# Gauss hypergeometric function

Symbol: Hypergeometric2F1 $\,{}_2F_1\!\left(a, b, c, z\right)$ Gauss hypergeometric function
Symbol: Hypergeometric2F1Regularized $\,{}_2{\textbf F}_1\!\left(a, b, c, z\right)$ Regularized Gauss hypergeometric function

## Hypergeometric series

$\,{}_2F_1\!\left(a, b, c, z\right) = \sum_{k=0}^{\infty} \frac{\left(a\right)_{k} \left(b\right)_{k}}{\left(c\right)_{k}} \frac{{z}^{k}}{k !}$
$\,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = \sum_{k=0}^{\infty} \frac{\left(a\right)_{k} \left(b\right)_{k}}{\Gamma\!\left(c + k\right)} \frac{{z}^{k}}{k !}$
$\,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = \frac{\,{}_2F_1\!\left(a, b, c, z\right)}{\Gamma\!\left(c\right)}$
$\,{}_2{\textbf F}_1\!\left(a, b, -n, z\right) = \frac{\left(a\right)_{n + 1} \left(b\right)_{n + 1} {z}^{n + 1}}{\left(n + 1\right)!} \,{}_2F_1\!\left(a + n + 1, b + n + 1, n + 2, z\right)$

## Differential equations

$z \left(1 - z\right) y''(z) + \left(c - \left(a + b + 1\right) z\right) y'(z) - a b y\!\left(z\right) = 0\; \text{ where } y\!\left(z\right) = \,{}_2F_1\!\left(a, b, c, z\right)$

## Specific values

$\,{}_2F_1\!\left(a, b, c, 0\right) = 1$
$\,{}_2F_1\!\left(a, b, c, 1\right) = \frac{\Gamma\!\left(c\right) \Gamma\!\left(c - a - b\right)}{\Gamma\!\left(c - a\right) \Gamma\!\left(c - b\right)}$
$\,{}_2F_1\!\left(1, 1, 2, z\right) = -\frac{\log\!\left(1 - z\right)}{z}$
$\,{}_2F_1\!\left(a, b, b, z\right) = {\left(1 - z\right)}^{-a}$

## Symmetries

$\,{}_2F_1\!\left(a, b, c, z\right) = \,{}_2F_1\!\left(b, a, c, z\right)$
$\,{}_2F_1\!\left(a, b, c, z\right) = \overline{\,{}_2F_1\!\left(\overline{a}, \overline{b}, \overline{c}, \overline{z}\right)}$

## Linear fractional transformations

$\,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = {\left(1 - z\right)}^{c - a - b} \,{}_2{\textbf F}_1\!\left(c - a, c - b, c, z\right)$
$\,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = {\left(1 - z\right)}^{-a} \,{}_2{\textbf F}_1\!\left(a, c - b, c, \frac{z}{z - 1}\right)$
$\,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = {\left(1 - z\right)}^{-b} \,{}_2{\textbf F}_1\!\left(c - a, b, c, \frac{z}{z - 1}\right)$
$\frac{\sin\!\left(\pi \left(b - a\right)\right)}{\pi} \,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = \frac{{\left(-z\right)}^{-a}}{\Gamma\!\left(b\right) \Gamma\!\left(c - a\right)} \,{}_2{\textbf F}_1\!\left(a, a - c + 1, a - b + 1, \frac{1}{z}\right) - \frac{{\left(-z\right)}^{-b}}{\Gamma\!\left(a\right) \Gamma\!\left(c - b\right)} \,{}_2{\textbf F}_1\!\left(b, b - c + 1, b - a + 1, \frac{1}{z}\right)$
$\frac{\sin\!\left(\pi \left(b - a\right)\right)}{\pi} \,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = \frac{{\left(1 - z\right)}^{-a}}{\Gamma\!\left(b\right) \Gamma\!\left(c - a\right)} \,{}_2{\textbf F}_1\!\left(a, c - b, a - b + 1, \frac{1}{1 - z}\right) - \frac{{\left(1 - z\right)}^{-b}}{\Gamma\!\left(a\right) \Gamma\!\left(c - b\right)} \,{}_2{\textbf F}_1\!\left(b, c - a, b - a + 1, \frac{1}{1 - z}\right)$
$\frac{\sin\!\left(\pi \left(c - a - b\right)\right)}{\pi} \,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = \frac{1}{\Gamma\!\left(c - a\right) \Gamma\!\left(c - b\right)} \,{}_2{\textbf F}_1\!\left(a, b, a + b - c + 1, 1 - z\right) - \frac{{\left(1 - z\right)}^{c - a - b}}{\Gamma\!\left(a\right) \Gamma\!\left(b\right)} \,{}_2{\textbf F}_1\!\left(c - a, c - b, c - a - b + 1, 1 - z\right)$
$\frac{\sin\!\left(\pi \left(c - a - b\right)\right)}{\pi} \,{}_2{\textbf F}_1\!\left(a, b, c, z\right) = \frac{{z}^{-a}}{\Gamma\!\left(c - a\right) \Gamma\!\left(c - b\right)} \,{}_2{\textbf F}_1\!\left(a, a - c + 1, a + b - c + 1, 1 - \frac{1}{z}\right) - \frac{{z}^{a - c} {\left(1 - z\right)}^{c - a - b}}{\Gamma\!\left(a\right) \Gamma\!\left(b\right)} \,{}_2{\textbf F}_1\!\left(c - a, 1 - a, c - a - b + 1, 1 - \frac{1}{z}\right)$

## Bounds and inequalities

$\left|\,{}_2F_1\!\left(a, b, c, z\right) - \sum_{k=0}^{N - 1} \frac{\left(a\right)_{k} \left(b\right)_{k}}{\left(c\right)_{k}} \frac{{z}^{k}}{k !}\right| \le \left|\frac{\left(a\right)_{N} \left(b\right)_{N}}{\left(c\right)_{N}} \frac{{z}^{N}}{N !}\right| \frac{1}{1 - D}\; \text{ where } D = \left|z\right| \left(1 + \frac{\left|a - c\right|}{\left|c + N\right|}\right) \left(1 + \frac{\left|b - 1\right|}{\left|1 + N\right|}\right)$
$\left|\frac{{f}^{(k)}(z)}{k !}\right| \le A {N + k \choose k} {\nu}^{k}\; \text{ where } f\!\left(z\right) = \,{}_2{\textbf F}_1\!\left(a, b, c, z\right),\,\nu = \max\!\left(\frac{1}{\left|z - 1\right|}, \frac{1}{\left|z\right|}\right),\,N = 2 \max\!\left(\sqrt{{\nu}^{-1} \left|a b\right|}, \left|a + b + 1\right| + 2 \left|c\right|\right),\,A = \max\!\left(\left|f\!\left(z\right)\right|, \frac{\left|f'(z)\right|}{\nu \left(N + 1\right)}\right)$

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2019-06-18 07:49:59.356594 UTC