Difference between revisions of "User:Tohline/Appendix/Ramblings/PPTori"

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<!-- __NOTOC__ will force TOC off -->
=Stability Analyses of PP Tori=
=Stability Analyses of PP Tori=
<font color="red"><b>[Comment by J. E. Tohline on 24 May 2016]</b></font> &nbsp; This chapter contains a set of technical notes and accompanying discussion that I put together several months ago as I was trying to gain a foundational understanding of the results of a large study of instabilities in self-gravitating tori published by the Imamura &amp; Hadley collaboration.  I have come to appreciate that some of the logic and interpretation of published results that are presented, below, has serious flaws.  Therefore, anyone reading this should be quite cautious in deciding what subsections provide useful insight.  I have written a separate chapter titled, "[[User:Tohline/Apps/ImamuraHadleyCollaboration#Characteristics_of_Unstable_Eigenvectors_in_Self-Gravitating_Tori|Characteristics of Unstable Eigenvectors in Self-Gravitating Tori]]," that contains a much more trustworthy analysis of this very interesting problem.
{{LSU_WorkInProgress}}
{{LSU_HBook_header}}
{{LSU_HBook_header}}


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</div>
</div>


<div align="center">
<div align="center" id="CubeRootImaginary">
<table border="1" width="60%" cellpadding="8">
<table border="1" width="60%" cellpadding="8">
<tr><td align="left">
<tr><td align="left">
Line 2,522: Line 2,528:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + xb ]^{-3/2} [4 + 9xb ] 
<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{
- [2b + 3xb^2 ]  + [6b + 6xb^2 ]
\biggr\} \frac{1}{2(1+xb)^{3/2}}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial \Lambda}{\partial \theta}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,541: Line 2,543:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\partial }{\partial \theta} \biggl\{
<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb]
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 2,556: Line 2,556:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\pm ~i~\frac{\partial }{\partial \theta} \biggl\{ \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2}   
<math>~\pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + xb ]^{-3/2} [4 + 3xb ]  
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<!-- SEPARATE EVALUATION OF 2nd DERIVATIVE wrt "x" -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">


<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{\partial^2\Lambda}{\partial x^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,571: Line 2,580:
   <td align="left">
   <td align="left">
<math>~
<math>~
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot x^3 \cdot  [\sin\theta (-3 + 3\cos^2\theta)]
2(n+1)[2^3(n+1)\cos^2\theta -3](1+3xb) ~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \, ,
+ [x^2 + x^3(3\cos\theta - \cos^3\theta)] \cdot [ -2^4(n+1)^2 \sin\theta \cos\theta ]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr><td colspan="3" align="left">which is the same.</td></tr>
</table>
</td></tr></table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{\partial \Lambda}{\partial \theta}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\frac{\partial }{\partial \theta} \biggl\{
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\frac{\partial }{\partial \theta} \biggl\{ \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} 
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot x^3 \cdot  [\sin\theta (-3 + 3\cos^2\theta)]
+ [x^2 + x^3(3\cos\theta - \cos^3\theta)] \cdot [ -2^4(n+1)^2 \sin\theta \cos\theta ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{-1/2}  \cdot  
~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{-1/2}  \cdot  
(-\sin\theta) [2x^2\cos\theta + x^3(9\cos^2\theta - 5\cos^4\theta) ]  
(-\sin\theta) [2x^2\cos\theta + x^3(9\cos^2\theta - 5\cos^4\theta) ]  
Line 2,956: Line 3,020:
</div>
</div>


=====Step 3=====
 
From our [[User:Tohline/Apps/PapaloizouPringle84#isolatingBlaes85|accompanying discussion of the Blaes85 derivation]], we expect the following equality to hold (see his equations 4.1 and 4.2):
<!-- SEPARATE EVALUATION OF 2nd DERIVATIVE wrt "theta" -->
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="1" cellpadding="8" align="center">
 
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\hat{L} (\delta W)</math>
<math>~\frac{\partial^2\Lambda}{\partial \theta^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,969: Line 3,036:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)(\delta W) \, ,</math>
<math>~
x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\hat{L} (\delta W)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\Theta_H x^2\cdot \frac{\partial^2(\delta W)}{\partial x^2}
+ x^3\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta
+\Theta_H \cdot \frac{\partial^2(\delta W)}{\partial\theta^2}
\biggr\}
+ \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x}
\biggr\} \cdot \frac{\partial (\delta W)}{\partial x}  
</math>
</math>
   </td>
   </td>
Line 3,005: Line 3,067:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } +  
\pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{
n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr]
(\beta\eta)\cos\theta + \frac{3x^3\sin^2\theta}{2(\beta\eta)}(5\cos^2\theta -2) + \frac{3^2x^6\sin^6\theta\cos\theta}{2^2(\beta\eta)^3}
\cdot \frac{\partial (\delta W)}{\partial\theta}
\biggr\} \, .
+ \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} -  \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]\delta W \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>
=====Step 3=====
From our [[User:Tohline/Apps/PapaloizouPringle84#isolatingBlaes85|accompanying discussion of the Blaes85 derivation]], we expect the following equality to hold (see his equations 4.1 and 4.2):
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~M</math>
<math>~\hat{L} (\delta W)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{x^2}{(1-\Theta_H)\beta^2} \, ,</math>
<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)(\delta W) \, ,</math>
  </td>
</tr>
</table>
</div>
where,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\hat{L} (\delta W)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\Theta_H x^2\cdot \frac{\partial^2(\delta W)}{\partial x^2}
+\Theta_H \cdot \frac{\partial^2(\delta W)}{\partial\theta^2}
+ \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x}
\biggr\} \cdot \frac{\partial (\delta W)}{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } +
n\cdot \frac{\partial \Theta_H}{\partial\theta}  \biggr]
\cdot \frac{\partial (\delta W)}{\partial\theta}
+ \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4}  -  \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]\delta W \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~M</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{x^2}{(1-\Theta_H)\beta^2} \, ,</math>
   </td>
   </td>
</tr>
</tr>
Line 4,233: Line 4,356:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\pm~~i~\{~~~\}</math>
<math>~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+x(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta
- x[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>
   </td>
   </td>
</tr>
</tr>


</table>
<tr>
 
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- x^2 \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ]
\biggr\}</math>
  </td>
</tr>
 
</table>
 
</td></tr></table>
</td></tr></table>
</div>
</div>
Line 4,278: Line 4,415:
</div>
</div>


Taken together, then, we have,


<!-- EVALUATE 2nd CROSS-TERM -->
<div align="center">
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{L}_{LHS} </math>
<math>~ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,292: Line 4,433:
   <td align="left">
   <td align="left">
<math>~
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 - \cancelto{0}{x^3b}] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
x\cdot 2^2(n+1)[2^3(n+1)\cos^2\theta -3]
</math>
</math>
   </td>
   </td>
Line 4,306: Line 4,447:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3b} ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
+ x^2\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \}
- n (1-x\cos\theta)\cdot \cancelto{0}{x^3} \biggl[ ( 2 +3xb )\cdot \frac{\partial  \Lambda }{\partial x} 
- 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\}  
</math>
</math>
   </td>
   </td>
Line 4,322: Line 4,461:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 - \cancelto{0}{x^3}b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
+x^3 \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \}
\cdot m^2 x^2\Lambda
</math>
</math>
   </td>
   </td>
Line 4,337: Line 4,475:
   <td align="left">
   <td align="left">
<math>~
<math>~
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot (1-x\cos\theta)^2  \Lambda
+x^3 (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \}
</math>
</math>
   </td>
   </td>
Line 4,347: Line 4,485:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\pm~~i~\beta~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+xb} \biggr]^{1/2} \biggl\{
(1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
4\cos\theta + 6x(2b\cos\theta + \sin^4\theta) + 3x^2(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta)
</math>
\biggr\} </math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>
Taken together, then, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{L}_{LHS} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
(1-x\cos\theta)^4 [ \beta^2 - x^2 - \cancelto{0}{x^3b}] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
\biggr\}  
</math>
</math>
   </td>
   </td>
Line 4,380: Line 4,529:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\}
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3b} ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
\cdot m^2 x^2\Lambda  
- n (1-x\cos\theta)\cdot \cancelto{0}{x^3} \biggl[ (  2 +3xb )\cdot \frac{\partial  \Lambda }{\partial x}  
- 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\}
</math>
</math>
   </td>
   </td>
Line 4,395: Line 4,545:
   <td align="left">
   <td align="left">
<math>~
<math>~
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 - \cancelto{0}{x^3}b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\}
\cdot m^2 x^2\Lambda  
</math>
</math>
   </td>
   </td>
Line 4,405: Line 4,556:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \Lambda
</math>
</math>
   </td>
   </td>
Line 4,419: Line 4,570:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta   
(1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
</math>
</math>
   </td>
   </td>
Line 4,436: Line 4,587:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}  
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta\biggr]
\biggr\}  
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,453: Line 4,603:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ x\beta^2(1-x\cos\theta)^3 \biggl\{
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
(1-2x \cos\theta )\cdot \biggl[ 2x(n+1)[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2\biggr]
\cdot m^2 x^2\Lambda
</math>
</math>
   </td>
   </td>
Line 4,468: Line 4,618:
   <td align="left">
   <td align="left">
<math>~
<math>~
~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2}  \cdot \sin^2\theta \biggr\}
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2  \Lambda
</math>
</math>
   </td>
   </td>
Line 4,478: Line 4,628:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ m^2 x^2\biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
(1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
\cdot \biggl\{- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,496: Line 4,645:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \cdot \biggl\{
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta   
- (4n+1)\beta^2  ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Let's further simplify:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{L}_{LHS} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
x^2\biggl\{ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]  
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)   -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta\biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,534: Line 4,676:
   <td align="left">
   <td align="left">
<math>~
<math>~
-~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-x\cos\theta)^2  \cdot
+ x\beta^2(1-x\cos\theta)^3 \biggl\{  
\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
(1-2x \cos\theta )\cdot \biggl[ 2x(n+1)[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2}   \biggr]
</math>
</math>
   </td>
   </td>
Line 4,549: Line 4,691:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \beta^2(1-x\cos\theta)^3
~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2}  \cdot \sin^2\theta  \biggr\}
(1-2x \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr]
</math>
</math>
   </td>
   </td>
Line 4,564: Line 4,705:
   <td align="left">
   <td align="left">
<math>~
<math>~
- 2n(4n+1)\beta^2   m^2  
+ m^2 x^2\biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n (1-x\cos\theta)^2 \biggr\}
+ (4n+1)\beta^2  m^2 [4n + \beta^2 - x^2 ] (1-x\cos\theta)^2
\cdot \biggl\{- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}\biggr\}
- (4n+1)\beta^2  m^2 n (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr]  
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,581: Line 4,720:
   <td align="left">
   <td align="left">
<math>~
<math>~
~~\pm ~~i x\beta^3 (1-x\cos\theta)^3
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \cdot \biggl\{
(1-2x \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2}
- (4n+1)\beta^2  ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Let's further simplify:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{L}_{LHS} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta (1-x\cos\theta)^4 [ \beta^2 - x^2 ]  
x^2\biggl\{ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]  
</math>
</math>
   </td>
   </td>
Line 4,610: Line 4,757:
   <td align="left">
   <td align="left">
<math>~
<math>~
\pm ~i~\beta m^2 x^2\biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n (1-x\cos\theta)^2 \biggr\}
-~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot
\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
</math>
   </td>
   </td>
Line 4,624: Line 4,771:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}
+ \beta^2(1-x\cos\theta)^3  
[2\cos^2\theta(15 - 7\cos^2\theta)    - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
(1-2x \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3]  \biggr]
</math>
</math>
   </td>
   </td>
Line 4,640: Line 4,787:
   <td align="left">
   <td align="left">
<math>~
<math>~
~~\pm ~~i~\beta x^2 (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-x\cos\theta)^[1 + xb ]^{-1/2} \cdot \sin^2\theta 
- 2n(4n+1)\beta^2   m^2
+ (4n+1)\beta^2  m^2  [4n + \beta^2 - x^2 ] (1-x\cos\theta)^2
- (4n+1)\beta^2   m^2 n (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,654: Line 4,804:
   <td align="left">
   <td align="left">
<math>~
<math>~
\pm i~x^2(-1) (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2   
~~\pm ~~i x\beta^3 (1-x\cos\theta)^3
(1-2x \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
=====Step 6=====
Hence, to lowest order we want to compare the following two expressions:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
2n(n+1) (n+1)(1-\cancelto{0}{x}\cos\theta)^2[ 4n + \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] + n(1-\cancelto{0}{x}\cos\theta)^4 [ 2(n+1) + 3\beta^2 ]  
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta (1-x\cos\theta)^4 [ \beta^2 - x^2 ]  
</math>
</math>
   </td>
   </td>
Line 4,689: Line 4,833:
   <td align="left">
   <td align="left">
<math>~
<math>~
\pm~i~  (1-\cancelto{0}{x}\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}  
\pm ~i~\beta m^2 x^2\biggl\{ 2n  - (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
</math>
   </td>
   </td>
Line 4,699: Line 4,844:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
2n(n+1) (n+1)[ 4n + \beta^2 ] + n [ 2(n+1) + 3\beta^2 ] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)   -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
</math>
</math>
   </td>
   </td>
Line 4,713: Line 4,859:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3n\beta^2 + (n+1)\biggl[
<math>~
2n -  4n - \beta^2  + 2n \biggr] \pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
~~\pm ~~i~\beta x^2 (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-x\cos\theta)^3  [1 + xb ]^{-1/2} \cdot \sin^2\theta 
</math>
</math>
   </td>
   </td>
Line 4,727: Line 4,873:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\beta^2 (2n-1) ~\pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
<math>~
\pm i~x^2(-1) (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-x\cos\theta)^2 
</math>
</math>
   </td>
   </td>
Line 4,737: Line 4,884:
</div>
</div>


