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Apr 16, 2021 at 21:31 vote accept Azad Tasan
Apr 16, 2021 at 20:44 comment added Ben McKay Sard's theorem proves Christian's and Carlo's wiggling argument.
Apr 16, 2021 at 20:34 answer added Iosif Pinelis timeline score: 1
Apr 16, 2021 at 20:11 comment added Christian Remling Just wiggling the constant term slightly will remove multiple zeros, so this probability is zero (if we avoid the exceptional scenarios that Carlo mentioned).
Apr 16, 2021 at 19:55 comment added Carlo Beenakker I would think that the set of roots with multiplicity $>1$ has measure zero, at least if $S_f$ contains both $s=0$ and $s=1$ (to exclude $x=0$)
Apr 16, 2021 at 19:24 history asked Azad Tasan CC BY-SA 4.0