Difference between revisions of "B(x,y)B(x+y,z)=B(y,z)B(y+z,x)"

From specialfunctionswiki
Jump to: navigation, search
(Created page with "==Theorem== The following formula holds: $$B(x,y)B(x+y,z)=B(y,z)B(y+z,x),$$ where $B$ denotes the beta function. ==Proof== ==References== * {{BookReference|Higher Transc...")
 
 
Line 7: Line 7:
  
 
==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=B(x,y+1)=(y/(x+y))B(x,y)|next=B(x,y)B(x+y,z)=B(z,x)B(x+z,y)}}: $\S 1.5 (7)$
+
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=B(x,y+1)=(y/(x+y))B(x,y)|next=B(x,y)B(x+y,z)=B(z,x)B(x+z,y)}}: $\S 1.5 (7)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 20:58, 3 March 2018

Theorem

The following formula holds: $$B(x,y)B(x+y,z)=B(y,z)B(y+z,x),$$ where $B$ denotes the beta function.

Proof

References