Editing Appendix/Ramblings/Bordeaux (section)

====Second (n  = 1) Term====
The second (n = 1) term in [[Apps/Wong1973Potential#D0andCn|Wong's (1973) expression for the exterior potential]] is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{a}{GM} \biggr) \Phi_\mathrm{W1}(\eta,\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-D_0
(\cosh\eta - \cos\theta)^{1 / 2} \cdot 2 \cos\theta \cdot  C_1(\cosh\eta_0)P_{+\frac{1}{2}}(\cosh\eta) \, ,
</math>
  </td>
</tr>
</table>
where, <math>~D_0</math> is the same as [[#Setup|above]], and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_1(\cosh\eta_0)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tfrac{3}{2} Q_{+\frac{3}{2}}(\cosh \eta_0) Q_{+\frac{1}{2}}^2(\cosh \eta_0) 
+ \tfrac{1}{2} Q_{+\frac{1}{2}}(\cosh \eta_0)~Q^2_{+ \frac{3}{2}}(\cosh \eta_0) \, .
</math>
  </td>
</tr>
</table>
Now, from our [[Appendix/SpecialFunctions#Toroidal_Function_Evaluations|accompanying table of "Toroidal Function Evaluations"]], it appears as though,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_{+\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sqrt{2}}{\pi} (\sinh\eta)^{+1 / 2} k^{-1} E(k) \, ,</math>
  </td>
</tr>
</table>

where, as above,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~k</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2}{\coth\eta+1} \biggr]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
Hence, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{a}{GM} \biggr) \Phi_\mathrm{W1}(\eta,\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} }{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] C_1(\cosh\eta_0) 
\biggl[ \cos\theta \cdot (\cosh\eta - \cos\theta)^{1 / 2}  (\sinh\eta)^{+1 / 2} \biggr] 
k^{-1} E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} }{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] C_1(\cosh\eta_0) \cdot \cos\theta
\biggl\{ \frac{a\sinh^2\eta}{\varpi}  \cdot \frac{\coth\eta + 1}{2} \biggr\}^{1 / 2} 
E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} }{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] C_1(\cosh\eta_0) \cdot \cos\theta
\biggl\{ \biggl( \frac{a}{2\varpi} \biggr) \biggl[ \frac{r_1^2 - r_2^2}{2r_1 r_2} \biggr]^2 \cdot \biggl[ \frac{2r_1^2}{r_1^2 - r_2^2} \biggr] \biggr\}^{1 / 2} 
E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} }{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] C_1(\cosh\eta_0) \cdot \cos\theta
\biggl\{ \biggl( \frac{a}{2} \biggr)\biggl[ \frac{4a}{r_1^2 - r_2^2} \biggr] \biggl[ \frac{r_1^2 - r_2^2}{2r_1 r_2} \biggr]^2 \cdot \biggl[ \frac{2r_1^2}{r_1^2 - r_2^2} \biggr] \biggr\}^{1 / 2} 
E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} a}{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] C_1(\cosh\eta_0) 
\biggl[ \frac{\cos\theta}{r_2} \biggr] 
E(k)
= 
- \frac{2^{3} a}{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr] C_1(\cosh\eta_0) 
\biggl[ \frac{\cos\theta}{\sqrt{ (\varpi - a)^2 + (z-Z_0)^2 }} \biggr] 
E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} a}{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr]  C_1(\cosh\eta_0) \biggl[\frac{ r_1^2 + r_2^2 - 4a^2}{2r_1 r_2^2} \biggr]
\boldsymbol{E}(k) \, .
</math>
  </td>
</tr>
</table>

<span id="Qrecurrence">&nbsp;</span>
<table border="1" align="center" width="80%" cellpadding="10">
<tr><td align="left">
From the [[#FirstEvaluations|above function tabulations &amp; evaluations]] &#8212; for example, <math>~ K(k_0) = 1.854074677</math> and <math>~ E(k_0) = 1.350643881</math> &#8212; and a [[Appendix/SpecialFunctions#Example_Recurrence_Relations|separate listing of ''Example Recurrence Relations'']], we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
[[Appendix/SpecialFunctions#Toroidal_Function_Evaluations|Appendix Expression:]] <math>~Q_{-\tfrac{1}{2}}(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~k_0 K(k_0)</math>
  </td>
</tr>

