Editing Appendix/Ramblings/NonlinarOscillation (section)

====Roots of Quartic Equation====

=====Analytic Solution=====
Here we will adopt the shorthand notation,
<div align="center">
<math>\chi \equiv  \frac{R_\mathrm{eq}}{R_\mathrm{SWS}}</math> &nbsp; &nbsp; &nbsp;and  &nbsp; &nbsp; &nbsp;<math>~m \equiv  \frac{M}{M_\mathrm{SWS}} \, .</math>
</div>

Drawing from the [https://en.wikipedia.org/wiki/Quartic_function Wikipedia discussion of the quartic function], we will think in terms of the generic quartic form,
<div align="center">
<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>a\chi^4 + b\chi^3 + c\chi^2 + d\chi + e \,.</math>
  </td>
</tr>
</table>
</div>
Relating this to our specific quartic function, we should make the following assignments:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>a</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2^2 \cdot 5 \pi}{3} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>b</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>c</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>d</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-5m</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>e</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>m^2</math>
  </td>
</tr>
</table>
</div>

We need to evaluate the following expressions:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>p</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\frac{8ac-3b^2}{8a^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>q</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\frac{b^3 - 4abc + 8a^2d}{8a^3} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{d}{a} = - \frac{3m}{2^2  \pi} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Delta_0</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>c^2 - 3bd + 12ae</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>12ae = 2^4 \cdot 5~\pi m^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Delta_1</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>2c^3 - 9bcd + 27b^2e+27ad^2 - 72ace</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>27ad^2 = 3^3 \cdot 5^2 m^2 \cdot \frac{2^2 \cdot 5 \pi}{3} = 2^2 \cdot 3^2 \cdot 5^3 ~\pi m^2</math>
  </td>
</tr>
</table>
</div>

Note that the discriminant is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Delta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
256a^3 e^3 - 192a^2bde^2 - 128a^2c^2e^2 + 144a^2 c d^2e - 27a^2d^4
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ 144 a b^2c e^2 - 6ab^2d^2e - 80 abc^2de + 18abcd^3 + 16ac^4e
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- 4ac^3 d^2 - 27b^4e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2c^2d^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
256a^3 e^3 - 27a^2d^4
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2^8\biggl( \frac{2^2 \cdot 5 \pi}{3} \biggr)^3 m^6 - 3^3\biggl( \frac{2^2 \cdot 5 \pi}{3} \biggr)^2 5^4 m^4
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2^{14} \cdot 5^3 \pi^3}{3^3} \biggr) m^6 - \biggl( 2^4 \cdot 3 \cdot 5^6 \pi^2 \biggr)  m^4
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl( 2^4 \cdot 3 \cdot 5^6 \pi^2 \biggr)  m^4 \biggl[1 - \biggl( \frac{1}{2^4 \cdot 3 \cdot 5^6 \pi^2 m^4} \biggr) \biggl( \frac{2^{14} \cdot 5^3 \pi^3}{3^3} \biggr) m^6 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl( 2^4 \cdot 3 \cdot 5^6 \pi^2 \biggr)  m^4 \biggl[1 - \biggl( \frac{2^{10}  \pi}{3^4\cdot 5^3} \biggr) m^2 \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and it will be negative (or, in the limit, zero) as long as <math>m \le m_\mathrm{max}</math>, where, as [[#Roots_of_Quadratic_Equation|defined above]], <math>m_\mathrm{max} \equiv [3^4\cdot 5^3/(2^{10}\pi)]^{1 / 2}</math>.  Because the discriminant is always negative (or, at most, zero), then our quartic equation has two distinct real roots and two complex conjugate non-real roots.

