- mathfun@mailman.xmission.com - mailman.xmission.com

Re: [math-fun] Packing problem
by Dan Asimov 22 Apr '19

22 Apr '19

These are very interesting questions! There's a useful concept of stability that's based on considering what happens under an arbitrarily small perturbation. Omitting mention of physics, we can define * for any set R = {r_j | j in J} of radii r_j > 0 (where J is some index set) with finite sum Sum_{j in J} r_j, let disk D_j in the plane have radius r_j. * Define an arrangement A = {D_j} in the plane of non-overlapping disks D_j in the plane, where the radius of D_j is r_j. This means that for each D_j, A is a choice of *center* = c_j in R^2. Non-overlapping means that for each pair of indices j ≠ k, we have ||c_j - c_k|| >= r_j + r_k. * Let D(A) be the smallest disk in R^2 containing each D_j in its interior. Then define diam(A) = diameter(D). * For an arrangement A and eps > 0, define an *eps-close perturbation* of A to be an arrangement A' with the same radii r_j but with each center c_j replaced by a new center c'_j, such that the sum of the center discrepancies ||c_j - c'_j|| is bounded above by eps: Sum_{j in J} ||c_j - c'_j|| < eps. * Finally, say an arrangement A is *stable* if there exists eps > 0 such that every eps-close perturbation A' of A satisfies diam(A') >= diam(A). In other words, small perturbations to A cannot decrease diam(A). —Dan ----- ... ... Is there a ... multiset of diameters ... that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler? My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches. ... PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. -----

1 0

[math-fun] Numerical fluke (?)
by Bill Gosper 22 Apr '19

22 Apr '19

(balter:=balanced ternary) Nine trits of even balanced ternary of π: In[329]:= 2 balter@Floor[3^9 π/2] Out[329]= {2, -2, -2, -2, 0, 2, 2, -2, 0, 2, -2} Check: In[330]:= FromDigits[{%, 2}, 3] // N Out[330]= 3.14149265863944 But In[337]:= Sum[4 (-1)^k/(1 - 2 k), {k, 10^4}] // N Out[337]= 3.14149265359004 ! —rwg

2 1

[math-fun] Berlekamp
by rcs＠xmission.com 20 Apr '19

20 Apr '19

Elwyn Berlekamp died April 9, age 78. I heard this from a G4G update. Obituary at https://news.berkeley.edu/2019/04/18/elwyn-berlekamp-game-theorist-and-codi… His web page https://math.berkeley.edu/~berlek/index.html Wikipedia https://en.wikipedia.org/wiki/Elwyn_Berlekamp --- Rich

1 0

[math-fun] Even balanced ternary
by Bill Gosper 20 Apr '19

20 Apr '19

Base 3, digits -2, 0, 2 would seem silly, representing only the even integers. But it is momentarily startling, and then trivially obvious that the pure fractions in this system completely cover [-1,1]. Digits {-1,0,1} only cover [-1/2,1/2]: In[306]:= Table[FromDigits[{{d1, d2, d3}, 0}, 3], {d1, -1, 1}, {d2, -1, 1}, {d3, -1, 1}] // Flatten Out[306]= {-(13/27), -(4/9), -(11/27), -(10/27), -(1/3), -(8/27), -(7/27), -(2/9), -(5/27), -(4/27), -(1/9), -(2/27), -(1/27), 0, 1/27, 2/27, 1/9, 4/27, 5/27, 2/9, 7/27, 8/27, 1/3, 10/27, 11/27, 4/9, 13/27} (stupid parentheses) But with {-2,0,2} In[307]:= Table[FromDigits[{{d1, d2, d3}, 0}, 3], {d1, -2, 2, 2}, {d2, -2, 2, 2}, {d3, -2, 2, 2}] // Flatten Out[307]= {-(26/27), -(8/9), -(22/27), -(20/27), -(2/3), -(16/27), -(14/27), -(4/9), -(10/27), -(8/27), -(2/9), -(4/27), -(2/27), 0, 2/27, 4/27, 2/9, 8/27, 10/27, 4/9, 14/27, 16/27, 2/3, 20/27, 22/27, 8/9, 26/27} You may object that there are no odd numerators. But .2222... is obviously 1, .2➋2➋2➋... = ½, and .20➋020➋020➋0... = ⅗, etc. And then 3 = 2.2222... = 2➋.➋➋➋... etc, and {-2,0,2} (more than) covers all the reals!—rwg

1 0

[math-fun] "maa.org" has been compromised?
by Henry Baker 19 Apr '19

19 Apr '19

I just received a "phishing" email whose header said it came from webwork.maa.org. This was a very good imitation of a Chase Bank email. There are some real pros out there. Be careful!

