Where is the fallacy here?

Fremen

This is a bit embarassing, but I can't quite find the gap in the logic behind this picture:



We've got this curve that's defined as the limit of an iterative procedure. It seems pretty clear to me that the curve converges pointwise to the circle, and also that it has some kind of fractal structure.

So where's the flaw in the argument? Is it that the iterative procedure converges pointwise but not uniformly?

I have a feeling this thread could open a can of worms in the way threads about Monty Hall and .9 recurring = 1 do. I'd just like to ask people to show a little restraint - please only try to answer if you actually think you understand what's going on. Don't just venture an opinion.

Thanks.

Edit: now I'm not even sure there's a sensible notion of pointwise convergence for the rectangular curve. First question is now "in what sense is convergence taking place here?".

PaulieBoy

I'll show my complete ignorance by saying that the square does not converge to a circle but is a deformed square that looks like it converges to a circle.

hivizman

It's to do with cardinality. In the diagram where the square is drawn round the circle, the perimeter touches the circle at 4 points. In the next diagram, the perimeter touches the circle at 8 points, in the fourth diagram at 16 points, and so on. As the process tends to infinity, the number of points where the perimeter touches the circle tends to infinity, but infinity in the sense of the size of the set of all natural numbers (the points making up the perimeter can be put into one-to-one correspondence with the natural numbers).

However, the circle, being a continuous line in two-dimensional space, consists of an infinite number of points where infinity has the sense of the size of the set of all real numbers (the points can be put into one-to-one correspondence with the real numbers).

As Georg Cantor showed, the size (cardinality) of the set of real numbers is strictly greater than the size (cardinality) of the set of the natural numbers. So, even at infinity, the perimeter touches the circle only at a strict subset of the circle's points, and hence the perimeter does not "converge" in a precise sense to the circle.

This implies a more subtle "paradox"; if the perimeter converges to a set of points on the circle that represents a strict subset of the set of all points making up the circle, how is the length of the perimeter greater than the circumference of the circle? I think I have an explanation, but I'll leave this for others to suggest.

Fremen

hivizman wrote: »

As Georg Cantor showed, the size (cardinality) of the set of real numbers is strictly greater than the size (cardinality) of the set of the natural numbers. So, even at infinity, the perimeter touches the circle only at a strict subset of the circle's points, and hence the perimeter does not "converge" in a precise sense to the circle.

I agree with what you've said up until here. I think there must be a sense in which the curve converges to a circle, though: the white area approaches 0 asymptotically, while all curves are guaranteed to enclose the blue area. It reminds me a bit of L^p convergence of functions.

In addition, the distance from the circle to the furthest point on the black curve tends to 0.

CJC86

Fremen wrote: »

I agree with what you've said up until here. I think there must be a sense in which the curve converges to a circle, though: the white area approaches 0 asymptotically, while all curves are guaranteed to enclose the blue area. It reminds me a bit of L^p convergence of functions.

In addition, the distance from the circle to the furthest point on the black curve tends to 0.

The (piecewise smooth) curve may converge pointwise to a circle, but it will have countably many singularities (ie. points where the dimension of the tangent space is greater than 1), so it will certainly never be a circle.

So, we have an approximation of a smooth curve by a piecewise smooth curve, that contains the exact same region of the plane. No reason to believe that they would be the same length... (I think that makes sense...)

bnt

hivizman wrote: »

This implies a more subtle "paradox"; if the perimeter converges to a set of points on the circle that represents a strict subset of the set of all points making up the circle, how is the length of the perimeter greater than the circumference of the circle? I think I have an explanation, but I'll leave this for others to suggest.

I thought I smelled a Fractal, but apparently not. Oh well.

ray giraffe

CJC86 wrote: »

The (piecewise smooth) curve may converge pointwise to a circle, but it will have countably many singularities (ie. points where the dimension of the tangent space is greater than 1), so it will certainly never be a circle.

So, we have an approximation of a smooth curve by a piecewise smooth curve, that contains the exact same region of the plane. No reason to believe that they would be the same length... (I think that makes sense...)

You could easily avoid singularities and still have the same problem.

[latex]z_n(\theta) = (1 + (1/n)\sin(n^2\theta))e^{i\theta}[/latex]

This is a smooth curve converging uniformly to the circle, but increasing in length to infinity!

I think the problem is that although the approximations converge pointwise to the circle, the derivatives (directions) of the approximations are not converging pointwise to the derivative (direction) of the circle. So when you find lengths of the approximations using their derivatives [latex]L_n = \int_{I}{x'_n(t)dt}[/latex] they won't converge to the length of the circle.

From wikipedia "even if the convergence is uniform,... the derivative of the limit need not be equal to the limit of the derivatives".

Therefore in my words "even if the convergence is uniform,... the length of the limit need not be equal to the limit of the lengths".

Quote: "If f_n converges uniformly to f, and if all the f_n are differentiable, and if derivatives f_n' converge uniformly to g, then f is differentiable and its derivative is g."

ray giraffe

hivizman wrote: »

So, even at infinity, the perimeter touches the circle only at a strict subset of the circle's points, and hence the perimeter does not "converge" in a precise sense to the circle.

So [latex] f_n:\mathbb{R} \to \mathbb{R}, f_n(x)= 1/n [/latex] doesn't converge to the zero function because it never "touches" the zero function?

Take a look at the definition of convergence of a function here

CJC86

ray giraffe wrote: »

You could easily avoid singularities and still have the same problem.

[latex]z_n(\theta) = (1 + 1/n) \sin(n^2\theta)e^{i\theta}[/latex]

This is a smooth curve converging uniformly to the circle, but increasing in length to infinity!

