Difference between revisions of "(c-a-b)2F1+a(1-z)2F1(a+1)-(c-b)2F1(b-1)=0"
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(Created page with "==Theorem== The following formula holds: $$(c-a-b){}_2F_1(a,b;c;z)+a(1-z){}_2F_1(a+1,b;c;z)-(c-b){}_2F_1(a,b-1;c;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1. ==Pr...") |
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Revision as of 03:16, 16 September 2016
Theorem
The following formula holds: $$(c-a-b){}_2F_1(a,b;c;z)+a(1-z){}_2F_1(a+1,b;c;z)-(c-b){}_2F_1(a,b-1;c;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 2.8 (33)$