Editing Apps/Wong1973Potential (section)

==Evaluate C<sub>n</sub> Coefficients==

From [[#D0andCn|above]], we have,
<div align="center">
<math>~D_0 \equiv \frac{2^{3/2} }{3\pi^2} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] \, ,</math>
</div>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_n(\cosh\eta_0)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(n+\tfrac{1}{2})Q_{n+\frac{1}{2}}(\cosh \eta_0) Q_{n - \frac{1}{2}}^2(\cosh \eta_0) 
- (n - \tfrac{3}{2}) Q_{n - \frac{1}{2}}(\cosh \eta_0)~Q^2_{n + \frac{1}{2}}(\cosh \eta_0) \, ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~\cosh\eta_0 \equiv \frac{1}{\epsilon} \, .</math>
</div>

Note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\sinh^3\eta_0}{\cosh\eta_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \, .</math>
  </td>
</tr>
</table>

===Case of n = 0===

====Phase 0A====

This specific case has already been examined, above.  But let's go through it again.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_0(\cosh\eta_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{1}{2}Q_{+\frac{1}{2}}(\cosh \eta_0) Q_{- \frac{1}{2}}^2(\cosh \eta_0) 
~+~\tfrac{3}{2} Q_{- \frac{1}{2}}(\cosh \eta_0)~Q^2_{+ \frac{1}{2}}(\cosh \eta_0) \, ,
</math>
  </td>
</tr>
</table>
where, from our [[Appendix/SpecialFunctions#Toroidal_Function_Evaluations|accompanying ''Equations'' appendix]] we have,
{{ Math/EQ_QminusHalf01 }}<br />
{{ Math/EQ_QplusHalf01 }}
and,
{{ Math/EQ_Q2minusHalf01 }}
and, finally,

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

<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}
\biggl\{ z k~K ( k )  ~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~2C_0(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
z k~K( k ) ~-~ [2(z+1)]^{1 / 2} E( k)
\biggr\} 
\biggl\{ 
\frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}} 
\biggr\}
</math>
  </td>
</tr>

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

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

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ 3z k^2~K \cdot K [(z-1) (z^2-1) ]^{1 / 2}  ~+~2^{1 / 2}\cdot 3 (z^2+3)  k K \cdot E
\biggr\}[(z-1) (z^2-1) ]^{- 1 / 2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~-~ \biggl[ 3z k^2~ [(z-1) (z^2-1) ]^{1 / 2}  
~+~ 2^{1 / 2} z(z-1) k~\biggr] K \cdot K
~-~ 8z(z+1)^{1 / 2} E\cdot E 
\biggr\}[(z-1)^2 (z+1) ]^{- 1 / 2} \, .
</math>
  </td>
</tr>
</table>


Notice that, in our case, <math>~z = \cosh\eta_0 = c/d</math>, while the argument of the elliptic integral functions is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~k_0^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2}{\cosh\eta_0 + 1}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2}{c/d + 1}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2d}{c + d}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{2d}{c}\biggr)\biggl[1 + \frac{d}{c}  \biggr]^{- 1 } \, .
</math>
  </td>
</tr>
</table>
This means, as well, that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~k_0^\prime</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{1 - k_0^2 \biggr\}^{1 / 2}
</math>
  </td>
</tr>

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

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

====Phase 0B====

Let's plow ahead, adopting the shorthand notation, <math>~\epsilon \equiv 1/z = d/c</math>.  The overall pre-factor (outside of the curly braces) is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{4(z-1) (z+1)^{1 / 2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4} \biggl[\frac{c}{d} - 1\biggr]^{-1 } \biggl[\frac{c}{d}+1\biggr]^{-1 / 2}</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-2} \epsilon^{3 / 2} (1 - \epsilon )^{-1 } (1 + \epsilon )^{-1 / 2} \ .</math> 
  </td>
</tr>
</table>

The coefficient of the function product, <math>~K\cdot E</math>, is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
2^{5 / 2} z^2 k~+~2 (z-1)(z+1)^{1 / 2} ~+~2^{1 / 2}\cdot 3 (z^2+3)  k 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^{1 / 2} \biggl(\frac{c}{d}\biggr)^2 k(4+3)
~+~2^{1 / 2}\cdot 3^2   k 
~+~2 \biggl(\frac{c}{d}-1 \biggr)\biggl(\frac{c}{d}+1 \biggr)^{1 / 2} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\epsilon^{-3/2} (1 + \epsilon)^{-1 / 2}\biggl[ 7  
~+~3^2  \epsilon^{2}  
~+~ (1 - \epsilon )( 1 + \epsilon ) \biggr]
</math>
  </td>
</tr>

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

The coefficient of the function product, <math>~E\cdot E</math>, is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>-~ 8z(z+1)^{1 / 2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~ \frac{8c}{d}\biggl(\frac{c}{d}+1 \biggr)^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~ 8 \biggl(\frac{c}{d}\biggr)^{3 / 2}\biggl(1 + \frac{d}{c}\biggr)^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~ 2^3 \epsilon^{-3 / 2} (1 + \epsilon )^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

The coefficient of the function product, <math>~K\cdot K</math>, is,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~6 (1 + \epsilon)^{- 1 } \epsilon^{-3 / 2}~ ( 1 - \epsilon ) ( 1 + \epsilon )^{1 / 2}  
~-~ 2 \epsilon^{-3 / 2} ( 1 - \epsilon ) (1 + \epsilon )^{- 1 /2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~2^3\epsilon^{-3 / 2} ( 1 - \epsilon ) ( 1 + \epsilon )^{- 1 / 2}  
</math>
  </td>
</tr>

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

Hence,

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

<tr>
  <td align="right">
<math>~2C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-2} \epsilon^{3 / 2} (1 - \epsilon )^{-1 } (1 + \epsilon )^{-1 / 2} \cdot 2^3 \epsilon^{-3 / 2} (1 + \epsilon )^{1 / 2} 
\biggl\{ 
\biggl[ \frac{2(1  ~+~\epsilon^{2} )}{(1 + \epsilon)} \biggr] K \cdot E
~-~ \biggl[\frac{ 1 - \epsilon}{1+\epsilon}\biggr] K \cdot K
~-~ E\cdot E 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~2 (1 - \epsilon )^{-1 } (1+\epsilon)^{-1} 
\biggl\{ (1 - \epsilon ) K \cdot K
~-~2(1  ~+~\epsilon^{2} ) K \cdot E
~+~ (1+\epsilon)E\cdot E 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1 - \epsilon^2 )^{-1 }  
\biggl\{(K - E)^2~+~ \epsilon (E\cdot E -K\cdot K)~-~2\epsilon^{2}  K \cdot E
\biggr\} \, .
</math>
  </td>
</tr>
</table>

