Polynomials – Prove or Disprove Single Peak of Positive Term Polynomial

inequalitiespolynomials

This is a question that a classmate asked me three years ago.

Let $P(x)=\sum_{i=0}^n a_ix^i$ be a polynomial such that each $a_i>0$. Prove or disprove that there exists a positive integer $r$ such that $P(x)^r=\sum_{i=0}^{nr} b_ix^i$ and there exists $0\le j\le nr$ such that $b_0\le b_1\le \dots\le b_{j}$ and $b_{j+1}\ge\dots\ge b_{nr}$.

This problem may have a probability problem as its original problem, since the classmate asked this while taking a probability class. What I could do is to try to apply CLT to claim that the "central terms" has only one peak and try to select some $r$ to make the first few terms and last few terms increasing/decreasing (that is, I proved that I can choose an $r$ to make $b_0\le b_1\le\dots \le b_k$ for any $k$, and same at the other side). But I failed to prove the problem… I also tried to factorize it into smaller quadratic polynomials but also failed… Also, I can't come up with a counterexample, too…

Note: if polynomials $p,q$ are single peak, this DOES NOT imply that $pq$ is single peak. For example: $p(x)=1+x+100x^2$ and $q(x)=10000+10000x+10100x^2+9000x^3+9000x^4$. But $p(x)q(x)=900000x^6+909000x^5+1028000x^4+1019100x^3+1020100x^2+20000x+10000$

P.S. This is exactly same as the problem in MSE. As @ChrisSanders in the comment pointed out, I just ask the question here…

Best Answer

This is answered affirmatively by Odlyzko and Richmond, On the unimodality of high convolutions of discrete distributions, Annals of probability (1985) 299--306: all sufficiently large powers of the polynomial (with positive coefficients and no gaps) are strongly unimodal, that is, the coefficients form a log concave sequence. The proof uses estimates of contour integrals over circles of just the right radius.

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