General differentials operators (Grothendieck definition) and polynomial rings

Question

Let $A$ be an algebra over some field $\mathbb{k}$. A linear map $f:A\to A$ is said to be a differential operator of an order $\le n$ if for all $a_0,a_1,\ldots a_n\in A$ we have $[a_n,[a_{n-1},[\ldots,[a_1,[a_0,D]]]=0$ (we identify each $a\in A$ with an operator of multiplication $x\mapsto ax$). This definition is general and works for every commutative algebras.
Let $\mathrm{Diff}^{\le n}$ be the set of all differential operators of order $\le n$. It is easy to verify that $\mathrm{Diff}^{\le 0}$ is just $A$ and each differential operator of an order $\le 1$ is a sum of some $a$ and some derivation.

It also can be easily proved, that $\mathrm{Diff}^{\le n}\circ \mathrm{Diff}^{\le m}\subset \mathrm{Diff}^{\le n+m}$ and $[\mathrm{Diff}^{\le n},\mathrm{Diff}^{\le m}]\subset \mathrm{Diff}^{\le n+m-1}$ (although I do not know wheather it is important for my question or not).

So, we see that that $f_k\frac{\partial^k}{\partial x^k}\in \mathrm{Diff}^{\le k}(\mathbb{k}[x])$ and also $\sum_{k\le N} f_k\frac{\partial^k}{\partial x^k}\in \mathrm{Diff}^{\le N}(\mathbb{k}[x])$ and, moreover, for $A=C^{\infty}(\mathbb{R})$ it is also true.

Could you please help me to prove the converse? I.e., we have an operator $D$ of an order $\le N$ and want to find polynomials $f_0,f_1,\ldots f_n$ such that $D=\sum_{0\le k\le N}f_k\frac{\partial^k}{\partial x^k}$. Perhaps we can get the coefficients? Yes, it is easy to write that $f_0=D(1)$, $f_1=D(z)-D(1)z$ but expressions for other coefficients seem to be very difficult... Thanks in advance!

Answer 1

Remark 2.1.9 in Differential operators and BV structures in noncommutative geometry by Ginzburg, Victor and Schedler, Travis deals with this.

http://link.springer.com/article/10.1007%2Fs00029-010-0029-8