center of mass of the Cantor set
Dave L. Renfro
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
Dave L. Renfro
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).
Ray Vickson
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
Robert Israel
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 isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Rob Johnson
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 <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
David C. Ullrich
The question is meaningless. Weight is a property of physical
objects, and that set is not a physical object.
I imagine that's an example of an answer that's not "convincing",
but it's true regardless.
>
>R.G. Vickson
William Hughes
If the set, call it
A, is not measurable, then it does not have a weight, [There is
a sense in which weight(A) <= 1.]
- William Hughes
Dave L. Renfro
> Looking back at the thread, I see that Robert Israel has
> come up with the same answer using a probabilistic argument.
Wow, two solutions between the time I left (Thur. afternoon)
and the time I returned (Fri. morning)!
For the record, here's one place (the only place?) the problem
can be found in the published literature:
Problem 116 (proposed by J. Wilker), Solution (by M. Shiffman and
S. Spital), Canadian Mathematical Bulletin 11 #2 (1968), 306-307.
http://books.google.com/books?id=mEpsils2t-0C&pg=PA306&lpg=PA306
The solution they arrive at is
-cos(2*pi/3)*cos(2*pi/3^2)*cos(2*pi/3^3)* ... *cos(2*pi/3^n)* ...
= 0.37143736...
Dave L. Renfro
Pubkeybreaker
<isr...@math.MyUniversitysInitials.ca> wrote:
>
> 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.
I do not see where Y comes from. Can you explain further?
Excuse my ignorance.
> 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)
Yes. This is clear.
Pubkeybreaker
> In article <2de3769a-6d5b-4c72-9e1f-43e877f41...@c11g2000vbe.googlegroups.com>,
>
> Looking back at the thread, I see that Robert Israel has come up
> with the same answer using a probabilistic argument.
>
> Rob Johnson <r...@trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a monospaced font
Your posts always impress me.
Did you write: JUST THE ESSENTIALS OF ELEMENTARY STATISTICS?
Rob Johnson
Pubkeybreaker <pubkey...@aol.com> wrote:
>On May 28, 6:40 am, r...@trash.whim.org (Rob Johnson) wrote:
>> In article <2de3769a-6d5b-4c72-9e1f-43e877f41...@c11g2000vbe.googlegroups.com>,
>
>>
>> Looking back at the thread, I see that Robert Israel has come up
>> with the same answer using a probabilistic argument.
>
Thank you. However, I hate it when posts leave an impression on my
bumper.
>Did you write: JUST THE ESSENTIALS OF ELEMENTARY STATISTICS?
It wasn't me. Nor did I sell my soul at the crossroads to play the
blues (enter "sell my soul at the crossroads to play the blues" into
Google).
Robert Israel
> On May 28, 3:45=A0am, Robert Israel
> <isr...@math.MyUniversitysInitials.ca> wrote:
> > "Dave L. Renfro" <renfr...@cmich.edu> writes:
> >
>
> >
> > 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. =A0Now if X_j are independent Bernoulli random
> > variables with p=3D1/2 (i.e. flips of a fair coin),
> > Y =3D sum_{j=3D1}^infty (2/3^j) X_j is uniformly distributed on
> > the Cantor set.
>
> I do not see where Y comes from. Can you explain further?
> Excuse my ignorance.
The Cantor set is the set of all members of [0,1] that have a base-3
expansion consisting only of 0's and 2's. 2 X_j is the j'th base-3
digit of Y.
Pubkeybreaker
<isr...@math.MyUniversitysInitials.ca> wrote:
> Pubkeybreaker <pubkeybrea...@aol.com> writes:
> > On May 28, 3:45=A0am, Robert Israel
> > <isr...@math.MyUniversitysInitials.ca> wrote:
> > > "Dave L. Renfro" <renfr...@cmich.edu> writes:
>
> > > 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. =A0Now if X_j are independent Bernoulli random
> > > variables with p=3D1/2 (i.e. flips of a fair coin),
> > > Y =3D sum_{j=3D1}^infty (2/3^j) X_j is uniformly distributed on
> > > the Cantor set.
