Calculation for negative integer order Associated Legendre Function

Question

I am currently engaging with the following hypergeometric function as a result of attempting to find a solution for this probability problem for $n$ number of dice:

$$_2F_1\left (\frac{n+k}{2}, \frac{n+k+1}{2};k+1;\frac{4b(d-e+1)}{d^2}\right ); \{n,k,b,d,e\}\in \mathbb{N}^+, z\in \mathbb{R},b < e \leq d$$

I'm trying to determine if there's way to calculate this function that doesn't require infinite sums (ideally, just with elementary functions and finite sums/products).

I first note that this function is of the pattern: $$_2F_1\left (a, a+\frac{1}{2};c;z\right )$$ DLMF 15.9.17 then states that:

$$_2\bar{F}_1\left (a, a+\frac{1}{2}; c; z \right )=2^{c-1}z^\frac{1-c}{2}(1-z)^{-a+\frac{c-1}{2}}P_{2a-c}^{1-c}\left ( \frac{1}{\sqrt{1-z}} \right )$$

where we're using DLMF's definition (14.7.14) of:

$$P_{n}^{m}(z)=\frac{(z^2-1)^{\frac{m}{2}}}{2^nn!}\frac{d^{m+n}}{dx^{m+n}}(z^2-1)^n$$

Okay, that seems reasonable enough to plug in, and if we do, we get:

$$_2\bar{F}_1\left (\frac{n+k}{2}, \frac{n+k+1}{2}; k+1; z \right )=2^{k}z^{-\frac{k}{2}}(1-z)^{-\frac{n}{2}}P_{n-1}^{-k}\left ( \frac{1}{\sqrt{1-z}} \right )$$

So that leaves us with an Associated Legendre Function of negative integer order. DLMF 14.9.3 gives us a recurrence relation that can convert this to an ALF of positive order:

$$P_{\nu}^{-m}\left ( z \right )=\frac{\Gamma(\nu-m+1)}{\Gamma(\nu+m+1)}P_{\nu}^{m}\left ( z \right )$$

But after a lot of work I've realised that this isn't actually suitable for the problem I'm looking at, because for ALFs of integer order and degree, $P_{n}^{m}=0$ where $m>n$, and the problem I'm looking at it's quite common for $|m|>n$, so the easy method is out. I've tried looking at as much of the literature as I can understand, and so far as I can tell, there doesn't seem to be a known identity for $P_{n}^{-m}$ for $|m|>n$ that doesn't involve infinite sums.

So the question I am asking is: Is there a known equation to calculate Associated Legendre Functions of arbitrarily large negative integer order and positive integer degree that doesn't require infinite sums (and, preferably, only requires elementary functions with finite sums/products)?

Answer 1

Sometimes you miss the easy solutions. It turns out that the Hypergeometric representation of the Associated Legendre Functions does, indeed, work with integer degrees and order, and will happily accomodate negative integer degree, even negative integer degrees where $|m|>n$ .

By DLMF 14.3.6:

$$P^\mu_\nu(x) = \left( \frac{x+1}{x-1}\right)^{\frac{\mu}{2}}\,_2\bar{F}_1\left(\nu+1,-\nu;1-\mu;\frac{1-x}{2}\right)$$ And by DLMF 15.4:

$$\,_2\bar{F}_1\left(b,-m;c;z\right)=\frac{1}{\Gamma(c)}\sum_{n=0}^{m}(-1)^n\binom{m}{n}\frac{(b)_n}{(c)_n}z^n$$

Since $\mu$ and $\nu$ are integers in our case, we can plug our figures in:

$$P^{-m}_{n}(x)=\left( \frac{x-1}{x+1}\right)^{\frac{m}{2}}\frac{1}{\Gamma(m+1)}\sum_{j=0}^{n}(-1)^j\binom{n}{j}\frac{(n+1)_j}{(m+1)_j}\left (\frac{1-x}{2} \right )^j$$

Since $m$ and $n$ are integers, we can go ahead and convert everything to factorials:

$$\begin{align*} P^{-m}_{n}(x)&=\left( \frac{x-1}{x+1}\right)^{\frac{m}{2}}\frac{1}{m!}\sum_{j=0}^{n}(-1)^j\frac{m!n!(n+j)!}{j!n!(n-j)!(m+j)!}\left (\frac{1-x}{2} \right )^j\\ &=\left( \frac{x-1}{x+1}\right)^{\frac{m}{2}}\sum_{j=0}^{n}\frac{(n+j)!}{j!(n-j)!(m+j)!}\left (\frac{x-1}{2} \right )^j \end{align*}$$

