Editing Appendix/Ramblings/PowerSeriesExpressions (section)

===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>