 
=====Step 6=====
 
Hence, to lowest order we want to compare the following two expressions:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 4,744: Line 4,891:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Re}\biggl[\frac{\mathcal{L}_{LHS}}{x^2}\biggr] </math>
<math>~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,751: Line 4,898:
   <td align="left">
   <td align="left">
<math>~
<math>~
(1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]  
2n(n+1) -  (n+1)(1-\cancelto{0}{x}\cos\theta)^2[ 4n + \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] + n(1-\cancelto{0}{x}\cos\theta)^4 [ 2(n+1) + 3\beta^2 ]  
</math>
</math>
   </td>
   </td>
Line 4,765: Line 4,912:
   <td align="left">
   <td align="left">
<math>~
<math>~
-~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-x\cos\theta)^2 \cdot
\pm~i~ (1-\cancelto{0}{x}\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}  
\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2}
</math>
</math>
   </td>
   </td>
Line 4,776: Line 4,922:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \beta^2(1-\cancelto{0}{x}\cos\theta)^3
2n(n+1) - (n+1)[ 4n + \beta^2  ] + n [ 2(n+1) + 3\beta^2 ] \pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
(1-\cancelto{0}{2x} \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3]  \biggr]
</math>
</math>
   </td>
   </td>
Line 4,791: Line 4,936:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~3n\beta^2 + (n+1)\biggl[
- 2n(4n+1)\beta^2   m^2
2n - 4n - \beta^2 + 2n \biggr] \pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
+ (4n+1)\beta^2  m^2 [4n + \beta^2 - \cancelto{0}{x^2} ] (1-\cancelto{0}{x}\cos\theta)^2
- (4n+1)\beta^2   m^2 n (1-\cancelto{0}{x}\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr]
</math>
</math>
   </td>
   </td>
Line 4,810: Line 4,953:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\beta^2 (2n-1) ~\pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]  
+ \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta  -6(n+1)\biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathrm{Re}\biggl[\frac{\mathcal{L}_{LHS}}{x^2}\biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~+ \beta^2\biggl\{- 2n(4n+1)  m^2+ (4n+1)m^2  [4n  ] - 2(4n+1) m^2 n
<math>~
\biggr\} + \beta^4\biggl\{(4n+1) m^2 - (4n+1) m^2 n \biggl[ \frac{3}{(n+1)} \biggr]
(1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]  
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,837: Line 4,984:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1)  \biggr]
-~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-x\cos\theta)^2 \cdot
+ \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta   -6(n+1)\biggr]
\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2}
</math>
</math>
   </td>
   </td>
Line 4,856: Line 5,003:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ m^2\beta^2\biggl[  -2^4(n+1)^2 (1 - 2\sin^2\theta ) \biggr]
+ \beta^2(1-\cancelto{0}{x}\cos\theta)^3
+ m^2\beta^2\biggl\{- 2n(4n+1)   + 4n(4n+1)  - 2n(4n+1) \biggr\}
(1-\cancelto{0}{2x} \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr]
</math>
</math>
   </td>
   </td>
Line 4,871: Line 5,018:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ (4n+1) m^2\beta^4\biggl[1 -  \frac{3n}{(n+1)} \biggr]
- 2n(4n+1)\beta^2  m^2
+ (4n+1)\beta^2  m^2 [4n + \beta^2 - \cancelto{0}{x^2} ] (1-\cancelto{0}{x}\cos\theta)^2
- (4n+1)\beta^2  m^2 n (1-\cancelto{0}{x}\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr]  
</math>
</math>
   </td>
   </td>
Line 4,885: Line 5,034:
   <td align="left">
   <td align="left">
<math>~
<math>~
2\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr]  
\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]  
+ m^2\beta^2\biggl[ 2^4(n+1)^2 (1 - 2\cos^2\theta ) \biggr]  
+ \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta   -6(n+1)\biggr]
</math>
</math>
   </td>
   </td>
Line 4,899: Line 5,048:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~+ \beta^2\biggl\{- 2n(4n+1)  m^2+ (4n+1)m^2 [4n  ] - 2(4n+1)  m^2 n
+ (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr]
\biggr\} + \beta^4\biggl\{(4n+1) m^2 - (4n+1) m^2 n \biggl[ \frac{3}{(n+1)} \biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,914: Line 5,064:
   <td align="left">
   <td align="left">
<math>~
<math>~
(1-m^2)2^5\beta^2(n+1)^2\cos^2\theta  
\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr]
+ 2^2(n+1)\beta^2\biggl[ 4m^2(n+1) -3\biggr]
+ \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta  -6(n+1)\biggr]
+ (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\pm~\mathrm{Im}\biggl[\mathcal{L}_{LHS} \biggr]</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
~x\biggl\{\beta^3 (1-\cancelto{0}{x}\cos\theta)^3
+ m^2\beta^2\biggl[  -2^4(n+1)^2 (1 - 2\sin^2\theta ) \biggr]
(1-\cancelto{0}{2x} \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + \cancelto{0}{xb} )^{-1/2}
+ m^2\beta^2\biggl\{- 2n(4n+1)   + 4n(4n+1) - 2n(4n+1) \biggr\}  
</math>
</math>
   </td>
   </td>
Line 4,949: Line 5,093:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
-~m^2\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + \cancelto{0}{xb} ]^{-3/2} \cos\theta  (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x}^2 ]  \biggr\}
+ (4n+1) m^2\beta^4\biggl[1 \frac{3n}{(n+1)} \biggr]
</math>
</math>
   </td>
   </td>
Line 4,960: Line 5,104:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+~x^2 \biggl\{ \beta m^2 \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
2\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr]  
\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2}
+ m^2\beta^2\biggl[ 2^4(n+1)^2 (1 - 2\cos^2\theta ) \biggr]  
</math>
</math>
   </td>
   </td>
Line 4,978: Line 5,122:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
+~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + \cancelto{0}{xb} )^{-3/2}  
+ (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr]
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
</math>
</math>
   </td>
   </td>
Line 4,990: Line 5,133:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
-~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-\cancelto{0}{x}\cos\theta)^3  [1 + \cancelto{0}{xb} ]^{-1/2} \cdot \sin^2\theta 
(1-m^2)2^5\beta^2(n+1)^2\cos^2\theta
+ 2^2(n+1)\beta^2\biggl[ 4m^2(n+1-3\biggr]
+ (4n+1) m^2\beta^4\biggl[1 \frac{3n}{(n+1)} \biggr] \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\pm~\mathrm{Im}\biggl[\mathcal{L}_{LHS} \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
-~ (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-\cancelto{0}{x}\cos\theta)^2  \biggr\}
~x\biggl\{\beta^3 (1-\cancelto{0}{x}\cos\theta)^3  
(1-\cancelto{0}{2x} \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + \cancelto{0}{xb} )^{-1/2}  
</math>
</math>
   </td>
   </td>
Line 5,018: Line 5,169:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
-~m^2\beta  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + \cancelto{0}{xb} ]^{-3/2} \cos\theta (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x}^2 ]  \biggr\}
</math>
</math>
   </td>
   </td>
Line 5,035: Line 5,186:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
+~~x^2 \beta^3 [ 2^3\cdot 3(n+1)^3 ]^{1/2}\biggl\{(12\cos^2\theta - 4\cos^4\theta) - 4\sin^2\theta
+~x^2 \biggl\{ \beta m^2 \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2}
</math>
</math>
   </td>
   </td>
Line 5,049: Line 5,201:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- m^2 \biggl[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta  
<math>
-~ \frac{n (4n+1)}{(n+1)^2} \biggr]\biggr\}
+~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + \cancelto{0}{xb} )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta\biggr]
</math>
</math>
   </td>
   </td>
Line 5,060: Line 5,213:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
-~\beta  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-\cancelto{0}{x}\cos\theta)^3  [1 + \cancelto{0}{xb} ]^{-1/2}  \cdot \sin^2\theta 
</math>
</math>
   </td>
   </td>
Line 5,077: Line 5,230:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
+~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{4\cos^2\theta - \cos^4\theta -1
-~ (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-\cancelto{0}{x}\cos\theta)^2 \biggr\}
</math>
</math>
   </td>
   </td>
Line 5,088: Line 5,241:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \frac{m^2}{2} \biggl[\cos^2\theta(15 - 7\cos^2\theta)  -12\sin^4\theta +6 \sin^2\theta
<math>~
-~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\}
~x(1-m^2)\cdot \beta^3  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
</math>
</math>
   </td>
   </td>
Line 5,102: Line 5,255:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
+~~x^2 \beta^3 [ 2^3\cdot 3(n+1)^3 ]^{1/2}\biggl\{(12\cos^2\theta - 4\cos^4\theta) - 4\sin^2\theta  
</math>
</math>
   </td>
   </td>
Line 5,119: Line 5,272:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~- m^2 \biggl[2\cos^2\theta(15 - 7\cos^2\theta)   - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta  
+~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{3-(1+\sin^2\theta)^2
-~ \frac{n (4n+1)}{(n+1)^2} \biggr]\biggr\}
- \frac{m^2}{2} \biggl[33\cos^2\theta-19 \cos^4\theta -6
-~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
=====Examples=====
Evaluate various expressions using the parameter set:&nbsp;&nbsp;
<math>~(n, \theta, x) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~b</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3}{2} - \frac{1}{8} = \frac{11}{8} </math>
<math>~
~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
</math>
   </td>
   </td>
</tr>
</tr>
Line 5,152: Line 5,294:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~(\beta\eta)^2</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl(\frac{1}{2^2}\biggr)^2\biggl[ 1 + \frac{11}{2^5} \biggr] = \frac{2^5 + 11}{2^9} = \frac{43}{2^9} </math>
<math>
+~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{4\cos^2\theta - \cos^4\theta -1
</math>
   </td>
   </td>
</tr>
</tr>
Line 5,164: Line 5,308:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Re}(\Lambda)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~- \frac{m^2}{2} \biggl[\cos^2\theta(15 - 7\cos^2\theta)  -12\sin^4\theta +6 \sin^2\theta
-5\beta^2 + \frac{43}{2^9} \biggl[ 2^3 - 6 \biggr]
-~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\}
= -5\beta^2 + \frac{43}{2^8}  
</math>
</math>
   </td>
   </td>
Line 5,179: Line 5,322:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Im}(\Lambda)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{\beta}{2}\biggl[ 2^{10}\cdot 3 \cdot \frac{43}{2^9} \biggr]^{1/2}
~x(1-m^2)\cdot \beta^3  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
= \beta \biggl[ \frac{3\cdot 43}{2} \biggr]^{1/2}
</math>
</math>
   </td>
   </td>
Line 5,194: Line 5,336:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{3-(1+\sin^2\theta)^2
- \frac{m^2}{2} \biggl[33\cos^2\theta-19 \cos^4\theta -6
-~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
 
