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 2 real fixpoints again ....... tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 02/23/2016, 10:17 PM (01/19/2016, 01:55 PM)sheldonison Wrote: tommy1729 Wrote:Yes , but the Superattracting fixpoint was just picked to make the example easy. The actual goal is to get helpers for a pair of fixed points that are not superattractive. (01/19/2016, 01:24 PM)tommy1729 Wrote: The Böttcher function does not really help much. .... His work seems not to help with matters like radius of convergeance , area of agreement with functional equation , agreement on fixpoints etc etc. ....It seems the super-attracting case is not essential to the discussion. For an attracting and a repelling fixed point, both fixed points have their own Schroeder function solutions, which don't agree with each other except in special cases (see http://math.stackexchange.com/questions/...-iteration for details). The biggest obstacle, is that there is no unique Abel/super function solution. For example, tetration b=sqrt(2) is no longer real valued, when viewed as a complex plane extension of tetration for real valued bases greater than exp(1/e). So tetration gives us two additional solutions to b=sqrt(2), which are complex conjugates of each other, depending on whether you rotate clockwise or counter-clockwise around the singularity at eta=exp(1/e). The result of this "tetration" theta mapping, is that the upper half of the complex plane looks like one of the fixed points, and the lower half of the complex plane looks like the other fixed point. There are an infinite number of other possible theta mappings of the two Abel or super-functions; some of them behave in other cool/interesting ways; but none of them seems particularly unique. There is also Gottfried's AIS sum solution. But there is no uniqueness, except when both fixed points are repelling, and then Henryk Trappmann's uniqueness proof is valid. About your first link. When i take z = 2 , i get a different result !? I use the velocity addition from relativity Tanh(2 arctan©) = ( a + b) / (1 + ab). Since a = b = c we get = 2c / (1+ c^2). This does not agree with your result ??? ( and neither for tan addition ?! ) --- Also how about results different from moebius ? Regards Tommy1729 « Next Oldest | Next Newest »

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