Editing Appendix/Mathematics/ToroidalConfusion (section)

=Cohl's Response to My (May 2018) Email Query=

==Proper Interpretation of DLMF Expression==
Most of the confusion expressed above stems from the DLMF's use of '''bold''' fonts, such as the function on the left-hand side of expression #3, above &#8212; that is, the Whipple formula from [https://dlmf.nist.gov/14.19.v &sect;14.19 of DLMF],

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

<tr>
  <td align="right">
<math>~\boldsymbol{Q}^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\Gamma\left(m-n+
\tfrac{1}{2}\right)}{\Gamma\left(m+n+\tfrac{1}{2}\right)}\left(\frac{\pi}{2
\sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, .
</math>
  </td>
</tr>
</table>

What has been missing in my discussion is an appreciation of the following relationship between [http://dlmf.nist.gov/14.3.E10 '''bold''' and plain-text function names],
<div align="center">
<math>
\boldsymbol{Q}^{\mu}_{\nu}\left(x\right)=e^{-\mu\pi i}\frac{Q^{\mu}_{\nu}\left(x\right)}{\Gamma\left(\nu+\mu+1\right)}.
</math>
</div>
After making the substitutions, <math>~\mu \rightarrow m</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, the Whipple formula displayed above as expression #3 becomes,

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

<tr>
  <td align="right">
<math>~e^{-m\pi i}\frac{Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(n+m+\tfrac{1}{2}\right)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\Gamma\left(m-n+
\tfrac{1}{2}\right)}{\Gamma\left(m+n+\tfrac{1}{2}\right)}\left(\frac{\pi}{2
\sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{m\pi i} 
\Gamma\left(m-n+\tfrac{1}{2}\right)\left(\frac{\pi}{2
\sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m 
\Gamma\left(m-n+\tfrac{1}{2}\right)\left(\frac{\pi}{2
\sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, ,
</math>
  </td>
</tr>
</table>

which matches expression #2, above.  But it does not ''appear'' to match expressions #1 or #4.

<table border="1" cellpadding="5" align="center" width="80%"><tr><td align="left">
The standard "Euler reflection formula for gamma functions" is usually presented in the form,
{{ Math/EQ_Gamma01 }}
If we make the association,
<div align="center">
<math>~z \leftrightarrow (m - n + \tfrac{1}{2}) \, ,</math>
</div>
with <math>~m</math> and <math>~n</math> both being either zero or a positive integer, then, this Euler reflection formula becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma(m - n + \tfrac{1}{2}) ~ \Gamma(n - m + \tfrac{1}{2} )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi \biggl\{ \sin\biggl[ \pi(m - n + \tfrac{1}{2})  \biggr] \biggr\}^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi (-1)^{m+n} \, .</math>
  </td>
</tr>
</table>

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

However, in our situation the so-called "Euler reflection formula for gamma functions" gives the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

Hence, we may also write,

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

<tr>
  <td align="right">
<math>~ Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m \biggl[ \frac{\pi (-1)^{m+n}}{\Gamma(n-m+\frac{1}{2}) } \biggr]
\left(\frac{\pi}{2
\sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{(-1)^n \pi }{\Gamma(n-m+\frac{1}{2}) }
\left(\frac{\pi}{2
\sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, ,
</math>
  </td>
</tr>
</table>
which matches expressions #1 and #4.  So everything appears to be in agreement!  Hooray!

==Derivation From Scratch==

Whenever he deals with these types of relations, Cohl usually begins with,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr><th align="center" colspan="3"><font color="maroon">Expression #5</font></th></tr>
<tr>
  <td align="right">
<math>~Q^\mu_\nu(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{\pi}{2}} ~\Gamma(\nu + \mu + 1) ~e^{i\mu\pi} \biggl[ \frac{1}{\sinh^2\eta} \biggr]^{1 / 4} P^{-\nu-\frac{1}{2}}_{-\mu - \frac{1}{2}} (\coth\eta)
</math>
  </td>
</tr>
</table>
</div>
Making the pair of substitutions,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~n - \frac{1}{2} \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~n ~~\in</math>
  </td>
  <td align="left">
<math>~\mathbb{N}_0 = \{ 0, 1, 2, \cdots\} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~m ~~\in</math>
  </td>
  <td align="left">
<math>~\mathbb{N}_0 = \{ 0, 1, 2, \cdots\} \, ,</math>
  </td>
</tr>
</table>
</div>
we also have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nu + \mu +1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~n - \frac{1}{2} + m + 1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~n + m + \frac{1}{2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~-\mu - \frac{1}{2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-m-\frac{1}{2} \, ,</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>~-\nu - \frac{1}{2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl(n - \frac{1}{2}\biggr)-\frac{1}{2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-n \, , </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~e^{i\mu\pi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{i m \pi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^{m} \, , </math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{-m - \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>
</table>
</div>

