Editing Appendix/Ramblings/PowerSeriesExpressions (section)

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> &#8212; expanded about the coordinate center, <math>~\xi = 0</math> &#8212; 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">
&nbsp;
  </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>:  &nbsp;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">
&nbsp;
  </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">
&nbsp;
  </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</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>
</table>
</td></tr></table>
</div>

NOTE:  &nbsp;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''.

===Isothermal Lane-Emden Function===

<!-- As we have discussed in [[SSC/Structure/IsothermalSphere#Governing_Relations|a separate chapter]], the 2<sup>nd</sup>-order ODE that governs the radial density distribution in an isothermal sphere is,
<div align="center" id="Chandrasekhar">
<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\psi}{d\xi}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{-\psi} \, .</math>
  </td>
</tr>
</table>
</div>
-->

Here we seek a power-series expression for the isothermal, Lane-Emden function &#8212; expanded about the coordinate center &#8212; 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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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 &sect;22 in Chapter IV of [[Appendix/References#C67|C67]].


NOTE:  &nbsp;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''.

===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">
&nbsp;
  </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>.  


====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>~\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>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<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{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</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{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
  </td>
</tr>
</table>
</td></tr></table>
</div>

===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,
<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 + ar + br^2 + cr^3 + dr^4 + \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( 1 + 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{\mathfrak{F} }{6}  \biggl( 1 + ar + br^2 \biggr) 
- \frac{\alpha}{3} \biggl(\frac{r^2}{10}  \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}}{6}</math>
  </td>
  <td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{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}{30}</math>
  </td>
  <td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] ~
\Rightarrow~
d = \frac{1}{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="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x(r)</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>
</td></tr></table>
</div>

===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>
&nbsp; &nbsp; &nbsp; (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>
&nbsp; &nbsp; &nbsp; for, &nbsp; &nbsp; &nbsp;
<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>
&nbsp; &nbsp;&nbsp; for <math>~(e^4 < 1)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;

  </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> &#8212; that is, in the case of an infinitesimally thin, axisymmetric disk &#8212; 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>
&nbsp; &nbsp; &nbsp; for, &nbsp; &nbsp; &nbsp;
<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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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">
&nbsp;
  </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>