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:

\(\ds \map {\Gamma_m} y\) \(=\) \(\ds \dfrac {m^y m!} {y \paren {y + 1} \paren {y + 2} \cdots \paren {y + m} }\)
\(\ds \) \(=\) \(\ds \dfrac {m^y m!} { y^{\overline {m + 1} } }\)


Proof 1

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \dfrac {x + y} y \map \Beta {x, y + 1}\) Beta Function of x with y+1 by x+y over y
\(\ds \) \(=\) \(\ds \paren {\dfrac {x + y} y } \paren {\dfrac {x + y + 1} {y + 1} } \map \Beta {x, y + 1 + 1}\) applying Beta Function of x with y+1 by x+y over y a second time
\(\ds \) \(=\) \(\ds \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}\) after $m + 1$ times: rising factorial: $y^{\overline {m + 1} } = y \paren {y + 1} \cdots \paren {y + m}$


Also:

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


and:

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


Hence:

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \dfrac {\map \Beta {y, m + 1} } {\map \Beta {x + y, m + 1} } \map \Beta {x, y + m + 1}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {m!} {y^{\overline {m + 1} } } } } {\paren {\dfrac {m!} {\paren {x + y}^{\overline {m + 1} } } } } \map \Beta {x, y + m + 1}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac {\map {\Gamma_m} y} {m^y} } \paren {\dfrac {m^{x + y} } {\map {\Gamma_m} {x + y} } } \map \Beta {x, y + m + 1}\) $\dfrac {m!} {y^{\overline {m + 1} } } = \dfrac {\map {\Gamma_m} y} {m^y}$ and $\dfrac {\paren {x + y}^{\overline {m + 1} } } {m! } = \dfrac {m^{x + y} } {\map {\Gamma_m} {x + y} }$
\(\ds \) \(=\) \(\ds \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}\) $m^y$ cancels

$\blacksquare$


Proof 2

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \dfrac {x + y} y \map \Beta {x, y + 1}\) Beta Function of x with y+1 by x+y over y
\(\ds \) \(=\) \(\ds \paren {\dfrac {x + y} y } \paren {\dfrac {x + y + 1} {y + 1} } \map \Beta {x, y + 1 + 1}\) applying Beta Function of x with y+1 by x+y over y a second time
\(\ds \) \(=\) \(\ds \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}\) after $m + 1$ times: rising factorial: $y^{\overline {m + 1} } = y \paren {y + 1} \cdots \paren {y + m}$


Hence:

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac {m^x m^y m!} {m^x m^y m!} } \dfrac {\paren {\dfrac 1 {y^{\overline {m + 1} } } } } {\paren {\dfrac 1 {\paren {x + y}^{\overline {m + 1} } } } } \map \Beta {x, y + m + 1}\) multiplying by $1$ and rearranging
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {m^x m^y m!} {y^{\overline {m + 1} } } } } {\paren {\dfrac {m^x m^y m!} {\paren {x + y}^{\overline {m + 1} } } } } \map \Beta {x, y + m + 1}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {m^y m!} {y^{\overline {m + 1} } } } m^x } {\paren {\dfrac {m^x m^y m!} {\paren {x + y}^{\overline {m + 1} } } } } \map \Beta {x, y + m + 1}\) isolating $m^x$
\(\ds \) \(=\) \(\ds \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}\) $\map {\Gamma_m} y = \dfrac {m^y m!} { y^{\overline {m + 1} } }$ and $\map {\Gamma_m} {x + y} = \dfrac {m^{x + y} m!} { \paren {x + y}^{\overline {m + 1} } }$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Beta_Function_of_x_with_y%2Bm%2B1&oldid=603784"