 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/17/d/solution

Termwise differentiation of a Fourier series, applied to the continuous periodic extension in part (c), gives for

- 1 < x < 1

2 x = - \frac{4}{π} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n} sin (nπ x) .

Hence the required Fourier sine series is

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

The endpoint deserves care: at

x = 1

every term is zero, so the Fourier series sums to

0

, the average of the periodic one-sided limits

1

and

- 1

. It therefore does not reproduce the specified endpoint value

g (1) = 1

. Changing a value at one point does not change any Fourier coefficient, so this is still the Fourier series of the stated function on

(- 1, 1]

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