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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ib/paper-3/15g/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 15G / a / Solution by

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

(0, 1)

, orthogonality gives

b_{n} = 2 \int_{0}^{1} x sin (nπ x) d x = 2 [- \frac{x c o s ( nπ x )}{nπ} + \frac{s i n ( nπ x )}{n ^{2} π ^{2}}]_{0}^{1} = \frac{2 ( - 1 ) ^{n + 1}}{nπ} .

(1)

Therefore

x = n = 1 \sum \infty \frac{2 ( - 1 ) ^{n + 1}}{nπ} sin (nπ x), 0 < x < 1.

(2)

The series is that of the odd, two-periodic extension. At

x = 0

its value is zero, agreeing with

f (0)

; at

x = 1

every summand is zero and the series converges to zero, the average of the two one-sided extension limits. Thus it represents the function in the interior and in the square-integrable sense, not the prescribed value one at the endpoint $x = 1$ . This endpoint issue matters when using sine expansions with homogeneous boundary conditions.

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