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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

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 ?

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

to

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

to

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

to

Evidently my posts are appearing, since I see replies.

I guess the question of why I don't see the posts themselves

\is ot here...

I guess the question of why I don't see the posts themselves

\is ot here...

On Thu, 18 Jun 2009 17:01:12 -0700 (PDT), Mark Dickinson

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

>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

>> >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".

>

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

>at least I'm not alone:

>

>"De?nition 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?

Perhaps not. I'm very surprised to see those definitions; I've

been a mathematician for 25 years and I've never seen a

curve defined a subset of the plane.

Hmm. You left out a bit in the first definition you cite:

"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. We assume that

each curve

comes with a fixed parametrization phi_J : R/Z ->� J. We call t in R/Z

the time

parameter. By abuse of notation, we write J(t) in R2 instead of phi_j

(t), using the

same notation for the function phi_J and its image J."

Close to sounding like he can't decide whether J is a set or a

function... Then later in the same paper

"Definition 2. A polygon is a Jordan curve that is a subset of a

finite union of

lines. A polygonal path is a continuous function P : [0, 1] ->� R2

that is a subset of

a finite union of lines. It is a polygonal arc, if it is 1 . 1."

By that definition a polygonal path is not a curve.

Worse: A polygonal path is a function from [0,1] to R^2

_that is a subset of a finite union of lines_. There's no

such thing - the _image_ of such a function can be a

subset of a finite union of lines.

Not that it matters, but his defintion of "polygonal path"

is, _if_ we're being very careful, self-contradictory.

So I don't think we can count that paper as a suitable

reference for what the _standard_ definitions are;

the standard definitions are not self-contradictory this way.

Then the second definition you cite: Amazon says the

prerequisites are two years of calculus. The stanard

meaning of log is log base e, even though it means

log base 10 in calculus.

>Mark

Jun 19, 2009, 3:13:08 PM6/19/09

to pytho...@python.org

I have.

>

>

> Hmm. You left out a bit in the first definition you cite:

>

> "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. We assume that

> each curve

> comes with a fixed parametrization phi_J : R/Z ->¨ J. We call t in R/Z

> the time

> parameter. By abuse of notation, we write J(t) in R2 instead of phi_j

> (t), using the

> same notation for the function phi_J and its image J."

>

>

> Close to sounding like he can't decide whether J is a set or a

> function...

On the contrary, I find this definition to be written with some care.

> Then later in the same paper

>

> "Definition 2. A polygon is a Jordan curve that is a subset of a

> finite union of

> lines. A polygonal path is a continuous function P : [0, 1] ->¨ R2

> that is a subset of

> a finite union of lines. It is a polygonal arc, if it is 1 . 1."

>

These are a bit too casual for me as well...

>

> By that definition a polygonal path is not a curve.

>

> Worse: A polygonal path is a function from [0,1] to R^2

> _that is a subset of a finite union of lines_. There's no

> such thing - the _image_ of such a function can be a

> subset of a finite union of lines.

>

> Not that it matters, but his defintion of "polygonal path"

> is, _if_ we're being very careful, self-contradictory.

> So I don't think we can count that paper as a suitable

> reference for what the _standard_ definitions are;

> the standard definitions are not self-contradictory this way.

Charles Yeomans

Jun 19, 2009, 3:40:36 PM6/19/09

to

On Jun 19, 7:43 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:

> Evidently my posts are appearing, since I see replies.

> I guess the question of why I don't see the posts themselves

> \is ot here...

> Evidently my posts are appearing, since I see replies.

> I guess the question of why I don't see the posts themselves

> \is ot here...

Judging by this thread, I'm not sure that much is off-topic

here. :-)

> Perhaps not. I'm very surprised to see those definitions; I've

> been a mathematician for 25 years and I've never seen a

> curve defined a subset of the plane.

That in turn surprises me. I've taught multivariable

calculus courses from at least three different texts

in three different US universities, and as far as I

recall a 'curve' was always thought of as a subset of

R^2 or R^3 in those courses (though not always with

explicit definitions, since that would be too much

to hope for with that sort of text). Here's Stewart's

'Calculus', p658:

"Examples 2 and 3 show that different sets of parametric

equations can represent the same curve. Thus we

distinguish between a *curve*, which is a set of points,

and a *parametric curve*, in which the points are

traced in a particular way."

Again as far as I remember, the rest of the language

in those courses (e.g., 'level curve', 'curve as the

intersection of two surfaces') involves thinking

of curves as subsets of R^2 or R^3. Certainly

I'll agree that it's then necessary to parameterize

the curve before being able to do anything useful

with it.

[Standard question when teaching multivariable

calculus: "Okay, so we've got a curve. What's

the first thing we do with it?" Answer, shouted

out from all the (awake) students: "PARAMETERIZE IT!"

(And then calculate its length/integrate the

given vector field along it/etc.)

Those were the days...]

