0F0(;;z)=exp(z)
From specialfunctionswiki
Theorem: The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric pFq and $e^z$ denotes the exponential.
Proof: █
Theorem: The following formula holds: $$e^z={}_0F_0(;;z),$$ where ${}_0F_0$ denotes the hypergeometric pFq and $e^z$ denotes the exponential.
Proof: █
