Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 4 / 5A (Methods)

The original PDF contains this subpart; the local TeX has omitted it. Write $a_n$ for the leading coefficient of $P_n$. The Legendre polynomial recurrence relation and the starting polynomials give $a_0=a_1=1$, and induction givesIndeed, $x{P}_n$ has degree of a polynomial $n+1$, while $P_{n-1}$ has smaller degree of a polynomial and cannot cancel its leading term. HenceAt $x=1$, the same linear recurrence relation gives $(n+1)P_{n+1}(1)=(2n+1)P_n(1)-n{P}_{n-1}(1)$. Starting from $P_0(1)=P_1(1)=1$, induction yields

This subpart is also present in the original PDF but missing from the local TeX. Put $p(x)=1-x^2$ and ${\lambda }_n=n(n+1)$. Multiply the Legendre differential equation for $P_n$ by $P_m$, multiply the equation for $P_m$ by $P_n$, and subtract. The product rule givesIntegrating from $-1$ to $1$, the boundary term vanishes: $p(\pm 1)=0$, and the Legendre polynomials and their derivatives are finite there. For distinct nonnegative integers $m,n$, ${\lambda }_m\ne {\lambda }_n$. Thus the Orthogonality of Legendre polynomials isThe vanishing coefficient at the endpoints causes no difficulty here because the solutions are polynomials.

Evaluate $R_n$ at $x=1$ using $(x^2-1{)}^n=(x-1{)}^n(x+1{)}^n$. In the Leibniz rule for the $n$th derivative, every term vanishes at $1$ except the term putting all $n$ derivatives on $(x-1{)}^n$. Therefore $R_n(1)=n!2^n$. Combining this with $P_n(1)=1$ gives the normalisation in the Rodrigues formula:For $n=0$, the formula gives ${\alpha }_0=1$, as required. As an independent check, the leading term of $R_n$ is $(2n)!x^n/n!$, so the resulting leading coefficient of $P_n$ is $(2n)!/(2^n(n!{)}^2)$, consistent with the Legendre polynomial recurrence relation.