Beta Function of x with y+m+1

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\map \Beta {x, y}$ denote the Beta function.

Then:

$\map \Beta {x, y} = \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}$

where $\Gamma_m$ is the partial Gamma function:

$\map {\Gamma_m} y := \dfrac {m^y m!} {y \paren {y + 1} \paren {y + 2} \cdots \paren {y + m} }$


Proof

\(\displaystyle \map \Beta {x, y}\) \(=\) \(\displaystyle \dfrac {x + y} y \map \Beta {x, y + 1}\) Beta Function of x with y+1 by x+y over y
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}\)


Also:

\(\displaystyle \map \Beta {y, m + 1}\) \(=\) \(\displaystyle \map \Beta {y, m} \dfrac m {y + m}\) Beta Function of x with y+1 by x+y over y
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m!} {\paren {y + 1}^{\overline m} } \map \Beta {y, 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m!} {y^{\overline {m + 1} } }\)


and:

\(\displaystyle \map \Beta {x + y, m + 1}\) \(=\) \(\displaystyle \map \Beta {x + y, m} \dfrac m {x + y + m}\) Beta Function of x with y+1 by x+y over y
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m!} {\paren {x + y}^{\overline {m + 1} } }\)


Hence:

\(\displaystyle \map \Beta {x, y}\) \(=\) \(\displaystyle \dfrac {\map \Beta {y, m + 1} } {\map \Beta {x + y, m + 1} } \map \Beta {x, y + m + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\map {\Gamma_m} y} {m^y} \dfrac {m^{x + y} } {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}\)

$\blacksquare$


Sources