====Phase 0C====

Referencing our [[2DStructure/ToroidalGreenFunction#Series_Expansions|separate listing of complete elliptic integral series expansions]] drawn from [http://www.mathtable.com/gr/ Gradshteyn &amp; Ryzhik (1965)], we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{2}{\pi} \biggl[K(k) - E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 + \biggl( \frac{1}{2} \biggr)^2k^2 
+ \biggl( \frac{1\cdot 3}{2\cdot 4}\biggr)^2 k^4 
+ \biggl( \frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2 k^6 
+ \cdots
+ \biggl[ \frac{(2n-1)!!}{2^n n!} \biggr]^2 k^{2n}
+ \cdots
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl( \frac{1}{2^2} \biggr)k^2 
+ \biggl( \frac{3^2}{2^6}\biggr)2 k^4 
+ \biggl( \frac{5^2}{2^8}\biggr) k^6 
\biggr\}
~+~
\biggl\{\frac{1}{2^2} ~k^2 
+ \frac{3}{2^6}~ k^4 
+ \biggl(\frac{5}{2^8}\biggr)~k^6
\biggr\} + \mathcal{O}(k^8)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl( \frac{1}{2^2} +\frac{1}{2^2} \biggr)k^2 
+ \biggl( \frac{3^2}{2^6} + \frac{3}{2^6}\biggr) k^4 
+ \biggl( \frac{5^2}{2^8} + \frac{5}{2^8} \biggr) k^6 
\biggr\} + \mathcal{O}(k^8)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl( \frac{1}{2} \biggr)k^2 
+ \biggl( \frac{3}{2^4} \biggr) k^4 
+ \biggl( \frac{3 \cdot 5}{2^7}  \biggr) k^6 
\biggr\} + \mathcal{O}(k^8)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{2^2}{\pi^2} \biggl[K(k) - E(k) \biggr]^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl( \frac{1}{2} \biggr)k^2 
+ \biggl( \frac{3}{2^4} \biggr) k^4 
+ \biggl( \frac{3 \cdot 5}{2^7}  \biggr) k^6 
\biggr\} 
\times
\biggl\{
\biggl( \frac{1}{2} \biggr)k^2 
+ \biggl( \frac{3}{2^4} \biggr) k^4 
+ \biggl( \frac{3 \cdot 5}{2^7}  \biggr) k^6 
\biggr\} 
+ \mathcal{O}(k^{10})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl( \frac{1}{2^2} \biggr)k^4 
+ \biggl( \frac{3}{2^5} \biggr) k^6 
+ \biggl( \frac{3 \cdot 5}{2^8}  \biggr) k^8 
\biggr\} 
~+~
\biggl\{
\biggl( \frac{3}{2^5} \biggr)k^6 
+ \biggl( \frac{3^2}{2^8} \biggr) k^8 
\biggr\} 
~+~\biggl\{
\biggl( \frac{3 \cdot 5}{2^8}  \biggr) k^8 
\biggr\} 
+ \mathcal{O}(k^{10})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{1}{2^2} \biggr)k^4 
+ \biggl( \frac{3}{2^4} \biggr) k^6 
+ \biggl( \frac{3 \cdot 13}{2^8}  \biggr) k^8 
+ \mathcal{O}(k^{10})
</math>
  </td>
</tr>
</table>


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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k) \cdot  E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 + \biggl( \frac{1}{2} \biggr)^2k^2 
+ \biggl( \frac{1\cdot 3}{2\cdot 4}\biggr)^2 k^4 
+ \biggl( \frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2 k^6 
+ \cdots
+ \biggl[ \frac{(2n-1)!!}{2^n n!} \biggr]^2 k^{2n}
+ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{1^2\cdot 3}{2^2\cdot 4^2}~ k^4 
- \biggl(\frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2~\frac{ k^6 }{5}
~-~ \cdots
 \biggl[ \frac{(2n-1)!!}{2^n n!} \biggr]^2 \frac{k^{2n}}{2n-1}
~-~ \cdots
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+~
\biggl\{
\biggl( \frac{1\cdot 3}{2\cdot 4}\biggr)^2 k^4 
\biggr\}
\times~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
\biggr\}
~+~
\biggl\{
\biggl( \frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2 k^6 
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{3}{2^6}~ k^4 
- \biggl(\frac{5}{2^8}\biggr)~k^6
\biggr\}
~+~
\biggl\{
\biggl( \frac{1}{2^2} \biggr)k^2 
- \frac{1}{2^4} ~k^4 
- \frac{3}{2^8}~ k^6 
\biggr\}
~+~
\biggl( \frac{3^2}{2^6}\biggr) k^4 
~-~
\biggl( \frac{3^2}{2^8}\biggr) k^6 
~+~
\biggl( \frac{5^2}{2^8}\biggr) k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 
~+~\biggl[ \frac{1}{2^2}  
- \frac{1}{2^2} \biggr] ~k^2 
~+~\biggl[ \frac{3^2}{2^6} 
- \frac{3}{2^6} 
- \frac{1}{2^4} \biggr]~k^4 
~+~\biggl[ \frac{5^2}{2^8}  
- \frac{5}{2^8}
- \frac{3}{2^8} 
~-~\frac{3^2}{2^8} \biggr]~ k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 ~+~\frac{1}{2^5} ~k^4 
~+~\frac{1}{2^5} ~ k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>
</table>


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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k) \cdot  K(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 + \biggl( \frac{1}{2} \biggr)^2k^2 
+ \biggl( \frac{1\cdot 3}{2\cdot 4}\biggr)^2 k^4 
+ \biggl( \frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2 k^6 
+ \cdots
+ \biggl[ \frac{(2n-1)!!}{2^n n!} \biggr]^2 k^{2n}
+ \cdots
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 
+ \frac{1}{2^2} k^2 
+ \frac{3^2}{2^6} k^4 
+ \frac{5^2}{2^8} k^6 
\biggr\}
\times~
\biggl\{
1 
+ \frac{1}{2^2} k^2 
+ \frac{3^2}{2^6} k^4 
+ \frac{5^2}{2^8} k^6 
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 
+ \frac{1}{2^2} k^2 
+ \frac{3^2}{2^6} k^4 
+ \frac{5^2}{2^8} k^6 
\biggr\}
~+~\frac{1}{2^2} k^2 
\biggl\{
1 
+ \frac{1}{2^2} k^2 
+ \frac{3^2}{2^6} k^4 
\biggr\}
~+~\frac{3^2}{2^6} k^4 
\biggl\{
1 
+ \frac{1}{2^2} k^2 
\biggr\}
~+~
\biggl\{
\frac{5^2}{2^8} k^6 
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \frac{1}{2^2} k^2 
+ \frac{3^2}{2^6} k^4 
+ \frac{5^2}{2^8} k^6 
~+~ 
\frac{1}{2^2} k^2
+ \frac{1}{2^4} k^4 
+ \frac{3^2}{2^8} k^6 
~+~\frac{3^2}{2^6} k^4 
~+~\frac{3^2}{2^8} k^6 
~+~
\frac{5^2}{2^8} k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \biggl[ \frac{1}{2^2} 
~+~ \frac{1}{2^2} \biggr] ~k^2
+ \biggl[ \frac{3^2}{2^6} 
+ \frac{1}{2^4} 
~+~\frac{3^2}{2^6} \biggr]~k^4 
+ \biggl[ \frac{5^2}{2^8} 
+ \frac{3^2}{2^8} 
~+~\frac{3^2}{2^8}  
~+~\frac{5^2}{2^8} \biggr]~ k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \frac{1}{2} k^2
+ \frac{11}{2^5} ~k^4 
+ \frac{17}{2^6} ~ k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>
</table>


