October 2013 - mathfun@mailman.xmission.com

[math-fun] Fwd: harder 7x7x7?
by Bill Gosper 03 Jun '15

03 Jun '15

I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314(a)gmail.com> Cc: Julian Ziegler Hunts <julianj.zh(a)gmail.com>, Jon Ziegler < jonathan.zh(a)gmail.com>, Corey Ziegler Hunts <corwin.zh(a)gmail.com>, Michael Beeler <mikebeeler(a)verizon.net>, rcs(a)xmission.com, stan(a)isaacs.com Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill

7 10

[math-fun] What is elliptic K and E closed form?
by Bill Gosper 25 May '15

25 May '15

I think the answer has changed in the last 50 yrs. A query from constantologist Steve Finch revived my interest in whether special value identities like EllipticK[(1 + Sqrt[2])^4] == ((1/4 - I/2)π^(3/2))/ ((2 + Sqrt[2]) Gamma[3/4]^2) and EllipticK[8 2^(1/4) (-1 + Sqrt[2])^3* E^(I ArcTan[(3 (3 + Sqrt[2]))/(4 2^(1/4))])] == (I (1 + Sqrt[2]) E^(-2 I ArcCot[2^(1/4)]) π^(3/2))/ (4 Sqrt[2] Gamma[3/4]^2) are already known, and where one might find tables much more extensive than, e.g., http://functions.wolfram.com/EllipticIntegrals/EllipticK/03/01/ or http://dlmf.nist.gov/19.6 . DLMF says "Exact values of K(k) and E(k) for various special values of k are given in Byrd and Friedman (1971, 111.10 and 111.11)" (1st English edition, 1954) Neil and his Mom were kind enough to fetch me this comprehensive (*Handbook* of *Elliptic* Integrals for Engineers and Physicists<http://www.amazon.com/Handbook-Elliptic-Integrals-Engineers-Physicists/dp/B…>), and to my surprise and disappointment, there was not a single identity of the above form, i.e. algebraics times Gammas(rationals). How could this be? My guess: In the 1950s K and E were considered closed forms while Gamma was not! It makes sense: K and E are of lower computational complexity. We Funsters once concluded that Gamma[k/24] was rapidly computable thanks to elliptic K identities. But now, in papers like "EVALUATION OF COMPLETE ELLIPTIC INTEGRALS OF THE FIRST KIND AT SINGULAR MODULI" (people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf*) "evaluation" means "determin[ation] explicitly in terms of Gamma values". Which is what I always thought. See, e.g., http://en.wikipedia.org/wiki/Elliptic_integral#Special_values --rwg

2 1

Re: [math-fun] parametrizing the Koch snowflake curve
by Bill Gosper 24 Apr '15

24 Apr '15

Fourier epicycles thru nth: (d16) sum(((product((1 - (( - 1)^j * tan(((%pi * (2 * k + (1/3)))/(2^j)))/(sqrt(3)))),j,1,inf)) * (%e)^(%i * (6 * k + 1) * t)/((2 * k + (1/3))^2)),k, - n,n) See http://www.tweedledum.com/rwg/cog.htm Also http://gosper.org/fst.dvi Pix: http://gosper.org/fst1.png ,http://gosper.org/fst2.png , http://gosper.org/fst3.png ,http://gosper.org/fst4.png . (For parametric, take Re and Im.) Rich's base 4 recursion loops for rational arg. Find the fixed point of the pending stack ops to simulate infinite recursion, and get an exact val on the curve. Equate to this Fourier series for a *non-obvious* identity. --rwg > If you view the starting triangle as running from 0 to 3, > and write the fraction base 4, you can use the > fraction digits to select which part of the (sub)edge > you are on. The orientation of the edge can be calculated > from the 1s and 2s of the high order portion of the fraction, > and that should give a somewhat kludgy sum for the (x,y) > coordinates of the image point. > RWG likely has some nice FFT clock sum expression. > > Rich > > > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com [mailto: math-fun-bounces(a)mailman.xmission.com] On > Behalf Of N. J. A. Sloane > Sent: Thursday, February 25, 2010 12:43 PM > To: math-fun(a)mailman.xmission.com > Cc: njas(a)research.att.com > Subject: Re: [math-fun] parametrizing the Koch snowflake curve > > Jim Propp asked: > >> Does anyone (Bill Gosper, maybe?) know an explicit parametrization > of the Koch snowflake curve? > > The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that > does not contain a 1. Is there a similar characterization of the Koch snowflake? > > Neil Sloane > >

