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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ib/paper-1/2h/solution

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Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 2H / Solution by

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

For the real even function, let

s_{N} = a_{0} /2 + \sum_{n = 1}^{N} a_{n} cos n x

. Fourier orthogonality gives

\int_{- π}^{π} cos n x d x = 0

for

n \geq 1

and

\int_{- π}^{π} cos n x cos m x d x = π δ_{nm}

. Consequently

\frac{1}{π} \int_{- π}^{π} s_{N}^{2} d x = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} .

For an even square-integrable function, the complete Fourier cosine series converges in the L2 norm; hence its norms converge. Taking

N \to \infty

proves Parseval's identity in the required normalization:

\frac{1}{π} \int_{- π}^{π} f (x)^{2} d x = \frac{a _{0}^{2}}{2} + n = 1 \sum \infty a_{n}^{2} .

This limit argument avoids multiplying an infinite series without a justification of L2 norm convergence.

For the quadratic function, the Fourier coefficients are

a_{0} = \frac{1}{π} \int_{- π}^{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}} .

The last equality follows from two integrations by parts, using

sin nπ = 0

. Thus

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

This Fourier series is absolutely convergent; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into Parseval's identity gives

\frac{2 π ^{4}}{5} = \frac{2 π ^{4}}{9} + 16 \sum_{n = 1}^{\infty} \frac{1}{n ^{4}}, n = 1 \sum \infty \frac{1}{n ^{4}} = \frac{π ^{4}}{90} .

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