Past exam of the options II course of the University of Oxford / 2011 / P / P3

A polynomial least-squares approximation $p_n\in {\Pi }_n$ minimizes the weighted residual norm:Equivalently it minimizes ${\int }_{-1}^1{x}^2(u(x)-q(x){)}^{2}dx$. This is a best approximation within a finite-dimensional polynomial space, without an interpolation requirement.

If $p$ is a minimizer, then for each $q\in {\Pi }_n$ the quadratichas a minimum at $t=0$, so $(u-p,q)=0$. Conversely, if this orthogonality holds, then for every $q\in {\Pi }_n$ the Pythagorean theorem in the inner product space givesThus the minimizers are exactly the orthogonal projections characterized byFor existence, write $p={\sum }_{j=0}^n{c}_j{x}^j$. The conditions form the normal equations $Gc=b$, with Gram matrix $G_{ij}=(x^i,x^j)$ and $b_i=(u,x^i)$. For every nonzero coefficient vector $v$,Indeed a nonzero polynomial is nonzero on some open interval away from $0$, where the weight is positive. Hence $G$ is a positive-definite matrix and is invertible, proving existence and uniqueness of $c$. Uniqueness also follows from the preceding Pythagorean theorem identity. The zero of the weight at one point therefore causes no degeneracy on continuous functions or polynomials.

Define the reflected polynomial $\tilde{p}(x)=-p(-x)$. It belongs to ${\Pi }_n$. Since $u$ is odd, a change of variable and the evenness of the weight giveThus $\tilde{p}$ is another polynomial least-squares approximation. By uniqueness, $\tilde{p}=p$, soThe conclusion includes the zero polynomial, which is both odd and even.

By part (c), the best polynomial of degree at most $2$ is odd, hence has the form $p_2(x)=cx$. The orthogonal projection condition with $q=x$ givesThe residual is odd, so its weighted inner products with the even polynomials $1$ and $x^2$ vanish automatically. Thus all three normal equations hold and