August 2019 - mathfun@mailman.xmission.com

[math-fun] Losing sleep because...
by bradklee＠gmail.com 08 Aug '19

08 Aug '19

For reals R, card(R^n)=card(R) yet R is measure zero in R^n. This is something of a paradox, yet: R^n=(R^(n-1)/p)*R + p*R, with p a lone point of R^(n-1), and p measure zero in R^(n-1). This sounds okay(?). What’s more troubling is reading on stack exchange and physics forum that Lebesgue integration allows, sometimes requires (?), 0*infinity=0? “I’m on the outside, looking inside / what do I see? / Much confusion, disillusion / all around me.” Arggggg.......

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Re: [math-fun] Random nonempty intersection of cube and plane, revisited
by Keith F. Lynch 08 Aug '19

08 Aug '19

James Propp <jamespropp(a)gmail.com> wrote: > Does your cube have side-length 1 or side-length 2? Sigh. Answered in the text of mine which you inexplicably quoted in full. > Since the volume of the cube of side-length 2 is 8, and since the > mean width is (a+b+c)/2 = (2+2+2)/2 = 3, the average cross-sectional > area should be 8/3 ~ 2.6667. Keith?s 2.8819 seems awfully far off > the mark (especially in comparison with his suggestively accurate > estimate 4.0004 from two years ago). I don't follow your reasoning as to why it should be 8/3. I have fairly high confidence in my program, and I would give ten-to-one odds that the correct value of average area is between 2.881 and 2.883. Allan Wechsler <acwacw(a)gmail.com> wrote: > My suspicion immediately falls on the definition of "random plane > intersecting a cube". Here are two possibilities: Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > But a subtler difficulty concerns the choice of integration weight: > ie. what probability distribution to assign to the phase space > (here, planes in 3-space). James Propp <jamespropp(a)gmail.com> wrote: > Geometric measure theory looks for measures that are invariant with > respect to symmetries of the underlying space. For instance, we > seek a continuous measure on the set of planes in 3-space that is > invariant under translation and rotation. And it turns out to be > unique (up to scaling)! That?s the measure I have in mind. Likewise. You beat me to it. And the same measure applies to planes through any other solid, including one with no symmetries. Invariant under reflections, too. Allan Wechsler <acwacw(a)gmail.com> wrote: > Rotating a plane around a line in that plane by a tiny angle ought > to be a tiny move, and invariance means that the choice of rotation > axis should not matter. But then, translating a plane perpendicular > to itself without changing its orientation can be shown to be a > smaller move than any rotation, and is therefore 0, which violates > positive definiteness. I was envisioning leaving the planes alone and rotating or translating the *cube*. I will describe how I generate random planes, and then let you guys shoot the planes down. :-) I use the formula ax + by + cz + d = 0 for the plane. (The cube is defined as the volume between -1 and +1 for all three coordinates.) I (pseudo-)randomly, uniformly, and independently choose each of a, b, and c to be real numbers between -1 and +1. That gives me a random point in a cube. I then set r to sqrt(a^2 + b^2 + c^2), and abort if r > 1, so as to carve the cube into a sphere. I then divide each of a, b, and c by r, so as to get a random point on the surface of the sphere. I then set d equal to a random number from 0 to d_max, where d_max is some number that exceeds the square root of 3, that being the maximum "radius" of the cube. (I've been using 1.75 for the first run, 2.0 for the second.) If the resulting plane does not intersect the cube, I don't count it. I keep going until I have the desired number of planes that intersect the cube. Most recently, one billion for each of the two runs. By this time tomorrow, if nothing goes wrong, I should have the results for two runs with ten billion planes each.

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[math-fun] Another step on the low road to Daitoku-Ji
by Brad Klee 08 Aug '19

