In the answer to this question the Lie Series result $$ \boxed{ \; e^{t \phi(x) \frac{d}{dx}} f(x) = f\left(e^{t \phi(x) \frac{d}{dx}} x\right) \; } $$ was mentioned. Now for $f(x)$ analytic and when $\phi(x)$ is a polynomial, this result makes sense.
Now for an arbitrary smooth $\phi(x)$ one can do a coordinate transformation (assume its one-one onto and therefore invertible) $$ s(x) = \int_{x_0}^{x}\phi(x')dx' $$ and get the result $$ e^{t \frac{d}{ds}}g(s) = g(s+t) = f(s^{-1}(s(x) +t)) $$
But now if I were to do a series expansion, $$ e^{t \frac{d}{ds}}g(s) = \sum_{0}^{\infty}\frac{t^n}{n!}\left[ \frac{d}{ds}\right]^n g(x) $$ and convert the series back to the original coordinates I get the series $$ e^{t \phi(x) \frac{d}{dx}} f(x) = \sum_{0}^{\infty}\frac{t^n}{n!}\left[ \phi(x) \frac{d}{dx}\right]^n f(x) $$ where I have used the relation \begin{align} \frac{ds}{dx} &= \phi \\ \frac{d}{ds} &= \frac{dx}{ds}\frac{d}{dx} \end{align} and used the result $$ \frac{dx}{ds} = \left( \frac{ds}{dx}\right)^{-1} $$ asuming $s(x)$ is invertible.
But this series seems to be divergent as the number of terms coming from the successive derivatives on $\phi$ in the $n$th series term grows asymptotically as $\sim \exp\left\{2^nlog(2)\right\}$.
In fact, if I were to Taylor expand the function $x(s)$ at $s=0$ as \begin{align} x(s) &= x_{0} + \sum_{0}^{\infty} \frac{dx}{ds} \Bigg |_{s = 0}\frac{s^n}{n!} \\ &=x_{0} + \sum_{1}^{\infty} \left[ \phi(x) \frac{d}{dx}\right]^{n-1}(\phi(x)) \Bigg |_{x = x_0}\frac{s^n}{n!}\\ &= e^{s\phi(x)\frac{d}{dx}} x \Bigg |_{x = x_0} \end{align} This also seems to have the same divergence issue.
If this was indeed Taylor expandable shouldn't it have been convergent?
How did the changing of coordinates change the convergence properties of the series?
Is there a more general condition on $\phi(x)$ for such a series to be convergent than it is a polynomial?
PS: Please forgive the abuse of notation.
