Difference between revisions of "User:Tohline/Appendix/Mathematics/ToroidalConfusion"

From VistrailsWiki
Jump to navigation Jump to search
 
(34 intermediate revisions by the same user not shown)
Line 11: Line 11:
From equation (34) of [http://adsabs.harvard.edu/abs/2000AN....321..363C H. S. Cohl, J. E. Tohline, A. R. P. Rau, & H. M. Srivastiva (2000, Astronomische Nachrichten, 321, no. 5, 363 - 372)] I find:
From equation (34) of [http://adsabs.harvard.edu/abs/2000AN....321..363C H. S. Cohl, J. E. Tohline, A. R. P. Rau, & H. M. Srivastiva (2000, Astronomische Nachrichten, 321, no. 5, 363 - 372)] I find:
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
 
<tr><th align="center" colspan="3"><font color="maroon">Expression #1</font></th></tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
Line 30: Line 30:
From [http://hcohl.sdf.org/WHIPPLE.html Howard Cohl's online overview] of toroidal functions, I find:
From [http://hcohl.sdf.org/WHIPPLE.html Howard Cohl's online overview] of toroidal functions, I find:
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
 
<tr><th align="center" colspan="3"><font color="maroon">Expression #2</font></th></tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
Line 48: Line 48:
Copying the Whipple's formula from [https://dlmf.nist.gov/14.19.v &sect;14.19 of DLMF],
Copying the Whipple's formula from [https://dlmf.nist.gov/14.19.v &sect;14.19 of DLMF],
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
 
<tr><th align="center" colspan="3"><font color="maroon">Expression #3</font></th></tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
Line 66: Line 66:
</table>
</table>


As per equation (8) in [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G A. Gil, J. Segura, &amp; N. M. Temme (2000, JCP, 161, 204 - 217)], the relationship is:
<span id="Gil">So far, this gives me three ''similar'' but not identical formulae for the same function mapping!</span>  As per equation (8) in (yet another source!) [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G A. Gil, J. Segura, &amp; N. M. Temme (2000, JCP, 161, 204 - 217)], the relationship is:
{{ User:Tohline/Math/EQ_Toroidal02 }}
 
This expression from Gil et al. (2000) means, for example, that by identifying <math>~x</math> with <math>~\coth\eta</math>, we have <math>~\lambda = \cosh\eta</math>, and,


<div align="center">
<div align="center">
Line 73: Line 76:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~Q_{n-1 / 2}^m (\lambda)</math>
<math>~Q_{n-1 / 2}^m (\cosh\eta)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 80: Line 83:
   <td align="left">
   <td align="left">
<math>~(-1)^n
<math>~(-1)^n
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(n-m+1 / 2)} (x^2-1)^{1 / 4} P_{m-1 / 2}^n(x) \, ,
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(n-m+1 / 2)} (\coth^2\eta-1)^{1 / 4} P_{m-1 / 2}^n(\coth\eta)
</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+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{\cosh^2\eta}{\sinh^2\eta}-1 \biggr]^{1 / 4} P_{m-1 / 2}^n(\coth\eta)
</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+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{1}{\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta) \, .
</math>
</math>
   </td>
   </td>
Line 86: Line 117:
</table>
</table>
</div>
</div>
where, <math>~\lambda \equiv x/\sqrt{x^2-1}</math>.  This expression from Gil et al. (2000) means, for example, that by identifying <math>~x</math> with <math>~\coth\eta</math>, we have <math>~\lambda = \cosh\eta</math>, and,
That is, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr><th align="center" colspan="3"><font color="maroon">Expression #4</font></th></tr>
<tr>
  <td align="right">
<math>~Q_{n-1 / 2}^m (\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl[\frac{\pi}{2\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta) \, ,
</math>
  </td>
</tr>
</table>
</div>
which matches the above expression #1 drawn from Cohl et al. (2000), but which appears ''not'' to match either of the other two "published" (online) formulae, expressions #2 or #3.


