Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V resit / Question 4

The Fourier cosine series and Fourier sine series use the respective formsTheir Fourier coefficients, obtained from Fourier orthogonality, areIn particular the constant term is $a_0/2={\pi }^{-1}{\int }_0^{\pi }f(t)dt$. The Fourier cosine series is the Fourier series of the even extension $f(|x|)$, made $2\pi$-periodic. The Fourier sine series is the Fourier series of the odd extension, equal to $f(x)$ for $0<x<\pi$ and $-f(-x)$ for $-\pi <x<0$, again made $2\pi$-periodic.

The theorem on pointwise convergence of a piecewise smooth Fourier series gives the average of the two one-sided values of the corresponding periodic extension. Under the usual piecewise smooth interpretation of the paper's derivative assumption, the even extension is continuous at $0$ and at the joined endpoints $\pm \pi$. ThusFor the odd extension, the values on either side of zero are $-f(0)$ and $f(0)$; those on either side of $\pi$ are $f(\pi )$ and $-f(\pi )$. ThereforeThe endpoint values are also immediate because each sine term vanishes there. If $f(0)$ or $f(\pi )$ is nonzero, the Fourier sine series does not reproduce that endpoint value. Merely assuming that the derivative is locally piecewise continuous in the open interval would be weaker than the usual piecewise smooth hypothesis; the convergence assertion uses the standard endpoint regularity implicit in that hypothesis.

For the Fourier cosine series, integration by parts givesand, for $n\ge 1$,Only odd Fourier harmonics remain, so

For the Fourier sine series, another integration by parts givesConsequentlyAt $x=0$ and $x=\pi$, the Fourier sine series instead has value zero.

To describe the requested graphs precisely, take $t\in [-\pi ,\pi )$ congruent to $x$ modulo $2\pi$. The Fourier cosine series converges to the continuous triangular periodic extensionIts minima occur at the even multiples of $\pi$, with value $1$, and its maxima at the odd multiples, with value $1+\pi$. The Fourier sine series converges toThe jump discontinuities at even multiples of $\pi$ go from $-1$ to $1$; those at odd multiples go from $1+\pi$ to $-(1+\pi )$. Each jump is assigned its midpoint, zero, by the Fourier series.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

Take $\boxed{f(x)=\sin x}$. This function is not a polynomial: it is nonzero and satisfies $f^{\prime \prime }=-f$, whereas a nonzero polynomial cannot satisfy that identity, since taking two derivatives reduces its degree or gives zero.

By Fourier orthogonality, the Fourier sine series consists of the single term $\sin x$. Thus every truncation with $N\ge 1$ is exact, including at the endpoints. In contrast, using the product-to-sum trigonometric identity in its Fourier cosine coefficients givesFor example, ${\int }_0^{\pi }\sin x\cos (2mx)dx=-2/(4m^2-1)$ follows by integrating the two resulting sine terms. Therefore the Fourier cosine series isEvery finite cosine truncation omits infinitely many nonzero Fourier coefficients, so Parseval's identity gives it strictly positive squared L2 norm error. The one-term sine representation has zero error in both the L2 norm and the supremum norm, proving the requested advantage.