September 2015 - mathfun@mailman.xmission.com

[math-fun] Non-real measures
by David Wilson 08 Sep '15

08 Sep '15

Is it possible to generalize measure on a set to non-real values, e.g. the surreal numbers? Could this allow us to measure more sets, perhaps distinguish sets with real measure 0?

2 1

Re: [math-fun] Verticals
by Keith F. Lynch 08 Sep '15

08 Sep '15

"Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> These ideas can of course be extended to lumpier worlds. Could a >> local vertical ever point directly away from the center? > Suppose the planet is a solid torus of uniform density. Then if you > stand on the inner Equator, gravity will pull you directly away from > the centre of the planet. Right. > (If you insist that the planet is homotopy-equivalent to a closed > ball, just include an infinitely thin membrane spanning the hole.) My solution was a hollow sphere with a hole drilled in the north pole. A plumbbob at the inner south pole would point directly away from the center. I'm now exploring two related ideas, each of which I may turn into a new post: It's well known that the Moon's gravity is too lumpy for there to be any long-term stable orbits. Similarly around Mercury, where the first and only Mercury orbiter did in fact recently crash into the planet. Of course an orbit can be adjusted as necessary by expending rocket fuel, but you would soon run out. Would it be possible to maintain an orbit indefinitely by having a satellite consisting of two masses on a tether, using solar energy to reel in and out the tether to circularize the orbit? A flywheel may also be necessary. What about the opposite extreme, a perfectly fluid planet? What shape would such a planet have? If it wasn't rotating, it would obviously be a sphere. With rotation, would be it be an ellipsoid, or is that only a good first-order approximation where gravity greatly exceeds centrifugal force? Would the shape depend on the distribution of mass, e.g. would it be one shape if nearly all the mass was near the center, and a different shape if the planet was of uniform density?

3 2

[math-fun] Defending fractions as a Countable Set
by Ed Pegg Jr 08 Sep '15

08 Sep '15

Over at Reddit, http://i.imgur.com/tDSX24E.jpg is getting a lot of attention. https://www.reddit.com/r/math/comments/3k1qe6/this_is_in_a_high_school_math… In short, the textbook Glencoe Algebra II claims that there is no relation between the integers and the rationals. Georg Cantor showed it in the 1870's. There is also Calkin-Wilf and Stern-Brocat CalkinWilfPosition[r_] := With[{cf = ContinuedFraction[r]}, First@FromDigits[ Flatten@MapIndexed[Table[Mod[#2[[1]], 2], {#1}] &, Reverse@#], {2}] &@ If[EvenQ[Length[cf]], MapAt[Sequence @@ {# - 1, 1} &, cf, -1], cf]]; CalkinWilfRational[n_Integer] := Module[{s0, s1}, s0 = IntegerDigits[n, 2]; s1 = If[EvenQ[Length[#]], MapAt[(Sequence @@ {#, {}}) &, #, -1], #] &@Split[s0]; FromContinuedFraction[Reverse[Length /@ s1]]]; AlmostBitReverse[n_] := FromDigits[ Prepend[Reverse[Drop[IntegerDigits[n, 2], 1]], 1], 2]; SternBrocatPosition[fraction_] := AlmostBitReverse[CalkinWilfPosition[fraction]]; SternBrocatRational[n_Integer] := CalkinWilfRational[AlmostBitReverse[n]]; I figured that the relationship between these would be in OEIS -- and they are. https://oeis.org/A059893 -- but this doesn't mention the relation. But it's how I came up with the Stern-Brocat code. If you order either set of fractions, the position in the other set is given by A059893 --Ed Pegg Jr.

3 2

[math-fun] Cantor set of measure 1?
by Warren D Smith 08 Sep '15

08 Sep '15

To ask whether these are homeomorphic, we need to know: what is a homeomorphism? https://en.wikipedia.org/wiki/Homeomorphism and yes, it seems to me all the Cantor sets I defined for each K>=3, including the K=3 case corresponding to Cantor's original defn, are homeomorphic. See, each point in each set is describable using a countably-infinitely-long bitstring. The Nth 0-or-1 specifies which of the two halves you went into at stage N of the construction. Simply make the bijection be: points with same bitstring correspond. Obviously it is a bijection. Then the only question is: is this bijection CONTINUOUS in both directions? It does not take long reviewing the epsilon delta defn of continuity, to claim the answer is "yes."

