Taylor Expansion for an Inverse function

Question

Is there a straightforward way to calculate the Taylor Expansion for the inverse of a function, for example $$ \ e^{\sin x} $$ by knowing the Taylor expansion for the function itself,

I think we use the formula $$ (f^{-1})'(x) = 1/f'\{f^{-1}(x)\} $$

Answer 1

Of course, if a function has a converging Taylor series, then it is identical to it, so anything to do with the function can also be realized with the series.

Assume that the Taylor series of the inverse $g_k=\sum_{j=0}^k b_jx^j$ is known up to degree $k$, $$ f(g_k(x))=x+O(x^{k+1}), $$ where the initial constant coefficient is $b_0=f^{-1}(0)$. Then the next coefficient can be obtained from the equation $$ f(g_k(x)+b_{k+1}x^{k+1})=f(g_k(x))+f'(g_k(x))b_{k+1}x^{k+1}+O(x^{k+2}) $$ where $f(g_k(x))$ is now evaluated up to degree $k+1$ and $f'(g_k(x))$ needs to be accurate up to degree $0$, i.e., has always the value $f'(f^{-1}(0))$.

One can also obtain more than one coefficient at once by employing a Newton-like scheme.

See also Hensel lifting.