Erdős Problem #1131

Erdős Problem 1131 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1131

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1131.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1131.tex.

FORMAL RESTATEMENT

Let $n\ge 1$ and let $x_1<\dots<x_n$ be distinct points in $[-1,1]$. Let $l_k$ be the associated Lagrange fundamental polynomials. Define \[ I(x_1,\dots,x_n):=\int_{-1}^1\sum_{k=1}^n |l_k(x)|^2\,dx. \] Since all quantities are real on $[-1,1]$, $|l_k(x)|^2=l_k(x)^2$. Problem: determine \[ \min_{x_1,\dots,x_n\in[-1,1]\ \text{distinct}} I(x_1,\dots,x_n) \] and in particular decide whether \[ \min I = 2-(1+o(1))\frac{1}{n}\qquad (n\to\infty) \] holds.

QUICK LITERATURE/CONTEXT CHECK

I only record context explicitly stated in the problem file. The file states bounds \[ 2-O\left(\frac{(\log n)^2}{n}\right)\le \min I\le 2-\frac{2}{2n-1}, \] with the upper bound witnessed by the roots of the integral of the Legendre polynomial.

ATTACK PLAN

Proof track ideas.

• Compute $I$ exactly for small $n$ to guess minimizers; derive Euler--Lagrange-type conditions.
• Use functional-analytic interpretation: $\sum l_k(x)^2$ is the diagonal of the reproducing kernel for the interpolation projector; attempt trace/minimax bounds.

Disproof track ideas.

• Try to find node sets for which $I$ is strictly smaller than $2-(1+o(1))/n$ or show that the conjectured asymptotic violates known lower bounds.

WORK

Lemma 1131.1 (partition of unity). For the Lagrange polynomials $l_k$ associated to any distinct nodes, \[ \sum_{k=1}^n l_k(x)=1\qquad\text{for all }x. \]

Proof. This is identical to Lemma 1129.1: $p(x)=\sum_k l_k(x)-1$ is a polynomial of degree $\le n-1$ vanishing at $n$ distinct points, hence identically zero. $\square$

Lemma 1131.2 (exact formula and minimizer for $n=2$). Let $n=2$ and let $x_1,x_2\in[-1,1]$ be distinct. Then \[ I(x_1,x_2)=\frac{\frac{4}{3}+2(x_1^2+x_2^2)}{(x_1-x_2)^2} =1+\frac{(x_1+x_2)^2+\frac{4}{3}}{(x_1-x_2)^2}. \] In particular, \[ \min I = \frac{4}{3}, \] achieved uniquely (up to swapping $x_1,x_2$) by $(x_1,x_2)=(-1,1)$.

Proof. For $n=2$ the Lagrange polynomials are \[ l_1(x)=\frac{x-x_2}{x_1-x_2},\qquad l_2(x)=\frac{x-x_1}{x_2-x_1}. \] Since $|l_k(x)|^2=l_k(x)^2$, we have \[ l_1(x)^2+l_2(x)^2 = \frac{(x-x_2)^2+(x-x_1)^2}{(x_1-x_2)^2}. \] Integrating termwise over $[-1,1]$ gives \[ I(x_1,x_2)=\frac{1}{(x_1-x_2)^2}\left(\int_{-1}^1 (x-x_1)^2\,dx+\int_{-1}^1 (x-x_2)^2\,dx\right). \] For any real $a$, \[ \int_{-1}^1 (x-a)^2\,dx=\int_{-1}^1 (x^2-2ax+a^2)\,dx=\frac{2}{3}+2a^2, \] because $\int_{-1}^1 x^2 dx=2/3$, $\int_{-1}^1 x\,dx=0$, and $\int_{-1}^1 1\,dx=2$. Hence \[ I(x_1,x_2)=\frac{(\frac{2}{3}+2x_1^2)+(\frac{2}{3}+2x_2^2)}{(x_1-x_2)^2} =\frac{\frac{4}{3}+2(x_1^2+x_2^2)}{(x_1-x_2)^2}. \] Writing $s=x_1+x_2$ and $d=x_1-x_2$, we have $x_1^2+x_2^2=(s^2+d^2)/2$, so \[ I(x_1,x_2)=\frac{\frac{4}{3}+s^2+d^2}{d^2}=1+\frac{s^2+\frac{4}{3}}{d^2}. \] To minimize $I$, we want to minimize $s^2$ and maximize $d^2$ subject to $x_1,x_2\in[-1,1]$. We always have $s^2\ge 0$, with equality iff $x_2=-x_1$. Also $|d|=|x_1-x_2|\le 2$, with equality iff $\{x_1,x_2\}=\{-1,1\}$. Choosing $(x_1,x_2)=(-1,1)$ gives $s=0$ and $d^2=4$, hence \[ I(-1,1)=1+\frac{0+\frac{4}{3}}{4}=\frac{4}{3}. \] For any other choice, either $s^2>0$ or $d^2<4$, making $(s^2+4/3)/d^2> (4/3)/4$ and hence $I>4/3$. This proves the claimed minimum and uniqueness up to swapping. $\square$