1 0

[math-fun] gravity on planet surface
by Warren D Smith 08 Sep '15

08 Sep '15

Perhaps more amusing puzzle than, e.g. asking for torus-shaped planets where gravity points "up" in places, is: can we devise planets where (a) gravity points "up" everywhere? (b) it points up on average over the whole surface? With Newtonian gravity where interior of planet has positive mass density, surface has a normal almost everywhere... because of the Gauss divergence theorem we see that the gravity field must, averaged over whole planet surface, always point "down" i.e. have negative dot product on average with the unit outward surface-normal. So both a and b are impossible. But you could by spinning a planet have a situation where the gravity+centrifugal force net pointed "up" everywhere except near the poles, and where the surface-average was "up." Is it possible to have that point up everywhere except a surface subset of measure 0? On 9/8/15, math-fun-request(a)mailman.xmission.com <math-fun-request(a)mailman.xmission.com> wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > Today's Topics: > > 1. Re: Cantor set of measure 1? (Thomas Colthurst) > 2. Re: Cantor set of measure 1? (Dan Asimov) > 3. Poincare dual of Penrose tiling? (Mike Stay) > 4. Cantor set of measure 1? (Warren D Smith) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Tue, 8 Sep 2015 10:51:43 -0400 > From: Thomas Colthurst <thomaswc(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Cantor set of measure 1? > Message-ID: > <CAHmrjv1Ch4CzzU_9LocrHnDa6NPT0RNCfv3FFTTTgBwWPRJMXQ(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Hi, Dan! > > For what it is worth, I get a different convex hull: [-2/3, 5/3]. The > left end point is the fixed point of x -> (1/2) x - 1/3 and the right end > point is the fixed point of x -> (1/2) x + 5/6. > > And actually, you are right about this not being a Cantor set; I was wrong > about it satisfying the open set condition. Specifically, the first > function maps [-2/3, 5/3] to [-2/3, 1/2] and the second function maps > [-2/3, 5/3] to [1/2, 5/3]. So the final attractor is just the line segment > [-2/3, 5/3]. > > Oh well, at least I was right about it having measure 1. :) > > -Thomas C > > On Tue, Sep 8, 2015 at 10:39 AM, Dan Asimov <asimov(a)msri.org> wrote: > >> ----- >> "It seems to me that at each stage we get 2^n disjoint closed intervals >> of >> length 1/2^n separated from one another by 1/3^n, and that every point >> (including the duplicate points created along the way) moves a finite >> distance and therefore has a well-defined location 'at time infinity'. So >> we seem to get a well-defined set 'at time infinity'." >> ----- >> >> If I understand the construction correctly, it can just as well go on >> within the limiting convex hull of all the points created,* namely [0-K, >> 1+K] where >> >> K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2, >> >> i.e., [-1/2, 3/2]. >> >> * It can just as well go on in [-1/2,3/2] because we can always dilate >> each stage about its midpoint by the appropriate factor so it exactly >> fits >> into [-1/2,3/2]. >> >> If that and the quoted paragraph above are correct, then IF this >> dilatation of each stage into [-1/2,3/2] is done, then the nth stage will >> consist of 2^n closed intervals of equal length L_n, separated by 2^n - 1 >> open intervals of length >> >> M_n = (2/3)^n * L_n, >> >> such that >> >> 2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n. >> >> (And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - >> 1)*(2/3)^n].) >> >> THEN it appears to me that, however the limiting set X is defined, there >> will (by symmetry) be a countable dense set of c in [0,2] for which the >> translations >> >> x |-> x + c >> >> will take a nonempty interval's worth of X to a different equal-size >> interval's worth of X. >> >> This implies that X contains points dense in some interval. But since the >> Cantor set is closed in R no matter how it is embedded, this implies it >> contains an entire interval, which is impossible for a Cantor set. >> >> Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely >> negating what I have hypothesized here. >> >> ?Dan >> >> >> > On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp(a)gmail.com> wrote: >> > >> > Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, >> > ...) >> as >> > the set obtained from S_{n-1} by replacing each of the 2^n closed >> intervals >> > [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >> > [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >> duplicate >> > the midpoint, shift the left half to the left by 1/3^n, and shift the >> right >> > half to the right by 1/3^n). >> >> > It seems to me that at each stage we get 2^n disjoint closed intervals >> > of >> > length 1/2^n separated from one another by 1/3^n, and that every point >> > (including the duplicate points created along the way) moves a finite >> > distance and therefore has a well-defined location "at time infinity". >> > So >> > we seem to get a well-defined set "at time infinity". >> >> > Am I right? And do we obtain a set of measure 1 that's homeomorphic to >> the >> > Cantor set? >> > >> > I've seen constructions of "fat Cantor sets" on the web, where one >> modifies >> > the culling procedure, but I've never seen one that uses sliding and >> > preserves the measure throughout the procedure. So I'm worried that I'm >> > overlooking something. >> > >> > Jim Propp >> > _______________________________________________ >> > math-fun mailing list >> > math-fun(a)mailman.xmission.com >> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > ------------------------------ > > Message: 2 > Date: Tue, 8 Sep 2015 07:58:25 -0700 > From: Dan Asimov <asimov(a)msri.org> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Cantor set of measure 1? > Message-ID: <59CD4A9E-A2D0-41D8-B717-1B9BB10B8D8B(a)msri.org> > Content-Type: text/plain; charset=utf-8 > >> On Sep 8, 2015, at 7:51 AM, Thomas Colthurst <thomaswc(a)gmail.com> wrote: >> >> Hi, Dan! >> >> For what it is worth, I get a different convex hull: [-2/3, 5/3]. The >> left end point is the fixed point of x -> (1/2) x - 1/3 and the right end >> point is the fixed point of x -> (1/2) x + 5/6. >> >> And actually, you are right about this not being a Cantor set; I was >> wrong >> about it satisfying the open set condition. Specifically, the first >> function maps [-2/3, 5/3] to [-2/3, 1/2] and the second function maps >> [-2/3, 5/3] to [1/2, 5/3]. So the final attractor is just the line >> segment >> [-2/3, 5/3]. >> >> Oh well, at least I was right about it having measure 1. :) > > Okay. (I also agreed with that.) > > ?Dan > > >> On Tue, Sep 8, 2015 at 10:39 AM, Dan Asimov <asimov(a)msri.org> wrote: >> >>> ----- >>> "It seems to me that at each stage we get 2^n disjoint closed intervals >>> of >>> length 1/2^n separated from one another by 1/3^n, and that every point >>> (including the duplicate points created along the way) moves a finite >>> distance and therefore has a well-defined location 'at time infinity'. >>> So >>> we seem to get a well-defined set 'at time infinity'." >>> ----- >>> >>> If I understand the construction correctly, it can just as well go on >>> within the limiting convex hull of all the points created,* namely [0-K, >>> 1+K] where >>> >>> K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2, >>> >>> i.e., [-1/2, 3/2]. >>> >>> * It can just as well go on in [-1/2,3/2] because we can always dilate >>> each stage about its midpoint by the appropriate factor so it exactly >>> fits >>> into [-1/2,3/2]. >>> >>> If that and the quoted paragraph above are correct, then IF this >>> dilatation of each stage into [-1/2,3/2] is done, then the nth stage >>> will >>> consist of 2^n closed intervals of equal length L_n, separated by 2^n - >>> 1 >>> open intervals of length >>> >>> M_n = (2/3)^n * L_n, >>> >>> such that >>> >>> 2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n. >>> >>> (And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - >>> 1)*(2/3)^n].) >>> >>> THEN it appears to me that, however the limiting set X is defined, there >>> will (by symmetry) be a countable dense set of c in [0,2] for which the >>> translations >>> >>> x |-> x + c >>> >>> will take a nonempty interval's worth of X to a different equal-size >>> interval's worth of X. >>> >>> This implies that X contains points dense in some interval. But since >>> the >>> Cantor set is closed in R no matter how it is embedded, this implies it >>> contains an entire interval, which is impossible for a Cantor set. >>> >>> Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely >>> negating what I have hypothesized here. >>> >>> ?Dan >>> >>> >>>> On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp(a)gmail.com> wrote: >>>> >>>> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, >>>> ...) >>> as >>>> the set obtained from S_{n-1} by replacing each of the 2^n closed >>> intervals >>>> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >>>> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >>> duplicate >>>> the midpoint, shift the left half to the left by 1/3^n, and shift the >>> right >>>> half to the right by 1/3^n). >>> >>>> It seems to me that at each stage we get 2^n disjoint closed intervals >>>> of >>>> length 1/2^n separated from one another by 1/3^n, and that every point >>>> (including the duplicate points created along the way) moves a finite >>>> distance and therefore has a well-defined location "at time infinity". >>>> So >>>> we seem to get a well-defined set "at time infinity". >>> >>>> Am I right? And do we obtain a set of measure 1 that's homeomorphic to >>> the >>>> Cantor set? >>>> >>>> I've seen constructions of "fat Cantor sets" on the web, where one >>> modifies >>>> the culling procedure, but I've never seen one that uses sliding and >>>> preserves the measure throughout the procedure. So I'm worried that I'm >>>> overlooking something. >>>> >>>> Jim Propp >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > ------------------------------ > > Message: 3 > Date: Tue, 8 Sep 2015 08:17:42 -0700 > From: Mike Stay <metaweta(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Poincare dual of Penrose tiling? > Message-ID: > <CAKQgqTa0GP2D0zb7NQAQ7De7oAN=WXNHdLTS-yPxr-UTt4apuA(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Anyone know where I can find a picture of the Poincare dual of a > Penrose tiling? Google images fails me. > -- > Mike Stay - metaweta(a)gmail.com > http://www.cs.auckland.ac.nz/~mike > http://reperiendi.wordpress.com > > > > ------------------------------ > > Message: 4 > Date: Tue, 8 Sep 2015 11:19:32 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Cantor set of measure 1? > Message-ID: > <CAAJP7Y3_F2tc3xSDmqY45Bp4aHbmpVv1NbGCvCxvkbMNkWOxuw(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > If we wanted to omit "sliding" from Propp's construction (to be more > in tune to what others have called "Cantor set" constructions) > then, fix K>2 and > > 1. start with unit interval [0,1] and N=1. > 2. from each of the 2^(N-1) intervals: > delete the open subinterval centered at the midpoint and > of width K^(-N). > 3. Increment N and go back to step 2. > > This construction begins with measure=1, and after the Nth iteration of > stage 2 > has deleted measure > 1/K + 2/K^2 + 4/K^3 + ... + 2^(N-1)/K^N > = > K^(-N) * (K^N-2^N) / (K-2) > --> > 1/(K-2) in the N-->infinity limit. > > Note if K=3, then the deleted measure is exactly 1, so we end up with > a zero-measure > nonempty set, which is, in fact, exactly Cantor's original "remove > middle third" construction (expressed differently, but it is the > same). > If K>3 then the deleted measure is less than 1, so we end up with a "Cantor" > set > whose measure is a fixed constant between 0 and 1 and which is totally > disconnected. > > > > On 9/8/15, math-fun-request(a)mailman.xmission.com > <math-fun-request(a)mailman.xmission.com> wrote: >> Send math-fun mailing list submissions to >> math-fun(a)mailman.xmission.com >> >> To subscribe or unsubscribe via the World Wide Web, visit >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> or, via email, send a message with subject or body 'help' to >> math-fun-request(a)mailman.xmission.com >> >> You can reach the person managing the list at >> math-fun-owner(a)mailman.xmission.com >> >> When replying, please edit your Subject line so it is more specific >> than "Re: Contents of math-fun digest..." >> >> >> Today's Topics: >> >> 1. Re: Where's the b? (rwg) >> 2. Re: Verticals (William R. Somsky) >> 3. Cantor set of measure 1? (James Propp) >> 4. Re: Cantor set of measure 1? (Thomas Colthurst) >> 5. Re: Cantor set of measure 1? (Dan Asimov) >> >> >> ---------------------------------------------------------------------- >> >> Message: 1 >> Date: Mon, 07 Sep 2015 18:45:49 -0700 >> From: rwg <rwg(a)sdf.org> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Cc: A.OldeDaalhuis(a)ed.ac.uk >> Subject: Re: [math-fun] Where's the b? >> Message-ID: <3b819eed1ec594f97bf227bef4918564(a)sdf.org> >> Content-Type: text/plain; charset=UTF-8; format=flowed >> >> OOPS!! Pofessor A. B. Olde Daalhuis, DLMF editor for Hypergeometric >> Functions >> (Chapters 13, 15, 16), kindly informs me that this transformation >> follows from >> switching the sides of http://dlmf.nist.gov/15.8.E22 and >> recanonicalizing the >> argument and parameters! I typed in all of 15.8, but I should have been >> more >> vigilant. Adding to my confusion, FullSimplify demurs due to the >> apparent >> failure of the reversed identity on the negative real axis. We may need >> one >> of Marichev's patented Sqrt[x]*Sqrt[1/x] fudge factors. --rwg >> >> On 2015-09-07 02:04, rwg wrote: >>> On 2015-09-07 01:34, Joerg Arndt wrote: >>>> * Bill Gosper <billgosper(a)gmail.com> [Sep 07. 