Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V resit / Question 4 / b / i / Solution

For the Fourier cosine series, integration by parts givesand, for $n\ge 1$,Only odd Fourier harmonics remain, so

For the Fourier sine series, another integration by parts givesConsequentlyAt $x=0$ and $x=\pi$, the Fourier sine series instead has value zero.

To describe the requested graphs precisely, take $t\in [-\pi ,\pi )$ congruent to $x$ modulo $2\pi$. The Fourier cosine series converges to the continuous triangular periodic extensionIts minima occur at the even multiples of $\pi$, with value $1$, and its maxima at the odd multiples, with value $1+\pi$. The Fourier sine series converges toThe jump discontinuities at even multiples of $\pi$ go from $-1$ to $1$; those at odd multiples go from $1+\pi$ to $-(1+\pi )$. Each jump is assigned its midpoint, zero, by the Fourier series.