Past exam of the numerical analysis course of the University of Oxford / 2014 / 2

The norm induced by the specified inner product is the L2 normA polynomial $p_n\in {\Pi }_n$ is a best polynomial least-squares approximation whenMinimizing this norm is equivalent to minimizing its square, the integrated squared error. Continuity ensures that a function with zero squared norm is identically zero: a nonzero value would give a neighbourhood on which the square has a positive integral.

Write $p_n(x)={\sum }_{j=0}^n{c}_j{x}^j$. The allowed orthogonal projection characterization, applied to the basis $1,x,\dots ,x^n$, gives the normal equationsExplicitly $G_{ij}=0$ if $i+j$ is odd and $G_{ij}=2/(i+j+1)$ if it is even. These equations can also be derived by differentiating $\int (u-\sum c_j{x}^j{)}^{2}dx$ with respect to each coefficient.

The Gram matrix $G$ is a symmetric positive-definite matrix. For any nonzero real coefficient vector $z$,The polynomial inside the square is nonzero, so its square is positive on some interval. Hence $G$ has zero kernel and is invertible. Thus the normal equations have exactly one solution, giving exactly one best least-squares polynomial. This proves both existence and uniqueness. For completeness, if the residual is orthogonal to ${\Pi }_n$, then for every $q\in {\Pi }_n$ the Pythagorean theorem gives $\Vert u-q{\Vert }^2=\Vert u-p_n{\Vert }^2+\Vert p_n-q{\Vert }^2$, directly verifying minimality and uniqueness.

Split the best polynomial into its even and odd parts, $p_n=p_e+p_o$. Both parts belong to ${\Pi }_n$. Since $u$ is odd, $u-p_o$ is odd and $p_e$ is even. Their product is odd, so its integral on the symmetric interval vanishes. The orthogonality givesThe candidate $p_o$ cannot have a strictly smaller error than a minimizer. Consequently $\Vert p_e{\Vert }^2=0$, and continuity gives $p_e=0$. Thus $\boxed{p_n(-x)=-p_n(x)}$. For $n=0$ this means $p_0=0$, which is indeed an odd function. More generally this is parity preservation by polynomial least-squares projection.

By parity preservation by polynomial least-squares projection, $p_2(x)=\beta x$: the constant and quadratic coefficients vanish. The remaining normal equation isThereforeThe residual is automatically orthogonal to $1$ and $x^2$ by parity and to $x$ by construction, so it is orthogonal to every element of ${\Pi }_2$. This checks the full linear least squares condition, rather than only a necessary coefficient equation.

Projection onto the constant polynomials requires the residual to be orthogonal to $1$. For the original inner product,For the weighted inner product the corresponding normal equation isHenceIt increases monotonically towards one. Although the weight vanishes at zero, it is positive elsewhere, and a nonzero continuous function is nonzero on an interval of positive weight. Thus the weighted expression really is an inner product and the constant minimizer is unique.

The normalized weight $(2n+1)x^{2n}/2$ has total mass one and assigns mass $r^{2n+1}$ to $[-r,r]$ for each $0<r<1$. Its mass therefore concentrates near $-1$ and $1$, where $x^2=1$. This explains why the weighted best constant tends to one rather than to the unweighted spatial average. It is endpoint concentration in weighted constant approximation.