What is the Taylor expansion of the function $f(x)$ near $x=0$? $$ f(x) = \frac{1}{1-g(x)} $$ where $g(x)$ is already a Taylor expansion at order $n>1$ (near $x=0$): $$ g(x) = a_1 x +...+a_n x^n + o(x^n) $$ with $a_1\neq 0$.
If we use the well known Taylor expansion of $\frac{1}{1-X}$ at order $m\geq1$ with $X=g(x)$ we obtain: $$ f(x) = 1 + g(x) + g(x)^2 + ... + g(x)^m + o\left(g(x)^m\right) $$
I feel like a lot of terms can be discarded, but I am not sure which ones. I am not comfortable with the expression $o\left(g(x)^m\right)$.
Is it true that $o\left(g(x)^m\right) = o\left(x^{\min(n,m)}\right)$?
To actually get the coefficients of $g(x)$, write it as $1 = f(1-g) =f-fg$ so $fg = f-1$.
Expand the left side by the standard formula for the product of polynomials to get a recurrence for the coefficients of $g$.
This will take about $n$ operations to get the coefficient of $x^n$.
I'd do it but I am on my phone.