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 Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/23/2008, 06:51 AM Hello, Henryk In Chaos Pro, is there a way to find coordinates [x,y] of an interesting region/point? So far I could only use zooming the are which was fine for e^pi/2, e^-e, e^(1/e) , e^pi/2 which are rather distinct points with interestingly different behaviour of iteration z_0=I, z=pixel^z in their neighborouhood, ( I have placed the pictures in the thread) but that is very time consuming if points are less obvious and e.g. off real axis. Imaginary axis is almost invisible. Also, could You look into my question in the thread, why (z^(z(^z(.............z^I) = h(z) (seems to be) where it converges on real axis? And would mapping the trajectory of this convergence give some additional information about the point compared to just computing it step by step on real axis? My interest would be to perform continuos iteration, of course, to see the full trajectory. I did discrete one for $a={1/e}$ leading to $h(a, I) = \Omega$ and the behaviour seems chaotic, but there are not enough points ( 10 integer iterations only). Perhaps any finite z on top of iterations of a^(a^(a^.....z) will lead to h(a) via different trajectories. Thank You in advance, Ivars bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 05/23/2008, 11:44 AM Ivars Wrote:In Chaos Pro, is there a way to find coordinates [x,y] of an interesting region/point? So far I could only use zooming the are which was fine for e^pi/2, e^-e, e^(1/e) , e^pi/2 which are rather distinct points with interestingly different behaviour of iteration z_0=I, z=pixel^z in their neighborouhood, ( I have placed the pictures in the thread) but that is very time consuming if points are less obvious and e.g. off real axis. Imaginary axis is almost invisible. Unfortunately I dont know either. Perhaps one can apply some tickmarks via the formula, if real and imaginary part are integer then draw a red dot or something like that. But it should be possible with intrinsic methods in ChaosPro. If you use Fractal Explorer the complex coordinates of the current mouse position are shown in the status bar, however fractal explorer is not that sophisticated as ChaosPro and magnitudes slower. If you consider the limit $a$ of $z_{n+1} = b^{z_n}$ then it is clear that $b^a=a$. However the sequence would only converge to an attracting fixed point and the only attracting fixed point of $b^z$ is the lower real fixed point, which exists only for $e^{-e}. Quote:Perhaps any finite z on top of iterations of a^(a^(a^.....z) will lead to h(a) via different trajectories. No, it does not converge for all starting values $z_0$, for example $\lim_{n\to\infty} z_n = \infty$ for $z_0>4$ and $b=\sqrt{2}$. But if it converges then indeed $\lim_{n\to\infty} z_n = h(b)$. Quote:And would mapping the trajectory of this convergence give some additional information about the point compared to just computing it step by step on real axis? *shrug* Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/24/2008, 08:13 AM I checked with the routine in PariTTY most kindly provided by Gottfried to me few trajectories of interesting points on real axis from fractal picture, $e^{-\pi/2}=I^I$. If we look for $h(e^{-\pi/2})$ without I on top, it is just real number 0.4745409995126511230174679441, while iterating with I on top obviosuly leads to oscillating behaviour since first iteration leads to $(I^I)^I=I^{-1}= -I.$ So I am looking for (a^(a^.........(a^I)) where $a= e^{-\pi/2}=0.2078795763507619085469556198$ mybase = exp(-Pi/2) %27 = 0.2078795763507619085469556198 ffrac_init(1.0*I,mybase) ffrac_init(1.0*I,mybase) %28 = 1.000000000000000000000000000*I for(k=1,20,print(ffrac)) for(k=1,20,print(ffrac)) -2.524354897 E-29 - 1.000000000000000000000000000*I -2.524354897 E-29 + 1.000000000000000000000000000*I -7.57306469 E-29 - 1.000000000000000000000000000*I -7.57306469 E-29 + 1.000000000000000000000000000*I -2.019483917 E-28 - 1.000000000000000000000000000*I -2.019483917 E-28 + 1.000000000000000000000000000*I -5.301145284 E-28 - 1.000000000000000000000000000*I -5.301145284 E-28 + 1.000000000000000000000000001*I -1.337908095 E-27 - 1.000000000000000000000000001*I -1.337908095 E-27 + 1.000000000000000000000000002*I -3.332148464 E-27 - 1.000000000000000000000000002*I -3.332148464 E-27 + 1.000000000000000000000000005*I -8.25464051 E-27 - 1.000000000000000000000000005*I -8.25464051 E-27 + 1.000000000000000000000000013*I -2.039678757 E-26 - 1.000000000000000000000000013*I -2.039678757 E-26 + 1.000000000000000000000000032*I -5.036088018 E-26 - 1.000000000000000000000000032*I -5.036088018 E-26 + 1.000000000000000000000000079*I -1.242739915 E-25 - 1.000000000000000000000000079*I -1.242739915 E-25 + 1.000000000000000000000000195*I But perhaps this corresponds to case You mentioned, that iterations with z on top of a does not converge. Ivars bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 05/24/2008, 08:55 AM (This post was last modified: 05/24/2008, 08:56 AM by bo198214.) But to see this oscillating behaviour between I and -I you dont need to do numerical computations, as you already mentioned: $b^I=(I^I)^I=I^{-1}=-I$ and the next step is to look at $b^{-I}=(I^I)^{-I}=I^1=I$ Hence this oscillating behaviour for base $b=I^I$. And yes this is one non-converging case. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/24/2008, 04:09 PM (This post was last modified: 05/24/2008, 04:13 PM by Ivars.) Then it seems, that for each real point in the region of convergence of h(a) we can check regions of z where convergence still holds. That should give a deeper view, perhaps split the region $(e^{-e} , e^{1/e})$ into subregions with similar behaviours? As You said, $\sqrt2$ does not converge if z>4, while I have checked that $e^{-\pi/2}$ does not converge at $z=I$. So how would it be possible to map such values of z = f(a) where h(a) with z on top diverges and where converges? I can imagine that requires some 3D picture, with a on vertical axis, and z regions on complex plane drawn at every a. But I do not know how to make such calculations for all a and all z simultaneously and plot it. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/25/2008, 08:28 AM While studying this, I have become little puzzled about the way $h(I)$ converges. When iterated from z_0=I with a any precision available in PARI, it shows a tri-cycle behaviour, as do many other imaginary numbers (e.g 2*I, 3*I etc. ), if I have made the calculations right. I know the analytic expression for convergence of $h(I)$ based on Lambert function, but the way $I^{I^{I^{I^{I...}}}}$ approaches fixed point seems not one that could ever stop having sharp angles . If it would be a spiral type behaviour, looping , than it would be easier to accept. Even 2 point oscillating approach would seem strange to accept as tending to limit value and truly stopping oscillations. ? Ivars Gottfried Ultimate Fellow Posts: 789 Threads: 121 Joined: Aug 2007 05/25/2008, 10:48 AM (This post was last modified: 05/25/2008, 10:50 AM by Gottfried.) Ivars Wrote:While studying this, I have become little puzzled about the way $h(I)$ converges. When iterated from z_0=I with a any precision available in PARI, it shows a tri-cycle behaviour, as do many other imaginary numbers (e.g 2*I, 3*I etc. ), if I have made the calculations right.Hi Ivars - I've discussed this in some initial state earlier here, but I'm currently with my head elsewhere so too lazy to find the thread. Here are some plots of that tri-furcation, see uncommented list below. I find the bi-,tri- and multifurcation an interesting subject, as we ask: can we assign an individual value if the iteration oscillates/is furcated - since this is somehow related to the partial evaluation of non-convergent oscillating series, to which we assign a value anyway. But I don't have a special conclusion for this matter, yet, which were worth to write it down here. Gottfried http://go.helms-net.de/math/tetdocs/traj..._165_I.png http://go.helms-net.de/math/tetdocs/traj..._175_I.png http://go.helms-net.de/math/tetdocs/trajectories_I.png http://go.helms-net.de/math/tetdocs/traj...rcated.png http://go.helms-net.de/math/tetdocs/trif...on_1_2.png http://go.helms-net.de/math/tetdocs/trif...on_1_8.png http://go.helms-net.de/math/tetdocs/trif...quator.png http://go.helms-net.de/math/tetdocs/trif..._bases.png http://go.helms-net.de/math/tetdocs/trif...detail.png