Attempt to find a limit point but each step needs doubling the precision...

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Attempt to find a limit point but each step needs doubling the precision... - Gottfried - 11/03/2014

I reread old posts and the Trappmann/Kousnetzov[1] - article about the tetration with base sqrt(2). The point around which I'm fiddling is the imaginary-height iteration, which allows to relate real values from the interval (-oo to 2) with values from (2 to 4), so for instance $x = -1$ gives
$Tet( x,\rho)=y \approx 2.66970618246$ where $\rho$ is $-{\pi \cdot i \over \log(\log(2))}$

It is now interesting, that the point $x_{-2}$ , which is usually assumed in tetration as "at -oo", has as well an existing relative to it, which can simply be computed if we compute the relative

$y_0$ to $x_0=1$ by $Tet(x_0,\rho)=y_0 \approx 2.46791405022$
and do integer Tetration to get
$Tet(y_0,-2) = y_{-2} \approx 2.76432104000$

Here is a picture, where I graphed the trajectories for the imaginary heights from $0 \dots \rho$ in steps of, say $0.1 \cdot \rho$ or $0.01 \cdot \rho$ starting at some example points on the line below +2.

Filename: LimitforY_4.png Size: 35.59 KB 11/03/2014, 12:18 PM

Even more interesting is now the question what happens for the points in the interval $y_{-2} \dots 4$ because they are related to somehow values larger than the negative infinity. There is the asymptotic vertical line at $x=y_{-2}$ ; the trajectories starting from values greater than $y_{-2}$ , say from $y_{-4}$ have a trajectory to the right side of that asymptote.

WHen I simply repeated my procedure for the computation of that trajectory from $y_{-4}$ it looked as if it arrives at some value with positive imaginary part and then, when the imaginary height goes over $\rho$ towards $2 \rho$ it does a jump to the conjugated values - with a discontinuity at $h = \rho$ exactly. This is the thicker red line at the right hand side of the image. But what is the exact limit-point where the jump occurs?

Increasing the precision, with which I compute the tetration shows, that I can extend the trajectory towards some expected limit. Such improved computations allow the grey circles which extend the red trajectory.

But here my problem occurs: I need to double the precision to increase the length of the trajectory, and the last point was already computed with precision of 1600 internal decimal digits... at $h= \rho \cdot (1 - 10^{-2^9})$ (and further approximations of $h \to 1$ do not change the computed values with that given precision)

So that calls for a better analytical consideration of that trajectory. How can we express the limit depending on the $h=\rho \cdot (1-\epsilon)$ where $\epsilon=10^{- 2^k}$ ? (Remark : in Henryk's/Dmitrii's article[1] at page 13(pg 1739 of the printed journal) there is a much nicer and richer picture of that trajectories, but I cannot relate anything in their pictures to *that* trajectory).

Updated: Ahh, I should also mention that I use the powerseries around the lower (attracting) fixpoint $t_0=2$ .

[1] PORTRAIT OF THE FOUR REGULAR SUPER-EXPONENTIALS TO BASE SQRT(2)
MATHEMATICS OF COMPUTATION
Volume 79, Number 271, July 2010, Pages 1727–1756
(Article electronically published on February 12, 2010)

RE: Attempt to find a limit point but each step needs doubling the precision... - tommy1729 - 11/03/2014

Dear Gottfried,

Its not completely clear to me what you are asking.

Maybe if I read it again , but its been a while since I considered base sqrt(2) tetration.

Maybe a few links will help.
( such as where this digital (freely available ?) paper occured , and which posts you are referring too )

Also perhaps express the problem in notation we all use here ( " standard notation ... at this forum " ) such as slog and sexp ?

If I find the time to read carefully and understand the OP I guess I will remove this post ... on the other hand these suggestions might improve readability.

I assume its a good question , and I like limits.

However some limits like the Paris constant are not expressible in closed form.

But once the question is clear to me , I will give it serious consideration.

***

EDIT : in particular your picture shows smooth curves , where is this jump point you talk about ??
I cant see it !

regards

tommy1729

RE: Attempt to find a limit point but each step needs doubling the precision... - Gottfried - 11/03/2014

(11/03/2014, 09:41 PM)tommy1729 Wrote: Dear Gottfried,

(... [1] ...)
Maybe a few links will help.
( such as where this digital (freely available ?) paper occured , and which posts you are referring too )

(... [2] ...)
Also perhaps express the problem in notation we all use here ( " standard notation ... at this forum " ) such as slog and sexp ?

If I find the time to read carefully and understand the OP I guess I will remove this post ... on the other hand these suggestions might improve readability.

(... [3] ...)
EDIT : in particular your picture shows smooth curves , where is this jump point you talk about ??
I cant see it !

regards

tommy1729

Tommy, thanks for your consideration.

Answering your three questions:

[1]: the paper is in our database (but requires user/password for access); also at http://www.ams.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/home.html Hmm. Seems to require also a password here; but I've the access to our database and can send the preprint to you. If you open the article you find pictures on the "physical" page 13 of the pdf.