==Specific Application==
I stumbled into this dilemma when I tried to explicitly demonstrate how <math>~Q_{-1 / 2}(\cosh\eta)</math> can be derived from <math>~P_{-1 / 2}(z)</math> where, from &sect;8.13 of [https://books.google.com/books?id=MtU8uP7XMvoC&printsec=frontcover&dq=Abramowitz+and+stegun&hl=en&sa=X&ved=0ahUKEwialra5xNbaAhWKna0KHcLAASAQ6AEILDAA#v=onepage&q=Abramowitz%20and%20stegun&f=false M. Abramowitz &amp; I. A. Stegun (1995)], we find,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{-1 / 2}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2 e^{- \eta / 2}
~K(e^{-\eta} ) \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">Abramowitz &amp; Stegun (1995), eq. (8.13.4)</td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~P_{-1 / 2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} \biggl[\frac{2}{z+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-1}{z+1}} \biggr) \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">Abramowitz &amp; Stegun (1995), eq. (8.13.1)</td>
</tr>
</table>
When I used the Whipple formula as defined in [https://dlmf.nist.gov/14.19.v &sect;14.19 of DLMF] (expression #3 reprinted above), the function mapping <font color="red">'''gave me the wrong result'''</font>; I was off by a factor of <math>~\Gamma(\tfrac{1}{2}) =\sqrt{\pi}</math>.  But, as demonstrated below, the Whipple formula provided by Cohl et al. (2000) and by Gil et al. (2000) &#8212; that is, expressions #1 and #4, above &#8212; ''does'' give the correct result.
<table border="1" cellpadding="5" align="center" width="80%">
<tr>
  <td align="center">
Demonstration that <math>~Q_{-\frac{1}{2}}</math> can be derived from <math>~P_{-\frac{1}{2}}</math>
  </td>
</tr>
<tr>
  <td align="left">
Copying equation (34) from [http://adsabs.harvard.edu/abs/2000AN....321..363C Cohl et al. (2000)], we begin with,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q^m_{n - 1 / 2}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(-1)^n \pi}{\Gamma(n - m + \tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P^n_{m - 1 / 2}(\coth\eta) \, ;
</math>
  </td>
</tr>
</table>
then setting <math>~m = n = 0</math>, we have,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\pi}{\Gamma(\tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P_{-\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}{\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta) \, .
</math>
  </td>
</tr>
</table>
Step #1: &nbsp; Associate &hellip; <math>z \leftrightarrow \cosh\eta</math>.  Then,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sqrt{z^2-1}} \biggr]^{1 / 2} P_{-\frac{1}{2}}\biggl(\frac{z}{\sqrt{z^2-1}} \biggr) \, .
</math>
  </td>
</tr>
</table>
Step #2: &nbsp; Now making the association &hellip; <math>\Lambda \leftrightarrow z/\sqrt{z^2-1}</math>, and drawing on eq. (8.13.1) from [https://books.google.com/books?id=MtU8uP7XMvoC&printsec=frontcover&dq=Abramowitz+and+stegun&hl=en&sa=X&ved=0ahUKEwialra5xNbaAhWKna0KHcLAASAQ6AEILDAA#v=onepage&q=Abramowitz%20and%20stegun&f=false Abramowitz &amp; Stegun (1995)], we can write,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~P_{-\frac{1}{2}}(\Lambda)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} \biggl[\frac{2}{\Lambda+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\Lambda-1}{\Lambda+1}} \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} \biggl[\frac{2\sqrt{z^2-1} }{z+\sqrt{z^2-1} }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-\sqrt{z^2-1} }{z+\sqrt{z^2-1} }} \biggr) \, .
</math>
  </td>
</tr>
</table>
Step #3: &nbsp; Again, making the association &hellip; <math>z \leftrightarrow \cosh\eta</math>, means,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~P_{-\frac{1}{2}}(\Lambda)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} \biggl[\frac{2\sinh\eta }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\eta-\sinh\eta }{\cosh\eta +\sinh\eta }} \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ Q_{-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sinh\eta} \biggr]^{1 / 2}
\frac{2}{\pi} \biggl[\frac{2\sinh\eta }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\eta-\sinh\eta }{\cosh\eta +\sinh\eta }} \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \biggl[\frac{1 }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh^2\eta-\sinh^2\eta }{[\cosh\eta +\sinh\eta ]^2}} ~\biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \biggl[\frac{1 }{e^\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{1 }{e^{2\eta}}} \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2 e^{-\eta/2} K(e^{-\eta}) \, .
</math>
  </td>
</tr>
</table>
This, indeed, matches eq. (8.13.4) from [https://books.google.com/books?id=MtU8uP7XMvoC&printsec=frontcover&dq=Abramowitz+and+stegun&hl=en&sa=X&ved=0ahUKEwialra5xNbaAhWKna0KHcLAASAQ6AEILDAA#v=onepage&q=Abramowitz%20and%20stegun&f=false Abramowitz &amp; Stegun (1995)].
  </td>
</tr>
</table>
=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,
{{ User:Tohline/Math/EQ_Gamma01 }}
If we make the association,
<div align="center">
<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">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~Q_{n-1 / 2}^m (\cosh\eta)</math>
<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>
   <td align="center">
   <td align="center">
Line 99: Line 555:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~(-1)^n
<math>~
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(n-m+1 / 2)} (\coth^2\eta-1)^{1 / 4} P_{m-1 / 2}^n(\coth\eta)  
\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>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <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;
&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>
   <td align="center">
   <td align="center">
Line 114: Line 778:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{\cosh^2\eta}{\sinh^2\eta}-1 \biggr]^{1 / 4} P_{m-1 / 2}^n(\coth\eta)  
\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>
</math>
   </td>
   </td>
Line 122: Line 825:
   <td align="right">
   <td align="right">
&nbsp;
&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>
   <td align="center">
   <td align="center">
Line 128: Line 854:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{1}{\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta)  
\sqrt{\frac{\pi}{2}} ~\Gamma(\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, .
</math>
</math>
   </td>
   </td>
Line 142: Line 868:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl[\frac{\pi}{2\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta) \, ,
\frac{\pi}{\sqrt{2}} ~ \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, .
</math>
</math>
   </td>
   </td>
Line 148: Line 874:
</table>
</table>
</div>
</div>
which matches the above expression drawn from Cohl et al. (2000), but which does ''not'' match either of the other two "published" (online) expressions.


