May 2014 - mathfun@mailman.xmission.com

[math-fun] Ramanujan, the movie (?)
by Warren D Smith 13 May '14

13 May '14

http://www.imdb.com/title/tt3037162/ pointed out by Dave Langer. Will be released sometime 2014. It perhaps might be available only in India, says it is Tamil+English language. No endorsement implied; in fact my total guess is, it'll be horrible. -- Warren D. Smith http://RangeVoting.org

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Re: [math-fun] Problem for high school algebra ?
by Henry Baker 13 May '14

13 May '14

You've got the problem transposed. We don't know the z coordinates, as we (like Plato) can only see the shadows (the projections on the XY plane). While we're currently seeking only the radius, it might also be nice to infer possible legal sets of zi's, as well. x1+x2+x3=0 (because the centroid is at the origin), but x1^2+y1^2+z1^2=R^2, but we don't know R or z1 (or z2 or z3). So we can't find the zi by simple linear algebra. So far, the simplest solution I've come up with is via the largest eigenvalue (or largest singular value) of the covariance matrix [see my later posting for this matrix]. But that's hardly high school stuff. It is cool, however, that PCA/SVD (Principal Component Analysis and Singular Value Decomposition) works so beautifully for such a small amount of data -- in this case, the bare minimum. I can see now why PCA/SVD has become the "goto" statistical method these days. At 05:24 PM 5/12/2014, Fred Lunnon wrote: > This ingeniously camouflaged problem had me bamboozled >for a day or two, off and on --- more red herrings than >a Soviet trawler! > > And that's before I started on the suggested solutions: >dunno how high-school kids would cope, 'cos I couldn't make >head nor tail of either of 'em ... > > So for the fainter-hearted amongst us, I propose the >following rather more pedestrian approach. > > Given some triangle P_1, P_2, P_3 in 3-space, >with centroid at the origin, and vertex x, y-components; >calculate circumradius R . > > x_i + y_i + z_i = 0 yields z-components; > (a_i)^2 = || P_(i+1) - P_(i+2) || yields square sides (a_i)^2 ; > Heron formula yields squared area A^2 ; > (a_1)^2 (a_2)^2 (a_3)^2 = 4 A^2 R^2 yields square radius R^2 ; > and a single square root yields R . > >While if also given that a_1 = a_2 = a_3 , then simply > z_2 = -(x_2 + y_2) , z_3 = -(x_3 + y_3) , > (a_1)^2 = (x_2 - x_3)^2 + (y_2 - y_3)^2 + (z_2 - z_3)^2 , > R = sqrt( (3/4)(a_1)^2 ) by Pythagoras. > > Mind you, as Knuth apparently remarked, I have only proved it >--- I haven't tested it. > >Fred Lunnon > > >Henry Baker (various dates) wrote ><< > >[High school kids could understand this, but probably >not do it as an exercise.] > >[Hints: least squares, Siebeck-Marden Theorem, >max/abs hack.] > >[Solution at bottom.] > >We have an _equilateral_ triangle in arbitrary position >in standard 3D space with coordinate axes x,y,z. > >Translate this triangle (without rotations) so that its >center becomes the origin. > >The 3 vertices are P1=(x1,y1,z1), P2=(x2,y2,z2), >P3=(x3,y3,z3). > >We are given x1,y1,x2,y2 (but _not_ the z coordinates). > >Obviously, x3=-x1-x2, y3=-y1-y2, due to the center being >the origin. > >Find a "simple" expression for the _radius_ of the >circumcircle. > >This problem is trivial in the case that x1^2+y1^2 = >x2^2+y2^2 = x3^2+y3^2 = R^2, in which case z1=z2=z3=0, >so we are interested in the more difficult case. > >(Hint: you don't actually need the z coordinates.) > >(Yes, I do have an answer, which is simpler than I >had expected.) >-------------------------------------------------- >The hardest part of this problem is convincing >yourself that it is even soluble. The 3 vertices >are equally spaced around a ring which is rotated >so that its projection onto the XY plane is an >ellipse whose semimajor axis is the radius we >seek. With only 3 points on the ring, it isn't >a priori obvious that we have enough data to >compute the parameters of the ellipse. > >Nevertheless, the radius can be calculated with >an elegant formula, thanks to a detour through >the complex numbers. > >Compute > >real sumx2 = x1^2+x2^2+x3^2 >real sumy2 = y1^2+y2^2+y3^2 >complex sumz2 = (x1+iy1)^2+(x2+iy2)^2+(x3+iy3)^2 > >real 3*R^2 = sumx2 + sumy2 + |sumz2| > >That's it! > >[snip] >>

