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May 8, 2023 at 15:06 comment added Peter Cordes Ah, I see now, thanks. So Q=P / R=1 is a valid solution for some P, but the interesting non-trivial cases to prove are P polynomials that have negative coefficients.
May 8, 2023 at 15:02 comment added Yaakov Baruch @PeterCordes I'm not sure I follow. $P$ is a "positive" polynomial and can have negative coefficients, while $Q$ is a non-zero polynomial with non-negative coefficients. So $R$ can be $1$ but it will only work for those $P$ that already have no negative coefficients. Does that clarify the question?
May 8, 2023 at 14:53 comment added Peter Cordes Isn't this trivially true because R = 0*x + 1 is a polynomial with non-negative coefficients, so Q=P, dividing by 1 doesn't change the value? Or does 1 not count as a polynomial in standard math terminology, since its value doesn't actually vary with x? I assume you want to rule that out trivial solution, but I don't know if you need any extra restriction in your definitions.
May 8, 2023 at 7:11 comment added Yaakov Baruch Thank you both for the great comments, pointing to results stronger than what I suspected. The connection with Chebyshev that I mentioned can at least be used to prove a small optimality result: $\exists P$ with degree $\le n$ and both $P$ and $P(x)(x^2-x+c)$ have coefficients $\ge 0 \Leftrightarrow c\ge 1/(2 \cos(\pi/(n+2))^2$.
May 7, 2023 at 21:38 history became hot network question
May 7, 2023 at 20:23 comment added David E Speyer There is also a cute way of doing this without using the fundamental theorem of algebra, math.stackexchange.com/a/1727591/448 . That one has the advantage of generalizing to multivariate polynomials.
May 7, 2023 at 15:59 vote accept Yaakov Baruch
May 7, 2023 at 15:52 answer added Fedor Petrov timeline score: 17
May 7, 2023 at 13:37 history asked Yaakov Baruch CC BY-SA 4.0