Re: [math-fun] Space-filling curves again
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com> wrote:
With much help from Julian, gosper.org/hellodoily.gif --rwg
On Fri, Sep 30, 2016 at 10:13 AM, Bill Gosper <billgosper@gmail.com <javascript:;>> wrote:
On Thu, Sep 29, 2016 at 8:33 PM, Bill Gosper <billgosper@gmail.com <javascript:;>> wrote:
The animation I intended was perhaps too computationally ambitious. Meanwhile, gosper.org/FHilbert.gif is just conventional plots of the first 288 approximations. I dislike such plots due to the finite line
thickness,
which someone might imagine contributed to spacefilling. If you read all the frames into Preview, say, you can see that the texture of frame n repeats at frame 4n. This gives us a way to interpolate the nonexistent frame n+1/4 from frame 4n+1.
Also uploaded: gosper.org/outcirc.gif made by flipping a sign in incirc.gif. --rwg Stay tuned for a Ptolemaic Hilbert sweep from a multi-hour computation.
Three hours. Foo, the Symbolics machine could do this nearly in real time. gosper.org/hilbert286.gif is the 287 rotor sweep. (288 would have been too gross.) It was to have been a single frame of the frequency expansion animation, analogous to gosper.org/FHilbert.gif , but the project would not have contributed much insight into the final "catastrophic" dimension bump.
I haven't tried very hard to telescope Julian's 3⨉3 matrix product for the Fourier coefficients. Success would provide a flood of strange identities like (d247) and (d246) in http://www.tweedledum.com/rwg/idents.htm , which come from the family that includes the Sierpinski Curve and Koch's Snowflake (http://gosper.org/fst.pdf pp137, 138) rather than Hilbert's curve. Failure to telescope would encumber the infinite products in the identities with matrices instead of scalars. --rwg The infinite Fourier series faithfully represents Hilbert's function, except it misses the top two corners (the endpoints of Hilbert's curve) and moves them instead to top center (where the animation begins and ends). This is a quadruple point if you think of the Fourier function on [0,1], but only a triple point if you regard it as a periodic function, since a period contains only one of its endpoints.
On 2016-09-26 18:42, James Propp wrote:
I hope Bill will write a lively document (or create a lively video)
that
explains what's wrong with so many accounts of spacefilling curves.
(I myself wish that more accounts started in a "Not Knot"-ish vein, explaining why the "kindergartner's space-filling curve" --- scribbling one's crayon back and forth until the square is filled --- isn't a solution to the mathematician's problem.)
> Is there a way to relax an approximation to a space-filling curve
in
> continuous time so that it works out its kinks and regresses to simpler > approximations? > > (No interim self-intersections please!) > > Jim Propp
http://gosper.org/FDrags128.mp4
If people really want to see it, Julian's recent Fourier matrix product can produce the analogous animation for Hilbert's "curve".
I would like to see it.
But it is a seductive thought crime to view a spacefilling function as some
kind of limit of "spacefilling curves"! Those curves Henry sketches are mere schematics, of no mathematical consequence. This leads inevitably to embarrassed hemming and hawing about how the area jumps from 0 to 1 at the very last moment, when both interior and exterior suddenly disappear and become boundary. Successive partial sums of the Fourier series are even more seductive. But no matter how many terms you take, you're still infinitely far from the end. --rwg
That's one way to look at it. But it depends on the notion of path-space you use and what metric you put on it, doesn't it? In particular, it's crucial to look at paths equipped with a parametrization. Then you really can get convergence to a limit. And the limiting object is a continuous function from a line segment onto a square. If you just look at the range of a path and not the parametrization, you can't describe the square as the limit in any meaningful way. Which I gather is part of Bill's point.
Jim Propp
See gosper.org/julianstrifle.png . Note the tiny self-crossings not visible in the cursive drawing. It is usually far clearer to portray a <2-dimensional curve as the boundary of a two-dimensional region. --rwg On Thu, Oct 20, 2016 at 4:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com>
wrote:
With much help from Julian, gosper.org/hellodoily.gif --rwg
On Fri, Sep 30, 2016 at 10:13 AM, Bill Gosper <billgosper@gmail.com <javascript:;>> wrote:
On Thu, Sep 29, 2016 at 8:33 PM, Bill Gosper <billgosper@gmail.com <javascript:;>> wrote:
The animation I intended was perhaps too computationally ambitious. Meanwhile, gosper.org/FHilbert.gif is just conventional plots of the first 288 approximations. I dislike such plots due to the finite line
thickness,
which someone might imagine contributed to spacefilling. If you read all the frames into Preview, say, you can see that the texture of frame n repeats at frame 4n. This gives us a way to interpolate the nonexistent frame n+1/4 from frame 4n+1.
Also uploaded: gosper.org/outcirc.gif made by flipping a sign in incirc.gif. --rwg Stay tuned for a Ptolemaic Hilbert sweep from a multi-hour computation.
Three hours. Foo, the Symbolics machine could do this nearly in real time. gosper.org/hilbert286.gif is the 287 rotor sweep. (288 would have been too gross.) It was to have been a single frame of the frequency expansion animation, analogous to gosper.org/FHilbert.gif , but the project would not have contributed much insight into the final "catastrophic" dimension bump.
I haven't tried very hard to telescope Julian's 3⨉3 matrix product for the Fourier coefficients. Success would provide a flood of strange identities like (d247) and (d246) in http://www.tweedledum.com/rwg/idents.htm , which come from the family that includes the Sierpinski Curve and Koch's Snowflake (http://gosper.org/fst.pdf pp137, 138) rather than Hilbert's curve. Failure to telescope would encumber the infinite products in the identities with matrices instead of scalars. --rwg The infinite Fourier series faithfully represents Hilbert's function, except it misses the top two corners (the endpoints of Hilbert's curve) and moves them instead to top center (where the animation begins and ends). This is a quadruple point if you think of the Fourier function on [0,1], but only a triple point if you regard it as a periodic function, since a period contains only one of its endpoints.
On 2016-09-26 18:42, James Propp wrote:
I hope Bill will write a lively document (or create a lively video)
that
explains what's wrong with so many accounts of spacefilling curves.
(I myself wish that more accounts started in a "Not Knot"-ish vein, explaining why the "kindergartner's space-filling curve" --- scribbling one's crayon back and forth until the square is filled --- isn't a solution to the mathematician's problem.)
> > > Is there a way to relax an approximation to a space-filling curve in > > continuous time so that it works out its kinks and regresses to simpler > > approximations? > > > > (No interim self-intersections please!) > > > > Jim Propp > > http://gosper.org/FDrags128.mp4 > > If people really want to see it, Julian's recent Fourier matrix product can > produce > the analogous animation for Hilbert's "curve".
