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Define $P(x)$ to be positive if $P(x)>0$ for $x>0$.

I can prove that a quadratic positive polynomial is the ratio of 2 polynomials with non negative coefficients, for example $\displaystyle x^2-x+1/3=\frac{x^6+1/27}{x^4+x^3+2/3 x^2+1/3 x+1/9}$, and similarly for every $x^2-x+c$ where $c>1/4$. The full proof is not hard and involves some recursive polynomials related to Chebyshev polynomials of the second kind.

This leads me to wonder

QUESTION: in general is it true that $P$ positive $\Leftrightarrow$ $\exists Q,R$ with non-negative coefficients, such that $\displaystyle P=Q/R$?

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    $\begingroup$ 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. $\endgroup$
    – David E Speyer
    Commented May 7, 2023 at 20:23
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    $\begingroup$ 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 $\ge 0 \Leftrightarrow c\ge 1/(2 \cos(\pi/(n+2))^2$. $\endgroup$
    – Yaakov Baruch
    Commented May 8, 2023 at 7:11
  • $\begingroup$ 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. $\endgroup$
    – Peter Cordes
    Commented May 8, 2023 at 14:53
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    $\begingroup$ @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? $\endgroup$
    – Yaakov Baruch
    Commented May 8, 2023 at 15:02
  • $\begingroup$ 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. $\endgroup$
    – Peter Cordes
    Commented May 8, 2023 at 15:06

1 Answer 1

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It is well-known. It is even known that you may take $R=(1+x)^m$ for large enough $m$. See, for example, John Scholes's solution to Problem 11 of the 38th IMO 1997 shortlist. Note that by the real fundamental theorem of algebra, the general case reduces to the case $\deg P\leqslant 2$.

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  • $\begingroup$ clickable version of your link: prase.cz/kalva/short/soln/sh9711.html (you forgot the HTTPS:// part when you pasted it: Stack Exchange will make it clickable if you do.) $\endgroup$
    – Peter Cordes
    Commented May 8, 2023 at 14:56
  • $\begingroup$ @PeterCordes fixed this $\endgroup$
    – Fedor Petrov
    Commented May 8, 2023 at 16:34

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