この記事の日本語版はまだありません。英語版を表示しています。

Erdős Problem #1132

Partial progress by Terence Tao with AlphaEvolve, Claude, Gemini Pro and GPT-5.4 Pro (24 Mar 2026): partial result.

問題
Erdős problem #1132 (analysis, polynomials); current status: open
貢献
Terence Tao with AlphaEvolve, Claude, Gemini Pro and GPT-5.4 Pro · AI working with humans
検証
Listed in Tao's wiki
経緯
First AI contribution 2026-03-24 · wiki data as of 2026-06-30
出典
Problem page (erdosproblems.com)Tao's AI-contributions wiki
部分的
結果
Per Tao's AI-contributions wiki

The problem

For x1,…,xn∈[−1,1]x_1,\ldots,x_n\in [-1,1] let

lk(x)=∏i≠k(x−xi)∏i≠k(xk−xi),l_k(x)=\frac{\prod_{i\neq k}(x-x_i)}{\prod_{i\neq k}(x_k-x_i)},

which are such that lk(xk)=1l_k(x_k)=1 and lk(xi)=0l_k(x_i)=0 for i≠ki\neq k.

Let x1,x2,…∈[−1,1]x_1,x_2,\ldots\in [-1,1] be an infinite sequence, and let

Ln(x)=∑1≤k≤n∣lk(x)∣,L_n(x) = \sum_{1\leq k\leq n}\lvert l_k(x)\rvert,

where each lk(x)l_k(x) is defined above with respect to x1,…,xnx_1,\ldots,x_n.

Must there exist x∈(−1,1)x\in (-1,1) such that

Ln(x)>2πlog⁡n−O(1)L_n(x) >\frac{2}{\pi}\log n-O(1)

for infinitely many nn?

Is it true that

lim sup⁡n→∞Ln(x)log⁡n≥2π\limsup_{n\to \infty}\frac{L_n(x)}{\log n}\geq \frac{2}{\pi}

for almost all x∈(−1,1)x\in (-1,1)?

Statement quoted from erdosproblems.com/1132, maintained by Thomas Bloom. Original source: [Er67,p.68][Va99,2.43]. Current status there: open.

AI contributions

As recorded in Terence Tao's wiki AI contributions to Erdős problems (data as of 30 Jun 2026). 🟢 full, 🟡 partial, ⚪ unverified, 🔴 incorrect.

Date Kind AI systems Humans Outcome
2 Feb-24 Mar, 2026 AI working with humans AlphaEvolve, Claude, Gemini Pro, GPT-5.4 Pro Terence Tao 🟡 Partial result

Caveats

The wiki's classifications are provisional and are not a benchmark. Many Erdős problems have never had a thorough literature review, so a result that looks new may later turn out to be known; some problems were also stated ambiguously or incorrectly. See the problem page for the full discussion.

∎