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Jan 3, 2016 at 14:00 vote accept Dongryul Kim
Jan 3, 2016 at 13:44 answer added Fedor Petrov timeline score: 14
Jan 3, 2016 at 12:50 comment added Peter Mueller @FedorPetrov You are right, my comment gives a weaker bound.
Jan 3, 2016 at 12:43 comment added Fedor Petrov @Peter it gives a twice worse upper bound in more general situation. The difference is seen already for $n=2$ (when there are indeed at most 2 roots, but direct application of Descartes rule does not prove this.)
Jan 3, 2016 at 12:31 comment added Peter Mueller I believe that math.stackexchange.com/questions/688606/… is an answer to a more general question. Probably also mathoverflow.net/questions/44443/… is relevant here.
Jan 3, 2016 at 12:30 history edited Dongryul Kim CC BY-SA 3.0
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Jan 3, 2016 at 12:27 comment added Dongryul Kim Yes. Actually I think one can give small perturbations to each of the $a_i$s and make all multiple roots into distinct simple roots.
Jan 3, 2016 at 11:52 comment added joro Do you count the multiplicity of $0$?
Jan 3, 2016 at 11:16 history asked Dongryul Kim CC BY-SA 3.0