June 2015 - mathfun@mailman.xmission.com

[math-fun] Fwd: What is the period of this integer sequence?
by Bill Gosper 07 Jun '15

07 Jun '15

I just noticed a typo in this item dated 2014-05-08 12:56: "Thus the only "restoring force" capable of pulling a[2n],a[2n+1] back to 0,1 ..." should say "back to (0,8)" Anyone who was confused by this is paying way too much attention. The item was primarily about the "Minsky Stock Index" restated toward the end. Last I heard, Tom Rokicki had chased it about 13 quadrillion steps without returning to (1,0). --rwg ----------------------------------------- [...] 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. "Shortly" 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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Re: [math-fun] theta[s](pi/6,q), or eta as theta again
by rwg 06 Jun '15

06 Jun '15

On 2009-07-05 01:35, Joerg Arndt wrote: > Hi, > [not on the list, > as unsure whether people \notin \{jj,rwg\} appreciate this stuff] List restored. Some of them do. This is just to remark that http://en.wikipedia.org/wiki/Dedekind_eta_function#Special_values gives the peculiar DedekindEta[\[Tau]] == E^((I \[Pi] \[Tau])/12) EllipticTheta[3, 1/2 \[Pi] (1 + \[Tau]), E^(3 I \[Pi] \[Tau])] instead of DedekindEta[\[Tau]] == EllipticTheta[1, \[Pi]/3, (E^(2 I \[Pi] \[Tau]))^(1/6)]/Sqrt[3] As a theta identity, this isn't obviously (to me) a quarter period or quarter quasiperiod identity, nor an imaginary transformation. The only thing I can guess is some sort of modular transformation of the eta. ? Which would suggest we add to our theta identities the modular eta relations, written as thetas? Will FullSimplify do these in our lifetime? --rwg > > * rwg(a)sdf.lonestar.org <rwg(a)sdf.lonestar.org> [Feb 16. 2009 19:33]: >> [...] > > > > Just used a few brain cycles on the equality between > third and last in this one: > >> 2 2 12 >> %pi %pi eta (q ) eta(q ) >> theta (---, q) = theta (---, q) = ----------------- >> 2 3 1 6 4 6 >> eta(q ) eta(q ) >> >> 3 9 >> theta (0, q ) theta (0, q) - 3 theta (0, q ) >> 2 2 1/3 2 2 >> = theta (0, q ) (---------------) = >> ------------------------------. >> 4 6 2 >> 2 theta (0, q ) >> 4 > > via relations 29.2-26a ... 29.2-26c on page 617 of the fxtbook > (version == 2009-June-30) > this is (warning: partition eta strikes again!) Somos' relation > t18_5_18 = +1*u1^3*u6*u9 +3*q*u1*u2*u3*u18^2 -1*u2^3*u3*u9 ; > (line 2278) of his file eta07.gp. > > Somos' file contains 250 such relations involving (a subset of) > u1, u2, u3, u6, u9, u18, and >4k relations in total. > So it may be feasible via some machinery to turn these > into relations involving thetas (at least theta[*](z=0,q)). > If so, _lots_ of relations would be obtained. > > >> >> This is a rich area of nonobvious(?) identities, e.g., >> >> 2 6 6 >> theta (0, q) theta (0, q ) theta (0, q ) >> 2 2 3 >> 2 2 2 3 >> = theta (0, q ) theta (0, q ) theta (0, q >> ), >> 2 3 2 >> or equivalently >> 2 6 >> theta (0, q) theta (0, q) theta (0, q ) >> 3 4 4 >> >> 2 2 3 3 >> = theta (0, q ) theta (0, q ) theta (0, q ) >> 4 3 4 > > All trivially verified via the referenced relations, finding them > may well be nontrivial. > > >> --rwg >> > > cheers, jj

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[math-fun] Alleged exact solution of Dirac eqn in a certain localized magnetic field
by Warren D Smith 06 Jun '15

