Exponential differential operator Consider the operator $D= e^{ax \frac{d}{dx}}$ operating on an infinitely differentiable function $f(x)$.
My approach:
$$
Df(x)= f(x) + ax \frac{df(x)}{dx} + (ax)^2\frac{d^2f(x)}{dx^2} + \cdots =f(x+ax)
$$
But this does not seem to be the answer. Can anyone tell me if and where did I go wrong?
 A: Sketch: Define
\begin{align}
u(t, x) = \exp\left(tx\frac{\partial}{\partial x}\right)f(x)
\end{align}
then we see that $u$ solves
\begin{align}
\partial_t u - x \partial_x u =0, \ \ \text{ with }\ \ u(0, x) = f(x).
\end{align}
By the method of characteristics, we see that
\begin{align}
u(t, x) = f(xe^{t})
\end{align}
which means
\begin{align}
\exp\left( ax\frac{\partial}{\partial x}\right) f = u(a, x) = f(xe^a).
\end{align}
A: Remember that
$$
e^x = 1 +x +\frac{1}{2!} x^2 + \cdots +\frac{1}{n!} x^n + \cdots
$$
Hence,
$$
\exp(ax \partial_x) f = f +ax\partial_xf +\frac{1}{2}ax\partial_x(ax\partial_x f)+\cdots
$$
Let $x=e^y$ so that $x\partial_x f(x) = \partial_y f(e^y)$. Therefore,
\begin{align}
\exp(ax \partial_x)f &= \sum_{n=0}^\infty \frac{1}{n!} (\partial_y^n f(e^y)) a^n \\
&= f(e^{y+a}) \\
&= f(xe^a)
\end{align}
A: 
It appears that the OP interpreted the operator $e^{\left((ax)\frac{d}{dx}\right)}$ on $f(x)$ incorrectly to mean $\sum_{n=0}^\infty \frac{(ax)^n}{n!}\frac{d^nf(x)}{dx^n}$.
This misinterpretation in the OP is due to applying incorrectly successive applications of the operator $ax\frac{d}{dx}$.  One application reveals $ \left((ax)\frac{d}{dx}\right)f(x)=axf'(x)$.  A subsequent application yields
$$\begin{align}
\left((ax)\frac{d}{dx}\right)^2 f(x)&=\left((ax)\frac{d}{dx}\right)\left(axf'(x)\right)\\\\
&=(ax)^2f''(x)+a^2xf'(x)\\\\
&\ne (ax)^2f''(x)
\end{align}$$
We will show in the following that $\sum_{n=0}^\infty \frac{(ax)^n}{n!}\frac{d^nf(x)}{dx^n}=f(ax+a)$ while $e^{\left((ax)\frac{d}{dx}\right)}=f(e^a x)$.  


We begin by analyzing the operator $e^{a\frac{d}{dx}}$ on $C^\infty$ as defined by
$$\left(e^{a\frac{d}{dx}}\right)\{f(x)\}=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}a^n\tag1$$
Note that the Taylor series of $f(x+a)$ around $x$ can be written
$$f(x+a)=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}a^n\tag2$$
Comparing $(1)$ and $(2)$ reveals
$$\left(e^{a\frac{d}{dx}}\right)\{f(x)\}=f(x+a)$$

Next, we transform the operator $x\frac{d}{dx}$ by enforcing the substitution $x=e^y$.  Then, denoting $f(x)=f(e^y)=g(y)$, we see that 
$$\begin{align}
\left(ax\frac{d}{dx}\right) f(x)&= \left(ae^y\frac{dy}{dx}\frac{d}{dy}\right) f(e^y)\\\\
&=\left(ae^ye^{-y}\frac{d}{dy}\right) f(e^y)\\\\
&=\left(a\frac{d}{dy}\right) g(y)\tag3
\end{align}$$

Finally, using $(2)$ and $(3)$ reveals
$$\begin{align}
\left(e^{(ax)\frac{d}{dx}}\right) f(x)&=\sum_{n=0}^\infty \frac1{n!}\left(\left(ax\frac{d}{dx}\right)^n \right)f(x)\\\\
&=\sum_{n=0}^\infty \frac{g^{(n)}(y)}{n!}a^n\\\\
&=g(y+a)\\\\
&=f(e^{y+a})\\\\
&=f(e^ae^y)\\\\
&=f(e^ax)
\end{align}$$
And we are done!
