A translation operator
The Taylor series of a function $f$ is
$$f(x)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(a)}{n!}(x-a)^n$$
where $\partial_x$ is the derivative operator. Expanding about $x+b$:
$$f(x+b)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(a)}{n!}(x+b-a)^n$$
Letting $a=x$:
$$f(x+b)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(x)}{n!}b^n=\sum_{n=0}^\infty\frac{((b\partial_x)^nf)(x)}{n!}$$
By definition
$$e^{b\partial_x}=\sum_{n=0}^\infty\frac{(b\partial_x)^n}{n!}$$
Hence
$$f(x+b)=(e^{b\partial_x}f)(x)$$
Hence $e^{\partial_x}=T$ where $T$ is the translation operator and $(Tf)(x)=f(x+1)$.
A scaling operator
We can also find a closed form for a scaling operator $S$ where $(Sf)(x)=f(ax)$.
$$f(xa)=f(e^{\log{xa}})=f(e^{\log x+\log a})=f(e^{y+\log a})$$
where $y=\log x$. Letting $g(z)=f(e^z)$:
$$f(xa)=g(y+\log a)$$
By our first theorem, $g(y+b)=(e^{b\partial_y}g)(y)$. Letting $b=\log a$:
$$f(xa)=(e^{(\log a)\partial_y}g)(y)=(a^{\partial_y}g)(y)=(a^{\partial_{\log x}}f)(e^y)$$
Since
$$\frac{\partial}{\partial \log x}=\frac{\partial x}{\partial \log x}\frac{\partial}{\partial x}=x\frac{\partial}{\partial x}$$
Then
$$f(xa)=(a^{x\partial_x}f)(e^{\log x})=(a^{x\partial_x}f)(x)$$
Therefore $S=a^{x\partial_x}$ defines our scaling operator.
A general operator
Suppose we want an operator $G$ such that $(Gf)(x)=f(g(x))$. Consider:
$$(e^{\partial_{h(x)}}f)(x)=(e^{\partial_y}f)(x)=(e^{\partial_y}f)(h^{-1}(h(x)))=(e^{\partial_y}f)(h^{-1}(y))$$
where $y=h(x)$. Letting $j=f\circ h^{-1}$ yields:
$$(e^{\partial_{h(x)}}f)(x)=(e^{\partial_y}j)(y)=j(y+1)=f(h^{-1}(y+1))=f(h^{-1}(h(x)+1))$$
Hence solving the functional equation
$$h^{-1}(h(x)+1)=g(x)$$
for $h(x)$ allows us to define our general operator $G=e^{\partial_{h(x)}}$.
For example, letting $g(x)=xa$, the function $h(x)=\frac{\log x}{\log a}$ is a solution:
$$h^{-1}(y)=a^y$$ $$h^{-1}(h(x)+1)=a^{h(x)+1}=a^{\frac{\log x}{\log a}+1}=a^{\frac{\log x}{\log a}}a=e^{\log x}a=xa$$
Hence the corresponding operator takes the form
$$e^{\partial_{h(x)}}=e^{\partial_{\frac{\log x}{\log a}}}=e^{\log a\partial_{\log x}}=a^{x\partial_x}$$
This is the scaling operator we derived before.
The case $e^{\partial_{h(x)}}=e^{x^n\partial_x}$ or equivalently $h(x)=\frac{x^{1-n}}{1-n}$ corresponds to the basis for the Witt algebra.
The question
My question is as follows: Can a similar procedure be used to find $e^{{\partial_x}^2}$, or $e^{{\partial_{h(x)}}^n}$ in general? Note that the commutator of $x$ and $\partial_x$ is nonzero and given by:
$$[\partial_x,x]=\partial_xx-x\partial_x=1$$
Furthermore, given the product rule $Dab=(Da)b+aDb$, it seems to be the case that
$$(x\partial_x)^2=x\partial_x+x^2\partial_x^2$$ $$(x\partial_x)^3=x\partial_x+3x^2\partial_x^2+x^3\partial_x^3$$ $$(x\partial_x)^4=x\partial_x+7x^2\partial_x^2+6x^3\partial_x^3+x^4\partial_x^4$$
and in general
$$(x\partial_x)^n=\sum_{k=1}^n \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{k} x^k\partial_x^k$$
where $\genfrac{\lbrace}{\rbrace}{0pt}{}{a}{b}$ are the Stirling numbers of the second kind. I have a strong suspicion the answer might have to do with properties of the Fourier and Laplace transforms, as seen here and here.
