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Re: [math-fun] Neptune-Pluto orbital resonance
by Keith F. Lynch 27 Apr '19

27 Apr '19

Bill Gosper <billgosper(a)gmail.com> wrote: > Wikipedia: > Under some circumstances, a resonant system can be stable and > self-correcting, so that the bodies remain in resonance. I'm reminded of one of my long term plans. One of my *very* long term plans. The sun will gradually become a red giant and incinerate our planet. To save Earth, it will be necessary to gradually move it into a larger orbit for the duration. And then into a smaller orbit once the sun becomes a white dwarf. There are at least three main ways Earth might be moved outwards: * Loop asteroids around it and Jupiter, pulling Earth up and pulling Jupiter down. (Of course this has problems once they approach each other.) * Place a giant mirror near Earth, such that sunlight will push the mirror forwards in Earth's orbit, and Earth's gravity will make Earth follow it. * Launch lots of stuff from Earth backwards along its orbit. Fortunately, we have time. We can spend the next hundred million years ignoring the problem, the hundred million years after that thinking about it, and the hundred million years after than gradually accelerating the Earth outwards such that at the end of that hundred million years its outward drift reaches the speed of a snail. However, it's not clear whether on its outward journey it would hit any bad resonances with the other planets that might cause planets to collide with each other or with the sun, or expel a planet from the solar system. (Of course once Earth reaches Mars's orbit, a collision is inevitable.) The obvious solution, at least with Mars, is to move the other planet outwards too. They will all move outwards even without our intervention, due to the sun losing mass to radiation and to the solar wind. Not enough to save us, unfortunately. Nor should it destabilize the solar system. If they retain their orbital velocity (and why wouldn't they?) the orbital radii and the orbital periods should all increase by the same proportion that the sun loses mass. (Kepler's law that says the square of the period is proportional to the cube of the radius doesn't apply here, as it implicitly assumes the sun's mass is unchanging.) I'll give details of my reasoning on request. Kepler's law most certainly *does* apply if we move all the planets outwards ourselves. So we can't just enlarge all the orbits keeping the proportions of their orbital radii constant. Can we get away with enlarging all the orbits keeping the proportions of their orbital *periods* constant? And is there a cheaper way, in which we can leave most of the larger planets alone? Has anyone studied this problem? Is there any known way to study it other than a massive amount of numerical simulation, i.e. a digital orrery? Urgent! Please reply within the next quarter billion years! Thanks.

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[math-fun] Neptune-Pluto orbital resonance
by Bill Gosper 27 Apr '19

27 Apr '19

Wikipedia: Under some circumstances, a resonant system can be stable and self-correcting, so that the bodies remain in resonance. Examples are the 1:2:4 resonance of Jupiter <https://en.wikipedia.org/wiki/Jupiter>'s moons Ganymede <https://en.wikipedia.org/wiki/Ganymede_(moon)>, Europa <https://en.wikipedia.org/wiki/Europa_(moon)> and Io <https://en.wikipedia.org/wiki/Io_(moon)>, and the 2:3 resonance between Pluto <https://en.wikipedia.org/wiki/Pluto> and Neptune <https://en.wikipedia.org/wiki/Neptune>. In[234]:= MinorPlanetData["Pluto"][EntityProperty["MinorPlanet", "OrbitPeriod"]] Out[234]= Quantity[247.92065, "JulianYears"] In[235]:= PlanetData["Neptune"][EntityProperty["Planet", "OrbitPeriod"]] Out[235]= Quantity[164.79132, "JulianYears"] In[236]:= ContinuedFraction[%%/%] Out[236]= {1, 1, 1, 55, 1, 1, 1, 7} That's a fairly lousy approximation to 2:3. What's going on? —rwg

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Re: [math-fun] Should I be surprised?
by Keith F. Lynch 27 Apr '19

27 Apr '19

James Propp <jamespropp(a)gmail.com> wrote: > The 91st Fibonacci number is > 4,660,046,610,375,530,309 > Should I be surprised that the first eight digits are even? > Should I be surprised that, if you strip out the zero-digits, this > number consists of six even digits followed by eight odd digits? No. This is called the Texas Sharpshooter Fallacy. The a priori odds of that exact pattern are small, but it's taken from vast population of equally unlikely patterns, e.g. that in base 17 the digits of the 23rd Catalan number are in ascending sequence if you exclude the digit "3." (I just made that up; I haven't checked. If by remarkable coincidence that actually *is* true, that's *not* an example of the fallacy, since I "claimed" it before checking.)

