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Jun 11, 2009, 11:32:26 PM6/11/09

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Greetings. Are there any modules, packages, whatever, that will

measure the fractal dimensions of a dataset, e.g. a time-series ?

Like the Correlation Dimension, the Information Dimension, etc...

measure the fractal dimensions of a dataset, e.g. a time-series ?

Like the Correlation Dimension, the Information Dimension, etc...

Peter

--

Peter Billam www.pjb.com.au www.pjb.com.au/comp/contact.html

Jun 14, 2009, 5:23:31 AM6/14/09

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In message <slrnh33j2b...@box8.pjb.com.au>, Peter Billam wrote:

> Are there any modules, packages, whatever, that will

> measure the fractal dimensions of a dataset, e.g. a time-series ?

I don't think any countable set, even a countably-infinite set, can have a

fractal dimension. It's got to be uncountably infinite, and therefore

uncomputable.

Jun 14, 2009, 8:04:39 AM6/14/09

to

I think there are attempts to estimate the fractal dimension of a set

using a finite sample from this set. But I can't remember where I got

this thought from!

--

Arnaud

Jun 14, 2009, 8:30:11 AM6/14/09

to

Arnaud Delobelle <arn...@googlemail.com> writes:

> I think there are attempts to estimate the fractal dimension of a set

> using a finite sample from this set. But I can't remember where I got

> this thought from!

> I think there are attempts to estimate the fractal dimension of a set

> using a finite sample from this set. But I can't remember where I got

> this thought from!

There are image data compression schemes that work like that, trying

to detect self-similarity in the data. It can go the reverse way too.

There was a program called Genuine Fractals that tried to increase the

apparent resolution of photographs by adding artificial detail

constructed from detected self-similarity. Its results were mixed, as

I remember.

Jun 14, 2009, 10:00:56 AM6/14/09

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Lawrence D'Oliveiro 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.

Strictly speaking, there's not one definition of "fractal dimension", there

are a number of them. One of the more useful is the "Hausdorf dimension",

which relates to the idea of how your measurement of the size of a thing

increases as you decrease the size of your yard-stick. The Hausdorf

dimension can be statistically estimated for finite objects, e.g. the

fractal dimension of the coast of Great Britain is approximately 1.25 while

that of Norway is 1.52; cauliflower has a fractal dimension of 2.33 and

crumpled balls of paper of 2.5; the surface of the human brain and lungs

have fractal dimensions of 2.79 and 2.97.

--

Steven

Jun 14, 2009, 5:29:04 PM6/14/09

to

On 14 Jun., 16:00, Steven D'Aprano

<st...@REMOVETHIS.cybersource.com.au> wrote:

<st...@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. 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.

Jun 14, 2009, 7:06:49 PM6/14/09

to

>> In message <slrnh33j2b...@box8.pjb.com.au>, Peter Billam wrote:

>>> Are there any modules, packages, whatever, that will

>>> measure the fractal dimensions of a dataset, e.g. a time-series ?

>>> Are there any modules, packages, whatever, that will

>>> measure the fractal dimensions of a dataset, e.g. a time-series ?

> Lawrence D'Oliveiro wrote:

>> I don't think any countable set, even a countably-infinite set, can

>> have a fractal dimension. It's got to be uncountably infinite, and

>> therefore uncomputable.

>> I don't think any countable set, even a countably-infinite set, can

>> have a fractal dimension. It's got to be uncountably infinite, and

>> therefore uncomputable.

You need a lot of data-points to get a trustworthy answer.

Of course edge-effects step in as you come up against the

spacing betwen the points; you'd have to weed those out.

On 2009-06-14, Steven D'Aprano <st...@REMOVETHIS.cybersource.com.au> wrote:

> Strictly speaking, there's not one definition of "fractal dimension", there

> are a number of them. One of the more useful is the "Hausdorf dimension",

They can be seen as special cases of Renyi's generalised entropy;

the Hausdorf dimension (D0) is easy to compute because of the

box-counting-algorithm:

http://en.wikipedia.org/wiki/Box-counting_dimension

Also easy to compute is the Correlation Dimension (D2):

http://en.wikipedia.org/wiki/Correlation_dimension

Between the two, but much slower, is the Information Dimension (D1)

http://en.wikipedia.org/wiki/Information_dimension

which most closely corresponds to physical entropy.

