Inversion of power series computational formula If I have a function $f(z)$ which is complex analytic on some neighbourhood of zero I know I can find a representation of the inverse as a series, and this series converges on some neighbourhood of $f(0)$. This is my understanding of the Lagrange inversion theorem. However, the formula for the coefficients of the series is not computationally friendly, expressed in terms of $n$-th derivatives and $n$-th powers of the variable $z$. In the same Wikipedia entry, a formula is given for formal power series which has an explicit formula for the coefficients of the inverse expressed in terms of the  Bell polynomials. Specifically, if $f_n$ and $g_n$ are the $n$-th coefficient of Taylor series of $f(z)$ and the inverse series respectively, and $f_0 = 0$, $f_1 \neq 0$, then
$$
g_n = \frac{1}{f^n_1}\sum_{k=1}^{n-1}(-1)^{k}n^{(k)}B_{n-1,k}(\hat{f}_1,\hat{f}_2,\ldots,\hat{f}_{n-k}),\quad n\geq 2,
$$
where,
$$
\hat{f}_k = \frac{f_{k+1}}{(k+1)f_1},\quad g_1 = \frac{1}{f_1}
,\quad \text{and } \quad n^{(k)}=n(n+1)\ldots(n+k-1),$$
and $B_{n-1,k}$ is the $(n-1,k)$-th partial exponential Bell polynomial. A similar formula, I think the same one actually although I haven't verified this, is used by Mathematica to invert series (see for example Series Reversion).
My question is: When can I assume this formal series actually represents the the inverse function? For me it would be enough to know that the formal series converges in some neighbourhood to the function, I don't need details about a specific radius of convergence.
 A: The Wronski's formula \eqref{f(t)g(t)=1=determ} below is a suitable answer to this question.
If $a_0\ne0$ and
$$
P(t)=a_0+a_1t+a_2t^2+\dotsm
$$
is a formal series, then the coefficients of the reciprocal series
$$
\frac{1}{P(t)}=b_0+b_1t+b_2t^2+\dotsm
$$
are given by
\begin{equation}\label{f(t)g(t)=1=determ}\tag{1}
b_r=\frac{(-1)^r}{a_0^{r+1}}
\begin{vmatrix}
a_1 & a_0 & 0 & 0 & \dotsm & 0& 0 & 0\\
a_2 & a_1 & a_0 & 0 & \dotsm & 0 & 0& 0\\
a_3 & a_2 & a_1 & a_0 & \dotsm & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots &\vdots & \vdots\\
a_{r-2} & a_{r-3} & a_{r-4} & a_{r-5} & \dotsm & a_1 & a_0 & 0\\
a_{r-1} & a_{r-2} & a_{r-3} & a_{r-4} & \dotsm & a_2 & a_1 & a_0\\
a_{r} & a_{r-1} & a_{r-2} & a_{r-3} & \dotsm & a_3 & a_2 & a_1
\end{vmatrix}, \quad r=1,2,\dotsc.
\end{equation}
There are alternative answers to this question and closely related information at the sites

*

*https://math.stackexchange.com/a/4262414/945479

*https://math.stackexchange.com/a/4262417/945479

*https://math.stackexchange.com/a/4262390/945479

*https://math.stackexchange.com/a/4262406/945479

*https://math.stackexchange.com/a/4261705/945479
