Editing SSC/StabilityConjecture/Bipolytrope51 (section)

===New Derivation===
====Expressions for q and &nu;====
Following through the numbered steps that we have used to [[SSC/Structure/BiPolytropes/Analytic51|construct a bipolytrope with]] <math>(n_c, n_e) = (5, 1)</math>, and adopting the substitute notation,
<div align="center">
<math>
\ell_i \equiv \frac{\xi_i}{\sqrt{3}}  \, ;
</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
<math>
m_3 \equiv 3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \, ,
</math>
</div>
we seek expressions for <math>\nu(m_3,\ell_i)</math> and  <math>q(m_3,\ell_i)</math>.  ['''Example #1''' numerical evaluation is for <math>\mu_e/\mu_c = 0.25</math> and <math>\xi_i = 0.5</math>, which implies that <math>m_3 = 0.75</math> and <math>\ell_i = (12)^{-1 / 2}</math>.] 

Focusing, first, on the core, we find,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(1 + \ell_i^2)^{-1 / 2} = 0.960768923</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{d\theta}{d\xi}\biggr)_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{\ell_i}{\sqrt{3}}(1 + \ell_i^2)^{-3 / 2} = -0.147810603</math>,
  </td>
</tr>
</table>

<!-- The radius and mass of the core -->
<table border="1" width="80%" align="center" cellpadding="8">
<tr><td align="center">
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1 / 2} r_\mathrm{core}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{3^2}{2\pi} \biggr)^{1 / 2} \ell_i  = 0.345494149</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl[ \frac{G^3 \rho_0^{2/5}}{K_c^3} \biggr]^{1 / 2} M_\mathrm{core}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3^2\biggl(\frac{2}{\pi} \biggr)^{1 / 2}\ell_i^3(1 + \ell_i^2)^{-3 / 2}  = 0.153203096</math>,
  </td>
</tr>
</table>
</td></tr>
</table>
Then moving across the interface, through the envelope, and ultimately to the surface of the configuration, we find,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\eta_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>m_3\ell_i \theta_i^2 = \frac{m_3 \ell_i}{(1+\ell_i^2)} = 0.199852016</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{d\phi}{d\eta}\biggr)_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3^{1 / 2} \theta_i^{-3} \biggl( \frac{d\theta}{d\xi}\biggr)_i = -0.288675135</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\Lambda_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta_i} + \biggl(\frac{d\phi}{d\eta}\biggr)_i
=
\frac{1}{m_3\theta_i^2 \ell_i} - \ell_i
=
\frac{1}{m_3\ell_i}\biggl[(1+\ell_i^2) - m_3\ell_i^2 \biggr]
=
\frac{1}{m_3\ell_i}\biggl[1 + (1 - m_3)\ell_i^2 \biggr] = 4.715027199\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\eta_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr) + \eta_i 
=
\biggl(\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr) +  \frac{m_3\ell_i }{(1 + \ell_i^2)} 
= 3.132453649\, , 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\eta_s - \pi
= -0.009139005\, , 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\eta_i(1+\Lambda_i^2)^{1 / 2}
=
\frac{m_3\ell_i}{(1+\ell_i^2)}\biggl\{
1 + \frac{1}{m_3^2 \ell_i^2}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{1 / 2}
=
0.963267676\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl( \frac{d\phi}{d\eta}\biggr)_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{A}{\eta_s^2} \biggl[\eta_s\cos(\pi) - \sin(\pi) \biggr]
=
-\frac{A}{\eta_s} 
=
-0.307512188\, ,
</math>
  </td>
</tr>
</table>

<!--  Total mass and radius -->
<table border="1" width="80%" align="center" cellpadding="8">
<tr><td align="center">
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1 / 2} R</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c}\biggr)^{-1} \theta_i^{-2} (2\pi)^{-1 / 2} \eta_s = 5.415228878</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~q \equiv \frac{r_\mathrm{core}}{R}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0.063800470 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl[ \frac{G^3 \rho_0^{2/5}}{K_c^3} \biggr]^{1 / 2} M_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \biggl(\frac{2}{\pi}\biggr)^{1 / 2} \biggl[- \frac{\eta_s^2}{\theta_i} \cdot \biggl(\frac{d\phi}{d\eta}\biggr)_s \biggr]
= 40.0934</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0.003821153 </math>
  </td>
</tr>
</table>
</td></tr>
</table>

