Erdős Problem #524 — For any t∈(0,1)t\in (0,1) let t=∑k=1∞ϵk(t)2−kt=\sum_{k=1}^\infty \epsilon_k(t)2^{-k} (where ϵk(t)∈{0,1}\epsilon_k(t)\in \{0,1\}).

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For any t∈(0,1)t\in (0,1) let t=∑k=1∞ϵk(t)2−kt=\sum_{k=1}^\infty \epsilon_k(t)2^{-k} (where ϵk(t)∈{0,1}\epsilon_k(t)\in \{0,1\}). What is the correct order of magnitude (for almost all t∈(0,1)t\in(0,1)) for Mn(t)=max⁡x∈[−1,1]∣∑k≤n(−1)ϵk(t)xk∣?M_n(t)=\max_{x\in [-1,1]}\left\lvert \sum_{k\leq n}(-1)^{\epsilon_k(t)}x^k\right\rvert?

References

Progress summary

Refreshed
Claimed solved

A 2026 paper settles the problem, giving both the largest recurring size and the precise scale of its unusually small values.

The problem asks for the almost-sure size of the maximum of a random polynomial on the interval from minus one to one. The lower-envelope question was raised by Salem and Zygmund and later reiterated by Erdős.

Known results

  • Salem and Zygmund: almost surely, the upper envelope satisfies
lim sup⁡n→∞Mnnlog⁡log⁡n=2.\limsup_{n\to\infty}\frac{M_n}{\sqrt{n\log\log n}}=\sqrt{2}.

Letwin–Sawhney result (2026)

Brayden Letwin and Mehtaab Sawhney determine the lower envelope through Gaussian-process small-ball estimates and obtain

lim inf⁡n→∞log⁡(Mn/n)(log⁡log⁡n)1/3=−3π41/3.\liminf_{n\to\infty}\frac{\log(M_n/\sqrt n)}{(\log\log n)^{1/3}}=-\frac{3\pi}{4^{1/3}}.

Thus the minimum values occurring infinitely often have scale nexp⁡ ⁣(−Θ((log⁡log⁡n)1/3))\sqrt n\exp\!\bigl(-\Theta((\log\log n)^{1/3})\bigr). The accompanying discussion says the argument was completed with assistance from GPT 5.2, with a small input from Gemini/Grok.

Current status (as of June 2026): The almost-sure upper and lower envelopes are settled by the Letwin–Sawhney paper.

Sources

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