# User:Tohline/Appendix/Mathematics/ToroidalConfusion

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# 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 $~Q^m_{n - 1 / 2}(\cosh\eta)$ in terms of $~P^n_{m - 1 / 2}(\coth\eta)$.

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## Published Expressions

Expression #1

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

$~=$

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

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

Expression #2

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

$~=$

$~(-1)^n ~\Gamma(n-m + \tfrac{1}{2}) \biggl[ \frac{\pi}{2\sinh\alpha} \biggr]^{1 / 2} P^m_{n- 1 / 2}(\coth\alpha)\, ,$

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

Expression #3

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

$~=$

$~ \frac{\Gamma\left(m-n+ \tfrac{1}{2}\right)}{\Gamma\left(m+n+\tfrac{1}{2}\right)}\left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, .$

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:

 $~Q_{n-1 / 2}^m (\lambda)$ $~=$ $~(-1)^n \frac{\pi^{3/2}}{\sqrt{2}~ \Gamma(n-m+1 / 2)} (x^2-1)^{1 / 4} P_{m-1 / 2}^n(x) \, ,$ Gil, Segura, & Temme (2000):  eq. (8)
 where: $~\lambda \equiv x/\sqrt{x^2-1}$

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

 $~Q_{n-1 / 2}^m (\cosh\eta)$ $~=$ $~(-1)^n \frac{\pi^{3/2}}{\sqrt{2} \Gamma(n-m+1 / 2)} (\coth^2\eta-1)^{1 / 4} P_{m-1 / 2}^n(\coth\eta)$ $~=$ $~ \frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{\cosh^2\eta}{\sinh^2\eta}-1 \biggr]^{1 / 4} P_{m-1 / 2}^n(\coth\eta)$ $~=$ $~ \frac{(-1)^n ~\pi}{\Gamma(n-m+1 / 2)} \biggl( \frac{\pi}{2}\biggr)^{1 / 2} \biggl[\frac{1}{\sinh\eta}\biggr]^{1 / 2} P_{m-1 / 2}^n(\coth\eta) \, .$

That is, we have,

Expression #4

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

$~=$

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

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 $~Q_{-1 / 2}(\cosh\eta)$ can be derived from $~P_{-1 / 2}(z)$ where, from §8.13 of M. Abramowitz & I. A. Stegun (1995), we find,

 $~Q_{-1 / 2}(\cosh\eta)$ $~=$ $~2 e^{- \eta / 2} ~K(e^{-\eta} ) \, ,$ Abramowitz & Stegun (1995), eq. (8.13.4)

and,

 $~P_{-1 / 2}(z)$ $~=$ $~ \frac{2}{\pi} \biggl[\frac{2}{z+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-1}{z+1}} \biggr) \, .$ 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 $~\Gamma(\tfrac{1}{2}) =\sqrt{\pi}$. 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 $~Q_{-\frac{1}{2}}$ can be derived from $~P_{-\frac{1}{2}}$

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

 $~Q^m_{n - 1 / 2}(\cosh\eta)$ $~=$ $~ \frac{(-1)^n \pi}{\Gamma(n - m + \tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P^n_{m - 1 / 2}(\coth\eta) \, ;$

then setting $~m = n = 0$, we have,

 $~Q_{-\frac{1}{2}}(\cosh\eta)$ $~=$ $~ \frac{\pi}{\Gamma(\tfrac{1}{2})} \biggl[ \frac{\pi}{2\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta)$ $~=$ $~ \frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sinh\eta} \biggr]^{1 / 2} P_{-\frac{1}{2}}(\coth\eta) \, .$

Step #1:   Associate … $z \leftrightarrow \cosh\eta$. Then,

 $~Q_{-\frac{1}{2}}(\cosh\eta)$ $~=$ $~ \frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sqrt{z^2-1}} \biggr]^{1 / 2} P_{-\frac{1}{2}}\biggl(\frac{z}{\sqrt{z^2-1}} \biggr) \, .$

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

 $~P_{-\frac{1}{2}}(\Lambda)$ $~=$ $~ \frac{2}{\pi} \biggl[\frac{2}{\Lambda+1}\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\Lambda-1}{\Lambda+1}} \biggr)$ $~=$ $~ \frac{2}{\pi} \biggl[\frac{2\sqrt{z^2-1} }{z+\sqrt{z^2-1} }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{z-\sqrt{z^2-1} }{z+\sqrt{z^2-1} }} \biggr) \, .$

