Difference between revisions of "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 pFq]] and $e^z$ denotes the [[exponential]].
+
where ${}_0F_0$ denotes the [[hypergeometric 0F0]] and $e^z$ denotes the [[exponential]].
  
 
==Proof==
 
==Proof==
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[[Category:Theorem]]
 
[[Category:Theorem]]
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[[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.

Proof

References