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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
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
18
answered Sufficient condition for integer Hausdorff dimension.
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.
May
17
comment Proving basic properties of Hausdorff dimension and measure
@AJY Again, my answer addresses exactly that.
May
17
revised Proving basic properties of Hausdorff dimension and measure
deleted 52 characters in body
May
17
comment Proving basic properties of Hausdorff dimension and measure
@AJY What? My counter example is stated on the line - i.e. $\mathbb R^1$. The salient point is that countable additivity must be with respect to some $\sigma$-algebra. The answer itself is certainly applicable on $\mathbb R^n$.
May
17
answered Proving basic properties of Hausdorff dimension and measure
May
17
comment Proving basic properties of Hausdorff dimension and measure
@AJY The purpose of an edit is to clarify, correct, or otherwise improve a question - not to change it's content. Your edit deleted one question entirely, thus making Daniel's comment superfluous.
May
17
revised Proving basic properties of Hausdorff dimension and measure
rolled back to a previous revision
May
12
comment How to compute a negative “Multibrot” set?
@DanielAllenLangdon I'm not too surprised about the disconnectedness. Orbit detection is an expensive operation, particularly near the boundary of the Mandelbrot set. Also, the point at infinity should be counted as an orbit as well. I placed some (not particularly fast) Javascript here that takes these things into account.
May
8
answered What is the integral of $e^{a \cdot x+b \cdot y}$ evaluated over the Koch Curve
May
8
comment Graph modeling using calculus
Must it be a polynomial? This would be very simple using a piecewise defined function.
May
7
comment What is the integral of $e^{a \cdot x+b \cdot y}$ evaluated over the Koch Curve
I would be very surprised if there is any known closed form expression for this integral. For what it's worth (which might not be much), I computed the integral numerically for $a=b=1$ and found a value of $1.8800692$. The inverse symbolic calculator didn't find any particularly promising results.
May
7
comment How to compute a negative “Multibrot” set?
@DanielAllenLangdon I made some updates to my answer, as well as to my Javascript application. Most importantly the Javascript contains code to return the detected period, as well as the detection time. Thus, the image of the bifurcation locus can be colored to indicate the various periods. I would love to know if you use this stuff to improve your own applications. I notice that you've been trying to create a fairly general bifurcation locus generator and I'd be happy to provide advice on that.
May
7
revised How to compute a negative “Multibrot” set?
added 1201 characters in body
May
6
comment What does a 3D periodic solution of a differential equation look like?
@2mkgz I thought so! Oh well. So, I know you've got upvotes available (it's early in the UTC day) but I guess my little response below was not worthy. I can certainly agree to disagree on this silly voting thing but, please, do let me know, if you ever find something I wrote upworthy. It would just make my day. :)