complex iteration (complex "height")

[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

[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 \( 1<b<\eta \):
   

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
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]

@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

[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]

@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]

[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]

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]

@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

x-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
   

[Ivars]

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]

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.