# How to prove ${}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};\frac{3}{2}; \frac{27}{4}z^2(1-z)\right) = \frac{1}{z}$

With reference to the following post

On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$

I used

$${}_{k}F_{k-1}\left(\frac{1}{k+1} ,\cdots ,\frac{k}{k+1};\frac{2}{k} \cdots ,\frac{k-1}{k},\frac{k+1}{k};\left( \frac{z(1-z^k)}{f_k}\right)^k \right) = \frac{1}{1-z^k}$$

Where

$$f_k \equiv \frac{k}{(1+k)^{1+1/k}}$$

This formula seems to be too complicated for the general case.

Let us look at the easiest case $k=2$ then we have

$${}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};\frac{3}{2}; \frac{27}{4}z^2(1-z^2)^2\right) = \frac{1}{1-z^2}$$

This is equivalent to proving
$${}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};\frac{3}{2}; \frac{27}{4}z^2(1-z)\right) = \frac{1}{z}\tag{1}$$

For some ranges of $z$.

Question:

Any idea how to prove $(1)$ ?

#### Solutions Collecting From Web of "How to prove ${}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};\frac{3}{2}; \frac{27}{4}z^2(1-z)\right) = \frac{1}{z}$"

Let’s first prove that $$_2F_{1} \left(a, 1-a; \frac{3}{2}; -z^{2} \right) =\frac{1}{2(2a-1)z} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} – \left(\sqrt{1+z^{2}}-z \right)^{2a-1}\right]$$ for $|z| <1$.

This identity is listed on DLMF but in a different form. (Conjugate multiplication shows that they’re equivalent.)

For evaluation purposes, let’s restrict $a$ to $0 < a < 1$. (For $a=1/2$, take the limit as $a\to \frac{1}{2}$.)

Using the generalized binomial theorem, we have

\begin{align} &\frac{1}{2(2a-1)z} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} – \left(\sqrt{1+z^{2}}-z \right)^{2a-1}\right] \\ &= \frac{1}{2(2a-1)z} \left[\sum_{k=0}^{\infty} \binom{2a-1}{k} \left(\sqrt{1+z^{2}} \right)^{2a-1-k}z^{k} – \sum_{k=0}^{\infty} \binom{2a-1}{k} \left(\sqrt{1+z^{2}} \right)^{2a-1-k}(-z)^{k}\right ] \\ &= \frac{1}{(2a-1)} \sum_{k=0}^{\infty} \binom{2a-1}{2k+1}(1+z^{2})^{a-1-k}z^{2k} \\ &= \frac{1}{2a-1} \sum_{k=0}^{\infty} \binom{2a-1}{2k+1}\sum_{j=0}^{\infty} \binom{a-1-k}{j}z^{2j}z^{2k} \\ &= \frac{1}{2a-1} \sum_{n=0}^{\infty} \sum_{m=0}^{n}\binom{2a-1}{2m+1} \binom{a-1-m}{n-m} z^{2n} \tag{1} \\ &=\frac{1}{2a-1} \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)}\sum_{m=0}^{n} \frac{\Gamma(a-m)}{\Gamma(2m+2)\Gamma(2a-2m-1) \Gamma(n-m+1)} \, z^{2n} \\ &= \frac{1}{2a-1} \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \sum_{m=0}^{n} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma(m+1) \Gamma(m+\frac{3}{2}) \Gamma(a-m-\frac{1}{2}) \Gamma(n-m+1)} \, z^{2n} \tag{2} \\ &= \frac{1}{2a-1} \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma(n+ \frac{3}{2}) \Gamma(a- \frac{1}{2})} \sum_{m=0}^{n} \binom{n+\frac{1}{2}}{n-m} \binom{a-\frac{3}{2}}{m} \, z^{2n} \\ &= \frac{1}{2a-1} \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}}\frac{1}{\Gamma(n+ \frac{3}{2}) \Gamma(a- \frac{1}{2})} \binom{a+n-1}{n} \, z^{2n} \tag{3} \\ &= \frac{1}{a-\frac{1}{2}} \sum_{n=0}^{\infty} \frac{\Gamma(a+\frac{1}{2})}{\Gamma(a- \frac{1}{2})} \frac{\Gamma(a+n)}{\Gamma(a)} \frac{\Gamma(a)}{\Gamma(a-n)} \frac{\frac{1}{2}\Gamma(\frac{1}{2})}{\Gamma(n+\frac{3}{2})} \, \frac{z^{2n}}{n!} \tag{4} \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a)} \frac{\Gamma(1-a+n)}{\Gamma(1-a)} \frac{\Gamma(\frac{3}{2})}{\Gamma(n+ \frac{3}{2})} \frac{(-z^{2})^{n}}{n!} \tag{5} \\ &= \, _2F_{1} \left(a, 1-a; \frac{3}{2}, -z^{2} \right) . \end{align}

