Past exam of the analysis course of the University of Oxford / 2009 / B / 4

A complete orthonormal sequence $(e_n)$ has $⟨e_n,e_m⟩={\delta }_{nm}$ and dense linear span; equivalently, the only vector orthogonal to every $e_n$ is zero. The Parseval identity says $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$, with expansion $x={\sum }_n⟨x,e_n⟩e_n$ in the Hilbert norm.

Put $G(t,z)=(1-2zt+z^2{)}^{-1/2}$ and $D=1-2zt+z^2$. Direct differentiation givesSubstitution yields $(1-t^2)G_{tt}-2t{G}_t+z^2{G}_{zz}+2z{G}_z=0$. Extracting the coefficient of $z^n$ in the generating function givesthe Legendre differential equation.

The generating function gives $P_n(-t)=(-1{)}^n{P}_n(t)$, so the odd function $f$ is orthogonal to every even-degree Legendre polynomial. For $n\ge 1$, integrate the self-adjoint form of the Legendre differential equation from zero to one:Thus ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. Differentiating the generating function in $t$ and then setting $t=0$ givesCombining these identities with the normalization of $p_n$ yieldsFor the Legendre expansion of the sign function, $\Vert f{\Vert }_2^2=2$, so the Parseval identity givesTherefore the requested sum equals

Expand both generating functions and use orthogonality of the Legendre polynomials, with ${\int }_{-1}^1{P}_n{P}_m=2{\delta }_{nm}/(2n+1)$. The allowed exchanges giveHence it depends only on $r=wz$. Explicitly,The series form alone already proves the required dependence.