Reputation
13,518
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
14 37
Newest
 Nice Answer
Impact
~174k people reached

16h
comment Apollonian gasket
I still don't see how your $C_4$ is well defined. If I understand your question correctly, though, then a somewhat simpler but still recursive formula for the curvatures $\kappa_n=1/r_n$ might be: $\kappa_n = 2 (\kappa_1 + \kappa_2 + \kappa_{n-1}) - \kappa_{n-2}$.
16h
comment Apollonian gasket
What's the $n^{\text{th}}$ circle? Keep in mind that, given three mutually tangent circles, we expect two more circles mutually tangent to all three. Not just one. So your $C_4$ is not uniquely determined.
Jun
27
comment Why does the boundary of the Mandelbrot set contain a cardioid?
One comment on the linked comment: I retract my statement that the boundary of the Mandelbrot set contains a line segment.
Jun
27
comment Why does the boundary of the Mandelbrot set contain a cardioid?
Your first paragraph is nonsense and I don't see how the remainder adds anything to Henning's response.
Jun
27
comment Why does the boundary of the Mandelbrot set contain a cardioid?
Yes, this is essentially how I would have responded. The fact that this cardioid lies on the boundary of the Mandelbrot set can be proved using rational external rays. It turns out that every point of the form $e^{2\pi i t}(2-e^{2\pi i t})$ for rational $t$ is the base of a bulb off of the main cardioid and that some external ray lands exactly at that point.
Jun
27
comment Can a fractal be a manifold? if so: will its boundary (if exists) be strictly one dimension lower?
@A.P. It can, in fact, be proved that every complex parameter on the cardioid $c=e^{i t} \left(2-e^{i t}\right)/4$ is on the boundary of the Mandelbrot set. I can't find a convenient pointer to that fact at the moment and proving it would take us a bit too far astray from this question. Perhaps you could raise the question on the site? I'd answer it this evening, if no one else has by then.
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
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
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
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
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.
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
19
comment How do you solve $\cos \pi z =0$?
Perhaps the downvotes arise because it's clear that you just copied and pasted the answer from a computer? In fact, your TeX can be produced exactly via the Mathematica command Reduce[Cos[Pi*z] == 0, z] // TeXForm.
May
17
comment Proving basic properties of Hausdorff dimension and measure
@AJY I'm not claiming a complete answer - just the idea and a reference, as you asked for. I suppose the key issue to address your present question is this: if you have disjoint closed sets (and they are bounded), then the they will be positively separated.