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Show that, for a positive integer $n$, $$F\left(-\frac{n}{2},-\frac{n}{2}+\frac{1}{2};n+\frac{3}{2};-\frac{1}{3}\right)=\left(\frac{8}{9}\right)^n\frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(n+\frac{4}{3}\right)}.$$

I can identify the right hand side (using the definition of the Pochhammer symbol) with: $$\left(\frac{8}{9}\right)^n\frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(n+\frac{4}{3}\right)}=\left(\frac{8}{9}\right)^n\frac{\left(\frac{3}{2}\right)_n}{\left(\frac{4}{3}\right)_n},$$ and get an expression for this in terms of factorials, but I'm not sure how to simplify the left hand side - it gets quite messy!

MrMazgari
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1 Answers1

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Your formula still holds if $n\notin \mathbb{Z}$. From $F(a,b;c;z)=(1-z)^{c-a-b}F(c-a,c-b;c;z)$, it suffices to calculate $$F(\frac{3n+3}{2},\frac{3n+2}{2};n+\frac{3}{2};-\frac{1}{3})$$ its value follows directly from a cubic transformation of $_2F_1$ due to Goursat: $$\tag{1}_2F_1(\frac{3a}{2}, \frac{3a-1}{2}; a+\frac{1}{2};-\frac{z^2}{3}) = {(1+z)^{1-3a}} {_2F_1}(a-\frac{1}{3}, a;2a;\frac{2z(3+z^2)}{(1+z)^3})$$ and the formula of $F(a,b;c;1)$.

Once an identity like $(1)$ has been conjectured, there is a mechanical way to prove it. One constructs a 2nd order ODE satisfied by RHS, then show LHS is also a solution. The calculation tends to be lengthy and not quite do-able by hand.

The Mathematica commands

DifferentialRootReduce[(1+z)^(1-3a)*Hypergeometric2F1[a-1/3,a,2a,2z(3+z^2)/(1+z)^3],z]

and

DifferentialRootReduce[Hypergeometric2F1[3a/2,(3a-1)/2,a+1/2,-z^2/3],z]

shows both sides of $(1)$ are satisfied by $$3a(3a-1)zy+6a(1+z^2)\frac{dy}{dz}+z(z^2+3)\frac{d^2y}{dz^2}=0$$ Therefore it suffices to show both sides have equal $0$th and $1$st derivative at $z=0$, this is even easier.

pisco
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