Consider the polynomial $P(X) =\prod_{i=1}^k (X-x_i)^s$ and let $M$ be the corresponding confluent Vandermonde matrix. Concretely, here is what I mean by that. Define $$ M^{(0)} = \begin{pmatrix} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots &x_k\\ x^2_1 & x^2_2 & \cdots &x^2_k\\ \vdots\\ x^{ks-1}_1 & x^{ks-1}_2 & \cdots &x^{ks-1}_k \end{pmatrix} $$ which is the $ks\times k$ Vandermonde matrix. Let $D=\sum_i \partial/\partial x_i$ be an operator. Define $$ M^{(n)} = \frac{1}{n!}D^n M^{(0)} $$ Then $M=(M^{(0)}, M^{(1)}, \cdots, M^{(s-1)})$ which is $ks\times ks$. With that definition, now here is my question:

Is there a clean formula for the solution $F$ of the system $MF=B$? Equivalently, is $M^{-1}$ known in terms of, say, symmetric functions? For all intents and purposes we can define $M^{(n)}=D^n M^{(0)}$ instead (as confluent Vandermonde is sometimes defined) if it helps.

Extra Detail: Since I have not managed to find much on this topic online, I'm fearful that there might not be a clean answer in general. In my specific application, however, both $F$ and $B$ have additional structure. Maybe these additional assumptions make life easier.

Structure of $F$: Let $F^{(n)}=(F^{(n)}_1, \cdots, F^{(n)}_k)^T$ so that $F=(F^{(0)}, \cdots, F^{(s-1)})$. Given a function $g(x_1, \cdots, x_k)$, consider the following of Dunkl operator $$ (Kg)(x_1, \cdots, x_k) = g(x_2, x_3, \cdots, x_k, x_1) $$ which rotates the variables. Then for each $0\leq n\leq s-1$ and $1\leq m\leq k$ I have $$ F^{(n)}_m(x_1, \cdots, x_k) = (K^{m-1} F^{(n)}_1)(x_1, \cdots, x_k) $$ Moreover, $F_1^{(n)}$ is symmetric (rational function) in variables $x_2, \cdots, x_k$.

Structure of B: Let $\Delta = \sum_{i=1}^k x_i^2\partial /\partial x_i$ and $f(x_1, \cdots, x_k)$ a symmetric homogeneous polynomial of degree $s$ such that $Df=\Delta^{(k-2)s+1}f=0$. Then $B=(b_1, \cdots, b_{ks})^T$ is obtained via the following $$ \sum_{l=1}^{ks} b_l t^l = t^{2s}\exp(t\Delta)f $$

All I really need is to find a formula for functions $F_1^{(0)}, F_1^{(1)}, \cdots, F_1^{(s-1)}$ in terms of $B$, or even better, $f$.

Consider the polynomial $P(X) =\prod_{i=1}^k (X-x_i)^s$ and let $M$ be the corresponding confluent Vandermonde matrix. Concretely, here is what I mean by that. Define $$ M^{(0)} = \begin{pmatrix} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots &x_k\\ x^2_1 & x^2_2 & \cdots &x^2_k\\ \vdots\\ x^{ks-1}_1 & x^{ks-1}_2 & \cdots &x^{ks-1}_k \end{pmatrix} $$ which is the $ks\times k$ Vandermonde matrix. Let $D=\sum_i \partial/\partial x_i$ be an operator. Define $$ M^{(n)} = \frac{1}{n!}D^n M^{(0)} $$ Then $M=(M^{(0)}, M^{(1)}, \cdots, M^{(s-1)})$ which is $ks\times ks$. With that definition, now here is my question:

Is there a clean formula for the solution $F$ of the system $MF=B$? Equivalently, is $M^{-1}$ known in terms of, say, symmetric functions? For all intents and purposes we can define $M^{(n)}=D^n M^{(0)}$ instead (as confluent Vandermonde is sometimes defined) if it helps.

Consider the polynomial $P(X) =\prod_{i=1}^k (X-x_i)^s$ and let $M$ be the corresponding confluent Vandermonde matrix. Concretely, here is what I mean by that. Define $$ M^{(0)} = \begin{pmatrix} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots &x_k\\ x^2_1 & x^2_2 & \cdots &x^2_k\\ \vdots\\ x^{ks-1}_1 & x^{ks-1}_2 & \cdots &x^{ks-1}_k \end{pmatrix} $$ which is the $ks\times k$ Vandermonde matrix. Let $D=\sum_i \partial/\partial x_i$ be an operator. Define $$ M^{(n)} = \frac{1}{n!}D^n M^{(0)} $$ Then $M=(M^{(0)}, M^{(1)}, \cdots, M^{(s-1)})$ which is $ks\times ks$. With that definition, now here is my question:

Is there a clean formula for the solution $F$ of the system $MF=B$? Equivalently, is $M^{-1}$ known in terms of, say, symmetric functions? For all intents and purposes we can define $M^{(n)}=D^n M^{(0)}$ instead (as confluent Vandermonde is sometimes defined) if it helps.

