Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 2 / 18F (Numerical Analysis)

Take $Q_j$ nonzero of exact degree $j$, so $Q_0,\dots ,Q_n$ are an orthogonal basis of the polynomials of degree at most $n$. Their orthogonal projection isIndeed its residual is orthogonal to every such polynomial. For another polynomial $q=p_n+r$ of degree at most $n$, the squared error issince the cross term vanishes. It is uniquely minimized at $r=0$. The approximation need not have exact degree $n$ when the highest coefficient happens to vanish.

Use first-kind Chebyshev polynomials $T_j$, characterized by $T_j(\cos \theta )=\cos (j\theta )$. Substituting $x=\cos \theta$ converts the weighted inner product into ${\int }_0^{\pi }g(\cos \theta )h(\cos \theta )d\theta$. Hence these polynomials are orthogonal, with squared norms $\pi$ for $j=0$ and $\pi /2$ for $j\ge 1$.

For the specified function, $f(\cos \theta )=\sin \theta$. Its constant projection coefficient is ${\pi }^{-1}{\int }_0^{\pi }\sin \theta d\theta =2/\pi$. For $j\ge 1$ the coefficient is $(2/\pi ){\int }_0^{\pi }\sin \theta \cos (j\theta )d\theta$. Symmetry about $\pi /2$ makes it zero for odd $j$. For $j=2m$, the product-to-sum identity givesTherefore the Chebyshev projection of a semicircle isIn particular the best degree-at-most-$n$ polynomial is even, so for odd $n$ its actual degree is at most $n-1$.