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Chebyshev–Lobatto interpolation folds polynomial modes

Statement

Let $N \geq 1$ , let $x_{j} = \cos (j π / N)$ , and let $I_{N}^{L}$ denote degree-at-most- $N$ interpolation on these nodes. For every integer $n \geq 0$ , let $ρ_{N} (n)$ be the unique integer $r \in {0, \dots, N}$ congruent to either $n$ or $- n$ modulo $2 N$ . Then

I_{N}^{L} T_{n} = T_{ρ_{N} (n)} .

In particular, sampling a Chebyshev series on the Lobatto grid can fold every degree $n > N$ to a degree at most $N$ .

Facts & Assumptions

Given: An integer $N \geq 1$ , its Chebyshev–Gauss–Lobatto grid, and an integer $n \geq 0$ .

[F1]

Chebyshev–Gauss–Lobatto collocation grid: $x_{j} = \cos (j π / N)$ for $0 \leq j \leq N$ , with distinct nodes.

[F2]

Chebyshev polynomials of the first and second kinds by their three-term recurrences defines the polynomials $T_{n}$ ; $T_{n} (\cos θ) = \cos (n θ)$ and $U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ)$ for every $n \in N$ gives $T_{n} (\cos θ) = \cos (n θ)$ .

[F3]

Existence and uniqueness of the cardinal-polynomial interpolant: a polynomial of degree at most $N$ that agrees with the data at the $N + 1$ distinct increasing nodes is their unique interpolant; reversing node order and data order preserves the same interpolation polynomial.

[F4]

Degrees and leading coefficients of the Chebyshev polynomials: $T_{r}$ has degree $r$ for every $r \geq 1$ , and $T_{0} = 1$ .

[F5]

Applying the cosine shift by $π$ in Quarter-turn values and shifts by pi/2 and pi twice gives $\cos (α + 2 q π) = \cos α$ for every integer $q$ ; Parity and the Pythagorean identity for sine and cosine gives $\cos (- α) = \cos α$ .

Proof

1.1givenF1choosealgebra

Reduce $n$ modulo $2 N$ to a residue $s \in {0, \dots, 2 N - 1}$ and put $r = \min (s, 2 N - s)$ . Then $0 \leq r \leq N$ , $r \equiv n$ or $- n (m o d 2 N)$ , and this $r$ is unique: the classes ${\pm r}$ are distinct for $0 < r < N$ , with singleton classes at $r = 0, N$ .

2.1F2F5step 1.1algebra

For every node $x_{j}$ , [F2] gives $T_{n} (x_{j}) = \cos (n j π / N)$ . Since $n \equiv \pm r (m o d 2 N)$ , [F5] gives $\cos (n j π / N) = \cos (r j π / N) = T_{r} (x_{j})$ . Thus $T_{n}$ and $T_{r}$ have the same nodal values.

3.1F1F3F4step 2.1∎

By [F4], $T_{r}$ has degree $r \leq N$ . Interpolation uniqueness [F3] therefore gives $I_{N}^{L} T_{n} = T_{r}$ .

Depends on

Used by

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Sources

Jie Shen and Tao Tang, Spectral and High-Order Methods with Applications, Ch. 2 (standard reference, not scraped)
Lloyd N. Trefethen, Finite Difference and Spectral Methods for Ordinary and Partial Differential Equations, Ch. 8 (standard reference, not scraped)