=====Examples=====
 
Evaluate various expressions using the parameter set:&nbsp;&nbsp;
<math>~(n, \theta, x) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,200: Line 5,369:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{3}{2} - \frac{1}{8} = \frac{11}{8} </math>
\biggl[ 2^3 - 6 \biggr]\frac{1}{2^2}\biggl( 2 + \frac{3\cdot 11}{2^2\cdot 2^3} \biggr)
  </td>
= \biggl( 1 + \frac{33}{2^6} \biggr)
  <td align="right">
</math>
1.375000000
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~(\beta\eta)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{1}{2^2}\biggr)^2\biggl[ 1 + \frac{11}{2^5} \biggr] = \frac{2^5 + 11}{2^9} = \frac{43}{2^9} </math>
  </td>
  <td align="right">
0.083984375
   </td>
   </td>
</tr>
</tr>
Line 5,209: Line 5,393:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>
<math>~\mathrm{Re}(\Lambda)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,216: Line 5,400:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{\beta}{2} \biggl[ \frac{2^9\cdot 2^8\cdot 3}{43} \biggr]^{1/2} \biggl[ \frac{1}{2^2}\cdot \biggl(2 + \frac{3\cdot 11}{2^2\cdot 2^3}\biggr) \biggr]
-5\beta^2 + \frac{43}{2^9} \biggl[ 2^3 - 6 \biggr]
= -5\beta^2 + \frac{43}{2^8}
</math>
  </td>
  <td align="right">
<math>~- 5\beta^2</math> + 0.167968750
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}(\Lambda)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\beta}{2}\biggl[ 2^{10}\cdot 3 \cdot \frac{43}{2^9} \biggr]^{1/2}
= \beta \biggl[ \frac{3\cdot 43}{2} \biggr]^{1/2}
</math>
  </td>
  <td align="right">
8.031189202 <math>~\beta</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2^3 - 6 \biggr]\frac{1}{2^2}\biggl( 2 + \frac{3\cdot 11}{2^2\cdot 2^3} \biggr)
= \biggl( 1 + \frac{33}{2^6} \biggr)
</math>
  </td>
  <td align="right">
1.515625000
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\beta}{2} \biggl[ \frac{2^9\cdot 2^8\cdot 3}{43} \biggr]^{1/2} \biggl[ \frac{1}{2^2}\cdot \biggl(2 + \frac{3\cdot 11}{2^2\cdot 2^3}\biggr) \biggr]
= \beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)  
= \beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)  
</math>
  </td>
  <td align="right">
36.23373732 <math>~\beta</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl(\frac{3}{4}\biggr)^{1/2} \biggl\{-2^4 \cdot \frac{43}{2^9} +\frac{3}{2^6}\biggl(\frac{3}{4}\biggr)\biggl[3-4\biggr]
\biggr\}
= ~-~\frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
-2.388335684
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot \biggl[ \frac{2^{10} \cdot 3\cdot 43}{2^9} \biggr]^{1/2}
\biggl\{1 + \frac{3\cdot 2^9}{2^7 \cdot 43} \biggl(\frac{3}{2^3}\biggr)\biggr\} =
(-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43  ]^{1/2}
\biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\}
</math>
  </td>
  <td align="right">
(-1) &times; 15.36617018 <math>~\beta</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^2[2^2-3]\biggl[1 + \frac{3}{2^2}\cdot \frac{11}{2^3}  \biggr] = 2^2 + \frac{33}{2^3} = \frac{65}{8}
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;  8.125000000
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta~
2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2}
=\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr] \beta
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;  30.76957507<math>~\beta</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl(\frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^4} \biggl\{2^6 \biggl(\frac{3}{2^2} - \frac{1}{2^2} \biggr)
\biggr\} + \frac{1}{2^6}\biggl\{-2^3\cdot 3  + 2 + \frac{3^3\cdot 5\cdot 7}{2^2}  -3\cdot 23
\biggr\} = 2 + \frac{1}{2^8}\biggl\{2^3 + 3^3\cdot 5\cdot 7  - 2^2\cdot 3\cdot 31
\biggr\}
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;  4.269531250
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)\beta [2^{10}\cdot 3]^{1/2} \biggl( \frac{43}{2^9} \biggr)^{1/2} \biggl\{
\frac{1}{2} + \frac{3}{2}\cdot \frac{2^9}{43} \cdot \frac{1}{2^6}\cdot \frac{3}{2^2}  \biggl(\frac{5}{2^2} -2 \biggr)
+ \biggl( \frac{3}{2}\cdot \frac{1}{2^6} \cdot \frac{2^9}{43}\biggr)^2 \biggl( \frac{3}{2^2} \biggr)^3 \frac{1}{2}
\biggr\}
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)\beta\biggl( \frac{3\cdot 43}{2} \biggr)^{1/2} \biggl\{
1 - \frac{3^3}{2\cdot 43} + \frac{3^5}{(2\cdot 43)^2}
\biggr\}
</math>
  </td>
  <td align="right">
(-1) &times; 5.773638858 <math>~\beta</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \mathrm{Re}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr)
-~\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~\frac{(2\cdot 3\cdot 97)-3\cdot (2^3\cdot 43 + 9)}{2^9}
</math>
  </td>
  <td align="right">
-0.931640625
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \mathrm{Im}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)
+~\frac{\sqrt{3}}{2}\cdot (-1) \beta~\frac{\sqrt{3}}{2}\cdot[ 2\cdot 3\cdot 43  ]^{1/2}
\biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\}
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
<!--
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl\{ \biggl[ \frac{3^3}{2^{3}\cdot 43} \biggr]^{1/2} (2^{6} + 3\cdot 11 )
-~\biggl[ \frac{3^4}{2^5\cdot 43} \biggr]^{1/2}
( 2\cdot 43 + 3^2 ) \biggr\}
</math>
  </td>
  <td align="right">
NEW:&nbsp;13.86780926 <math>~\beta</math>
  </td>
</tr>
-->
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl[ \frac{3^3}{2^5\cdot 43} \biggr]^{1/2} [2 (2^6 + 33) - (2\cdot 43 + 3^2) ]
</math>
  </td>
  <td align="right">
13.86780926 <math>~\beta</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \mathrm{Re}\biggl\{ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr)
+~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{97^2}{2^{11}} +~\frac{3^3}{2^{11}}\cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
9.248046874
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \mathrm{Im}\biggl\{ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl\{ \biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr) \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)
+~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43  ]^{1/2}
\biggl[1 + \frac{3^2}{2\cdot 43} \biggr]\biggr\}
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl\{ \biggl( \frac{3}{2^9\cdot 43} \biggr)^{1/2}  \biggl[ ( 2^6 + 33)^2
+~3^3(2\cdot 43 +3^2 ) \biggr]\biggr\}
</math>
  </td>
  <td align="right">
139.7753772
  </td>
</tr>
</table>
</div>
=====Step 7=====
Let's begin by slightly redefining the LHS and RHS collections of terms.
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{RHS}_4</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\mathrm{RHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~x^2 ~ [ 2^2(n+1)^2  - m^2 \Lambda ] \cdot \mathcal{A} \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathrm{LHS}_4</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\mathrm{LHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
</math>
  </td>
</tr>
</table>
<!-- Early attempt at lowest order ****************
======To Lowest Order======
Now, let's gather together only the lowest order components of all the LHS terms.
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{LHS}_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta)
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n \cancelto{0}{x^3} (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\beta^2~\biggl\{
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{1}{\beta^2} \cdot \mathrm{LHS}_4</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3\cancelto{0}{x}b)}{(1+\cancelto{0}{x}b)^{3/2}} \biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
~~\pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{
\cancelto{x}{(\beta\eta)}\cos\theta \biggr\} 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x\biggl\{
x(n+1)[-6 + 2^4(n+1)\cos^2\theta]
~~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl[ 2\cos\theta
- \cancelto{0}{x}(2 - 7\cos^2\theta + 3\cos^4\theta )\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3] + 2^4(n+1)^2(\sin^2\theta - \cos^2\theta) + (n+1)[-6 + 2^4(n+1)\cos^2\theta] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\pm~~i~\biggl\{(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \cdot x\cos\theta
+\beta x [ 2^5\cdot 3 (n+1)^3 ]^{1/2} \cdot 2\cos\theta
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \cdot 2(n+1)\biggl\{ [2^3(n+1)\cos^2\theta -3] + 2^3(n+1)(\sin^2\theta - \cos^2\theta) + [-3 + 2^3(n+1)\cos^2\theta] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\pm~~i~x\beta \cos\theta\biggl\{
[ 2^5\cdot 3 (n+1)^3 ]^{1/2} \cdot 2 - [2^7\cdot 3 (n+1)^3 ]^{1/2}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \cdot 2(n+1)\biggl\{2^3(n+1)-6\biggr\}
\pm~~i~x\beta \cos\theta\biggl\{ ~0~ \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \cdot 4(n+1)(4n+1) \, .
</math>
  </td>
</tr>
</table>
Contrast this with,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{RHS}_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2 ~ [m^2 \Lambda - 2^2(n+1)^2 ] \cdot \mathcal{A}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\beta^2 ~~\pm~i~\cancelto{0}{x}\beta \cos\theta [2^7\cdot 3(n+1)^3]^{1/2}\biggr]
- 2^2(n+1)^2 \biggr\} \cdot \mathcal{A}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\beta^2 \biggr]
- 2^2(n+1)^2 \biggr\} \cdot \mathcal{A}
</math>
  </td>
</tr>
</table>
Now, to lowest order [<font color="red">Case B</font>],
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(n+1)\cdot \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)(1-\cancelto{0}{x}\cos\theta)^2 [ \cancelto{0}{ x^2}(1+xb) -  \beta^2 - 4n ]
+ (1-\cancelto{0}{x}\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~\cancelto{0}{x}\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2n  (n+1) + (n+1)[ -  \beta^2 - 4n ]+ [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-  \beta^2(4n+1)
</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{\beta^2}\cdot \mathrm{RHS}_4</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\cancelto{0}{\beta^2} \biggr]
- 2^2(n+1)^2 \biggr\} \cdot \biggl[ \frac{-(4n+1)}{(n+1)} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ 
2^2(n+1) \biggr\} \cdot (4n+1) \, ,
</math>
  </td>
</tr>
</table>
and we see that, to lowest order, the two sides do match.  Notice that the mode number, <math>~m</math>, drops out in this lowest order approximation.
***************** -->
======Next Lowest Order======
Let's begin with the RHS (<font color="red">Case B</font>).
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(n+1)\cdot \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)(1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [  x^2(1+xb) -  \beta^2 - 4n ]
+ [1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3) ] [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x\cos\theta [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^0\biggl\{2n  (n+1) -4n(n+1) + 2n(n+1) \biggr\}
+ x^1\biggl\{8n(n+1)\cos\theta - 8n(n+1)\cos\theta\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x^2\biggl\{-4n(n+1)\cos^2\theta + (n+1)\biggl[  1 -  \biggl(\frac{\beta}{x}\biggr)^2 \biggr] + 12n(n+1)\cos^2\theta - 3n \biggl(\frac{\beta}{x}\biggr)^2 \biggr\}
+ \mathcal{O}(x^3)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^0\biggl\{0 \biggr\}
+ x^1\biggl\{0\biggr\}
+ x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2  \biggr\}
+ \mathcal{O}(x^3)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] 
</math>
  </td>
</tr>
</table>
</div>
where,
<table border="0" align="center" cellpadding="8">
<tr><td align="center">
<math>b_0 \equiv [ 2^7\cdot 3 (n+1)^3 \cos^2\theta ]^{1/2} \, .</math>
</td></tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathrm{RHS}_4}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 2^2(n+1)^2\cdot \mathcal{A}  + m^2 \cdot \mathcal{A} \biggl\{
-(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta)
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 2^2(n+1)^2\cdot \mathcal{A}  + m^2 \cdot \mathcal{A} (n+1)\biggl\{
-x^2\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 + x^2(1+\cancelto{0}{x}b)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~x^2 \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} (1+\cancelto{0}{x}b)^{1/2}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\mathcal{A} (n+1)\biggl\{-~ 2^2(n+1)  + m^2 \cancelto{0}{x^2} \biggl[
[2^3(n+1)\cos^2\theta - 3] -\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm ~~i~ \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} 
\biggr] \biggl\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-2^2(n+1)x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 
~~~\pm~i~\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2\cancelto{0}{x}\cos\theta + \cancelto{0}{x^2}\cos^2\theta + \cancelto{0}{\mathcal{O}}(x^3) \biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~\frac{\mathrm{RHS}_4}{x^4}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>2^2(n+1)(4n+1)\biggl(\frac{\beta}{x}\biggr)^2  ~-~2^2(n+1)^2[8n\cos^2\theta + 1]
~~\pm~i~(-1)\biggl(\frac{\beta}{x}\biggr) nb_0  \, .
</math>
  </td>
</tr>
</table>
This should be compared to,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathrm{LHS}_4}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - xb \biggr] (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \, .
</math>
  </td>
</tr>
</table>
Now, from above, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2\biggl\{
2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~\pm~~i~\beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]
\biggr\} + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \cancelto{0}{x^3}\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~(-1)\beta b_0 ~x(1+\cancelto{0}{x}b)^{1/2}\biggl\{
1 + \frac{3\cancelto{0}{x}\sin^2\theta (5\cos^2\theta -2)}{2(1+xb)\cos\theta } + \frac{3^2\cancelto{0}{x^2}\sin^6\theta}{2^2(1+xb)^2}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2\biggl\{
2(n+1)[2^3(n+1)\cos^2\theta -3]
+ 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
~\pm~~i~x \biggl\{ \cancelto{0}{x\beta} \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- \beta b_0\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\}
~\pm~~i~x^2\biggl\{ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
Also,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ x\biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2(n+1)[-6 + 2^4(n+1)\cos^2\theta]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \cancelto{0}{x^3}(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\cancelto{0}{x^4}(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta
\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta
- \cancelto{0}{x}[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- \cancelto{0}{x^2} \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ]
\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \, .
</math>
  </td>
</tr>
</table>
Finally,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ nx\biggl[ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ n \cancelto{0}{x^3}\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+n \cancelto{0}{x^4} \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ n\cancelto{0}{x^4} (n+1)\sin^4\theta \{  -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm~~i~nx^2\biggl(\frac{\beta}{x}\biggr)~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+\cancelto{0}{x}b} \biggr]^{1/2} \biggl\{
4\cos\theta + 6\cancelto{0}{x}(2b\cos\theta + \sin^4\theta) + 3\cancelto{0}{x^2}(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta)
\biggr\} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 \, .
</math>
  </td>
</tr>
</table>
Inserting these three approximate expressions into the LHS_4 ensemble gives,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathrm{LHS}_4}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - \cancelto{0}{xb} \biggr] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta)
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x (1-\cancelto{0}{x}\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr]  \biggl\{
~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr]
