Measuring Fractal Dimension ?

Mark Dickinson dickinsm at gmail.com
Wed Jun 17 16:23:33 CEST 2009


On Jun 17, 2:18 pm, pdpi <pdpinhe... at gmail.com> wrote:
> On Jun 17, 1:26 pm, Jaime Fernandez del Rio <jaime.f... at gmail.com>
> wrote:
>
> > P.S. The snowflake curve, on the other hand, is uniformly continuous, right?
>
> The definition of uniform continuity is that, for any epsilon > 0,
> there is a delta > 0 such that, for any x and y, if x-y < delta, f(x)-f
> (y) < epsilon. Given that Koch's curve is shaped as recursion over the
> transformation from ___ to _/\_, it's immediately obvious that, for a
> delta of at most the length of ____, epsilon will be at most the
> height of /. It follows that, inversely, for any arbitrary epsilon,
> you find the smallest / that's still taller than epsilon, and delta is
> bound by the respective ____. (hooray for ascii demonstrations)

I think I'm too stupid to follow this.  It looks as though
you're treating (a portion of?) the Koch curve as the graph
of a function f from R -> R and claiming that f is uniformly
continuous.  But the Koch curve isn't such a graph (it fails
the 'vertical line test', in the language of precalculus 101),
so I'm confused.

Here's an alternative proof:

Let K_0, K_1, K_2, ... be the successive generations of the Koch
curve, so that K_0 is the closed line segment from (0, 0) to
(1, 0), K_1 looks like _/\_, etc.

Parameterize each Kn by arc length, scaled so that the domain
of the parametrization is always [0, 1] and oriented so that
the parametrizing function fn has fn(0) = (0,0) and fn(1) = (1, 0).

Let d = ||f1 - f0||, a positive real constant whose exact value
I can't be bothered to calculate[*] (where ||f1 - f0|| means
the maximum over all x in [0, 1] of the distance from
f0(x) to f1(x)).

Then from the self-similarity we get ||f2 - f1|| = d/3,
||f3 - f2|| = d/9, ||f4 - f3|| = d/27, etc.

Hence, since sum_{i >= 0} d/(3^i) converges absolutely,
the sequence f0, f1, f2, ... converges *uniformly* to
a limiting function f : [0, 1] -> R^2 that parametrizes the
Koch curve.  And since a uniform limit of uniformly continuous
function is uniformly continuous, it follows that f is
uniformly continuous.

Mark

[*] I'm guessing 1/sqrt(12).



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