Considering: sum_n(z^n)/(n!)=e^z

Zaiden Soto

Zaiden Soto

Open question

2022-08-22

Considering:
n z n n ! = e z

Answer & Explanation

Javier Mathis

Javier Mathis

Beginner2022-08-23Added 7 answers

n = 0 z n n ! ! = e z 2 2 ( 1 + π 2 erf ( z 2 ) )
where appears the error function
n = 0 z 2 n ( 2 n ) ! ! = e z 2 2
n = 0 z 2 n + 1 ( 2 n + 1 ) ! ! = π 2 e z 2 2 erf ( z 2 )

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