January 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] sqrt(pi)
by Bill Gosper 24 Mar '13

24 Mar '13

Pairwise combining the terms of A&S 6.5.31 (upper Gamma CF), then adding last week's lower gamma CF: In[185]:= Equal[Gamma[ z], ((y^z*((1/(cfk[n*(z - n), -z + y + 2*n + 1, List[n, Infinity]] - z + y + 1)) + (1/(z + cfk[n*y, z - y + n, List[n, Infinity]] - y))))/(E^y))] z 1 Gamma[z] == (y (------------------------------------------- + -y + z + cfk[n y, n - y + z, {n, Infinity}] 1 y -----------------------------------------------------------)) / E 1 + y - z + cfk[n (-n + z), 1 + 2 n + y - z, {n, Infinity}] where cfk:=ContinuedFractionK and y>0 can be chosen to optimize convergence. Specializing, In[186]:= % /. z -> 1/2 /. y -> 1 Out[186]= 1 Sqrt[Pi] == (-------------------------------------- + 1 1 -(-) + cfk[n, -(-) + n, {n, Infinity}] 2 2 1 ------------------------------------------) / E 3 1 3 - + cfk[(- - n) n, - + 2 n, {n, Infinity}] 2 2 2 In[192]:= N[List @@ %186 /. cfk[num_, den_, _] :> ContinuedFractionK[num, den, {n, 9}]] Out[192]= {1.77245, 1.77244} This is the first non-asymptotic sum or CF I've seen for sqrt(pi). But the sum behind the lowergamma cf has been there all along, so I must not have been paying attention. --rwg

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Re: [math-fun] poisoned wine bottles
by Veit Elser 22 Mar '13

22 Mar '13

I have no improvement to report on the 10-rat case, but here's a surprise development we owe to my former student Mike Quist. Shortly after I explained the poisoned bottle problem to him he pointed out that one shouldn't make the assumption that a uniform partition of the bottles is always going to be optimal. To prove his point, Mike constructed, for 7 rats, a 4-unique code of size 19. When the bottles are partitioned evenly this would give a discard fraction of 4/19, inferior to what one gets with Michael's size 21 4-unique code reported Nov. 28. However, by unequally partitioning the bottles you can get a discard fraction of 2/11 with Mike's code! Rather than show you Mike's code I'll give you the simplest case of this phenomenon which requires 6 rats. Using a mixed-integer program I checked that for less than 6 rats the equal partition is optimal. The same linear programming method found the following code for 6 rats: Construct the 2+2+2 product code from the 2-rat code {00,01,10}. This has 27 codewords. Now remove all the codewords of weight 3; that gives a code having 1 word of weight 0, 6 words of weight 1, and 12 words of weight 2, for a total of 19 codewords. Partition the wine bottles so 7/49 are assigned to the zero word, 1/49 to each of the weight 1 words, and 3/49 to each of the weight 2 words. You can check that for any union of codeword pairs you never have to discard more than the fraction 12/49 -- a modest improvement over 1/4. For example, there are some unions where 6 codewords are implicated (it is a 6-unique code), but in all these cases the bottle partition has the form (1+1+1+3+3+3)/49=12/49. Are unequal bottle partitions going to be the rule for optimal codes? So far we only know it is true for 6 and 7 rats. When I relax the equal partition condition and optimize the good codes we found in the past with respect to this I get no improvement (the optimal partition is the uniform one). So it seems we need a new approach to get the truly optimal codes. I can send the *.lp file for 7 or more rats to anyone interested with access to a big computer. At this point we only have optimal solutions up to 6 rats: n 1 2 3 4 5 6 f 1 2/3 1/2 2/5 1/3 12/49 The gurobi solver finds the last entry in 2 sec when I restrict the codewords to weight 2 and 15 minutes when I allow weight 3. For 7 rats it finds the 4/21 solution, that is optimal for weight < 3 , in 15 sec. But finding Quist's 2/11 solution, which includes weight 3, is not going to happen on my laptop! -Veit On Jan 6, 2013, at 4:46 PM, Veit Elser <ve10(a)cornell.edu> wrote: > Wow -- that was fast! > > wax & wane has found several more 50/4's, all with the same > (quaternion unit) symmetry group. I'll try a deeper search to > see if something else turns up. > > -Veit

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[math-fun] Sums of two cubes
by Allan Wechsler 16 Feb '13

16 Feb '13

I am sure I am very late to this party, but I have been looking at positive sums of two (not necessarily positive) cubes, catalogued on OEIS as A045980. The corresponding set for squares is easily characterized in terms of prime factorizations: the exponents of primes of the form 4k-1 must be even. (I hope I have that characterization right; if it's not, it's close.) The experience for sums of two squares inspires me to factorize the first few dozen entries in A045980. The result is very suggestive, about as suggestive as it's possible to be with actually, you know, *suggesting* anything. The most striking feature is a marked preference for primes of the form 6k+1. Of course if a number n is the sum of two cubes, then nm^3 is also, for all integers m. But the converse is not true. For example, 1^3 + 7^3 = 344 = 2^3*43, but 43 is not the sum of two cubes. The property is not multiplicative: for example, 2 and 7 are sums of two cubes, but 14 isn't. That having been said, it displays frustrating *hints* of multiplicativity. After staring at these numbers for a couple of days, I have come up with a hard question related to this spooky para-multiplicativity. For any k, let Ck be the set of integers n such that kn is the sum of two squares. C1, for instance, is A045980 itself. C2 is {1, 4, 8, 13, 14 ...}, which is not in OEIS. (In fact, none of the Ck's I looked at were in OEIS except C1.) Here is my question: do there exist *any* two integers j and k such that Cj = Ck?

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[math-fun] Lissajous illusion
by Bill Gosper 07 Feb '13

07 Feb '13

This simple fact might make a nice math objet d'art. Labeled[Manipulate[ Labeled[ParametricPlot[{{t - 2*\[Pi] - 1, Cos[a*(t + v) + \[Phi]]}, {Sin[b*(t + v)], t - 2*\[Pi] - 1}, {Cos[t], Sin[t]}/11 + {Sin[b*v], Cos[a*v + \[Phi]]}, {Sin[b*v], t/\[Pi] - 1}, {t/\[Pi] - 1, Cos[a*v + \[Phi]]}, {Sin[b*t], Cos[a*t + \[Phi]]}}, {t, 0, 2*\[Pi]}, PlotRange -> {{-4 - \[Pi], 9/8}, {-4 - \[Pi], 9/8}}], {StringJoin[ "Vertical frequency a=", ToString[a]], StringJoin["Horizontal frequency b=", ToString[b]], StringJoin["Phase \[Phi]=", ToString[\[Phi]/\[Pi]], "\[Pi]"]}, {Top, Right, Bottom}, RotateLabel -> True], {v, -\[Pi], \[Pi]}, {\[Phi], 0, 2 \[Pi]}, {a, Range[5]}, {b, Range[5]}], "Lissajous Mechanism", Top] Click frequencies a to 3, b to 4, say. Pop open the + sign on the phase (φ) parameter and run it slowly. Imagine the precessing Lissajous figure to be drawn on a rotating transparent cylinder. Is it vertical or horizontal? So make two metal "Lissajous hoops", one for each axis and rotate them side by side on mutually perpendicular turntables. --rwg

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