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Appendix/Ramblings/BiPolytrope51AnalyticStability

2026-11-06T00:34:45Z

Joel2: /* Is There an Analytic Expression for the Eigenfunction? */

__FORCETOC__

=Marginally Unstable (n<sub>c</sub>,n<sub>e</sub>) = (5,1) Bipolytropes=

This ''Ramblings Appendix'' chapter provides some detailed trial derivations — mostly blind alleyways — in support of the [[SSC/Stability/BiPolytropes#Eigenvectors_for_Marginally_Unstable_Models_with_.28.CE.B3c.2C_.CE.B3e.29_.3D_.286.2F5.2C_2.29|accompanying, thorough discussion of this topic]].

==Fundamental Modes==

We decided to examine, first, whether any model along each sequence marks a transition from dynamically stable to dynamically unstable configurations. We accomplished this by setting <math>~\sigma_c^2</math> = 0, then integrating the relevant LAWE from the center toward the surface for many different ''guesses'' of the core-envelope interface radius until an eigenfunction with no radial nodes — ''i.e.,'' an eigenfunction associated with the fundamental mode of radial oscillation — was found whose behavior at the surface matched with high precision the physically desired surface boundary condition. We were successful in this endeavor. A marginally unstable model was identified on each of the six separate equilibrium sequences.

===Equilibrium Properties of Marginally Unstable Models===
Table 2 summarizes some of the equilibrium properties of these six models. For example, the second column of the table gives the value of the core-envelope interface radius, <math>~\xi_i</math>, associated with each marginally unstable model. The table also lists: the value of the model's dimensionless radius, <math>~R^*_\mathrm{surf}</math>, the key structural parameters, <math>~q</math> & <math>~\nu</math>, and the central-to-mean density associated with each model; and in each case the dimensionless thermal energy <math>~(\mathfrak{s})</math> and dimensionless gravitational potential energy <math>~(\mathfrak{w})</math> associated, separately, with the core and the envelope. Note that, once the pair of parameters, <math>~(\mu_e/\mu_c, \xi_i)</math>, has been specified, we can legitimately assign high-precision values to all of the other model parameters because they are [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|analytically prescribed]].

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center" colspan="10">
'''Table 2:''' Properties of Marginally Unstable Bipolytropes Having<br /><br /><math>~(n_c, n_e) = (5, 1)</math> and <math>~(\gamma_c, \gamma_e) = (\tfrac{6}{5}, 2)</math><br /><br />Determined from Integration of the LAWE
</th>
</tr>
<tr>
<td align="center"><math>~\frac{\mu_e}{\mu_c}</math></td>
<td align="center"><math>~\xi_i</math></td>
<td align="center"><math>~R^*_\mathrm{surf}</math></td>
<td align="center"><math>~q \equiv \frac{r_\mathrm{core}}{R_\mathrm{surf}}</math></td>
<td align="center"><math>~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math></td>
<td align="center"><math>~\frac{\rho_c}{\bar\rho}</math></td>
<td align="center"><math>~\mathfrak{s}_\mathrm{core}</math></td>
<td align="center"><math>~\mathfrak{w}_\mathrm{core}</math></td>
<td align="center"><math>~\mathfrak{s}_\mathrm{env}</math></td>
<td align="center"><math>~\mathfrak{w}_\mathrm{env}</math></td>
</tr>
<tr>
<td align="center">1</td>
<td align="left">1.6686460157</td>
<td align="right">2.139737</td>
<td align="center">0.53885819</td>
<td align="center">0.497747626</td>
<td align="center">8.51704656</td>
<td align="center">3.021916335</td>
<td align="center">-3.356583022</td>
<td align="center">1.47780476</td>
<td align="center">-5.642859167</td>
</tr>
<tr>
<td align="center"><math>~\tfrac{1}{2}</math></td>
<td align="left">2.27925811317</td>
<td align="right">5.146499</td>
<td align="center">0.306021732</td>
<td align="center">0.401776274</td>
<td align="center">63.29514949</td>
<td align="center">4.241287117</td>
<td align="center">-6.074241035</td>
<td align="center">4.284931508</td>
<td align="center">-10.97819621</td>
</tr>
<tr>
<td align="center">0.345</td>
<td align="left">2.560146865247</td>
<td align="right">9.554041</td>
<td align="center">0.185160563</td>
<td align="center">0.234302525</td>
<td align="center">209.7739052</td>
<td align="center">4.639705843</td>
<td align="center">-7.125754184</td>
<td align="center">11.72861751</td>
<td align="center">-25.61089252</td>
</tr>
<tr>
<td align="center"><math>~\tfrac{1}{3}</math></td>
<td align="left">2.582007485476</td>
<td align="right">10.120558</td>
<td align="center">0.176288391</td>
<td align="center">0.218241608</td>
<td align="center">230.4125398</td>
<td align="center">4.667042505</td>
<td align="center">-7.200966267</td>
<td align="center">13.15887139</td>
<td align="center">-28.45086152</td>
</tr>
<tr>
<td align="center">0.309</td>
<td align="left">2.6274239687695</td>
<td align="center">11.464303</td>
<td align="center">0.158362807</td>
<td align="center">0.184796947</td>
<td align="center">279.0788798</td>
<td align="center">4.722277318</td>
<td align="center">-7.354156963</td>
<td align="center">17.1374434</td>
<td align="center">-36.36528446</td>
</tr>
<tr>
<td align="center"><math>~\tfrac{1}{4}</math></td>
<td align="left">2.7357711469398</td>
<td align="center">15.895632</td>
<td align="center">0.118924863</td>
<td align="center">0.11071211</td>
<td align="center">430.0444648</td>
<td align="center">4.84592201</td>
<td align="center">-7.70305421</td>
<td align="center">37.84289623</td>
<td align="center">-77.67458196</td>
</tr>
</table>
</div>



As was expected from our [[#What_to_Expect_for_Equilibrium_Configurations|above discussion of virial equilibrium conditions]], we found that to high precision for each of these equilibrium models,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
(\mathfrak{w}_\mathrm{core}
~+~\mathfrak{w}_\mathrm{env})
~+~2(\mathfrak{s}_\mathrm{core}
~+~\mathfrak{s}_\mathrm{env})
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~0 \, .</math>
</td>
</tr>
</table>
<span id="Figure4">However, contrary to expectations,</span> in no case did we find that <math>~\mathfrak{s}_\mathrm{core}/\mathfrak{s}_\mathrm{env} = 5</math>. That is to say, we found that ''none'' of the models lies on the (red-dashed) curve in the <math>~q-\nu</math> parameter space that separates stable from unstable models as defined by our [[#What_to_Expect_for_Equilibrium_Configurations|above free-energy-based stability analysis]]. The left-hand panel of Figure 4 shows this (red-dashed) demarcation curve; for all intents and purposes, it is a reproduction of the right-hand panel of [[#Virial_Stability_Evaluation|Figure 3, above]] — turning-point markers have been removed to minimize clutter, the equilibrium sequences have been labeled, and the horizontal axis has been extended to unity in order to include a longer portion of the <math>~\mu_e/\mu_c = 1</math> sequence. The orange triangular markers that appear in the right-hand panel of Figure 4 pinpoint where each of the Table 2 "marginally unstable" models resides in this <math>~q-\nu</math> plane. Clearly, all six of the orange triangles lie well off of — and to the ''stable'' side of — the red-dashed demarcation curve. This discrepancy, which has resulted from our use of two separate approaches to stability analysis, will be discussed further and gratifyingly resolved, below.

<table border="0" cellpadding="5" align="center">

<tr>
<th align="center">Figure 4</th>
</tr>
<tr>
<td align="center" colspan="10">[[File:NEWCompositeDlabeled.png|800px|Marginally unstable models]]</td>
</tr>
</table>

===Eigenfunction Details===

Here we examine some of the properties of the fundamental-mode eigenfunctions that we have found are associated with marginally unstable, <math>~(n_c, n_e) = (5,1)</math> bipolytropes.
<table border="0" align="right" width="40%">
<tr>
<th align="center">Figure 5</th>
</tr>
<tr><td align="center">
[[File:Mod0MuRatio100.png|450px|Example eigenvector]]
</td></tr>
</table>
Consider the model on the <math>~\mu_e/\mu_c = 1</math> sequence for which <math>~\sigma_c^2=0~</math>; key properties of this specific equilibrium model are enumerated in the first row of numbers provided in [[#Equilibrium_Properties_of_Marginally_Unstable_Models|Table 2, above]]. Figure 5 shows how our numerically derived, fundamental-mode eigenfunction, <math>~x = \delta r/r_0</math>, varies with the fractional radius over the entire range, <math>~0 \le r/R \le 1</math>. By prescription, the eigenfunction has a value of unity and a slope of zero at the center <math>~(r/R = 0)</math>. Integrating the LAWE outward from the center, through the model's core (blue curve segment), <math>~x</math> drops smoothly to the value <math>~x_i = 0.81437</math> at the interface <math>~(\xi_i = 1.6686460157 ~\Rightarrow~ q = r_\mathrm{core}/R_\mathrm{surf} = 0.53885819)</math>. Our numerical integration of the LAWE showed that, at the interface, the logarithmic slope of the ''core'' (blue) segment of the eigenfunction is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i \biggr\}_\mathrm{core}
=
\biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_i \biggr\}_\mathrm{core}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- 0.455872 \, .</math>
</td>
</tr>
</table>
Next, following the [[#Interface|above discussion of matching conditions at the interface]], we determined that, from the perspective of the envelope, the slope of the eigenfunction at the interface must therefore be,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i \biggr\}_\mathrm{env}
=
\biggl\{ \frac{d\ln x}{d\ln \eta}\biggr|_i \biggr\}_\mathrm{env}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~3\biggl(\frac{\gamma_c}{\gamma_e} -1\biggr) + \frac{\gamma_c}{\gamma_e} \biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_i \biggr\}_\mathrm{core} = -1.47352 \, .</math>
</td>
</tr>
</table>

Adopting this "env" slope along with the amplitude, <math>~x_i = 0.81437</math>, as the appropriate ''interface'' boundary conditions, we integrated the LAWE from the interface to the surface, obtaining the green-colored segment of the eigenfunction that is shown in Figure 5. The amplitude continued to steadily decrease, reaching a value of <math>~x_s = 0.38203</math>, at the model's surface <math>~(r/R = 1)</math>. At the surface, this ''envelope'' (green) segment of the eigenfunction exhibits a logarithmic slope that matches to eight significant digits the value that is [[#SurfaceCondition|expected from astrophysical arguments]] for this marginally unstable <math>~(\sigma_c^2=0)</math> model, namely,
<div align="center">
<math>~
\frac{d\ln x}{d\ln \eta}\biggr|_s = \biggl[ \biggl( \frac{\rho_c}{\bar\rho} \biggr)\frac{\cancelto{0}{\sigma_c^2}}{2\gamma_e} - \biggl(3 - \frac{4}{\gamma_e}\biggr)\biggr] = -1 \, .
</math>
</div>

<table border="1" cellpadding="10" width="80%" align="center"><tr><td align="left">
<font color="red">'''Key Reminder:'''</font> We were able to find an eigenfunction whose surface boundary condition matched the desired value — in this particular case, a logarithmic slope of negative one — to this high level of precision only by iterating many times and, at each step, fine-tuning our choice of the equilibrium model's radial interface location, <math>~\xi_i</math> before performing a numerical integration of the LAWE.
</td></tr></table>

The discontinuous jump that occurs in the ''slope'' of the eigenfunction at the interface results from our assumption that the effective adiabatic index of material in the core <math>~(\gamma_c = 6/5)</math> is different from the effective adiabatic index of the envelope material <math>~(\gamma_e = 2)</math>. In an effort to emphasize and more clearly illustrate the behavior of this fundamental-mode eigenfunction as it crosses the core/envelope interface, we have added a pair of dashed line segments to the Figure 5 plot. The red-dashed line segment touches, and is tangent to, the blue segment of the eigenfunction at the location of the core/envelope interface; it has a slope,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{dx}{d(r/R)}\biggr|_i
=
\frac{x_i}{(r_i/R)}\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i\biggr\}_\mathrm{core}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- 0.455872 \biggl(\frac{ 0.81437 }{ 0.53885819 }\biggr) = - 0.68895\, .</math>
</td>
</tr>
</table>
On the other hand, the purple-dashed line segment touches, and is tangent to, the green segment of the eigenfunction at the location of the core/envelope interface; it has a slope,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{dx}{d(r/R)}\biggr|_i
=
\frac{x_i}{(r_i/R)}\biggl\{ \frac{d\ln x}{d\ln r}\biggr|_i\biggr\}_\mathrm{env}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- 1.47352 \biggl(\frac{ 0.81437 }{ 0.53885819 }\biggr) = - 2.22691\, .</math>
</td>
</tr>
</table>
For comparison purposes, the eigenfunction shown in Figure 5 has been presented again in Figure 6, along with several other of our numerically derived eigenfunctions, but in Figure 6 the plotted amplitude has been renormalized to give a surface value — rather than a central value — of unity.

In Figure 6 we show the behavior of the fundamental-mode eigenfunction for each of the marginally unstable models identified in Table 2. In the top figure panel, each curve shows — on a linear-linear plot — how the amplitude varies with radius; in the bottom figure panel, the amplitude is plotted on a logarithmic scale. On each curve, the black plus sign marks the radial location of the core-envelope interface; in the bottom panel, these markers are accompanied by the values of <math>~\xi_i</math> that are associated with each corresponding model (see also the second column of Table 2). Each eigenfunction has been normalized such that the surface amplitude is unity. In the top panel, the value of the central amplitude of the eigenfunction that results from this normalization is recorded near the point where each eigenfunction touches the vertical axis. (In each case, the value provided on the plot is simply the inverse of the value of <math>~x_s</math> given in Table 3, below.)

<table border="0" align="center" cellpadding="8">
<tr>
<th align="center">
'''Figure 6: Eigenfunctions Associated with the Fundamental-Mode of Radial Oscillation'''<br />
'''in Marginally Unstable Models having Various''' <math>~\mu_e/\mu_c</math>
</th>
</tr>
<tr>
<td align="center">
[[File:Mode0EigenfunctionsCombinedSmall.png|800px|Eigenfunctions for Marginally Unstable Models]]
</td>
</tr>
</table>

Notice that, especially as they approach the surface, the "envelope" segments of these six marginally unstable eigenfunction appear to merge into the same curve, irrespective of their value of the ratio of mean molecular weights. Note as well that the discontinuous jump that occurs in the ''slope'' of each eigenfunction at the radial location of the core/envelope interface — resulting from our choice to adopt a different adiabatic index, <math>~\gamma_g</math>, in the core from the one in the envelope — becomes less and less noticeable for smaller and smaller values of the ratio of mean molecular weights.

===Is There an Analytic Expression for the Eigenfunction?===
After noticing that, in Figure 6, the ''envelope'' segments of all of the marginally unstable eigenfunctions merge into the same curve, we began to wonder whether a single expression — and, even better, an ''analytically defined'' expression — would perfectly describe the eigenfunction. We had reason to believe that this might actually be possible because, in [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|pressure-truncated polytropic configurations, we have derived analytic expressions for the marginally unstable, fundamental-mode eigenfunctions]] of both <math>~n = 5</math> and <math>~n=1</math> systems.

Very quickly, we convinced ourselves that a parabolic function does indeed perfectly match the "core" segment of each displayed eigenfunction. Specifically, throughout the core <math>~(0 \le \xi \le \xi_i)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_P\biggr|_\mathrm{core}</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~1 - \frac{\xi^2}{15}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~\frac{dx_P}{d\xi}\biggr|_\mathrm{core}</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~- \frac{2\xi}{15} </math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~\frac{d\ln x_P}{d\ln \xi}\biggr|_\mathrm{core}</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~- \frac{2\xi^2}{15} \biggl[ \frac{(15 - \xi^2)}{15} \biggr]^{-1} = - \frac{2\xi^2}{(15 - \xi^2)} \, .</math>
</td>
</tr>
</table>

The envelope segment posed a much greater challenge.

=====Attempt 1=====
Building on our [[SSC/Stability/InstabilityOnsetOverview#Configurations_Having_an_Index_Less_Than_Three|accompanying discussion of ''Pressure-Truncated Configurations Having a Polytropic Index less than Three'']] — see, for example, a relevant [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|''succinct demonstration'']] — a promising analytic expression is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_P\biggr|_\mathrm{env}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1 - \eta\cot(\eta - C) \biggr] \, ,
</math>
</td>
</tr>
</table>
where the values of the pair of coefficients, <math>~b</math> and <math>~C</math>, is to be determined. Most likely, we should set <math>~C = B</math>.

Focusing on the case of <math>~\mu_e/\mu_c = 1</math>, <math>~\gamma_c = 6/5</math>, and <math>~\gamma_e = 2</math>, here are some parameters that we ''think'' we know.
<table border="1" align="center" cellpadding="10">
<tr>
<th align="center" colspan="5">Relevant Parameters for<br />Marginally Unstable Model <math>~(\Omega^2 = 0)</math> <br />with <math>~\mu_e/\mu_c = 1</math> and <math>~(\gamma_c, \gamma_e) = (\tfrac{6}{5}, 2)</math></th>
</tr>
<tr>
<td align="right"><math>~\xi_i</math></td>
<td align="center"> </td>
<td align="center">numerically<br />determined</td>
<td align="center"><math>~\approx</math></td>
<td align="left">1.6686460157</td>
</tr>
<tr>
<td align="right"><math>~x_i</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~1 - \frac{\xi_i^2}{15} </math></td>
<td align="center"><math>~=</math></td>
<td align="left">0.814374699</td>
</tr>
<tr>
<td align="right"><math>~\biggl[\frac{d\ln x}{d\ln \xi}\biggr]_i</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~- \frac{2\xi_i^2}{(15 - \xi_i^2)}</math></td>
<td align="center"><math>~=</math></td>
<td align="left">-0.455871976</td>
</tr>
<tr>
<td align="right"><math>~\eta_i</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~\biggl(\frac{\mu_e}{\mu_c}\biggr) \sqrt{3} \xi_i \biggl[1 + \frac{\xi_i^2}{3}\biggr]^{-1 } </math></td>
<td align="center"><math>~=</math></td>
<td align="left">1.498957494</td>
</tr>
<tr>
<td align="right"><math>~B</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~\eta_i - \frac{\pi}{2} + \tan^{-1}\biggl[ \frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}} \biggr] </math></td>
<td align="center"><math>~=</math></td>
<td align="left">- 0.359863579</td>
</tr>
<tr>
<td align="right"><math>~\eta_s</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~\pi + B </math></td>
<td align="center"><math>~=</math></td>
<td align="left">2.781729075</td>
</tr>
<tr>
<td align="right"><math>~\biggl[\frac{d\ln x}{d\ln \eta}\biggr]_i</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~3\biggl( \frac{\gamma_c}{\gamma_e}-1\biggr) + \frac{\gamma_c}{\gamma_e} \biggl[\frac{d\ln x}{d\ln \xi}\biggr]_i</math></td>
<td align="center"><math>~=</math></td>
<td align="left">-1.473523186</td>
</tr>
<tr>
<td align="right"><math>~\alpha_e</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~3 - \frac{4}{\gamma_e} </math></td>
<td align="center"><math>~=</math></td>
<td align="left">+ 1</td>
</tr>
<tr>
<td align="right"><math>~\biggl[\frac{d\ln x}{d\ln \eta}\biggr]_\mathrm{surf}</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~\frac{\cancelto{0}{\Omega^2}}{\gamma_e} - \alpha_e </math></td>
<td align="center"><math>~=</math></td>
<td align="left">- 1</td>
</tr>
<tr>
<td align="right"><math>~b</math></td>
<td align="center"><math>~\equiv</math></td>
<td align="left"><math>~\frac{3}{5}\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{15-\xi_i^2}{3+\xi_i^2}\biggr]</math></td>
<td align="center"><math>~=</math></td>
<td align="left">1.26097406</td>
</tr>
</table>

Notice that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\cot(\eta_i - B)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan\biggl[ \frac{\pi}{2} -(\eta_i - B ) \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[ \frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}} \biggr] </math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \eta_i \cot(\eta_i - B)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left" colspan="3">
<math>~\biggl[ 1 - \frac{\xi_i \eta_i}{\sqrt{3}} \biggr] </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left" colspan="3">
<math>~1 - \xi_i^2 \biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[1 + \frac{\xi_i^2}{3}\biggr]^{-1 } </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left" colspan="3">
<math>~1 - \biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3\xi_i^2 }{3 + \xi_i^2}\biggr] \, .</math>
</td>
</tr>
</table>

So if we adopt the expression for <math>~x_P</math> as given above (with C = B), then we can evaluate the leading ''b'' factor by examining the function at the interface, that is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~b</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
x_i \eta_i^2 \biggl[1 - \eta_i \cot(\eta_i - B) \biggr]^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
x_i \eta_i^2 \biggl[\frac{\xi_i \eta_i }{\sqrt{3}} \biggr]^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\eta_i \biggl[\frac{\sqrt{3}}{\xi_i} \biggr]x_i
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl(\frac{\mu_e}{\mu_c}\biggr) \sqrt{3} \xi_i \biggl[1 + \frac{\xi_i^2}{3}\biggr]^{-1 } \biggl[\frac{\sqrt{3}}{\xi_i} \biggr] \biggl[1 - \frac{\xi_i^2}{15} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
3\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3}{3+\xi_i^2}\biggr]\biggl[\frac{15-\xi_i^2}{15} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3}{5}\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{15-\xi_i^2}{3+\xi_i^2}\biggr] \, .
</math>
</td>
</tr>
</table>

<table border="0" align="right">
<tr>
<th align="center">Figure 7: Analytic Trial</th>
</tr>
<tr><td align="center">[[File:AnalyticAttempt1.png|400px|First Analytic Trial]]</td></tr>
</table>
An even clearer way of looking at this is to realize that, quite generally,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(\eta - B)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\eta - \eta_i + \frac{\pi}{2} - \tan^{-1} f \, ,</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~\frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}} \, .</math>
</td>
</tr>
</table>
Hence, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\cot(\eta - B)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan\biggl[ \frac{\pi}{2} - (\eta - B) \biggr]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan\biggl[ (\eta_i - \eta) + \tan^{-1}f \biggr]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{\tan(\eta_i - \eta) + f}{1 - f\tan(\eta_i - \eta)} \, .</math>
</td>
</tr>
</table>
As a result, we can rewrite the expression for our ''guess'' of the envelope segment of the eigenfunction in the form,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_P\biggr|_\mathrm{env}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{b}{\eta^2} \biggl\{ 1 - \eta \biggl[ \frac{\tan(\eta_i - \eta) + f}{1 - f\tan(\eta_i - \eta)} \biggr] \biggr\} \, .
</math>
</td>
</tr>
</table>
This blows up when <math>\eta \rightarrow \eta_s</math> because, as it turns out, <math>~f = 1/\tan(\eta_i - \eta_s)</math>. We should point out, as well, that the expression for ''b'' can be rewritten in the form,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~b</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{\eta_i^2 (x_P)_i}{(1 - \eta_i f)} \, .
</math>
</td>
</tr>
</table>

In Figure7 we have reprinted the numerically determined fundamental-mode eigenfunction that was first displayed in [[#Eigenfunction_Details|Figure 5, above]]. We have added to this plot the eigenfunction segments that are defined by our trial analytic functions: The ''core'' segment, <math>~x_P|_\mathrm{core}</math>, matches the numerically determined segment with sufficient precision that the two curve segments are indiscernible from one another. However, our analytically defined "env" segment, <math>~x_P|_\mathrm{env}</math> — identified in Figure 7 by the solid black, small circular markers — does not match the numerically determined envelope segment at all. We therefore have more work to do!

=====Attempt 2=====

Using an Excel spreadsheet as a sandbox, we employed crude, brute force iterations in an effort to fit the numerically constructed envelope eigenfunction. Here is a trial function that works pretty well. Using <math>~\eta_\mathrm{F}</math> to represent the envelope's dimensionless radial coordinate, over the range,
<div align="center">
<math>~\eta_i \le \eta_\mathrm{F} \le \eta_s \, ,</math>
</div>
and defining the parameter,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~g_\mathrm{F}</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
\frac{\pi}{8(\eta_s - \eta_i)}
\, ,</math>
</td>
</tr>
</table>

<table border="1" cellpadding="5" align="right">
<tr>
<th align="center" colspan="4">Limiting Parameter Values</th>
</tr>
<tr>
<td align="center"> </td>
<td align="center">min</td>
<td align="center">max</td>
<td align="center"><math>~\alpha = \alpha_s</math>
</tr>
<tr>
<td align="center"><math>~\eta_\mathrm{F}</math></td>
<td align="center"><math>~\eta_i</math></td>
<td align="center"><math>~\eta_s</math></td>
<td align="center"><math>~\frac{8}{\pi} ( \eta_s - \eta_i )^2 + 2\eta_s - \eta_i</math></td>
</tr>
<tr>
<td align="center"><math>~\alpha</math></td>
<td align="center"><math>~-\frac{\pi}{2}</math></td>
<td align="center"><math>~-\frac{5\pi}{8}</math></td>
<td align="center"><math>~\eta_i - \eta_s - \frac{3\pi}{4}</math></td>
</tr>
<tr>
<td align="center"><math>~\Lambda</math></td>
<td align="center"><math>~\eta_i - \frac{\pi}{4}</math></td>
<td align="center"><math>~\eta_i - \frac{\pi}{8}</math></td>
<td align="center"><math>~\eta_s</math></td>
</tr>
</table>
we propose,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_\mathrm{trial}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \biggl[ \frac{\tan(\alpha) + f_\alpha}{1 - f_\alpha \cdot \tan(\alpha)} \biggr] \biggr\} - a_0 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{f_\alpha} = \tan(\alpha_s)</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
\tan[ - (\eta_s - \eta_i + \tfrac{3\pi}{4}) ]
\, ,</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\alpha(\eta_\mathrm{F})</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~g_\mathrm{F} \biggl[ 5\eta_i - 4\eta_s - \eta_\mathrm{F}\biggr] \, ,</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Lambda(\eta_\mathrm{F})</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
\eta_i + g_\mathrm{F} \biggl[ \eta_i - 2\eta_s + \eta_\mathrm{F} \biggr]
\, .</math>
</td>
</tr>
</table>

<table border="0" align="right">
<tr>
<th align="center">Figure 8: Another Analytic Trial</th>
</tr>
<tr><td align="center">[[File:AnalyticAttempt2.png|400px|Second Analytic Trial]]</td></tr>
</table>
This function, <math>~x_\mathrm{trial}</math>, is displayed as the black-dotted curve segment in Figure 8 with the tuning/scaling parameters set to the values, <math>~(a_0, b_0) = (0.31, 0.96)</math>. We should point out that, when plotting this curve segment in Figure 8, the dimensionless radial coordinate has been defined by the relation, <math>~r^*/R^* = \eta_F/\eta_s</math>.

<font color="red">'''[16 February 2019: Comment by Tohline]'''</font> When assessed visually, this ''trial'' function appears to match pretty well the numerically derived eigenfunction for the envelope. We have not yet critically assessed whether or not the function satisfies the LAWE or whether it satisfies either one (or both) of the required boundary values. This work is still to be done.

A couple of days after inserting this <font color="red">''Comment''</font>, we recognized for the first time that, quite generally,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\alpha</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\eta_i - \biggl(\Lambda + \frac{3\pi}{4} \biggr) \, .</math>
</td>
</tr>
</table>
Hence, the parameter, <math>~\alpha</math>, can be straightforwardly removed from the expression for the trial eigenfunction to give,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_\mathrm{trial}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \biggl[ \frac{\tan(\eta_i - \Lambda - 3\pi/4) + f_\alpha}{1 - f_\alpha \cdot \tan(\eta_i - \Lambda - 3\pi/4)} \biggr] \biggr\} - a_0 \, .
</math>
</td>
</tr>
</table>

Drawing from our [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|accompanying discussion of pressure-truncated polytropes]], we need the eigenfunction to satisfy the,
<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>
(Note that, in order to bring the notation of this ''Key Equation'' in line with the notation used elsewhere in this chapter, we will hereafter adopt the variable mapping <math>~\xi \rightarrow \eta</math> and <math>~\theta \rightarrow \phi</math>.) Here we are especially focused on finding a solution in the case where <math>~\sigma_c^2 = 0</math> and <math>~n = 1</math>, that is — see also our [[#Envelope:|above discussion]] — the relevant envelope LAWE is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl[ 2 - Q \biggr] \frac{2}{\eta}\cdot \frac{dx}{d\eta} -2\alpha_g Q \cdot \frac{x}{\eta^2} \, ,
</math>
</td>
</tr>
</table>
where, drawing from our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|discussion of the n = 1 envelope's equilibrium structure]],
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[ - \frac{d\ln\phi}{d\ln\eta} \biggr]_\mathrm{n=1} = - \frac{\eta}{\phi}\biggl[ \frac{d\phi}{d\eta} \biggr]_\mathrm{n=1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \biggl[ \frac{1}{\sin(\eta - B)} \biggr]\biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[1 - \eta\cot(\eta-B) \biggr] \, .</math>
</td>
</tr>
</table>

In an [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|accompanying discussion]] — see also a [[SSC/Stability/InstabilityOnsetOverview#Configurations_Having_an_Index_Less_Than_Three|short summary of the same]] — we have shown that an analytically specified displacement function that precisely satisfies this LAWE for pressure-truncated configurations (i.e., when B = 0) is,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_P\biggr|_{n=1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3}{\eta^2}\biggl[ 1- \eta \cot\eta \biggr] \, .
</math>
</td>
</tr>
</table>
</div>

In still [[SSC/Stability/n1PolytropeLAWE#What_About_Bipolytropes.3F|another related discussion]], we have attempted to construct an analytic eigenfunction expression that satisfies the LAWE when <math>~B \ne 0</math>.

