TITLE: A sum involving derivatives of Vandermonde
QUESTION [12 upvotes]: Consider the standard Vandermonde $V(x_1, \ldots, x_n) = \prod_{i < j} (x_i - x_j)$.
I am intersted in the calculation of the following expression for fixed $k$:
$$\sum_i (x_i)^k (d/dx_i)^k V(x_1 , \ldots , x_n).$$
My guess is that it equals  $c \cdot V(x_1, \ldots, x_n)$
where $c$ is an expression depending on $k$ and $n$ but not on the $x_i$'s. Is it true ? If yes, what is this constant $c$?
I think, if it is true, then it is pretty well-known.
Would you be so kind to provide with the answer and/or proof and/or references?
What can be the context people study it? Symmetric functions? Quantum Calogero-Moser?

REPLY [6 votes]: The Vandermonde Matrix is $V_{ij}=x^{i-1}_j$. Notice that only the $k-th$ column depends on the $k-th$ variable so we can use the Laplace expansion for the determinant of $V_{ij}$ over such column
\begin{eqnarray}
\sum_kx^n_k\frac{\partial^n}{\partial x_k^n}\underset{1 \leq i,j \leq N}{\det}(V_{ij}) & = & \sum_kx^n_k\frac{\partial^n}{\partial x_k^n}\sum_p x_k^{p-1}C_{pk} \\
 & = & \sum_kx^n_k\sum_p \frac{(p-1)!}{(p-1-n)!} x_k^{p-1-n}C_{pk} \\
 & = & \sum_p \frac{(p-1)!}{(p-1-n)!}\sum_k x_k^{p-1}C_{pk} \\
 & = & \sum_p \frac{(p-1)!}{(p-1-n)!}\underset{1 \leq i,j \leq N}{\det}(V_{ij})
\end{eqnarray}
were $C_{ij}$ are the cofactors of $V_{ij}$.
After applying the operator, the sum over $k$ yield back a determinant as an expansion over the $p-th$ row.