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

[tommy1729]

[quote='sheldonison' pid='6885' dateline='1398794535']

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I do find it interesting that it seems there are no known entire functions with fractional exponential growth. I would define the exponential growth by the following equation....
\( \text{growth}_f = \lim_{n \to\infty}\frac{\text{slog}(f^{o n})}{n} \)
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Lets call that "conjecture entire 1".
Your intuition is correct.
And also it is strongly related to another recent conjecture of you ( or a repost of it ) :

\( \lim_{z \to \infty} \text{slog}(f^z)=\text{slog}(f^{z+1})-1 \)

(made in the slog_b(sexp_b(z)) thread)

Lets call that " conjecture entire 2"

Im very optimistic about both conjectures and dare to say a proof is 99% complete.
I have believed them for over 25 years.
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As for " conjecture entire 1" I feel forced to notice
http://en.wikipedia.org/wiki/Weierstrass_product

Yes the famous weierstrass factorization theorem.

Together with induction that should be a strong tool in a proof.
....
you ask for an entire function that grows like exp^[1/2].
How about F(F^[-1](z)+1/2) as a solution to your problem ? Clearly F^[-1] cannot be entire ( it must have branches , being the inverse of a nontrivial entire function ).
Hence F(F^[-1](z)+1/2) is NOT the solution we want.

Maybe that helps.

regards

tommy1729

Tommy, Interesting, thanks for your response...
- Sheldon

Even more intresting, I have a proof of "conjecture entire 1".
ON the other side its a shame we missed it because its almost written before here.

Its a bit sketchy but the proof goes something like this :

Lets say exp*(z) is a function with order 1 and 2 conjugate repelling fixpoints.
sexp*(z) is its superfunction.
If we want an entire function close to exp^[1/2](z) then it should be exp*^[1/2](z).

So exp*^[1/2] is not entire because sexp*(z) is not entire.
sexp*(z) is not entire => sexp*(slog*(z)+1/2) is not entire.

So for the same reasons exp^[1/2](z) is not entire , exp*^[1/2](z) is not entire.

we know sexp*(z) is not entire because it has to have log branches or algebraic branches ( coming from sexp*(z-1) = log*(sexp*(z)) ).

regards

tommy1729