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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/2/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 2 / b / Solution by

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

Multiply by the

(m, n)

sine product and integrate over the square. Fourier orthogonality gives squared norm

ℓ^{2} /4

, and thus

c_{mn} = \frac{4}{ℓ ^{2}} (\int_{0}^{ℓ} sin \frac{mπ x}{ℓ} d x) (\int_{0}^{ℓ} sin \frac{nπ y}{ℓ} d y) = \frac{4 [ 1 - ( - 1 ) ^{m} ] [ 1 - ( - 1 ) ^{n} ]}{π ^{2} mn} .

(1)

Therefore

c_{mn} = {16/ (π^{2} mn), 0, m, n both odd, otherwise .

(2)

This double Fourier sine series equals one at interior points. At boundary points every summand vanishes, consistent with the endpoint value of the sine extension.

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