# Tetration Forum

Full Version: sexp(strip) is winding around the fixed points
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I plot the image of the halfstrip $S=\{z\in\mathbb{C}: -1\le\Re(z)\le0, \Im(z)>0\}$ at the mapping with sexp. The $\mathrm{sexp}(S)$ is marked with colored lines. Let $s\in S$, and $z=\mathrm{sexp}(s)$. The red lines indicate $\Re(s)=-1,-0.8,-0.6,-0.4,-0.2,0$. The blue lines which indicate $\Im(z)=0,1,2,3, ... 22$; these numbers are marked at the graphics. The halff of the yellow sickle indicates the "basic range" G for the slog.
The resolution of my screen happened to be 10 orders of magnitude out of range required to see the details in one figure. Therefore, I plot the sequence of zooms; each next represents the piece corresponding to the smallest sell of the previous one.
[attachment=431]
The figire confirms that the slog(strip) is winding around the fixed point. In particular, at $z=-0.5+20i$, we have $t=\mathrm{sexp}(z)\approx 1.3161315052205+1.338235701329i$, which is again within the "basic range" G for the slog function. G and sexp(S) have many intersecitons. (we need to evaluate the tetration with at least 12 decimal digits in order to see these intersections).

P.S. I have the fast slog and fast sexp. I just finished the tests and I begin the description. Now it is implemented in C++, but I plan to translate it to other languages. If anyone can help with the translation? Has anyone experience with automatic translation from C++ to Mathematica? Is the result of such a translation workable?
Dmitrii, thats thorough work!
Thank you for this diligent elaboration.

But now it would be really interesting to see slog(G) (with non-winding cut) with image range up to 20i on the imaginary axis. Whether it intersects the strip there or not.
I am a bit confused how slog(G) looks globally. It seems to be impossible if slog is biholomorphic on G.
bo198214 Wrote:.. would be really interesting to see slog(G) (with non-winding cut) with image range up to 20i on the imaginary axis. Whether it intersects the strip there or not.
I have implemented the funciton fsexp, that approximates sexp, and fslog, that approximates slog. I plot $f= |\mathrm{fslog}(\mathrm{fsexp}(z))-z|$.
The level $f=10^{-2}$ is shown with dark yellow.
The level $f=10^{-4}$ is shown with thick yellow.
The level $f=10^{-6}$ is shown with thin yellow.
The level $f=10^{-8}$ is shown with black.
The level $f=10^{-10}$ is shown with red.
The level $f=10^{-12}$ is shown with pink.
The level $f=10^{-14}$ is shown with green.
[attachment=432]
The plot below indicates the range of values of slog. The $\mathrm{slog}(G)$ is well inside the range of values; it intersects the strip $S=\{z\in \mathbb{C} : -1<\Re(z)\le 0 \}$ only once. In particular, the slog never has value 20i; but it may have value -4+20i.
The levels indicate the precision of the fast numerical implementaitons. While the imaginary part does not exceed 5, the errors are regular, and, perhaps, are due to the approximation fslog. Currently fslog is made of 2 elementary functions (expansion in vicinity of the fixed point and expansion in vicinity of unity); fsexp uses 3 functions (expansion at $i \infty$, expansion at 3i and expansion at zero). I estimate, in the worst case (distance of order of 0.1 from the fixed point), fslog returns at least 10 significant figures; in vicinity of the real axis, each of functions fslog and fsexp return values what deviate from slog and sexp in 15th decimal digit.
While the imaginary part exceeds 8, the deviation of $f= |\mathrm{fslog}(\mathrm{fsexp}(z))-z|$ from zero is mainly due to the smallness of derivatives of sexp; it is difficult to distinguish values of the sexp from the fixed point.
Thank you for being so patient with me, Dmitrii. Now it finally came through that if sexp maps $A$ biholomorphically to $C$ then it can also map $B$ biholomorphically to $C$ though $A\cap B=\emptyset$ may be.

