Difference between revisions of "0F1(;r;z)0F1(;s;z)=2F1(r/2+s/2, r/2+s/2-1/2;r,s,r+s-1;4z)"
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(Created page with "==Theorem== The following formula holds: $${}_0F_1(;r;z){}_0F_1(;s;z)={}_2F_1 \left( \dfrac{r}{2} + \dfrac{s}{2}, \dfrac{r}{2}+\dfrac{s}{2}-\dfrac{1}{2};r,s,r+s-1;4z \right),$...") |
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Revision as of 19:58, 17 June 2017
Theorem
The following formula holds: $${}_0F_1(;r;z){}_0F_1(;s;z)={}_2F_1 \left( \dfrac{r}{2} + \dfrac{s}{2}, \dfrac{r}{2}+\dfrac{s}{2}-\dfrac{1}{2};r,s,r+s-1;4z \right),$$ where ${}_0F_1$ denotes hypergeometric 0F1 and ${}_2F_1$ denotes hypergeometric 2F1.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.2 (2)$
