Reduction of order of generalized hypergeometric function

Question

If a generalized hypergeometric series has $b_i=a_i+n$ for a positive integer $n$, is there a simple or straightforward method to reduce the order? It seems like there should be, and there are some results on the internet regarding this, but in general they seem needlessly complicated for the case I am interested in, and I am wondering if these is something simpler.

In particular I'm interested in evaluating $$ {_4}F_3\left( \frac{1}{2}, \frac{1}{2}, \ell, m ; a, b, 1 ; 1 \right) $$ where $\ell$, $m$, $a$, and $b$ are all positive integers with the constraints that $\ell \le a$, $m \le b$, and equality cannot hold in both cases (implying that the result is guaranteed to be finite).

I think that this case is simple enough that it should be amenable to reduction. (Based on particular values, I am guessing that the result should be expressible in terms of $\Gamma$ of integers and half-integers.) But I'm just checking whether there is a standard result or standard method for this.

Answer 1

I managed to figure this out, but was too busy to post the solution until now.

Notation: Let me introduce some notation to make this answer more general and also simpler. I will call the arguments that I am trying to eliminate $\alpha$ and $\beta$, where both $\alpha$ and $\beta$ must be natural numbers and $\beta > \alpha$. For simplicity of notation, I will use $F$ to represent $_pF_q$ for a any $p$ and $q$, and I will write $a$ as a shorthand for all of the $a_i$'s that I am not trying to eliminate, and $b$ for all of the $b_i$'s that I am not trying to eliminate. That is, I will write

$$ F\left( \begin{matrix} a & \alpha \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) $$

instead of

$$ _{p+1}F_{q+1}\left( \begin{matrix} a_1 & ... & a_p & \alpha \\ b_1 & ... & b_q & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) $$

When I write $a$ as a factor in a product, I actually mean $a_1 ... a_p$, and when I write the rising factorial $(a)_n$, I actually mean $(a_1)_n (a_2)_n ... (a_p)_n$, etc.

Solution: The key to the solution is recognizing that when one of the numerator arguments is equal to one (so that the series has no $n!$ factor), then the series satisfies the following:

$$ \begin{align} F\left( \begin{matrix} a & 1 \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) &= \sum_{n=0}^\infty \frac{(a)_n (\alpha)_n}{(b)_n (\beta)_n} z^n \\ &= 1 + \frac{a \alpha}{b \beta} z \sum_{n=0}^\infty \frac{(a+1)_n (\alpha+1)_n}{(b+1)_n (\beta+1)_n} z^n \\ &= 1 + \frac{a \alpha}{b \beta} z F \left( \begin{matrix} a + 1 & 1 \\ b + 1 & \beta + 1 \end{matrix} \right) \end{align} $$

That is, all but one of the arguments can be incremented in the relation. Repeating this gives

$$ \begin{align} F\left( \begin{matrix} a - \beta + 1 \\ b - \beta + 1 \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) &= F\left( \begin{matrix} a - \beta + 1 & 1 \\ b - \beta + 1 & 1 \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) \\ &= \sum_{n=0}^{\beta-1} \frac{(a - \beta + 1)_n z^n}{(b - \beta + 1)_n n!} + \frac{(a - \beta + 1)_{\beta-1} z^{\beta-1}}{(b-\beta+1)_{\beta-1}} F\left( \begin{matrix} a & 1 \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) \end{align} \tag{1}$$

This can be solved for $$ F\left( \begin{matrix} a & 1 \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) $$

This is where the reduction of order actually occurs. The only missing piece is to actually write $F$ in such a way that one of the numeratorial arguments is 1.

If there is only one pair of integer arguments to reduce, then the easiest way to do this is to write

$$ F\left( \begin{matrix} a & \alpha \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) = \frac{(b-\alpha+1)_{\alpha-1} (\beta-\alpha+1)_{\alpha-1}}{(a-\alpha+1)_{\alpha-1}} \left( \frac{d}{dz} \right)^{\alpha-1} F\left( \begin{matrix} a - \alpha + 1 & 1 \\ b - \alpha + 1 & \beta - \alpha + 1 \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) \tag{2} $$ Combining $(1)$ and $(2)$ accomplishes the reduction of order.

