Measuring Fractal Dimension ?
David C. Ullrich
ullrich at math.okstate.edu
Tue Jun 16 20:57:57 CEST 2009
On 15 Jun 2009 04:55:03 GMT, Steven D'Aprano
<steven 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).
No, the length of the perimeter is infinity, period. Calling it
"countably infinite" makes no sense.
You're confusing two different sorts of "infinity". A set has a
cardinality - "countably infinite" is the smallest infinite
cardinality.
Limits, as in calculus, as in that limit above, are not
cardinailities.
>
>> 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.
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