GPT 6 Astra Ultra attempt on Math-5714

FORMAL RESTATEMENT

Erdős #525; the counting statement from minimum-modulus universality, 2026-09-24. Among the $2^{n+1}$ polynomials $F_n(z)=\sum_{j=0}^n\epsilon_jz^j$, $\epsilon_j\in\{-1,1\}$, show that all but $o(2^n)$ have $\min_{|z|=1}|F_n(z)|<1$.

QUICK LITERATURE AND CONTEXT CHECK

Cook--Nguyen prove an exponential limiting distribution for the minimum modulus, extending Yakir--Zeitouni's Gaussian result. We use their stated universality theorem as the external input and derive the exact counting conclusion, keeping normalization and probability quantifiers explicit.

OBJECTIVE AND ATTACK PLAN

Translate the normalized trigonometric-polynomial theorem to ordinary sign polynomials, then use a fixed-threshold limit followed by a second limit. A distributional theorem alone should not be silently promoted to almost-sure eventual behavior for a coupled infinite sign sequence.

WORK

The external theorem and normalization. For independent symmetric signs $\xi_{-m},\ldots,\xi_m$, define \[ P_m(t)=\frac1{\sqrt{2m+1}}\sum_{j=-m}^m\xi_j e^{ijt}, \qquad \mu_m=\min_t|P_m(t)|. \] Cook--Nguyen, Corollary 1.5, gives, for every fixed $\tau>0$, \[ \lim_{m\to\infty}\mathbb P(m\mu_m>\tau)=e^{-\lambda\tau}, \qquad \lambda=2\sqrt{\pi/3}>0. \tag{1} \] Their discussion immediately following $1.4$ explicitly extends the arguments to odd polynomial degrees as well. Symmetric signs have mean zero, variance one, and a sub-Gaussian tail, so they meet the hypotheses of the theorem. Multiplication by $e^{imt}$ does not affect modulus. Thus, for degree $n=2m$ and $M_n=\min_{|z|=1}|F_n(z)|$, \[ \sqrt n\,M_n =\frac{\sqrt{2m(2m+1)}}{m}\,(m\mu_m). \] The multiplier tends to $2$. Equation $1$, and its odd-degree extension, therefore yield the full-sequence limit \[ \lim_{n\to\infty}\mathbb P(\sqrt n\,M_n>T) =e^{-\lambda T/2}\qquad(T>0). \tag{2} \] The continuous exponential limit justifies using either strict or non-strict inequalities at a fixed positive threshold.

The required counting estimate. Fix $T>0$. Once $\sqrt n>T$, \[ \{M_n\geq1\}\subseteq\{\sqrt n\,M_n>T\}. \] Taking the upper limit in probability and using $2$ gives $\limsup_n\mathbb P(M_n\geq1)\leq e^{-\lambda T/2}$. Now let $T\to\infty$. The right side tends to zero, so $\mathbb P(M_n\geq1)=o(1)$. Uniform independent signs assign probability $2^{-(n+1)}$ to each coefficient vector. Consequently \[ \#\{(\epsilon_0,\ldots,\epsilon_n):M_n\geq1\} =2^{n+1}o(1)=o(2^n), \] exactly as required.

What the theorem additionally says. For any fixed $0<a<b$, the limiting probability that $a<\sqrt n\,M_n\leq b$ is $e^{-\lambda a/2}-e^{-\lambda b/2}$. Thus the natural scale is $n^{-1/2}$. The argument also proves $M_n\to0$ in probability, by replacing $1$ with any fixed positive threshold. It supplies no summable upper bound on exceptional probabilities, so no almost-sure eventual assertion is inferred here.

RESULT AND LIMITATIONS

Attempt status: solved. The counting conclusion and all normalization and limiting steps are complete, conditional only on the explicitly stated Cook--Nguyen universality theorem and its odd-degree extension. The deep small-ball and local central-limit estimates proving that theorem are not reconstructed in this attempt.

SOURCES

https://www.erdosproblems.com/525; N. Cook and H. Nguyen, Universality of the minimum modulus for random trigonometric polynomials, Theorem 1.2 and Corollary 1.5, https://arxiv.org/abs/2101.07203, published in Discrete Analysis 2021:20.

Completion rate estimate: 100%.