Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 5D (Methods)

Use the real Fourier series conventionThe constant coefficient is $a_0=2{\int }_0^{1}x(1-x)dx=1/3$. Reflection about $x=1/2$ changes the sine factor's sign and preserves the quadratic, giving $b_n=0$. Two integrations by parts, with $q=2\pi n$, giveConsequently the Fourier series of a periodically extended quadratic arch isThe periodic function is continuous, including at integers where its value is zero. Its piecewise smoothness gives pointwise Fourier series convergence everywhere, and the summable coefficient bound gives uniform and absolute convergence.

The endpoint agreement $f(1)=f(0)=0$ makes the boundary terms vanish when integrating the Fourier coefficients of $f^{\prime }$ by parts. If $a_n^{\prime },b_n^{\prime }$ denote these coefficients, thenThese are exactly the coefficients obtained by differentiating the preceding Fourier series:On every compact subinterval of $(0,1)$ the differentiated series converges uniformly by the Dirichlet test, since partial sums of the sine factors are uniformly bounded there. The Termwise differentiation of a Fourier series is therefore valid away from integers and gives $1-2x$ on $(0,1)$. At an integer the periodic original has a corner, so the differentiated series is not asserting the existence of an ordinary derivative there.

Every sine term vanishes at zero, so $G(0)=0$. The one-sided values of the periodic extension of $f^{\prime }$ are $f_{\mathrm{per}}^{\prime }(0-)=-1$ and $f_{\mathrm{per}}^{\prime }(0+)=1$. The Fourier series converges at this jump to their midpoint,This midpoint need not equal the arbitrarily assigned value of the periodic derivative at the endpoint.