==Specific Application==
==Joel's Additional Manipulations==
I stumbled into this dilemma when I tried to explicitly demonstrate how <math>~Q_{-1 / 2}(\cosh\eta)</math> can be derived from <math>~P_{-1 / 2}(z)</math> where, from &sect;8.13 of [https://books.google.com/books?id=MtU8uP7XMvoC&printsec=frontcover&dq=Abramowitz+and+stegun&hl=en&sa=X&ved=0ahUKEwialra5xNbaAhWKna0KHcLAASAQ6AEILDAA#v=onepage&q=Abramowitz%20and%20stegun&f=false M. Abramowitz &amp; I. A. Stegun (1995)], we find,
 
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">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~Q_{-1 / 2}(\cosh\eta)</math>
<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>
   <td align="center">
   <td align="center">
Line 163: Line 907:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2 e^{- \eta / 2}
<math>~
~K(e^{-\eta} ) \, ,
\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>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="center" colspan="3">Abramowitz &amp; Stegun (1995), eq. (8.13.4)</td>
  <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>
</tr>
</table>
</table>
and,
</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">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~P_{-1 / 2}(z)</math>
<math>~Q_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 183: Line 954:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \biggl[
\frac{2}{\pi} \biggl[\frac{2}{z+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-1}{z+1}} \biggr) \, .
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
But this can be rewritten in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="center" colspan="3">Abramowitz &amp; Stegun (1995), eq. (8.13.1)</td>
  <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>
</tr>
</table>
</table>
</div>


=See Also=
=See Also=

Latest revision as of 04:43, 13 June 2018


Confusion Regarding Whipple Formulae

May, 2018 (J.E.Tohline): I am trying to figure out what the correct relationship is between half-integer degree, associated Legendre functions of the first and second kinds. In order to illustrate my current confusion, here I will restrict my presentation to expressions that give <math>~Q^m_{n - 1 / 2}(\cosh\eta)</math> in terms of <math>~P^n_{m - 1 / 2}(\coth\eta)</math>.


