February 2008 - mathfun@mailman.xmission.com

Re: [math-fun] CORRECTED Product formula puzzle
by Michael Reid 15 Feb '08

15 Feb '08

> Let N = 2K+1 for some positive integer K. Evaluate explicitly > > f(N) := Product (5 - 4*cos(2*j*pi/N)) > > where j runs through 1,2,...,K, and prove your answer. over R , the polynomial x^(2K) + x^(2K-1) + ... + x^2 + x + 1 factors as Product (x^2 - 2 cos(2 j pi / (2K + 1)) x + 1) , the product being for j = 1, 2, ... , K . this is easy to see by factoring the polynomial completely over C , and then pairing up complex conjugate factors. now take x = 2 to get the answer: 2^(2K+1) - 1 . mike

1 0

Re: [math-fun] Homotheties of the hyperbolic plane
by Dan Asimov 15 Feb '08

15 Feb '08

Fred writes: << On 2/14/08, [I] wrote: << ... In any case, the impossibility of a bijection H^2 -> H^2 that increases all distances by a factor of (say) 2 brings up an interesting question: Can one quantify just how close one can get to such a map? >> In particular, can one choose constants c1, c2 both > 1 such that c1 <= |f(x) - f(y)| / |x - y| <= c2 ?>> >> Good question. (My guess is no, because lengths pushed a distance R away from a point increase in an exponential manner as a function of R.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

[math-fun] How many passphrases?
by Bernie Cosell 15 Feb '08

15 Feb '08

I'm trying to get a *rough* feel for how many simple passphrases there are in (say) twenty characters. Nothing fancy with punctuation and capitalization, and I don't want to worry that "doghouse" and "dog" followed by "house" will actually result in the same passphrase. Including one-letter-words it is easy to see that there are 2^19 different ways to divide the 20-character phrase into "words". But beyond that I'm stuck for getting even a vague-order-of-magnitude of how many passphrases you can get... But I know the number is going to be *HUGE*. The division 7/7/6, using just a small simple wordlist, gets me nearly 10^12 possibilities. Five four-letter words is >10^17 (four five letter words gets you a mere 10^15). My very rough intuitive guess is that since the different combinations will add together, the order of magnitude will be dominated by the tall pole in the tent and I bet that my guess of 5 four-letter-words [at >10^17] will probably be the answer. (actually, for my word list, 6 three-letter words and a two-letter word is about a hundred times larger). [for my not-very-big wordlist, 6 three-lets and a two-let came up with 83,631,769,001,984,000,000 pass phrases, nearly 10^20] [I'm trying to argue that if you make your passphrase 20 characters long, it doesn't make *ANY* difference what it is, it'll be massively hard to brute-force crack [compared, for example, to *anything* you do with an eight-character password -- even if you allow utterly random character strings, there are only about 6*10^15 8-char passwords], and so just pick something easy to remember and easy to type, but make it LONG. /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

1 0

[math-fun] Product formula puzzle
by Dan Asimov 15 Feb '08

15 Feb '08

Let N = 2K+1 for some positive integer K. Evaluate explicitly f(N) := Product (5 - 2*cos(2*j*pi/N)) over 1 <= j <= K, and prove your answer. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

1 0

Re: [math-fun] Homotheties of the hyperbolic plane
by Eugene Salamin 14 Feb '08

14 Feb '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Wednesday, February 13, 2008 4:03:45 PM Subject: Re: [math-fun] Homotheties of the hyperbolic plane ... Which reminds me of another similarly off-the-wall query. In the Poincar\'e model ("circle" including line and point in Euclidean plane) circles within the disc correspond to Circles; circles orthogonal to the disc correspond to Lines; circles tangent to the disc correspond to Horocycles; circles meeting to the disc correspond to Equidistant Curves; but circles outside the disc correspond to what in Hyperbolic space? WFL _______________________________________________ There is one more circle, the boundary of the disk. It is infinity, the set of directions in the hyperbolic plane. As for the "outside", it seems to be a parallel universe, another hyperbolic plane. On the Poincare sphere, pick a circle, call it "infinity". It separates the sphere into two caps, which can be made into two hyperbolic planes. Conformal maps on the sphere which preserve infinity become congruences in the hyperbolic planes. Reflection through infinity exchanges the two universes and establishes a correspondence between their objects. A circle orthogonal to infinity reflects onto itself, so its two hyperbolic lines correspond. On the other hand, a circle intersecting infinity nonorthogonally reflects onto a different circle, and so its two equidistant curves are noncorresponding. No doubt much more can be said. In particular, many parallel universes can coexist on the Poincare sphere, and be related via groups of conformal mappings. Gene ____________________________________________________________________________________ Looking for last minute shopping deals? Find them fast with Yahoo! Search. http://tools.search.yahoo.com/newsearch/category.php?category=shopping