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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[E(k) \cdot  E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{1^2\cdot 3}{2^2\cdot 4^2}~ k^4 
- \biggl(\frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2~\frac{ k^6 }{5}
~-~ \cdots
 \biggl[ \frac{(2n-1)!!}{2^n n!} \biggr]^2 \frac{k^{2n}}{2n-1}
~-~ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{1^2\cdot 3}{2^2\cdot 4^2}~ k^4 
- \biggl(\frac{1\cdot 3\cdot 5}{2^4\cdot 3}\biggr)^2~\frac{ k^6 }{5}
~-~ \cdots
 \biggl[ \frac{(2n-1)!!}{2^n n!} \biggr]^2 \frac{k^{2n}}{2n-1}
~-~ \cdots
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{3}{2^6}~ k^4 
- \frac{5}{2^8}~ k^6
\biggr\}
\times
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{3}{2^6}~ k^4 
- \frac{5}{2^8}~ k^6
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{3}{2^6}~ k^4 
- \frac{5}{2^8}~ k^6
\biggr\}
~+~
\biggl\{- \frac{1}{2^2} ~k^2 
\biggr\}
\times
\biggl\{
1 - \frac{1}{2^2} ~k^2 
- \frac{3}{2^6}~ k^4 
\biggr\}
~+~
\biggl\{
- \frac{3}{2^6}~ k^4 
\biggr\}
\times
\biggl\{
1 - \frac{1}{2^2} ~k^2 
\biggr\}
~+~
\biggl\{
- \frac{5}{2^8}~ k^6
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{1}{2^2} ~k^2 
- \frac{3}{2^6}~ k^4 
- \frac{5}{2^8}~ k^6
- \frac{1}{2^2} ~k^2  + \frac{1}{2^4} ~k^4 
+ \frac{3}{2^8}~ k^6 
~-~ \frac{3}{2^6}~ k^4 
~+~ \frac{3}{2^8}~ k^6 
~-~\frac{5}{2^8}~ k^6
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{1}{2} ~k^2 
+ \biggl[ \frac{1}{2^4}  - \frac{3}{2^6} ~-~ \frac{3}{2^6} \biggr]~ k^4 
+ \biggl[ \frac{3}{2^8} ~+~ \frac{3}{2^8} - \frac{5}{2^8}~-~\frac{5}{2^8} \biggr] ~ k^6
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{1}{2} ~k^2 ~-~ \frac{1}{2^5} ~ k^4 ~-~\frac{1}{2^6} ~ k^6
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{2^2}{\pi^2} \biggl[E(k) \cdot  E(k) - K(k) \cdot  K(k)\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 - \frac{1}{2} ~k^2 ~-~ \frac{1}{2^5} ~ k^4 ~-~\frac{1}{2^6} ~ k^6
\biggr\}
~-~
\biggl\{
1 + \frac{1}{2} k^2+ \frac{11}{2^5} ~k^4 + \frac{17}{2^6} ~ k^6 
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~\biggl\{
k^2 ~+~ \frac{3}{2^3} ~ k^4 ~+~\frac{3^2}{2^5} ~ k^6
\biggr\}
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>
</table>


<table border="1" align="center" cellpadding="8">
<tr>
  <th align="center">Summary</th>
</tr>
<tr>
  <td align="left">

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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k) \cdot  E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 ~+~\frac{1}{2^5} ~k^4 
~+~\frac{1}{2^5} ~ k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k) \cdot  K(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \frac{1}{2} k^2
+ \frac{11}{2^5} ~k^4 
+ \frac{17}{2^6} ~ k^6 
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[E(k) \cdot  E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{1}{2} ~k^2 ~-~ \frac{1}{2^5} ~ k^4 ~-~\frac{1}{2^6} ~ k^6
+ \mathcal{O}(k^{8})
</math>
  </td>
</tr>
</table>

  </td>
</tr>
</table>

====Phase 0D====

We therefore have,

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

<tr>
  <td align="right">
<math>~(1 - \epsilon^2) \biggl[ \frac{2^2}{\pi^2} \biggr] C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  
\biggl[ \frac{2^2}{\pi^2} \biggr]\biggl\{ (K - E)^2 \biggr\}
~-~ \epsilon \biggl[ \frac{2^2}{\pi^2} \biggr]\biggl\{ E\cdot E -K\cdot K\biggr\}
~+~2\epsilon^{2}  \biggl[ \frac{2^2}{\pi^2} \biggr]\biggl\{ K \cdot E \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  
\biggl\{ \biggl( \frac{1}{2^2} \biggr)k^4 
+ \biggl( \frac{3}{2^4} \biggr) k^6 
+ \biggl( \frac{3 \cdot 13}{2^8}  \biggr) k^8 + \mathcal{O}(k^{10})
\biggr\}
~+~ \epsilon \biggl\{ k^2 ~+~ \frac{3}{2^3} ~ k^4 ~+~\frac{3^2}{2^5} ~ k^6 + \mathcal{O}(k^{8})\biggr\}
~+~2\epsilon^{2} \biggl\{ 1 ~+~\frac{1}{2^5} ~k^4 
~+~\frac{1}{2^5} ~ k^6 + \mathcal{O}(k^{8}) \biggr\} \, .
</math>
  </td>
</tr>
</table>

Now, given that,
<div align="center">
<math>~k_0^2 = 2\epsilon (1+\epsilon)^{-1} \, ,</math>
</div>
we furthermore have,

