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 Bifurcation of tetration below E^-E Gottfried Ultimate Fellow Posts: 754 Threads: 114 Joined: Aug 2007 01/18/2008, 06:56 AM (This post was last modified: 01/18/2008, 07:10 AM by Gottfried.) andydude Wrote:@Gottfried: Your graph confuses me, why is the 1-tick for b=... and the 1-tick for f=... not aligned? and why do you have X's for points, do you have any options for boxes? diamonds? circles? no dots? That would make the graph much nicer to look at. But anyway, I got the general idea of the graph, I just think it could use some work. I'm not sure what the benefit of such a change-of-variables would be, unless it would simplify the series for b^b^x.Hi Andrew - the line for b is just for completeness. It could have had the same y-scale as the other lines, but I wanted to prevent overlay of the curves, so the y-scale for the b-curve is at the right side of the graph. I'm copy&pasting the data computed by Pari/gp into Excel and do the standard graph with some editing. Not always optimal, though... ;-( Quote:@Gottfried: As for your series expansion, I really, really like your presentation of the series for (b^x-1), it shows patterns in the coefficients, where I never saw patterns before Good job. Well It's quite a time ago, early eighties, when I sat night after night in the campus' computing center and hacked the PL/I-runtime-system by analyzing the pointers and data-structures from memory-dumps. The greatest success was to implement a routine, which was able to modify the file-parameters at runtime so that I could open-and reopen any file and assign record-length and block-size-parameters dynamically. I could then provide *very* versatile programs for file-backup, file-restructuring and transformation in dialog-programs; usually with PL/I you needed to define fixed file-types in the source-code of an application. So sometimes I get nice associations to this inspired time of pattern-detection... For the given coefficients a_k you may observe, that in each of that the numerical coefficients of the last column contains the Stirling numbers 1'st kind and the first column that of 2'nd kind, scaled by factorials. The condition, that with integer h the a_k-polynomials produce multiples of the denominator seem to make the other coefficients unique, so I manually solved for the numerical coefficients in a_4, to get an example of the path for a general solution This is in principle possible for all other a_k, so this description allows to compute each single term a_k independently of the matrix-eigensystem-computation. I think, I don't need to explain, why this is a remarkable achievement - in principle. I say "in principle" because the computation-requirements seem again to be huge, and I don't have a good recursive algorithm yet. But the possibility to do the computation for as many terms of the powerseries of x as wished sequentially and independently from each other, frees us principally from the impossible memory management for matrices of sizes for even some dozen powerseries-terms only and of the computation of their symbolical eigensystem-decomposition, and allows, for instance, to compute term a_96 this week, term a_97 next week and so on, if that computation needs any such amount of time. --- ... and I love series, too :-) ("Euler - the master of us all...") Gottfried Ps. your explanation of the role of x0 and x1 are good, I understand now. Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/18/2008, 08:53 AM (This post was last modified: 01/18/2008, 09:26 AM by andydude.) To recap, we know a simple parameterization of $x = b(x)^{b(x)^x}$ on the middle branch is just $b(x) = x^{1/x}$, and the other branches can be described by the differential equation given in my previous post, along with a given initial condition. The initial condition used for the previous series expansion was the point of intersection: $(x=\frac{1}{e}, b=e^{-e})$. I just found another point (probably found before) on the outer branches, one that has many advantages over the point of intersection. This point is $(x=\frac{1}{2}, b=\frac{1}{16})$ which is 2-periodic in the sense: $(1/16)^{(1/2)} = (1/4)$ $(1/16)^{(1/4)} = (1/2)$ as can be seen in this plot:     The problem with the point $(x=\frac{1}{e}, b=e^{-e})$ is that its really not a periodic point at all, its a fixed point, so it doesn't qualify as a 2-periodic point. Another problem with the this is that when it is used as an