July 2011 - mathfun@mailman.xmission.com

Re: [math-fun] tennis and randomness
by Dan Asimov 07 Jul '11

07 Jul '11

It's common knowledge in tennis that (other things being equal) the worse your opponent is at predicting where you'll hit the ball, the better you'll do. The best players don't merely randomize; they deceive. --Dan << No, I don't think so. Judging from the abstract, that article is about the inherent tactical value of various different serves, not the importance of using mixed as opposed to pure strategies (i.e., not always serving the same way). >> Sometimes the brain has a mind of its own.

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Re: [math-fun] tennis and randomness
by James Propp 07 Jul '11

07 Jul '11

No, I don't think so. Judging from the abstract, that article is about the inherent tactical value of various different serves, not the importance of using mixed as opposed to pure strategies (i.e., not always serving the same way). On Tue, Jul 5, 2011 at 3:54 PM, Seb Perez-D <sbprzd(a)gmail.com> wrote: > Could it be this one? > > http://www.ncbi.nlm.nih.gov/pubmed/19197212 > > Best, > > Seb > > On Thu, Jun 30, 2011 at 16:38, James Propp <jamespropp(a)gmail.com> wrote: > >> Someone (I've forgotten who, alas!) mentioned to me a few weeks ago that >> researchers have studied patterns in tennis service, and have found that >> when it comes to deciding where to serve, top-ranked tennis players do a >> better job of "acting like a coin" than lower-ranked players. >> >> Can anyone provide information on this? >> >> Thanks! >> >> Jim Propp >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > >

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[math-fun] Minsky reversible circle hack, again
by Henry Baker 07 Jul '11

07 Jul '11

Steve Witham shows how even the integer (floor) version of Minsky's circle hack is invertible/reversible: http://www.tiac.net/~sw/2005/03/Mandala/index.html The Minsky steps are: x := x - floor(delta*y) y := y + floor(epsilon*x) This means, starting with (x0,y0): x1 = x0 - floor(delta*y0) y1 = y0 + floor(epsilon*x1) Suppose now that we have (x1,y1): Since y1 = y0 + floor(epsilon*x1), we can solve for y0: y0 = y1 - floor(epsilon*x1) Since x1 = x0 - floor(delta*y0), we can solve for x0: x0 = x1 + floor(delta*y0) Notice that "+"/"-" doesn't even have to be the usual integers, but could be some group -- e.g., Z_2^n, Z_p, or something even wilder. --- Neil Bickford comments that when (delta,epsilon)=(9/17,15/2), the initial point (1,0) goes through at least 104 _trillion_ points without looping. http://nbickford.wordpress.com/2011/04/03/the-minsky-circle-algorithm/ Given delta, epsilon, let us define an equivalence relation for lattice points on the X-Y plane: (x1,y1) ~ (x0,y0) if (x1,y1) is a finite number of forward or backwards Minsky steps from (x0,y0). Since the Minsky algorithm is reversible, each of the equivalence classes is either a finite cycle or an infinite chain (isomorphic to Z). We want to place some bounds on the sizes of these equivalence classes. Under "normal" conditions, the Minsky circle algorithm starting with (x,y) (x,y integers) should explore no more than 2*pi*sqrt(x^2+y^2)*C points before looping, where C>0 is some small constant. Rationale: this _is_, after all, a _circle_ drawing algorithm, so if it colors very many more points than are on the circumference of the circle, it can't possibly be a circle drawing algorithm. We already know that some difference equation approximations to differential equations don't converge to the differential equation if the "step size" is too big. Something analogous to this behavior seems to be going on here. 1. Can we get tight bounds on C, if delta & epsilon are bounded in some way? 2. For the less bounded case: Is the Collatz Conjecture hiding in the Minsky algorithm somewhere?

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[math-fun] more polynomial factor heights
by Bill Gosper 07 Jul '11

07 Jul '11

>From 19 Aug '05 math-fun: "Apropos the cyclotomic discussion, I assume everybody knows about the 2s in factor(x^105-1). What is the lowest degree polynomial with coeffs in {-1,0,1} with a coeff of 2 in its irreducible factorization? --rwg" >From 8 May 09:"rwg>> Puzzle: find a trinomial with maximal |coefficient| < some coefficient of >> a(n irreducible) factor. > Edwin Clark> > x^70-x^35+1 has irreducible factor > > x^48-x^47+x^46+x^43-x^42+2*x^41-x^40+x^39+x^36-x^35+x^34-x^33+x^32 > -x^31-x^28-x^26-x^24-x^22-x^20-x^17+x^16-x^15+x^14-x^13+x^12+x^9 > -x^8+2*x^7-x^6+x^5+x^2-x+1 > > Note the coefficient of x^41 and also of x^7 is 2. Right, equivalently x^70+x^35+1 = (x^(3*5*7)-1)/(x^(5*7)-1). More surprising (to me), your factor/.x->-x, 1+x+x^2-x^5-x^6 -2*x^7 -x^8-x^9+x^12+x^13+x^14+x^15+x^16+x^17-x^20-x^22-x^24-x^26 -x^28+x^31+x^32+x^33+x^34+x^35+x^36-x^39-x^40 -2*x^41 -x^42-x^43+x^46+x^47+x^48, divides x^245 + x^70 + 1. I.e., (x^2+x+1)*(x^5-x^4+x^2-x+1) = x^7+x^2+1. > JamesB> > 2*x^5 - 5*x^2 + 3 = (2*x^3 + 4*x^2 + 6*x + 3)*(x-1)^2 > > 5 < 6. > > Jim Buddenhagen > Wow, I missed that one! --rwg" My searchware refuses to find Michael Kleber's solution of a related puzzle. Anyway, here's another: Find the height 1 polynomial of least degree d with an irreducible factor of height > d. And >=d? I'll start the bidding with d = 37, height 41. Whose value I'll sequester for now just to make things harder. --rwg

