= Solution
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 $\binom nj(-1)^je^{-x}(n!/j!)x^j$. Cancelling the prefactor gives
$$
\boxed{L_n(x)=\sum_{j=0}^n(-1)^j\binom nj\frac{x^j}{j!}}.
$$
The leading coefficient $(-1)^n/n!$ is nonzero, so this is a <polynomial> of degree exactly $n$. In particular, the requested coefficients in $L_{10}$ are
$$
\boxed{[x^{10}]L_{10}=\frac1{10!},\qquad [x^9]L_{10}=-\frac{10!}{(9!)^2}}.
$$
These are the ordinary <Laguerre polynomials>, with normalization $L_n(0)=1$.
Back to article page