Multifractals are very common in nature

(like stock exchanges, if that counts as nature :-))

http://en.wikipedia.org/wiki/Multifractal_analysis

but there you really need _huge_ datasets to get useful answers ...

There have been lots of papers published (these are some refs I have:

G. Meyer-Kress, "Application of dimension algorithms to experimental

chaos," in "Directions in Chaos", Hao Bai-Lin ed., (World Scientific,

Singapore, 1987) p. 122

S. Ellner, "Estmating attractor dimensions for limited data: a new

method, with error estimates" Physi. Lettr. A 113,128 (1988)

P. Grassberger, "Estimating the fractal dimensions and entropies

of strange attractors", in "Chaos", A.V. Holden, ed. (Princeton

University Press, 1986, Chap 14)

G. Meyer-Kress, ed. "Dimensions and Entropies in Chaotic Systems -

Quantification of Complex Behaviour", vol 32 of Springer series

in Synergetics (Springer Verlag, Berlin, 1986)

N.B. Abraham, J.P. Gollub and H.L. Swinney, "Testing nonlinear

dynamics," Physica 11D, 252 (1984)

) but I haven't chased these up and I don't think they contain

any working code. But the work has been done, so the code must

be there still, on some computer somwhere...

Regards, Peter

Jun 15, 2009, 12:55:03 AM6/15/09

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On Sun, 14 Jun 2009 14:29:04 -0700, Kay Schluehr wrote:

> On 14 Jun., 16:00, Steven D'Aprano

> <st...@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

Jun 15, 2009, 7:14:14 AM6/15/09

to

On Jun 15, 5:55 am, Steven D'Aprano

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).

Jun 16, 2009, 2:22:46 PM6/16/09

to

Lawrence D'Oliveiro <l...@geek-central.gen.new_zealand> writes:

> I don't think any countable set, even a countably-infinite set, can have a

> fractal dimension. It's got to be uncountably infinite, and therefore

> uncomputable.

> fractal dimension. It's got to be uncountably infinite, and therefore

> uncomputable.

I think the idea is you assume uniform continuity of the set (as

expressed by a parametrized curve). That should let you approximate

the fractal dimension.

As for countability, remember that the reals are a separable metric

space, so the value of a continuous function any dense subset of the

reals (e.g. on the rationals, which are countable) completely

determines the function, iirc.

Jun 16, 2009, 2:57:57 PM6/16/09

to

On 15 Jun 2009 04:55:03 GMT, Steven D'Aprano

<ste...@REMOVE.THIS.cybersource.com.au> wrote:

<ste...@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...@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 Beno�t Mandelbrot, it's good

>enough for me.

Jun 16, 2009, 10:50:28 PM6/16/09

to

In message <7x63ew3...@ruckus.brouhaha.com>, wrote:

> Lawrence D'Oliveiro <l...@geek-central.gen.new_zealand> writes:

>

>> I don't think any countable set, even a countably-infinite set, can have

>> a fractal dimension. It's got to be uncountably infinite, and therefore

>> uncomputable.

>

> I think the idea is you assume uniform continuity of the set (as

> expressed by a parametrized curve). That should let you approximate

> the fractal dimension.

Fractals are, by definition, not uniform in that sense.

Jun 17, 2009, 1:35:35 AM6/17/09

to pytho...@python.org

I had my doubts on this statement being true, so I've gone to my copy

of Gerald Edgar's "Measure, Topology and Fractal Geometry" and

Proposition 2.4.10 on page 69 states: "The sequence (gk), in the

dragon construction of the Koch curve converges uniformly." And

uniform continuity is a very well defined concept, so there really

shouldn't be an interpretation issue here either. Would not stick my

head out for it, but I am pretty sure that a continuous sequence of

curves that converges to a continuous curve, will do so uniformly.

