Derivative of Gaussian Hypergeometric Function

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Theorem

$\map {\dfrac \d {\d x} } {\map F {a, b; c; x} } = \dfrac {a b} c \map F {a + 1, b + 1; c + 1; x} $

where:

$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
$x^{\overline k}$ denotes the $k$th rising factorial power of $x$


Proof

\(\ds \map F {a, b; c; x}\) \(=\) \(\ds 1 + \dfrac {a b} c x + \dfrac {a \paren {a + 1} b \paren {b + 1} } {c \paren {c + 1} } \dfrac {x^2} {2!} + \cdots + \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {x^k} {k!}\) Definition of Gaussian Hypergeometric Function
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac \d {\d x} } {\map F {a, b; c; x} }\) \(=\) \(\ds \dfrac {a b} c + \dfrac {a \paren {a + 1} b \paren {b + 1} } {c \paren {c + 1} } x + \cdots + \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {x^{k - 1} } {\paren {k - 1}!}\) Derivative of Power of Function
\(\ds \) \(=\) \(\ds \dfrac {a b} c \paren {1 + \dfrac {\paren {a + 1} \paren {b + 1} } {\paren {c + 1} } x + \cdots + \dfrac {\paren {a + 1}^{\overline k} \paren {b + 1}^{\overline k} } {\paren {c + 1}^{\overline k} } \dfrac {x^k } {k!} }\) factor out $\dfrac {a b} c$
\(\ds \) \(=\) \(\ds \dfrac {a b} c \map F {a + 1, b + 1; c + 1; x}\)

$\blacksquare$


Also see


Sources

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