Editing SSC/Structure/OtherAnalyticModels (section)

===Prasad's Work===

====Overview====
{{ Prasad49 }} performed a semi-analytic analysis of the radial oscillations and stability of structures having a parabolic density distribution.  Let's examine his tabulated results to see if they help us understand more fully whether or not our analysis is on the right track.  For example, from his Table I, we see that <math>~\mathfrak{F} = 0</math> when <math>~\alpha = 0</math>, where, according to his equation (3),

<div align="center">
<math>~\mathfrak{F} \equiv \sigma^2 - 5\alpha \, .</math>
</div>

This means that, also, <math>~\sigma^2 = 0</math>.  Now, from our derived [[#Second_Constraint|second constraint]], we deduce that,

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

<tr>
  <td align="right">
<math>~\mathfrak{F} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~5[s_{nm} -n(1  - 2\lambda)] \, .</math>
  </td>
</tr>
</table>
</div>
Hence, since <math>~\mathfrak{F} = 0</math>, we conclude that,

<div align="center">
<math>~s_{nm} = n(1  - 2\lambda) \, .</math>
</div>

Also, since by definition <math>~s_{nm} = n + m</math>, we conclude that,

<div align="center">
<math>~\frac{m}{n} = - 2\lambda \, .</math>
</div>

Next, given that <math>~\alpha = 0</math>, we conclude from our derived [[#First_Constraint|first constraint]], that
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~s_{nm} = \frac{3^2}{2^2}\biggl[ -1 \pm 1\biggr] </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow</math>&nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~s_{nm}^{+} =0 </math>&nbsp; &nbsp; and &nbsp; &nbsp; <math>~s_{nm}^{-} = -\frac{9}{2} \, .</math>
  </td>
</tr>
</table>
</div>

=====The Minus Root=====

Combining these two results for the "minus" solution, we furthermore conclude that, for this ''specific'' mode, the relationship between the two exponents and <math>~\lambda</math> are,


<div align="center">
<math>~n^- = - \frac{9}{2(1-2\lambda)}</math> &nbsp;  &nbsp;  &nbsp; and  &nbsp;  &nbsp;  &nbsp;  <math>~m^- = (s_{nm} - n^-) = \frac{9\lambda}{(1-2\lambda)} \, .</math>
</div>

=====The Plus Root=====
Next, let's examine the "plus" solution.  Because <math>~s_{nm}^{+} =0 </math>, this solution implies that,


<div align="center">
<math>~m^+ = -n^+</math> &nbsp; &nbsp; &nbsp;<math>~\Rightarrow</math>&nbsp; &nbsp; &nbsp;<math>~\frac{m}{n} = -1</math>.
</div>

In this case, then, we deduce that,


<div align="center">
<math>~\lambda = -\frac{1}{2}\biggl(\frac{m}{n}\biggr) = +\frac{1}{2}</math>.
</div>

So, even though these first two constraints have not revealed the ''value'' of either of the exponents, <math>~n</math> and <math>~m</math>, we see that the resulting trial eigenfunction must be,
<div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="center">
<math>~\mathcal{G}_\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0^n(1 + \lambda x^2)^n \cdot (2 - x^2)^m </math>
  </td>
</tr>

<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_0^n\biggl[\frac{(1 + \tfrac{1}{2} x^2)}{(2 - x^2)}\biggr]^n </math>
  </td>
</tr>

<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{a_0}{2}\biggr)^n\biggl[\frac{(2 + x^2)}{(2 - x^2)}\biggr]^n \, .</math>
  </td>
</tr>
</table>

Interesting!

====Third Constraint====

=====The Minus Root=====

Let's insert all of these relations into the algebraic expression that we have derived from the <math>~x^2</math> coefficient:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2s_{nm}(9 + 2 s_{nm})+ 5(1 - 4 \lambda)s_{nm} -5n(1 - 4 \lambda)(1  - 2\lambda) + 60n \lambda  -20n \lambda^2 - 8n(n-1)\lambda^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (s_{nm}-n)[- 23  + 20 \lambda + 8n \lambda]   - 2(s_{nm}-n)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0
-\frac{45}{2}\biggl(1 - 4 \lambda\biggr) +\frac{45}{2}\biggl(1 - 4 \lambda\biggr) + n\lambda \biggl\{ 60  -20 \lambda + 8\biggl[\frac{11-4\lambda}{2(1-2\lambda)}\biggr]\lambda \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{9\lambda}{(1-2\lambda)} \biggl\{- 23  + 20 \lambda - \biggl[ \frac{36\lambda}{(1-2\lambda)} \biggr] \lambda \biggr\}   - 2\biggl[ \frac{9\lambda}{(1-2\lambda)}\biggr]^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)} \biggl\{ -30  +10 \lambda - 2\biggl[\frac{11-4\lambda}{(1-2\lambda)}\biggr]\lambda 
- 23  + 20 \lambda - \biggl[ \frac{36\lambda}{(1-2\lambda)} \biggr] \lambda  - \biggl[ \frac{18\lambda}{(1-2\lambda)}\biggr]\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)} \biggl\{ -53  +30 \lambda - \biggl[\frac{58-8\lambda}{(1-2\lambda)}\biggr]\lambda 
- \biggl[ \frac{18\lambda}{(1-2\lambda)}\biggr]\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)^2} \biggl\{( -53  +30 \lambda)(1-2\lambda) - (58-8\lambda)\lambda - 18\lambda  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)^2} \biggl[ -53  +30 \lambda  + 106\lambda -60\lambda^2 - 58\lambda + 8\lambda^2 - 18\lambda  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{9\lambda}{(1-2\lambda)^2} \biggl[ -53   + 60\lambda -52\lambda^2   \biggr]
</math>
  </td>
</tr>