Jaime

--

(\__/)

( O.o)

( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus

planes de dominación mundial.

Jun 17, 2009, 2:04:11 AM6/17/09

to

Jaime Fernandez del Rio <jaime...@gmail.com> writes:

> I am pretty sure that a continuous sequence of

> curves that converges to a continuous curve, will do so uniformly.

> I am pretty sure that a continuous sequence of

> curves that converges to a continuous curve, will do so uniformly.

I think a typical example of a curve that's continuous but not

uniformly continuous is

f(t) = sin(1/t), defined when t > 0

It is continuous at every t>0 but wiggles violently as you get closer

to t=0. You wouldn't be able to approximate it by sampling a finite

number of points. A sequence like

g_n(t) = sin((1+1/n)/ t) for n=1,2,...

obviously converges to f, but not uniformly. On a closed interval,

any continuous function is uniformly continuous.

Jun 17, 2009, 7:37:32 AM6/17/09

to pytho...@python.org

Isn't (-∞, ∞) closed?

Charles Yeomans

Jun 17, 2009, 7:52:29 AM6/17/09

to

On Jun 17, 7:04 am, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:

> I think a typical example of a curve that's continuous but not

> uniformly continuous is

>

> f(t) = sin(1/t), defined when t > 0

>

> It is continuous at every t>0 but wiggles violently as you get closer

> to t=0. You wouldn't be able to approximate it by sampling a finite

> number of points. A sequence like

>

> g_n(t) = sin((1+1/n)/ t) for n=1,2,...

>

> obviously converges to f, but not uniformly. On a closed interval,

> any continuous function is uniformly continuous.

> I think a typical example of a curve that's continuous but not

> uniformly continuous is

>

> f(t) = sin(1/t), defined when t > 0

>

> It is continuous at every t>0 but wiggles violently as you get closer

> to t=0. You wouldn't be able to approximate it by sampling a finite

> number of points. A sequence like

>

> g_n(t) = sin((1+1/n)/ t) for n=1,2,...

>

> obviously converges to f, but not uniformly. On a closed interval,

> any continuous function is uniformly continuous.

Right, but pointwise convergence doesn't imply uniform

convergence even with continuous functions on a closed

bounded interval. For an example, take the sequence

g_n (n >= 0), of continuous real-valued functions on

[0, 1] defined by:

g_n(t) = nt if 0 <= t <= 1/n else 1

Then for any 0 <= t <= 1, g_n(t) -> 0 as n -> infinity.

But the convergence isn't uniform: max_t(g_n(t)-0) = 1

for all n.

Maybe James is thinking of the standard theorem

that says that if a sequence of continuous functions

on an interval converges uniformly then its limit

is continuous?

Mark

Jun 17, 2009, 7:56:20 AM6/17/09

to

On Jun 17, 12:52 pm, Mark Dickinson <dicki...@gmail.com> wrote:

> g_n(t) = nt if 0 <= t <= 1/n else 1

> g_n(t) = nt if 0 <= t <= 1/n else 1

Whoops. Wrong definition. That should be:

g_n(t) = nt if 0 <= t <= 1/n else

n(2/n-t) if 1/n <= t <= 2/n else 0

Then my claim that g_n(t) -> 0 for all t might

actually make sense...

Jun 17, 2009, 8:26:27 AM6/17/09

to Mark Dickinson, pytho...@python.org

On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinson<dick...@gmail.com> wrote:

> Maybe James is thinking of the standard theorem

> that says that if a sequence of continuous functions

> on an interval converges uniformly then its limit

> is continuous?

> Maybe James is thinking of the standard theorem

> that says that if a sequence of continuous functions

> on an interval converges uniformly then its limit

> is continuous?