2015 09:28]: >>>>> The elliptic K transformation that NeilB & I are investigating >>>>> generalizes >>>>> (from a=1/2) at least to >>>>> (*) Hypergeometric2F1[a,1/2, 2 a, z] == >>>>> Hypergeometric2F1[2 a - 1/2, 1/2, 1/2 + a, -((-1 + Sqrt[1 - z])^2/(4 >>>>> Sqrt[1 - z]))]/(1 - z)^(1/4), >>>> >>>> In >>>> George E.\ Andrews, Richard Askey, Ranjan Roy: >>>> {Special functions}, >>>> Cambridge University Press, (1999). >>>> On page 176, "Exercises", relation (1.c) is >>>> F([a,b],[2b],x) = (1-x)^{-a/2} *F([a, 2b-a],[b+1/2], - >>>> (1-sqrt(1-x))^2 / (4*sqrt(1-x)) ) >>>> Swap a and b: >>>> F([b,a],[2a],x) = (1-x)^{-b/2} *F([b, 2a-b],[a+1/2], - >>>> (1-sqrt(1-x))^2 / (4*sqrt(1-x)) ) >>>> Set b = 1/2: >>>> F([1/2,a],[2a],x) = (1-x)^{-1/4} *F([1/2, 2a-1/2],[a+1/2], - >>>> (1-sqrt(1-x))^2 / (4*sqrt(1-x)) ) >>>> This is your transformation. >>>> >>>> Best regards, jj >>> >>> WOW, problem solved! I wonder if you can get this by combining >>> quadratic >>> and linear transformations in http://dlmf.nist.gov/15.8 . Thanks! >>> --rwg >>> (I should check why my methodology that introduced a (matching series >>> coefficients) failed to introduce b.) >>>> >>>> >>>>> which strongly resembles published 2F1 quadratic transformations >>>>> (e.g. http://dlmf.nist.gov/15.8) except these generally have two >>>>> degrees >>>>> of freedom, >>>>> e.g. 2F1[a,b,c(a,b),z] = f(z,a,b) 2F1[A(a,b),B(a,b),C(a,b),g(z)], >>>>> where f >>>>> and g are >>>>> algebraic in z. DLMF lists an exception: >>>>> >>>>> "When the intersection of two groups in Table 15.8.1 >>>>> <http://dlmf.nist.gov/15.8#T1> is not empty there exist special >>>>> quadratic >>>>> transformations, with only one free parameter, between two >>>>> hypergeometric >>>>> functions in the same group. . . . >>>>> >>>>> For further examples see Andrews et al. (1999 >>>>> <http://dlmf.nist.gov/bib/#bib102>, pp. 130?132 and 176?177)." >>>>> Can someone with access peek at >>>>> http://www.ams.org/mathscinet-getitem?mr=1688958 >>>>> >>>>> (http://ebooks.cambridge.org/ebook.jsf?bid=CBO9781107325937) and tell >>>>> us if >>>>> (*) appears there? >>>>> >>>>> Or do you know of other such single degree of freedom formul?? Or >>>>> best of >>>>> all, can you find >>>>> >>>>> the 2F1[a,b,...] generalization of (*) that would finally put this >>>>> question >>>>> to rest? --rwg >>> >> >> >> >> ------------------------------ >> >> Message: 2 >> Date: Tue, 8 Sep 2015 02:40:06 -0700 >> From: "William R. Somsky" <wrsomsky(a)gmail.com> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Subject: Re: [math-fun] Verticals >> Message-ID: >> <CAOnbAQBZWZur0kzcLu9S=i=Nvaux7Kru18nuJhfAVT=PACf30A(a)mail.gmail.com> >> Content-Type: text/plain; charset=UTF-8 >> >> Within a hollow spherical shell (uniform thickness and density), the >> gravitational forces cancel. The plumb Bob floats. >> On Sep 6, 2015 2:18 PM, "Keith F. Lynch" <kfl(a)keithlynch.net> wrote: >> >>> "Adam P. Goucher" <apgoucher(a)gmx.com> wrote: >>> > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >>> >> These ideas can of course be extended to lumpier worlds. Could a >>> >> local vertical ever point directly away from the center? >>> >>> > Suppose the planet is a solid torus of uniform density. Then if you >>> > stand on the inner Equator, gravity will pull you directly away from >>> > the centre of the planet. >>> >>> Right. >>> >>> > (If you insist that the planet is homotopy-equivalent to a closed >>> > ball, just include an infinitely thin membrane spanning the hole.) >>> >>> My solution was a hollow sphere with a hole drilled in the north pole. >>> A plumbbob at the inner south pole would point directly away from the >>> center. >>> >>> I'm now exploring two related ideas, each of which I may turn into a >>> new post: >>> >>> It's well known that the Moon's gravity is too lumpy for there to be >>> any long-term stable orbits. Similarly around Mercury, where the >>> first and only Mercury orbiter did in fact recently crash into the >>> planet. Of course an orbit can be adjusted as necessary by expending >>> rocket fuel, but you would soon run out. Would it be possible to >>> maintain an orbit indefinitely by having a satellite consisting of >>> two masses on a tether, using solar energy to reel in and out the >>> tether to circularize the orbit? A flywheel may also be necessary. >>> >>> What about the opposite extreme, a perfectly fluid planet? What shape >>> would such a planet have? If it wasn't rotating, it would obviously >>> be a sphere. With rotation, would be it be an ellipsoid, or is that >>> only a good first-order approximation where gravity greatly exceeds >>> centrifugal force? Would the shape depend on the distribution of >>> mass, e.g. would it be one shape if nearly all the mass was near the >>> center, and a different shape if the planet was of uniform density? >>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> >> >> ------------------------------ >> >> Message: 3 >> Date: Tue, 8 Sep 2015 09:42:33 -0400 >> From: James Propp <jamespropp(a)gmail.com> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Subject: [math-fun] Cantor set of measure 1? >> Message-ID: >> <CA+G9J-eANS_Zg9QtJBfsJ4C7taD43ePQbwgxF0CHe5ZxW3pLEw(a)mail.gmail.com> >> Content-Type: text/plain; charset=UTF-8 >> >> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) >> as >> the set obtained from S_{n-1} by replacing each of the 2^n closed >> intervals >> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >> duplicate >> the midpoint, shift the left half to the left by 1/3^n, and shift the >> right >> half to the right by 1/3^n). >> >> It seems to me that at each stage we get 2^n disjoint closed intervals of >> length 1/2^n separated from one another by 1/3^n, and that every