Kummer's Hypergeometric Theorem/Proof 2
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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 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 |
$\blacksquare$
Source of Name
This entry was named for Ernst Eduard Kummer.