April 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Eisenstein representation
by Dan Asimov 17 Apr '18

17 Apr '18

Although I didn't know offhand what the equivalence classes of binary quadratic forms over Z is, I still meant to point out that: ----- This could be used to simplify the process of determining which numbers are represented by a given quadratic form f. If say f is equivalent over Z to the standard representative g_k, then it represents the same integers as g_k, which can be known in advance. ----- where "this" means: a) the presumed classification, plus b) knowing what integers each equivalence class represents, plus c) how to determine, in terms of (a,b,c), which equivalence class the quadratic form a x^2 + b x y + y^2 belongs to. My comment is just that once those things are established they might (emphasis on might) provide a shortcut to determining the set of integers that the given quadratic form represents. —Dan

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Re: [math-fun] Physics: Longest possible solid object?
by Keith F. Lynch 16 Apr '18

16 Apr '18

Brent Meeker <meekerdb(a)verizon.net> wrote: > I don't follow that? Are you calculating that, because of the > expansion rate, a place 5e4lyr (0.015megaparsec) away will be moving > away from us at about 1km/sec /and therefore the CMB won't look > isotropic from there?/ No, I was saying the exact opposite of that. Since the CMB looks isotropic from everywhere as long as you're at rest relative to galaxies that are close to you, and since places at rest relative to galaxies that are close to them are moving away from each other at a rate proportional to their distance apart, the CMB's rest frame must be different in different places. So if there's a ring a constant 1 megaparsec in radius (the ring is not expanding), and its center is at rest relative to the CMB at that location, then everywhere on the ring will be moving 70 km/sec toward the center relative to the CMB at that point. But of course no part of the ring ever gets any closer to the center. > The expansion is uniform and so looks the same from any point. Yes, but only if you're participating in the expansion. > Otherwise you could imagine a point 14e9lyr away at which the CMB > was stationary relative to you in one direction. The ring's radius obviously can't exceed the distance to the cosmic horizon. If it was slightly less than that, and of course made of some impossibly strong material so that it wouldn't be torn apart, everywhere on the ring the CMB would look very asymmetrical. Perhaps it would be the same brightness and color temperature as the sun in the direction of the center of the ring, and completely undetectable at all directions more than a degree away from that direction. (Speaking of which, I've often wondered what it would be like on an Earth-like planet in deep space going at close to the speeed of light relative to the CMB. Obviously there's some velocity at which the planet would get the same flux as Earth gets from the sun, but would the bright spot look anything like the sun, and would the peak be in the visible part of the spectrum? Also, how long would it last (by the planet's clocks) before either CMB-drag or the expansion of the universe made the faux sun cold and dark?)

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Re: [math-fun] Eisenstein representation
by Dan Asimov 16 Apr '18

16 Apr '18

I imagine the quadratic form f in Fred's post is over the integers. If Z = the integers then f : Z x Z —> Z via f(x, y) = a x^2 + b x y + c y^2 with a, b, c in Z. And that "binary" quadratic form just means that it's in 2 variables — right? Question (probably for Victor): What are appropriate definitions of equivalence of such quadratic forms? I would think that we would say that the quadratic forms f and g over Z are "equivalent over Z" if there is an invertible matrix (P Q) M = ( ) of integers P, Q, R, S such that by abuse of notation, (R S) f(M(x, y)') = g(x, y), for all column vectors (x, y)' in Z. Or non-abusively f(Px+Qy, Rx+Sy) = g(x, y). Then: What is a set containing one simple representative for each equivalence class? Is it finite? And: Given an arbitrary quadratic form f as above: What is a simple way to determine which of these representatives it is equivalent to? This could be used to simplify the process of determining which numbers are represented by a given quadratic form f. If say f is equivalent over Z to the standard representative g_k, then it represents the same integers as g_k, which can be known in advance. How does the equivalence relation, and standard representatives, change if the coefficients {P, Q, R, S} can be any rational numbers? —Dan ----- I spent most of the weekend failing to adapt Will Jagy's algorithm, encouraged by a delusion that I understood better why that should work. Once that notion had been dispelled, Victor's method succeeded at once, and generalises to an arbitrary quadratic form as follows. (A) To represent a prime number p by a quadratic form f(x, y) : (1) Solve quadratic equation f(w, 1) = 0 over |F_p ; (2) Reduce the lattice basis { [w, 1], [p, 0] } to a minimal basis; (3) Now either reduced vector yields [s, t] such that f(s, t) = p . The second quadratic root actually applies to the conjugate form, so that signs in step (3) may be incorrect. Efficient polynomial-time algorithms are eg. Tonelli-Shanks for step (1), and enhanced Euclidean for step (2). However, one's favourite CAS these days conceals such drudgery behind clean facades, such as Magma's Sqrt() and LLL() . Useful test info is available at http://oeis.org/wiki/User:Peter_Luschny/BinaryQuadraticForms There remain two supplementary questions. (B) Exactly why does method (A) work? (C) How do representations of prime factors combine into representations of a composite target? All these seem bread-and-butter questions to my innocent mind, and it is mystifying that they are not part-and-parcel of standard treatments. Or are they? Fred Lunnon On 4/14/18, Victor Miller <victorsmiller(a)gmail.com> wrote: > Yes, See my answer to the question about the sum of two squares: > > > https://mathoverflow.net/questions/70208/system-of-diophantine-equations?rq… > > For your problem you need to find a primitive 6th root of unity (really > just finding sqrt(-3)), then set up the lattice and apply LLL. > > > Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > >> There exists an algorithm for finding integers m,n such that >> x = m^2 + n^2 , in amortised time proportional to (log x)^3 , >> where x is prime with x mod 4 = 1 . See >> Richard E. Crandall, Carl Pomerance "Prime Numbers: >> A Computational Perspective" Springer (2001), sect 2.3.2 pp 93--96 . >> >> Is there an analogous efficient algorithm for x = m^2 + m n + n^2 ? >> See >> https://en.wikipedia.org/wiki/Eisenstein_integer >> https://oeis.org/A034017 , https://oeis.org/A000086 >> >> What about more general binary quadratic forms? -----