expansion point, the corresponding series has a radius of convergence of 1/e. The reason why is because there are singularities at (x=0,b=0) and (x=1,b=0), if you don't believe me, then do the math, you'll see the singularities. This means that we can't evaluate the series near x=1, because this is outside the radius of convergence. The advantage of the point $(x=\frac{1}{2}, b=\frac{1}{16})$ is that not only is it a true 2-periodic point, but it is half way between x=0 and x=1 (obviously), which means the radius of convergence would be 1/2. Another advantage of this point is that the derivative of b is non-zero, which means the series is invertible. The problem here is that since it is so close to the other point (where the series is not invertible) the inverse series would have a radius of convergence of $|(e^{-e})-\frac{1}{16}| \approx 0.00349$ which is pretty useless. Despite these short-commings, it still converges between x=0 and x=1, so it is still useful for this. Here is the series expansion that I found about the second point: $ \begin{tabular}{rl} b(x) & = \frac{1}{16} \\ & +\ (-2 + 4 \ln(2)^2) \frac{(x-\frac{1}{2})}{4(2\ln(2)-1)} \\ & +\ (1-8\ln(2)+28\ln(2)^2-20\ln(2)^3-32\ln(2)^4+32\ln(2)^5) \frac{(x-\frac{1}{2})^2}{2(2\ln(2)-1)^3} \\ & +\ (24-192\ln(2)+2176\ln(2)^3-5568\ln(2)^4+4544\ln(2)^5+2048\ln(2)^6-5120\ln(2)^7+2048\ln(2)^ \frac{(x-\frac{1}{2})^3}{12(2\ln(2)-1)^5} \\ & +\ \cdots \end{tabular} " align="middle" /> It might be interesting to note that the highest power of $\ln(2)$ displays a pattern in the numerator (3n-1) and the denominator (2n-1). Which means a closed form might be possible some day. Andrew Robbins Attached Files   bump4.pdf (Size: 33.49 KB / Downloads: 238) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/18/2008, 09:13 AM GFR Wrote:Hi IVARS! Therefore, I agree with you (IVARS) thet we need an extra dimension, but I think that it should be represented by the imaginary axis itself. In this case, it will be the "fourth dimension" of the diagram. GFR But if we use only infinite tetration heights, 3 dimensions would be enough? There is one puzzle I am having with the circle since complex plane are actually 2 half circles, how do we get them divided... One more thing is that there is also negative region on the left that havent been accessible to tetration , I understand. Is it not accesible also to superroots or x^(1/x) - since any root has as many values as x real , imaginary, positive - where do we see these many values? what if x= e? x=i? How many roots should we have? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/18/2008, 09:19 AM andydude Wrote:@Ivars: First of all those are not my definitions, they are Knoebel's. His paper, called "Exponentials Reiterated" is an amazing exposition of the infinitely iterated exponential. It covers almost all aspects of it, and mentions many aspects of tetration as well. Andrew Robbins I can not access that one. I have made an oath never to pay for an article even if i am in dire need for it. Makes investigations more chalenging. I just can not afford it. I buy books, but never online articles. Sometimes, I would mail the author asking for a copy, but usuallly those guys are not very responsive understandably, or already dead. Ivars GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/20/2008, 04:30 PM (This post was last modified: 01/20/2008, 04:32 PM by GFR.) andydude Wrote:...I just found another point (probably found before) on the outer branches, one that has many advantages over the point of intersection. This point is $(x=\frac{1}{2}, b=\frac{1}{16})$ which is 2-periodic in the sense: $(1/16)^{(1/2)} = (1/4)$ $(1/16)^{(1/4)} = (1/2)$ as can be seen in this plot: [attachment=198] Andrew Robbins I found a similar result with "my" approximation, including your two 2-peridic points, obviously referred to a different base. Interesting! @ IVARS Please see the attachment, in which I try to imagine the situation represented in a 3-dimensional environment (y,b,x). The plots of y=b#x are the (appromimated and simulated) real projectionns on the yx plane of the y=d#x complex helicoidal "functions". The transition (yellow, off-limit) zone on the yb diagram is (in my ... simple opinion) the graph of the max-min span of the undetermined y values, when x -> oo. Persintent oscillations. Difficult to describe, ... see the plots. GFR Attached Files   Approx.pdf (Size: 91.6 KB / Downloads: 280) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/22/2008, 