Prev: Precompactness
Next: Cosmology, Pain, and Esoteric Mathematics Merit Million Dollar ShawPrizes
From: Dave L. Renfro on 27 May 2010 14:19
Here's an interesting problem I recently saw.

Take the usual middle thirds Cantor set, constructed on the
closed interval [0, 2*pi] instead of the closed interval
[0,1], and bend it without stretching into a circle of
radius 1 centered at the origin of the xy-coordinate plane
so that the points 0 and 2*pi are glued together at (0,1).
What are the xy-coordinates for the center of mass of the
resulting circular Cantor set, assuming a uniform density
for the Cantor set?

Dave L. Renfro
From: Dave L. Renfro on 27 May 2010 14:24
Ooops, I intended the glued points to be at (1,0),
not at (0,1), not that it really matters.

----------------------------------------------------

Here's an interesting problem I recently saw.

Take the usual middle thirds Cantor set, constructed on the
closed interval [0, 2*pi] instead of the closed interval
[0,1], and bend it without stretching into a circle of
radius 1 centered at the origin of the xy-coordinate plane
so that the points 0 and 2*pi are glued together at (1,0).
What are the xy-coordinates for the center of mass of the
resulting circular Cantor set, assuming a uniform density
for the Cantor set?


Dave L. Renfro
From: Ray Vickson on 27 May 2010 15:24
On May 27, 11:19 am, "Dave L. Renfro" <renfr...(a)cmich.edu> wrote:
> Here's an interesting problem I recently saw.
>
> Take the usual middle thirds Cantor set, constructed on the
> closed interval [0, 2*pi] instead of the closed interval
> [0,1], and bend it without stretching into a circle of
> radius 1 centered at the origin of the xy-coordinate plane
> so that the points 0 and 2*pi are glued together at (0,1).
> What are the xy-coordinates for the center of mass of the
> resulting circular Cantor set, assuming a uniform density
> for the Cantor set?
>
> Dave L. Renfro

This reminds me of a problem I posted to this group several years ago,
and did not receive a convincing answer. Take [0,1]^2 with a uniform
distribution of mass whose total weight = 1. Cut out a Lebsegue non-
measurable set in [0,1]^2. How much does it weigh?

R.G. Vickson
From: Robert Israel on 28 May 2010 03:45
"Dave L. Renfro" <renfr1dl(a)cmich.edu> writes:

> Ooops, I intended the glued points to be at (1,0),
> not at (0,1), not that it really matters.
>
> ----------------------------------------------------
>
> Here's an interesting problem I recently saw.
>
> Take the usual middle thirds Cantor set, constructed on the
> closed interval [0, 2*pi] instead of the closed interval
> [0,1], and bend it without stretching into a circle of
> radius 1 centered at the origin of the xy-coordinate plane
> so that the points 0 and 2*pi are glued together at (1,0).
> What are the xy-coordinates for the center of mass of the
> resulting circular Cantor set, assuming a uniform density
> for the Cantor set?

It's most convenient to consider the circular Cantor set as the image
of the usual Cantor set under the map f: t -> exp(2 pi i t) from [0,1]
into the complex plane. Now if X_j are independent Bernoulli random
variables with p=1/2 (i.e. flips of a fair coin),
Y = sum_{j=1}^infty (2/3^j) X_j is uniformly distributed on
the Cantor set. Thus you want

E[exp(2 pi i Y)] = product_{j=1}^infty E[exp((4/3^j) pi i X_j)]
= product_{j=1}^infty (1 + exp(4 pi i/3^j))/2
= - product_{j=1}^infty cos(2 pi/3^j)

Numerically the value is approximately
0.37143735670876563505338187851509444203474586752320

but I doubt that there is a closed form.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Rob Johnson on 28 May 2010 06:40
In article <2de3769a-6d5b-4c72-9e1f-43e877f4194a(a)c11g2000vbe.googlegroups.com>,
"Dave L. Renfro" <renfr1dl(a)cmich.edu> wrote:
>Ooops, I intended the glued points to be at (1,0),
>not at (0,1), not that it really matters.
>
>----------------------------------------------------
>
>Here's an interesting problem I recently saw.
>
>Take the usual middle thirds Cantor set, constructed on the
>closed interval [0, 2*pi] instead of the closed interval
>[0,1], and bend it without stretching into a circle of
>radius 1 centered at the origin of the xy-coordinate plane
>so that the points 0 and 2*pi are glued together at (1,0).
>What are the xy-coordinates for the center of mass of the
>resulting circular Cantor set, assuming a uniform density
>for the Cantor set?

I apologize to those with news readers that do not handle UTF-8.
The character π is pi.

The uniform measure concentrated on the Cantor set is

d_0 + d_{2/3} d_0 + d_{2/9} d_0 + d_{2/27}
u = ------------- * ------------- * -------------- * ...
2 2 2

where d_t is the dirac delta centered at t and * is convolution.

The center of mass of the Cantor set, in the complex plane, is then

|\1
| exp(2πix) u(x) dx
\| 0

Since this is the Fourier Transform of a convolution of measures
evaluated at 1, it is also the product of the Fourier Transforms of
the measures evaluated at 1. That is, the product of

|\1 d_0 + d_{2/3^k}
| exp(2πix) --------------- dx
\| 0 2

1 + exp(4πi/3^k)
= ----------------
2

= exp(2πi/3^k) cos(2π/3^k)

Thus, the point in question is

oo oo
--- ---
| | exp(2πi/3^k) | | cos(2π/3^k)
k=1 k=1

oo
---
= exp(πi) | | cos(2π/3^k)
k=1

oo
---
= - | | cos(2π/3^k)
k=1

= .371437356708765635053381878515094442034745867523203332293...

Looking back at the thread, I see that Robert Israel has come up
with the same answer using a probabilistic argument.

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
 |  Next  |  Last
Pages: 1 2 3 4 5
Prev: Precompactness
Next: Cosmology, Pain, and Esoteric Mathematics Merit Million Dollar ShawPrizes