point >> (including the duplicate points created along the way) moves a finite >> distance and therefore has a well-defined location "at time infinity". So >> we seem to get a well-defined set "at time infinity". >> >> Am I right? And do we obtain a set of measure 1 that's homeomorphic to >> the >> Cantor set? >> >> I've seen constructions of "fat Cantor sets" on the web, where one >> modifies >> the culling procedure, but I've never seen one that uses sliding and >> preserves the measure throughout the procedure. So I'm worried that I'm >> overlooking something. >> >> Jim Propp >> >> >> ------------------------------ >> >> Message: 4 >> Date: Tue, 8 Sep 2015 10:14:03 -0400 >> From: Thomas Colthurst <thomaswc(a)gmail.com> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Subject: Re: [math-fun] Cantor set of measure 1? >> Message-ID: >> <CAHmrjv0o+w=2AgNRTFadZxu9fRoHtU3rLyaSQp03wwXg8sv4bA(a)mail.gmail.com> >> Content-Type: text/plain; charset=UTF-8 >> >> This is just the IFS { x -> (1/2) x - 1/3, x -> (1/2) x + 5/6 }. It >> satisfies the open set condition, so yes, it is homeomorphic to a Cantor >> set and has Hausdorff and similarity dimension = 1. >> >> -Thomas C >> >> On Tue, Sep 8, 2015 at 9:42 AM, James Propp <jamespropp(a)gmail.com> wrote: >> >>> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) >>> as >>> the set obtained from S_{n-1} by replacing each of the 2^n closed >>> intervals >>> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >>> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >>> duplicate >>> the midpoint, shift the left half to the left by 1/3^n, and shift the >>> right >>> half to the right by 1/3^n). >>> >>> It seems to me that at each stage we get 2^n disjoint closed intervals >>> of >>> length 1/2^n separated from one another by 1/3^n, and that every point >>> (including the duplicate points created along the way) moves a finite >>> distance and therefore has a well-defined location "at time infinity". >>> So >>> we seem to get a well-defined set "at time infinity". >>> >>> Am I right? And do we obtain a set of measure 1 that's homeomorphic to >>> the >>> Cantor set? >>> >>> I've seen constructions of "fat Cantor sets" on the web, where one >>> modifies >>> the culling procedure, but I've never seen one that uses sliding and >>> preserves the measure throughout the procedure. So I'm worried that I'm >>> overlooking something. >>> >>> Jim Propp >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> >> >> ------------------------------ >> >> Message: 5 >> Date: Tue, 8 Sep 2015 07:39:59 -0700 >> From: Dan Asimov <asimov(a)msri.org> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Subject: Re: [math-fun] Cantor set of measure 1? >> Message-ID: <82035032-32EA-4BAB-8F36-08A1EC86AC90(a)msri.org> >> Content-Type: text/plain; charset=utf-8 >> >> ----- >> "It seems to me that at each stage we get 2^n disjoint closed intervals >> of >> length 1/2^n separated from one another by 1/3^n, and that every point >> (including the duplicate points created along the way) moves a finite >> distance and therefore has a well-defined location 'at time infinity'. So >> we seem to get a well-defined set 'at time infinity'." >> ----- >> >> If I understand the construction correctly, it can just as well go on >> within >> the limiting convex hull of all the points created,* namely [0-K, 1+K] >> where >> >> K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2, >> >> i.e., [-1/2, 3/2]. >> >> * It can just as well go on in [-1/2,3/2] because we can always dilate >> each >> stage about its midpoint by the appropriate factor so it exactly fits >> into >> [-1/2,3/2]. >> >> If that and the quoted paragraph above are correct, then IF this >> dilatation >> of each stage into [-1/2,3/2] is done, then the nth stage will consist of >> 2^n closed intervals of equal length L_n, separated by 2^n - 1 open >> intervals of length >> >> M_n = (2/3)^n * L_n, >> >> such that >> >> 2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n. >> >> (And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - >> 1)*(2/3)^n].) >> >> THEN it appears to me that, however the limiting set X is defined, there >> will (by symmetry) be a countable dense set of c in [0,2] for which the >> translations >> >> x |-> x + c >> >> will take a nonempty interval's worth of X to a different equal-size >> interval's worth of X. >> >> This implies that X contains points dense in some interval. But since the >> Cantor set is closed in R no matter how it is embedded, this implies it >> contains an entire interval, which is impossible for a Cantor set. >> >> Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely >> negating what I have hypothesized here. >> >> ?Dan >> >> >>> On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp(a)gmail.com> wrote: >>> >>> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) >>> as >>> the set obtained from S_{n-1} by replacing each of the 2^n closed >>> intervals >>> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >>> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >>> duplicate >>> the midpoint, shift the left half to the left by 1/3^n, and shift the >>> right >>> half to the right by 1/3^n). >> >>> It seems to me that at each stage we get 2^n disjoint closed intervals >>> of >>> length 1/2^n separated from one another by 1/3^n, and that every point >>> (including the duplicate points created along the way) moves a finite >>> distance and therefore has a well-defined location "at time infinity". >>> So >>> we seem to get a well-defined set "at time infinity". >> >>> Am I right? And do we obtain a set of measure 1 that's homeomorphic to >>> the >>> Cantor set? >>> >>> I've seen constructions of "fat Cantor sets" on the web, where one >>> modifies >>> the culling procedure, but I've never seen one that uses sliding and >>> preserves the measure throughout the procedure. So I'm worried that I'm >>> overlooking something. >>> >>> Jim Propp >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> >> >> ------------------------------ >> >> Subject: Digest Footer >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> ------------------------------ >> >> End of math-fun Digest, Vol 151, Issue 8 >> **************************************** >> > > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step) > > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > ------------------------------ > > End of math-fun Digest, Vol 151, Issue 9 > **************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