~\pm~~i~x^2\biggl[ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\mathrm{LHS}_4}{x^4}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr]  \biggl\{
~2(n+1)\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] + 2(n+1)[2^3(n+1)\cos^2\theta -3] \biggr\} -~ 2^2n(n+1)[2^3(n+1)\cos^2\theta -3]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2  [2^2(n+1)  - 3 ]
-(n+1) \biggl\{ [2^4(n+1)  -12]
+~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\}
~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2  [4n+1 ]
- 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
\, .
</math>
  </td>
</tr>
</table>
======Assessment======
The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression.  But the imaginary differ by a factor of 2.  So, let's repeat the steps leading to the imaginary parts.
'''<font color="red" size="+1">Case B:</font>'''
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^2} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~2^2(n+1)\cdot \mathrm{Im}[(n+1)\mathcal{A}]
+\frac{m^2}{(n+1)}\biggl\{ \mathrm{Im}[(n+1)\mathcal{A}]\cdot \mathrm{Re}[\Lambda] + \mathrm{Re}[(n+1)\mathcal{A}]\cdot \mathrm{Im}[\Lambda]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~2^2(n+1)\cdot x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{m^2}{(n+1)}\biggl\{ x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2} \biggr\} \cdot \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{m^2}{(n+1)}\biggl\{ 2n  (n+1)
+ (n+1)(1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
\biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~x^2 \biggl(\frac{\beta}{x}\biggr) \biggl\{ (1-x\cos\theta)^2 nb_0 \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+m^2 x^4\biggl(\frac{\beta}{x}\biggr)\biggl\{ (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] \biggr\} \cdot
\biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \biggl(\frac{\beta}{x}\biggr)^2 + (1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2\biggl(\frac{\beta}{x}\biggr) \biggl\{ 2n
+ (1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)}  \biggr]
\biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\}
</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow~~~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^4} \biggr]\biggl(\frac{x}{\beta}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1-x\cos\theta)^2 nb_0
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2  \biggl\{ 2n
+ (1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)}  \biggr]
\biggr\} \cdot (1+xb)^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+m^2
\biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \beta^2 + x^2(1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\}
\cdot (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1-x\cos\theta)^2 nb_0
+ m^2  \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4 \biggr\} \cdot (1+xb)^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)^{3/2} -  \beta^2\cdot (1+xb)^{1/2} \biggl[ 1
+ \frac{3n(1-x\cos\theta)^2}{(n+1)}  \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+m^2  (1-x\cos\theta)^2
\biggl\{ x^2(1+xb)[2^3(n+1)\cos^2\theta - 3]\cdot\biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] -\biggl[ \frac{nb_0(4n+1)}{2^2(n+1)^3}\biggr] \beta^2 \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1-x\cos\theta)^2 nb_0
+ m^2  \biggl\{ 2n - 4n [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)]
+ 2n[ 1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3)] \biggr\} \cdot (1+xb)^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2 (1-x\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+xb)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0
\biggr\} \cdot \frac{(1+xb)}{2^2(n+1)^2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- m^2 \beta^2 (1-x\cos\theta)^2
\biggl\{ nb_0(4n+1)  +  2^2(n+1)^2(1+xb)^{1/2} [ (n+1)
+ 3n(1-x\cos\theta)^2  ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ nb_0 [1 -2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)]
+ m^2  \biggl\{
8n x^2\cos^2\theta + \mathcal{O}(x^3)\biggr\} \cdot (1+\cancelto{0}{x}b)^{1/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0
\biggr\} \cdot \frac{(1+\cancelto{0}{x}b)}{2^2(n+1)^2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- m^2 \beta^2 (1-\cancelto{0}{x}\cos\theta)^2
\biggl\{ nb_0(4n+1)  +  2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} [ (n+1)
+ 3n(1-\cancelto{0}{x}\cos\theta)^2  ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-~ nb_0 [1 -2x\cos\theta  ]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~-nb_0x^2\cos^2\theta
+ m^2 x^2 \biggl\{2^5n(n+1)^2 \cos^2\theta + 2^2(n+1)^2 + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0
\biggr\} \cdot \frac{1}{2^2(n+1)^2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- m^2 \beta^2 
\biggl\{ nb_0(4n+1)  +  2^2(n+1)^2 [ (n+1)
+ 3n ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
</math>
</math>
   </td>
   </td>

Latest revision as of 23:18, 20 December 2016

Stability Analyses of PP Tori

[Comment by J. E. Tohline on 24 May 2016]   This chapter contains a set of technical notes and accompanying discussion that I put together several months ago as I was trying to gain a foundational understanding of the results of a large study of instabilities in self-gravitating tori published by the Imamura & Hadley collaboration. I have come to appreciate that some of the logic and interpretation of published results that are presented, below, has serious flaws. Therefore, anyone reading this should be quite cautious in deciding what subsections provide useful insight. I have written a separate chapter titled, "Characteristics of Unstable Eigenvectors in Self-Gravitating Tori," that contains a much more trustworthy analysis of this very interesting problem.


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may contain incorrect mathematical equations and/or physical misinterpretations.
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As has been summarized in an accompanying chapter — also see our related detailed notes — we have been trying to understand why unstable nonaxisymmetric eigenvectors have the shapes that they do in rotating toroidal configurations. For any azimuthal mode, <math>~m</math>, we are referring both to the radial dependence of the distortion amplitude, <math>~f_m(\varpi)</math>, and the radial dependence of the phase function, <math>~\phi_\mathrm{max}(\varpi)</math> — the latter is what the Imamura and Hadley collaboration refer to as a "constant phase locus." Some old videos showing the development over time of various self-gravitating "constant phase loci" can be found here; these videos supplement the published work of Woodward, Tohline & Hachisu (1994).


Here, we focus specifically on instabilities that arise in so-called (non-self-gravitating) Papaloizou-Pringle tori and will draw heavily from three publications:

PP III

Figure 2 extracted without modification from p. 274 of J. C. B. Papaloizou & J. E. Pringle (1987)

"The Dynamical Stability of Differentially Rotating Discs.   III"

MNRAS, vol. 225, pp. 267-283 © The Royal Astronomical Society

Figure 2 from PP III

Blaes (1985)

Equilibrium Configuration

In our separate discussion of PP84, we showed that the equilibrium structure of a PP-torus is defined by the enthalpy distribution,

<math> H = \frac{GM_\mathrm{pt}}{\varpi_0} \biggl[ (\chi^2 + \zeta^2)^{-1/2} - \frac{1}{2}\chi^{-2} - C_B^' \biggr] . </math>

Normalizing this expression by the enthalpy at the "center" — i.e., at the pressure maximum — of the torus which, as we have already shown, is

<math> H_0 = \frac{GM_\mathrm{pt}}{2\varpi_0} [1-2C_B^' ] \, </math>

gives,

<math> [1-2C_B^' ]\biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1 + [1 - 2C_B^' ]. </math>

Now, in our review of Kojima's (1986) work, we showed that the square of the Mach number at the "center" of the torus is,

<math>~\mathfrak{M}_0^2 \equiv \frac{(v_\varphi|_0)^2}{(c_s|_0)^2}</math>

<math>~=</math>

<math>~\frac{2(n+1)}{\gamma}\biggl[ \frac{1}{\chi_-} - 1 \biggr]^{-2}</math>

 

<math>~=</math>

<math>~2n [ 1- 2C_B^' ]^{-1} </math>

<math>~\Rightarrow ~~~~ [1 - 2C_B^'] </math>

<math>~=</math>

<math>~\frac{2n}{\mathfrak{M}_0^2} \, , </math>

where, in obtaining this last expression we have related the adiabatic exponent to the polytropic index via the relation, <math>~\gamma = (n+1)/n</math>. Instead of specifying the system's Mach number, Blaes (1985) defines the dimensionless parameter,

<math>~\beta^2 </math>

<math>~\equiv</math>

<math>~\frac{2n}{\mathfrak{M}_0^2} \, .</math>

Implementing this parameter swap, the equilibrium expression becomes,

<math> \beta^2 \biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1 + \beta^2 \, , </math>

or,

<math>~\frac{H}{H_0} </math>

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl[\chi^{-2} - 2(\chi^2 + \zeta^2)^{-1/2} + 1 \biggr] \, .</math>

Looking at Figure 1 of Blaes85 — see also the coordinate definitions given in his equation (2.1) — I conclude that,

<math>~\chi = 1 - x\cos\theta</math>       and         <math>\zeta = x\sin\theta \, .</math>

Hence,

<math>~\frac{H}{H_0} </math>

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - x\cos\theta)^2 + x^2\sin^2\theta]^{-1/2} + 1 \biggr\} </math>

 

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - 2x\cos\theta + x^2\cos^2\theta) + x^2(1-\cos^2\theta)]^{-1/2} + 1 \biggr\} </math>

 

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl[ (1 - x\cos\theta)^{-2} - 2(1 - 2x\cos\theta + x^2)^{-1/2} + 1 \biggr] \, .</math>

This matches equation (2.2) of Blaes85, if we acknowledge that Blaes uses <math>~f</math> — instead of the parameter notation, <math>~\Theta_H</math>, found in our discussion of equilibrium polytropic configurations — to denote the normalized enthalpy; that is,

<math>~f_\mathrm{Blaes85} = \Theta_H \equiv \frac{H}{H_0} \, .</math>

This expression for the enthalpy throughout a Papaloizou-Pringle torus is valid for tori of arbitrary thickness <math>~(0 < \beta < 1)</math>. When considering only slim tori, Blaes (1985) points out that this expression can be written in terms of the following power series in <math>~x</math> (see his equation 1.3):

<math>~\Theta_H</math>

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(x^4) \biggr] \, .</math>

Blaes then adopts a related parameter that is constant on isobaric surfaces, namely,

<math>\eta^2 \equiv 1 - \Theta_H \, ,</math>

which is unity at the surface of the torus and which goes to zero at the cross-sectional center of the torus. Notice that <math>~\eta</math> tracks the "radial" coordinate that measures the distance from the center of the torus; in particular, keeping only the leading-order term in <math>~x</math>, there is a simple linear relationship between <math>~\eta</math> and <math>~x</math>, namely,

<math>~\eta</math>

<math>~=</math>

<math>~[1 - \Theta_H]^{1/2} \approx \frac{x}{\beta} \, .</math>

Cubic Equation Solution

For later use, let's invert the cubic relation to obtain a more broadly applicable <math>~x(\eta)</math> function. Because we are only interested in radial profiles in the equatorial plane — that is, only for the values of <math>~\theta = 0</math> or <math>~\theta=\pi</math> — the relation to be inverted is,

<math>~x^2 \pm 2 x^3</math>

<math>~=</math>

<math>~(\beta\eta)^2</math>

<math>~\Rightarrow ~~~~ x^3 \pm \tfrac{1}{2}x^2 \mp \tfrac{1}{2}(\beta\eta)^2</math>

<math>~=</math>

<math>~0 \, .</math>


Table 1:  Example Parameter Values

determined by iterative solution for <math>~\beta = \tfrac{1}{10}</math>
<math>~\eta</math> <math>~\Gamma^2 = 54\beta^2\eta^2</math> Inner solution <math>~(\theta = 0)</math>

[Superior sign in cubic eq.]
Outer solution <math>~(\theta = \pi)</math>

[Inferior sign in cubic eq.]
<math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math> <math>~6(S + T)</math> <math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math> <math>~6(S + T)</math>
0.25 0.03375 0.244112 1.14647 0.256675 -0.84600
1.0 0.54 0.91909 1.55145 1.1378 -0.31732

Here, <math>~x_\mathrm{root}</math> has been determined via a brute-force, iterative technique.


Following Wolfram's discussion of the cubic formula, we should view this expression as a specific case of the general formula,

<math>~x^3 + a_2x^2 + a_1x + a_0 = 0 \, ,</math>

in which case, as is detailed in equations (54) - (56) of Wolfram's discussion of the cubic formula, the three roots of any cubic equation are:

<math>~x_1</math>

<math>~=</math>

<math>~ -\frac{1}{3}a_2 + (S + T) \, , </math>

<math>~x_2</math>

<math>~=</math>

<math>~ -\frac{1}{3}a_2 - \frac{1}{2} (S + T) + \frac{1}{2} \it{i} \sqrt{3} (S-T)\, , </math>

<math>~x_3</math>

<math>~=</math>

<math>~ -\frac{1}{3}a_2 - \frac{1}{2} (S + T) - \frac{1}{2} \it{i} \sqrt{3} (S-T)\, , </math>

where,

<math>~S</math>

<math>~\equiv</math>

<math>~[R + \sqrt{D}]^{1/3} \, ,</math>

<math>~T</math>

<math>~\equiv</math>

<math>~[R - \sqrt{D}]^{1/3} \, ,</math>

<math>~D</math>

<math>~\equiv</math>

<math>~Q^3 + R^2 \, ,</math>

<math>~Q</math>

<math>~\equiv</math>

<math>~\frac{3a_1 - a_2^2}{3^2} \, ,</math>

<math>~R</math>

<math>~\equiv</math>

<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} \, . </math>

Outer [inferior sign] Solution

Focusing, first, on the inferior sign convention, which corresponds to the "outer" solution <math>~(\theta = \pi)</math>, we see that the coefficients that lead to our specific cubic equation are:

<math>~a_2</math>

<math>~=</math>

<math>~- \tfrac{1}{2} \, ,</math>

<math>~a_1</math>

<math>~=</math>

<math>~0 \, ,</math>

<math>~a_0</math>

<math>~=</math>

<math>~\tfrac{1}{2}(\beta\eta)^2 \, .</math>

Applying Wolfram's definitions of the <math>~Q</math> and <math>~R</math> parameters to our particular problem gives,

<math>~Q</math>

<math>~\equiv</math>

<math>~\frac{3a_1 - a_2^2}{3^2} = -\biggl(\frac{a_2}{3}\biggr)^2 = - \frac{1}{2^2\cdot 3^2} \, ;</math>

<math>~R</math>

<math>~\equiv</math>

<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} = \frac{1}{2\cdot 3^3} \biggl[ -\frac{ 3^3}{2}(\beta\eta)^2 + \frac{1}{2^2}\biggr] </math>

 

<math>~\equiv</math>

<math>~\frac{1}{2^3\cdot 3^3} \biggl[ 1 - 2\cdot 3^3(\beta\eta)^2\biggr] \, . </math>

Defining the parameter,

<math>~\Gamma^2</math>

<math>~\equiv</math>

<math>~ 2\cdot 3^3(\beta\eta)^2 \, ,</math>

we therefore have,

<math>~(2\cdot 3)^6 D</math>

<math>~=</math>

<math>~( 1 - \Gamma^2 )^2-1 \, ,</math>

<math>~(2\cdot 3)^3S^3</math>

<math>~\equiv</math>

<math>~(2\cdot 3)^3 R + \sqrt{(2\cdot 3)^6D} </math>

 

<math>~\equiv</math>

<math>~(1-\Gamma^2) + \sqrt{( 1 - \Gamma^2 )^2-1}</math>

 

<math>~\equiv</math>

<math>~(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \, ;</math>

<math>~(2\cdot 3)^3T^3</math>

<math>~\equiv</math>

<math>~(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \, .</math>

ASIDE:  The cube root of an imaginary number …

<math>~\ell^3</math>

<math>~=</math>

<math>~A \pm i \sqrt{1-A^2}</math>

 

<math>~=</math>

<math>~r_\ell e^{i\theta_\ell} \, ,</math>

where,

<math>~r_\ell</math>

<math>~=</math>

<math>~( A^2 + 1-A^2 )^{1/2} = 1 \, ,</math>

and,

<math>~\theta_\ell</math>

<math>~=</math>

<math>~\pm \tan^{-1}\biggl( \frac{\sqrt{1-A^2}}{A} \biggr) = \pm \cos^{-1}A \, .</math>

Now, according to this online resource, the three roots <math>~(j=0,1,2)</math> of <math>~\ell^3</math> are,

<math>~\ell_j = r_\ell^{1/3}e^{i(\theta_\ell + 2j\pi)/3)} \, ,</math>

which, for our specific problem gives,

<math>~\ell_j</math>

<math>~=</math>

<math>~e^{i\theta_\pm/3} \cdot e^{i(2j\pi/3)} \, ,</math>

where the subscript on <math>~\theta</math> refers to the <math>~\pm</math> in our original expression for <math>~\ell</math>.


In our particular case, after associating <math>~A \leftrightarrow (1-\Gamma^2)</math>, we can write,

<math>~ 2\cdot 3(S + T) </math>

<math>~=</math>

<math>~ \biggl[(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} + \biggl[(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} </math>

 

<math>~=</math>