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[math-fun] Dual tic-tac-toe?
by Mike Stay 13 May '14

13 May '14

Where can I find a write-up of the dual version of tic-tac-toe (i.e. exchanging lines and points)? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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Re: [math-fun] Problem for high school algebra ?
by Henry Baker 13 May '14

13 May '14

[High school kids could understand this, but probably not do it as an exercise.] [Hints: least squares, Siebeck-Marden Theorem, max/abs hack.] [Solution at bottom.] We have an _equilateral_ triangle in arbitrary position in standard 3D space with coordinate axes x,y,z. Translate this triangle (without rotations) so that its center becomes the origin. The 3 vertices are P1=(x1,y1,z1), P2=(x2,y2,z2), P3=(x3,y3,z3). We are given x1,y1,x2,y2 (but _not_ the z coordinates). Obviously, x3=-x1-x2, y3=-y1-y2, due to the center being the origin. Find a "simple" expression for the _radius_ of the circumcircle. This problem is trivial in the case that x1^2+y1^2 = x2^2+y2^2 = x3^2+y3^2 = R^2, in which case z1=z2=z3=0, so we are interested in the more difficult case. (Hint: you don't actually need the z coordinates.) (Yes, I do have an answer, which is simpler than I had expected.) -------------------------------------------------- The hardest part of this problem is convincing yourself that it is even soluble. The 3 vertices are equally spaced around a ring which is rotated so that its projection onto the XY plane is an ellipse whose semimajor axis is the radius we seek. With only 3 points on the ring, it isn't a priori obvious that we have enough data to compute the parameters of the ellipse. Nevertheless, the radius can be calculated with an elegant formula, thanks to a detour through the complex numbers. Compute real sumx2 = x1^2+x2^2+x3^2 real sumy2 = y1^2+y2^2+y3^2 complex sumz2 = (x1+iy1)^2+(x2+iy2)^2+(x3+iy3)^2 real 3*R^2 = sumx2 + sumy2 + |sumz2| That's it! [ If you want to work only with reals, real sumx2 = x1^2+x2^2+x3^2 real sumy2 = y1^2+y2^2+y3^2 real sumxy = x1*y1+x2*y2+x3*y3 real |sumz2| = sqrt((sumx2-sumy2)^2+4*sumxy^2) real 3*R^2 = sumx2+sumy2+|sumz2| ] Here's the math: real a,b // The semi axes of the projected ellipse complex r1 = exp(i*phi)*(a*cos(theta)+b*i*sin(theta)) complex r2 = exp(i*phi)*(a*cos(theta+2*pi/3)+b*i*sin(theta+2*pi/3)) complex r3 = exp(i*phi)*(a*cos(theta-2*pi/3)+b*i*sin(theta-2*pi/3)) real sumx2=realpart(r1)^2+realpart(r2)^2+realpart(r3)^2 real sumy2=imagpart(r1)^2+imagpart(r2)^2+imagpart(r3)^2 real sumx2+sumy2 = (3/2) (a^2+b^2) complex sumz2 = r1^2+r2^2+r3^2 complex sumz2 = (3/2) exp(2*i*phi) (a^2-b^2) real |sumz2| = (3/2) |a^2-b^2| real sumx2+sumy2+|sumz2|=(3/2)(a^2+b^2+|a^2-b^2|)=3/2*2*max(a^2,b^2) = 3*max(a^2,b^2) [This is my first _natural_ use of the max/abs hack in >50 years!] The Marden Theorem tells us the major & minor axes of the ellipse formed by the 3 given points. If p(z) is the monic polynomial with r1,r2,r3 as roots, then p'(0) = (3/4)exp(2*i*phi)(a^2-b^2). complex r1*r2+r1*r3+r2*r3 = (3/4) exp(2*i*phi) (a^2-b^2) which is another way to compute |a^2-b^2|. Does anyone know the earliest reference to the max/abs hack? I tried Googling, but wasn't able to come up with anything.