[2]: It's also on my side, the difficulties with the notion of something (in this case with "superfunction", and how to be applied...)
My approach to tetration as iterated exponention is always: take a initial value $x_0$ assign a base b as an exponential expression to it and evaluate: so you have tetration of "height 1" from $x_0$ : $x_1 = b^{x_0}$ . Assign b as base again, so you have tetration of height 2. Generalize this to negative, then fractional and then complex heights. Notation $x_h = \exp_b^{\circ h}(x_0)$ (For whatever reason we discuss superfunctions only using $x_0=1$ and reduce by this default the notation to b^^h ...) The generalization to fractional and complex heights is possible for bases between 1 and exp(exp(-1)) because that bases allow convergent power series in the computation and namely allow the Schröder-function-mechanism on real numbers with power series with real coefficients.

[3]: In the curve at the right hand there is the smooth thicker red curve. That is the trajectory beginning from $y_{-4}$ in the near of 3.1 . Tetrating with increasing purely imaginary height it first ascends and then descends further to the right side, but does not arrive at the real axis (which is however expected to happen when the height is exactly $h_1 = \pi*I/\log(\log(2))$ . But the proceeding of the trajectory becomes radically stuffed: you are at the final height h_1 by about 99.9 % and still one cm away from the real axis. And if you go to 100.1 % you find the same value but below the real axis. So you conclude, the value at the final hight might be just in the middle. But that's not true. If you go nearer to $h_1$ to 100%-1e-6% there is another remarkable step with a horizontal component. After that, the default internal precision of the software (I do with 200 digits by default) does not suffice to improve the computation. In fact, one can proceed when 400 digits are used to come nearer to the 100%-level for the height, then 800 digits, then 1600 digits precision and the height can then approach $h_1 - 1e-400$ - and one can see the tiny black circles with which I've marked that results. But obviously this type of computation cannot be stretched far more to see only a shadow of the true limit ... - conclusion: one needs an analytical approach (but I don't know how...)

If this explanations do not suffice, please ask for more.

Gottfried

RE: Attempt to find a limit point but each step needs doubling the precision... - jaydfox - 11/03/2014

Gottfried,

Have you looked at this thread before?
http://math.eretrandre.org/tetrationforum/showthread.php?tid=89

I think, if you look at the pictures, that you will see that there are singularities, fanning out from the fixed point at real 4. To get to the real axis, you will have to navigate a course between the singularities. This is likely the cause of the slow convergence, as well as the need for high precision.

I've never actually tried to reach the real axis, and indeed I assumed it was impossible. But without actually trying, I can't be sure. If you do find a path to the real axis, I'd be curious to know the details.

RE: Attempt to find a limit point but each step needs doubling the precision... - tommy1729 - 11/03/2014

http://www.ils.uec.ac.jp/~dima/PAPERS/2009sqrt2.pdf

That seems to be the free paper ...

If that is legal ...

Hope its ok to post it here.

regards

tommy1729

RE: Attempt to find a limit point but each step needs doubling the precision... - tommy1729 - 11/04/2014

(11/03/2014, 10:46 PM)Gottfried Wrote: [3]: In the curve at the right hand there is the smooth thicker red curve. That is the trajectory beginning from $y_{-4}$ in the near of 3.1 . Tetrating with increasing purely imaginary height it first ascends and then descends further to the right side, but does not arrive at the real axis (which is however expected to happen when the height is exactly $h_1 = \pi*I/\log(\log(2))$ . But the proceeding of the trajectory becomes radically stuffed: you are at the final height h_1 by about 99.9 % and still one cm away from the real axis. And if you go to 100.1 % you find the same value but below the real axis. So you conclude, the value at the final hight might be just in the middle. But that's not true. If you go nearer to $h_1$ to 100%-1e-6% there is another remarkable step with a horizontal component. After that, the default internal precision of the software (I do with 200 digits by default) does not suffice to improve the computation. In fact, one can proceed when 400 digits are used to come nearer to the 100%-level for the height, then 800 digits, then 1600 digits precision and the height can then approach $h_1 - 1e-400$ - and one can see the tiny black circles with which I've marked that results. But obviously this type of computation cannot be stretched far more to see only a shadow of the true limit ... - conclusion: one needs an analytical approach (but I don't know how...)

If this explanations do not suffice, please ask for more.

Gottfried

Oh quantum tunneling $Smile$

First guess is this relates to the radius of convergeance of the Taylor series expanded at 2 for the half-iterate.

But thats a wild guess.

[2] was clear to me after reading a second time, just saying to give you back some confidence. There was only 1 logical interpretation $Smile$
Although the formal way of math does not need those kind of puzzles of course.

Seems we have work to do once again.
It seems the amount of problems about tetration grow like sexp and the number of solutions like slog , however both are rising functions and I like both questions and answers $Smile$

regards

tommy1729

RE: Attempt to find a limit point but each step needs doubling the precision... - tommy1729 - 11/04/2014

Perhaps a silly question but assuming Gottfried and Jay plot iterations of type sexp^[k](x) for x values near the real line ...

Then why do these contours of iterations not intersect ?

I mean why do we not get contours like here :

http://math.eretrandre.org/tetrationforum/showthread.php?tid=499

Afterall iterations of sqrt(2)^x are just as chaotic as those of exp(x) or not ?

