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7h
comment Is the function $f(x) = |x|$ convex?
There was a very similar discussion in this discussion.
1d
comment If the Chaos Game result is a Sierpinski attractor when the random seed is a sequence (Möbius function), does it imply that the sequence is random?
Why don't you use the ternary digits of $\pi$ or some other interesting number to determine your sequence. Or you might try a sequence generated by an interesting cellular automaton or a gray code or the iteration of a chaotic function. These are all deterministic but likely to generate the full Sierpinski triangle.
1d
comment If the Chaos Game result is a Sierpinski attractor when the random seed is a sequence (Möbius function), does it imply that the sequence is random?
Neat construction! However, I don't think that the set of limit points of the sequence being the same as the Sierpinski triangle implies that the process generates a nice image of the Sierpinski triangle. It's quite likely that points will not be uniformly distributed over the attractor. In technical terms, the self-similar measure generated by your procedure is very different from that generated by a random procedure. It might very well be concentrated on a line segment and that's image the procedure would generate.
1d
comment Are there any other non-differentiable that came be constructed from summation besides the Weierstrass function?
If $g(n,t)$ are the terms in any sequence that sum to the zero function, then you can add them to the $f$s and maintain the sum. Certainly, you can choose the $g$s to destroy any periodicity that might have been present in the original terms.
Jul
23
comment Wolfram Alpha formula not working
@GregVoit The question is not at all about Mathematica; it's about WolframAlpha which is quite different and definitely not a Mathematica interpreter.
Jul
22
comment Box-Counting Dimension with finite resolution
@Bob Well, you lost me there. I'd say an $n$-dimensional grid looks, well, $n$-dimensional regardless of the resolution. Now, if you have an image that consists of a large number of line segments, then it might appear to have a fractal dimension at large resolution while, at small scales, the one-dimensional behavior of those segments might dominate. Is that the point you're trying to make? Regardless, we certainly don't expect $\log(N_{\varepsilon})/\log(1/\varepsilon)$ to be constant. If it's close to constant over a wide range, then that constant could be called the fractal dimension.
Jul
22
comment Box-Counting Dimension with finite resolution
@Bob Right - I'm not sure I see much difference between your statement and mine. I guess the main point that I should really emphasize is that the relationship should hold over many orders of magnitude. Thus, we measure $\log(N_{a^k})$ for some constant $a$ larger than $0$ and (much) less than $1$ as well as for a large number of $k$s. If the relationship between $\log(N_{a^k})$ and $a^k$ holds up over that range, then we see that the fractal object behaves the same on a wide range of scales. This is somewhat like self-similarity.
Jul
22
comment Example of polynomial in dynamics
@JimBelk Have you noticed this in V10: JuliaSetPlot[2 z^3 - 3 z^2 + 1/2, z]?
Jul
20
comment Example of polynomial in dynamics
You can rule out the mulitbrot family, since they all have a single critical point, namely the origin.
Jul
20
comment Example of polynomial in dynamics
Very nice! It might be worth pointing out that it's easy to show that $1$ is in the Julia set using the characterization that the Julia set is the closure of the repelling fixed points. First, $-1/2$ is in the Julia set since it's a repulsive fixed point (i.e. a point of period $1$). Then, $1$ is in the Julia set since it's the inverse image of $-1/2$ and the Julia set is backward invariant.
Jul
15
comment Why do we say “radius” of convergence?
@ElliotG Your question in the title is 'Why do we say “radius” of convergence?' This response is an important part of that answer.
Jul
10
comment Estimating the value of an improper integral numerically
@TanMath When you say that you "can choose what values of x I want to evaluate f(x) at", that certainly sounds like a routine. If you have only samples, then you can probably interpolate first. Have a look at scipy.interpolate.
Jul
10
comment Estimating the value of an improper integral numerically
If you're using python, why don't you use the quad command defined in scipy's integrate package? You don't need a mathematical formula, just a python function (possibly, defined as a routine) that returns a real number. Though, this only works over bounded intervals, you can likely use joriki's excellent advice, which you should upvote.
Jul
4
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: $\kappa_n = 2 (\kappa_1 + \kappa_2 + \kappa_{n-1}) - \kappa_{n-2}$.
Jul
4
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.