Difference between revisions of "(b-a)2F1+a2F1(a+1)-b2F1(b+1)=0"
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(Created page with "==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== ==Ref...") |
(No difference)
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Revision as of 03:10, 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.
