I'm working with Laguerre polynomials for numerically solving a differential equation, and I've stumbled upon an identity that I feel should be documented somewhere (e.g., https://en.wikipedia.org/wiki/Laguerre_polynomials). I can't seem to find anything related to this however. The identity is $$\frac{d L_n}{dx}=-\sum_{k=0}^{n-1}L_k(x).$$
I stumbled upon this numerically and couldn't find it documented anywhere, so I generated a proof as follows. Note that by the closed form expression $$L_n(x)=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{k!}x^k$$ we have $$\frac{d L_n}{dx}=\sum_{k=1}^n\binom{n}{k}\frac{(-1)^k}{(k-1)!}x^{k-1}.$$
Now, consider the sum
$$\sum_{k=0}^{n-1}L_k(x)=\sum_{k=0}^{n-1}\sum_{j=0}^k\binom{k}{j}\frac{(-1)^j}{j!}x^j=\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}\binom{k}{j}\frac{(-1)^j}{j!}x^j.$$
By the Hockey-Stick identity $$\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}\binom{k}{j}\frac{(-1)^j}{j!}x^j=\sum_{j=0}^{n-1}\binom{n}{j+1}\frac{(-1)^j}{j!}x^j=\sum_{j=1}^{n-1}\binom{n}{j}\frac{(-1)^{j-1}}{(j-1)!}x^{j-1}.$$ One can observe the similarity to the closed form for the derivative. If we simply adjust by a factor of $-1$, we have $$\sum_{k=0}^{n-1}L_k(x)=-\sum_{j=1}^{n-1}\binom{n}{j}\frac{(-1)^{j}}{(j-1)!}x^{j-1}=-\frac{d L_n}{dx}.$$ Have I simply missed an equivalent identity listed somewhere? This seems like a very useful relation (especially for solving diff eqs) that I can't find documented anywhere.