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Jun
27
comment Can a fractal be a manifold? if so: will its boundary (if exists) be strictly one dimension lower?
@iadvd I guess you mean a smooth manifold, which is certainly a common assumption. Otherwise, the graph of a nowhere differentiable function can be both a manifold and a fractal. The boundary of the Mandelbrot set contains a line segment, a circle, and a cardioid. So I guess a fractal can contain a smooth manifold.
Jun
26
comment Is :$\frac{\Bbb d}{\Bbb d x}$ a chaotic operator in infinite-dimensional Hilbert space?
@user51189 The question is on hold still because it still doesn't have enough reopen votes; I was just the fourth to vote to reopen so it needs one more. On the other hand, I think that Noah's comment contains a fine pointer.
Jun
19
revised What is the difference between $f(f^{-1}(U))$ and $f^{-1}(f(U))$?
deleted 3 characters in body
Jun
19
comment How to solve this integral and have a result in term of arccos?
@SimonS I believe he used Mathematica, actually. In fact, the exact TeX in the answer is produced by the Mathematica command TeXForm[Integrate[Sqrt[R/r - 1], r]]. I've told him before that this is not acceptable and he should at least acknowledge the source. I have no idea why he refuses.
Jun
15
comment How many vertices are in the Koch Snowflake?
@LinusS. We have now carefully defined a vertex as a vertex of one of the polygonal approximations. These correspond to the (finite) addresses. If you allow the branches to have infinite length, you can then determine other points on the Koch curve. This is analogous to the fact that the terminating decimal expansions determine only countably many real numbers. You need non-terminating expansions get the remaining reals.
Jun
14
answered How many vertices are in the Koch Snowflake?
Jun
12
comment How many vertices are in the Koch Snowflake?
Supposing that by "vertex" you mean one of the vertices of the standard, polygonal approximations to the Koch curve, then perhaps the answer to this question might help? That indicates a technique to place the vertices in 1-1 correspondence with finite strings of symbols so, if you can enumerate those, then you've enumerated the vertices.
Jun
7
comment What does “points spanned by powers” mean in the Goffinet dragon definition?
@trichoplax Put another way - it's the set of all linear combinations of the numbers $p^k$ where the coefficients can be either zero or one. If you consider that these coefficients form a finite field, then this is exactly the linear the span of the $p^k$s.
Jun
6
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Jun
5
comment what is the parametric form for “mystery curve”?
It appears to be taken verbatim from Frank Farris's Wheels on Wheels on Wheels, in the June 1996 edition of Math Mag.
Jun
4
answered Categories of fractals
Jun
3
comment Existence of a recurrent point
No biggie - my comment was just informative. :)
Jun
3
comment Existence of a recurrent point
I voted to close this since it's exactly the same as your more recent (and now answered) question here.
Jun
3
comment Subsets of set satisfying open set condition
@skipdivided If $S_i(O)\cap S_j(O)$ is empty and $U\subset O$, then certainly $S_i(U)\cap S_j(U)$ will be empty as well. So, if $U$ happens to satisfy $S_i(U)\subset U$ for each $i$, then $U$ will verify OSC as well. But, again, I think this is quite unusual and certainly not likely to happen for all open $U\subset O$. On the contrary, if you have at least two different contractions with distinct fixed points, then you will always be able to find an open ball $B$ of sufficiently small radius so that $B\not\supset S_i(B)\cup S_j(B)$.
Jun
3
comment Subsets of set satisfying open set condition
@skipdivided You want $S_i(A) \subset A$ for any set $A$ and independent of $i$? Seems like quite a strong assumption. In fact, if you have at least two different contractions with distinct fixed points, then I don't think this condition can be arranged.
Jun
3
answered Subsets of set satisfying open set condition
May
25
comment Mandelbrot set of $c \cdot \cos(z)$
@Unit Thank you. Your observation is correct, of course. I guess that entire is a broader class of functions, though. Somethings could be true of entire functions in general that are not true of polynomials. Importantly, $c\cos(z)$ is transcendental but, really, I was just following the nomenclature that is common in the research literature.
May
25
revised Mandelbrot set of $c \cdot \cos(z)$
edited tags
May
25
answered Mandelbrot set of $c \cdot \cos(z)$
May
22
answered Calculate moment of inertia of Koch snowflake