06 Jun '15

Alonso Contreras-Astorga, David J. Fernandez C., Javier Negro: Solutions of the Dirac Equation in a Magnetic Field and Intertwining Operators http://arxiv.org/abs/1210.7416 They claim to exactly solve the Dirac equation associated to a charged particle of spin 1/2 immersed in a magnetic field with cylindrical symmetry generated by the vector potential [Ax,Ay,Az] = [0, 0, c*k / (e*r)] where c=lightspeed, e=electron charge, r=distance to symmetry axis, and k is an arbitrary field-strength parameter. This magnetic field wraps round the z-axis and has strength proportional to r^(-2). The energy levels E are (middle of page 8): E = +- sqrt( (m*c/hbar)^2 + d0^2 + n*(2a+n)*b^2 / (a^2 * (a+n)^2) ) where m = electron mass hbar = (planck const)/(2pi) pz = momentum in z-direction d0 = pz * L / (hbar*sqrt(L^2+k^2)) a = sqrt(L^2+k^2) / hbar b = pz * k / hbar^2 n = 0,1,2,3,... L = angular momentum quantum number You may object that this is obviously dimensionally incorrect. True. It's always amazing how physicists publish papers like that. Perhaps they have some magic secret undisclosed units that repair that. You may also object that there is no spin parameter. Also true. Perhaps it is absorbed inside their L parameter somehow. Anyhow, assuming they are not completely full of shit, the important thing to observe is that E when n=0 (and pz and L are nonzero and fixed) is a DECREASING function of the field strength parameter k. Therefore, we here have an alleged exact Dirac solution, where the energy of a state is smaller in higher magnetic field, so that, as I wanted, pair creation should occur if the magnetic field is strong enough and |pz| is large enough. Frankly, though, I have reached the point where I feel one has to assume that every physicist that ever publishes about this topic, is publishing a result which you should assume as your default assumption, is utterly false garbage. That is because that definitely has been the case in over 50% of the papers I've looked at. Under that assumption, life is difficult since you have to do everything yourself from the ground up.

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[math-fun] *OLD* proof sought
by Bill Gosper 05 Jun '15

05 Jun '15

Way back, when, I think, ACW moderated, NJAS requested more terms of A003459, and rcs posted an amazing proof that all the rest were repunits. Can anyone retrieve it? --rwg

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[math-fun] Exhaustive garage door opener using De Bruijn sequences
by Henry Baker 05 Jun '15

05 Jun '15

FYI -- Brute force garage door opener using De Bruijn sequences. https://threatpost.com/using-a-toy-to-open-a-fixed-code-garage-door-in-10-s… Using a Toy to Open a Fixed-Code Garage Door in 10 Seconds by Dennis Fisher Follow @dennisf June 4, 2015 , 10:42 am It may be time to upgrade your garage door opener. Security researcher Samy Kamkar has developed a new technique that enables him to open almost any garage door that uses a fixed codeÂand he implemented it on a $12 childÂs toy. The attack Kamkar devised, known as OpenSesame, reduces the amount of time it takes to guess the fixed code for a garage door from several minutes down to less than 10 seconds. Most openers in commercially available garage door openers have a set of 12 dip switches, which are binary, and provide a total of 4,096 possible code combinations. This is a highly limited keyspace and is open to brute-force attacks. But even on such a small keyspace, those attacks take some time. ÂNow in a common garage and clicker, weÂre going to be using between an 8-12 bit code, and we see a single click sends the same code 5 times, and we see each ÂbitÂ takes 2ms to send, with a 2ms wait period per bit after the entire code is sent. So a single 12-bit combination takes (12 bits * 2ms transmit * 2ms wait * 5 times = 240ms),Â Kamkar said in a post explaining his new attack. https://www.youtube.com/watch?v=iSSRaIU9_Vc With a simple brute-force attack, that would take 29 minutes, Kamkar said. To begin reducing that time, he eliminated the retransmission of each code, bringing the time down to about six minutes. He then removed the wait period after each code is sent, which reduced the time even further, to about three minutes. Looking to further reduce the time, Kamkar discovered that many garage door openers use a technique known as a bit shift register. This means that when the opener receives a 12-bit code, it will test that code, and if itÂs incorrect, the opener will then shift out one bit and pull in one bit of the next code transmitted. ÂSo the garage actually tests: 011111100000 (incorrect) (chops off the first bit, then pulls in the next bit) 111111000000 (correct!) Meaning we sent 13 bits to test two 12-bit codes instead of sending a full 24 bits. Incredible!Â Kamkar said. ÂWhatÂs even more beautiful is that since the garage is not clearing an attempted code, a 12 bit code also tests five 8 bit codes, four 9 bit codes, three 10 bit codes, four 11 bit codes, and of course one 12 bit code! As long as we send every 12 bit code, the 8-11 bit codes will all be tested simultaneously.Â Kamkar implemented an algorithm known as the De Bruijn sequence to automate this process and then loaded his code onto a now-discontinued toy called the Mattel IM-ME. The toy was designed as a short-range texting device for kids, but Kamkar reprogrammed it using the GoodFET adapter built by Travis Goodspeed. Once that was done, Kamkar tested the device against a variety of garage door openers and discovered that the technique worked on systems manufactured by several companies, including Nortek and NSCD. It also works on older systems made by Chamberlain, Liftmaster, Stanley, Delta-3, and Moore-O-Matic. ÂI donÂt think thereÂs any solution to having a really small key space. Upgrade is the only solution. Unfortunately the upgrades have their own set of problems, but no where nearly as bad,Â Kamkar said via email. Kamkar has released the source code for the OpenSesame attack, but he has modified it slightly so that it wonÂt actually work, to prevent criminals from using it. Kamkar has a long history of creative projects like OpenSesame. Earlier this year he released a tool called KeySweeper, which is a $10 USB wall charger that can record keystrokes from wireless keyboards. And last year he released a project called SkyJack that could forcibly disconnect drones from their controllers and make them connect to his drone. Kamkar said it took him quite a bit of time to get this attack in working order, but said he wasnÂt working on it full time. ÂI got all of this working late last year (2014). I just kept improving the attack over time and honestly, getting the pink toy working was the longest part. Using existing RF hardware to employ the attack takes seconds or minutes if you know what youÂre doing, but making a mobile version on a very small hardware budget (very little memory to produce a lot of data) took many nights, on and off,Â Kamkar said. About Dennis Fisher Dennis Fisher is a journalist with more than 13 years of experience covering information security.