<math>~ e^{i\theta_+/3} \cdot e^{i(2j\pi/3)} + e^{i\theta_-/3} \cdot e^{i(2j\pi/3)} </math>

 

<math>~=</math>

<math>~e^{i(2j\pi/3)} \biggl\{ e^{i[\cos^{-1}(1-\Gamma^2)]/3} + e^{-i[\cos^{-1}(1-\Gamma^2)]/3} \biggr\} </math>

 

<math>~=</math>

<math>~e^{i(2j\pi/3)} \biggl\{\cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] + i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] </math>

 

 

<math>~+ \cos\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] - i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~2 e^{i(2j\pi/3)} \cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, .</math>

Similarly, we can write,

<math>~ 2\cdot 3(S - T) </math>

<math>~=</math>

<math>~ \biggl[(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} - \biggl[(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} </math>

 

<math>~=</math>

<math>~ e^{i\theta_+/3} \cdot e^{i(2j\pi/3)} - e^{i\theta_-/3} \cdot e^{i(2j\pi/3)} </math>

 

<math>~=</math>

<math>~e^{i(2j\pi/3)} \biggl\{ e^{i[\cos^{-1}(1-\Gamma^2)]/3} - e^{-i[\cos^{-1}(1-\Gamma^2)]/3} \biggr\} </math>

 

<math>~=</math>

<math>~e^{i(2j\pi/3)} \biggl\{\cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] + i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] </math>

 

 

<math>~- \cos\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] + i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~2i e^{i(2j\pi/3)} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, . </math>

Focusing specifically on the "j=0" root, and setting <math>~a_2 = -\tfrac{1}{2}</math>, we therefore have,

<math>~6x_1-1</math>

<math>~=</math>

<math>~ 6(S + T) </math>

 

<math>~=</math>

<math>~2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math>

<math>~6x_2-1</math>

<math>~=</math>

<math>~ - \frac{1}{2} \biggl\{ 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] - i\sqrt{3} \cdot 2i \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ -2\biggl\{ \frac{1}{2}\cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] +\frac{\sqrt{3}}{2} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr] \biggr\} \, ; </math>

<math>~6x_3-1</math>

<math>~=</math>

<math>~ - \frac{1}{2} \biggl\{ 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] + i\sqrt{3} \cdot 2i \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ -2\biggl\{ \frac{1}{2}\cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] -\frac{\sqrt{3}}{2} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr] \biggr\} \, . </math>


Table 1:  Analytically Evaluated Roots

determined for <math>~\beta = \tfrac{1}{10}</math>
<math>~\eta</math> <math>~\Gamma^2 = 54\beta^2\eta^2</math> Inner solution <math>~(\theta = 0)</math>

[Superior sign in cubic eq.]
Outer solution <math>~(\theta = \pi)</math>

[Inferior sign in cubic eq.]
<math>~x_1/\beta</math> <math>~x_2/\beta</math> <math>~x_3/\beta</math> <math>~x_1/\beta</math> <math>~x_2/\beta</math> <math>~x_3/\beta</math>
0.25 0.03375 -4.98744 0.24411 -0.25667 4.98744 -0.24411 0.25667
1.0 0.54 -4.78128 0.91909 -1.1378 4.78128 -0.91909 1.1378
  CONFIRMATION: In all cases,

<math>~x^2 + 2x^3 = (\beta\eta)^2</math>
CONFIRMATION: In all cases,

<math>~x^2 - 2x^3 = (\beta\eta)^2</math>


Inner [superior sign] Solution

Next, examing the superior sign convention, which corresponds to the "inner" solution <math>~(\theta = 0)</math>, we see that the coefficients that lead to our specific cubic equation are:

<math>~a_2</math>

<math>~=</math>

<math>~\tfrac{1}{2} \, ,</math>

<math>~a_1</math>

<math>~=</math>

<math>~0 \, ,</math>

<math>~a_0</math>

<math>~=</math>

<math>~- \tfrac{1}{2}(\beta\eta)^2 \, .</math>

Following the same set of steps that were followed in determining the "outer" solution, here we find: <math>~Q</math> remains the same; <math>~R</math> has the same magnitude, but changes sign; and, hence, <math>~D</math> remains the same. We therefore have,

<math>~(2\cdot 3)^3S^3</math>

<math>~=</math>

<math>~- (1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \, ;</math>

<math>~(2\cdot 3)^3T^3</math>

<math>~=</math>

<math>~- (1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \, ,</math>

which leads to the following expressions for the three "inner" roots:

<math>~6x_1+1</math>

<math>~=</math>

<math>~- 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math>

<math>~6x_2+1</math>

<math>~=</math>

<math>~ - 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr] \biggr\} \, ; </math>

<math>~6x_3+1</math>

<math>~=</math>

<math>~ - 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr] \biggr\} \, . </math>

Analytically Prescribed Eigenvector

Our Notation

As is explicitly defined in Figure 1 of our accompanying detailed notes, we have chosen to represent the spatial structure of an eigenfunction in the equatorial-plane of toroidal-like configurations via the expression,

<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>

<math>~=</math>

<math>~\biggl\{ f_m(\varpi)e^{-im\phi_m} \biggr\} \, .</math>

In general, we should assume that the function that delineates the radial dependence of the eigenfunction has both a real and an imaginary component, that is, we should assume that,

<math>~f_m(\varpi)</math>

<math>~=</math>

<math>~\mathcal{A}(\varpi) + i\mathcal{B}(\varpi) \, ,</math>

in which case the square of the modulus of the function is,

<math>~|f_m|^2 \equiv f_m \cdot f^*_m </math>

<math>~=</math>

<math>~\mathcal{A}^2 + \mathcal{B}^2 \, .</math>

We can rewrite this complex function in the form,

<math>~f_m(\varpi)</math>

<math>~=</math>

<math>~|f_m|e^{-i[\alpha(\varpi) + \pi/2]} \, ,</math>

if the angle, <math>~\alpha(\varpi)</math> is defined such that,

<math>~\sin\alpha = \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>

        and        

<math>~\cos\alpha = \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>

<math>~\Rightarrow ~~~~ \alpha</math>

<math>~\equiv</math>

<math>~\tan^{-1}\biggl(\frac{\mathcal{A}}{\mathcal{B}}\biggr) = \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] \, .</math>

Hence, the spatial structure of the eigenfunction can be rewritten as,

<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>

<math>~=</math>

<math>~|f_m(\varpi)|e^{-i[\alpha(\varpi) + \pi/2+ m\phi_m]} \, . </math>

From this representation we can see that, at each radial location, <math>~\varpi</math>, the phase angle(s) at which the fractional perturbation exhibits its maximum amplitude, <math>~|f_m|</math>, is identified by setting the exponent of the exponential to zero. That is,

<math>~\phi_m = \phi_\mathrm{max}(\varpi)</math>

<math>~\equiv</math>

<math>~-\frac{1}{m}\biggl[\alpha(\varpi) +\frac{\pi}{2}\biggr] = -\frac{1}{m}\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} \biggr\} \, .</math>

An equatorial-plane plot of <math>~\phi_\mathrm{max}(\varpi)</math> should produce the "constant phase locus" referenced, for example, in recent papers from the Imamura & Hadley collaboration.


General Formulation

From my initial focused reading of the analysis presented by Blaes (1985), I conclude that, in the infinitely slender torus case, unstable modes in PP tori exhibit eigenvectors of the form,

<math>~\frac{\delta W}{W_0} \equiv \biggl[ \frac{W(\eta,\theta)}{C} - 1 \biggr]e^{im\Omega_p t}e^{-y_2 (\Omega_0 t)} </math>

<math>~=</math>

<math>~\biggl\{ f_m(\eta,\theta)e^{-i[m\phi_m + k\theta]} \biggr\} \, ,</math>

where we have written the perturbation amplitude in a manner that conforms with the notation that we have used in Figure 1 of a related, but more general discussion. As is summarized in §1.3 of Blaes (1985), for "thick" (but, actually, still quite thin) tori, "exactly one exponentially growing mode exists for each value of the azimuthal wavenumber <math>~m</math>," and its complex amplitude takes the following form (see his equation 1.10):

<math>~f_m(\eta,\theta)</math>

<math>~=</math>

<math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta\cos\theta\biggr] + \mathcal{O}(\beta^3) \, . </math>

Aside from an arbitrary leading scale factor, we should therefore find that the amplitude (modulus) of the enthalpy perturbation is,

<math>~\biggl|\frac{\delta W}{W_0} \biggr|</math>

<math>~=</math>

<math>~\sqrt{[\mathrm{Re}(f_m)]^2+ [\mathrm{Im}(f_m)]^2} \, ;</math>

and the associated phase function is,

<math>~m\phi_\mathrm{max}(\varpi)</math>

<math>~=</math>

<math>~ -\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} +k\theta \biggr\} \, .</math>

Now, keeping in mind that, for the time being, we are only interested in examining the shape of the unstable eigenvector in the equatorial plane of the torus, we can set,

<math>~\cos\theta ~~ \rightarrow ~~ \pm 1 \, .</math>

Hence, we have,

<math>~\frac{1}{\beta^4 m^4}\biggl|\frac{\delta W}{W_0} \biggr|^2</math>

<math>~=</math>

<math>~\biggl[2\eta^2 - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}\biggr]^2 + 16\biggl[\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta \biggr]^2 </math>

 

<math>~=</math>

<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[2^3(n+1)^2\eta^2 - 3(n+1)\eta^2 - (4n+1) \biggr]^2 + \frac{2^3 \cdot 3\eta^2}{(n+1)} </math>

 

<math>~=</math>

<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2 + \frac{2^3 \cdot 3\eta^2}{(n+1)} </math>

<math>~\Rightarrow ~~~~ \biggl[\frac{2(n+1)}{\beta m} \biggr]^4 \biggl|\frac{\delta W}{W_0} \biggr|^2</math>

<math>~=</math>

<math>~\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2 + 2^7 \cdot 3(n+1)^3\eta^2 \, . </math>

Also, keeping in mind that, because of the <math>~\cos\theta</math> factor, the sign on the imaginary term flips its sign when switching from the "inner" region to the "outer" region of the torus,

 

<math>~m\phi_\mathrm{max}</math>

<math>~=</math>

<math>~- \biggl\{ \tan^{-1}\biggl[ \frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta} \biggr]+\frac{\pi}{2}\biggr\}</math>

        over        

inner <math>~(\theta=0)</math> region of the torus;

while
 

<math>~m\phi_\mathrm{max}</math>

<math>~=</math>

<math>~- \biggl\{ \tan^{-1}\biggl[- \frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}\biggr\} +\frac{3\pi}{2} \biggr\}</math>

        over        

outer <math>~(\theta=\pi)</math> region of torus.

Incompressible Slim Tori

If we specifically consider geometrically slim, incompressible tori — that is, if we set the polytropic index, <math>~n=0</math> — to lowest order the eigenfunction derived by Blaes (1985) takes the form,

<math>~f_m(\eta,\theta)</math>

<math>~=</math>

<math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} - \frac{1}{4} \pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr] + \cancelto{0}{\mathcal{O}(\beta^3)} \, . </math>

Check Validity of Blaes85 Eigenvector

Step 1

Equation (2.6) of Blaes85 states that,

<math>~\beta^2 \eta^2 = [x^2 + x^3(3\cos\theta - \cos^3\theta) ]</math>

        <math>~\Rightarrow</math>        

<math>~ \beta^2(1 - \eta^2) = [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \, .</math>

This means that,

<math>~\frac{\partial \eta^2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{\beta^2}\biggl[ 2x +3x^2(3\cos\theta - \cos^3\theta) \biggr] \, ; </math>

and,

<math>~\frac{\partial \eta^2}{\partial\theta}</math>

<math>~=</math>

<math>~\frac{x^3}{\beta^2}\biggl[ -3\sin\theta + 3\sin\theta \cos^2\theta \biggr] </math>

 

<math>~=</math>

<math>~\frac{3x^3 \sin\theta}{\beta^2}\biggl[\cos^2\theta -1\biggr] \, .</math>

Furthermore,

<math>~\frac{\partial \eta}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ 2x +3x^2(3\cos\theta - \cos^3\theta) \biggr] \, ; </math>

and,

<math>~\frac{\partial \eta}{\partial\theta}</math>

<math>~=</math>

<math>~\frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ -3\sin\theta + 3\sin\theta \cos^2\theta \biggr] \, .</math>

Step 2

Equation (4.14) of Blaes85 states that,

<math>~\nu + m</math>

<math>~=</math>

<math>~ \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, ; </math>

and equation (4.13) of Blaes85 states that,

<math>~\frac{\delta W}{A_{00}}</math>

<math>~=</math>

<math>~1 + \beta^2 m^2\biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl( \frac{3}{2n+2} \biggr)^{1/2} \eta\cos\theta \biggr] </math>

<math>~\Rightarrow~~~~\frac{1}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math>

<math>~=</math>

<math>~\beta^2 \biggl\{- \frac{(4n+1)}{4(n+1)^2} + \eta^2\biggl[ 2\cos^2\theta - \frac{3}{4(n+1)}\biggr] \pm i\biggl( \frac{2^3\cdot 3}{n+1} \biggr)^{1/2} \eta\cos\theta \biggr\} </math>

 

<math>~=</math>

<math>~\frac{\beta^2}{2^2(n+1)^2} \biggl\{- (4n+1) + \eta^2 [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \pm i ~[ 2^7\cdot 3(n+1)^3 ]^{1/2} \eta\cos\theta \biggr\} </math>

<math>~\Rightarrow~~~~\Lambda \equiv \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math>

<math>~=</math>

<math>~- (4n+1)\beta^2 + [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)] </math>

 

 

<math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} </math>

 

<math>~=</math>

<math>~- (4n+1)\beta^2 + (n+1)x^2[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

Through a separate white-board derivation I have obtained …

<math>~\Lambda </math>

<math>~\equiv</math>

<math>~ \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr] </math>

 

<math>~=</math>

<math>~ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \, , </math>

where,

<math>~(\beta\eta)^2</math>

<math>~=</math>

<math>~x^2(1+xb) \, ,</math>

<math>~b</math>

<math>~\equiv</math>

<math>~3\cos\theta - \cos^3\theta \, .</math>

Note that, differentiating the left-hand-side with respect to either coordinate <math>~(x</math> or <math>~\theta)</math> gives,

<math>~\frac{\partial \Lambda}{\partial x_i} </math>

<math>~=</math>

<math>~\frac{2^2(n+1)^2}{m^2}\cdot \frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math>

<math>~\Rightarrow~~~~\frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math>

<math>~=</math>

<math>~\biggl[\frac{m}{2(n+1)}\biggr]^2 \cdot \frac{\partial \Lambda}{\partial x_i} \, .</math>


Given that <math>~\nu</math> has both real and imaginary parts, presumably,

<math>~\nu^2 \equiv \nu \cdot \nu^*</math>

<math>~=</math>

<math>~ \biggl\{ -m \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} \biggl\{ -m \mp im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} </math>

 

<math>~=</math>

<math>~m^2 + m^2\biggl[\frac{3}{2(n+1)}\biggr]\beta^2 \, . </math>

For later reference, let's take the relevant partial derivatives of the function, <math>~\Lambda(x,\theta)</math>. Adopting the shorthand notation,

<math>~b \equiv (3\cos\theta-\cos^3\theta) \, ,</math>

we have,

<math>~\frac{\partial \Lambda}{\partial x}</math>

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b] \biggr\} </math>

 

 

<math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b ]^{1/2} \biggr\} </math>

 

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + 3x^2b] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + 3xb ] </math>

Through a separate white-board derivation I have obtained …

<math>~\frac{\partial\Lambda}{\partial x}</math>

<math>~=</math>

<math>~ (n+1)[2^3(n+1)\cos^2\theta -3]x(2+3xb) ~~~\pm ~~i~\beta\cos\theta [2^5\cdot 3 (n+1)^3]^{1/2} \cdot \frac{x(2+3xb)}{(\beta\eta)} \, , </math>

which is the same.


<math>~\frac{\partial^2 \Lambda}{\partial x^2}</math>

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x}[2x + 3x^2b] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x}\biggl\{[1 + xb ]^{-1/2} \cdot [2 + 3xb ] \biggr\} </math>

 

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2 + 6xb] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{ -\tfrac{b}{2}[1 + xb ]^{-3/2} \cdot [2 + 3xb ] + [1 + xb ]^{-1/2} \cdot [3b ] \biggr\} </math>

 

<math>~=</math>

<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{ - [2b + 3xb^2 ] + [6b + 6xb^2 ] \biggr\} \frac{1}{2(1+xb)^{3/2}} </math>

 

<math>~=</math>

<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb] </math>

 

 