=====Attempt 3=====
======Straightforward Trial======
Let's adopt a trial eigenfunction of the form
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_\mathrm{trial}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{b_0}{\Lambda^2} \biggl[
1 - \Lambda \cot(\Lambda - E)
\biggr] - a_0 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Lambda</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\Lambda_0 + g_\mathrm{F} \eta \, .</math>
</td>
</tr>
</table>

<table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left">
<font color="red">'''NOTE:'''</font>  We can retrieve the empirical expression for <math>~x_\mathrm{trial}</math> obtained above in '''Attempt 2''' if we eventually set,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Lambda_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\eta_i + g_\mathrm{F}(\eta_i - 2\eta_s)</math>
</td>
</tr>

<tr>
<td align="right">
<math>~g_\mathrm{F}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{\pi}{8(\eta_s - \eta_i)} \, ,</math>       and,
</td>
</tr>

<tr>
<td align="right">
<math>~E</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\eta_i - \frac{5\pi}{4} + \tan^{-1} f_\alpha \, .</math>
</td>
</tr>
</table>
The last of these expressions arises because,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\cot(\Lambda - E)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan \biggl[ \frac{\pi}{2} - (\Lambda - E)\biggr]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan\biggl[ \frac{\pi}{2} - \Lambda + \eta_i - \frac{5\pi}{4} + \tan^{-1} f_\alpha \biggr]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan\biggl[ (\eta_i - \Lambda - \tfrac{3\pi}{4}) + \tan^{-1}f_\alpha \biggr]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{\tan(\eta_i - \Lambda - \tfrac{3\pi}{4}) + f_\alpha }{1 - f_\alpha \tan(\eta_i - \Lambda - \tfrac{3\pi}{4} )} \, .</math>
</td>
</tr>
</table>

</td></tr></table>

Because the LAWE requires derivatives of <math>~x_\mathrm{trial}</math> with respect to <math>~\eta</math>, we will often need to recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d}{d\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d\Lambda}{d\eta} \cdot \frac{d}{d\Lambda} = g_\mathrm{F} \cdot \frac{d}{d\Lambda} \, .</math>
</td>
</tr>
</table>
Hence, in particular,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\Lambda - E) \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- g_\mathrm{F} [1 + \cot^2(\Lambda - E) ] \, .</math>
</td>
</tr>
</table>
The first derivative gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d}{d\eta}\biggl[ x_\mathrm{trial} \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{d}{d\eta} \biggl\{
\frac{b_0}{\Lambda^2} \biggl[
1 - \Lambda \cot(\Lambda - E)
\biggr] - a_0
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
-~\frac{2b_0 g_\mathrm{F}}{\Lambda^3} \biggl[1 - \Lambda \cot(\Lambda - E)\biggr]
+ \frac{b_0}{\Lambda^2} \biggl[- g_\mathrm{F} \cot(\Lambda - E)\biggr]
+ \frac{b_0}{\Lambda^2} \biggl[g_\mathrm{F} \Lambda [1 + \cot^2(\Lambda - E) ] \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~ g_\mathrm{F} b_0 \biggl\{
-~\frac{2}{\Lambda^3} \biggl[1 - \Lambda \cot(\Lambda - E)\biggr]
- \frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr]
+ \frac{1}{\Lambda} \biggl[1 + \cot^2(\Lambda - E) \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~ g_\mathrm{F} b_0 \biggl\{
\frac{1}{\Lambda}
-\frac{2}{\Lambda^3}
+\frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr]
+ \frac{1}{\Lambda} \biggl[\cot^2(\Lambda - E) \biggr]
\biggr\} \, .
</math>
</td>
</tr>
</table>

The second derivative gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{g_\mathrm{F}b_0} \cdot \frac{d^2}{d\eta^2}\biggl[ x_\mathrm{trial} \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d}{d\eta}\biggl\{
\frac{1}{\Lambda}
-\frac{2}{\Lambda^3}
+\frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr]
+ \frac{1}{\Lambda} \biggl[\cot^2(\Lambda - E) \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
-\frac{g_\mathrm{F}}{\Lambda^2}
+\frac{6 g_\mathrm{F}}{\Lambda^4}
-\frac{2g_\mathrm{F}}{\Lambda^3} \biggl[\cot(\Lambda - E)\biggr]
- \frac{g_\mathrm{F}}{\Lambda^2} \biggl[1 + \cot^2(\Lambda - E)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{g_\mathrm{F}}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr]
- \frac{2g_\mathrm{F}}{\Lambda} \biggl[\cot(\Lambda - E) \biggr] \biggl[1 + \cot^2(\Lambda - E) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{g^2_\mathrm{F}b_0} \cdot \frac{d^2}{d\eta^2}\biggl[ x_\mathrm{trial} \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
-\frac{1}{\Lambda^2}
+\frac{6 }{\Lambda^4}
-\frac{2}{\Lambda^3} \biggl[\cot(\Lambda - E)\biggr]
- \frac{1}{\Lambda^2}
- \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr]
- \frac{2}{\Lambda} \biggl[\cot(\Lambda - E) + \cot^3(\Lambda - E) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
-\frac{1}{\Lambda^2}
- \frac{1}{\Lambda^2}
+\frac{6 }{\Lambda^4}
-\frac{2}{\Lambda^3} \biggl[\cot(\Lambda - E)\biggr]
- \frac{2}{\Lambda} \biggl[\cot(\Lambda - E) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr]
- \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E)\biggr]
- \frac{2}{\Lambda} \biggl[ \cot^3(\Lambda - E) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
-\frac{2}{\Lambda^2}
+\frac{6 }{\Lambda^4}
- \biggl[ \frac{2}{\Lambda^3} + \frac{2}{\Lambda}\biggr]\biggl[\cot(\Lambda - E)\biggr]
- \frac{2}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr]
- \frac{2}{\Lambda} \biggl[ \cot^3(\Lambda - E) \biggr] \, .
</math>
</td>
</tr>
</table>
So, appreciating that, <math>~\eta = (\Lambda - \Lambda_0 )/g_\mathrm{F}</math>, and dividing the relevant LAWE through by <math>~b_0 g_\mathrm{F}^2</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{b_0 g_\mathrm{F}^2} \cdot \frac{d^2x_\mathrm{trial} }{d\eta^2}
+ \frac{1}{b_0 g_\mathrm{F}^2} \biggl[ 2 - Q \biggr] \frac{2g_\mathrm{F}}{(\Lambda - \Lambda_0 )}\cdot \frac{dx_\mathrm{trial}}{d\eta}
- \frac{2Q}{b_0 g_\mathrm{F}^2} \cdot \frac{g^2_\mathrm{F} x_\mathrm{trial}}{(\Lambda - \Lambda_0 )^2}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{1}{(\Lambda - \Lambda_0 )^2} \biggl\{
\frac{(\Lambda - \Lambda_0 )^2}{b_0 g_\mathrm{F}^2} \cdot \frac{d^2x}{d\eta^2}
+ \frac{ (\Lambda - \Lambda_0 )}{b_0 g_\mathrm{F}} \biggl[ 2 - Q \biggr] \frac{dx}{d\eta}
- \frac{2Q}{b_0 } \cdot x
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~ \frac{ (\Lambda - \Lambda_0 )^2}{b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
(\Lambda - \Lambda_0 )^2 \biggl\{ \frac{1}{b_0 g_\mathrm{F}^2} \cdot \frac{d^2x}{d\eta^2} \biggr\}
+ (\Lambda - \Lambda_0 ) \biggl[ 4 - 2Q \biggr] \biggl\{ \frac{1}{b_0 g_\mathrm{F}} \frac{dx}{d\eta} \biggr\}
- 2Q \biggl\{ \frac{x}{b_0} \biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
(\Lambda - \Lambda_0 )^2 \biggl\{
-\frac{2}{\Lambda^2}
+\frac{6 }{\Lambda^4}
- \biggl[ \frac{2}{\Lambda^3} + \frac{2}{\Lambda}\biggr]\biggl[\cot(\Lambda - E)\biggr]
- \frac{2}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr]
- \frac{2}{\Lambda} \biggl[ \cot^3(\Lambda - E) \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ (\Lambda - \Lambda_0 ) \biggl[ 4 - 2Q \biggr] \biggl\{ \frac{1}{\Lambda}
-\frac{2}{\Lambda^3}
+\frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr]
+ \frac{1}{\Lambda} \biggl[\cot^2(\Lambda - E) \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- 2Q \biggl\{ \frac{1}{\Lambda^2} \biggl[1 - \Lambda \cot(\Lambda - E) \biggr] - \frac{ a_0 }{b_0}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
(\Lambda - \Lambda_0 )^2 \biggl\{
(6-2\Lambda^2)
- (2\Lambda + 2\Lambda^3 ) \cot(\Lambda - E)
- 2\Lambda^2 \cot^2(\Lambda - E)
- 2\Lambda^3 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ (\Lambda - \Lambda_0 ) (4 - 2Q ) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2 \cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- 2Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
(\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)(\Lambda - \Lambda_0 )
- (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot(\Lambda - E)
- \Lambda^2 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
- \Lambda^3 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3
-4\Lambda
+2\Lambda^2 \cot(\Lambda - E)
+ 2\Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-Q \biggl\{ \Lambda^3(\Lambda - \Lambda_0 )
-2\Lambda (\Lambda - \Lambda_0 )
+\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E)
+ \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\} \, .
</math>
</td>
</tr>
</table>

----

The right-hand-side of this expression should simplify considerably if we let <math>~E \rightarrow B</math>, if we set <math>~(g_\mathrm{F}, \Lambda_0) = (1, 0) ~\Rightarrow~ \Lambda = \eta</math>, and if <math>~(a_0, b_0) = (0, 3)</math>. Let's see.

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
RHS
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda \biggl\{
(3-\Lambda^2)\Lambda
- (\Lambda + \Lambda^3 )\Lambda \cot(\Lambda - E)
- \Lambda^3 \cot^2(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
+2\Lambda^3
-4\Lambda
+2\Lambda^2 \cot(\Lambda - E)
+ 2\Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-Q \biggl\{ \Lambda^4
-2\Lambda^2
+\Lambda^3 \cot(\Lambda - E)
+ \Lambda^4 \cot^2(\Lambda - E)
+\Lambda^2 - \Lambda^3 \cot(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda \biggl\{
(-\Lambda +\Lambda^3)
+ (\Lambda^2 - \Lambda^4) \cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
-\Lambda^3
+ \Lambda
- \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+\Lambda\cot(\lambda-E) \biggl\{ \Lambda^4
-2\Lambda^2
+\Lambda^3 \cot(\Lambda - E)
+ \Lambda^4 \cot^2(\Lambda - E)
+\Lambda^2 - \Lambda^3 \cot(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda \biggl\{
(\Lambda^2 - \Lambda^4) \cot(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
\biggr\}
+\Lambda\cot(\lambda-E) \biggl\{ \Lambda^4
-\Lambda^2
+ \Lambda^4 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
0 \, .
</math>
</td>
</tr>
</table>

EXCELLENT !!

----

Let's return to the more general expression and see if it can be simplified.

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda (\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \Lambda (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^2
-4
+2\Lambda \cot(\Lambda - E)
+ 2\Lambda^2 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-Q(\Lambda - \Lambda_0 ) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
- Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda (\Lambda - \Lambda_0 ) \biggl\{
(\Lambda^2-1)
+ (\Lambda - \Lambda^3 ) \cot(\Lambda - E)
+ \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
- \Lambda (\Lambda - \Lambda_0 ) \biggl\{ \Lambda^2
-2
+\Lambda \cot(\Lambda - E)
+ \Lambda^2 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
- \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta\cot(\eta-B) (\Lambda - \Lambda_0 ) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
+ \eta\cot(\eta-B) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda (\Lambda - \Lambda_0 ) \biggl\{
1
- \Lambda^3 \cot(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
- \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta\cot(\eta-B) (\Lambda - \Lambda_0 ) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
+ \eta\cot(\eta-B) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>
</table>

Simplifying again produces,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
RHS
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda^2 \biggl\{
1
- \Lambda^3 \cot(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
- \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \Lambda^2 \cot(\Lambda-E) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
+ \Lambda \cot(\Lambda-E) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Lambda^3 \cot(\Lambda - E)
- \Lambda^5 \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+\Lambda^5 \cot(\Lambda-E) - 2\Lambda^3 \cot(\Lambda-E)
+\Lambda^4\cot^2(\Lambda - E)
+ \Lambda^5 \cot^3(\Lambda - E)
+ \Lambda^3 \cot(\Lambda-E) - \Lambda^4 \cot^2(\Lambda - E)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
0\, .
</math>
</td>
</tr>
</table>

EXCELLENT !! Keep trying to simplify the more general expression …

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl\{ \Lambda^2
- \Lambda^5 \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
- \Lambda^2 + \Lambda^3 \cot(\Lambda - E) + \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\}
- \Lambda_0 \biggl\{ \Lambda
- \Lambda^4 \cot(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta\cot(\eta-B) (\Lambda ) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
+ \eta\cot(\eta-B) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \eta\cot(\eta-B) (\Lambda_0) \biggl\{ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \Lambda^2
- \Lambda^5 \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
- \Lambda^2 + \Lambda^3 \cot(\Lambda - E) + \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[ \Lambda^4
-2\Lambda^2
+\Lambda^3 \cot(\Lambda - E)
+ \Lambda^4 \cot^2(\Lambda - E)
+ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 \biggl\{ (\Lambda - \Lambda_0 ) \biggl[
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr]
+ \biggl[ \Lambda
- \Lambda^4 \cot(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E) \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[
\biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
+ (\Lambda^3 - \Lambda^5) \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
\biggr]
+ \eta\cot(\eta-B) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4
- \Lambda^2
+ \Lambda^4 \cot^2(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 \biggl\{ (\Lambda - \Lambda_0 ) \biggl[
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr]
+ \biggl[ \Lambda
- \Lambda^4 \cot(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta\cot(\eta-B) \biggl[ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E) \biggr]
\biggr\} \, .
</math>
</td>
</tr>
</table>

----

Simplify again …

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
RHS
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[
\biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
+ (\Lambda^3 - \Lambda^5) \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
\biggr]
+ \eta\cot(\eta-B) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4
- \Lambda^2
+ \Lambda^4 \cot^2(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[
\biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
+ (\Lambda^3 - \Lambda^5) \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
\biggr]
+ \biggl[
(\Lambda^5
- \Lambda^3) \cot(\Lambda-E)
- \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^5\cot(\Lambda-E)
+ \Lambda^5 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl( \frac{ a_0 }{b_0}\biggr) \biggl[ \Lambda^4
- \Lambda^5\cot(\Lambda-E)\biggr] \, .
</math>
</td>
</tr>
</table>

======First Argument Relationships Guess======

Let's try the relationship,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(\eta - B)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~m(\Lambda - E) \, .</math>
</td>
</tr>
</table>
Then, for example, if <math>~m = 2</math>, we can make the replacement,
<div align="center">
<math>~\cot(\eta - B) = \cot[2(\Lambda - E)] = \frac{\cot^2(\Lambda - E) - 1}{2\cot(\Lambda - E)} \, .</math>
</div>
And, alternatively, if <math>~m = 3</math>, we can make the replacement,
<div align="center">
<math>~\cot(\eta - B) = \cot[3(\Lambda - E)] = \frac{\cot^3(\Lambda - E) - 3\cot(\Lambda - E)}{3\cot^2(\Lambda - E)-1} \, .</math>
</div>

<table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left">
<div align="center"><font color="red">'''IMPLICATIONS'''</font></div>

Given that we also are assuming the relationship,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\eta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{1}{g_\mathrm{F}} (\Lambda - \Lambda_0) \, ,</math>
</td>
</tr>
</table>
we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~m\Lambda - \eta </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~mE - B</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ m\Lambda - \frac{1}{g_\mathrm{F}} (\Lambda - \Lambda_0)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~mE - B</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \Lambda \biggl[m - \frac{1}{g_\mathrm{F}} \biggr] </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~mE - B - \frac{\Lambda_0}{g_\mathrm{F}} \, . </math>
</td>
</tr>
</table>
In order for this statement to be true for all <math>~\Lambda</math>, the RHS and the LHS must independently be zero. Hence, we require,
<div align="center">
<math>~g_\mathrm{F} = \frac{1}{m}</math>       and      <math>~\Lambda_0 = E - \frac{B}{m} \, .</math>
</div>
</td></tr></table>

Let's try <math>~m=2</math>:
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~ 2\cot(\Lambda - E) \cdot \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2\cot(\Lambda - E)\biggl[
\biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
+ (\Lambda^3 - \Lambda^5) \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2(\Lambda - \Lambda_0) [\cot^2(\Lambda - E) - 1]\biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4
- \Lambda^2
+ \Lambda^4 \cot^2(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 \biggl\{ 2(\Lambda - \Lambda_0 )\cot(\Lambda - E) \biggl[
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2(\Lambda - \Lambda_0)[\cot^2(\Lambda - E) - 1] \biggl[ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2\cot(\Lambda - E)\biggl[ \Lambda
- \Lambda^4 \cot(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~ \Rightarrow ~~~\cot(\Lambda - E) \cdot \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \Lambda_0 (\Lambda - \Lambda_0) \biggl[ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- (\Lambda - \Lambda_0) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4
- \Lambda^2
+ \Lambda^4 \cot^2(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\cot(\Lambda - E)\biggl[
\biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
+ (\Lambda^3 - \Lambda^5) \cot(\Lambda - E)
- \Lambda^5 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-\Lambda_0 (\Lambda - \Lambda_0 )\cot(\Lambda - E) \biggl[
(3-\Lambda^2)
- (\Lambda + \Lambda^3 ) \cot(\Lambda - E)
- \Lambda^2 \cot^2(\Lambda - E)
- \Lambda^3 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 \cot(\Lambda - E)\biggl[ \Lambda
- \Lambda^4 \cot(\Lambda - E)
- \Lambda^4 \cot^3(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ (\Lambda - \Lambda_0) \cot^2(\Lambda - E) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4
- \Lambda^2
+ \Lambda^4 \cot^2(\Lambda - E)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \Lambda_0 (\Lambda - \Lambda_0)\cot^2(\Lambda - E) \biggl[ \Lambda^3
-2\Lambda
+\Lambda^2\cot(\Lambda - E)
+ \Lambda^3 \cot^2(\Lambda - E) \biggr]
</math>
</td>
</tr>
</table>

======Second Argument Relationships Guess======
Let's go back up to the general expression,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Rightarrow~~~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
(\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)(\Lambda - \Lambda_0 )
- (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot(\Lambda - E)
- \Lambda^2 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
- \Lambda^3 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3
-4\Lambda
+2\Lambda^2 \cot(\Lambda - E)
+ 2\Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-Q \biggl\{ \Lambda^3(\Lambda - \Lambda_0 )
-2\Lambda (\Lambda - \Lambda_0 )
+\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E)
+ \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\} \, ,
</math>
</td>
</tr>
</table>
where,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[ 1 - \eta \cot(\eta - B) \biggr] \, .</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - \biggl[\frac{(\Lambda - \Lambda_0)}{g_\mathrm{F}}\biggr] \cot[ (\Lambda - \Lambda_0)/g_\mathrm{F} - B] \, . </math>
</td>
</tr>
</table>

Then, let's try setting,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\cot[ (\Lambda - \Lambda_0)/g_\mathrm{F} - B]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan(\Lambda - E)</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \cot\biggl\{ \frac{\pi}{2} - \biggl[ \frac{\pi}{2} + B - (\Lambda - \Lambda_0)/g_\mathrm{F} \biggr] \biggr\}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\tan(\Lambda - E)</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{\pi}{2} + B - \frac{ (\Lambda - \Lambda_0) }{g_\mathrm{F}} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\Lambda - E \, .</math>
</td>
</tr>
</table>
This will only work for all <math>~\Lambda</math> if, <math>~g_\mathrm{F} = -1</math>; in which case,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{\pi}{2} + B + (\Lambda - \Lambda_0) </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\Lambda - E </math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ E </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\Lambda_0 - \frac{\pi}{2} - B \, . </math>
</td>
</tr>
</table>
Hence, we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 + (\Lambda - \Lambda_0)\tan(\Lambda - E) </math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ Q \cot(\Lambda-E)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\cot(\Lambda-E) + (\Lambda - \Lambda_0) \, .</math>
</td>
</tr>
</table>

This means that,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\cot(\Lambda-E) \cdot \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\cot(\Lambda-E)(\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)(\Lambda - \Lambda_0 )
- (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot(\Lambda - E)
- \Lambda^2 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
- \Lambda^3 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \cot(\Lambda-E)(\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3
-4\Lambda
+2\Lambda^2 \cot(\Lambda - E)
+ 2\Lambda^3 \cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-[\cot(\Lambda-E) + (\Lambda - \Lambda_0)] \biggl\{ \Lambda^3(\Lambda - \Lambda_0 )
-2\Lambda (\Lambda - \Lambda_0 )
+\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E)
+ \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- [\cot(\Lambda-E) + (\Lambda - \Lambda_0)] \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
(\Lambda - \Lambda_0 ) \biggl\{
(3-\Lambda^2)(\Lambda - \Lambda_0 )\cot(\Lambda-E)
- (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot^2(\Lambda - E)
- \Lambda^2 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E)
- \Lambda^3 (\Lambda - \Lambda_0 )\cot^4(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3\cot(\Lambda-E)
-4\Lambda \cot(\Lambda-E)
+2\Lambda^2 \cot^2(\Lambda - E)
+ 2\Lambda^3 \cot^3(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-(\Lambda - \Lambda_0) \biggl\{ \Lambda^3(\Lambda - \Lambda_0 )
-2\Lambda (\Lambda - \Lambda_0 )
+\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E)
+ \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- (\Lambda - \Lambda_0) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-[\cot(\Lambda-E) ] \biggl\{ \Lambda^3(\Lambda - \Lambda_0 )
-2\Lambda (\Lambda - \Lambda_0 )
+\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E)
+ \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E)
+ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4
\biggr\} \, .
</math>
</td>
</tr>
</table>

====Attempt 4A====

Try, the ''structural'' function,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{1}{\xi} \biggl[A\sin\xi - B\cos\xi \biggr] </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{a_0}{\xi} \cdot \sin(\xi - b_0) \, ,</math>
</td>
</tr>
</table>
where, recognizing that, <math>~f \equiv B/A</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~a_0 \equiv A \sqrt{1 + f^2}</math>
</td>
<td align="center">
      and      
</td>
<td align="left">
<math>~b_0 \equiv \tan^{-1}f \, .</math>
</td>
</tr>
</table>

Does this ''structural'' function satisfy the Lane-Emden equation, namely …
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\xi^2 \frac{d^2\theta}{d\xi^2} + 2\xi \frac{d\theta}{d\xi} + \xi^2\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~0 \, .</math>
</td>
</tr>
</table>

First, recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a_0}{\xi} \cdot \cos(\xi - b_0)
-
\frac{a_0}{\xi^2} \cdot \sin(\xi - b_0) \, ,
</math>
</td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2\theta}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{a_0}{\xi} \cdot \sin(\xi - b_0)
-
\frac{a_0}{\xi^2} \cdot \cos(\xi - b_0)
-
\frac{a_0}{\xi^2} \cdot \cos(\xi - b_0)
+
\frac{2a_0}{\xi^3} \cdot \sin(\xi - b_0) \, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\xi^2 \frac{d^2\theta}{d\xi^2} + 2\xi \frac{d\theta}{d\xi} + \xi^2\theta
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\xi^2 \biggl\{
- \frac{a_0}{\xi} \cdot \sin(\xi - b_0)
-
\frac{a_0}{\xi^2} \cdot \cos(\xi - b_0)
-
\frac{a_0}{\xi^2} \cdot \cos(\xi - b_0)
+
\frac{2a_0}{\xi^3} \cdot \sin(\xi - b_0)
\biggr\}
+ 2\xi \biggl\{
\frac{a_0}{\xi} \cdot \cos(\xi - b_0)
-
\frac{a_0}{\xi^2} \cdot \sin(\xi - b_0)
\biggr\}
+ \xi^2\biggl\{ \frac{a_0}{\xi}\cdot \sin(\xi - b_0) \biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl\{
-a_0 \xi \cdot \sin(\xi - b_0)
-
2a_0 \cdot \cos(\xi - b_0)
+
\frac{2a_0}{\xi} \cdot \sin(\xi - b_0)
\biggr\}
+ \biggl\{
2a_0\cdot \cos(\xi - b_0)
-
\frac{2a_0}{\xi} \cdot \sin(\xi - b_0)
\biggr\}
+ a_0 \xi\cdot \sin(\xi - b_0)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
0 \, .
</math>
</td>
</tr>
</table>
Good!

====Attempt 4B====
Let's continue with the same '''4A''' approach, but shift to the variable notation that we have used in the [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|accompanying description of how to build an equilibrium, (n<sub>c</sub>, n<sub>e</sub>) = (5, 1) bipolytrope]]. For example, the envelope's ''structural function'' will be referred to as <math>~\phi(\eta)</math> instead of <math>~\theta(\xi)</math>. Specifically, we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\phi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~a_0 \biggl[ \frac{\sin(\eta - b_0)}{\eta} \biggr] \, ,</math>
</td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\phi}{d\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{a_0}{\eta^2} \biggl[ \eta \cos(\eta - b_0) - \sin(\eta - b_0) \biggr] \, ,</math>
</td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2\phi}{d\eta^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{a_0}{\eta} \cdot \sin(\eta - b_0)
-
\frac{2a_0}{\eta^2} \cdot \cos(\eta - b_0)
+
\frac{2a_0}{\eta^3} \cdot \sin(\eta - b_0) \, .
</math>
</td>
</tr>
</table>
This satisfies the Lane-Emden equation for any values of the parameter pair, <math>~a_0</math> and <math>~b_0</math>. Note that,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q \equiv - \frac{d\ln \phi}{d\ln\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{a_0}{\eta^2} \biggl[\sin(\eta - b_0) - \eta \cos(\eta - b_0) \biggr] \cdot \frac{\eta}{\phi}</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[1 - \eta \cot(\eta - b_0) \biggr] </math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~ \eta \cot(\eta - b_0) </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~(1 - Q ) \, .</math>
</td>
</tr>
</table>

<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left">
Note that for the <math>~\mu_e/\mu_c = 1</math> model having a core-envelope interface of <math>~\xi_i = 1.668646016</math> (and an adopted normalization, φ<sub>i</sub> = 1), we also have,
<table border="0" cellpadding="8" align="center">
<tr>
<td align="right"><math>~\theta_i =</math></td>
<td align="right">0.720165</td>
</tr>
<tr>
<td align="right"><math>~-\biggl(\frac{d\theta}{d\xi}\biggr)_i =</math></td>
<td align="right">0.207749</td>
</tr>
<tr>
<td align="right"><math>~\eta_i =</math></td>
<td align="right">1.498956</td>
</tr>
<tr>
<td align="right"><math>~-\biggl(\frac{d\phi}{d\eta}\biggr)_i =</math></td>
<td align="right">0.963393</td>
</tr>
<tr>
<td align="right"><math>~\Lambda_i =</math></td>
<td align="right">- 0.296262</td>
</tr>
<tr>
<td align="right"><math>~b_0 =</math></td>
<td align="right">- 0.359864 (or +2.78173)</td>
</tr>
<tr>
<td align="right"><math>~a_0 =</math></td>
<td align="right">1.563355</td>
</tr>
<tr>
<td align="right"><math>~\eta_s = \pi + b_0 =</math></td>
<td align="right">+2.78173 (or - 0.359864 = 5.92332)</td>
</tr>
</table>
Some of this is also detailed [[#Attempt_1|above in the Attempt 1 subsection]].
</td></tr></table>

Now, guided by a [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|separate parallel discussion]], we want to see whether, in the case of a bipolytropic configuration for which <math>n_e=1</math>, the
<table border="0" cellpadding="5" align="center">
<tr>
<td align="center" colspan="4"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
      and      
</td>
<td align="left">
<math>~x_P </math>
</td>
<td align="left">
<math>~ \equiv \frac{3c_0 (n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr] </math>
</td>
</tr>
<tr>
<td align="right" colspan="3">
 
</td>
<td align="left">
<math>~= -\biggl( \frac{3c_0}{\eta \phi}\biggr) \frac{d\phi}{d\eta} = \frac{3c_0}{\eta^2} \cdot Q \, , </math>
</td>
</tr>
</table>

satisfies the governing LAWE, namely,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{d^2x_P}{d\eta^2} + \biggl[4 - 2Q\biggr]\frac{1}{\eta}\cdot \frac{dx_P}{d\eta} - 2Q\cdot \frac{x_P}{\eta^2} \, .
</math>
</td>
</tr>
</table>

Well … recognizing that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d}{d\eta} \biggl[\cot(\eta - b_0)\biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \biggl[ 1 + \cot^2(\eta - b_0)\biggr] \, ,</math>
</td>
</tr>
</table>
we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dQ}{d\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\eta -\cot(\eta - b_0)
+\eta\cot^2(\eta - b_0)
\, ,
</math>
</td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2Q}{d\eta^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d}{d\eta}\biggl[
\eta -\cot(\eta - b_0)
+\eta\cot^2(\eta - b_0) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \biggl[ 1 + \cot^2(\eta - b_0)\biggr]
+\cot^2(\eta - b_0)
-2\eta \cot(\eta - b_0)\biggl[ 1 + \cot^2(\eta - b_0)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2
-2\eta \cot(\eta - b_0)
+ 2\cot^2(\eta - b_0)
-2\eta \cot^3(\eta - b_0)
\, .
</math>
</td>
</tr>
</table>

Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{3c_0}\cdot \frac{dx_P}{d\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{1}{\eta^2} \frac{dQ}{d\eta} - \frac{2Q}{\eta^3} </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{\eta^2}\biggl[\eta -\cot(\eta - b_0)
+\eta\cot^2(\eta - b_0)
\biggr]
- \frac{2Q}{\eta^3}
\, , </math>
</td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{3c_0}\cdot \frac{d^2x_P}{d\eta^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{\eta^2} \frac{d^2Q}{d\eta^2}
- \frac{2}{\eta^3} \frac{dQ}{d\eta}
+ \frac{6Q}{\eta^4}
- \frac{2}{\eta^3} \frac{dQ}{d\eta}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{\eta^2} \biggl[2
-2\eta \cot(\eta - b_0)
+ 2\cot^2(\eta - b_0)
-2\eta \cot^3(\eta - b_0)
\biggr]
- \frac{4}{\eta^3} \biggl[ \eta -\cot(\eta - b_0)
+\eta\cot^2(\eta - b_0)
\biggr]
+ \frac{6Q}{\eta^4} \, .
</math>
</td>
</tr>
</table>
<span id="proof">So the relevant LAWE becomes,</span>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( \frac{\eta^4}{3c_0} \biggr) \cdot</math> LAWE
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\eta^2 \biggl[2
-2\eta \cot(\eta - b_0)
+ 2\cot^2(\eta - b_0)
-2\eta \cot^3(\eta - b_0)
\biggr]
- 4\eta \biggl[ \eta -\cot(\eta - b_0)
+\eta\cot^2(\eta - b_0)
\biggr]
+ 6Q
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta^3 \biggl[4 - 2Q\biggr]\biggl\{ \frac{1}{\eta^2}\biggl[\eta -\cot(\eta - b_0)
+\eta\cot^2(\eta - b_0)
\biggr]
- \frac{2Q}{\eta^3}
\biggr\} - 2Q^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2\eta^2 \biggl[1
- \eta \cot(\eta - b_0)
+ \cot^2(\eta - b_0)
- \eta \cot^3(\eta - b_0)
\biggr]
- 4 \biggl[ \eta^2 - \eta\cot(\eta - b_0)
+\eta^2\cot^2(\eta - b_0)
\biggr]
+ 6Q
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 4\biggl[\eta^2 - \eta\cot(\eta - b_0) + \eta^2\cot^2(\eta - b_0) - 2Q \biggr]
- 2Q \biggl[\eta^2 - \eta\cot(\eta - b_0) + \eta^2\cot^2(\eta - b_0) - 2Q \biggr]
- 2Q^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2\eta^2 \biggl[Q
+ \cot^2(\eta - b_0)
- \eta \cot^3(\eta - b_0)
\biggr]
- 2Q
- 2Q \biggl[\eta^2 - \eta\cot(\eta - b_0) + \eta^2\cot^2(\eta - b_0) - 2Q \biggr]
- 2Q^2
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~\biggl( \frac{\eta^4}{6c_0} \biggr) \cdot</math> LAWE
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\eta^2 \cot^2(\eta - b_0)
- \eta^3 \cot^3(\eta - b_0)
\biggr]
+ \biggl[ 1-\eta\cot(\eta - b_0)\biggr] \biggl[\eta\cot(\eta - b_0) - \eta^2\cot^2(\eta - b_0) \biggr]
- Q(1-Q)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\eta\cot(\eta - b_0) - \eta^2\cot^2(\eta - b_0) \biggr]
- Q(1-Q)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
0\, .
</math>
</td>
</tr>
</table>

====Attempt 5====

=====New Strategy=====
In the vast majority of our prior attempts to derive an analytic expression for the envelope's eigenvector, we have started with the presumption — as voiced by [http://adsabs.harvard.edu/abs/1988Ap%26SS.147..219B Beech (1988)] and repeated in [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|our accompanying derivation of the structure of the <math>~(n_c, n_e) = (5, 1)</math> bipolytrope's structure]] — that the most general solution to the n = 1 Lane-Emden equation can be written in the form,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\phi</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, .</math>
</td>
</tr>
</table>
But, as we have [[SSC/Structure/Polytropes#Primary_E-Type_Solution|emphasized in a separate context]], another expression that satisfies the relevant Lane-Emden equation has the form,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\phi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- A \biggl[ \frac{\cos(\eta - B)}{\eta} \biggr] \, .</math>
</td>
</tr>
</table>

<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
We fully appreciate that for appropriately chosen and ''different'' values of the parameter, <math>~B</math>, these two functions can be made equal to one another. But, for now, let's work through our analysis pretending that they are different functions.
</td></tr></table>

This alternate "cosine" expression was not the solution of choice when we were seeking a mathematical description of the structure of an ''isolated'', n = 1 polytrope because it does not satisfy the relevant central boundary conditions. However, it occurs to us that this alternate expression might work in the context we are considering now <font color="red">[20 April 2019]</font>, which deals with properties of the envelope of a bipolytrope. In this case,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q \equiv - \frac{d\ln\phi}{d\ln\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~A \biggl[\frac{\cos(\eta-B)}{\eta^2} + \frac{\sin(\eta-B)}{\eta} \biggr]\cdot \frac{\eta^2}{A\cos(\eta-B)}</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[\cos(\eta-B) + \eta\sin(\eta-B) \biggr]\cdot \frac{1}{\cos(\eta-B)}</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 + \eta\tan(\eta-B) \, ;</math>
</td>
</tr>
</table>
and, following along the lines of our earlier '''[[#Attempt_4B|Attempt 4B]]''' discussion, a reasonable ''guess'' for the dimensionless displacement function is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_P</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1 Q}{\eta^2}</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] \, .</math>
</td>
</tr>
</table>

What are the first and second derivatives of this trial eigenfunction?