08 Aug '19

Some time ago I wrote a totally crazy article about how I thought that the aliens from Villeneuve's "Arrival" had extra-sensory perception of manifold geometry: https://community.wolfram.com/groups/-/m/t/1034626 ( with over 13,000 views !? What? How can this be? ) Tonight it occurred to me that this theory could be improved and advanced by physical construction of a curved surface sculpture using wire and //black thread only//. It took more than ten yards of winding, but here it is: https://0x0.st/zO6H.JPG An Enso is supposed to be drawn in one fell swoop. This is not possible without ink... At the end of the process, leave extra thread, and it can be wrapped around cyclically for added texture, even in only one breath. Of course, in terms of topology, the sculpture is a an orientable annulus, with a 2*Pi cycle on either side. Seen this way, yes it's foolhardy to rush on to Mobius loops, Klein bottles, and anything else the Germans might have invented. And at last: "The creation of enso was continued by Daitoku-ji masters during the early Edo period (1600-1868) such as Takuan Soho (1573-1645) and Kosetsu Soryu (1595-166), and then became increasingly popular with Zen masters in the Hakuin lineage such as Torei Enji (1721-1792) and Shunso Joshu (1751-1839)" I found a picture of "Drunk on the Moon", by Dairyo, abbot number 341. Ha ha ha, me too, but don't forget Hotaru! Cheers --Brad

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[math-fun] Homology of YGB Sculpture
by Brad Klee 07 Aug '19

07 Aug '19

Consider a Mobius strip, colored symmetrically: Yellow-Green-Blue-Green. (Actually, there are two shades of blue and accents of orange, purple, and magenta, but these extra colors do not contribute to analysis.) If the Mobius strip were non-self-intersecting, a 4*PI cycle could be written as a word, Y1G1B1g1Y2G2B2g2. Instead require that B2 is bisected by the inner Y edge. The path through B2 is obstructed, and the 4*Pi cycle can not be completed, see for example: https://0x0.st/zpTf.JPG https://0x0.st/zpT9.JPG https://0x0.st/zpTp.JPG https://0x0.st/zpTL.JPG Fix the topology by allowing that Y1 and Y2 connect directly to the left and right parts of B2. PROBLEM: How many homology classes does the YGB sculpture have? ANSWER: Three paths are valid from Y1 to Y2: Y1G1B1g1Y2, Y1B2G2Y2, and Y1g2B2Y2. An entire cycle involves one half path plus the reverse of another. Three choose two equals three. Thus the geometry has three homology classes, two of length 7, one of length 6. Explicitly they are: Y1G1B1g1Y2G2B2, Y1G1B1g1Y2B2g2, and Y1B2G2Y2B2g2. A Mobius strip has only one homology class, so this YGB Surface must be something else entirely. Is it known already? If yes, does it also have a vanity name after a famous mathematician? Cheers, Brad

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Re: [math-fun] Wanted/needed: video of morphing dissection of the plane
by Keith F. Lynch 07 Aug '19

07 Aug '19

Tom Karzes <karzes(a)sonic.net> wrote: > I think Fred's point is that the maximum (or minimum) may be > a limit case, where the limit polygon is not a pentagon, even > though the polygons leading to the limit are. Thanks. I mostly take a computational view of math, so that's not an issue for me. I'm interested in what my random-plane program should get as largest and smallest area and perimeter for each polygon. For instance the lower limits are obviously 0 for triangles. That doesn't mean I expect to find a triangle with that exact area or perimeter. It means that the more random planes I generate, the closer both numbers should get to 0. And indeed that's what I observe. It's only a problem if I do something stupid, such as take the geometric or harmonic mean, since those are dominated by values close to 0. I don't doubt that the intersection areas have well-defined geometric and harmonic means, and that they aren't zero. But they can't be found by a random plane program, not even with a googolplex planes. (They could be found with hairy triple integrals.) Fortunately, James Propp asked for the better-behaved arithmetic mean. I could also give him the quadratic mean (RMS) if he likes.

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[math-fun] Bill Gosper's famous 1989 MIT talk now scanned
by Neil Sloane 06 Aug '19

06 Aug '19

It is scanned and stored on the OEIS server. Direct URL to it: https://oeis.org/A309204/a309204.pdf It is linked to from 6 sequences right now, e.g. https://oeis.org/A309204 There are a lot of "intentionally blank" pages, because about half the originals were two-sided and half were not. It is a composite of two different paper copies, each of which had about half the pages unnumbered, and in a different order in each case. Here you get one copy of each side, and pages after 15 have been numbered in a fairly random order. What is really needed is 40 more pages from Bill explaining the formulas and the illustrations! Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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[math-fun] KFL pentagonal perimeter problem
by Fred Lunnon 05 Aug '19

