December 2013 - mathfun@mailman.xmission.com

[math-fun] tightening knots
by Warren D Smith 31 Dec '13

31 Dec '13

I've posted before about knots are really machines and should be analysed as such. This idea turns out not to be new. Here is a paper http://arxiv.org/abs/1002.1723 with computer simulations of knots trying to find the "tightest" form of each knot type. You can see movies of the tightening process for many knots & links here: http://www.jasoncantarella.com/movs/ Many questions suggest themselves... -- Warren D. Smith http://RangeVoting.org

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[math-fun] new climate change prediction: 4C rise (from CO2 doubling vs prehuman) by year 2100
by Warren D Smith 31 Dec '13

31 Dec '13

http://www.nature.com/nature/journal/v505/n7481/full/nature12829.html 4+-1 C, more precisely. That's 6.4+-1.6 Fahrenheit. Previous estimates had been 1.5 to 5C. The reduction in the error bar is because the errors among the previous estimates are about half explained by a mechanism involving low versus high clouds, and water-vs-altitude mixing. So upon fitting that to observations this error mechanism can be reduced. And note, this error reduction now unfortunately for humanity takes us to the high end of the previous prediction range.

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[math-fun] knots and algorithms
by Warren D Smith 31 Dec '13

31 Dec '13

In what would seem to be a fantastic result, Marc Lackenby: A polynomial upper bound on Reidemeister moves, http://arxiv.org/abs/1302.0180 claims to prove that any diagram of the unknot with c crossings may be reduced to the trivial diagram using at most (231*c)^11 Reidemeister moves. He depends heavily on http://arxiv.org/abs/math/0208153 For two knots or links which are the same, what is an upper bound on the number of Reidemeister moves needed? Alexander Coward, Marc Lackenby: An upper bound on Reidemeister moves http://arxiv.org/abs/1104.1882 give a rather worse upper bound, namely, if n is the sum of the crossing numbers, then 2^(2^(2^(...(2^(n))...))) moves suffice, where the number of "2^" is 10^(1000000*n). Now that's what I call an upper bound, baby. It is known that the equivalence problem for manifolds is undecidable in all sufficiently large dimensions (in fact I think for 4D manifolds?) Kuperberg's claim that knottedness is in NP (under GRH) is here: http://arxiv.org/abs/1112.0845 and he says in 2002 Ian Agol claimed he'd proved this without requiring the GRH, but says his paper was never completely written up. But see thorem 2 in this: Ian Agol, Joel Hass, William P. Thurston: The Computational Complexity of Knot Genus and Spanning Area http://arxiv.org/abs/math/0205057 where it is claimed to be proven without need for GRH. Chad Musick in http://arxiv.org/abs/1110.2871 claimed unknottedness and unlinkedness was in P, but this paper "has been withdrawn until the proof can be clarified and verified."

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Re: [math-fun] tightening knots
by Bill Gosper 31 Dec '13

31 Dec '13

Sorry about the missing Subject line(s). On Tue, Dec 31, 2013 at 4:23 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > DanA> > > A few years ago I tried to solve for the tightest (smallest constant slope) helical rope about the z-axis in R^3. > > I.e., the lowest-slope helix about the z-axis such that if each point P of the helix has a flat, closed, perpendicular disk of radius = 1 centered at P that touches the z-axis, the interiors of all such disks are disjoint. > > This results in a transcendental equation that can only be solved numerically, but it was a fun endeavor and is in some sense the simplest "tightest-knot" type question I can think of. > > --Dan > It might be easier to think in terms of a moving sphere. > > Was your equation amenable to Lambert-W, or equivalent to Kepler's? > (http://www.tweedledum.com/rwg/pizza.html) > > Did you grab 20 digits and ask > http://isc.carma.newcastle.edu.au ? > In any event, I'd count this as resoundingly solved. All we need > is *any* system of equations admitting routine numerical methods. > > --rwg > > The pizza page ends with a sequence of polynomials in (Lambert) w, 1, 2 + 3 w, 9 + 26 w + 20 w^2, 64 + 273 w + 404 w^2 + 210 w^3, 625 + 3524 w + 7672 w^2 + 7692 w^3 + 3024 w^4,... whose "Rodrigues-like" formula can be considerably simplified: (D[t^(2*k)/w[t]^(2*k), {t, k}]*w[t]^k*(1 + w[t])^(-1 + 2*k))/(t^k*(2*k)) (none of which is a multiple of 1+w; they all seem irreducible). The high order coefficients are (2 n - 3)!/(n - 1)! (A006963). The constant terms are n^(n-1) (A000169). So why not edit the page instead of muttering here? Because the underlying notebook reads into NeilB's Student Edition but not my Mma 9.0 ! --rwg

