 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-4/5a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The function is even, so its Fourier series has

b_{n} = 0

. In the convention

f \sim a_{0} /2 + \sum_{n \geq 1} (a_{n} cos n x + b_{n} sin n x)

a_{0} = \frac{2}{π} \int_{0}^{π} x^{2} d x = \frac{2 π ^{2}}{3} .

Twice applying integration by parts gives, for

n \geq 1

\int_{0}^{π} x^{2} cos (n x) d x = - \frac{2}{n} \int_{0}^{π} x sin (n x) d x = \frac{2 π ( - 1 ) ^{n}}{n ^{2}}, a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2}} .

Therefore the periodic Fourier representation is

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

with periodic continuation. The periodic extension is continuous, including the identified endpoints, and piecewise smooth, so the Fourier series converges to it. The coefficients also give absolute convergence and uniform convergence by comparison with

\sum n^{- 2}

x = π

, the cosine cancels the alternating coefficient sign; at

x = 0

, it is identically one. These evaluations yield

n = 1 \sum \infty \frac{1}{n ^{2}} = \frac{π ^{2}}{6}, n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n ^{2}} = \frac{π ^{2}}{12} .

The first identity solves the Basel problem.

 New to topics? Read the docs here!