Multiple exp^[1/2](z) by same sexp ?

[tommy1729]

What about f(z)=sexp(2*z)? We can generate the function
\( g(z) = f(f^{-1}(z)+1/2)=\exp(z) \), which is entire... I guess that's not fair since the multiple branches as you circle the fixed point of L all agree at integer values of n.

Your f(z) has order 2 and your g(z) has order 1.
In my example it was order 1 and order 1/2.
The clue is that a point on sexp*(a) that is analytic cannot be mapped to a singularity at sexp*(a+b) while assuming that exp*^[b] is entire.

But what if the fixed point is real? Can we say that \( g(z) = f(f^{-1}(z)+1/2)=\exp(z) \) is never entire?

If the only fixpoint is real then it resembles iterations of eta^z (or exp(z)-1).
But eta^z (or exp(z)-1) has a parabolic fixpoint ( you are long enough into tetration to know what that implies ).

---

It seems that things get complicated if we assume more than 1 fixpoint pair.

However this means sexp* has more branches than sexp.
And It seems therefore we get the same result.

This lacks some formal formulation I admit. But I think you understand what I mean.

---

regards

tommy1729