I would like to see it.
But it is a seductive thought crime to view a spacefilling function as some > kind of > limit of "spacefilling curves"! Those curves Henry sketches are mere > schematics, > of no mathematical consequence. This leads inevitably to embarrassed > hemming > and hawing about how the area jumps from 0 to 1 at the very last moment, > when > both interior and exterior suddenly disappear and become boundary. > Successive partial sums of the Fourier series are even more seductive. But > no matter how many terms you take, you're still infinitely far from the > end. > --rwg
That's one way to look at it. But it depends on the notion of path-space you use and what metric you put on it, doesn't it? In particular, it's crucial to look at paths equipped with a parametrization. Then you really can get convergence to a limit. And the limiting object is a continuous function from a line segment onto a square. If you just look at the range of a path and not the parametrization, you can't describe the square as the limit in any meaningful way. Which I gather is part of Bill's point.
Jim Propp
On Fri, Oct 21, 2016 at 7:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
See gosper.org/julianstrifle.png . Note the tiny self-crossings not visible in the cursive drawing. It is usually far clearer to portray a <2-dimensional curve as the boundary of a two-dimensional region. --rwg
Julian's Fourier expander takes an optional argument to arrange m copies of a curve around a regular m-gon. gosper.org/6trifil.png . Instead of spacefilling an equilateral triangle divided into quarters, I tried dividing it into five little ones plus a double size one. Arranged around a hexagon, gosper.org/ringnoring1234.png (one of the few cases where the outline looks maybe better than the filled area.) (Shades of https://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane ?) Note that with 2468 rotors, the Fourier approximation gets nowhere near the middle. So is this a spacefill, or not? Julian: "Of course it is ..." Proof that it reaches all three vertices of each triangle: In[422]:= hollow /@ {0, 4/5, 1} // Expand
Out[422]= {{0}, {3/2 + (3 I Sqrt[3])/2}, {3}} via (Julian's ever amazing) piecewiserecursivefractal Clear[hollow]; hollow[t_] := piecewiserecursivefractal[t, Identity, {Min[6, 1 + Floor[6*#]]} &, {6 # &, 6 # - 1 &, 3 - 6 # &, 6 # - 3 &, 6 # - 4 &, 6 - 6 # &}, {#/3 &, #/3 + 1 &, #/3/I^(2/3) + 3/2 + I Sqrt[3]/2 &, 3/2 + I Sqrt[3]/2 - #/3 &, 2 #/3 + 1/2 + I Sqrt[3]/2 &, #/3*I^(4/3) + 3 &}] A polygonal sampling: gosper.org/hollowfill150.png (Reminder: True spacefills self-contact like crazy.) --rwg
On Thu, Oct 20, 2016 at 4:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com>
wrote:
With much help from Julian, gosper.org/hellodoily.gif --rwg
On Fri, Sep 30, 2016 at 10:13 AM, Bill Gosper <billgosper@gmail.com <javascript:;>> wrote:
On Thu, Sep 29, 2016 at 8:33 PM, Bill Gosper <billgosper@gmail.com <javascript:;>> wrote:
The animation I intended was perhaps too computationally ambitious. Meanwhile, gosper.org/FHilbert.gif is just conventional plots of
the
first 288 approximations. I dislike such plots due to the finite line thickness, which someone might imagine contributed to spacefilling. If you read all the frames into Preview, say, you can see that the texture of frame n repeats at frame 4n. This gives us a way to interpolate the nonexistent frame n+1/4 from frame 4n+1.
Also uploaded: gosper.org/outcirc.gif made by flipping a sign in incirc.gif. --rwg Stay tuned for a Ptolemaic Hilbert sweep from a multi-hour computation.
Three hours. Foo, the Symbolics machine could do this nearly in real time. gosper.org/hilbert286.gif is the 287 rotor sweep. (288 would have been too gross.) It was to have been a single frame of the frequency expansion animation, analogous to gosper.org/FHilbert.gif , but the project would not have contributed much insight into the final "catastrophic" dimension bump.
I haven't tried very hard to telescope Julian's 3⨉3 matrix product for the Fourier coefficients. Success would provide a flood of strange identities like (d247) and (d246) in http://www.tweedledum.com/rwg/idents.htm , which come from the family that includes the Sierpinski Curve and Koch's Snowflake (http://gosper.org/fst.pdf pp137, 138) rather than Hilbert's curve. Failure to telescope would encumber the infinite products in the identities with matrices instead of scalars. --rwg The infinite Fourier series faithfully represents Hilbert's function, except it misses the top two corners (the endpoints of Hilbert's curve) and moves them instead to top center (where the animation begins and ends). This is a quadruple point if you think of the Fourier function on [0,1], but only a triple point if you regard it as a periodic function, since a period contains only one of its endpoints.
[chop]
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers. True or false: There is a countable set of grunches such that every grunch is a subset of one of them. Prove your answer. —Dan
Nice question, Dan! If I didn't have to grade midterm exams for the next several hours, I'd definitely play with this puzzle right now. Maybe I'll think about it as I drift off to hard-earned sleep tonight... Jim Propp On Thu, Oct 27, 2016 at 3:04 PM, Dan Asimov <asimov@msri.org> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Yes, it is a nice question. I wish I'd thought of it (but it wasn't me). —Dan
On Oct 27, 2016, at 5:13 PM, James Propp <jamespropp@gmail.com> wrote:
Nice question, Dan! If I didn't have to grade midterm exams for the next several hours, I'd definitely play with this puzzle right now. Maybe I'll think about it as I drift off to hard-earned sleep tonight...
Jim Propp
On Thu, Oct 27, 2016 at 3:04 PM, Dan Asimov <asimov@msri.org> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
SPOILER below ... On 10/27/16, Dan Asimov <asimov@msri.org> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
—Dan _______________________________________________
A bounded closed subset of reals is compact, so a grunch contains all its accumulation points, which must also be rational; conversely any bounded set of rationals containing its rational accumulation points is a grunch. Suppose [H_i] is some denumerable sequence of grunches such that for each grunch S there exists i such that S is a subset of H_i . Let G = {0} union {x_i} , where x_i is rational, x_i notin Union_{j <= i} H_j , and 0 < x_i < x_{i-1} . G contains its sole accumulation point 0 , so G is a grunch; however for all i , G notin H_i --- contradiction. Hence there is no such sequence [H_i] . QED Fred Lunnon
Er, spoiler and clarification below ... On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
SPOILER below ...