09:46 PM GFR Wrote:@ IVARS Please see the attachment, in which I try to imagine the situation represented in a 3-dimensional environment (y,b,x). The plots of y=b#x are the (appromimated and simulated) real projectionns on the yx plane of the y=d#x complex helicoidal "functions". The transition (yellow, off-limit) zone on the yb diagram is (in my ... simple opinion) the graph of the max-min span of the undetermined y values, when x -> oo. Persintent oscillations. Difficult to describe, ... see the plots. GFR I was studying it so long but its still above my head. I wanted to see oscillating behaviour as n=infinity (both odd and even) in region x infinity are continuous functions, if we infinitely backtetrate them , there has to be a distance on x axis between the results, but its infinitesimal. So the spiral ,or circle in perpendicular plane will consist of infinitely close planes , but nevertheless will have distinctly separate endpoints in plane xy as points of continuos functions h odd (x) h even x . So in real plane xy we do not see any difference between x for 2 neighbouring values of infinite tetration 2n-> infinity and 2n-1 -> infinity, but we know there must be an infinitesimal difference so perhaps it is visible in perpendicular space as a direction of rotation of each of these spirals - i can not imagine any other way left to differentiate them. I guess i was not clear enough, but do not have the skills to make or post a picture. In the end , it should be possible to explain such things in plain words though it might take a lot of space GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/26/2008, 03:52 PM Dear Ivars! Concerning: Ivars Wrote:Ref.: GFR Wrote:@ IVARS Please see the attachment, in which I try to imagine the situation represented in a 3-dimensional environment (y,b,x). The plots of y=b#x are the (appromimated and simulated) real projections on the yx plane of the y=d#x complex helicoidal "functions". The transition (yellow, off-limit) zone on the yb diagram is (in my ... simple opinion) the graph of the max-min span of the undetermined y values, when x -> oo. Persintent oscillations. Difficult to describe, ... see the plots.I was studying it so long but its still above my head. I wanted to see oscillating behaviour as n=infinity (both odd and even) in region x infinity are continuous functions, if we infinitely backtetrate them , there has to be a distance on x axis between the results, but its infinitesimal. .....Ops, ... sorry again! I shall try now to be as clear as possible, but I believe that the "plots" are more explicit than any linguistic explanation, particularly when the available terminology is not final and agreed upon, as in our present situation. Nevertheless, let's try it, hoping not to create furher confusion. :< In my comments, I chose to start defining y = b # x as the tetrational function to be analyzed (b-tetra-x), for 0 < b < e^(-e) [i.e.: in 0 ..... Beta, where Beta is 0.06598036..]. Generally, it is a ternary relation among three variables (y, b, x) or a binary operation (y, x) with parameter b. Moreover, if parameter b is a positive constant number (for ... any constant b > 0), y = b # x is (or .. we hope that it is) a function y = f(x), which we expect to be ... continuous and smooth. :> Having said that, let me try to fix our ideas in examining the y = b#x "function": - x: is the independent variable, the "prototype" of which could be ... the independently running "time"; - b: is the tetration "base", provisionally considered as a constant, similarly to what we admit when we study y = b ^ x; - y: is the result of the operation (or the "function"); - h: is used in the literature to indicate what I am calling "the infinite tower height" i.e.: h = (x->+oo)lim(b # x) = b # oo; we must observe that b # +oo can be real or complex, finite or infinite (variable with b), while b ^ +oo is always +oo, as far as I know, for b>0. ;> In some cases, as we may see, it can be non-determinated, between two extreme values. Well, the latter case describes exactly the situation when 0 < b < Beta [b = e^(-e)]. In this framework, we discovered that h = b#oo is itself a "function" of b, inverse of b = y-root(y), with complex solutions for b > Eta, double real solutions (a two-valued real function) for 1 < b < Eta, one real solution for Beta < b < 1 and undetermined values in 0 < b < Beta. Actually, in 0 < b < Beta, we know three values of h, which we might