 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/4b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

f (k) = \int_{- \infty}^{\infty} f (x) e^{- ik x} d x .

Then

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

with the continuous value

f (0) = 2 A

(f * f) (x)

A^{2}

times the length of the overlap of the intervals

[- 1, 1]

and

[x - 1, x + 1]

. Thus

(f * f) (x) = {A^{2} (2 - ∣ x ∣), 0, ∣ x ∣ \leq 2, ∣ x ∣ > 2.

The convolution theorem states, for this convention, that

u * v (k) = u (k) v (k) .

Since

g = (B / A^{2}) (f * f)

g (k) = 4 B (\frac{sin k}{k})^{2},

again interpreted continuously at

k = 0

, where

g (0) = 4 B

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