2 1

[math-fun] Antipodal New Zealand
by Hans Havermann 19 Apr '19

19 Apr '19

2001 vs 2019, in Mathematica. http://gladhoboexpress.blogspot.com/2019/04/antipodal-new-zealand.html

3 4

[math-fun] One base, two digit sets
by Bill Gosper 19 Apr '19

19 Apr '19

On 2019-04-17 16:27, Adam P. Goucher wrote: > Interesting! Are there any reasons to prefer quater-imaginary over > base i-1? Rectangular vs fractal pure fractions region? Julian's magic piecewiserecursivefractal function, seems to solve membership questions. E.g., it can give the exact real preimages of any complex value of the Heighway Dragon function.Heighway triple point <http://gosper.org/dragtrip!.png>> > As mentioned before, base 3 + omega (with digits 0 and the 6 roots > of unity) is my favourite positional number system for the complex > numbers. The reason being that: > > (a) the sum of any 3 one-digit numbers is a two-digit number; > (b) the product of any 2 one-digit numbers is a one-digit number; > > which together make it possible to build a ripple carry adder out > of 'full adders'. (The same is, of course, true for binary.) > > But balanced ternary, which Knuth himself describes as the most > beautiful of all positional number systems (which I agree with), > satisfies a strengthened version, where '3' is replaced with '4' > in (a). > > So a balanced ternary adder can sum *three* numbers together, > instead of (as is the case in binary) requiring sums to be > performed pairwise. (That's actually quite cute: in binary, the > adder is a binary function; in balanced ternary, it's a ternary > function.) > The magicd function in gasket Fourier <http://gosper.org/gaskettalk.pdf> is a fairly exotic application of balanced ternary. balter[x_Integer] := Sign[x]* (IntegerDigits[Abs[x] + (1/2)* (3^Ceiling[Log[3, 1 + Abs[2*x]]] - 1), 3] - 1) That paper also illustrates an interesting alternative to convergent sums, products, and continued fractions: Infinite products of idempotent matrices with vanishing perturbations. —rwg > > Best wishes, > > > Adam P. Goucher > > >> Sent: Thursday, April 18, 2019 at 12:14 AM >> From: "Bill Gosper" <billgosper(a)gmail.com> >> To: math-fun(a)mailman.xmission.com >> Subject: [math-fun] One base, two digit sets >> >> Knuth has proposed two positional base + digits schemes for representing >> complex numbers. >> Base i-1, digits {0,1}: >> https://en.wikipedia.org/wiki/Complex-base_system#/media/File:ComplexTwindr… >> , >> gosper.org/basei-1.png >> and Base 2i, digits {0,1,2,3}: >> https://en.wikipedia.org/wiki/Quater-imaginary_base >> These systems are closely related, because (i-1)² = -2i . >> So in base 2i, you can get the mirror image of the twindragon system by >> using the digits {0, 1, - i - 1, -i} instead of {0,1,2,3}! —rwg >> ______

1 0

[math-fun] Double normal puzzle + question
by Dan Asimov 17 Apr '19

17 Apr '19

Some time ago I wondered if there could be a smooth simple closed curve in the plane without a double normal — a chord of the curve that is perpendicular at both ends. This is a fun puzzle which I won't spoil for anyone who hasn't seen it. But today I wondered about the same question for a smooth surface in R^3 topologically equivalent to the 2-sphere S^2. Now for many smooth S^2 surfaces M — like convex ones — following the internal normal N(p) at point p of M until it first hits the surface, say at point Q(p) — the map Q : M —> M taking p to Q(p) is a continuous. If so, then having obtained Q(p) from p we now follow the internal normal N(Q(p)) at Q(p) until it first intersects the tangent plane T_p(M) of M at p. Let this point be f(p). Assuming no double normal, we know f has no fixed points: for all p in M, f(p) ≠ p. But then f(p) - p provides a vector field on S^2 continuous in p, known to be impossible (the hairy ball theorem), contradiction. But if that map Q above is *not* continuous, this argument doesn't work. So, Question: Does there exist a smooth S^2 in R^3 without a double normal? —Dan

1 0

[math-fun] Mathematica CDF player
by Fred Lunnon 17 Apr '19

17 Apr '19

I'm considering implementing a solid honeycomb / 3-D tiling demo for the CDF player, and soliciting feedback concerning how well it works for author and user. I also wonder how in practice casual users might feel about going to the trouble of downloading this player to view interactive video material. Fred Lunnon

3 3

[math-fun] One base, two digit sets
by Bill Gosper 17 Apr '19

17 Apr '19

Knuth has proposed two positional base + digits schemes for representing complex numbers. Base i-1, digits {0,1}: https://en.wikipedia.org/wiki/Complex-base_system#/media/File:ComplexTwindr… , gosper.org/basei-1.png and Base 2i, digits {0,1,2,3}: https://en.wikipedia.org/wiki/Quater-imaginary_base These systems are closely related, because (i-1)² = -2i . So in base 2i, you can get the mirror image of the twindragon system by using the digits {0, 1, - i - 1, -i} instead of {0,1,2,3}! —rwg

3 2