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Aug
21
comment Did Wolfram|Alpha mess up $\int 1 - \frac{1}{1-e^{-x}} \mathrm{d}x$ or did I?
I see you just removed the complex-integration tag yet, I would say, that's a major source of the discrepancy!
Aug
21
comment Did Wolfram|Alpha mess up $\int 1 - \frac{1}{1-e^{-x}} \mathrm{d}x$ or did I?
+1 For showing how to obtain the WA result, with the exception of the absolute value bars. In fact, those absolute bars (contrary to the assertion made by @wythagoras) are not quite correct here. This is actually where key source of the discrepancy arises, since the symbolic functionality in Mathematica (and, therfore, WA) makes the implicit assumption that all variables are complex. In that context, the absolute value bars are simply incorrect.
Aug
20
comment Interesting and unexpected applications of $\pi$
@wltrup I think you forgot to multiply by $m^2$, but yeah.
Aug
15
comment Advantages to learning Sage?
@kcrisman Thanks - it's been some time and I've moved on, though. It was an infinite group, incidentally - the Baumslag-Solitar group.
Aug
14
comment Advantages to learning Sage?
@kcrisman Well, that's rather the point isn't it? Maxima is incorporated into Sage with a clunky pexpect interface (or, at least that's how it used to be) and Sage is ill prepared to deal with this type of interaction. That's not the only example, though. Some years ago, I was working Cayley graphs - i.e. graphs associated with groups. I figured that Sage would be perfect, as it contained Gap and NetworkX. The group computations and the graph computations went both individually fine, but the interaction between the two was very difficult.
Aug
10
comment Relation between Power Laws and Fractals
There is a book, Chaos, Fractals, and Power Laws, devoted to this topic. In that context, a power law is used to relate the dimension $d$ of the set to the number $N_r$ of sets of size $r$ required to cover the set via $N_r\sim 1/r^d$. This leads to the so-called box-counting dimension. There is no assertion that the graph of a power function is itself a fractal.
Aug
4
revised Prove that the Mandelbrot Set Is A Closed Set
edited tags
Aug
4
comment Question About Filled Julia and Julia Sets
@Overachiever I completed the argument more explicitly, rather than just providing hints. To address your specific questions, it seems that you have one point of fundamental confusion. Your original question does not concern the boundary of the unit disk. That's just a specific case, namely when $c=0$; most Julia sets are much more complicated than this. So there's no reason to suppose that $|Q_c(z_0)|=1$ when $Q_c(z_0)\in J_c$. Rather, you should draw conclusions on points in a neighborhood of $z_0$ from assumptions involving corresponding points in a neighborhood of $Q_c(z_0)$.
Aug
4
revised Question About Filled Julia and Julia Sets
added 109 characters in body
Aug
4
answered Question About Filled Julia and Julia Sets
Aug
1
comment Is the function $f(x) = |x|$ convex?
There was a very similar discussion in this discussion.
Aug
1
answered Advantages to learning Sage?
Jul
31
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.
Jul
31
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.
Jul
30
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
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
answered Box-Counting Dimension with finite resolution
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
21
answered Construction of Rauzy Fractals with substitutions without a fixed point