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Lets consider this method of finding inverse function:

$$f^{-1}(x) = \sum_{k=0}^\infty A_k(x) \frac{(x-f(x))^k}{k!}$$

where coefficients $A_k(x)$ recursively defined as

$$\begin{cases} A_0(x)=x \\ A_{n+1}(x)=\frac{A_n'(x)}{f'(x)}\end{cases}$$

It is evident that for some classes of functions starting from some point $A_k(x)$ becomes zero and thus the inverse function can be expressed in closed form.

For example, the expression has limited number of terms for any function of the following form:

$$f(x)=a \sqrt[n]{x+b}+c$$

where $n\in \mathbb{N}$. It is also evident that there are other classes of functions for which the series have limited number of terms.

So my question is for which other classes of functions this series give inverse function is closed form?

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I can see the way to derive formally this formula. It seems to be meaningful when $f$ is some perturbation of the identity function. Is there any convergence/equality result ? Do you have any reference ? – Denis Serre Nov 2 '10 at 10:40
up vote 6 down vote accepted

The answer is not that difficult. Assume that $A_{n+1}\equiv0$. This means $A_n={\rm cst}$, that is $A_{n-1}'=cf'$. Integrating, $A_{n-1}=cf+d$, that is $A_{n-2}'=(cf+d)f'$. Integrating again, $A_{n-2}=\frac{c}{2}f^2+df+e$. And so on. By induction, we find that $A_{n-k}=P_k(f)$, where $P_\ell$ is a polynomial of degree $\ell$. When $k=n$, we find the necessary and sufficient condition that $x=P_n(f)$.

In conclusion, the functions $f$ for which the series in the formula has finitely many terms are the roots of equations $P(f)=x$, where $P$ is some polynomial.

Edit. This answer suggests that the formula is nothing but Taylor expansion for $f^{-1}$, which turns out to be finite if and only if $f^{-1}$ is a polynomial. This also suggests that the series converges in a non-trivial interval whenever $f^{-1}$ is analytic.

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Thank you very much. Interesting how such class of functions (inverted polynomials) is called? P.S. Probably there is a typo: $A{n−k}=P_k(f)$. – Anixx Nov 2 '10 at 11:19

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