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[Update] Comparision of 5 methods of interpolation to continuous tetration
#21
(10/28/2013, 11:29 PM)tommy1729 Wrote: The only thing I can come up with is that :

1) we simply take the principal schroeder function near the fixpoint c.

2) in that domain we use analytic continuation ( monodromy ) so that by small radiuses recentered we can reach any value in the upper plane.

3) In that upper plane we find ( after many continuations ) the values equalling the reals between 0 and 1. ( and those are connected )
( for reasons yet to be explained !)
.....
10) we have sexp analytic near the positive real line.
....
And if I got It right , there is still alot to be explained !!

Tommy,

Is your question about using Kneser's method to get numerical results, or is it about Kneser's proof of a real valued sexp(z) solution? It is not practical to use Kneser's method to generate usable results for sexp(z), though it is a proof of the existence of such a real valued solution, even though numerically, Kneser's Riemann mapping is very difficult to work with. Also, Kneser's proof is hard to follow, and I am not good at math to recreate it.

On the other hand, I wrote a pari-gp program that is numerically equivalent to Kneser's method that is practical, though I cannot rigorously prove my method converges. But assuming my method converges, than I can show it converges to the same solution as would be given by Kneser's Riemann mapping solution. Obviously, I could explain my method, and its equivalence to Kneser's solution, but that would not be relevant to your questions.
- Sheldon
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#22
I simply asked if I am correct with my 10 step way to do kneser.

I do not ask for the Riemann mapping so I guess its more of a proof question.

But actually I'd say its a " construction " question.

Is the Kneser proof/solution constructed in the 10 steps I posted or is one or more steps wrong ?

I assumed sheldon was an expert on kneser.

Maybe bo is.

I must say Im not so good with code.

And actually I wonder , if you cannot explain kneser completely , how can you find your own method based on kneser and equivalent to kneser ??

That is a mystery to me.

Also apart from the fact that i can read only math - not so good with code - I also wonder how Im suppose to fully understand an equivalent method if I do not understand the Original completely ?!

ON the other hand , Im convinved that once I understand Kneser , I will be able to prove convergeance of your method ....

Maybe Bo or Jay can help us out here.

Regards

tommy1729
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#23
(10/29/2013, 01:11 PM)tommy1729 Wrote: I simply asked if I am correct with my 10 step way to do kneser.

I do not ask for the Riemann mapping so I guess its more of a proof question.

But actually I'd say its a " construction " question.

Is the Kneser proof/solution constructed in the 10 steps I posted or is one or more steps wrong ?
Tommy, the sequence is good, but the Riemann mapping step 5) has a lot of sub-steps. For me, the biggest complexity hurdle in Kneser's construction, besides the fact that I don't have a formal math degree, is taking the Abel function, from Schroeder function, of the real axis, which is after the step where he generates the chi-star, but still one or two steps before the Riemann mapping. I don't read German, so have no idea how he proved the infinite region is simply connected, and I wouldn't know how to do so, since the region is increasingly recursively complex.

Here is the rough Abel function of the real axis, showing the repeating pattern; here . Kneser multiplies this repeating pattern by and then takes the exponent of that; . That is the contour that gets wrapped around a unit circle for the Riemann mapping.
   

Here we zoom in on one of the singularities, where sexp(z)=0. The singularity gets ever more complicated as we super-exponentially approach zero. Here, I show what the contour looks like if .
   

My algorithm has a mathematical description, as well as pari-gp code. I don't want to side track too much, but it does something different but equivalent to generate , via a 1-cyclic mapping from the inverse Abel function, , as well as an sexp(z) Taylor series representation at the real axis. This is because the function has a singularity at the real axis, so adequate convergence is not possible with a reasonable number of terms. So my algorithm actually has to iteratively generate two different equivalent representations of sexp(z).
- Sheldon
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#24
Thank you sheldon.

At this point I wonder most about how the riemann mapping of the abel function maintains its functional equation.

I recall knowing this but I forgot ... if I recall well. ( how much doubt about memory can one sentence have ? :p )

Regards

tommy1729
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#25
An application of the "polynomial method/Diagonalization" (matrix-size 64x64) which was in the instance of the article shown with base and in that article likely asymptotic to the Kneser method.

Here I provide a picture with focus on complex iteration from one real starting point with base . (I've not yet checked against Sheldon's Kneser-implementation ).

The picture shows roughly circles: along the circumferences the iteration-height is purely imaginary; one revolving means where v is the log of the log of the fixpoint


   

It is still surprising to me that we can proceed from one real point below the fixpoint to some other real point above the fixpoint - which means to avoid/surpass the infinite height-iteration: just by using imaginary heights...


Gottfried

Here I added two more (hopefully instructive) views;

Here the base-point is z0=0
   

and here is the iteration to one more negative height, where I had to leave out the infinitely distant point z0 (-> - infinity)

   

Now I've got my matrices for base b=1.44, near eta. What a mess!
I don't have any idea - there is no obvious divergence in the power series with 64 terms. I also took z0=b^b ~ 1.69 as initial point; the curves originating from z0=1 were even more messed up.
   

This is the picture base b=1.44 z0=1+0î - I've no explanation so far for the messed curves.

   


<hr>


P.s.: A 3-D picture with colors indicating height, and "isobares"-grid and the fixpoint shown as peak of infinite height were nicer but I do not know how to draw one(which software). If someone else likes to play with this I can provide the coefficients of the diagonalization matrices in Pari/GP-convention: the computation of that matrices is extremely costly (it needed 6000 secs to be computed and more than 3000 digits decimal precision;I chose then 4000 digits) so it might be interesting to get the ready-made numbers by download instead by a new computation.
Gottfried Helms, Kassel
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#26
Very nice pic Gottfried.

