Difference between revisions of "(b-a)2F1+a2F1(a+1)-b2F1(b+1)=0"

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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=(c-2a-(b-a)z)2F1+a(1-z)2F1(a+1)-(c-a)2F1(a-1)=0|next=findme}}: $\S 2.8 (32)$
+
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=(c-2a-(b-a)z)2F1+a(1-z)2F1(a+1)-(c-a)2F1(a-1)=0|next=(c-a-b)2F1+a(1-z)2F1(a+1)-(c-b)2F1(b-1)=0}}: $\S 2.8 (32)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Revision as of 03:14, 16 September 2016

Theorem

The following formula holds: $$(b-1){}_2F_1(a,b;c;z)+a{}_2F_1(a+1,b;c;z)-b{}_2F_1(a,b+1;c;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1.

Proof

References