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Three-point Clenshaw–Curtis integration of a quadratic

Example

For the three-point rule ( $n = 2$ ) on $[- 1, 1]$ , the nodes are $1, 0, - 1$ and the quadrature weights in that order are $1 / 3, 4 / 3, 1 / 3$ . Hence it integrates $1$ , $x$ , and $x^{2}$ exactly, with respective values $2$ , $0$ , and $2 / 3$ .

Facts & Assumptions

Given: The degree $n = 2$ Clenshaw–Curtis rule on $[- 1, 1]$ and the sample data of the polynomials $1, x, x^{2}$ at its three nodes.

[F1]

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

[F2]

DCT-I coefficients of the Chebyshev–Lobatto interpolant: $a_{0} = X_{0} / 4$ , $a_{1} = X_{1} / 2$ , and $a_{2} = X_{2} / 4$ .

[F3]

Exact integrals of Chebyshev polynomials: $μ_{0} = 2$ , $μ_{1} = 0$ , and $μ_{2} = - 2 / 3$ .

[F4]

Clenshaw–Curtis quadrature with the midpoint zero-degree case: $Q_{2, [- 1, 1]} (f) = \sum_{k = 0}^{2} a_{k} μ_{k}$ since $d = 1$ .

[F5]

Chebyshev polynomials of the first and second kinds by their three-term recurrences: $T_{0} (x) = 1$ , $T_{1} (x) = x$ , and $T_{2} (x) = 2 x^{2} - 1$ .

Proof

1.1givenF1F2F4algebra

For sample values $(y_{0}, y_{1}, y_{2})$ at $(1, 0, - 1)$ , [F1] and the exact values $\cos (0) = 1$ , $\cos (π / 2) = 0$ , $\cos (π) = - 1$ give $X_{0} = y_{0} + 2 y_{1} + y_{2}$ , $X_{1} = y_{0} - y_{2}$ , and $X_{2} = y_{0} - 2 y_{1} + y_{2}$ . Substituting [F2] into [F4] gives $Q_{2} (f) = \frac{y_{0} + 2 y_{1} + y_{2}}{2} - \frac{y_{0} - 2 y_{1} + y_{2}}{6} = \frac{y_{0}}{3} + \frac{4 y_{1}}{3} + \frac{y_{2}}{3} .$

2.1givenF3F5step 1.1algebra∎

For $1$ , the sample vector is $(1, 1, 1)$ and the rule gives $2$ . For $x$ , the vector is $(1, 0, - 1)$ and the rule gives $0$ . For $x^{2}$ , it is $(1, 0, 1)$ and the rule gives $2 / 3$ . Direct integration gives these same values, respectively $2$ , $0$ , and $[x^{3} / 3]_{- 1}^{1} = 2 / 3$ . Equivalently, [F5] gives $x^{2} = (T_{0} + T_{2}) / 2$ , and [F3] gives integral $μ_{0} / 2 + μ_{2} / 2 = 2 / 3$ . Thus all three evaluations agree with the exact integrals.

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)