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](https://math.stackexchange.com/questions/4399939/prove-or-disprove-that-the-power-of-positive-term-polynomial-will-be-eventually). As @ChrisSanders in the comment pointed out, I just ask the question here...



Prove or disprove that the power of positive term polynomial will be eventually single peak
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.


