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)) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 2: | Line 2: | ||
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== | ||
| Line 9: | Line 9: | ||
[[Category:Theorem]] | [[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.
