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: $${}_2F_0(a,b;;z){}_2F_0(a,b;;-z)={}_4F_1 \left(a, b, \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+b; 4z...") |
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Revision as of 20:07, 17 June 2017
Theorem
The following formula holds: $${}_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
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.2 (4)$