Extra Detail: Since I have not managed to find much on this topic online, I'm fearful that there might not be a clean answer in general. In my specific application, however, both $F$ and $B$ have additional structure. Maybe these additional assumptions make life easier.

Structure of $F$: Let $F^{(n)}=(F^{(n)}_1, \cdots, F^{(n)}_k)^T$ so that $F=(F^{(0)}, \cdots, F^{(s-1)})$. Given a function $g(x_1, \cdots, x_k)$, consider the following of Dunkl operator $$ (Kg)(x_1, \cdots, x_k) = g(x_2, x_3, \cdots, x_k, x_1) $$ which rotates the variables. Then for each $0\leq n\leq s-1$ I have $$ F^{(n)}_m(x_1, \cdots, x_k) = (K^{m-1} F^{(n)}_1)(x_1, \cdots, x_k) $$ Moreover, $F_1^{(n)}$ is symmetric in variables $x_2, \cdots, x_k$.

Structure of B: Let $\Delta = \sum_{i=1}^k x_i^2\partial /\partial x_i$ and $f(x_1, \cdots, x_k)$ a symmetric homogeneous polynomial of degree $s$ such that $Df=\Delta^{(k-2)s+1}f=0$. Then $B=(b_1, \cdots, b_{ks})^T$ is obtained via the following $$ \sum_{l=1}^{ks} b_l t^l = t^{2s}\exp(t\Delta)f $$

All I really need is to find a formula for functions $F_1^{(0)}, F_1^{(1)}, \cdots, F_1^{(s-1)}$ in terms of $B$, or even better, $f$.

Consider the polynomial $P(X) =\prod_{i=1}^k (X-x_i)^s$ and let $M$ be the corresponding confluent Vandermonde matrix. Concretely, here is what I mean by that. Define $$ M^{(0)} = \begin{pmatrix} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots &x_k\\ x^2_1 & x^2_2 & \cdots &x^2_k\\ \vdots\\ x^{ks-1}_1 & x^{ks-1}_2 & \cdots &x^{ks-1}_k \end{pmatrix} $$ which is the $ks\times k$ Vandermonde matrix. Let $D=\sum_i \partial/\partial x_i$ be an operator. Define $$ M^{(n)} = \frac{1}{n!}D^n M^{(0)} $$ Then $M=(M^{(0)}, M^{(1)}, \cdots, M^{(s-1)})$ which is $ks\times ks$. With that definition, now here is my question:

Is there a clean formula for the solution $F$ of the system $MF=B$? Equivalently, is $M^{-1}$ known in terms of, say, symmetric functions? For all intents and purposes we can define $M^{(n)}=D^n M^{(0)}$ instead (as confluent Vandermonde is sometimes defined) if it helps.

Extra Detail: Since I have not managed to find much on this topic online, I'm fearful that there might not be a clean answer in general. In my specific application, however, both $F$ and $B$ have additional structure. Maybe these additional assumptions make life easier.

Structure of $F$: Let $F^{(n)}=(F^{(n)}_1, \cdots, F^{(n)}_k)^T$ so that $F=(F^{(0)}, \cdots, F^{(s-1)})$. Given a function $g(x_1, \cdots, x_k)$, consider the following of Dunkl operator $$ (Kg)(x_1, \cdots, x_k) = g(x_2, x_3, \cdots, x_k, x_1) $$ which rotates the variables. Then for each $0\leq n\leq s-1$ and $1\leq m\leq k$ I have $$ F^{(n)}_m(x_1, \cdots, x_k) = (K^{m-1} F^{(n)}_1)(x_1, \cdots, x_k) $$ Moreover, $F_1^{(n)}$ is symmetric (rational function) in variables $x_2, \cdots, x_k$.

Structure of B: Let $\Delta = \sum_{i=1}^k x_i^2\partial /\partial x_i$ and $f(x_1, \cdots, x_k)$ a symmetric homogeneous polynomial of degree $s$ such that $Df=\Delta^{(k-2)s+1}f=0$. Then $B=(b_1, \cdots, b_{ks})^T$ is obtained via the following $$ \sum_{l=1}^{ks} b_l t^l = t^{2s}\exp(t\Delta)f $$

All I really need is to find a formula for functions $F_1^{(0)}, F_1^{(1)}, \cdots, F_1^{(s-1)}$ in terms of $B$, or even better, $f$.