Exponential of the product between $x$ the derivative operator of $x$ acting in a $f(x)$ The question
I'm stuck here trying to figure out how to compute and prove, the following operator action in a function:
$\exp(\varepsilon x \partial_x) f(x) = f(x \exp(\varepsilon) )$
where $\varepsilon$ is a constant. 
I saw this result and I failed in my attempt to reproduced it. What I did was to expand $\exp(\varepsilon x \partial_x)$ in Taylor's series as:
$\begin{align*}
\exp(\varepsilon x \partial_x)f(x)& = \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon x \partial_x)^m f(x)\\ 
 &= \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon x)^m\frac{\partial^m}{\partial x^m}f(x) \\  
\end{align*}$
I took this procedure because I already know how to compute $e^{\partial_x}f(x)$. Let me show you what I did in this case:
The translation operator
The Taylor series of a function f is
\begin{equation}
f(x)=\sum_{n=0}^\infty\frac{(\partial_x^nf)(a)}{n!}(x-a)^n
\end{equation}
Expanding about $x+b$ and letting $a=x$:
\begin{equation}
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!}
\end{equation}
By definition:
\begin{equation}
e^{b\partial_x}=\sum_{n=0}^\infty\frac{(b\partial_x)^n}{n!}
\end{equation}
Hence
\begin{equation}
f(x+b)=(e^{b\partial_x}f)(x)
\end{equation}
Returning to my question
I tried to generalize or make something similar for the previous case I discussed but I didn't get anywhere. Anyone can give me a tip or recommend a book or paper?
Thanks!!
 A: You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$.  If $f(x)=x^k$, then 
$$
[M_x\partial_x f](x)=kx^k,\ \ [(M_x\partial_x)^2f](x)=k^2x^k,\ \ \cdots\ ,\ \ [(M_x\partial_x)^mf](x)=k^mx^k.
$$
Then
\begin{align*}
[\exp(\varepsilon x \partial_x)f](x)& = \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon x \partial_x)^m f(x)\\ 
 &= \sum_{m=0}^{\infty}\frac{1}{m!}(\varepsilon )^mk^mx^k \\  
&=x^k\,\exp(\varepsilon k)=x^k (\exp(\varepsilon)^k\\
&=f(x\exp(\varepsilon)). 
\end{align*}
Thus we get by linearity that, for any polynomial $p$, 
$$\tag1
[\exp(\varepsilon x \partial_x)p](x)=p(x\exp(\varepsilon)). 
$$
Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic. 
A: Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.
If we assume that $f(x)$ has a Fourier Transform, we can write:
\begin{align*}
e^{\varepsilon x \partial_x}f(x) &= \int d^3k \, f(k)e^{\varepsilon x \partial_x}e^{ikx}\\  
&= \int d^3k\, f(k) \sum_{n,m} \frac{(\varepsilon x \partial_x)}{m!}\frac{ikx}{n!}
\end{align*}
Notice that
\begin{align*}
(x \partial_x)^m x^n = n^m x^n \, \, ,
\end{align*}
where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:
\begin{align*}
e^{\varepsilon x \partial_x}f(x) &= \int d^3k\, f(k) \sum_{n} e^{(\varepsilon)^n}\frac{(ikx)^n}{n!}\\  
&= \int d^3k\, f(k) \sum_{n} \frac{(e^{\varepsilon}ikx)^n}{n!} \\
&= \int d^3k \, f(k) e^{ike^\varepsilon x} = f(e^\epsilon x)
\end{align*}
It's a physicist solution because we're assuming that every operation is well defined. 
