Question about the radius of convergence in the Lagrange Inversion Theorem

Question

Suppose that $f(x)$ has an inverse function $g(x)$. Suppose that both functions have a power series representations: \begin{align} f(x)= \sum_{k=0}^\infty f_k\frac{x^k}{k!} \text{ and } g(x)= \sum_{k=0}^\infty g_k\frac{x^k}{k!} \end{align} with $f_0=0$ and $f_1 \neq 0$. Then, the Lagrange inversion theorem say that we can coefficients of $f$ and $g$ as follows: \begin{align} g_1&=\frac{1}{f_1},\\ g_n&= \frac{1}{f_1^n} \sum_{k=1}^{n-1} (-1)^k n^{(k)} B_{n-1,k}(\hat{f}_1,\ldots, \hat{f}_{n-k}), n \ge 2 \end{align} where $\hat{f}_k=\frac{f_{k+1}}{(k+1) f_1}$ and $n^{(k)}$ is rising factorial.

Questions: I have the following question about this result

1. Suppose that the representation of $f$ holds only for $|x|<r$ (i.e., has a finite radius of convergence). Does this the result still hold on $|x|<r$?
2. What can we say about the radius of convergence of $g$? Can it be found? Or do we have to do a root test on the new coefficients to find it?
3. How does this formula change if I have an expansion of $f$ around $x=a$ (i.e., $f(x)= \sum_{k=0}^\infty f_k\frac{(x-a)^k}{k!}$)

Answer 1

Think about $f(x) = \log(1+x)$ and $g(x) = e^x - 1$. The radii of convergence of their power series are really quite unrelated.