----

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

<tr>
  <td align="right">
<math>~P^\mu_\nu(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~P^\mu_{-\nu-1}(z) \, ,</math>
  </td>
</tr>
</table>
</div>
if we make the substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-(\nu + 1)</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~-(m+\tfrac{1}{2})</math>
  </td>
  <td align="center">&nbsp; &nbsp; <math>~\Rightarrow</math>&nbsp; &nbsp;</td>
  <td align="right">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~m - \tfrac{1}{2} \, ,</math>
  </td>
</tr>
</table>
</div>
we also know that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P^\mu_{m-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~P^\mu_{-m-\frac{1}{2}}(z) \, .</math>
  </td>
</tr>
</table>
</div>
<span id="QPrelation">Hence, we can write,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{m - \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>
</table>
</div>

----

Finally, another relation states that, for <math>~n \in \mathbb{N}_0</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P^{-n}_{m-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\Gamma(m-n+\frac{1}{2})}{\Gamma(m+n+\frac{1}{2})} \biggr] P^n_{m-\frac{1}{2}}(z) \, .</math>
  </td>
</tr>
</table>
</div>
So, we obtain,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] \biggl[ \frac{\Gamma(m-n+\frac{1}{2})}{\Gamma(m+n+\frac{1}{2})} \biggr]P^{n}_{m - \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\sqrt{\frac{\pi}{2}} ~\Gamma(m-n+\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{n}_{m - \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>
</table>
</div>
This matches expressions #2 and #3, above.

==Index Values of Zero==
Setting  <math>~n = m = 0</math> gives the following sought-for relationship:

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

<tr>
  <td align="right">
<math>~Q^0_{-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{\pi}{2}} ~\Gamma(\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\pi}{\sqrt{2}} ~ \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>
</table>
</div>

==Joel's Additional Manipulations==

From [https://dlmf.nist.gov/14.19.iv &sect;14.19.6 of DLMF], we find the following summation expression:

<div align="center">
<math>~\boldsymbol{Q}^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)+2\sum_{n=1}^{\infty}
\frac{\Gamma\left(\mu+n+\tfrac{1}{2}\right)}{\Gamma\left(\mu+\tfrac{1}{2}
\right)}\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right)\cos\left(n
\phi\right)=\dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu
}}{\left(\cosh\xi-\cos\phi\right)^{\mu+(1/2)}}
</math></div>

Then, if we again employ the [http://dlmf.nist.gov/14.3.E10 DLMF relationship between '''bold''' and plain-text function names], namely,
<div align="center">
<math>
\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(x\right)
=
e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(x\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \, ,
</math>
</div>
where we have made the substitution, <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, the ''Sums'' expression becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~e^{-\mu\pi i}\frac{Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+ \tfrac{1}{2} \right)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu
}}{\left(\cosh\xi-\cos\phi\right)^{\mu+(1/2)}}
-
2\sum_{n=1}^{\infty}
\frac{\Gamma\left(\mu+n+\tfrac{1}{2}\right)}{\Gamma\left(\mu+\tfrac{1}{2}
\right)}
\biggl[ e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \biggr] \cos\left(n
\phi\right)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ e^{\mu\pi i} \Gamma\left(\mu+ \tfrac{1}{2} \right) \biggl[
\dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu
}}{\left(\cosh\xi-\cos\phi\right)^{\mu+(1/2)}}\biggr]
-
2\sum_{n=1}^{\infty}
Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
\phi\right) \, .
</math>
  </td>
</tr>
</table>
</div>

When dealing with Dyson-Wong tori, we will set <math>~\mu = 0</math>, in which case the ''Sums'' expression becomes,

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

<tr>
  <td align="right">
<math>~Q_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl[
\dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}}  }\biggr]
-
2\sum_{n=1}^{\infty}
Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
\phi\right) \, .
</math>
  </td>
</tr>
</table>
</div>

But this can be rewritten in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\sum_{n=0}^{\infty} \epsilon_n
Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
\phi\right) 

</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl[
\dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}}  }\biggr]
</math>
  </td>
</tr>
</table>
</div>