Difference between revisions of "(b-a)2F1+a2F1(a+1)-b2F1(b+1)=0"
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $$(b- | + | $$(b-a){}_2F_1(a,b;c;z)+a{}_2F_1(a+1,b;c;z)-b{}_2F_1(a,b+1;c;z)=0,$$ |
where ${}_2F_1$ denotes [[hypergeometric 2F1]]. | where ${}_2F_1$ denotes [[hypergeometric 2F1]]. | ||
Revision as of 03:30, 16 September 2016
Theorem
The following formula holds: $$(b-a){}_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
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 2.8 (32)$