Whitworth's (1981) Isothermal Free-Energy Surface
|   Tiled Menu   |   Tables of Content   |  Banner Video   |  Tohline Home Page   |

Published Expressions

From equation (34) of H. S. Cohl, J. E. Tohline, A. R. P. Rau, & H. M. Srivastiva (2000, Astronomische Nachrichten, 321, no. 5, 363 - 372) I find:

Expression #1

<math>~Q^m_{n - 1 / 2}(\cosh\eta)</math>

<math>~=</math>

<math>~ \frac{(-1)^n \pi}{\Gamma(n - m + \tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P^n_{m - 1 / 2}(\coth\eta) \, . </math>


From Howard Cohl's online overview of toroidal functions, I find:

Expression #2

<math>~Q^n_{m- 1 / 2}(\cosh\alpha)</math>

<math>~=</math>

<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>

Copying the Whipple's formula from §14.19 of DLMF,

Expression #3

<math>~\boldsymbol{Q}^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>

<math>~=</math>

<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>

So far, this gives me three similar but not identical formulae for the same function mapping! As per equation (8) in (yet another source!) A. Gil, J. Segura, & N. M. Temme (2000, JCP, 161, 204 - 217), the relationship is:

LSU Key.png

<math>~Q_{n-1 / 2}^m (\lambda)</math>

<math>~=</math>

<math>~(-1)^n \frac{\pi^{3/2}}{\sqrt{2}~ \Gamma(n-m+1 / 2)} (x^2-1)^{1 / 4} P_{m-1 / 2}^n(x) \, , </math>

Gil, Segura, & Temme (2000):  eq. (8)

where:    

<math>~\lambda \equiv x/\sqrt{x^2-1}</math>

This expression from Gil et al. (2000) means, for example, that by identifying <math>~x</math> with <math>~\coth\eta</math>, we have <math>~\lambda = \cosh\eta</math>, and,

<math>~Q_{n-1 / 2}^m (\cosh\eta)</math>

<math>~=</math>

<math>~(-1)^n \frac{\pi^{3/2}}{\sqrt{2} \Gamma(n-m+1 / 2)} (\coth^2\eta-1)^{1 / 4} P_{m-1 / 2}^n(\coth\eta) </math>

 

<math>~=</math>

<math>~ \frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{\cosh^2\eta}{\sinh^2\eta}-1 \biggr]^{1 / 4} P_{m-1 / 2}^n(\coth\eta) </math>

 

<math>~=</math>

<math>~ \frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{1}{\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta) \, . </math>

That is, we have,

Expression #4

<math>~Q_{n-1 / 2}^m (\cosh\eta)</math>

<math>~=</math>

<math>~ \frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl[\frac{\pi}{2\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta) \, , </math>

which matches the above expression #1 drawn from Cohl et al. (2000), but which appears not to match either of the other two "published" (online) formulae, expressions #2 or #3.

Specific Application

I stumbled into this dilemma when I tried to explicitly demonstrate how <math>~Q_{-1 / 2}(\cosh\eta)</math> can be derived from <math>~P_{-1 / 2}(z)</math> where, from §8.13 of M. Abramowitz & I. A. Stegun (1995), we find,

<math>~Q_{-1 / 2}(\cosh\eta)</math>

<math>~=</math>

<math>~2 e^{- \eta / 2} ~K(e^{-\eta} ) \, , </math>

Abramowitz & Stegun (1995), eq. (8.13.4)

and,

<math>~P_{-1 / 2}(z)</math>

<math>~=</math>

<math>~ \frac{2}{\pi} \biggl[\frac{2}{z+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-1}{z+1}} \biggr) \, . </math>

Abramowitz & Stegun (1995), eq. (8.13.1)

When I used the Whipple formula as defined in §14.19 of DLMF (expression #3 reprinted above), the function mapping gave me the wrong result; I was off by a factor of <math>~\Gamma(\tfrac{1}{2}) =\sqrt{\pi}</math>. But, as demonstrated below, the Whipple formula provided by Cohl et al. (2000) and by Gil et al. (2000) — that is, expressions #1 and #4, above — does give the correct result.