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[math-fun] (no subject)
by Bill Gosper 31 Dec '13

31 Dec '13

DanA> A few years ago I tried to solve for the tightest (smallest constant slope) helical rope about the z-axis in R^3. I.e., the lowest-slope helix about the z-axis such that if each point P of the helix has a flat, closed, perpendicular disk of radius = 1 centered at P that touches the z-axis, the interiors of all such disks are disjoint. This results in a transcendental equation that can only be solved numerically, but it was a fun endeavor and is in some sense the simplest "tightest-knot" type question I can think of. --Dan It might be easier to think in terms of a moving sphere. Was your equation amenable to Lambert-W, or equivalent to Kepler's? (http://www.tweedledum.com/rwg/pizza.html) Did you grab 20 digits and ask http://isc.carma.newcastle.edu.au ? In any event, I'd count this as resoundingly solved. All we need is *any* system of equations admitting routine numerical methods. --rwg

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Re: [math-fun] tightening knots
by Bill Gosper 31 Dec '13

31 Dec '13

wds>I've posted before about knots are really machines and should be analysed as such. This idea turns out not to be new. Here is a paper http://arxiv.org/abs/1002.1723 with computer simulations of knots trying to find the "tightest" form of each knot type. You can see movies of the tightening process for many knots & links here: http://www.jasoncantarella.com/movs/ Many questions suggest themselves... -- Warren D. Smith "Because their strut sets break into a particularly simple form (see pages 36 and 37), these knots may be better candidates for an explicit solution than seemingly simpler knots such as the trefoil." http://www.jasoncantarella.com/movs/3_1ac.mov (note the glitch at 8 o'clock.) Back in the day, IBM hired me to write a demo of Mathematica 1.0(?) for their RT workstation rollout. I produced a fairly tight trefoil as a helix on a torus. WRI wanted the image for a poster, but IBM refused. WRI reverse-engineered the image, but kludgily. They sent me one nicely embroidered on a T-shirt. Just rendering *any* trefoil was a big deal in those days. Are there still no explicitly solved tightenings of it (or any other true knot)? I expect the trefoil to get sorted fairly soon. A crude program of mine came up with 17 and 13 as the minimal "lengths" of the trefoil, and the subarc of the trefoil that could not spontaneously "untie itself". Evidently, I used tube diameter where the convention seems to be radius. Fred> The player spontaneously jumps back in time (Apple iMac / OS10.9.1 / Safari). Anyone else have this problem? I'd email him, if I could decipher his address! < WFL This happened for a while on my MAC, but dragging the player progress indicator seems to force it to finish downloading, and play smoothly. --rwg

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[math-fun] (no subject)
by Bill Gosper 28 Dec '13

