Inverse Function for a Taylor Series?

I have the feeling that, for truncated series, you are looking for series reversion which is well described in the given link.

Using a simple example $$y=\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{3 9916800}+O\left(x^{13}\right)$$ Aplying the method you will get $$x=\sin^{-1}(y)=y+\frac{y^3}{6}+\frac{3 y^5}{40}+\frac{5 y^7}{112}+\frac{35 y^9}{1152}+\frac{63 y^{11}}{2816}+O\left(y^{13}\right)$$ You can check that the result is the Taylor series of the inverse function.

A more complex example (but the same method applies) $$y=1+e^x \log (x+1) \sin (x)=1+x^2+\frac{x^3}{2}+\frac{x^4}{6}-\frac{x^5}{12}+\frac{x^6}{36}-\frac{2 x^7}{45}+\frac{x^8}{28}+O\left(x^{9}\right)$$ would lead to $$x=\sqrt{y-1}-\frac{y-1}{4}+\frac{7(y-1)^{3/2}}{96} +\frac{(y-1)^2}{24} -\frac{233 (y-1)^{5/2}}{2048}+\frac{449 (y-1)^3}{2880}-\frac{3442697 (y-1)^{7/2}}{20643840}+O\left((y-1)^4\right)$$ The problem is that you need to solve quite large sets of equations.

We have $e^0=1$. From the Taylor series for $e^x$ centered at $x=0$, we may obtain the Taylor series for $\ln y$ centered at $y=1$.

Another quirk here. We get the terms of the inverse only one at a time, recursively. So, for example, the Taylor series for $\arctan x$ at $x=0$ has a nice formula for the terms, $$ \arctan x = x -\frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots $$ but the Taylor series for $\tan y$ at $y=0$ doesn't... those terms involve Bernoulli numbers.