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My question is focused on the King's Dream fractal, which can be defined as follows (nice pictures can be found here) : $$ \Omega = \{f^n(0.1,0.1) \;\vert\; n \in \mathbb N \} \quad f(x,y)=(\sin(ax)+b\sin(ay),\sin(cx)+d \sin(cy))$$ where $-3 < a,c < 3$ and $-0.5<b,d<1.5$ are given.

Then, I was wondering if $K=\overline \Omega$ is symmetric w.r.t. $(0,0)$, i.e. $(x,y) \in K \implies (-x,-y)\in K$ (at least for some values of $a,b,c$ and $d$, e.g. those given below).

I don't think that $\Omega$ itself is symmetric w.r.t $(0,0)$. Looking at the picture below, it seems that it could be the case for $K$, but it doesn't appear obvious when looking at the definition of $f$.

I was also wondering whether $K$ is connected. If these properties hold, I don't know how this could be proved...

Here is a picture of $\{f^n(0.1,0.1) \;\vert\; 0 \leq n \leq N = 9000 \}$ for $$a=2.879879,b=−0.765145,c=−0.966918,d=0.744728.$$ King's dream fractal for N = 9000, a=2.879879,b=−0.765145,c=−0.966918,d=0.744728

Thank you in advance for your answers !

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  • $\begingroup$ it may depend on the parameters, obviously the case $a=b=c=d=0$ results in $\Omega=\{(0.1,0.1),(0,0)\}$ which is not connected (and uninteresting). I suspect that if $a,b,c,d$ each is close to $0$, then $f^n(0.1,0.1)\to0$ so $K$ would be a converging sequence of points and won't be connected. If $b=d=0$ but $a=c>0$ is small, then all iterations are positive, small, and converge monotonically to $0$, so they remain in the first quadrant and cannot be symmetric with respect to the origin. (Use that $0<\sin(x)<x$ for $x>0$.)This is not an answer but you might want to clarify the interesting cases $\endgroup$ – Mirko Dec 15 '15 at 23:19
  • $\begingroup$ Thank you for your comment. I was indeed interested in non trivial cases such as $a=2.879879, b=-0.765145,c=-0.966918,d=0.744728$ which seems to be symmetric w.r.t. the origin. $\endgroup$ – Watson Dec 16 '15 at 8:57
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    $\begingroup$ I bet if you plot a 2-dimensional map over two of the four variables (fix the other two), and plotted a pixel black if the attractor is connected, the resulting plot could also contain fractal properties. $\endgroup$ – Mark Jeronimus Jan 3 '16 at 11:07
  • $\begingroup$ @Mark Jeronimus : so there is "no hope" (i.e. it may be very difficult) to prove that $\overline{\Omega}$ is symmetric w.r.t to the origin (for the values I gave above for instance) ? $\endgroup$ – Watson Jan 9 '16 at 15:11
  • $\begingroup$ Related: math.stackexchange.com/questions/2506900 $\endgroup$ – Watson Nov 25 '18 at 13:44

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