Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 18

For period two, the Fourier series convention isSince $f$ is an even function, every $b_n$ is zero. Also $a_0={\int }_{-1}^1\cosh xdx=2\sinh 1$. For $n\ge 1$, an antiderivative useful for the cosine coefficient isEvaluation at $\pm 1$ gives $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. ThusThe periodic extension is continuous at $x=1$, so its Fourier series converges to $\cosh 1$ there. Since $\cos (n\pi )=(-1{)}^n$,Therefore $\boxed{p=2,q=\pi }$, with $q=-\pi$ giving the same expression because only $q^2$ occurs. This is the $\beta =1$ case of Fourier series of periodic hyperbolic sine and cosine.

On the open interval, $g=f^{\prime }$. It is an odd function there, so its constant and cosine coefficients vanish; assigned values at the endpoints do not affect these integrals. For its sine coefficient, integration by parts givesConsequentlyThe endpoint qualification matters. The periodic $g$ has a jump discontinuity at every odd integer, with one-sided limits $\pm \sinh 1$. By the Dirichlet-Jordan convergence theorem, the Fourier series converges to their average, zero, there, not to the assigned value $g(1)=\sinh 1$. The integration by parts argument also avoids assuming unrestricted termwise differentiation of the first series.

For the period-two Fourier series of a square-integrable function $w$, Parseval's identity statesFor $w=f-g$, write $s=\sinh 1$. Its coefficients are $a_0=2s$, $a_n=2(-1{)}^{n}s/(1+n^2{\pi }^2)$ and $b_n=n\pi a_n$, because $g$ is subtracted. ThereforeThe sum was obtained in part (a). As a separate check, $f-g=e^{-x}$ in the interval, whose square has integral ${\int }_{-1}^1{e}^{-2x}dx=\sinh 2$. Endpoint jumps have no effect on this integral.