Lemma 1131.3 (a crude universal lower bound). For any $n\ge 1$ and any distinct nodes $x_1,\dots,x_n\in[-1,1]$, \[ I(x_1,\dots,x_n)\ge \frac{2}{n}. \]

Proof. Fix $x\in[-1,1]$. By Cauchy--Schwarz, \[ \sum_{k=1}^n l_k(x)^2\ \ge\ \frac{\left(\sum_{k=1}^n l_k(x)\right)^2}{n}. \] By Lemma 1131.1, $\sum_k l_k(x)=1$. Therefore $\sum_k l_k(x)^2\ge 1/n$ for all $x\in[-1,1]$. Integrating over $[-1,1]$ yields $I\ge \int_{-1}^1 (1/n)\,dx = 2/n$. $\square$

FAST REALITY CHECK (small $n$ numerics). Using Gauss--Legendre quadrature, I computed $I$ for a few node choices:

n=1 nodes [0]: I 2.0000000000000004
n=2 nodes [-1,1]: I 1.3333333333333293
n=2 nodes [-0.5,0.5]: I 2.333333333333319
n=3 nodes [-1,0,1]: I 1.59999999999999
n=3 nodes [-0.5,0,0.5]: I 7.599999999999338

A coarse grid search with step $0.1$ over ordered triples $(x_1,x_2,x_3)$ suggested the smallest value occurs at $(-1,0,1)$ with $I\approx 1.6$ on that grid.

VERIFICATION

• Lemma 1131.2 is a complete calculation; the optimization step uses only the bounds $s^2\ge 0$ and $|d|\le 2$ in $[-1,1]$.
• Lemma 1131.3 is correct even though it is far from sharp for large $n$.
• Numerical values are consistent with the exact $n=2$ minimum $4/3$ and show that bringing nodes away from endpoints can increase $I$ dramatically.

FINAL

UNRESOLVED

(i) Strongest proved partial result. The exact minimizer for $n=2$ is $\{-1,1\}$ with $\min I=4/3$ (Lemma 1131.2). Universally, $I\ge 2/n$ for all $n$ (Lemma 1131.3).

(ii) First gap (crisp). Determine the minimising node sets and asymptotics of $\min I(x_1,\dots,x_n)$ as $n\to\infty$, in particular whether $\min I = 2-(1+o(1))/n$.

(iii) Top 3 next moves.

• For fixed small $n$ (e.g. $n=3,4,5$), perform rigorous (possibly computer-assisted) constrained optimization to identify minimizers and exact minimal values.
• Develop a variational formula for $I$ in terms of the Gram matrix $G_{ij}=\int_{-1}^1 l_i(x)l_j(x)\,dx$ and attempt to lower bound $\operatorname{tr}(G)$ under interpolation constraints.
• Try to connect $\sum l_k(x)^2$ to Christoffel functions/orthogonal polynomials to access asymptotics.