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

<tr>
  <td align="right">
<math>~(1 - \epsilon^2) \biggl[ \frac{2^2}{\pi^2} \biggr] C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  
\biggl\{ \biggl( \frac{1}{2^2} \biggr)[2\epsilon (1+\epsilon)^{-1}]^2 
+ \biggl( \frac{3}{2^4} \biggr) [2\epsilon (1+\epsilon)^{-1}]^3 
+ \biggl( \frac{3 \cdot 13}{2^8}  \biggr) [2\epsilon (1+\epsilon)^{-1}]^4 + \mathcal{O}(\epsilon^{5})
\biggr\}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  \biggl\{
\epsilon^2 (1+\epsilon)^{-2} 
+ \biggl( \frac{3}{2} \biggr) \epsilon^3 (1+\epsilon)^{-3} 
+ \biggl( \frac{3 \cdot 13}{2^4}  \biggr) \epsilon^4 (1+\epsilon)^{-4} 
\biggr\}
~+~ \biggl\{ 2\epsilon^2 (1+\epsilon)^{-1} ~+~ \frac{3}{2} ~ \epsilon^3 (1+\epsilon)^{-2} ~+~\frac{3^2}{2^2} ~ \epsilon^4 (1+\epsilon)^{-3} \biggr\}
~+~ \biggl\{ 2\epsilon^{2} ~+~\frac{1}{2^2} ~\epsilon^4 (1+\epsilon)^{-2}  \biggr\} 
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 2 + 2(1+\epsilon)^{-1} - (1+\epsilon)^{-2} \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (1+\epsilon)^{-2} - \biggl( \frac{3}{2} \biggr) (1+\epsilon)^{-3}\biggr\}
~+~ \epsilon^4 \biggl\{ \frac{1}{2^2} (1+\epsilon)^{-2} + \frac{3^2}{2^2} (1+\epsilon)^{-3} - \biggl( \frac{3 \cdot 13}{2^4}  \biggr) (1+\epsilon)^{-4}\biggr\}
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>
</table>

Recalling that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~D_0 \equiv \frac{2^{3 / 2}}{3\pi^2} \biggl[\frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^{3 / 2}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr] \, ,</math>
  </td>
</tr>
</table>
to lowest order we see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2^{3 / 2} D_0 \cdot C_0</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{2^{3}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr] \epsilon^2 \biggl\{ 2 + 2(1+\epsilon)^{-1} - (1+\epsilon)^{-2} \biggr\} (1-\epsilon^2)^{-1} \biggl[ \frac{\pi^2}{2^2}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{2}{3} \biggl\{ 3 \biggr\} = 2 \, ,</math>
  </td>
</tr>
</table>
as [[#Speculation|speculated above]].

What about the next order?  Using the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial expansion]] we have,

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

<tr>
  <td align="right">
<math>~(1 - \epsilon^2) \biggl[ \frac{2^2}{\pi^2} \biggr] C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 2 + 2(1-\epsilon + \epsilon^2) - (1-2\epsilon +3\epsilon^2) \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (1-2\epsilon) - \biggl( \frac{3}{2} \biggr) (1-3\epsilon)\biggr\}
~+~ \epsilon^4 \biggl\{ \frac{1}{2^2}  + \frac{3^2}{2^2}  - \biggl( \frac{3 \cdot 13}{2^4}  \biggr) \biggr\}
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 3 + 2(-\epsilon + \epsilon^2) - (-2\epsilon +3\epsilon^2) \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (-2\epsilon) - \biggl( \frac{3}{2} \biggr) (-3\epsilon)\biggr\}
~+~ \frac{1}{2^4}\epsilon^4 \biggl\{ 40  - 3 \cdot 13  \biggr\}
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
<math>\Rightarrow~~~2^{3 / 2} D_0 \cdot C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~2^{3 / 2}D_0
\cdot (3\epsilon^2 )  \biggl[ 1  
~+~ \frac{3}{2^4}  \epsilon^2\biggr](1 - \epsilon^2)^{-1} \biggl[ \frac{\pi^2}{2^2} \biggr] 
</math>
  </td>
</tr>

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2  ~+~ \frac{3 + 2^3\cdot 5}{2^3}  \epsilon^2 \, .
</math>
  </td>
</tr>
</table>

This does ''not'' match the <math>~\epsilon^2</math> term, as [[#Speculation|speculated above]].  But keep in mind that Wong's coordinate system is slightly shifted from the one used by Dyson, so perhaps this difference can be entirely reconciled via the proper coordinate-system mapping.

===Case of n = 1===

====Phase 1A====

In this case we want to rewrite in terms of complete elliptic integrals the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2C_1(\cosh\eta_0)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~3 Q_{+\frac{3}{2}}(\cosh \eta_0) Q_{+ \frac{1}{2}}^2(\cosh \eta_0) 
+ Q_{+ \frac{1}{2}}(\cosh \eta_0)~Q^2_{+ \frac{3}{2}}(\cosh \eta_0) \, ,
</math>
  </td>
</tr>
</table>
where, to start with, [[#Phase_0A|from above]] we have,
{{ Math/EQ_QplusHalf01 }}
and, 

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

<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}
\biggl\{ z k~K ( k )  ~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} \, ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~k \equiv \sqrt{\frac{2}{z+1}} \, .</math>
</div>

An expression for <math>~Q_{+\frac{3}{2}}(z)</math> can be obtained by setting <math>~m=2</math> in the recurrence relation (see [[Appendix/SpecialFunctions#Example_Recurrence_Relations|accompanying appendix]]),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^0_{m-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4 \biggl[ \frac{m-1}{2m-1} \biggr] z Q^0_{m-\frac{3}{2}}(z) - \biggl[ \frac{2m-3}{2m-1}\biggr]Q^0_{m-\frac{5}{2}}(z) \, ,</math>
  </td>
</tr>
</table>

and inserting the expressions for <math>~Q_{+\frac{1}{2}}(z)</math> and <math>~Q_{-\frac{1}{2}}(z)</math> (again, [[#Phase_0A|see above]]).  We obtain,

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

<tr>
  <td align="right">
<math>~Q_{+\frac{3}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3} \biggl[ 4z Q_{+\frac{1}{2}}(z) - Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
</tr>

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

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

An expression for <math>~Q^2_{+\frac{3}{2}}(z)</math> can be obtained by setting <math>~\mu = 2</math>, <math>~\nu = +\tfrac{1}{2}</math>, and by inserting the expressions for <math>~Q^2_{+\frac{1}{2}}(z)</math> and <math>~Q^2_{-\frac{1}{2}}(z)</math> ([[#Phase_0A|see above]]) into the Key recurrence relation,

