 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/18/b/solution

The function is odd, so its constant and cosine coefficients are zero. Its sine coefficient is

b_{n} = \frac{2}{π} I_{n}, I_{n} = \int_{0}^{π} (x^{3} - π^{2} x) sin n x d x .

The polynomial vanishes at both integration endpoints. Two integrations by parts therefore give

I_{n} = \frac{1}{n} \int_{0}^{π} (3 x^{2} - π^{2}) cos n x d x = - \frac{1}{n ^{2}} \int_{0}^{π} 6 x sin n x d x .

One more integration yields

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

Consequently

I_{n} = 6 π (- 1)^{n} / n^{3}

and

b_{n} = 12 (- 1)^{n} / n^{3}

. Hence

g (x) = 12 n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{3}} sin n x, p = 3 .

The periodic continuation is continuous at

\pm π

, where its values are zero, so the series represents it there as well as in the open interval.

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