</table>
</div>

Let's repeat this step, but start from an earlier expression for the <math>~x^2</math> coefficient, namely,

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

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + 60n \lambda  -20n \lambda^2- 25m    
+ 20m \lambda + 8n m \lambda  - [ 8n(n-1)\lambda^2 + 2m(m-1) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + n\biggl[60 \lambda  -20 \lambda^2 + \frac{m}{n}\biggl(- 25    
+ 20\lambda \biggr)\biggr] + 8n m \lambda  - [ 8n(n-1)\lambda^2 + 2m(m-1) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + n\biggl[60 \lambda  -12 \lambda^2 + \frac{m}{n}\biggl(- 23    
+ 20\lambda \biggr)\biggr] + 2n^2\biggl[- 4\lambda^2 +  4 \biggl(\frac{m}{n}\biggr) \lambda  -  \biggl(\frac{m}{n}\biggr) ^2  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

The first two terms on the RHS immediately go to zero because, for this ''specific'' eigenfunction, both <math>~\alpha</math> and <math>~\sigma^2</math> are zero.  Plugging in our determined expressions for <math>~n^-</math> and <math>~(m^-/n^-)</math> gives,


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

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{9}{2(1-2\lambda)}\biggl[60 \lambda  -12 \lambda^2 -2\lambda\biggl(- 23    
+ 20\lambda \biggr)\biggr] + 2\biggl[- \frac{9}{2(1-2\lambda)}\biggr]^2\biggl[- 4\lambda^2 +  4 \biggl(-2\lambda\biggr) \lambda  -  \biggl(-2\lambda\biggr) ^2  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{9\lambda}{(1-2\lambda)}\biggl[53   -26 \lambda\biggr] 
- \frac{8\cdot 81 \lambda^2}{(1-2\lambda)^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{9\lambda}{(1-2\lambda)^2}\biggl[ (1-2\lambda)(53   -26 \lambda) + 72 \lambda \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{9\lambda}{(1-2\lambda)^2}\biggl[53 - 60 \lambda + 52\lambda^2 \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
which exactly matches the previous, but messier, derivation.  Now, the two roots of the quadratic expression inside the square brackets are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3\cdot 13} \biggl[ 2^2\cdot 3\cdot 5 \pm \sqrt{ -2^8\cdot 29 }\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2\cdot 13} \biggl[ 3\cdot 5 \pm \sqrt{ -2^4\cdot 29 }\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Both roots are imaginary numbers and therefore not of interest in the context of this astrophysical problem.

=====The Plus Root=====
Next, in addition to setting <math>~\alpha = \sigma^2 = 0</math>, we'll plug <math>~\lambda = \tfrac{1}{2}</math> and <math>~m^+/n^+ = -1</math> into the third constraint expression as follows:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha(- 11 +20 \lambda) + \sigma^2(1 - 4 \lambda) + n\biggl[ 60 \lambda  -20 \lambda^2- 25\biggl(\frac{m}{n}\biggr)    
+ 20\biggl(\frac{m}{n}\biggr) \lambda + 8\lambda^2 + 2\biggl(\frac{m}{n}\biggr)  \biggr] + n^2\biggl[ 8\biggl(\frac{m}{n}\biggr) \lambda  - 8\lambda^2 - 2\biggl(\frac{m}{n}\biggr)^2  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n\biggl[ 60 \lambda  -20 \lambda^2+ 25    
- 20 \lambda + 8\lambda^2 - 2  \biggr] + n^2\biggl[ -8 \lambda  - 8\lambda^2 - 2  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n[ 40 \lambda  - 12 \lambda^2+ 23 ] -2 n^2[ 4 \lambda  + 4\lambda^2 +1 ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n[ 20  - 3+ 23 ] -2 n^2[ 2  + 1 +1 ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8n(5-n) \, .
</math>
  </td>
</tr>

