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Statement

Problem 4.15 (p. 76, quoted). "If again $\varepsilon_k=\mp1$, is it true that, for large $n$, all but $o(2^n)$ polynomials $P(z)=\sum_{k=1}^{n}\varepsilon_kz^k$ have just $n/2+o(n)$ roots in $\mathbb{D}$?"

The word "again" refers to Problem 4.13 (p. 75), which introduces the polynomials $P(z)=\sum_{k=1}^n\varepsilon_kz^k$ with each $\varepsilon_k=\mp1$. There are $2^n$ such polynomials, so the question asks whether, with the signs chosen independently and uniformly, the number of roots in the unit disc $\mathbb{D}$ divided by $n/2$ tends to $1$ in probability. Chapter 4 does not define $\mathbb{D}$; the notation of Chapter 5 (p. 83) takes $\mathbb{D}$ to be the open unit disc $|z|<1$, and the book writes $\mathbb{D}=\{|\omega|<1\}$ in Problem 6.35 (p. 130). The problem carries no attribution line, and Table 2 (p. 253) lists it among the problems of the 1967 edition.

Update 4.15 (p. 76). No progress had been reported to the authors.

Source. W. K. Hayman and E. F. Lingham, Research Problems in Function Theory, arXiv:1809.07200v2 (21 September 2018), Chapter 4, p. 76. The edition read is identified on the source card.

Read depth. Claims checked: the problem, its update and the definition in Problem 4.13 were read clause by clause on the printed page. The book proves nothing; it poses and reports.

Proof pointer

None; a problem. The Yakir card records Yakir's 2021 Theorem 1 as an affirmative answer, for the polynomials $\sum_{k<n}X_kz^k$ with independent uniform signs; the 2018 update predates it. Since $P(z)=z\sum_{k=1}^n\varepsilon_kz^{k-1}$, the two root counts in the disc differ by the root at $0$.

Dependencies

None.

Bears on

• Problem 522: Problem 4.15 is the in-probability form of the problem's almost-sure question. The book's sum starts at $k=1$ and counts roots in $\mathbb{D}$; the problem's starts at $k=0$ and counts roots in the closed disc. Whether $\mathbb{D}$ is read as the open or the closed disc, an affirmative answer to #522 gives one to Problem 4.15 (observation made here): almost-sure convergence implies convergence in probability, the factor $z$ adds one root at $0$, and the reflection $z\mapsto1/z$ maps the uniform sign polynomials of each degree onto themselves and the roots outside the closed disc onto those inside the open one. The converse does not follow.