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S Apr 9, 2023 at 13:10 vote accept Oleksandr Kulkov
Apr 1, 2023 at 18:10 answer added Oleksandr Kulkov timeline score: 1
Nov 11, 2017 at 20:35 vote accept Oleksandr Kulkov
S Apr 9, 2023 at 13:10
Nov 10, 2017 at 23:40 answer added Pietro Majer timeline score: 1
Nov 5, 2017 at 20:45 answer added Terry Tao timeline score: 16
Nov 3, 2017 at 16:46 answer added Pietro Majer timeline score: 6
Nov 3, 2017 at 12:11 answer added Wolfgang timeline score: 2
Nov 3, 2017 at 7:48 answer added მამუკა ჯიბლაძე timeline score: 9
Nov 3, 2017 at 6:20 history edited Oleksandr Kulkov CC BY-SA 3.0
added 7 characters in body
Nov 3, 2017 at 4:12 history edited Oleksandr Kulkov CC BY-SA 3.0
added 55 characters in body
Nov 3, 2017 at 4:10 comment added Oleksandr Kulkov You got it wrong, $a_n$ is the polynomial from $x$. So $\forall k \hookrightarrow |[x^k]a_n(x)| \leq 1$.
Nov 3, 2017 at 3:50 comment added Anthony Quas What is $x$? It seems hard to decide if $|a_n|>1$ without knowing what $x$ is.
Nov 3, 2017 at 2:37 review First posts
Nov 3, 2017 at 2:55
Nov 3, 2017 at 2:34 history asked Oleksandr Kulkov CC BY-SA 3.0