Editing 2DStructure/ToroidalGreenFunction (section)

=Appendix B: Elliptic Integrals=
==Summary Table==
If the default argument of a [[Appendix/EquationTemplates#Complete_Elliptic_Integrals|complete elliptic integral function]] is the parameter, <math>~k</math>, then we should also appreciate that the following additional arguments may be useful:

<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" rowspan="2" width="10%">parameter</td>
  <td align="center" colspan="4">In terms of &hellip;</td>
</tr>
<tr>
  <td align="center" width="20%"><math>~k</math><br />
<font size="-1">([http://dlmf.nist.gov/22.7.i Descending Landen Transformation])</font>
  </td>
  <td align="center" width="20%"><math>~k_1</math><br />
<font size="-1">([http://dlmf.nist.gov/22.7.ii Ascending Landen Transformation])</font>
  </td>
  <td align="center" width="20%><math>~\frac{R}{R_1} = e^{-\eta}</math></td>
  <td align="center"><math>~\eta \equiv \ln\biggl(\frac{R_1}{R}\biggr)</math></td>
</tr>
<tr>
  <td align="center"><math>~k</math></td>
  <td align="center"><math>~k</math></td>
  <td align="center"><math>~\frac{2\sqrt{k_1}}{1+k_1}</math></td>
  <td align="center"><math>~\biggl[1 - \biggl( \frac{R}{R_1}\biggr)^2 \biggr]^{1 / 2}</math></td>
  <td align="center"><math>~\biggl[ \frac{2}{\coth\eta + 1} \biggr]^{1 / 2}= \biggl[ 1 - e^{-2\eta} \biggr]^{1 / 2}</math></td>
</tr>
<tr>
  <td align="center"><math>~k^{\prime}</math></td>
  <td align="center"><math>~\sqrt{1 - k^2}</math></td>
  <td align="center"><math>~\frac{1-k_1}{1+k_1}</math></td>
  <td align="center"><math>~\frac{R}{R_1}</math></td>
  <td align="center"><math>~\frac{1}{\cosh\eta + \sinh\eta} = e^{-\eta}</math></td>
</tr>
<tr>
  <td align="center"><math>~k_1 = \frac{1-k^{\prime}}{1+k^{\prime}}</math></td>
  <td align="center"><math>~\frac{1-\sqrt{1 - k^2}}{1+\sqrt{1 - k^2}}</math></td>
  <td align="center"><math>~k_1</math></td>
  <td align="center"><math>~\frac{R_1 - R}{R_1 +R}</math></td>
  <td align="center"><math>~\tanh\biggl(\frac{\eta}{2}\biggr) = \frac{1 - e^{-\eta}}{1 + e^{-\eta}}</math></td>
</tr>
</table>


<span id="GandR65">As published, for example,</span> in &sect;8.126 of [http://www.mathtable.com/gr/ Gradshteyn &amp; Ryzhik (1965)], the following mappings are valid:

<table border="1" cellpadding="8" width="85%" align="center">
<tr>
<th>
Explicitly Published in &sect;8.126 of [http://www.mathtable.com/gr/ G &amp; R (1965)]
</th>
<th>
Additional Implications
</th>
</tr>
<tr>
  <td>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
K\biggl( \frac{1-k^\prime}{1+k^\prime} \biggr)
=
K(k_1)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} (1+k^{\prime}) K(k)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
E\biggl( \frac{1-k^\prime}{1+k^\prime} \biggr)
=
E(k_1)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{1+k^\prime} \biggl[E(k) + k^\prime K(k) \biggr]</math>
  </td>
</tr>
</table>
  </td>
  <td>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\frac{4K(k_1)}{R_1 + R}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2K(k)}{R_1}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~E(k)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1+k^\prime)E(k_1) - k^\prime K(k)</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">[http://dlmf.nist.gov/19.8.E12 DLMS &sect;19.8.12]</td>
</tr>
</table>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~
K\biggl( \frac{2\sqrt{k}}{1+k}  \biggr)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1+k)K(k)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
E\biggl( \frac{2\sqrt{k}}{1+k} \biggr)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{1+k} \biggl[2E(k) - (k^\prime)^2 K(k) \biggr]</math>
  </td>
</tr>
</table>
  </td>
  <td>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
K( k )
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1+k_1)K(k_1)</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">[http://dlmf.nist.gov/19.8.E12 DLMS &sect;19.8.12]</td>
</tr>