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dx_P}{d\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr]
- \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3c_1}{\eta^3} \biggl[\frac{\eta^2}{\cos^2(\eta-B)} -2 - \eta\tan(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]\, .
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{d^2x_P}{d\eta^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
3c_1 \frac{d}{d\eta}\biggl[\eta^{-3}\cos^{-2}(\eta-B) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{3c_1}{\eta^3\cos^2(\eta-B)}\cdot \frac{d}{d\eta}\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
3c_1 \biggl[2\eta^{-3}\sin(\eta-B)\cos^{-3}(\eta-B) -3\eta^{-4}\cos^{-2}(\eta-B) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[2\eta + 4\sin(\eta-B)\cos(\eta-B)
- \sin(\eta-B)\cos(\eta-B)- \eta \cos^2(\eta-B)+ \eta\sin^2(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)}
\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
\cdot \biggl[2\eta\sin(\eta-B) -3\cos(\eta-B) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[2\eta^2\cos(\eta-B) + 4\eta \sin(\eta-B)\cos^2(\eta-B)
- \eta \sin(\eta-B)\cos^2(\eta-B)- \eta^2 \cos^3(\eta-B)+ \eta^2\sin^2(\eta-B)\cos(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{
2\eta^3 \sin(\eta-B) - 4\eta\sin(\eta-B)\cos^2(\eta-B) - 2\eta^2 \sin^2(\eta-B)\cos(\eta-B)
</math>
</td>
</tr>

<tr>
<td align="right">
  </td>
<td align="center">
 
</td>
<td align="left">
<math>~
-3\eta^2 \cos(\eta-B) + 6\cos^3(\eta-B) +3 \eta\sin(\eta-B)\cos^2(\eta-B)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta \biggl[ 4\sin(\eta-B)\cos^2(\eta-B) - \sin(\eta-B)\cos^2(\eta-B)\biggr]
+ \eta^2 \biggl[ 2\cos(\eta-B) - \cos^3(\eta-B)+ \sin^2(\eta-B)\cos(\eta-B)\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{6\cos^3(\eta-B)
+ \eta^3 \biggl[2\sin(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta \biggl[ 4\sin(\eta-B)\cos^2(\eta-B) - \sin(\eta-B)\cos^2(\eta-B) - 4\sin(\eta-B)\cos^2(\eta-B) +3 \sin(\eta-B)\cos^2(\eta-B) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \eta^2 \biggl[ 2\cos(\eta-B) - \cos^3(\eta-B)+ \sin^2(\eta-B)\cos(\eta-B) - 2 \sin^2(\eta-B)\cos(\eta-B) -3 \cos(\eta-B) \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B)
+ 2\eta \sin(\eta-B)\cos^2(\eta-B)
- 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B)
\biggr] \, .
</math>
</td>
</tr>
</table>

=====Illustration=====

<table border="1" align="center" width="710px" cellpadding="8">
<tr><td align="center">
[[File:Frame06good.png|700px|Attempt5 Trial Eigenfunction]]
</td></tr>
<tr><td align="left">
''Above & Below:''  The solid, light-blue circular markers trace how the function, <math>~x_P(\eta)</math>, varies over the "radial" range, <math>~- \pi \le \eta \le \pi</math>, when <math>~c_1 = 1</math>, for various values of the parameter, <math>~B</math>; in the above panel as well as in each frame of the animation, the chosen value of <math>~B</math> is recorded in the upper-right corner of the image. (These values are also recorded in the table immediately below the animation.) The red vertical dashed line segment identifies the value of <math>~\eta</math> at which the argument of the tangent function goes to <math>~\tfrac{\pi}{2}</math> and, hence, where the function <math>~x_P(\eta)</math> flips discontinuously from plus- to minus-infinity. As explained further, below, the solid purple curve shows how the x<sub>P</sub>-''intercept'' function varies with <math>~\eta</math>; as defined, this curve is independent of the parameter, <math>~B</math>, so it is unchanging in the animation sequence. The single larger yellow-circular marker (with a black border) shows where this "intercept" curve intersects the <math>~x_P</math> function and, therefore, where along this trial eigenfunction <math>~d\ln x_P/d\ln\eta = -1</math>; the coordinates (abscissa & ordinate) of this yellow marker are recorded in the accompanying table, for each illustrative value of <math>~B</math>.
</td></tr>
<tr><td align="center">
[[File:TangentTrialAnalytic51movie.gif|700px|Animation of Attempt5 Trial Eigenfunction]]
</td></tr>
<tr><td align="center">
<table border="0" cellpadding="8">
<tr>
<th align="center" colspan="2"> </th>
<th align="center" colspan="2"><math>~\frac{d\ln x_P}{d\ln\eta} = -1</math><br /><hr /></th>
</tr>
<tr>
<th align="center">Animation<br />Frame</th>
<th align="center"><math>~B</math></th>
<th align="center">Abscissa<br /><math>~\eta</math></th>
<th align="center">Ordinate<br /><math>~x_P(\eta)</math></th>
</tr>
<tr>
<td align="center"><math>~1</math></td>
<td align="center"><math>~1.02 \times \tfrac{\pi}{2} = 1.6022</math></td>
<td align="center"><math>~0.014</math></td>
<td align="center"><math>~27,165</math></td>
</tr>
<tr>
<td align="center"><math>~2</math></td>
<td align="center"><math>~1.2 \times \tfrac{\pi}{2} = 1.8850</math></td>
<td align="center"><math>~0.157</math></td>
<td align="center"><math>~242.7</math></td>
</tr>
<tr>
<td align="center"><math>~3</math></td>
<td align="center"><math>~1.4 \times \tfrac{\pi}{2} = 2.1991</math></td>
<td align="center"><math>~0.311</math></td>
<td align="center"><math>~60.31</math></td>
</tr>
<tr>
<td align="center"><math>~4</math></td>
<td align="center"><math>~1.6 \times \tfrac{\pi}{2} = 2.5133</math></td>
<td align="center"><math>~0.462</math></td>
<td align="center"><math>~26.52</math></td>
</tr>
<tr>
<td align="center"><math>~5</math></td>
<td align="center"><math>~1.8 \times \tfrac{\pi}{2} = 2.8274</math></td>
<td align="center"><math>~0.606</math></td>
<td align="center"><math>~14.68</math></td>
</tr>
<tr>
<td align="center"><math>~6</math></td>
<td align="center"><math>~2 \times \tfrac{\pi}{2} = 3.1416</math></td>
<td align="center"><math>~0.739</math></td>
<td align="center"><math>~9.187</math></td>
</tr>
<tr>
<td align="center"><math>~7</math></td>
<td align="center"><math>~0.2 \times \tfrac{\pi}{2} = 0.3142</math></td>
<td align="center"><math>~0.856</math></td>
<td align="center"><math>~6.200</math></td>
</tr>
<tr>
<td align="center"><math>~8</math></td>
<td align="center"><math>~0.4 \times \tfrac{\pi}{2} = 0.6283</math></td>
<td align="center"><math>~0.949</math></td>
<td align="center"><math>~4.381</math></td>
</tr>
<tr>
<td align="center"><math>~9</math></td>
<td align="center"><math>~0.5 \times \tfrac{\pi}{2} = 0.7854</math></td>
<td align="center"><math>~0.981</math></td>
<td align="center"><math>~3.724</math></td>
</tr>
<tr>
<td align="center"><math>~10</math></td>
<td align="center"><math>~0.5 \times \tfrac{\pi}{2} = 0.9425</math></td>
<td align="center"><math>~0.998</math></td>
<td align="center"><math>~3.178</math></td>
</tr>
<tr>
<td align="center"><math>~11</math></td>
<td align="center"><math>~0.8 \times \tfrac{\pi}{2} = 1.2566</math></td>
<td align="center"><math>~0.955</math></td>
<td align="center"><math>~2.313</math></td>
</tr>
<tr>
<td align="center"><math>~12</math></td>
<td align="center"><math>~0.9 \times \tfrac{\pi}{2} = 1.4137</math></td>
<td align="center"><math>~0.840</math></td>
<td align="center"><math>~1.944</math></td>
</tr>
<tr>
<td align="center"><math>~13</math></td>
<td align="center"><math>~0.98 \times \tfrac{\pi}{2} = 1.5394</math></td>
<td align="center"><math>~0.545</math></td>
<td align="center"><math>~1.632</math></td>
</tr>
</table>
</td></tr>
</table>

What is the expression for the logarithmic derivative of this eigenfunction guess?

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\ln x_P}{d\ln\eta} = \frac{\eta}{x_P} \cdot \frac{dx_P}{d\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{\eta^3}{3c_1} \biggl[1 + \eta\tan(\eta-B)\biggr]^{-1} \biggl\{
\frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr]
- \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{\cos (\eta-B)}{\cos(\eta-B) + \eta\sin(\eta-B)}\biggr] \biggl\{
\frac{\eta}{\cos^2(\eta-B)} \biggl[\sin(\eta - B)\cos(\eta-B) +\eta\biggr]
- \frac{2}{\cos^2(\eta-B)} \biggl[\cos^2(\eta-B) + \eta\sin(\eta-B)\cos(\eta-B)\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{\eta^2 - \eta\sin(\eta - B)\cos(\eta-B) - 2\cos^2(\eta-B)}{\cos^2(\eta-B) + \eta\sin(\eta-B) \cos(\eta-B)} \, .
</math>
</td>
</tr>
</table>

At the surface of the bipolytropic configuration — that is, presumably when <math>~\eta = \eta_s</math> — we must find that this logarithmic derivative is negative one. So, for a given value of the parameter, <math>~B</math>, what is the value of <math>~\eta_s</math>? Well …

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\ln x_P}{d\ln\eta}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~-1</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ -\cos^2(\eta-B) - \eta\sin(\eta-B) \cos(\eta-B)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\eta^2 - \eta\sin(\eta - B)\cos(\eta-B) - 2\cos^2(\eta-B)</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ 0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\eta^2 - \cos^2(\eta-B)</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \eta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\pm \cos(\eta-B)\, .</math>
</td>
</tr>
</table>

Now, given that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\tan(\eta-B) </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\pm \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \, ,</math>
</td>
</tr>
</table>
we see that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x_P\biggr|_\mathrm{intercept}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \biggr\} </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \biggr\} </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\eta^2}}{\eta} \biggr] \biggr\} </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \sqrt{1-\eta^2}\biggr\} \, .</math>
</td>
</tr>
</table>

The solid-purple, x<sub>P</sub>-''intercept'' curve that appears in the above figure/animation is defined by this function. Notice that this function never exceeds unity. This presumably means that if the tangent-based <math>~x_P</math> eigenfunction is the correct solution to the envelope's LAWE, then the dimensionless radius, <math>~\eta_s</math>, of the bipolytrope must never exceed unity.

The following diagram — the original of which appears in our [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying discussion of the equilibrium properties of bipolytropic configurations having]] <math>~(n_c, n_e) = (5, 1)</math> — shows how <math>~\eta_i</math> (purple curve) and <math>~\eta_s</math> (green curve) vary with the interface location <math>~\xi_i</math> (ordinate). The solid yellow circular markers (with black edges) identify where the logarithmic derivative of the dimensionless displacement function, <math>~d\ln x_P/d\ln\eta</math>, equals negative one. If <math>~x_P</math> is the correct eigenfunction for the marginally unstable bipolytropic configuration, one of these yellow circular markers should coincide with the green curve, that is, it should be associated with the configuration's surface. Since the curve identified by the yellow circular markers does not appear to intersect the green curve, we conclude that we have not yet identified the correct eigenfunction.

<div align="center">
[[File:Bipolytrope51Boundaries03b.png|500px|Bipolytrope Properties]]
</div>

=====Is This Compatible With LAWE=====

In an effort to track the two <math>~Q(\eta)</math> functions separately, we will add a subscript zero to the one that applies to the ''structural'' properties of the underlying equilibrium configuration. Again, we will be focused on finding a solution in the case where <math>~\sigma_c^2 = 0</math> and <math>~n = 1</math>, that is — see also our [[#Envelope:|above discussion]] — the relevant envelope LAWE is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2\alpha_g Q_0 \cdot \frac{x_P}{\eta^2} \, ,
</math>
</td>
</tr>
</table>
where, drawing from our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|discussion of the n = 1 envelope's equilibrium structure]],
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[ - \frac{d\ln\phi}{d\ln\eta} \biggr]_\mathrm{n=1} = - \frac{\eta}{\phi}\biggl[ \frac{d\phi}{d\eta} \biggr]_\mathrm{n=1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \biggl[ \frac{1}{\sin(\eta - B_0)} \biggr]\biggl[ \eta\cos(\eta-B_0) - \sin(\eta-B_0) \biggr]</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[1 - \eta\cot(\eta-B_0) \biggr] \, .</math>
</td>
</tr>
</table>

Hence, after recognizing that for this specific case, <math>~\alpha_g = +1</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
LAWE
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2 Q_0 \cdot \frac{x_P}{\eta^2}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~</math>  LAWE
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B)
+ 2\eta \sin(\eta-B)\cos^2(\eta-B)
- 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
- \frac{2 Q_0}{\eta^2} \cdot \frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>  LAWE
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 3\cos^3(\eta-B)
+ \eta \sin(\eta-B)\cos^2(\eta-B)
- \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ 2 - Q_0 \biggr]\cos(\eta - B)\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]
- Q_0 \cos^3(\eta - B) \biggl[1 + \eta\tan(\eta-B)\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
3\cos^3(\eta-B)
+ \eta \sin(\eta-B)\cos^2(\eta-B)
- \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
-4\cos^3(\eta-B) - 2\eta\sin(\eta-B)\cos^2(\eta-B) + 2\eta^2\cos(\eta - B)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- Q_0\biggl\{ \biggl[\eta^2\cos(\eta - B) -2\cos^3(\eta-B) - \eta\sin(\eta-B)\cos^2(\eta-B)\biggr]
+ \biggl[\cos^3(\eta - B) + \eta\sin(\eta-B)\cos^2(\eta - B)\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
-\cos^3(\eta-B)
- \eta \sin(\eta-B)\cos^2(\eta-B)
+ \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
+ Q_0\biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta - B) \biggr]
</math>
</td>
</tr>
</table>

<table border="1" width="85%" cellpadding="10" align="center"><tr><td align="left">
<font color="red">'''Quick Check:'''</font>   Now, if we set <math>~Q_0 = 1 + \eta\tan(\eta-B)</math>, these RHS terms should sum to zero. Let's check.

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>  LAWE
</td>
<td align="center">
<math>~\rightarrow</math>
</td>
<td align="left">
<math>~
-\cos^3(\eta-B)
- \eta \sin(\eta-B)\cos^2(\eta-B)
+ \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\cos(\eta - B)+ \eta\sin(\eta-B) \biggr] \cdot \biggl[ \cos^2(\eta-B) - \eta^2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
0 \, .
</math>
</td>
</tr>
</table>

Excellent!
</td></tr></table>

Now, let's plug in the expression for the ''structural'' <math>~Q_0</math>. Specifically, we want,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~Q_0</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~\biggl[ 1 - \eta\cot(\eta-B_0) \biggr]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~\biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr] \, ,</math>
</td>
</tr>
</table>
in which case we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>  LAWE
</td>
<td align="center">
<math>~\rightarrow</math>
</td>
<td align="left">
<math>~
-\cos^3(\eta-B)
- \eta \sin(\eta-B)\cos^2(\eta-B)
+ \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr] \cdot \biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta-B) \biggr] \, .
</math>
</td>
</tr>
</table>
As we have just shown, above, in the context of a "<font color="red">'''Quick Check'''</font>", the expression on the RHS will go to zero if we adopt the transformation, <math>~[B_0 + \tfrac{\pi}{2} \mp m\pi] \rightarrow B</math>. Does this help shift the coordinate, <math>~\eta</math>?

====Attempt 6 (5 November 2026)====

=See Also=

* [http://adsabs.harvard.edu/abs/2018Sci...362..201D K. De et al. (12 October 2018, Science, Vol. 362, No. 6411, pp. 201 - 206)], ''A Hot and Fast Ultra-stripped Supernova that likely formed a Compact Neutron Star Binary.''

{{ SGFfooter }}

SSC/Structure/BiPolytropes/Analytic51/Pt2

2026-11-06T00:29:07Z

Joel2: /* UV Plane */

__FORCETOC__
=BiPolytrope with n<sub>c</sub> = 5 and n<sub>e</sub> = 1 (Pt 2)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
<td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51|Part I:  (n<sub>c</sub>,n<sub>e</sub>) = (5,1) BiPolytrope]]
 
</td>
<td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt2|Part II:  Example Models]]
 
</td>
<td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt3|Part III:  Limiting Mass]]
 
</td>
<td align="center" bgcolor="lightblue"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt4|Part IV:  Free Energy]]
 
</td>
</tr>
</table>

==Examples==

===Normalization===
The dimensionless variables used in Tables 1 & 2 are defined as follows:
<div align="center">
<table border="0" cellpadding="3">
<tr>
<td align="right">
<math>\rho^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{\rho}{\rho_0}</math>
</td>

<td align="center">;    </td>

<td align="right">
<math>r^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{r}{[K_c^{1/2}/(G^{1/2}\rho_0^{2/5})]}</math>
</td>
</tr>

<tr>
<td align="right">
<math>P^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{P}{K_c\rho_0^{6/5}}</math>
</td>

<td align="center">;    </td>

<td align="right">
<math>M_r^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{M_r}{[K_c^{3/2}/(G^{3/2}\rho_0^{1/5})]}</math>
</td>
</tr>

<tr>
<td align="right">
<math>H^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{H}{K_c\rho_0^{1/5}}</math>
</td>

<td align="center">.    </td>

<td align="right" colspan="3">
 
</td>
</tr>

</table>
</div>

===Parameter Values===

The <math>2^\mathrm{nd}</math> column of Table 1 catalogues the analytic expressions that define various parameters and physical properties (as identified, respectively, in column 1) of the <math>(n_c, n_e) = (5, 1)</math> bipolytrope. We have evaluated these expressions for various choices of the dimensionless interface radius, <math>\xi_i</math>, and have tabulated the results in the last few columns of the table. The tabulated values have been derived assuming <math>\mu_e/\mu_c = 1</math>, that is, assuming that the core and the envelope have the same mean molecular weights.


<div align="center">
<b>Table 1: Properties of <math>(n_c, n_e) = (5, 1)</math> BiPolytrope Having Various Interface Locations, <math>\xi_i</math></b><br>
[[File:BiPolytropeParametersV01.xml|Accompanying spreadsheet with parameter values]]

<table border="1" cellpadding="5" width="80%">
<tr>
<td align="center">
Parameter
</td>
<td align="center">
<math>\xi_i</math>
</td>
<td align="center">
0.5
</td>
<td align="center">
1.0
</td>
<td align="center">
1.66864602<sup>&dagger;</sup>
</td>
<td align="center">
3.0
</td>
<td rowspan="21">
[[File:Bipolytrope51Boundaries02.png|500px|Examples]]<br />
<table border="0" align="left" cellpadding="10"><tr><td align="left">
For bipolytropic models having <math>\mu_e/\mu_c = 1.0</math>, this figure shows how the interface location, <math>\eta_i</math> (solid purple curve), the surface radius, <math>\eta_s</math> (green circular markers), and the parameter, <math>\tan^{-1}\Lambda_i</math> (orange circular markers), vary with <math>\xi_i</math> (ordinate) over the range, <math>0 \le \xi_i \le 12</math>. The three horizontal, red-dashed line segments identify the values of <math>\xi_i</math> for which numerical values of these (and other) parameters have been listed in the table shown here on the left.</td></tr></table>
</td>
</tr>

<tr>
<td align="center">
<math>\theta_i</math>
</td>
<td align="center">
<math>\biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-1/2}</math>
</td>
<td align="center">
0.96077
</td>

<td align="center">
0.86603
</td>
<td align="center">
0.72016538
</td>
<td align="center">
0.50000
</td>
</tr>

<tr>
<td align="center">
<math>-\biggl(\frac{d\theta_i}{d\xi}\biggr)_i</math>
</td>
<td align="center">
<math>\frac{1}{3} \xi_i \biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-3/2}</math>
</td>
<td align="center">
0.14781
</td>

<td align="center">
0.21651
</td>
<td align="center">
0.20774935
</td>
<td align="center">
0.12500
</td>
</tr>

<tr>
<td align="center">
<math>r^*_\mathrm{core} \equiv r^*_i</math>
</td>
<td align="center">
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i</math>
</td>
<td align="center">
0.34549
</td>

<td align="center">
0.69099
</td>
<td align="center">
1.15301487
</td>
<td align="center">
2.07297
</td>
</tr>

<tr>
<td align="center">
<math>\rho^*_i \biggr|_c = \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \rho^*_i \biggr|_e</math>
</td>
<td align="center">
<math>\biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-5/2}</math>
</td>
<td align="center">
0.81864
</td>

<td align="center">
0.48714
</td>
<td align="center">
0.19371408
</td>
<td align="center">
0.03125
</td>
</tr>

<tr>
<td align="center">
<math>P^*_i</math>
</td>
<td align="center">
<math>\biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-3}</math>
</td>
<td align="center">
0.78653
</td>

<td align="center">
0.42188
</td>
<td align="center">
0.13950617
</td>
<td align="center">
0.01563
</td>
</tr>

<tr>
<td align="center">
<math>H^*_i \biggr|_c = \frac{n_c+1}{n_e+1} \biggl( \frac{\mu_e}{\mu_c} \biggr) H^*_i \biggr|_e</math>
</td>
<td align="center">
<math>6 \biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-1/2}</math>
</td>
<td align="center">
5.76461
</td>

<td align="center">
5.19615
</td>
<td align="center">
4.32099225
</td>
<td align="center">
3.00000
</td>
</tr>

<tr>
<td align="center">
<math>M^*_\mathrm{core}</math>
</td>
<td align="center">
<math>\biggl( \frac{6}{\pi}\biggr)^{1/2} (\xi_i \theta_i)^3</math>
</td>
<td align="center">
0.15320
</td>

<td align="center">
0.89762
</td>
<td align="center">
2.39822567
</td>
<td align="center">
4.66417
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}\eta_i</math>
</td>
<td align="center">
<math>\sqrt{3} ~\theta_i^2 \xi_i</math>
</td>
<td align="center">
0.79941
</td>

<td align="center">
1.29904
</td>
<td align="center">
1.49895749
</td>
<td align="center">
1.29904
</td>
</tr>

<tr>
<td align="center">
<math>-\biggl( \frac{d\phi}{d\eta} \biggr)_i</math>
</td>
<td align="center">
<math>\sqrt{3} ~\theta_i^{-3} \biggl( - \frac{d\theta}{d\xi} \biggr)_i = \frac{\xi_i}{\sqrt{3}}</math>
</td>
<td align="center">
0.28868
</td>

<td align="center">
0.57735
</td>
<td align="center">
0.96339323
</td>
<td align="center">
1.73205
</td>
</tr>

<tr>
<td align="center">
<math>\Lambda_i</math>
</td>
<td align="center">
<math>\frac{1}{\eta_i} + \biggl( \frac{d\phi}{d\eta} \biggr)_i</math>
</td>
<td align="center">
0.96225
</td>

<td align="center">
0.19245
</td>
<td align="center">
-  0.2962629
</td>
<td align="center">
-0.96225
</td>
</tr>

<tr>
<td align="center">
<math>A</math>
</td>
<td align="center">
<math>\eta_i (1 + \Lambda_i^2)^{1/2}</math>
</td>
<td align="center">
1.10940
</td>

<td align="center">
1.32288
</td>
<td align="center">
1.56335712
</td>
<td align="center">
1.80278
</td>
</tr>

<tr>
<td align="center">
<math>B</math>
</td>
<td align="center">
<math>\eta_i - \frac{\pi}{2} + \tan^{-1}( \Lambda_i)</math>
</td>
<td align="center">
- 0.00523
</td>

<td align="center">
-0.08163
</td>
<td align="center">
-0.359863583
</td>
<td align="center">
-1.03792
</td>
</tr>

<tr>
<td align="center">
<math>\eta_s</math>
</td>
<td align="center">
<math>\pi + B</math>
</td>
<td align="center">
3.13637
</td>

<td align="center">
3.05996
</td>
<td align="center">
2.781729071
</td>
<td align="center">
2.10367
</td>
</tr>

<tr>
<td align="center">
<math>- \biggl( \frac{d\phi}{d\eta} \biggr)_s</math>
</td>
<td align="center">
<math>\frac{A}{\eta_s}</math>
</td>
<td align="center">
0.35372
</td>

<td align="center">
0.43232
</td>
<td align="center">
0.562009126
</td>
<td align="center">
0.85697
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \cdot \biggl[ R^* \equiv r^*_s \biggr]</math>
</td>
<td align="center">
<math>\frac{\eta_s}{\sqrt{2\pi} ~\theta_i^2}</math>
</td>
<td align="center">
1.35550
</td>

<td align="center">
1.62766
</td>
<td align="center">
2.139737125
</td>
<td align="center">
3.35697
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^2 M^*_\mathrm{tot}</math>
</td>
<td align="center">
<math>\biggl(\frac{2}{\pi}\biggr)^{1/2} \theta_i^{-1} \biggl( -\eta^2 \frac{d\phi}{d\eta} \biggr)_s = \biggl(\frac{2}{\pi}\biggr)^{1/2} \frac{A\eta_s}{\theta_i}</math>
</td>
<td align="center">
2.88959
</td>

<td align="center">
3.72945
</td>
<td align="center">
4.818155932
</td>
<td align="center">
6.05187
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\rho_c}{\bar\rho}</math>
</td>
<td align="center">
<math>~\frac{\eta_s^2}{3A\theta_i^5}</math>
</td>
<td align="center">
3.61035
</td>

<td align="center">
4.84326
</td>
<td align="center">
8.517046605
</td>
<td align="center">
26.1844
</td>
</tr>

<tr>
<td align="center">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \cdot \biggl[ \nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}} \biggr]</math>
</td>
<td align="center">
<math>\sqrt{3} ~\biggl( \frac{\xi_i^3 \theta_i^4}{A\eta_s} \biggr)</math>
</td>
<td align="center">
0.05302
</td>

<td align="center">
0.24068
</td>
<td align="center">
0.497747627
</td>
<td align="center">
0.77070
</td>
</tr>

<tr>
<td align="center">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} \cdot \biggl[ q \equiv \frac{r_\mathrm{core}}{R} \biggr]</math>
</td>
<td align="center">
<math>\sqrt{3}~\biggl[\frac{\xi_i \theta_i^2}{\eta_s}\biggr]</math>
</td>
<td align="center">
0.25488
</td>

<td align="center">
0.42453
</td>
<td align="center">
0.538858190
</td>
<td align="center">
0.61751
</td>
</tr>

<tr>
<td align="left" colspan="6">
<sup>&dagger;</sup>This choice of the value of <math>\xi_i = 1.66864602</math> is motivated by our [[SSC/Stability/BiPolytropes/Pt3#Fundamental_Modes|discussion of the fundamental mode of oscillation]] of the marginally unstable model that has <math>\mu_e/\mu_c = 1.0</math>; see especially row 1 of Table 2 in that associated chapter.
</tr>

</table>
</div>


Alternatively, if given <math>\mu_e/\mu_c</math> and the value of the parameter, <math>\eta_i</math>, then we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(\frac{\mu_e}{\mu_c} \biggr)^{-1}\eta_i</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\frac{3^{3/2}\xi_i}{3 + \xi_i^2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ 0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\xi_i^2 - \biggl[ \biggl(\frac{\mu_e}{\mu_c} \biggr) \frac{3^{3/2}}{\eta_i }\biggr] \xi_i + 3</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~\xi_i </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{3}
\biggl(\frac{\mu_e}{\mu_c} \biggr) \frac{3}{2\eta_i } \biggl\{ 1 \pm
\sqrt{1 - \biggl[ \biggl(\frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{2 \eta_i }{3}\biggr]^2 }
\biggr\} \, .
</math>
</td>
</tr>
</table>

It must be understood, therefore, that the interface location is restricted to the range,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0</math>
</td>
<td align="center">
<math>~\le \eta_i \le</math>
</td>
<td align="left">
<math>~\frac{3}{2}\biggl(\frac{\mu_e}{\mu_c} \biggr)\, ,</math>
</td>
</tr>
</table>

and that this upper limit on <math>\eta_i</math> is associated with a model whose core radius is, <math>\xi_i = \sqrt{3}</math>. Also,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\Lambda_i </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{1}{\eta_i} -
\biggl(\frac{\mu_e}{\mu_c} \biggr) \frac{3}{2\eta_i } \biggl\{ 1 \pm
\sqrt{1 - \biggl[ \biggl(\frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{2 \eta_i }{3}\biggr]^2 }
\biggr\} \, .
</math>
</td>
</tr>
</table>

===Profile===

Once the values of the key set of parameters have been determined as illustrated in Table 1, the radial profile of various physical variables can be determined throughout the bipolytrope as detailed in [[#Steps_2_.26_3|step #4]] and [[#Step_7|step #8]], above. Table 2 summarizes the mathematical expressions that define the profile throughout the core (column 2) and throughout the envelope (column 3) of the normalized mass density, <math>\rho^*(r^*)</math>, the normalized gas pressure, <math>P^*(r^*)</math>, and the normalized mass interior to <math>r^*</math>, <math>M_r^*(r^*)</math>. For all profiles, the relevant normalized radial coordinate is <math>r^*</math>, as defined in the <math>2^\mathrm{nd}</math> row of Table 2. Graphical illustrations of these resulting profiles can be viewed by clicking on the thumbnail images posted in the last few columns of Table 2.

