Uniqueness of tetration. Small lemma.

[sheldonison]

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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?
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Otherwise it would seem likely that there would be other singularities, outside of the real axis.

What does it mean, "based on the fixed point"? Do you mean the case of other base?

I was refering to the base e solution.

Khoustenov earlier wrote:

\( \mathrm{tem}(z)=\mathrm{tet}(J(z)) \)
where \( J(z)=z+a\cdot\sin(2 \pi z) \)

In your equation, you have the original tet(z) and you also have tem(z)=tet(J(z)). When I said "solution based on a fixed point", I was referring to tet(z), the "correct" solution, rather than the modified tem(z) equation, with the extra singularities.

For tet(z), is the only contour line where \( \Im(\text{tet}(z))=0 \) at the real axis, for z>-2?
Just realized this cannot be true, since \( \Im(e^{(\pi*k*i)})=0 \). So when the imaginary part >=pi, contours where the imaginary component is zero arise.


For the modified equation tem(z)=tet(J(z)), the fractal copies of the real axis (including the singularities) occur where \( \Im(\text{tem}(z))=0 \), and these occur where there is a contour of \( \Im(J(z))=0 \). Wherever there is a contour line for \( \Im(J(z))=0 \) outside of the real axis, a distorted copy of the original tet(z) real axis gets generated.

All of this seems to imply that the contours of \( \Im(\text{tem}(z))=0 \) (outside the real axis) have much more to do with the particular one-cyclic \( \theta(z) \) function, where \( J(z)=z+\theta(z) \), then with tet(z) itself.
- Sheldon