LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02

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Degrees and leading coefficients of the Chebyshev polynomials

Statement

For $n \geq 1$ , $T_{n}$ has degree $n$ and leading coefficient $2^{n - 1}$ , while $U_{n}$ has degree $n$ and leading coefficient $2^{n}$ . The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, The principle of mathematical induction, Integer powers $a^{m}$ .

Facts & Assumptions

Given: A natural $n$ .

[L1]

Chebyshev polynomials of the first and second kinds by their three-term recurrences defines $T_{0} = 1$ , $T_{1} = x$ , $T_{n + 1} = 2 x T_{n} - T_{n - 1}$ and $U_{0} = 1$ , $U_{1} = 2 x$ , $U_{n + 1} = 2 x U_{n} - U_{n - 1}$ .

Proof

technique · induction

1.1

The initial values in [L1] give the asserted degrees and leading coefficients at $n = 1$ (and the recurrence needs the consecutive base indices $0, 1$ ).

L1base

1.2

Assume the degree and leading-coefficient assertions at consecutive indices.

2.1

In each recurrence of [L1], $2 x$ times the degree- $n$ term has degree $n + 1$ , whereas the subtracted predecessor has degree $n - 1$ . Thus no leading-term cancellation is possible, and the next leading coefficients are $2 \cdot 2^{n - 1} = 2^{n}$ for $T_{n + 1}$ and $2 \cdot 2^{n} = 2^{n + 1}$ for $U_{n + 1}$ .

L1step 1.2algebra

3.1

This proves the stated degree and leading-coefficient formulas at every index.

discharge-induction∎

Depends on

Used by

Dependency tree · two levels

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Sources

NIST Digital Library of Mathematical Functions, Chapter 18 (standard reference, not scraped)