28 Dec '13

VM> W[]oops, for (1) I meant that for -pi/2 <= x <= pi/2, that |sin(x)| >= |x| or that |sin(x)| >= pi || x/pi || for *all* real x. The rest of what I said follows without change. <VM Isn't this *still* backwards? I.e., For -pi/2 <= x <= pi/2, |sin(x)| <= |x|. WDS> Returning to pi, http://mathworld.wolfram.com/IrrationalityMeasure.html indicates pi has a most a finite set of rational approxs p/q with |pi-p/q| < q^(-7.7). Therefore, in view of limit(N-->oo) N^(7.7/N)=1, the limit exists and is 1. <WDS Can someone confirm or correct this rephrasing?: "For every r>1 and q> 7.7, the nth term of π's continued fraction exceeds r^q^n at most finitely often." In any event, this is pathetically lame, since we probably only need q>1. --rwg REPHRASING SPRINGHARE RANGERSHIP

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[math-fun] limit puzzle
by Warren D Smith 28 Dec '13

28 Dec '13

As VM said, the only way the limit can fail to be 1 is if pi has occasional incredibly close rational approximations. If the role of pi were replaced by some evilly constructed "Liouville-like" transcendental number like SUM 10^(-2^(2^N)) then the limit would fail to exist. However, I think one can show that if pi is replaced by X then for almost all reals X, the limit exists and is 1. Also, for X any algebraic irrational, ditto. I.e. the set of "bad" real numbers X which have got an infinity of unbelievably close rational approximations, is a set of measure zero. To prove this, consider the standard continued fraction expansion of X and use the Wirsing/Khinchin probability distribution argue to that it is probability 0 that it has an infinite subsequence of large-enough partial quotients. For X algebraic you can use Roth theorem http://mathworld.wolfram.com/RothsTheorem.html Returning to pi, http://mathworld.wolfram.com/IrrationalityMeasure.html indicates pi has a most a finite set of rational approxs p/q with |pi-p/q| < q^(-7.7). Therefore, in view of limit(N-->oo) N^(7.7/N)=1, the limit exists and is 1. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Mathematica: Can this be?
by Bill Gosper 28 Dec '13

28 Dec '13

Asimov> In[1]:= Limit[Abs[Sin[x]]^(1/x), x -> Infinity] Out[1]= 1 But on the RHS of In[1] inside the limit, it takes the value 0 for arbitrarily large x. --Dan Just poaching your irresistable (despite spellcheck resistince) subject line, In[365]:= N[Sum[Sin[n]/Log[n], {n, 2, ∞}]] <brown disclaimers> Out[365]= 0.683914 In[366]:= N[Sum[Sin[n]/Log[n], {n, 2, ∞}], 21] <longer brown disclaimers, but sooner> Out[366]= -2.90772260714548440205*10^28 In[369]:= N[Integrate[Sin[x]/Log[x], {x, 2, ∞}], 26] Out[369]= -0.0964377 In[370]:= N[Integrate[Sin[x]/Log[x], {x, 2, ∞}], 27] <frantic brown diclaimers> Out[370]= -8.24193564681189931453106429*10^34 Boy, them low order digits can sneak up on you. --rwg

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Re: [math-fun] Turing -- world's fastest mathematician?
by Henry Baker 27 Dec '13

27 Dec '13

Be extremely careful about such analyses; the mere existence of Usain Bolt obliterated 100 years of scientific speculation about what makes a good human sprinter. I think that I will see (at least on TV!) someone (probably a Kenyan) do a sub-2 hour marathon in my lifetime (I'm currently 66). (Turing was 1 hour faster in the marathon, but he also had 30 years on me!) At 10:01 AM 12/26/2013, Warren D Smith wrote: >This page by Brian Skinner gives an interesting statistical analysis >which concludes the fastest achievable mile time is 3min 39.6sec. > http://gravityandlevity.wordpress.com/2009/04/22/the-fastest-possible-mile/ > >He then redid it for marathon finding 2hr 02min 43sec: > http://gravityandlevity.wordpress.com/2011/04/11/the-fastest-possible-marat… >In the comments he says he thinks this estimate has an error bar of >+-15 seconds, >i.e. making the bold prediction that 2hr 2min will never happen. > >if these true, the best athletes so far were already quite near to >best possible performances.

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