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[math-fun] Landau levels (explanation of what they are)
by Warren D Smith 05 Jun '15

05 Jun '15

The Landau energy levels E of an electron in a constant magnetic field B (pointed in z direction) are as follows. E = +- sqrt( m^2*c^4 + pz^2*c^2 + e*B*c^2*hbar*(2*n+1-s) ) Here n=0,1,2,3,... is an integer; +- is for positron and electron states; pz is an arbitrary real momentum in z-direction; s is +1 or -1 representing electron spin times 2, e is the charge of the positron, m is mass of electron, c = lightspeed, hbar = planck const/2pi, B=magnetic field. Source: see EQ 21 of http://xxx.tau.ac.il/pdf/0705.4275v2.pdf or exercise 7.6 in W.Greiner+J.Reinhardt: Quantum electrodynamics 2nd ed. Springer 1994, or EQ13 of http://merlin.phys.uni.lodz.pl/concepts/2007_1/2007_1_141.pdf . The lattermost source has the advantage it derives it using a cylindrically symmetric wavefunction, which is just what I want to think about re also imposing a "cylindrical box." But you pay the price of a more complicated derivation. Also, the lattermost source claims the Landau levels are countably-infinitely degenerate, not merely 2-fold like the other two sources claim (More precisely, these states are 2-fold degenerate except when n=0 there is only 1). All these sources derived the energy levels by exactly solving Dirac equation in various ways. If the cylindrical box has a radius that is just right (exactly equal to something arising from a zero of a confluent hypergeometric) then their unboxed wavefunction still should be valid even if the box is there. More generally by taking a linear combination of the 2 degenerate landau states you should be able to choose the coefficients to make it be a solution in any-radius box (at least if that radius is large enough) i.,e. meet the boundary condition that the wavefn by 0 at the box wall. Consequently, I conclude that the landau levels in an aligned-oriented cylindrical box are exactly the same as the unboxed levels, except that: (a) the pz must be chosen to fit an integer number of wavelengths in 2*(box height). (b) for a fixed-radius box, only certain landau level n-parameters may be allowed? (c) the n=0 ground landau level, is not permitted to exist with box confinement. (d) the 2-fold degeneracy for unboxed landau levels reduces to 1 with boxing.

