Editing SSC/Stability/n1PolytropeLAWE/Pt4 (section)

==Establish Relevant (n=1) LAWE==
From [[SSC/Stability/n1PolytropeLAWE#WorkInProgress|a related discussion]] &#8212; or [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|a broader overview of Instability Onset]] &#8212; we find the

<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>

Furthermore &#8212; see, for example, [[SSC/Stability/InstabilityOnsetOverview#Pressure_and_Density_Displacement_Functions|here]],
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="1"><font color="maroon"><b>Exact Solution to the Polytropic LAWE</b></font></td>
</tr>
<tr>
  <td align="left">
<math>x_P \equiv \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>
in which case for <math>n=1</math>,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="left">
<math>
x_P = -3 \biggl( \frac{1}{\xi \theta}\biggr) \frac{d\theta}{d\xi}
=
-3 \biggl( \frac{1}{\xi^2}\biggr) \frac{d\ln\theta}{d\ln\xi}
=
\frac{3}{\xi^2} Q \, .
</math>
  </td>
</tr>
</table>


===Isolated Sphere===
For an isolated n = 1 <math>(\gamma_g = 2, \alpha = 1)</math> polytrope, we know that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\frac{\sin\xi}{\xi}</math>
  </td>
  <td align="right">
&nbsp; &nbsp; <math>\Rightarrow ~~~ Q(\xi) \equiv - \frac{d \ln \theta}{d\ln \xi}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\biggl[1 - \xi\cot\xi\biggr] \, .</math>
  </td>
</tr>
</table>
Hence, the relevant LAWE is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{d^2 x}{d\xi^2} 
+ 
\biggl[4 - 2(1 - \xi\cot\xi)\biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+
2 \biggl[\biggl( \frac{\sigma_c^2}{12}\biggr)\frac{\xi^3}{\sin\xi} - (1 - \xi\cot\xi)  \biggr]\frac{x}{\xi^2}
</math>
  </td>
</tr>
</table>

<div align="center">
<font color="maroon"><b>LAWE for n = 1 Polytrope</b></font><br />
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{d^2 x}{d\xi^2} 
+ 
\frac{2}{\xi} \biggl[1 +  \xi\cot\xi\biggr] \frac{dx}{d\xi}
+
\frac{1}{2} \biggl[\biggl( \frac{\sigma_c^2}{3}\biggr)\frac{\xi}{\sin\xi} - \frac{4}{\xi^2} \biggl(1 - \xi\cot\xi \biggr)  \biggr]x
</math>
  </td>
</tr>
</table>
</div>

[[SSC/Stability/Polytropes#Boundary_Conditions|Surface boundary condition]]:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>- \frac{d\ln x}{d\ln \xi} \biggr|_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl( \frac{3-n}{n+1}\biggr) + \frac{n\sigma_c^2}{6(n+1)}
\biggl[ \frac{\xi}{\theta'}\biggr]_\mathrm{surf}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ - \frac{d\ln x}{d\ln \xi} \biggr|_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
1 + \frac{\sigma_c^2}{12}
\biggl[ \frac{\xi^3}{(\xi \cos\xi - \sin\xi)}\biggr]_{\xi=\pi}
=
1 - \frac{\pi^2 \sigma_c^2}{12}
</math>
  </td>
</tr>
</table>
</td></tr></table>

===Spherical Shell===

In the context of a spherically symmetric n = 1 <math>(\gamma_g = 2, \alpha = 1)</math> shell (''envelope'') outside of a spherically symmetric bipolytropic ''core'', we should adopt the more general Lane-Emden structural solution,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>A \biggl[\frac{\sin\xi}{\xi}\biggr] - B \biggl[\frac{\cos\xi}{\xi}\biggr]</math>
  </td>
  <td align="right">
&nbsp; &nbsp; <math>\Rightarrow ~~~ Q(\xi) \equiv - \frac{d \ln \theta}{d\ln \xi}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl[1 - \xi\cot(\xi-\beta)  \biggr]
=
\frac{\xi^2 x_P}{3} \, .
</math>
  </td>
</tr>
</table>

<table border="5" bordercolor="purple" align="center" cellpadding="10" width="80%">
<tr>
  <td align="left">
Reminder: the expression for <math>x_P</math> is, 
<div align="center"><math>x_P = \frac{3}{\xi^2}\biggl[1 - \xi\cot(\xi-\beta)\biggr]</math>.</div>

Playing around a bit, we find that,

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

<tr>
  <td align="right"><math>\frac{\xi^2}{3} \cdot x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
1 - \xi \cdot \frac{\cos(\xi-\beta)}{\sin(\xi-\beta)}
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
1 - \xi 
\biggl[\frac{\cos(\xi-\beta)}{\sin(\xi-\beta)}\biggr] \cdot \frac{(\xi-\beta)}{(\xi-\beta)} 
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
1 - \xi 
\biggl[\frac{\cos(\xi-\beta)}{(\xi-\beta)}\biggr] 
\biggl[\frac{(\xi-\beta)}{\sin(\xi-\beta)}\biggr] 
</math></td>
</tr>
</table>

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

As a result, the governing LAWE becomes,
<!--
<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>
-->

