an operator question I know how the derivative operator $\Big(\frac{d}{dx}\Big)^n$ works. But then how does it work if I have $$\exp{\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)}f(x)$$ I thought to use $$\exp z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$ to have 
\begin{align}
\exp{\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)}f(x)&=f(x)\\
&+\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)f(x)\\
&+\frac{1}{2!}\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)^2f(x)\\
&+\frac{1}{3!}\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)^3f(x)\\
&+\cdots
\end{align}
but then this looks strange ... I am not surer if I could do something like this. 
I appreciate hints, references, ...
Many thanks in advance.
 A: $e^{a∂_x}$ is the Taylor shift operator, if $f$ is analytical, then 
$$
(e^{a∂_x}f)(x)=\sum_{k=0}^\infty \frac{f^{(k)}(x)}{k!}a^k=f(x+a).
$$
I'm afraid the exponential of the second derivative does not have such a nice interpretation.

One can do more, as is routinely done in quantum physics: Set
$$
u(a,b,x)=(e^{a∂_x+b∂_x^2}f)(x)
$$
then
$$
∂_au=∂_xu\text{ and }∂_bu=∂_x^2u
$$
with $u(0,0,x)=f(x)$ and from the first with the method of characteristics one gets again that $u(a,0,x)=f(x+a)$. The development in direction of growing $b$ and constant $a$ is then a heat equation, where the solution is obtained via convolution with the heat kernel.
A: For the operators $a \partial_{x} + b \partial_{x}^{2}$ it is seen that
\begin{align}
(a \partial_{x} + b \partial_{x}^{2})^{1} &= a \partial_{x} + b \partial_{x}^{2} \\
(a \partial_{x} + b \partial_{x}^{2})^{2} &= (a \partial_{x} + b \partial_{x}^{2})(a \partial_{x} + b \partial_{x}^{2}) \\
&= a^{2} \partial_{x}^{2} + 2 a b \partial_{x}^{3} + b^{2} \partial_{x}^{4} = \sum_{k=0}^{2} \binom{2}{k} \, a^{2-k} b^{k} \, \partial_{x}^{2+k} \\
(a \partial_{x} + b \partial_{x}^{2})^{3} &= a^{3} \partial_{x}^{3} + 3 a^{2} b \partial_{x}^{4} + 3 a b^{2} \partial_{x}^{5} + b^{3} \partial_{x}^{6}  = \sum_{k=0}^{3} \binom{3}{k} \, a^{3-k} b^{k} \, \partial_{x}^{3+k} \\
\cdots &= \cdots \\
(a \partial_{x} + b \partial_{x}^{2})^{n} &= \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \, \partial_{x}^{n+k}
\end{align}
Now, for the exponential case,
\begin{align}
e^{a \partial_{x} + b \partial_{x}^{2}} \, f(x) &= \sum_{n=0}^{\infty} \frac{1}{n!} \, 
\left( \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \, \partial_{x}^{n+k} \right) \, f(x) 
\end{align}
This may also be obtained as follows:
\begin{align}
a \partial_{x} + b \partial_{x}^{2} = (a \partial_{x}) ( 1 + \frac{b}{a} \partial_{x})
\end{align}
for which
\begin{align}
e^{a \partial_{x} + b \partial_{x}^{2}} f(x) &= \sum_{n=0}^{\infty} \frac{1}{n!} [ (a \partial_{x}) (1 + \frac{b}{a} \partial_{x}) ] f(x) \\
&= \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \partial_{x}^{n+k} \, f(x)
\end{align}