I don't get it...

ray giraffe

CJC86 wrote: »

The (piecewise smooth) curve may converge pointwise to a circle, but it will have countably many singularities (ie. points where the dimension of the tangent space is greater than 1)

Maybe I don't understand this. What will have countably many singularities? Each approximation has finitely many singularities, the pointwise limit is the circle which has no singularities, am i right?

ray giraffe

CJC86 wrote: »

I don't get it...

Length is not preserved in taking a uniform limit of smooth functions. (i.e. even with no non-differentiable points)
[latex]f_n: [0,1] \to \mathbb{R}, f_n(x)=(1/n)\sin(nx) [/latex] does not approach length 1.

Enkidu

hivizman wrote: »

It's to do with cardinality....

I think the cardinality is only countable for the points that converge in a finite number of steps, however all the points do converge. Your arguement would go through for almost any construction in real analysis if this wasn't the case.

My opinion is that this is just the break down of Lebesgue's dominated convergence theorem. The lines converge to the circle, but the limit of the length (given by an integral) is not the length of the limiting line, i.e.
if [latex]x_n(t)[/latex] are the functions parameterising the lines, the length
is given by
[latex]L_n = \int_{\Gamma}{x_n(t)dt}[/latex]
with [latex]\Gamma[/latex] just the path of integration, then
[latex]lim_{n\rightarrow\infty} L_n \neq \int_{\Gamma}{x(t)dt}[/latex]
with [latex]x(t)[/latex] the limiting curve, i.e. the circle.

CJC86

ray giraffe wrote: »

Length is not preserved in taking a uniform limit of smooth functions. (i.e. even with no non-differentiable points)
[latex]f_n: [0,1] \to \mathbb{R}, f_n(x)=(1/n)\sin(nx) [/latex] does not approach length 1.

What I don't get is how the function you described above converges uniformly to the circle. It could be "obvious", but I just don't see it.

I messed up my language in my first post. It does seem that the pointwise limit of the piecewise smooth curve is the circle, hence smooth. What I should then have said is that for each n, there are finitely many singularities, hence I think it would be foolish to think that the limit of the lengths of these curves would converge to the length of the circle.

As you pointed out in your edited post above, the reason the lengths don't converge is due to there being no convergence of the derivatives to the derivative of the limit curve. This is largely saying the same thing, replace "derivative" with "tangent space" and there you go.

ray giraffe

CJC86 wrote: »

What I don't get is how the function you described above converges uniformly to the circle.

Sorry, I made an error in bracketing. It's fixed now!

Remember the circle is just [latex]z(\theta) = e^{i\theta}[/latex]

LeixlipRed

I need to brush up on my analysis, this is all going over my head! Anyone got a link to a good video tutorial, intro to real analysis type thing? Really is the biggest hole in my mathematical knowledge and it's about the worst hole one could have :pac:

Fremen

bnt wrote: »

I thought I smelled a Fractal, but apparently not. Oh well.

Yes, that's a good explanation. So even though the curve is converging uniformly, you can't necessarily exchange limit and integral, as Enkidu pointed out.

LR, Terry Tao's recent book on measure theory might be a good place to learn this stuff. I find his explanations really enlightening. He has a PDF online here. I'd love to work through the thing myself at some point. There's some material on convergence of functions from p.114. Don't know of any videos though.

ray giraffe

Enkidu wrote: »

I think the cardinality is only countable for the points that converge in a finite number of steps, however all the points do converge. Your arguement would go through for almost any construction in real analysis if this wasn't the case.

My opinion is that this is just the break down of Lebesgue's dominated convergence theorem. The lines converge to the circle, but the limit of the length (given by an integral) is not the length of the limiting line, i.e.
if [latex]x_n(t)[/latex] are the functions parameterising the lines, the length
is given by
[latex]L_n = \int_{\Gamma}{x_n(t)dt}[/latex]
with [latex]\Gamma[/latex] just the path of integration, then
[latex]lim_{n\rightarrow\infty} L_n \neq \int_{\Gamma}{x(t)dt}[/latex]
with [latex]x(t)[/latex] the limiting curve, i.e. the circle.

Surely the length of the curve is [latex]L_n = \int_{I}{\|x'_n(t)\|dt}[/latex] where [latex]x_n(t)[/latex] is the function parameterising the line on the interval I?

See wikipedia

ray giraffe

LeixlipRed wrote: »

I need to brush up on my analysis, this is all going over my head! Anyone got a link to a good video tutorial, intro to real analysis type thing? Really is the biggest hole in my mathematical knowledge and it's about the worst hole one could have :pac:

I remember this book used to be in the Maynooth library. Tony recommended it to me in a second year module. The module started with proof of "A metric space is compact iff it is complete and totally bounded iff it is sequentially compact". This was 12 months before we started the module on basic pointset topology.

LeixlipRed

Cheers guys!

Enkidu

ray giraffe wrote: »

Surely the length of the curve is [latex]L_n = \int_{I}{x'_n(t)dt}[/latex] where [latex]x_n(t)[/latex] is the function parameterising the line on the interval I?

See wikipedia

What a silly error, thanks ray giraffe.:)

equivariant

I think the key issue with this example is the type of convergence (or the choice of norm). Essentially what you have is a sequence of functions that converge point wise to a given function (in fact the convergence is uniform even). There is no reason to expect a corresponding convergence of a property (ie arc length) that is defined in terms of the derivatives of the functions. To get that type of convergence you would probably require convergence wrt some sort of sobolev norm

See http://en.wikipedia.org/wiki/Sobolev_space