Difference between revisions of "0F0(;;z)=exp(z)"
From specialfunctionswiki
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[[Category:Theorem]] | [[Category:Theorem]] | ||
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Latest revision as of 04:11, 30 June 2016
Theorem
The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric 0F0 and $e^z$ denotes the exponential.