Step #3:   Again, making the association … $z \leftrightarrow \cosh\eta$, means,

 $~P_{-\frac{1}{2}}(\Lambda)$ $~=$ $~ \frac{2}{\pi} \biggl[\frac{2\sinh\eta }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\eta-\sinh\eta }{\cosh\eta +\sinh\eta }} \biggr)$ $~\Rightarrow ~~~ Q_{-\frac{1}{2}}(\cosh\eta)$ $~=$ $~ \frac{\pi}{\sqrt{2}} \biggl[ \frac{1}{\sinh\eta} \biggr]^{1 / 2} \frac{2}{\pi} \biggl[\frac{2\sinh\eta }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh\eta-\sinh\eta }{\cosh\eta +\sinh\eta }} \biggr)$ $~=$ $~ 2 \biggl[\frac{1 }{\cosh\eta +\sinh\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{\cosh^2\eta-\sinh^2\eta }{[\cosh\eta +\sinh\eta ]^2}} ~\biggr)$ $~=$ $~ 2 \biggl[\frac{1 }{e^\eta }\biggr]^{1 / 2} ~K\biggl( \sqrt{ \frac{1 }{e^{2\eta}}} \biggr)$ $~=$ $~2 e^{-\eta/2} K(e^{-\eta}) \, .$

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,

 $~\boldsymbol{Q}^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)$ $~=$ $~ \frac{\Gamma\left(m-n+ \tfrac{1}{2}\right)}{\Gamma\left(m+n+\tfrac{1}{2}\right)}\left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, .$

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

$\boldsymbol{Q}^{\mu}_{\nu}\left(x\right)=e^{-\mu\pi i}\frac{Q^{\mu}_{\nu}\left(x\right)}{\Gamma\left(\nu+\mu+1\right)}.$

After making the substitutions, $~\mu \rightarrow m$ and $~\nu \rightarrow (n-\tfrac{1}{2})$, the Whipple formula displayed above as expression #3 becomes,

 $~e^{-m\pi i}\frac{Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(n+m+\tfrac{1}{2}\right)}$ $~=$ $~ \frac{\Gamma\left(m-n+ \tfrac{1}{2}\right)}{\Gamma\left(m+n+\tfrac{1}{2}\right)}\left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right)$ $~\Rightarrow ~~~ Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)$ $~=$ $~e^{m\pi i} \Gamma\left(m-n+\tfrac{1}{2}\right)\left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right)$ $~=$ $~(-1)^m \Gamma\left(m-n+\tfrac{1}{2}\right)\left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, ,$

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,

$~ \Gamma(z) ~\Gamma(1-z)$

$~=$

$~ \frac{\pi}{\sin(\pi z)}$

$~\biggl|$

for example, if
$~z \rightarrow (m-n + \tfrac{1}{2})$

 $~\Rightarrow ~~~\Gamma(m-n+\tfrac{1}{2})~\Gamma(n-m+\tfrac{1}{2})$ $~=$ $~\pi \biggl\{\sin\biggl[ \frac{\pi}{2} + \pi(m-n) \biggr] \biggr\}^{-1}$ $~=$ $~\pi (-1)^{m-n}$
$~\biggl|$
Valid for:

$~z \ne0, \pm 1, \pm 2,$

$~\biggl|$

If we make the association,

$~z \leftrightarrow (m - n + \tfrac{1}{2}) \, ,$

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

 $~\Gamma(m - n + \tfrac{1}{2}) ~ \Gamma(n - m + \tfrac{1}{2} )$ $~=$ $~\pi \biggl\{ \sin\biggl[ \pi(m - n + \tfrac{1}{2}) \biggr] \biggr\}^{-1}$ $~=$ $~\pi (-1)^{m+n} \, .$

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

 $~\frac{\pi (-1)^{m+n}}{\Gamma(n-m+\frac{1}{2}) }$ $~=$ $~\Gamma(m-n+\tfrac{1}{2}) \, .$

Hence, we may also write,

 $~ Q^{m}_{n-\frac{1}{2}}\left(\cosh\xi\right)$ $~=$ $~(-1)^m \biggl[ \frac{\pi (-1)^{m+n}}{\Gamma(n-m+\frac{1}{2}) } \biggr] \left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right)$ $~=$ $~ \frac{(-1)^n \pi }{\Gamma(n-m+\frac{1}{2}) } \left(\frac{\pi}{2 \sinh\xi}\right)^{1 / 2}P^{n}_{m-\frac{1}{2}}\left(\coth\xi\right) \, ,$

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

$~Q^\mu_\nu(\cosh\eta)$

$~=$

$~ \sqrt{\frac{\pi}{2}} ~\Gamma(\nu + \mu + 1) ~e^{i\mu\pi} \biggl[ \frac{1}{\sinh^2\eta} \biggr]^{1 / 4} P^{-\nu-\frac{1}{2}}_{-\mu - \frac{1}{2}} (\coth\eta)$

