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First i want to say that the equation
Slog(ln(x)) = slog(x) - 1 is sometimes better then
Slog(exp(x)) = slog(x) + 1.
Basically because slog is NOT periodic as the exp suggests.
Also because we almost always only consider the fixpoint x = ln(x) rather then all the fixpoints x = exp(x).

Let the fixpoints be L and L*.

First question

Im fascinated by the fact that f(g(x)+1) can be periodic while g is not.
Are there elementary nonpolynomial f,g that satisfy this ?

Second question

How does slog behave around the singularities at L and L* ??

3rd question

Does there exist An entire function E such that,

Slog(x) = E( ln(x - L) + ln(x - L*) ) ?

Or something similar ?

Regards

tommy1729
Notice that the 3rd question seems to relate to the Tommy-kouznetsov expansion.

Not sure what that implies.

Regards

Tommy1729
I think |slog(z)| - slog(|z|) is bounded for |z|>|L|+1.

I think most agree.

Regards

Tommy1729
(05/12/2015, 09:55 PM)tommy1729 Wrote: [ -> ]First i want to say that the equation
Slog(ln(x)) = slog(x) - 1 is sometimes better then
Slog(exp(x)) = slog(x) + 1.
Basically because slog is NOT periodic as the exp suggests.
....
Let the fixpoints be L and L*.
...
How does slog behave around the singularities at L and L* ??

What is slog(-5)? What is the slog(z) as z goes to minus infinity?
slog(-5)=slog(exp(-5))-1=slog(0.0067)-1 = -1.993817....
as z goes to minus infinity slog(z) goes to -2. Since sexp(z) has a fairly straightforward logarithmic singularity at z=-2, , then slog(z) as z goes to minus infinity is approximated by the inverse:

So if the cutpoints are drawn correctly, then for real(z)<real(L), slog is 2pi i periodic! The question becomes how do you draw the slog cutpoints; which Dimitrii Kouznetsov investigated. The main singularity is at L, L*, which is a complicated singularity since it involves both the singularity of the Koenig solution for the Abel function at the complex fixed point, as well as the perturbation due to the theta mapping. But the Koenig abel function singularity is dominant, and as you circle around the fixed point, you approximately add multiples of the sexp pseudo period which is ~= 4.4470 + 1.0579i. It is only approximate because there is a second singularity at the fixed point of L. if is the Koenig Abel function at the fixed point then:
where theta(z) is a 1-cyclic function which goes to a constant as imag(z) goes to infinity
An important related question : is slog univalent near the cutlines at L and L* ?

If not then question 3 has answer no i guess.

Regards

Tommy1729
Sheldon i Cannot simply agree.

If slog was periodic even in the limit, then

1) sexp(slog(x)+1/2) is periodic too.

2) the cutlines and singularities due to L,L* are COPIED infinitely often because of the periodicity. Resulting in slog having infinitely many singularities.

3) the fake analogues of 1),2).

These are all contradictions !?

Maybe a weaker form , pseudo pseudo semi periodic or so I take my cutlines paralell to the real axis pointing to - oo.

Maybe your truncation is too brutal ?
Or An unusual branch. But even on other branches im skeptical.
In particular when |im| > |re|.

Tommy-sheldon disputes are always intresting.
Students of tetration should take notes.

Regards

Tommy1729
(05/13/2015, 11:49 PM)tommy1729 Wrote: [ -> ]If slog was periodic even in the limit, then

1) sexp(slog(x)+1/2) is periodic too.

2) the cutlines and singularities due to L,L* are COPIED infinitely often because of the periodicity. Resulting in slog having infinitely many singularities.

By definition; slog(z) = slog(exp(z))-1. So if exp(z) is a small number, close to zero, than the behavior of slog(x) is governed by he behavior of slog(z) near zero, where slog is a nicely defined analytic function in the neighborhood of z=0; where slog(z)~=-1+0.915946x. So now consider a path for slog(z) for z=-5 to z=-5+pi i. slog(-5+pi i) ~= -2.0062. Notice that the pi i values for slog(z) are real valued again, numbers between -2 and -3. So we can do a Schwarz reflection about the pi i line as well....

If you look in the right places, than exp^{1/2}=sexp(slog(z)+1/2) has 2pi i periodicity as well, and exp^{1/2}(z+pi*I) for real(z)~<-0.3624, than exp^{1/2}(z) is real valued, and tends to sexp(-1.5)~=-0.696 as z goes to minus infinity; There is a singularity for slog(z)=-2.5. And exp^{1/2} has singularities at L,L*

Based on these arguments, additional singularities would be expected for slog(z) for z=L+2n pi i, L*+2n pi i

Cake is something to turn Tommy into conjectures.
( i dont drink coffee )

Enough jokes.

I conjecture there is a nonzero constant c such that

C f(x) = f(exp(x))

Where f is entire and Clog f is real-analytic.

Some motivation
Take Clog on both sides:
We then get a NEW F :
F(x) + 1 = log(f(exp(x)) / log C.
This is similar to the slog or slog equation.
We know the branch cut difference should be approx. the pseudoperiod P.
So 2pi i / log C = P i assume.
Then C = exp(2 pi i / P).

That Logic is a bit handwaving but it was the motivation.

Related is the idea - not to be confused with question 3 from the OP - ;

There exist nontrivial entire functions E_1,E_2 such that

E_1(slog(x)) = E_2(x).

If the exp only had one primary fixpoint these questions would all be easy.
It inspires to ask these question anyway as if we were unaware of the other fixpoint.

Regards

Tommy1729
Indeed Sheldon , if sexp(z) is never equal to L + 2 pi n i then all these values are singularities of slog.

Unless the functional equations no longer hold there.

Hmm.

Reminds me of An old question i asked :

Are all solutions to exp(exp(s)) = s singularities for slog ?

I wonder where things are periodic , I bet plots would be Nice.
I do not recall the fake semi-exp to mimic periodicity but i might be wrong, or that may be normal for fakes.

Regards

Tommy1729
If there are additional singularities then question 3 needs revision.

Not sure how , it seems complicated due to not exactly periodic.

Suggestions ?

Regards

Tommy1729
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