1 0

[math-fun] Cantor set of measure 1?
by Warren D Smith 08 Sep '15

08 Sep '15

If we wanted to omit "sliding" from Propp's construction (to be more in tune to what others have called "Cantor set" constructions) then, fix K>2 and 1. start with unit interval [0,1] and N=1. 2. from each of the 2^(N-1) intervals: delete the open subinterval centered at the midpoint and of width K^(-N). 3. Increment N and go back to step 2. This construction begins with measure=1, and after the Nth iteration of stage 2 has deleted measure 1/K + 2/K^2 + 4/K^3 + ... + 2^(N-1)/K^N = K^(-N) * (K^N-2^N) / (K-2) --> 1/(K-2) in the N-->infinity limit. Note if K=3, then the deleted measure is exactly 1, so we end up with a zero-measure nonempty set, which is, in fact, exactly Cantor's original "remove middle third" construction (expressed differently, but it is the same). If K>3 then the deleted measure is less than 1, so we end up with a "Cantor" set whose measure is a fixed constant between 0 and 1 and which is totally disconnected. On 9/8/15, math-fun-request(a)mailman.xmission.com <math-fun-request(a)mailman.xmission.com> wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > Today's Topics: > > 1. Re: Where's the b? (rwg) > 2. Re: Verticals (William R. Somsky) > 3. Cantor set of measure 1? (James Propp) > 4. Re: Cantor set of measure 1? (Thomas Colthurst) > 5. Re: Cantor set of measure 1? (Dan Asimov) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Mon, 07 Sep 2015 18:45:49 -0700 > From: rwg <rwg(a)sdf.org> > To: math-fun <math-fun(a)mailman.xmission.com> > Cc: A.OldeDaalhuis(a)ed.ac.uk > Subject: Re: [math-fun] Where's the b? > Message-ID: <3b819eed1ec594f97bf227bef4918564(a)sdf.org> > Content-Type: text/plain; charset=UTF-8; format=flowed > > OOPS!! Pofessor A. B. Olde Daalhuis, DLMF editor for Hypergeometric > Functions > (Chapters 13, 15, 16), kindly informs me that this transformation > follows from > switching the sides of http://dlmf.nist.gov/15.8.E22 and > recanonicalizing the > argument and parameters! I typed in all of 15.8, but I should have been > more > vigilant. Adding to my confusion, FullSimplify demurs due to the > apparent > failure of the reversed identity on the negative real axis. We may need > one > of Marichev's patented Sqrt[x]*Sqrt[1/x] fudge factors. --rwg > > On 2015-09-07 02:04, rwg wrote: >> On 2015-09-07 01:34, Joerg Arndt wrote: >>> * Bill Gosper <billgosper(a)gmail.com> [Sep 07. 2015 09:28]: >>>> The elliptic K transformation that NeilB & I are investigating >>>> generalizes >>>> (from a=1/2) at least to >>>> (*) Hypergeometric2F1[a,1/2, 2 a, z] == >>>> Hypergeometric2F1[2 a - 1/2, 1/2, 1/2 + a, -((-1 + Sqrt[1 - z])^2/(4 >>>> Sqrt[1 - z]))]/(1 - z)^(1/4), >>> >>> In >>> George E.\ Andrews, Richard Askey, Ranjan Roy: >>> {Special functions}, >>> Cambridge University Press, (1999). >>> On page 176, "Exercises", relation (1.c) is >>> F([a,b],[2b],x) = (1-x)^{-a/2} *F([a, 2b-a],[b+1/2], - >>> (1-sqrt(1-x))^2 / (4*sqrt(1-x)) ) >>> Swap a and b: >>> F([b,a],[2a],x) = (1-x)^{-b/2} *F([b, 2a-b],[a+1/2], - >>> (1-sqrt(1-x))^2 / (4*sqrt(1-x)) ) >>> Set b = 1/2: >>> F([1/2,a],[2a],x) = (1-x)^{-1/4} *F([1/2, 2a-1/2],[a+1/2], - >>> (1-sqrt(1-x))^2 / (4*sqrt(1-x)) ) >>> This is your transformation. >>> >>> Best regards, jj >> >> WOW, problem solved! I wonder if you can get this by combining >> quadratic >> and linear transformations in http://dlmf.nist.gov/15.8 . Thanks! >> --rwg >> (I should check why my methodology that introduced a (matching series >> coefficients) failed to introduce b.) >>> >>> >>>> which strongly resembles published 2F1 quadratic transformations >>>> (e.g. http://dlmf.nist.gov/15.8) except these generally have two >>>> degrees >>>> of freedom, >>>> e.g. 2F1[a,b,c(a,b),z] = f(z,a,b) 2F1[A(a,b),B(a,b),C(a,b),g(z)], >>>> where f >>>> and g are >>>> algebraic in z. DLMF lists an exception: >>>> >>>> "When the intersection of two groups in Table 15.8.1 >>>> <http://dlmf.nist.gov/15.8#T1> is not empty there exist special >>>> quadratic >>>> transformations, with only one free parameter, between two >>>> hypergeometric >>>> functions in the same group. . . . >>>> >>>> For further examples see Andrews et al. (1999 >>>> <http://dlmf.nist.gov/bib/#bib102>, pp. 130?132 and 176?177)." >>>> Can someone with access peek at >>>> http://www.ams.org/mathscinet-getitem?mr=1688958 >>>> >>>> (http://ebooks.cambridge.org/ebook.jsf?bid=CBO9781107325937) and tell >>>> us if >>>> (*) appears there? >>>> >>>> Or do you know of other such single degree of freedom formul?? Or >>>> best of >>>> all, can you find >>>> >>>> the 2F1[a,b,...] generalization of (*) that would finally put this >>>> question >>>> to rest? --rwg >> > > > > ------------------------------ > > Message: 2 > Date: Tue, 8 Sep 2015 02:40:06 -0700 > From: "William R. Somsky" <wrsomsky(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Verticals > Message-ID: > <CAOnbAQBZWZur0kzcLu9S=i=Nvaux7Kru18nuJhfAVT=PACf30A(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Within a hollow spherical shell (uniform thickness and density), the > gravitational forces cancel. The plumb Bob floats. > On Sep 6, 2015 2:18 PM, "Keith F. Lynch" <kfl(a)keithlynch.net> wrote: > >> "Adam P. Goucher" <apgoucher(a)gmx.com> wrote: >> > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> >> These ideas can of course be extended to lumpier worlds. Could a >> >> local vertical ever point directly away from the center? >> >> > Suppose the planet is a solid torus of uniform density. Then if you >> > stand on the inner Equator, gravity will pull you directly away from >> > the centre of the planet. >> >> Right. >> >> > (If you insist that the planet is homotopy-equivalent to a closed >> > ball, just include an infinitely thin membrane spanning the hole.) >> >> My solution was a hollow sphere with a hole drilled in the north pole. >> A plumbbob at the inner south pole would point directly away from the >> center. >> >> I'm now exploring two related ideas, each of which I may turn into a >> new post: >> >> It's well known that the Moon's gravity is too lumpy for there to be >> any long-term stable orbits. Similarly around Mercury, where the >> first and only Mercury orbiter did in fact recently crash into the >> planet. Of course an orbit can be adjusted as necessary by expending >> rocket fuel, but you would soon run out. Would it be possible to >> maintain an orbit indefinitely by having a satellite consisting of >> two masses on a tether, using solar energy to reel in and out the >> tether to circularize the orbit? A flywheel may also be necessary. >> >> What about the opposite extreme, a perfectly fluid planet? What shape >> would such a planet have? If it wasn't rotating, it would obviously >> be a sphere. With rotation, would be it be an ellipsoid, or is that >> only a good first-order approximation where gravity greatly