Difference between revisions of "0F0(;;z)=exp(z)"
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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 | + | where ${}_0F_0$ denotes the [[hypergeometric 0F0]] and $e^z$ denotes the [[exponential]]. |
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
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.
