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 [Update] Comparision of 5 methods of interpolation to continuous tetration sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 10/29/2013, 09:37 AM (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 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 10/29/2013, 01:11 PM 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 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 10/29/2013, 02:52 PM (This post was last modified: 10/29/2013, 06:26 PM by sheldonison.) (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 $\alpha(z)$ 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 $\Im(z)=\frac{1}{sexp(3.5)}\approx 10^{-78}$. Kneser multiplies this repeating pattern by $2\pi i$ and then takes the exponent of that; $\exp(2\pi i z)$. 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 $\Im(z)=\frac{1}{\text{sexp}(8.5)}$.     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 $\text{sexp}(z)=\alpha^{-1}(z+\theta(z))$, via a 1-cyclic mapping from the inverse Abel function, $\alpha^{-1}(z)$, as well as an sexp(z) Taylor series representation at the real axis. This is because the $\theta(z)$ 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 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/01/2013, 02:16 PM 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 Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 02/01/2014, 12:09 PM (This post was last modified: 02/04/2014, 12:29 AM by Gottfried.) An application of the "polynomial method/Diagonalization" (matrix-size 64x64) which was in the instance of the article shown with base $b=4$ 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 $z_0 = 1$ with base $b=1.3$. (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 $2 \pi i / v$ where v is the log of the log of the fixpoint $t \approx 1.48$     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.
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 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 02/03/2014, 01:13 PM 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 Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 02/03/2014, 01:44 PM (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 Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 02/03/2014, 01:59 PM (This post was last modified: 02/03/2014, 04:15 PM by Gottfried.) (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 $z_0=1$ 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 ( $\text{reflexion}(z_0) = y_0 \approx 1.86300282587$ the "zero-reflexion" $\text{reflexion} (z_{-1}) = y_{-1} \approx 2.37147238675802$ and then "neg infinity-reflexion" $\text{reflexion}(z_{-2})= y_{-2} \approx 3.29126767409758$). The left part of the curve has then the horizontal lines +- as limiting asymptote. Well, I'm really curious too what the shape is with bases nearer to eta... Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 02/03/2014, 10:35 PM (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 ( $\text{reflexion}(z_0) = y_0 \approx 1.86300282587$ the "zero-reflexion" $\text{reflexion} (z_{-1}) = y_{-1} \approx 2.37147238675802$ and then "neg infinity-reflexion" $\text{reflexion}(z_{-2})= y_{-2} \approx 3.29126767409758$). The left part of the curve has then the horizontal lines +- as limiting asymptote. GottfriedIm 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 +- is actually 1.3 2 pi i ? From the formula b 2pi i where b is the base for 1< 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 « Next Oldest | Next Newest »

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