Rewrite operator $(\beta \partial/\partial\beta)^n$ I am trying to calculate the following thing
$$e^{A \cdot \beta \frac{\partial}{\partial \beta}}e^{i\eta \beta},$$
where $A,\eta$ are constants. Any idea how to write $e^{A \beta \partial/\partial\beta}$ in series?
In general
$$e^{A \cdot \beta \frac{\partial}{\partial \beta}} = \sum\limits_{n=0}^{\infty} \frac{A^n}{n!}\left(\beta \frac{\partial}{\partial \beta} \right)^n$$
and for $n=2$ I get
$$\left(\beta \frac{\partial}{\partial \beta} \right)^2 = \beta \frac{\partial}{\partial \beta} \beta \frac{\partial}{\partial \beta} = \beta \frac{\partial}{\partial \beta} + \beta^2 \frac{\partial^2}{\partial\beta^2}$$
 A: For any operator $T$,
$e^{AT}$ formally satisfies the differential equation
$$ \dfrac{\partial}{\partial A} e^{AT} = T e^{AT}$$
If $$v(A,\beta) = \exp\left(A \beta \dfrac{\partial}{\partial \beta}\right) e^{i\eta\beta}$$ this says
$$ \dfrac{\partial}{\partial A} v(A,\beta) = \beta \dfrac{\partial}{\partial \beta} v(A,\beta) $$
We have boundary condition $v(0,\beta) = e^{i\eta \beta}$.
The solution, obtained by the method of characteristics, is
$$ v(A,\beta) = \exp(i \eta \beta e^{A})$$
A: It is well known that the stirling numbers of the second kind $S(n,i)$ occur in the formula
\begin{eqnarray*}
\left( \beta \frac{ \partial}{\partial \beta} \right)^n =\sum_{i=j}^{n} S(n,j) \beta^{j} \frac{ \partial^j}{\partial \beta^j}.
\end{eqnarray*}
So your quantity becomes
\begin{eqnarray*}
e^{A \cdot \beta \frac{\partial}{\partial \beta}} e^{i\eta \beta} &=& \sum\limits_{n=0}^{\infty} \frac{A^n}{n!}\left(\beta \frac{\partial}{\partial \beta} \right)^n  e^{i\eta \beta}
&=& \sum\limits_{n=0}^{\infty} \frac{A^n}{n!} \sum_{j=1}^{n} S(n,j) \beta^{j}  \eta^j i^j e^{i\eta \beta} =\color{red}{\operatorname{exp}( i\beta \eta(e^A))}.
\end{eqnarray*}
See here for more info on the Stirling numbers of the second kind https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
