recover by taking $l=0$):

[@several2 Thm. 1]\[weiergen\] Let $\Delta: R \rightarrow R$ be a differential operator of the form $\Delta = \sum_{i=0}^{l} r_i \partial^i$ where $r_i \in R$ for all $i$ and $r_l$ is $x_n$-regular. Then $R/\Delta(R)$ is a finitely generated $R_{n-1}$-module.

We state and prove a special case of Proposition \[weiergen\], due to Malgrange, separately:

[@malgrange Prop. 1.3]\[malglem\] Let $R = k[[x]]$ and let $\Delta: R \rightarrow R$ be a nonzero differential operator. Then $R/\Delta(R)$ is a finite-dimensional $k$-space.

For any formal power series $r = \sum \alpha_i x^i \in R$, we let $\nu(r) = \min{\{i | \alpha_i \neq 0\}}$. The differential operator $\Delta$ takes the form $\Delta = \sum_{i=0}^{l} r_i \partial^i$ where $r_l \neq 0$. Set $s = \max{\{i - \nu(r_i)\}}$, and let $I \subset \{0, \ldots, l\}$ be the set of indices for which this maximum is attained, that is, for which $s = i - \nu(r_i)$. For each $i \in I$, $r_i = x^{i-s}\rho_i$ for some $\rho_i \in R$ such that $\rho_i(0) \neq 0$. For any integer $t \geq s$, we have $$\Delta(x^t) = \sum_{i \in I} t(t-1)\cdots (t-i+1) \rho_i(0) x^{t-s} + \textrm{higher order terms}.$$ The coefficient of $x^{t-s}$ in the above expression, namely $\sum_{i \in 