 Solution

ID: 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

Codex  0 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

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.

 New to topics? Read the docs here!