Vector Field Exponential Map I've got ${\bf v} = x^2\partial_x$, and I'm trying to find $\exp(\varepsilon{\bf v})$, but I'm having some trouble.
If I define ${\bf v}^{n+1} = {\bf v}{\bf v}^n$ then I get a different outcome to ${\bf v}^{n+1} = {\bf v}^n{\bf v}$.
For example:
$${\bf v}^2 = {\bf vv} = (x^2\partial_x)(x^2\partial_x) = x^2(\partial_xx^2)\partial_x = 
x^2(2x)\partial_x = 2x^3\partial_x$$
$${\bf v}^3 = {\bf v}{\bf v}^2 =(x^2\partial_x)(2x^3\partial_x)=x^2(\partial_x2x^3)\partial_x = x^2(6x^2)\partial_x = 6x^4\partial_x$$
$${\bf v}^3 = {\bf v}^2{\bf v} = (2x^3\partial_x)(x^2\partial_x) = 2x^3(\partial_xx^2)\partial_x = 2x^3(2x)\partial_x = 4x^4\partial_x$$
Using ${\bf v}^{n+1} = {\bf v}{\bf v}^n$ gives 
$$\exp(\varepsilon {\bf v})x = \frac{x}{1-\varepsilon x}$$
While using ${\bf v}^{n+1} = {\bf v}^n{\bf v}$ gives
$$\exp(\varepsilon {\bf v})x = \frac{x}{2}(1+\mathrm e^{2\varepsilon x})$$
In both cases, when $\varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$\lim_{\varepsilon \to 0} \frac{\mathrm d}{\mathrm d\varepsilon} \exp(\varepsilon {\bf v})x = {\bf v}$$
The same ${\bf v} \in \mathfrak g$ can't possible generate two different flows, can it?
 A: If you apply the product rule carefully, you don't get two different answers.  In your example, \begin{align*}
\mathbf{v}(f(x)) &= (x^2 \partial_x)(f(x))  \\
    &= x^2 f'(x)  \text{, and }\\
\mathbf{v}^2(f(x)) &= (x^2 \partial_x)\left( (x^2 \partial_x)(f(x)) \right)  \\
    &= (x^2 \partial_x)\left( x^2 f'(x) \right)  \\
    &= x^2(2x f'(x) + x^2 f''(x) )  \text{, or } \\
\mathbf{v}^2(f(x)) &= \left((x^2 \partial_x) (x^2 \partial_x) \right) (f(x))  \\
    &= (x^2(2x \partial_x + x^2 \partial_x^2))(f(x))  \\
    &= x^2(2x f'(x) + x^2 f''(x) )
\end{align*}
so we see $\mathbf{v}^2 = 2x^3 \partial_x + x^4 \partial_x^2$.  Similarly, $\mathbf{v}^3 = 6 x^4 \partial_x + 6 x^5 \partial_x^2 + x^6 \partial_x^3$.
(Of course, if we're linearizing, we project onto the first term in both of those.)
A: You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.
v generates a shift operator, and it pays to define suitable canonical coordinates,
$$
y=-1/x, \qquad \Longrightarrow \qquad x^2 \partial_x=\partial_y ,
$$ 
so that you are shifting y by $\epsilon$,
$$
e^{\epsilon \partial_y} ~~f(y)= f(y+\epsilon),
$$
which reads
$$
e^{\epsilon x^2\partial_x} ~~g(x)= g\left(\frac{-1}{y+\epsilon}\right )=g\left (\frac{x}{1-\epsilon x}\right ), 
$$
a standard formula in the RG advection of QFT.
Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.
A: The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because ${\bf v}(f)$ is a function and so ${\bf v}[{\bf v}(f)]$ is also a function, while $[{\bf v}(f)]{\bf v}$ is a function times a vector field, i.e. a vector field.
The differential operator ${\bf v} = x^2\partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.
Applied to $x$, we get $x \mapsto 1 \mapsto x^2$, meaning ${\bf v}(x) = x^2$.
Applied to $x^2$, we get $x^2 \mapsto 2x \mapsto 2x^3$, meaning ${\bf v^2}(x) = 2x^3$.
Applied to $2x^3$, we get $2x^3 \mapsto 6x^2 \mapsto 6x^4$, meaning ${\bf v}^3(x) = 6x^4$.
In general, ${\bf v}^n(x) = n!x^{n+1}$, and so for all $|\varepsilon x|<1$
\begin{eqnarray*}
\sum_{n \ge 0} \frac{\varepsilon^n}{n!}{\bf v}^n(x) &=& \sum_{n \ge 0} \frac{\varepsilon^n}{n!}n!x^{n+1} \\ \\
&=& x\sum_{n \ge 0} (\varepsilon x)^n \\ \\
&=& \frac{x}{1-\varepsilon x}
\end{eqnarray*}
