For any $t\in (0,1)$ let $t={\sum }_{k=1}^{\infty }{ϵ}_k(t)2^{-k}$ (where ${ϵ}_k(t)\in \{0,1\}$). What is the correct order of magnitude (for almost all $t\in (0,1)$) for$$M_n(t)=\underset{x\in [-1,1]}{\max }\left|\sum _{k\le n}(-1{)}^{{ϵ}_k(t)}{x}^k\right|?$$

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Let [ a_k(t):=(-1)^{\epsilon_k(t)}\in{\pm1},\qquad P_n(x):=\sum_{k=1}^n a_k(t),x^k, \qquad M_n(t)=\sup_{x\in[-1,1]}|P_n(x)|. ] For Lebesgue-a.e. (t\in(0,1)), the bits (\epsilon_k(t)) are i.i.d. Bernoulli($1/2$), hence ((a_k(t))) are i.i.d. Rademacher [[nomath]]($\pm1$ with prob. $1/2$)[[/nomath]]. (We ignore the dyadic…

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.