March 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Physicists Discover 13 New Solutions To Three-Body Problem
by Henry Baker 16 Mar '13

16 Mar '13

I don't see a problem with this at all. Some processes -- e.g., Newton's iteration of the square root -- pushes _all of the information_ (except for the sign) into lower and lower order bits. You can recover these bits by running the Newton process backwards. This backwards Newton process will reach further & further into the "innards" of the initial number -- i.e., pulling out lower & lower order bits. The only way you can get this process to cycle is if you started with a rational number (which has a repeating expansion). A simple "bounded" system which is composed of a finite set of real numbers, will have an unbounded number of bits to fool with. I can envision such a system which recycles larger & larger numbers of bits from the initial condition real numbers, but stays bounded because all the real numbers are a priori bounded by some fixed constant. For example, take some real numbers whose integer parts are all equal to 3. These real numbers remain bounded in the range (3,4), but we simply cycle larger & larger numbers of the fractional bits. This system is chaotic, non-periodic and bounded. At 05:04 PM 3/15/2013, Bill Gosper wrote: >The idea that such a simple, bounded system can produce endless novelty is a little hard to swallow.

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Re: [math-fun] a certain formula of Euler, primes and Pi
by Bill Gosper 14 Mar '13

14 Mar '13

This somewhat explains it: In[59]:= FullSimplify[FunctionExpand[LerchPhi[1, s, a]/Zeta[s, a]], s > 1 && a > 0] Out[59]= LerchPhi[1, s, a]/Zeta[s, a] In[60]:= Table[%, {s, 2, 3, 1/2}] Out[60]= {Zeta[2, a, IncludeSingularTerm -> False]/Zeta[2, a], Zeta[5/2, a, IncludeSingularTerm -> False]/Zeta[5/2, a], Zeta[3, a, IncludeSingularTerm -> False]/Zeta[3, a]} In[61]:= FullSimplify[FunctionExpand[%], a > 0] Out[61]= {1, 1, 1} but I don't see why I didn't get this for %59. --rwg Wouter> It is quite irresistible to play around with these formulae until one finds a simple expression that even Mma 9.0.0 can't simplify: try this: LerchPhi[1, 2*k, 1/2]/2^(2*k) /( -(-1)^k BernoulliB[2k] (2 Pi)^(2k)/(2k)!/2) == 1-1/2^(2k) or shorter: FullSimplify[LerchPhi[1, 2*k, 1/2]/2^(2*k)/Zeta[2 k]] just cute, no? Wouter -----Original Message----- From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Simon Plouffe Sent: woensdag 13 maart 2013 23:02 To: math-fun Subject: [math-fun] a certain formula of Euler, primes and Pi Hello, I have a friend in Moncton (NB, canada) a math professor, Paul Deguire, browsing around with the history of math and came accros this formula (well known ?) of pi : http://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80 see the center of the page with prime numbers, also this one http://mathworld.wolfram.com/PiFormulas.html formulas 60 and 61. It deals with prime numbers , an infinite product and pi. Is there someone that knows a reference for this formula, when it was found by Euler ? Any information that would enlight this almost pi day, Best regards, Simon Plouffe

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Re: [math-fun] A129935
by Phil Carmody 14 Mar '13

14 Mar '13

> Neil B. points out that Phil's PARI code misses eleven entries in > http://oeis.org/A129935/b129935.txt, and suggests > 13*cand (or larger) instead of 2*cand in > > default(realprecision, 500); c=contfrac(log(2)/2); for(n=2,#c, > cand=contfracpnqn(vecextract(c, 2^n-1))[1, 1];forstep(m=cand, 2*cand, > cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)),print(m)))) > > 13, and even 20, is in turn insufficient for longer tables. 20 gives > 638215591788186030688409<<5153>>7190350518178849343531561 > as A(1520), but 26 gives it as A(1527). Is there any finite > alternative to 26 that works indefinitely? If not, what is a > correct algorithm here? This may relate to the startling > interludes of slow, arithmetic growth at, e.g., A(56)-A(68). Would it depend on the next term in the CF? Could the numerator and denominator ever exceed the values attained by the next convergent? Of not, then that should provide an upper limit. I can't say I was ever happy with that 2, it's just that I knew I was missing some even numbers in the listed terms, and the 2 would catch them. Phil -- () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani]

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[math-fun] a certain formula of Euler, primes and Pi
by Simon Plouffe 14 Mar '13

14 Mar '13

Hello, I have a friend in Moncton (NB, canada) a math professor, Paul Deguire, browsing around with the history of math and came accros this formula (well known ?) of pi : http://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80 see the center of the page with prime numbers, also this one http://mathworld.wolfram.com/PiFormulas.html formulas 60 and 61. It deals with prime numbers , an infinite product and pi. Is there someone that knows a reference for this formula, when it was found by Euler ? Any information that would enlight this almost pi day, Best regards, Simon Plouffe

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[math-fun] A129935
by Bill Gosper 14 Mar '13

