 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/2/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The periodically extended

x^{2}

is even, so every sine coefficient is zero. With the convention

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

For the second integral, two applications of integration by parts give

\int_{0}^{π} x^{2} cos n x d x = 2 π (- 1)^{n} / n^{2}

. Consequently

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

The periodic extension is continuous even at the joined endpoints, where both one-sided values equal

π^{2}

. The pointwise convergence of a piecewise smooth Fourier series therefore gives the displayed values throughout the interval; the coefficient bound also gives uniform convergence by the Weierstrass M-test. Setting

x = π

and then

x = 0

yields

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} .

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