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DCT-I coefficients of the Chebyshev–Lobatto interpolant

Statement

Let $n \in N$ with $n \geq 1$ , and let $g \in C [- 1, 1]$ . Put $y_{j} = g (\cos (j π / n))$ for $0 \leq j \leq n$ , and let $X_{k}$ be the DCT-I of Type-I cosine transform for Chebyshev–Lobatto data. Define $a_{0} : = \frac{X_{0}}{2 n}, a_{n} : = \frac{X_{n}}{2 n}, a_{k} : = \frac{X_{k}}{n} (1 \leq k < n) .$ Then $p_{n} (t) : = \sum_{k = 0}^{n} a_{k} T_{k} (t)$ is the unique polynomial of degree at most $n$ interpolating $g$ at the Chebyshev–Lobatto nodes $\cos (j π / n)$ ; in particular $p_{n} = I_{n}^{L} g$ in the convention of Best uniform polynomial error and Chebyshev interpolation operators.

Facts & Assumptions

Given: An integer $n \geq 1$ and a continuous $g : [- 1, 1] \to R$ .

[F1]

Type-I cosine transform for Chebyshev–Lobatto data: $X_{k} = y_{0} + (- 1)^{k} y_{n} + 2 \sum_{j = 1}^{n - 1} y_{j} \cos (j k π / n)$ .

[F2]

Weighted cosine orthogonality on the Lobatto grid: the half-endpoint weighted cosine products sum to $n$ at matching endpoint frequencies, $n / 2$ at matching interior frequencies, and $0$ at distinct frequencies.

[F3]

The Chebyshev recurrence agrees with cosine multiple angles: $T_{k} (\cos θ) = \cos (k θ)$ for every real $θ$ .

[F4]

Best uniform polynomial error and Chebyshev interpolation operators: the Lobatto nodes are $x_{j}^{L, n} = \cos (j π / n)$ and $I_{n}^{L} g$ is the Lagrange-cardinal interpolant at those nodes.

[F5]

Chebyshev cardinal interpolation reproduces polynomials and its norm is the cardinal sum: $I_{n}^{L} g$ is a polynomial of degree at most $n$ and has values $I_{n}^{L} g (x_{j}^{L, n}) = g (x_{j}^{L, n})$ at every node.

[F6]

Cardinal-polynomial interpolation and Existence and uniqueness of the cardinal-polynomial interpolant: at $n + 1$ strictly increasing distinct real nodes, the degree-at-most- $n$ Lagrange interpolant is the unique polynomial with the given node values.

[F7]

The Chebyshev recurrence gives degree at most its index: each $T_{k}$ has degree at most $k$ .

Proof

1.1givenF1F3algebra

Use the half-endpoint sum notation $\sum_{j = 0}^{n}^{'} v_{j} : = v_{0} / 2 + \sum_{j = 1}^{n - 1} v_{j} + v_{n} / 2$ . By [F1], $X_{k} = 2 \sum_{j = 0}^{n}^{'} y_{j} \cos (j k π / n)$ . The recurrence-angle identity [F3] gives $T_{k} (\cos (j π / n)) = \cos (j k π / n)$ , so the definition of $a_{k}$ then gives $p_{n} (\cos (j π / n)) = \frac{1}{n} \sum_{k = 0}^{n}^{'} X_{k} \cos (j k π / n) .$

2.1givenF1F2step 1.1algebra

Substitute the expression for $X_{k}$ from step 1.1 into the displayed reconstruction sum and interchange the two finite sums. By [F2], every frequency $m \neq j$ contributes zero. If $j$ is an endpoint, its outer half-weight and the diagonal sum $n$ contribute $(1 / n) \cdot 2 \cdot (1 / 2) n y_{j} = y_{j}$ ; if $j$ is interior, the diagonal sum $n / 2$ contributes $(1 / n) \cdot 2 \cdot (n / 2) y_{j} = y_{j}$ . Thus $p_{n} (\cos (j π / n)) = y_{j}$ for each $j$ .

3.1givenF4F5F6F7step 2.1algebra∎

By [F7], $p_{n}$ has degree at most $n$ . The interpolant $I_{n}^{L} g$ has the same degree bound and node values by [F4]–[F5]. Reorder the common node set increasingly; it remains $n + 1$ distinct nodes because $\cos$ is strictly decreasing on $[0, π]$ . Apply the uniqueness statement [F6] to the two polynomials with the same values from step 2.1. Hence $p_{n} = I_{n}^{L} g$ .

Depends on

Used by

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Sources

Lloyd N. Trefethen, Is Gauss Quadrature Better than Clenshaw–Curtis?, §2 (standard reference, not scraped)
L. Ridgway Scott, Numerical Analysis, 2nd ed., §13.4.1 (standard reference, not scraped)