On 10/27/16, Dan Asimov <asimov@msri.org> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
—Dan _______________________________________________
A bounded closed subset of reals is compact, so a grunch contains all its accumulation points, which must also be rational; conversely any bounded set of rationals containing its rational accumulation points is a grunch.
Suppose [H_i] is some denumerable sequence of grunches such that
for each grunch S there exists i such that S is a subset of H_i .
Let
G = {0} union {x_i} ,
where x_i is rational,
x_i notin Union_{j <= i} H_j ,
and
0 < x_i < x_{i-1} .
G contains its sole accumulation point 0 , so G is a grunch; however
for all i , G notin H_i
--- contradiction. Hence there is no such sequence [H_i] . QED
Fred Lunnon
Following a walk with the dog, not to mention a cryptic memo from Dan, it dawns on me that I failed to establish above that there exists some x_i with 0 < x_i < x_{i-1} and x notin Union_{j <= i} H_j . Firstly, it was unneccesary to take that union anyway: x notin H_i suffices. Now if every rational in (0, x_{i-1}) were a member of H_i , then all real numbers in that interval would be accumulation points, hence via compactness also members of H_i , which could therefore not be a grunch. By a second contradiction then, there must be some rational x_i satisfying the (weakened) constraint. Fred Lunnon
Nice! Jim On Thursday, October 27, 2016, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Er, spoiler and clarification below ...
On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com <javascript:;>> wrote:
SPOILER below ...
On 10/27/16, Dan Asimov <asimov@msri.org <javascript:;>> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
—Dan _______________________________________________
A bounded closed subset of reals is compact, so a grunch contains all its accumulation points, which must also be rational; conversely any bounded set of rationals containing its rational accumulation points is a grunch.
Suppose [H_i] is some denumerable sequence of grunches such that
for each grunch S there exists i such that S is a subset of H_i .
Let
G = {0} union {x_i} ,
where x_i is rational,
x_i notin Union_{j <= i} H_j ,
and
0 < x_i < x_{i-1} .
G contains its sole accumulation point 0 , so G is a grunch; however
for all i , G notin H_i
--- contradiction. Hence there is no such sequence [H_i] . QED
Fred Lunnon
Following a walk with the dog, not to mention a cryptic memo from Dan, it dawns on me that I failed to establish above that there exists some x_i with 0 < x_i < x_{i-1} and x notin Union_{j <= i} H_j .
Firstly, it was unneccesary to take that union anyway: x notin H_i suffices. Now if every rational in (0, x_{i-1}) were a member of H_i , then all real numbers in that interval would be accumulation points, hence via compactness also members of H_i , which could therefore not be a grunch. By a second contradiction then, there must be some rational x_i satisfying the (weakened) constraint.
Fred Lunnon
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
So far, no one has submitted a correct solution. —Dan
On Oct 28, 2016, at 7:18 AM, James Propp <jamespropp@gmail.com> wrote:
On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org> <javascript:;>> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thin I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0. But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i. Andy
—Dan
On Oct 28, 2016, at 7:18 AM, James Propp <jamespropp@gmail.com> wrote:
On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org> <javascript:;>> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
Yes, that's it. Nice solving! Fred ... and Andy. The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered. —Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
Some physics questions I don't know the answer to: 1) An old Car Talk puzzle of the week is about a truck carrying a load of chickens. As it approaches a mandatory truck scale, the driver realizes that its weight is just over the legal limit. To lighten the truck, he or she makes a loud noise, frightening the chickens so that they fly into the air just in time for the truck to be weighed. Question: Does this succeed in lightening the load? 1a) Also, does it matter whether the truck is airtight or open? ----- 2) In a related question, people on a sailboat realize that it needs to get back to shore in a hurry, so they attach a powerful fan to a powerful battery and aim it at the sail. Can this help propel the boat? 2a) Likewise, suppose this picture is scaled down to become a small toy boat. Does it matter if all takes place in an airtight enclosed case? ----- —Dan
Regarding question 2, Mythbusters did it. Probably doesn't match the "in a hurry" criterion, but it moved. https://www.youtube.com/watch?v=uKXMTzMQWjo On Fri, Oct 28, 2016 at 11:52 AM, Dan Asimov <asimov@msri.org> wrote:
Some physics questions I don't know the answer to:
1) An old Car Talk puzzle of the week is about a truck carrying a load of chickens.
As it approaches a mandatory truck scale, the driver realizes that its weight is just over the legal limit. To lighten the truck, he or she makes a loud noise, frightening the chickens so that they fly into the air just in time for the truck to be weighed.
Question: Does this succeed in lightening the load?
1a) Also, does it matter whether the truck is airtight or open? -----
2) In a related question, people on a sailboat realize that it needs to get back to shore in a hurry, so they attach a powerful fan to a powerful battery and aim it at the sail. Can this help propel the boat?
2a) Likewise, suppose this picture is scaled down to become a small toy boat. Does it matter if all takes place in an airtight enclosed case? -----
—Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Turns out Mythbusters is your friend for question 1, too. https://www.youtube.com/watch?v=lVeP6oqH-Qo On Fri, Oct 28, 2016 at 12:44 PM, Kerry Mitchell <lkmitch@gmail.com> wrote:
Regarding question 2, Mythbusters did it. Probably doesn't match the "in a hurry" criterion, but it moved.
https://www.youtube.com/watch?v=uKXMTzMQWjo
On Fri, Oct 28, 2016 at 11:52 AM, Dan Asimov <asimov@msri.org> wrote:
Some physics questions I don't know the answer to:
1) An old Car Talk puzzle of the week is about a truck carrying a load of chickens.
As it approaches a mandatory truck scale, the driver realizes that its weight is just over the legal limit. To lighten the truck, he or she makes a loud noise, frightening the chickens so that they fly into the air just in time for the truck to be weighed.
Question: Does this succeed in lightening the load?
1a) Also, does it matter whether the truck is airtight or open? -----
2) In a related question, people on a sailboat realize that it needs to get back to shore in a hurry, so they attach a powerful fan to a powerful battery and aim it at the sail. Can this help propel the boat?
2a) Likewise, suppose this picture is scaled down to become a small toy boat. Does it matter if all takes place in an airtight enclosed case? -----
—Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 10/28/2016 11:52 AM, Dan Asimov wrote:
Some physics questions I don't know the answer to:
1) An old Car Talk puzzle of the week is about a truck carrying a load of chickens.
As it approaches a mandatory truck scale, the driver realizes that its weight is just over the legal limit. To lighten the truck, he or she makes a loud noise, frightening the chickens so that they fly into the air just in time for the truck to be weighed. The chickens can only fly by pushing air down with sufficient momentum flux to equal their weight. So it couldn't have the desired effect in a closed truck. If would even be very hard for all the chickens to fly at once since they would create a circulation flow inside the truck that would work against them. In an open truck a small part of the flying chicken's pressure 'footprint' might fall outside the truck bed
Question: Does this succeed in lightening the load?
