# inverse function and maclaurin series coefficients.

i dunno if this is asked before, and i am not sure where to find this on the web or in textbooks.

we are given a function (that is too hard to invert by solving for $x$):

$$y = f(x)$$

which has an unknown inverse:

$$x = g(y)$$

so $y = f(g(y))$ and $x = g(f(x))$ and let’s say that $f(\cdot)$ is an odd-symmetry function

$$f(-x) = -f(x) \quad \quad \forall x \in \mathbb{R}$$

i think that means that $g(\cdot)$ must also be an odd-symmetry function.

since we know $f(x)$, we can compute derivatives of it around $x=0$. we can represent $f(x)$ as a Maclaurin series. and we also know that all of the even-power terms of the series are zero.

$$y = f(x) = x \cdot \sum\limits_{n=0}^{\infty} a_n \ x^{2n}$$

the same can be said about $g(y)$

$$x = g(y) = y \cdot \sum\limits_{n=0}^{\infty} b_n \ y^{2n}$$

i know coefficients $a_n$ because i know $f(x)$ and all of its derivatives. now by slugging through this manually, i can compute coefficients $b_n$ for the Maclaurin expansion of $g(y)$. the first two are

$$b_0 = \frac{1}{a_0}$$

$$b_1 = \frac{-a_1}{a_0^4}$$

now, is there a more general method of computing these inverse coefficients? is there a paper or online reference that deals with this, ostensibly quite practical, calculus problem? or am i sentenced to just slog through this manually and stop at the $y^5$ or $y^7$ term?