Difference between revisions of "(b-a)2F1+a2F1(a+1)-b2F1(b+1)=0"
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==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=(c-2a-(b-a)z)2F1+a(1-z)2F1(a+1)-(c-a)2F1(a-1)=0 |next=}}: $\S 2.8 (32)$ | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=(c-2a-(b-a)z)2F1+a(1-z)2F1(a+1)-(c-a)2F1(a-1)=0|next=findme}}: $\S 2.8 (32)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Revision as of 03:12, 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.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 2.8 (32)$
