complex iteration (complex "height") - Printable Version +- Tetration Forum (https://tetrationforum.org) +-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3) +--- Thread: complex iteration (complex "height") (/showthread.php?tid=87) |
complex iteration (complex "height") - Gottfried - 11/14/2007 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 RE: complex iteration (complex "height") - bo198214 - 03/21/2008 @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 \( 1<b<\eta \): 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 There is also this interesting base \( b>1 \) for which \( b[4]I \) is real. RE: complex iteration (complex "height") - Gottfried - 03/22/2008 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 RE: complex iteration (complex "height") - Ivars - 03/22/2008 @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 RE: complex iteration (complex "height") - bo198214 - 03/22/2008 Ivars Wrote:@Henryk 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: \( b[4]z=\exp_b^{\circ z}(1)=\lim_{n\to\infty} \log_b^{\circ n}(a(1-\ln(a)^z) + \ln(a)^z \exp_b^{\circ n}(1)) \) RE: complex iteration (complex "height") - bo198214 - 03/22/2008 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 \( \sqrt{2}[4]I=1.210309011+.5058275611*I \) RE: complex iteration (complex "height") - Gottfried - 03/22/2008 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 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: ´ 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 RE: complex iteration (complex "height") - Gottfried - 03/22/2008 Ivars Wrote:@Gottfriedx-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 RE: complex iteration (complex "height") - Ivars - 03/22/2008 bo198214 Wrote:As I am not so interested in particular values perhaps you need to stretch your own brain. The formula I used was: My brain is already overstretched... I think I need some software that can calculate infinitely many logarithms Ivars RE: complex iteration (complex "height") - bo198214 - 03/22/2008 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. |