December 2010 - mathfun@mailman.xmission.com

[math-fun] q. re Farey-ish fractions with weighted mediants
by N. J. A. Sloane 15 Dec '10

15 Dec '10

Rich Schroeppel asked me: > Neil, have you looked at "Farey-depth"? > I.e., 1. the number of mediant steps required to reach a particular rational? > Or 2. the number of steps needed to reach the earliest and latest k/N > (as a function of N = 1, 3, 5, 7, ...). This was in response to my comment that > I added the triangles of numerators and denominators > of these fractions to the OEIS: see http://oeis.org/A177405 and A177407. I haven't looked at "Farey-depth". Maybe someone on this list could work out the initial terms of the two sequences in 2.? Neil

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[math-fun] Hitler struggles with point-set topology
by James Propp 14 Dec '10

14 Dec '10

http://www.youtube.com/watch?v=SyD4p8_y8Kw Jim Propp

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Re: [math-fun] Farey-ish fractions with weighted mediants
by James Propp 14 Dec '10

14 Dec '10

I also tried the same construction using the four weighted mediants (4a+c)/(4b+d), (3a+2c)/(3b+2d), (2a+3c)/(2b+3d), and (a+4c)/(b+4d) (removing common factors as they arise). At the 10th level, I find that every fraction in the list has denominator congruent to 1 mod 4, and most fractions of this form with small denominator do occur althought a few don't; specifically, 4/13, 6/13, 7/13, and 9/13 haven't occurred by level 10 even though 1/13, 2/13, 3/13, 5/13, 8/13, 10/13, 11/13, and 12/13 all occurred back in level 4 or earlier. The only other proper fractions with denominator less than 30 and congruent to 1 mod 4 that haven't occurred by level 10 are 4/29 and 25/29. I'll probably post something about these questions to MathOverflow unless someone on this list says they're working on them. Jim Propp

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[math-fun] Newton's cradle
by Bill Gosper 13 Dec '10

13 Dec '10

Veit>Here's a fun application of the pairwise model that you can try at home. Hold a basketball while resting a tennis ball on top of it. Now drop the basketball (the tennis ball should be co-moving and nearly in contact). You'll find that after all the collisions (pairwise!) have finished the tennis ball is shot up to a surprising height. It took this to remind me of the short-lived AstroBlaster toy, an unopened copy of which I may still own. It was a tapered sequence of (very) bouncy balls on a stem firmly embedded in the bottom one. The middle ones could slide on the stem, but were trapped by a thickening near the top. The topmost, tiny ball had a large enough hole to slide off. Dangle it by the stem, drop it, and instantly lose the tiny ball, or your eye if it was above the impact point. I wonder if Tim Rowett owns one. --rwg

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[math-fun] Cubic sections
by Bill Gosper 13 Dec '10

13 Dec '10

(Simple stuff I couldn't visualize w/o plotting.) Rotate a unit cube about a major diagonal. Its intersection with a plane containing that diagonal varies (with period pi/3) between a rhombus (of side rt(5)/2) and a 1 by rt2 rectangle. --rwg Who might be able to tell me that "my" Fourier expansion of Jacobi am is a) Centuries old, or b) Just what we've been looking for. ?

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Re: [math-fun] Farey-ish fractions with weighted mediants
by N. J. A. Sloane 13 Dec '10

13 Dec '10

I added the triangles of numerators and denominators of these fractions to the OEIS: see http://oeis.org/A177405 and A177407. The analogous entries for the classical Farey series are A006842 and A006843. Updates to the entries would be welcomed. Neil

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Re: [math-fun] possible?
by Bill Gosper 13 Dec '10

13 Dec '10

Bill Thurston> It's known (and not hard to prove, using C^\infty bump functions --- I > think this may have been originally due to Hassler Whitney) that every > closed set in $R^n$ is the zero set of a C^\infty function. So, you can > have two C^\infty surfaces intesecting in R^3 even intersect in much > wilder sets than a simple arc --- e.g. they can intersect in a set that is > connected but not locally connected, or whatever. If you're concerned > about how the two surfaces cross, > the construction allows you to independently choose the sign of the > function in any component of the complement. > On Dec 8, 2010, at 3:05 PM, Bill Gosper wrote: > >> Good grief, tutor Julian has plotted two infinitely differentiable >> surfaces >> intersecting in a simple arc that is not differentiable. >> --rwg >> __________ Thanks! This was probably a special case: The intersection became momentarily tangential, whereat the arc took an abrupt turn. It's a little hard to believe, even as you tumble the Mathematica plot, whose code I'll send soon. --rwg

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Re: [math-fun] Newton's cradle
by rcs＠xmission.com 13 Dec '10

