Problem Statement
For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). What is the correct order of magnitude (for almost all $t\in(0,1)$) for\[M_n(t)=\max_{x\in [-1,1]}\left\lvert \sum_{k\leq n}(-1)^{\epsilon_k(t)}x^k\right\rvert?\]
Categories:
Analysis Probability Polynomials
Progress
A problem of Salem and Zygmund [SaZy54]. Chung showed that, for almost all $t$, there exist infinitely many $n$ such that\[M_n(t) \ll \left(\frac{n}{\log\log n}\right)^{1/2}.\]Erdős (unpublished) showed that for almost all $t$ and every $\epsilon>0$ we have $\lim_{n\to \infty}M_n(t)/n^{1/2-\epsilon}=\infty$.Source: erdosproblems.com/524 | Last verified: January 15, 2026