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>0</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} + \biggl[4 - 2Q\biggr]\frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
+ 2\biggl[\cancelto{0}{\biggl(\frac{\sigma_c^2}{12}\biggr)}\frac{\xi^2}{\theta} - Q\biggr]\frac{x_P}{\xi^2}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{dx_P}{d\xi} 
- \biggl[2Q\biggr]\frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
- \biggl[2Q\biggr]\frac{x_P}{\xi^2}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{dx_P}{d\xi} 
- 2Q \biggl\{  \frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
+\frac{x_P}{\xi^2}\biggr\}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{dx_P}{d\xi} 
- \frac{2\xi^2 x_P}{3} \biggl\{  \frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
+\frac{x_P}{\xi^2}\biggr\}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \biggl[ \frac{4}{\xi}- \frac{2\xi x_P}{3}  \biggr] \cdot \frac{dx_P}{d\xi} 
- \frac{2 x_P^2}{3} 
\, .
</math></td>
</tr>
</table>

Let's plug in the expression for <math>x_P</math>, namely, <math>x_P = 3[1 - \xi\cot(\xi-\beta)]/\xi^2</math>.  We have, first of all,

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

<tr>
  <td align="right"><math>
x_p^2
  </math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{3^2}{\xi^4}\biggl[1 - \xi\cot(\xi-\beta) \biggr]^2
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{3^2}{\xi^4}\biggl[1 - 2\xi\cot(\xi-\beta) + \xi^2\cot^2 (\xi-\beta)\biggr] \, ;
</math></td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl[1 - \xi\cot(\xi-\beta)\biggr]
\frac{d}{d\xi}\biggl\{
\frac{3}{\xi^2}
\biggr\}
-
\frac{3}{\xi}
\frac{d}{d\xi}
\biggl[\cot(\xi-\beta)\biggr]
-
\frac{3}{\xi^2}
\biggl[\cot(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{6}{\xi^3}\biggl[1 - \xi\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi}
\biggl[1 + \cot^2(\xi-\beta)\biggr]
-
\frac{3}{\xi^2}
\biggl[\cot(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr] \, .
</math></td>
</tr>
</table>


<table border="5" bordercolor="purple" align="center" cellpadding="10" width="80%">
<tr>
  <td align="left">
Note for later use that,

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

<tr>
  <td align="right">
<math>
\frac{x_P^2}{r^2} \biggl(\frac{d\ln x_P}{d \ln r}\biggr)^2 = \biggl(\frac{dx_P}{dr} \biggr)^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr] \biggr\}^2

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

Recognize that we have used the trigonometric relations, 

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

<tr>
  <td align="right">
<math>\frac{d}{d\xi}\biggl[\cot(u)\biggr]</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- ~\frac{1}{\sin^2(u)} ~\frac{du}{d\xi}
=
- ~ \biggl[1 + \cot^2(u)\biggr] \frac{du}{d\xi}
\, .
</math></td>
</tr>
</table>
And,

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

<tr>
  <td align="right">
<math>
\frac{d^2x_p}{d\xi^2}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d}{d\xi}\biggl\{
- \frac{6}{\xi^3}
+
\frac{3}{\xi} 
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+ 
\frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\}
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{3}{\xi^2} 
- \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{3}{\xi^2}\biggl[1 + \cot^2(\xi-\beta)\biggr]
- \frac{3}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
- \frac{6}{\xi}\biggl[\cot(\xi-\beta)\biggr]\biggl[1 + \cot^2(\xi-\beta) \biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] - \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
\, .
</math></td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>
\frac{d^2x_P}{d\xi^2} 
+ \biggl[ \frac{4}{\xi}- \frac{2\xi x_P}{3}  \biggr] \cdot \frac{dx_P}{d\xi} 
- \frac{2 x_P^2}{3} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl\{
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] - \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
\biggr\}
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+
\biggl[ \frac{4}{\xi}- \frac{2\xi x_P}{3}  \biggr] \cdot \biggl\{
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
-
\biggl\{
\frac{6}{\xi^4}\biggl[1 - 2\xi\cot(\xi-\beta) + \xi^2\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+
\frac{4}{\xi} \biggl\{
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+
\biggl[ \frac{2\xi x_P}{3}  \biggr] \cdot \biggl\{
 \frac{6}{\xi^3}
- \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
-
\frac{3}{\xi} - \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{6}{\xi^4}\biggl[-1 + 2\xi\cot(\xi-\beta) - \xi^2\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
-
\frac{24}{\xi^4}
+ \frac{12}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{12}{\xi^2} + \frac{12}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr] 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{2}{\xi}\biggl[1-\xi\cot(\xi-\beta)\biggr]\cdot \biggl\{
 \frac{6}{\xi^3}
- \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
-
\frac{3}{\xi} - \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{6}{\xi^4}\biggl[-1 + 2\xi\cot(\xi-\beta) - \xi^2\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
-\frac{6}{\xi^4} + \frac{6}{\xi^2}  + \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
+ \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^3}\cdot \cot(\xi-\beta) 
+ \frac{6}{\xi} \cdot\cot(\xi-\beta) 
+ \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr] 
+ \frac{6}{\xi}\biggl[\cot^3(\xi-\beta)\biggr] 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{12}{\xi^4}
- \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
-
\frac{6}{\xi^2} - \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{6}{\xi^4} + \frac{12}{\xi^3}\cdot \cot(\xi-\beta) 
- \frac{6}{\xi^2}\cdot \cot^2(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
+ \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{6}{\xi} \cdot\cot(\xi-\beta) 
+ \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr] 
+ \frac{6}{\xi}\biggl[\cot^3(\xi-\beta)\biggr] 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

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

<hr>

<b>Debugging LaTeX layout:</b>
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^2}\biggl[\cot^3(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ \biggl[\cot^1(\xi-\beta)\biggr]
+ \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)_{m}\biggr] 
+ \frac{6}{\xi}\biggl[\cot^3(\xi-\beta)\biggr] 
</math>
  </td>
</tr>
</table>