Editing Appendix/Ramblings/PPToriPt1A (section)

=====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>
  </td>
</tr>
</table>
</div>