Difference between revisions of "B(x,y)B(x+y,z)=B(z,x)B(x+z,y)"
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(Created page with "==Theorem== The following formula holds: $$B(x,y)B(x+y,z)=B(z,x)B(x+z,y),$$ where $B$ denotes the beta function. ==Proof== ==References== * {{BookReference|Higher Transc...") |
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==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=B(x,y)B(x+y,z)=B(y,z)B(y+z,x)|next=B(x,y)B(x+y,z)B(x+y+z,u)=Gamma(x)Gamma(y)Gamma(z)Gamma(u)/Gamma(x+y+z+u)}}: $\S 1.5 (7)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 20:58, 3 March 2018
Theorem
The following formula holds: $$B(x,y)B(x+y,z)=B(z,x)B(x+z,y),$$ where $B$ denotes the beta function.
Proof
References
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (7)$
