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Composite midpoint error bound for a twice continuously differentiable function

Statement

Let $a < b$ , let $n \geq 1$ be an integer, and put $h = (b - a) / n$ . If $f \in C^{2} ([a, b])$ , define $M_{2} = \max_{x \in [a, b]} ∣ f^{''} (x) ∣$ . The composite midpoint value $Q_{M} (f; a, b, n)$ of Composite midpoint rule on a uniform grid satisfies

$∣ \int_{a}^{b} f (x) d x - Q_{M} (f; a, b, n) ∣ \leq \frac{(b - a) h^{2}}{24} M_{2} .$

The integral is the Riemann integral. This is an exact-arithmetic quadrature bound; evaluation and summation roundoff are not included.

Facts & Assumptions

Given: Reals $a < b$ , an integer $n \geq 1$ , $h = (b - a) / n$ , and $f \in C^{2} ([a, b])$ .

[F1]

Composite midpoint rule on a uniform grid: $Q_{M}$ is the sum of the values at the $n$ panel midpoints, each multiplied by $h$ .

[F2]

Midpoint panel error kernel and second-derivative bound: on a panel of length $h$ , the difference between its integral and its midpoint value times $h$ has absolute value at most $M_{2} h^{3} / 24$ .

[F3]

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ : the integral over a finite partition is the sum of the integrals over its adjacent panels.

[F4]

Triangle inequality for finite sums: the absolute value of a finite sum is at most the sum of the absolute values.

Proof

1.1givenF2construct

Put $x_{j} = a + j h$ and $m_{j} = (x_{j} + x_{j + 1}) / 2$ for $0 \leq j < n$ . Each panel $[x_{j}, x_{j + 1}]$ has length $h$ and inherits the $C^{2}$ restriction of $f$ ; $\max_{[x_{j}, x_{j + 1}]} ∣ f^{''} ∣ \leq M_{2}$ . By [F2], its midpoint error $E_{j}$ obeys $∣ E_{j} ∣ \leq M_{2} h^{3} / 24$ .

2.1step 1.1F1F3algebra

Repeated application of [F3] gives $\int_{a}^{b} f = \sum_{j = 0}^{n - 1} \int_{x_{j}}^{x_{j + 1}} f$ . By [F1], $Q_{M} = h \sum_{j = 0}^{n - 1} f (m_{j})$ , so subtraction gives $\int_{a}^{b} f - Q_{M} = \sum_{j = 0}^{n - 1} E_{j}$ .

3.1step 1.1step 2.1F4algebra∎

The finite-sum triangle inequality [F4] and step 1.1 yield $∣ \int_{a}^{b} f - Q_{M} ∣ \leq \sum_{j = 0}^{n - 1} ∣ E_{j} ∣ \leq n M_{2} h^{3} / 24$ . Since $n h = b - a$ , this is the claimed bound.

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Sources

MATH 373 lecture notes, §8 Approximation of Integrals (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §5.6 (standard reference, not scraped)