June 2011 - mathfun@mailman.xmission.com

[math-fun] Exact N-body simulations?
by Bill Gosper 10 Jun '11

10 Jun '11

Henry Baker> All this talk of Gaussian integers reminded me of a problem that I'd dearly love to see solved. Consider N bodies, each with integer masses. Start them in initial positions in a 2D (for the moment, there's plenty of complexity in 2D) which are on integer coordinates. The attractive Newtonian gravitation force is proportional to m*m'/(dx^2+dy^2), so this force is rational. We need to compute new positions for each of the bodies in such a way that all the conservation laws are preserved exactly (if this is possible), yet the number of bits required doesn't grow too quickly (see below). We need to preserve the total energy, we need to preserve momentum & angular momentum, & we might like to preserve information (the system is reversible). Note that we don't necessarily require that delta-t be constant, although if it isn't, we may need to have another counter to represent the "clock" -- i.e., the global time. I don't know how to do this even in the 2-body case, but the Minsky circle hack may provide some guidance -- if we could get the two bodies to each draw a Minsky circle while revolving around one another, then we'd have a good start on the problem. In addition to solving the problem of finite arithmetic, we also need to figure out what happens in collisions. Taking a cue from Fredkin, we should probably have 100% elastic collisions in order to ensure reversibility. How does this square with Poincaré ? Well, I'm assuming that my N-body problem for finite N is Turing complete, so with enough bodies and enough "tape" (empty space), we should be able to compute any computable function. Minsky already provides one potential representation with his "counter machines". So we want a reasonably "efficient" Turing Machine (or Counter Machine) simulation so that we aren't wasting more than a fixed fraction of the bits required to represent the state (N pairs of integers or N gaussian integers). Has anyone attempted this sort of exact simulation before? The variable speed of the Kepler solution is pretty unMinskylike, although there might be a transformation through which a nonphysical but errorless simulation can appear physical, analogous to the circularization methods in Minskys and Trinskys <http://www.blurb.com/books/2172660>. If http://www.freewebz.com/vitaliy/triApplet/triGrav3.html is any indication, the n=3 case in 2D is unstable with p ~ 1. (Unlike the Windows version, the Apple version seems capable of occasional gross integration errors. But both simulations seem always to blow up.) --rwg

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Re: [math-fun] Some Sum
by Henry Baker 10 Jun '11

10 Jun '11

Isn't this problem somewhat akin to defining mod(p,q), p,q complex numbers? The problem is that almost every definition is arbitrary: besides |mod(p,q)|<|q|, and the fact that mod(p+n*q,q)=mod(p,q) (n an integer), and perhaps mod(p+i*n*q,q)=mod(p,q), what other criteria should we use to pin down this function? We could require the image of mod(p,q) to be a parallelogram in (1,i) space, but that might be far too confining: we might want to use one of Gosper's dragons as the basic "rep-tile". BTW, I assume that there is some sort of connection between mod(p,q) and elliptic functions? I've never seen such a connection written down, but the basic 2-D periodicity would seem to scream it out. At 02:04 PM 6/8/2011, Dan Asimov wrote: >I was going to ask the same question. > >It seems the canonicalest choice would be elements z of Z[i] >with Re(z) > 0 and Im(z) >= 0. But there's something >arbitrary about this. > >But even more natural might be to define the sum of aliquot >parts in something like Z[i] / z ~ iz. I would hope this >object would retain some kind of useful algebraic structure, >that would be related to the ring structure of Z[i] as Z+ >is related to the group structure of Z. > >--Dan > >Gene wrote: > ><< >In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i. Thus if d is a divisor of 3+i, so are -d, id, -id. In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor? If you take them all, the sum is zero. In the rational integers, one chooses the positive divisor. >>>

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Re: [math-fun] Some Sum
by Dan Asimov 10 Jun '11

10 Jun '11

At the risk of repeating what I wrote here years ago: What finally got me interested in algebra and number theory as I approached my 50th birthday is the theorem (conjectured by Gauss, proved by Heegner, reproved by Stark) -- that among imaginary quadratic fields Q(sqrt(-d), their rings of algebraic integers are unique factorization domains precisely for 9 of these fields, and these 9 rings can be expressed as Z[sqrt(-d)] for d = 1,2,3,7,11,19,43,67,163. (Gauss already knew that these were UFD's, and conjectured that there were no others; that's what Heegner proved in 1952; see < http://en.wikipedia.org/wiki/Stark%E2%80%93Heegner_theorem#History >.) As may have been mentioned here, Conway & Guy give a good heuristic idea of how the proof proceeds in their book Numbers (thanks again for that, Rich & Hilarie). (I love theorems with a strange set of exceptions like that. (E.g., the automorphism group Aut(S_n) of the symmetric group S_n is isomorphic to S_n itself for precisely all n not equal to 2 or 6.) --Dan Hans writes: << I'm going to guess that the "positive" Gaussian quadrant illustrated by Conway and Guy can be rotated into the *top-right* quadrant without loss of the unique-factorization principle. >> Sometimes the brain has a mind of its own.

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Re: [math-fun] Some Sum
by Dan Asimov 10 Jun '11

10 Jun '11

<< But x=1-i has no "positive" Gaussian integer divisors at all. 1-i = (-i)(1+i). --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] Some Sum
by Dan Asimov 08 Jun '11

08 Jun '11

I was going to ask the same question. It seems the canonicalest choice would be elements z of Z[i] with Re(z) > 0 and Im(z) >= 0. But there's something arbitrary about this. But even more natural might be to define the sum of aliquot parts in something like Z[i] / z ~ iz. I would hope this object would retain some kind of useful algebraic structure, that would be related to the ring structure of Z[i] as Z+ is related to the group structure of Z. --Dan Gene wrote: << In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i. Thus if d is a divisor of 3+i, so are -d, id, -id. In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor? If you take them all, the sum is zero. In the rational integers, one chooses the positive divisor. >> Sometimes the brain has a mind of its own.

