Editing Apps/Wong1973Potential (section)

===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>