Editing Appendix/Mathematics/ToroidalFunctions (section)

=Our Mucking Around=
Begins on p. 332 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)]

==Recurrence Relations==

According to [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)], both <math>~P_\nu^\mu</math> and <math>~Q_\nu^\mu</math> satisfy the same recurrence relations.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_\nu^{\mu+1}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1)^{- 1 / 2} \biggl[ (\nu - \mu)z P_\nu^\mu(z) - (\nu + \mu)P_{\nu - 1}^\mu(z) \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(z^2-1) \frac{dP_\nu^{\mu}(z)}{dz}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\nu + \mu)(\nu - \mu +1_(z^2-1)^{1 / 2} P_\nu^{\mu - 1}(z) - \mu zP_\nu^\mu(z) \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(\nu - \mu + 1)P_{\nu + 1}^{\mu}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\nu+1)zP_\nu^\mu (z) -(\nu + \mu)P_{\nu-1}^\mu(z) \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(z^2-1) \frac{dP_\nu^{\mu}(z)}{dz}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\nu z P_\nu^{\mu }(z) - (\nu + \mu)P_{\nu-1}^\mu(z) \, .
</math>
  </td>
</tr>
</table>
</div>

<table border="1" cellpadding="5" align="center" width="90%"><tr><td align="left">
According to equation (14) of [https://www.jstor.org/stable/2369515?seq=1#page_scan_tab_contents A. B. Basset (1893, American Journal of Mathematics, vol. 15, No. 4, pp. 287 - 302)],
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~
(n-m+\tfrac{1}{2})P_{n+1}^m(\nu)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n\nu P_n^m(\nu) 
+ \biggl[ \frac{m^2}{n-m-\tfrac{1}{2}} \biggr]P_{n-1}^m(\nu)
- \frac{(n-\tfrac{1}{2})^2}{(n-m-\tfrac{1}{2})}P_{n-1}^m(\nu)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n\nu P_n^m(\nu) 
+ \biggl[ \frac{m^2 - (n-\tfrac{1}{2})^2}{n-m-\tfrac{1}{2}} \biggr]P_{n-1}^m(\nu)
</math>
  </td>
</tr>
</table>

After replacing, <math>~n</math>, with <math>~(n + \tfrac{1}{2})</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~
(n-m+1)P_{n+3 / 2}^m(\nu)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2n+1)\nu P_{n+1 / 2}^m(\nu) 
+ \biggl[ \frac{m^2 - n^2}{n-m} \biggr]P_{n-1 / 2}^m(\nu)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2n+1)\nu P_{n+1 / 2}^m(\nu) 
- (n+m)P_{n-1 / 2}^m(\nu) \, .
</math>
  </td>
</tr>
</table>

The ''coefficients'' of this last expression precisely match the coefficients in the above expression provided by [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)], but the ''subscript'' notation is off by <math>~\tfrac{1}{2}</math>.  This inconsistency most likely should be blamed on the notation adopted by Basset (1893).  At the top of his p. 289 &#8212; which is a couple of pages before his equation (14) &#8212; Basset says: &nbsp; <font color="#009999">A toroidal function is an associated function of degree <math>~n - \tfrac{1}{2}</math> and order <math>~m</math>; and the notation which ought in strictness to be adopted for the two kinds of toroidal functions is <math>~P_{n-1 / 2}^m</math> and <math>~Q_{n-1 / 2}^m</math>; but as these functions rarely if ever occur in an investigation which also involves associated functions of integral degree <math>~n</math>, it will be generally sufficient to employ the suffix <math>~n</math> instead of <math>~n - \tfrac{1}{2}</math></font>.  Thus, we probably should have shifted the ''subscript'' notation in his equation (14) by "-&frac12;" before incorporating our additional replacement everywhere of <math>~n</math> by <math>~(n + \tfrac{1}{2})</math>.
<!--
&#8212; after replacing, <math>~n</math>, with <math>~(n - \tfrac{1}{2})</math>,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~[n-m]P_{n+1 / 2}^m(\nu)
-(2n-1)\nu P_{n-1 / 2}^m(\nu)
+\frac{ [n-1]^2 }{ [n-m-1] }P_{n-3 / 2}^m(\nu)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{m^2}{n-m-1} \biggr]P_{n-3 / 2}^m(\nu)</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~[n-m]P_{n+1 / 2}^m(\nu)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2n-1)\nu P_{n-1 / 2}^m(\nu)
+ \biggl[ \frac{m^2 -(n-1)^2}{n-m-1} \biggr]P_{n-3 / 2}^m(\nu)
</math>
  </td>
</tr>
</table>
-->


