Timeline for A finite alternating sum

12 events

when toggle format	what		by	license	comment
Nov 1, 2019 at 23:21	review	Suggested edits
Nov 2, 2019 at 9:37
Oct 30, 2019 at 16:24	comment	added	esg		For $j=0$ the equality $\sum_{n\geq 0}n^j t^j=\frac{t A_j(t)}{(1-t)^{j+1}}$ uses the convention $0^0:=0$. With $0^0:=1$ you arrive at Fedor Petrov's result.
Oct 30, 2019 at 7:45	comment	added	Fedor Petrov		You do not actually need Euler polynomials: for fixed $n$ we get $\sum_j (-e)^{-j}n^j t^j/j!=\exp(-nt/e)$, then summing by $n$ we have $\sum_n \exp(-nt/e)t^n=\sum_n (t\exp(-t/e))^n=1/(1-t\exp(-t/e))$, hm, this slightly differs from yours formula. Let's check the coefficients of $t^0$.
Oct 30, 2019 at 5:29	vote	accept	Francisco
Oct 30, 2019 at 5:28	vote	accept	Francisco
Oct 30, 2019 at 5:28
Oct 30, 2019 at 5:27	comment	added	Francisco		Thank you! I was not familiarized with Eulerian polynomials. I really enjoyed learning about this and the approach.
Oct 28, 2019 at 0:52	history	edited	WhatsUp	CC BY-SA 4.0	added 357 characters in body
Oct 28, 2019 at 0:46	history	edited	WhatsUp	CC BY-SA 4.0	added 357 characters in body
Oct 28, 2019 at 0:40	history	edited	WhatsUp	CC BY-SA 4.0	added 357 characters in body
Oct 28, 2019 at 0:29	history	edited	WhatsUp	CC BY-SA 4.0	added 357 characters in body
Oct 28, 2019 at 0:22	history	edited	WhatsUp	CC BY-SA 4.0	added 357 characters in body
Oct 27, 2019 at 23:59	history	answered	WhatsUp	CC BY-SA 4.0