Second derivative of a Laguerre polynomial: Show that $L_n''(0)=\frac{1}{2}n(n-1).$

Question

Show that $$L_n''(0)=\frac{1}{2}n(n-1)$$

My first thought was to use the recurrence relation $$L_n^k(x)=L_{n-1}^k(x)+L_n^{k-1}(x)$$ and the orthogonality property $$\int_0^\infty e^{-x} x^kL_n^k(x)L_m^k(x)dx=\frac{(n+k)1}{n!},$$ for $k=2$, but doing so didn't seem to lead anywhere.

Answer 1

We know $L_n(0)=1$ and $L'_n(0)=-n$, so by derivation of Laguerre equation $$xy''+(1-x)y'+ny=0$$ again and set $x=0$ you will find it.

Answer 2

$$L_n(x)=\Sigma_{k=0}^{n}\frac{(-1)^k}{k!}\left ( \begin{align*}n \\k \end{align*} \right )x^k.$$ With $k=2$, we have $\left ( \begin{align*} n \\ \ 2 \end{align*} \right)=\frac{n(n-1)}{2}, (-1)^2=1, k!=2, x^2.$ So $$\frac{d}{dx}\left[ \frac{n(n-1)}{4}x^2\right ]=\frac{n(n-1)}{2}x,$$ and $$\frac{d^2}{dx^2}\left [ \frac{n(n-1)}{2}x\right ] = \frac{1}{2}n(n-1).$$