Difference between revisions of "1F1(a;2a;z)1F1(b;2b;-z)=2F3(a/2+b/2,a/2+b/2+1/2;a+1/2,b+1/2,a+b;z^2/4)"
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(Created page with "==Theorem== The following formula holds: $${}_1F_1(a;2a;z){}_1F_1(b;2b;-z)={}_2F_3 \left( \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+\dfrac{1}{...") |
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Revision as of 20:18, 17 June 2017
Theorem
The following formula holds: $${}_1F_1(a;2a;z){}_1F_1(b;2b;-z)={}_2F_3 \left( \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+\dfrac{1}{2}, b+\dfrac{1}{2}, a+b; \dfrac{z^2}{4} \right),$$ where ${}_1F_1$ denotes hypergeometric 1F1 and ${}_2F_3$ denotes hypergeometric 2F3.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.2 (6)$
