Difference between revisions of "0F0(;;z)=exp(z)"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$e^z={}_0F_0(;;z),$$ | $$e^z={}_0F_0(;;z),$$ | ||
where ${}_0F_0$ denotes the [[hypergeometric pFq]] and $e^z$ denotes the [[exponential]]. | where ${}_0F_0$ denotes the [[hypergeometric pFq]] and $e^z$ denotes the [[exponential]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
Revision as of 03:46, 6 June 2016
Theorem
The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric pFq and $e^z$ denotes the exponential.