<div align="center">
<b>Table 2: Radial Profile of Various Physical Variables</b>
<table border="1" cellpadding="6">
<tr>
<td align="center" rowspan="2">
Variable
</td>
<td align="center" rowspan="2">
Throughout the Core<br>
<math>0 \le \xi \le \xi_i</math>
</td>
<td align="center" rowspan="2">
Throughout the Envelope<sup>&dagger;</sup><br>
<math>\eta_i \le \eta \le \eta_s</math>
</td>

<td align="center" colspan="3">
Plotted Profiles
</td>
</tr>

<tr>
<td align="center">
<math>\xi_i = 0.5</math>
</td>
<td align="center">
<math>\xi_i = 1.0</math>
</td>
<td align="center">
<math>\xi_i = 3.0</math>
</td>
</tr>

<tr>
<td align="center">
<math>r^*</math>
</td>
<td align="center">
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi</math>
</td>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta</math>
</td>

<td align="center" colspan="3">
 
</td>
</tr>

<tr>
<td align="center">
<math>\rho^*</math>
</td>
<td align="center">
<math>\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}</math>
</td>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \phi(\eta)</math>
</td>

<td align="center">

[[Image:DenXi05.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:DenXi10.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:DenXi30.jpg|thumb|75px]]
</td>
</tr>

<tr>
<td align="center">
<math>P^*</math>
</td>
<td align="center">
<math>\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3}</math>
</td>
<td align="center">
<math>\theta^{6}_i [\phi(\eta)]^{2}</math>
</td>

<td align="center">

[[Image:PresXi05.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:PresXi10.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:PresXi30.jpg|thumb|75px]]
</td>
</tr>

<tr>
<td align="center">
<math>M_r^*</math>
</td>
<td align="center">
<math>\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]</math>
</td>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>
</td>

<td align="center">

[[Image:MassXi05.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:MassXi10.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:MassXi30.jpg|thumb|75px]]
</td>
</tr>

<tr>
<td align="left" colspan="6">
<sup>&dagger;</sup>In order to obtain the various envelope profiles, it is necessary to evaluate <math>\phi(\eta)</math> and its first derivative using the information presented in Step 6, above.
</td>
</tr>
</table>
</div>

[As of 28 April 2013] For the interface locations <math>\xi_i = 0.5, 1.0,\mathrm{and}~3.0</math>, Table 2 provides profiles for three values of the molecular weight ratio: <math>\mu_e/\mu_c = 1.0, 1/2,\mathrm{and}~1/4</math>. In all nine graphs, blue diamonds trace the structure of the <math>n_c=5</math> core; the core extends to a radius, <math>r^*_\mathrm{core}</math>, that is independent of molecular weight ratio but varies in direct proportion to the choice of <math>\xi_i</math>. Specifically, as tabulated in the fourth row of Table 1, <math>r^*_\mathrm{core} = 0.34549, ~0.69099, \mathrm{and} ~2.07297</math> for, respectively, <math>\xi_i = 0.5,~1,~\mathrm{and}~3</math>. Notice that, while the pressure profile and mass profile are continuous at the interface for all choices of the molecular weight ratio, the density profile exhibits a discontinuous jump that is in direct proportion to the chosen value of <math>\mu_e/\mu_c</math>.

Throughout the <math>n_e = 1</math> envelope, the profile of all physical variables varies with the choice of the molecular weight ratio. In the Table 2 graphs, red squares trace the envelope profile for <math>\mu_e/\mu_c = 1.0</math>; green triangles trace the envelope profile for <math>\mu_e/\mu_c = 1/2</math>; and purple crosses trace the envelope profile for <math>\mu_e/\mu_c = 1/4</math>. The surface of the bipolytropic configuration is defined by the (normalized) radius, <math>R^*</math>, at which the envelope density and pressure drop to zero; the values tabulated in row 16 of Table 1 — <math>1.35550, ~1.62766, ~\mathrm{and} ~3.35697</math> for, respectively, <math>\xi_i = 0.5,~1,~\mathrm{and}~3</math> — correspond to a molecular weight ratio of unity and, hence also, to the envelope profiles traced by red squares in the Table 2 graphs. As the molecular weight ratio is decreased from unity to <math>1/2</math> and, then, <math>1/4</math> for a given choice of <math>\xi_i</math>, the (normalized) radius of the bipolytrope increases roughly in inverse proportion to <math>\mu_e/\mu_c</math> as suggested by the formula for <math>R^*</math> shown in Table 1. This proportional relation is not exact, however, because the parameter <math>\eta_s</math>, which also appears in the formula for <math>R^*</math>, contains an implicit dependence on the chosen value of the molecular weight ratio through the parameter <math>\eta_i</math>.

For a given choice of the interface parameter, <math>\xi_i</math>, the (normalized) mass that is contained in the core is independent of the choice of the molecular weight ratio. However, the (normalized) total mass, <math>M_\mathrm{tot}^*</math>, varies significantly with the choice of <math>\mu_e/\mu_c</math>; as suggested by the expression provided in row 17 of Table 1, the variation is in rough proportion to <math>(\mu_e/\mu_c)^{-2}</math> but, as with <math>R^*</math>, this proportional relation is not exact because the parameters <math>\eta_s</math> and <math>A</math> which also appear in the formula for <math>M_\mathrm{tot}^*</math> harbor an implicit dependence on the molecular weight ratio.

===Model Sequences===

For a given choice of <math>\mu_e/\mu_c</math> a physically relevant sequence of models can be constructed by steadily increasing the value of <math>\xi_i</math> from zero to infinity — or at least to some value, <math>\xi_i \gg 1</math>. Figure 1 shows how the fractional core mass, <math>\nu \equiv M_\mathrm{core}/M_\mathrm{tot}</math>, varies with the fractional core radius, <math>q \equiv r_\mathrm{core}/R</math>, along sequences having six different values of <math>\mu_e/\mu_c</math>, as detailed in the figure caption. The natural expectation is that an increase in <math>\xi_i</math> along a given sequence will correspond to an increase in the relative size — both the radius and the mass — of the core. This expectation is realized along the sequences marked by blue diamonds (<math>\mu_e/\mu_c = 1</math>) and by red squares (<math>\mu_e/\mu_c = </math>½). But the behavior is different along the other four illustrated sequences. For sufficiently large <math>\xi_i</math>, the relative radius of the core begins to decrease; then, as <math>\xi_i</math> is pushed to even larger values, eventually the relative core mass begins to decrease. Additional properties of these equilibrium sequences are discussed in [[SSC/FreeEnergy/PolytropesEmbedded#Behavior_of_Equilibrium_Sequence|an accompanying chapter]].

<div align="center">
<table border="0" cellpadding="5" width="85%">

<tr>
<td align="center" colspan="2" bgcolor="white">
[[Image:PlotSequencesBest02.png|500px|center]]
</td>
</tr>

<tr>
<td align="left" colspan="2">
'''Figure 1:''' Analytically determined plot of fractional core mass (<math>\nu</math>) versus fractional core radius (<math>q</math>) for <math>(n_c, n_e) = (5, 1)</math> bipolytrope model sequences having six different values of <math>\mu_e/\mu_c</math>: 1 (blue diamonds), ½ (red squares), 0.345 (dark purple crosses), ⅓ (pink triangles), 0.309 (light green dashes), and ¼ (purple asterisks). Along each of the model sequences, points marked by solid-colored circles correspond to models whose interface parameter, <math>\xi_i</math>, has one of three values: 0.5 (green circles), 1 (dark blue circles), or 3 (orange circles); the images linked to Table 2 provide plots of the density, pressure and mass profiles for nine of these identified models.
</td>
</tr>
</table>
</div>

The variation of <math>\nu</math> with <math>q</math> for a seventh analytically determined model sequence — one for which <math>\mu_e/\mu_c = 1/5</math> — is mapped out by a string of blue diamond symbols in the left-hand side of Figure 2. It behaves in an analogous fashion to the <math>\mu_e/\mu_c = </math>¼ (purple asterisks) sequence displayed in Figure 1. It also quantitatively, as well as qualitatively, resembles the sequence that was numerically constructed by {{ SC42full }} for models with an isothermal core (<math>n_c = \infty</math>) and an <math>n_e=3/2</math> envelope; Fig. 1 from their paper has been reproduced here on the right-hand side of Figure 2.

<div align="center">
<table border="1" cellpadding="5" width="80%">
<tr>
<td align="center" colspan="3">
<b>
Figure 2: Relationship to Schönberg-Chandrasekhar Mass Limit
</b>
</td>
</tr>

<tr>
<td align="center">
Analytic BiPolytrope with <math>n_c=5</math>, <math>n_e = 1</math>, and <math>\mu_e/\mu_c = 1/5</math>
</td>
<td align="center">
Edited excerpt from [http://adsabs.harvard.edu/abs/1942ApJ....96..161S Schönberg & Chandrasekhar (1942)]
</td>
<td align="center">
Figure from [http://adsabs.harvard.edu/abs/1941ApJ....94..525H Henrich & Chandraskhar (1941)]
</td>
</tr>
<tr>
<td align="center">
[[Image:SC_42Comparison.jpg|400px|center]]
</td>
<td align="center" rowspan="2">
[[Image:SC42_Fig1.jpg|400px|center]]

</td>
<td align="center" rowspan="2">
[[Image:HenrichChandra41b.jpg|200px|center]]

</td>
</tr>

<tr>
<td align="left">
(''Above'') Plot of fractional core mass (<math>\nu</math>) versus fractional core radius (<math>q</math>) for the analytic bipolytrope having <math>\mu_e/\mu_c = 1/5</math>. The behavior of this analytically defined model sequence resembles the behavior of the numerically constructed isothermal core models presented by (''center'') {{ SC42 }} and by (''far right'') {{ HC41full }}.
</td>
</tr>
</table>
</div>

==UV Plane==

Pulling from, for example, p. 109 of Schwarzschild's "Structure and Evolution of the Stars", let's examine these equilibrium structures using a "UV-plane" analysis, where,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>U</math></td>
<td align="center"><math>\equiv</math></td>
<td align="left">
<math>
+ \frac{d\ln m_0}{d\ln r} = \frac{4\pi r^3 \rho}{M_r} \, ,
</math>
</td>
</tr>

<tr>
<td align="right"><math>V</math></td>
<td align="center"><math>\equiv</math></td>
<td align="left">
<math>
+ \frac{d\ln P}{d\ln r} = \frac{G \rho M_r}{Pr} \, .
</math>
</td>
</tr>
</table>

===Core===

<table border="0" align="center" cellpadding="8">

<tr>
<td align="right"><math>V</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{G \rho M_r}{Pr}
</math>
</td>
</tr>
</table>

===Envelope===

=Related Discussions=
* [[SSC/Structure/PolytropesEmbedded#Embedded_Polytropic_Spheres|Polytropes emdeded in an external medium]]
* [[SSC/Structure/BiPolytropes#BiPolytropes|Constructing BiPolytropes]]
* [[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert spheres]]
** [http://en.wikipedia.org/wiki/Bonnor-Ebert_mass Bonnor-Ebert Mass] according to Wikipedia
** <font color="red">Link has disappeared:</font> [http://www.astro.umd.edu/~cychen/MATLAB/ASTR320/matlabFrom320spring2011/Bonnor-EbertSphere/html/BonnorEbert.html A MATLAB script to determine the Bonnor-Ebert Mass coefficient] developed by [http://www.astro.umd.edu/people/cychen.html Che-Yu Chen] as a graduate student in the University of Maryland Department of Astronomy
* [[SSC/Structure/LimitingMasses#Sch.C3.B6nberg-Chandrasekhar_Mass|Schönberg-Chandrasekhar limiting mass]]
* [[SSC/Structure/LimitingMasses#Relationship_Between_the_Bonnor-Ebert_and_Sch.C3.B6nberg-Chandrasekhar_Critical_Masses|Relationship between Bonnor-Ebert and Schönberg-Chandrasekhar limiting masses]]

{{ SGFfooter }}

SSC/Structure/BiPolytropes/Analytic51/Pt2

2026-11-03T22:42:30Z

Joel2:

__FORCETOC__
=BiPolytrope with n<sub>c</sub> = 5 and n<sub>e</sub> = 1 (Pt 2)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
<td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51|Part I:  (n<sub>c</sub>,n<sub>e</sub>) = (5,1) BiPolytrope]]
 
</td>
<td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt2|Part II:  Example Models]]
 
</td>
<td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt3|Part III:  Limiting Mass]]
 
</td>
<td align="center" bgcolor="lightblue"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt4|Part IV:  Free Energy]]
 
</td>
</tr>
</table>

==Examples==

===Normalization===
The dimensionless variables used in Tables 1 & 2 are defined as follows:
<div align="center">
<table border="0" cellpadding="3">
<tr>
<td align="right">
<math>\rho^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{\rho}{\rho_0}</math>
</td>

<td align="center">;    </td>

<td align="right">
<math>r^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{r}{[K_c^{1/2}/(G^{1/2}\rho_0^{2/5})]}</math>
</td>
</tr>

<tr>
<td align="right">
<math>P^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{P}{K_c\rho_0^{6/5}}</math>
</td>

<td align="center">;    </td>

<td align="right">
<math>M_r^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{M_r}{[K_c^{3/2}/(G^{3/2}\rho_0^{1/5})]}</math>
</td>
</tr>

<tr>
<td align="right">
<math>H^*</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{H}{K_c\rho_0^{1/5}}</math>
</td>

<td align="center">.    </td>

<td align="right" colspan="3">
 
</td>
</tr>

</table>
</div>

===Parameter Values===

The <math>2^\mathrm{nd}</math> column of Table 1 catalogues the analytic expressions that define various parameters and physical properties (as identified, respectively, in column 1) of the <math>(n_c, n_e) = (5, 1)</math> bipolytrope. We have evaluated these expressions for various choices of the dimensionless interface radius, <math>\xi_i</math>, and have tabulated the results in the last few columns of the table. The tabulated values have been derived assuming <math>\mu_e/\mu_c = 1</math>, that is, assuming that the core and the envelope have the same mean molecular weights.


<div align="center">
<b>Table 1: Properties of <math>(n_c, n_e) = (5, 1)</math> BiPolytrope Having Various Interface Locations, <math>\xi_i</math></b><br>
[[File:BiPolytropeParametersV01.xml|Accompanying spreadsheet with parameter values]]

<table border="1" cellpadding="5" width="80%">
<tr>
<td align="center">
Parameter
</td>
<td align="center">
<math>\xi_i</math>
</td>
<td align="center">
0.5
</td>
<td align="center">
1.0
</td>
<td align="center">
1.66864602<sup>&dagger;</sup>
</td>
<td align="center">
3.0
</td>
<td rowspan="21">
[[File:Bipolytrope51Boundaries02.png|500px|Examples]]<br />
<table border="0" align="left" cellpadding="10"><tr><td align="left">
For bipolytropic models having <math>\mu_e/\mu_c = 1.0</math>, this figure shows how the interface location, <math>\eta_i</math> (solid purple curve), the surface radius, <math>\eta_s</math> (green circular markers), and the parameter, <math>\tan^{-1}\Lambda_i</math> (orange circular markers), vary with <math>\xi_i</math> (ordinate) over the range, <math>0 \le \xi_i \le 12</math>. The three horizontal, red-dashed line segments identify the values of <math>\xi_i</math> for which numerical values of these (and other) parameters have been listed in the table shown here on the left.</td></tr></table>
</td>
</tr>

<tr>
<td align="center">
<math>\theta_i</math>
</td>
<td align="center">
<math>\biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-1/2}</math>
</td>
<td align="center">
0.96077
</td>

<td align="center">
0.86603
</td>
<td align="center">
0.72016538
</td>
<td align="center">
0.50000
</td>
</tr>

<tr>
<td align="center">
<math>-\biggl(\frac{d\theta_i}{d\xi}\biggr)_i</math>
</td>
<td align="center">
<math>\frac{1}{3} \xi_i \biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-3/2}</math>
</td>
<td align="center">
0.14781
</td>

<td align="center">
0.21651
</td>
<td align="center">
0.20774935
</td>
<td align="center">
0.12500
</td>
</tr>

<tr>
<td align="center">
<math>r^*_\mathrm{core} \equiv r^*_i</math>
</td>
<td align="center">
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i</math>
</td>
<td align="center">
0.34549
</td>

<td align="center">
0.69099
</td>
<td align="center">
1.15301487
</td>
<td align="center">
2.07297
</td>
</tr>

<tr>
<td align="center">
<math>\rho^*_i \biggr|_c = \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \rho^*_i \biggr|_e</math>
</td>
<td align="center">
<math>\biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-5/2}</math>
</td>
<td align="center">
0.81864
</td>

<td align="center">
0.48714
</td>
<td align="center">
0.19371408
</td>
<td align="center">
0.03125
</td>
</tr>

<tr>
<td align="center">
<math>P^*_i</math>
</td>
<td align="center">
<math>\biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-3}</math>
</td>
<td align="center">
0.78653
</td>

<td align="center">
0.42188
</td>
<td align="center">
0.13950617
</td>
<td align="center">
0.01563
</td>
</tr>

<tr>
<td align="center">
<math>H^*_i \biggr|_c = \frac{n_c+1}{n_e+1} \biggl( \frac{\mu_e}{\mu_c} \biggr) H^*_i \biggr|_e</math>
</td>
<td align="center">
<math>6 \biggl( 1+\frac{1}{3}\xi_i^2 \biggr)^{-1/2}</math>
</td>
<td align="center">
5.76461
</td>

<td align="center">
5.19615
</td>
<td align="center">
4.32099225
</td>
<td align="center">
3.00000
</td>
</tr>

<tr>
<td align="center">
<math>M^*_\mathrm{core}</math>
</td>
<td align="center">
<math>\biggl( \frac{6}{\pi}\biggr)^{1/2} (\xi_i \theta_i)^3</math>
</td>
<td align="center">
0.15320
</td>

<td align="center">
0.89762
</td>
<td align="center">
2.39822567
</td>
<td align="center">
4.66417
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}\eta_i</math>
</td>
<td align="center">
<math>\sqrt{3} ~\theta_i^2 \xi_i</math>
</td>
<td align="center">
0.79941
</td>

<td align="center">
1.29904
</td>
<td align="center">
1.49895749
</td>
<td align="center">
1.29904
</td>
</tr>

<tr>
<td align="center">
<math>-\biggl( \frac{d\phi}{d\eta} \biggr)_i</math>
</td>
<td align="center">
<math>\sqrt{3} ~\theta_i^{-3} \biggl( - \frac{d\theta}{d\xi} \biggr)_i = \frac{\xi_i}{\sqrt{3}}</math>
</td>
<td align="center">
0.28868
</td>

<td align="center">
0.57735
</td>
<td align="center">
0.96339323
</td>
<td align="center">
1.73205
</td>
</tr>

<tr>
<td align="center">
<math>\Lambda_i</math>
</td>
<td align="center">
<math>\frac{1}{\eta_i} + \biggl( \frac{d\phi}{d\eta} \biggr)_i</math>
</td>
<td align="center">
0.96225
</td>

<td align="center">
0.19245
</td>
<td align="center">
-  0.2962629
</td>
<td align="center">
-0.96225
</td>
</tr>

<tr>
<td align="center">
<math>A</math>
</td>
<td align="center">
<math>\eta_i (1 + \Lambda_i^2)^{1/2}</math>
</td>
<td align="center">
1.10940
</td>

<td align="center">
1.32288
</td>
<td align="center">
1.56335712
</td>
<td align="center">
1.80278
</td>
</tr>

<tr>
<td align="center">
<math>B</math>
</td>
<td align="center">
<math>\eta_i - \frac{\pi}{2} + \tan^{-1}( \Lambda_i)</math>
</td>
<td align="center">
- 0.00523
</td>

<td align="center">
-0.08163
</td>
<td align="center">
-0.359863583
</td>
<td align="center">
-1.03792
</td>
</tr>

<tr>
<td align="center">
<math>\eta_s</math>
</td>
<td align="center">
<math>\pi + B</math>
</td>
<td align="center">
3.13637
</td>

<td align="center">
3.05996
</td>
<td align="center">
2.781729071
</td>
<td align="center">
2.10367
</td>
</tr>

<tr>
<td align="center">
<math>- \biggl( \frac{d\phi}{d\eta} \biggr)_s</math>
</td>
<td align="center">
<math>\frac{A}{\eta_s}</math>
</td>
<td align="center">
0.35372
</td>

<td align="center">
0.43232
</td>
<td align="center">
0.562009126
</td>
<td align="center">
0.85697
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \cdot \biggl[ R^* \equiv r^*_s \biggr]</math>
</td>
<td align="center">
<math>\frac{\eta_s}{\sqrt{2\pi} ~\theta_i^2}</math>
</td>
<td align="center">
1.35550
</td>

<td align="center">
1.62766
</td>
<td align="center">
2.139737125
</td>
<td align="center">
3.35697
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^2 M^*_\mathrm{tot}</math>
</td>
<td align="center">
<math>\biggl(\frac{2}{\pi}\biggr)^{1/2} \theta_i^{-1} \biggl( -\eta^2 \frac{d\phi}{d\eta} \biggr)_s = \biggl(\frac{2}{\pi}\biggr)^{1/2} \frac{A\eta_s}{\theta_i}</math>
</td>
<td align="center">
2.88959
</td>

<td align="center">
3.72945
</td>
<td align="center">
4.818155932
</td>
<td align="center">
6.05187
</td>
</tr>

<tr>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\rho_c}{\bar\rho}</math>
</td>
<td align="center">
<math>~\frac{\eta_s^2}{3A\theta_i^5}</math>
</td>
<td align="center">
3.61035
</td>

<td align="center">
4.84326
</td>
<td align="center">
8.517046605
</td>
<td align="center">
26.1844
</td>
</tr>

<tr>
<td align="center">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \cdot \biggl[ \nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}} \biggr]</math>
</td>
<td align="center">
<math>\sqrt{3} ~\biggl( \frac{\xi_i^3 \theta_i^4}{A\eta_s} \biggr)</math>
</td>
<td align="center">
0.05302
</td>

<td align="center">
0.24068
</td>
<td align="center">
0.497747627
</td>
<td align="center">
0.77070
</td>
</tr>

<tr>
<td align="center">
<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} \cdot \biggl[ q \equiv \frac{r_\mathrm{core}}{R} \biggr]</math>
</td>
<td align="center">
<math>\sqrt{3}~\biggl[\frac{\xi_i \theta_i^2}{\eta_s}\biggr]</math>
</td>
<td align="center">
0.25488
</td>

<td align="center">
0.42453
</td>
<td align="center">
0.538858190
</td>
<td align="center">
0.61751
</td>
</tr>

<tr>
<td align="left" colspan="6">
<sup>&dagger;</sup>This choice of the value of <math>\xi_i = 1.66864602</math> is motivated by our [[SSC/Stability/BiPolytropes/Pt3#Fundamental_Modes|discussion of the fundamental mode of oscillation]] of the marginally unstable model that has <math>\mu_e/\mu_c = 1.0</math>; see especially row 1 of Table 2 in that associated chapter.
</tr>

</table>
</div>


Alternatively, if given <math>\mu_e/\mu_c</math> and the value of the parameter, <math>\eta_i</math>, then we have,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(\frac{\mu_e}{\mu_c} \biggr)^{-1}\eta_i</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\frac{3^{3/2}\xi_i}{3 + \xi_i^2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ 0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\xi_i^2 - \biggl[ \biggl(\frac{\mu_e}{\mu_c} \biggr) \frac{3^{3/2}}{\eta_i }\biggr] \xi_i + 3</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~\xi_i </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{3}
\biggl(\frac{\mu_e}{\mu_c} \biggr) \frac{3}{2\eta_i } \biggl\{ 1 \pm
\sqrt{1 - \biggl[ \biggl(\frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{2 \eta_i }{3}\biggr]^2 }
\biggr\} \, .
</math>
</td>
</tr>
</table>

It must be understood, therefore, that the interface location is restricted to the range,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0</math>
</td>
<td align="center">
<math>~\le \eta_i \le</math>
</td>
<td align="left">
<math>~\frac{3}{2}\biggl(\frac{\mu_e}{\mu_c} \biggr)\, ,</math>
</td>
</tr>
</table>

and that this upper limit on <math>\eta_i</math> is associated with a model whose core radius is, <math>\xi_i = \sqrt{3}</math>. Also,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\Lambda_i </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{1}{\eta_i} -
\biggl(\frac{\mu_e}{\mu_c} \biggr) \frac{3}{2\eta_i } \biggl\{ 1 \pm
\sqrt{1 - \biggl[ \biggl(\frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{2 \eta_i }{3}\biggr]^2 }
\biggr\} \, .
</math>
</td>
</tr>
</table>

===Profile===

Once the values of the key set of parameters have been determined as illustrated in Table 1, the radial profile of various physical variables can be determined throughout the bipolytrope as detailed in [[#Steps_2_.26_3|step #4]] and [[#Step_7|step #8]], above. Table 2 summarizes the mathematical expressions that define the profile throughout the core (column 2) and throughout the envelope (column 3) of the normalized mass density, <math>\rho^*(r^*)</math>, the normalized gas pressure, <math>P^*(r^*)</math>, and the normalized mass interior to <math>r^*</math>, <math>M_r^*(r^*)</math>. For all profiles, the relevant normalized radial coordinate is <math>r^*</math>, as defined in the <math>2^\mathrm{nd}</math> row of Table 2. Graphical illustrations of these resulting profiles can be viewed by clicking on the thumbnail images posted in the last few columns of Table 2.

<div align="center">
<b>Table 2: Radial Profile of Various Physical Variables</b>
<table border="1" cellpadding="6">
<tr>
<td align="center" rowspan="2">
Variable
</td>
<td align="center" rowspan="2">
Throughout the Core<br>
<math>0 \le \xi \le \xi_i</math>
</td>
<td align="center" rowspan="2">
Throughout the Envelope<sup>&dagger;</sup><br>
<math>\eta_i \le \eta \le \eta_s</math>
</td>

<td align="center" colspan="3">
Plotted Profiles
</td>
</tr>

<tr>
<td align="center">
<math>\xi_i = 0.5</math>
</td>
<td align="center">
<math>\xi_i = 1.0</math>
</td>
<td align="center">
<math>\xi_i = 3.0</math>
</td>
</tr>

<tr>
<td align="center">
<math>r^*</math>
</td>
<td align="center">
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi</math>
</td>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta</math>
</td>

<td align="center" colspan="3">
 
</td>
</tr>

<tr>
<td align="center">
<math>\rho^*</math>
</td>
<td align="center">
<math>\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}</math>
</td>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \phi(\eta)</math>
</td>

<td align="center">

[[Image:DenXi05.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:DenXi10.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:DenXi30.jpg|thumb|75px]]
</td>
</tr>

<tr>
<td align="center">
<math>P^*</math>
</td>
<td align="center">
<math>\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3}</math>
</td>
<td align="center">
<math>\theta^{6}_i [\phi(\eta)]^{2}</math>
</td>

<td align="center">

[[Image:PresXi05.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:PresXi10.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:PresXi30.jpg|thumb|75px]]
</td>
</tr>

<tr>
<td align="center">
<math>M_r^*</math>
</td>
<td align="center">
<math>\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]</math>
</td>
<td align="center">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>
</td>

<td align="center">

[[Image:MassXi05.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:MassXi10.jpg|thumb|75px]]
</td>
<td align="center">
[[Image:MassXi30.jpg|thumb|75px]]
</td>
</tr>

<tr>
<td align="left" colspan="6">
<sup>&dagger;</sup>In order to obtain the various envelope profiles, it is necessary to evaluate <math>\phi(\eta)</math> and its first derivative using the information presented in Step 6, above.
</td>
</tr>
</table>
</div>

[As of 28 April 2013] For the interface locations <math>\xi_i = 0.5, 1.0,\mathrm{and}~3.0</math>, Table 2 provides profiles for three values of the molecular weight ratio: <math>\mu_e/\mu_c = 1.0, 1/2,\mathrm{and}~1/4</math>. In all nine graphs, blue diamonds trace the structure of the <math>n_c=5</math> core; the core extends to a radius, <math>r^*_\mathrm{core}</math>, that is independent of molecular weight ratio but varies in direct proportion to the choice of <math>\xi_i</math>. Specifically, as tabulated in the fourth row of Table 1, <math>r^*_\mathrm{core} = 0.34549, ~0.69099, \mathrm{and} ~2.07297</math> for, respectively, <math>\xi_i = 0.5,~1,~\mathrm{and}~3</math>. Notice that, while the pressure profile and mass profile are continuous at the interface for all choices of the molecular weight ratio, the density profile exhibits a discontinuous jump that is in direct proportion to the chosen value of <math>\mu_e/\mu_c</math>.