Making the pair of substitutions,

 $~\nu$ $~=$ $~n - \frac{1}{2} \, ,$ $~n ~~\in$ $~\mathbb{N}_0 = \{ 0, 1, 2, \cdots\} \, ,$ $~\mu$ $~=$ $~m \, ,$ $~m ~~\in$ $~\mathbb{N}_0 = \{ 0, 1, 2, \cdots\} \, ,$

we also have,

 $~\nu + \mu +1$ $~=$ $~n - \frac{1}{2} + m + 1$ $~=$ $~n + m + \frac{1}{2} \, ,$ $~-\mu - \frac{1}{2}$ $~=$ $~-m-\frac{1}{2} \, ,$ $~-\nu - \frac{1}{2}$ $~=$ $~-\biggl(n - \frac{1}{2}\biggr)-\frac{1}{2}$ $~=$ $~-n \, ,$ $~e^{i\mu\pi}$ $~=$ $~e^{i m \pi}$ $~=$ $~(-1)^{m} \, ,$

in which case,

 $~Q^m_{n-\frac{1}{2}}(\cosh\eta)$ $~=$ $~ \sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{-m - \frac{1}{2}} (\coth\eta) \, .$

Now, since,

 $~P^\mu_\nu(z)$ $~=$ $~P^\mu_{-\nu-1}(z) \, ,$

if we make the substitution,

 $~-(\nu + 1)$ $~\rightarrow$ $~-(m+\tfrac{1}{2})$ $~\Rightarrow$ $~\nu$ $~\rightarrow$ $~m - \tfrac{1}{2} \, ,$

we also know that,

 $~P^\mu_{m-\frac{1}{2}}(z)$ $~=$ $~P^\mu_{-m-\frac{1}{2}}(z) \, .$

Hence, we can write,

 $~Q^m_{n-\frac{1}{2}}(\cosh\eta)$ $~=$ $~ \sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{m - \frac{1}{2}} (\coth\eta) \, .$

Finally, another relation states that, for $~n \in \mathbb{N}_0$,

 $~P^{-n}_{m-\frac{1}{2}}(z)$ $~=$ $~\biggl[ \frac{\Gamma(m-n+\frac{1}{2})}{\Gamma(m+n+\frac{1}{2})} \biggr] P^n_{m-\frac{1}{2}}(z) \, .$

So, we obtain,

 $~Q^m_{n-\frac{1}{2}}(\cosh\eta)$ $~=$ $~(-1)^m \sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] \biggl[ \frac{\Gamma(m-n+\frac{1}{2})}{\Gamma(m+n+\frac{1}{2})} \biggr]P^{n}_{m - \frac{1}{2}} (\coth\eta) \, .$ $~=$ $~(-1)^m \sqrt{\frac{\pi}{2}} ~\Gamma(m-n+\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{n}_{m - \frac{1}{2}} (\coth\eta) \, .$

This matches expressions #2 and #3, above.

## Index Values of Zero

Setting $~n = m = 0$ gives the following sought-for relationship:

 $~Q^0_{-\frac{1}{2}}(\cosh\eta)$ $~=$ $~ \sqrt{\frac{\pi}{2}} ~\Gamma(\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, .$ $~=$ $~ \frac{\pi}{\sqrt{2}} ~ \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{0}_{- \frac{1}{2}} (\coth\eta) \, .$

## Joel's Additional Manipulations

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

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

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

$\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(x\right) = e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(x\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \, ,$

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

 $~e^{-\mu\pi i}\frac{Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+ \tfrac{1}{2} \right)}$ $~=$ $~ \dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu }}{\left(\cosh\xi-\cos\phi\right)^{\mu+(1/2)}} - 2\sum_{n=1}^{\infty} \frac{\Gamma\left(\mu+n+\tfrac{1}{2}\right)}{\Gamma\left(\mu+\tfrac{1}{2} \right)} \biggl[ e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \biggr] \cos\left(n \phi\right)$ $~\Rightarrow ~~~Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)$ $~=$ $~ e^{\mu\pi i} \Gamma\left(\mu+ \tfrac{1}{2} \right) \biggl[ \dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu }}{\left(\cosh\xi-\cos\phi\right)^{\mu+(1/2)}}\biggr] - 2\sum_{n=1}^{\infty} Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n \phi\right) \, .$

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

 $~Q_{-\frac{1}{2}}\left(\cosh\xi\right)$ $~=$ $~ \biggl[ \dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr] - 2\sum_{n=1}^{\infty} Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n \phi\right) \, .$

But this can be rewritten in the form,

 $~ \sum_{n=0}^{\infty} \epsilon_n Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n \phi\right)$ $~=$ $~ \biggl[ \dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr]$

# See Also

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