</table>
</div>

So the nontrivial solution is <math>~n^+ = 5</math> &#8212; and, hence, <math>~m^+ = -5</math> &#8212; in which case the trial eigenfunction is,
<div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="center">
<math>~\mathcal{G}_\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{a_0}{2}\biggr)^5\biggl[\frac{(2 + x^2)}{(2 - x^2)}\biggr]^5 \, .</math>
  </td>
</tr>
</table>



====Fifth Constraint====
=====The Minus Root=====
In a similar vein, let's insert all of the deduced relations into the algebraic expression that we have derived from the <math>~x^6</math> coefficient:

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + 11 n \lambda  +22 m \lambda  - 47n \lambda^2 - 25m \lambda^2  -12 n m \lambda^2 + 4n m \lambda
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 10 n(n-1)\lambda^2 -2m(m-1)\lambda^2  +4m(m-1)\lambda 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + n\biggl[ 11 \lambda - 47\lambda^2  
+22 \biggl(\frac{m}{n}\biggr) \lambda  - 25\biggl(\frac{m}{n}\biggr) \lambda^2
+ 10 \lambda^2 + 2\biggl(\frac{m}{n}\biggr)\lambda^2  - 4\biggl(\frac{m}{n}\biggr)\lambda\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+n^2\biggl[ 
-12 \biggl(\frac{m}{n}\biggr) \lambda^2 + 4\biggl(\frac{m}{n}\biggr) \lambda - 10 \lambda^2 -2\biggl(\frac{m}{n}\biggr)^2\lambda^2  +4\biggl(\frac{m}{n}\biggr)^2\lambda \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + n\biggl[ 11 \lambda - 37\lambda^2  
+ \lambda\biggl(\frac{m}{n}\biggr)(18   - 23\lambda)
\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+n^2\biggl[ - 10 \lambda^2 
+ 4\lambda\biggl(\frac{m}{n}\biggr)(1-3 \lambda^2 ) + 2\lambda\biggl(\frac{m}{n}\biggr)^2 ( 2 -\lambda ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

As above, the first two terms on the RHS immediately go to zero because, for this ''specific'' eigenfunction, both <math>~\alpha</math> and <math>~\sigma^2</math> are zero.  Plugging in our determined expressions for <math>~n^-</math> and <math>~(m^-/n^-)</math> gives,


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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9}{2(1-2\lambda)}
\biggl[ 11 \lambda - 37\lambda^2  -2 \lambda^2(18   - 23\lambda)\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+2\biggl[  -\frac{9}{2(1-2\lambda)} \biggr]^2\biggl[ - 5 \lambda^2 -4\lambda^2(1-3 \lambda^2 ) + 4\lambda^3 ( 2 -\lambda ) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9\lambda}{2(1-2\lambda)}
\biggl[ 11 - 73\lambda  +46 \lambda^2\biggr]  
+\biggl[ \frac{9^2\lambda^2}{2(1-2\lambda)^2} \biggr]\biggl[ - 9 + 8\lambda  +8\lambda^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9\lambda}{2(1-2\lambda)^2} \biggl\{(1-2\lambda)
[ 11 - 73\lambda  +46 \lambda^2]  
-9\lambda [ - 9 + 8\lambda  +8\lambda^2]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\frac{9\lambda}{2(1-2\lambda)^2} \biggl[11 - 14\lambda  +120 \lambda^2 -164 \lambda^3
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

=====The Plus Root=====
Next, in addition to setting <math>~\alpha = \sigma^2 = 0</math>, we'll plug <math>~\lambda = \tfrac{1}{2}</math> and <math>~m^+/n^+ = -1</math> into the fifth constraint expression as follows:

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
\alpha(6\lambda - 11\lambda^2)  + \sigma^2\lambda^2 + n\biggl[ 11 \lambda - 37\lambda^2  
+ \lambda\biggl(\frac{m}{n}\biggr)(18   - 23\lambda)
\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+n^2\biggl[ - 10 \lambda^2 
+ 4\lambda\biggl(\frac{m}{n}\biggr)(1-3 \lambda^2 ) + 2\lambda\biggl(\frac{m}{n}\biggr)^2 ( 2 -\lambda ) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
n\lambda [ 11 - 37\lambda  - (18   - 23\lambda) ]  
+n^2\lambda [ - 10 \lambda - 4 (1-3 \lambda^2 ) + 2 ( 2 -\lambda ) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
n\lambda [ -7 - 14\lambda ]  +n^2\lambda [ - 10 \lambda -4 + 12 \lambda^2  + 4 - 2\lambda  ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
- 7n  - \biggl( \frac{3}{2} \biggr) n^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td>
<math>~ 
- 7n\biggl[1  + \biggl( \frac{3}{14} \biggr) n \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>
From this constraint, it appears that the nontrivial result is, <math>~n = -14/3</math>.

