Appendix/Mathematics/ToroidalConfusion

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 $Q_{n-1/2}^m(\cosh \eta )$ in terms of $P_{m-1/2}^n(\coth \eta )$.

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
$Q_{n-1/2}^m(\cosh \eta )$	$=$	$\frac{(-1{)}^n\pi }{\mathrm{\Gamma }(n-m+\frac{1}{2})}[\frac{\pi }{2\sinh \eta }{]}^{1/2}{P}_{m-1/2}^n(\coth \eta ).$

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

Expression #2
$Q_{m-1/2}^n(\cosh \alpha )$	$=$	$(-1{)}^n\mathrm{\Gamma }(n-m+\frac{1}{2})[\frac{\pi }{2\sinh \alpha }{]}^{1/2}{P}_{n-1/2}^m(\coth \alpha ),$

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

Expression #3
${\mathit{Q}}_{n-\frac{1}{2}}^m(\cosh \xi )$	$=$	$\frac{\mathrm{\Gamma }(m-n+\frac{1}{2})}{\mathrm{\Gamma }(m+n+\frac{1}{2})}{\left(\frac{\pi }{2\sinh \xi }\right)}^{1/2}{P}_{m-\frac{1}{2}}^n(\coth \xi ).$

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:

	$Q_{n-1/2}^m(\lambda )$	$=$	$(-1{)}^n\frac{{\pi }^{3/2}}{\sqrt{2}\mathrm{\Gamma }(n-m+1/2)}(x^2-1{)}^{1/4}{P}_{m-1/2}^n(x),$
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,

$Q_{n-1/2}^m(\cosh \eta )$	$=$	$(-1{)}^n\frac{{\pi }^{3/2}}{\sqrt{2}\mathrm{\Gamma }(n-m+1/2)}({\coth }^2\eta -1{)}^{1/4}{P}_{m-1/2}^n(\coth \eta )$
	$=$	$\frac{(-1{)}^n\pi }{\mathrm{\Gamma }(n-m+1/2)}\left(\frac{\pi }{2}{)}^{1/2}\right[\frac{{\cosh }^2\eta }{{\sinh }^2\eta }-1{]}^{1/4}{P}_{m-1/2}^n(\coth \eta )$
	$=$	$\frac{(-1{)}^n\pi }{\mathrm{\Gamma }(n-m+1/2)}\left(\frac{\pi }{2}{)}^{1/2}\right[\frac{1}{\sinh \eta }{]}^{1/2}{P}_{m-1/2}^n(\coth \eta ).$

That is, we have,

Expression #4
$Q_{n-1/2}^m(\cosh \eta )$	$=$	$\frac{(-1{)}^n\pi }{\mathrm{\Gamma }(n-m+1/2)}[\frac{\pi }{2\sinh \eta }{]}^{1/2}{P}_{m-1/2}^n(\coth \eta ),$