2 1

Re: [math-fun] Homotheties of the hyperbolic plane
by Dan Asimov 14 Feb '08

14 Feb '08

Gene writes [of the hyperbolic plane H^2]: << That no nontrivial homothety exists is clear from the observation that a space of constant nonzero curvature possesses an intrinsic length scale, namely the inverse of its curvature. >> Perhaps this is the best short description of why this is impossible. In any case, the impossibility of a bijection H^2 -> H^2 that increases all distances by a factor of (say) 2 brings up an interesting question: Can one quantify just how close one can get to such a map? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

Re: [math-fun] Homotheties of the hyperbolic plane
by Eugene Salamin 14 Feb '08

14 Feb '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: Dan Asimov <dasimov(a)earthlink.net>; math-fun <math-fun(a)mailman.xmission.com> Sent: Wednesday, February 13, 2008 10:28:15 AM Subject: Re: [math-fun] Homotheties of the hyperbolic plane On 2/13/08, Dan Asimov <dasimov(a)earthlink.net> wrote: > ... > But any bijection of a manifold that multiplies distance by a fixed constant > c must be angle-preserving ... Is that a theorem? I couldn't justify it, and that's why I resorted to the computation. Even then, I assumed that the transformation had a fixed point, and furthermore was isotropic at that point. Neither of these assumptions has currently been justified, however plausible they might seem. WFL ______________________________________________ Yes, it's a theorem. If ABC is an infinitesimal triangle, then by the law of cosines, c^2 = a^2 + b^2 - 2 a b cos C, etc. So if all lengths are multiplied by a fixed constant the angles stay the same. That no nontrivial homothety exists is clear from the observation that a space of constant nonzero curvature possesses an intrinsic length scale, namely the inverse of its curvature. Gene ____________________________________________________________________________________ Never miss a thing. Make Yahoo your home page. http://www.yahoo.com/r/hs

2 1

Re: [math-fun] Homotheties of the hyperbolic plane
by Dan Asimov 13 Feb '08

13 Feb '08

Fred wrote: << On 2/13/08, [I] wrote: << I don't think I ever considered this before, but it seems I just figured out all possible ways such that, given a real number c > 0, one can construct a bijection of the hyperbolic plane H^2 to itself that multiplies all distances by c: f: H^2 -> H^2 such that dist(f(x),f(y)) = c dist(x,y) for all x, y in H^2. (If the word isn't "homothety", then I withdraw it.) I wonder if anyone else would care to tackle the same question. >> ... [Fred's argument goes here] ... The conclusion is that there are no nontrivial hyperbolic homotheties --- which is not too surprising, since (a) we would otherwise surely have heard of them already; and (b) there are none in spherical space. >> That's the conclusion I reached. My reasoning wasn't entirely from scratch: It's well known that the isometries of the hyperbolic plane H^2 are precisely its angle-preserving bijections. (I.e., bijections that are conformal or anti-conformal.) But any bijection of a manifold that multiplies distance by a fixed constant c must be angle-preserving, and so in this case is an isometry of H^2. Thus c = 1. --Dan P.S. I agree with Fred's sociological reasoning that if non-isometric homotheties of H^2 exzisted, we would have heard of them. But I was somewhat surprised that they don't, since if the Euclidean plane R^2 is infinite, H^2 seems "more" infinite, in that there is so much extra area. With my naive intuition, it felt like there ought to be more flexibility to arrange the points as desired. But the nonexsitence is entirely consistent with and also follows from the fact that a regular hyperbolic polygon that tiles H^2 with a fixed number per vertex -- e.g., say a hyperbolic square that occurs 5 per vertex -- can occur only in one exact size and shape: its side length is uniquely determined. If the fictional homotheties existed, this would not be true. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

