Editing SSC/Stability/MoreGeneralApproach (section)

==Additional Setup==
Benefitting from our [[SSC/Structure/OtherAnalyticModels#Promising_Avenue_of_Exploration|earlier exploration of this problem]], let's divide through by the product, <math>~(a_0 b_0)</math>, and introduce the new variable notations,

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<math>~\lambda \equiv \frac{a_2}{a_0} \, ,</math> &nbsp; &nbsp; &nbsp; and  &nbsp; &nbsp; &nbsp;
<math>~\eta \equiv \frac{b_2}{b_0} \, .</math>
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The LAWE becomes,

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

<tr>
  <td align="right">
<math>~ 0
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \alpha(1 + \lambda x^2) (1 + \eta x^2) + 2x^2(n \lambda  + m\eta  )  + 2x^4 (n\lambda  \eta + m\eta  \lambda ) \biggr](5-3x^2) 
-\sigma^2 (1 + \lambda x^2) (1 + \eta x^2) 
</math>
  </td>
</tr>

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

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

Multiplying through by the denominator of the last  term(s) &#8212; that is, multiplying through by <math>~(1 + \lambda x^2)(1 + \eta x^2)</math> &#8212; will give us a polynomial with coefficient expressions for 6 terms <math>~(x^0, x^2, x^4, x^6, x^8, x^{10})</math> expressed in terms of 5 unknowns <math>~(\sigma^2, n, m, \lambda, \eta)</math>.

Wouldn't a better strategy be to insert yet another quadratic factor &#8212; specifically, <math>~(1+\beta x^2)^\ell</math> &#8212; which will introduce two additional unknowns but only add one more term into the polynomial expression?  This would bring the total number of coefficient expressions to 7 while simultaneously raising the number of unknowns to 7.  It will be tedious and messy, but worth the try.