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"Kneser"-like mapping also for complex base?
#3
(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
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RE: "Kneser"-like mapping also for complex base? - by sheldonison - 11/07/2010, 01:37 PM

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