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[math-fun] Landau levels
by Warren D Smith 05 Jun '15

05 Jun '15

to ES: The lowest landau level has energy m*c^2 independent of field, and there is therefore no force repelling electrons from high magnetic field in your reasoning. Further, the landau states are totally delocalized, which kind of "repels" reasoning based on speaking about localized regions...

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[math-fun] Can a strong magnetic field create electron-positron pairs?
by Warren D Smith 04 Jun '15

04 Jun '15

Here is an argument, which seems absolutely indisputable, that an INHOMOGENOUS magnetic field can and will create pairs. Let the field be of form (Bx,By,Bz) = (0, 0, K*x) incidentally arising from vector potential (Ax,Ay,Az) = (0, K*x*x/2, 0). We know from the Stern-Gerlach effect that an electron (positron also) in such a field will be sucked in either the +x or -x direction depending on its spin. And the force sucking it will be F = mu_e * K. And this is experimentally confirmed, not mere theorizing. Therefore, if the distance L we can go in the x direction obeys L*F > m*c^2 then the total energy from all that sucking is enough to create an electron. A pair could be created at x=0, then suck distance L, and then have regained enough energy to pay for the pair creation. Therefore, pairs will be created. And if a distance L <= hbar/(m*c) suffices, then this creation process should be very fast. I don't see how anybody can possibly argue with this. Now any real-world magnetic field is NOT uniform, and it must go to 0 someplace. Therefore, any real magnetic field with enough strength, specifically if max strength B>m*c^2/mu_e, will generate pairs at a nonzero rate. Then the only question, it seems to me, is "what is that rate?" And how does the rate depend on the specific geometry of the field, for example on grad|B|? Optimally, we would express the pair generation rate (pairs/second) in terms of some triple integral of some function (or functional) of the magnetic field. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Schwinger pair creation rate
by Warren D Smith 04 Jun '15

04 Jun '15

> [I happen to have found 3 different published papers which state 3 > mutually disagreeing formulas for the Schwinger creation rate, and I > suspect all 3 wrong... but they all agree it is nonzero :)... ] --No -- make that 4 papers with 4 mutually disagreeing formulas when we add EQ1 of http://arxiv.org/pdf/hep-th/0507174v2.pdf to the list. No wait, here is a 5th paper with a 5th disagreeing formula: http://arxiv.org/pdf/hep-th/0107135v2.pdf EQ10 (see also EQ 2,3,4)! This last formula (which is claimed to agree with EQ18 in [Sree ram Valluri, Darrell R. Lamm, Julian Mielniczuk: Applications of the representation of the Heisenberg–Euler Lagrangian by means of special functions, Canadian Journal of Physics 71,7-8 (1993) 389-397; erratum 72,9-10 (1994) 683-685] claims to be valid in any uniform E and B combination-field and always yields a positive pair-creation rate if |E.B|>0. (Incidentally, this paper also uses a formula from Ramanujan notebooks.) Neato, except the Valluri-L-M paper claims the E field must be above a critical threshold to produce pairs, which is SIMPLY FALSE, so why should I believe anything else they say? These guys are unbelievable! How can they do this? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Reply to Meeker re Kinetic energy
by Warren D Smith 04 Jun '15

04 Jun '15

> First, the linear Bz field is only an approximation of the field in a SG > device. You > can't really have a quadratically increasing potential out to infinity. --it would be fine if it only kept going for a finite range of x, if wide enough (as I'd said). > Second, if you > did all you've shown is that the electron could be accelerated to high > energy by the > magnetic field acting on it's magnetic moment. But an electron won't decay > all by itself > just because it has lots of kinetic energy. The kinetic energy is frame > dependent. A > suitable Lorentz transformation to the electron's rest frame reduces that > energy to zero --sorry, by that reasoning you would conclude a large E-field would not create pairs, because E-fields can merely accelerate electrons to high kinetic energies, which means nothing. Wrong: The "Schwinger limit" is the well-accepted statement that an E-field creates pairs if E*(available length) > 1.022 MVolts. The Schwinger creation is rapid if E * e*hbar/(m*c) > 1.022 MeV = m*c^2. And the pairs actually will not have high kinetic energy, if created at threshold, basically since they arise from "tunneling." [I happen to have found 3 different published papers which state 3 mutually disagreeing formulas for the Schwinger creation rate, and I suspect all 3 wrong... but they all agree it is nonzero :)... ]

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