Corollary 1. For $\frac{1}{2} < z < 2$, $$ {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-\frac{1}{z}\right) = \sqrt{z} {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-z\right). $$ |
Proof. We recall Bailey’s Formula ((i) in [2]) for $w\in\mathbb R$:
\begin{equation}\tag{1}
(1-w)^{-a} {}_2 F_1 \left( a,b;c; – \frac{w}{1-w}\right) = {}_2 F_1 (a,c-b;c;w), \qquad -1< w <\frac{1}{2}
\end{equation}
Setting $a=b=\frac{1}{2}$, $c=1$ and $z=1-w$ at (1), we obtain
$$
z^{-1/2} {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-\frac{1}{z}\right)
=
{}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-z\right), \qquad \frac{1}{2} < z < 2.
$$
$\square$
Corollary 2. $$ {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1 – \left( \frac{1-z}{1+z}\right)^2\right) = (1+z) {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2}; 1; z^2\right) $$ |
Proof. Replacing
$$
z \to \left(\frac{1-z}{1+z}\right)^2
$$
in Corollary 1 above, we have
\begin{equation}
{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1 – \left( \frac{1-z}{1+z}\right)^2\right)
=
\frac{1+z}{1-z} {}_2 F_1\left( \frac{1}{2}, \frac{1}{2};1; – \frac{4z}{(1-z)^2}\right).
\end{equation}
We remember that Gauss’s Quadratic Transformation ([3]) with $a=b=\frac{1}{2}$ and $z \to -z$ is
\begin{equation}
{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; – \frac{4z}{(1-z)^2}\right)
=
(1-z) {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; z^2 \right).
\end{equation}
Piecing (2) and (3) together, we obtain Corollary 2.$\square$
References.
[1] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, pp.92-93.
[2] Leun Kim, [Calculation 7] Bailey’s Formulas for Hypergeometric Functions.
[3] Leun Kim, [Calculation 8] Gauss’s Quadratic Trasformation.
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