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?

  • $\begingroup$ In "Higher Trancendental Functions", Vol 1, by A. Erdelyi (ed.), on page 86 equation (4) says $_2F_0(\alpha,\beta;z) \ _2F_0(\alpha,\beta;z) = _4F_1(\alpha, \beta, \frac{1}{2} (\alpha+\beta), \frac{1}{2} (\alpha+\beta+1); \alpha+\beta; 4 z^2)$, from which your formula can be derived by setting $\alpha=-n$, $\beta=n+1$. $\endgroup$ – Johannes Trost Jun 28 '19 at 8:53
  • $\begingroup$ typo in my comment : the second $_2F_0$ has argument $-z$. $\endgroup$ – Johannes Trost Jun 28 '19 at 9:09
  • $\begingroup$ @JohannesTrost: Nice catch! Please add this as an answer - I will accept it. Do they give any proof idea? $\endgroup$ – Max Alekseyev Jun 28 '19 at 11:32

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.


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.

  • $\begingroup$ Nice! Thank you! $\endgroup$ – Max Alekseyev Jun 29 '19 at 12:27

$a(x) = {}_2F_0(-n,n+2; x/2)$ 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.

  • $\begingroup$ I like this proof although I'd not call it straightforward. Also, it essentially verifies the identity rather than derives it. And it's not surprising to see such differential equations here given the Bessel origin of the identity. I still wonder if it can be generalized. $\endgroup$ – Max Alekseyev Jun 27 '19 at 22:00
  • $\begingroup$ Hmm, well it looks like ${}_2F_0(-n,n+2; x/2) {}_2F_0(-n,n+2;-x/2) = {}_4F_1(1,3/2,-n,n+2; 2; x^2)$. $\endgroup$ – Robert Israel Jun 28 '19 at 3:20
  • $\begingroup$ Indeed, this is another instance of the same general identity -- see Johannes Trost's comment above. $\endgroup$ – Max Alekseyev Jun 28 '19 at 11:34

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