Kummer's Hypergeometric Theorem

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Theorem

Let $x, n \in \C$.

Let $n \notin \Z_{< 0}$.

Let $\map \Re {x + 1} > 0$.


Then:

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

where:

$\map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$
$\map \Gamma {n + 1} = n!$ is the Gamma function.


Proof 1

First we note the definition of Gaussian hypergeometric function:

$\map F {n, -x; x + n + 1; -1} = \ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } {\paren {x + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}$

where $x^{\overline k}$ denotes the $k$th rising factorial power of $x$.


Two lemmata:

Lemma 1

$\ds \lim_{y \mathop \to \infty} \dfrac {y^{\underline k} } {\paren {y + n + 1}^{\overline k} } = 1$

$\Box$


Lemma 2

$\ds \lim_{y \mathop \to \infty} \dfrac {\paren {y + \dfrac n 2 + 1}^{\overline x} } {\paren {y + n + 1}^{\overline x} } = 1$

$\Box$


We use Dixon's Hypergeometric Theorem:

$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }$

where $\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$.


So:

\(\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }\) Dixon's Hypergeometric Theorem
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} } \dfrac {1^k} {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }\) Definition of Generalized Hypergeometric Function
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {\paren {-y}^{\overline k} } {\paren {y + n + 1}^{\overline k} } \dfrac {1^k} {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + \dfrac n 2 + 1} } \times \dfrac {\map \Gamma {y + n + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {x + y + n + 1} \map \Gamma {y + \dfrac n 2 + 1} }\) reorganizing both sides: isolating $y$
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren{-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {y^{\underline k} } {\paren {y + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + \dfrac n 2 + 1} } \times \dfrac {\map \Gamma {y + n + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {x + y + n + 1} \map \Gamma {y + \dfrac n 2 + 1} }\) on the left hand side: Rising Factorial in terms of Falling Factorial of Negative
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren{-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {y^{\underline k} } {\paren {y + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + \dfrac n 2 + 1} } \times \dfrac {\paren {y + \dfrac n 2 + 1}^{\overline x} } {\paren {y + n + 1}^{\overline x} }\) on the right hand side: Rising Factorial as Quotient of Factorials
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {n^{\overline k} \paren{-x}^{\overline k} } {\paren {x + n + 1}^{\overline k} } \paren 1 \dfrac {\paren {-1}^k} {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {n + 1} \map \Gamma {x + \dfrac n 2 + 1} } \times 1\) Lemma 1 and Lemma 2: letting $y \to \infty$
\(\ds \leadsto \ \ \) \(\ds \map F {n, -x; x + n + 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }\) Definition of Gaussian Hypergeometric Function

$\blacksquare$


Proof 2

From Euler's Integral Representation of Hypergeometric Function, we have:

$\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{- a} \rd t$

Where $a, b, c \in \C$.

and $\size x < 1$

and $\map \Re c > \map \Re b > 0$.

Since Euler's Integral Representation only applies where $\size x < 1$, we will determine the limit of the integral as $x \to -1$.


By symmetry, we have:

$\ds \map F {n, -x; x + n + 1; -1} = \ds \map F {-x, n; x + n + 1; -1}$

Therefore:

\(\ds \map F {-x, n; x + n + 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + n + 1 - n } } \int_0^1 t^{n - 1} \paren {1 - t}^{x + n + 1 - n - 1} \paren {1 - \paren {-1} t}^{- \paren {-x} } \rd t\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 t^{n - 1} \paren {1 - t}^x \paren {1 + t}^x \rd t\) simplifying
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 t^{n - 1} \paren {1 - t^2}^x \rd t\) simplifying further: $\paren {1 - t^2} = \paren {1 - t}\paren {1 + t}$

We now apply a u-substitution: Let $u = t^2$

\(\ds u\) \(=\) \(\ds t^2\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \sqrt u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d t}\) \(=\) \(\ds 2 t\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \d t\) \(=\) \(\ds \frac {\d u} {2 t}\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \frac {\d u} {2 \sqrt u}\)

Substituting back into our equation, we have:

\(\ds \map F {-x, n; x + n + 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 \paren {\sqrt u}^{n - 1} \paren {1 - u}^x \frac {\d u} {2 \sqrt u}\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 u^{\frac n 2 - 1} \paren {1 - u}^x \d u\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac 1 2 \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \dfrac {\map \Gamma {\dfrac n 2 } \map \Gamma {x + 1 } } {\map \Gamma {\dfrac n 2 + x + 1 } }\) Definition of Beta Function
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\dfrac 1 2 \map \Gamma {x + n + 1} } {\map \Gamma n } \dfrac {\map \Gamma {\dfrac n 2 } } {\map \Gamma {\dfrac n 2 + x + 1 } }\) simplifying and canceling $\map \Gamma {x + 1 }$
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\dfrac n 2 \map \Gamma {x + n + 1} } {n \map \Gamma n } \dfrac {\map \Gamma {\dfrac n 2 } } {\map \Gamma {\dfrac n 2 + x + 1 } }\) multiplying by $1$
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }\) Definition of Gamma Function

$\blacksquare$


Proof 3

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

$\blacksquare$


Examples

Example: $\map F {\dfrac 1 2, \dfrac 1 2; 1; -1}$

$1 - \paren {\dfrac 1 2}^2 + \paren {\dfrac {1 \times 3} {2 \times 4} }^2 - \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\sqrt \pi} {\sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }$


Example: $\map F {\dfrac 1 3, \dfrac 1 3; 1; -1}$

$1 - \paren {\dfrac 1 3}^2 + \paren {\dfrac {1 \times 4} {3 \times 6} }^2 - \paren {\dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} }^2 + \cdots = \dfrac {\pi} {\paren {\map \Gamma {\dfrac 5 6} }^2 \map \Gamma {\dfrac 1 3} }$


Example: $\map F {\dfrac 2 5, \dfrac 1 {10}; \dfrac {13} {10}; -1}$

$1 - \paren {\dfrac {\paren {2} \paren {1} } {\paren {5} \paren {13} } } + \paren {\dfrac {\paren {2 \times 7} \paren {1 \times 11} } {\paren {5 \times 10} \paren {13 \times 23} } } - \paren {\dfrac {\paren {2 \times 7 \times 12} \paren {1 \times 11 \times 21} } {\paren {5 \times 10 \times 15} \paren {13 \times 23 \times 33} } } + \cdots = \dfrac {1944^{\frac 1 5} \pi^{\frac 3 2} } {\phi \map \Gamma {\dfrac 1 {10} } \paren {\map \Gamma {\dfrac 7 {10} } }^2 }$


Also see


Source of Name

This entry was named for Ernst Eduard Kummer.


Sources

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