Difference between revisions of "Beta is symmetric"

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==Proof==
 
==Proof==
 +
Using [[beta in terms of gamma]], we may calculate
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$$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \dfrac{\Gamma(y)\Gamma(x)}{\Gamma(y+x)} = B(y,x),$$
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as was to be shown.
  
 
==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=findme|next=Beta as product of gamma functions}}: $\S 1.5 (4)$
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* {{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}}: $6.2.2$
 
* {{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:Unproven]]
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[[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