0F0(;;z)=exp(z)
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Revision as of 21:36, 26 June 2016 by Tom (talk | contribs) (Tom moved page Exponential in terms of hypergeometric 0F0 to 0F0(;;z)=exp(z))
Theorem
The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric pFq and $e^z$ denotes the exponential.