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[math-fun] Some Sum
by Hans Havermann 08 Jun '11

08 Jun '11

What is the sum of the aliquot (proper) divisors of 3+i? a) The divisors are {1, 1+i, 1+2i, 3+i}. The sum of the first three numbers is 3+3i. b) Robert Spira (1961) defined a "complex sum of divisors" that is incorporated into Mathematica's DivisorSigma function: In[1]:= DivisorSigma[1,3+I] Out[1]= 2+6 I Of course that answer includes the divisor 3+i, so we subtract that from 2+6i to get -1+5i. c) Neither of the above. Spira's sum-of-divisors function is (I think) a product. So subtracting the 3+i after calculating the "sum" is a leap of faith.

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Re: [math-fun] Paradox
by Dan Asimov 08 Jun '11

08 Jun '11

Thank you for the clarification, Andy and Bill. I believe there are subsets of the integers that really should not be assigned a density. I'm interested in giving a density to a maximum -- or at least a maximal -- collection of subsets of Z that *ought* to have a density. Such a thing should of course be an extension of the usual density to a wider class of subsets. I'm not sure such a thing exists, but it would be nice to think it did. (Certainly subsets of Z whose density can be defined by iterated Cesaro summation should be included.) Auxiliary question: If the power set of the integers is identified with {0,1}^Z and given its product measure, then what is the measure of the collection of those subsets of Z that "ought" to have a density? --Dan Andy L. wrote: << I believe that what Bill is saying is that density can be extended to a finitely additive translation-invariant measure defined on all sets of integers. But there's no canonical way to do this, and no such way can be explicitly specified, and proving the existence of this extension requires use of AC. Bill T. wrote: << That is correct. . . . >> >> Sometimes the brain has a mind of its own.

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[math-fun] iterated insertion of mediants
by James Propp 07 Jun '11

07 Jun '11

If you start with two arbitrary (positive) rational numbers p,q and start iteratively interpolating mediants, and you reduce every unreduced fraction you encounter to lowest terms, do you get all the rationals between p and q? Example: Start with 1/2 and 3/1. 1/2 3/1 gives 1/2 4/3 3/1 gives 1/2 1/1 4/3 7/4 3/1 gives 1/2 2/3 1/1 5/4 4/3 11/7 7/4 2/1 3/1 etc. After n rounds of mediant interpolation, you have 2^n+1 fractions. Jim

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Re: [math-fun] Paradox
by Dan Asimov 06 Jun '11

06 Jun '11

About 10 years ago I devised a construction, based on the Axiom of Choice, that results in the selection of a random integer under the assumption that it makes sense to pick a random point from the circle R/Z -- or equally, the interval [0,1). Of course Bill is referring to the well-known finitely-additive measure on the integers known as density, which of course is defined only on certain subsets. And speaking of density, it seems that using special averaging methods like Cesaro summation, one can extend the collection of subsets of Z on which density is defined. My question is,\: Is there a well-defined maximum collection of such subsets? Or at least maximal ones? --Dan Bill T. wrote: << On Jun 6, 2011, at 1:41 AM, Andy Latto wrote: > On Sun, Jun 5, 2011 at 9:32 PM, Bill Thurston <wpthurston(a)mac.com> wrote: > >> ====== >> Non countably-additive measures are legitimate structures, and provide >> one of the ways to define amenability for a group, >> but they are totally weird and I don't think you really want to go into that territory. In theory, there are >> translation-invariant additive but non-countaby-additive measures on Z, but it is known to be impossible to >> actually define any particular instance of one. > > How can this be? > > Let the measure of {1} be x > 0, and choose an integer p such that x > 1/p. > > If the measure is translation invariant, then the measure of any singleton is x. > > But if the measure is finitely additive, then the measure of > {1,2,3,...p} is xp > 1, so the measure can't be a probability measure. You've given a correct deduction that the measure of any singleton, and in fact the measure of every finite set is 0. The measure must be 1 for every set whose complement is finite. The measure must be 1/2 for all even integers, and 1/2 for all odd integers, etc --- there are certain similar things you can deduce. The craziness comes in extending these definitions to *all* subsets of the integers. >> Sometimes the brain has a mind of its own.

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Re: [math-fun] DARPA Starship study
by Henry Baker 06 Jun '11

06 Jun '11

(DARPA should have asked for time travel, as well. Proposals were due at noon last Friday, June 3rd.) Step 1: Talk the Chinese into converting $1T of their holdings of U.S. debt to fund this effort. Their Terra Cotta Warriors show that previous Chinese govts have spent similar fractions of their GDP for attempts at immortality. At 01:08 PM 6/6/2011, Michael Kleber wrote: >On Mon, Jun 6, 2011 at 4:04 PM, <rcs(a)xmission.com> wrote: > >> The physicist Robert Forward proposed anti-hydrogen as fuel for a >> starship engine. He needed 4g of anti-H2, to be annihilated and >> mixed with tons of regular matter (reaction mass), to travel to >> Alpha Centauri at a respectable fraction of c. The fuel is >> expensive, but has a wonderful specific impulse. > >A timely observation, given DARPA's recent request for ideas on how to set >up a self-sufficient agency to work on "interstellar flight in 100 years"! > >https://www.fbo.gov/index?s=opportunity&mode=form&id=ad1728884c2dc05a344e33… > >--Michael

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