<tr>
  <td align="right">
[[Appendix/SpecialFunctions#Toroidal_Function_Evaluations|Appendix Expression:]] <math>~Q_{+\tfrac{1}{2}}(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z~k_0 K(k_0) - [2(z+1)]^{1 / 2} E(k_0)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q^0_{m-\tfrac{1}{2}}</math> [[Appendix/SpecialFunctions#Example_Recurrence_Relations|recurrence]] with m = 2:  
<math>~Q_{+\tfrac{3}{2}}(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4}{3} z~Q_{+\tfrac{1}{2}}(z_0) - \frac{1}{3} Q_{-\tfrac{1}{2}}(z_0)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4}{3} z \{z~k_0 K(k_0) - [2(z+1)]^{1 / 2} E(k_0)\} - \frac{1}{3}k_0 K(k_0)</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_0 K(k_0) - 4 z[2(z+1)]^{1 / 2} E(k_0) \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
Hence, <math>~Q_{+\tfrac{3}{2}}(3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0.014544576 \, .</math>
  </td>
</tr>
</table>

----

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
[[Appendix/SpecialFunctions#Toroidal_Function_Evaluations|Appendix Expression:]] <math>~Q^2_{-\tfrac{1}{2}}(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[2^3(z-1)(z^2-1)]^{-1 / 2} [4zE(k_0) - (z-1)K(k_0)]</math>
  </td>
</tr>

<tr>
  <td align="right">
[[Appendix/Mathematics/ToroidalSynopsis01#Evaluating_Q2.CE.BD|Additional derivation:]] <math>~Q^2_{+\tfrac{1}{2}}(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}
\biggl\{ z k_0~K ( k_0 )  
~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k_0)\biggr\} 
</math>
  </td>
</tr>
</table>

Then, letting <math>~\mu \rightarrow 2</math> and, for all m &ge; 2, letting <math>~\nu \rightarrow (m - \tfrac{1}{2})</math> in the "Key Equation," 

{{ Math/EQ_Toroidal04 }}

we have,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(m - \tfrac{3}{2})Q^{2}_{m+\tfrac{1}{2}} (z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2m)z Q_{m-\tfrac{1}{2}}^{2}(z) - (m + \tfrac{3}{2})Q^{2}_{m - \tfrac{3}{2}}(z) \, .
</math>
  </td>
</tr>
</table>

Therefore, specifically for m = 1, we obtain the recurrence relation,

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

<tr>
  <td align="right">
<math>~Q^{2}_{+\tfrac{3}{2}} (z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5 Q^{2}_{- \tfrac{1}{2}}(z_0)-4z Q_{+\tfrac{1}{2}}^{2}(z_0)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5 \biggl\{ [2^3(z-1)(z^2-1)]^{-1 / 2} [4zE(k_0) - (z-1)K(k_0)] \biggr\}
+ z \biggl\{ z k_0~K ( k_0 )  
~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k_0)\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^{1 / 2}[ (z-1)(z^2-1) ]^{- 1 / 2} \biggl\{  [5 z] 
~-~z (z^2+3) \biggr\} E(k_0) 
+ \biggl\{ z^2 k_0~  
- [(z-1)(z^2-1)]^{-1 / 2} [ 2^{-3 / 2} \cdot 5 (z-1)] \biggr\}K(k_0)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^{-3 / 2}(z+1)^{-1 / 2} [4 z^2  -  5  ]K(k_0)
-~2^{1 / 2}[ (z-1)(z^2-1) ]^{- 1 / 2} (z^2 - 2)z  E(k_0) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
Hence, <math>~Q^{2}_{+\tfrac{3}{2}} (3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0.132453829 \, . 
</math>
  </td>
</tr>
</table>

----

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ C_1(3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2}~Q_{+\frac{3}{2}}(3) \cdot Q_{+ \frac{1}{2}}^2(3) 
+ \frac{1}{2}~ Q_{+ \frac{1}{2}}(3)\cdot Q^2_{+ \frac{3}{2}}(3) 
= 0.017278633 \, .
</math>
  </td>
</tr>
</table>

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


While keeping in mind that, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cosh\eta_0 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~k_0^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{\cosh\eta_0 + 1} 
=
\frac{2}{z_0 + 1}
\, ,</math>
  </td>
</tr>
</table>