(iv) Minimal counterexample structure. A counterexample to the asymptotic $2-(1+o(1))/n$ would be a family of node sets for which $I$ stays bounded away from $2$ by more than a constant multiple of $1/n$, or conversely an explicit construction driving $I$ below $2- c/n$ for some $c>1$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

FORMAL RESTATEMENT

Let $n\ge 1$ and let $x_1<\dots<x_n$ be distinct points in $[-1,1]$. Let $l_k$ be the associated Lagrange fundamental polynomials. Define \[ I(x_1,\dots,x_n):=\int_{-1}^1\sum_{k=1}^n |l_k(x)|^2\,dx. \] Since all quantities are real on $[-1,1]$, $|l_k(x)|^2=l_k(x)^2$. Problem: determine \[ \min_{x_1,\dots,x_n\in[-1,1]\ \text{distinct}} I(x_1,\dots,x_n) \] and in particular decide whether \[ \min I = 2-(1+o(1))\frac{1}{n}\qquad (n\to\infty) \] holds.

QUICK LITERATURE/CONTEXT CHECK

I only record context explicitly stated in the problem file. The file states bounds \[ 2-O\left(\frac{(\log n)^2}{n}\right)\le \min I\le 2-\frac{2}{2n-1}, \] with the upper bound witnessed by the roots of the integral of the Legendre polynomial.

ATTACK PLAN

Proof track ideas.

• Compute $I$ exactly for small $n$ to guess minimizers; derive Euler--Lagrange-type conditions.
• Use functional-analytic interpretation: $\sum l_k(x)^2$ is the diagonal of the reproducing kernel for the interpolation projector; attempt trace/minimax bounds.

Disproof track ideas.

• Try to find node sets for which $I$ is strictly smaller than $2-(1+o(1))/n$ or show that the conjectured asymptotic violates known lower bounds.

WORK

Lemma 1131.1 (partition of unity). For the Lagrange polynomials $l_k$ associated to any distinct nodes, \[ \sum_{k=1}^n l_k(x)=1\qquad\text{for all }x. \]

Proof. This is identical to Lemma 1129.1: $p(x)=\sum_k l_k(x)-1$ is a polynomial of degree $\le n-1$ vanishing at $n$ distinct points, hence identically zero. $\square$

Lemma 1131.2 (exact formula and minimizer for $n=2$). Let $n=2$ and let $x_1,x_2\in[-1,1]$ be distinct. Then \[ I(x_1,x_2)=\frac{\frac{4}{3}+2(x_1^2+x_2^2)}{(x_1-x_2)^2} =1+\frac{(x_1+x_2)^2+\frac{4}{3}}{(x_1-x_2)^2}. \] In particular, \[ \min I = \frac{4}{3}, \] achieved uniquely (up to swapping $x_1,x_2$) by $(x_1,x_2)=(-1,1)$.

Proof. For $n=2$ the Lagrange polynomials are \[ l_1(x)=\frac{x-x_2}{x_1-x_2},\qquad l_2(x)=\frac{x-x_1}{x_2-x_1}. \] Since $|l_k(x)|^2=l_k(x)^2$, we have \[ l_1(x)^2+l_2(x)^2 = \frac{(x-x_2)^2+(x-x_1)^2}{(x_1-x_2)^2}. \] Integrating termwise over $[-1,1]$ gives \[ I(x_1,x_2)=\frac{1}{(x_1-x_2)^2}\left(\int_{-1}^1 (x-x_1)^2\,dx+\int_{-1}^1 (x-x_2)^2\,dx\right). \] For any real $a$, \[ \int_{-1}^1 (x-a)^2\,dx=\int_{-1}^1 (x^2-2ax+a^2)\,dx=\frac{2}{3}+2a^2, \] because $\int_{-1}^1 x^2 dx=2/3$, $\int_{-1}^1 x\,dx=0$, and $\int_{-1}^1 1\,dx=2$. Hence \[ I(x_1,x_2)=\frac{(\frac{2}{3}+2x_1^2)+(\frac{2}{3}+2x_2^2)}{(x_1-x_2)^2} =\frac{\frac{4}{3}+2(x_1^2+x_2^2)}{(x_1-x_2)^2}. \] Writing $s=x_1+x_2$ and $d=x_1-x_2$, we have $x_1^2+x_2^2=(s^2+d^2)/2$, so \[ I(x_1,x_2)=\frac{\frac{4}{3}+s^2+d^2}{d^2}=1+\frac{s^2+\frac{4}{3}}{d^2}. \] To minimize $I$, we want to minimize $s^2$ and maximize $d^2$ subject to $x_1,x_2\in[-1,1]$. We always have $s^2\ge 0$, with equality iff $x_2=-x_1$. Also $|d|=|x_1-x_2|\le 2$, with equality iff $\{x_1,x_2\}=\{-1,1\}$. Choosing $(x_1,x_2)=(-1,1)$ gives $s=0$ and $d^2=4$, hence \[ I(-1,1)=1+\frac{0+\frac{4}{3}}{4}=\frac{4}{3}. \] For any other choice, either $s^2>0$ or $d^2<4$, making $(s^2+4/3)/d^2> (4/3)/4$ and hence $I>4/3$. This proves the claimed minimum and uniqueness up to swapping. $\square$