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[math-fun] Closed form for series?
by Bill Gosper 07 Jul '11

07 Jul '11

>Is there a closed form for the series sum(n^p z^n / n!, n=1..infinity) , where p is >complex, or at least p > 0, or even just for p = 1/2 ? > -- Gene Gene, this would require an analytic interpolation of the Bell polynomials analogous to Hurwitz zeta interpolating the Bernoulli polynomials. --Bill

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[math-fun] Closed form for series?
by Eugene Salamin 06 Jul '11

06 Jul '11

Is there a closed form for the series sum(n^p z^n / n!, n=1..infinity) , where p is complex, or at least p > 0, or even just for p = 1/2 ? -- Gene

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Re: [math-fun] tennis and randomness
by Dan Asimov 06 Jul '11

06 Jul '11

I would be surprised if there were any advantageous tennis behavior that didn't occur more often among top-ranked players, on average. --Dan << Someone (I've forgotten who, alas!) mentioned to me a few weeks ago that researchers have studied patterns in tennis service, and have found that when it comes to deciding where to serve, top-ranked tennis players do a better job of "acting like a coin" than lower-ranked players. Can anyone provide information on this? >> Sometimes the brain has a mind of its own.

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Re: [math-fun] Minsky Harmonic Motion
by Henry Baker 05 Jul '11

