Difference between revisions of "Beta is symmetric"
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==Proof== | ==Proof== | ||
| + | Using [[beta in terms of gamma]], we may calculate | ||
| + | $$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \dfrac{\Gamma(y)\Gamma(x)}{\Gamma(y+x)} = B(y,x),$$ | ||
| + | as was to be shown. | ||
==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)=integral (t^(x-1)+t^(y-1))(1+t)^(-x-y) dt|next=Beta as product of gamma functions}}: $\S 1.5 (4)$ |
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Beta in terms of gamma|next=Digamma}}: $6.2.2$ | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proven]] |
Latest revision as of 20:57, 3 March 2018
Theorem
The following formula holds: $$B(x,y)=B(y,x),$$ where $B$ denotes the beta function.
Proof
Using beta in terms of gamma, we may calculate $$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \dfrac{\Gamma(y)\Gamma(x)}{\Gamma(y+x)} = B(y,x),$$ as was to be shown.
References
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (4)$
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $6.2.2$
