Difference between revisions of "2F0(a,b;;z)2F0(a,b;;-z)=4F1(a,b,a/2+b/2,a/2+b/2+1/2;a+b;4z^2)"

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(Created page with "==Theorem== The following formula holds: $${}_2F_0(a,b;;z){}_2F_0(a,b;;-z)={}_4F_1 \left(a, b, \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+b; 4z...")
 
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=0F1(;r;z)0F1(;r;-z)=0F3(r,r/2,r/2+1/2;-z^2/4)|next=findme}}: $4.2 (4)$  
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=0F1(;r;z)0F1(;r;-z)=0F3(r,r/2,r/2+1/2;-z^2/4)|next=1F1(a;r;z)1F1(a;r;-z)=2F3(a,r-a;r,r/2,r/2+1/2;z^2/4)}}: $4.2 (4)$
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=2F0(a,b;;z)2F0(a,b;;-z)=4F1(a,b,a/2+b/2,a/2+b/2+1/2;a+b;4z^2)|next=}}: $4.2 (4)$  
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Revision as of 20:08, 17 June 2017

Theorem

The following formula holds: $${}_2F_0(a,b;;z){}_2F_0(a,b;;-z)={}_4F_1 \left(a, b, \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+b; 4z^2 \right),$$ where ${}_2F_0$ denotes hypergeometric 2F0 and ${}_4F_1$ denotes hypergeometric 4F1.

Proof

References