I think the phenomenon is not typical for tetration but for most functions with a limited amount of fixpoints.

What surprises me is that the shapes are so close to perfect circles.

I wonder how it looks like if the real fixpoints approaches its limits.

In other words what happens if the base gets close to eta ?

Do we get figure 8 shapes instead of circles because of the anticipation of the pair of conjugate fixpoints ?

What happens with functions with 2 or more fixpoints ? do they also have these circles around their fixpoints ? In other words, do they locally ( around the fixpoints ) behave the same as the plot given by gottfried here ?

Why not ellipses ??

And what if there is no fixpoint ?

Maybe we need riemann surfaces to understand this better ?

Many questions as usual.

Regards

Tommy1729
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#27
(02/03/2014, 01:13 PM)tommy1729 Wrote: Very nice pic Gottfried.

I wonder how it looks like if the real fixpoints approaches its limits.

In other words what happens if the base gets close to eta ?

Hmm, I'm trying hard to get pictures with such bases, but for some reason (which I do not yet understand and thus cannot compensate for), this becomes extremely difficult - for instance: perhaps I'll need power series with thousands of terms, I don't know. I'm still fiddling with it to find the key to access the problem in a reasonable generality/range of bases - so this might take some time...

Gottfried
Gottfried Helms, Kassel
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#28
(02/03/2014, 01:13 PM)tommy1729 Wrote: What surprises me is that the shapes are so close to perfect circles.

I wonder how it looks like if the real fixpoints approaches its limits.

In other words what happens if the base gets close to eta ?

Do we get figure 8 shapes instead of circles because of the anticipation of the pair of conjugate fixpoints ?

What happens with functions with 2 or more fixpoints ? do they also have these circles around their fixpoints ? In other words, do they locally ( around the fixpoints ) behave the same as the plot given by gottfried here ?

Why not ellipses ??
Concerning the shape:[please see my updated previous post, I've inserted two new pictures] as the base point goes to the left then the shape becomes first a distorted oval (picture 2) and later a -left-open ellipsoidic looking shape going to negative infinity to the left (picture 3).

Note also on the real axis, righthand of the fixpoint, the occurence of some "mirror"- or "Reflexion"-points: just for the two additional iterations to the right(=negative height) from the currrently most right point ( the "zero-reflexion" and then "neg infinity-reflexion" ).

The left part of the curve has then the horizontal lines +-<not-yet-determined-imaginary-value> as limiting asymptote.

Well, I'm really curious too what the shape is with bases nearer to eta...

Gottfried
Gottfried Helms, Kassel
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#29
(02/03/2014, 01:59 PM)Gottfried Wrote: Note also on the real axis, righthand of the fixpoint, the occurence of some "mirror"- or "Reflexion"-points: just for the two additional iterations to the right(=negative height) from the currrently most right point ( the "zero-reflexion" and then "neg infinity-reflexion" ).

The left part of the curve has then the horizontal lines +-<not-yet-determined-imaginary-value> as limiting asymptote.

Gottfried
Im not sure what you meant by reflexion. I noticed the fixpoint is a red x in your plots and there are also green x's. I guess that relates.
Maybe its about pseudocenters of the pseudocircles or the analogue for ellipses ( pseudofocal points) . It seems these centers move away from the fixpoints. These centers seem to follow the real iterations although probably not linearly.

The idea of circles becoming other shapes gradually , reminds me of pseudoperiodic functions. Afterall its like f(a+2pi i) = f(a) but f(a+1+2pi i) =/= f(a+1). Kinda.

As for the asymptotic to the left I wonder - based on the idea of pseudoperiodic somewhat too - if that +-<not-yet-determined-imaginary-value> is actually 1.3 2 pi i ? From the formula b 2pi i where b is the base for 1<<b<<eta.

Perhaps a bit naive. I need more training.

As for these shapes *very* near the fixpoints I believe they are mainly determined by the first 2 derivatives of the fixpoint.

The shapes *far* away are then just the ln or exp iterations of the others (ln and exp in base b of course) . This seems to make sense.

For instance *near* the fixpoints if the function is well approximated by ( recenter fixpoint at 0 for simplicity ) :

0 + x + (x^2)/15

,then we get a circle approximation since the derivative is 1.

This is because the complex iterations of x around x=0 are also a circle.

Maybe a bit naive but my first impression.

Thanks for the updates.

regards

tommy1729
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#30
(02/03/2014, 10:35 PM)tommy1729 Wrote: Im not sure what you meant by reflexion. I noticed the fixpoint is a red x in your plots and there are also green x's. I guess that relates.
The green x is the position of the base. With "reflexion" I hope to introduce a meaningful expression for the impression, that the two points on the opposite sides of the "circle" - opposite with respect to the fixpoint- look like mirrored images of each other - and are also just looking like half-circle rotations of each other (which would then be a very nice and significant property).

With the <not-yet-determined-imaginary-value> yes, that 1.3*2*pi is perhaps a good idea. I think it should have a simple solution; I just don't get it at the moment... [update] It should be 2*Pi*I/v where v= log(log(LH(1.3))) where the function t=LH(b) gives the fixpoint t such that b^t=t and v is about 1/v~-1.0503


With the other questions - I do not yet know...



Gottfried
Gottfried Helms, Kassel
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