Jaime was simply plain wrong... The example that always comes to mind

when figuring out uniform convergence (or lack of it), is the step

function , i.e. f(x)= 0 if x in [0,1), x(x)=1 if x >= 1, being

approximated by the sequence f_n(x) = x**n if x in [0,1), f_n(x) = 1

if x>=1, where uniform convergence is broken mostly due to the

limiting function not being continuous.

I simply was too quick with my extrapolations, and have realized I

have a looooot of work to do for my "real and functional analysis"

exam coming in three weeks...

Jaime

P.S. The snowflake curve, on the other hand, is uniformly continuous, right?

Jun 17, 2009, 8:46:22 AM6/17/09

to

On Jun 17, 1:26 pm, Jaime Fernandez del Rio <jaime.f...@gmail.com>

wrote:

> On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinson<dicki...@gmail.com> wrote:

> > Maybe James is thinking of the standard theorem

> > that says that if a sequence of continuous functions

> > on an interval converges uniformly then its limit

> > is continuous?

wrote:

> On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinson<dicki...@gmail.com> wrote:

> > Maybe James is thinking of the standard theorem

> > that says that if a sequence of continuous functions

> > on an interval converges uniformly then its limit

> > is continuous?

s/James/Jaime. Apologies.

> P.S. The snowflake curve, on the other hand, is uniformly continuous, right?

Yes, at least in the sense that it can be parametrized

by a uniformly continuous function from [0, 1] to the

Euclidean plane. I'm not sure that it makes a priori

sense to describe the curve itself (thought of simply

as a subset of the plane) as uniformly continuous.

Mark

Jun 17, 2009, 9:18:17 AM6/17/09

to

> 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)

Curiously enough, it's the recursive/self-similar nature of the Koch

curve so easy to prove as uniformly continuous.

Jun 17, 2009, 10:23:33 AM6/17/09

to

On Jun 17, 2:18 pm, pdpi <pdpinhe...@gmail.com> wrote:

> On Jun 17, 1:26 pm, Jaime Fernandez del Rio <jaime.f...@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)

> On Jun 17, 1:26 pm, Jaime Fernandez del Rio <jaime.f...@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).

Jun 17, 2009, 10:46:12 AM6/17/09

to

Mark Dickinson <dick...@gmail.com> 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',

> 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.

Jun 17, 2009, 11:18:52 AM6/17/09

to

On Jun 17, 3:46 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:

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

Jun 17, 2009, 11:58:48 AM6/17/09

to

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".

Jun 18, 2009, 2:13:42 PM6/18/09

to

I won't ask where I can find this definition. That Koch thing is a

closed curve in R^2. That means _by definition_ that it is a

continuous function from [0,1] to R^2 (with the same value

at the endpoints). And any continuous fu

Jun 18, 2009, 2:14:20 PM6/18/09

to

On Wed, 17 Jun 2009 14:50:28 +1200, Lawrence D'Oliveiro

<l...@geek-central.gen.new_zealand> wrote:

<l...@geek-central.gen.new_zealand> wrote:

Jun 18, 2009, 2:16:06 PM6/18/09

to

On Wed, 17 Jun 2009 14:50:28 +1200, Lawrence D'Oliveiro

<l...@geek-central.gen.new_zealand> wrote:

<l...@geek-central.gen.new_zealand> wrote:

Sorry if I've already posted half of this - having troubles hitting

the toushpad on this little machine by accident.

The fractal in question is a curve in R^2. By definition that

means it is a continuous function from [a,b] to R^2 (with

the same value at the two endpoints). Hence it's

uniformly continuous.

Jun 18, 2009, 2:17:44 PM6/18/09

to

Nope. Not that I see the relvance here - the g_k _do_

converge uniformly.

>Jaime

Jun 18, 2009, 2:19:23 PM6/18/09

to

>Isn't (-?, ?) closed?

What is your version of the definition of "closed"?

>Charles Yeomans

Jun 18, 2009, 2:21:50 PM6/18/09

to

As long as people are throwing around all this math stuff:

Officially, by definition a curve _is_ a parametrization.

Ie, a curve in the plane _is_ a continuous function from

an interval to the plane, and a subset of the plane is

not a curve.