A Google Books search even turned up some complex

analysis texts where the word 'curve' is used to

mean a subset of the plane; check out

the book by Ian Stewart and David Orme Tall,

'Complex Analysis: a Hitchhiker's Guide to the

Plane': they distinguish 'curves' (subset of the

complex plane) from 'paths' (functions from a

closed bounded interval to the complex plane).

> "Definition 2. A polygon is a Jordan curve that is a subset of a

> finite union of

> lines. A polygonal path is a continuous function P : [0, 1] ->¨ R2

> that is a subset of

> a finite union of lines. It is a polygonal arc, if it is 1 . 1."

>

> By that definition a polygonal path is not a curve.

Right. I'm much more willing to accept 'path' as standard

terminology for a function [a, b] -> (insert_favourite_space_here).

> Not that it matters, but his defintion of "polygonal path"

> is, _if_ we're being very careful, self-contradictory.

Agreed. Surprising, coming from Hales; he must surely rank

amongst the more careful mathematicians out there.

> So I don't think we can count that paper as a suitable

> reference for what the _standard_ definitions are;

> the standard definitions are not self-contradictory this way.

I just don't believe there's any such thing as

'the standard definition' of a curve. I'm happy

to believe that in complex analysis and differential

geometry it's common to define a curve to be a

function. But in general I'd suggest that it's one

of those terms that largely depends on context, and

should be defined clearly when it's not totally

obvious from the context which definition is

intended. For example, for me, more often than not,

a curve is a 1-dimensional scheme over a field k

(usually *not* algebraically closed), that's at

least some of {geometrically reduced, geometrically

irreducible, proper, smooth}. That's a far cry either

from a subset of an affine space or from a

parametrization by an interval.

> Then the second definition you cite: Amazon says the

> prerequisites are two years of calculus. The stanard

> meaning of log is log base e, even though means

> log base 10 in calculus.

Sorry, I've lost context for this comment. Why

are logs relevant? (I'm very well aware of the

debates over the meaning of log, having frequently

had to help students 'unlearn' their "log=log10"

mindset when starting a first post-calculus course).

Mark

Jun 20, 2009, 7:36:01 PM6/20/09

to pytho...@python.org

This simply isn't true.

Charles Yeomans

Jun 22, 2009, 8:46:55 AM6/22/09

to

I find the usage of image slightly ambiguous (as it suggests the image

set defines the curve), but that's my only qualm with it as well.

Thinking pragmatically, you can't have non-simple curves unless you

use multisets, and you also completely lose the notion of curve

orientation and even continuity without making it a poset. At this

point in time, parsimony says that you want to ditch your multiposet

thingie (and God knows what else you want to tack in there to preserve

other interesting curve properties) and really just want to define the

curve as a freaking function and be done with it.

Jun 22, 2009, 10:31:26 AM6/22/09

to pytho...@python.org

On Jun 22, 2009, at 8:46 AM, pdpi wrote:

> On Jun 19, 8:13 pm, Charles Yeomans <char...@declareSub.com> wrote:

>> On Jun 19, 2009, at 2:43 PM, David C. Ullrich wrote:

>>

>>

>> <snick>

>>

>>

>>

>>> Hmm. You left out a bit in the first definition you cite:

>>

>>> "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. We assume that

>>> each curve

>>> comes with a fixed parametrization phi_J : R/Z ->¨ J. We call t in

>>> R/Z

>>> the time

>>> parameter. By abuse of notation, we write J(t) in R2 instead of

>>> phi_j

>>> (t), using the

>>> same notation for the function phi_J and its image J."

>>

>>> Close to sounding like he can't decide whether J is a set or a

>>> function...

>>

>> On the contrary, I find this definition to be written with some care.

>

> I find the usage of image slightly ambiguous (as it suggests the image

> set defines the curve), but that's my only qualm with it as well.

>

> Thinking pragmatically, you can't have non-simple curves unless you

> use multisets, and you also completely lose the notion of curve

> orientation and even continuity without making it a poset. At this

> point in time, parsimony says that you want to ditch your multiposet

> thingie (and God knows what else you want to tack in there to preserve

> other interesting curve properties) and really just want to define the

> curve as a freaking function and be done with it.

> --

But certainly the image set does define the curve, if you want to view

it that way -- all parameterizations of a curve should satisfy the

same equation f(x, y) = 0.

Charles Yeomans

Jun 22, 2009, 2:11:02 PM6/22/09

to

On Mon, 22 Jun 2009 05:46:55 -0700 (PDT), pdpi <pdpin...@gmail.com>

wrote:

wrote:

>On Jun 19, 8:13�ｿｽpm, Charles Yeomans <char...@declareSub.com> wrote:

>> On Jun 19, 2009, at 2:43 PM, David C. Ullrich wrote:

>>

>>

>>

>>

>>

>> > Evidently my posts are appearing, since I see replies.

>> > I guess the question of why I don't see the posts themselves

>> > \is ot here...