Which means we can now use this to determine a finite sum for our hypergeometric function of interest:

$$\begin{align*} _2F_1\left ( \tfrac{n+k}{2}, \tfrac{n+k+1}{2};k+1;z\right )&= 2^{k}k!z^{-\frac{k}{2}}(1-z)^{-\frac{n}{2}}P^{-k}_{n-1}\left(\frac{1}{\sqrt{1-z}}\right)\\ &=2^{k}k!z^{-\frac{k}{2}}(1-z)^{-\frac{n}{2}}\left( \frac{\frac{1}{\sqrt{1-z}}-1}{\frac{1}{\sqrt{1-z}}+1}\right)^{\frac{k}{2}}\sum_{j=0}^{n-1}\frac{(n+j-1)!}{j!(n-j-1)!(k+j)!}\left (\frac{\frac{1}{\sqrt{1-z}}-1}{2} \right )^j\\ &=2^{k}k!z^{-\frac{k}{2}}(1-z)^{-\frac{n}{2}}\left( \frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\right)^{\frac{k}{2}}\sum_{j=0}^{n-1}\frac{(n+j-1)!}{j!(n-j-1)!(k+j)!}\left (\frac{1-\sqrt{1-z}}{2\sqrt{1-z}} \right )^j \end{align*}$$

And this will work even if $k>n$, and thus can be used in my particular application.

Answer 2

To evaluate \begin{equation} G={}_2F_1\left (\frac{n+k}{2}, \frac{n+k+1}{2};k+1;x \right) \end{equation} one can recognize the pattern \begin{equation} G={}_2F_1\left (s, s+1/2;2s+1-n;x \right) \end{equation} where $s=(n+k)/2$. Under this form, it can be interpreted as the $n$-th derivative of a simpler hypergeometric function by this relation: \begin{equation} \frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\left(z^{c-1}{}_2F_1\left(a,b;c;z\right)% \right)={\left(c-n\right)_{n}}z^{c-n-1}{}_2F_1\left(a,b;c-n;z\right) \end{equation} ($(.)_n$ correspond to the Pochhamer symbols). Here $a=s,b=s+1/2,c=2s+1$, then \begin{equation} G=\frac{x^{n-2s}}{(2s+1-n)_n}\frac{{\mathrm{d}}^{n}}{{\mathrm{d}x}^{n}}\left(z^{2s}F\left(s,s+1/2;2s+1;x\right)% \right) \end{equation} But this special hypergeometric function is tabulated: \begin{equation} {}_2F_{1}(s,s+{\frac{1}{2}};2\,s+1;z)=2^{2s}\left({\sqrt{1-z}}+1\right)^{-2s} \end{equation} then \begin{equation} G=\frac{2^{n+k}x^{-k}}{(k+1)_n}\frac{{\mathrm{d}}^{n}}{{\mathrm{d}x}^{n}}\left(z^{n+k}\left({1+\sqrt{1-x}}\right)^{-n-k} \right) \end{equation} The inner function \begin{align} z^{n+k}\left({1+\sqrt{1-z}}\right)^{-n-k}&=\left( \frac{z}{1+\sqrt{1-z}} \right)^{n+k}\\ &=\left( 1-\sqrt{1-x} \right)^{n+k}\\ &=\sum_{m=0}^{n+k}\binom{n+k}m (-1)^m\left( 1-x \right)^{m/2} \end{align} As \begin{equation} \frac{{\mathrm{d}}^{n}}{{\mathrm{d}x}^{n}}\left( \left( 1-x \right)^{m/2} \right)=(-1)^n\left( m/2+1-n \right)_n(1-x)^{m/2-n} \end{equation} one obtains \begin{equation} G=\frac{2^{n+k}(-1)^nx^{-k}}{(k+1)_n}\sum_{m=0}^{n+k}(-1)^m\binom{n+k}m \left( m/2+1-n \right)_n(1-x)^{m/2-n} \end{equation} which seems to be numerically correct.