Attempt to find a limit point but each step needs doubling the precision...

[tommy1729]

[3]: In the curve at the right hand there is the smooth thicker red curve. That is the trajectory beginning from \( y_{-4} \) in the near of 3.1 . Tetrating with increasing purely imaginary height it first ascends and then descends further to the right side, but does not arrive at the real axis (which is however expected to happen when the height is exactly \( h_1 = \pi*I/\log(\log(2)) \). But the proceeding of the trajectory becomes radically stuffed: you are at the final height h_1 by about 99.9 % and still one cm away from the real axis. And if you go to 100.1 % you find the same value but below the real axis. So you conclude, the value at the final hight might be just in the middle. But that's not true. If you go nearer to \( h_1 \) to 100%-1e-6% there is another remarkable step with a horizontal component. After that, the default internal precision of the software (I do with 200 digits by default) does not suffice to improve the computation. In fact, one can proceed when 400 digits are used to come nearer to the 100%-level for the height, then 800 digits, then 1600 digits precision and the height can then approach \( h_1 - 1e-400 \) - and one can see the tiny black circles with which I've marked that results. But obviously this type of computation cannot be stretched far more to see only a shadow of the true limit ... - conclusion: one needs an analytical approach (but I don't know how...)

If this explanations do not suffice, please ask for more.

Gottfried

Oh quantum tunneling

First guess is this relates to the radius of convergeance of the Taylor series expanded at 2 for the half-iterate.

But thats a wild guess.

[2] was clear to me after reading a second time, just saying to give you back some confidence. There was only 1 logical interpretation
Although the formal way of math does not need those kind of puzzles of course.

Seems we have work to do once again.
It seems the amount of problems about tetration grow like sexp and the number of solutions like slog , however both are rising functions and I like both questions and answers

regards

tommy1729