Throughout the <math>n_e = 1</math> envelope, the profile of all physical variables varies with the choice of the molecular weight ratio. In the Table 2 graphs, red squares trace the envelope profile for <math>\mu_e/\mu_c = 1.0</math>; green triangles trace the envelope profile for <math>\mu_e/\mu_c = 1/2</math>; and purple crosses trace the envelope profile for <math>\mu_e/\mu_c = 1/4</math>. The surface of the bipolytropic configuration is defined by the (normalized) radius, <math>R^*</math>, at which the envelope density and pressure drop to zero; the values tabulated in row 16 of Table 1 — <math>1.35550, ~1.62766, ~\mathrm{and} ~3.35697</math> for, respectively, <math>\xi_i = 0.5,~1,~\mathrm{and}~3</math> — correspond to a molecular weight ratio of unity and, hence also, to the envelope profiles traced by red squares in the Table 2 graphs. As the molecular weight ratio is decreased from unity to <math>1/2</math> and, then, <math>1/4</math> for a given choice of <math>\xi_i</math>, the (normalized) radius of the bipolytrope increases roughly in inverse proportion to <math>\mu_e/\mu_c</math> as suggested by the formula for <math>R^*</math> shown in Table 1. This proportional relation is not exact, however, because the parameter <math>\eta_s</math>, which also appears in the formula for <math>R^*</math>, contains an implicit dependence on the chosen value of the molecular weight ratio through the parameter <math>\eta_i</math>.

For a given choice of the interface parameter, <math>\xi_i</math>, the (normalized) mass that is contained in the core is independent of the choice of the molecular weight ratio. However, the (normalized) total mass, <math>M_\mathrm{tot}^*</math>, varies significantly with the choice of <math>\mu_e/\mu_c</math>; as suggested by the expression provided in row 17 of Table 1, the variation is in rough proportion to <math>(\mu_e/\mu_c)^{-2}</math> but, as with <math>R^*</math>, this proportional relation is not exact because the parameters <math>\eta_s</math> and <math>A</math> which also appear in the formula for <math>M_\mathrm{tot}^*</math> harbor an implicit dependence on the molecular weight ratio.

===Model Sequences===

For a given choice of <math>\mu_e/\mu_c</math> a physically relevant sequence of models can be constructed by steadily increasing the value of <math>\xi_i</math> from zero to infinity — or at least to some value, <math>\xi_i \gg 1</math>. Figure 1 shows how the fractional core mass, <math>\nu \equiv M_\mathrm{core}/M_\mathrm{tot}</math>, varies with the fractional core radius, <math>q \equiv r_\mathrm{core}/R</math>, along sequences having six different values of <math>\mu_e/\mu_c</math>, as detailed in the figure caption. The natural expectation is that an increase in <math>\xi_i</math> along a given sequence will correspond to an increase in the relative size — both the radius and the mass — of the core. This expectation is realized along the sequences marked by blue diamonds (<math>\mu_e/\mu_c = 1</math>) and by red squares (<math>\mu_e/\mu_c = </math>½). But the behavior is different along the other four illustrated sequences. For sufficiently large <math>\xi_i</math>, the relative radius of the core begins to decrease; then, as <math>\xi_i</math> is pushed to even larger values, eventually the relative core mass begins to decrease. Additional properties of these equilibrium sequences are discussed in [[SSC/FreeEnergy/PolytropesEmbedded#Behavior_of_Equilibrium_Sequence|an accompanying chapter]].

<div align="center">
<table border="0" cellpadding="5" width="85%">

<tr>
<td align="center" colspan="2" bgcolor="white">
[[Image:PlotSequencesBest02.png|500px|center]]
</td>
</tr>

<tr>
<td align="left" colspan="2">
'''Figure 1:''' Analytically determined plot of fractional core mass (<math>\nu</math>) versus fractional core radius (<math>q</math>) for <math>(n_c, n_e) = (5, 1)</math> bipolytrope model sequences having six different values of <math>\mu_e/\mu_c</math>: 1 (blue diamonds), ½ (red squares), 0.345 (dark purple crosses), ⅓ (pink triangles), 0.309 (light green dashes), and ¼ (purple asterisks). Along each of the model sequences, points marked by solid-colored circles correspond to models whose interface parameter, <math>\xi_i</math>, has one of three values: 0.5 (green circles), 1 (dark blue circles), or 3 (orange circles); the images linked to Table 2 provide plots of the density, pressure and mass profiles for nine of these identified models.
</td>
</tr>
</table>
</div>

The variation of <math>\nu</math> with <math>q</math> for a seventh analytically determined model sequence — one for which <math>\mu_e/\mu_c = 1/5</math> — is mapped out by a string of blue diamond symbols in the left-hand side of Figure 2. It behaves in an analogous fashion to the <math>\mu_e/\mu_c = </math>¼ (purple asterisks) sequence displayed in Figure 1. It also quantitatively, as well as qualitatively, resembles the sequence that was numerically constructed by {{ SC42full }} for models with an isothermal core (<math>n_c = \infty</math>) and an <math>n_e=3/2</math> envelope; Fig. 1 from their paper has been reproduced here on the right-hand side of Figure 2.

<div align="center">
<table border="1" cellpadding="5" width="80%">
<tr>
<td align="center" colspan="3">
<b>
Figure 2: Relationship to Schönberg-Chandrasekhar Mass Limit
</b>
</td>
</tr>

<tr>
<td align="center">
Analytic BiPolytrope with <math>n_c=5</math>, <math>n_e = 1</math>, and <math>\mu_e/\mu_c = 1/5</math>
</td>
<td align="center">
Edited excerpt from [http://adsabs.harvard.edu/abs/1942ApJ....96..161S Schönberg & Chandrasekhar (1942)]
</td>
<td align="center">
Figure from [http://adsabs.harvard.edu/abs/1941ApJ....94..525H Henrich & Chandraskhar (1941)]
</td>
</tr>
<tr>
<td align="center">
[[Image:SC_42Comparison.jpg|400px|center]]
</td>
<td align="center" rowspan="2">
[[Image:SC42_Fig1.jpg|400px|center]]

</td>
<td align="center" rowspan="2">
[[Image:HenrichChandra41b.jpg|200px|center]]

</td>
</tr>

<tr>
<td align="left">
(''Above'') Plot of fractional core mass (<math>\nu</math>) versus fractional core radius (<math>q</math>) for the analytic bipolytrope having <math>\mu_e/\mu_c = 1/5</math>. The behavior of this analytically defined model sequence resembles the behavior of the numerically constructed isothermal core models presented by (''center'') {{ SC42 }} and by (''far right'') {{ HC41full }}.
</td>
</tr>
</table>
</div>

==UV Plane==

Pulling from, for example, p. 109 of Schwarzschild's "Structure and Evolution of the Stars", let's examine these equilibrium structures using a "UV-plane" analysis, where,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>U</math></td>
<td align="center"><math>\equiv</math></td>
<td align="left">
<math>
+ \frac{d\ln m_0}{d\ln r} = \frac{4\pi r^3 \rho}{M_r} \, ,
</math>
</td>
</tr>

<tr>
<td align="right"><math>V</math></td>
<td align="center"><math>\equiv</math></td>
<td align="left">
<math>
+ \frac{d\ln P}{d\ln r} = \frac{G \rho M_r}{Pr} \, .
</math>
</td>
</tr>
</table>

=Related Discussions=
* [[SSC/Structure/PolytropesEmbedded#Embedded_Polytropic_Spheres|Polytropes emdeded in an external medium]]
* [[SSC/Structure/BiPolytropes#BiPolytropes|Constructing BiPolytropes]]
* [[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert spheres]]
** [http://en.wikipedia.org/wiki/Bonnor-Ebert_mass Bonnor-Ebert Mass] according to Wikipedia
** <font color="red">Link has disappeared:</font> [http://www.astro.umd.edu/~cychen/MATLAB/ASTR320/matlabFrom320spring2011/Bonnor-EbertSphere/html/BonnorEbert.html A MATLAB script to determine the Bonnor-Ebert Mass coefficient] developed by [http://www.astro.umd.edu/people/cychen.html Che-Yu Chen] as a graduate student in the University of Maryland Department of Astronomy
* [[SSC/Structure/LimitingMasses#Sch.C3.B6nberg-Chandrasekhar_Mass|Schönberg-Chandrasekhar limiting mass]]
* [[SSC/Structure/LimitingMasses#Relationship_Between_the_Bonnor-Ebert_and_Sch.C3.B6nberg-Chandrasekhar_Critical_Masses|Relationship between Bonnor-Ebert and Schönberg-Chandrasekhar limiting masses]]

{{ SGFfooter }}

SSC/Stability/BiPolytropes/RedGiantToPN/TryAgain

2026-10-31T00:33:52Z

Joel2: /* Review */

__FORCETOC__
=BiPolytropes with n<sub>c</sub> = 5 and n<sub>e</sub> = 1=

==Review==

(October 30, 2026): A significant portion of the information that we have gathered regarding the stability of (n<sub>c</sub>, n<sub>e</sub>) = (5, 1) bipolytropes is [[SSC/Stability/BiPolytropes/51Models|presented on the following page]] ...

The most important result is associated with the case of <math>\mu_e/\mu_c = 1</math>. With our first numerical integration of the LAWE, we found that the interface location for which <math>\sigma_c^2 = 0</math> is <math>\xi = 1.6686460157</math>; a separate numerical integration gave the result: <font color="darkgreen"><math>\xi_i = 1.6639103365</math></font>.

Separately, we asked where along the equilibrium sequence we will find that <math>d M_\mathrm{tot}/d\xi = 0</math>. For the case of unity molecular-weight ratio, we found analytically that <math>\xi_i = 1.6684629814</math>. These are virtually identical coordinate positions; it seems likely that they should be identically the same.

=See Also=

{{ SGFfooter }}

SSC/Stability/BiPolytropes/RedGiantToPN/TryAgain

2026-10-31T00:12:05Z

Joel2: /* BiPolytropes with nc = 5 and ne = 1 */

SSC/Stability/BiPolytropes/RedGiantToPN/TryAgain

2026-10-30T23:52:44Z

Joel2:

SSC/Stability/BiPolytropes/RedGiantToPN/TryAgain

2026-10-30T23:31:38Z

Joel2: Created page with "__FORCETOC__ =BiPolytrope with n<sub>c</sub> = 5 and n<sub>e</sub> = 1 (Pt 1)= =Related Discussions= * Polytropes emdeded in an external medium * Constructing BiPolytropes * Bonnor-Ebert spheres ** [http://en.wikipedia.org/wiki/Bonnor-Ebert_mass Bonnor-Ebert Mass] according to Wikipedia ** <font c..."

__FORCETOC__
=BiPolytrope with n<sub>c</sub> = 5 and n<sub>e</sub> = 1 (Pt 1)=

=Related Discussions=
* [[SSC/Structure/PolytropesEmbedded#Embedded_Polytropic_Spheres|Polytropes emdeded in an external medium]]
* [[SSC/Structure/BiPolytropes#BiPolytropes|Constructing BiPolytropes]]
* [[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert spheres]]
** [http://en.wikipedia.org/wiki/Bonnor-Ebert_mass Bonnor-Ebert Mass] according to Wikipedia
** <font color="red">Link has disappeared:</font> [http://www.astro.umd.edu/~cychen/MATLAB/ASTR320/matlabFrom320spring2011/Bonnor-EbertSphere/html/BonnorEbert.html A MATLAB script to determine the Bonnor-Ebert Mass coefficient] developed by [http://www.astro.umd.edu/people/cychen.html Che-Yu Chen] as a graduate student in the University of Maryland Department of Astronomy
* [[SSC/Structure/LimitingMasses#Sch.C3.B6nberg-Chandrasekhar_Mass|Schönberg-Chandrasekhar limiting mass]]
* [[SSC/Structure/LimitingMasses#Relationship_Between_the_Bonnor-Ebert_and_Sch.C3.B6nberg-Chandrasekhar_Critical_Masses|Relationship between Bonnor-Ebert and Schönberg-Chandrasekhar limiting masses]]

{{ SGFfooter }}

H BookTiledMenu

2026-10-30T23:23:55Z

Joel2: /* Stability Analysis */

__NOTOC__


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==Context==

{| class="Chap1B" style="float:right; margin-left: 150px; border-style: solid; border-color:black;"
|-
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[VE|Global Energy<br />Considerations]]
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{| class="Chap1A" style="border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:solid black;" |[[PGE|<b>Principal<br />Governing<br />Equations</b>]]<br />(PGEs)
|
! style="height: 150px; width: 150px; background-color:white; border-right:dashed black;" |[[PGE/ConservingMass|Continuity]]
|
! style="height: 150px; width: 150px; border-right:dashed black;" |[[PGE/Euler|Euler (inertial)]] <br /> <hr /><br /> [[PGE/RotatingFrame|Euler (rotating frame)]]
|
! style="height: 150px; width: 150px; border-right:dashed black;" |[[PGE/FirstLawOfThermodynamics|1<sup>st</sup> Law of<br />Thermodynamics]]
|
! style="height: 150px; width: 150px;"|[[PGE/PoissonOrigin|Poisson]]
|}
<p> </p>
{| class="Chap1C" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SR|<b>Equation<br />of State</b>]]<br />(EOS)
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SR/IdealGas|Ideal Gas]]
|
! style="height: 150px; width: 150px; " |[[SR/PressureCombinations|Total Pressure]] <br /> <hr /><br />[[Apps/SMS|Bond, Arnett, & Carr<br />(1984)]]
|}
<p> </p>
{| class="Chap1KW94" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width:150px; background-color:#ffff99;" |[[SR/Piston|<b>KW94<br />Piston Model</b>]]
|}

==Spherically Symmetric Configurations==
{| class="Chap2A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">(Initially) Spherically Symmetric Configurations</font>
|}
<p> </p>

{| class="Chap2B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px solid black;"|[[SphericallySymmetricConfigurations/IndexFreeEnergy|Free-Energy<br />Index]]
|
! style="height: 150px; width: 150px;"|[[File:FreeNRGpressureRadiusIsothermal.png|150px|link=SSCpt1/Virial/PolytropesEmbeddedOutline#Virial_Equilibrium_of_Embedded_Polytropic_Spheres|Whitworth's (1981) Isothermal Free-Energy Surface]]
|
! style="height: 150px; width: 150px; border-left:2px dashed black;"|[[SSCpt1/Virial/FormFactors#Synopsis|Structural<br />Form<br />Factors]]
|
! style="height: 150px; width: 150px; background-color:#9390DB; border-left:2px solid black;"|[[SSCpt1/Virial#Free_Energy_Expression|Free-Energy<br />of<br />Spherical<br />Systems]]
|}
{| class="Chap2C" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px solid black; background-color:lightgreen; " |[[SSCpt1/PGE|One-Dimensional<br /> PGEs]]
|
! style="height: 150px; width: 150px; background-color:white; border-right:2px; " |[[SSC/Index|SSC<br />Index]]
|}

===Equilibrium Structures===
{| class="Chap3A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">1D STRUCTURE</font>
|}
<p> </p>
{| class="Chap3B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:SSC_SynopsisImage1.png|150px|link=SSC/SynopsisStyleSheet#Structure|Spherical Structures Synopsis]]
|
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[VE#Scalar_Virial_Theorem|Scalar<br />Virial<br />Theorem]]
|}
{| class="Chap3C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:2px solid black;" |[[SSCpt2/IntroductorySummary#Applications|<b>Hydrostatic<br />Balance<br />Equation</b>]]
|
! style="height: 150px; width: 310px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Template:Math/EQ_SShydrostaticBalance01 }}</div>
|
! style="height: 150px; width: 150px;" |[[SSCpt2/SolutionStrategies#Solution_Strategies|Solution<br />Strategies]]
|}

{| class="Chap3D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; " |[[SSC/Structure/UniformDensity#Isolated_Uniform-Density_Sphere|<b>Uniform-Density<br />Sphere</b>]]
|}
<p> </p>
{| class="Chap3E" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Structure/IsothermalSphere#Isothermal_Sphere|<b>Isothermal<br />Sphere</b>]]
|
! style="height: 150px; width: 310px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_SSLaneEmden02 }}</div>
|
! style="height: 150px; width: 150px;" |[[SSC/Structure/IsothermalSphere#Our_Numerical_Integration|via<br />Direct<br />Numerical<br />Integration]]
|}

{| class="Chap3F" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Structure/Polytropes#Polytropic_Spheres|<b>Isolated<br />Polytropes</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[SSC/Structure/Lane1870#Lane.27s_1870_Work|<b>Lane<br />(1870)</b>]]
|
! style="height: 150px; width: 310px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_SSLaneEmden01 }}</div>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/Polytropes/Analytic|Known<br />Analytic<br />Solutions]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/Polytropes/Numerical#Straight_Numerical_Integration|via<br />Direct<br />Numerical<br />Integration]]
|
! style="height: 150px; width: 150px; " |[[SSC/Structure/Polytropes/Numerical#HSCF_Technique|via<br />Self-Consistent<br />Field (SCF)<br />Technique]]
|}

<span id="MoreModels"> </span>
{| class="Chap3G" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Zero-Temperature<br />White Dwarf</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee;" |[[SSC/Structure/WhiteDwarfs#Chandrasekhar_mass|Chandrasekhar<br />Limiting<br />Mass<br />(1935)]]
|}

{| class="Chap3H" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; "|[[SSC/Structure/Polytropes/VirialSummary|Virial Equilibrium<br />of<br />Pressure-Truncated<br />Polytropes]]
|}
{| class="Chap3I" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |<b>Pressure-Truncated<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert<br />(Isothermal)<br />Spheres<br />(1955 - 56)]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/PolytropesEmbedded#Embedded_Polytropic_Spheres|Embedded<br />Polytropes]]
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/InstabilityOnsetOverview#Marginally_Unstable_Pressure-Truncated_Gas_Clouds|Equilibrium<br />Sequence<br />Turning-Points]]<br /><font color="green">&hearts;</font>
|
! style="height: 150px; width: 150px; border-right:2px dashed black; " |[[File:MassVsRadiusCombined02.png|130px|link=SSC/Stability/InstabilityOnsetOverview#Turning_Points_along_Sequences_of_Pressure-Truncated_Polytropes|Equilibrium sequences of Pressure-Truncated Polytropes]]
|
! style="height: 150px; width: 150px; " |[[Appendix/Ramblings/TurningPoints#Turning_Points|Turning-Points<br />(Broader Context)]]
|}

{| class="Chap3J" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; "|[[SSC/Structure/BiiPolytropes/FreeEnergy51#Free_Energy_of_BiPolytrope_with_(nc,_ne)_=_(5,_1)|Free Energy<br />of<br />Bipolytropes]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}
{| class="Chap3K" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Structure/BiPolytropes#BiPolytropes|<b>Composite<br />Polytropes</b>]]<br />(Bipolytropes)
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/BiPolytropes/Analytic1.53#BiPolytrope_with_(nc,_ne)_=_(3/2,_3)|Milne<br />(1930)]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[SSC/Structure/LimitingMasses#Sch.C3.B6nberg-Chandrasekhar_Mass|Schönberg-<br />Chandrasekhar<br />Mass<br />(1942)]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/BiPolytropes/Analytic15#BiPolytrope_with_nc_.3D_1_and_ne_.3D_5|Murphy (1983)<br /><br /><br />Analytic]]<br />(n<sub>c</sub>, n<sub>e</sub>) = (1, 5)
|
! style="height: 150px; width: 150px; border-right:2px dashed black; " |[[SSC/Structure/BiPolytropes/Analytic51#BiPolytrope_with_nc_=_5_and_ne_=_1_(Pt_1)|Eggleton, Faulkner<br />& Cannon (1998)<br /><br />Analytic]]<br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|
! style="height: 150px; width: 150px; " |[[Image:TurningPoints51Bipolytropes.png|150px|link=SSC/Stability/BiPolytropes#Organizational_Index|Equilibrium sequences of (5, 1) Bipolytropes]]
|}
<br />

===Stability Analysis===

{| class="Chap4A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">1D STABILITY</font>
|}

<font color="darkgreen"><span id="BKB74pt1">Three different approaches are used in the study of hydrodynamical stability of stars</span> and other gravitating objects …  
<ul><li>The first approach is based on the use of the equations of small oscillations. In that case the problem is reduced to a search for the solution of the boundary-value problem of the Stourme-Liuville type for the linearised system of equations of small oscillations. The solutions consist of a set of eigenfrequencies and eigenfunctions.</font> The following set of menu tiles include links to chapters where this approach has been applied to: (a) uniform-density configurations, (b) pressure-truncated isothermal spheres, (c) an isolated n = 3 polytrope, (d) pressure-truncated n = 5 configurations, and (e) bipolytropes having <math>(n_c, n_e) = (1, 5)</math>.</li>

<li>Second, one can derive <font color="darkgreen">a variational principle from the equations of small oscillations.</font> Below, an appropriately labeled (purple) menu tile links to a chapter in which the foundation for this approach is developed.  <font color="darkgreen">With the aid of the variational principle, the problem is reduced to the search of the best trial functions; this leads to approximate eigenvalues of oscillations. In spite of the simplifications introduced by the use of the variational principle and by not solving the equations of motion exactly, the problem still remains complicated …</font> One menu tile, below, links to a chapter in which an analytic (''exact'') demonstration of the variational principle's utility is provided in the context pressure-truncated n = 5 polytropes.</li>

<li>The third approach is what we have already referred to as a free-energy — and associated virial theorem — analysis. <font color="darkgreen">When this method is used, it is not necessary to use the equations of small oscillations but, instead, the functional expression for the total energy of the momentarily stationary (but not necessarily in equilibrium) star is sufficient. The condition that the first variation of the energy vanishes, determines the state of equilibrium of the star and the positiveness of a second variation indicates stability.</font> </li>
</ul>

<span id="BKB74pt2"><font color="darkgreen">If one wants to know from a stability analysis the answer to only one question — whether the model is stable or not — then the most straightforward procedure is to use the third, static method … For the application of this method, one needs to construct only equilibrium, stationary models, with no further calculation. Generally the static analysis gives no information about the shape of the modes of oscillation, but, in the vicinity of critical points, where instability sets in, this method makes it possible to find the eigenfunction of the mode which becomes unstable at the critical point.</font></span> Generally in what follows, this will be referred to as the "B-KB74 conjecture;" a menu tile carrying this label is linked to a chapter in which this approach is used to analyze the onset of a dynamical instability along the equilibrium sequence of pressure-truncated n = 5 polytropes.
<div align="right">--- Text in ''green'' taken directly from [http://adsabs.harvard.edu/abs/1974A%26A....31..391B G. S. Bisnovatyi-Kogan & S. I. Blinnikov (1974)]; B-KB74, for short.</div>
 <br />
 <br />

{| class="Chap4B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:SSC_SynopsisImage2.png|150px|link=SSC/SynopsisStyleSheet#Stability|Synopsis: Stability of Spherical Structures]]
|
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[SSC/VariationalPrinciple#Ledoux.27s_Variational_Principle|Variational<br />Principle]]
|}
{| class="Chap4C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:2px solid black;" |<b>Radial<br />Pulsation<br />Equation</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Perturbations#Spherically_Symmetric_Configurations_.28Stability_.E2.80.94_Part_II.29|Example<br />Derivations<br />&<br />Statement of<br />Eigenvalue<br />Problem]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/PerspectiveReconciliation#Reconciling_Eulerian_versus_Lagrangian_Perspectives|(poor attempt at)<br />Reconciliation]]
|
! style="height: 150px; width: 150px;" |[[SSC/SoundWaves#Sound_Waves|Relationship<br />to<br />Sound Waves]]
|}
<p> </p>
{| class="Chap4D" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:ImageOfDerivations06GoodJeansBonnor.png|120px|thumb|center|Jeans (1928) or Bonnor (1957)]]
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:ImageOfDerivations07GoodLedouxWalraven.png|120px|thumb|center|Ledoux & Walraven (1958)]]
|
! style="height: 150px; width: 150px; " |[[File:ImageOfDerivations08GoodRosseland.png|120px|thumb|center|Rosseland (1969)]]
|}
<p> </p>
{| class="Chap4E" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Uniform-Density<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black; " |[[SSC/Stability/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Sterne's<br />Analytic Sol'n<br />of Eigenvalue<br />Problem<br />(1937)]]
|
! style="height: 150px; width: 150px; " |[[File:Sterne1937SolutionPlot1.png|150px|link=SSC/Stability/UniformDensity#Properties_of_Eigenfunction_Solutions|Sterne's (1937) Solution to the Eigenvalue Problem for Uniform-Density Spheres]]
|}
<p> </p>
{| class="Chap4F" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Stability/Isothermal|<b>Pressure-Truncated<br />Isothermal<br />Spheres</b>]]
|
! style="height: 150px; width: 620px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_RadialPulsation03 }}</div>
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/Isothermal#Our_Numerical_Integration|via<br />Direct<br />Numerical<br />Integration]]
|
! style="height: 150px; width: 376px;" |[[File:TaffVanHorn1974Fundamental.gif|376px|Fundamental-Mode Eigenvectors]]
|}
<span id="MoreStabilityAnalyses"> </span>

{| class="Chap5A" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black; " |[[SSC/Stability/InstabilityOnsetOverview#Yabushita.27s_Insight_Regarding_Stability|Yabushita's<br />Analytic Sol'n for<br />Marginally Unstable<br />Configurations<br />(1974)]]
|
! style="height: 150px; width: 310px;"|<table border="0" cellpadding="2" align="center">
<tr>
<td align="center">
<math>\sigma_c^2 = 0 \, , ~~~~\gamma_\mathrm{g} = 1</math>
</td>
</tr>
<tr>
<td align="center">
 and  
</td>
</tr>
<tr>
<td align="center">
<math>x = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} </math>
</td>
</tr>
</table>
|}
<p> </p>
{| class="Chap5B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Stability/Polytropes#Radial_Oscillations_of_Polytropic_Spheres|<b>Polytropes</b>]]
|
! style="height: 150px; width: 620px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_RadialPulsation02 }}</div>
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/n3PolytropeLAWE#Radial_Oscillations_of_n_.3D_3_Polytropic_Spheres|Isolated<br />n = 3<br />Polytrope]]
|
! style="height: 150px; width: 150px; " |[[File:Schwarzschild1941movie.gif|300px|link=SSC/Stability/n3PolytropeLAWE#SchwarzschildMovie|Schwarzschild's Modal Analysis]]
|}
<br />
{| class="Chap5C" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px;"|[[Appendix/Ramblings/NonlinarOscillation#Radial_Oscillations_in_Pressure-Truncated_n_.3D_5_Polytropes|''Exact''<br />Demonstration<br />of<br />B-KB74<br />Conjecture]]
|
! style="height: 150px; width: 150px; border-left:2px dashed black;"|[[SSC/VariationalPrinciple#Directly_to_n_.3D_5_Polytropic_Configurations|''Exact''<br />Demonstration<br />of<br />Variational<br />Principle]]
|}
{| class="Chap5D" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/n5PolytropeLAWE#Radial_Oscillations_of_n_.3D_5_Polytropic_Spheres|Pressure-Truncated<br />n = 5<br />Configurations]]
|
! style="height: 150px; width: 150px; " |[[File:OutputC.gif|270px|n5 Truncated Movie]]
|}
<br />
{| class="Chap5E" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black; " |[[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|Our (2017)<br />Analytic Sol'n for<br />Marginally Unstable<br />Configurations<br /><font color="green">&hearts;</font>]]
|
! style="height: 150px; width: 465px;"|<table border="0" cellpadding="2" align="center">
<tr>
<td align="center">
<math>~\sigma_c^2 = 0 \, , ~~~~\gamma_\mathrm{g} = (n+1)/n</math>
</td>
</tr>
<tr>
<td align="center">
 and  
</td>
</tr>
<tr>
<td align="center">
<math>~x = \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] </math>
</td>
</tr>
</table>
|}
<br />
{| class="Chap5F" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px;"|[[SSC/StabilityConjecture/Bipolytrope51|B-KB74<br />Conjecture<br />RE: Bipolytrope]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}
{| class="Chap5G" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>BiPolytropes</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black; " |[[SSC/Stability/Yabushita75|Yabushita<br />(1975)]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (∞, 3/2)
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black; " |[[SSC/Stability/MurphyFiedler85|Murphy & Fiedler<br />(1985b)]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (1,5)
|
! style="height: 150px; width: 150px; " |[[SSC/Stability/BiPolytropes|Stability Analysis <br />of BiPolytropes with]]<br /> <br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}
<br />
{| class="Chap5Y" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px solid black; " |[[SSC/Stability/BiPolytropes/RedGiantToPN|Eureka!<br>(October 2025)]]
|
! style="height: 150px; width: 150px; |[[SSC/Stability/BiPolytropes/RedGiantToPN/TryAgain|Try Again<br>(November 2026)]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}

