Difference between revisions of "(c-a-b)2F1-(c-a)2F1(a-1)+b(1-z)2F1(b+1)=0"
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(Created page with "==Theorem== The following formula holds: $$(c-a-b){}_2F_1(a,b;c;z)-(c-a){}_2F_1(a-1,b;c;z)+b(1-z){}_2F_1(a,b+1;c;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1. ==Pro...") |
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Revision as of 03:25, 16 September 2016
Theorem
The following formula holds: $$(c-a-b){}_2F_1(a,b;c;z)-(c-a){}_2F_1(a-1,b;c;z)+b(1-z){}_2F_1(a,b+1;c;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous): $\S 2.8 (36)$