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Negative weights amplify sample perturbations in the nine-point Newton–Cotes rule

Example

For a sample vector $s = (s_{0}, \dots, s_{8}) \in R^{9}$ , write ${\hat{Q}}_{8} (s) : = \sum_{j = 0}^{8} w_{j} s_{j}$ . If a perturbed sample vector is $s + δ$ with $∣ δ_{j} ∣ \leq ε$ , then $∣ {\hat{Q}}_{8} (s + δ) - {\hat{Q}}_{8} (s) ∣ \leq \frac{13714}{4725} ε .$ This constant is sharp and exceeds $2$ , the signed sum of the weights and the total weight of any positive rule exact on constants over $[- 1, 1]$ . Independently, for $f \in C^{10} [- 1, 1]$ the quadrature error satisfies the Peano bound $(\int_{- 1}^{1} ∣ K_{10} (t) ∣ d t) ∥ f^{(10)} ∥_{\infty}$ of Peano remainder bound for the eight-panel rule.

Facts & Assumptions

Given: A sample vector $s \in R^{9}$ , the nine weights of $Q_{8}$ , and real perturbations $δ_{0}, \dots, δ_{8}$ satisfying $∣ δ_{j} ∣ \leq ε$ .

[A1]

$ε \geq 0$ and $∣ δ_{j} ∣ \leq ε$ for each node.

[F1]

Closed Newton–Cotes rule on eight equal panels: the rule has weights with numerators $(989, 5888, - 928, 10496, - 4540, 10496, - 928, 5888, 989)$ and common denominator $14175$ .

[F2]

Exact weights and degree-nine exactness of the eight-panel rule: these coefficients are the integrated cardinal weights and the rule is exact through degree $9$ .

[F3]

Peano remainder bound for the eight-panel rule: for $f \in C^{10}$ , its integral error is bounded by $∥ K_{10} ∥_{1} ∥ f^{(10)} ∥_{\infty}$ with the stated explicit kernel.

Proof

1.1givenA1F1algebra

The positive weight numerators sum to $34746$ and the absolute values of the negative numerators sum to $928 + 4540 + 928 = 6396$ . Thus $\sum_{j} ∣ w_{j} ∣ = (34746 + 6396) / 14175 = 41142 / 14175 = 13714 / 4725$ , while $\sum_{j} w_{j} = 2$ . The triangle inequality and [A1] give $∣ {\hat{Q}}_{8} (s + δ) - {\hat{Q}}_{8} (s) ∣ = ∣ \sum_{j} w_{j} δ_{j} ∣ \leq ε \sum_{j} ∣ w_{j} ∣$ . Equality holds by taking $δ_{j} = ε$ when $w_{j} > 0$ and $δ_{j} = - ε$ when $w_{j} < 0$ (with either value when $w_{j} = 0$ ), so the bound is sharp.

2.1givenF2F3step 1.1∎

For any separately supplied $f \in C^{10} [- 1, 1]$ , [F2] and [F3] bound its deterministic integration error as stated. That smooth-function bound does not remove the sample-perturbation amplification in step 1.1.

Depends on

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Sources

L. Ridgway Scott, Numerical Analysis, 2nd ed., §13.1.1 and §13.2 (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, §3.5(iv), Interpolatory Quadrature Rules (standard reference, not scraped)