LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

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A barycentric denominator polynomial with nonzero weights is nonzero

Statement

Let $k \geq 1$ , let $z_{1}, \dots, z_{k} \in C$ be distinct, and let $w_{1}, \dots, w_{k} \in C$ be nonzero. Then the polynomial

Q (z) = \sum_{j = 1}^{k} w_{j} \prod_{ℓ \neq j} (z - z_{ℓ})

is not identically zero.

Facts & Assumptions

Given: A positive integer $k$ , distinct complex nodes $z_{1}, \dots, z_{k}$ , and nonzero complex weights $w_{1}, \dots, w_{k}$ .

Proof

1.1givenchoosealgebra

Fix an index $j \in {1, \dots, k}$ . At $z = z_{j}$ , each summand with index different from $j$ contains the factor $z_{j} - z_{j} = 0$ , so $Q (z_{j}) = w_{j} \prod_{ℓ \neq j} (z_{j} - z_{ℓ}) .$ This is the support-node evaluation of $Q$ .

2.1givenstep 1.1algebra∎

Every factor in the product in step 1.1 is nonzero because the nodes are distinct, and $w_{j} \neq 0$ . Thus $Q (z_{j}) \neq 0$ ; an identically zero polynomial would vanish at $z_{j}$ , so $Q$ is not identically zero.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Yuji Nakatsukasa, Olivier Sète, and Lloyd N. Trefethen, The AAA algorithm for rational approximation, Section 2 (standard reference, not scraped)
Jean-Paul Berrut and Lloyd N. Trefethen, Barycentric Lagrange Interpolation, Section 9.2 (standard reference, not scraped)