----


If we set <math>~\mu = 0</math> in the [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)] recurrence relation, then replace <math>~\nu</math> everywhere with <math>~\nu - \tfrac{1}{2}</math>, we obtain,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\nu + 1)P_{\nu + 1}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\nu+1)zP_\nu (z) -(\nu )P_{\nu-1}(z) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\nu \rightarrow \nu - \tfrac{1}{2} ~~~\Rightarrow ~~~ (\nu + \tfrac{1}{2})P_{\nu + 1 / 2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\nu)zP_{\nu - 1 / 2} (z) -(\nu - \tfrac{1}{2})P_{\nu-3 / 2}(z) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
Mult. thru by 2 &nbsp; &nbsp; &nbsp;<math>~~~\Rightarrow ~~~ (2\nu + 1)P_{\nu + 1 / 2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4\nu)zP_{\nu - 1 / 2} (z) -(2\nu - 1)P_{\nu-3 / 2}(z) 
</math>
  </td>
</tr>
</table>

Independently, from equation (56) of [https://archive.org/stream/atreatiseonhydr02bassgoog#page/n44/mode/2up Basset's (1888, Cambridge: Beighton, Bell and Co.) ''A Treatise on Hydrodynamics''], we have,

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

<tr>
  <td align="right">
<math>~(2n+1)P_{n+1}(\nu) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4nCP_n(\nu) - (2n-1)P_{n-1}(\nu) \, .
</math>
  </td>
</tr>
</table>
This matches the Abramowitz &amp; Stegun expression if, as before, we employ the mapping, <math>~n \rightarrow n-\tfrac{1}{2}</math>, ''in the subscripts only''; also, note that, due to what must have been a typesetting error, the coefficient, <math>~C</math>, in Basset's expression must be replaced by the independent variable, <math>~\nu</math>.

From equations (57) - (60) of [https://archive.org/stream/atreatiseonhydr02bassgoog#page/n44/mode/2up Basset's (1888) ''Hydrodynamics''], we also obtain,

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

<tr>
  <td align="right">
<math>~P_{-1 / 2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \sqrt{k} ~F \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P_{+1 / 2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{ \sqrt{k}}~ E \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q_{-1 / 2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \sqrt{k} ~F \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q_{+1 / 2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{ \sqrt{k}}~[ F - E] \, ;
</math>
  </td>
</tr>
</table>
where,

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

<tr>
  <td align="right">
<math>~k^2 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
e^{-2\eta} \, ,
</math>
  </td>
  <td align="center">&nbsp; &nbsp; and &nbsp; &nbsp;</td>
  <td align="right">
<math>~(k^')^2 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
1 - e^{-2\eta} \, .
</math>
  </td>
</tr>
</table>

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

==Toroidal Functions==
Relationship between one another, 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)]:

{{ Math/EQ_Toroidal02 }}

Note that the relationship between <math>~\lambda</math> and <math>~x</math> is the same as the relationship between <math>~\cosh\alpha</math> and <math>~\coth\alpha</math>, that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\coth\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pm \cosh\alpha}{ \sqrt{\cosh^2\alpha - 1}} \, ;</math>
  </td>
<td align="center">&nbsp; &nbsp; or &nbsp; &nbsp; </td>
  <td align="right">
<math>~\cosh\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\sqrt{1-\tanh^2\alpha}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pm \coth\alpha}{ \sqrt{\coth^2\alpha - 1 }} \, .</math>
  </td>
</tr>
</table>
</div>


Relation to Elliptic Integrals

===PminusHalf01===

{{ Math/EQ_PminusHalf01 }}

<table border="1" cellpadding="5" align="center" width="80%">
<tr>
  <td align="center">
Proof that these are the same expressions:
  </td>
</tr>
<tr>
  <td align="left">
From standard relationships between hyperbolic functions, we know that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\cosh u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 1 - \tanh^2u  \biggr]^{1 / 2}
</math>
  </td>
</tr>
</table>
So, if we let <math>~u \equiv \eta/2</math> and make the association,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tanh u</math>
  </td>
  <td align="center">
<math>~~~\leftrightarrow~~~</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{z-1}{z+1}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\cosh u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 - \frac{z-1}{z+1} \biggr]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2}{z+1} \biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
Also,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cosh \eta = \cosh(2u) = 2\cosh^2 u - 1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\biggl[\frac{z+1}{2}\biggr] - 1 = z \, .</math>
  </td>
</tr>
</table>
Q.E.D.
  </td>
</tr>
</table>


===QminusHalf01===

{{ Math/EQ_QminusHalf01 }}

<div align="center">
<table border="1" cellpadding="5" align="center" width="80%">
<tr>
  <td align="center">
Proof that these are the same expressions:
  </td>
</tr>
<tr>
  <td align="left">
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">

