# Expressing the hypergeometric function as an infinite sum of squares

The hypergeometric function evaluated on paramaters of a specific form, can be expressed as an infinite sum of squares. I'm wondering if this gives it any interesting closed form expressions.

So in general the hypergeometric function is defined as $$_2F_1(a,b;c;z)=\sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$$

But for parameters of the form $a=b, c=1$ the function can be expressed as $$_2F_1(a,a;1;z^2)=\sum_{n=0}^\infty \frac{(a)_n(a)_n}{(1)_n}\frac{z^{2n}}{n!}=\sum_{n=0}^\infty \Big(\frac{(a)_nz^n}{n!}\Big)^2$$ Or equivalently $$_2F_1(a,a;1;z^2)=\frac{1}{\Gamma(a)^2}\sum_{n=0}^\infty \Big(\frac{\Gamma(a+n)z^n}{\Gamma(n+1)}\Big)^2$$ Does the fact that we have an infinite sum of squares give rise to a closed form expression for any $a$?

In particular I'm interested in $a=\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{1}{6},\frac{5}{6}$

• @Masacroso What would $n$-choose-$k$ be interpreted as for fractional values? – Christian Woll Jul 27 '17 at 16:45
• Ramanujan' theories of elliptic functions to alternative bases is what you want to look for. – Somos Jul 27 '17 at 18:18

In case of particular values of the parameters, the hypergeometric2F1 function reduces to functions of lower level : $$_2F_1(a,a;1;x)=\frac{1}{(1+x)^a} \mathrm{P}_\nu(X) \quad\text{with}\quad \nu=-a \quad\text{and}\quad X=\frac{1+x}{1-x}$$ where $\mathrm{P}_\nu(X)$ is the Legendre function.
Moreover, in some particular cases of $a$, it reduces to even simpler functions :
$$_2F_1(\frac{1}{2},\frac{1}{2};1;x)=\frac{2}{\pi}\mathrm{K}(x)$$
$$_2F_1(\frac{1}{4},\frac{1}{4};1;x)=\frac{2}{\pi}\mathrm{K}\left(\frac{1-\sqrt{1-x}}{2}\right)$$
$$_2F_1(\frac{3}{4},\frac{3}{4};1;x)=\frac{2}{\pi\sqrt{1-x}}\mathrm{K}\left(\frac{1-\sqrt{1-x}}{2}\right)$$ Function $\mathrm{K}$ is the complete elliptic integral of first kind.