B(x,y+1)=(y/(x+y))B(x,y)
From specialfunctionswiki
Theorem
The following formula holds: $$B(x,y+1)=\dfrac{y}{x+y} B(x,y),$$ where $B$ denotes the beta function.
Proof
References
- 1953: {{ #if: |{{{2}}}|Arthur Erdélyi}}{{#if: Wilhelm Magnus|{{#if: Fritz Oberhettinger|, {{ #if: |{{{2}}}|Wilhelm Magnus}}{{#if: Francesco G. Tricomi|, {{ #if: |{{{2}}}|Fritz Oberhettinger}}{{#if: |, {{ #if: |{{{2}}}|Francesco G. Tricomi}}{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and {{ #if: |{{{2}}}|Francesco G. Tricomi}}}}| and {{ #if: |{{{2}}}|Fritz Oberhettinger}}}}| and {{ #if: |{{{2}}}|Wilhelm Magnus}}}}|}}: [[Book:Arthur Erdélyi/Higher Transcendental Functions Volume I{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Higher Transcendental Functions Volume I{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}}|{{#if: | ({{{ed}}} ed.)}}}}]]{{#if: | (translated by [[Mathematician:{{{translated}}}|{{ #if: |{{{2}}}|{{{translated}}}}}]])}}{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: B(x,y+1)=(y/x)B(x+1,y) | ... (previous)|}}{{#if: B(x,y)B(x+y,z)=B(y,z)B(y+z,x) | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: $\S 1.5 (6)$
