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

[sheldonison]

At the real axis near the neighborhood of 3, the asum itself has a very small amplitude, of approximately 7.66E-12, as compared to an amplitude of 0.012 for Gottfried's asum(1.3^z-1) example. typo: off by 2x, the amplitude is approximately 1.533E-11

Here is an earlier plot for the decreasing of the amplitude as the base b goes to 2:

I have an equation for the approximate log(amplitude), for the alternating sum, iterating b^z-1. The approximation works for bases<e, and becomes more accurate as the base gets closer to e, with the adjust factor seeming to have a value of about 1.685 (log base e).
\( \log(\text{amplitude})=\frac{\pi^2}{\log(\log(b))}+\text{constant} \)edit, I shouldn't have used "constant". adjust(b) seems ok ....
\( \log(\text{amplitude})=\frac{\pi^2}{\log(\log(b))}+\text{adjust}(b) \)

First, I reproduced Gottfried's graph,
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And then I extended the graph to b=2.65, with the graph here shown to b=2.5, with both the calculated values and the approximation.
   

The key to the approximation, and calculating the asum for such tiny amplitudes lies in the complex plane. I prefer to work with the upper fixed point because it is entire, rather than the lower fixed point. The following Fourier series works for \( f^{[z]}(m) \), where \( f(z)=b^z-1 \) is the decremented exponential, m is the midpoint between the upper fixed point and the lower fixed point of zero, and the superfunction \( f^{[z]}(m) \) was generated from the upper fixed point.

\( \text{asum}(f^{[z]}(m)) = \sum_{n=1,3,5,7...} a_n\exp(n\pi i z) + \overline{a_n}\exp(-n\pi i z) \)

At the real axis, this is a 2-periodic real valued fourier series, and only has odd terms in the series, since asum(z)=-asum(z+1). The asum may have very tiny values at the real axis, as the example for b=2 shows, but it grows exponentially as imag(z) increases or decreases away from the real axis. This graph is for b=2, with the upper fixed point=1, and m=0.5. The graph is the asum of the upper fixed point superfunction, generated at 1/2 the period from the lower fixed point where \( \Im(z)=\frac{-\pi i}{\log(\log(2))}\approx8.572i \). On the left is the asum(z), and on the right is the superfunction(z), \( f^{[z-\pi i/\log(\log(2))]}(0.5) \). Real is graphed in red, imaginary is in green.

   

The ideal asum(z)=amplitude*cos(pi z) which grows exponentially as imaginary of z increases. \( \cos(\pi z)=0.5\times(\exp(\pi iz) + \exp(\pi -iz)) \) If it has some defined limiting behavior as the base approaches e, for \( \Im(z)=\frac{\pi i}{\log(\log(b))} \), then the value at the real axis must be scaled from that value by \( \exp(\frac{\pi^2}{\log(\log(b))}) \). That is where the approximation comes form. At base=e, the period is infinite, but for the upper fixed point, there is still a defined point where the upper fixed point best approximates the lower fixed point, see http://math.eretrandre.org/tetrationforu....php?fid=3. You can generate the asum of this function, which goes to 0 at both +/- real infinity. I'll post more when I calculate it.


One final plot. Eventually the \( f^z(0.5) \) ceases converges towards the upper fixed point and instead goes to infinity. This is a singularity for the asum(f^z), where f^z is generated from the upper fixed point. The analytic boundary of the asum, as expressed by the Fourier series, occurs when \( f^z=1+\frac{2\pi i}{\log(2)} \), because then f(z+1)=1, which is a the upper fixed point for f(z) since 2^1-1=1. Arbitrarily close to this point (but above it), is a point where f^(z+n) goes to infinity. It is fairly straightforward to calculate the value for z for this value, which occurs at 0.882829631453880483797+8.92366740700108685788i. This is about +0.35i bigger than the Period/2. Here is a graph of the asum(f(z)), at its analytic limit. Here, the asum is still continuous, but you can clearly see the singularity. On the left is the asum, and on the right is the superfunction, \( f^z \). Real is graphed in red, imaginary is in green.
   

Maybe I'll post more later. I also figured out some approximations for the singularities of the asum based superfunction, which eventually hits a wall of fractal singularities as |imag(z)| increases. The analytic limit for the asum based superfunction is a little bit less than the analytic limit of the asum from the upper fixed point, which was shown here.
- Sheldon