If you have two pairs to reduce, as I do in the original post, then this is problematic because $a-\alpha+1$ could have an element that is zero or a negative integer, in which case the preceding equation is not true. In this case, another strategy is to use the contiguous relation

$$ (\alpha - \beta + 1) F\left( \begin{matrix} a & \alpha \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) = \alpha F\left( \begin{matrix} a & \alpha + 1 \\ b & \beta \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) - (\beta - 1) F\left( \begin{matrix} a & \alpha \\ b & \beta - 1 \end{matrix} \hspace{3pt} ; \hspace{3pt} z \right) $$

repeatedly until all integer pairs differ only by one, and then apply the previous technique on the smallest pair. Note that this contiguous relation cannot be used for the actual reduction of order since when $\beta=\alpha+1$, it gives the trivial relation $0=0$. An iterated version of this relation is in Eq. (A2) of this useful paper.

Note that after both reductions are performed in my original problem and $z=1$ is plugged in, the result will be of the form

$$ c + \sum_n d_n \ {_2}F_1\left( \begin{matrix} \frac{1}{2} & \frac{1}{2} \\ n \end{matrix} \right) $$

where $c$ and $d_n$ are rational constants. The $_2F_1$ can be evaluated using Gauss's hypergeometric theorem and is a rational multiple of $1/\pi$. This explains why the values all satisfy the form noted by the commenter above and suggested in the original post.

If there is any additional interest (or when I get to the point where I am actually using this in a paper) then I will try to combine all of this into a closed-form result.

Answer 2

$$ F(x) = {_4}F_3\left( \frac{1}{2}, \frac{1}{2}, \ell, m ; a, b, 1 ; -x \right) $$ Take the Mellin transform from the Barnes integral representation $$ \mathcal{M}[F]=\frac{\Gamma (a) \Gamma (b) \Gamma \left(\frac{1}{2}-s\right)^2 \Gamma (s) \Gamma (l-s) \Gamma (m-s)}{\pi \Gamma (l) \Gamma (m) \Gamma (1-s) \Gamma (a-s) \Gamma (b-s)} $$ now we can tinker with $a\to l+ n_1$ and $b \to m+n_2$ for positive $n_1$ and $n_2$ that aren't both $0$. $$ \mathcal{M}[F]=\frac{\Gamma (l+n_1) }{\Gamma (l)}\frac{\Gamma (m+n_2)}{\Gamma (m)}\frac{ \Gamma \left(\frac{1}{2}-s\right)^2 \Gamma (s) \Gamma (l-s) \Gamma (m-s)}{\pi \Gamma (1-s) \Gamma (a-s) \Gamma (b-s)} $$ $$ \mathcal{M}[F]=\prod_{k=0}^{n_1-1}(l+k)\prod_{k=0}^{n_2-1}(m+k)\frac{ \Gamma \left(\frac{1}{2}-s\right)^2 \Gamma (s) \Gamma (l-s) \Gamma (m-s)}{\pi \Gamma (1-s) \Gamma (l+n_1-s) \Gamma (m+n_2-s)} $$ $$ \mathcal{M}[F]=\prod_{k=0}^{n_1-1}(l+k)\prod_{k=0}^{n_2-1}(m+k)\frac{ \Gamma \left(\frac{1}{2}-s\right)^2 \Gamma (s) }{\pi \Gamma (1-s)}\prod_{k=0}^{n_1-1}(l-s+k)^{-1}\prod_{k=0}^{n_2-1}(m-s+k)^{-1} $$ retake the Mellin transform and see what kind of monster $M(x;n_1,n_2)$ we have created $$ F(x)=\prod_{k=0}^{n_1-1}(l+k)\prod_{k=0}^{n_2-1}(m+k) M(x;n_1,n_2) $$ by Ramanujan's master theorem $$ M(x;n_1,n_2)=\sum_{s=0}^\infty \frac{ \Gamma \left(\frac{1}{2}+s\right)^2 }{\pi \Gamma (1+s)}\prod_{k=0}^{n_1-1}(l+s+k)^{-1}\prod_{k=0}^{n_2-1}(m+s+k)^{-1}\frac{(-1)^s}{s!}x^s $$ which when evaluated in Mathematica for given $n_1,n_2$ pairs produces a big mess of $\;_3F_2(\cdots,-x)$ functions. I can't produce much more than this formula. Hope it is helpful.