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 "Kneser"-like mapping also for complex base? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/05/2010, 09:56 AM (This post was last modified: 11/05/2010, 10:00 AM by mike3.) Hi. I was wondering about this. I noticed that the tetrational constructed with the continuum sum for the base $1.33 + 1.28i$, shown here: http://math.eretrandre.org/tetrationforu...hp?tid=514 looks in the upper half-plane like the regular iteration at the attracting fixed point, while in the lower half-plane it looks different. The lower half-plane actually looks like the regular iteration at a repelling fixed point:     Compare to: Both fixed points are fixed points of the associated logarithm. This behavior suggests that it may be possible to form the tetrational at any complex base out of the two regular iterations of the two fixed points of logarithm. Could it be possible that some kind of Kneser-like 1-cyclic transform, or a pair of such transforms, be applied to "bend" the two regular iterations so they flow together in a holomorphic manner on the right half-plane to yield the tetrational function? sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 11/06/2010, 02:59 PM (This post was last modified: 11/07/2010, 12:54 PM by sheldonison.) (11/05/2010, 09:56 AM)mike3 Wrote: Both fixed points are fixed points of the associated logarithm. This behavior suggests that it may be possible to form the tetrational at any complex base out of the two regular iterations of the two fixed points of logarithm. Could it be possible that some kind of Kneser-like 1-cyclic transform, or a pair of such transforms, be applied to "bend" the two regular iterations so they flow together in a holomorphic manner on the right half-plane to yield the tetrational function? I highlighted pair of transforms, and I like the phrase "flow together". I don't know if its possible or not. But your ideas are starting to make sense to me. I like having a pair of kneser mappings, with two different theta functions, one for each repelling fixed point, that have equal sexp(z) for z>-2 at the real axis. The first theta function maps the upper half of the complex plane, and the second theta function maps the lower half of the complex plane. Both theta functions would have singularities at the real axis for integer values of z, corresponding to sexp(0)=1, sexp(1)=B, sexp(2)=B^B. For z>-2, the singularities gets canceled by $\text{sexp}(z)=\text{superfunction}(z+\theta(z))$, so that sexp(z) is analytic at the real axis. For real valued bases, the sexp(z) defining boundary condition is that sexp(z) is real valued at the real axis after the 1-cyclic theta function, and then we use the Schwarz reflection theorem to generate the values for imag(z)<0. For your complex base, the defining condition would be that the complex contour at the real axis for z>-2 be the same for the upper theta(z) mapping from one fixed point, as the lower theta(z) from the other fixed point. This actually seems like a reasonable and probably unique definition, and probably also works for real valued bases. I think you should pursue it further. Also, you could graph the sexp(z) value at the real axis, for z>-2. Your example looks like it may have singularities roughly corresponding to z=-2,-3,-4 .... etc, but the singularities look a little off kilter, in that the imag(z) for the singularities should be in a horizontal line, spaced apart by a unit length. It would be interesting to find some way of computing such a pair of kneser $\theta(z)$ mappings, and their corresponding analytic complex tetration. - Sheldon sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 11/07/2010, 01:37 PM (11/05/2010, 09:56 AM)mike3 Wrote: Both fixed points are fixed points of the associated logarithm. This behavior suggests that it may be possible to form the tetrational at any complex base out of the two regular iterations of the two fixed points of logarithm.....I think the case Mike brings up involves both an attracting fixed point and a repelling fixed point. In my earlier post, I was implicitly assuming two repelling fixed points. Repelling fixed point -1.135218577668813217726900190 - 1.989607927210830803666191570i Attracting fixed point 0.5793702071827351250349231804 + 0.6472265454524960204764987313i If we start with a solution from the repelling fixed point (which is guaranteed to be entire), and then we do a superfunction(z+theta(z)) kneser mapping, to allow sexp(0)=1, but the mapping would only be defined for 1/2 of the complex plane, due to the singularity in theta(z). There are infinitely many such mappings. So I think the question is, can this mapping equal another theta(z) mapping of the solution from the attracting fixed point? Where the attracting fixed point mapping is defined on the other half of the complex plane, and the two mappings are equal at the real axis, with sexp(0)=1, sexp(1)=B..... - Sheldon « Next Oldest | Next Newest »

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