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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...

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  • $\begingroup$ Not only can the product of two single-peak polynomials be non-single-peak, but also powers of a single-peak polynomial can be non-single-peak. E.g., the polynomial $(1 + x + 10 x^2)^r$ is non-single-peak for $r=2,\dots,9$. $\endgroup$ Commented Mar 11, 2022 at 2:46
  • $\begingroup$ Even a greater violation of the monotonicity of the single-peakedness in $r$: For $r\in\{1,\dots,40\}$, the polynomial $(10+x+x^2+10x^3)^r$ is single-peak only for $r\in\{30,32,34,36,37,38,39,40\}$. $\endgroup$ Commented Mar 11, 2022 at 4:25
  • $\begingroup$ @IosifPinelis I wonder whether your examples can be a counterexample... or not? Since I am not sure how to prove a polynomial to be a counterexample... I have thought the examples of $Ax^2+Bx+C$ where $B\sim 0$, but failed to do so... $\endgroup$
    – JetfiRex
    Commented Mar 11, 2022 at 4:40
  • $\begingroup$ I think your conjecture is true and have a very vague idea of how it could be proved, but at this point am far from an actual proof. $\endgroup$ Commented Mar 11, 2022 at 6:21

2 Answers 2

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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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  • $\begingroup$ Thank you! Since I can't find the correct key word, I was not able to find it... $\endgroup$
    – JetfiRex
    Commented Mar 13, 2022 at 1:22
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A rough sketch.

Each polynomial $P_n(x)$ with positive coefficients $a_i>0$, for $0\leq i\leq n$ induces a probability distribution on the set $\{0,1,\dots,n\}$ with probability weights specified by the coefficients of $P_n(x)$ Namely, a random variable $X$ is distributed according to $P_n(x)$, if, for $i=0,1,\dots,n$, $$\text{Prob}(X = i)=\frac{a_i}{P_n(1)}.$$ If $X$ and $Y$ are random variables distributed according to $P(x)$ and $Q(x)$, respectively, then $X + Y$ is distributed according to the product $P(x)Q(x)$. In particular, let $X_1,\dots,X_r$ be independent random variables distributed according to $P(x)$. Then their sum $X_1+\cdots+X_r$ is distributed according to $P(x)^r$. The central limit theorem therefore suggests that the coefficients of $P(x)^r$ should eventually become “more log-concave” as $r$ increases.

We know this: if a log-concave polynomial has no internal zeroes then it is unimodal. This applies to your polynomials $P_n(x)$ here, due to the hypothesis $a_i>0$. Hence, your polynomials $P_n(x)^r$ will eventually settle down to unimodality.

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  • $\begingroup$ Thank you, that is a nice extension to what I am thinking for ''apply CLT'' in the statement to.''claim that the middle ones are ''somewhat'' single peak.'' Unimodel is kind of close to single peak though... $\endgroup$
    – JetfiRex
    Commented Mar 11, 2022 at 22:06
  • $\begingroup$ @JetfiRex: Isn't "single peak" exactly the same as "unimodal"? If not, you should explain the difference. $\endgroup$ Commented Mar 12, 2022 at 18:35
  • $\begingroup$ @SamHopkins: JetfiRex was misled by my remark (deleted now) that said "no guarantee for unique peak". My mistake it was. $\endgroup$ Commented Mar 12, 2022 at 18:38
  • $\begingroup$ @SamHopkins When I tried to read wiki, it is said that "unimodal" is kind-of ambiguous... One interpretation is that unimodal means that "there is exactly one maximum". $\endgroup$
    – JetfiRex
    Commented Mar 12, 2022 at 23:27
  • $\begingroup$ @T.Amdeberhan For the deleted remark, when I tried to read wiki, it is said that "unimodal" is kind-of ambiguous... One interpretation is that unimodal means that "there is exactly one maximum", and the other is "single-peak" or so... Also, could you elaborate more about the content (from "we know this") there...? Could you give a reference (if possible) and how this proved the problem (since... I don't know what is "internal zero")... And also log-concave is really strong result though, at least stronger than single peak. But... I am not sure whether log-concave is a thing that can be proved. $\endgroup$
    – JetfiRex
    Commented Mar 12, 2022 at 23:27

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