[AIS] (alternating) Iteration series: Half-iterate using the AIS?

[sheldonison]

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Theta(z) has a radius of convergence that is slightly smaller, |imag(z)|<8.779i. The nearest singularity occurs when the asum(f^z) has a derivative of zero, where the acos(asum(z)/amplitude) has a singularity, but has a defined value. Here, where f'(z)=0, the acos(f^z/amplitude)= 0.445180168159228295119296 +8.77930110300507236428937*I. This is a simple branch singularity, but the function is starting to behave pretty poorly at this point, and there are many other singularities nearby, where the acos has singularities, or where the asum(f^z) has singularities, so it seems impossible to extend the function in any meaningful way much beyond this point in the complex plane....

My primary interest in the asum function was understanding its analytic limits, and how it compares to the other sueprfunctions. I chose b=2 as an example, since b^2-1 is analogous to tetration b=sqrt(2), which has been previously studied, including a paper published by Henryk and Dimitrii.

Here are two graphs. On the left, is the 1-cyclic theta(z) function at its analytic limit, 8.779i, showing the cusp of the branch singularity. On the right, is the corresponding asum of the superfunction, which is defined to be the \( \text{asum}(f^{[z+\theta(z)]})=\text{amplitude}*\cos(\pi\times z), \Im(z)=8.77930110300507236428937i \)
Notice, that the asum is a perfect 2-cyclic cosine, even though theta has a singularity. Real is graphed in red, imaginary is in green.
   

Above this point, if imag(z) is any larger, than the singularity becomes a discontinuity branch. At the left, is the asum superfunction over the same range as above, and on the right is the asum superfunction over a larger range, from -4 to 4. You can see the singularities, corresponding to the singularities in the theta(z) function.
   

Let f^z be the superfunction from the upper fixed point. As noted, the singularity occurs whenever the asum(f^z) has a zero derivative not at the real axis. Let a(z)=asum(f^z). Let s=singularity, where a'(z)=0, and s=a(z), imag(z)<>0. The singularity is caused since a^-1(s) is a branch point due to the zero derivative. There are an infinite number of other similar zero derivative singularities nearby, as the asum(f^z) function approaches the singularity posted earlier. In fact, these occur within im(z)=0.0153 of this closest singularity to the real axis, with the next singularity at approximately 0.44325 - 8.7899i. So this is as far as one can go in defining a continuous asum based superfunction for b=2, whose asum is exactly a multiple of the cosine function. Above this point, the asum(z) becomes an increasingly discontinous function with an infinite number of branch jumps required to maintain the asum(f^z)=cos(z) definition.

Traditionally, for f=2^z-1, there are two superfunctions defined from the two schroder functions for the fixed point of zero and the fixed point of 1. These two functions are imaginary periodic. The f^z from the upper fixed point of one is entire, the f^z from the fixed point of zero has logarithmic singularities for negative integer values, at half the period~=8.571i = \( \frac{\pi i}{-\log(\log(2))} \). At the real axis, between zero and one, the two functions differ by a fractional part in 10^-25, as published by Henryk and Dimitrii. The asum superfunction is closer to upper fixed point superfunction. As calculated here, the asum superfunction's analytic range is slightly larger than the analytic range of the lower fixed point function, =~8.779i, but it is not periodic, and encounters an infinite number of other chaotic discontinuities above this singularity.
- Sheldon