Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / P / P3

The best polynomial minimizes the weighted squared error among all polynomials of degree at most $n$:The given inner product is positive definite on continuous functions despite the weight vanishing at zero: a continuous nonzero function is nonzero on an interval containing nonzero points, so its weighted square integral is strictly positive.

If the error is orthogonal to ${\Pi }_n$, then for any $q\in {\Pi }_n$,proving sufficiency. Conversely, if $p_n$ minimizes, perturb it to $p_n+tq$ for any real $t$. The squared error becomesA minimum at $t=0$ forces the linear coefficient to vanish, proving necessity.

For existence, choose the basis $1,x,\dots ,x^n$. Its Gram matrix $G$ is positive definite because $c^{T}Gc$ is the squared norm of the nonzero polynomial with coefficients $c$. Thus $G$ is invertible, and the normal equations $Gc=((u,x^j){)}_{j=0}^n$ have a solution, giving an orthogonal residual and therefore a minimizer. The Pythagorean identity shows any other minimizer differs by a polynomial of zero norm, so it is equal to the first. Hence the best polynomial exists and is unique.

For odd $u$, define $q(x)=-p_n(-x)$. It is another polynomial in ${\Pi }_n$. Changing variables $x\mapsto -x$, and using evenness of the weight $x^2$, gives $\Vert u-q\Vert =\Vert u-p_n\Vert$. Thus $q$ is also a best polynomial. Uniqueness from part (b) gives $p_n(x)=-p_n(-x)$, provingEquivalently, the even component of a candidate is orthogonal to the odd residual and can only increase the squared error.

By part (c), the best degree-at-most-two polynomial is odd, so it must be $p_2(x)=bx$. Orthogonality to $x$ givesThereforeIts error is automatically orthogonal to the even basis functions $1,x^2$ by parity, and the chosen coefficient makes it orthogonal to $x$, so part (b) verifies optimality.