Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 4 / d / Solution

Evenness means the negative half of the function is determined as well: this is the triangular function

q (x) = 1 - ∣ x ∣

on

∣ x ∣ \leq 1

, zero outside. Its transform is real and even:

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

Integration by parts, or integrating the two terms separately, gives

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

The apparent singularity at zero is removable, and its limiting value is the area under

q

. This is the Fourier transform of a triangular function.

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