11/27/2015, 03:41 PM
(This post was last modified: 11/27/2015, 07:34 PM by sheldonison.)
(11/26/2015, 11:56 PM)tommy1729 Wrote: There are many ways to solve f(f(x)) = exp(x) or g(g(x)) = 2sinh(x).
If g < f then g is uniquely so.
Also using the 2sinh method on g gives us f.
....
But one wonders if f and g related as above must
1) both be analytic or both not.
If g=2sinh(z), and f=exp(z),
One can start with the analytic Abel function \( \alpha_g(z)\;\; \) and the superfunction for g. \( \alpha^{-1}_g(z) \). For example, the abel and superfunction for 2sinh(z) can be developed using Koenig's method.
Now this gives us \( g^{[0.5]}(z) = \alpha_g^{-1}(\alpha_g(z)+0.5) \)
We wish to generate the superfunction for f(z) from \( \alpha_g\;\alpha_g^{-1} \), with the equation below. In the case at hand this limit for the superfunction of f converges at the real axis, but it is not analytic; it does not converge in the complex plane. So far, this seems always be the case for arbitrary exponentially growing functions, f and g.
\( \alpha^{-1}_f(x)\; = \; \lim_{n \to \infty} \; f^{[-n]}( \alpha^{-1}_g(x+n))\;\;\; \)the superfunction for f(z)
So the resulting half iterate of f developed in this way is not analytic either, even thought it converges at the real axis
\( f^{[0.5]}(z) = \alpha_f^{-1}(\alpha_f(z)+0.5) \)
- Sheldon