Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 41B (Numerical Analysis)

With $x=\cos \theta$, $T_k(x)=\cos k\theta$, proving orthogonality. In the convention $f\sim c_0+{\sum }_{k\ge 1}{c}_k{T}_k$, $c_0={\pi }^{-1}{\int }_0^{\pi }f(\cos \theta )d\theta$ and $c_k=2{\pi }^{-1}{\int }_0^{\pi }f(\cos \theta )\cos k\theta d\theta$. For $f=\sqrt{1-x^2}$, $c_0=2/\pi$, odd coefficients vanish, and $c_{2m}=-4/[\pi (4m^2-1)]$.

The substitution $x=\cos \theta$ identifies the weighted norm with the $L^2(0,\pi )$ norm of $f(\cos \theta )$. Completeness of cosine Fourier series therefore makes the Chebyshev partial sums converge in $\Vert \cdot {\Vert }_w$.

Parseval's identity for the cosine series, with $\Vert T_0{\Vert }_w^2=\pi$ and $\Vert T_k{\Vert }_w^2=\pi /2$ for $k\ge 1$, gives the claimed equality.

Repeated integration by parts in the cosine-coefficient integral shows $|c_k|\le C_r{k}^{-r}$ for every $r$ when $f$ is smooth (endpoint terms vanish after using the even periodic extension). Parseval then bounds the squared tail by $C_r{\sum }_{k>N}{k}^{-2r}$, which is $O(N^{-2r+1})$. Choosing $r$ arbitrarily large proves spectral convergence.