<math>~\pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + xb ]^{-3/2} [4 + 3xb ] </math>


Through a separate white-board derivation I have obtained …

<math>~\frac{\partial^2\Lambda}{\partial x^2}</math>

<math>~=</math>

<math>~ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3xb) ~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \, , </math>

which is the same.


<math>~\frac{\partial \Lambda}{\partial \theta}</math>

<math>~=</math>

<math>~\frac{\partial }{\partial \theta} \biggl\{ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)] \biggr\} </math>

 

 

<math>~\pm ~i~\frac{\partial }{\partial \theta} \biggl\{ \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} \biggr\} </math>

 

<math>~=</math>

<math>~ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot x^3 \cdot [\sin\theta (-3 + 3\cos^2\theta)] + [x^2 + x^3(3\cos\theta - \cos^3\theta)] \cdot [ -2^4(n+1)^2 \sin\theta \cos\theta ] </math>

 

 

<math> ~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{-1/2} \cdot (-\sin\theta) [2x^2\cos\theta + x^3(9\cos^2\theta - 5\cos^4\theta) ] </math>

 

<math>~=</math>

<math>~ -3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] -2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] </math>

 

 

<math> ~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] </math>

 

<math>~=</math>

<math>~ -3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] -2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + xb ] </math>

 

 

<math> ~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] </math>


Through a separate white-board derivation I have obtained …

<math>~\frac{\partial\Lambda}{\partial \theta}</math>

<math>~=</math>

<math>~ 9(n+1)x^3\sin^3\theta - 2^3\cdot 3(n+1)^2x^3\sin^3\theta \cos^2\theta -2^4 (n+1)^2 (\beta\eta)^2 \sin\theta\cos\theta </math>

 

 

<math>~ \pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2}\biggl\{ (\beta\eta)\sin\theta +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^3\theta \cos\theta}{(\beta\eta)} \biggr]\biggr\} </math>

 

<math>~=</math>

<math>~(n+1)\sin\theta \biggl\{ -2^4 (n+1) (\beta\eta)^2 \cos\theta + 3x^3 \sin^2\theta \biggl[3 - 2^3(n+1)\cos^2\theta \biggr] \biggr\} </math>

 

 

<math>~ \pm~~i~(-1)\beta \sin\theta [2^7\cdot 3 (n+1)^3 (\beta\eta)^2]^{1/2}\biggl\{ 1 +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^2\theta \cos\theta}{(\beta\eta)^2} \biggr]\biggr\} \, , </math>

which is the same.


<math>~\frac{\partial^2 \Lambda}{\partial \theta^2}</math>

<math>~=</math>

<math>~ -3 x^3 \frac{\partial }{\partial \theta} \cdot \biggl\{\sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \biggr\} -2^4(n+1)^2 x^2 \cdot \frac{\partial }{\partial \theta} \biggl\{\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] \biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \frac{\partial }{\partial \theta} \biggl\{ \sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \biggr\} </math>

 

<math>~=</math>

<math>~ -3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ] \biggr\} </math>

 

 

<math>~ -2^4(n+1)^2 x^2 \cdot \biggl\{ (\cos^2\theta - \sin^2\theta ) [1 + x(3\cos\theta-\cos^3\theta) ] - 3x\sin^4\theta \cos\theta \biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl\{ \cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} </math>

 

 

<math> - 3x\sin^2\theta [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot (3 - 5\cos^2\theta) </math>

 

 

<math> + \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-3/2} \biggr\} </math>

 

<math>~=</math>

<math>~ -3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ] \biggr\} </math>

 

 

<math>~ -2^4(n+1)^2 x^2 \cdot \biggl\{ (1 - 2\sin^2\theta ) + x (3\cos\theta -\cos^3\theta - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta ) \biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{ \cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + xb ] </math>

 

 

<math> - 3x\sin^2\theta [1 + xb ] \cdot (3 - 5\cos^2\theta) + \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math>

 

<math>~=</math>

<math>~-2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) - x^3 \cdot \biggl\{ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta </math>

 

 

<math>~ + 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta )\biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{ 2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + x^2 b(9\cos\theta - 5\cos^3\theta)] </math>

 

 

<math> - 6x\sin^2\theta (1 + xb ) \cdot (3 - 5\cos^2\theta) + 3 \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math>


Through a separate white-board derivation I have obtained …

<math>~\frac{\partial^2\Lambda}{\partial \theta^2}</math>

<math>~=</math>

<math>~ x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta) \biggr\} </math>

 

 

<math>~ + x^3\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta \biggr\} </math>

 

 

<math>~ \pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{ (\beta\eta)\cos\theta + \frac{3x^3\sin^2\theta}{2(\beta\eta)}(5\cos^2\theta -2) + \frac{3^2x^6\sin^6\theta\cos\theta}{2^2(\beta\eta)^3} \biggr\} \, . </math>

Step 3

From our accompanying discussion of the Blaes85 derivation, we expect the following equality to hold (see his equations 4.1 and 4.2):

<math>~\hat{L} (\delta W)</math>

<math>~=</math>

<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)(\delta W) \, ,</math>

where,

<math>~\hat{L} (\delta W)</math>

<math>~\equiv</math>

<math>~ \Theta_H x^2\cdot \frac{\partial^2(\delta W)}{\partial x^2} +\Theta_H \cdot \frac{\partial^2(\delta W)}{\partial\theta^2} + \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x} \biggr\} \cdot \frac{\partial (\delta W)}{\partial x} </math>

 

 

<math>~ + \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr] \cdot \frac{\partial (\delta W)}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]\delta W \, , </math>

<math>~M</math>

<math>~\equiv</math>

<math>~\frac{x^2}{(1-\Theta_H)\beta^2} \, ,</math>

<math>~N</math>

<math>~\equiv</math>

<math>~\frac{2mx^2}{(1-\Theta_H)\beta^2(1-x\cos\theta)^2} \, .</math>

Immediately evaluating the right-hand-side (RHS) of the equality, we have,

<math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{RHS}{A_{00}}</math>

<math>~=</math>

<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)\cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math>

 

<math>~=</math>

<math>~-\frac{2nx^2}{\beta^2}\biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math>

 

<math>~=</math>

<math>~-\frac{2nx^2}{\beta^2}\biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} \, .</math>

And the similarly modified LHS is:

<math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{LHS}{A_{00}}</math>

<math>~=</math>

<math>~ \Theta_H x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +\Theta_H \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}} </math>

 

<math>~=</math>

<math>~ (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial (1-\eta^2) }{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial (1-\eta^2) }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math>

 

<math>~=</math>

<math>~ (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] - nx^2 \cdot \frac{\partial \eta^2 }{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } - n\cdot \frac{\partial \eta^2 }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math>

Now multiply both sides by …

<math>~\beta^2 m^2 (1-x\cos\theta)^4 \, .</math>


We have,

<math>~m^2\mathcal{L}_{RHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}}</math>

<math>~=</math>

<math>~- 2n m^2 x^2(1-x\cos\theta)^4 \biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math>

 

<math>~=</math>

<math>~- 2n x^2 (1-x\cos\theta)^2 \biggl[ (1-x\cos\theta)^2 \nu^2 + (2m\nu) \biggr] \cdot \{m^2 \Lambda + [ 2(n+1) ]^2 \} </math>

 

<math>~=</math>

<math>~- 2n x^2 (1-x\cos\theta)^2 \{m^2 \Lambda + [ 2(n+1) ]^2 \} \biggl\{ m^2 (1-x\cos\theta)^2 \biggl[ 1+\frac{3\beta^2}{2(n+1)} \biggr] + 2m^2 \biggl[ -1 ~\pm~ i~\biggl[ \frac{3\beta^2}{2(n+1)} \biggr]^{1/2} \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ - \frac{n m^2 x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{m^2 \Lambda + [ 2(n+1) ]^2 \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] </math>

 

 

<math>~ - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} \, . </math>

And,

<math>~m^2\mathcal{L}_{LHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}}</math>

<math>~=</math>

<math>~ \beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +\beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

 

 

<math>~ + \beta^2 m^2 (1-x\cos\theta)^3 \biggl[ (1-\eta^2) x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial x} \biggr] \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \beta^2 m^2 (1-x\cos\theta)^3 \biggl[(1-\eta^2) x\sin\theta - n (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} </math>

 

 

<math>~ + m^2 \biggl[ 2n x^2 - (1-x\cos\theta)^2 \beta^2 x^2(1-\eta^2) \biggr] \cdot \biggl\{m^2 \Lambda + \biggl[ 2(n+1) \biggr]^2 \biggr\} </math>

 

<math>~=</math>

<math>~ m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

 

 

<math>~ + m^2 (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot [ 2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + m^2 (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta - 3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta \biggr\} \cdot \frac{\partial\Lambda}{\partial\theta} </math>

 

 

<math>~ + m^2 \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[m^2 \Lambda + 2^2(n+1)^2 \biggr] \, . </math>

Step 4

Let's divide both sides by <math>~m^2</math> and swap a couple of terms between the sides in order to group, on the right, terms with no explicit mention of <math>~\Lambda</math>.

<math>~\mathcal{L}_{RHS} </math>

<math>~\equiv</math>

<math>~ \frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}} ~\pm~\mathrm{swaps} </math>

 

<math>~=</math>

<math>~ - \frac{n x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{[ 2(n+1) ]^2 \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math>

 

 

<math>~ - \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[2^2(n+1)^2 \biggr] </math>

<math>~\Rightarrow ~~~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>

<math>~=</math>

<math>~ n \cdot (1-x\cos\theta)^2 \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math>

 

 

<math>~ + (n+1)\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} </math>

 

<math>~=</math>

<math>~ 2n(n+1) - (1-x\cos\theta)^2 (n+1)[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] </math>

 

 

<math>~ + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] - 4 n(n+1) (1-x\cos\theta)^2 </math>

 

 

<math>~ \pm~i~ (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~=</math>

<math>~ 2n(n+1) - (n+1)(1-x\cos\theta)^2[ 4n + \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] </math>

 

 

<math>~ \pm~i~ (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} \, . </math>


Through a separate white-board derivation I have obtained …

<math>~\mathrm{RHS}_3</math>

<math>~\equiv</math>

<math>~ \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{RHS}_1 ~~\pm~~ \mathrm{swaps} \biggr\} = \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{RHS}_0 ~~\pm~~ \mathrm{swaps} \biggr\} </math>

 

<math>~=</math>

<math>~-~ 2^2(n+1)^2 x^2 \cdot \mathcal{A} \, , </math>

where,

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~ 2n + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) - \beta^2 + x^2(1+xb)\biggr] + 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2 \, , </math>

<math>~\frac{\nu}{m}</math>

<math>~=</math>

<math>~-~1~~ \pm~~ i~\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, . </math>

Case A:       <math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu \cdot \nu^*}{m^2} = 1+\frac{3\beta^2}{2(n+1)} ~~~\Rightarrow</math>

<math>~(n+1)\cdot \mathcal{A}</math>

<math>~=</math>

<math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) + 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~(1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \, . </math>

Case B:       <math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu }{m} \cdot \frac{\nu }{m} = 1 - \frac{3\beta^2}{2(n+1)} ~~\pm~i~(-1)\biggl[\frac{2\cdot 3\beta^2}{(n+1)}\biggr]^{1/2} ~~~\Rightarrow</math>

<math>~(n+1)\cdot \mathcal{A}</math>

<math>~=</math>

<math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \, . </math>


And,

<math>~\mathcal{L}_{LHS} </math>

<math>~\equiv</math>

<math>~ \frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}} ~\mp~\mathrm{swaps} </math>

 

<math>~=</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot [ 2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta - 3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta \biggr\} \cdot \frac{\partial\Lambda}{\partial\theta} </math>

 

 

<math>~ + \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[m^2 \Lambda \biggr] </math>

 

 

<math>~ + \frac{n x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{m^2 \Lambda \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math>

 

<math>~=</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3b] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - x^2 - x^3b ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] - n (1-x\cos\theta)\cdot x^3 \biggl[ ( 2 +3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math>


Through a separate white-board derivation I have obtained …

<math>~\mathrm{LHS}_3</math>

<math>~\equiv</math>

<math>~ \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{LHS}_1 ~~\mp~~ \mathrm{swaps} \biggr\} = \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{LHS}_0 ~~\mp~~ \mathrm{swaps} \biggr\} </math>

 

<math>~=</math>

<math>~ \beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] + m^2 x^2 \Lambda \cdot \mathcal{A} \, ,</math>

where, as above in the definition of <math>~\mathrm{RHS_3} \, ,</math>

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~ 2n + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) - \beta^2 + x^2(1+xb)\biggr] + 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2 \, . </math>


The remaining question is, does <math>~\mathcal{L}_{LHS} = \mathcal{L}_{RHS}</math> — at least to lowest order(s) in <math>~x</math> — after the Blaes85 expression for the eigenfunction, <math>~\Lambda</math> (and its derivatives), is inserted into the LHS expression?

Step 5

Now let's evaluate the LHS terms, keeping only leading-orders in <math>~x</math> before plugging derivatives of <math>~\Lambda</math> into each term. For example,

<math>~ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

<math>~=</math>

<math>~ x^2 \biggl\{2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + \cancelto{0}{3xb}] \pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + \cancelto{0}{xb} ]^{-3/2} [4 + \cancelto{0}{9xb} ] \biggr\} </math>

 

 

<math>~m^2\cdot \biggl\{ -2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) - \cancelto{0}{x^3 }\cdot \biggl[ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta </math>

 

 

<math>~ + 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta )\biggr] </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl[ 2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + \cancelto{0}{x^2 b}(9\cos\theta - 5\cos^3\theta)] </math>

 

 

<math> - 6x\sin^2\theta (1 + \cancelto{0}{xb }) \cdot (3 - 5\cos^2\theta) + 3 \cdot x\sin^4\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] \biggr] \biggr\} </math>

 

<math>~\approx</math>

<math>~ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta </math>

 

 

<math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math>

Next,

<math>~ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>

<math>~=</math>

<math>~ (1-2x \cos\theta )\cdot \biggl\{ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + \cancelto{0}{3x^2b}] ~~\pm ~~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + \cancelto{0}{3xb }] \biggr\} </math>

 

 