===Nonlinear Dynamical Evolution===

{| class="Chap6A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">1D DYNAMICS</font>
|}
<p> </p>
{| class="Chap6B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99;;" |[[SSC/Dynamics/FreeFall#Free-Fall|<b>Free-Fall<br />Collapse</b>]]
|}
<p> </p>
{| class="Chap6C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Dynamics/IsothermalCollapse#Collapse_of_Isothermal_Spheres|<b>Collapse of<br />Isothermal<br />Spheres</b>]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[http://adsabs.harvard.edu/abs/1993ApJ...416..303F via<br/>Direct<br />Numerical<br />Integration]
|
! style="height: 150px; width: 150px;" |[[SSC/Dynamics/IsothermalSimilaritySolution#Similarity_Solution|Similarity<br />Solution]]
|}
<p> </p>
{| class="Chap6D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99;" |[[Apps/GoldreichWeber80#Homologously_Collapsing_Stellar_Cores|<b>Collapse of<br />an Isolated<br />n = 3<br />Polytrope</b>]]
|}
<p> </p>

==Two-Dimensional Configurations (Axisymmetric)==
{| class="Chap7A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">(Initially) Axisymmetric Configurations</font>
|}
<p> </p>
{| class="Chap7B" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:black;" |[[AxisymmetricConfigurations/Storyline|<font color="white">Storyline</font>]]
|}
<p> </p>
{| class="Chap7C" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen;" |[[AxisymmetricConfigurations/PGE#Axisymmetric_Configurations_.28Part_I.29|PGEs<br />for<br />Axisymmetric<br />Systems]]
|}

===Axisymmetric Equilibrium Structures===

{| class="Chap8A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D STRUCTURE</font>
|}
<p> </p>
{| class="Chap8B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:2px solid black;" |[[AxisymmetricConfigurations/Equilibria|<b>Constructing<br />Steady-State<br />Axisymmetric<br />Configurations</b>]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[2DStructure/AxisymmetricInstabilities|Axisymmetric<br />Instabilities<br />to Avoid]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[AxisymmetricConfigurations/SolutionStrategies#Simple_Rotation_Profile_and_Centrifugal_Potential|''Simple''<br />Rotation<br />Profiles]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[AxisymmetricConfigurations/HSCF|Hachisu Self-Consistent-Field<br />[HSCF]<br />Technique]]
|
! style="height: 150px; width: 150px; " |[[AxisymmetricConfigurations/SolvingPE#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Solving the<br />Poisson Equation]]
|}
<p> </p>
{| class="Chap8C" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[2DStructure/UsingTC#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Using<br />Toroidal Coordinates<br />to Determine the<br />Gravitational<br /> Potential]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[File:Apollonian_myway4.png|150px|link=2DStructure/ToroidalCoordinateIntegrationLimits#Mapping_from_Cylindrical_to_Toroidal_Coordinates|Apollonian Circles]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[2DStructure/TCsimplification#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Attempt at<br />Simplification<br /><br /><font color="green" size="+2">&hearts;</font>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/WongAP#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Wong's<br />Analytic Potential<br />(1973)]]
|
! style="height: 150px; width: 150px; background-color:#D0FFFF;" |[[File:MovieWongN4.gif|130px|link=Apps/DysonWongTori#The_Coulomb_Potential|n = 3 contribution to potential]]
|}

====Spheroidal & Spheroidal-Like====
<br />
{| class="Chap9A" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |[[Apps/MaclaurinSpheroids#Maclaurin_Spheroids_.28axisymmetric_structure.29|<b>Uniform-Density<br />(Maclaurin)<br />Spheroids</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/MaclaurinSpheroids/GoogleBooks#Excerpts_from_A_Treatise_of_Fluxions|Maclaurin's<br />Original Text<br />&<br />Analysis<br />(1742)]]
|
! style="height: 200px; width: 200px; border-right:2px dashed black;" |[[File:Maclaurin01.gif|282px|link=Apps/MaclaurinSpheroids/GoogleBooks#Prolate_Spheroid|Our Construction of Maclaurin's Figure 291Pt2]]
|
! style="height: 150px; width: 150px; border-right:2px solid black; background-color:maroon;" |[[Apps/MaclaurinSpheroidSequence|<font color="white">Maclaurin<br />Spheroid<br />Sequence</font>]]
|}
<p> </p>
{| class="Chap9B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Rotationally<br />Flattened<br />Isothermal<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[Apps/HayashiNaritaMiyama82#Rotationally_Flattened_Isothermal_Structures|Hayashi, Narita<br /> & Miyama's<br />Analytic Sol'n<br />(1982)]]
|
! style="height: 150px; width: 150px;" |[[Apps/ReviewStahler83|Review of<br /> Stahler's (1983)<br />Sol'n Technique]]
|}
<p> </p>
{| class="Chap9C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Rotationally<br />Flattened<br />Polytropes</b>
|
! style="height: 150px; width: 150px;" |[[Apps/RotatingPolytropes|Example<br />Equilibria]]
|}
<p> </p>
{| class="Chap9D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Rotationally<br />Flattened<br />White Dwarfs</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[Apps/OstrikerBodenheimerLyndenBell66|Ostriker<br />Bodenheimer<br />& Lynden-Bell<br />(1966)]]
|
! style="height: 150px; width: 150px;" |[[Apps/RotatingWhiteDwarfs|Example<br />Equilibria]]
|}

====Toroidal & Toroidal-Like====
 <br />

{| class="Chap10B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Massless<br />Polytropic<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/PapaloizouPringleTori#Massless_Polytropic_Tori|Papaloizou-Pringle<br />Tori<br />(1984)]]
|
! style="height: 150px; width: 250px; background-color:#D0FFFF;" |[[File:TorusMovie1.gif|250px|link=Apps/PapaloizouPringleTori#Boundary_Conditions|Pivoting PP Torus]]
|}
<p> </p>
{| class="Chap10C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Self-Gravitating<br />Incompressible<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[Apps/DysonPotential|Dyson<br />(1893)]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[Apps/DWT#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Dyson-Wong<br />Tori]]
|
! style="height: 150px; width: 150px; " |[[Apps/MaclaurinToroid|Maclaurin<br />Toroid Sequence]]<br />{{ MPT77hereafter }}
|}
<p> </p>
{| class="Chap10D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Self-Gravitating<br />Compressible<br />Configurations</b>
|
! style="height: 150px; width: 150px;" |[[Apps/Ostriker64|Ostriker<br />(1964)]]
|}
<p> </p>

===Stability Analysis===
{| class="Chap11A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D STABILITY</font>
|}
<p> </p>
====Sheroidal & Spheroidal-Like====

{| class="Chap11B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Incompressible<br />Sequences</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[Appendix/Ramblings/MacSphCriticalPoints|Critical Points<br />Along<br />Maclaurin Spheroid<br />Sequence]]
|
! style="height: 150px; width: 150px;" |[[Apps/EriguchiHachisu/Models|Eriguchi, Hachisu<br />& Colleagues]]
|}
<p> </p>
{| class="Chap11B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Linear<br />Analysis<br /> of<br />Bar-Mode<br />Instability</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[Apps/RotatingPolytropes/BarmodeIncompressible|Bifurcation<br />from<br />Maclaurin<br />Sequence]]
|
! style="height: 150px; width: 150px;" |[[Apps/RotatingPolytropes/BarmodeEigenvector|Traditional<br />Analyses]]
|}
<p> </p>
* [https://ui.adsabs.harvard.edu/abs/1949ApJ...109..149C/abstract T. G. Cowling & R. A. Newing (1949)], ApJ, 109, 149: ''The Oscillations of a Rotating Star''
* [https://ui.adsabs.harvard.edu/abs/1965ApJ...141..210C/abstract M. J. Clement (1965)], ApJ, 141, 210: ''The Radial and Non-Radial Oscillations of Slowly Rotating Gaseous Masses''
* [https://ui.adsabs.harvard.edu/abs/1963ApJ...137..777R/abstract P. H. Roberts & K. Stewartson (1963)], ApJ, 137, 777: ''On the Stability of a Maclaurin spheroid of small viscosity''
* [https://ui.adsabs.harvard.edu/abs/1967ApJ...148..825R/abstract C. E. Rosenkilde (1967)], ApJ, 148, 825: ''The tensor virial-theorem including viscous stress and the oscillations of a Maclaurin spheroid''
* [https://ui.adsabs.harvard.edu/abs/1968ApJ...152..267C/abstract S. Chandrasekhar & N. R. Lebovitz (1968)], ApJ, 152, 267: ''The Pulsations and the Dynamical Stability of Gaseous Masses in Uniform Rotation''
* [https://ui.adsabs.harvard.edu/abs/1977ApJ...213..497H/abstract C. Hunter (1977)], ApJ, 213, 497: ''On Secular Stability, Secular Instability, and Points of Bifurcation of Rotating Gaseous Masses''
* [https://ui.adsabs.harvard.edu/abs/1985ApJ...294..474I/abstract J. N. Imamura, J. L. Friedman & R. H. Durisen (1985)], ApJ, 294, 474: ''Secular stability limits for rotating polytropic stars''
<table border="0" align="center" width="100%" cellpadding="1"><tr>
<td align="center" width="5%"> </td><td align="left">
<font color="green">The equilibrium models are calculated using the polytrope version (Bodenheimer & Ostriker 1973) of the Ostriker and Mark (1968) self-consistent field (SCF) code … the equilibrium models rotate on cylinders and are completely specified by <math>~n</math>, the total angular momentum, and the specific angular momentum distribution <math>~j(m_\varpi)</math>. Here <math>~m_\varpi</math> is the mass contained within a cylinder of radius <math>~\varpi</math> centered on the rotation axis. The angular momentum distribution is prescribed in several ways: (1) imposing strict uniform rotation; (2) using the same <math>~j(m_\varpi)</math> as that of a uniformly rotating spherical polybrope of index <math>~n^'</math> (see Bodenheimer and Ostriker 1973); and (3) using <math>~j(m_\varpi) \propto m_\varpi</math>, which we refer to as <math>~n^' = L</math>, <math>~L</math> for "linear."</font>
</td></tr></table>
* [https://ui.adsabs.harvard.edu/abs/1990ApJ...355..226I/abstract J. R. Ipser & L. Lindblom (1990)], ApJ, 355, 226: ''The Oscillations of Rapidly Rotating Newtonian Stellar Models''
* [https://ui.adsabs.harvard.edu/abs/1991ApJ...373..213I/abstract J. R. Ipser & L. Lindblom (1991)], ApJ, 373, 213: ''The Oscillations of Rapidly Rotating Newtonian Stellar Models. II. Dissipative Effects''
* [https://ui.adsabs.harvard.edu/abs/2000ApJ...528..946I/abstract J. N. Imamura, J. L. Friedman & R. H. Durisen (2000)], ApJ, 528, 946: ''Nonaxisymmetric Dynamic Instabilities of Rotating Polytropes. II. Torques, Bars, and Mode Saturation with Applications to Protostars and Fizzlers''
* [https://ui.adsabs.harvard.edu/abs/2003MNRAS.343..619S/abstract M. Shibata, S. Karino, & Y. Eriguchi (2003)], MNRAS, 343, 619 - 626: ''Dynamical bar-mode instability of differentially rotating stars: effects of equations of state and velocity profiles''
* [https://ui.adsabs.harvard.edu/abs/2019ApJ...877....9H/abstract G. P. Horedt (2019)], ApJ, 877, 9: ''On the Instability of Polytropic Maclaurin and Roche ellipsoids''
<p> </p>

====Toroidal & Toroidal-Like====
{| class="Chap11C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"

|-
! style="height: 150px; width: 150px; background-color:lightgreen; " |[[Apps/PapaloizouPringle84#Formulation_of_Eigenvalue_Problem|<b>Defining the<br />Eigenvalue Problem</b>]]
|}
<p> </p>
{| class="Chap11D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>(Massless)<br />Papaloizou-Pringle<br />Tori</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/ImamuraHadleyCollaboration#Analytic_Solution|Analytic Analysis<br />by<br />Blaes<br />(1985)]]
|
! style="height: 150px; width: 150px; " |[[File:N1.5j2_Combinedsmall.png|450px|center|link=Apps/ImamuraHadleyCollaboration#Plots_of_a_Few_Example_Eigenvectors|j2 Eigenfunction from Blaes85|]]
|}
<p> </p>
{| class="Chap11E" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Self-Gravitating<br />Polytropic<br />Rings</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[https://ui.adsabs.harvard.edu/abs/1990ApJ...361..394T/abstract Numerical Analysis<br />by<br />Tohline & Hachisu<br />(1990)]
|
! style="height: 150px; width: 150px; background-color:black; border-right:2px dashed black; " |[[File:Minitorus.animated.gif|150px|center|PP torus instability]]
|
! style="height: 150px; width: 150px; background-color:white; border-right:2px dashed black; " |[[Apps/WoodwardTohlineHachisu94#Online_Movies|Thick<br />Accretion<br />Disks]]<br />(WTH94)
|
! style="height: 150px; width: 150px;" |[[Apps/ImamuraHadleyCollaboration|Hadley &<br />Imamura<br />Collaboration]]
|}
<p> </p>

===Nonlinear Dynamical Evolution===

====Sheroidal & Spheroidal-Like====

{| class="Chap12A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D DYNAMICS</font>
|}
<p> </p>

{| class="Chap12B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99;" |[[Aps/MaclaurinSpheroidFreeFall|<b>Free-Fall<br />Collapse<br />of an<br />Homogeneous<br />Spheroid</b>]]
|}
<p> </p>
{| class="Chap12C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Nonlinear<br />Development of<br />Bar-Mode</b>
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[Apps/RotatingPolytropes/BarmodeLinearTimeDependent|Initially<br />Axisymmetric<br /> & Differentially<br />Rotating<br />Polytropes]]
|
! style="height: 150px; width: 150px; background-color: black;" |[[File:Dissertation.fig3cropped.png|112px|link=Apps/RotatingPolytropes/BarmodeLinearTimeDependent|Cazes Model A Simulation]]
|}
<p> </p>

==Two-Dimensional Configurations (Nonaxisymmetric Disks)==
{| class="Chap13A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">Infinitesimally Thin, Nonaxisymmetric Disks</font>
|}
<p> </p>

{| class="Chap13B" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D STRUCTURE</font>
|}
<p> </p>
{| class="Chap13C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"

|-
! style="height: 150px; width: 150px; background-color:lightgreen; " |[[Apps/Korycansky_Papaloizou_1996#Korycansky_and_Papaloizou_.281996.29|<b>Constructing<br />Infinitesimally Thin<br />Nonaxisymmetric<br />Disks</b>]]
|}
<p> </p>

==Three-Dimensional Configurations==
{| class="Chap14A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">(Initially) Three-Dimensional Configurations</font>
|}
<p> </p>

===Equilibrium Structures===
{| class="Chap14B" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">3D STRUCTURE</font>
|}


<table border="0" cellpadding="3" align="center" width="60%">
<tr><td align="left">
Special numerical techniques must be developed <font color="darkgreen">"to build three-dimensional compressible equilibrium models with complicated flows."</font> To date … <font color="darkgreen">"techniques have only been developed to build compressible equilibrium models of nonaxisymmetric configurations for a few systems with simplified rotational profiles, e.g., rigidly rotating systems ([https://ui.adsabs.harvard.edu/abs/1984PASJ...36..239H/abstract Hachisu & Eriguchi 1984]; [https://ui.adsabs.harvard.edu/abs/1986ApJS...62..461H/abstract Hachisu 1986)], irrotational systems ([https://ui.adsabs.harvard.edu/abs/1998ApJS..118..563U/abstract Uryū & Eriguchi 1998]), and configurations that are stationary in the inertial frame ([https://ui.adsabs.harvard.edu/abs/1996MNRAS.282..653U/abstract Uryū & Eriguchi 1996])."</font>
</td></tr>
<tr><td align="right">
— Drawn from §1 of {{ Ou2006full }}
</td></tr></table>
====Ellipsoidal & Ellipsoidal-Like====
 

{| class="Chap15F" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[VE/RiemannEllipsoids|Steady-State<br />2<sup>nd</sup>-Order<br />Tensor Virial<br />Equations]]
|}
{| class="Chap15G" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right: 2px solid black; " |<b>Uniform-Density<br />Incompressible<br />Ellipsoids</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee;" |[[3Dconfigurations/RiemannEllipsoids|<b>Bernhard<br />Riemann<br />(1861)</b>]]
|}
<p> </p>

{| class="Chap15C" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:white; border-right: 2px dashed black; " |[[ThreeDimensionalConfigurations/HomogeneousEllipsoids|<b>The<br />Gravitational<br />Potential</b>]]<br />(A<sub>i</sub> coefficients)
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black; " |[[ThreeDimensionalConfigurations/JacobiEllipsoids#Jacobi_Ellipsoids|<b>Jacobi<br />Ellipsoids</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black;" |[[ThreeDimensionalConfigurations/RiemannStype#Riemann_S-type_Ellipsoids|<b>Riemann<br />S-Type<br />Ellipsoids</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black;" |[[3Dconfigurations/DescriptionOfRiemannTypeI|<b>Type I<br />Riemann<br />Ellipsoids</b>]]
|
! style="height: 150px; width: 150px; background-color:white;" |[[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS|<b>Riemann<br />meets<br />COLLADA<br />&<br /> Oculus Rift S</b>]]
|}
<p> </p>

{| class="Chap15D" style="margin-right: auto; margin-left: 550px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:white; border-right: 2px dashed black; " |[https://www.aimspress.com/article/10.3934/math.2019.2.215/fulltext.html <b>A Gauge Theory<br />of<br />Riemann Ellipsoids</b>]
|
! style="height: 150px; width: 150px; background-color:white; " |[https://ui.adsabs.harvard.edu/abs/2020PhRvL.124e2501S/abstract <b>Nuclear<br />Wobbling Motion</b>]
|}
<p> </p>

{| class="Chap16A" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right: 2px solid black; " |<b>Compressible<br />Analogs of<br />Riemann Ellipsoids</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black;" |[[ThreeDimensionalConfigurations/FerrersPotential|<b>Ferrers<br />Potential<br />(1877)</b>]]
|
! style="height: 150px; width: 150px; background-color:white; border-right: 2px dashed black;" |[https://ui.adsabs.harvard.edu/abs/2006ApJ...639..549O/abstract <b>Constructing<br />Ellipsoidal<br /> & Ellipsoidal-Like<br />Configurations</b>]
|
! style="height: 150px; width: 150px; background-color:white;" |[[ThreeDimensionalConfigurations/CAREs|<b>Thoughts<br />&<br />Challenges</b>]]
|}
<p> </p>
* [https://ui.adsabs.harvard.edu/abs/1985Ap.....23..654K/abstract B. P. Kondrat'ev (1985)], Astrophysics, 23, 654: ''Irrotational and zero angular momentum ellipsoids in the Dirichlet problem''
* {{ LRS93bfull }}: ''Ellipsoidal Figures of Equilibrium: Compressible models''

====Binary Systems====

* [https://ui.adsabs.harvard.edu/abs/1933MNRAS..93..539C/abstract S. Chandrasekhar (1933)], MNRAS, 93, 539: ''The equilibrium of distorted polytropes. IV. the rotational and the tidal distortions as functions of the density distribution''
* {{ Chandrasekhar63_XIXfull }}: ''The Equilibrium and the Stability of the Roche Ellipsoids''
<table border="0" align="center" width="100%" cellpadding="1"><tr>
<td align="center" width="5%"> </td><td align="left">
<font color="green">Roche's problem is concerned with the equilibrium and the stability of rotating homogeneous masses which are, further, distorted by the constant tidal action of an attendant rigid spherical mass.</font>
</td></tr></table>

===Stability Analysis===
{| class="Chap17A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">3D STABILITY</font>
|}

====Ellipsoidal & Ellipsoidal-Like====
{| class="Chap11C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"

|-
! style="height: 150px; width: 150px; background-color:white; " |[[ThreeDimensionalConfigurations/Stability/RiemannEllipsoids|<b>Lebovitz & Lifschitz (1996)</b>]]
|}

<p> </p>

====Binary Systems====

* {{ Chandrasekhar63_XIXfull }}: ''The Equilibrium and the Stability of the Roche Ellipsoids''
* [https://ui.adsabs.harvard.edu/abs/2019ApJ...877....9H/abstract G. P. Horedt (2019)], ApJ, 877, 9: ''On the Instability of Polytropic Maclaurin and Roche Ellipsoids''

===Nonlinear Evolution===
{| class="Chap18A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">3D DYNAMICS</font>
|}
<p> </p>
{| class="Chap18B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 320px; background-color:#D0FFFF;"|[[File:JacobiMaclaurin2.gif|320px|link=ThreeDimensionalConfigurations/EFE_Energies#Animation|Animation related to Fig. 3 from Christodoulou1995]]
|
! style="height: 150px; width: 150px; background-color:#9390DB; border-left:2px solid black;"|[[ThreeDimensionalConfigurations/EFE_Energies#Properties_of_Homogeneous_Ellipsoids_.282.29|Free-Energy<br />Evolution<br />from the Maclaurin<br />to the Jacobi<br />Sequence]]
|}
{| class="Chap18C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right: 2px solid black; " |[[ThreeDimensionalConfigurations/BinaryFission#Fission_Hypothesis_of_Binary_Star_Formation|<b>Fission<br />Hypothesis</b>]]
|
! style="height: 150px; width: 150px; background-color:white;" |[http://www.phys.lsu.edu/~tohline/fission.movies.html <b>"Fission"<br />Simulations<br />at LSU</b>]
|}
<p> </p>

====Secular====
* [https://ui.adsabs.harvard.edu/abs/1971ApJ...170..143F/abstract M. Fujimoto (1971)], ApJ, 170, 143: ''Nonlinear Motions of Rotating Gaseous Ellipsoids''
* [https://ui.adsabs.harvard.edu/abs/1973ApJ...181..513P/abstract W. H. Press & S. A. Teukolsky (1973)], ApJ, 181, 513: ''On the Evolution of the Secularly Unstable, Viscous Maclaurin Spheroids''
* [https://ui.adsabs.harvard.edu/abs/1977ApJ...213..193D/abstract S. L. Detweiler & L. Lindblom (1977)], ApJ, 213, 193: ''On the evolution of the homogeneous ellipsoidal figures.''

====Dynamical====

=See Also=

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2026-10-30T23:21:29Z

Joel2: /* Stability Analysis */

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==Context==

{| class="Chap1B" style="float:right; margin-left: 150px; border-style: solid; border-color:black;"
|-
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[VE|Global Energy<br />Considerations]]
|}
{| class="Chap1A" style="border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:solid black;" |[[PGE|<b>Principal<br />Governing<br />Equations</b>]]<br />(PGEs)
|
! style="height: 150px; width: 150px; background-color:white; border-right:dashed black;" |[[PGE/ConservingMass|Continuity]]
|
! style="height: 150px; width: 150px; border-right:dashed black;" |[[PGE/Euler|Euler (inertial)]] <br /> <hr /><br /> [[PGE/RotatingFrame|Euler (rotating frame)]]
|
! style="height: 150px; width: 150px; border-right:dashed black;" |[[PGE/FirstLawOfThermodynamics|1<sup>st</sup> Law of<br />Thermodynamics]]
|
! style="height: 150px; width: 150px;"|[[PGE/PoissonOrigin|Poisson]]
|}
<p> </p>
{| class="Chap1C" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SR|<b>Equation<br />of State</b>]]<br />(EOS)
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SR/IdealGas|Ideal Gas]]
|
! style="height: 150px; width: 150px; " |[[SR/PressureCombinations|Total Pressure]] <br /> <hr /><br />[[Apps/SMS|Bond, Arnett, & Carr<br />(1984)]]
|}
<p> </p>
{| class="Chap1KW94" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width:150px; background-color:#ffff99;" |[[SR/Piston|<b>KW94<br />Piston Model</b>]]
|}

==Spherically Symmetric Configurations==
{| class="Chap2A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">(Initially) Spherically Symmetric Configurations</font>
|}
<p> </p>

{| class="Chap2B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px solid black;"|[[SphericallySymmetricConfigurations/IndexFreeEnergy|Free-Energy<br />Index]]
|
! style="height: 150px; width: 150px;"|[[File:FreeNRGpressureRadiusIsothermal.png|150px|link=SSCpt1/Virial/PolytropesEmbeddedOutline#Virial_Equilibrium_of_Embedded_Polytropic_Spheres|Whitworth's (1981) Isothermal Free-Energy Surface]]
|
! style="height: 150px; width: 150px; border-left:2px dashed black;"|[[SSCpt1/Virial/FormFactors#Synopsis|Structural<br />Form<br />Factors]]
|
! style="height: 150px; width: 150px; background-color:#9390DB; border-left:2px solid black;"|[[SSCpt1/Virial#Free_Energy_Expression|Free-Energy<br />of<br />Spherical<br />Systems]]
|}
{| class="Chap2C" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px solid black; background-color:lightgreen; " |[[SSCpt1/PGE|One-Dimensional<br /> PGEs]]
|
! style="height: 150px; width: 150px; background-color:white; border-right:2px; " |[[SSC/Index|SSC<br />Index]]
|}

===Equilibrium Structures===
{| class="Chap3A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">1D STRUCTURE</font>
|}
<p> </p>
{| class="Chap3B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:SSC_SynopsisImage1.png|150px|link=SSC/SynopsisStyleSheet#Structure|Spherical Structures Synopsis]]
|
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[VE#Scalar_Virial_Theorem|Scalar<br />Virial<br />Theorem]]
|}
{| class="Chap3C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:2px solid black;" |[[SSCpt2/IntroductorySummary#Applications|<b>Hydrostatic<br />Balance<br />Equation</b>]]
|
! style="height: 150px; width: 310px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Template:Math/EQ_SShydrostaticBalance01 }}</div>
|
! style="height: 150px; width: 150px;" |[[SSCpt2/SolutionStrategies#Solution_Strategies|Solution<br />Strategies]]
|}

{| class="Chap3D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; " |[[SSC/Structure/UniformDensity#Isolated_Uniform-Density_Sphere|<b>Uniform-Density<br />Sphere</b>]]
|}
<p> </p>
{| class="Chap3E" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Structure/IsothermalSphere#Isothermal_Sphere|<b>Isothermal<br />Sphere</b>]]
|
! style="height: 150px; width: 310px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_SSLaneEmden02 }}</div>
|
! style="height: 150px; width: 150px;" |[[SSC/Structure/IsothermalSphere#Our_Numerical_Integration|via<br />Direct<br />Numerical<br />Integration]]
|}

{| class="Chap3F" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Structure/Polytropes#Polytropic_Spheres|<b>Isolated<br />Polytropes</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[SSC/Structure/Lane1870#Lane.27s_1870_Work|<b>Lane<br />(1870)</b>]]
|
! style="height: 150px; width: 310px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_SSLaneEmden01 }}</div>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/Polytropes/Analytic|Known<br />Analytic<br />Solutions]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/Polytropes/Numerical#Straight_Numerical_Integration|via<br />Direct<br />Numerical<br />Integration]]
|
! style="height: 150px; width: 150px; " |[[SSC/Structure/Polytropes/Numerical#HSCF_Technique|via<br />Self-Consistent<br />Field (SCF)<br />Technique]]
|}

<span id="MoreModels"> </span>
{| class="Chap3G" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Zero-Temperature<br />White Dwarf</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee;" |[[SSC/Structure/WhiteDwarfs#Chandrasekhar_mass|Chandrasekhar<br />Limiting<br />Mass<br />(1935)]]
|}

{| class="Chap3H" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; "|[[SSC/Structure/Polytropes/VirialSummary|Virial Equilibrium<br />of<br />Pressure-Truncated<br />Polytropes]]
|}
{| class="Chap3I" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |<b>Pressure-Truncated<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert<br />(Isothermal)<br />Spheres<br />(1955 - 56)]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/PolytropesEmbedded#Embedded_Polytropic_Spheres|Embedded<br />Polytropes]]
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/InstabilityOnsetOverview#Marginally_Unstable_Pressure-Truncated_Gas_Clouds|Equilibrium<br />Sequence<br />Turning-Points]]<br /><font color="green">&hearts;</font>
|
! style="height: 150px; width: 150px; border-right:2px dashed black; " |[[File:MassVsRadiusCombined02.png|130px|link=SSC/Stability/InstabilityOnsetOverview#Turning_Points_along_Sequences_of_Pressure-Truncated_Polytropes|Equilibrium sequences of Pressure-Truncated Polytropes]]
|
! style="height: 150px; width: 150px; " |[[Appendix/Ramblings/TurningPoints#Turning_Points|Turning-Points<br />(Broader Context)]]
|}

{| class="Chap3J" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; "|[[SSC/Structure/BiiPolytropes/FreeEnergy51#Free_Energy_of_BiPolytrope_with_(nc,_ne)_=_(5,_1)|Free Energy<br />of<br />Bipolytropes]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}
{| class="Chap3K" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Structure/BiPolytropes#BiPolytropes|<b>Composite<br />Polytropes</b>]]<br />(Bipolytropes)
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/BiPolytropes/Analytic1.53#BiPolytrope_with_(nc,_ne)_=_(3/2,_3)|Milne<br />(1930)]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[SSC/Structure/LimitingMasses#Sch.C3.B6nberg-Chandrasekhar_Mass|Schönberg-<br />Chandrasekhar<br />Mass<br />(1942)]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Structure/BiPolytropes/Analytic15#BiPolytrope_with_nc_.3D_1_and_ne_.3D_5|Murphy (1983)<br /><br /><br />Analytic]]<br />(n<sub>c</sub>, n<sub>e</sub>) = (1, 5)
|
! style="height: 150px; width: 150px; border-right:2px dashed black; " |[[SSC/Structure/BiPolytropes/Analytic51#BiPolytrope_with_nc_=_5_and_ne_=_1_(Pt_1)|Eggleton, Faulkner<br />& Cannon (1998)<br /><br />Analytic]]<br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|
! style="height: 150px; width: 150px; " |[[Image:TurningPoints51Bipolytropes.png|150px|link=SSC/Stability/BiPolytropes#Organizational_Index|Equilibrium sequences of (5, 1) Bipolytropes]]
|}
<br />

===Stability Analysis===

{| class="Chap4A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">1D STABILITY</font>
|}

<font color="darkgreen"><span id="BKB74pt1">Three different approaches are used in the study of hydrodynamical stability of stars</span> and other gravitating objects …  
<ul><li>The first approach is based on the use of the equations of small oscillations. In that case the problem is reduced to a search for the solution of the boundary-value problem of the Stourme-Liuville type for the linearised system of equations of small oscillations. The solutions consist of a set of eigenfrequencies and eigenfunctions.</font> The following set of menu tiles include links to chapters where this approach has been applied to: (a) uniform-density configurations, (b) pressure-truncated isothermal spheres, (c) an isolated n = 3 polytrope, (d) pressure-truncated n = 5 configurations, and (e) bipolytropes having <math>(n_c, n_e) = (1, 5)</math>.</li>

