Difference between revisions of "0F1(;r;z)0F1(;r;-z)=0F3(r,r/2,r/2+1/2;-z^2/4)"
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(Created page with "==Theorem== The following formula holds: $${}_0F_1(;r;z){}_0F_1(;r;-z)={}_0F_3\left( ;r, \dfrac{r}{2}, \dfrac{r}{2} + \dfrac{1}{2}; - \dfrac{z^2}{4} \right),$$ where ${}_0F_1$...") |
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Revision as of 20:04, 17 June 2017
Theorem
The following formula holds: $${}_0F_1(;r;z){}_0F_1(;r;-z)={}_0F_3\left( ;r, \dfrac{r}{2}, \dfrac{r}{2} + \dfrac{1}{2}; - \dfrac{z^2}{4} \right),$$ where ${}_0F_1$ denotes hypergeometric 0F1 and ${}_0F_3$ denotes hypergeometric 0F3.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.2 (3)$