<math>~ + \sin\theta \cdot \biggl\{ -2^4(n+1)^2 \cancelto{0}{x^2}\sin\theta \cos\theta [1 + xb ] -3 \cancelto{0}{x^3} \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] </math>

 

 

<math>~ ~~\pm ~~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot x~\sin\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math>

 

<math>~\approx</math>

<math>~ (1-2x \cos\theta )\cdot \biggl[ 2x(n+1) [ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} \biggr] </math>

 

 

<math>~ ~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot \sin^2\theta </math>


Through a separate white-board derivation I have obtained …

<math>~ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>

<math>~=</math>

<math>~ x(n+1)[-6 + 2^4(n+1)\cos^2\theta] </math>

 

 

<math>~ - x^2(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \} </math>

 

 

<math>~ +x^3(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+x(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta - x[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>

 

 

<math>~- x^2 \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ] \biggr\}</math>


Also, from above,

<math>~\Lambda </math>

<math>~=</math>

<math>~- (4n+1)\beta^2 + (n+1)\cancelto{0}{x^2}[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

 

<math>~\approx</math>

<math>~- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>


Through a separate white-board derivation I have obtained …

<math>~ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>

<math>~=</math>

<math>~ x\cdot 2^2(n+1)[2^3(n+1)\cos^2\theta -3] </math>

 

 

<math>~ + x^2\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \} </math>

 

 

<math>~ +x^3 \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~ +x^3 (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \} </math>

 

 

<math>~\pm~~i~\beta~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+xb} \biggr]^{1/2} \biggl\{ 4\cos\theta + 6x(2b\cos\theta + \sin^4\theta) + 3x^2(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta) \biggr\} </math>


Taken together, then, we have,

<math>~\mathcal{L}_{LHS} </math>

<math>~=</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - \cancelto{0}{x^3b}] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3b} ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] - n (1-x\cos\theta)\cdot \cancelto{0}{x^3} \biggl[ ( 2 +3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - \cancelto{0}{x^3}b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math>

 

<math>~\approx</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] \biggr\} </math>

 

 

<math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math>

 

<math>~\approx</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta </math>

 

 

<math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] \biggr\} </math>

 

 

<math>~ + x\beta^2(1-x\cos\theta)^3 \biggl\{ (1-2x \cos\theta )\cdot \biggl[ 2x(n+1)[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} \biggr] </math>

 

 

<math>~ ~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot \sin^2\theta \biggr\} </math>

 

 

<math>~ + m^2 x^2\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot \biggl\{- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}\biggr\} </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \cdot \biggl\{ - (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} \biggr\} </math>

Let's further simplify:

<math>~\mathcal{L}_{LHS} </math>

<math>~\approx</math>

<math>~ x^2\biggl\{ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math>~ -~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math>~ + \beta^2(1-x\cos\theta)^3 (1-2x \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr] </math>

 

 

<math>~ - 2n(4n+1)\beta^2 m^2 + (4n+1)\beta^2 m^2 [4n + \beta^2 - x^2 ] (1-x\cos\theta)^2 - (4n+1)\beta^2 m^2 n (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] \biggr\} </math>

 

 

<math>~ ~~\pm ~~i x\beta^3 (1-x\cos\theta)^3 (1-2x \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} </math>

 

 

<math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta (1-x\cos\theta)^4 [ \beta^2 - x^2 ] </math>

 

 

<math>~ \pm ~i~\beta m^2 x^2\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math>

 

 

<math>~ ~~\pm ~~i~\beta x^2 (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-x\cos\theta)^3 [1 + xb ]^{-1/2} \cdot \sin^2\theta </math>

 

 

<math>~ \pm i~x^2(-1) (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 </math>

Step 6

Hence, to lowest order we want to compare the following two expressions:

<math>~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>

<math>~\approx</math>

<math>~ 2n(n+1) - (n+1)(1-\cancelto{0}{x}\cos\theta)^2[ 4n + \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] + n(1-\cancelto{0}{x}\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] </math>

 

 

<math>~ \pm~i~ (1-\cancelto{0}{x}\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~\approx</math>

<math>~ 2n(n+1) - (n+1)[ 4n + \beta^2 ] + n [ 2(n+1) + 3\beta^2 ] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~\approx</math>

<math>~3n\beta^2 + (n+1)\biggl[ 2n - 4n - \beta^2 + 2n \biggr] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~\approx</math>

<math>~\beta^2 (2n-1) ~\pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>


<math>~\mathrm{Re}\biggl[\frac{\mathcal{L}_{LHS}}{x^2}\biggr] </math>

<math>~\approx</math>

<math>~ (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math>~ -~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math>~ + \beta^2(1-\cancelto{0}{x}\cos\theta)^3 (1-\cancelto{0}{2x} \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr] </math>

 

 

<math>~ - 2n(4n+1)\beta^2 m^2 + (4n+1)\beta^2 m^2 [4n + \beta^2 - \cancelto{0}{x^2} ] (1-\cancelto{0}{x}\cos\theta)^2 - (4n+1)\beta^2 m^2 n (1-\cancelto{0}{x}\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] </math>

 

<math>~\approx</math>

<math>~ \beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] + \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta -6(n+1)\biggr] </math>

 

 

<math>~+ \beta^2\biggl\{- 2n(4n+1) m^2+ (4n+1)m^2 [4n ] - 2(4n+1) m^2 n \biggr\} + \beta^4\biggl\{(4n+1) m^2 - (4n+1) m^2 n \biggl[ \frac{3}{(n+1)} \biggr] \biggr\} </math>

 

<math>~\approx</math>

<math>~ \beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr] + \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta -6(n+1)\biggr] </math>

 

 

<math>~ + m^2\beta^2\biggl[ -2^4(n+1)^2 (1 - 2\sin^2\theta ) \biggr] + m^2\beta^2\biggl\{- 2n(4n+1) + 4n(4n+1) - 2n(4n+1) \biggr\} </math>

 

 

<math>~ + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] </math>

 

<math>~\approx</math>

<math>~ 2\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr] + m^2\beta^2\biggl[ 2^4(n+1)^2 (1 - 2\cos^2\theta ) \biggr] </math>

 

 

<math>~ + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] </math>

 

<math>~\approx</math>

<math>~ (1-m^2)2^5\beta^2(n+1)^2\cos^2\theta + 2^2(n+1)\beta^2\biggl[ 4m^2(n+1) -3\biggr] + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] \, . </math>

<math>~\pm~\mathrm{Im}\biggl[\mathcal{L}_{LHS} \biggr]</math>

<math>~\approx</math>

<math>~ ~x\biggl\{\beta^3 (1-\cancelto{0}{x}\cos\theta)^3 (1-\cancelto{0}{2x} \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + \cancelto{0}{xb} )^{-1/2} </math>

 

 

<math> -~m^2\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + \cancelto{0}{xb} ]^{-3/2} \cos\theta (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x}^2 ] \biggr\} </math>

 

 

<math>~ +~x^2 \biggl\{ \beta m^2 \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math> +~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + \cancelto{0}{xb} )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math>

 

 

<math>~ -~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-\cancelto{0}{x}\cos\theta)^3 [1 + \cancelto{0}{xb} ]^{-1/2} \cdot \sin^2\theta </math>

 

 

<math>~ -~ (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-\cancelto{0}{x}\cos\theta)^2 \biggr\} </math>

 

<math>~\approx</math>

<math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

 

 

<math> +~~x^2 \beta^3 [ 2^3\cdot 3(n+1)^3 ]^{1/2}\biggl\{(12\cos^2\theta - 4\cos^4\theta) - 4\sin^2\theta </math>

 

 

<math>~- m^2 \biggl[2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta -~ \frac{n (4n+1)}{(n+1)^2} \biggr]\biggr\} </math>

 

<math>~\approx</math>

<math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

 

 

<math> +~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{4\cos^2\theta - \cos^4\theta -1 </math>

 

 

<math>~- \frac{m^2}{2} \biggl[\cos^2\theta(15 - 7\cos^2\theta) -12\sin^4\theta +6 \sin^2\theta -~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\} </math>

 

<math>~\approx</math>

<math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

 

 

<math> +~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{3-(1+\sin^2\theta)^2 - \frac{m^2}{2} \biggl[33\cos^2\theta-19 \cos^4\theta -6 -~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\} </math>


Examples

Evaluate various expressions using the parameter set:   <math>~(n, \theta, x) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math>

<math>~b</math>

<math>~=</math>

<math>~\frac{3}{2} - \frac{1}{8} = \frac{11}{8} </math>

1.375000000

<math>~(\beta\eta)^2</math>

<math>~=</math>

<math>~\biggl(\frac{1}{2^2}\biggr)^2\biggl[ 1 + \frac{11}{2^5} \biggr] = \frac{2^5 + 11}{2^9} = \frac{43}{2^9} </math>

0.083984375

<math>~\mathrm{Re}(\Lambda)</math>

<math>~=</math>

<math>~ -5\beta^2 + \frac{43}{2^9} \biggl[ 2^3 - 6 \biggr] = -5\beta^2 + \frac{43}{2^8} </math>

<math>~- 5\beta^2</math> + 0.167968750

<math>~\mathrm{Im}(\Lambda)</math>

<math>~=</math>

<math>~ \frac{\beta}{2}\biggl[ 2^{10}\cdot 3 \cdot \frac{43}{2^9} \biggr]^{1/2} = \beta \biggl[ \frac{3\cdot 43}{2} \biggr]^{1/2} </math>

8.031189202 <math>~\beta</math>

<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>

<math>~=</math>

<math>~ \biggl[ 2^3 - 6 \biggr]\frac{1}{2^2}\biggl( 2 + \frac{3\cdot 11}{2^2\cdot 2^3} \biggr) = \biggl( 1 + \frac{33}{2^6} \biggr) </math>

1.515625000

<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>

<math>~=</math>

<math>~ \frac{\beta}{2} \biggl[ \frac{2^9\cdot 2^8\cdot 3}{43} \biggr]^{1/2} \biggl[ \frac{1}{2^2}\cdot \biggl(2 + \frac{3\cdot 11}{2^2\cdot 2^3}\biggr) \biggr] = \beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) </math>

36.23373732 <math>~\beta</math>

<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>

<math>~=</math>

<math>~ 2\biggl(\frac{3}{4}\biggr)^{1/2} \biggl\{-2^4 \cdot \frac{43}{2^9} +\frac{3}{2^6}\biggl(\frac{3}{4}\biggr)\biggl[3-4\biggr] \biggr\} = ~-~\frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math>

-2.388335684

<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>

<math>~=</math>

<math>~ (-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot \biggl[ \frac{2^{10} \cdot 3\cdot 43}{2^9} \biggr]^{1/2} \biggl\{1 + \frac{3\cdot 2^9}{2^7 \cdot 43} \biggl(\frac{3}{2^3}\biggr)\biggr\} = (-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43 ]^{1/2} \biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\} </math>

(-1) × 15.36617018 <math>~\beta</math>


<math>~\mathrm{Re}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>

<math>~=</math>

<math>~ 2^2[2^2-3]\biggl[1 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr] = 2^2 + \frac{33}{2^3} = \frac{65}{8} </math>

          8.125000000

<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>

<math>~=</math>

<math>~\beta~ 2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2} =\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr] \beta </math>

          30.76957507<math>~\beta</math>

<math>~\mathrm{Re}\biggl(\frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>

<math>~=</math>

<math>~ \frac{1}{2^4} \biggl\{2^6 \biggl(\frac{3}{2^2} - \frac{1}{2^2} \biggr) \biggr\} + \frac{1}{2^6}\biggl\{-2^3\cdot 3 + 2 + \frac{3^3\cdot 5\cdot 7}{2^2} -3\cdot 23 \biggr\} = 2 + \frac{1}{2^8}\biggl\{2^3 + 3^3\cdot 5\cdot 7 - 2^2\cdot 3\cdot 31 \biggr\} </math>

          4.269531250

<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>

<math>~=</math>

<math>~(-1)\beta [2^{10}\cdot 3]^{1/2} \biggl( \frac{43}{2^9} \biggr)^{1/2} \biggl\{ \frac{1}{2} + \frac{3}{2}\cdot \frac{2^9}{43} \cdot \frac{1}{2^6}\cdot \frac{3}{2^2} \biggl(\frac{5}{2^2} -2 \biggr) + \biggl( \frac{3}{2}\cdot \frac{1}{2^6} \cdot \frac{2^9}{43}\biggr)^2 \biggl( \frac{3}{2^2} \biggr)^3 \frac{1}{2} \biggr\} </math>

         

 

<math>~=</math>

<math>~(-1)\beta\biggl( \frac{3\cdot 43}{2} \biggr)^{1/2} \biggl\{ 1 - \frac{3^3}{2\cdot 43} + \frac{3^5}{(2\cdot 43)^2} \biggr\} </math>

(-1) × 5.773638858 <math>~\beta</math>


<math>~ \mathrm{Re}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr) -~\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math>

 

 

<math>~=</math>

<math>~ ~\frac{(2\cdot 3\cdot 97)-3\cdot (2^3\cdot 43 + 9)}{2^9} </math>

-0.931640625

<math>~ \mathrm{Im}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) +~\frac{\sqrt{3}}{2}\cdot (-1) \beta~\frac{\sqrt{3}}{2}\cdot[ 2\cdot 3\cdot 43 ]^{1/2} \biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\} </math>

 

 

<math>~=</math>

<math>~\beta \biggl[ \frac{3^3}{2^5\cdot 43} \biggr]^{1/2} [2 (2^6 + 33) - (2\cdot 43 + 3^2) ] </math>

13.86780926 <math>~\beta</math>

<math>~ \mathrm{Re}\biggl\{ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr) +~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math>

 

 

<math>~=</math>

<math>~\frac{97^2}{2^{11}} +~\frac{3^3}{2^{11}}\cdot (2^3\cdot 43 + 9) </math>

9.248046874

<math>~ \mathrm{Im}\biggl\{ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\beta \biggl\{ \biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr) \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) +~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43 ]^{1/2} \biggl[1 + \frac{3^2}{2\cdot 43} \biggr]\biggr\} </math>

 

 

<math>~=</math>

<math>~\beta \biggl\{ \biggl( \frac{3}{2^9\cdot 43} \biggr)^{1/2} \biggl[ ( 2^6 + 33)^2 +~3^3(2\cdot 43 +3^2 ) \biggr]\biggr\} </math>

139.7753772

Step 7

Let's begin by slightly redefining the LHS and RHS collections of terms.

<math>~\mathrm{RHS}_4</math>

<math>~\equiv</math>

<math>~\mathrm{RHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>

 

<math>~=</math>

<math>~-~x^2 ~ [ 2^2(n+1)^2 - m^2 \Lambda ] \cdot \mathcal{A} \, , </math>

<math>~\mathrm{LHS}_4</math>

<math>~\equiv</math>

<math>~\mathrm{LHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>

 

<math>~=</math>

<math>~ \beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math>


Next Lowest Order

Let's begin with the RHS (Case B).

<math>~(n+1)\cdot \mathcal{A}</math>

<math>~=</math>

<math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math>

 

<math>~=</math>

<math>~ 2n (n+1) + (n+1)[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ x^2(1+xb) - \beta^2 - 4n ] + [1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3) ] [2n(n+1) - 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~x\cos\theta [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math>

 

<math>~=</math>

<math>~x^0\biggl\{2n (n+1) -4n(n+1) + 2n(n+1) \biggr\} + x^1\biggl\{8n(n+1)\cos\theta - 8n(n+1)\cos\theta\biggr\} </math>

 

 

<math>~ + x^2\biggl\{-4n(n+1)\cos^2\theta + (n+1)\biggl[ 1 - \biggl(\frac{\beta}{x}\biggr)^2 \biggr] + 12n(n+1)\cos^2\theta - 3n \biggl(\frac{\beta}{x}\biggr)^2 \biggr\} + \mathcal{O}(x^3) </math>

 

 

<math>~ \pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] </math>

 

<math>~=</math>

<math>~x^0\biggl\{0 \biggr\} + x^1\biggl\{0\biggr\} + x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 \biggr\} + \mathcal{O}(x^3) </math>

 

 

<math>~ \pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] </math>

where,

<math>b_0 \equiv [ 2^7\cdot 3 (n+1)^3 \cos^2\theta ]^{1/2} \, .</math>

Hence,

<math>~\frac{\mathrm{RHS}_4}{x^2}</math>

<math>~=</math>

<math>~-~ 2^2(n+1)^2\cdot \mathcal{A} + m^2 \cdot \mathcal{A} \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\} </math>

 