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[math-fun] Should I be surprised?
by James Propp 26 Apr '19

26 Apr '19

The 91st Fibonacci number is 4,660,046,610,375,530,309 Should I be surprised that the first eight digits are even? Should I be surprised that, if you strip out the zero-digits, this number consists of six even digits followed by eight odd digits? I know that the initial digits of the Fibonacci numbers have an asymptotic equidistribution property, but 91 is not that big... Jim Propp

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[math-fun] The Lost Code
by Fred Lunnon 26 Apr '19

26 Apr '19

Seated one day at an analogue of the aperiodic Penrose tiling, I struck one 10-dimensional rank-4 null space of beauty with the (redundant) weight-4 basis: [ 1, 0, 0, 0, -1, -1, -1, 0, 0, 0] [ 0, 1, 0, 0, -1, 0, 0, 0, -1, 1] [ 0, 0, 1, 0, 0, -1, 0, 1, 0, -1] [ 0, 0, 0, 1, 0, 0, -1, -1, 1, 0] [ 1, -1, -1, -1, 0, 0, 0, 0, 0, 0] ; and on dropping signs it struck me (back!) that this was case n,m = 5,2 of an elementary binary block code with: number = n words, weight = (n-1)_C_(m-1) , length = n_C_m , distance = 2 (n-2)_C_(m-1) , where n_C_m denotes binomial coefficient. I have sought, but I seek it vainly --- along with its name and rank. Can somebody please jog my memory? Fred Lunnon

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Re: [math-fun] Hot tea sucks.
by Dan Asimov 26 Apr '19

26 Apr '19

Drop the bottle on the sidewalk from 3 ft. above. The tea will come out immediately. —Dan ----- I brew it in a 46oz pickle jar that easily tips over, and leaks if the lid isn't snug. But if I forget to momentarily loosen the lid within a few minutes, a powerful vacuum develops that makes it virtually impossible to unscrew. What is the trivial solution? -----

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Re: [math-fun] Square hex numbers?
by Dan Asimov 25 Apr '19

25 Apr '19

Thanks, Adam! I thought there should be some general method for solving P(n) = Q(K) where P, Q are quadratic polynomials over the integers, but didn't know where to go from there. * * * When there is an integer like 169 that belongs to two figurate numbers (H_7 = 169 = 13^2), I wonder if there is a set of 169 points in R^4 = R^2 x R^2 whose projection by p_12 onto one R^2 factor is the first figure and onto the other by p_34 is the second figure. I seem to recall asking this here some years ago for T_8 = 6^2, and that Michael Kleber showed that there's no set X of 36 points in R^4 with p_12(X) = triangle of side 8 and p_34(X) = square of side 6. Maybe there's a general statement about when such a thing can exist? —Dan —Dan Adam Goucher wrote: ----- Your equation is of the form: (quadratic in n) = (quadratic in K) ... ... -----

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Re: [math-fun] Square hex numbers?
by Dan Asimov 25 Apr '19

25 Apr '19

This suggests the question: ----- P and Q be integer polynomials of degrees >= 2. Find the solution set {(N,K) in Z^2 | P(N) = Q(K)}. ----- What's known about the general case when max(deg(P), deg(Q)} = 2, 3, 4, or >= 5 ??? I would try googling this but I'm not sure how to search for it. —Dan Adam Goucher wrote: ----- Your equation is of the form: (quadratic in n) = (quadratic in K) so you can complete the square on both sides and ..... -----

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Re: [math-fun] Packing problem
by Dan Asimov 25 Apr '19

25 Apr '19

Rigorous posts don't get no props. My previous attemptedly rigorous post on the subject was just to say, I guess opaquely, that it seems to capture the essence of a stable arrangement — like one forced by an elastic collar — to merely require: any sufficiently small perturbation of the arrangement (modify the centers but not the radii) cannot decrease the *diameter* of the arrangement (= the diameter of the smallest disk containing all disks of the arrangement). It seems clear that a rosette of 7 equal disks, but with the central one removed, is stable in this sense (despite the hole). —Dan ----- On 4/23/19, Allan Wechsler <acwacw(a)gmail.com> wrote: > I don't mind any particular model, but I think I just don't understand the > no-slip condition. Can you rigorize it? -----

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Re: [math-fun] "Non-orientable gyroid" question from John Graham
by Dan Asimov 24 Apr '19

24 Apr '19

Apologies for sending a private e-mail to math-fun under this subject. —Dan

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