Or does this occur in the white space ?

What happens to secondary and higher fixpoints ... are they the " cut-in's " of the white space / fractal / branch ?

I note that b^x = b^(x+period)

but I do not see that at the plots.

My guess is it ( b^(x+period) is replaced by b^x + period , and this explains some missing contour intersections / merges / discontinuity.

I think f ' (2) / f ' (4) is involved.

EDIT : yes I understand the iterations are of IMAGINARY height and that PARTIALLY explains some things like the circle shapes and locally not intersecting ... but still.

regards

tommy1729

RE: Attempt to find a limit point but each step needs doubling the precision... - tommy1729 - 11/04/2014

Looking at the pics I can't help wonder :

How different is imaginary height for functions with 2 fixpoints from

functions with 2 parabolic fixpoints whose derivatives (at the fixpoints) are nonreal.

??

regards

tommy1729

RE: Attempt to find a limit point but each step needs doubling the precision... - Gottfried - 11/04/2014

(11/04/2014, 12:48 AM)tommy1729 Wrote: Looking at the pics I can't help wonder :

How different is imaginary height for functions with 2 fixpoints from

functions with 2 parabolic fixpoints whose derivatives (at the fixpoints) are nonreal.

??

regards

tommy1729

Well the iteration of real x with real height (base b=sqrt(2)):

1) take any point x at the half line -oo to 2: forward-iteration y=(b^x) then y is again on the real halfline but nearer to the fixpoint 2; backwards iteration (y = log(x)/log(b)) if x>0 then y is still on the half line but farther away from 2; if x<0 then y becomes complex; if x=0 then y becomes negative infinity . Result for forward iteration: point t=2 is limit point and cannot be overstepped by as many forward iterations as you want.

2) take any point x at the real interval 2 to 4 (4 not included): forward-iteration y=(b^x) then y is again on that interval but nearer to the fixpoint 2; backwards iteration (y = log(x)/log(b)) y is nearer to the upper fixpoint 4. Result for forward iteration: point t=2 is limit point and cannot be overstepped by as many forward iterations as you want. For backward iterations: point t=4 is limit point and cannot be overstepped by as many forward iterations as you want.

3) take any point x at the half line 4 to +oo : forward-iteration y=(b^x) then y is again on the real halfline but farther away from the fixpoint 4; backwards iteration (y = log(x)/log(b)): y is still on the half line but nearer to 4; Result for backward iteration: point t=4 is limit point and cannot be overstepped by as many forward iterations as you want.

Iterations with complex height:
x is a point on the halfline -infinity to 2: y is in general complex and if h=Pi*I/log(log(2)) then y is real and in the interval 2..4. Thus with complex heights you can "overstep" the fixpoint to the other side.
That's what I'm interested in here.

But this gives an interesting relation: we choose some point y in the interval 2..4 and find its relative x in the negative halfline, call it y_0 when x_0 = 1, call it y_(-1) when x_(-1)=0 , call it y_(-2) when x_(-2) = - infinity. But there is also y_(-3) ; now what is x_(-3)? Is it log(-infinity)/log(log(2)) ? Anyway: we have some y_(-3) just by y_(-3) = log(y_(-2))/log(log(2)) and so on. Numerical computation usually shall give a *complex* value near the upper real halfline >4 and the software does not allow to find the expected, exactly real, value x_(-3).

And here is now my problem: if you can approximate the complex height very near to h = \rho = Pi*I/log(log(2)) then you come a bit nearer to the real axis, and when you get even nearer to \rho then cou come also nearer to the real axis - but to what rate? In the Trappmann/Kousnetzov-article at best we find some white area in the graph for F_(2,-1) . And my hope is that by some analytical formula we find that the true limit is something on the real axis, possibly positive infinity ( but I don't know).

Well, I'm not (yet) aware of some analytical expression for tetrations with imaginary or complex heights so at the moment I/we can only try to extend the precision to come nearer to the result and see the tendency.

Gottfried

RE: Attempt to find a limit point but each step needs doubling the precision... - Gottfried - 11/04/2014

(11/04/2014, 12:33 AM)tommy1729 Wrote: Perhaps a silly question but assuming Gottfried and Jay plot iterations of type sexp^[k](x) for x values near the real line ...

Then why do these contours of iterations not intersect ?

I mean why do we not get contours like here :

http://math.eretrandre.org/tetrationforum/showthread.php?tid=499

Afterall iterations of sqrt(2)^x are just as chaotic as those of exp(x) or not ?

Just to that question: no, it's different. Because b=sqrt(2) is in the interval 1 < b < exp(exp(-1)) and has two fixpoints on the real line (so all iterates of values on the real line >0 are on the real line again and we have even finite values for the iterates when h goes to infinite height (namely the two real fixpoints) while when b=exp(1) then it is outside the Euler-interval 1<b<exp(exp(-1)) and we cannot iterate infinitely on the real axis except with the result going to positive infinity as well.
For the iterations with complex heights one can consider Dmitrii's further plots in his articles on tetration with base b=exp(1)...