<tr>
  <td align="right">
<math>~
E\biggl( \frac{2\sqrt{k_1}}{1+k_1} \biggr) = E(k)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{1+k_1} \biggl[2E(k_1) - (1-k_1^2) K(k_1) \biggr]</math>
  </td>
</tr>
</table>
  </td>
</tr>
</table>

==One Set of Details==
Following advice that we have received from Howard Cohl (private communication), we can demonstrate this most efficiently by employing the [https://dlmf.nist.gov/19.8#ii ''Descending Landen Transformation'' for the complete elliptic integral of the first kind], that is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~K(k)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
(1 + k_1)K(k_1) \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; where, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~k_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 
\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}} \, .
</math>
  </td>
</tr>
</table>
Given our just-stated definition of the parameter, <math>~k</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sqrt{1-k^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 1 - \frac{2}{\coth\eta + 1} \biggr]^{1 / 2} = \biggl[ \frac{\coth\eta - 1 }{\coth\eta + 1} \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 1 \pm \sqrt{1-k^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 \pm  \biggl[ \frac{\coth\eta - 1 }{\coth\eta + 1} \biggr]^{1 / 2} = \frac{ [\coth\eta + 1]^{1 / 2} \pm  [\coth\eta - 1]^{1 / 2}  }{ [\coth\eta + 1]^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~k_1 \equiv \frac{1 - \sqrt{1-k^2}}{1 + \sqrt{1-k^2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ [\coth\eta + 1]^{1 / 2} -  [\coth\eta - 1]^{1 / 2}  }{ [\coth\eta + 1]^{1 / 2} +  [\coth\eta - 1]^{1 / 2}  }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~k_1^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ [\coth\eta + 1] -2[\coth\eta + 1]^{1 / 2}[\coth\eta - 1]^{1 / 2} +  [\coth\eta - 1]  }{ [\coth\eta + 1] + 2[\coth\eta + 1]^{1 / 2}[\coth\eta - 1]^{1 / 2} +  [\coth\eta - 1] }
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\cosh\eta - 1}{\cosh\eta + 1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tanh^2\biggl(\frac{\eta}{2}\biggr) \, .</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~1 + k_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1+ \frac{1 - \sqrt{1-k^2}}{1 + \sqrt{1-k^2}} = \frac{2}{1 + \sqrt{1-k^2}} \, .</math>
  </td>
</tr>
</table>

Hence, the "second attempt" expression for <math>~P_{1 / 2}(z)</math> becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_{-1 / 2}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\sqrt{2}}{\pi}~ (\sinh\eta)^{-1 / 2} \biggl[ \frac{2k}{1 + \sqrt{1-k^2}} \biggr] K\biggl(\tanh \frac{\eta}{2} \biggr) 
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\pi}{2} \cdot \cosh\frac{\eta}{2}  \biggr]^{-1} K\biggl(\tanh \frac{\eta}{2} \biggr) \, .
</math>
  </td>
</tr>
</table>
This is, indeed, identical to the "Attempt #1" expression for  <math>~P_{-1 / 2}(z)</math> that was used by Wong.

'''Q.E.D.'''

==Series Expansions==

From our [[Appendix/EquationTemplates#Complete_Elliptic_Integrals|accompanying set of Key mathematical relations]] we find the following series expansions for complete elliptic integrals of, respectively, the first and second kind:

<div align="center">
{{ Math/EQ_EllipticIntegral01 }}<br />
{{ Math/EQ_EllipticIntegral02 }}
</div>
If you don't know how to handle a ''double factorial'' &#8212; also referred to as a ''semifactorial'' &#8212; see, for example, [https://en.wikipedia.org/wiki/Double_factorial this wikipedia page].