exceeds >> centrifugal force? Would the shape depend on the distribution of >> mass, e.g. would it be one shape if nearly all the mass was near the >> center, and a different shape if the planet was of uniform density? >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > ------------------------------ > > Message: 3 > Date: Tue, 8 Sep 2015 09:42:33 -0400 > From: James Propp <jamespropp(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Cantor set of measure 1? > Message-ID: > <CA+G9J-eANS_Zg9QtJBfsJ4C7taD43ePQbwgxF0CHe5ZxW3pLEw(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as > the set obtained from S_{n-1} by replacing each of the 2^n closed intervals > [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and > [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate > the midpoint, shift the left half to the left by 1/3^n, and shift the right > half to the right by 1/3^n). > > It seems to me that at each stage we get 2^n disjoint closed intervals of > length 1/2^n separated from one another by 1/3^n, and that every point > (including the duplicate points created along the way) moves a finite > distance and therefore has a well-defined location "at time infinity". So > we seem to get a well-defined set "at time infinity". > > Am I right? And do we obtain a set of measure 1 that's homeomorphic to the > Cantor set? > > I've seen constructions of "fat Cantor sets" on the web, where one modifies > the culling procedure, but I've never seen one that uses sliding and > preserves the measure throughout the procedure. So I'm worried that I'm > overlooking something. > > Jim Propp > > > ------------------------------ > > Message: 4 > Date: Tue, 8 Sep 2015 10:14:03 -0400 > From: Thomas Colthurst <thomaswc(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Cantor set of measure 1? > Message-ID: > <CAHmrjv0o+w=2AgNRTFadZxu9fRoHtU3rLyaSQp03wwXg8sv4bA(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > This is just the IFS { x -> (1/2) x - 1/3, x -> (1/2) x + 5/6 }. It > satisfies the open set condition, so yes, it is homeomorphic to a Cantor > set and has Hausdorff and similarity dimension = 1. > > -Thomas C > > On Tue, Sep 8, 2015 at 9:42 AM, James Propp <jamespropp(a)gmail.com> wrote: > >> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) >> as >> the set obtained from S_{n-1} by replacing each of the 2^n closed >> intervals >> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >> duplicate >> the midpoint, shift the left half to the left by 1/3^n, and shift the >> right >> half to the right by 1/3^n). >> >> It seems to me that at each stage we get 2^n disjoint closed intervals of >> length 1/2^n separated from one another by 1/3^n, and that every point >> (including the duplicate points created along the way) moves a finite >> distance and therefore has a well-defined location "at time infinity". So >> we seem to get a well-defined set "at time infinity". >> >> Am I right? And do we obtain a set of measure 1 that's homeomorphic to >> the >> Cantor set? >> >> I've seen constructions of "fat Cantor sets" on the web, where one >> modifies >> the culling procedure, but I've never seen one that uses sliding and >> preserves the measure throughout the procedure. So I'm worried that I'm >> overlooking something. >> >> Jim Propp >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > ------------------------------ > > Message: 5 > Date: Tue, 8 Sep 2015 07:39:59 -0700 > From: Dan Asimov <asimov(a)msri.org> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Cantor set of measure 1? > Message-ID: <82035032-32EA-4BAB-8F36-08A1EC86AC90(a)msri.org> > Content-Type: text/plain; charset=utf-8 > > ----- > "It seems to me that at each stage we get 2^n disjoint closed intervals of > length 1/2^n separated from one another by 1/3^n, and that every point > (including the duplicate points created along the way) moves a finite > distance and therefore has a well-defined location 'at time infinity'. So > we seem to get a well-defined set 'at time infinity'." > ----- > > If I understand the construction correctly, it can just as well go on within > the limiting convex hull of all the points created,* namely [0-K, 1+K] > where > > K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2, > > i.e., [-1/2, 3/2]. > > * It can just as well go on in [-1/2,3/2] because we can always dilate each > stage about its midpoint by the appropriate factor so it exactly fits into > [-1/2,3/2]. > > If that and the quoted paragraph above are correct, then IF this dilatation > of each stage into [-1/2,3/2] is done, then the nth stage will consist of > 2^n closed intervals of equal length L_n, separated by 2^n - 1 open > intervals of length > > M_n = (2/3)^n * L_n, > > such that > > 2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n. > > (And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - > 1)*(2/3)^n].) > > THEN it appears to me that, however the limiting set X is defined, there > will (by symmetry) be a countable dense set of c in [0,2] for which the > translations > > x |-> x + c > > will take a nonempty interval's worth of X to a different equal-size > interval's worth of X. > > This implies that X contains points dense in some interval. But since the > Cantor set is closed in R no matter how it is embedded, this implies it > contains an entire interval, which is impossible for a Cantor set. > > Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely > negating what I have hypothesized here. > > ?Dan > > >> On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp(a)gmail.com> wrote: >> >> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) >> as >> the set obtained from S_{n-1} by replacing each of the 2^n closed >> intervals >> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and >> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, >> duplicate >> the midpoint, shift the left half to the left by 1/3^n, and shift the >> right >> half to the right by 1/3^n). > >> It seems to me that at each stage we get 2^n disjoint closed intervals of >> length 1/2^n separated from one another by 1/3^n, and that every point >> (including the duplicate points created along the way) moves a finite >> distance and therefore has a well-defined location "at time infinity". So >> we seem to get a well-defined set "at time infinity". > >> Am I right? And do we obtain a set of measure 1 that's homeomorphic to >> the >> Cantor set? >> >> I've seen constructions of "fat Cantor sets" on the web, where one >> modifies >> the culling procedure, but I've never seen one that uses sliding and >> preserves the measure throughout the procedure. So I'm worried that I'm >> overlooking something. >> >> Jim Propp >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > ------------------------------ > > End of math-fun Digest, Vol 151, Issue 8 > **************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

