Editing Appendix/Mathematics/ToroidalConfusion (section)

==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,

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<math>~Q_{-1 / 2}(\cosh\eta)</math>
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<math>~=</math>
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<math>~2 e^{- \eta / 2}
~K(e^{-\eta} ) \, ,
</math>
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  <td align="center" colspan="3">Abramowitz &amp; Stegun (1995), eq. (8.13.4)</td>
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and,

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<math>~P_{-1 / 2}(z)</math>
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<math>~=</math>
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<math>~
\frac{2}{\pi} \biggl[\frac{2}{z+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-1}{z+1}} \biggr) \, .
</math>
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  <td align="center" colspan="3">Abramowitz &amp; Stegun (1995), eq. (8.13.1)</td>
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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.

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Demonstration that <math>~Q_{-\frac{1}{2}}</math> can be derived from <math>~P_{-\frac{1}{2}}</math>
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Copying equation (34) from [http://adsabs.harvard.edu/abs/2000AN....321..363C Cohl et al. (2000)], we begin with,
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<math>~Q^m_{n - 1 / 2}(\cosh\eta)</math>
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<math>~=</math>
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<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>
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then setting <math>~m = n = 0</math>, we have,
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<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>
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<math>~=</math>
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<math>~
\frac{\pi}{\Gamma(\tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta) 
</math>
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&nbsp;
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<math>~=</math>
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<math>~
\frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta) \, .
</math>
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Step #1: &nbsp; Associate &hellip; <math>z \leftrightarrow \cosh\eta</math>.  Then,
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<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>
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<math>~=</math>
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<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>
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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,

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<math>~P_{-\frac{1}{2}}(\Lambda)</math>
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<math>~=</math>
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<math>~
\frac{2}{\pi} \biggl[\frac{2}{\Lambda+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\Lambda-1}{\Lambda+1}} \biggr) 
</math>
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&nbsp;
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<math>~=</math>
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<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>
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Step #3: &nbsp; Again, making the association &hellip; <math>z \leftrightarrow \cosh\eta</math>, means,
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<math>~P_{-\frac{1}{2}}(\Lambda)</math>
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<math>~=</math>
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<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>
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<math>~\Rightarrow ~~~ Q_{-\frac{1}{2}}(\cosh\eta)</math>
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<math>~=</math>
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<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>
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&nbsp;
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<math>~=</math>
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<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>
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&nbsp;
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<math>~=</math>
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<math>~
2 \biggl[\frac{1 }{e^\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{1 }{e^{2\eta}}} \biggr)
</math>
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&nbsp;
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<math>~=</math>
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<math>~2 e^{-\eta/2} K(e^{-\eta}) \, .
</math>
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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)].
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