A hypergeometric identity related to Bessel functions

Question

The identity in my recent answer can be stated in a particularly neat form: $${}_2F_0\left({-n, n+1\atop{}};\frac{x}{2}\right) ~\cdot~ {}_2F_0\left({-n, n+1\atop{}};-\frac{x}{2}\right) ~=~ {}_3F_0\left({\frac{1}{2}, -n, n+1\atop{}};x^2\right).$$ Is this by any chance a partial case of something more general and/or has a straightforward proof?

Answer 1

In "Higher Trancendental Functions", Vol. 1, by A. Erdelyi (ed.), on page 86, equation (4) says $$ _2F_0(α,β;z)\ _2F_0(α,β;-z) = \, _4F_1(α,β,\frac{1}{2}(α+β),\frac{1}{2}(α+β+1);α+β;4z^2), $$ from which your formula can be derived by setting $\alpha=−n$, $\beta=n+1$.

Unfortunately Erdelyi, et al., do not give a proof only references, which I reproduce here:

W.N. Bailey (1928), Proc. London Math. Soc. (2) 28, 242-254

Preece, C.T. (1924), Proc. London Math. Soc. (2) 22, 370-380,

but I cannot say more about these papers and their content.

Answer 2

The coefficient of $x^n$ in $_2F_0(\alpha,\beta;z) {}_2F_0(\alpha,\beta;-z)$, multiplied by $n!$, is $$\begin{aligned} \sum_{k=0}^n (-1)^{n-k}\binom nk &(\alpha)_k(\beta)_k (\alpha)_{n-k}(\beta)_{n-k}\hfill\\ &=(-1)^n (\alpha)_n(\beta)_n\, {}_3F_2\left({-n,\atop}\!{\alpha,\atop 1-\alpha -n,}\, {\beta\atop1-\beta-n}\biggm | 1\right). \end{aligned} $$ The right side can be evaluated by Dixon's theorem to give the identity cited by Johannes Trost.

Answer 3

$a(x) = {}_2F_0(-n,n+2; x/2)$ satisfies the differential equation $$ \left( -{n}^{2}-n \right) a \left( x \right) + \left( 2\,x-2 \right) {\frac {\rm d}{{\rm d}x}}a \left( x \right) +{x}^{2}{\frac { {\rm d}^{2}}{{\rm d}{x}^{2}}}a \left( x \right) =0$$ $b(x) = a(-x)$ satisfies $$ \left( -{n}^{2}-n \right) b \left( x \right) + \left( 2\,x+2 \right) {\frac {\rm d}{{\rm d}x}}b \left( x \right) +{x}^{2}{\frac { {\rm d}^{2}}{{\rm d}{x}^{2}}}b \left( x \right)=0$$ $c(x) = {}_3F_0(1/2,-n,n+1; x^2)$ satisfies $$ \left( -4\,{n}^{2}x-4\,nx \right) c \left( x \right) + \left( -4\,{n}^{2}{x}^{2}-4\,n{x}^{2}+6\,{x}^{2}-4 \right) {\frac {\rm d}{{\rm d}x}}c \left( x \right) +6\,{x}^{3}{\frac {{\rm d}^{2}}{ {\rm d}{x}^{2}}}c \left( x \right) +{x}^{4}{\frac {{\rm d}^{3}}{ {\rm d}{x}^{3}}}c \left( x \right)=0$$ Substitute $c(x) = a(x) b(x)$, use the differential equations for $a(x)$ and $b(x)$, and this simplifies to $0=0$. That is, $c(x)$ and $a(x)b(x)$ satisfy the same third order linear differential equation. There is only a one-dimensional family of analytic solutions of this equation ($0$ being an irregular singular point), and the fact that $a(0) b(0) = c(0) = 1$ completes the proof.