Difference between revisions of "(b-a)2F1+a2F1(a+1)-b2F1(b+1)=0"

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(Created page with "==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== ==Ref...")
 
 
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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$(b-1){}_2F_1(a,b;c;z)+a{}_2F_1(a+1,b;c;z)-b{}_2F_1(a,b+1;c;z)=0,$$
+
$$(b-a){}_2F_1(a,b;c;z)+a{}_2F_1(a+1,b;c;z)-b{}_2F_1(a,b+1;c;z)=0,$$
 
where ${}_2F_1$ denotes [[hypergeometric 2F1]].
 
where ${}_2F_1$ denotes [[hypergeometric 2F1]].
  
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==References==
 
==References==
 +
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=(c-2a-(b-a)z)2F1+a(1-z)2F1(a+1)-(c-a)2F1(a-1)=0|next=(c-a-b)2F1+a(1-z)2F1(a+1)-(c-b)2F1(b-1)=0}}: $\S 2.8 (32)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 23:23, 3 March 2018

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