Which (approximative) methods are there to compute the inverse of a complicated function?

Asked Jul 27 '14 at 08:57

Active Aug 03 '14 at 18:43

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I have a complicated function $f(x)$ for which I want to compute the inverse $f^{-1}$ over a certain range $R(f): a \leq f(x) \leq b$.

The only way to find the inverse I can think of is power series reversion at multiple points within the interval $(a,b)$, so that the radii of convergence of the inverse series overlap. One can try to improve the fit by using Newton's method on $f(y)-x = 0$.
Lagrange's theorem offers another method (see [Dominici]).
[Dominici] gives a third using nested derivatives.
Using Newton's method on the equation $f(y)-x = 0$ is probably the first choice. It gives the values of the inverse function at discrete points $x_i$. See StackExchange (thanks Antonio Vargas) and [Koepf]

Are there other methods to compute the inverse of a complicated function $f$?

edited Aug 03 '14 at 18:43

asked Jul 27 '14 at 08:57

jjack

You could try to approximate $f$ by an analytically invertible function then use Newton's method to correct the approximation as in this question. – Antonio Vargas Jul 27 '14 at 10:11
Is there a method to choose such a function from a certain class of functions? – jjack Jul 27 '14 at 12:34
Not in general. – Antonio Vargas Jul 27 '14 at 12:37
Is the function monotone? (Otherwise the inverse isn't defined). Also, if you are looking for a numerical approximation, then I don't understand what "choose from a certain class of functions" means. – Jul 27 '14 at 17:28
Of course, the function is strictly monotone. – jjack Jul 27 '14 at 17:32
What do you mean by "compute" $f^{-1}$? If you want to solve $f(x) = y$ for $x$, given $y$, there are numerical methods for that. – littleO Jul 27 '14 at 17:34
I would like to have a power series representation for taking derivatives and integrating. How would you go about numerically computing $f^{-1}$ for a known function $f(x)$? – jjack Jul 27 '14 at 17:57
This answer of mine may give you an idea of how one could numerically compute an inverse. – Antonio Vargas Jul 27 '14 at 18:02

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