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[math-fun] Fibonacci's contest problem
by Bill Gosper 12 May '14

12 May '14

All I found was a slow but amusing Mma solution. E.g. changing 5 to 14, In[838]:= Intersection @@ Transpose[ Reap[NestWhile[1/(2*Floor[#] + 1 - #) &,1/3, (If[NumberQ[#[[1]]], Sow[#]]; True) &[{1, #}/Sqrt[(#^2 - 1)/14]] &, 1, 9999]][[2, 1]]] Out[838]= {-((65 I)/12), 65/12} Mystery: Why doesn't this find 337/120 for 7? --rwg NeilB just privately sent a big improvement to the search technique, plus some interesting results. I'll leave it up to him to post it here. ---------- Forwarded message ---------- From: Neil Bickford <techie314(a)gmail.com> Date: Thu, May 8, 2014 at 10:01 PM Subject: Question for tomorrow To: Bill Gosper <billgosper(a)gmail.com> I came across this problem in a recreational mathematics book allegedly by H.E. Licks (apparently the I.J. Matrix of the late 19th century), who attributes it to Fibonacci- specifically, that Fibonacci solved it in a mathematical competition of the time: Find a rational number n such that: n^2+5 is a square of a rational, and n^2-5 is a square of a rational. I've been working on this problem for some time, and while it's almost trivial to find a solution by brute force, I haven't been able to find an elegant solution. So far, the best observation I've been able to make is that it's Pellish. Any thoughts? --Neil

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[math-fun] Mathematical Model Suggests That Human Consciousness Is Noncomputable
by Ray Tayek 11 May '14

11 May '14

http://science-beta.slashdot.org/story/14/05/08/1957225/mathematical-model-… <http://beta.slashdot.org/%7Etimothy>timothy posted 9 hours ago | from the opposite-would-be-more-suprising dept. <http://science-beta.slashdot.org/story/14/05/08/1957225/mathematical-model-…>297 <http://beta.slashdot.org/%7EKentuckyFC>KentuckyFC (1144503) writes "One of the most profound advances in science in recent years is the way researchers from a variety of fields are beginning to formulate the problem of consciousness in mathematical terms, in particular using information theory. That's largely thanks to a relatively <http://en.wikipedia.org/wiki/Integrated_Information_Theory>new theory that consciousness is a phenomenon which integrates information in the brain in a way that cannot be broken down. Now a group of researchers has taken this idea further using algorithmic theory to study whether this kind of integrated information is computable. They say that the process of integrating information is equivalent to compressing it. That allows memories to be retrieved but it also loses information in the process. But they point out that this cannot be how real memory works; otherwise, retrieving memories repeatedly would cause them to gradually decay. By assuming that the process of memory is non-lossy, they use algorithmic theory to show that the <https://medium.com/the-physics-arxiv-blog/898b104158d>process of integrating information must noncomputable. In other words, your PC can never be conscious in the way you are. That's likely to be a controversial finding but the bigger picture is that the problem of consciousness is finally opening up to mathematical scrutiny for the first time." --- co-chair http://ocjug.org/

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Re: [math-fun] Heavy ice D2O sinks in light water H2O
by Bill Gosper 10 May '14