1a) Also, does it matter whether the truck is airtight or open? -----
2) In a related question, people on a sailboat realize that it needs to get back to shore in a hurry, so they attach a powerful fan to a powerful battery and aim it at the sail. Can this help propel the boat?
Not if they put the sail out for a reach and blow into it. But if they trim the sail as for a tack and direct the fan back over the outboard surface of the sail they would get some propulsion. But simpler is just furl the sail and point the fan astern.
2a) Likewise, suppose this picture is scaled down to become a small toy boat. Does it matter if all takes place in an airtight enclosed case?
If would make it less effective as the circulation of the air tends to be a counterflow. But here's a more interesting, and real world, question. If you build a landsailer (c.f. nalsa.org) but instead of a mast and sail you provide it with a big rear facing, variable pitch propeller (like an air boat) and you gear the propeller to the rear axle so it turns with the wheels, can you use it to sail directly into the wind? Brent Meeker
Rats --- I did get there, but too late (reinvigorated by a post-prandial snooze, combined with further editorial sanction ...) To restore my bruised ego, I shall propose a supplementary. Denote by Lim(S) the set of accumulation points of set S ; and define the height h(S) to be the minimum n such that Lim^n(S) = { } is empty. Question: is h(S) necessarily finite for every grunch S ? Fred Lunnon On 10/28/16, Dan Asimov <asimov@msri.org> wrote:
Yes, that's it. Nice solving! Fred ... and Andy.
The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered.
—Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> wrote: > Puzzle: Define a grunch to be any closed and bounded > subset of the real line R consisting entirely of rational numbers. > > True or false: There is a countable set of grunches such that > every grunch is a subset of one of them. > > Prove your answer.
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Strictly speaking, height should have been defined by h(S) = sup( n | Lim^n(S) /= { } ) --- otherwise it fails if the answer is "no" ! WFL On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Rats --- I did get there, but too late (reinvigorated by a post-prandial snooze, combined with further editorial sanction ...) To restore my bruised ego, I shall propose a supplementary.
Denote by Lim(S) the set of accumulation points of set S ; and define the height h(S) to be the minimum n such that Lim^n(S) = { } is empty.
Question: is h(S) necessarily finite for every grunch S ?
Fred Lunnon
On 10/28/16, Dan Asimov <asimov@msri.org> wrote:
Yes, that's it. Nice solving! Fred ... and Andy.
The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered.
—Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
> On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> > wrote: >> Puzzle: Define a grunch to be any closed and bounded >> subset of the real line R consisting entirely of rational numbers. >> >> True or false: There is a countable set of grunches such that >> every grunch is a subset of one of them. >> >> Prove your answer.
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<< Denote by Lim(S) the set of accumulation points of set S ; and define the height ht(S) to be the minimum n such that Lim^n(S) = { } is empty. Question: is ht(G) necessarily finite for every grunch G ? >> Three weeks or so ago, I posed this supplementary to Dan's grunch puzzle; la grippe promptly gripped, and has only just relented sufficiently to permit preparing the answer (11) below. The argument is surprisingly delicate --- perhaps it might be simplified --- so paragraphs are numbered for ease of reference, elision, refutation, general derision etc. The discussion employs set operators ∈ (member of), ⋃ (union), ⋂ (intersection), ⊂, ⊆ (proper/subset), since such symbols are apparently becoming more widely available to mail browsers. Please complain if still unable to decipher! Fred lunnon ____________________________________________________________ (0) Given some real set S , define Lim(S) to be the set of accumulation points of S . (1) Every x ∈ S can be classified as just one of: `ambidextrous' if all neighbourhoods of x contain y,z ∈ S with y < x < z ; else `left-handed' if all neighbourhoods of x contain y ∈ S with y < x ; else `right-handed' if all neighbourhoods of x contain z ∈ S with x < z ; else `isolated' if none of the above. (2) Define a grunch to be any closed and bounded set of rational numbers. For any grunch G : G is denumerable since its points are rational; G is compact since closed and bounded; Lim(G) is also a grunch; Lim(G) ⊆ G . *(3)* Required to prove that Lim(G) ⊂ G properly for nonempty grunch G . Assume that on the contrary G = Lim(G) ; then --- (4) Via (2) every x ∈ G is an accumulation point of G , so no x is isolated. (5) Gaps in G occur only between consecutive left- and right-handed points; otherwise some neighbour intrudes between them. Therefore G may be decomposed uniquely into an unrefinable, denumerable union of separated subsets G' with distinct right- and left-handed boundary points. (6) Each G' is a grunch such that G' = Lim(G') also. (7) Having no gaps, G' is dense: every x ∈ G' is ambidextrous except its boundaries! (8) Via (6), G' is rational; but via (7) and compactness, G' includes all real numbers between its bounds. Via contradiction G' does not exist, so G is empty, proving (3) QED. (9) Define the `omega-limit' of G , H == G ⋂ Lim(G) ⋂ ... ⋂ Lim^n(G) ⋂ ... For grunch G , via (2) H = Lim(H) ; so via (3), H = { } is empty. (10) Assume G is a grunch with Lim^n(G) nonempty for all natural n . Using (3), for each n choose some x_n ∈ Lim^n(G) - Lim^{n+1}(G) ; then let x be some accumulation point of the infinite set {x_n} ⊆ G . Via compactness, x ∈ Lim^n(G) for all n , so x ∈ H ; and H is nonempty, contradicting (9). *(11)* Hence assumption (10) was false: for any grunch G , the height ht(G) == min( n | Lim^n(G) = { } ) is finite. (12) The following sequence of examples is instructive. For each n , denote by G(n) the dyadic rationals in the unit interval with at most n bits nonzero: formally, F(n) = { 2^(i_1) + ... + 2^(i_n) | 0 > i_1 > ... > i_n } ; G(n) = { 0 } ⋃ F(1) ⋃ ... ⋃ F(n) . (13) Easily G(n) is a grunch, with Lim(G(n+1)) = G(n) , ht(G(n)) = n+1 . (14) As n -> oo , the omega-limit H is the entire unit interval, expressed in binary; so as indicated by the proof of (3), extending its height to infinity crunches the grunch into continuity. Fred lunnon, Maynooth 20/11/16 ____________________________________________________________ On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Rats --- I did get there, but too late (reinvigorated by a post-prandial snooze, combined with further editorial sanction ...) To restore my bruised ego, I shall propose a supplementary.
