 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/15/a/solution

Use the real Fourier series convention

g = a_{0} /2 + \sum_{n \geq 1} [a_{n} cos (nπ x) + b_{n} sin (nπ x)]

for period two. Since

g

b_{n} = 0

, and

a_{0} = \int_{- 1}^{1} x^{2} d x = 2/3

. Twice applying integration by parts, with

k = nπ

, gives

a_{n} = 2 \int_{0}^{1} x^{2} cos (k x) d x = 2 [\frac{x ^{2} s i n ( k x )}{k} + \frac{2 x c o s ( k x )}{k ^{2}} - \frac{2 s i n ( k x )}{k ^{3}}]_{0}^{1} = \frac{4 ( - 1 ) ^{n}}{π ^{2} n ^{2}} .

Hence

g (x) = \frac{1}{3} + \frac{4}{π ^{2}} n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}} cos (nπ x) .

The periodic extension is continuous at its endpoints, so the series takes the specified endpoint values as well as the interior values.

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