Editing Appendix/Mathematics/ToroidalFunctions (section)

=Overview by Howard Cohl=
This subsection is drawn verbatim from [http://hcohl.sdf.org/WHIPPLE.html Howard Cohl's online overview] of toroidal functions.

<font color="#009999">
&hellip; These last two expressions allow us to express toroidal functions of a certain kind (first or second, respectively) with argument hyperbolic cosine, as a direct proportionality in terms of the toroidal function of the other kind (second or first, respectively) with argument hyperbolic cotangent. The Whipple formulae may also be expressed as follows:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^n_{m- 1 / 2}(\cosh\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^n
~\Gamma(n-m + \tfrac{1}{2}) \biggl[ \frac{\pi}{2\sinh\alpha} \biggr]^{1 / 2} P^m_{n- 1 / 2}(\coth\alpha)\, ,
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">&hellip; and &hellip;</td></tr>
<tr>
  <td align="right">
<math>~Q^n_{m- 1 / 2}(\coth\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
~\frac{\pi}{\Gamma(m-n + \tfrac{1}{2})}  \biggl[ \frac{\pi \sinh\alpha}{2} \biggr]^{1 / 2} P^m_{n- 1 / 2}(\cosh\alpha) \, .
</math>
  </td>
</tr>
</table>


These interesting formulae have the property that they can relate Legendre functions of the first and second kinds directly in terms of each other. The only hitch is that you need a different argument to relate them. The way it works is as such. The Legendre functions of the first kind generally are well-behaved near the origin and blow up at positive infinity. Consequently the Legendre functions of the second kind blow up at unity and exponentially converges towards zero for large values of the argument. The relevant domain for toroidal functions is from 1 to infinity. The standard hyperbolic argument for these functions are naturally chosen to be the hyperbolic cosine since it ranges from 1 to infinity. The Whipple formulae relate the Legendre functions with argument 1 to infinity, cosh, to a reversed range given by the hyperbolic cotangent function. the hyperbolic cotangent function ranges from infinity at unity to unity at infinity. At what point alpha does cosh alpha equal coth alpha? The point alpha is given by
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ln(1+\sqrt{2}) \cong 0.88137359 \, .
</math>
  </td>
</tr>
</table>
Therefore, <math>~e^\alpha</math> and <math>~e^{-\alpha}</math> are given, respectively, by
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~e^\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{2} + 1 \cong 2.41421356 \, ,
</math>
  </td>
  <td align="center">&nbsp; &nbsp; and &nbsp; &nbsp;</td>
  <td align="right">
<math>~e^{-\alpha}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{2} - 1 \cong 0.41421356 \, .
</math>
  </td>
</tr>
</table>

The value that <math>~\cosh \alpha</math> and <math>~\coth \alpha</math> obtain at <math>~\alpha</math> is given by
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cosh\alpha = \coth\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{2} \cong 1.41421356 \, .
</math>
  </td>
</tr>
</table>

The value that <math>~1/\cosh\alpha</math> and <math>~\tanh \alpha</math> obtain at <math>~\alpha</math> is given by
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\cosh\alpha} = \tanh\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\sqrt{2}} \cong 0.70710678 \, .
</math>
  </td>
</tr>
</table>

Finally, <math>~\sinh \alpha</math> and it's inverse are given respectively by unity,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sinh\alpha = \frac{1}{\sinh\alpha} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 \, .
</math>
  </td>
</tr>
</table>

We now see that the value at which the argument of the Legendre functions inversely maps the entire domain is given by cosh alpha = coth alpha ~ 1.4142356. By using the Whipple formulae for ring functions, we can inversely map the entire domain from 1 to infinity about this point cosh alpha, the square root of 2, and take full advantage of this new symmetry for Legendre functions. There being previously more definite and indefinite integrals tabulated for the Legendre function of the first kind than for the Legendre function of the second kind. In fact, this new transformation, when applied to toroidal functions yields distinct expressions which relate correspondingly the complete elliptic integrals of the first and second kind, which don't seem to be related to the linear and quadratic transformations of hypergeometric functions.

</font>


Note that, as shown above, [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G Gil, ''et al.''  (2000)] state:

{{ Math/EQ_Toroidal02 }}

Hence, if we swap the indexes, <math>~m\leftrightarrow n</math>, and make the assignments, <math>~x \leftrightarrow \cosh\alpha</math> and <math>~\lambda \leftrightarrow \coth\alpha</math>, this last relation becomes,

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

<tr>
  <td align="right">
<math>~Q_{m-1 / 2}^n (\coth\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(m-n+1 / 2)} (\cosh^2\alpha-1)^{1 / 4} P_{n-1 / 2}^m(\cosh\alpha) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\frac{\pi}{ \Gamma(m-n+1 / 2)} \biggl[ \frac{\pi\sinh\alpha }{2} \biggr]^{1 / 2} P_{n-1 / 2}^m(\cosh\alpha) 
</math>
  </td>
</tr>
</table>
</div>

Alternatively, if we swap the indexes, <math>~m\leftrightarrow n</math>, and make the assignments, <math>~x \leftrightarrow \coth\alpha</math> and <math>~\lambda \leftrightarrow \cosh\alpha</math>, the [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G Gil, ''et al.''  (2000)] relation becomes,

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

<tr>
  <td align="right">
<math>~Q_{m-1 / 2}^n (\cosh\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(m-n+1 / 2)} (\coth^2\alpha-1)^{1 / 4} P_{n-1 / 2}^m(\coth\alpha) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\frac{\pi}{ \Gamma(m-n+1 / 2)} \biggl[ \frac{ \pi}{2\sinh\alpha} \biggr]^{1 / 2} P_{n-1 / 2}^m(\coth\alpha) 
</math>
  </td>
</tr>
</table>
</div>

However, in our situation the so-called "Euler reflection formula for gamma functions" gives the relation,

{{ Math/EQ_Gamma01 }}

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

<tr>
  <td align="right">
<math>~\frac{\pi (-1)^{m}}{\Gamma(m-n+\frac{1}{2}) }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^{n}\Gamma(n-m+\tfrac{1}{2}) \, .</math>
  </td>
</tr>
</table>
</div>
So,

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

<tr>
  <td align="right">
<math>~Q_{m-1 / 2}^n (\cosh\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-1)^{n}\Gamma(n-m+\tfrac{1}{2})\biggl[ \frac{ \pi}{2\sinh\alpha} \biggr]^{1 / 2} P_{n-1 / 2}^m(\coth\alpha) 
</math>
  </td>
</tr>
</table>
</div>
We see, then, that we are able to generate both of Cohl's relations from the Gil ''et al.'' relation.  Yeah!