Demonstration that <math>~Q_{-\frac{1}{2}}</math> can be derived from <math>~P_{-\frac{1}{2}}</math>

Copying equation (34) from Cohl et al. (2000), we begin with,

<math>~Q^m_{n - 1 / 2}(\cosh\eta)</math>

<math>~=</math>

<math>~ \frac{(-1)^n \pi}{\Gamma(n - m + \tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P^n_{m - 1 / 2}(\coth\eta) \, ; </math>

then setting <math>~m = n = 0</math>, we have,

<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<math>~ \frac{\pi}{\Gamma(\tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta) </math>

 

<math>~=</math>

<math>~ \frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta) \, . </math>

Step #1:   Associate … <math>z \leftrightarrow \cosh\eta</math>. Then,

<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<math>~ \frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sqrt{z^2-1}} \biggr]^{1 / 2} P_{-\frac{1}{2}}\biggl(\frac{z}{\sqrt{z^2-1}} \biggr) \, . </math>

Step #2:   Now making the association … <math>\Lambda \leftrightarrow z/\sqrt{z^2-1}</math>, and drawing on eq. (8.13.1) from Abramowitz & Stegun (1995), we can write,

<math>~P_{-\frac{1}{2}}(\Lambda)</math>

<math>~=</math>

<math>~ \frac{2}{\pi} \biggl[\frac{2}{\Lambda+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\Lambda-1}{\Lambda+1}} \biggr) </math>

 

<math>~=</math>

<math>~ \frac{2}{\pi} \biggl[\frac{2\sqrt{z^2-1} }{z+\sqrt{z^2-1} }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-\sqrt{z^2-1} }{z+\sqrt{z^2-1} }} \biggr) \, . </math>

Step #3:   Again, making the association … <math>z \leftrightarrow \cosh\eta</math>, means,

<math>~P_{-\frac{1}{2}}(\Lambda)</math>

<math>~=</math>

<math>~ \frac{2}{\pi} \biggl[\frac{2\sinh\eta }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\eta-\sinh\eta }{\cosh\eta +\sinh\eta }} \biggr) </math>

<math>~\Rightarrow ~~~ Q_{-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<math>~ \frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sinh\eta} \biggr]^{1 / 2} \frac{2}{\pi} \biggl[\frac{2\sinh\eta }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\eta-\sinh\eta }{\cosh\eta +\sinh\eta }} \biggr) </math>

 

<math>~=</math>

<math>~ 2 \biggl[\frac{1 }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh^2\eta-\sinh^2\eta }{[\cosh\eta +\sinh\eta ]^2}} ~\biggr) </math>

 

<math>~=</math>

<math>~ 2 \biggl[\frac{1 }{e^\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{1 }{e^{2\eta}}} \biggr) </math>

 

<math>~=</math>

<math>~2 e^{-\eta/2} K(e^{-\eta}) \, . </math>

This, indeed, matches eq. (8.13.4) from Abramowitz & Stegun (1995).

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 — that is, the Whipple formula from §14.19 of DLMF,

<math>~\boldsymbol{Q}^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>

<math>~=</math>

<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>

What has been missing in my discussion is an appreciation of the following relationship between bold and plain-text function names,

<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>

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,

<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>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~ Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

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

The standard "Euler reflection formula for gamma functions" is usually presented in the form,

LSU Key.png

<math>~ \Gamma(z) ~\Gamma(1-z) </math>

<math>~=</math>

<math>~ \frac{\pi}{\sin(\pi z)} </math>

<math>~\biggl|</math>

for example, if
<math>~z \rightarrow (m-n + \tfrac{1}{2})</math>

<math>~\Rightarrow ~~~\Gamma(m-n+\tfrac{1}{2})~\Gamma(n-m+\tfrac{1}{2})</math>

<math>~=</math>

<math>~\pi \biggl\{\sin\biggl[ \frac{\pi}{2} + \pi(m-n) \biggr] \biggr\}^{-1}</math>

 

<math>~=</math>

<math>~\pi (-1)^{m-n} </math>

DLMF §5.5(ii)

<math>~\biggl|</math>
Valid for:

   <math>~z \ne0, \pm 1, \pm 2, </math> …

<math>~\biggl|</math>

If we make the association,

<math>~z \leftrightarrow (m - n + \tfrac{1}{2}) \, ,</math>

with <math>~m</math> and <math>~n</math> both being either zero or a positive integer, then, this Euler reflection formula becomes,

