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.
