Editing Apps/Wong1973Potential (section)

===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.