05 Jul '11

I should have pointed out that the Trinsky analysis of Minsky essentially produces Pythagorean triples/Gaussian primes for certain rational numbers; e.g., if d=1, delta*eps = 4/(1+4*d^2) = 4/(1+4*1^2)=4/5, and w = 2*asin(sqrt(4/5)/2) = 2*asin(1/sqrt(5)) = asin(4/5) -- i.e., we get a 3,4,5 triangle. When we compute arbitrarily high powers ((3+4*i)/5)^n, we conserve the length (i.e., energy) of the system, but we don't actually get an "orbit", since Q(i) is a field of characteristic 0. I now need to analyze the much harder case where we execute the _floor(x)_ and/or _round(x)_ function at each Minsky step. This system is the one actually discovered by Minsky, and produces a slightly blurry circle/ellipse, because it touches on a number of close by integer points and may require several "cycles" before it literally repeats. >From: Henry Baker <hbaker1(a)pipeline.com> > >Here's one way to do "Minsky Motion", aka Minsky Physics. I haven't >tried it with inverse square orbits, but here's a version for harmonic orbits. > >The Minsky method of drawing a "circle", a la Minskys & Trinskys: > >http://nbickford.wordpress.com/2011/04/03/the-minsky-circle-algorithm/ > >http://www.blurb.com/bookstore/detail/2172660?alt=Minskys+%26+Trinskys+3rd+… > >x := x - delta * y ; >y := y + eps * x >/* _sequential_, not parallel, updates */ > >or equivalently, > >[x] [ 1 -delta ][x] >[y] := [eps 1-eps*delta][y] > >or > >[x] [x] >[y] := M [y], where > > [ 1 -delta ] >M = [eps 1-eps*delta]. > >But we can diagonalize M with P, s.t. >P^-1 M P = diag(lambda_1,lambda_2), > >where lambda_1, lambda_2 are the eigenvalues of M. > >Since det(M) = 1 = lambda_1*lambda_2, >lambda_2 = 1/lambda_1. Also, > >trace(M) = 2-eps*delta = lambda_1+lambda_2 > = lambda_1+1/lambda_1 > >If 0 < eps*delta < 4, then >-4 < -eps*delta < 0, and >-2 < 2-eps*delta < 2, and >-1 < 1-eps*delta/2 < 1. > >So we can choose real w=acos(1-eps*delta/2), and the >eigenvalues will be complex conjugates >lambda_1 = cos(w)-i*sin(w) = exp(-iw), >lambda_2 = cos(w)+i*sin(w) = exp(iw). > >However, it will be slightly more convenient to choose >angle phi=w/2. > >Since cos(w) = 1-eps*delta/2, > >sin(phi) = sin(w/2) = sqrt(1-cos(w))/sqrt(2) > = sqrt(1-(1-eps*delta/2))/sqrt(2) > = sqrt(eps*delta/2)/sqrt(2) > = sqrt(eps*delta)/2 > >So, phi = w/2 = asin(sqrt(eps*delta)/2), and >w = 2*asin(sqrt(eps*delta)/2). > >Using Maxima's eigenvectors command and a lot of manipulation, > > [ delta delta ] >P = [1-exp(-iw) 1-exp(iw)] > >We can now construct a real matrix > >[cos(-w) sin(-w)] >[-sin(-w) cos(-w)] > >with the same eigenvalues exp(-iw), exp(iw), and > > [ cos(-w) sin(-w)] [exp(-iw) 0 ] >Q^-1 [-sin(-w) cos(-w)] Q = [ 0 exp(iw)] > > [1 1] >with Q = [i -i]. > >Combining these two efforts, we get > > [ 1 -delta ] [ cos(-w) sin(-w)] >Q P^-1 [eps 1-delta*eps] P Q^-1 = [-sin(-w) cos(-w)] > >So if with compute (Q P^-1 M P Q^-1)^n, we can get the >effect of n steps, i.e., > >[ cos(-n*w) sin(-n*w)] >[-sin(-n*w) cos(-n*w)] > >Thus, the effect of the n-th Minsky step is > >M^n = P Q^-1 (Q P^-1 M P Q^-1))^n Q P^-1 > >In effect, we slightly skew the space, perform n iterations, >and then reskew back. > >Slightly more perspicuously, > >x_n = x_0*cos(-n*w)+(x_0/2-y_0/eps)*sin(-n*w)/d and >y_n = y_0*cos(-n*w)+(x_0/delta-y_0/2)*sin(-n*w)/d, > >where d=sqrt(1/(delta*eps)-1/4) is a real number. > >--------- > >Now instead of considering two spatial coordinates x,y, >we consider instead a spatial coordinate x and its velocity -- >i.e., we consider the circle in _phase space_: > >[x] [1 -delta][x] >[v] := [eps 1-eps*delta][v] > >This phase plot can be interpreted as a simple harmonic >oscillator consisting of a weight moving forward & back >on a spring. > >Let k be the spring constant, and m be the mass >on the end of the spring. > >Then m*x'' = m*v' = m*a = F = -kx, where x'' is d^2 x/dt^2. > >Also, conservation of energy requires that > >E = (1/2)*k*x^2 + (1/2)*m*v^2 = (1/2)*(k*x^2+m*v^2) > >If k=m in suitable units, then the phase plot of such an >harmonic oscillator will be a perfect circle. > >Following Minsky/Trinsky, as above, we slightly skew our >(digital) phase space, compute n iterations, and then >reskew back to our digital phase space. > >Note that since the the harmonic oscillator in two dimensions >is separable, we can do exactly the same thing for y, v_y, and >simply include these 2 additional dimensions in our expanded >phase space. > >For a differential equation which _isn't_ separable, we could >(for n=2 objects), utilize quaternions (2 position coordinates, >2 velocity coordinates), and try to keep the _length_ of the >quaternion (~ total energy) constant.

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Re: [math-fun] Minsky Harmonic Motion
by Henry Baker 05 Jul '11

05 Jul '11

You are right; my rational rationale was wrong. But I still don't want to extend to Q(sqrt(3)), because therein lies madness (or at least a slippery slope). --- Someone over the years _must_ have looked at the "orbits" of the Minsky algorithm (including floor(x)). I hope this doesn't degenerate into something like the 3n+1 problem (Collatz conjecture)... At 03:22 PM 7/5/2011, Andy Latto wrote: >On Tue, Jul 5, 2011 at 3:14 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> I should have pointed out that the Trinsky analysis of Minsky essentially >> produces Pythagorean triples/Gaussian primes for certain rational numbers; >> >> e.g., if d=1, delta*eps = 4/(1+4*d^2) = 4/(1+4*1^2)=4/5, >> >> and w = 2*asin(sqrt(4/5)/2) = 2*asin(1/sqrt(5)) = asin(4/5) >> >> -- i.e., we get a 3,4,5 triangle. >> >> When we compute arbitrarily high powers ((3+4*i)/5)^n, we conserve the >> length (i.e., energy) of the system, but we don't actually get an "orbit", >> since Q(i) is a field of characteristic 0. > >While I agree we don't get an orbit, because tan(3/4)/tau is not >rational, I don't see the connection with the characteristic of the >field. Q((sqrt(3)) is a field of characteristic 0, but it includes >(sqrt(3) + i)/2, and the twelfth power of this is 1. > >Andy

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[math-fun] tennis and randomness
by James Propp 05 Jul '11

05 Jul '11

Someone (I've forgotten who, alas!) mentioned to me a few weeks ago that researchers have studied patterns in tennis service, and have found that when it comes to deciding where to serve, top-ranked tennis players do a better job of "acting like a coin" than lower-ranked players. Can anyone provide information on this? Thanks! Jim Propp

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