Furthermore note that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Delta_1^2 - 4\Delta_0^3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(2^2 \cdot 3^2 \cdot 5^3 ~\pi m^2)^2 - 2^2(2^4 \cdot 5~\pi m^2  )^3
= (2^4 \cdot 3^4 \cdot 5^6 ~\pi^2 m^4) \biggl[1 - \frac{(2^{14} \cdot 5^3~\pi^3 m^6  )}{2^4 \cdot 3^4 \cdot 5^6 ~\pi^2 m^4}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(2^4 \cdot 3^4 \cdot 5^6 ~\pi^2 m^4) \biggl[1 - \biggl(\frac{2^{10} ~\pi}{3^4 \cdot 5^3 }\biggr) m^2\biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and it will never be negative, as long as <math>m \le m_\mathrm{max}</math>.


For a given value of <math>m</math>, then, the pair of real roots is:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\chi_\pm</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{b}{4a} + S \pm \frac{1}{2}\biggl[ -4S^2 - 2p - \frac{q}{S} \biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>S</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl[- \frac{2p}{3} + \frac{1}{3a}\biggl(Q + \frac{\Delta_0}{Q}\biggr)  \biggr]^{1/2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>Q</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2} \biggr]^{1/3} \, .
</math>
  </td>
</tr>
</table>
</div>

Let's work through these expressions.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>Q</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(2 \cdot 3^2 \cdot 5^3 ~\pi m^2)^{1 / 3}
\biggl\{ 1 +  \biggl[1 - \biggl( \frac{m}{m_\mathrm{max}}\biggr)^2\biggr]^{1 / 2} 
\biggr\}^{1/3} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>S^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2^2 \cdot 3a}\biggl(Q + \frac{\Delta_0}{Q}\biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{Q}{2^4 \cdot 5 \pi} + \frac{m^2}{Q} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\chi_\pm</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
S \pm \frac{1}{2}\biggl[\frac{1}{S}\biggl( \frac{3m}{2^2\pi} \biggr) -4S^2   \biggr]^{1/2} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
S \biggl\{1 \pm \biggl[\frac{3m}{2^4\pi ~S^3}  - 1  \biggr]^{1/2} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
<!--
We have used an Excel spreadsheet to evaluate these expressions.  The following table identifies <math>~\epsilon_\pm</math> pairs (the middle two columns of numbers) for twenty different values of the external pressure; more specifically, for twenty values of <math>~0 \le \delta \le 0.69938</math>, equally spaced between the two limits.  The corresponding pairs of <math>~\tilde\xi_\pm</math> are also listed (rightmost pair of columns).
-->

=====Check at Maximum Mass=====
As a check, recognize that the two roots should be identical, and given by (see [[#Roots_of_Quadratic_Equation|above definition]]) <math>~r_\mathrm{crit} = [3^2 \cdot 5/(2^6\pi)]^{1 / 2}</math>, when <math>~m = m_\mathrm{max}</math>.  Let's see &hellip;
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q\biggr|_{m_\mathrm{max}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 2 \cdot 3^2 \cdot 5^3 ~\pi \biggl( \frac{3^4\cdot 5^3}{2^{10}\pi}\biggr)\biggr]^{1 / 3} 
= \biggl[ \frac{3^6\cdot 5^6}{2^{9}} \biggr]^{1 / 3}
= \frac{3^2\cdot 5^2}{2^{3}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ S^2 \biggr|_{m_\mathrm{max}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^4 \cdot 5 \pi} \biggl(\frac{3^2\cdot 5^2}{2^{3}}\biggr) + \biggl( \frac{3^4\cdot 5^3}{2^{10}\pi}\biggr)\biggl(\frac{2^{3}}{3^2\cdot 5^2}\biggr) 
= \frac{3^2\cdot 5}{2^{6} \pi} = r_\mathrm{crit}^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \chi_\pm\biggr|_{m_\mathrm{max}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r_\mathrm{crit} \biggl\{1 \pm \biggl[\frac{3}{2^4\pi }\biggl( \frac{3^4\cdot 5^3}{2^{10}\pi}\biggr)^{1 / 2}\biggl( \frac{2^{6} \pi}{3^2\cdot 5} \biggr)^{3 / 2}  - 1  \biggr]^{1/2} \biggr\} 
= r_\mathrm{crit} \, , 
</math>
  </td>
</tr>

</table>
</div>
Q.E.D.