Denote by Lim(S) the set of accumulation points of set S ; and define the height h(S) to be the minimum n such that Lim^n(S) = { } is empty.
Question: is h(S) necessarily finite for every grunch S ?
Fred Lunnon
On 10/28/16, Dan Asimov <asimov@msri.org> wrote:
Yes, that's it. Nice solving! Fred ... and Andy.
The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered.
—Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
> On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> > wrote: >> Puzzle: Define a grunch to be any closed and bounded >> subset of the real line R consisting entirely of rational numbers. >> >> True or false: There is a countable set of grunches such that >> every grunch is a subset of one of them. >> >> Prove your answer.
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The crux of the matter is in step (9), where you state that Lim(H) = H. I don't see why this is true, and the justification you give "via (2)" is not illuminating to me. Why does x ∈ H imply x ∈ lim(H)? You use a complex argument, that I don't fully understand, to conclude that " Therefore G may be decomposed uniquely into an unrefinable, denumerable union of separated subsets G' with distinct right- and left-handed boundary points." Isn't the decomposition of G into sets with a single element such a decomposition? If not, what do you mean by saying a collection of sets is "separated"? Andy On Sun, Nov 20, 2016 at 1:27 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Denote by Lim(S) the set of accumulation points of set S ; and define the height ht(S) to be the minimum n such that Lim^n(S) = { } is empty. Question: is ht(G) necessarily finite for every grunch G ? >>
Three weeks or so ago, I posed this supplementary to Dan's grunch puzzle; la grippe promptly gripped, and has only just relented sufficiently to permit preparing the answer (11) below. The argument is surprisingly delicate --- perhaps it might be simplified --- so paragraphs are numbered for ease of reference, elision, refutation, general derision etc.
The discussion employs set operators ∈ (member of), ⋃ (union), ⋂ (intersection), ⊂, ⊆ (proper/subset), since such symbols are apparently becoming more widely available to mail browsers. Please complain if still unable to decipher!
Fred lunnon
____________________________________________________________
(0) Given some real set S , define Lim(S) to be the set of accumulation points of S .
(1) Every x ∈ S can be classified as just one of: `ambidextrous' if all neighbourhoods of x contain y,z ∈ S with y < x < z ; else `left-handed' if all neighbourhoods of x contain y ∈ S with y < x ; else `right-handed' if all neighbourhoods of x contain z ∈ S with x < z ; else `isolated' if none of the above.
(2) Define a grunch to be any closed and bounded set of rational numbers. For any grunch G : G is denumerable since its points are rational; G is compact since closed and bounded; Lim(G) is also a grunch; Lim(G) ⊆ G .
*(3)* Required to prove that Lim(G) ⊂ G properly for nonempty grunch G . Assume that on the contrary G = Lim(G) ; then ---
(4) Via (2) every x ∈ G is an accumulation point of G , so no x is isolated.
(5) Gaps in G occur only between consecutive left- and right-handed points; otherwise some neighbour intrudes between them. Therefore G may be decomposed uniquely into an unrefinable, denumerable union of separated subsets G' with distinct right- and left-handed boundary points.
(6) Each G' is a grunch such that G' = Lim(G') also.
(7) Having no gaps, G' is dense: every x ∈ G' is ambidextrous except its boundaries!
(8) Via (6), G' is rational; but via (7) and compactness, G' includes all real numbers between its bounds. Via contradiction G' does not exist, so G is empty, proving (3) QED.
(9) Define the `omega-limit' of G , H == G ⋂ Lim(G) ⋂ ... ⋂ Lim^n(G) ⋂ ... For grunch G , via (2) H = Lim(H) ; so via (3), H = { } is empty.
(10) Assume G is a grunch with Lim^n(G) nonempty for all natural n . Using (3), for each n choose some x_n ∈ Lim^n(G) - Lim^{n+1}(G) ; then let x be some accumulation point of the infinite set {x_n} ⊆ G . Via compactness, x ∈ Lim^n(G) for all n , so x ∈ H ; and H is nonempty, contradicting (9).
*(11)* Hence assumption (10) was false: for any grunch G , the height ht(G) == min( n | Lim^n(G) = { } ) is finite.
(12) The following sequence of examples is instructive. For each n , denote by G(n) the dyadic rationals in the unit interval with at most n bits nonzero: formally, F(n) = { 2^(i_1) + ... + 2^(i_n) | 0 > i_1 > ... > i_n } ; G(n) = { 0 } ⋃ F(1) ⋃ ... ⋃ F(n) .
(13) Easily G(n) is a grunch, with Lim(G(n+1)) = G(n) , ht(G(n)) = n+1 .
(14) As n -> oo , the omega-limit H is the entire unit interval, expressed in binary; so as indicated by the proof of (3), extending its height to infinity crunches the grunch into continuity.
Fred lunnon, Maynooth 20/11/16
____________________________________________________________
On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Rats --- I did get there, but too late (reinvigorated by a post-prandial snooze, combined with further editorial sanction ...) To restore my bruised ego, I shall propose a supplementary.
Denote by Lim(S) the set of accumulation points of set S ; and define the height h(S) to be the minimum n such that Lim^n(S) = { } is empty.
Question: is h(S) necessarily finite for every grunch S ?
Fred Lunnon
On 10/28/16, Dan Asimov <asimov@msri.org> wrote:
Yes, that's it. Nice solving! Fred ... and Andy.
The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered.
—Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
>> On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> >> wrote: >>> Puzzle: Define a grunch to be any closed and bounded >>> subset of the real line R consisting entirely of rational numbers. >>> >>> True or false: There is a countable set of grunches such that >>> every grunch is a subset of one of them. >>> >>> Prove your answer.
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-- Andy.Latto@pobox.com
Regarding (9) I attempted to avoid introducing ordinals so came to grief, failing to take into account that 1+𝜔 = 𝜔 ≠ 𝜔+1 (where 𝜔 denotes the first infinite ordinal). Although using Lim(G) ⊆ G etc via (2), easily H = Lim^𝜔(G) = G ⋂ Lim(G) ⋂ ... ⋂ Lim^n(G) ⋂ ... and Lim^𝜔(Lim(G)) = Lim^{1+𝜔}(G) = Lim(G) ⋂ ... ⋂ Lim^n(G) ⋂ ... = H , what I actually needed was instead Lim(H) = Lim^{𝜔+1}(G) . There is no obvious reason why H might not include eg. an isolated point; so my entire strategy fails irreparably at this point! Regarding (5) the structure of G can be more complicated than I foresaw. Eg. if an accumulation point x of left-handed points is left-handed, then x has no matching smaller right-hand point delimiting an interval G' separated (on the left) by a gap from its complement in G ; and similar situations arise involving right-handed and ambidextrous x instead. It seems to me that such iterated structures demand some interval G' to occur somewhere "at bottom level", but I cannot at this stage see why. Both barrels ... anyway, thanks Andy! Fred Lunnon On 11/20/16, Andy Latto <andy.latto@pobox.com> wrote:
The crux of the matter is in step (9), where you state that Lim(H) = H. I don't see why this is true, and the justification you give "via (2)" is not illuminating to me. Why does x ∈ H imply x ∈ lim(H)?
