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Given the hypergeometric function $\,_2F_1$, Pochhammer symbol $(m)_n$, and $0, anybody knows how to prove that,

$\,_2F_1(a,1-a;1;z) = \sqrt{\sum_{n=0}^\infty \frac{(a)_n (1-a)_n (\tfrac{1}{2})_n}{n!^3} \,\big(4z(1-z)\big)^n}$

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The identity is usually written as $ \, _2F_1(1-a,a\,;\,1\,;\,z){}^2 = \sum_{n=0}^\infty \frac{(a)_n (1-a)_n (\tfrac{1}{2})_n}{n!^3} \,\big(4z(1-z)\big)^n = \, _3F_2\left(1-a,a,\tfrac{1}{2} \,;\,1,1\,;\, 4 (1-z) z\right) $

This is a special case of a more general (two-parameter) identity, combined with this quadratic transformation: $ {}_2F_1\left(1-a,a\,;\,1\,;\,z\right) = {}_2F_1\left(\frac{1-a}{2},\frac{a}{2}\,;\,1\,;\,4 (1-z)z\right) $

To prove it, it suffices to compare first few series coefficients of the Taylor expansion at the origin (1st, 2nd, 3rd, 4th would do it) to determine that rhs and lhs satisfy the same linear ODE with polynomial coefficients.

Then one also needs to compare main terms at $z=1$ and at $z=\infty$ to make sure that not only do the rhs and lhs belong to the same fundamental solution, but are actually the same globally. This may not be a very illuminating proof, but a proof nonetheless.

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    Thanks, Sasha. I found the relation empirically while looking at Ramanujan's pi formulas. It is good to know it is really a valid identity.2011-12-24