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,

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

and,

$P_{-1/2}(z)$	$=$	$\frac{2}{\pi }\left[\frac{2}{z+1}{]}^{1/2}K\right(\sqrt{\frac{z-1}{z+1}}).$
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 $\mathrm{\Gamma }(\frac{1}{2})=\sqrt{\pi }$. 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, $Q_{n-1/2}^m(\cosh \eta )$ $=$ $\frac{(-1{)}^n\pi }{\mathrm{\Gamma }(n-m+\frac{1}{2})}[\frac{\pi }{2\sinh \eta }{]}^{1/2}{P}_{m-1/2}^n(\coth \eta );$ then setting $m=n=0$, we have, $Q_{-\frac{1}{2}}(\cosh \eta )$ $=$ $\frac{\pi }{\mathrm{\Gamma }(\frac{1}{2})}[\frac{\pi }{2\sinh \eta }{]}^{1/2}{P}_{-\frac{1}{2}}(\coth \eta )$   $=$ $\frac{\pi }{\sqrt{2}}[\frac{1}{\sinh \eta }{]}^{1/2}{P}_{-\frac{1}{2}}(\coth \eta ).$ Step #1:   Associate … $z\leftrightarrow \cosh \eta$. Then, $Q_{-\frac{1}{2}}(\cosh \eta )$ $=$ $\frac{\pi }{\sqrt{2}}\left[\frac{1}{\sqrt{z^2-1}}{]}^{1/2}{P}_{-\frac{1}{2}}\right(\frac{z}{\sqrt{z^2-1}}).$ Step #2:   Now making the association … $\mathrm{\Lambda }\leftrightarrow z/\sqrt{z^2-1}$, and drawing on eq. (8.13.1) from Abramowitz & Stegun (1995), we can write, $P_{-\frac{1}{2}}(\mathrm{\Lambda })$ $=$ $\frac{2}{\pi }\left[\frac{2}{\mathrm{\Lambda }+1}{]}^{1/2}K\right(\sqrt{\frac{\mathrm{\Lambda }-1}{\mathrm{\Lambda }+1}})$   $=$ $\frac{2}{\pi }\left[\frac{2\sqrt{z^2-1}}{z+\sqrt{z^2-1}}{]}^{1/2}K\right(\sqrt{\frac{z-\sqrt{z^2-1}}{z+\sqrt{z^2-1}}}).$ Step #3:   Again, making the association … $z\leftrightarrow \cosh \eta$, means, $P_{-\frac{1}{2}}(\mathrm{\Lambda })$ $=$ $\frac{2}{\pi }\left[\frac{2\sinh \eta }{\cosh \eta +\sinh \eta }{]}^{1/2}K\right(\sqrt{\frac{\cosh \eta -\sinh \eta }{\cosh \eta +\sinh \eta }})$ $\Rightarrow Q_{-\frac{1}{2}}(\cosh \eta )$ $=$ $\frac{\pi }{\sqrt{2}}\left[\frac{1}{\sinh \eta }{]}^{1/2}\frac{2}{\pi }\right[\frac{2\sinh \eta }{\cosh \eta +\sinh \eta }{]}^{1/2}K\left(\sqrt{\frac{\cosh \eta -\sinh \eta }{\cosh \eta +\sinh \eta }}\right)$   $=$ $2\left[\frac{1}{\cosh \eta +\sinh \eta }{]}^{1/2}K\right(\sqrt{\frac{{\cosh }^2\eta -{\sinh }^2\eta }{[\cosh \eta +\sinh \eta {]}^2}})$   $=$ $2\left[\frac{1}{e^{\eta }}{]}^{1/2}K\right(\sqrt{\frac{1}{e^{2\eta }}})$   $=$ $2e^{-\eta /2}K(e^{-\eta }).$ 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,

${\mathit{Q}}_{n-\frac{1}{2}}^m(\cosh \xi )$

$=$

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

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

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

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

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

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, $\mathrm{\Gamma }(z)\mathrm{\Gamma }(1-z)$ $=$ $\frac{\pi }{\sin (\pi z)}$ $|$ for example, if $z\to (m-n+\frac{1}{2})$ $\Rightarrow \mathrm{\Gamma }(m-n+\frac{1}{2})\mathrm{\Gamma }(n-m+\frac{1}{2})$ $=$ $\pi \{\sin [\frac{\pi }{2}+\pi (m-n)]{\}}^{-1}$   $=$ $\pi (-1{)}^{m-n}$ DLMF §5.5(ii) $|$ Valid for:    $z\ne 0,\pm 1,\pm 2,$ … $|$ If we make the association, $z\leftrightarrow (m-n+\frac{1}{2}),$ with $m$ and $n$ both being either zero or a positive integer, then, this Euler reflection formula becomes, $\mathrm{\Gamma }(m-n+\frac{1}{2})\mathrm{\Gamma }(n-m+\frac{1}{2})$ $=$ $\pi \{\sin [\pi (m-n+\frac{1}{2})]{\}}^{-1}$   $=$ $\pi (-1{)}^{m+n}.$

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

$\frac{\pi (-1{)}^{m+n}}{\mathrm{\Gamma }(n-m+\frac{1}{2})}$

$=$

$\mathrm{\Gamma }(m-n+\frac{1}{2}).$

Hence, we may also write,

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

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

Making the pair of substitutions,

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

we also have,

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

in which case,

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

$=$

$\sqrt{\frac{\pi }{2}}\mathrm{\Gamma }(n+m+\frac{1}{2})(-1{)}^m[\frac{1}{\sqrt{\sinh \eta }}]P_{-m-\frac{1}{2}}^{-n}(\coth \eta ).$

---

Now, since,

$P_{\nu }^{\mu }(z)$

$=$

$P_{-\nu -1}^{\mu }(z),$

if we make the substitution,

$-(\nu +1)$

$\to$

$-(m+\frac{1}{2})$

$\Rightarrow$

$\nu$

$\to$

$m-\frac{1}{2},$

we also know that,

$P_{m-\frac{1}{2}}^{\mu }(z)$

$=$

$P_{-m-\frac{1}{2}}^{\mu }(z).$

Hence, we can write,

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

$=$

$\sqrt{\frac{\pi }{2}}\mathrm{\Gamma }(n+m+\frac{1}{2})(-1{)}^m[\frac{1}{\sqrt{\sinh \eta }}]P_{m-\frac{1}{2}}^{-n}(\coth \eta ).$

---

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

$P_{m-\frac{1}{2}}^{-n}(z)$

$=$

$\left[\frac{\mathrm{\Gamma }(m-n+\frac{1}{2})}{\mathrm{\Gamma }(m+n+\frac{1}{2})}\right]P_{m-\frac{1}{2}}^n(z).$

So, we obtain,

$Q_{n-\frac{1}{2}}^m(\cosh \eta )$	$=$	$(-1{)}^m\sqrt{\frac{\pi }{2}}\mathrm{\Gamma }(n+m+\frac{1}{2})[\frac{1}{\sqrt{\sinh \eta }}\left]\right[\frac{\mathrm{\Gamma }(m-n+\frac{1}{2})}{\mathrm{\Gamma }(m+n+\frac{1}{2})}]P_{m-\frac{1}{2}}^n(\coth \eta ).$
	$=$	$(-1{)}^m\sqrt{\frac{\pi }{2}}\mathrm{\Gamma }(m-n+\frac{1}{2})[\frac{1}{\sqrt{\sinh \eta }}]P_{m-\frac{1}{2}}^n(\coth \eta ).$

This matches expressions #2 and #3, above.

Index Values of Zero

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

$Q_{-\frac{1}{2}}^0(\cosh \eta )$	$=$	$\sqrt{\frac{\pi }{2}}\mathrm{\Gamma }(\frac{1}{2})\left[\frac{1}{\sqrt{\sinh \eta }}\right]P_{-\frac{1}{2}}^0(\coth \eta ).$
	$=$	$\frac{\pi }{\sqrt{2}}\left[\frac{1}{\sqrt{\sinh \eta }}\right]P_{-\frac{1}{2}}^0(\coth \eta ).$

Joel's Additional Manipulations

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

${\mathit{Q}}_{-\frac{1}{2}}^{\mu }(\cosh \xi )+2\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(\mu +n+\frac{1}{2})}{\mathrm{\Gamma }(\mu +\frac{1}{2})}{\mathit{Q}}_{n-\frac{1}{2}}^{\mu }(\cosh \xi )\cos \left(n\varphi \right)=\frac{{\left(\frac{1}{2}\pi \right)}^{1/2}{(\sinh \xi )}^{\mu }}{{(\cosh \xi -\cos \varphi )}^{\mu +(1/2)}}$

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

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

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

$e^{-\mu \pi i}\frac{Q_{-\frac{1}{2}}^{\mu }(\cosh \xi )}{\mathrm{\Gamma }(\mu +\frac{1}{2})}$	$=$	$\frac{{\left(\frac{1}{2}\pi \right)}^{1/2}{(\sinh \xi )}^{\mu }}{{(\cosh \xi -\cos \varphi )}^{\mu +(1/2)}}-2\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(\mu +n+\frac{1}{2})}{\mathrm{\Gamma }(\mu +\frac{1}{2})}\left[e^{-\mu \pi i}\frac{Q_{n-\frac{1}{2}}^{\mu }(\cosh \xi )}{\mathrm{\Gamma }(\mu +n+\frac{1}{2})}\right]\cos \left(n\varphi \right)$
$\Rightarrow Q_{-\frac{1}{2}}^{\mu }(\cosh \xi )$	$=$	$e^{\mu \pi i}\mathrm{\Gamma }(\mu +\frac{1}{2})\left[\frac{{\left(\frac{1}{2}\pi \right)}^{1/2}{(\sinh \xi )}^{\mu }}{{(\cosh \xi -\cos \varphi )}^{\mu +(1/2)}}\right]-2\sum _{n=1}^{\mathrm{\infty }}{Q}_{n-\frac{1}{2}}^{\mu }(\cosh \xi )\cos \left(n\varphi \right).$

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

$Q_{-\frac{1}{2}}(\cosh \xi )$

$=$

$\left[\frac{\pi /\sqrt{2}}{{(\cosh \xi -\cos \varphi )}^{\frac{1}{2}}}\right]-2\sum _{n=1}^{\mathrm{\infty }}{Q}_{n-\frac{1}{2}}(\cosh \xi )\cos \left(n\varphi \right).$

But this can be rewritten in the form,

$\sum _{n=0}^{\mathrm{\infty }}{ϵ}_n{Q}_{n-\frac{1}{2}}(\cosh \xi )\cos \left(n\varphi \right)$

$=$

$\left[\frac{\pi /\sqrt{2}}{{(\cosh \xi -\cos \varphi )}^{\frac{1}{2}}}\right]$

See Also