You use a complex argument, that I don't fully understand, to conclude that " Therefore G may be decomposed uniquely into an unrefinable, denumerable union of separated subsets G' with distinct right- and left-handed boundary points."
Isn't the decomposition of G into sets with a single element such a decomposition? If not, what do you mean by saying a collection of sets is "separated"?
Andy
On Sun, Nov 20, 2016 at 1:27 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Denote by Lim(S) the set of accumulation points of set S ; and define the height ht(S) to be the minimum n such that Lim^n(S) = { } is empty. Question: is ht(G) necessarily finite for every grunch G ? >>
Three weeks or so ago, I posed this supplementary to Dan's grunch puzzle; la grippe promptly gripped, and has only just relented sufficiently to permit preparing the answer (11) below. The argument is surprisingly delicate --- perhaps it might be simplified --- so paragraphs are numbered for ease of reference, elision, refutation, general derision etc.
The discussion employs set operators ∈ (member of), ⋃ (union), ⋂ (intersection), ⊂, ⊆ (proper/subset), since such symbols are apparently becoming more widely available to mail browsers. Please complain if still unable to decipher!
Fred lunnon
____________________________________________________________
(0) Given some real set S , define Lim(S) to be the set of accumulation points of S .
(1) Every x ∈ S can be classified as just one of: `ambidextrous' if all neighbourhoods of x contain y,z ∈ S with y < x < z ; else `left-handed' if all neighbourhoods of x contain y ∈ S with y < x ; else `right-handed' if all neighbourhoods of x contain z ∈ S with x < z ; else `isolated' if none of the above.
(2) Define a grunch to be any closed and bounded set of rational numbers. For any grunch G : G is denumerable since its points are rational; G is compact since closed and bounded; Lim(G) is also a grunch; Lim(G) ⊆ G .
*(3)* Required to prove that Lim(G) ⊂ G properly for nonempty grunch G . Assume that on the contrary G = Lim(G) ; then ---
(4) Via (2) every x ∈ G is an accumulation point of G , so no x is isolated.
(5) Gaps in G occur only between consecutive left- and right-handed points; otherwise some neighbour intrudes between them. Therefore G may be decomposed uniquely into an unrefinable, denumerable union of separated subsets G' with distinct right- and left-handed boundary points.
(6) Each G' is a grunch such that G' = Lim(G') also.
(7) Having no gaps, G' is dense: every x ∈ G' is ambidextrous except its boundaries!
(8) Via (6), G' is rational; but via (7) and compactness, G' includes all real numbers between its bounds. Via contradiction G' does not exist, so G is empty, proving (3) QED.
(9) Define the `omega-limit' of G , H == G ⋂ Lim(G) ⋂ ... ⋂ Lim^n(G) ⋂ ... For grunch G , via (2) H = Lim(H) ; so via (3), H = { } is empty.
(10) Assume G is a grunch with Lim^n(G) nonempty for all natural n . Using (3), for each n choose some x_n ∈ Lim^n(G) - Lim^{n+1}(G) ; then let x be some accumulation point of the infinite set {x_n} ⊆ G . Via compactness, x ∈ Lim^n(G) for all n , so x ∈ H ; and H is nonempty, contradicting (9).
*(11)* Hence assumption (10) was false: for any grunch G , the height ht(G) == min( n | Lim^n(G) = { } ) is finite.
(12) The following sequence of examples is instructive. For each n , denote by G(n) the dyadic rationals in the unit interval with at most n bits nonzero: formally, F(n) = { 2^(i_1) + ... + 2^(i_n) | 0 > i_1 > ... > i_n } ; G(n) = { 0 } ⋃ F(1) ⋃ ... ⋃ F(n) .
(13) Easily G(n) is a grunch, with Lim(G(n+1)) = G(n) , ht(G(n)) = n+1 .
(14) As n -> oo , the omega-limit H is the entire unit interval, expressed in binary; so as indicated by the proof of (3), extending its height to infinity crunches the grunch into continuity.
Fred lunnon, Maynooth 20/11/16
____________________________________________________________
On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Rats --- I did get there, but too late (reinvigorated by a post-prandial snooze, combined with further editorial sanction ...) To restore my bruised ego, I shall propose a supplementary.
Denote by Lim(S) the set of accumulation points of set S ; and define the height h(S) to be the minimum n such that Lim^n(S) = { } is empty.
Question: is h(S) necessarily finite for every grunch S ?
Fred Lunnon
On 10/28/16, Dan Asimov <asimov@msri.org> wrote:
Yes, that's it. Nice solving! Fred ... and Andy.
The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered.
—Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
>>> On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> >>> wrote: >>>> Puzzle: Define a grunch to be any closed and bounded >>>> subset of the real line R consisting entirely of rational numbers. >>>> >>>> True or false: There is a countable set of grunches such that >>>> every grunch is a subset of one of them. >>>> >>>> Prove your answer.
