Proving a quadratic transformation of ${}_2F_1$

Question

For context, please see this MSE post

I want to prove that $${}_2F_1\left(a,b;a−b+1;z\right)=(1−z)^{1−2b}(1+z)^{2b−a−1}{}_2F_1\left[\frac{a−2b+1}{2},\frac{a−2b+2}{2};a−b+1;\frac{4z}{(1+z)^2}\right]$$ The comment said I can prove this by showing that both sides satisfy the same hypergeometric differential equation. However, when I actually tried it out the expression became extremely tedious and I failed to prove the identity using such method. Furthermore, using hypergeometric differential equation to prove quadratic transformations feel forced and unnatural way.

I would like a proof, preferably that doesn't use the hypergeometric differential equation. If there isn't one, a proof using hypergeometric differential equation would suffice.

Any help is appreciated

Answer 1

Here is a brute force proof using the definition of the ${}_2F_1$. Move the factor $(1-z)^{1-2b}$ to the LHS and expand both sides as series in $z$. On the left, we get $$\sum_{k=0}^\infty\frac{(1-2b)_k}{k!}\,z^k\sum_{m=0}^\infty\frac{(a)_m(b)_m}{m!(a-b+1)_m}\,z^m. $$ Put $k+m=n$ and pull the sum in $m$ inside. You get $$\sum_{n=0}^\infty\frac{(1-2b)_n}{n!}\,z^n\sum_{m=0}^n\frac{(a)_m(b)_m(-n)_m}{m!(a-b+1)_m(2b-n)_m}.$$ The inner sum is computed by the Pfaff-Saalschütz summation and we end up with $$\sum_{n=0}^\infty\frac{(1-b)_n(a-2b+1)_n}{n!(a-b+1)_n}\,z^n.$$ The right-hand side can be written \begin{multline*}\sum_{m=0}^\infty\frac{(a-2b+1)_{2m}}{m!(a-b+1)_m}\frac{z^m}{(1+z)^{a+1-2b+2m}}\\ =\sum_{m=0}^\infty\frac{(a-2b+1)_{2m}}{m!(a-b+1)_m}\, z^m\sum_{k=0}^\infty\frac{(a-2b+1+2m)_k}{k!}\,(-z)^k. \end{multline*} Again we put $k+m=n$ and pull $m$ inside. This gives $$ \sum_{n=0}^\infty \frac{(a-2b+1)_n}{n!}(-z)^n \sum_{m=0}^n\frac{(a-2b+1+n)_{m}(-n)_m}{m!(a-b+1)_m}. $$ The inner sum is computed by the Chu-Vandermonde summation and we get the same expression $$\sum_{n=0}^\infty\frac{(1-b)_n(a-2b+1)_n}{n!(a-b+1)_n}\,z^n$$ as before.