Lemma 1131.3 (a crude universal lower bound). For any $n\ge 1$ and any distinct nodes $x_1,\dots,x_n\in[-1,1]$, \[ I(x_1,\dots,x_n)\ge \frac{2}{n}. \]

Proof. Fix $x\in[-1,1]$. By Cauchy--Schwarz, \[ \sum_{k=1}^n l_k(x)^2\ \ge\ \frac{\left(\sum_{k=1}^n l_k(x)\right)^2}{n}. \] By Lemma 1131.1, $\sum_k l_k(x)=1$. Therefore $\sum_k l_k(x)^2\ge 1/n$ for all $x\in[-1,1]$. Integrating over $[-1,1]$ yields $I\ge \int_{-1}^1 (1/n)\,dx = 2/n$. $\square$

FAST REALITY CHECK (small $n$ numerics). Using Gauss--Legendre quadrature, I computed $I$ for a few node choices:

n=1 nodes [0]: I 2.0000000000000004
n=2 nodes [-1,1]: I 1.3333333333333293
n=2 nodes [-0.5,0.5]: I 2.333333333333319
n=3 nodes [-1,0,1]: I 1.59999999999999
n=3 nodes [-0.5,0,0.5]: I 7.599999999999338

A coarse grid search with step $0.1$ over ordered triples $(x_1,x_2,x_3)$ suggested the smallest value occurs at $(-1,0,1)$ with $I\approx 1.6$ on that grid.

VERIFICATION

• Lemma 1131.2 is a complete calculation; the optimization step uses only the bounds $s^2\ge 0$ and $|d|\le 2$ in $[-1,1]$.
• Lemma 1131.3 is correct even though it is far from sharp for large $n$.
• Numerical values are consistent with the exact $n=2$ minimum $4/3$ and show that bringing nodes away from endpoints can increase $I$ dramatically.

FINAL

UNRESOLVED

(i) Strongest proved partial result. The exact minimizer for $n=2$ is $\{-1,1\}$ with $\min I=4/3$ (Lemma 1131.2). Universally, $I\ge 2/n$ for all $n$ (Lemma 1131.3).

(ii) First gap (crisp). Determine the minimising node sets and asymptotics of $\min I(x_1,\dots,x_n)$ as $n\to\infty$, in particular whether $\min I = 2-(1+o(1))/n$.

(iii) Top 3 next moves.

• For fixed small $n$ (e.g. $n=3,4,5$), perform rigorous (possibly computer-assisted) constrained optimization to identify minimizers and exact minimal values.
• Develop a variational formula for $I$ in terms of the Gram matrix $G_{ij}=\int_{-1}^1 l_i(x)l_j(x)\,dx$ and attempt to lower bound $\operatorname{tr}(G)$ under interpolation constraints.
• Try to connect $\sum l_k(x)^2$ to Christoffel functions/orthogonal polynomials to access asymptotics.

(iv) Minimal counterexample structure. A counterexample to the asymptotic $2-(1+o(1))/n$ would be a family of node sets for which $I$ stays bounded away from $2$ by more than a constant multiple of $1/n$, or conversely an explicit construction driving $I$ below $2- c/n$ for some $c>1$.