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-- Andy.Latto@pobox.com
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There's an interesting difference between the uniform triangle fill (by quarters) vs the nonuniform five ninths plus a quadruple ninth version. It appears that the former maps the rationals in [0,1] onto the "Eisenstein rationals" (r₁ + r₂ i √3) in the (equilateral) triangle based on [0,1]. So a (= the) preimage of the center Out[416]= 1/2 + I/(2 Sqrt[3]) is In[417]:= untrifil@% Out[417]= {2/5} Check: In[418]:= trifil @@ % // Simplify Out[418]= {1/6 (3 + I Sqrt[3])} More convincingly, the preimage of Out[405]= 113/355+(7 I Sqrt[3])/22 is In[406]:= untrifil@%//tim During evaluation of In[406]:= 188.873607 secs Out[406]= {66067177350394731417555234130689770663588400360845847462258734629314942043718190483511 /124330809102446660538845562036705210025114037699336929360115994223289874253133343883328} Check: In[407]:= trifil@@%//tim During evaluation of In[407]:= 42.187317 secs Out[407]= {113/355+(7 I Sqrt[3])/22} Interestingly, the centroid of the nonuniform filler (based on [0,3] instead of [0,1]) is a triple point: In[420]:= wolloh[3 %416] (triple size triangle) Out[420]= {3/10, 1/2, 7/10} Check: In[422]:= Simplify[hollow /@ %%]/3 Out[422]= {{1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}} Likewise, the preimage of Out[408]= 13/9 + (13 I)/(6 Sqrt[3]) is In[423]:= wolloh@%408 Out[423]= {1204367/1680912} Check: In[424]:= Expand@Simplify[hollow @@ %] Out[424]= {13/9 + (13 I)/(6 Sqrt[3])} But wolloh[3/2 + I 3 Sqrt[3]/4] apparently recurses forever. I don't even have a good enough approximation to guess if this preimage is algebraic. --rwg On Wed, Oct 26, 2016 at 7:22 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Oct 21, 2016 at 7:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
See gosper.org/julianstrifle.png . Note the tiny self-crossings not visible in the cursive drawing. It is usually far clearer to portray a <2-dimensional curve as the boundary of a two-dimensional region. --rwg
Julian's Fourier expander takes an optional argument to arrange m copies of a curve around a regular m-gon. gosper.org/6trifil.png . Instead of spacefilling an equilateral triangle divided into quarters, I tried dividing it into five little ones plus a double size one. Arranged around a hexagon, gosper.org/ringnoring1234.png (one of the few cases where the outline looks maybe better than the filled area.) (Shades of https://en.wikipedia.org/wiki/Uniform_tilings_in_ hyperbolic_plane ?) Note that with 2468 rotors, the Fourier approximation gets nowhere near the middle. So is this a spacefill, or not? Julian: "Of course it is ..." Proof that it reaches all three vertices of each triangle: In[422]:= hollow /@ {0, 4/5, 1} // Expand
Out[422]= {{0}, {3/2 + (3 I Sqrt[3])/2}, {3}}
via (Julian's ever amazing) piecewiserecursivefractal Clear[hollow]; hollow[t_] := piecewiserecursivefractal[t, Identity, {Min[6, 1 + Floor[6*#]]} &, {6 # &, 6 # - 1 &, 3 - 6 # &, 6 # - 3 &, 6 # - 4 &, 6 - 6 # &}, {#/3 &, #/3 + 1 &, #/3/I^(2/3) + 3/2 + I Sqrt[3]/2 &, 3/2 + I Sqrt[3]/2 - #/3 &, 2 #/3 + 1/2 + I Sqrt[3]/2 &, #/3*I^(4/3) + 3 &}] A polygonal sampling: gosper.org/hollowfill150.png (Reminder: True spacefills self-contact like crazy.) --rwg
On Thu, Oct 20, 2016 at 4:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com>
wrote:
With much help from Julian, gosper.org/hellodoily.gif --rwg
>[chop]
It turns out the preimage of 3/2 + i √3/4 under the nonuniform triangle-filling function is BaseForm[401474566812783980411091190154813161/ 1403266106709659827505450695237042176., 6] (0.14144414144141141144444441444414414444411411414...)₆, so we "know" it isn't algebraic. For lack of understanding Julian's remark, "You can invert it just fine and it will always converge, though good luck finding points with explicit inverses except by going forward.", it took me a week to figure out that there's a straightforward process for graphically determining two or three base 6 digits at a time. So we have the incautious conjecture that the preimages of the "Eisenstein rationals" are either rational, or "Cantorish", base 6. --rwg On Sun, Oct 30, 2016 at 5:15 PM, Bill Gosper <billgosper@gmail.com> wrote:
There's an interesting difference between the uniform triangle fill (by quarters) vs the nonuniform five ninths plus a quadruple ninth version. It appears that the former maps the rationals in [0,1] onto the "Eisenstein rationals" (r₁ + r₂ i √3) in the (equilateral) triangle based on [0,1]. So a (= the) preimage of the center
Out[416]= 1/2 + I/(2 Sqrt[3])
is In[417]:= untrifil@%
Out[417]= {2/5} Check: In[418]:= trifil @@ % // Simplify
Out[418]= {1/6 (3 + I Sqrt[3])}
More convincingly, the preimage of
Out[405]= 113/355+(7 I Sqrt[3])/22
is In[406]:= untrifil@%//tim
During evaluation of In[406]:= 188.873607 secs Out[406]= {660671773503947314175552341306897706635884003608458474622587 34629314942043718190483511 /124330809102446660538845562036705210025114037699336929360115 994223289874253133343883328} Check: In[407]:= trifil@@%//tim During evaluation of In[407]:= 42.187317 secs Out[407]= {113/355+(7 I Sqrt[3])/22}
Interestingly, the centroid of the nonuniform filler (based on [0,3] instead of [0,1]) is a triple point: In[420]:= wolloh[3 %416] (triple size triangle)
Out[420]= {3/10, 1/2, 7/10} Check: In[422]:= Simplify[hollow /@ %%]/3
Out[422]= {{1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}}
Likewise, the preimage of Out[408]= 13/9 + (13 I)/(6 Sqrt[3])
is In[423]:= wolloh@%408
Out[423]= {1204367/1680912} Check: In[424]:= Expand@Simplify[hollow @@ %]
Out[424]= {13/9 + (13 I)/(6 Sqrt[3])}
But wolloh[3/2 + I 3 Sqrt[3]/4] apparently recurses forever. I don't even have a good enough approximation to guess if this preimage is algebraic. --rwg
On Wed, Oct 26, 2016 at 7:22 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Oct 21, 2016 at 7:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
See gosper.org/julianstrifle.png . Note the tiny self-crossings not visible in the cursive drawing. It is usually far clearer to portray a <2-dimensional curve as the boundary of a two-dimensional region. --rwg
