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@mailman.xmission.com <math-fun-request@mailman.xmission.com> wrote:
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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)
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Message: 1 Date: Tue, 8 Sep 2015 10:51:43 -0400 From: Thomas Colthurst <thomaswc@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Cantor set of measure 1? Message-ID: <CAHmrjv1Ch4CzzU_9LocrHnDa6NPT0RNCfv3FFTTTgBwWPRJMXQ@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@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@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@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Message: 2 Date: Tue, 8 Sep 2015 07:58:25 -0700 From: Dan Asimov <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Cantor set of measure 1? Message-ID: <59CD4A9E-A2D0-41D8-B717-1B9BB10B8D8B@msri.org> Content-Type: text/plain; charset=utf-8
On Sep 8, 2015, at 7:51 AM, Thomas Colthurst <thomaswc@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@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@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@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Message: 3 Date: Tue, 8 Sep 2015 08:17:42 -0700 From: Mike Stay <metaweta@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Poincare dual of Penrose tiling? Message-ID: <CAKQgqTa0GP2D0zb7NQAQ7De7oAN=WXNHdLTS-yPxr-UTt4apuA@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@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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Message: 4 Date: Tue, 8 Sep 2015 11:19:32 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] Cantor set of measure 1? Message-ID: <CAAJP7Y3_F2tc3xSDmqY45Bp4aHbmpVv1NbGCvCxvkbMNkWOxuw@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@mailman.xmission.com <math-fun-request@mailman.xmission.com> wrote:
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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)
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Message: 1 Date: Mon, 07 Sep 2015 18:45:49 -0700 From: rwg <rwg@sdf.org> To: math-fun <math-fun@mailman.xmission.com> Cc: A.OldeDaalhuis@ed.ac.uk Subject: Re: [math-fun] Where's the b? Message-ID: <3b819eed1ec594f97bf227bef4918564@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@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
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Message: 2 Date: Tue, 8 Sep 2015 02:40:06 -0700 From: "William R. Somsky" <wrsomsky@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Verticals Message-ID: <CAOnbAQBZWZur0kzcLu9S=i=Nvaux7Kru18nuJhfAVT=PACf30A@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@keithlynch.net> wrote:
"Adam P. Goucher" <apgoucher@gmx.com> wrote:
"Keith F. Lynch" <kfl@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?
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Message: 3 Date: Tue, 8 Sep 2015 09:42:33 -0400 From: James Propp <jamespropp@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Cantor set of measure 1? Message-ID: <CA+G9J-eANS_Zg9QtJBfsJ4C7taD43ePQbwgxF0CHe5ZxW3pLEw@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@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Cantor set of measure 1? Message-ID: <CAHmrjv0o+w=2AgNRTFadZxu9fRoHtU3rLyaSQp03wwXg8sv4bA@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@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@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@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Cantor set of measure 1? Message-ID: <82035032-32EA-4BAB-8F36-08A1EC86AC90@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@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@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
------------------------------
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