let's attempt to express this leading coefficient, <math>~C_1(\cosh\eta_0)</math>, entirely in terms of the pair of complete elliptic integral functions.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2C_1(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 \biggl[ Q_{+\frac{3}{2}}(z_0) \biggr]\times \biggl[ Q_{+\frac{1}{2}}^2(z_0) \biggr]
+ \biggl[ Q_{+\frac{1}{2}}(z_0) \biggr]\times \biggl[ 5 Q^{2}_{- \tfrac{1}{2}}(z_0)-4z Q_{+\tfrac{1}{2}}^{2}(z_0) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[3 Q_{+\frac{3}{2}}(z_0)  -4z Q_{+\frac{1}{2}}(z_0) \biggr]\times \biggl[ Q_{+\frac{1}{2}}^2(z_0) \biggr]
+ \biggl[ 5Q_{+\frac{1}{2}}(z_0) \biggr]\times \biggl[ Q^{2}_{- \tfrac{1}{2}}(z_0) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{1}{2^2}\biggl\{  (4z^2 - 1 )k_0 K(k_0) - 4 z[2(z+1)]^{1 / 2} E(k_0)    
-4z \biggl[ z~k_0 K(k_0) - [2(z+1)]^{1 / 2} E(k_0) \biggr] \biggr\}
\times 
\biggl\{ z k_0~K ( k_0 )  
~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k_0) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 5\biggl[ ~z~k_0 K(k_0) - [2(z+1)]^{1 / 2} E(k_0) \biggr]
\times 
\biggl\{ [2^3(z-1)(z^2-1)]^{-1 / 2} [4zE(k_0) - (z-1)K(k_0)] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^2} \cdot k_0 K(k_0) 
\times 
\biggl\{ z k_0~K ( k_0 )  
~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k_0) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 5[2^3(z-1)(z^2-1)]^{-1 / 2} \biggl[ ~z~k_0 K(k_0) - [2(z+1)]^{1 / 2} E(k_0) \biggr]
\times 
\biggl\{ 4zE(k_0) - (z-1)K(k_0) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
K(k_0)\cdot K(k_0) \biggl\{ \frac{z k_0^2}{2^2} - 5[2^3(z-1)(z^2-1)]^{-1 / 2} \cdot zk_0(z-1)\biggr\}
+
E(k_0)\cdot E(k_0) \biggl\{ -5[2^3(z-1)(z^2-1)]^{-1 / 2}\cdot [2(z+1)]^{1 / 2} \cdot 4z  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~K(k_0)\cdot E(k_0) \biggl\{ 
-~\frac{1}{2^2} \cdot k_0(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2}  
+ 5[2^3(z-1)(z^2-1)]^{-1 / 2} \cdot 4z^2k_0
+ 5[2^3(z-1)(z^2-1)]^{-1 / 2}\cdot [2(z+1)]^{1 / 2} \cdot (z-1)
\biggr\} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~2[(z-1)(z^2-1)]^{1 / 2} C_1(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
z k_0 \cdot K(k_0)\cdot K(k_0) \biggl\{ \frac{k_0}{2^2} \biggl[(z-1)(z^2-1) \biggr]^{1 / 2}  - \frac{5(z-1)}{2^{3/2}} \biggr\}
-~10 z(z+1)^{1 / 2}  \cdot E(k_0)\cdot E(k_0) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~2^{-3/2} K(k_0)\cdot E(k_0) \biggl\{ 
k_0[19z^2 - 3 ]   
+ 5(z-1) [2(z+1)]^{1 / 2}  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~2^{3/2}\biggl[ \frac{(z-1)}{k_0} \biggr] C_1(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
z k_0 \cdot K(k_0)\cdot K(k_0) \biggl\{ \frac{k_0}{2^2} \biggl[\frac{2^{1 / 2}(z-1)}{k_0} \biggr]  - \frac{5(z-1)}{2^{3/2}} \biggr\}
-~\biggl[ \frac{2^{3 / 2} \cdot 5z}{k_0}  \biggr] E(k_0)\cdot E(k_0) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~2^{-3/2} K(k_0)\cdot E(k_0) \biggl\{ 
k_0[19z^2 - 3 ]   
+ \frac{10 (z-1)}{k_0}   
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~C_1(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2(3z^2 - 1)}{(z^2-1)}      \biggr]K(k_0)\cdot E(k_0) 
-~\biggl[ \frac{z}{(z+1)} \biggr] K(k_0)\cdot K(k_0) 
-~\biggl[ \frac{ 5z}{(z-1)}  \biggr] E(k_0)\cdot E(k_0) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~(z_0^2-1)C_1(z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2(3z^2 - 1) K(k_0)\cdot E(k_0) 
-~z_0(z_0-1) K(k_0)\cdot K(k_0) 
-~5z_0(z_0+1) E(k_0)\cdot E(k_0) \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\biggl( \frac{a}{GM} \biggr) \Phi_\mathrm{W1}(\eta,\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^{3} a}{3\pi^3} \biggl[ \frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr]  C_1(\cosh\eta_0) \biggl[\frac{ r_1^2 + r_2^2 - 4a^2}{2r_1 r_2^2} \biggr]
\boldsymbol{E}(k) \, .
</math>
  </td>
</tr>
</table>