center of mass of the Cantor set
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From: Niels Diepeveen on 31 May 2010 08:14 Niels Diepeveen wrote: > Rob Johnson wrote: > >> In article <4c008003$0$22920$e4fe514c(a)news.xs4all.nl>, >> Niels Diepeveen <n659474(a)dv1.demon.nl> wrote: >>>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. >> >> 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. > > 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. > 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
From: Robert Israel on 31 May 2010 12:01 Niels Diepeveen <n659474(a)dv1.demon.nl> writes: > Niels Diepeveen wrote: > > > Rob Johnson wrote: > > > >> In article <4c008003$0$22920$e4fe514c(a)news.xs4all.nl>, > >> Niels Diepeveen <n659474(a)dv1.demon.nl> wrote: > >>>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. > >> > >> 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. > > > > 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. > > > > 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. 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. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Niels Diepeveen on 4 Jun 2010 20:45 Robert Israel wrote: > Niels Diepeveen <n659474(a)dv1.demon.nl> writes: > >> Niels Diepeveen wrote: >> >> > Rob Johnson wrote: >> > >> >> In article <4c008003$0$22920$e4fe514c(a)news.xs4all.nl>, >> >> Niels Diepeveen <n659474(a)dv1.demon.nl> wrote: >> >>>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. >> >> >> >> 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. >> > >> > 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. >> > >> >> 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. > > True: in general a compact infinite set of reals has no natural "uniform" > probability measure. 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
From: Dave L. Renfro on 6 Jun 2010 09:17 Niels Diepeveen wrote: > 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
From: Dave L. Renfro on 6 Jun 2010 09:24 Dave L. Renfro wrote (in part): > 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
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