Kummer's Hypergeometric Theorem/Proof 3

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Theorem

$\map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }$


Proof

From Kummer's Quadratic Transformation, we have:

$\ds \map F {a, b; 1 + a - b; z} = \paren {1 - z}^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; \dfrac {-4 z} {\paren {1 - z }^2} }$

Let $z \to -1$ and we have:

$\ds \map F {a, b; 1 + a - b; -1} = 2^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; 1 }$

From Gauss's Hypergeometric Theorem, we have:

$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$

Therefore, the right hand side becomes:

\(\ds 2^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; 1 }\) \(=\) \(\ds 2^{-a} \dfrac {\map \Gamma {1 + a - b} \map \Gamma {\paren {1 + a - b} - \dfrac a 2 - \paren {\dfrac {1 + a} 2 - b} } } {\map \Gamma {\paren {1 + a - b} - \dfrac a 2} \map \Gamma {\paren {1 + a - b} - \paren {\dfrac {1 + a} 2 - b} } }\) Gauss's Hypergeometric Theorem
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {1 + a - b} } {\map \Gamma {1 + \dfrac a 2 - b} } \paren {2^{-a} \dfrac {\map \Gamma {\dfrac 1 2 } } {\map \Gamma {\dfrac {1 + a} 2 } } }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {1 + a - b} } {\map \Gamma {1 + \dfrac a 2 - b} } \paren {\dfrac {\map \Gamma {\dfrac a 2 + 1 } } {\map \Gamma {a + 1 } } }\) Legendre's Duplication Formula
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {1 + a - b} \map \Gamma {\dfrac a 2 + 1 } } {\map \Gamma {1 + \dfrac a 2 - b} \map \Gamma {a + 1 } }\) simplifying

Substituting $a = n$ and $b = -x$, we obtain:

\(\ds \map F {a, b; 1 + a - b; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {1 + a - b} \map \Gamma {\dfrac a 2 + 1 } } {\map \Gamma {1 + \dfrac a 2 - b} \map \Gamma {a + 1 } }\) before substitution
\(\ds \leadsto \ \ \) \(\ds \map F {n, -x; 1 + n + x; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {1 + n + x} \map \Gamma {\dfrac n 2 + 1 } } {\map \Gamma {1 + \dfrac n 2 + x} \map \Gamma {n + 1 } }\) after substitution

$\blacksquare$


Source of Name

This entry was named for Ernst Eduard Kummer.


Sources