13 Dec '10

A message from Steve Gray .... --- Date: Sun, 12 Dec 2010 17:16:36 -0800 Subject: Re: [math-fun] Newton's cradle From: "Stephen B. Gray" <stevebg(a)roadrunner.com> I don't think anyone has mentioned conservation of energy or momentum. ?? Steve Gray On 12/11/2010 11:07 AM, Gary Antonick wrote: > first of all, we have a serious problem in that one ball appears to be > kicking out only one ball. so pretty much everyone might be wrong.. or > at least a new group is a bit more right. and now completely > perplexed. am epoxying together a larger cradle (1" and 1.25" balls) > this moment. > > second... in defense of Fred (from very much the layperson view on > this) it does seem that given three colliding bodies all you know is > that the ending velocities (along with the beginning velocities) will > lie on a circle at the intersection of a sphere (kinetic energy) and a > plane (momentum). I don't see how this endpoint could ever be > determinate without another constraint. > > On Sat, Dec 11, 2010 at 10:45 AM, Robert Munafo <mrob27(a)gmail.com> wrote: > >> How ill-defined? >> >> For example, take the case where all three have equal mass, the central body >> starts out motionless and the outer two are approaching at the same speed >> (from opposite directions). The midpoint between the positions of the outer >> two is X, and isn't changing until the first impact because of the initial >> conditions I just stated. >> >> If the central body starts out at position "X+epsilon", I get two >> non-simultaneous collisions after which the central body ends up at position >> "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous >> collisions, the central body remains motionless and the outer two >> simultaneously bounce off it. >> >> If I decompose in the other order (starting with a negative "epsilon" so the >> other impact happens first) the limit still gives me the same answer. >> >> Why is the limit ill-defined? Because the velocity of either particle is not >> defined (and discontinuous) at the moment of impact? Wouldn't that make all >> individual collision events ill-defined? >> >> (Of course this is all assuming the bodies are ideal perfectly rigid, i.e. >> the speed of sound in their material is infinite and similar idealizations) >> >> - Robert >> >> On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon(a)gmail.com> wrote: >> >> >>> [...] >>> I've never been convinced that these problems are even well-defined, >>> for the same reason that a simultaneous collision between three bodies >>> in the plane is ill-defined. In that case, it's immediately obvious that >>> you >>> get completely different results according to how you decompose into >>> a sequence of 2-body collisions. >>> >>> The resemblance to the preceding discussion is striking! WFL >>> >

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[math-fun] Correction; Analemma
by Bill Gosper 12 Dec '10

12 Dec '10

rwg>Based on [Veit's] earlier analysis, I withdraw my claim about hypercubes. >The effective center varies (slightly) depending on what plane the centroid >oscillates in. Wrong again. The moment of inertia of a cube about its center is orientation- independent. The center of oscillation, when hung from a vertex, always divides the major diagonal in proportion 11:7. (Off list, Veit pointed out that the center of oscillation of a hoop swinging in its plane about a point on its rim is indeed its "south pole", as if all its mass were concentrated there. Absurd. But apparently irrefutable. Slightly less absurd: Turning an ordinary disk shaped pendulum 90 degrees to swing like a paddle shortens its period. (Assuming zero air resistance.) ---------- Doug Dodds at Symbolics used to send the following every year at this time. If you haven't seen one, check out "analemma" in Google images for one of those 365-exposure sky pix. There's even one for Mars!.--rwg --- Received: from ELEPHANT-BUTTE.SCRC.Symbolics.COM by RUSSIAN.SPA.Symbolics.COM via CHAOS with CHAOS-MAIL id 52312; Tue 8-Dec-87 13:52:55 PST Received: from YUKON.SCRC.Symbolics.COM by ELEPHANT-BUTTE.SCRC.Symbolics.COMvia CHAOS with CHAOS-MAIL id 244027; Tue 8-Dec-87 16:18:32 EST Received: from ANNISQUAM.SCRC.Symbolics.COM by YUKON.SCRC.Symbolics.COM via INTERNET with SMTP id 303771; 8 Dec 87 16:11:04 EST Date: Tue, 8 Dec 87 16:11 EST From: Doug Dodds <DODDS(a)YUKON.SCRC.Symbolics.COM> Subject: Rejoice, the light returneth! To: BBoard(a)YUKON.SCRC.Symbolics.COM, Fun(a)YUKON.SCRC.Symbolics.COM File-References: SCRC|Y:>Dodds>analemma.mss.newest, SCRC|Y:>Dodds>analemma.text.newest Message-ID: <19871208211112.0.DODDS(a)ANNISQUAM.SCRC.Symbolics.COM> The sunset today in Boston is at 16:11 EST, which is Plenty Early. But take heart, today's sunset is the earliest of the year! (Everywhere in the northern hemisphere, regardless of the actual sunset time.) The sunset starts to become later tomorrow, even though the shortest day is still two weeks off. If you haven't yet learned of this fascinating phenomenon, take a look at one of the referenced files. The first is formattable (m-X Format File); if formatting is not available to you, then print the second referenced file. Enjoy the increasing afternoon light! 0,, *** EOOH ***

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[math-fun] Circle of differences
by Joshua Zucker 11 Dec '10

11 Dec '10

I suspect this is well-known but I am still excited that I figured it out for myself last night. Start with four numbers written around a circle. At each moment, replace the four numbers with the absolute value of the difference of each adjacent pair. For example 1 9 4 7 would be replaced by 8 5 3 6 and then 3 2 3 2 and then 1 1 1 1 and then 0 0 0 0 and thereafter it's pretty boring. The problem: 1) Prove that given any four integers, it eventually reaches 0 0 0 0. (I had seen this result before). Also give a reasonable upper bound relating the size of the largest integer to the number of steps to reach 0 0 0 0. (A a reasonable bound that at least has the right proportionality in big O terms but maybe with a slightly too generous constant by a factor of 2 or so). 2) Prove that there exist lists of four integers that take as many steps as you want to reach 0 0 0 0. Let's make it a constructive proof that actually gives an algorithm for constructing a set of integers that, say, lasts 1000 steps. 3) What happens if you start with irrationals? Say, things of the form a + b sqrt(2)? Does it still always reach 0 0 0 0 for them? 4) Prove that there exists a list of four irrationals that never reach 0 0 0 0. Last night I solved 2 and 4 and that made me happy. I have what looks to me like an answer to 3 but I am not totally convinced that my proof is really solid. I'm sure that in the late 1990s I heard some version of this problem somewhere, but I don't know where nor exactly which form I saw it in. I think the place I found it only asked for the proof that any four integers eventually reached 0 0 0 0, and that the other parts here are my invention, but I'm not sure of that either. --Joshua Zucker

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