Julian's Fourier expander takes an optional argument to arrange m copies of a curve around a regular m-gon. gosper.org/6trifil.png . Instead of spacefilling an equilateral triangle divided into quarters, I tried dividing it into five little ones plus a double size one. Arranged around a hexagon, gosper.org/ringnoring1234.png (one of the few cases where the outline looks maybe better than the filled area.) (Shades of https://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbol ic_plane ?) Note that with 2468 rotors, the Fourier approximation gets nowhere near the middle. So is this a spacefill, or not? Julian: "Of course it is ..." Proof that it reaches all three vertices of each triangle: In[422]:= hollow /@ {0, 4/5, 1} // Expand
Out[422]= {{0}, {3/2 + (3 I Sqrt[3])/2}, {3}}
via (Julian's ever amazing) piecewiserecursivefractal Clear[hollow]; hollow[t_] := piecewiserecursivefractal[t, Identity, {Min[6, 1 + Floor[6*#]]} &, {6 # &, 6 # - 1 &, 3 - 6 # &, 6 # - 3 &, 6 # - 4 &, 6 - 6 # &}, {#/3 &, #/3 + 1 &, #/3/I^(2/3) + 3/2 + I Sqrt[3]/2 &, 3/2 + I Sqrt[3]/2 - #/3 &, 2 #/3 + 1/2 + I Sqrt[3]/2 &, #/3*I^(4/3) + 3 &}] A polygonal sampling: gosper.org/hollowfill150.png (Reminder: True spacefills self-contact like crazy.) --rwg
On Thu, Oct 20, 2016 at 4:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com>
wrote:
With much help from Julian, gosper.org/hellodoily.gif --rwg
>>[chop]
With this filling function, the triangle's centroid is a triple point: In[434]:= wolloh[1/2 (3 + I Sqrt[3])] // tim During evaluation of In[434]:= 1.413051,3 Out[434]= {3/10, 1/2, 7/10} But there is a tidier order of filling the one large + five small subtriangles: http://gosper.org/tameavoid2592.png http://gosper.org/tamedoily1728.png and Julian makes the remarkable observation that this function likewise has (≥) three preimages of the centroid, but one of them is irratonal! In[432]:= unTrifill[1/2 (3 + I Sqrt[3])] // tim During evaluation of In[432]:= 166.861545,2 Out[432]= {4/15, 2/3} The evidence for the irrational is the timing: 167 sec (vs 1.4), which would have been "$RecursionLimit exceeded" but for a safety valve tripping for denominators near a googol. Irrationality could be proven with a bound on preimage denominators, which should be feasible. Anybody need something to work on? --rwg On Tue, Nov 1, 2016 at 2:29 AM, Bill Gosper <billgosper@gmail.com> wrote:
It turns out the preimage of 3/2 + i √3/4 under the nonuniform triangle-filling function is BaseForm[401474566812783980411091190154813161/ 1403266106709659827505450695237042176., 6]
(0.14144414144141141144444441444414414444411411414...)₆, so we "know" it isn't algebraic. For lack of understanding Julian's remark, "You can invert it just fine and it will always converge, though good luck finding points with explicit inverses except by going forward.", it took me a week to figure out that there's a straightforward process for graphically determining two or three base 6 digits at a time. So we have the incautious conjecture that the preimages of the "Eisenstein rationals" are either rational, or "Cantorish", base 6. --rwg
On Sun, Oct 30, 2016 at 5:15 PM, Bill Gosper <billgosper@gmail.com> wrote:
There's an interesting difference between the uniform triangle fill (by quarters) vs the nonuniform five ninths plus a quadruple ninth version. It appears that the former maps the rationals in [0,1] onto the "Eisenstein rationals" (r₁ + r₂ i √3) in the (equilateral) triangle based on [0,1]. So a (= the) preimage of the center
Out[416]= 1/2 + I/(2 Sqrt[3])
is In[417]:= untrifil@%
Out[417]= {2/5} Check: In[418]:= trifil @@ % // Simplify
Out[418]= {1/6 (3 + I Sqrt[3])}
More convincingly, the preimage of
Out[405]= 113/355+(7 I Sqrt[3])/22
is In[406]:= untrifil@%//tim
During evaluation of In[406]:= 188.873607 secs Out[406]= {66067177350394731417555234130689770663588400360845847462258 734629314942043718190483511 /12433080910244666053884556203670521002511403769933692936011 5994223289874253133343883328} Check: In[407]:= trifil@@%//tim During evaluation of In[407]:= 42.187317 secs Out[407]= {113/355+(7 I Sqrt[3])/22}
Interestingly, the centroid of the nonuniform filler (based on [0,3] instead of [0,1]) is a triple point: In[420]:= wolloh[3 %416] (triple size triangle)
Out[420]= {3/10, 1/2, 7/10} Check: In[422]:= Simplify[hollow /@ %%]/3
Out[422]= {{1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}}
Likewise, the preimage of Out[408]= 13/9 + (13 I)/(6 Sqrt[3])
is In[423]:= wolloh@%408
Out[423]= {1204367/1680912} Check: In[424]:= Expand@Simplify[hollow @@ %]
Out[424]= {13/9 + (13 I)/(6 Sqrt[3])}
But wolloh[3/2 + I 3 Sqrt[3]/4] apparently recurses forever. I don't even have a good enough approximation to guess if this preimage is algebraic. --rwg
On Wed, Oct 26, 2016 at 7:22 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Oct 21, 2016 at 7:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
See gosper.org/julianstrifle.png . Note the tiny self-crossings not visible in the cursive drawing. It is usually far clearer to portray a <2-dimensional curve as the boundary of a two-dimensional region. --rwg
Julian's Fourier expander takes an optional argument to arrange m copies of a curve around a regular m-gon. gosper.org/6trifil.png . Instead of spacefilling an equilateral triangle divided into quarters, I tried dividing it into five little ones plus a double size one. Arranged around a hexagon, gosper.org/ringnoring1234.png (one of the few cases where the outline looks maybe better than the filled area.) (Shades of https://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbol ic_plane ?) Note that with 2468 rotors, the Fourier approximation gets nowhere near the middle. So is this a spacefill, or not? Julian: "Of course it is ..." Proof that it reaches all three vertices of each triangle: In[422]:= hollow /@ {0, 4/5, 1} // Expand
Out[422]= {{0}, {3/2 + (3 I Sqrt[3])/2}, {3}}
via (Julian's ever amazing) piecewiserecursivefractal Clear[hollow]; hollow[t_] := piecewiserecursivefractal[t, Identity, {Min[6, 1 + Floor[6*#]]} &, {6 # &, 6 # - 1 &, 3 - 6 # &, 6 # - 3 &, 6 # - 4 &, 6 - 6 # &}, {#/3 &, #/3 + 1 &, #/3/I^(2/3) + 3/2 + I Sqrt[3]/2 &, 3/2 + I Sqrt[3]/2 - #/3 &, 2 #/3 + 1/2 + I Sqrt[3]/2 &, #/3*I^(4/3) + 3 &}] A polygonal sampling: gosper.org/hollowfill150.png (Reminder: True spacefills self-contact like crazy.) --rwg
On Thu, Oct 20, 2016 at 4:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com>
wrote:
> With much help from Julian, gosper.org/hellodoily.gif > --rwg >
> >>[chop]
participants (8)
-
Andy Latto -
Bill Gosper -
Brent Meeker -
Dan Asimov -
Dan Asimov -
Fred Lunnon -
James Propp -
Kerry Mitchell