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

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.