Measuring Fractal Dimension ?
pdpi
pdpinheiro at gmail.com
Mon Jun 15 07:14:14 EDT 2009
On Jun 15, 5:55 am, Steven D'Aprano
<ste... at REMOVE.THIS.cybersource.com.au> wrote:
> On Sun, 14 Jun 2009 14:29:04 -0700, Kay Schluehr wrote:
> > On 14 Jun., 16:00, Steven D'Aprano
> > <st... at REMOVETHIS.cybersource.com.au> wrote:
>
> >> Incorrect. Koch's snowflake, for example, has a fractal dimension of
> >> log 4/log 3 ≈ 1.26, a finite area of 8/5 times that of the initial
> >> triangle, and a perimeter given by lim n->inf (4/3)**n. Although the
> >> perimeter is infinite, it is countably infinite and computable.
>
> > No, the Koch curve is continuous in R^2 and uncountable.
>
> I think we're talking about different things. The *number of points* in
> the Koch curve is uncountably infinite, but that's nothing surprising,
> the number of points in the unit interval [0, 1] is uncountably infinite.
> But the *length* of the Koch curve is not, it's given by the above limit,
> which is countably infinite (it's a rational number for all n).
>
> > Lawrence is
> > right and one can trivially cover a countable infinite set with disks of
> > the diameter 0, namely by itself. The sum of those diameters to an
> > arbitrary power is also 0 and this yields that the Hausdorff dimension
> > of any countable set is 0.
>
> Nevertheless, the Hausdorff dimension (or a close approximation thereof)
> can be calculated from the scaling properties of even *finite* objects.
> To say that self-similar objects like broccoli or the inner surface of
> the human lungs fails to nest at all scales is pedantically correct but
> utterly pointless. If it's good enough for Benoît Mandelbrot, it's good
> enough for me.
>
> --
> Steven
You're mixing up the notion of countability. It only applies to set
sizes. Unless you're saying that there an infinite series has a
countable number of terms (a completely trivial statement), to say
that the length is "countably finite" simply does not parse correctly
(let alone being semantically correct or not). This said, I agree with
you: I reckon that the Koch curve, while composed of uncountable
cardinality, is completely described by the vertices, so a countable
set of points. It follows that you must be able to correctly calculate
the Hausdorff dimension of the curve from those control points alone,
so you should also be able to estimate it from a finite sample (you
can arguably infer self-similarity from a limited number of self-
similar generations).
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