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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/14a/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 14A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

By the Fourier inversion theorem and Fubini's theorem,

h (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} f (k) g (k) e^{ik x} d k = \int_{- \infty}^{\infty} f (y) [\frac{1}{2 π} \int_{- \infty}^{\infty} g (k) e^{ik (x - y)} d k] d y = \int_{- \infty}^{\infty} f (y) g (x - y) d y .

(1)

Thus multiplication of Fourier transforms corresponds to the convolution $h = f * g$ , which is the convolution theorem in the convention used here.

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