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Statement

Setting (p. 1). For $n\ge2$ let $P(z)=\sum_{k=0}^{n-1}X_kz^k$, where the $X_k$ are independent with $\mathbb P(X_k=1)=\mathbb P(X_k=-1)=\tfrac12$, so that $P$ is uniform among the $2^n$ Littlewood polynomials of degree $n-1$. Let $\nu_n$ be the counting measure of the roots of $P$ and $\mathbb D$ the unit disk.

Theorem 1 (pp. 1--2). As $n\to\infty$,

$$\mathbb P\Bigl(\Bigl|\nu_n(\mathbb D)-\frac n2\Bigr|\ge n^{9/10}\Bigr)\to0 .$$

In particular $\nu_n(\mathbb D)/n\to1/2$ in probability as $n\to\infty$.

Equivalently, all but $o(2^n)$ of the $2^n$ Littlewood polynomials of degree $n-1$ have $n/2+o(n)$ roots in $\mathbb D$ (the abstract, p. 1). The paper presents this as an affirmative answer to Problem 4.15 of Hayman's problem book, which it quotes for polynomials $\sum_{k=1}^n\varepsilon_kz^k$ with $\varepsilon_k=\pm1$ and which the fiftieth anniversary reprint lists with no progress reported, and to the same question asked by Borwein, Choi, Ferguson and Jankauskas (p. 1). The author says the exponent $9/10$ is not optimal and that the deviations are probably of order $\sqrt n$, which the paper's methods do not reach (p. 2).

Source. Oren Yakir, Approximately half of the roots of a random Littlewood polynomial are inside the disk, arXiv:2011.06234v2 (2022); published in Studia Math. 261 (2021), 227--240. Labels and pages here are those of arXiv v2: the setting and Theorem 1 on pp. 1--2, the proof in Section 2 on pp. 3--5. The edition read is identified on the source card.

Read depth. Claims checked: the setting and the statement were read clause by clause on the printed pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Section 2, pp. 3--5, assuming Lemma 1.3. Take $\tau=n^{-11/10}$. For the upper bound, Jensen's formula on the circles of radius $1$ and $1+\tau$ bounds $\nu_n(\mathbb D)$ by the difference of the two logarithmic integrals of $P$ divided by $\log(1+\tau)$. Normalizing $P$ by $\sigma(r)=(\mathbb E|P(re^{i\theta})|^2)^{1/2}$ splits off the deterministic part $(\log\sigma(1+\tau)-\log\sigma(1))/\log(1+\tau)$, which a second-order Taylor bound puts within $2\tau n^2$ of $n/2$. An excess of $n^{9/10}$ over $n/2$ then forces one of the two normalized logarithmic integrals to differ from $-\gamma/2$ by at least $n^{-1/5}$, and Chebyshev's inequality with the first two moments from Lemma 1.3 bounds that probability by a constant times $n^{2/5-1/2}(\log n)^2$. The lower bound runs the same way on the circles of radius $1-\tau$ and $1$. A remark on p. 5 gives a second route to the lower bound: by a result of Konyagin and Schlag, $P$ has no root on the unit circle with probability tending to 1, and the reversed polynomial $z^{n-1}P(1/z)$ has the same distribution as $P$.

Dependencies

Lemma 1.3 (p. 2); Jensen's formula (the paper's (2.3), p. 3).

Bears on

• Problem 522: the problem asks whether the number $R_n$ of roots of a random $\pm1$ polynomial of degree $n$ in the closed disk $|z|\le1$ satisfies $R_n/(n/2)\to1$ almost surely. Theorem 1 gives $\nu_n(\mathbb D)/n\to1/2$ in probability for degree $n-1$, with deviation below $n^{9/10}$ with probability tending to 1; it proves convergence in probability, not almost sure convergence.