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 complex iteration (complex "height") Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 11/14/2007, 09:32 AM (This post was last modified: 11/14/2007, 02:42 PM by Gottfried.) Here I add two plots, which show bases s=2 and base s=sqrt(2), when tetrated to complex heights h. The height-parameter h follows the border of the complex unit-disk, so passes the coordinates (1,0),(0,i),(-1,0),(0,-i ) and is exp(2*Pi*I*x) . Also I have added 4 different scalings of h by the zoom-factor g, where g=1 (h is on the complexunit-circle), g=0.75 (h has radius 0.75), g=0.5 and g=0.25. This is a small picture for s=2:     and this a small picture for s=sqrt(2):     The curves for value g=1.25 (radius of h=1.25) are partly erratic; a better approximation-method should be able to smooth them a bit. Gottfried Attached Files Image(s)         Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/21/2008, 05:21 PM (This post was last modified: 03/21/2008, 05:23 PM by bo198214.) @Gottfried: By which method did you compute your values? Just triggered by the question about b[4]I, I computed it by $b[4]t=\exp_b^{\circ t}(1)$ with $\exp_b^{\circ t}$ being the regular iteration at the lower fixed point of $b^x$ (via the formula given here) for various bases $1:     This picture has also the disadvantage that you dont see which base $b$ is associated to which point on the curve. So I add a list of values: $b$ , $b[4]I={exp_b}^{\circ I}(1)$ Code:1.01, 1.011233887 - 0.01003728120 I 1.02, 1.035202770 - 0.01405751590 I 1.03, 1.059991495 - 0.01057562160 I 1.04, 1.083268505 - 0.002332372600 I 1.05, 1.104485443 + 0.009021821700 I 1.06, 1.123581670 + 0.02245648730 I 1.07, 1.140655009 + 0.03729317620 I 1.08, 1.155854058 + 0.05306607940 I 1.09, 1.169338195 + 0.06944512180 I 1.10, 1.181261981 + 0.08619082190 I 1.11, 1.191769271 + 0.1031263661 I 1.12, 1.200991438 + 0.1201195506 I 1.13, 1.209047362 + 0.1370706818 I 1.14, 1.216044168 + 0.1539042223 I 1.15, 1.222078233 + 0.1705628566 I 1.16, 1.227236250 + 0.1870032102 I 1.17, 1.231596256 + 0.2031926760 I 1.18, 1.235228576 + 0.2191070431 I 1.19, 1.238196668 + 0.2347287000 I 1.20, 1.240557891 + 0.2500452354 I 1.21, 1.242364161 + 0.2650483724 I 1.22, 1.243662552 + 0.2797331122 I 1.23, 1.244495805 + 0.2940970757 I 1.24, 1.244902798 + 0.3081399627 I 1.25, 1.244918930 + 0.3218631346 I 1.26, 1.244576497 + 0.3352692637 I 1.27, 1.243904987 + 0.3483620634 I 1.28, 1.242931367 + 0.3611460599 I 1.29, 1.241680334 + 0.3736264086 I 1.30, 1.240174520 + 0.3858087482 I 1.31, 1.238434699 + 0.3976990769 I 1.32, 1.236479952 + 0.4093036533 I 1.33, 1.234327829 + 0.4206289147 I 1.34, 1.231994485 + 0.4316814110 I 1.35, 1.229494806 + 0.4424677512 I 1.36, 1.226842509 + 0.4529945543 I 1.37, 1.224050270 + 0.4632684218 I 1.38, 1.221129787 + 0.4732959017 I 1.39, 1.218091867 + 0.4830834635 I 1.40, 1.214946518 + 0.4926374870 I 1.41, 1.211702993 + 0.5019642400 I 1.42, 1.208369868 + 0.5110698727 I 1.43, 1.204955087 + 0.5199604031 I 1.44, 1.201466009 + 0.5286417061 I There is also this interesting base $b>1$ for which $b[4]I$ is real. Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 03/22/2008, 01:01 AM bo198214 Wrote:@Gottfried: By which method did you compute your values? Hmm, for whatever reason I didn't find the pari.gp-syntax. I used the diagonalization and took complex powers for the diagonal. Maybe I did it using the square-bell-matrices directly or already using the shifting-to-triangular method, don't remember. What I have are the data for the plots in Excel. Since I'm busy tomorrow (tibet-action) I'll not be able to reproduce the syntax, however I could send you the data if this helps. Everything else I can do on sat evening or sunday, let's see. Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/22/2008, 09:15 AM (This post was last modified: 03/22/2008, 09:32 AM by Ivars.) @Henryk The next questions to provoke further thinking is : b[4](i*pi/2)=b[4]ln(i)? b[4](i/2)=? b[4](-i)=? and at what exact value of b is b[4]i purely real by Your method? @Gottfried What is on axis? What is k?The relation between k and x? Why there is k/64 in expression for h? ... Interesting anyway , Your pictures always seem to contain a lot of information. Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/22/2008, 11:59 AM Ivars Wrote:@Henryk The next questions to provoke further thinking is : b[4](i*pi/2)=b[4]ln(i)? b[4](i/2)=? b[4](-i)=? As I am not so interested in particular values perhaps you need to stretch your own brain. The formula I used was: $f^{\circ t}(x)=\lim_{n\to\infty} f^{\circ -n}(a(1-q^t) + q^t f^{\circ n}(x))$, $q=f'(a)$. where $a$ is the attracting fixed point and $f$ is the function to be iterated. Specialized to our case using the lower fixed point $a$ we get the base by $b=a^{1/a}$ and $f'(a)=(b^x)'(a)=\ln(b)b^a=\ln(b)a=\ln(b^a)=\ln(a)$, so: $b[4]z=\exp_b^{\circ z}(1)=\lim_{n\to\infty} \log_b^{\circ n}(a(1-\ln(a)^z) + \ln(a)^z \exp_b^{\circ n}(1))$ bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/22/2008, 12:10 PM Gottfried Wrote:[What I have are the data for the plots in Excel. Since I'm busy tomorrow (tibet-action) I'll not be able to reproduce the syntax, however I could send you the data if this helps. Everything else I can do on sat evening or sunday, let's see. I was only interested whether our results match. But one couldnt see from the graph what $\sqrt{2}[4]I$ is, so I would be glad if you could compare this with my result of $\sqrt{2}[4]I=1.210309011+.5058275611*I$ Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 03/22/2008, 03:25 PM (This post was last modified: 03/22/2008, 11:45 PM by Gottfried.) bo198214 Wrote:I was only interested whether our results match. But one couldnt see from the graph what $\sqrt{2}[4]I$ is, so I would be glad if you could compare this with my result of $\sqrt{2}[4]I=1.210309011+.5058275611*I$ Yepp, that's exactly the value that I've got for height h=I (up to 7'th digit) In my excel-file it is real=1.2103090 imag = 0.50582757 [update] with higher precision I recomputed the value and got in the 92..96 partial-sums the following approximations: Code:´   1.210309025559961+0.5058275713618201*I   1.210309025559961+0.5058275713618201*I   1.210309025559961+0.5058275713618201*I   1.210309025559961+0.5058275713618201*I   1.210309025559961+0.5058275713618201*I ---------------------------------------------   1.210309011      + .5058275611          *I  <--- yours So this differs from the 7'th digit; anyway - I just computed this with the function U_t (x) = t^x - 1 T_b(x) = b^x U_t°h(x) and T_b°h(x) as their iterates of general height h and the shift T_b°h(x) = (U_t°h(x/t-1) +1)*t and the height-iteration by applying powers to the diagonal of the diagonalized operator for U_t(x) with 96 terms, using b=sqrt(2),t=2,x=1 If your values have full accuracy, then there must be a methodical difference, which I would like to find; I'll check today with other fixpoint-shifts. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 03/22/2008, 03:32 PM (This post was last modified: 03/22/2008, 03:47 PM by Gottfried.) Ivars Wrote:@Gottfried What is on axis? What is k?The relation between k and x? Why there is k/64 in expression for h? ... Interesting anyway , Your pictures always seem to contain a lot of information. Ivarsx-Axis is real, y-axis imaginary value of the result. The complex circle of radius |1| was divided in 64 steps. indicated by k as the number, so each k represents 1/64 of the circumference of the complex unit-circle. But this is only using g=1. I added the other results for equivalent parameters on the complex circles of smaller radii too. So the blue line shows the values for sqrt(2)^^(exp(2*Pi*I*k/64)), k=0..64 the green line for sqrt(2)^^(0.75*exp(2*Pi*I*k/64)), k=0..64 and so on. Here is an updated plot, where the points for h=I (the 16'th point on the blue line) , h=-1 (32'th point) , h=-I (48'th point) are marked     Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/22/2008, 07:45 PM bo198214 Wrote:As I am not so interested in particular values perhaps you need to stretch your own brain. The formula I used was: $f^{\circ t}(x)=\lim_{n\to\infty} f^{\circ -n}(a(1-q^t) + q^t f^{\circ n}(x))$, $q=f'(a)$. where $a$ is the attracting fixed point and $f$ is the function to be iterated. Specialized to our case using the lower fixed point $a$ we get the base by $b=a^{1/a}$ and $f'(a)=(b^x)'(a)=\ln(b)b^a=\ln(b)a=\ln(b^a)=\ln(a)$, so: $b[4]z=\exp_b^{\circ z}(1)=\lim_{n\to\infty} \log_b^{\circ n}(a(1-\ln(a)^z) + \ln(a)^z \exp_b^{\circ n}(1))$ My brain is already overstretched... I think I need some software that can calculate infinitely many logarithms Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/22/2008, 07:49 PM Ivars Wrote:My brain is already overstretched... I think I need some software that can calculate infinitely many logarithms Ya, software might be quite helpful, however no software will compute infinitely many logarithms. Its just a question of approximation. « Next Oldest | Next Newest »

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