04/14/2009, 01:20 AM

sheldonison Wrote:In the tetration curve, the imaginary value for f(z) at the real axis is equal to zero for all z>=-2. For the solution based on the fixed point, is the imaginary value of f(z) nonzero everywhere outside of the real axis?No.

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I was refering to the base e solution.

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For tet(z), is the only contour line where at the real axis, for z>-2?

The part of the real axis is not the only contour where .

There are many of them in the right hand side of the complex plane.

Both real and imaginary parts of have huge values in vicinity of the real axis. Many times the imaginary part passes through the values that are integer factors of

Each line , at the unity translation , produces the line ,

These lines form the complicated structure.

The fractal of lines is shown in fig.2 at

http://www.ils.uec.ac.jp/~dima/PAPERS/2009fractal.pdf

sheldonison Wrote:Just realized this cannot be true, since . So when the imaginary part >=pi, contours where the imaginary component is zero arise.Yes.

sheldonison Wrote:For the modified equation tem(z)=tet(J(z)), the fractal copies of the real axis (including the singularities) occur where , and these occur where there is a contour of . Wherever there is a contour line for outside of the real axis, a distorted copy of the original tet(z) real axis gets generated.Yes. In the paper mentioned, the example with modified tetration is also considered.

However, we have no need to modify the tetration in order to have lines

outside the real axis, but these lines are at .

sheldonison Wrote:... the contours of (outside the real axis) have much more to do with the particular one-cyclic function, where , then with tet(z) itself.Yes.

The multiple reproduction of reduced and deformed patterns is not a specific property of a tetration.

Take any other function with beautiful pattern in the complex plane, modify its argument with function J , and you get the fractal of the modified patterns at the plot.