 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/18/b/solution

Evenness makes every sine coefficient zero. For the constant coefficient,

a_{0} = \frac{2}{L} \int_{0}^{L} x d x = L .

n \geq 1

a_{n} = \frac{2}{L} \int_{0}^{L} x cos \frac{nπ x}{L} d x = \frac{2 L}{n ^{2} π ^{2}} [(- 1)^{n} - 1] .

The even coefficients vanish and the odd coefficients are

- 4 L / (n^{2} π^{2})

. Therefore

f (x) = \frac{L}{2} - \frac{4 L}{π ^{2}} k = 0 \sum \infty \frac{cos (( 2 k + 1 ) π x / L )}{( 2 k + 1 ) ^{2}} .

f

at every point, including the endpoints. The absolute summability of the coefficients also gives uniform convergence.

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