Measuring Fractal Dimension ?

pdpi pdpinheiro at
Wed Jun 17 11:58:48 EDT 2009

On Jun 17, 4:18 pm, Mark Dickinson <dicki... at> wrote:
> On Jun 17, 3:46 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
> > Mark Dickinson <dicki... at> writes:
> > > 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',
> > I think you treat it as a function f: R -> R**2 with the usual
> > distance metric on R**2.
> Right.  Or rather, you treat it as the image of such a function,
> if you're being careful to distinguish the curve (a subset
> of R^2) from its parametrization (a continuous function
> R -> R**2).  It's the parametrization that's uniformly
> continuous, not the curve, and since any curve can be
> parametrized in many different ways any proof of uniform
> continuity should specify exactly which parametrization is
> in use.
> Mark

I was being incredibly lazy and using loads of handwaving, seeing as I
posted that (and this!) while procrastinating at work.

an even lazier argument: given the _/\_ construct, you prove that its
vertical growth is bound: the height of / is less than 1/3 (given a
length of 1 for ___), so, even if you were to build _-_ with the
middle segment at height = 1/3, the maximum vertical growth would be
sum 1/3^n from 1 to infinity, so 0.5. Sideways growth has a similar
upper bound. 0.5 < 1, so the chebyshev distance between any two points
on the curve is <= 1. Ergo, for any x,y, f(x) is at most at chebyshev
distance 1 of (y). Induce the argument for "smaller values of one".

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