Now, putting all these steps together, we can generate the pair of desired model-parameter expressions:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>q(m_3, \ell_i)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{m_3 \ell_i \theta_i^2}{\eta_s}
=
\frac{m_3 \ell_i }{(1 + \ell_i^2)} \biggl\{
\biggl(\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr) +  \frac{m_3\ell_i }{(1 + \ell_i^2)}
\biggr\}^{-1} 
=
\biggl\{
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr]\frac{(1 + \ell_i^2)}{m_3\ell_i} +  1 
\biggr\}^{-1} 
= 0.063800470\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\nu(m_3, \ell_i)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sqrt{3}\biggl[ \frac{\xi_i^3 \theta_i^4}{A\eta_s} \biggr]\frac{m_3^2}{3^2}
=
\frac{m_3 \ell_i \theta_i^2}{\eta_s}\biggl[ \frac{\xi_i^2 \theta_i^2}{A} \biggr]\frac{m_3}{3}
=
~q\ell_i \biggl[ \frac{m_3 \ell_i }{(1+\ell_i^2)A} \biggr]
=
~q\ell_i \biggl\{
1 + \frac{1}{m_3^2 \ell_i^2}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{-1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
~q m_3 \ell_i^2 \biggl\{
m_3^2 \ell_i^2 + \biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{-1 / 2} 
=
(0.059892291)q = 0.00382116\, .
</math>
  </td>
</tr>
</table>
<!-- UNNECESSARY CALCULATION 
'''Example #1:'''  Trying, <math>\xi_i = 0.5 ~~\Rightarrow~~ \ell_i = (12)^{-1 / 2}</math>, and, <math>\mu_e/\mu_c = 0.25  ~~\Rightarrow~~ m_3 = 3/4</math>, we expect from [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|Table 1 of our accompanying discussion]] that <math>(q, \nu) = (0.063720, 0.0033138)</math>.  Using our just-derived expressions, we obtain, <math>(\Lambda_i, q, \nu) = (4.71503, 0.063800, 0.0038211)</math>.
-->

Let's fully spell out the final <math>\nu(m_3, \ell_i)</math> function by incorporating the "q" function:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\nu(m_3, \ell_i)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m_3^2 \ell_i^3 \biggl\{
m_3^2 \ell_i^2 + \biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{-1 / 2} 
\biggl\{
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) +  m_3\ell_i 
\biggr\}^{-1} = 0.003821156\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\Lambda_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{m_3\ell_i}\biggl[1 + (1 - m_3)\ell_i^2 \biggr] = 4.715027198 \, .
</math>
  </td>
</tr>
</table>
For later use, note that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>m_3^2 \ell_i^2 (1 + \Lambda_i^2)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
[1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{1}{(1 + \Lambda_i^2)}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m_3^2 \ell_i^2 \{[1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2\}^{-1}
\, .
</math>
  </td>
</tr>
</table>

====Differentiate &nu; with Respect to &ell;<sub>I</sub>====
In order to determine the maximum value of the fractional core mass, we next need to determine the derivative of <math>\nu</math> with respect to <math>\ell_i</math>. ['''Example #2:''' Borrowing from [[#Table1|Table 1, above]], in this case our numerical evaluation is for <math>\mu_e/\mu_c = 0.25</math> and <math>\xi_i = 4.93827</math>, for which the expected maximum mass-fraction is, <math>\nu_\mathrm{max} = 0.1394</math>.  This implies that <math>m_3 = 0.75</math> and <math>\ell_i = 2.85111</math>.]

