12/17/2012, 10:19 PM
(This post was last modified: 12/17/2012, 11:01 PM by sheldonison.)
I decided to compare the asum function for tetration base sqrt(2), to the bummer thread on this forum. \( \sqrt{2}^{z}(3) \) has an upper fixed point of 4 and a lower fixed point of 2. You can compare this image to the last image in the post I linked to. The base(sqrt(2)) has been discussed a great deal on this forum, so I thought it would be a good comparison study. The differences between these functions are extremely tiny, so the differences are scaled by 10^25 so as to fit on the same plot which shows a little bit of the superfunction going from the upper fixed point of 4 down towards the lower fixed point of 2, with the graph centered at 3.
Once again, the asum super function function is not an average of the superfunctions from the two fixed points. As you can see from the graph, the asum superfunction is much closer to the upper fixed point superfunction, which is entire, than to the lower fixed point superfunction, which is also imaginary periodic in the complex plane, but has logarithmic singularities.
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. Also, the asum needs a lot more iterations for the sqrt(2) base than for Gottfried's example because of slower convergence towards the fixed points. If you take the asum of the upper fixed point superfunction, than the asum has a period of 2, and the third harmonic harmonic has a magnitude of 1.186E-36. The even harmonics of the asum of the superfunction from the fixed point are zero since by definition asum(z)=-asum(z+1), and this is only true for odd harmonics, not odd harmonics.
As imag(z) increases towards >8.5*I, the differences between these three functions becomes larger and large in amplitude, becoming macroscopic instead of microscopic. The difference increase with a scaling factor of \( \exp(2\pi \Im z) \). At some point, the higher harmonics become visible too, and one expects the three functions to diverge once the upper fixed point superfunction is no longer converging towards 2 as real of z increases. Eventually the asum superfunction is probably dominated by singulariies; but I conjecture the asum superfunction is not imaginary periodic like the two fixed point superfunctions.
- Sheldon
Once again, the asum super function function is not an average of the superfunctions from the two fixed points. As you can see from the graph, the asum superfunction is much closer to the upper fixed point superfunction, which is entire, than to the lower fixed point superfunction, which is also imaginary periodic in the complex plane, but has logarithmic singularities.
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. Also, the asum needs a lot more iterations for the sqrt(2) base than for Gottfried's example because of slower convergence towards the fixed points. If you take the asum of the upper fixed point superfunction, than the asum has a period of 2, and the third harmonic harmonic has a magnitude of 1.186E-36. The even harmonics of the asum of the superfunction from the fixed point are zero since by definition asum(z)=-asum(z+1), and this is only true for odd harmonics, not odd harmonics.
As imag(z) increases towards >8.5*I, the differences between these three functions becomes larger and large in amplitude, becoming macroscopic instead of microscopic. The difference increase with a scaling factor of \( \exp(2\pi \Im z) \). At some point, the higher harmonics become visible too, and one expects the three functions to diverge once the upper fixed point superfunction is no longer converging towards 2 as real of z increases. Eventually the asum superfunction is probably dominated by singulariies; but I conjecture the asum superfunction is not imaginary periodic like the two fixed point superfunctions.
- Sheldon