$(1)$: Cauchy product

$(2)$: Duplication formula for the gamma function

$(3)$: Chu-Vandermonde identity

$(4)$: Multiply and divide by $\frac{1}{2} \Gamma(a)$.

$(5)$ $\frac{\Gamma(a)}{\Gamma(a-n)}= (-1)^{n} \frac{\Gamma(1-a+n)}{\Gamma(1-a)}$.

Replacing $z$ with $i \sin x$, $- \frac{\pi}{2}<x <\frac{\pi}{2}$, we get \begin{align}\, _2F_{1} \left(a, 1-a; \frac{3}{2}, \sin^{2}(x) \right) &= \frac{(\cos x + i \sin x)^{2a-1}-(\cos x – i \sin x)^{2a-1}}{2i(2a-1) \sin x} \\ &= \frac{e^{ix(2a-1)}-e^{-ix(2a-1)}}{2i(2a-1) \sin x} \\ &= \frac{\sin[(2a-1)x]}{(2a-1) \sin x}. \end{align}

So proving that $${}_{2}F_{1}\left(\frac{1}{3};\frac{2}{3};\frac{3}{2}; \frac{27}{4}x^2(1-x^2)^2\right) = \frac{1}{1-x^2}, \quad -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}},$$ is equivalent to proving that $$\frac{\sin\left[\frac{1}{3} \, \arcsin \left(\frac{3 \sqrt{3}}{2} x(1-x^{2}) \right) \right]}{\frac{1}{3} \frac{3 \sqrt{3}}{2} x(1-x^{2})} = \frac{1}{1-x^{2}} , \quad – \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}, \tag{6}$$ which in turn is equivalent to proving that $$\frac{1}{3} \, \arcsin \left(\frac{3 \sqrt{3}}{2} x(1-x^{2}) \right) = \arcsin \left(\frac{\sqrt{3}}{2}x \right) , \quad – \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}. \tag{7}$$

(Identity $(6)$ is entry (27) on the Wolfram MathWorld page about hypergeometric functions.)

But under the restriction on $x$, \begin{align} \frac{1}{3} \frac{\mathrm d}{\mathrm dx} \arcsin \left(\frac{3 \sqrt{3}}{2} x(1-x^{2}) \right) &= \frac{1}{3} \frac{\frac{3\sqrt{3}}{2}(1-3x^{2})}{\sqrt{1- \frac{27}{4} x^{2}(1-x^{2})^{2}}} \\ &= \frac{\sqrt{3}}{2} \frac{1-3x^{2}}{\sqrt{(1- \frac{3}{4}z^{2})(1-3x^{2})^{2}}} \\ &= \frac{\sqrt{3}}{2} \frac{1}{\sqrt{1- \frac{3}{4} z^{2}}} \\ &= \frac{\mathrm d}{\mathrm dx} \arcsin \left(\frac{\sqrt{3}}{2} z \right). \end{align}

And both sides of $(7)$ agree at $x=0$.