Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 17

Every harmonic in this Fourier series has period $2L$. Since the first-harmonic coefficient is nonzero, the fundamental period cannot be smaller: translating by a period $T$ must multiply that nonzero complex coefficient by $e^{i\pi T/L}=1$. Thus, taking $L>0$,

Write $h_n=a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L)$. For the partial sum $S_N=a_0/2+{\sum }_{n=1}^N{h}_n$, ordinary multiplication givesOn $[-L,L]$, the integrals of every nonconstant sine or cosine vanish. Fourier orthogonality further gives $\int \cos (m\pi x/L)\cos (n\pi x/L)dx=L{\delta }_{mn}$ and the same identity for two sines, while every sine-cosine integral is zero. ThusFor square-integrable $f$, its Fourier partial sums converge to $f$ in L2 space, so their squared norms converge to the squared norm of $f$. Passing to the limit proves Parseval's identity:Working with finite partial sums before taking the norm limit avoids an unjustified interchange of an infinite double series and integration.

The periodic ramp is an odd function, so $a_0=a_n=0$. Integration by parts givesThus the Fourier series of a periodic sawtooth isThe periodic function is smooth at $x=\pi /2$, so the Dirichlet-Jordan convergence theorem permits evaluation there. Even-index terms vanish and odd-index sines alternate in sign, giving