Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 18

For a real $2\pi$-periodic square-integrable function, use the Fourier series conventionwith Fourier coefficientsThe trigonometric orthogonality relations are ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$ for positive $n,m$, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral $2\pi$ and is orthogonal to every positive harmonic. Thus the squared norm of the $N$th partial sum isTaking the limit in the L2 space gives the Fourier-series Parseval identity:Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.

The function is odd, so its constant and cosine coefficients are zero. Its sine coefficient isThe polynomial vanishes at both integration endpoints. Two integrations by parts therefore giveOne more integration yieldsConsequently $I_n=6\pi (-1{)}^n/n^3$ and $b_n=12(-1{)}^n/n^3$. HenceThe periodic continuation is continuous at $\pm \pi$, where its values are zero, so the series represents it there as well as in the open interval.

For this odd function, the Parseval identity from part (a) givesThe integrand is even. Expanding and integrating it givesDividing by $144$ therefore provesThis evaluates the Riemann zeta function at six by using the norm of a simple polynomial.