[math-fun] Homotheties of the hyperbolic plane
by Dan Asimov 13 Feb '08

13 Feb '08

I don't think I ever considered this before, but it seems I just figured out all possible ways such that, given a real number c > 0, one can construct a bijection of the hyperbolic plane H^2 to itself that multiplies all distances by c: f: H^2 -> H^2 such that dist(f(x),f(y)) = c dist(x,y) for all x, y in H^2. (If the word isn't "homothety", then I withdraw it.) I wonder if anyone else would care to tackle the same question. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

[math-fun] The Kiss Precise
by Fred lunnon 13 Feb '08

13 Feb '08

I know some of this has appeared in math-fun before, but don't recall the final generalisation making a bow at the time --- apologies if it's old stuff! There is a very neat proof in terms of Cayley-Menger type determinants ("distance geometry"). Quoted (more or less) from Pat Ballew at URL www.pballew.net/soddy.html << The Kiss Precise by Frederick Soddy For pairs of lips to kiss maybe Involves no trigonometry. 'Tis not so when four circles kiss Each one the other three. To bring this off the four must be As three in one or one in three. If one in three, beyond a doubt Each gets three kisses from without. If three in one, then is that one Thrice kissed internally. Four circles to the kissing come. The smaller are the benter. The bend is just the inverse of The distance from the centre. Though their intrigue left Euclid dumb There's now no need for rule of thumb. Since zero bend's a dead straight line And concave bends have minus sign, The sum of the squares of all four bends Is half the square of their sum. To spy out spherical affairs An oscular surveyor Might find the task laborious, The sphere is much the gayer, And now besides the pair of pairs A fifth sphere in the kissing shares. Yet, signs and zero as before, For each to kiss the other four The square of the sum of all five bends Is thrice the sum of their squares. In _Nature_, June 20, 1936 Later another verse was written by Thorold Gosset to describe the even more general case in n dimensions for n+2 hyperspheres of the n-th dimension. The Kiss Precise (Generalised) by Thorold Gosset And let us not confine our cares To simple circles, planes and spheres, But rise to hyper flats and bends Where kissing multiple appears, In n-ic space the kissing pairs Are hyperspheres, and Truth declares - As n + 2 such osculate Each with an n + 1 fold mate The square of the sum of all the bends Is n times the sum of their squares. In _Nature_ January 9, 1937. >> While I can't claim to have published the next result, I did in fact discover it independently. So I feel justified in attempting an ode in the traditional vein. The Kiss Precise (Further Generalised) by Fred Lunnon How frightfully pedestrian My predecessors were To pose in space Euclidean Each fraternising sphere! Let Gauss' k squared be positive When space becomes elliptic, And conversely turn negative For spaces hyperbolic: Squared sum of bends is sum times n Of twice k squared plus squares of bends. In more conventional notation, let n+2 spheres be mutually tangent at distinct points in uniform space of n dimensions with (Gaussian, second) curvature k^2. The Descartes / Lachlan / Soddy/ Gosset relation between their curvatures x_i generalises to (\sum_i x_i)^2 = n (\sum_i x_i^2 + 2 k^2) . Examples: 4 circles touching at 6 points in plane n = 2: Euclidean k^2 = 0: x = [-1,2,2,3], 36 = 2(18 + 0); spherical k^2 = 1: x = [0,1,1,2], 16 = 2(6 + 2); hyperbolic k^2 = -1: x = [7,1,1,1], 100 = 2(52 - 2). See Ivars Petersen "Circle Game" in Science News (2001) \bf 159 (16) p.254 http://ww.sciencenews.org/articles/20010421/bob18.asp Result (rather obscurely) implicit in Lagarias, Mallows, Wilks \sl Beyond the Descartes Circle Theorem \rm, Amer. Math. Monthly also at http://xxx.lanl.gov/abs/math.MG/0101066. [apollo/0101066.pdf] Graham et al \sl Apollonian Circle Packings: Number Theory \rm http://xxx.lanl.gov/abs/math.NT/0009113. Websites: Graham at http://math.ucsd.edu/~fan/ron Lagarias, Wilks at http://www.research.att.com/~jcl, ~alan Fred Lunnon

1 0