>>

>> > On Thu, 18 Jun 2009 17:01:12 -0700 (PDT), Mark Dickinson

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

>>

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

>> >> wrote:

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

>> >>>> Right. �ｿｽ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). �ｿｽ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. �ｿｽSo I've been getting it wrong all this time. �ｿｽOh well,

>> >> at least I'm not alone:

>>

>> >> "De?nition 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?

>>

>> > Perhaps not. I'm very surprised to see those definitions; I've

>> > been a mathematician for 25 years and I've never seen a

>> > curve defined a subset of the plane.

>>

>> I have.

>>

>>

>>

>>

>>

>>

>>

>> > Hmm. You left out a bit in the first definition you cite:

>>

>> > "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. We assume that

>> > each curve

>> > comes with a fixed parametrization phi_J : R/Z ->�ｿｽ J. We call t in R/Z

>> > the time

>> > parameter. By abuse of notation, we write J(t) in R2 instead of phi_j

>> > (t), using the

>> > same notation for the function phi_J and its image J."

>>

>> > Close to sounding like he can't decide whether J is a set or a

>> > function...

>>

>> On the contrary, I find this definition to be written with some care.

>

>I find the usage of image slightly ambiguous (as it suggests the image

>set defines the curve), but that's my only qualm with it as well.

>

>Thinking pragmatically, you can't have non-simple curves unless you

>use multisets, and you also completely lose the notion of curve

>orientation and even continuity without making it a poset. At this

>point in time, parsimony says that you want to ditch your multiposet

>thingie (and God knows what else you want to tack in there to preserve

>other interesting curve properties) and really just want to define the

>curve as a freaking function and be done with it.

Precisely.

Jun 22, 2009, 2:16:56 PM6/22/09

to

This sounds like you didn't read his post, or totally missed the

point.

Say S is the set of (x,y) in the plane such that x^2 + y^2 = 1.

What's the "index", or "winding number", of that curve about the

origin?

(Hint: The curve c defined by c(t) = (cos(t), sin(t)) for

0 <= t <= 2pi has index 1 about the origin. The curve

d(t) = (cos(-t), sin(-t)) (0 <= t <= 2pi) has index -1.

The curve (cos(2t), sin(2t)) (same t) has index 2.)

>Charles Yeomans

Jun 22, 2009, 2:43:19 PM6/22/09

to

Surely you don't say a curve is a subset of the plane and

also talk about the integrals of verctor fields over _curves_?

This is getting close to the point someone else made,

before I had a chance to: We need a parametriztion of

that subset of the plane before we can do most interesting

things with it. The parametrization determines the set,

but the set does not determine the parametrization

(not even "up to" some sort of isomorphism; the

set does not determine multiplicity, orientation, etc.)

So if the definition of "curve" is not as I claim then

in some sense it _should_ be.

Hales defines a curve to be a set C and then says he assumes

that there is a parametrization phi_C. Does he ever talk

about things like the orientation of a curve a about a point?

Seems likely. If so then his use of the word "curve" is

simply not consistent with his definition.

>A Google Books search even turned up some complex

>analysis texts where the word 'curve' is used to

>mean a subset of the plane; check out

>the book by Ian Stewart and David Orme Tall,

>'Complex Analysis: a Hitchhiker's Guide to the

>Plane': they distinguish 'curves' (subset of the

>complex plane) from 'paths' (functions from a

>closed bounded interval to the complex plane).

Hmm. I of all people am in no position to judge a book

on complex analysis by the silliness if its title...

Ok.

>> Then the second definition you cite: Amazon says the

>> prerequisites are two years of calculus. The stanard

>> meaning of log is log base e, even though means

>> log base 10 in calculus.

>

>Sorry, I've lost context for this comment. Why

>are logs relevant? (I'm very well aware of the

>debates over the meaning of log, having frequently

>had to help students 'unlearn' their "log=log10"

>mindset when starting a first post-calculus course).

The point is that a calculus class is not mathematics.

In my universe the standard definition of "log" is different

froim what log means in a calculus class, and my point

was that a definition of "curve" in a book that specifies

it's supposed to be accessible to calculus students

doesn't seem to me like much evidence regarding

the standard definition in mathematics.

>Mark

Jun 22, 2009, 3:38:51 PM6/22/09

to

On Jun 22, 7:43 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:

> Surely you don't say a curve is a subset of the plane and

> also talk about the integrals of verctor fields over _curves_?

> [snip rest of long response that needs a decent reply, but

> possibly not here... ]

I wonder whether we can find a better place to have this

discussion; I think there are still plenty of interesting

things to say, but I fear we're rather abusing the hospitality

of comp.lang.python at the moment.

I'd suggest moving it to sci.math, except that I've seen the

noise/signal ratio over there...

Mark

Jun 22, 2009, 5:03:14 PM6/22/09

to pytho...@python.org

That is to say, the "winding number" is a property of both the curve

and a parameterization of it. Or, in other words, the winding number

is a property of a function from S1 to C.

Charles Yeomans

Jun 22, 2009, 10:52:59 PM6/22/09