<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>
then setting <math>~m = n = 0</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{Q}^{0}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\Gamma\left(\tfrac{1}{2}\right)}{\Gamma\left(\tfrac{1}{2}\right)}\left(\frac{\pi}{2\sinh\xi}\right)^{1 / 2}P^{0}_{-\frac{1}{2}}\left(\coth\xi\right) \, .
</math>
  </td>
</tr>
</table>
Step #1: &nbsp; Associate &hellip; <math>z \leftrightarrow \cosh\xi</math>.  Then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{Q}^{0}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\pi}{2} \biggr)^{1/2}
\left[\frac{1}{\sqrt{z^2-1}}\right]^{1 / 2}
P^{0}_{-\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>, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_{-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\xi</math>, means,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \boldsymbol{Q}^{0}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2\sinh\xi} \biggr]^{ 1 / 2}
\frac{2}{\pi} \biggl[\frac{2\sinh\xi }{\cosh\xi+\sinh\xi }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\xi-\sinh\xi }{\cosh\xi+\sinh\xi }} \biggr) 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\sqrt{\pi}} ~e^{-\xi/2} ~K( e^{-\xi}) \, ,
</math>
  </td>
</tr>
</table>
which, apart from the leading factor of <math>~\pi^{-1 / 2}</math>, exactly matches the above expression.

----

Note:  From [http://hcohl.sdf.org/WHIPPLE.html Howard Cohl's online overview] &#8212; see, also, [[#Overview_by_Howard_Cohl|below]] &#8212; we find that the Whipple formula is slightly different from the one (quoted above) drawn from DLMF.  According to Cohl the Whipple formula should be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{n- 1 / 2}^m(\cosh\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-1)^m \Gamma (m - n + \tfrac{1}{2} )\biggl( \frac{\pi}{2\sinh\alpha} \biggr)^{1 / 2} P^{n}_{m - 1 / 2}(\coth\alpha) \, .
</math>
  </td>
</tr>
</table>
The DLMF expression needs to be multiplied by <math>~(-1)^m\Gamma (m + n + \tfrac{1}{2} )</math> in order to match the expression provided by Cohl; for the case being considered here of <math>~m=n=0</math>,  this factor is precisely <math>~\Gamma(\tfrac{1}{2}) = \sqrt{\pi}</math> &#8212; [https://en.wikipedia.org/wiki/Gamma_function#Properties see, for example, Wikipedia's discussion of the gamma function] &#8212; which cancels this confusing factor of <math>~\pi^{-1 / 2}</math>.
  </td>
</tr>
</table>
</div>


===PplusHalf01===

{{ Math/EQ_PplusHalf01 }}

<div align="center">
<table border="1" cellpadding="5" align="center" width="80%">
<tr>
  <td align="center">
Proof that these are the same expressions:
  </td>
</tr>
<tr>
  <td align="left">
If we associate,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~e^\eta</math>
  </td>
  <td align="center">
<math>~~~\leftrightarrow~~~</math>
  </td>
  <td align="left">
<math>~
z + \sqrt{z^2-1}
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\sqrt{z^2-1}}{[z + \sqrt{z^2-1}]} \, .
</math>
  </td>
</tr>
</table>

It also means that,
<table border="0" cellpadding="5" align="center">

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
z \, .
</math>
  </td>
</tr>
</table>
Q.E.D.
  </td>
</tr>
</table>
</div>


===QplusHalf01===

{{ Math/EQ_QplusHalf01 }}


===Other===
When the argument, <math>~x</math>, lies in the range, <math>~-1 < x < 1</math>:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_{-1 / 2}(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} ~K\biggl( \sqrt{ \frac{1-x}{2} } \biggr) \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P_{-1 / 2}(\cos\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} ~K\biggl( \sin \frac{\theta}{2}\biggr) \, ;
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~P_{+1 / 2}(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi} \biggl[2E\biggl( \sqrt{ \frac{1-x}{2} } \biggr) - ~K\biggl( \sqrt{ \frac{1-x}{2} } \biggr) \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q_{+ 1 / 2}(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
K\biggl( \sqrt{ \frac{1+x}{2} } \biggr) - 2E\biggl( \sqrt{ \frac{1+x}{2} } \biggr)\, ;
</math>
  </td>
</tr>
</table>
</div>

==Piece Together==

When <math>~\mu = 0</math>, and <math>~\nu = (m- 3/ 2)</math>, the recurrence relation should be &hellip;

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

<tr>
  <td align="right">
<math>~(m - \tfrac{1}{2})P_{m-1 / 2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[2m-2]zP_{m-3 / 2} (z) -(m - \tfrac{3}{2} )P_{m - 5 / 2} (z) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~(2m -1)P_{m - 1 / 2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4(m-1)zP_{m - 3 /2 } (z) - (2m -3)P_{m-5 / 2} (z) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~P_{m - 1 / 2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{
4(m-1)zP_{m - 3 /2 } (z) - (2m -3)P_{m-5 / 2} (z) }{(2m -1)} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
for all <math>~m \ge 2</math>.