The nature of g(exp(f(s))

[JmsNxn]

Also i want to point out for iterations of exp ;
(...)
However I believe it is true that for nonreal s :

In the infinitesimal neigbourhood of any nonreal s there is a point with some period B. (where B is not the period of s if s is periodic).
(...)

Hmm, it seems that the n-periodic points are very "dense" (in  the visual sense) in the right halfplane, but I don't know about the left half plane. At least I didn't come across of 2-,3- or 4-periodic points (for the exp()-function to base e) in the left halfplane (don't have proof -positive or negative- so far)

Hey, Gottfried.


You will not get periodic points in the left half plane, because necessarily they would be attracting, and \( \exp \) has no attracting periodic points. Suppose that,

\(
\Re(z_j) < 0\\
\exp(z_j) = z_{j+1}\\
z_{n} = z_0\\
\)

Then,

\(
|\frac{d}{dz}\exp^{\circ n}(z_j)| = \prod_{j=0}^{n-1} |\exp(z_j)| < 1\\
\)

And

\(
\exp^{\circ n}(z_j) = z_j\\
\)

The function

\(
\exp^{\circ n}(z)\\
\)

has no attracting fixed points. So this is impossible.

The only way you can have a cycle in the left half-plane is if some elements of the cycle are in the right half plane. This won't really happen though, and if it does, they'll be far less numerous than periodic points in the right half plane. Which yes, look almost dense. They aren't though. They're actually a Lebesgue measure zero set. Quite the opposite of dense. But they are kind of numerous.

Regards, James