{{ Math/EQ_Toroidal04 }}

Specifically, we obtain,

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

<tr>
  <td align="right">
<math>~(\nu - \mu + 1)Q^2_{+\frac{3}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>(2\nu + 1)z Q_{+\frac{1}{2}}^2(z) - (\nu + \mu)Q^2_{-\frac{1}{2}}(z)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ -\biggl(\frac{1}{2}\biggr)Q^2_{+\frac{3}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>2z Q_{+\frac{1}{2}}^2(z) - \biggl( \frac{5}{2} \biggr)Q^2_{-\frac{1}{2}}(z)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ Q^2_{+\frac{3}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>5 Q^2_{-\frac{1}{2}}(z) -4z Q_{+\frac{1}{2}}^2(z) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 20 z E(k) - 5(z-1) K(k) }{ [2^{3} (z-1) (z^2-1) ]^{1 / 2}} 
~+~z 
\biggl\{ z k~K ( k )  ~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ 2^3(z-1) (z^2-1) ]^{- 1 / 2} \biggl\{
\biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] K(k)  
~+~ \biggl[ 20 z  
~-~4z(z^2+3) \biggr] E(k) \biggr\} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~2^3 C_1(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl[ (4z^2 - 1) k ~K( k  ) ~-~ 4z[2(z+1)]^{1 / 2} E( k ) \biggr] 
\times 
\biggl( -~\frac{1}{2^2} \biggr)\biggl\{ z k~K ( k )  ~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~4\biggl[z k~K( k ) ~-~ [2(z+1)]^{1 / 2} E(k )\biggr]
\times ~
[ 2^3(z-1) (z^2-1) ]^{- 1 / 2} \biggl\{
\biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] K(k)  
~+~ \biggl[ 20 z  
~-~4z(z^2+3) \biggr] E(k) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ (z-1) (z^2-1) ]^{- 1 / 2} 
\biggl[ (4z^2 - 1) k ~K( k  ) ~-~ 4z[2(z+1)]^{1 / 2} E( k ) \biggr] 
\times 
\biggl\{  2^{1 / 2}(z^2+3) E(k) ~-~z k~K ( k )[ (z-1) (z^2-1) ]^{ 1 / 2}  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~2^{1 / 2}[ (z-1) (z^2-1) ]^{- 1 / 2} \biggl[z k~K( k ) ~-~ [2(z+1)]^{1 / 2} E(k )\biggr]
\times ~
\biggl\{
\biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] K(k)  
~+~ \biggl[ 20 z  
~-~4z(z^2+3) \biggr] E(k) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2^3 [ (z-1) (z^2-1) ]^{1 / 2}C_1(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ (4z^2 - 1) k ~K( k  ) ~-~ 4z[2(z+1)]^{1 / 2} E( k ) \biggr] 
\times 
\biggl\{  2^{1 / 2}(z^2+3) E(k) ~-~z k~K ( k )[ (z-1) (z^2-1) ]^{ 1 / 2}  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl[2^{1 / 2} z k~K( k ) ~-~ 2[(z+1)]^{1 / 2} E(k )\biggr]
\times ~
\biggl\{
\biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] K(k)  
~+~ \biggl[ 20 z  
~-~4z(z^2+3) \biggr] E(k) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ (4z^2 - 1) k ~K( k  )\biggr] 
\times 
\biggl\{  2^{1 / 2}(z^2+3) E(k) \biggr\} 
~-~\biggl[ 4z[2(z+1)]^{1 / 2} E( k ) \biggr] 
\times 
\biggl\{  2^{1 / 2}(z^2+3) E(k) \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\biggl[ (4z^2 - 1) k ~K( k  ) \biggr] 
\times 
\biggl\{  z k~K ( k )[ (z-1) (z^2-1) ]^{ 1 / 2}  \biggr\} 
~+~\biggl[ 4z[2(z+1)]^{1 / 2} E( k ) \biggr] 
\times 
\biggl\{  z k~K ( k )[ (z-1) (z^2-1) ]^{ 1 / 2}  \biggr\} 
</math>
  </td>
</tr>

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

~-~\biggl[2[(z+1)]^{1 / 2} E(k )\biggr]
\times 
\biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] K(k)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl[2^{1 / 2} z k~K( k ) \biggr]
\times 
\biggl[ 20 z  - 4z(z^2+3) \biggr] E(k) 
~-~\biggl[2[(z+1)]^{1 / 2} E(k )\biggr]
\times 
\biggl[ 20 z  - 4z(z^2+3) \biggr] E(k) 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~(4z^2 - 1) k \biggl[  z k~[ (z-1) (z^2-1) ]^{ 1 / 2}  \biggr] K\cdot K
~+~ 4z[2(z+1)]^{1 / 2} \biggl[  z k~[ (z-1) (z^2-1) ]^{ 1 / 2}  \biggr] K\cdot E
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~2^{1 / 2} z k~\biggl[ 20 z  - 4z(z^2+3) \biggr] K\cdot E 
~-~2[(z+1)]^{1 / 2} \biggl[ 20 z  - 4z(z^2+3) \biggr] E\cdot E 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~2^{1 / 2} z k~\biggl[ 20 z  - 4z(z^2+3) \biggr] 
~-~2[(z+1)]^{1 / 2} \biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] 
\biggr\} K \cdot E  
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ \biggl\{ 2[(z+1)]^{1 / 2} \biggl[ 20 z  - 4z(z^2+3) \biggr] 
~+~4z[2(z+1)]^{1 / 2}  \biggl[  2^{1 / 2}(z^2+3) \biggr] 
\biggr\} E\cdot E
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ 2^{5 / 2}z^2[ (z+1)(z-1) (z^2-1) ]^{1 / 2}  k  
~+~2 \cdot 5[(z^2-1)(z-1)]^{1 / 2}   
\biggr\} K \cdot E  
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl\{ 
zk^2[ (z-1) (z^2-1) ]^{ 1 / 2}  
~-~ 2^{1 / 2} \cdot 5 z k~(z-1)   
\biggr\} K\cdot K
~-~ \biggl\{ 2^3 \cdot 5 z[(z+1)]^{1 / 2} 
\biggr\} E\cdot E
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl\{ 
zk (z-1) \bigg[k(z+1)^{ 1 / 2}  
~-~ 2^{1 / 2}\cdot 5 \biggr]   
\biggr\} K\cdot K
~-~ \biggl\{ 2^3 \cdot 5 z(z+1)^{1 / 2} 
\biggr\} E\cdot E
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl\{ 2^{1 / 2} k (19z^2 -3 )
~+~2 \cdot 5(z-1)(z+1)^{1 / 2}   
\biggr\} K \cdot E  
~+~\biggl\{ 
zk (z-1) \bigg[k(z+1)^{ 1 / 2}  
~-~ 2^{1 / 2}\cdot 5 \biggr]   
\biggr\} K\cdot K
~-~ \biggl\{ 2^3 \cdot 5 z(z+1)^{1 / 2} 
\biggr\} E\cdot E
</math>
  </td>
</tr>
</table>