Let's rewrite the function as,

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

<tr>
  <td align="right">
<math>\nu</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{m_3^2 \ell_i^3}{F^{1 / 2} \cdot H} 
= 0.139370157 (8)\, ,
</math>
  </td>
</tr>
</table>
where,

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

<tr>
  <td align="right">
<math>\Lambda_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{m_3\ell_i}\biggl[1 + (1 - m_3)\ell_i^2 \biggr] = 1.418024375 (7) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>F</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
m_3^2 \ell_i^2 + [1 + (1 - m_3)\ell_i^2 ]^2
= 13.76676346 (4)
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>H</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) +  m_3\ell_i 
= 25.21038191 (5)
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
NOTE: &nbsp;<math>q</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{m_3 \ell_i}{H} 
= 0.084820
\, .
</math>
  </td>
</tr>
</table>
Then we have,

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

<tr>
  <td align="right">
<math>\frac{1}{m_3^2} \cdot \frac{d\nu}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3\ell_i^2}{F^{1 / 2} \cdot H} 
-\frac{1}{2}\biggl[ \frac{\ell_i^3}{F^{3 / 2} \cdot H}\biggr] \frac{dF}{d\ell_i}
- \biggl[ \frac{\ell_i^3}{F^{1 / 2} \cdot H^2} \biggr] \frac{dH}{d\ell_i}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{
3FH - \frac{\ell_i H}{2}\cdot \frac{dF}{d\ell_i}  - \ell_i F \cdot \frac{dH}{d\ell_i}
\biggr\} 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{ 1041.196093 - 425.9706908 - 615.2543231 \biggr\} 
= \frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{-0.028921\biggr\}\, .
</math>
&nbsp; &nbsp; <font color="red">EXCELLENT!</font>  
  </td>
</tr>
</table>
Furthermore,

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

<tr>
  <td align="right">
<math>\frac{dF}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2m_3^2 \ell_i + 4[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i
= 11.85266706 (6)\, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{dH}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
+ 2\ell_i\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + m_3
= 15.67503863 (9)\, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{1}{1 + \Lambda_i^2} \biggr] \frac{d\Lambda_i}{d\ell_i}
=
\biggl[\frac{1}{1 + \Lambda_i^2} \biggr] \frac{d}{d\ell_i}\biggl\{
\frac{1}{m_3\ell_i}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \frac{m_3^2 \ell_i^2 }{[1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2} \biggr\} \biggl\{
-\frac{1}{m_3\ell_i^2}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]
+
\frac{2(1-m_3)\ell_i}{m_3\ell_i}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ [1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2 \biggr\}^{-1} \biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\} 
= 0.056233763 (8)\, .
</math>
  </td>
</tr>
</table>
Now, along an equilibrium sequence of fixed <math>m_3</math>, the point of maximum core mass is located at the point where,

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

<tr>
  <td align="right">
<math>\frac{d\nu}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{
3FH - \frac{\ell_i H}{2}\cdot \frac{dF}{d\ell_i}  - \ell_i F \cdot \frac{dH}{d\ell_i}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ 0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
H \biggl\{ 3F - \frac{\ell_i}{2} \cdot \frac{dF}{d\ell_i}\biggr\}
-\ell_i F \cdot \frac{dH}{d\ell_i}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1041.196093 - 425.9706908 - 615.2543231 = -0.0289209
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) +  m_3\ell_i \biggr\}
\biggl\{ 3F - \ell_i \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \ell_i F \biggl\{ 
(1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
+ 2\ell_i\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + m_3
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) \biggr\}
\biggl\{ 3F - \ell_i \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] \biggr\}

- 2\ell_i^2 F \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- m_3\ell_i F
+ m_3\ell_i 
\biggl\{ 3F - \ell_i \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] \biggr\} 
- \ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
\biggl\{ 
3F(1 + \ell_i^2) 
- 2\ell_i^2 F 
- m_3^2 \ell_i \cdot \ell_i(1 + \ell_i^2)  
- 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i^2 (1 + \ell_i^2)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+  
3Fm_3\ell_i 
- m_3\ell_i F
- \ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
- m_3\ell_i^2 \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
\biggl\{ 
F(3 + \ell_i^2) 
- m_3^2 \ell_i^2(1 + \ell_i^2)  
- 2\biggl[1 + (1 - m_3)\ell_i^2 \biggr](1-m_3)\ell_i^2 (1 + \ell_i^2)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+  
2Fm_3\ell_i 
- \ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
- m_3\ell_i^2 \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2.527380938 
\biggl\{ 111.4667252 -112.5053101 \biggr\} + 38.72662783  -36.13064883 
=
-2.624899679 + 2.595979 = -0.028920679 \, .
</math>
&nbsp; &nbsp; <font color="red">EXCELLENT!</font>
  </td>
</tr>
</table>