Officially, anyway.

>Mark

Jun 18, 2009, 2:26:56 PM6/18/09

to

On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson

<dick...@gmail.com> wrote:

<dick...@gmail.com> wrote:

>On Jun 17, 3:46�pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:

>> Mark Dickinson <dicki...@gmail.com> 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,

Again, it doesn't really matter, but since you use the phrase

"if you're being careful": In fact what you say is exactly

backwards - if you're being careful that subset of the plane

is _not_ a curve (it's sometimes called the "trace" of 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.

Any _closed_ curve must have [a,b] as its parameter

interval, and hence is uniformly continuous since any

continuous function on [a,b] is uniformly continuous.

>Mark

Jun 18, 2009, 2:32:13 PM6/18/09

to pytho...@python.org

My version of a closed interval is one that contains its limit points.

Charles Yeomans

Jun 18, 2009, 4:56:57 PM6/18/09

to

David C. Ullrich <ull...@math.okstate.edu> writes:

> On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson

> <dick...@gmail.com> wrote:

>

>>On Jun 17, 3:46 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:

>>> Mark Dickinson <dicki...@gmail.com> 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,

>

> Again, it doesn't really matter, but since you use the phrase

> "if you're being careful": In fact what you say is exactly

> backwards - if you're being careful that subset of the plane

> is _not_ a curve (it's sometimes called the "trace" of the curve".

I think it is quite common to refer to call 'curve' the image of its

parametrization. Anyway there is a representation theorem somewhere

that I believe says for subsets of R^2 something like:

A subset of R^2 is the image of a continuous function [0,1] -> R^2

iff it is compact, connected and locally connected.

(I might be a bit -or a lot- wrong here, I'm not a practising

mathematician) Which means that there is no need to find a

parametrization of a plane curve to know that it is a curve.

To add to this, the usual definition of the Koch curve is not as a

function [0,1] -> R^2, and I wonder how hard it is to find such a

function for it. It doesn't seem that easy at all to me - but I've

never looked into fractals.

--

Arnaud

Jun 18, 2009, 8:01:12 PM6/18/09

to

On Jun 18, 7:26 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:

> On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson

> On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson

> >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,

>

> Again, it doesn't really matter, but since you use the phrase

> "if you're being careful": In fact what you say is exactly

> backwards - if you're being careful that subset of the plane

> is _not_ a curve (it's sometimes called the "trace" of the curve".

> >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,

>

> Again, it doesn't really matter, but since you use the phrase

> "if you're being careful": In fact what you say is exactly

> backwards - if you're being careful that subset of the plane

> is _not_ a curve (it's sometimes called the "trace" of the curve".

Darn. So I've been getting it wrong all this time. Oh well,

at least I'm not alone:

"Deﬁnition 1. A simple closed curve J, also called a

Jordan curve, is the image of a continuous one-to-one

function from R/Z to R2. [...]"

- Tom Hales, in 'Jordan's Proof of the Jordan Curve Theorem'.

"We say that Gamma is a curve if it is the image in

the plane or in space of an interval [a, b] of real

numbers of a continuous function gamma."

- Claude Tricot, 'Curves and Fractal Dimension' (Springer, 1995).

Perhaps your definition of curve isn't as universal or

'official' as you seem to think it is?

Mark

Jun 18, 2009, 8:40:55 PM6/18/09

to

David C. Ullrich <ull...@math.okstate.edu> writes:

> >> obviously converges to f, but not uniformly. On a closed interval,

> >> any continuous function is uniformly continuous.

> >

> >Isn't (-?, ?) closed?

>

> What is your version of the definition of "closed"?

> >> any continuous function is uniformly continuous.

> >

> >Isn't (-?, ?) closed?

>

> What is your version of the definition of "closed"?

I think the whole line is closed, but I hadn't realized anyone

considered the whole line to be an "interval". Apparently they do.

So that the proper statement specifies compactness (= closed and

bounded) rather than just "closed".

Jun 19, 2009, 2:43:11 PM6/19/09