[AIS] (alternating) Iteration series: Half-iterate using the AIS? - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: [AIS] (alternating) Iteration series: Half-iterate using the AIS? ( /showthread.php?tid=760) |

RE: Iteration series: Half-iterate using the infinite iteration-series? - sheldonison - 01/03/2013
(01/03/2013, 07:07 AM)Gottfried Wrote: Hmm, perhaps I have made some basic error now, or something is around which I did not understand correctly from the beginning.In general, asum(z+x) <> -asum(f(z+x)) <> asum(f(f(z))), for arbitrary values of x. This is sort of obvious since the difference of adjacent iterations goes to zero as you approach the fixed point. This explains what Gottfried is seeing, which matches expectations. The derivatives are not the same for the asum of the different points in the sequence. asum(f^z) is 2-periodic, where f^z is a superfunction of b^z-1 developed from either fixed point, or even developed from the asum itself. If f^z is developed from the asum itself than asum(f^z) simplifies to following, where the amplitude is |a1|. This is exactly a scale/shift factor of the cosine function, since that is the definition of the asum super function. For example, take b=2. f(z)=2^z-1. This has two fixed points, zero and one. Take m=0.5, halfway in between. From the upper fixed point, z=1, generate f^z(0.5), which is entire. I choose to use the upper fixed point instead of the lower fixed point of zero, because the upper fixed point f^z is entire. Then asum(f^z(0.5)) is a 2-periodic function. Approximations for the first 16 a_n values for this function are listed below; this is the 2-periodic function I posted graphs for earlier in this thread. Carried to an infinite number of terms, this series converges if |imag(z)|<= 8.9236674*I ...because the singularity closest to the real axis is approximately 0.882829631453880483797 - 8.92366740700108685788*I. See my previous post. Code: `a1= 3.78754977991134100255081 E-12 - 6.66245137733325455039148 E-12*I` Now the next step in my algorithm is to generate the asum superfunction itself. Again, starting from the upper fixed point entire function, f^z(0.5), I calculate a 1-cyclic theta(z) function, which also has very small coefficients. But like any periodic function, the terms in theta(z) grow exponentially as imag(z) increases, as an*exp(2n pi). Theta(z) has a radius of convergence that is slightly smaller, |imag(z)|<8.7793i. 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. The pari-gp code I have works for b in the range of 1.1 to around 2.5. With no attempt at optimizations, it takes about 8 seconds to generate the asum and theta fourier series results accurate to 32 decimal digits, for b=2, accurate in the range |imag(z)|<8.57. For 64 digits accuracy, triple the time required. I could post the code (after cleaning it up a little). Here are the theta(z) coefficients, normalized so that fa(z)=amplitude*cos(z). This is the best way to accurately represent the asum superfunction over fairly wide range of values, approaching within 0.3i of the nearest singularity to the real axis. At the real axis, theta(z) has very small values, and the two superfunctions are nearly identical. So theta(z) provides one clear way to document the small differences between the functions, in a way that also shoes how those small differences grow as imag(z) increases. - Sheldon Code: `b0= 0.335456219829438730039` - Sheldon RE: Iteration series: Half-iterate using the infinite iteration-series? - sheldonison - 01/04/2013
(01/03/2013, 09:58 PM)sheldonison Wrote: ....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 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. [attachment=991] 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. [attachment=992] 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 = . 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 RE: [AIS] (alternating) Iteration series: Half-iterate using the AIS? - Gottfried - 03/26/2015
The AIS-method for finding the fractional iteratives seems to be still insufficient. In a discussion in mathoverflow (http://mathoverflow.net/questions/201098/oscillation-aspects-of-two-way-infinite-alternating-series-a-followup-from-the ) I study the two-way-infinite alternating iteration series on the base function r(x)=x^b, so the left-associative power tower. I applied the scheme as I did here in this thread where the base function is decremented exponentiation (r(x)= b^x-1) and also got an oscillating function asum. However, that base function allows a natural explicte fractional iterate, so we can compare the asum-computed and the naturally computed funcion values. I found that the asum for this function differs from the sine-wave by tiny differences, in the sixth digit or even below. So the consequence is, that I should distrust the apparently nice solution for the determination of the fractional iterate for the tetration based on the AIS. Maybe someone else here of the forum is able to determine the characteristic of the d(x)-function in my linked MO-post, perhaps by some fourier-decomposition, with the hope, that such an axpression could then also be introduced as a correction for the asum() for the tetration, but I don't really know whether it shall lead to somewhere. Gottfried RE: [AIS] (alternating) Iteration series: Half-iterate using the AIS? - tommy1729 - 03/27/2015
The alternating reminds me of ramanujan's master theorem ... Maybe that helps here. regards tommy1729 |