March 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

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[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

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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 > >

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[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

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[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.

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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)

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[math-fun] Size of the Cherbakul Lake Meteorite Fragment
by Hans Havermann 16 Oct '13

16 Oct '13

From a National Post (newspaper) article today: "Viktor Grokhovsky, who led the expedition from Urals Federal University, said Monday the meteorites plucked from the ice-covered Chebarkul Lake so far are less than a centimeter and had an iron content of about 10 percent. Locals saw a big meteorite fall into the lake on Friday, leaving a six-meter-wide hole in the ice. Grokhovsky said a meteorite up to 50-60 centimeters could eventually be found in the lake." When the initial reports of the 6 m hole surfaced, I thought (looking at a picture of the hole in the ice): That must be a 4 m (or more) chunk of rock. But .5 m? An impact in a surface ice layer surely is not like an impact on ground where a wider hole might be expected.

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[math-fun] Modulus Nine
by Hans Havermann 24 Apr '13

24 Apr '13

Start with 0,0,...,0,1 (n-1 0's and a single 1); thereafter the next term is the sum of the previous n terms, modulus 9. The sequence will repeat; for example, n=2: 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, … (period 24); and n=3: 0, 0, 1, 1, 2, 4, 7, 4, 6, 8, 0, 5, 4, 0, 0, 4, 4, 8, 7, 1, 7, 6, 5, 0, 2, 7, 0, 0, 7, 7, 5, 1, 4, 1, 6, 2, 0, 8, 1, 0, 0, 1, … (period 39). Periods can get quite large; for instance, n=24 has period 421900912158 but, empirically, periods for n= one-less-than-powers-of-three & powers-of-three appear to be not only more manageable but surprisingly regular: {2,3} -> {24,39} {8,9} -> {240,273} {26,27} -> {2184,2271} {80,81} -> {19680,19929} {242,243} -> {177144,177879} {728,729} -> {1594320,1596513} {2186,2187} -> {14348904,14355471} {6560,6561} -> {129140160,129159849} {19682,19683} -> {1162261464,1162320519} Can anyone come up with an exact formulation for these?

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[math-fun] Futility Closet
by Jason 04 Apr '13

04 Apr '13

This blog is thoroughly delightful, with brain teasers and historical oddities. In particular, I thought the solution to this tiling problem was quite lovely, and generalizes to other shapes: http://www.futilitycloset.com/2012/09/30/tiling-task/ Upon further reading, it appears to be related to the mutilated chessboard problem.

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[math-fun] direction of fastest convergence
by Bill Gosper 03 Apr '13

03 Apr '13

Using path-invariance, we can, in polynomial vs exponential time, exhaustively search a taxicab-metric "sphere" for the direction of fastest convergence. Starting with the matrices i: {{(i + k) (i + n)/((1 + i) (i + j + k + n)), -j k n}, {0, 1}}, j: {{(j + k) (j + n)/((1 + j) (i + j + k + n)), -i k n}, {0, 1}}, k: {((i + k) (j + k)/((1 + k) (i + j + k + n)), i j n}, {0, 1}}, n: {{(i + n) (j + n)/((1 + n) (i + j + k + n)), i j k}, {0, 1}} which in the {n,k} plane give Gauß's identity 2F1(a,b,c;1)=Gamma Gamma/Gamma/Gamma, and adding the h dimension via Sunday's "cheap trick", h: {{(h (h + i - j))/((h + k) (h + i + n)), -1 + j + k + n}, {0, 1}}}, i: {{((h + i - j) (i + n))/((1 + i - j - k) (h + i + n)), ((-1 + h + k) (-1 + j + k + n))/(-1 - i + j + k)}, {0, 1}}}, j: {{((i - j - k) (j + n))/((-1 + h + i - j) (j + k + n)), ((-1 + h + k) (-1 + h + i + n))/(-1 + h + i - j)}, {0, 1}}}, k: {{(k (-i + j + k))/((h + k) (j + k + n)), -1 + h + i + n}, {0, 1}}}, n: {{((i + n) (j + n))/((h + i + n) (j + k + n)), -1 + h + k}, {0, 1}}} then searching the neighborhood of starting point n=1,i=0,j=0,h=1,k=1 at a radius of 18 says that "best" is to take 6 n steps, 6 h steps, and 6 k steps, yielding the 6 bits/term identity (combining just one step of n, h, and k) π^2/6==Sum[(13 + 21*k)*π^(3/2)*k!^3/(64^(k + 1)*(1/2 + k)!^3), {k, 0, ∞}] which Mathematica converts to a useless pair of 4F3s. Notice how notationally and computationally inferior are both Sum and pFq notations vs this teensy matrix product In[893]:= {{(2/j + 4)^-3, 21 j/8 - 1}, {0, 1}} In[894]:= Dot @@ Table[%, {j, 3}] Out[894]= {{1/2744000, 23687/14400}, {0, 1}} In[895]:= 23687/14400 - π^2/6. Out[895]= -3.51129*10^-6 (~3 ppm from 3 terms). The Sum identity appears to express π^2 in terms of π^(3/2) ! But of course you don't need π to evaluate the summand. I quoted "best" above because, e.g., we could have combined the n, h, and k steps with a coordinate transformation, yielding fewer steps with messier matrices. Subjectivity persists. --rwg Soon: the "best" matrices (from this system) for csc πx, sin πx, π, and 1/π .

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