<li>Second, one can derive <font color="darkgreen">a variational principle from the equations of small oscillations.</font> Below, an appropriately labeled (purple) menu tile links to a chapter in which the foundation for this approach is developed.  <font color="darkgreen">With the aid of the variational principle, the problem is reduced to the search of the best trial functions; this leads to approximate eigenvalues of oscillations. In spite of the simplifications introduced by the use of the variational principle and by not solving the equations of motion exactly, the problem still remains complicated …</font> One menu tile, below, links to a chapter in which an analytic (''exact'') demonstration of the variational principle's utility is provided in the context pressure-truncated n = 5 polytropes.</li>

<li>The third approach is what we have already referred to as a free-energy — and associated virial theorem — analysis. <font color="darkgreen">When this method is used, it is not necessary to use the equations of small oscillations but, instead, the functional expression for the total energy of the momentarily stationary (but not necessarily in equilibrium) star is sufficient. The condition that the first variation of the energy vanishes, determines the state of equilibrium of the star and the positiveness of a second variation indicates stability.</font> </li>
</ul>

<span id="BKB74pt2"><font color="darkgreen">If one wants to know from a stability analysis the answer to only one question — whether the model is stable or not — then the most straightforward procedure is to use the third, static method … For the application of this method, one needs to construct only equilibrium, stationary models, with no further calculation. Generally the static analysis gives no information about the shape of the modes of oscillation, but, in the vicinity of critical points, where instability sets in, this method makes it possible to find the eigenfunction of the mode which becomes unstable at the critical point.</font></span> Generally in what follows, this will be referred to as the "B-KB74 conjecture;" a menu tile carrying this label is linked to a chapter in which this approach is used to analyze the onset of a dynamical instability along the equilibrium sequence of pressure-truncated n = 5 polytropes.
<div align="right">--- Text in ''green'' taken directly from [http://adsabs.harvard.edu/abs/1974A%26A....31..391B G. S. Bisnovatyi-Kogan & S. I. Blinnikov (1974)]; B-KB74, for short.</div>
 <br />
 <br />

{| class="Chap4B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:SSC_SynopsisImage2.png|150px|link=SSC/SynopsisStyleSheet#Stability|Synopsis: Stability of Spherical Structures]]
|
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[SSC/VariationalPrinciple#Ledoux.27s_Variational_Principle|Variational<br />Principle]]
|}
{| class="Chap4C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:2px solid black;" |<b>Radial<br />Pulsation<br />Equation</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/Perturbations#Spherically_Symmetric_Configurations_.28Stability_.E2.80.94_Part_II.29|Example<br />Derivations<br />&<br />Statement of<br />Eigenvalue<br />Problem]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[SSC/PerspectiveReconciliation#Reconciling_Eulerian_versus_Lagrangian_Perspectives|(poor attempt at)<br />Reconciliation]]
|
! style="height: 150px; width: 150px;" |[[SSC/SoundWaves#Sound_Waves|Relationship<br />to<br />Sound Waves]]
|}
<p> </p>
{| class="Chap4D" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:ImageOfDerivations06GoodJeansBonnor.png|120px|thumb|center|Jeans (1928) or Bonnor (1957)]]
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[File:ImageOfDerivations07GoodLedouxWalraven.png|120px|thumb|center|Ledoux & Walraven (1958)]]
|
! style="height: 150px; width: 150px; " |[[File:ImageOfDerivations08GoodRosseland.png|120px|thumb|center|Rosseland (1969)]]
|}
<p> </p>
{| class="Chap4E" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Uniform-Density<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black; " |[[SSC/Stability/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Sterne's<br />Analytic Sol'n<br />of Eigenvalue<br />Problem<br />(1937)]]
|
! style="height: 150px; width: 150px; " |[[File:Sterne1937SolutionPlot1.png|150px|link=SSC/Stability/UniformDensity#Properties_of_Eigenfunction_Solutions|Sterne's (1937) Solution to the Eigenvalue Problem for Uniform-Density Spheres]]
|}
<p> </p>
{| class="Chap4F" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Stability/Isothermal|<b>Pressure-Truncated<br />Isothermal<br />Spheres</b>]]
|
! style="height: 150px; width: 620px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_RadialPulsation03 }}</div>
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/Isothermal#Our_Numerical_Integration|via<br />Direct<br />Numerical<br />Integration]]
|
! style="height: 150px; width: 376px;" |[[File:TaffVanHorn1974Fundamental.gif|376px|Fundamental-Mode Eigenvectors]]
|}
<span id="MoreStabilityAnalyses"> </span>

{| class="Chap5A" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black; " |[[SSC/Stability/InstabilityOnsetOverview#Yabushita.27s_Insight_Regarding_Stability|Yabushita's<br />Analytic Sol'n for<br />Marginally Unstable<br />Configurations<br />(1974)]]
|
! style="height: 150px; width: 310px;"|<table border="0" cellpadding="2" align="center">
<tr>
<td align="center">
<math>\sigma_c^2 = 0 \, , ~~~~\gamma_\mathrm{g} = 1</math>
</td>
</tr>
<tr>
<td align="center">
 and  
</td>
</tr>
<tr>
<td align="center">
<math>x = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} </math>
</td>
</tr>
</table>
|}
<p> </p>
{| class="Chap5B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Stability/Polytropes#Radial_Oscillations_of_Polytropic_Spheres|<b>Polytropes</b>]]
|
! style="height: 150px; width: 620px; text-align:center; border-right:2px dashed black;" |<div align="center">{{ Math/EQ_RadialPulsation02 }}</div>
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/n3PolytropeLAWE#Radial_Oscillations_of_n_.3D_3_Polytropic_Spheres|Isolated<br />n = 3<br />Polytrope]]
|
! style="height: 150px; width: 150px; " |[[File:Schwarzschild1941movie.gif|300px|link=SSC/Stability/n3PolytropeLAWE#SchwarzschildMovie|Schwarzschild's Modal Analysis]]
|}
<br />
{| class="Chap5C" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px;"|[[Appendix/Ramblings/NonlinarOscillation#Radial_Oscillations_in_Pressure-Truncated_n_.3D_5_Polytropes|''Exact''<br />Demonstration<br />of<br />B-KB74<br />Conjecture]]
|
! style="height: 150px; width: 150px; border-left:2px dashed black;"|[[SSC/VariationalPrinciple#Directly_to_n_.3D_5_Polytropic_Configurations|''Exact''<br />Demonstration<br />of<br />Variational<br />Principle]]
|}
{| class="Chap5D" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[SSC/Stability/n5PolytropeLAWE#Radial_Oscillations_of_n_.3D_5_Polytropic_Spheres|Pressure-Truncated<br />n = 5<br />Configurations]]
|
! style="height: 150px; width: 150px; " |[[File:OutputC.gif|270px|n5 Truncated Movie]]
|}
<br />
{| class="Chap5E" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black; " |[[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|Our (2017)<br />Analytic Sol'n for<br />Marginally Unstable<br />Configurations<br /><font color="green">&hearts;</font>]]
|
! style="height: 150px; width: 465px;"|<table border="0" cellpadding="2" align="center">
<tr>
<td align="center">
<math>~\sigma_c^2 = 0 \, , ~~~~\gamma_\mathrm{g} = (n+1)/n</math>
</td>
</tr>
<tr>
<td align="center">
 and  
</td>
</tr>
<tr>
<td align="center">
<math>~x = \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] </math>
</td>
</tr>
</table>
|}
<br />
{| class="Chap5F" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px;"|[[SSC/StabilityConjecture/Bipolytrope51|B-KB74<br />Conjecture<br />RE: Bipolytrope]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}
{| class="Chap5G" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>BiPolytropes</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black; " |[[SSC/Stability/Yabushita75|Yabushita<br />(1975)]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (∞, 3/2)
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black; " |[[SSC/Stability/MurphyFiedler85|Murphy & Fiedler<br />(1985b)]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (1,5)
|
! style="height: 150px; width: 150px; " |[[SSC/Stability/BiPolytropes|Stability Analysis <br />of BiPolytropes with]]<br /> <br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}
<br />
{| class="Chap5Y" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px solid black; " |[[SSC/Stability/BiPolytropes/RedGiantToPN|Eureka!<br>(October 2025)]]
|
! style="height: 150px; width: 150px; |[[SSC/Stability/BiPolytropes/RedGiantToPN|Try Again<br>(November 2026)]]<br /><br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)
|}

===Nonlinear Dynamical Evolution===

{| class="Chap6A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">1D DYNAMICS</font>
|}
<p> </p>
{| class="Chap6B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99;;" |[[SSC/Dynamics/FreeFall#Free-Fall|<b>Free-Fall<br />Collapse</b>]]
|}
<p> </p>
{| class="Chap6C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black;" |[[SSC/Dynamics/IsothermalCollapse#Collapse_of_Isothermal_Spheres|<b>Collapse of<br />Isothermal<br />Spheres</b>]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[http://adsabs.harvard.edu/abs/1993ApJ...416..303F via<br/>Direct<br />Numerical<br />Integration]
|
! style="height: 150px; width: 150px;" |[[SSC/Dynamics/IsothermalSimilaritySolution#Similarity_Solution|Similarity<br />Solution]]
|}
<p> </p>
{| class="Chap6D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99;" |[[Apps/GoldreichWeber80#Homologously_Collapsing_Stellar_Cores|<b>Collapse of<br />an Isolated<br />n = 3<br />Polytrope</b>]]
|}
<p> </p>

==Two-Dimensional Configurations (Axisymmetric)==
{| class="Chap7A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">(Initially) Axisymmetric Configurations</font>
|}
<p> </p>
{| class="Chap7B" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:black;" |[[AxisymmetricConfigurations/Storyline|<font color="white">Storyline</font>]]
|}
<p> </p>
{| class="Chap7C" style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen;" |[[AxisymmetricConfigurations/PGE#Axisymmetric_Configurations_.28Part_I.29|PGEs<br />for<br />Axisymmetric<br />Systems]]
|}

===Axisymmetric Equilibrium Structures===

{| class="Chap8A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D STRUCTURE</font>
|}
<p> </p>
{| class="Chap8B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:lightgreen; border-right:2px solid black;" |[[AxisymmetricConfigurations/Equilibria|<b>Constructing<br />Steady-State<br />Axisymmetric<br />Configurations</b>]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[2DStructure/AxisymmetricInstabilities|Axisymmetric<br />Instabilities<br />to Avoid]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[AxisymmetricConfigurations/SolutionStrategies#Simple_Rotation_Profile_and_Centrifugal_Potential|''Simple''<br />Rotation<br />Profiles]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[AxisymmetricConfigurations/HSCF|Hachisu Self-Consistent-Field<br />[HSCF]<br />Technique]]
|
! style="height: 150px; width: 150px; " |[[AxisymmetricConfigurations/SolvingPE#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Solving the<br />Poisson Equation]]
|}
<p> </p>
{| class="Chap8C" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[2DStructure/UsingTC#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Using<br />Toroidal Coordinates<br />to Determine the<br />Gravitational<br /> Potential]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[File:Apollonian_myway4.png|150px|link=2DStructure/ToroidalCoordinateIntegrationLimits#Mapping_from_Cylindrical_to_Toroidal_Coordinates|Apollonian Circles]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[2DStructure/TCsimplification#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Attempt at<br />Simplification<br /><br /><font color="green" size="+2">&hearts;</font>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/WongAP#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Wong's<br />Analytic Potential<br />(1973)]]
|
! style="height: 150px; width: 150px; background-color:#D0FFFF;" |[[File:MovieWongN4.gif|130px|link=Apps/DysonWongTori#The_Coulomb_Potential|n = 3 contribution to potential]]
|}

====Spheroidal & Spheroidal-Like====
<br />
{| class="Chap9A" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |[[Apps/MaclaurinSpheroids#Maclaurin_Spheroids_.28axisymmetric_structure.29|<b>Uniform-Density<br />(Maclaurin)<br />Spheroids</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/MaclaurinSpheroids/GoogleBooks#Excerpts_from_A_Treatise_of_Fluxions|Maclaurin's<br />Original Text<br />&<br />Analysis<br />(1742)]]
|
! style="height: 200px; width: 200px; border-right:2px dashed black;" |[[File:Maclaurin01.gif|282px|link=Apps/MaclaurinSpheroids/GoogleBooks#Prolate_Spheroid|Our Construction of Maclaurin's Figure 291Pt2]]
|
! style="height: 150px; width: 150px; border-right:2px solid black; background-color:maroon;" |[[Apps/MaclaurinSpheroidSequence|<font color="white">Maclaurin<br />Spheroid<br />Sequence</font>]]
|}
<p> </p>
{| class="Chap9B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Rotationally<br />Flattened<br />Isothermal<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[Apps/HayashiNaritaMiyama82#Rotationally_Flattened_Isothermal_Structures|Hayashi, Narita<br /> & Miyama's<br />Analytic Sol'n<br />(1982)]]
|
! style="height: 150px; width: 150px;" |[[Apps/ReviewStahler83|Review of<br /> Stahler's (1983)<br />Sol'n Technique]]
|}
<p> </p>
{| class="Chap9C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Rotationally<br />Flattened<br />Polytropes</b>
|
! style="height: 150px; width: 150px;" |[[Apps/RotatingPolytropes|Example<br />Equilibria]]
|}
<p> </p>
{| class="Chap9D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Rotationally<br />Flattened<br />White Dwarfs</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[Apps/OstrikerBodenheimerLyndenBell66|Ostriker<br />Bodenheimer<br />& Lynden-Bell<br />(1966)]]
|
! style="height: 150px; width: 150px;" |[[Apps/RotatingWhiteDwarfs|Example<br />Equilibria]]
|}

====Toroidal & Toroidal-Like====
 <br />

{| class="Chap10B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Massless<br />Polytropic<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/PapaloizouPringleTori#Massless_Polytropic_Tori|Papaloizou-Pringle<br />Tori<br />(1984)]]
|
! style="height: 150px; width: 250px; background-color:#D0FFFF;" |[[File:TorusMovie1.gif|250px|link=Apps/PapaloizouPringleTori#Boundary_Conditions|Pivoting PP Torus]]
|}
<p> </p>
{| class="Chap10C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Self-Gravitating<br />Incompressible<br />Configurations</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dashed black;" |[[Apps/DysonPotential|Dyson<br />(1893)]]
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[Apps/DWT#Common_Theme:_Determining_the_Gravitational_Potential_for_Axisymmetric_Mass_Distributions|Dyson-Wong<br />Tori]]
|
! style="height: 150px; width: 150px; " |[[Apps/MaclaurinToroid|Maclaurin<br />Toroid Sequence]]<br />{{ MPT77hereafter }}
|}
<p> </p>
{| class="Chap10D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Self-Gravitating<br />Compressible<br />Configurations</b>
|
! style="height: 150px; width: 150px;" |[[Apps/Ostriker64|Ostriker<br />(1964)]]
|}
<p> </p>

===Stability Analysis===
{| class="Chap11A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D STABILITY</font>
|}
<p> </p>
====Sheroidal & Spheroidal-Like====

{| class="Chap11B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Incompressible<br />Sequences</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[Appendix/Ramblings/MacSphCriticalPoints|Critical Points<br />Along<br />Maclaurin Spheroid<br />Sequence]]
|
! style="height: 150px; width: 150px;" |[[Apps/EriguchiHachisu/Models|Eriguchi, Hachisu<br />& Colleagues]]
|}
<p> </p>
{| class="Chap11B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Linear<br />Analysis<br /> of<br />Bar-Mode<br />Instability</b>
|
! style="height: 150px; width: 150px; border-right:2px dashed black;" |[[Apps/RotatingPolytropes/BarmodeIncompressible|Bifurcation<br />from<br />Maclaurin<br />Sequence]]
|
! style="height: 150px; width: 150px;" |[[Apps/RotatingPolytropes/BarmodeEigenvector|Traditional<br />Analyses]]
|}
<p> </p>
* [https://ui.adsabs.harvard.edu/abs/1949ApJ...109..149C/abstract T. G. Cowling & R. A. Newing (1949)], ApJ, 109, 149: ''The Oscillations of a Rotating Star''
* [https://ui.adsabs.harvard.edu/abs/1965ApJ...141..210C/abstract M. J. Clement (1965)], ApJ, 141, 210: ''The Radial and Non-Radial Oscillations of Slowly Rotating Gaseous Masses''
* [https://ui.adsabs.harvard.edu/abs/1963ApJ...137..777R/abstract P. H. Roberts & K. Stewartson (1963)], ApJ, 137, 777: ''On the Stability of a Maclaurin spheroid of small viscosity''
* [https://ui.adsabs.harvard.edu/abs/1967ApJ...148..825R/abstract C. E. Rosenkilde (1967)], ApJ, 148, 825: ''The tensor virial-theorem including viscous stress and the oscillations of a Maclaurin spheroid''
* [https://ui.adsabs.harvard.edu/abs/1968ApJ...152..267C/abstract S. Chandrasekhar & N. R. Lebovitz (1968)], ApJ, 152, 267: ''The Pulsations and the Dynamical Stability of Gaseous Masses in Uniform Rotation''
* [https://ui.adsabs.harvard.edu/abs/1977ApJ...213..497H/abstract C. Hunter (1977)], ApJ, 213, 497: ''On Secular Stability, Secular Instability, and Points of Bifurcation of Rotating Gaseous Masses''
* [https://ui.adsabs.harvard.edu/abs/1985ApJ...294..474I/abstract J. N. Imamura, J. L. Friedman & R. H. Durisen (1985)], ApJ, 294, 474: ''Secular stability limits for rotating polytropic stars''
<table border="0" align="center" width="100%" cellpadding="1"><tr>
<td align="center" width="5%"> </td><td align="left">
<font color="green">The equilibrium models are calculated using the polytrope version (Bodenheimer & Ostriker 1973) of the Ostriker and Mark (1968) self-consistent field (SCF) code … the equilibrium models rotate on cylinders and are completely specified by <math>~n</math>, the total angular momentum, and the specific angular momentum distribution <math>~j(m_\varpi)</math>. Here <math>~m_\varpi</math> is the mass contained within a cylinder of radius <math>~\varpi</math> centered on the rotation axis. The angular momentum distribution is prescribed in several ways: (1) imposing strict uniform rotation; (2) using the same <math>~j(m_\varpi)</math> as that of a uniformly rotating spherical polybrope of index <math>~n^'</math> (see Bodenheimer and Ostriker 1973); and (3) using <math>~j(m_\varpi) \propto m_\varpi</math>, which we refer to as <math>~n^' = L</math>, <math>~L</math> for "linear."</font>
</td></tr></table>
* [https://ui.adsabs.harvard.edu/abs/1990ApJ...355..226I/abstract J. R. Ipser & L. Lindblom (1990)], ApJ, 355, 226: ''The Oscillations of Rapidly Rotating Newtonian Stellar Models''
* [https://ui.adsabs.harvard.edu/abs/1991ApJ...373..213I/abstract J. R. Ipser & L. Lindblom (1991)], ApJ, 373, 213: ''The Oscillations of Rapidly Rotating Newtonian Stellar Models. II. Dissipative Effects''
* [https://ui.adsabs.harvard.edu/abs/2000ApJ...528..946I/abstract J. N. Imamura, J. L. Friedman & R. H. Durisen (2000)], ApJ, 528, 946: ''Nonaxisymmetric Dynamic Instabilities of Rotating Polytropes. II. Torques, Bars, and Mode Saturation with Applications to Protostars and Fizzlers''
* [https://ui.adsabs.harvard.edu/abs/2003MNRAS.343..619S/abstract M. Shibata, S. Karino, & Y. Eriguchi (2003)], MNRAS, 343, 619 - 626: ''Dynamical bar-mode instability of differentially rotating stars: effects of equations of state and velocity profiles''
* [https://ui.adsabs.harvard.edu/abs/2019ApJ...877....9H/abstract G. P. Horedt (2019)], ApJ, 877, 9: ''On the Instability of Polytropic Maclaurin and Roche ellipsoids''
<p> </p>

====Toroidal & Toroidal-Like====
{| class="Chap11C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"

|-
! style="height: 150px; width: 150px; background-color:lightgreen; " |[[Apps/PapaloizouPringle84#Formulation_of_Eigenvalue_Problem|<b>Defining the<br />Eigenvalue Problem</b>]]
|}
<p> </p>
{| class="Chap11D" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>(Massless)<br />Papaloizou-Pringle<br />Tori</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[[Apps/ImamuraHadleyCollaboration#Analytic_Solution|Analytic Analysis<br />by<br />Blaes<br />(1985)]]
|
! style="height: 150px; width: 150px; " |[[File:N1.5j2_Combinedsmall.png|450px|center|link=Apps/ImamuraHadleyCollaboration#Plots_of_a_Few_Example_Eigenvectors|j2 Eigenfunction from Blaes85|]]
|}
<p> </p>
{| class="Chap11E" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Self-Gravitating<br />Polytropic<br />Rings</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right:2px dotted black;" |[https://ui.adsabs.harvard.edu/abs/1990ApJ...361..394T/abstract Numerical Analysis<br />by<br />Tohline & Hachisu<br />(1990)]
|
! style="height: 150px; width: 150px; background-color:black; border-right:2px dashed black; " |[[File:Minitorus.animated.gif|150px|center|PP torus instability]]
|
! style="height: 150px; width: 150px; background-color:white; border-right:2px dashed black; " |[[Apps/WoodwardTohlineHachisu94#Online_Movies|Thick<br />Accretion<br />Disks]]<br />(WTH94)
|
! style="height: 150px; width: 150px;" |[[Apps/ImamuraHadleyCollaboration|Hadley &<br />Imamura<br />Collaboration]]
|}
<p> </p>

===Nonlinear Dynamical Evolution===

====Sheroidal & Spheroidal-Like====

{| class="Chap12A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D DYNAMICS</font>
|}
<p> </p>

{| class="Chap12B" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99;" |[[Aps/MaclaurinSpheroidFreeFall|<b>Free-Fall<br />Collapse<br />of an<br />Homogeneous<br />Spheroid</b>]]
|}
<p> </p>
{| class="Chap12C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right:2px solid black; " |<b>Nonlinear<br />Development of<br />Bar-Mode</b>
|
! style="height: 150px; width: 150px; border-right:2px dotted black;" |[[Apps/RotatingPolytropes/BarmodeLinearTimeDependent|Initially<br />Axisymmetric<br /> & Differentially<br />Rotating<br />Polytropes]]
|
! style="height: 150px; width: 150px; background-color: black;" |[[File:Dissertation.fig3cropped.png|112px|link=Apps/RotatingPolytropes/BarmodeLinearTimeDependent|Cazes Model A Simulation]]
|}
<p> </p>

==Two-Dimensional Configurations (Nonaxisymmetric Disks)==
{| class="Chap13A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">Infinitesimally Thin, Nonaxisymmetric Disks</font>
|}
<p> </p>

{| class="Chap13B" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">2D STRUCTURE</font>
|}
<p> </p>
{| class="Chap13C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"

|-
! style="height: 150px; width: 150px; background-color:lightgreen; " |[[Apps/Korycansky_Papaloizou_1996#Korycansky_and_Papaloizou_.281996.29|<b>Constructing<br />Infinitesimally Thin<br />Nonaxisymmetric<br />Disks</b>]]
|}
<p> </p>

==Three-Dimensional Configurations==
{| class="Chap14A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:navy;"
|-
! style="height: 50px; width: 800px; background-color:lightgrey;"|<font color="navy" size="+2">(Initially) Three-Dimensional Configurations</font>
|}
<p> </p>

===Equilibrium Structures===
{| class="Chap14B" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">3D STRUCTURE</font>
|}


<table border="0" cellpadding="3" align="center" width="60%">
<tr><td align="left">
Special numerical techniques must be developed <font color="darkgreen">"to build three-dimensional compressible equilibrium models with complicated flows."</font> To date … <font color="darkgreen">"techniques have only been developed to build compressible equilibrium models of nonaxisymmetric configurations for a few systems with simplified rotational profiles, e.g., rigidly rotating systems ([https://ui.adsabs.harvard.edu/abs/1984PASJ...36..239H/abstract Hachisu & Eriguchi 1984]; [https://ui.adsabs.harvard.edu/abs/1986ApJS...62..461H/abstract Hachisu 1986)], irrotational systems ([https://ui.adsabs.harvard.edu/abs/1998ApJS..118..563U/abstract Uryū & Eriguchi 1998]), and configurations that are stationary in the inertial frame ([https://ui.adsabs.harvard.edu/abs/1996MNRAS.282..653U/abstract Uryū & Eriguchi 1996])."</font>
</td></tr>
<tr><td align="right">
— Drawn from §1 of {{ Ou2006full }}
</td></tr></table>
====Ellipsoidal & Ellipsoidal-Like====
 

{| class="Chap15F" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 150px; background-color:#9390DB;"|[[VE/RiemannEllipsoids|Steady-State<br />2<sup>nd</sup>-Order<br />Tensor Virial<br />Equations]]
|}
{| class="Chap15G" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right: 2px solid black; " |<b>Uniform-Density<br />Incompressible<br />Ellipsoids</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee;" |[[3Dconfigurations/RiemannEllipsoids|<b>Bernhard<br />Riemann<br />(1861)</b>]]
|}
<p> </p>

{| class="Chap15C" style="margin-right: auto; margin-left: 230px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:white; border-right: 2px dashed black; " |[[ThreeDimensionalConfigurations/HomogeneousEllipsoids|<b>The<br />Gravitational<br />Potential</b>]]<br />(A<sub>i</sub> coefficients)
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black; " |[[ThreeDimensionalConfigurations/JacobiEllipsoids#Jacobi_Ellipsoids|<b>Jacobi<br />Ellipsoids</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black;" |[[ThreeDimensionalConfigurations/RiemannStype#Riemann_S-type_Ellipsoids|<b>Riemann<br />S-Type<br />Ellipsoids</b>]]
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black;" |[[3Dconfigurations/DescriptionOfRiemannTypeI|<b>Type I<br />Riemann<br />Ellipsoids</b>]]
|
! style="height: 150px; width: 150px; background-color:white;" |[[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS|<b>Riemann<br />meets<br />COLLADA<br />&<br /> Oculus Rift S</b>]]
|}
<p> </p>

{| class="Chap15D" style="margin-right: auto; margin-left: 550px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:white; border-right: 2px dashed black; " |[https://www.aimspress.com/article/10.3934/math.2019.2.215/fulltext.html <b>A Gauge Theory<br />of<br />Riemann Ellipsoids</b>]
|
! style="height: 150px; width: 150px; background-color:white; " |[https://ui.adsabs.harvard.edu/abs/2020PhRvL.124e2501S/abstract <b>Nuclear<br />Wobbling Motion</b>]
|}
<p> </p>

{| class="Chap16A" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right: 2px solid black; " |<b>Compressible<br />Analogs of<br />Riemann Ellipsoids</b>
|
! style="height: 150px; width: 150px; background-color:#ffeeee; border-right: 2px dashed black;" |[[ThreeDimensionalConfigurations/FerrersPotential|<b>Ferrers<br />Potential<br />(1877)</b>]]
|
! style="height: 150px; width: 150px; background-color:white; border-right: 2px dashed black;" |[https://ui.adsabs.harvard.edu/abs/2006ApJ...639..549O/abstract <b>Constructing<br />Ellipsoidal<br /> & Ellipsoidal-Like<br />Configurations</b>]
|
! style="height: 150px; width: 150px; background-color:white;" |[[ThreeDimensionalConfigurations/CAREs|<b>Thoughts<br />&<br />Challenges</b>]]
|}
<p> </p>
* [https://ui.adsabs.harvard.edu/abs/1985Ap.....23..654K/abstract B. P. Kondrat'ev (1985)], Astrophysics, 23, 654: ''Irrotational and zero angular momentum ellipsoids in the Dirichlet problem''
* {{ LRS93bfull }}: ''Ellipsoidal Figures of Equilibrium: Compressible models''

====Binary Systems====

* [https://ui.adsabs.harvard.edu/abs/1933MNRAS..93..539C/abstract S. Chandrasekhar (1933)], MNRAS, 93, 539: ''The equilibrium of distorted polytropes. IV. the rotational and the tidal distortions as functions of the density distribution''
* {{ Chandrasekhar63_XIXfull }}: ''The Equilibrium and the Stability of the Roche Ellipsoids''
<table border="0" align="center" width="100%" cellpadding="1"><tr>
<td align="center" width="5%"> </td><td align="left">
<font color="green">Roche's problem is concerned with the equilibrium and the stability of rotating homogeneous masses which are, further, distorted by the constant tidal action of an attendant rigid spherical mass.</font>
</td></tr></table>

===Stability Analysis===
{| class="Chap17A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">3D STABILITY</font>
|}

====Ellipsoidal & Ellipsoidal-Like====
{| class="Chap11C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"

|-
! style="height: 150px; width: 150px; background-color:white; " |[[ThreeDimensionalConfigurations/Stability/RiemannEllipsoids|<b>Lebovitz & Lifschitz (1996)</b>]]
|}

<p> </p>

====Binary Systems====

* {{ Chandrasekhar63_XIXfull }}: ''The Equilibrium and the Stability of the Roche Ellipsoids''
* [https://ui.adsabs.harvard.edu/abs/2019ApJ...877....9H/abstract G. P. Horedt (2019)], ApJ, 877, 9: ''On the Instability of Polytropic Maclaurin and Roche Ellipsoids''

===Nonlinear Evolution===
{| class="Chap18A" width=100% style="margin-right: auto; margin-left: 0px; border-style: solid; border-width: 3px; border-color:darkgrey;"
|-
! style="height: 30px; width: 800px; background-color:lightgrey;"|<font color="white" size="+1">3D DYNAMICS</font>
|}
<p> </p>
{| class="Chap18B" style="float:right; margin-left: 150px; border-style: solid; border-width: 3px; border-color:black;"
|-
! style="height: 150px; width: 320px; background-color:#D0FFFF;"|[[File:JacobiMaclaurin2.gif|320px|link=ThreeDimensionalConfigurations/EFE_Energies#Animation|Animation related to Fig. 3 from Christodoulou1995]]
|
! style="height: 150px; width: 150px; background-color:#9390DB; border-left:2px solid black;"|[[ThreeDimensionalConfigurations/EFE_Energies#Properties_of_Homogeneous_Ellipsoids_.282.29|Free-Energy<br />Evolution<br />from the Maclaurin<br />to the Jacobi<br />Sequence]]
|}
{| class="Chap18C" style="margin-right: auto; margin-left: 75px; border-style: solid; border-width: 3px; border-color: black;"
|-
! style="height: 150px; width: 150px; background-color:#ffff99; border-right: 2px solid black; " |[[ThreeDimensionalConfigurations/BinaryFission#Fission_Hypothesis_of_Binary_Star_Formation|<b>Fission<br />Hypothesis</b>]]
|
! style="height: 150px; width: 150px; background-color:white;" |[http://www.phys.lsu.edu/~tohline/fission.movies.html <b>"Fission"<br />Simulations<br />at LSU</b>]
|}
<p> </p>

====Secular====
* [https://ui.adsabs.harvard.edu/abs/1971ApJ...170..143F/abstract M. Fujimoto (1971)], ApJ, 170, 143: ''Nonlinear Motions of Rotating Gaseous Ellipsoids''
* [https://ui.adsabs.harvard.edu/abs/1973ApJ...181..513P/abstract W. H. Press & S. A. Teukolsky (1973)], ApJ, 181, 513: ''On the Evolution of the Secularly Unstable, Viscous Maclaurin Spheroids''
* [https://ui.adsabs.harvard.edu/abs/1977ApJ...213..193D/abstract S. L. Detweiler & L. Lindblom (1977)], ApJ, 213, 193: ''On the evolution of the homogeneous ellipsoidal figures.''

