Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 4 / b

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- Solution b

Solution

Direct evaluation gives the first Fourier transform:

f (k) = \int_{- 1/2}^{1/2} e^{- ik x} d x = \frac{2 sin ( k /2 )}{k}, f (0) = 1 .

For the even triangular function, the odd sine part integrates to zero. Two elementary integrations by parts or one evaluation of the cosine integral give

g (k) = 2 \int_{0}^{1} (1 - x) cos (k x) d x = \frac{2 ( 1 - c o s k )}{k ^{2}} = \frac{4 sin ^{2} ( k /2 )}{k ^{2}}, g (0) = 1 .

The values at zero are the respective areas, agreeing with the removable limits. Therefore

g = f^{2}

. By the convolution theorem and uniqueness of the Fourier transform,

(f * f) (x) = g (x) = {1 - ∣ x ∣, 0, ∣ x ∣ < 1, ∣ x ∣ \geq 1.

There is also a direct interpretation: the convolution is the length of overlap of the intervals

(- 1/2, 1/2)

and

(x - 1/2, x + 1/2)

, giving exactly that triangular profile.

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