Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 20

Apply the Leibniz rule to the $n$th derivative in Rodrigues' formula. If $j$ derivatives fall on $e^{-x}$ and $n-j$ on $x^n$, the corresponding term is $\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-1{)}^j{e}^{-x}(n!/j!)x^j$. Cancelling the prefactor givesThe leading coefficient $(-1{)}^n/n!$ is nonzero, so this is a polynomial of degree exactly $n$. In particular, the requested coefficients in $L_{10}$ areThese are the ordinary Laguerre polynomials, with normalization $L_n(0)=1$.

Differentiating $v=x^n{e}^{-x}$ gives $x{v}^{\prime }=(n-x)v$. Applying the Leibniz rule $n+1$ times to this identity givesOnly two terms survive on each side because the multiplying factors are linear. Rearranging yields $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. Since $v^{(n)}=n!e^{-x}{L}_n$, the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substituting and cancelling $n!e^{-x}$ givesThis is deriving the Laguerre equation from Rodrigues' formula, exhibiting $L_n$ as an eigenfunction of the Laguerre differential operator on polynomials.

Multiply the two displayed equations by $e^{-x}$ and subtract the one with indices reversed. The product terms combine into the Lagrange identity for a Sturm-Liouville operator:Integrating from zero to infinity givesAll factors $L_j,L_j^{\prime }$ are polynomials. The endpoint term at zero vanishes because of the factor $x$, and the one at infinity vanishes because the exponential function dominates every polynomial. Therefore, if $m\ne n$,The integral is absolutely convergent. This is orthogonality in the weighted inner product with weight $e^{-x}$; the boundary cancellation is the self-adjoint mechanism behind the Laguerre differential operator on polynomials.