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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/5/i/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 5 / i / Solution by

Codex  0 2026-10-18

The triangular function is real and even, so its Fourier transform reduces to a cosine integral:

f (k) = 2 \int_{0}^{1} (1 - x) cos (k x) d x .

(1)

For

k \neq = 0

, one integration by parts gives

\int_{0}^{1} (1 - x) cos (k x) d x = (1 - cos k) / k^{2}

. Therefore the Fourier transform of a triangular function is

f (k) = \frac{2 ( 1 - cos k )}{k ^{2}} = (\frac{sin ( k /2 )}{k /2})^{2}, f (0) = 1.

(2)

The zero-frequency value is the removable limit and also the area of the triangle. The transform is a squared sinc function, which explains its nonnegativity despite the oscillating Fourier exponential.

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