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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 14, 2014 at 2:33 comment added Qiaochu Yuan So if $n = 2k$ is even we can get small eigenvalue $1 + e^{2 \pi i (k+1)/2k} = 1 - e^{\pi i / k}$ which has absolute value $O \left( \frac{1}{k} \right)$. By comparison, Denis Serre's argument in the linked answer gives $O \left( \frac{1}{k^{O(1) \log k + O(1)}} \right)$, I think. (Really not a fan of constants today.)
Feb 13, 2014 at 22:34 history answered Gerry Myerson CC BY-SA 3.0