10 May '14

TomKnight>That is correct. I've personally done the experiment. On Nov 29, 2013, at 1:57 PM, Eugene Salamin wrote: A while ago we had a discussion about whether heavy ice floats or sinks in light water. It sinks. I found a picture and tried to post it, but it exceeded the 40 KB limit for math-fun postings. It's on page 6 of this document. http://mragheb.com/NPRE%20402%20ME%20405%20Nuclear%20Power%20Engineering/Is… people should don their tinfoil hat before viewing. -- Gene Now if you titrate to neutral buoyancy, you'd almost have something practical. I just learned that the whole reason to brominate vegetable oils is to raise their specific gravity so they don't separate from aqueous solution. (So you probably need to tweak for salt or sugar content.) A more direct (and costly?) approach might be simply to deuterate the (citrus or whatever) oil. The catch is that the D will wander around replacing your H. So what? Apparently this can derange the relative efficiencies of various enzymes, leading to chemical imbalances. Some cells (and some organisms) are better than others at adjusting the enzyme concentrations to compensate. (This from a long ago talk I attended.) --rwg The .0156% natural abundance suggests that the effects of dietary deuterium deprivation would be subtle. I read somewhere that willow roots concentrate D, but now all I can find is that they concentrate heavy metals. And aspirin. (Well, actually the bark. Well, salicylic acid, of which aspirin is the precursor.)

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Re: [math-fun] Mathematical Model Suggests That Human Consciousness Is Noncomputable
by Bill Gosper 10 May '14

10 May '14

Mike Speciner: "But there's a lot of evidence that real memories DO decay." Apparently "the continuous addition of new neurons both degrades existing information stored in hippocampal circuits and simultaneously provides substrates for new learning". HansH> http://www.sciencemag.org/content/344/6184/598 "Throughout life, new neurons are continuously added to the dentate gyrus. As this continuous addition remodels hippocampal circuits, computational models predict that neurogenesis leads to degradation or forgetting of established memories."<Hans I was taught that the only cells you can generate postnatally were blood, bone, and liver. When did this change? Or are they referring to the conversion to neurons of glia? Or maybe the ancients were right and the liver really *is* the seat of consciousness! --rwg

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Re: [math-fun] What is the period of this integer sequence?
by Bill Gosper 08 May '14

