Difference between revisions of "1F1(a;r;z)1F1(a;r;-z)=2F3(a,r-a;r,r/2,r/2+1/2;z^2/4)"
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(Created page with "==Theorem== The following formula holds: $${}_1F_1(a;r;z){}_1F_1(a;r;-z)={}_2F_3\left(a, r-a; r, \dfrac{r}{2}, \dfrac{r}{2}+\dfrac{1}{2}; \dfrac{z^2}{4} \right),$$ where ${}_1...") |
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Revision as of 20:10, 17 June 2017
Theorem
The following formula holds: $${}_1F_1(a;r;z){}_1F_1(a;r;-z)={}_2F_3\left(a, r-a; r, \dfrac{r}{2}, \dfrac{r}{2}+\dfrac{1}{2}; \dfrac{z^2}{4} \right),$$ where ${}_1F_1$ denotes hypergeometric 1F1 and ${}_2F_3$ denotes hypergeometric 2F3.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous): $4.2 (5)$
