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 The tangent approximation sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/11/2017, 11:36 PM (This post was last modified: 02/11/2017, 11:48 PM by sheldonison.) Much of the math behind Perturbed Fatou coordinates is inaccessible to me.  I can read some of the papers, but I don't have the background to understand them.  This is the first of several posts, which probably has some connection to perturbed fatou coordinates.  In this post, I will present the tangent approximation for iterating the function f which is congruent to tetration base b, where $k=\ln(\ln(b))+1$   The details are left to the reader.  if k=0, then we have the parabolic case.  Iterating $z\mapsto \exp(z)-1$ is congruent to iterating $z \mapsto \eta^z \;\; \eta=exp(1/e)$.   $f(x) = \exp(x)-1 + k\;\;\;$  This post is about iterating f(x) When iterating f(x), there is a singularity at k=0; x=0.   Then the first conjecture is that this tangent approximation nonetheless gets arbitrarily good as k approaches 0.  I'm pretty sure the approximation works and gets arbitrarily good for complex values of k as the absolute value of k approaches 0.  This approximation is for the superfunction of f(x), iterating from k/2.  You might notice that k/2 is very close to the inflection point of the iterated function for f(x), where at the inflection point, the 2nd derivative goes to zero.   The tangent function has an inflection point at zero.  This is a bit like the linear approximation for sexp(z) base(e) between sexp(-1)=0, and sexp(0)=1, where the inflection point is close to 0.5. $f^{[\circ z]}(k/2) = \tan(z \sqrt{\frac{k}{2}}) \cdot sqrt{2k} + \frac{k}{2}\;\;\;$ Tangent approximation for iterating f(x) This equation started from the observation that iterating f(x) as k approaches zero, the Period for the complex fixed points is proportional to $\sqrt{1/k}$.  But the iterated superfunction is just about linear as k approaches 0, with a slope of 1/k.  So let's define $g=f^{[\circ z]$, so g is the superfunction.  Then my hunch is this limit as k approaches zero might exist.    $g(\frac{z}{\sqrt{k}}) \cdot \sqrt{\frac{1}{k}}$ Then I looked at how g(z) behaved, and centered g at it's inflection point, and observed the connection with the tangent function.  This soon lead to the equation above, along with the beginnings of an asymptotic series which I'm still working on.  The asymptotic series includes the Tangent approximation, as well as an asymptotic perturbation of the tangent approximation. - Sheldon « Next Oldest | Next Newest »

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