 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-2/16a/a/solution

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

The function is an even function, so every sine Fourier coefficient vanishes. Direct integration gives

a_{0} = \frac{2}{π} \int_{0}^{π} x d x = π, a_{n} = \frac{2}{π} \int_{0}^{π} x cos (n x) d x = \frac{2}{π n ^{2}} ((- 1)^{n} - 1) .

Thus

∣ x ∣ = \frac{π}{2} - \frac{4}{π} k = 0 \sum \infty \frac{cos (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}}

for the

2 π

-periodic extension.

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