====Phase 1B====

Drawing from our [[#Phase_0B|above, "Phase 0B"]] derivation, we recall that, <math>~z = 1/\epsilon</math>,
<div align="center">
<math>~k = 2^{1 / 2} \epsilon^{1 / 2} (1+\epsilon)^{-1 / 2} \, ,</math>
</div>

and, in this case the overall pre-factor coming from the LHS is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{2^3(z-1) (z+1)^{1 / 2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-3} \epsilon^{3 / 2} (1 - \epsilon )^{-1 } (1 + \epsilon )^{-1 / 2} \ .</math> 
  </td>
</tr>
</table>

The coefficient of the function product, <math>~K\cdot E</math>, is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
2^{1 / 2} k (19z^2 -3 )
~+~2 \cdot 5(z-1)(z+1)^{1 / 2}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\epsilon^{1 / 2}(1+\epsilon)^{-1 / 2} \epsilon^{-2}(19 -3\epsilon^2 )
~+~2 \cdot 5 \epsilon^{-3 / 2} (1 - \epsilon)(1 + \epsilon)^{1 / 2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\epsilon^{-3 / 2}(1+\epsilon)^{-1 / 2} [ 
(19 -3\epsilon^2 )
~+~5  (1 - \epsilon^2) ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^4\epsilon^{-3 / 2}(1+\epsilon)^{-1 / 2} ( 3 -\epsilon^2 ) \, .
</math>
  </td>
</tr>
</table>

The coefficient of the function product, <math>~K\cdot K</math>, is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
zk (z-1) \bigg[k(z+1)^{ 1 / 2}  
~-~ 2^{1 / 2}\cdot 5 \biggr]   
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{1 / 2} \epsilon^{1 / 2} (1+\epsilon)^{-1 / 2} 
\epsilon^{-2} (1 - \epsilon) \bigg[2^{1 / 2} \epsilon^{1 / 2} (1+\epsilon)^{-1 / 2} \epsilon^{-1 / 2}(1 + \epsilon)^{ 1 / 2}  
~-~ 2^{1 / 2}\cdot 5 \biggr]   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~2^3 \epsilon^{- 3 / 2} (1+\epsilon)^{-1 / 2} (1 - \epsilon)   
\, .
</math>
  </td>
</tr>
</table>

The coefficient of the function product, <math>~E\cdot E</math>, is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
-~2^3 \cdot 5 z(z+1)^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~2^3 \cdot 5 \epsilon^{-3 / 2} (1+\epsilon)^{- 1 / 2} (1+\epsilon) \, .
</math>
  </td>
</tr>
</table>

Hence,

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

<tr>
  <td align="right">
<math>~C_1(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-3} \epsilon^{3 / 2} (1 - \epsilon )^{-1 } (1 + \epsilon )^{-1 / 2} \biggl\{
\biggl[ 2^4\epsilon^{-3 / 2}(1+\epsilon)^{-1 / 2} ( 3 -\epsilon^2 )   
\biggr] K \cdot E  
~-~\biggl[ 
2^3 \epsilon^{- 3 / 2} (1+\epsilon)^{-1 / 2} (1 - \epsilon)\biggr] K\cdot K
~-~ \biggl[ 2^3 \cdot 5 \epsilon^{-3 / 2} (1+\epsilon)^{- 1 / 2} (1+\epsilon) 
\biggr] E\cdot E
\biggr\}
</math>
  </td>
</tr>

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

====Phase 1C====
From [[#Phase_0C|Phase 0C, above]], we have,

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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k) \cdot  E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 ~+~\frac{1}{2^5} ~k^4 ~+~\frac{1}{2^5} ~ k^6 + \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k) \cdot  K(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \frac{1}{2} k^2 + \frac{11}{2^5} ~k^4 + \frac{17}{2^6} ~ k^6 + \mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[E(k) \cdot  E(k) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{1}{2} ~k^2 ~-~ \frac{1}{2^5} ~ k^4 ~-~\frac{1}{2^6} ~ k^6 + \mathcal{O}(k^{8})
</math>
  </td>
</tr>
</table>

Hence,

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

<tr>
  <td align="right">
<math>~\biggl[ \frac{2^2(1 - \epsilon^2 )}{\pi^2} \biggr] C_1(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
2 ( 3 -\epsilon^2 )  \biggl[ 1 ~+~\frac{1}{2^5} ~k^4 ~+~\frac{1}{2^5} ~ k^6 + \mathcal{O}(k^{8})  \biggr]
~-~ (1 - \epsilon) \biggl[ 1 + \frac{1}{2} k^2 + \frac{11}{2^5} ~k^4 + \frac{17}{2^6} ~ k^6 + \mathcal{O}(k^{8}) \biggr]
~-~ 5  (1+\epsilon) \biggl[  1 - \frac{1}{2} ~k^2 ~-~ \frac{1}{2^5} ~ k^4 ~-~\frac{1}{2^6} ~ k^6 + \mathcal{O}(k^{8})\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2 ( 3 -\epsilon^2 ) -(1-\epsilon) - 5(1+\epsilon) \biggr]
~+~ 
\biggl[ ~-~ \frac{1}{2}(1 - \epsilon)  + \frac{5}{2}(1+\epsilon)\biggr]k^2
~+~ 
\biggl[ 2^{-4} ( 3 -\epsilon^2 ) - \frac{11}{2^5}(1-\epsilon) + \frac{5}{2^5}(1+\epsilon) \biggr]k^4
~+~ 
\biggl[ 2^{-4} ( 3 -\epsilon^2 ) - \frac{17}{2^6}(1-\epsilon) + \frac{5}{2^6}(1+\epsilon) \biggr]k^6
~+~ 
\mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 6 -2\epsilon^2  -1+\epsilon - 5 - 5\epsilon \biggr]
~+~ 
\frac{1}{2}\biggl[(\epsilon-1)  + 5(1+\epsilon)\biggr]k^2
~+~ 
\frac{1}{2^5}\biggl[ 2( 3 -\epsilon^2 ) - 11(1-\epsilon) + 5(1+\epsilon) \biggr]k^4
~+~ 
\frac{1}{2^6}\biggl[ 4 ( 3 -\epsilon^2 ) -17(1-\epsilon) + 5(1+\epsilon) \biggr]k^6
~+~ 
\mathcal{O}(k^{8})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~\epsilon \biggl[ 4  + 2\epsilon \biggr]
~+~ 
\biggl[2+3\epsilon \biggr]k^2
~+~ 
\frac{\epsilon}{2^4}\biggl[8  -\epsilon  \biggr]k^4
~+~ 
\frac{\epsilon}{2^5}\biggl[11  - 2\epsilon  \biggr]k^6
~+~ 
\mathcal{O}(k^{8})
</math>
  </td>
</tr>
</table>

