Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 20

Direct differentiation of $v=(x^2-1{)}^l$ gives $(x^2-1)v^{\prime }=2lxv$; the identity also holds for $l=0$. Differentiate it $l+1$ times using the Leibniz rule. The quadratic factor has only its first two derivatives nonzero, soWriting $y=v^{(l)}$ and collecting terms gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$. Multiplication by $-1$ gives the Legendre differential equation:

The polynomial $v$ has leading term $x^{2l}$. Its $l$th derivative has leading term $(2l)!x^l/l!$, and all other terms have lower degree. Thus the Rodrigues formula for Legendre polynomials defines a polynomial with nonzero leading coefficientIt therefore has degree exactly $l$, including the constant case $P_0=1$.

Write $v=(x+1{)}^l(x-1{)}^l$. In the Leibniz rule for its $l$th derivative at one, any term with fewer than $l$ derivatives on $(x-1{)}^l$ vanishes. The sole surviving term leaves $(x+1{)}^l$ undifferentiated and differentiates the second factor $l$ times, givingThe normalization in Rodrigues' formula consequently givesThe same argument covers $l=0$, with both zeroth powers equal to one.