I now also see that the image of slog has a width of roughly $\frac{2\pi}{\arg(L)}$ because an increment by 1 causes the image of sexp to revolve $\arg(L)=\Im(L)$ around the fixed point (near the fixed point). And after $2\pi$ revolution it reaches the cut.
I have new doubts that sexp(S) (where S is the strip: 0<Re(z)<1) winds infinitely around L.
When we consider T=slog(G) (where G is the crescent: |z|<|L| and Re(z)>Re(L)) then T is also a strip of width 1.

Let the z-plane contain T and S, and the w-plane being the image under sexp, i.e. containing L, G and sexp(S).
If we slowly continuously transform T into S in the z-plane, then G slowly transforms into sexp(S) in the w-plane.
But continuous deformations of G always wind finitely many around L.
So we would never reach sexp(S) if it winds infinitely around L.
(06/27/2009, 09:55 AM)bo198214 Wrote: [ -> ]I have new doubts that sexp(S) (where S is the strip: 0<Re(z)<1) winds infinitely around L.
When we consider T=slog(G) (where G is the crescent: |z|<|L| and Re(z)>Re(L)) then T is also a strip of width 1.

Let the z-plane contain T and S, and the w-plane being the image under sexp, i.e. containing L, G and sexp(S).
If we slowly continuously transform T into S in the z-plane, then G slowly transforms into sexp(S) in the w-plane.
But continuous deformations of G always wind finitely many around L.
So we would never reach sexp(S) if it winds infinitely around L.

You have no need to deal with so exotic objects like sexp, in order to reproduce the same paradox. Consider the logatirhmic spiral x=log(t)cos(pt), y=log(t)sin(pt) .
For posiive p, at real t, this spiral winds infinitely around zero.
"If we slowly continuously transform" this spiral to the straight line (just reduce p to zero), then you may wander, how the infinite number of turns of this spiral arond zero becomes finite (no windings at all).
(06/27/2009, 11:50 PM)Kouznetsov Wrote: [ -> ]You have no need to deal with so exotic objects like sexp, in order to reproduce the same paradox. Consider the logatirhmic spiral x=log(t)cos(pt), y=log(t)sin(pt) .
For posiive p and real t, this spiral winds infinitely around sero.
"If we slowly continuously transform" this spiral to the straight line (just reduce t to zero), then you may wander, how the infinite number of turns of this spiral arond zero becomes finite (no windings at all).

Well there must be a critical point where it changes from infinite winding to finite winding.
we have a continuous transform h:[0,1]->region of CC, with h(0)=T and h(1)=S.
Where is this point t in the transform of T into S (can it visibly recognized?) that sexp(h(t)) is infintely winding while sexp(h(t')) is only finitely winding for t'<t.
(06/28/2009, 02:31 PM)bo198214 Wrote: [ -> ]
(06/27/2009, 11:50 PM)Kouznetsov Wrote: [ -> ]You have no need to deal with so exotic objects like sexp, in order to reproduce the same paradox. Consider the logatirhmic spiral x=log(t)cos(pt), y=log(t)sin(pt) .
For posiive p and real t, this spiral winds infinitely around sero.
"If we slowly continuously transform" this spiral to the straight line (just reduce t to zero), then you may wander, how the infinite number of turns of this spiral arond zero becomes finite (no windings at all).

Well there must be a critical point where it changes from infinite winding to finite winding.
we have a continuous transform h:[0,1]->region of CC, with h(0)=T and h(1)=S.
Where is this point t in the transform of T into S (can it visibly recognized?) that sexp(h(t)) is infintely winding while sexp(h(t')) is only finitely winding for t'<t.

Henryk, you reveal the misprint in my post. Sorry. I just corrected there
"(just reduce t to zero)" to "(just reduce p to zero)".
Now t parametrizes the set (curve),
and p is parameter we play in order to modify the "speed" of winding. Only at p=0, there is no winding, although at small p, it is difficult to count many turns, we need strong zooming in.
Oh now I see what you mean. In this case the critical point is $p=0$ where it changes from finite winding into infinite winding.