2 1

[math-fun] That fractal pentagram again
by Bill Gosper 09 Apr '15

09 Apr '15

For some pictorial. mildly trick questions: http://gosper.org/foupent.pdf (1.6M) --rwg

1 4

[math-fun] Alan Adler, of Aerobie fame,
by Bill Gosper 14 Sep '14

14 Sep '14

finds Euler's product over primes for zeta(s) somewhat magical, and wonders if we funsters know other really neat sums or products involving primes. Hence yesterday's "Minor twist". Otherwise, all I could remember was our Valentine's Day 2011 discussion of "Euler's crazy pi product". --rwg

5 11

[math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Gary Antonick 19 Jun '14

19 Jun '14

Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.

4 7

Re: [math-fun] Iridium & Indium (inside future analogue of light bulbs?)
by Henry Baker 13 May '14

13 May '14

The good news is that "quantum dots" don't depend upon normal valence electron energy levels in order to achieve their properties. Their properties are primarily a result of their size & configuration, which is in the 5-20 nm range -- almost achievable in the next 1-2 generations of lithography. So many of the needs for exotic materials with very specific bandgaps will be replaced by these engineered quantum dots. --- It was my impression that most, if not all, iridium comes from outer space; there is very little indigenous iridium on the Earth's surface. There may be a lot in the Earth's core, but it's going to be a little difficult to mine it from there. At 04:42 PM 12/4/2012, Warren Smith wrote: >Iridium is 40x rarer than gold and total world production is only >about 10 tons annually. > >Indium is also pretty rare, but not as rare: 1000 ton annual production. >So I think if iridium is a substantial component of solid plastic >lighting source then that will not be a useful technology for Average >Joe Consumer. I would think >you'd want ingredients common enough for million-ton production. >If they can remove the need for Iridium and other rare elements, the >general concept sounds excellent as a future lighting source. > >-- >Warren D. Smith >http://RangeVoting.org <-- add your endorsement (by clicking >"endorse" as 1st step)

10 15

Re: [math-fun] Julian's dodecahedron
by Bill Gosper 20 Feb '14

20 Feb '14

On 2013-09-02 15:15, Adam P. Goucher wrote: > The legendary Bill Gosper wrote: I'd rather be young. > >> is made by replacing ten of the regular pentagons of a dodecahedron with >> concave equilateral pentagons, preserving the topology: >> http://gosper.org/dodohedron.png >> ( !#%&*@! was deleted again. Apologies to any of you who looked for it.) >> >> http://en.wikipedia.org/wiki/Dodecahedron glaringly omits it: > > Even worse, the Wikipedia article omits the much more well-known > _endo-dodecahedron_ (the Symmetries of Things, page 328), which has > concave regular pentagonal faces and is *face-transitive*. And it's equilateral! > Its symmetry group is pyritohedral (3*2), and it is defined as: > > "The spaces left over by positioning dodecahedra of maximum size in > the natural orientation, centred on the points of the FCC lattice" > > The faces are different from the concave pentagons of the > tympanohedron, having internal angles of {7pi/5, pi/5, 3pi/5, 3pi/5, > pi/5} in cyclic order. > > > Sincerely, > > > Adam P. Goucher > > http://cp4space.wordpress.com Actually the concave figure in http://en.wikipedia.org/wiki/Pyritohedron is probably your endododecahedron, but they don't say. It would appear in their stupid throbbing animation, if the author wasn't so hung up on convexiity. Here's my stupid throbbing animation: ListAnimate[ Table[Graphics3D[ Polygon[Join[#, -#] &@ Join[#, Map[RotateLeft, #, {2}], Map[RotateRight, #, {2}]] &@{{{-(1/2) + 2 z^2, -(1/2) + z, 0}, {-(1/2), -(1/2), -(1/2)}, {0, -(1/2) + 2 z^2, -(1/2) + z}, {1/2, -(1/2), -(1/2)}, {1/2 - 2 z^2, -(1/2) + z, 0}}, {{1/2 - 2 z^2, -(1/2) + z, 0}, {1/2, -(1/2), 1/ 2}, {0, -(1/2) + 2 z^2, 1/2 - z}, {-(1/2), -(1/2), 1/ 2}, {-(1/2) + 2 z^2, -(1/2) + z, 0}}}]] /. z -> (5*Sin[t] + Sin[5*t])/12, {t, -\[Pi]/2, \[Pi]/2, \[Pi]/30}], AnimationRunning -> True, AnimationDirection -> ForwardBackward] The mgif: gosper.org/dodex.gif --rwg BTW, yesterday Neil predictively sketched the key frames of this on a small piece of paper while riding in a car in the Santa Cruz Mountains.

