Difference between revisions of "Beta is symmetric"
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==Proof== | ==Proof== | ||
| − | + | Using [[beta in terms of gamma]], we may calculate | |
$$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \dfrac{\Gamma(y)\Gamma(x)}{\Gamma(y+x)} = B(y,x),$$ | $$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \dfrac{\Gamma(y)\Gamma(x)}{\Gamma(y+x)} = B(y,x),$$ | ||
as was to be shown. | as was to be shown. | ||
Revision as of 15:12, 6 October 2016
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: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (4)$
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $6.2.2$
