Asymptotic Zero Distribution of Random Polynomials Spanned by General Bases

Abstract

Zeros of Kac polynomials spanned by monomials with i.i.d. random coefficients are asymptotically uniformly distributed near the unit circumference. We give estimates of the expected discrepancy between the zero counting measure and the normalized arclength on the unit circle. Similar results are established for polynomials with random coefficients spanned by different bases, e.g., by orthogonal polynomials. We show almost sure convergence of the zero counting measures to the corresponding equilibrium measures, and quantify this convergence, relying on the potential theoretic methods developed for deterministic polynomials. Applications include estimates of the expected number of zeros in various sets. Random coefficients may be dependent and need not have identical distributions in our results.

Zero distribution of random polynomials

“…for a fixed t > 1. Detailed proofs of these statements may be found in [25], and we confine ourselves to an outline of the necessary arguments in this paper.…”

Equidistribution of zeros of random polynomials

“…If the random coefficients satisfy mild assumptions such as in Theorem 2.1, then the zero counting measures of random polynomials P n (z) = n k=0 A k B k (z) converge almost surely to µ E for very general sets E and associated bases {B k } ∞ k=0 . We direct the reader to the recent papers [8], [3,4] and [22,23], and to references found therein. However, the necessity part of Theorem 2.1 seems to be open in such general setting.…”

“…Sufficient conditions for almost sure equidistribution of zeros of random orthogonal polynomials were considered by Shiffman and Zelditch [26] and [27], Bloom [6] and [7], Bloom and Shiffman [9], Bloom and Levenberg [8], Bayraktar [3] and [4], and others. Pritsker [22] and [23] considered zero distribution for random polynomials spanned by general bases.…”

Random Bernstein–Markov factors

“…Continuing with the proof of Theorem 2, we apply Lemma 8 with X k = |A k | 2 , a k = k 2 , and S n = n k=1 k 2 to obtain see, e.g., [7]. Therefore, with probability one, both series Hence the first part of this theorem follows.…”