identify as: - h/mid = y/mid = - plog[-ln(x)/ln(x); - h/sup = (x->+oo)lim[y/sup]; - h/inf = (x->+oo)lim[y/inf]. Then, attention, please! What I mean is that: I believe that function y = b # x, for b in the 0 .... Beta domain, oscillates, with oscillations between y/inf and y/sup, asymptotically decreasing towards h/inf and h/sup. These decreasing values are always and only verified for integer values of x (odd and even, respectively). The "jumps" among them are justified by the reationship: y(x+1) = b^y(x) and are detectable at a minimum distance of one unit in the x axis. Also: the h/inf and h/sup are the asymptotic values of y/inf and y/sup, respectively, for x -> +oo. Think of a similar situation of a sinus function having, for x -> +oo, a residual oscillation around 0, between h/inf and h/sup. Its asymptotic behaviour, for x -< +oo, could be described as being undeterminad between h/inf and h/sup, with an average at y = 0. Non-orthodox, but true ! As you see, no infinitesimal distance on x axis between the results should, in any case, appear. Great Euler should have said: "Functio non facit saltus". But, probably he never did it. This I what I was trying to say. As somebody once said: "Please pay attention to it, because it might be interesting, but dont take it for granted, because ... I might be wrong!" GFR GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 03/02/2008, 04:18 PM andydude Wrote:About GFR's "yellow zone" (YZ) or Ivars' "phase transition region" I was very interested to see the formula $x = (y^{-y} -1)\frac{e^{-e}}{e^{1/e}-1}$. I think it is a very good approximation, but I think it is not accurate. Sorry, Andydude, you are right! My formula is wrong, because it is only a very approximated .... approximation. The only way, I think, is to take into consideration [/b]log x = b^x. I shall come back to that asap. Sorry again! GFR bo198214 Administrator Posts: 1,380 Threads: 90 Joined: Aug 2007 03/03/2008, 09:01 PM (This post was last modified: 03/03/2008, 09:21 PM by bo198214.) GFR Wrote:- h: is used in the literature to indicate what I am calling "the infinite tower height" i.e.: h = (x->+oo)lim(b # x) = b # oo; we must observe that b # +oo can be real or complex, finite or infinite (variable with b), while b ^ +oo is always +oo, as far as I know, for b>0. Hey Gianfranco, forgive me, but I have to correct: $b^\infty=0$ for $-1 $1^\infty=1$ $b^\infty=\infty$ for $b>1$. And that we have two limit points in the case $b=-1$: $(-1)^\infty=\pm 1$; and two limit "points" for $b<-1$: $b^\infty=\pm\infty$. And there is really some similarity between exponentiation and tetration. The first similarity is when considering odd/even exponents. $x^n$ and ${^n}x$ is bijective exactly for odd exponents $n$ (assuming that the domain is $\mathbb{R}$ for $x^n$ and $\mathbb{R}_+$ for ${^nx}$, i.e. we map $-\infty\mapsto 0$, $0\mapsto 1$). And this similarity continues to the limits, where we would map $-\infty\mapsto 0$, $-1\mapsto e^{-e}$, $0\mapsto 1$, $1\mapsto e^{1/e}$: ${^\infty b}$ has two limit points for $0 ${^\infty b}$ has one limit for $e^{-e}\le b\le e^{1/e}$ ${^\infty b}=\infty$ for $b>e^{1/e}$. The difference in this analogy is that the behaviour directly at the borders is different. So ${^\infty \left(e^{-e}\right)}$ has only one limit while $(-1)^{\infty}$ has two limit points and the function ${^\infty}x$ is continuous in $x=e^{1/e}$, i.e. $\lim_{x\uparrow e^{1/e}}{^\infty x}={^\infty(e^{1/e})}$ while the function $x^{\infty}$ is not continuous in $x=1$, i.e. $\lim_{x\uparrow 1} x^\infty =0\neq 1= 1^\infty$. Quote:Then, attention, please! What I mean is that: I believe that function y = b # x, for b in the 0 .... Beta domain, oscillates, with oscillations between y/inf and y/sup, asymptotically decreasing towards h/inf and h/sup. These decreasing values are always and only verified for integer values of x (odd and even, respectively). If we extend this analogy and asking ourselves why $b^x$ has two limit points for $b< -1$ then the answer is not that $b^x$ oscillates for $b\le -1$ but that it takes complex values there which just happen to be alternating real for natural $x$. I pointed out a similar phenomenon for the function ${^x}b$ somewhere already. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/16/2008, 08:40 AM (This post was last modified: 03/16/2008, 08:41 AM by Ivars.) This value, if correct, seems to fall in the region b

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