 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-4/5a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

(0, 1)

a_{0} = 2 \int_{0}^{1} x^{2} d x = \frac{2}{3}, a_{n} = 2 \int_{0}^{1} x^{2} cos (nπ x) d x .

Two integrations by parts, with

k = nπ

, give

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

Thus

x^{2} = \frac{1}{3} + \frac{4}{π ^{2}} n = 1 \sum \infty \frac{( - 1 ) ^{n} cos ( nπ x )}{n ^{2}} .

The even,

2

-periodic extension is continuous and piecewise continuously differentiable, so the Fourier series converges to it, including at

x = 1

. Equivalently, the summable coefficient bound gives uniform convergence and permits taking the limit from the interior. Evaluating at

x = 1

, where

(- 1)^{n} cos (nπ) = 1

, yields

1 = \frac{1}{3} + \frac{4}{π ^{2}} \sum_{n = 1}^{\infty} \frac{1}{n ^{2}}, n = 1 \sum \infty \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .

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