<math>~=</math>

<math>~-~ 2^2(n+1)^2\cdot \mathcal{A} + m^2 \cdot \mathcal{A} (n+1)\biggl\{ -x^2\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 + x^2(1+\cancelto{0}{x}b)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~x^2 \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} (1+\cancelto{0}{x}b)^{1/2} \biggr\} </math>

 

<math>~\approx</math>

<math>~\mathcal{A} (n+1)\biggl\{-~ 2^2(n+1) + m^2 \cancelto{0}{x^2} \biggl[ [2^3(n+1)\cos^2\theta - 3] -\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm ~~i~ \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} \biggr] \biggl\} </math>

 

<math>~\approx</math>

<math>~-2^2(n+1)x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm~i~\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2\cancelto{0}{x}\cos\theta + \cancelto{0}{x^2}\cos^2\theta + \cancelto{0}{\mathcal{O}}(x^3) \biggr] \biggr\} </math>

<math>~\Rightarrow ~~~~\frac{\mathrm{RHS}_4}{x^4}</math>

<math>~\approx</math>

<math>2^2(n+1)(4n+1)\biggl(\frac{\beta}{x}\biggr)^2 ~-~2^2(n+1)^2[8n\cos^2\theta + 1] ~~\pm~i~(-1)\biggl(\frac{\beta}{x}\biggr) nb_0 \, . </math>

This should be compared to,

<math>~\frac{\mathrm{LHS}_4}{x^2}</math>

<math>~=</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - xb \biggr] (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \, . </math>

Now, from above, we can write,

<math>~\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]</math>

<math>~=</math>

<math>~x^2\biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~\pm~~i~\beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \biggr\} + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} </math>

 

 

<math>~ + \cancelto{0}{x^3}\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta \biggr\} </math>

 

 

<math>~ \pm~~i~(-1)\beta b_0 ~x(1+\cancelto{0}{x}b)^{1/2}\biggl\{ 1 + \frac{3\cancelto{0}{x}\sin^2\theta (5\cos^2\theta -2)}{2(1+xb)\cos\theta } + \frac{3^2\cancelto{0}{x^2}\sin^6\theta}{2^2(1+xb)^2} \biggr\} </math>

 

<math>~\approx</math>

<math>~x^2\biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3] + 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} ~\pm~~i~x \biggl\{ \cancelto{0}{x\beta} \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- \beta b_0\biggr\}</math>

 

<math>~\approx</math>

<math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\} ~\pm~~i~x^2\biggl\{ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math>

Also,

<math>~ x\biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]</math>

<math>~=</math>

<math>~ x^2(n+1)[-6 + 2^4(n+1)\cos^2\theta] </math>

 

 

<math>~ - \cancelto{0}{x^3}(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \} </math>

 

 

<math>~ +\cancelto{0}{x^4}(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta - \cancelto{0}{x}[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>

 

 

<math>~- \cancelto{0}{x^2} \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ] \biggr\}</math>

 

<math>~\approx</math>

<math>~ 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \, . </math>

Finally,

<math>~ nx\biggl[ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr]</math>

<math>~=</math>

<math>~ x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] </math>

 

 

<math>~ + n \cancelto{0}{x^3}\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \} </math>

 

 

<math>~ +n \cancelto{0}{x^4} \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~ + n\cancelto{0}{x^4} (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \} </math>

 

 

<math>~\pm~~i~nx^2\biggl(\frac{\beta}{x}\biggr)~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+\cancelto{0}{x}b} \biggr]^{1/2} \biggl\{ 4\cos\theta + 6\cancelto{0}{x}(2b\cos\theta + \sin^4\theta) + 3\cancelto{0}{x^2}(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta) \biggr\} </math>

 

<math>~\approx</math>

<math>~ x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 \, . </math>

Inserting these three approximate expressions into the LHS_4 ensemble gives,

<math>~\frac{\mathrm{LHS}_4}{x^2}</math>

<math>~=</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - \cancelto{0}{xb} \biggr] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x (1-\cancelto{0}{x}\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math>

<math>~</math>

<math>~\approx</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{ ~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] ~\pm~~i~x^2\biggl[ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr] </math>

<math>~</math>

 

<math>~ + 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \biggr\} </math>

 

 

<math>~ -~x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 </math>

<math>~\Rightarrow~~~\frac{\mathrm{LHS}_4}{x^4}</math>

<math>~\approx</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{ ~2(n+1)\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] + 2(n+1)[2^3(n+1)\cos^2\theta -3] \biggr\} -~ 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] </math>

<math>~</math>

 

<math>~ ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 </math>

 

<math>~\approx</math>

<math>~ 2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [2^2(n+1) - 3 ] -(n+1) \biggl\{ [2^4(n+1) -12] +~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\} ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 </math>

 

<math>~\approx</math>

<math>~ 2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [4n+1 ] - 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 \, . </math>

Assessment

The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression. But the imaginary differ by a factor of 2. So, let's repeat the steps leading to the imaginary parts.

Case B:

<math>~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^2} \biggr]</math>

<math>~=</math>

<math>~-~2^2(n+1)\cdot \mathrm{Im}[(n+1)\mathcal{A}] +\frac{m^2}{(n+1)}\biggl\{ \mathrm{Im}[(n+1)\mathcal{A}]\cdot \mathrm{Re}[\Lambda] + \mathrm{Re}[(n+1)\mathcal{A}]\cdot \mathrm{Im}[\Lambda] \biggr\} </math>

 

<math>~=</math>

<math>~-~2^2(n+1)\cdot x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math>

 

 

<math>~ +\frac{m^2}{(n+1)}\biggl\{ x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \biggr\} \cdot \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] \biggr\} </math>

 

 

<math>~ +\frac{m^2}{(n+1)}\biggl\{ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] \biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\} </math>

 

<math>~=</math>

<math>~-~x^2 \biggl(\frac{\beta}{x}\biggr) \biggl\{ (1-x\cos\theta)^2 nb_0 \biggr\} </math>

 

 

<math>~ +m^2 x^4\biggl(\frac{\beta}{x}\biggr)\biggl\{ (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] \biggr\} \cdot \biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \biggl(\frac{\beta}{x}\biggr)^2 + (1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\} </math>

 

 

<math>~ + m^2 x^2\biggl(\frac{\beta}{x}\biggr) \biggl\{ 2n + (1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr] \biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\} </math>

<math>~\Rightarrow~~~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^4} \biggr]\biggl(\frac{x}{\beta}\biggr) </math>

<math>~=</math>

<math>~-~(1-x\cos\theta)^2 nb_0 </math>

 

 

<math>~ + m^2 \biggl\{ 2n + (1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr] \biggr\} \cdot (1+xb)^{1/2} </math>

 

 

<math>~ +m^2 \biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \beta^2 + x^2(1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\} \cdot (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] </math>

 

<math>~=</math>

<math>~-~(1-x\cos\theta)^2 nb_0 + m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4 \biggr\} \cdot (1+xb)^{1/2} </math>

 

 

<math>~ + m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)^{3/2} - \beta^2\cdot (1+xb)^{1/2} \biggl[ 1 + \frac{3n(1-x\cos\theta)^2}{(n+1)} \biggr] \biggr\} </math>

 

 

<math>~ +m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)[2^3(n+1)\cos^2\theta - 3]\cdot\biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] -\biggl[ \frac{nb_0(4n+1)}{2^2(n+1)^3}\biggr] \beta^2 \biggr\} </math>

 

<math>~=</math>

<math>~-~(1-x\cos\theta)^2 nb_0 + m^2 \biggl\{ 2n - 4n [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)] + 2n[ 1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3)] \biggr\} \cdot (1+xb)^{1/2} </math>

 

 

<math>~ + m^2 x^2 (1-x\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+xb)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{(1+xb)}{2^2(n+1)^2} </math>

 

 

<math>~ - m^2 \beta^2 (1-x\cos\theta)^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+xb)^{1/2} [ (n+1) + 3n(1-x\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math>

 

<math>~=</math>

<math>~-~ nb_0 [1 -2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)] + m^2 \biggl\{ 8n x^2\cos^2\theta + \mathcal{O}(x^3)\biggr\} \cdot (1+\cancelto{0}{x}b)^{1/2} </math>

 

 

<math>~ + m^2 x^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{(1+\cancelto{0}{x}b)}{2^2(n+1)^2} </math>

 

 

<math>~ - m^2 \beta^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} [ (n+1) + 3n(1-\cancelto{0}{x}\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math>

 

<math>~\approx</math>

<math>~-~ nb_0 [1 -2x\cos\theta ] </math>

 

 

<math>~-nb_0x^2\cos^2\theta + m^2 x^2 \biggl\{2^5n(n+1)^2 \cos^2\theta + 2^2(n+1)^2 + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{1}{2^2(n+1)^2} </math>

 

 

<math>~ - m^2 \beta^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2 [ (n+1) + 3n ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math>

Goldreich, Goodman and Narayan (1986)

Unperturbed Slim Torus Structure

Goldreich, Goodman & Narayan (1986, MNRAS, 221, 339) — hereafter, GGN86 — also used analytic techniques to analyze the properties of unstable, nonaxisymmetric eigenmodes in Papaloizou-Pringle tori. They restricted their discussion to only the slimmest tori, so overlap between the GGN86 and Blaes85 work is easiest to recognize if we begin with the enthalpy distribution prescribed for a "slim torus" by Blaes (1985), as discussed above, namely,

<math>~H = H_0\Theta_H</math>

<math>~=</math>

<math>~H_0 - \frac{H_0}{\beta^2}\biggl[r^2 + r^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(r^4) \biggr] \, .</math>

[Note:   Here we have replaced the variable name, <math>~x</math>, as used in Blaes85, with the variable name, <math>~r</math>, in order (1) to emphasize that the variable represents a dimensionless radial coordinate, and (2) to avoid conflict with the GGN86 variable, <math>~x</math>, which is a Cartesian coordinate with the standard dimension of length.]

Now, from our above discussion of equilibrium PP tori and recognizing that the Keplerian angular frequency at the location of the enthalpy maximum is,

<math>\Omega_K \equiv \frac{GM_\mathrm{pt}}{\varpi_0^3} \, ,</math>

we can set,

<math>~H_0</math>

<math>~=</math>

<math>~\frac{GM_\mathrm{pt}\beta^2}{2\varpi_0} = \tfrac{1}{2} \Omega_K^2 \varpi_0^2 \beta^2 \, .</math>

Hence, for the slimmest tori — that is, keeping only the lowest order term in <math>~r</math> — the enthalpy distribution becomes,

<math>~H </math>

<math>~=</math>

<math>~\tfrac{1}{2} \Omega_K^2 \varpi_0^2 \beta^2 - \tfrac{1}{2} \Omega_K^2 \varpi_0^2\biggl[r^2 + \cancelto{0}{r^3}(3\cos\theta - \cos^3\theta) + \cancelto{0}{\mathcal{O}(r^4)} ~~\biggr] </math>

 

<math>~\approx</math>

<math>~\frac{\Omega_K^2}{2} [\varpi_0^2 \beta^2 - \varpi_0^2 r^2] \, .</math>

Following GGN86, the surface of the torus — where the enthalpy drops to zero — occurs at <math>~r = a/\varpi_0</math>. Hence, we recognize that,

<math>\beta = \frac{a}{\varpi_0} \, ,</math>

and we can rewrite the expression for the unperturbed enthalpy distribution as,

<math>~H </math>

<math>~=</math>

<math>~\frac{\Omega_K^2}{2} [a^2 - \varpi_0^2 r^2 ] \, .</math>

This expression exactly matches equation (2.13) of GGN86 — which, is,

<math>~Q_0(x,z)</math>

<math>~=</math>

<math>~\frac{\Omega^2}{2} \biggl[ (2q-3)(a^2 - x^2) - z^2 \biggr] \, ,</math>

once it is appreciated that, in moving from the Blaes85 discussion to the GGN86 discussion, <math>~\varpi_0^2 r^2 \rightarrow (x^2 + z^2)</math>, and it is recognized that Blaes85 restricted his investigation to tori that have uniform specific angular momentum <math>~(q = 2)</math>.

Additional Notation

<math>~(ky)_\mathrm{GGN} = \biggl( \frac{my}{\varpi_0} \biggr)_\mathrm{GGN} ~~\leftrightarrow ~~ (m\phi)_\mathrm{Blaes}</math>

<math>~\beta_\mathrm{GGN} \equiv \biggl( \frac{ma}{\varpi_0} \biggr)_\mathrm{GGN} ~~\leftrightarrow ~~ m\beta_\mathrm{Blaes}</math>

From equation (5.16) of GGN86 we obtain "the lowest order [complex] expression for the [perturbed] velocity potential," namely,

<math>~\psi </math>

<math>~=</math>

<math>~1+\tfrac{1}{4} k^2(5x^2 - 3z^2) \mp 4i\biggl(\frac{3}{2}\biggr)^{1/2} k x \beta_\mathrm{GGN} \, .</math>

Working on the imaginary part of this expression to put it in the terminology of Blaes85, we find,

<math>~\mathrm{Im}(\psi)</math>

<math>~=</math>

<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} k x \beta_\mathrm{GGN} </math>

 

<math>~=</math>

<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} \biggl(\frac{m}{\varpi_0}\biggr) [\varpi_0 (\eta\beta_\mathrm{Blaes})\cos\theta ](m\beta_\mathrm{Blaes}) </math>

 

<math>~=</math>

<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} m^2\beta^2_\mathrm{Blaes} \eta\cos\theta \, ,</math>

which exactly matches <math>~\mathrm{Im}(f_m)</math> as derived by Blaes85 and summarized above. Similarly,

<math>~\mathrm{Re}(\psi)</math>

<math>~=</math>

<math>~1+\tfrac{1}{4} k^2(5x^2 - 3z^2) </math>

 

<math>~=</math>

<math>~1+\frac{1}{4} \biggl(\frac{m}{\varpi_0}\biggr)^2[\varpi_0^2 r^2(5\cos^2\theta - 3\sin^2\theta)] </math>

 

<math>~=</math>

<math>~1+\frac{1}{4} \eta^2 m^2 \beta^2_\mathrm{Blaes}[8\cos^2\theta - 3] </math>

 

<math>~=</math>

<math>~1+m^2 \beta^2_\mathrm{Blaes}\biggl[2\eta^2\cos^2\theta - \frac{3\eta^2}{4}\biggr] \, .</math>

This exactly matches <math>~\mathrm{Re}(f_m)</math> as derived by Blaes85 and summarized above. This is in line with the following statement that appears in the acknowledgement section of GGN86: "We note that Omar Blaes … [has] independently derived many of the results reported in this paper."

Summary Comparison

For slim, incompressible tori with uniform specific angular momentum, Blaes85 gives, to lowest order:

<math>~f_m(\eta,\theta) + \tfrac{1}{4} \beta^2 m^2</math>

<math>~=</math>

<math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} \pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr] \, . </math>

By comparison, from GGN86 we obtain:

<math>~\Psi(\eta,\theta) - 1</math>

<math>~=</math>

<math>~ m^2 \beta^2_\mathrm{Blaes}\biggl[2\eta^2\cos^2\theta - \frac{3\eta^2}{4} \mp 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta \biggr] \, . </math>

To within an additive constant, these two functions are identical. The resulting amplitude function is (to within an overall scale factor and to within an arbitrary additive constant),

<math>~F(\eta,\theta)</math>

<math>~=</math>

<math>~\biggl\{ \biggl[ 2\eta^2\cos^2\theta - \frac{3\eta^2}{4} \biggr]^2 + \biggl[ 4\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta \biggr]^2 \biggr\}^{1/2}</math>

 

<math>~=</math>

<math>~\biggl\{ \eta^2 \biggl[ 4^2\biggl(\frac{3}{2}\biggr) \cos^2\theta \biggr] + \frac{1}{4^2}\eta^4\biggl[ 8\cos^2\theta - 3\biggr]^2 \biggr\}^{1/2}</math>

 

<math>~=</math>

<math>~\biggl\{ \biggl(\frac{\eta}{2}\biggr)^2 \biggl[ 2^5\cdot 3 \cos^2\theta \biggr] + \biggl( \frac{\eta}{2}\biggr)^4\biggl[ 8\cos^2\theta - 3\biggr]^2 \biggr\}^{1/2} \, ;</math>

and the associate phase function is,

<math>~m\phi_m </math>

<math>~=</math>

<math>~\tan^{-1} \biggl[ \frac{ 2^7 \cdot 3 \cos^2\theta }{\eta^2 ( 8\cos^2\theta - 3)^2} \biggr] + k\theta \, .</math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

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