I was playing with commuting the derivative operator $D$ and the scalar multiplication of $x$ applying to a function $f$. If $[\cdot,\cdot]$ denote de commutator, then:
$$[D,x]f=D(xf)-xDf=f+xDf-xDf=f$$
If we abstract for applying to a function, $[D,x]=1$. I notice that when you choose an arbitrary $n$-grade of $D$, denoted as $D^n$, the commutation with $x$ gives a "power rule", similar to the derivation of just $x^n$, the last exponent meaning the usual $n$-power of $x$:
$$[D^n,x]=nD^{n-1}$$
For example:
$$[D^2,x]f=D^2(xf)-xD^2f=D(f+xDf)-xD^2f$$ $$=Df+xD^2f+Df-xD^2f=2Df $$
$$\Rightarrow [D^2,x]=2D $$ If we go further and generalize to $[D^n,x^m]\dots$
$$[D^n,x^m]=\sum_{k=1}^n {n \choose k}D^k(x^m)D^{n-k} $$
which resemble to the binomial formula.
Is there a meaning for this result?