====Dynamical====

=See Also=

{{SGFfooter}}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-15T22:44:11Z

Joel2: /* Power-Series Derivation */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (62) from §5 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots ~~~~~\Rightarrow ~~ \frac{2}{3} x_Y = \frac{2}{3} - \frac{\xi^2}{45} + \frac{\xi^4}{1890} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>, or <font color="green">0.322</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>, or <font color="green">0.017191666</font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>, or <font color="green">0.329910995</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>, or <font color="green">0.335631748</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>, or <font color="green">2.955536 × 10<sup>-04</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> , or <font color="green">2.955533 × 10<sup>-04</sup></font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>, or <font color="green">5.080120 × 10<sup>-06</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font>, or <font color="green">5.081054 × 10<sup>-06</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>, or <font color="green">8.917476 × 10<sup>-08</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font>, or <font color="green">8.735179 × 10<sup>-08</sup></font>)
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212 ~~ or ~~ - 0.329910995}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167 ~ or ~ 1.017340}
\approx </math>   <font color="orange">-0.333587444</font>, or <font color="green">-0.3356316</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   (<font color="orange">-0.333587685</font>, or <font color="green">-0.3356304</font>)
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ x_Y</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{2}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>
</td>
</tr>
</table>
<b><font color="red">YES!!</font></b>
</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-15T22:33:42Z

Joel2: /* Displacement Function for Isothermal LAWE */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots ~~~~~\Rightarrow ~~ \frac{2}{3} x_Y = \frac{2}{3} - \frac{\xi^2}{45} + \frac{\xi^4}{1890} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>, or <font color="green">0.322</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>, or <font color="green">0.017191666</font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>, or <font color="green">0.329910995</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>, or <font color="green">0.335631748</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>, or <font color="green">2.955536 × 10<sup>-04</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> , or <font color="green">2.955533 × 10<sup>-04</sup></font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>, or <font color="green">5.080120 × 10<sup>-06</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font>, or <font color="green">5.081054 × 10<sup>-06</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>, or <font color="green">8.917476 × 10<sup>-08</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font>, or <font color="green">8.735179 × 10<sup>-08</sup></font>)
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212 ~~ or ~~ - 0.329910995}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167 ~ or ~ 1.017340}
\approx </math>   <font color="orange">-0.333587444</font>, or <font color="green">-0.3356316</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   (<font color="orange">-0.333587685</font>, or <font color="green">-0.3356304</font>)
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ x_Y</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{2}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>
</td>
</tr>
</table>
<b><font color="red">YES!!</font></b>
</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-15T21:31:18Z

Joel2: /* Displacement Function for Isothermal LAWE */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>, or <font color="green">0.322</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>, or <font color="green">0.017191666</font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>, or <font color="green">0.329910995</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>, or <font color="green">0.335631748</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>, or <font color="green">2.955536 × 10<sup>-04</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> , or <font color="green">2.955533 × 10<sup>-04</sup></font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>, or <font color="green">5.080120 × 10<sup>-06</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font>, or <font color="green">5.081054 × 10<sup>-06</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>, or <font color="green">8.917476 × 10<sup>-08</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font>, or <font color="green">8.735179 × 10<sup>-08</sup></font>)
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212 ~~ or ~~ - 0.329910995}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167 ~ or ~ 1.017340}
\approx </math>   <font color="orange">-0.333587444</font>, or <font color="green">-0.3356316</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   (<font color="orange">-0.333587685</font>, or <font color="green">-0.3356304</font>)
</td>
</tr>
</table>

</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-15T21:15:48Z

Joel2: /* Displacement Function for Isothermal LAWE */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>, or <font color="green">0.322</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>, or <font color="green">0.017191666</font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>, or <font color="green">0.329910995</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>, or <font color="green">0.335631748</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>, or <font color="green">2.955536 × 10<sup>-04</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> , or <font color="green">2.955533 × 10<sup>-04</sup></font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>, or <font color="green">5.080120 × 10<sup>-06</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font>, or <font color="green">5.081054 × 10<sup>-06</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>, or <font color="green">8.917476 × 10<sup>-08</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font>, or <font color="green">8.735179 × 10<sup>-08</sup></font>)
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>, or <font color="green">1.017340</font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167}
\approx </math>   <font color="orange">-0.333587444</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   <font color="orange">(-0.333587685)</font>
</td>
</tr>
</table>

</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-15T20:58:49Z

Joel2: /* Displacement Function for Isothermal LAWE */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>, or <font color="green">0.322</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>, or <font color="green">0.017191666</font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>, or <font color="green">0.329910995</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>, or <font color="green">0.335631748</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>, or <font color="green">2.955536 × 10<sup>-04</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> , or <font color="green">2.955533 × 10<sup>-04</sup></font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>, or <font color="green">5.080120 × 10<sup>-06</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font>, or <font color="green">5.081054 × 10<sup>-06</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>, or <font color="green">8.917476 × 10<sup>-08</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font>, or <font color="green">8.735179 × 10<sup>-08</sup></font>)
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167}
\approx </math>   <font color="orange">-0.333587444</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   <font color="orange">(-0.333587685)</font>
</td>
</tr>
</table>

</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-15T19:00:45Z

Joel2: /* Displacement Function for Isothermal LAWE */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>, or <font color="green">0.322</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>, or <font color="green">0.017191666</font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>, or <font color="green">1.017340293</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>, or <font color="green">0.329910995</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>, or <font color="green">0.335631748</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>, or <font color="green">2.955536 × 10<sup>-04</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> , or <font color="green">2.955533 × 10<sup>-04</sup></font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font> )
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167}
\approx </math>   <font color="orange">-0.333587444</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   <font color="orange">(-0.333587685)</font>
</td>
</tr>
</table>

</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2026-10-13T23:28:30Z

Joel2: /* Displacement Function for Isothermal LAWE */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===

====Power-Series Derivation====
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_{H}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6
+ \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====Theta-Function Examples====

<font color="darkgreen"><b>When <math>n=0</math></b></font>, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>\theta_{n=0}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 - \frac{\xi^2}{6} \, .</math>
</td>
</tr>
</table>

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=1</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{6} + \frac{\xi^4}{120} - \frac{\xi^6}{5040}
+ \frac{\xi^8}{362880} + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{5!} - \frac{\xi^6}{7!}
+ \frac{\xi^8}{9!} + \cdots
</math>
</td>
</tr>
</table>
which is consistent with the power-series expression for <math>(\sin \xi)/\xi</math>.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=5</math></b></font>, the first few terms in the power-series expression are,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \biggl(\frac{n}{5!}\biggr) \xi^4 - \frac{n}{7!} \biggl( \frac{8n-5}{3} \biggr) \xi^6
+ \frac{n}{9!}\biggl[ \frac{(122n^2 -183n + 70)}{9} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!} - \frac{\xi^6}{6!} \biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl[ \frac{5^2}{3} \cdot \frac{7^2 }{3} \biggr] + \cdots
</math>
</td>
</tr>
</table>
This should be compared with the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution_2|known analytic solution]], which is
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>(\Theta_H)_{n=5}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-1/2} \, .</math>
</td>
</tr>
</table>

From the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]], we start with,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{m}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ mb + \biggl[ \frac{m(m-1)}{2!}\biggr] b^{2} + \biggl[ \frac{m(m-1)(m-2)}{3!} \biggr] b^{3} + \biggl[ \frac{m(m-1)(m-2)(m-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>
then set <math>b=\xi^2/3</math> and <math>m = -1/2</math> to obtain:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\theta_{n=5}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ m\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ m\biggl(m-1 \biggr)\biggr] \biggl( \frac{\xi^2}{3}\biggr)^{2}
+ \frac{1}{3!}\biggl[ m \biggl(m-1 \biggr)\biggl(m-2\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{3}
+ \frac{1}{4!}\biggl[ m \biggl(m-1\biggr)\biggl(m-2\biggr)\biggl(m-3\biggr) \biggr] \biggl( \frac{\xi^2}{3}\biggr)^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ \biggl(-\frac{1}{2}\biggr)\biggl( \frac{\xi^2}{3}\biggr)
+ \frac{1}{2!}\biggl[ -\frac{1}{2}\biggl(-\frac{1}{2}-1 \biggr)\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1 \biggr)\biggl(-\frac{1}{2}-2\biggr) \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ -\frac{1}{2} \biggl(-\frac{1}{2}-1\biggr)\biggl(-\frac{1}{2}-2\biggr)\biggl(-\frac{1}{2}-3\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{\xi^2}{2\cdot 3}\biggr)
+ \frac{1}{2!}\biggl[ \frac{3}{4}\biggr] \biggl( \frac{\xi^4}{3^2}\biggr)
+ \frac{1}{3!}\biggl[ -\frac{15}{8} \biggr] \biggl( \frac{\xi^6}{3^3}\biggr)
+ \frac{1}{4!}\biggl[ \biggl(\frac{105}{16}\biggr) \biggr]
\biggl( \frac{\xi^8}{3^4}\biggr) + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{3!} + \frac{\xi^4}{4!}
- \frac{\xi^6}{6!}\biggl( \frac{5^2}{3} \biggr)
+ \frac{\xi^8}{8!}\biggl(\frac{5^2 \cdot 7^2}{3^2}\biggr) + \dots
</math>
</td>
</tr>
</table>

<b>QED</b>

===Isothermal Lane-Emden Function===

====Derivation====



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

====For Later Use====

We have just determined that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, ,
</math>
</td>
</tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>a = c = e = g</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
0 \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>b</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>d</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
- \frac{1}{2^3\cdot 3\cdot 5} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>f</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
+ \frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{2^3\cdot 3^2} + \frac{1}{2^3\cdot 3\cdot 5} \biggr)
+r^6\biggl( - \frac{1}{2^4 \cdot 3^2\cdot 5} - \frac{1}{2^4\cdot 3^4} - \frac{1}{2\cdot 3^3\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{5}{2^3\cdot 3^2\cdot 5} + \frac{3}{2^3\cdot 3^2\cdot 5} \biggr)
+r^6\biggl( - \frac{3^2\cdot 7}{2^4 \cdot 3^4\cdot 5\cdot 7} - \frac{5\cdot 7}{2^4\cdot 3^4 \cdot 5 \cdot 7}
- \frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7} \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, .
</math>
</td>
</tr>
</table>
Therefore, also,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} = (e^{-w})^{-1}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
(1 + \epsilon )^{-1}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\epsilon</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
- \frac{r^2}{6}
\cdot
\biggl[
- \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\biggr]
+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
\cdot
\biggl[
- \frac{r^2}{6}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\epsilon^3</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-\frac{r^2}{2\cdot 3}\cdot
\frac{r^4}{2^2\cdot 3^2} = - \frac{r^6}{2^3\cdot 3^3}\, .
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{+w} </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \epsilon + \epsilon^2 - \epsilon^3 + \cancelto{0}{\epsilon^4} - \cancelto{0}{\epsilon^5}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 - \biggl[ - \frac{r^2}{6}+ r^4 \biggl(\frac{1}{3^2\cdot 5} \biggr)
- r^6\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) \biggr]
+ \biggl[ \frac{r^4}{2^2\cdot 3^2} - r^6 \biggl(\frac{1}{ 3^3\cdot 5} \biggr) \biggr]
- \biggl[ - \frac{r^6}{2^3\cdot 3^3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2} - \frac{1}{3^2\cdot 5} \biggr]r^4
+ \biggl[\biggl( \frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} \biggr) - \biggl(\frac{1}{ 3^3\cdot 5} \biggr) + \frac{1}{2^3\cdot 3^3} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
+ \biggl[\frac{61}{2^3 \cdot 3^4\cdot 5\cdot 7} - \frac{2^3 \cdot 3\cdot 7}{2^3 \cdot 3^4\cdot 5 \cdot 7}
+ \frac{3 \cdot 5 \cdot 7}{2^3\cdot 3^4 \cdot 5 \cdot 7} \biggr]r^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6}
+ \biggl[ \frac{1}{2^2\cdot 3^2 \cdot 5} \biggr]r^4
- \biggl[\frac{1}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr]r^6 \, .
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>
1 + \frac{r^2}{6} + \biggl[ \frac{1}{180} \biggr]r^4 - \biggl[\frac{1}{11,340} \biggr]r^6
\, .
</math>
</td>
</tr>
</table>

We also have shown that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{1}{r} \frac{dw}{dr}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\frac{1}{3} - \frac{4 r^2}{120} + \frac{6 r^4}{1890} - \frac{8\cdot 61 r^6}{1,632,960} \, .</math>
</td>
</tr>
</table>

<span id="xYraw">Therefore, together we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>-\biggl[ \frac{1}{r} \frac{dw}{dr} \biggr]e^{+w}
</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>\biggl\{ - \frac{1}{3} + \frac{r^2}{2\cdot 3\cdot 5} - \frac{r^4}{3^2\cdot 5\cdot 7} + \frac{61 r^6}{204,120} \biggr\}
\times
\biggl\{ 1 + \frac{r^2}{2\cdot 3}
+ \frac{r^4}{2^2\cdot 3^2 \cdot 5}
- \frac{r^6}{2^2 \cdot 3^4\cdot 5\cdot 7} \biggr\}
</math>
</td>
</tr>
</table>

===Displacement Function for Polytropic LAWE===

The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...
</font>

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>
\biggl[ 6\theta \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ \theta \biggr]
- 6(n+1)\xi \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl[- \frac{d\theta}{d\xi}\biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
\biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 24 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4
- \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots\biggr]
- 2(n+1)\xi \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{1}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[\xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\} x
</math>
</td>
</tr>
</table>

Try <math>\cdots</math>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{2}{\xi} \cdot \frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>

in which case,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center"><math>\approx</math></td>
<td align="left">
<math>
\biggl[ 6 - \xi^2 + \frac{n}{20} \xi^4 - \frac{n}{63} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 \biggr] \biggl[2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~ \biggl\{ 12 - (n+3)\xi^2 + \biggl[\frac{n(n+2)}{10}\biggr] \xi^4 - \biggl[
\frac{n(n+19)}{7\cdot 27} \biggr] \biggl( \frac{8n-5}{40} \biggr)\xi^6 \biggr\}
\biggl[ \frac{2a}{\xi} + 4b + 6c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"> </td>
<td align="left">
<math>
+~(n+1) \biggl\{ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha }{21} \biggl( \frac{8n-5}{40} \biggr) \xi^4 \biggr\}
\biggl[ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 \biggr]
</math>
</td>
</tr>
</table>

<table align="center" width="80%">
<tr>
<td align="left" width="20%">
<math>\xi^{-1}:</math>
</td>
<td align="left">
<math>0 = 24a</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ a = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{0}:</math>
</td>
<td align="left">
<math>0 = 12b + 48b + (n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{1}:</math>
</td>
<td align="left">
<math>0 = 36c + 72c + \cancelto{0}{a}(n+1)\mathfrak{F}</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ c = 0</math>
</td>
</tr>
<tr>
<td align="left" width="20%">
<math>\xi^{2}:</math>
</td>
<td align="left">
<math>
0 = 72d - 2b + 96d - 4b(n+3) + b(n+1)\mathfrak{F} + \frac{n(n+1)\alpha}{5}
=
168d + b[-14 - 4n + (n+1)\mathfrak{F}] + \frac{n(n+1)\alpha}{5}
</math>
</td>
<td align="right">
<math>\Rightarrow ~~~ d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>

</td></tr></table>

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{12n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>

<tr>
<td colspan="4">
<font color="red">NOTE:</font>  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, <math>d</math>, should be <font color="red"><math>12n\alpha</math></font> instead of just <math>n\alpha</math>.
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{12 n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

====Polytropic Displacement-Function Examples====

We will only consider situations in which <math>\sigma_c^2 = 0</math> and <math>\gamma = (n+1)/n</math>, in which case, <math>\alpha = (3-n)/(n+1)</math> and <math>\mathfrak{F} = 2(n-3)/(n+1)</math>.

<font color="darkgreen"><b>When <math>n=5</math></b></font> <math>\Rightarrow \alpha = - 1 / 3</math> and <math>\mathfrak{F} = 2 / 3 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{6}{60}\cdot \frac{2}{3}\biggr) \xi^2- (6)\biggl\{ \frac{-20 + (2/3)[(34)- 4 ]}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( \frac{1}{15}\biggr) \xi^2 - 0 + \cdots
</math>
</td>
</tr>
</table>
Compare this result with the expression for <math>x_P\biggr|_{n=5}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.
<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=3</math></b></font> <math>\Rightarrow \alpha = 0</math> and <math>\mathfrak{F} = 0 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \biggl( 0\biggr) \xi^2- (6)\biggl\{ \frac{0}{2^5\cdot 3^2 \cdot 5\cdot 7} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \cdots
</math>
</td>
</tr>
</table>

Compare this result with the expression for <math>x_P\biggr|_{n=3}</math> found in an [[SSC/Stability/InstabilityOnsetOverview#Déjà_Vu|an accompanying "related thoughts"]] paragraph.

<hr width="100%" align="center">

<font color="darkgreen"><b>When <math>n=0</math></b></font> <math>\Rightarrow \alpha = +3</math> and <math>\mathfrak{F} = -6 \ldots</math> we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \frac{1}{10} \xi^2 + \frac{1 }{84} \xi^4 + \cdots
</math>
</td>
</tr>
</table>

[[SSC/Stability/UniformDensity#Validity_Check|Sterne (1937)]] derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x\biggr|_{\mathrm{Sterne}}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 \, .
</math>
</td>
</tr>
</table>
But in his analysis, the accompanying value of <math>\mathfrak{F}</math> is zero; and the accompanying (square of the) dimensionless eigenfrequency is <math>\sigma_c^2 = 2(\gamma-4/3)</math>. We cannot adopt the requirement that <math>\gamma = (n+1)/n</math> because <math>n = 0</math>.

<hr width="100%" align="center">

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form <font color="red">[modified leading term on 9/29/2026]</font>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>x</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
x_0 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
a + 2br + 3cr^2 + 4dr^3 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr]
\biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \mathfrak{F} + \frac{\alpha r^2}{5} - \frac{2\alpha r^4}{105} \biggr]
\biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
+ \frac{\alpha r^4}{315} \biggl( x_0 + ar + br^2 + cr^3 + dr^4 \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F}}{6} \biggl( x_0 + ar + br^2 \biggr)
- \frac{\alpha r^2}{30} \biggl( x_0 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}x_0}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}x_0}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha x_0}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 20b - 5\mathfrak{F} b - \alpha x_0 \biggr] ~
\Rightarrow~
d = \frac{x_0}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{x(r)}{x_0}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
Then, in the particular case where <math>\gamma = 1</math> and <math>\sigma_c^2=0</math> — which implies that <math>\alpha = (3-4/\gamma)=-1</math> and <math>\mathfrak{F} = (\sigma_c^2/\gamma - 2\alpha) = +2</math> — we have,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>x_Y(\xi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \frac{\xi^2}{30} + \frac{\xi^4}{1260} + \cdots
</math>
</td>

</tr>
</table>

</td></tr></table>
</div>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<font color="red">
This is a derivation check ...</font>      (Numerical values in parentheses assume <math>\xi = 0.056</math>, or <font color="orange">0.107</font>)
</font>

As was [[SSC/Stability/InstabilityOnsetOverview#Yabushita's_Insight_Regarding_Stability|discovered by Yabushita]], the analytically prescribed isothermal displacement function is,
<table border="0" cellpadding="5" align="center">
<tr>
<td align="right">
<math>~\sigma_c^2 = 0</math>
</td>
<td align="center">
 and  
</td>
<td align="left">
<math>~x_Y = 1 - \biggl( \frac{1}{\xi e^{-\psi}}\biggr) \frac{d\psi}{d\xi} \, .</math>
</td>
</tr>
</table>
[[Appendix/Ramblings/PowerSeriesExpressions#Isothermal_Lane-Emden_Function|Given that]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi(\xi)</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots
</math>      (5.22585 × 10<sup>-4</sup>, or <font color="orange">1.907075 × 10<sup>-03</sup></font>)<math>~~~\Rightarrow ~~~ e^\psi =</math>(1.000522, or <font color="orange">1.001908895</font>)
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{1}{\xi}\cdot\frac{d\psi}{d\xi}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{1}{3} - \frac{\xi^2}{30} + \frac{6\xi^4}{1890} - \frac{8\cdot 61 \xi^6}{1,632,960} + \cdots
</math>      (3.33229× 10<sup>-1</sup>, or <font color="orange">0.332952116</font>)
<math>~~~\Rightarrow ~~~ x_Y = </math>(1 -  0.465010   =   0.34990)
    <math>(1-x_Y) = </math> (xxx, or <font color="orange">0.333587080</font>)
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^2</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^2}{6} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+
\biggl[ - \frac{\xi^4}{120} \biggr]
\times \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} \biggr]
+ \biggl[ \frac{\xi^6}{1890} \biggr]
\times \biggl[ \frac{\xi^2}{6} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{\xi^4}{6 \cdot 6} - \frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{6\cdot 1890} \biggr]
+
\biggl[ -\frac{\xi^6}{6\cdot 120} + \frac{\xi^8}{120\cdot 120} \biggr]
+ \biggl[ \frac{\xi^8}{6\cdot 1890} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2} - \frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
-\frac{\xi^6}{2^4\cdot 3^2\cdot 5} + \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+ \frac{\xi^8}{2^2\cdot 3^4\cdot 5 \cdot 7}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2\cdot 3^4\cdot 5 \cdot 7}
+ \frac{\xi^8}{2^6\cdot 3^2\cdot 5^2}
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
+\mathcal{O}(\xi^{10})
</math>      (2.73095 × 10<sup>-07</sup>, or <font color="orange">3.636931 × 10<sup>-06</sup></font>  
vs. from above  2.73095 × 10<sup>-7</sup>, or <font color="orange">3.636935 × 10<sup>-06</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{
\frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl[2^5\cdot 5 + 3^2\cdot 7 \biggr]
\biggr\} \times
\biggl[\frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} + \cdots\biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^2 \cdot 3^2 \cdot 120}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5 \cdot 3^3\cdot 5}
-
\biggl[
\frac{\xi^8}{2^4\cdot 3^3\cdot 5} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
+\mathcal{O}(\xi^{10})
</math>      (1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935893 × 10<sup>-09</sup></font>  
vs. from above  1.42715 × 10<sup>-10</sup>  , or <font color="orange">6.935909 × 10<sup>-09</sup></font> )
</td>
</tr>
</table>

and that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right"><math>\psi^4</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{\xi^8}{2^4 \cdot 3^4}
+\mathcal{O}(\xi^{10})
</math>      (7.46276 × 10<sup>-14</sup>  , or <font color="orange">1.325760 × 10<sup>-11</sup></font>
vs. from above  7.45808 × 10<sup>-14</sup>  , or <font color="orange">1.322730 × 10<sup>-11</sup></font> )
</td>
</tr>
</table>

we appreciate that the [[Appendix/Ramblings/PowerSeriesExpressions#Exponential|exponential function gives]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>~e^\psi</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + \psi + \frac{\psi^2}{2!} + \frac{\psi^3}{3!} + \frac{\psi^4}{4!} + \cdots
</math>      (1 + 5.227213 × 10<sup>-4</sup>  
 , or <font color="orange">1+1.908894 × 10<sup>-03</sup></font>
vs. from above   1.000522 × 10<sup>0</sup>  , or <font color="orange">1+1.908895 × 10<sup>-03</sup></font>)
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} - \frac{61 \xi^8}{1,632,960} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5}
+ \frac{\xi^8}{2^6\cdot 3^4 \cdot 5^2 \cdot 7}\biggl(2^5\cdot 5 + 3^2\cdot 7 \biggr)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
-
\frac{\xi^8}{2^5\cdot 3^2 \cdot 5}
\biggr]
+ \frac{1}{4!}\biggl[ \frac{\xi^8}{2^4 \cdot 3^4} \biggr]
+\mathcal{O}(\xi^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \frac{1}{2!}\biggl[ \frac{\xi^4}{2^2 \cdot 3^2}
- \frac{\xi^6}{2^3\cdot 3^2\cdot 5} \biggr]
+ \frac{1}{3!}\biggl[ \frac{\xi^6}{2^3 \cdot 3^3}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1
+ \biggl[ \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} \biggr]
+ \biggl[ \frac{\xi^4}{2^3 \cdot 3^2}
- \frac{\xi^6}{2^4\cdot 3^2\cdot 5} \biggr]
+ \biggl[ \frac{\xi^6}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^3 \cdot 3^2}
- \frac{1}{2^3\cdot 3\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{1}{2\cdot 3^3\cdot 5\cdot 7}
- \frac{1}{2^4\cdot 3^2\cdot 5}
+ \frac{1}{2^4 \cdot 3^4}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] + \xi^6 \biggl[
\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}
- \frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5 \cdot 7}
+ \frac{5\cdot 7}{2^4 \cdot 3^4 \cdot 5\cdot 7}
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>~
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{180}
\biggr] + \xi^6 \biggl[
\frac{-4}{45360}
\biggr]
</math>      (1 + 5.22721 × 10<sup>-4</sup>, or <font color="orange">1 + 1.908894 × 10<sup>-03</sup></font>)
</td>
</tr>
</table>

As a result, the approximate expression for <math>x_Y</math> is ([[#xYraw|also see above)]],
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>x_Y -1= - \biggl[\frac{1}{\xi} \frac{d\psi}{d\xi} \biggr] e^{\psi}</math>
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
\overbrace{\biggl[- \frac{1}{3} + \frac{\xi^2}{2\cdot 3\cdot 5} - \frac{\xi^4}{3^2 \cdot 5\cdot 7} + \frac{61 \xi^6}{ 2^3 \cdot 3^6\cdot 5\cdot 7}\biggr]}^{-0.33295212}
\times \underbrace{\biggl\{
1 + \frac{\xi^2}{6}
+ \xi^4\biggl[
\frac{1}{2^2 \cdot 3^2\cdot 5}
\biggr] - \xi^6 \biggl[
\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}
\biggr]
\biggr\}}_{1.001908167}
\approx </math>   <font color="orange">-0.333587444</font>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
-
\biggl[\frac{1}{3} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr] - \xi^6 \biggl[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}+ \xi^4\biggl[\frac{1}{2^2 \cdot 3^2\cdot 5} \biggr]
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-
\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr]
\times
\biggl\{1 + \frac{\xi^2}{6}
\biggr\}
+
\biggl[\frac{61 \xi^6}{2^3\cdot 3^6\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \frac{\xi^2}{2\cdot 3^2} - \xi^4\biggl[\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr] + \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr] + \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\frac{\xi^2}{6}
+ \biggl[\frac{\xi^2}{2\cdot 3\cdot 5} \biggr]\biggl[\frac{\xi^4}{2^2 \cdot 3^2\cdot 5} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-\biggl[\frac{\xi^4}{3^2 \cdot 5\cdot 7} \biggr] - \biggl[\frac{\xi^6}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+
\biggl[\frac{61 \xi^6}{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2}\biggr] + \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\biggr]
+ \xi^2\biggl[\frac{1}{2\cdot 3\cdot 5} \biggr] + \xi^4\biggl[\frac{1}{2^2\cdot 3^2\cdot 5} \biggr]
+ \xi^6\biggl[\frac{1}{2^3 \cdot 3^3\cdot 5^2} \biggr]
+ \xi^4\biggl[-\frac{1}{3^2 \cdot 5\cdot 7} \biggr] + \xi^6\biggl[-\frac{1}{2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6\biggl[\frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{1}{2\cdot 3^2} + \frac{1}{2\cdot 3\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{1}{2^2 \cdot 3^3\cdot 5} + \frac{1}{2^2\cdot 3^2\cdot 5} -\frac{1}{3^2 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7} + \frac{1}{2^3 \cdot 3^3\cdot 5^2} + \frac{61 }{2^3 \cdot 3^5 \cdot 5 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} + \xi^2\biggl[- \frac{5}{2\cdot 3^2 \cdot 5} + \frac{3}{2\cdot 3^2\cdot 5} \biggr]
+ \xi^4\biggl[-\frac{7}{2^2 \cdot 3^3\cdot 5 \cdot 7} + \frac{3\cdot 7}{2^2\cdot 3^3\cdot 5 \cdot 7} -\frac{2^2\cdot 3}{2^2\cdot 3^3 \cdot 5\cdot 7} \biggr]
+ \xi^6 \biggl[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7} + \frac{3^2 \cdot 7}{2^3 \cdot 3^5 \cdot 5^2}
+ \frac{5\cdot 61 }{2^3 \cdot 3^5 \cdot 5^2 \cdot 7}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \xi^2\biggl[\frac{1}{3^2 \cdot 5} \biggr]
+ \xi^4\biggl[ \frac{1}{2 \cdot 3^3\cdot 5 \cdot 7} \biggr]
+ \xi^6 \biggl[\frac{1}{2^2\cdot 3^2\cdot 5^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>\approx</math>
</td>
<td align="left">
<math>
- \frac{1}{3} - \biggl[\frac{\xi^2}{45} \biggr]
+ \biggl[ \frac{\xi^4}{1890} \biggr]
+ \biggl[\frac{\xi^6}{900} \biggr]
</math>   <font color="orange">(-0.333587685)</font>
</td>
</tr>
</table>

</td></tr></table>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

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