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 Incoming links: Fourier cosine coefficients of a sine mode

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 11 / b / Solution Created 2026-10-18 Updated 2026-10-20

Because the function is even, all sine Fourier coefficients vanish. Its average is zero, so

a_{0} = 0

. The Fourier cosine coefficients of a sine mode follow from the product-to-sum identity:

a_{n} = \frac{2}{π} \int_{0}^{π} sin 2 x cos n x d x = ⎩ ⎨ ⎧ \frac{8}{π ( 4 - n ^{2} )}, 0, n odd, n even .

For

n = 2

, the integral is directly zero rather than an undefined substitution into a denominator. Thus

f (x) = \frac{8}{π} m = 0 \sum \infty \frac{cos (( 2 m + 1 ) x )}{4 - ( 2 m + 1 ) ^{2}} .

The Fourier series equals the continuous function at every point, including the corners. Its coefficients decay quadratically, so the series is also absolutely convergent.

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