=====In Terms of the Limiting Mass=====

Because it will be useful to us later, let's rewrite the expression for <math>~\chi_\pm</math> in terms of the new parameter, 
<div align="center">
<math>~\mu \equiv \biggl[ 1 - \biggl( \frac{m}{m_\mathrm{max}} \biggr)^2 \biggr]^{1 / 2}</math> &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; <math>~m^2 =
 m_\mathrm{max}^2(1-\mu^2) = \frac{3^4\cdot 5^3}{2^{10} \pi}(1-\mu^2) \, ,</math>
</div>
which will be small &#8212; and always positive &#8212; in the vicinity of the limiting mass.  For <math>~S</math> and <math>~Q</math> we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(2 \cdot 3^2 \cdot 5^3 ~\pi m^2)(1+\mu)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~S^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{Q}{2^4 \cdot 5 \pi}\biggl[1 + \frac{2^4\cdot 5\pi ~m^2}{Q^2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^4 \cdot 5 \pi}\biggl[ \frac{3^2\cdot 5^2}{2^{3}} \biggr] (1-\mu^2)^{1 / 3}(1+\mu)^{1 / 3} 
\biggl[1 + \frac{2^4\cdot 5\pi }{Q^2} \cdot \frac{3^4\cdot 5^3}{2^{10} \pi}(1-\mu^2)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{3^2\cdot 5}{2^{7}~\pi} \biggr] (1-\mu^2)^{1 / 3}(1+\mu)^{1 / 3} 
\biggl[1 + \frac{3^4\cdot 5^4}{2^{6} } \cdot \frac{(1-\mu^2)}{Q^2}\biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{S}{r_\mathrm{crit}}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\mu^2)^{1 / 3}(1+\mu)^{1 / 3} 
\biggl[ \frac{1 + (1-\mu^2)^{1 / 3} (1+\mu)^{- 2 / 3}}{2} \biggr] 
</math>
  </td>
</tr>

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

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

This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{3m}{2^4\pi ~S^3} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3m}{2^4\pi } \biggl\{
\biggl( \frac{r_\mathrm{crit}^2}{2} \biggr) (1-\mu^2)^{1 / 3}(1+\mu)^{1 / 3} 
\biggl[ 1 + (1-\mu^2)^{1 / 3} (1+\mu)^{- 2 / 3} \biggr] 
\biggr\}^{-3 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{2^4\pi } \biggl[ \frac{3^4\cdot 5^3}{2^{10} \pi} \biggr]^{ 1 / 2} \biggl( \frac{2}{r_\mathrm{crit}^2} \biggr)^{3 / 2} (1-\mu^2)^{1 / 2} \biggl\{
(1-\mu^2)^{-1 / 2}(1+\mu)^{- 1 / 2} 
\biggl[ 1 + (1-\mu^2)^{1 / 3} (1+\mu)^{- 2 / 3} \biggr]^{-3 / 2} 
\biggr\}
</math>
  </td>
</tr>

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

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

Hence, utilizing the shorthand notation,
<div align="center">
<math>~u^2(\mu) \equiv \biggl[ \frac{(1+\mu)^{1 / 3} + (1-\mu)^{1 / 3} }{2} \biggr]^{-1} \, ,</math>
</div>
we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl(\frac{\chi}{r_\mathrm{crit}}\biggr)_\pm</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{S}{r_\mathrm{crit}} \biggl\{1 \pm \biggl[\frac{3m}{2^4\pi ~S^3}  - 1  \biggr]^{1/2} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-\mu^2)^{1 / 6}}{u}  \biggl[1 \pm (u^3 - 1 )^{1/2} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

