# Manipulating the left-hand side into a right-hand side of a Hypergeometric sequence expansion

How do you show that$$F\left(\frac 13,\frac 23;\frac 32;\frac {27}4x^2(1-x^2)^2\right)=\frac {2\sin\left[\tfrac 13\arcsin\left(\tfrac {3\sqrt3}2x(x^2-1)\right)\right]}{x\sqrt3(x^2-1)}$$

I'm not sure where to begin. I thought about using the expansion of $\sin^{-1}$ since its expansion is equivalent to$$\arcsin z=z\, F\left(\frac 12,\frac 12;\frac 32;z^2\right)$$But am not sure how to incorporate that. I also thought about using some identity to make a transformation, but I'm not sure which specific one to use.

The series expansion, about $x=0$, of $\sin\left(\frac{1}{3} \, \sin^{-1}(x)\right)$ is given by $$\sin\left(\frac{1}{3} \, \sin^{-1}(x)\right) = 3 \, x \, \sum_{n=0}^{\infty} \frac{(3n)!}{n! \, (2n+1)!} \, \left(\frac{4 \, x^{2}}{27}\right)^{n}.$$
Now, by comparison it is determined that $${}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right) = \frac{1}{3 \, t} \, \sin\left(\frac{1}{3} \, \sin^{-1}(t)\right).$$ By setting $2 \, t = 3 \, \sqrt{3} \, x \, (1-x^{2})$ leads to the desired expression. The region of convergence is $0 \leq x \leq \sqrt{3}/3$.
With much magic, or confusion, one may show that $${}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right) = \frac{1}{3 \, t} \, \sin\left(\frac{1}{3} \, \sin^{-1}(t)\right) = \frac{1}{1-x^{2}},$$ where $2 \, t = 3 \, \sqrt{3} \, x \, (1-x^{2})$.
• I made a mistake and your answer is very nice $\to +1$. – Claude Leibovici Oct 31 '17 at 9:21