<math>~\Gamma(m - n + \tfrac{1}{2}) ~ \Gamma(n - m + \tfrac{1}{2} )</math>

<math>~=</math>

<math>~\pi \biggl\{ \sin\biggl[ \pi(m - n + \tfrac{1}{2}) \biggr] \biggr\}^{-1}</math>

 

<math>~=</math>

<math>~\pi (-1)^{m+n} \, .</math>

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

<math>~\frac{\pi (-1)^{m+n}}{\Gamma(n-m+\frac{1}{2}) }</math>

<math>~=</math>

<math>~\Gamma(m-n+\tfrac{1}{2}) \, .</math>

Hence, we may also write,

<math>~ Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

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,

Expression #5

<math>~Q^\mu_\nu(\cosh\eta)</math>

<math>~=</math>

<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>

Making the pair of substitutions,

<math>~\nu</math>

<math>~=</math>

<math>~n - \frac{1}{2} \, ,</math>

     

<math>~n ~~\in</math>

<math>~\mathbb{N}_0 = \{ 0, 1, 2, \cdots\} \, ,</math>

<math>~\mu</math>

<math>~=</math>

<math>~m \, ,</math>

     

<math>~m ~~\in</math>

<math>~\mathbb{N}_0 = \{ 0, 1, 2, \cdots\} \, ,</math>

we also have,

<math>~\nu + \mu +1</math>

<math>~=</math>

<math>~n - \frac{1}{2} + m + 1</math>

<math>~=</math>

<math>~n + m + \frac{1}{2} \, ,</math>

<math>~-\mu - \frac{1}{2}</math>

<math>~=</math>

<math>~-m-\frac{1}{2} \, ,</math>

 

 

<math>~-\nu - \frac{1}{2}</math>

<math>~=</math>

<math>~-\biggl(n - \frac{1}{2}\biggr)-\frac{1}{2} </math>

<math>~=</math>

<math>~-n \, , </math>

<math>~e^{i\mu\pi}</math>

<math>~=</math>

<math>~e^{i m \pi}</math>

<math>~=</math>

<math>~(-1)^{m} \, , </math>

in which case,

<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<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>


Now, since,

<math>~P^\mu_\nu(z)</math>

<math>~=</math>

<math>~P^\mu_{-\nu-1}(z) \, ,</math>

if we make the substitution,

<math>~-(\nu + 1)</math>

<math>~\rightarrow</math>

<math>~-(m+\tfrac{1}{2})</math>

    <math>~\Rightarrow</math>   

<math>~\nu</math>

<math>~\rightarrow</math>

<math>~m - \tfrac{1}{2} \, ,</math>

we also know that,

<math>~P^\mu_{m-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~P^\mu_{-m-\frac{1}{2}}(z) \, .</math>

Hence, we can write,

<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<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>


Finally, another relation states that, for <math>~n \in \mathbb{N}_0</math>,

<math>~P^{-n}_{m-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~\biggl[ \frac{\Gamma(m-n+\frac{1}{2})}{\Gamma(m+n+\frac{1}{2})} \biggr] P^n_{m-\frac{1}{2}}(z) \, .</math>

So, we obtain,

<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

This matches expressions #2 and #3, above.

Index Values of Zero

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

<math>~Q^0_{-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<math>~ \sqrt{\frac{\pi}{2}} ~\Gamma(\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, . </math>

 

<math>~=</math>

<math>~ \frac{\pi}{\sqrt{2}} ~ \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, . </math>

Joel's Additional Manipulations

From §14.19.6 of DLMF, we find the following summation expression:

<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>

Then, if we again employ the DLMF relationship between bold and plain-text function names, namely,

<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>

where we have made the substitution, <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, the Sums expression becomes,

<math>~e^{-\mu\pi i}\frac{Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+ \tfrac{1}{2} \right)}</math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>

<math>~=</math>

<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>

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

<math>~Q_{-\frac{1}{2}}\left(\cosh\xi\right)</math>

<math>~=</math>

<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>

But this can be rewritten in the form,

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

</math>

<math>~=</math>

<math>~ \biggl[ \dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr] </math>

See Also

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
|   H_Book Home   |   YouTube   |
Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS |
Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation