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When using the hypergeometric representation for a Legendre polynomial, I encounter, for integer n and l, the following ratio: $$\frac{\Gamma(n-l)}{\Gamma(-l)}$$

Where $n \leq l$ (the quantity is definitely zero for $n > l$, as it should be in the definition of a Legendre polynomial). I am unsure as to how to evaluate this ratio; as it stands, it is indeterminate. My original idea was to use: $$\Gamma(k+1) = (k+1)\Gamma(k)$$

Multiple times to reduce $\Gamma(n-l)$ to: $$(n-l)(n-l-1)(...)(-l+1) \ \Gamma(-l)$$

Then the $\Gamma(-l)$ terms would cancel and I'd be left only with some sensible terms. Unfortunately, I do not think that this approach is valid, as it does not yield the correct representation for the Legendre polynomial. Secondly, the above can be written as: $$(-1)^n\frac{(l-1)!}{(l-n-1)!} \ \Gamma(-l)$$

Which now no longer permits us to set $n=l$ as is necessary to obtain a polynomial of order $l$. I'm trying to remain brief on the references to Legendre polynomials as it is specifically the ratio of the Gamma functions listed at the start of this post that I am interested in evaluating.

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  • $\begingroup$ I know your question is about gamma ratios, but you can use hypergeometric functions for the (associated) Legendre polynomials without taking limits if you work with the regularized Gauss function $_2\tilde{F}_1(a,b,c,x) = \frac{1}{\Gamma(c)}\;{_2}F_1(a,b,c,x)$ see functions.wolfram.com/HypergeometricFunctions/…. The formula for $P^m_n(z)$ functions.wolfram.com/Polynomials/LegendreP2/26/01/01/0002 reduces to $P^0_n(z)$ functions.wolfram.com/Polynomials/LegendreP/26/01/01/0001. $\endgroup$ – gammatester Sep 16 '16 at 12:32
  • $\begingroup$ In the link you have provided; the expression for $P^0_n$, when expanded out, is: $$P^0_l(x) = \frac{1}{\Gamma(l+1) \Gamma(-l)}\sum^{\infty}_{n=0}\frac{\Gamma(l+1+n) \Gamma(n-l)}{\Gamma (1-n)\ n!}(\frac{1-x}{2})^n$$ Which is identical to the result given in my comment - one still must still end up taking that tricky ratio $\frac{\Gamma(n-l)}{\Gamma(-l)}$ - As such, I can't see any way out without having to calculate that quantity. $\endgroup$ – CrossProduct Sep 16 '16 at 13:39
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I am not sure where you obtain your equation but the correct recursive formula for gamma function should be $$\Gamma(k+1)=k\Gamma(k)$$

If I proceed in a similar way as yours, then $\Gamma(n-l)$ can easily be reduced to $$\begin{aligned}\Gamma(n-l)&=(n-l-1)(n-l-2)\cdots (-l)\Gamma(-l)\\ &=(-1)^{n}\frac{l!}{(l-n)!}\Gamma(-l)\end{aligned}$$

The Legendre polynomial defined in terms of hypergeometric function can be written as $$\begin{aligned} P_l(z)&={}_2F_1(-l,l+1;1;\frac{1-z}{2})\\ &=\sum^{\infty}_{n=0}\frac{(-l)_n(l+1)_n}{(1)_n}\frac{1}{n!}\left(\frac{1-z}{2}\right)^2 \end{aligned}$$ Using $(a)_n\equiv \frac{\Gamma(a+n)}{\Gamma(a)}$, $$\begin{aligned} P_l(z)&=\sum^{\infty}_{n=0}\frac{\Gamma(n-l)}{\Gamma(-l)}\frac{\Gamma(l+n+1)}{\Gamma(l+1)\Gamma(n+1)n!}\left(\frac{1-z}{2}\right)^2\\ &=\sum^{\infty}_{n=0}\frac{(-1)^{n}l!}{n!(l-n)!}\frac{\Gamma(l+n+1)}{\Gamma(l+1)\Gamma(n+1)}\left(\frac{1-z}{2}\right)^2 \end{aligned}$$ (Btw, I believe the $\Gamma(1-n)$ in the denominator inside the summation series of the your comment is a mistake) The binomial coefficient for complex argument is related to gamma function via $$ \begin{pmatrix}x \\y\end{pmatrix}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)} $$ and using the identity $$ \begin{pmatrix}x \\y\end{pmatrix}=(-1)^y\begin{pmatrix}y-x-1 \\y\end{pmatrix} $$, we can establish the relation $$\begin{aligned} \begin{pmatrix}-l-1 \\n\end{pmatrix}&=(-1)^n\begin{pmatrix}l+n \\n\end{pmatrix}\\ &=(-1)^n\frac{\Gamma(l+n+1)}{\Gamma(n+1)\Gamma(l+1)} \end{aligned}$$ Finally, we can substitute the above relation into the Legendre polynomial equation to obtain $$P_l(z)=\sum^{l}_{n=0}\begin{pmatrix}l \\n\end{pmatrix}\begin{pmatrix}-l-1 \\n\end{pmatrix}\left(\frac{1-z}{2}\right)^2$$ where the series terminates after $n=l$. This is exactly the other form of Legendre polynomial expression, therefore the ratio I showed here $$\frac{\Gamma(n-l)}{\Gamma(-l)}=(-1)^{n}\frac{l!}{(l-n)!}$$ is consistent with the formulas even if the arguments are negative integers.

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We have for $k\in\mathbf{N}$ and $x\rightarrow 0$ $$\Gamma(-k+x) \sim \frac{1}{k!x} + O(1)$$ and therefore $$\lim_{x\rightarrow 0}\frac{\Gamma(n-l+x)}{\Gamma(-l+x)} = \frac{1}{(l-n)!}/\frac{1}{l!} = \frac{l!}{(l-n)!}$$ Of course this is only a very special limit. It is finite for $n=l$ but I do not know what you expect for the value. If the above does reproduce it, you will at least have a simple arugument.

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  • $\begingroup$ Thanks very much for your response. I made a small error in my original working which I have now updated, but it does not affect the answer you have given (just what I got for the limit was wrong). I am, for $\frac{\Gamma(n-l)}{\Gamma(-l)}$, obtaining: $(-1)^n\frac{(l-1)!}{(l-n-1)!} \ \Gamma(-l)$ - This is not exactly identical to what you get but it does actually yield the same answer for $n=l$ (zero). My problem is that this does indeed kill off the $x^l$ term in the expression, which should not be the case for a Legendre polynomial. $\endgroup$ – CrossProduct Sep 16 '16 at 12:02
  • $\begingroup$ For completeness, here is the actual original expression. You can see where the $\frac{\Gamma(n-l)}{\Gamma(-l)}$ term comes from: $$P^0_l(x) = \frac{1}{\Gamma(-l) \Gamma(l+1)}\sum^{\infty}_{n=0}\frac{\Gamma(l+1+n) \Gamma(n-l)}{(n!)^2}(\frac{1-x}{2})^n$$ This expression can be found in many places in the literature, e.g: dlmf.nist.gov/14.3 - eqn. 14.3.6 (set $m=0$, and note that $l=\nu$, $m=\mu$ on this page). If this expression is indeed to hold for all $l$, $m$, then one should obtain a polynomial of degree $l$ for the case of integer $l$ (i.e the $x^{l+1}$ term onwards is killed). $\endgroup$ – CrossProduct Sep 16 '16 at 12:06

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