Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 18 / b

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- Solution b

Solution

Separate the already finite sine terms from the even part

g (x) = x^{2} - π ∣ x ∣

. Since

g

is an even function, it has no sine coefficients. Its constant coefficient is

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

For

n \geq 1

, integration by parts gives

\int_{0}^{π} x cos n x d x = \frac{( - 1 ) ^{n} - 1}{n ^{2}}, \int_{0}^{π} x^{2} cos n x d x = \frac{2 π ( - 1 ) ^{n}}{n ^{2}} .

Therefore the cosine Fourier coefficients are

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

Only even frequencies survive:

a_{2 m} = 1/ m^{2}

. The sine Fourier coefficients are

b_{1} = 1

,

b_{2} = - 2

, with all others zero. Hence

f (x) = sin x - 2 sin 2 x - \frac{π ^{2}}{6} + m = 1 \sum \infty \frac{cos 2 m x}{m ^{2}} .

The periodic function is continuous at zero and at its endpoint joins, and its coefficients are absolutely summable. Thus this Fourier series represents it everywhere, including those joins.

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