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Re: [math-fun] Physics: Longest possible solid object?
by Keith F. Lynch 16 Apr '18

16 Apr '18

Brent Meeker <meekerdb(a)verizon.net> wrote: > Why do you think the CMB defines a different rest frame at different > points in the universe? Because of the expansion of the universe. The Hubble constant is usually given in units of kilometers per second per megaparsec. (A megaparsec is about 3 million light years, or about 3E+22 meters.) In those units its current value (it's not really a constant) is about 70. In other words, if every civilization in the universe were to erect a survey marker in their neighborhood at rest relative to the CMB, two such markers a megaparsec apart would be receding from each other at 70 kilometers per second. Earth's motion relative to the CMB is given as 371 kilometers per second. (Presumably that's averaged over the year, as Earth's orbital speed is about 30 kilometers per second.) So with today's technology it can apparently be measured to a precision of 1 kilometer per second. That means the CMB rest frame ought to be detectably different if we measure it from about 50,000 light years from here. Obviously not practical. But if the CWB rest frame can be measured to a precision 8 orders of magnitude better -- and I can't see any reason why that should be impossible -- it will become possible to notice the differences from opposite ends of the solar system. More interesting is whether the expansion of the universe exerts a force, and if so whether it could be detected and measured with anything close to today's technology. The most distant man-made object, Voyager 1, is about 2E+13 meters away, moving at about 17 kilometers per second relative to the sun. If the cosmic expansion is pushing it via "linear frame dragging," it ought to have added a speed of 4E-5 meters per second to it. That would obviously be lost in the noise, given that the mass of the solar system is not known to a sufficient precision that its deceleration of Voyager can be subtracted out. On the other hand, the total additional distance to Voyager due to that hypothetical acceleration would be 24 kilometers. And the distance to Voyager can presumably be measured to within about 1 meter by timing the return signal. So if the experiment were to be repeated in deep space, it would be perfectly feasible. Another way to think about it is that there's a distance from the sun at which Hubble's constant times the distance to the sun is equal to solar escape velocity at that location. Would an object placed there at rest relative to the sun just hang there? I wrote a lot more here, but it was very confused and muddled, so I've deleted it. I may get back to it, especially if others are interested. Better yet, a real GR expert will speak up. Getting back to the enormously long loop of steel cable, it occurs to me that the CMB itself would exert pressure on it. If it's at rest relative to the CMB, that pressure would balance out (assuming the minor anisotropies in the CMB are negligible on that scale). But if the loop is large enough, not all of it can be at rest relative to the CMB. The CMB would induce an outward pressure on the loop, placing it under tension. This is a purely classical effect, no GR needed. Unlike gravitational and other GR forces, the cable thickness doesn't cancel out for this, since a cable of twice the diameter has eight times the strength, but only twice the CMB-intercepting surface area.

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[math-fun] Physics: Longest possible solid object?
by Keith F. Lynch 14 Apr '18

14 Apr '18

What's the longest a solid object can be in our universe, i.e. the greatest possible distance between its parts? If you had a lot of stuff in one place, solid or otherwise, it would collapse into a black hole. But what about an I-beam in deep space? How long could it be before it collapsed due to its own gravity? You could partially counteract the gravity by slowly spinning it end over end, but the forces wouldn't balance everywhere. Then I realized that in a ring the forces would balance. I'm envisioning a narrow steel cable (or carbon nanotube if you prefer) in a circular loop, slowly rotating around the loop's axis. Gravity would point toward the center, and centrifugal force would point away from the center. The forces should balance everywhere so long as it's in very deep space where tidal stresses are negligible. If it's steel, you have the advantage that most metals, including steel, get stronger as they get colder, and in deep space it would be very cold. (The reason why the insulation on the Space Shuttle's main tank was on the outside, despite all the trouble that caused, was so that the tank's aluminum fuselage would be very cold, hence could be thinner and lighter for the required strength.) You'd give it a slight surplus of spin so that it's under slight tension everywhere, as steel is stronger under tension than under compression. Especially if it's a cable rather than a beam. I think the size would only be limited by stresses from the expansion of the universe. But that raises a bunch of questions. If you build it with all the parts at rest relative to each other (other than the slow rotation) rather than at rest relative to the cosmological microwave background (whose rest frame is different in different places) would the cosmic expansion have any effect? I think it would have to, since distant parts of the universe are receding from us faster than light. (The speed limit of c is strictly local.) The radius of the ring obviously couldn't exceed the cosmic horizon distance. Even if I'm wrong on that, there's also the *acceleration* of the expansion of the universe to account for. And nobody knows how that varies with time. One dire conjecture is that eventually we'll all be literally torn limb from limb by it. Dyson found a possible way to live forever in an expanding universe, and Tipler found a possible way to live a subjective forever in a collapsing universe, but as far as I know nobody has yet found a way to survive a "Big Rip" universe.