Hence,

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

<tr>
  <td align="right">
<math>
20.14923887 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2.624899679 + 22.74521787
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
\biggl\{ 
F(3 + \ell_i^2) 
- \ell_i^2(1 + \ell_i^2) \biggl[ m_3^2   
+ 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr]  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+  
2Fm_3\ell_i 
- m_3\ell_i^3 \biggl[ m_3^2 + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
\ell_i F (1 + \ell_i^2) 
\biggl\{ [1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2 \biggr\}^{-1} \biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
F\biggl\{ \biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
(3 + \ell_i^2) 
+  
2m_3\ell_i \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \ell_i^2\biggl[ m_3^2   + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr]  
\biggl\{
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
(1 + \ell_i^2) +m_3\ell_i \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
\ell_i F (1 + \ell_i^2)^2 
\biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
F \cdot J \biggl\{ (H - m_3\ell_i) 
(3 + \ell_i^2) 
+  
2m_3\ell_i (1+\ell_i^2)\biggr\}
~- ~\ell_i^2(1 + \ell_i^2) H\cdot J\biggl[ m_3^2   + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
2532.246281
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
56062.28134 -53533.66963 = 2528.61171
</math>
  </td>
</tr>
</table>
The difference between the LHS and RHS &#8212; <math>(2528.61171 - 2532.246281) = -3.634571 </math> &#8212; is larger than our previously obtained "difference" <math>(-0.028920679)</math> by the factor, <math>(1 + \ell_i^2)J = 125.6745377</math>.  We are therefore satisfied that, for a given value of <math>m_3</math>, the value of <math>\ell_i</math> associated with the model that has the maximum core mass-fraction is identified when the LHS and RHS of this final expression match.

Note that, in reaching this final expression, we have recognized that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{H - m_3\ell_i}{1+\ell_i^2}\biggr]
\, ,
</math>
  </td>
</tr>
</table>
and have introduced the short-hand notation,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>
J 
</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2 + m_3^2 \ell_i^2
= 13.76676346 (3)\, .
</math>
  </td>
</tr>
</table>