4 8

[math-fun] Slinkiest slinky
by Bill Gosper 31 Oct '13

31 Oct '13

~17 yrs ago I mentioned here making a giant slinky by taping together about sixty 13" Aerobies on alternate edges. Regrettably, I can't find my videos, either digital or analog. The best featured it descending the roof of the Cherry Chase School and then stepping all the way to the ground near some unimpressed kids playing basketball. It was also able to go a few steps down a particularly steep section of a street near my house. The device exhibited some "momentum" by sometimes going an extra step after reaching a level surface. Anyway, I think a neat engineering class challenge would be to design the slinky needing the least inclination. Helpful might be if the last few coils on either end were heavier and stiffer: those on the upper end lying not quite flat, and those on the lower end releasing stored energy near the end of each step. --rwg

2 2

[math-fun] FCC "HAAT" calculations
by Henry Baker 30 Oct '13

30 Oct '13

The FCC uses its "HAAT" calculation to figure out how high an antenna should be above the surrounding terrain. http://transition.fcc.gov/mb/audio/bickel/haat_calculator.html >From a mathematical perspective, HAAT appears to be trying to compute the average elevation for circles at 2,4,6,8,10 mile radii from the antenna. Each of these 5 circles are sampled at 8 or more equal-spaced angles -- e.g., (2*pi*i)/8, i=0..7. The 8 (at least) points from each of these 5 circles are averaged together with equal weights. Suppose you wanted your antenna to cover the maximum _area_ with the _minimum HAAT_. How do you choose the position of your (omnidirectional) antenna? Although the sample points at 2,4,6,8 miles weigh as much in the formula as those at 10 miles, there is more area a 10 miles (and beyond). So, ideally, you position your antenna on a plateau 8 miles in diameter. Alternatively, you position your antenna in the middle of a parabolic peak so that the contribution of the more distant samples scales with the additional area. It isn't clear who gets to choose where the sample points are located: you or the FCC. For example, the terrain could be "crumpled" in radial directions in such a way that the answer could vary widely depending upon the angle of the original baseline relative to the equator. So perhaps the FCC might interpret this as the infimum or supremum over all choices of original baseline. Similarly, the terrain could have thin "walls" at exactly 2,4,6,8,10 miles from the antenna, in which case the average terrain would be lower than the tops of these walls. Of course, in reality, these wall would also create _valleys_ behind them, which would obscure the sight of the antenna. So, taking account of both the FCC and line-of-sight wave propagation, what would be the optimum location? An antenna at the top of a gradually-sloping _conical_ hill would provide the most area covered by line-of-sight of the antenna; the steeper the conical hill, the better (to maximize area and minimize the HAAT number).

1 0