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[math-fun] Headlights (was Re: Miss those IBM/DEC blinking lights?)
by Keith F. Lynch 14 Apr '18

14 Apr '18

Henry Baker <hbaker1(a)pipeline.com> wrote: > I'm surprised that auto LED headlights don't already broadcast their > vehicle VIN# (maybe they do?). I don't know, but many of them do rapidly flash on and off. So do LED taillights. > Apparently, one of the countries in the Middle East is now requiring > some sort of RFID IFF for automobiles. Sure makes the CIA drone > program a lot more accurate... RFID doesn't work over long distances. I'd worry more about RFID tags in US passports, which make it possible to build a IED that will detonate only when a US citizen is in range. > I seem to recall that someone at MIT had developed an LED display > for bicycle spokes that was time-multiplexed in such a way as to > provide a full circular bit-mapped display -- kinda like those WWII > radar displays. Synchronized clock displays, and the like, on things like spinning fan blades can be found in any novelty shop. But I think that would be a bad idea on bicycle wheels, since cyclists don't want motorists to be the least bit confused as to what they are seeing. Speaking of which, traditional headlights are a very bad way to illuminate your path, as they make it difficult for other road users to see anything except your headlights. Car headlights are designed to only light what's directly in front or below, so as to not shine directly in the eyes of motorists, cyclists, or pedestrians. And maybe that works well on perfectly flat terrain, but it doesn't work well anywhere else. Gyroscopically stabilizing them wouldn't help, since other road users follow the terrain. Decades ago, Edwin Land proposed that all headlights and all windshields be diagonally polarized. Pedestrians and cyclists could wear diagonally polarized glasses. You'd see the reflected light from your own headlights, but you wouldn't be dazzled by oncoming headlights as you are now. Since polarizers aren't perfect, you could still *see* the oncoming headlights, but they'd no longer be blindingly bright. The disadvantage is that all lights would have to be twice as bright to compensate. Anyhow, it never caught on, and is unlikely to at this late date. Today's technology gives more options. I suggest that car headlights use something like Google's face-detection algorithm to keep bright light out of everyone's eyes. Some light would be sent toward eyes, but not enough to keep people from seeing anything else. Bike headlights, at least in brightly lit urban and suburban areas, are mostly for being seen rather than for seeing. The same algorithm could be used in reverse for them, aiming all their much dimmer light toward eyes. That would greatly increase the battery life, assuming the computer chips don't use more power than the lights. The light could also be directed toward anything dangerous that the cyclist might not have noticed, such as a fallen branch across the road.

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[math-fun] Paper-cutters
by James Propp 14 Apr '18

14 Apr '18

Why do paper-cutters have curved blades? Is the usual design intended to optimize something? Jim Propp

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Re: [math-fun] More information begets less?
by Dan Asimov 13 Apr '18

13 Apr '18

----- > At first, "turn of the century" was unambiguous, but now even the > augmented phrase "turn of the 20th century" is ambiguous. According to the Wikipedia article on it, the British English version of the augmented phrase is still unambiguous. ----- Interesting! (I have little knowledge of other languages.) —Dan

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[math-fun] More information begets less?
by Dan Asimov 13 Apr '18

13 Apr '18

I don't know from information theory, but consider this: (As a post-W.W.II baby boomer) I had always heard the phrase "turn of the century" used unambiguously to mean around the time that 1899 became 1900. But as someone pointed out to me, nowadays the same phrase "turn of the century" is rarely mentioned without *which century* being added ... evidently because *without* the added info, the bare phrase "turn of the century" could mean the recent one or the earlier one. However, the called-for information is not necessarily the "which century" that people tend to assume is what's missing. For most people, the phrase "turn of the 20th century" is about as likely to mean 1899 —> 1900 as 1999 —> 2000 (or for pedants, add one). At first, "turn of the century" was unambiguous, but now even the augmented phrase "turn of the 20th century" is ambiguous. Clearly, more information is less. —Dan

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Re: [math-fun] "Random" cards
by Henry Baker 13 Apr '18

13 Apr '18

I claim that "1" is the *least-picked*/*least-random?* number. Three Dog Night even had a popular song about this phenomenon: "One is the Loneliest Number" At 06:06 PM 4/12/2018, James Propp wrote: >Cf. the observation that when people are asked to pick a number between 1 >and 20, the most common answer is 17.

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