=====Approximation Near the Limiting Mass=====
When <math>m</math> is ''near'' <math>m_\mathrm{max}</math>, we know that <math>0 \le \mu \ll 1</math>.  So, drawing on the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial expansion]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{(1+\mu)^{1 / 3} + (1-\mu)^{1 / 3} }{2} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{2} \biggl[ 
1+ \frac{1}{3}\mu + \frac{1}{2}\biggl(\frac{1}{3}\biggr) \biggl(\frac{1}{3}-1\biggr) \mu^2 + \cdots
+ 1 - \frac{1}{3}\mu + \frac{1}{2}\biggl(\frac{1}{3}\biggr) \biggl(\frac{1}{3}-1\biggr) \mu^2 + \cdots
\biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1  - \biggl(\frac{1}{3^2}\biggr) \mu^2 + \biggl[ \frac{1}{2^3 \cdot 3^2 } \biggl(-\frac{2}{3}\biggr) \biggl(- \frac{5}{3}\biggr) \biggl(- \frac{8}{3}\biggr)\biggr]\mu^4 + \cdots
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1  - \biggl(\frac{\mu}{3}\biggr)^2  - \biggl[ \frac{2\cdot 5}{3^5 } \biggr]\mu^4 + \cdots
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{(1-\mu^2)^{1 / 6}}{u}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 - \frac{\mu^2}{6} + \frac{1}{2}\biggl(\frac{1}{6}\biggr) \biggl(\frac{1}{6}-1\biggr)\mu^4 + \mathcal{O}(\mu^6)\biggr]
\biggl[ 1  - \biggl( \frac{\mu^2}{3^2}  + \frac{2\cdot 5 \mu^4}{3^5 } \biggr) + \mathcal{O}(\mu^6) \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 - \frac{\mu^2}{2\cdot 3} - \biggl(\frac{5}{2^3\cdot 3^2}\biggr)\mu^4 + \mathcal{O}(\mu^6) \biggr]
\biggl[ 1  - \frac{1}{2}\biggl( \frac{\mu^2}{3^2}  + \frac{2\cdot 5 \mu^4}{3^5 } \biggr) - \biggl( \frac{\mu^4}{2^3\cdot 3^4}  \biggr)+ \mathcal{O}(\mu^6) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 - \frac{\mu^2}{2\cdot 3} - \biggl(\frac{5}{2^3\cdot 3^2}\biggr)\mu^4 + \mathcal{O}(\mu^6) \biggr]
\biggl[ 1  - \frac{\mu^2}{2 \cdot 3^2}  - \biggl( \frac{43}{2^3 \cdot 3^5 }  \biggr)\mu^4  + \mathcal{O}(\mu^6) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \frac{\mu^2}{2\cdot 3} - \biggl(\frac{5}{2^3\cdot 3^2}\biggr)\mu^4 
- \biggl[ 1 - \frac{\mu^2}{2\cdot 3} \biggr]
\biggl[ \frac{\mu^2}{2 \cdot 3^2}   \biggr]
- \biggl( \frac{43}{2^3 \cdot 3^5 }  \biggr)\mu^4  
+ \mathcal{O}(\mu^6)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \biggl[ \frac{\mu^2}{2\cdot 3} + \frac{\mu^2}{2 \cdot 3^2}   \biggr]
+ \biggl[ \frac{1}{2^2 \cdot 3^3} 
- \frac{5}{2^3\cdot 3^2} 
- \frac{43}{2^3 \cdot 3^5 }  \biggr]\mu^4  
+ \mathcal{O}(\mu^6)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \biggl(\frac{2}{3^2}\biggr) \mu^2
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
+ \mathcal{O}(\mu^6) \, ;
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>u^3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1  - \biggl( \frac{\mu^2}{3^2}  + \frac{2\cdot 5 \mu^4}{3^5 } \biggr) + \mathcal{O}(\mu^6) \biggr]^{-3 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1  + \frac{3}{2} \biggl( \frac{\mu^2}{3^2}  + \frac{2\cdot 5 \mu^4}{3^5 } \biggr) 
+ \frac{1}{2} \biggl( -\frac{3}{2}\biggr) \biggl( -\frac{3}{2} - 1\biggr) \biggl( \frac{\mu^2}{3^2}  \biggr)^2
+ \mathcal{O}(\mu^6) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1  + \biggl( \frac{\mu^2}{2 \cdot 3}\biggr)  + \biggl(\frac{5 }{3^4 }  + \frac{5}{2^3 \cdot 3^3}  \biggr)\mu^4
+ \mathcal{O}(\mu^6) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1  + \biggl( \frac{\mu^2}{2 \cdot 3}\biggr)  + \biggl( \frac{55}{2^3 \cdot 3^4} \biggr) \mu^4
+ \mathcal{O}(\mu^6) \, .
</math>
  </td>
</tr>
</table>
</div>

