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Erdős #521 · analysis / polynomials

Let (ϵk)k≥0(\epsilon_k)_{k\geq 0} be independently uniformly chosen at random from {−1,1}\{-1,1\}. If RnR_n counts the number of real roots of fn(z)=∑0≤k≤nϵkzkf_n(z)=\sum_{0\leq k\leq n}\epsilon_k z^k then is it true that, almost surely,

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Problem statement

Let (ϵk)k≥0(\epsilon_k)_{k\geq 0} be independently uniformly chosen at random from {−1,1}\{-1,1\}. If RnR_n counts the number of real roots of fn(z)=∑0≤k≤nϵkzkf_n(z)=\sum_{0\leq k\leq n}\epsilon_k z^k then is it true that, almost surely,

lim⁡n→∞Rnlog⁡n=2π?\lim_{n\to \infty}\frac{R_n}{\log n}=\frac{2}{\pi}?

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Background

Erdős and Offord [EO56] showed that the number of real roots of a random degree nn polynomial with ±1\pm 1 coefficients is (2π+o(1))log⁡n(\frac{2}{\pi}+o(1))\log n.

It is ambiguous in [Er61] whether Erdős intended the coefficients to be uniformly chosen from {−1,1}\{-1,1\} or {0,1}\{0,1\}. In the latter case, the constant 2π\frac{2}{\pi} should be 1π\frac{1}{\pi} (see the discussion in the comments).

In the case of {−1,1}\{-1,1\} Do [Do24] proved that, if Rn[−1,1]R_n[-1,1] counts the number of roots in [−1,1][-1,1], then, almost surely,

lim⁡n→∞Rn[−1,1]log⁡n=1π.\lim_{n\to \infty}\frac{R_n[-1,1]}{\log n}=\frac{1}{\pi}.

See also [522].

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #521 link
  2. [2]Upstream Lean 4 statement for Erdős #521 linkExternal statement reference. Not compiled or verified by this platform; upstream files may contain open proof obligations.

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