Kummer's Hypergeometric Theorem
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
- Dixon's Hypergeometric Theorem
- Gauss's Hypergeometric Theorem
- Properties of Generalized Hypergeometric Function
Source of Name
This entry was named for Ernst Eduard Kummer.
Sources
- 1989: Bruce C. Berndt: Ramanujan's Notebooks: Part II: Chapter $\text {10}$. Hypergeometric Series: $\text I$
- Weisstein, Eric W. "Kummer's Theorem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/KummersTheorem.html