Again, given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~k^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\epsilon (1+\epsilon)^{-1} \, ,</math>
  </td>
</tr>
</table>
and, drawing from the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial series expansion]], we can write,

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

<tr>
  <td align="right">
<math>~\biggl[ \frac{2^2(1 - \epsilon^2 )}{\pi^2} \biggr] C_1(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~\epsilon \biggl[ 4  + 2\epsilon \biggr]
~+~ \biggl[2+3\epsilon \biggr]2\epsilon (1+\epsilon)^{-1}
~+~ 
\frac{\epsilon}{2^4}\biggl[8  -\epsilon  \biggr]2^2 \epsilon^2 (1+\epsilon)^{-2}
~+~ 
\frac{\epsilon}{2^5}\biggl[11  - 2\epsilon  \biggr]2^3\epsilon^3 (1+\epsilon)^{-3}
~+~ 
\mathcal{O}(\epsilon^{5})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~4\epsilon  - 2\epsilon^2 
~+~ (4\epsilon + 6\epsilon^2 ) (1 - \epsilon +\epsilon^2 - \epsilon^3 + \epsilon^4 - \epsilon^5 + \cdots)
~+~ \frac{\epsilon^3}{2^2} (8  -\epsilon ) (1 - 2\epsilon + 3\epsilon^2 - 4\epsilon^3 + 5\epsilon^4 - 6\epsilon^5 + \cdots)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~+~ \frac{\epsilon^4}{2^2} ( 11  - 2\epsilon ) (1 - 3\epsilon + 6\epsilon^2 - 10\epsilon^3 + 15\epsilon^4 - 21\epsilon^5 + \cdots)
~+~ 
\mathcal{O}(\epsilon^{5})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~4\epsilon  - 2\epsilon^2 
~+~ (4\epsilon + 6\epsilon^2 ) (1 - \epsilon +\epsilon^2 - \epsilon^3 + \epsilon^4 )
~+~ \frac{\epsilon^3}{2^2} (8  -\epsilon ) (1 - 2\epsilon +3\epsilon^2)
~+~ \frac{\epsilon^4}{2^2} ( 11  - 2\epsilon ) (1 - 3\epsilon )
~+~ 
\mathcal{O}(\epsilon^{5})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~4\epsilon  - 2\epsilon^2 
~+~ (4\epsilon - 4\epsilon^2 +4\epsilon^3 - 4\epsilon^4 )
~+~ (6\epsilon^2 - 6\epsilon^3 +6\epsilon^4 )
~+~ \epsilon^3 (2 - 4\epsilon +6\epsilon^2)
~-~ \frac{\epsilon^4}{2^2} (1 - 2\epsilon +3\epsilon^2)
~+~ 11~\cdot \frac{\epsilon^4}{2^2} 
~+~ \mathcal{O}(\epsilon^{5})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(- 4\epsilon^4 )
~+~ (6\epsilon^4 )
~-~ 4\epsilon^4 
~-~ \frac{\epsilon^4}{2^2} 
~+~ 11~\cdot \frac{\epsilon^4}{2^2} 
~+~ \mathcal{O}(\epsilon^{5})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{\epsilon^4}{2}
~+~ \mathcal{O}(\epsilon^{5}) \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~2^{3/2}D_0 \cdot C_1</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3\pi^2} \biggl[\epsilon^{-2}(1+\epsilon^2)^{3 / 2}\biggr] \biggl[ \frac{\pi^2}{2^2(1 - \epsilon^2 )} \biggr] \frac{\epsilon^4}{2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\epsilon^2}{3}  ~+~ \mathcal{O}(\epsilon^{3})\, .
</math>
  </td>
</tr>
</table>

===Summary===

<table border="1" align="center" width="85%" cellpadding="8">
<tr>
  <th align="center">
Summary of Relevant Toroidal Function Expressions<br />
(see also an [[Appendix/SpecialFunctions#Toroidal_Function_Evaluations|associated ''Special Functions'' appendix]])
  </th>
</tr>
<tr>
  <td align="left">
<table border="0" cellpadding="5" align="center">
<tr>
  <th align="left" colspan="3">Function Expression</th>
  <th align="center" colspan="1"><sup>&dagger;</sup>Case A</th>
  <th align="center" colspan="1">&nbsp; &nbsp; &nbsp;</th>
  <th align="center" colspan="1"><sup>&dagger;</sup>Case B</th>
</tr>

<tr>
  <td align="right">
<math>~Q^0_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
k~K( k )  \, ;
</math>
  </td>
  <td align="right"><math>~1.419337751 \times 10^0</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~5.54081487 \times 10^{-1}</math></td>
</tr>

<tr>
  <td align="right">
<math>~Q^0_{+\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
z k~K( k ) ~-~ [2(z+1)]^{1 / 2} E(k ) \, ;
</math>
  </td>
  <td align="right"><math>~1.426580119 \times 10^{-1}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~8.614495 \times 10^{-3}</math></td>
</tr>

<tr>
  <td align="right">
<math>~Q^0_{+\frac{3}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{3} \biggl[ 4zQ^0_{+\frac{1}{2}}(z) - Q^0_{-\frac{1}{2}}(z) \biggr] 
</math>
  </td>
  <td align="right"><math>~2.143519083 \times 10^{-2}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~1.99242 \times 10^{-4}</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{3} \biggl[ (4z^2 - 1) k ~K( k  ) ~-~ 4z[2(z+1)]^{1 / 2} E( k ) \biggr] \, ;
</math>
  </td>
  <td align="center" colspan="3">&nbsp; &nbsp; &nbsp;</td>
</tr>

<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z-1) (z^2-1) ]^{1 / 2}} \, ;
</math>
  </td>
  <td align="right"><math>~1.246521876 \times 10^{0}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~4.171704 \times 10^{-1}</math></td>
</tr>

<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}
\biggl\{ z k~K ( k )  ~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} \, ;
</math>
  </td>
  <td align="right"><math>~5.80241772 \times 10^{-1}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~3.236720 \times 10^{-2}</math></td>
</tr>

<tr>
  <td align="right">
<math>~ Q^2_{+\frac{3}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5Q^2_{-\frac{1}{2}}(z) - 4zQ^2_{+\frac{1}{2}}(z)
</math>
  </td>
  <td align="right"><math>~1.98094951 \times 10^{-1}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~1.758501 \times 10^{-3}</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ 2^3(z-1) (z^2-1) ]^{- 1 / 2} \biggl\{
\biggl[ 2^{3 / 2}z^2[ (z-1) (z^2-1) ]^{1 / 2}  k 
~-~ 5(z-1) \biggr] K(k)  
</math>
  </td>
  <td align="right">&nbsp;</td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~+~ \biggl[ 20 z  
~-~4z(z^2+3) \biggr] E(k) \biggr\} \, ;
</math>
  </td>
  <td align="right">&nbsp;</td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">&nbsp;</td>
</tr>