====Examples====
By trial-and-error, we have searched for accurate <math>(m_3, \ell_i)</math> pairs; this, of course gives us the desired <math>(\mu_e/\mu_c, \xi_i)</math> pairs.  When an accurate pair has been discovered, we should find that the LHS and RHS of the following expression should be equal to one another, to a very high degree of precision.
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>
\ell_i F (1 + \ell_i^2)^2 
\biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
F \cdot J \biggl\{ (H - m_3\ell_i) 
(3 + \ell_i^2) 
+  
2m_3\ell_i (1+\ell_i^2)\biggr\}
~- ~\ell_i^2(1 + \ell_i^2) H\cdot J\biggl[ m_3^2   + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
In the following table, the first row of numbers (associated with <math>\mu_e/\mu_c = 1/4</math>) shows results from the relatively crude "trial" Example #2 that we used, above as we debugged our derivation of this analytic expression.  The second row of numbers improves on this initial guess, while the other rows give high-precision results for other selected values of <math>\mu_e/\mu_c</math>.
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="9">
[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/Bipolytrope/Stability/qAndNuMax.xlsx --- worksheet = K-BK74]]<br />'''Example High-Precision Determinations of <math>\nu_\mathrm{max}(\mu_e/\mu_c)</math>'''
  </td>
</tr>
<tr>
  <td align="center"><math>\frac{\mu_e}{\mu_c}</math></td>
  <td align="center"><math>\xi_i</math></td>
  <td align="center">LHS</td>
  <td align="center">RHS:TERM1</td>
  <td align="center">RHS:TERM2</td>
  <td align="center">error<br />(TERM1 - TERM2 - LHS)</td>
  <td align="center"><math>q</math></td>
  <td align="center"><math>\nu_\mathrm{max}</math></td>
  <td align="center">earlier<br />fractional error</td>
</tr>
<tr>
  <td align="center"><math>\frac{1}{4}</math></td>
  <td align="center"><math>4.93827</math></td>
  <td align="center">2532.246285</td>
  <td align="center">56062.281392</td>
  <td align="center">53533.669713</td>
  <td align="center"><math>-3.6346</math></td>
  <td align="center"><math>0.084820</math></td>
  <td align="center"><math>0.1393701568</math></td>
  <td align="center"><math>-1.1 \times 10^{-2}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>\frac{1}{4}</math></td>
  <td align="center"><math>4.9379256</math></td>
  <td align="center">2530.312408401</td>
  <td align="center">56030.44257523</td>
  <td align="center">53500.13020660</td>
  <td align="center"><math>-0.0000398</math></td>
  <td align="center"><math>0.0848241365</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.1393701572</math></td>
  <td align="center"><math>-1.2 \times 10^{-7}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>0.295</math></td>
  <td align="center"><math>7.07531489</math></td>
  <td align="center">22437.37085296</td>
  <td align="center">424789.588653918</td>
  <td align="center">402352.217777713</td>
  <td align="center"><math>+0.0000232</math></td>
  <td align="center"><math>0.0832775611</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.2646775149</math></td>
  <td align="center"><math>6.3 \times 10^{-9}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>0.3</math></td>
  <td align="center"><math>7.569605936</math></td>
  <td align="center">34614.27130158</td>
  <td align="center">652591.38554202</td>
  <td align="center">617977.11415666</td>
  <td align="center"><math>+0.0000838</math></td>
  <td align="center"><math>0.0814202240</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.2860557405</math></td>
  <td align="center"><math>1.5 \times 10^{-8}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>0.305</math></td>
  <td align="center"><math>8.193828507</math></td>
  <td align="center">57980.93749506</td>
  <td align="center">1095371.3718054</td>
  <td align="center">1037390.4343464</td>
  <td align="center"><math>-0.0000361</math></td>
  <td align="center"><math>0.0788994904</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.3100155910</math></td>
  <td align="center"><math>9.6 \times 10^{-9}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>0.310</math></td>
  <td align="center"><math>9.014959766</math></td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center"><math>+0.000169</math></td>
  <td align="center"><math>0.0755022550</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.3372170065</math></td>
  <td align="center"><math>-3.8 \times 10^{-9}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>0.320</math></td>
  <td align="center"><math>11.914571350</math></td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center"><math>-0.0000119</math></td>
  <td align="center"><math>0.0644564059</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.4061310924</math></td>
  <td align="center"><math>1.4 \times 10^{-9}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>0.325</math></td>
  <td align="center"><math>15.0964057345</math></td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center"><math>-0.000216</math></td>
  <td align="center"><math>0.0549312331</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.4531316008</math></td>
  <td align="center"><math>-4.7 \times 10^{-10}</math></td>
</tr>
<tr>
  <td align="center" bgcolor="lightgreen"><math>\frac{1}{3}</math></td>
  <td align="center"><math>\infty</math></td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center">-</td>
  <td align="center">--</td>
  <td align="center"><math>0.0</math></td>
  <td align="center" bgcolor="lightgreen"><math>0.63661977</math></td>
  <td align="center">-</td>
</tr>
</table>

In a [[SSC/Structure/BiPolytropes/Analytic51#Derivation|separate earlier derivation]], we determined that the analytic expression from which the value of <math>\nu_\mathrm{max}</math> can be derived is,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>
\biggl(\frac{\pi}{2} + \tan^{-1} \Lambda_i\biggr) (1+\ell_i^2) [ 3 + (1-m_3)^2(2-\ell_i^2)\ell_i^2] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m_3 \ell_i [(1-m_3)\ell_i^4 - (m_3^2 - m_3 +2)\ell_i^2 - 3] \, .
</math> 
  </td>
</tr>
</table>

The last column of the table &#8212; labeled "earlier fractional error" &#8212; shows the result of subtracting the LHS of this earlier expression from its RHS, then dividing by the LHS.  Because our derived "earlier fractional error" values are tiny, we are convinced that these two separately derived expressions are indeed identical.