Velocity Reviews > Re: Measuring Fractal Dimension ?

# Re: Measuring Fractal Dimension ?

pdpi
Guest
Posts: n/a

 06-17-2009
On Jun 17, 4:18*pm, Mark Dickinson <(E-Mail Removed)> wrote:
> On Jun 17, 3:46*pm, Paul Rubin <http://(E-Mail Removed)> wrote:
>
> > Mark Dickinson <(E-Mail Removed)> 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".

Charles Yeomans
Guest
Posts: n/a

 06-18-2009

On Jun 18, 2009, at 2:19 PM, David C. Ullrich wrote:

> On Wed, 17 Jun 2009 07:37:32 -0400, Charles Yeomans
> <(E-Mail Removed)> wrote:
>
>>
>> On Jun 17, 2009, at 2:04 AM, Paul Rubin wrote:
>>
>>> Jaime Fernandez del Rio <(E-Mail Removed)> writes:
>>>> 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.

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

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

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

Charles Yeomans

Arnaud Delobelle
Guest
Posts: n/a

 06-18-2009
David C. Ullrich <(E-Mail Removed)> writes:

> On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson
> <(E-Mail Removed)> wrote:
>
>>On Jun 17, 3:46*pm, Paul Rubin <http://(E-Mail Removed)> wrote:
>>> Mark Dickinson <(E-Mail Removed)> 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

Mark Dickinson
Guest
Posts: n/a

 06-19-2009
On Jun 18, 7:26*pm, David C. Ullrich <(E-Mail Removed)> 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:

"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

Paul Rubin
Guest
Posts: n/a

 06-19-2009
David C. Ullrich <(E-Mail Removed)> 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"?

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".

Charles Yeomans
Guest
Posts: n/a

 06-19-2009

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
> <(E-Mail Removed)> wrote:
>
>> On Jun 18, 7:26 pm, David C. Ullrich <(E-Mail Removed)>
>> 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.

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

Mark Dickinson
Guest
Posts: n/a

 06-19-2009
On Jun 19, 7:43*pm, David C. Ullrich <(E-Mail Removed)> 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...

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

Charles Yeomans
Guest
Posts: n/a

 06-20-2009

On Jun 18, 2009, at 2:21 PM, David C. Ullrich wrote:

> On Wed, 17 Jun 2009 05:46:22 -0700 (PDT), Mark Dickinson
> <(E-Mail Removed)> wrote:
>
>> On Jun 17, 1:26 pm, Jaime Fernandez del Rio <(E-Mail Removed)>
>> wrote:
>>> On Wed, Jun 17, 2009 at 1:52 PM, Mark
>>> Dickinson<(E-Mail Removed)> 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.

>
> 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.

This simply isn't true.

Charles Yeomans

pdpi
Guest
Posts: n/a

 06-22-2009
On Jun 19, 8:13*pm, Charles Yeomans <(E-Mail Removed)> 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
> > <(E-Mail Removed)> wrote:

>
> >> On Jun 18, 7:26 pm, David C. Ullrich <(E-Mail Removed)> *
> >> 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.

>
> 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.

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.

Charles Yeomans
Guest
Posts: n/a

 06-22-2009

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

> On Jun 19, 8:13 pm, Charles Yeomans <(E-Mail Removed)> 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