Multiple exp^[1/2](z) by same sexp ?
#4
(04/29/2014, 10:04 PM)tommy1729 Wrote: [quote='sheldonison' pid='6885' dateline='1398794535']
Quote:.....
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....
\( \text{growth}_f = \lim_{n \to\infty}\frac{\text{slog}(f^{o n})}{n} \)
.....

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.
.....
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


Messages In This Thread
RE: Multiple exp^[1/2](z) by same sexp ? - by sheldonison - 04/30/2014, 10:47 AM

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