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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 3A / Solution

The function is odd, so only sine coefficients occur. For

n \geq 1

,

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

Integration by parts gives

\int_{0}^{π} x sin (n x) d x = - \frac{π ( - 1 ) ^{n}}{n},

and

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

The terms of order

1/ n

cancel, leaving

b_{n} = \frac{12 ( - 1 ) ^{n}}{n ^{3}} .

Thus

x^{3} - π^{2} x = 12 \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n ^{3}} sin (n x), - π < x < π .

The Parseval identity gives

\frac{1}{π} \int_{- π}^{π} (x^{3} - π^{2} x)^{2} d x = 144 \sum_{n = 1}^{\infty} \frac{1}{n ^{6}} .

Direct integration yields

\frac{1}{π} \int_{- π}^{π} (x^{6} - 2 π^{2} x^{4} + π^{4} x^{2}) d x = \frac{16 π ^{6}}{105} .

Therefore

n = 1 \sum \infty \frac{1}{n ^{6}} = \frac{π ^{6}}{945},

as recorded in the Fourier series of x cubed minus pi squared x.

Solved by gpt-5.6-sol high.

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