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revised How can the Hausdorff measure be nonzero?
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Apr
23
comment how do i prove that a collection of contractions does not satisfy the open set condition?
To be more blunt, could you please present the IFS that you are working with? The question is interesting and potentially challenging. Given the IFS, someone might very well be interested in the challenge. Without it, certainly no one can answer your question.
Apr
22
comment how do i prove that a collection of contractions does not satisfy the open set condition?
I believe this is a very difficult problem from a purely algorithmic perspective and that there is no known, general algorithm. It might be possible to address your question specifically, if you could present the four functions that comprise the IFS.
Apr
22
comment This one wierd trick integrates fractals. But does it deliver the correct results?
@Zach466920 Cool! I agree with your $2/3$ computation. I guess I'm assuming that your set looks something like this. If you're a Mathematica user, I have Mathematica code that automates self-similar integration procedure.
Apr
22
comment Hausdorf dimension of fractal iterates
If $S_0$ is point-wise of dimension one (which it might or might not be when constructing the Koch curve), then yes each $S_n$ will be pointwise of dimension 1. The limit (if we're talking about a strictly self-similar set) will certainly have constant dimension. Topological properties of self-similar sets is quite a tricky topic!
Apr
22
comment Hausdorf dimension of fractal iterates
Again, the dimension of $S_n$ need not be integral. If $S_0$ is the Cantor set, then each $S_n$ will have dimension $\log(2)/\log(3)$. You can certainly make reasonable notion of local dimension regardless - just look at $$\lim_{r\rightarrow0}\dim(B_r(x)\cap S).$$ You might prefer a $\limsup$ or $\liminf$ in some contexts. I don't believe there's any reason to expect the local dimension to be continuous, though.
Apr
22
comment Hausdorf dimension of fractal iterates
Your statement that "each $S_n$ is constructed by finite operations" is quite imprecise. If you're talking about the iterative procedure where we apply an IFS $\{f_i\}_{i=1}^m$ recursively from an initial compact set $S_0$ to obtain a sequence via the formula $$S_{n+1}=\bigcup_{i=1}^m f_i(S_n),$$ then the answer is - each $S_n$ has the same dimension as $S_0$ since it's just a countable collection of copies of $S_n$. Of course, what you've discovered is that the limit in the Hutchinson metric does not respect the Hausdorff dimension.
Apr
22
comment 3-cycles of the iterated quadratic maps and the Galois Theory of $x^3 = a(x-1)$
Well, that factorization is exactly what gives you the result that you need - is it not? Your second question seems quite separate. Perhaps it should be a separate question?
Apr
22
comment This one wierd trick integrates fractals. But does it deliver the correct results?
@columbus8myhw It's worked out in the paper I referred to.
Apr
22
answered This one wierd trick integrates fractals. But does it deliver the correct results?
Apr
18
answered The image of a specific Mobius transformation
Apr
17
answered an example of when Hausdorff and box-counting dimensions are equal?
Apr
17
answered At how many points will $\lfloor(sin x + cos x )\rfloor$ be discontinuous in the interval [0,2$\pi$]
Apr
17
revised Contractions and finding Fixed Points
added 136 characters in body
Apr
17
revised Contractions and finding Fixed Points
added 1082 characters in body
Apr
17
answered Contractions and finding Fixed Points
Apr
16
comment Newton Iteration Function
Very nice analysis. It might be interesting to note that this is unexpected. Typically, we figure that $N(x)=x-f(x)/f'(x)$ has a fixed point at $x_0$ only if $f(x_0)=0$. I guess this example also shows that we can also have a fixed point at $x_0$ if $f'(x_0)=\infty$. Another example would be $\sqrt[3]{x}$.
Apr
16
answered Sierpinski (Triangle) for Other Polygons
Apr
15
comment Parametrization by arclength
The question asks for a re-parametrization yielding the same path but with constant speed. Does your answer yield a technique to find that re-parametrization?
Apr
15
comment Parametrization by arclength
Use NIntegrate[...] rather than N[Integrate[...]]. The first goes straight to numerical techniques while the second attempts symbolic evaluation before following back on numerical techniques.