In the answering of another question in MSE I've dealt with the iteration of $x=b^x$ where the base $b=i$. If I reversed that iteration $x=log(x)/log(i)$ then I run into a cycle of 3 periodic fixpoints. Another complex base on the unit circle however, $b = \sqrt{0.5}(1+i)$ gives only one fixpoint. I asked myself, whether there is a point on the complex unit circle between that values, where the "switching" occurs, but could not recognize a meaningful interpretation. By binary search I narrowed the range to the following boundaries.

Let $\varphi_1=0.345008597051431179704966664178 $ and $b_1=\exp(\pi i \varphi_1) $ and
Let $\varphi_2= 0.345008597051431179704966665794$ and $b_2=\exp(\pi i \varphi_2) $
such that the bases $b_1,b_2$ lay on the complex unit circle.

Then consider the iteration to the fixpoint beginning at some arbitrary $x_0$ in the near of the guessed fixpoint, say $ x_0=1+I$, and iterate $ x_k= \log(x_{k-1})/\log(b_1) $. We shall approximate one fixpoint.

Now we use the other base $b_2$, begin again at $ x_0=1+I$, and iterate $ x_k= \log(x_{k-1})/\log(b_2) $. We get a 3-periodic cycle of fixpoints, one of them very near at $ t_\infty \sim - \pi/2 + \delta i $ where $ \delta$ goes to zero if we let go $\varphi_2 $into the direction of $ \varphi_1$.

Q1: What is the exact/symbolic value of $\varphi_x$ where the splitting from one into three fixpoints occurs? Wolfram-alpha and Ries at Robert Munafo's site were not helpful so far...
Q2: And what happens at that base? Do we see a fixpoint-cycle or one single fixpoint?

  • $\begingroup$ What do you mean by $\log$ here? A particular branch or the multi-valued logarithm? $\endgroup$ – mrf Jan 17 '13 at 23:42
  • $\begingroup$ @mrf: the principal branch (my writing might be spoiled by regular and heavy use of the software Pari/GP... ;-) ) $\endgroup$ – Gottfried Helms Jan 18 '13 at 0:03

The iteration $x \to f(x) = \log_b(x)$ (using the principal branch of log) has a fixed point at $x = x^*$ where $x^* = -W(-\ln(b))/\ln(b)$, $W$ being the Lambert W function. The fixed point is stable if $|f'(x^*)| < 1$ and unstable if $|f'(x^*)| > 1$, where $f'(x^*) = \dfrac{-1}{W(-\ln(b))}$. For $b = e^{it}$, $-\pi < t < \pi$, the plot of $|f'(x^*)|$ looks like this:

enter image description here

The transition between stable and unstable occurs at approximately $t = t_0 = \pm 1.96130884645945594019766830511$.
I doubt that this has a closed form.

EDIT: That is for the "0" branch of LambertW. The other branches also produce fixed points, some of which will be stable. Which fixed point or cycle a particular initial point will go to might not be easy to predict.

EDIT: For $b=i$, there is an attracting $3$-cycle $[-1.14012370875626+.706579007931335 i, 1.646798829-.1869454749 i, -0.07196134893-.3216429665 i]$, and there is an attracting fixed point $-1.861743075-.4107999686 i$ (from the "$1$" branch of Lambert W). These both have big basins of attraction, as shown in the plot below. Points in the white region are attracted to the fixed point, those in gray or black are attracted to the $3$-cycle.

enter image description here

EDIT: Not all branches of Lambert W produce fixed points: you do need $\text{Im}(x^* \ln(b)) \in (-\pi,\pi]$. At $b = (1+i)/\sqrt{2}$ the three-cycle seems to have disappeared, and nearly everything is attracted to the stable fixed point from $i=1$.

EDIT: It seems that the three-cycle disappears at approximately $b = e^{1.083876408 i}$ as one of its points hits the negative real axis.


Note: I wrote this from memory but didn't write it in the form of a proof. So at best I can only say this appears to be true, but it hasn't been proven yet. Sorry.

The question can be solved with an exact algebraic answer. Let $a^A=A$, with $A$ is a fixed point of $a$. Consider that for an infinitesimal $dz$

$a^{A+dz}=a^{dz} a^A =a^{dz} A = (1+ \ln(a){dz})A = A+ \ln(a^A){dz} = A+ \ln(A){dz}$.

So exponentiation maps ${A+dz}$ into $A+ \ln(A){dz}$ or more simply ${dz}$ into $\ln(A){dz}$. But this means that $\ln(A)$ is the multiplier or Lyapunov characteristic at $A$ of the fixed point of $a$.

Set the Lyapunov exponent to $\frac{1}{3}$ giving a multiplier of $e^\frac{2 \pi i}{3}$. Let the two multipliers equal, thus $e^\frac{2 \pi i}{3}=\ln(A)$ and $e^{e^\frac{2 \pi i}{3}}=A$. But $A^\frac{1}{A} = a $.

$\ln(A)=-0.5 + 0.866025 i $

$A=0.392947 + 0.462031 i$

At $a =0.0309536 + 1.73922 i $ the dynamics of tetration switches from one fixed point to the 3-periodic fixed point cycle.

See Projective Fractals.

  • $\begingroup$ (+1)Daniel, thank you very much. I'll go through this later. $\endgroup$ – Gottfried Helms Mar 17 '13 at 19:15

Hmm, I've seen, that the point $\varphi_x$ is also depending on the selection of the starting points $x_0$ of the iteration. So this is a much more complicated problem than I thought and I leave this (a bit casual) question better aside.

I thought of deleting it, but assume now it's a better policy on MSE to delete only completely off or meaningless questions. If anyone thinks it should be deleted: just feel free to flag so.

  • 1
    $\begingroup$ Gottfried, since you're familiar with the shell thron region, restrict ourselves to base=b inside that region; all of the example bases you gave, b=I, and the other two, are inside that region. To get the attracting fixed point, iterate, $z \gets b^z$. But the logarithm is multi-valued, so getting the primary repelling fixed point may require that the first iteration, starting with z=1+i, use $z \gets (\log(z)-2\pi i)/\log(b)$. Then continue iterating $z \gets \log(z)/\log(b)$ For the example b=I, the repelling fixed point is -1.8617-0.4108i. $\endgroup$ – Sheldon L Jan 18 '13 at 18:20
  • $\begingroup$ @SheldonL : true; my confusion began, when I found, that if we simply begin at another starting point $x_0$, for instance $x_0=0.1-0.1 i$ we run into a 3-periodic cycle of fixpoints $ \small -1.14012+0.706579 i, 1.64680-0.186945 i, -0.0719613-0.321643 i $. This means, I've jumped into the problem of different fixpoints - and I think I'd to consider the "basins of attraction" - but this all is much more than I expected after that small casual observation... So I don't think I'll continue this here. Thanks anyway for the comment! $\endgroup$ – Gottfried Helms Jan 18 '13 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.