1 0

[math-fun] Poincare dual of Penrose tiling?
by Mike Stay 08 Sep '15

08 Sep '15

Anyone know where I can find a picture of the Poincare dual of a Penrose tiling? Google images fails me. -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

1 0

[math-fun] Where's the b?
by Bill Gosper 07 Sep '15

07 Sep '15

The elliptic K transformation that NeilB & I are investigating generalizes (from a=1/2) at least to (*) Hypergeometric2F1[a,1/2, 2 a, z] == Hypergeometric2F1[2 a - 1/2, 1/2, 1/2 + a, -((-1 + Sqrt[1 - z])^2/(4 Sqrt[1 - z]))]/(1 - z)^(1/4), which strongly resembles published 2F1 quadratic transformations (e.g. http://dlmf.nist.gov/15.8) except these generally have two degrees of freedom, e.g. 2F1[a,b,c(a,b),z] = f(z,a,b) 2F1[A(a,b),B(a,b),C(a,b),g(z)], where f and g are algebraic in z. DLMF lists an exception: "When the intersection of two groups in Table 15.8.1 <http://dlmf.nist.gov/15.8#T1> is not empty there exist special quadratic transformations, with only one free parameter, between two hypergeometric functions in the same group. . . . For further examples see Andrews et al. (1999 <http://dlmf.nist.gov/bib/#bib102>, pp. 130–132 and 176–177)." Can someone with access peek at http://www.ams.org/mathscinet-getitem?mr=1688958 (http://ebooks.cambridge.org/ebook.jsf?bid=CBO9781107325937) and tell us if (*) appears there? Or do you know of other such single degree of freedom formulæ? Or best of all, can you find the 2F1[a,b,...] generalization of (*) that would finally put this question to rest? --rwg

3 3

[math-fun] Elliptic integrals: expressions for dk/dq, DK/dq, DE/dq etc.
by Joerg Arndt 07 Sep '15

07 Sep '15

Computing dq/dk by using q = exp( -Pi K'/K ) and \diff{K}{k} = \frac{E - k'^2 K}{k k'^2}, \diff{K'}{k} = \frac{k^2 K'-E'}{k k'^2} (LaTeX-note: read \diff{y}{x} as \frac{dy}{dx}) one obtains a somewhat lengthy expression. Simplifying using k^2+k'^2 = 1 and then (surprise!) using E K' + E' K + K K' = Pi/2 (Legendre's relation) one can obtain (setting T := (2 K^2)/(\pi^2 q) = \Theta_3(q)^4 / (2 q) ) \diff{k}{q} = + T k k'^2 \diff{k'}{q} = - T k^2 k' \diff{K}{q} = + T (E - k'^2 K) \diff{K'}{q} = - T (E' - k^2 K') \diff{E}{q} = + T k'^2 (E - K) \diff{E'}{q} = - T k^2 (E' - K') I have not seen any of these before (pointers, anyone?). >From these follow relations involving theta derivatives such as \frac{2 E}{\pi} = \frac{1}{\Theta_3^3(q)} ( \Theta_4^4(q) \Theta_3(q) + 4 q \diff{\Theta_3(q)}{q} ) (Cf. http://oeis.org/A194094 ) (End of bore). Best regards, jj

1 0

Re: [math-fun] animated music
by Henry Baker 06 Sep '15

06 Sep '15

Cool! This reminds me of a piece of music from the 1960's which came from stock market movements. If I recall correctly, IBM may have had something to do with this composition. Tymoczko himself in one of his YouTube videos took a random number generator & applied more & more rules to it until it started to sound like actual music. He demonstrates each "musical composition" in the sequence in the video, and he calls it his "first year composition class" sequence, as he claims that during the first 6 weeks of this composition class, his students make the same progression. At 04:52 PM 9/6/2015, Neil Sloane wrote: >Very interesting to see the Malinowski video. > >On a somewhat similar note, any sequence in the OEIS >can be converted to "music" by clicking the "listen" >button (next to the History tab). Two recent entries >that are almost worth listening to are https://oeis.org/A261684 >and https://oeis.org/A261422. (The latter arose from trying to prove that >every number is the sum of 3 palindromes. Its graph is quite remarkable.) > >(On my machine I have to click listen then save then I click the saved file) > >You can also click the "graph" button (to the left >of the "history" button) to see a plot of the notes.

1 0