Therefore the two real roots of the quartic equation are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl(\frac{\chi}{r_\mathrm{crit}}\biggr)_\pm</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{(1-\mu^2)^{1 / 6}}{u}  
\biggl\{1 \pm \biggl[u^3 
- 1 \biggr]^{1/2} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[
1 - \biggl(\frac{2}{3^2}\biggr) \mu^2
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
+ \mathcal{O}(\mu^6) 
\biggr]
\biggl\{1 \pm \biggl[\biggl( \frac{\mu^2}{2 \cdot 3}\biggr)  + \biggl( \frac{55}{2^3 \cdot 3^4} \biggr) \mu^4
+ \mathcal{O}(\mu^6) \biggr]^{1/2} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[
1 - \biggl(\frac{2}{3^2}\biggr) \mu^2
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
\biggr]
\biggl\{1 \pm \biggl( \frac{\mu^2}{2 \cdot 3}\biggr)^{1 / 2}
\biggl[1  + \biggl( \frac{5\cdot 11}{2^2 \cdot 3^3} \biggr) \mu^2 + \mathcal{O}(\mu^4) \biggr]^{1/2} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[
1 - \biggl(\frac{2}{3^2}\biggr) \mu^2
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
\biggr]
\biggl\{1 \pm \biggl( \frac{\mu^2}{2 \cdot 3}\biggr)^{1 / 2}
\biggl[1  + \biggl( \frac{5\cdot 11}{2^3 \cdot 3^3} \biggr) \mu^2 + \mathcal{O}(\mu^4) \biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[
1 - \biggl(\frac{2}{3^2}\biggr) \mu^2
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
\biggr]
\biggl[ 1 
\pm \biggl( \frac{1}{2 \cdot 3}\biggr)^{1 / 2}\mu  
\pm \biggl( \frac{5^2 \cdot 11^2}{2^7 \cdot 3^7} \biggr)^{1 / 2} \mu^3   
+ \mathcal{O}(\mu^5)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \biggl(\frac{2}{3^2}\biggr) \mu^2- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  \biggr]
~\pm~ \biggl( \frac{1}{2 \cdot 3}\biggr)^{1 / 2}\mu  
\biggl[1 - \biggl(\frac{2}{3^2}\biggr) \mu^2 \biggr]
~\pm~ \biggl( \frac{5^2 \cdot 11^2}{2^7 \cdot 3^7} \biggr)^{1 / 2} \mu^3   
+ \mathcal{O}(\mu^5)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 ~\pm~ \biggl( \frac{1}{2 \cdot 3}\biggr)^{1 / 2}\mu  
- \biggl(\frac{2}{3^2}\biggr) \mu^2
~\pm~ \biggl[ \biggl( \frac{5^2 \cdot 11^2}{2^7 \cdot 3^7} \biggr)^{1 / 2}   
- \biggl(\frac{2}{3^5}\biggr)^{1 / 2} \biggr]\mu^3 
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
+ \mathcal{O}(\mu^5)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 ~\pm~ \biggl( \frac{1}{2 \cdot 3}\biggr)^{1 / 2}\mu  
- \biggl(\frac{2}{3^2}\biggr) \mu^2
~\pm~ \biggl( \frac{7^2}{2^7 \cdot 3^7} \biggr)^{1 / 2}\mu^3 
- \biggl( \frac{2^2 \cdot 5}{3^5 } \biggr)\mu^4  
+ \mathcal{O}(\mu^5) \, .
</math>
  </td>
</tr>
</table>
</div>

An Excel-spreadsheet check appears to indicate that this expansion is correct, to the specified order.