Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 11

The function is even and continuous, with period $2\pi$. On $0<x<\pi$ it is $\sin 2x$, so it rises from zero to one at $\pi /4$, crosses zero at $\pi /2$, reaches minus one at $3\pi /4$, and returns to zero at $\pi$. Reflect this curve about the vertical axis and repeat periodically. The joins at integer multiples of $\pi$ have corners, but no jumps.

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:For $n=2$, the integral is directly zero rather than an undefined substitution into a denominator. ThusThe Fourier series equals the continuous function at every point, including the corners. Its coefficients decay quadratically, so the series is also absolutely convergent.

At $x=0$, the Fourier series equals $f(0)=0$. Isolating the first term givesChanging the sign of the remaining denominators provesThe absolute convergence justifies the evaluation and rearrangement.