Littlewood in [L] states several conjectures regarding asymptotics of polynomials with $\pm1$ coefficients. He considers the class $\mathscr F$ of polynomials of form $\sum^n\pm z^m$ and asks whether there exists a sequence $f_n$ in $\mathscr F$ such that $$A\leq\frac{|f_n(\theta)|}{\sqrt{n+1}}\leq B$$ for large n's.

1. I'd like to know if this is still open and would be grateful if someone could provide references for recent survey papers.

2. If the conjecture is not settled yet, I was thinking if one can apply integer programming methods to tackle the problem?

Update 1 As pointed out by Fedor, J-P. Kahane's work gives an elegant proof for polynomials with unimodular coefficients. His result is as follows(I didn't translate it to English as seems rather easy to understand ):

Theoreme. Il existe une suite de polynomes $$P_n(z)=\sum_{m=1}^{n}a_{m,n}z^m,\,(|a_{m,n}| = 1; \, n=1,2,\ldots,m=1,\ldots,n)$$ et une suite $\epsilon_n$ positive tendant vers 0 telles que pour tout z de module 1 on ait$$(1-\epsilon_n)\sqrt{n}\leq|P_n(z)|\leq(1+\epsilon_n)\sqrt{n}.$$

Numerical experiments suggest there might be hope to construct a sequence of $P_n\in\mathbb {Z}_2[x]$ which satisfies Kahane's asymptotics.

[L]Littlewood, J. E., On polynomials $\sum^n\pm z^m, \sum^n e^{\alpha_mi}z^m, z = e^{\theta i}$, J. Lond. Math. Soc. 41, 367-376 (1966). ZBL0142.32603.

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    J.-P. Kahane (Bull. London Math. Soc. 12 (1980), 321–342) answers this question in positive for polynomials with coefficients on the unit circle, and even proves that $A$ and $B$ may be arbitrarily close to 1. At least in this strength the question looks to be open. It was called open in a recent paper by A. Prikhodko (in which such polynomials are applied to spectral ergodic theory.) – Fedor Petrov Nov 4 at 21:40
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    What does $\mathbb{Z}_2[x]$ stand for in the edit - polynomials with $\pm 1$ coefficients? You may also look up Bombieri and Bourgain's 2009 paper in J. European Math. Soc. (Volume 11, Issue 3, 2009, pp. 627–703) for some significant quantitative improvement on Kahane's result. – Vesselin Dimitrov Nov 6 at 17:38

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