08 May '14

On 2014-05-05 07:01, Fred Lunnon wrote: > Very nice pics in the book! Thanks! > > (Neckties? T-shirts? Bedspreads? Wallpaper? > Um, maybe not wallpaper ...) > > WFL I am hoping the silence from the rest of you means at least a few took a crack at the sequence. By way of couragement, here is a similar sequence we did track. a[-2], a[-1], a[0], a[1],... = 8, 17, 8, 0, -17, -17, 12, 21, 4, -5,... a[k_?OddQ] == a[k - 2] - Floor[d*a[k - 1]], a[k_?EvenQ] == a[k - 2] + Floor[e*a[k - 1]] where the exotic rationals d := 2450365/1120052, e := 12497/8798 arose from a hopeless attempt to completely map the periods in a subpatch of (d,e) space in http://nbickford.files.wordpress.com/2011/03/neighborhood.png . Instead of the scalars a[n], we ran the pairs {x[n],y[n]} :={a[2n+1],a[2n]}, which lie on a very eccentric ellipse, which can be circularized by {x',y'}:= {(x - (d*y)/2)/(d^2*(-(1/4) + 1/(d*e)))^(1/4), (d* Sqrt[-(1/4) + 1/(d*e)]*y)/(d^2*(-(1/4) + 1/(d*e)))^(1/4)} This is the period 51284705405843 (times 2 for a[n]) "monster" mentioned on p13. The radius plot of (x',y') vs time is http://gosper.org/51TRadiusHistory.png , where the time axis is compressed a billionfold. Note that it is much trendier than a random walk. The sudden market crash at the end took 3 CPU days. Casual observers may have missed the crucial fact that, despite the Floor functions, the a[n] sequence is exactly reversible, so it must return to 8,0,-17 or blow up. Likewise for the subject sequence and all similarly defined. "Shortlly" after time 104T, the circularized radius is triple that of 51T's maximum. At such a large radius the Floors are very nearly no-ops, leaving the circularized radius very nearly constant. Thus the only "restoring force" capable of pulling a[2n],a[2n+1] back to 0,1 is that, until then, it must self-avoid. Somehow. --rwg > > > On 5/4/14, Bill Gosper <billgosper(a)gmail.com> wrote: >> 0, 1, 7, -2, -8, 3, 14, -4, -16, 5, 21, -6, -24, 7, 28, -7, -25, 7, >> 27, -7, -26, 7, 26, -6, -19, 5, 18, -4, -12, 3, 10, -2, -5, 1, 2, 0, >> 2, -1, -6, 3, 16, -5, -22, 7, 30, -8, -30, 8, 30, -7, -23, 6, 22, -5, >> -16, 4, 14, -3, -9, 2, 6, -1, -2, 1, 5, -1, -3, 1, 4, -1, ... >> >> a[0] = 0; a[1] = 1; >> a[k_?OddQ] := a[k] = a[k - 2] - Floor[9*a[k - 1]/17]; >> a[k_?EvenQ] := a[k] = a[k - 2] + Floor[15*a[k - 1]/2] >> >> This is the "Minsky Stock Index" that Corey and Julian ran fruitlessly >> from a[-10^14] to a[10^14] a few years ago. "If you haven't looked at a >> problem in the last few years, you haven't looked at it." --Ed Pegg, Jr. >> There is a slight chance that the period is infinite. >> (When it reached 18 trillion, we exclaimed AT&T!) >> >> The above recursive definition crashes Mathematica in under a million, >> even with memoizing. Instead use the iteration >> In[115]:= NestList[Function[xy, {#, xy[[2]] + Floor[15*#/2]} & >> [xy[[1]] - Floor[9*xy[[2]]/17]]], {1, 0}, 35] >> e.g., for {a[-1],a[0}, {a[1],a[2]},...{a[69],a[70]} >> Out[115]= {{1, 0}, {1, 7}, ..., {-1, -4}} >> >> Background: >> http://www.blurb.com/books/2172660-minskys-trinskys-3rd-edition >> Unlike Collatz or twin prime searching, this is a very specific question >> about >> very specific quantities, rather than a question about the infinitude of >> integers. >> --rwg

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[math-fun] othello puzzles...
by Warren D Smith 08 May '14

08 May '14

As a lower bound, this real (as opposed to artificial) othello game, a 64-0 wipeout by white (who makes lowercase moves), involves 23 passes by black (uppercase moves): F5d6C4g5D7c7H5d8B8b7B6a7b5A6a5b4B3c6E8f8C8a8e7a4b2A2c2D2d1e2F2c5c3a1a3e6d3f6F7f1g2H2e3F4b1g1c1e1g8g7f3h1h3h4h6 count of passes so far: 1 2 3 4 5 6 7 8 9 a b c d e f 0 1 2 3 4 5 6 7 assuming I did not screw up. The maximum possible number of passes is clearly <=61 since two consecutive passes end the game (by rules). And if one cannot pass during first 10 moves, upper bound shrinks to 51. Concerning the question of the shortest possible non-wipeout game, here's an 11-move nonwipeout game (15 disk final position): 1.f5 2.f4 3.c3 4.e6 5.f7 6.g6 7.g4 8.e8 9.g8 10.g5 11.h5 ends B14, W1: ........ ........ ..B..... ...BBBB. ...BBBBB ....B.B. .....B.. ....W.B. Re the max-mobility question: I found a position with mobility=25 arising in a real game, but it is certainly increasable. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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