>
> > I do not see where Y comes from. Can you explain further?
> > Excuse my ignorance.
>
> The Cantor set is the set of all members of [0,1] that have a base-3
> expansion consisting only of 0's and 2's. 2 X_j is the j'th base-3
> digit of Y.
Thanks.
Niels Diepeveen
I'm curious. When I first saw this post, I imagined the answer was
"Right next to the centre of mass of the rationals".
Usually, calculations of centres of mass are based on counting measure
for finite sets or Lebesue measure for infinite sets. Neither applies in
this case, yet everyone who replied seemed to make to make
(essentially) the same assumptions. Is there some general definition
that I'm not aware of, or is it really an ad hoc solution based on "it
stands to reason" given the symmetries.
--
Niels Diepeveen
Rob Johnson
There are several ways to characterize the uniform measure supported
on the Cantor set. One is defined in parallel to the way the Cantor
set is defined. For each stage of the middle thirds set, define the
measure to be the usual uniform measure on the remaining closed
intervals divided by their total measure. For example,
u_0 = X_[0,1]
u_1 = 3/2 X_{[0,1/3]U[2/3,1]}
u_2 = (3/2)^2 X_{[0,1/9]U[2/91/3]U[2/3,7/9]U[8/9,1]}
etc.
Each measure is absolutely continuous and as measures, they converge
to the Cantor measure.
Another way is using delta measures, as I did in my solution. The
following singular measures put delta masses at the left end of each
of the intervals of the previous measure
u_0 = d_0
d_0 + d_{2/3}
u_1 = -------------
2
d_0 + d_{2/3} d_0 + d_{2/9}
u_2 = ------------- * -------------
2 2
d_0 + d_{2/3} d_0 + d_{2/9} d_0 + d_{2/27}
u_3 = ------------- * ------------- * --------------
2 2 2
etc.
Using these measures and looking at the center of mass for each,
leads us to the center of mass of the limit measure, the Cantor
measure.
Thus the Cantor measure is the limit of absolutely continuous
measures and the limit of singular measures, as well.
David R Tribble
>> Did you write: JUST THE ESSENTIALS OF ELEMENTARY STATISTICS?
>
Rob Johnson wrote:
> It wasn't me. Nor did I sell my soul at the crossroads to play the
> blues (enter "sell my soul at the crossroads to play the blues" into
> Google).
But of course you still enjoy the blues, yes?
"The Blues ain't nothin' but a good man feelin' bad."
Rob Johnson
David R Tribble <da...@tribble.com> wrote:
>Pubkeybreaker wrote:
>>> Did you write: JUST THE ESSENTIALS OF ELEMENTARY STATISTICS?
>>
>
>Rob Johnson wrote:
>> It wasn't me. Nor did I sell my soul at the crossroads to play the
>> blues (enter "sell my soul at the crossroads to play the blues" into
>> Google).
>
>But of course you still enjoy the blues, yes?
I would say that I have a fond appreciation of the blues.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
Who put the tribbles in the quadrotriticale?
Robert Israel
The Cantor set has a "natural" measure, which corresponds to Haar measure
on {0,1}^N (N = natural numbers), and this is pretty clearly what was meant,
although "uniform density for the Cantor set" is not good terminology.
Niels Diepeveen
> Niels Diepeveen <n65...@dv1.demon.nl> writes:
>
>> Dave L. Renfro 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
>>
>> I'm curious. When I first saw this post, I imagined the answer was
>> "Right next to the centre of mass of the rationals".
>> Usually, calculations of centres of mass are based on counting measure
>> for finite sets or Lebesue measure for infinite sets. Neither applies in
>> this case, yet everyone who replied seemed to make to make
>> (essentially) the same assumptions. Is there some general definition
>> that I'm not aware of, or is it really an ad hoc solution based on "it
>> stands to reason" given the symmetries.
>
> The Cantor set has a "natural" measure, which corresponds to Haar measure
> on {0,1}^N (N = natural numbers), and this is pretty clearly what was meant,
> although "uniform density for the Cantor set" is not good terminology.