14 Mar '13

On 2013-02-27, at 3:05 PM, Bill Gosper wrote: Sum[2^k/(1 + z^2^k), {k, -Infinity, Infinity}] == 1/Log[z] Can somebody tell me where? --rwg And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1) MRob>This reminds me of something Neil Sloane mentions in his OEIS talks, namely that the function f(n) = floor(2n/log(2)) has the same value as the function g(n) = ceiling(2/(2^(1/n)-1)) (Sloane's sequence A78608) for all integer n from 1 to [7]77451915729367, but differs at n=77451915729368 (and at some other larger values, seeoeis.org/A129935) -------------------------- Neil B. points out that Phil's PARI code misses eleven entries in http://oeis.org/A129935/b129935.txt, and suggests 13*cand (or larger) instead of 2*cand in default(realprecision, 500); c=contfrac(log(2)/2); for(n=2, #c, cand=contfracpnqn(vecextract(c, 2^n-1))[1, 1]; forstep(m=cand, 2*cand, cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)), print(m)))) 13, and even 20, is in turn insufficient for longer tables. 20 gives 638215591788186030688409<<5153>>7190350518178849343531561 as A(1520), but 26 gives it as A(1527). Is there any finite alternative to 26 that works indefinitely? If not, what is a correct algorithm here? This may relate to the startling interludes of slow, arithmetic growth at, e.g., A(56)-A(68). Here is equivalent Mma code: $MaxExtraPrecision = 999999999; Select[Flatten[ Transpose[Outer[Times, Range[26], Denominator@Convergents[2/Log[2], 9999]]]], Floor[2*#/Log[2]] != Ceiling[2/(2^(1/#) - 1)] &]; This should really say $MaxExtraPrecision = ∞ but that tickles a numerics bug in the current version (9.0.1). --rwg

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Re: [math-fun] Discretized disk-packing
by Warren D Smith 13 Mar '13

13 Mar '13

It is trivial to solve your discrete ball-packing problem optimally, IF we subject ourselves to the constraint that the solution be a LATTICE. (Get computer to try every lattice. Lattice if D dimensions defined by D basis vectors, the lattice is their integer linear combinations.) In 3 and 4 and 8 dimensions the best ball packing lattices in fact have integer coordinates (if suitably scaled & rotated). In 2 dimensions this is also true if we cheat by using integer coordinates in 3-space, and agree to stay on the plane x+y+z=0. In 2D one way to get a very close approximation to the perfect beehive packing is to use lattice defined by basis vectors (0, 2A) and (B, A) where B/A is a good rational approximation to sqrt(3), got from the continued fraction expansion of sqrt(3)...

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[math-fun] Discretized disk-packing
by James Propp 13 Mar '13

13 Mar '13

Let d_n be the greatest density achieved by any subset of Z^2, no two points of which are closer in Euclidean distance than sqrt(n). What is known about d_n? Is it known to be always rational? I can use a compactness argument to show that the supremum density is achieved (that's why I felt free to write "the greatest density" above) but I can't prove that it's achieved by a doubly-periodic set (which would imply rational density). I get n d_n approx 2/sqrt(3) for large n using a back-of-the-envelope calculation, but I may have mis-programmed the envelope, especially since it's not an envelope but rather a bag for a pizza-slice-to-go --- a medium with no track-record as a calculation-aid. Jim Propp

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Re: [math-fun] "memoization"
by Phil Carmody 11 Mar '13

11 Mar '13

Cris Moore <moore(a)santafe.edu> wrote: > Of course, "memoization" is a silly neologism, redolent of > 1950's office work ;-) when I teach this subject, I use the > perfectly good word "memorization". Indeed. "Memoization" has never achieved full cromulence, in my book. I've never had a problem with just "caching" results. Phil -- () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani]

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[math-fun] Cris Moore action-minimizing technique: Implies stability? Apparently not.
by Warren D Smith 11 Mar '13

11 Mar '13

Moore's technique suggested the following QUESTION: If it can be shown that some N-body (or other Hamiltonian system) periodic solution is an action minimum, does that prove a stability claim? And by minimum I mean strict local minimum within some space of small-perturbations. Actually Moore's numerical technique I would think only shows it is an action min within the space of periodic solutions (and maybe even that was only shown nonrigorously and within some subspace, but it is probably possible to show it rigorously using second variations and some bounding argument about high frequencies). BEGINNINGS OF AN ANSWER: I'm a little confused because the space we are dealing with in variational problems is infinite dimensional, but I'm used to finite-dimensional. In the finite-dim'l case, certainly action-min implies stability, because 1. every Newtonian mechanics solution is an action-min (unlike, e.g. Einstein GR where the action is merely stationarized, nobody said minimized...) and 2. it is known (at least in the 3-body case, but I presume in N-body case for any N if we stay away from collisions) that solutions are analytic... and 3. there cannot be another min arbitrarily close to your min for the same reason analytic functions cannot have unboundedly many zeros within a neighborhood of a point; zeros always must be isolated points (any "analytic" function disobeying this would have an essential singularity there, not analytic) and this is also true for analytic functions of any finite number of variables. QED. BUT that reasoning does not work in the infinite dimensional case, because the following entire analytic function of infinity variables x1,x2,x3,... SUM F(x[n]*n)/n^9 where F(x) is a polynomial with two mins, one at x=0 and the other at x=1, has (x1,x2,x3...)=(0,0,0,...) as a non-isolated min. MY USUAL NASTY COUNTEREXAMPLE: is a point mass moving along a geodesic on a compact hyperbolic manifold (boundaryless surface of constant negative curvature). As I mentioned before, all such solutions are unstable. There are plenty of closed orbits (closed geodesics). These are not *strict* action mins since the action is the same for every solution with same kinetic energy. But given any particular closed geodesic, you could modify the problem by adding a smooth potential energy function which was 0 on that geodesic and slightly positive off it (say proportional to the squared distance to the geodesic, to be concrete) , in which case that orbit would then become a strict action min, indeed the unique global action min. So would this orbit then become stable to infinitesimally small perturbations? I believe the answer is NO, it still (if the point has enough mass and kinetic energy and the manifold's |curvature| is large enough, versus the magnitude of the potential energy) is unstable. So if I'm right about that (and I think I am) that answers my original question in the negative. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] best tilt angle for solar panel
by Warren D Smith 11 Mar '13

11 Mar '13

To be clear, I regard this problem as simply a cute calculus puzzle. It seems to be a pretty good exercise in calculus. I think it can be solved in more-or-less closed form, although I have not fully done so at present. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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