# Computing hypergeometric function at 1

I'm looking to compute

$${}_ 3F_ 2\biggl(\begin{matrix} -m-1/2,\ -m,\ k-m+1/2 \cr 1/2-m,\ k-m+3/2\end{matrix};1\biggr)$$

for $$m,k > 0$$ are positive integers and $$0 < k < m$$. I'm wondering if there is any reduction that can be applied, such as Thomae's theorem, which is mentioned in this post (though I don't think I can apply it here).

I would also be interested in any reductions, or if this evaluation was $$0$$ or had a definite sign like $$(-1)^m$$. Apologies in advance as I am a geometric analyst looking to compute things with eigenfunctions of the laplacian on the hyperbolic disk - any references there would also be useful!

For questions like this, Mathematica is your friend:

$$\, _3F_2\left(-m-\tfrac{1}{2},-m,k-m+\tfrac{1}{2};\tfrac{1}{2}-m,k-m+\tfrac{3}{2};1\right)$$ $$=\tfrac{1}{2}(k+1)^{-1}\Gamma (m+1) \left(\frac{(2 k-2 m+1) \Gamma \left(\frac{1}{2}-m\right)}{\sqrt{\pi }}+\frac{(2 m+1) \Gamma \left(k-m+\frac{3}{2}\right)}{\Gamma \left(k+\frac{3}{2}\right)}\right).$$

• Thank you so much! I was woefully unaware that Mathematica could do these computations
– JMK
Commented Dec 18, 2023 at 21:09
• it is always smart to not blindly trust the output, I typically insert a few values to check the identity in the parameter range where it is needed; in the past I would use reference books, which may have typo's, so the situation is not that different. Commented Dec 18, 2023 at 21:22
• Interestingly, you can even get this from the online free version here just entering the expression "HypergeometricPFQ[{-m-1/2, -m, k-m+1/2}, {1/2-m,k-m+3/2}, 1]" Commented Dec 19, 2023 at 18:20
• certainly, the functionality is the same, the restrictions are mainly on the time needed for the computation. Commented Dec 19, 2023 at 18:21

Carlo's answer is correct but doesn't show why the identity holds, so let me explain how to do this easily by hand.

Look more generally at $$S={}_3F_2\left(\begin{matrix}-m,a,b\\a+1,b+1\end{matrix};1\right).$$ The terms involve the factor $$\frac{(a)_k(b)_k}{(a+1)_k(b+1)_k}=\frac{ab}{(a+k)(b+k)}=\frac{ab}{b-a}\left(\frac{1}{a+k}-\frac{1}{b+k}\right) =\frac{1}{b-a}\left(\frac{b(a)_k}{(a+1)_k}-\frac{a(b)_k}{(b+1)_k}\right).$$ Hence, $$S=\frac{b}{b-a}\,{}_2F_1\left(\begin{matrix}-m,a\\a+1\end{matrix};1\right)- \frac{a}{b-a}\,{}_2F_1\left(\begin{matrix}-m,b\\b+1\end{matrix};1\right).$$ These sums are computed by the Chu-Vandermonde summation and we obtain $$S=\frac{m!}{b-a}\left(\frac{b}{(a+1)_m}-\frac{a}{(b+1)_m}\right).$$ Your sum is a special case of this. Since your sum is finite, it is of course a rational expression, you really shouldn't write it in terms of $$\sqrt{\pi}$$ etc. like Mathematica.

• Thank you for the derivation - it is quite clever! It also seems that to convert to Carlos' answer I could use the Gauss summation from the second to last equation.
– JMK
Commented Dec 19, 2023 at 7:02
• @JMK Yes, and it also follows that if we drop the condition that $m$ is a non-negative integer, then the sum of the series is $$\frac{\Gamma(m+1)}{b-a}\left(\frac{b\,\Gamma(a+1)}{\Gamma(a+1+m)}-\frac{a\,\Gamma(b+1)}{\Gamma(b+1+m)}\right),$$ subject to convergence conditions. Commented Dec 19, 2023 at 8:24