Two Attracting Fixed Points Catullus Fellow   Posts: 213 Threads: 47 Joined: Jun 2022 07/01/2022, 10:04 AM Does there exist a number a, such that both of the two primary fixed points of a to the x are attracting? If so, what could a be? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 07/01/2022, 08:49 PM (07/01/2022, 10:04 AM)Catullus Wrote: Does there exist a number a, such that both of the two primary fixed points of a to the x are attracting? If so, what could a be? The general form for the multiplier of a fixed point $x_0$ of an exponential $a^x$ is $\log(a)x_0$. Therefore you are asking if for a given $a$ there exists two values $x_0$ and $x_1$ such that each is less than $1/\log(a)$. By which we would have: \begin{align} |x_0| &< \frac{1}{|\log(a)|}\\ |x_1| &< \frac{1}{|\log(a)|}\\ \end{align} These points also satisfy: \begin{align} \log(x_0)/\log(a) &= x_0\\ \log(x_1)/\log(a) &= x_1\\ \end{align} Therefore $|\log(x_1)| < 1$ and $|\log(x_0)| < 1$. Therefore these two solutions would be rather close together. One can check for $a \in \mathbb{R}^+$ and $x \in \mathbb{R}^+$ this never happens. One can also show that within the Shell-thron region this cannot happen, as there is a unique attracting fixed point. Outside of the Shell-Thron region this doesn't happen either, as there are only repelling fixed points. On the boundary this can't happen because the fixed points are neutral or repelling. I believe the answer to your question is no, pretty sure this is standard. It's miraculous enough $a^x$ even has one attracting fixed point, for it to have two would be incredible. General rule of thumb when dealing with exponentials, is that the Julia set is the entire complex plane, or there's one attracting fixed point/neutral fixed point. Catullus Fellow   Posts: 213 Threads: 47 Joined: Jun 2022 07/02/2022, 10:23 AM Is eta the only number a, such that a to the power of x has a parabolic fixed point? If not, what else could a be? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 07/03/2022, 07:21 AM (07/02/2022, 10:23 AM)Catullus Wrote: Is eta the only number a, such that a to the power of x has a parabolic fixed point? If not, what else could a be? The boundary of shell thron always has a parabolic fixed point. Let $a \in \mathcal{S} = \{a \in \mathbb{C}\mid |\log(a)| = 1\}$. The function: $f(z) = e^{\log(a)z/a}\\$ Which is an exponential; always has a parabolic fixed point at $a$. Essentially to answer your question, the boundary of the Shell-thron region is the only place you get parabolic fixed points. The value $\eta$ is just one particular value. tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 07/04/2022, 01:04 PM (07/03/2022, 07:21 AM)JmsNxn Wrote: (07/02/2022, 10:23 AM)Catullus Wrote: Is eta the only number a, such that a to the power of x has a parabolic fixed point? If not, what else could a be? The boundary of shell thron always has a parabolic fixed point. Let $a \in \mathcal{S} = \{a \in \mathbb{C}\mid |\log(a)| = 1\}$. The function: $f(z) = e^{\log(a)z/a}\\$ Which is an exponential; always has a parabolic fixed point at $a$. Essentially to answer your question, the boundary of the Shell-thron region is the only place you get parabolic fixed points. The value $\eta$ is just one particular value. yes that is correct. I sometimes wonder about the analogue shell-thron for the double exponential functions. In particular because there are fixpoints outside the shell-thron such as z^z = z. ( we studies that number , I posted a thread about it with title simply the complex value of the smallest nonreal z ) And also because there are half-iterates of exp(exp(x)) that are not exp(x). I could say more , as could most regular posters here. We all have considered such ideas. So I will leave it at that for now. regards tommy1729 « Next Oldest | Next Newest »

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