Difference between revisions of "0F0(;;z)=exp(z)"
From specialfunctionswiki
m (Tom moved page Exponential in terms of hypergeometric 0F0 to 0F0(;;z)=exp(z)) |
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The following formula holds: | 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== | ==Proof== | ||
Revision as of 21:37, 26 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.
