Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options questions only not deleted user 1056

Fractals deal with special sets that exhibit complicated patterns in every scale. Fractal sets usually have a Hausdorff dimension different from its topological dimension. Examples include Julia sets, the Sierpinski triangle, the Cantor set. Fractals naturally appear in dynamical system, such as iterations in the complex plane, or as strange attractors to continuous dynamical systems, (see Lorentz attractor).

10 votes
1 answer
415 views

Convex Julia sets

Consider the classical Julia set $J_f$ associated with $f(z)=z^2+c$. Since $J_c$ is completely invariant, we know that $f^{-1}(J_f) \subseteq J_f$. Now, let $H_f$ be the convex hull of $J_f$. Is it t …
Per Alexandersson's user avatar
19 votes
2 answers
3k views

Is there some known way to create the Mandelbrot set (the boundary), with an iterated functi...

Is there some known way to create the Mandelbrot set (the boundary), with an iterated function system (IFS)? Julia sets can be formed by iterating the two functions $z \mapsto \pm \sqrt{z-c},$ and th …
Per Alexandersson's user avatar
13 votes
5 answers
3k views

Hausdorff dimension for invariant measure?

A fractal set has a Hausdorff dimension. In some cases, we may generate a fractal by iterating $f,$ and let the fractal be the set of starting points $x$ such that $|f^{\circ n}(x)|$ is bounded as $ …
Per Alexandersson's user avatar