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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 7, 2015 at 12:57 comment added Will Sawin The bound cannot actually be super exponential, as there is a fixed vector such that the probability that it is in the kernel is $2^{-m}$.
Jul 7, 2015 at 3:05 comment added Igor Rivin @WillSawin Well, very special $v$ is not random...
Jul 7, 2015 at 2:43 comment added Will Sawin There are several complexities here, like that the bound is not right for short $v$, and especially for very special $v$ like a $v$ with only a couple nonzero entries.
Jul 7, 2015 at 2:11 history answered Igor Rivin CC BY-SA 3.0