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A Fourier integral missed by one sample but captured by Filon moments

Example

Let $N$ be a positive odd integer, $ω = 2 π N$ , and $g (x) = x$ on $[0, 1]$ . For $f (x) = 1$ , the exact integral is $0$ and the endpoint Filon rule returns $0$ , while the one-sample midpoint value $M_{ω} (f) = f (1 / 2) e^{i ω / 2}$ is $- 1$ . For $f (x) = x^{2}$ , the same endpoint Filon rule has the exact error $2 / ω^{2}$ .

Facts & Assumptions

Given: The positive odd integer $N$ , $ω = 2 π N$ , $g (x) = x$ , and the two amplitudes $f (x) = 1$ and $f (x) = x^{2}$ on $[0, 1]$ .

[F1]

Filon quadrature from nodal interpolation and phase moments: the Filon output is $\sum_{j} f (c_{j}) {\hat{μ}}_{j}$ , using the Lagrange interpolant's phase moments.

[F2]

Exact recurrence for polynomial Fourier moments: for endpoints $a, b$ , the recurrence is $μ_{0} = (e^{i ω b} - e^{i ω a}) / (i ω)$ and $μ_{k} = (b^{k} e^{i ω b} - a^{k} e^{i ω a}) / (i ω) - k μ_{k - 1} / (i ω)$ .

[F3]

Exact polynomial moments for the linear Fourier phase: each output $μ_{k}$ is the exact integral $\int_{a}^{b} x^{k} e^{i ω x} d x$ .

[F4]

Nonstationary error bound for endpoint Filon interpolation: endpoint Filon with exact moments has error bounded by an explicit constant times $∣ ω ∣^{- 2}$ under $f \in C^{2}$ and nonstationary $g \in C^{3}$ .

[F5]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential: $e^{z + w} = e^{z} e^{w}$ for complex $z, w$ .

[F6]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : $e^{i π} = - 1$ .

[F7]

Laws of integer exponents: for a nonzero field element and natural exponents, $(a^{m})^{n} = a^{m n}$ and $a^{m + n} = a^{m} a^{n}$ .

Verification

1.1givenF1F2F3F5F6F7algebra

Since $e^{i 2 π N} = (e^{i π})^{2 N} = (- 1)^{2 N} = 1$ and $e^{i π N} = (- 1)^{N} = - 1$ for odd $N$ , by [F5]--[F7], [F2] and [F3] give $μ_{0} = 0$ . For $f = 1$ the endpoint interpolant is constant, so [F1] returns the exact integral $0$ , while the one-sample value is $M_{ω} (1) = e^{i π N} = - 1$ .

2.1givenF1F2F3F4step 1.1algebra∎

For $f = x^{2}$ , the degree-one endpoint interpolant is $p (x) = x$ . Step 1.1 gives $e^{i ω} = 1$ , so [F2] gives $μ_{1} = 1 / (i ω)$ and $μ_{2} = 1 / (i ω) - 2 / (i ω)^{2}$ . Thus $I_{ω} (f, g) - Q_{1} = μ_{2} - μ_{1} = - 2 / (i ω)^{2} = 2 / ω^{2}$ . The hypotheses of [F4] hold, so this exact error has the asserted second-order frequency decay.

Depends on

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Sources

A. Iserles, S. P. Nørsett, and S. Olver, Highly oscillatory quadrature: The story so far, §3.2 (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, §2.3(i) (standard reference, not scraped)