====Fourth Constraint====

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(10 \lambda^2  - 22 \lambda +3) - \sigma^2(2\lambda^2- 2\lambda ) - 47n \lambda+ 60n \lambda^2  - 50m \lambda    +11m + 10m \lambda^2-12 n m \lambda    + 8n m \lambda^2
</math>
  </td>
</tr>

<tr>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
  <td align="left">
<math>
~ + 16n(n-1)\lambda^2 - 4m(m-1)\lambda + 2m(m-1) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(10 \lambda^2  - 22 \lambda +3) - \sigma^2(2\lambda^2- 2\lambda ) + n\biggl[- 47\lambda+ 60 \lambda^2  - 50\biggl(\frac{m}{n}\biggr) \lambda    
+11\biggl(\frac{m}{n}\biggr) + 10\biggl(\frac{m}{n}\biggr) \lambda^2 - 16\lambda^2 + 4\biggl(\frac{m}{n}\biggr) \lambda - 2\biggl(\frac{m}{n}\biggr)\biggr] 
</math>
  </td>
</tr>

<tr>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
  <td align="left">
<math>
~+n^2 \biggl[ -12 \biggl(\frac{m}{n}\biggr) \lambda  + 8\biggl(\frac{m}{n}\biggr) \lambda^2 + 16\lambda^2 - 4\biggl(\frac{m}{n}\biggr)^2\lambda + 2\biggl(\frac{m}{n}\biggr)^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\alpha(10 \lambda^2  - 22 \lambda +3) - \sigma^2(2\lambda^2- 2\lambda ) + n\biggl[- 47\lambda+ 44 \lambda^2  - 37\biggl(\frac{m}{n}\biggr) \lambda    
+ 10\biggl(\frac{m}{n}\biggr) \lambda^2 \biggr] 
</math>
  </td>
</tr>

<tr>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
 <td align="right">&nbsp;</td>
  <td align="left">
<math>
~+n^2 \biggl[ + 16\lambda^2 -12 \biggl(\frac{m}{n}\biggr) \lambda  + 8\biggl(\frac{m}{n}\biggr) \lambda^2 - 4\biggl(\frac{m}{n}\biggr)^2\lambda + 2\biggl(\frac{m}{n}\biggr)^2 \biggr]
</math>
  </td>
</tr>
</table>
</div>

=====The Minus Root=====
In addition to setting <math>~\alpha = \sigma^2 = 0</math>, here we plug <math>~n^- = -9/[2(1-2\lambda)]</math> and <math>~m^+/n^+ = -2\lambda</math> into the fourth constraint expression as follows:


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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
n\lambda [- 47+ 118 \lambda    -20 \lambda^2 ] +n^2 \lambda^2[ 16 +24   -16 \lambda - 16 \lambda + 8 ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\biggl[\frac{9}{2(1-2\lambda)}\biggr]
\lambda [- 47+ 118 \lambda    -20 \lambda^2 ] +\biggl[\frac{9}{2(1-2\lambda)}\biggr]^2 \lambda^2[ 48   -32 \lambda]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{9\lambda}{2(1-2\lambda)^2}\biggr]\biggl\{9 \lambda[ 24   -16 \lambda]  -
(1-2\lambda) [- 47+ 118 \lambda    -20 \lambda^2 ] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{9\lambda}{2(1-2\lambda)^2}\biggr]\biggl\{216\lambda - 144\lambda^2  
+ [47- 118 \lambda    +20 \lambda^2 ] + [- 94\lambda + 236 \lambda^2    -40 \lambda^3 ]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{9\lambda}{2(1-2\lambda)^2}\biggr]\biggl\{47 -4\lambda + 112\lambda^2   -40 \lambda^3 
\biggr\}
</math>
  </td>
</tr>
</table>
</div>


=====The Plus Root=====
Next, in addition to setting <math>~\alpha = \sigma^2 = 0</math>, we'll plug <math>~\lambda = \tfrac{1}{2}</math> and <math>~m^+/n^+ = -1</math> into the fourth constraint expression as follows:

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

<tr>
  <td align="right">
<math>~x^6:</math>&nbsp; &nbsp;
  </td>
  <td align="right">
RHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
n\biggl[- 47\lambda+ 44 \lambda^2  + 37 \lambda    - 10 \lambda^2 \biggr] +n^2 \biggl[ + 16\lambda^2 + 12 \lambda  - 8 \lambda^2 - 4 \lambda + 2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{7n}{2}\biggl[1+ n\biggl(\frac{16}{7}\biggr)  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

From this constraint, it appears that the nontrivial result is, <math>~n^+ = -7/16</math>.