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 tetration from alternative fixed point JmsNxn Ultimate Fellow Posts: 921 Threads: 111 Joined: Dec 2010 12/05/2011, 07:58 PM That's a very weird looking sexp! Does this mean we're going to have to clarify uniqueness based on the infinite list of fix points? Already, by looking at the new sexp it doesn't seem as fitting as the one based off of the primary fix point. But I don't know how you'd express this aesthetic requirement in mathematical notation. Perhaps merely because the derivative goes to zero at integers? And, I wonder if we'll see more erratic behaviour across a third fix point or a fourth! Is it likely that they'll get more erratic as we get further from the primary fixpoint? Or will it be more chaotic, as to which ones look nice and which ones don't? Very very interesting. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 12/06/2011, 02:43 PM (This post was last modified: 12/06/2011, 04:13 PM by sheldonison.) (12/05/2011, 07:58 PM)JmsNxn Wrote: That's a very weird looking sexp! Does this mean we're going to have to clarify uniqueness based on the infinite list of fix points?James, Thanks for your comments. The primary fixed point solution obviously looks best at the real axis, but the secondary fixed point solution is very beautiful in the complex plane (see my previous post). I haven't tried generating the nth fixed point of e^z, beyond n=2, but that would be my intuition, that there are an infinite number of other analytic sexp(z) solutions, one for each fixed point. The functions are hypothesized to all be analytic in the complex plane, except for singularities at the real axis for negative integers less than or equal to -2. As $\Im(z)$ goes to +/- infinity, the $\text{sexp}_n(z)$ function converges to the nth fixed point of exp(z). Also, for the nth fixed point, each loop around the singularity at z=-2 increments by (2n-1)2pi i. Correspondingly, at z=-1, the first 2n-1 taylor series terms are zero. The existence of such solutions depends on the topology of the Schroder function of the real number line and how the real axis unfolds in the complex plane for the superfunction so that the singularities can cancel out for integers>-2, after the Rieman mapping. Unfortunately, I'm over my head.... - Shel Daniel Fellow Posts: 188 Threads: 59 Joined: Aug 2007 12/24/2019, 06:26 AM I asked this question in 1993 in the following manner,  Consider two fixed points $\alpha_1, \alpha_2$ of the complex exponential map $a^z$ where $a^{\alpha_1}=\alpha_1$ and $a^{\alpha_2}=\alpha_2$ and their Lyapunov multipliers $\lambda_1, \lambda_2$.  Can the forward orbit of $a^z$ traverse a region of space dominated by $\alpha_1,\lambda_1$ to the region dominated by $\alpha_2,\lambda_2$? My numerical research indicated that the answer can be affirmative. The main problem I faced what the chaotic region between $\alpha_1, \alpha_2$. By adjusting $a\rightarrow 1$ the chaotic region becomes arbitrarily thin and several hundred iterations can map $\alpha_1,\lambda_1$ to $\alpha_2,\lambda_2$. Daniel « Next Oldest | Next Newest »

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