LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Lagrange interpolation at distinct real nodes

Statement

Let $n \geq 1$ and let $c_{0}, \dots, c_{n}$ be distinct real numbers. For arbitrary values $y_{0}, \dots, y_{n} \in C$ , define $ℓ_{j} (x) = \prod_{\begin{array}{c} 0 \leq k \leq n \\ k \neq j \end{array}} \frac{x - c_{k}}{c_{j} - c_{k}}, p (x) = \sum_{j = 0}^{n} y_{j} ℓ_{j} (x) .$ Then $p$ is the unique polynomial of degree at most $n$ satisfying $p (c_{j}) = y_{j}$ . In particular, interpolation reproduces every polynomial of degree at most $n$ .

Facts & Assumptions

Given: The stated distinct nodes and complex data.

[F1]

A nonzero real polynomial of degree $n$ has no more than $n$ distinct real roots: a nonzero real polynomial of degree at most $n$ has at most $n$ distinct real roots. Applied to real and imaginary parts, the same root bound holds for complex polynomials.

Proof

1.1givenconstructalgebra

Since $c_{j} - c_{k} \neq 0$ for $j \neq k$ , every denominator in $ℓ_{j}$ is nonzero. At $x = c_{m}$ , the factor for $k = m$ makes $ℓ_{j} (c_{m}) = 0$ when $j \neq m$ , while every factor in $ℓ_{m} (c_{m})$ equals $1$ ; hence $ℓ_{j} (c_{m}) = 1$ exactly when $j = m$ . The displayed $p$ has degree at most $n$ and $p (c_{m}) = y_{m}$ .

2.1givenF1algebra∎

If $q$ is another such polynomial, then $q - p$ has degree at most $n$ and vanishes at the $n + 1$ distinct nodes. Its real and imaginary parts each have those roots, so [F1] forces both parts to be zero polynomials. Thus $q = p$ . If the data are values of a polynomial of degree at most $n$ , uniqueness identifies that polynomial with $p$ .

Depends on

A nonzero real polynomial of degree $n$ has no more than $n$ distinct real roots

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Iserles, S. P. Nørsett, and S. Olver, Highly oscillatory quadrature: The story so far, §3.2 (standard reference, not scraped)
D. Levin, Procedures for computing one- and two-dimensional integrals of functions with rapid irregular oscillations, §1 (standard reference, not scraped)