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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/18/c/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 18 / c / Solution by

Codex  0 2026-10-18

Put

s = ω t

, so the nonzero half-cycle is

0 \leq s \leq π

. The constant coefficient is

a_{0} = \frac{V _{0}}{π} \int_{0}^{π} sin s d s = \frac{2 V _{0}}{π} .

The sine coefficients satisfy

b_{n} = \frac{V _{0}}{π} \int_{0}^{π} sin s sin n s d s = {V_{0} /2, 0, n = 1, n \neq = 1,

by sine orthogonality. For the cosine coefficients, the product-to-sum identity gives, when

n \neq = 1

,

\int_{0}^{π} sin s cos n s d s = \frac{1}{2} \int_{0}^{π} [sin ((1 + n) s) + sin ((1 - n) s)] d s = \frac{1 + ( - 1 ) ^{n}}{1 - n ^{2}} .

The case

n = 1

is zero directly. Hence

a_{2 m} = - \frac{2 V _{0}}{π ( 4 m ^{2} - 1 )}, a_{2 m - 1} = 0, b_{1} = \frac{V _{0}}{2}, b_{n} = 0 (n \neq = 1) .

The first five nonzero terms, counting the constant as the zero-frequency term, are

V_{1} (t) = \frac{V _{0}}{π} + \frac{V _{0}}{2} sin ω t - \frac{2 V _{0}}{3 π} cos 2 ω t - \frac{2 V _{0}}{15 π} cos 4 ω t - \frac{2 V _{0}}{35 π} cos 6 ω t + \dots .

The general coefficient formula specifies every remaining term. This is the Fourier series of a half-wave rectified sinusoid; the waveform is continuous at the joins, so the series gives its actual value there.

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