http://go.helms-net.de/math/tetdocs/trif...bases1.png Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/25/2008, 06:08 PM (This post was last modified: 05/25/2008, 09:38 PM by Ivars.) Gottfried Wrote:Here are some plots of that tri-furcation, see uncommented list below. I find the bi-,tri- and multifurcation an interesting subject, as we ask: can we assign an individual value if the iteration oscillates/is furcated - since this is somehow related to the partial evaluation of non-convergent oscillating series, to which we assign a value anyway. Yes, that is an interesting idea, that these seemingly convergent iterations are actually divergent but get the value in the same way like e.g. series 1-1+1-1..........= 1/2. What then, one would assign to the point $e^{-\pi/2}$ which when iterated with $z=I$ on top oscillates between $+I$ and $- I$ ? From complex geometric series , would we have to assign value to that Iteration by analogy with divergent (?) sum: $1/(1+I)=I{^0}+I{^1}-I{^2}+I{^3}-I{^4}+..............=1+I+1-I-1...$ ${1/(1+I)} =1/2-I/2$ whose module is ${\sqrt2/2}$ and argument $-\pi/4$, so value would be: ${(\sqrt2/2)}*e^{-I*\pi/4}$ No, the sum is not the same as $I-I+I-I...$. First I have to generate such sum where only odd powers of I are present. Ivars Gottfried Ultimate Fellow Posts: 789 Threads: 121 Joined: Aug 2007 05/25/2008, 10:12 PM Ivars Wrote:Gottfried Wrote:Here are some plots of that tri-furcation, see uncommented list below. I find the bi-,tri- and multifurcation an interesting subject, as we ask: can we assign an individual value if the iteration oscillates/is furcated - since this is somehow related to the partial evaluation of non-convergent oscillating series, to which we assign a value anyway. Yes, that is an interesting idea, that these seemingly convergent iterations are actually divergent but get the value in the same way like e.g. series 1-1+1-1..........= 1/2. What then, one would assign to the point $e^{-\pi/2}$ which when iterated with $z=I$ on top oscillates between $+I$ and $- I$ ? From complex geometric series , would we have to assign value to that Iteration by analogy with divergent (?) sum: $1/(1+I)=I{^0}+I{^1}-I{^2}+I{^3}-I{^4}+..............=1+I+1-I-1...$ ${1/(1+I)} =1/2-I/2$ whose module is ${\sqrt2/2}$ and argument $-\pi/4$, so value would be: ${(\sqrt2/2)}*e^{-I*\pi/4}$ No, the sum is not the same as $I-I+I-I...$. First I have to generate such sum where only odd powers of I are present. IvarsHmm, for 1+I-1-I+1+I-1-I+.... I get 1/(1-I) = 1/2 + 1/2I by the formula just invoking Pari/GP on that, and also 1/2 + 1/2I by conventional Euler-summation. For the second, to clarify the series you must assign an indexing-scheme, for instance powers of x. Did you mean: 0 + Ix +0x^2 - Ix^3 + 0 x^4 + I x^5 .... = Ix *(1/(1+x^2)) -> 1/2 I or I - Ix + Ix^2 - Ix^3 .... = I*(1/(1+x)) -> 1/2 I The limits should evaluate this way then (good to see in in the view of matrix-operations on formal powerseries, then the index is always clear under any transformation... ) Fortunately, they come out to be the same, but such equality may not be concluded only by the non-indexed (perhaps compressed by eliminating intermediate zeros) notation. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 789 Threads: 121 Joined: Aug 2007 05/25/2008, 10:51 PM (This post was last modified: 04/12/2014, 06:40 AM by Gottfried.) Refering to the header of this thread and your obvious love to iteration and trajectories, I challenge you to compute the alternating sum of all the intermediate steps of the trajectory I, 2^I, 2^(2^I), 2^(2^(2^I)... AS(2,I) = I - 2^I + 2^2^I - 2^2^2^I + ... -... = ?? My proposal is AS(2,I) = -0.440033096027+0.928380628227*I What do you think? Gottfried [update 4'2014] The value was computed using "regular tetration" which includes a fixpoint-shift of the exponential-series (which also leads to triangular matrices). When I wrote this posting I was not aware of the deviance of the evaluation when repelling or attracting fixpoints were chosen. A short explanation of this and a better/more meaningful estimate for this sum one can find in the mail from yesterday a couple of post below this one. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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