05 Aug '19

Keith F. Lynch <kfl(a)keithlynch.net> asked << I have a related question, in support of testing my program that finds the average perimeter and area of random intersections of cubes and planes. I know what the maximum perimeter and maximum area are when the intersection is a triangle, a quadrilateral, or a hexagon. (Or at least I think I do.) But I don't know what it is when the intersection is a pentagon. Does anyone know? Thanks. >> In principle such questions can definitively be answered by the method of Lagrange multipliers, followed by inspection of the (lower-dimensional) semi-algebraic boundary; but in practice I prefer to start by trying to get a provisional feel for the shape of solutions via a (quick-and-dirty) numerical Monte-Carlo attack. [ Maybe not so quick but certainly dirty, since convergence to any critical point is necessarily ill-conditioned! ] The cube will have vertices with components in {+1,-1} . For this problem we get off to a good start by observing that, modulo symmetries of the cube, the interior of the phase space of potential plane sections is connected: for example, plane x + y + 3 z = 2 is continuously connected to every possible pentagonal cross-section modulo congruence. Our luck quickly runs out when it transpires that any solutions have at least one pentagon corner with a zero component, causing large cancellation errors in the generated approximate solutions. But, after a number of runs with different settings of convergence parameters and initial planes, a hazy picture starts to emerge. Apparently all solution pentagons are degenerate: no proper maximal pentagons exist, only quadrilaterals yielding perimeter suprema. We find the following provisional solutions (there may be others): A local interior maximum perimeter 8.24621125 at quadrilateral on plane x + y + 4 z = 2 ; A global(?) boundary maximum perimeter 4 sqrt(5) ~ 8.94427191 at quadrilateral on plane x + y + 2 z = 0 . Notice that the latter is NOT the globally maximal quadrilateral with perimeter 4(1 + rt2) ~ 9.65685425 , which it seems cannot tilt to a pentagon while maintaining planarity. The area problem remains pending. Further details on request. Fred Lunnon

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[math-fun] Wanted/needed: video of morphing dissection of the plane
by James Propp 05 Aug '19

05 Aug '19

I just realized that, to illustrate Warren Smith's way of proving the Wall of Fire theorem at my August 7 talk, it'd be cool to have a video or GIF showing how the intersection between the 2-skeleton of a moving cubical network and a fixed plane evolves in time. For instance, say the plane is {(x,y,z): x+y+z=0} and the cubical network is the standard one in Z^3 moving at constant speed in the (1,1,1) direction, which one can write as {(x,y,z); x≡t (mod 1) or y≡t (mod 1) or z≡t (mod 1)}. We see a dynamic dissection of the plane in which equilateral triangles grow and turn into hexagons and then turn into shrinking triangles pointing the other way. Can anyone dash off such a video? If I use it in my talk I will of course give credit. Thanks, Jim Propp

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[math-fun] Cross-sections of regular dodecahedron
by James Propp 05 Aug '19

05 Aug '19

It appears that no cross-section of a regular dodecahedron has more than ten sides, but how does one prove it? (I did five minutes of Googling but didn't find relevant literature.) Thanks, Jim Propp

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Re: [math-fun] Random nonempty intersection of cube and plane, revisited
by Keith F. Lynch 02 Aug '19

02 Aug '19

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > Someone can collaborate right now by giving me some suggestion on > how to figure out the order of the intersections between the plane > and the cube edges. Without knowing the order, I don't know which > of the pairwise distances to add to get the perimeter (except, of > course, in the case of the triangle). Okay, I think I figured it out. Where n is the order of the intersection polygon, 3 through 6, there are n intersection points between the cube and the plane, hence n! possible sequences of intersection points. Since the polygon is convex, I'm pretty sure that the correct order to visit the n intersection points, returning to my starting point (i.e. a Hamiltonian circuit) will simply be the one with the shortest total distance. Someone please politely tell me if I'm wrong. Since it doesn't matter which intersection point I start on, I only have to try (n-1)! sequences. Make that (n-1)!-1, since there are two correct solutions (the same solution in opposite directions) and they can't both be in the last place. So for the largest possible n, 6, 119 tries. Of course it scales very badly for larger polygons, even worse than exponential in n, but that's not an issue here. For larger convex polygons maybe always go to the nearest neighbor not yet visited? No, that can fail. Maybe that always succeeds for *some* starting point, just not necessarily for all of them. That would scale with n^2. > Areas are even tougher. There's no analog of Heron's formula for > quadrilaterals, pentagons, or hexagons. Certainly there is. Once I know the order of the intersection points, all I have to do is run a line from one intersection point to each of the other intersection points, to divide the polygon into triangles, then Heron each of the triangles and add their areas together. If I get the correct average area, that will give me confidence that I also have the correct average perimeter.

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