I reread old posts and the Trappmann/Kousnetzov[1] - article about the tetration with base sqrt(2). The point around which I'm fiddling is the imaginary-height iteration, which allows to relate real values from the interval (-oo to 2) with values from (2 to 4), so for instance \( x = -1 \) gives

\( Tet( x,\rho)=y \approx 2.66970618246 \) where \( \rho \) is \( -{\pi \cdot i \over \log(\log(2))} \)

It is now interesting, that the point \( x_{-2} \) , which is usually assumed in tetration as "at -oo", has as well an existing relative to it, which can simply be computed if we compute the relative

\( y_0 \) to \( x_0=1 \) by \( Tet(x_0,\rho)=y_0 \approx 2.46791405022 \)

and do integer Tetration to get

\( Tet(y_0,-2) = y_{-2} \approx 2.76432104000 \)

Here is a picture, where I graphed the trajectories for the imaginary heights from \( 0 \dots \rho \) in steps of, say \( 0.1 \cdot \rho \) or \( 0.01 \cdot \rho \) starting at some example points on the line below +2.

Even more interesting is now the question what happens for the points in the interval \( y_{-2} \dots 4 \) because they are related to somehow values larger than the negative infinity. There is the asymptotic vertical line at \( x=y_{-2} \) ; the trajectories starting from values greater than \( y_{-2} \), say from \( y_{-4} \) have a trajectory to the right side of that asymptote.

WHen I simply repeated my procedure for the computation of that trajectory from \( y_{-4} \) it looked as if it arrives at some value with positive imaginary part and then, when the imaginary height goes over \( \rho \) towards \( 2 \rho \) it does a jump to the conjugated values - with a discontinuity at \( h = \rho \) exactly. This is the thicker red line at the right hand side of the image. But what is the exact limit-point where the jump occurs?

Increasing the precision, with which I compute the tetration shows, that I can extend the trajectory towards some expected limit. Such improved computations allow the grey circles which extend the red trajectory.

But here my problem occurs: I need to double the precision to increase the length of the trajectory, and the last point was already computed with precision of 1600 internal decimal digits... at \( h= \rho \cdot (1 - 10^{-2^9}) \) (and further approximations of \( h \to 1 \) do not change the computed values with that given precision)

So that calls for a better analytical consideration of that trajectory. How can we express the limit depending on the \( h=\rho \cdot (1-\epsilon) \) where \( \epsilon=10^{- 2^k} \) ? (Remark : in Henryk's/Dmitrii's article[1] at page 13(pg 1739 of the printed journal) there is a much nicer and richer picture of that trajectories, but I cannot relate anything in their pictures to *that* trajectory).

Updated: Ahh, I should also mention that I use the powerseries around the lower (attracting) fixpoint \( t_0=2 \).

[1] PORTRAIT OF THE FOUR REGULAR SUPER-EXPONENTIALS TO BASE SQRT(2)

MATHEMATICS OF COMPUTATION

Volume 79, Number 271, July 2010, Pages 1727–1756

(Article electronically published on February 12, 2010)

\( Tet( x,\rho)=y \approx 2.66970618246 \) where \( \rho \) is \( -{\pi \cdot i \over \log(\log(2))} \)

It is now interesting, that the point \( x_{-2} \) , which is usually assumed in tetration as "at -oo", has as well an existing relative to it, which can simply be computed if we compute the relative

\( y_0 \) to \( x_0=1 \) by \( Tet(x_0,\rho)=y_0 \approx 2.46791405022 \)

and do integer Tetration to get

\( Tet(y_0,-2) = y_{-2} \approx 2.76432104000 \)

Here is a picture, where I graphed the trajectories for the imaginary heights from \( 0 \dots \rho \) in steps of, say \( 0.1 \cdot \rho \) or \( 0.01 \cdot \rho \) starting at some example points on the line below +2.

Even more interesting is now the question what happens for the points in the interval \( y_{-2} \dots 4 \) because they are related to somehow values larger than the negative infinity. There is the asymptotic vertical line at \( x=y_{-2} \) ; the trajectories starting from values greater than \( y_{-2} \), say from \( y_{-4} \) have a trajectory to the right side of that asymptote.

WHen I simply repeated my procedure for the computation of that trajectory from \( y_{-4} \) it looked as if it arrives at some value with positive imaginary part and then, when the imaginary height goes over \( \rho \) towards \( 2 \rho \) it does a jump to the conjugated values - with a discontinuity at \( h = \rho \) exactly. This is the thicker red line at the right hand side of the image. But what is the exact limit-point where the jump occurs?

Increasing the precision, with which I compute the tetration shows, that I can extend the trajectory towards some expected limit. Such improved computations allow the grey circles which extend the red trajectory.

But here my problem occurs: I need to double the precision to increase the length of the trajectory, and the last point was already computed with precision of 1600 internal decimal digits... at \( h= \rho \cdot (1 - 10^{-2^9}) \) (and further approximations of \( h \to 1 \) do not change the computed values with that given precision)

So that calls for a better analytical consideration of that trajectory. How can we express the limit depending on the \( h=\rho \cdot (1-\epsilon) \) where \( \epsilon=10^{- 2^k} \) ? (Remark : in Henryk's/Dmitrii's article[1] at page 13(pg 1739 of the printed journal) there is a much nicer and richer picture of that trajectories, but I cannot relate anything in their pictures to *that* trajectory).

Updated: Ahh, I should also mention that I use the powerseries around the lower (attracting) fixpoint \( t_0=2 \).

[1] PORTRAIT OF THE FOUR REGULAR SUPER-EXPONENTIALS TO BASE SQRT(2)

MATHEMATICS OF COMPUTATION

Volume 79, Number 271, July 2010, Pages 1727–1756

(Article electronically published on February 12, 2010)

Gottfried Helms, Kassel