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 Permeable Shell-Thron-Boundary? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 02/24/2013, 01:35 PM It seems possible to flawlessly (i.e. base-holomorphic tetration) transform a real fixpoint pair (one attracting below $\eta$, one repelling above $\eta$) into a conjugated fixpoint pair with same absolute value of the repelling multiplier. There seems no disturbance of passing the attractive fixpoint through the STB turning it into a repelling fixpoint. I conclude that from that there is no disturbance in the sickles between the fixpoints while transforming (Well this all really far from being conclusive, its just some good guessing). To visualize this I took our famous (2,4) fixpoint pair. Just as clarification some symbols and formulas. We consider the function $f(z)=exp{c*z}$ (i.e. $b=e^c$ is our base). $L=exp{A}$ is our fixpoint, then we have the relations $c = A*exp{-A}$ $A = - W_k ( -c )$ $f'(L) = A$ (which is called multiplier) So for a each fixpoint with $|A|=\log(4)$, we can calculate "the other" fixpoint by $A_2=-W_k(- A * exp{-A})$ by choosing $k$ suitably. And this is what I did in the next picture. At the left side I go through $A=\log(4)*exp^{i\phi}$ via the red line, and connect it with the corresponding $A_2$ (blue line) via straight lines. On the right side I apply $f$ to the straight line on the left side. So the left picture shows the left side of the sickles and the right picture shows the right sides of the sickles. The black curve is the fixpoints with |A|=1 (their base corresponds to the STB).         And on the next two pictures, this is shown with the values $L$ instead of $A$.         For a given value of the repelling real fixpoint we can calculate the base, for which they are transformed into a conjugate fixpoint pair with same absolute value of the multiplier. For this we are seeking two fixpoints which are conjugate having the same base corresponding to $c$ and same $|A|=r$. Say the upper fixpoint is $A=r exp{i\phi}=a+ib$. Same base: $r exp{i\phi} exp{-a-ib} = r exp{-i\phi} exp{-a+ib}$ $exp{i 2\phi } = exp{i 2b}$ $\phi = b + \pi k$ $\phi = r sin(\phi) + \pi k$ In our case for $r=log(4)$ we get $\phi\approx 1.354$. So the base is: $c = r exp{i\phi} exp{-a-i\phi} = r exp{-a}$ $a=r\cos(\phi)$ $c = r exp{-r\cos(\phi)}$ $b = exp{r e^{-r\cos(\phi)}} \approx 2.794$ In the brown line we follow the base $b$ corresponding to our red fixpoints (black is STB): « Next Oldest | Next Newest »

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