Difference between revisions of "0F0(;;z)=exp(z)"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric...") |
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| − | <strong>Theorem:</strong> The following formula holds: | + | <strong>[[Exponential in terms of hypergeometric 0F0|Theorem]]:</strong> 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]]. | ||
Revision as of 05:41, 8 February 2016
Theorem: The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric pFq and $e^z$ denotes the exponential.
Proof: █
