Is sexp(z) pseudounivalent for Re(z) > 0 ?

Is that a uniqueness condition ?

The difference between univalent and pseudounivalent is :

speudounivalent is weaker : f(z+k) = f(z) only possible if k is real.

regards

tommy1729

I have to add : psuedounivalent must be a uniqueness criterion but without existance proven.

This follows from the earlier proven fact that sexp(z) analytic for Re(z)>0 is already a uniqueness criterion.

Hence the question becomes : is this analytic function sexp(z) pseudounivalent ?

We know that sexp(z) has singularies. In the thread " the fermat superfunction " I noticed that most entire functions are not pseudounivalent. The follows from the nonbijective nature of most entire functions and the many amount of fixpoints/zero's of most entire functions.

This motivated me to wonder about functions that are not entire hence returning to sexp(z).

Naturally I also wonder about other superfunctions ofcourse.

I do not recall the exact location of all previous posts regarding these issues but on MSE mick made a related post.

The case which mick answer is actually about exp, but the same applies by analogue to sexp for Re(z)>0.

http://math.stackexchange.com/questions/...-equations
Btw his score of -5 is rediculous.

This reminds me of the reasons why I am not or no longer on places such as MSE , sci.math, wiki etc.

Tetration forum rules

But other tend not too imho.

regards

tommy1729

Let a,b > 0. Let a',b' > 0.

Because exponential iterations of a + b i usually get arbitrarily close to any other a' + b' i , I am tempted to say that sexp(z) is NOT psuedounivalent.

So entire superfunctions are usually not pseudounivalent and since many iterations of analytic functions behave chaotic (like exp) , neither are most nonentire superfunctions.

So, what is an example of a nontrivial pseudounivalent superfunction ?

If they exist at all ?

Together with my friend mick im working on this.

Im pessimistic though.

And you ?

regards

tommy1729