<tr>
  <td align="right">
<math>~C_0(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{1}{2}Q^0_{+\frac{1}{2}}(z) Q_{- \frac{1}{2}}^2(z) 
~+~\tfrac{3}{2} Q^0_{- \frac{1}{2}}(z)~Q^2_{+ \frac{1}{2}}(z) \, ,
</math>
  </td>
  <td align="right"><math>~1.324251744 \times 10^{0}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~2.869795 \times 10^{-2}</math></td>
</tr>

<tr>
  <td align="right">
<math>~C_1(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{3}{2} Q^0_{+\frac{3}{2}}(z) Q_{+ \frac{1}{2}}^2(z) 
+ \tfrac{1}{2} Q^0_{+ \frac{1}{2}}(z)~Q^2_{+ \frac{3}{2}}(z) \, ;
</math>
  </td>
  <td align="right"><math>~3.278631 \times 10^{-2}</math></td>
  <td align="center" colspan="1">&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~1.724765 \times 10^{-5}</math></td>
</tr>
</table>

  </td>
</tr>
<tr>
  <td align="left">
<sup>&dagger;</sup>''Example'' values are given here in an effort to illustrate agreement with &#8212; and partial extension of &#8212; toroidal function evaluations that we have [[Appendix/SpecialFunctions#Comparison_with_Table_IX_from_MF53|tabulated elsewhere for comparison with MF53]].  Tabulated function values are for the argument:
* '''Case A:''' &nbsp; <math>~z = 13/5 = 2.6</math>, in which case, <math>~k = \sqrt{2/(z+1)} = \sqrt{5}/3</math>, <math>~K(k) = 1.904241417</math>, and <math>~E(k) = 1.322119966</math>. 
* '''Case B:''' &nbsp; <math>~k = \sin(\pi/9)</math>, in which case, <math>~z = (2/k^2 - 1) = 16.09726435</math>, <math>~K(k) = 1.62002589</math>, and <math>~E(k) = 1.52379921</math>. 

See, also, a separate [[2DStructure/ToroidalCoordinateIntegrationLimits#Evaluation_of_Elliptic_Integrals|table giving example evaluations of elliptic integrals]], and other [[2DStructure/ToroidalGreenFunction#Appendix_B:_Elliptic_Integrals|useful elliptic integral expressions/relations]].
  </td>
</tr>
</table>


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

<tr>
  <td align="right">
<math>~D_0 \equiv \frac{2^{3 / 2}}{3\pi^2} \biggl[\frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^{3 / 2}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr] \, ,</math>
  </td>
</tr>
</table>


From [[#ThreeTermsAdded|above]], when added together, the three leading terms in Wong's expression for the exterior potential give,

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

<tr>
  <td align="right">
<math>~\frac{\pi}{a}\biggl[\Phi_{\mathrm{W}0} + \Phi_{\mathrm{W}1} + \Phi_{\mathrm{W}2}\biggr]_\mathrm{exterior} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2^{3 / 2} D_0 \biggl[ \frac{2K(\mu)}{R_1 + R} \biggr] \biggl\{ C_0(z) 
-~\biggl( \frac{2}{3} \biggr)
C_2(z)~\cos(2\theta)  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 2^{5 / 2}D_0 \biggl[ \frac{E(k)}{R}  \biggr]   \biggl\{
~ \biggl[ C_1(z) + \frac{2}{3} \cdot C_2(z) \biggr]\cos(\theta)  
+~\biggl( \frac{2}{3} \biggr)\biggl( \frac{4c^2 }{RR_1} \biggr)
C_2(z)~\cos(2\theta)  
+~\biggl( \frac{2}{3} \biggr)
C_2(z)~\cos (3\theta )       
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2^{3 / 2} \cdot \frac{2^{3 / 2}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  \biggl[ \frac{2K(\mu)}{R_1 + R} \biggr] \biggl\{ 
\biggl[ \frac{\pi^2}{2^2(1-\epsilon^2)} \biggr] 3\epsilon^2 \biggl[1+\frac{3}{2^4}~\epsilon^2 \biggr]
-~\biggl( \frac{2}{3} \biggr) C_2(z)~\cos(2\theta)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 2^{5 / 2} \cdot \frac{2^{3 / 2}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  \biggl[ \frac{E(k)}{R}  \biggr]   \biggl\{
~ \frac{\epsilon^4}{2}\biggl[ \frac{\pi^2}{2^2(1-\epsilon^2)} \biggr]\cos\theta + \frac{2}{3} \cdot C_2(z) \cos(\theta)  
+~\biggl( \frac{2}{3} \biggr)\biggl( \frac{4c^2 }{RR_1} \biggr)
C_2(z)~\cos(2\theta)  
+~\biggl( \frac{2}{3} \biggr)
C_2(z)~\cos (3\theta )       
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{2K(\mu)}{R_1 + R} \biggr] \biggl\{ 
\biggl[ \frac{\pi^2}{2^2(1-\epsilon^2)} \biggr] 3\epsilon^2 \biggl[1+\frac{3}{2^4}~\epsilon^2 \biggr]\frac{2^{3}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  
-~\frac{2^{4}}{3^2\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  C_2(z)~\cos(2\theta)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{E(k)}{R}  \biggr]   \biggl\{
~ \frac{\epsilon^4}{2}\biggl[ \frac{\pi^2}{2^2(1-\epsilon^2)} \biggr] \frac{2^{4}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  \cos\theta 
~+~ C_2(z) \cdot \frac{2^{5}}{3^2\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  \biggl[ \cos(\theta)  
~+~\biggl( \frac{4c^2 }{RR_1} \biggr)~\cos(2\theta)  
~+~\cos (3\theta )  \biggr]     
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{2K(\mu)}{R_1 + R} \biggr] \biggl\{ 
\biggl[2+\frac{3}{2^3}~\epsilon^2 \biggr] (1 + \epsilon^2)^{3 / 2}  (1-\epsilon^2)^{-1}
-~\frac{2^{4}}{3^2\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]  C_2(z)~\cos(2\theta)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{E(k)}{R}  \biggr]   \biggl\{
~ \frac{2\epsilon^2}{3} (1 + \epsilon^2)^{3 / 2} (1-\epsilon^2)^{-1}  \cos\theta 
~+~\frac{2^{5}}{3^2\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr] C_2(z)  \biggl[ \cos(\theta)  
~+~\biggl( \frac{4c^2 }{RR_1} \biggr)~\cos(2\theta)  
~+~\cos (3\theta )  \biggr]     
\biggr\} \, .
</math>
  </td>
</tr>
</table>