# 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?

• This looks like a special case of the Lagrange Inversion theorem, but I could be wrong. In any case, that theorem does what you want in more generality. – solstafir Jul 20 '16 at 20:33
• You will find a very large number of hits under reversion of series. The Wikipedia article has information, and there are implementations galore. A little known fact is that this is where the original Newton's Method comes from. At the beginning it was not about numerical solution of equations, it was about reversion of series. – André Nicolas Jul 20 '16 at 20:42
• thanks. it is a special case of the Lagrange inversion theorem. the special case being odd-symmetry functions and the "analytic point" $a=0$. so it looks like: $$g(y) = y \cdot \sum_{n=0}^{\infty} \frac{y^{2n}}{(2n+1)!} \lim_{x \to 0} \left( \frac{\mathrm{d}^{\,2n}}{\mathrm{d}x^{\,2n}} \left( \frac{x}{f(x)} \right)^{(2n+1)}\right)$$. – robert bristow-johnson Jul 21 '16 at 3:38
• here is a link link giving a more generic formula. – user115350 Jul 21 '16 at 3:39