Difference between revisions of "(c-a-1)2F1+a2F1(a+1)-(c-1)2F1(c-1)=0"
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(Created page with "==Theorem== The following formula holds: $$(c-a-1){}_2F_1(a,b;c;z)+a{}_2F_1(a+1,b;c;z)-(c-1){}_2F_1(a,b;c-1;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1. ==Proof==...") |
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Revision as of 03:22, 16 September 2016
Theorem
The following formula holds: $$(c-a-1){}_2F_1(a,b;c;z)+a{}_2F_1(a+1,b;c;z)-(c-1){}_2F_1(a,b;c-1;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous): $\S 2.8 (35)$