Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 4

For the Fourier cosine series, first take the even extension $F_e(x)=f(|x|)$ on $[-\pi ,\pi ]$, and then its $2\pi$-periodic extension. Evenness eliminates the sine terms. With the constant term written as $a_0/2$, the Fourier coefficients and series areFor the Fourier sine series, take the odd extension $F_o(x)=f(x)$ for $0<x<\pi$ and $F_o(x)=-f(-x)$ for $-\pi <x<0$, and extend with period $2\pi$. Values assigned at isolated endpoints do not change the Fourier coefficients. Oddness eliminates the constant and cosine terms, leavingBoth formulas follow by orthogonality on $[-\pi ,\pi ]$ and the parity of the extensions.

Using the usual piecewise smooth interpretation of the derivative hypothesis, Dirichlet's theorem on Fourier series gives the average of the two one-sided limits of the appropriate periodic extension. The even extension is continuous at $0$ and at the identified endpoints $\pm \pi$, so the Fourier cosine series converges toFor the odd extension, the one-sided limits at zero are $-f(0)$ and $f(0)$, while at $\pi$ the periodic one-sided limits are $f(\pi )$ and $-f(\pi )$. Hence the Fourier sine series converges toAt an endpoint this equals $f$ precisely when $f$ is zero there. One can also see the zero endpoint values directly: every partial sum consists of sines that vanish at both endpoints.

Put $a=\pi /2$. The constant Fourier coefficient isFor $n\ge 1$, integration by parts givesThe $n^{-1}$ terms cancel. ThusThis Fourier cosine series has $a_n=0$ when $n$ is divisible by four, $a_n=-4/(\pi n^2)$ when $n\equiv 2(\mathrm{mod}4)$, and $a_n=-2/(\pi n^2)$ when $n$ is odd.

Similarly,ThereforeThe Fourier sine series also equals $f(0)=0$ at zero, but has value zero at $\pi$, where $f(\pi )=\pi /2$.

The Fourier cosine series converges to the continuous even $2\pi$-periodic extension. If $r\in [-\pi ,\pi ]$ represents $x$ modulo $2\pi$, its value isIt has value zero at $x=2j\pi$, rises linearly with distance from these points until height $\pi /2$, and then stays flat until the next descending ramp. Thus the sketch on $[-3\pi ,3\pi ]$ has minima at $-2\pi ,0,2\pi$, and plateaux centred at the odd multiples of $\pi$. Its large partial sums round the corners and closely follow the whole graph.

The Fourier sine series converges, away from the odd multiples of $\pi$, to the odd $2\pi$-periodic extensionAt every odd multiple of $\pi$ its value is zero, the midpoint of the jump from $\pi /2$ to $-\pi /2$. At even multiples of $\pi$ it passes continuously through zero. Thus on every interval $((2j-1)\pi ,(2j+1)\pi )$ it has a negative plateau, a rising linear segment through $2j\pi$, and a positive plateau. The jump size is $\pi$.

The blue curves are the partial sums through $n=60$. The open endpoint circles and filled zero dots in the lower panel distinguish the one-sided limits from the value of the Fourier sine series at each jump. Near these jumps the partial sums exhibit the Gibbs phenomenon: the oscillations become concentrated in narrower regions as the number of terms grows, while their limiting overshoot remains about $0.08949$ times the jump, here about $0.281$.

The truncated Fourier cosine series gives the better approximation over the whole closed interval. Its Fourier coefficients obey $|a_n|\le 4/(\pi n^2)$, so absolute convergence and the tail boundgive uniform convergence. The even periodic extension is continuous; its slope changes only cause corners.

By contrast $b_n=(-1{)}^{n+1}/n+O(n^{-2})$. The odd periodic extension has a jump at $\pi$, giving slower coefficient decay and the Gibbs phenomenon near that endpoint. Every truncated Fourier sine series satisfies $S_N(\pi )=0$, soIt therefore cannot converge uniformly to $f$ on $[0,\pi ]$, although it converges at every point of $[0,\pi )$. This comparison concerns approximation across the interval; it does not assert that the cosine truncation has smaller error at every individual interior point for every $N$.