Past exam of the options II course of the University of Oxford / 2013 / P3

With the weighted inner product $⟨p,q{⟩}_w={\int }_{-1}^{1}p(x)q(x)w(x)dx$, a family of orthogonal polynomials has $\mathrm{deg}{p}_j=j$, each $p_j$ nonzero, andContinuity and positivity of the weight make the integral finite and give $⟨p_j,p_j{⟩}_w>0$. Unit norm or monic leading coefficient is an additional optional normalization.

The weight is even, so opposite-parity polynomials are automatically orthogonal. Set $p_0=1$, $p_1=x$, then subtract lower-degree components within each parity class. The needed moments areThus $p_2=x^2-{\mu }_2/{\mu }_0=x^2-2/5$ is orthogonal to $p_0$, and $p_3=x^3-({\mu }_4/{\mu }_2)x=x^3-9x/14$ is orthogonal to $p_1$. Parity supplies every other pair. One valid answer isAll have the required degree and positive norm; arbitrary nonzero rescalings are equally valid.

The degree-zero polynomial is constant and therefore even. Assume inductively that $p_j(-x)=(-1{)}^j{p}_j(x)$ for $j<k$. Define $q_k(x)=(-1{)}^k{p}_k(-x)$, which has the same leading coefficient as $p_k$.

Changing variable $x\mapsto -x$ and using the even weight and inductive hypothesis gives $⟨q_k,p_j{⟩}_w=(-1{)}^{k+j}⟨p_k,p_j{⟩}_w=0$ for $j<k$. Hence $q_k-p_k$ has degree below $k$ and is orthogonal to every $p_0,\dots ,p_{k-1}$. Those polynomials form a basis for the polynomials of degree below $k$, so the difference is orthogonal to itself. Positivity of the weighted inner product forces it to be zero.

Thereforecompleting the induction and proving parity of orthogonal polynomials for an even weight. Thus even indices give even functions and odd indices give odd functions, regardless of the nonzero normalization chosen.