I had heard about Haar measure, but I hadn't realized that it could be
applied to this. Some more reading to do. It sounds like the sort of
justification I was looking for.
Thanks for the pointer.
--
Niels Diepeveen
Niels Diepeveen
I had dismissed this approach earlier, because I didn't see how it could
be generalized in a meaningful way. Now that you made me look at it
again, I'm starting to believe that it generalizes neatly to any compact
inifinite set of reals.
--
Niels Diepeveen
Niels Diepeveen
No, I was right the first time. The convergence to the Cantor
distribution depends on taking away pieces in a nice symmetric way.
Taking away one interval at a time doesn't necessarily give a convergent
sequence, and in carefully chosen cases a different distribution.
Thus, the distribution depends not on the set itself, but on a
particular way of contructing it.
The problem, as far as I am concerned, with these symmetry-based
arguments is that they only apply to a handful of sets. As such they
don't seem to lead to a useful extension of already existing
definitions of uniform distribution.
--
Niels Diepeveen
Robert Israel
True: in general a compact infinite set of reals has no natural "uniform"
probability measure.
You might look at Hausdorff measures, though. The Cantor measure is Hausdorff
measure of dimension ln(2)/ln(3) on the Cantor set. In general,
for a set of Hausdorff dimension d you might try looking at d-dimensional
Hausdorff measure restricted to the set. If the measure of the set
is nonzero, I think this gives a reasonable interpretation of a "uniform
distribution" on the set.
Niels Diepeveen
And many of them are probably "unsalvageable", such as the countable
ones.
>
> You might look at Hausdorff measures, though. The Cantor measure is Hausdorff
> measure of dimension ln(2)/ln(3) on the Cantor set. In general,
> for a set of Hausdorff dimension d you might try looking at d-dimensional
> Hausdorff measure restricted to the set. If the measure of the set
> is nonzero, I think this gives a reasonable interpretation of a "uniform
> distribution" on the set.
I knew most of this, but somehow I had never thought of applying it to
probability theory. It makes perfect sense though. At dimension zero,
Hausdorff measure reduces naturally to counting measure and at higher
integral dimensions it seems consistent with Lebesgue measure. (I
haven't checked the details, but for simple examples it matches.)
So, in a way, it's a natural "interpolation" between the usual
notions of "uniform".
--
Niels Diepeveen
Dave L. Renfro
> I knew most of this, but somehow I had never thought of applying
> it to probability theory. It makes perfect sense though. At
> dimension zero, Hausdorff measure reduces naturally to counting
> measure and at higher integral dimensions it seems consistent
> with Lebesgue measure. (I haven't checked the details, but for
> simple examples it matches.) So, in a way, it's a natural
> "interpolation" between the usual notions of "uniform".
Actually, there exist uncountable sets (even closed sets -- hence
fairly nice sets -- with cardinality continuum) having zero Hausdorff
dimension. Indeed, there is quite a bit of "room" at the zero
Hausdorff dimension realm for all sorts of Hausdorff h-dimension
possibilities, where h represents a Hausdorff measure function
whose graph is steeper, as you approach the origin from the right,
than is the case for any power function's graph. A "natural
example" of an uncountable set having zero Hausdorff dimension
is the set of Liouville numbers. Although this set is not closed
(it's a G_delta set), it has the added feature of being the complemnt
of a first category set (a notion of "almost all" that is different
from the "almost all" notion from measure theory), which has the
consequence that continuum many points in common with every open
interval.
Dave L. Renfro
Dave L. Renfro
> is the set of Liouville numbers. Although this set is not closed
> (it's a G_delta set), it has the added feature of being the complemnt
Better would have been to write "... is not closed (however, it is
a G_delta set), it has ...", since the way I first wrote this makes
it sound as if being a G_delta set is a reason for not being closed,
which is like saying a polygon is a rectangle is a reason for the
polygon to not be a square. The actual point I wanted to make, to
continue with the polygon analogy, is that while the polygon isn't
a square, it's still a rectangle.
Dave L. Renfro