LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge 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.

The Chebyshev recurrence agrees with cosine multiple angles

Statement

For every integer $n \geq 0$ and every real $θ$ , the first-kind Chebyshev polynomial from Chebyshev polynomials of the first and second kinds by their three-term recurrences satisfies $T_{n} (\cos θ) = \cos (n θ) .$

Facts & Assumptions

Given: An integer $n \geq 0$ and a real angle $θ$ .

[A1]

$n$ is a nonnegative integer and $θ \in R$ .

[F1]

Chebyshev polynomials of the first and second kinds by their three-term recurrences: $T_{0} = 1$ , $T_{1} = X$ , and $T_{k + 2} = 2 X T_{k + 1} - T_{k}$ for every $k \geq 0$ .

[F2]

The addition formulas for sine and cosine: $\cos ((k + 2) θ) = 2 \cos θ \cos ((k + 1) θ) - \cos (k θ)$ for every integer $k \geq 0$ and real $θ$ .

[F3]

The principle of mathematical induction: an identity true at the initial indices and preserved by the stated recurrence holds for every natural index.

Proof

1.1givenA1F1basealgebra

For the fixed real $θ$ in [A1], define $P (k)$ to assert the identity at both indices $k$ and $k + 1$ . At $k = 0$ , the identities for $T_{0}$ and $T_{1}$ give $P (0)$ .

1.2givenA1F1F2IHalgebra

Fix $θ \in R$ as in [A1] and suppose $P (k)$ . By the polynomial recurrence [F1], $T_{k + 2} (\cos θ) = 2 \cos θ T_{k + 1} (\cos θ) - T_{k} (\cos θ)$ . Substituting the two induction hypotheses and then [F2] gives $T_{k + 2} (\cos θ) = \cos ((k + 2) θ)$ , so $P (k + 1)$ holds.

2.1A1F3step 1.1step 1.2inductiondischarge-induction∎

By [F3], steps 1.1 and 1.2 prove $P (k)$ for every natural $k$ ; its first clause is the identity for every $n$ in [A1].

Depends on

Used by

Dependency tree · two levels

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Sources

Lloyd N. Trefethen, Is Gauss Quadrature Better than Clenshaw–Curtis?, §2 (standard reference, not scraped)