March 2012 - mathfun@mailman.xmission.com

Re: [math-fun] Disk polyominoes
by Allan Wechsler 13 Apr '12

13 Apr '12

A funster who might or might not wish to remain anonymous has corrected me in private. A3 does not equal 2; it equals 1. The corrected sequence is: 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 3, 3 ... As before, I'm not entirely confident of the last couple of values. Thank you for the correction.

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Re: [math-fun] Topology puzzle
by Dan Asimov 31 Mar '12

31 Mar '12

I wrote: << In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2. >> My solution a few lines below Gareth's solution is correct, of course, and essentially the same as mine, which looks a little more symmetrical then the brick arrangement. Namely, consider the tiling of the plane by regular hexagons. The group G of translations preserving this tiling must be a lattice in the plane. Hence the quotient of the plane by the lattice must be a torus, topologically. Call the torus T. Let one of the hexagons be called H. Since every point of the plane is equivalent to some point of H, we can obtain the same quotient torus T by just identifying the points of H that are equivalent by the group G of translations. This means: Take the (closed) hexagon H and identify opposite edges in the simplest way -- identify each point of each edge with its closest poiwwnt on the opposite edge. This identification turns H into the torus T. It's easy to check that the boundary of H becomes a theta curve C on T. It's also clear that identifying the boundary of H to a point gives a topological sphere S^2 as the quotient space. Which means that taking the theta curve C on the torus T, and identifying it to a point, will also give the same topological sphere S^2 as *its* quotient space. Puzzle solved. * * * ((( Interestingly, it's easy to check (from orientability and Euler characteristic) that taking any regular polygon with an even number 2n of sides, and identifying its opposite sides (in the simplest way), has as quotient space the surface having its genus = floor(n/2). I.e., 2n-gon where 2n = 4 6 8 10 12 14 16 18 20 22 24 26 28 30 . . . -------------------------------------------------------------------------------- genus 1 1 2 2 3 3 4 4 6 6 8 8 10 10 . . . To get a highly symmetric *metric* surface of constant curvature, this already works for the torus using a square or regular hexagon in the plane. But for surfaces of higher genus, we need to start with the regular 2n-gon of constant negative curvature (say = -1) whose vertex angles are each pi/n if n is even, and 2pi/n if n is odd (with geodesic edges). These are unique up to isometry. ))) --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] sqrt(x*x+y*y)+3*cos(sqrt(x*x+y*y))+5
by Ray Tayek 30 Mar '12

30 Mar '12

https://www.google.com/search?btnG=1&pws=0&q=sqrt%28x*x%2By*y%29%2B3*cos%28… --- co-chair http://ocjug.org/

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[math-fun] Bessel at High Precision?
by quad 30 Mar '12

30 Mar '12

Hey all. Does anyone have any ideas on computing Bessel functions to high precision (over the complex plane)? Googling gives [1] and [2]. Sorry, [1] is behind a paywall. I am wondering if one of you has a secret magical sum or product that just magically works well for high precision computation. If not, maybe we can develop some strategies! Thanks, Robert Smith [1] dl.acm.org/citation.cfm?id=1466254 [2] web.cs.dal.ca/~jborwein/bessel.pdf

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[math-fun] Topology puzzle
by Dan Asimov 30 Mar '12

30 Mar '12

In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2. --Dan ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] Topology puzzle
by Dan Asimov 30 Mar '12

30 Mar '12

So far, two messages (one directed only to me) have questioned whether this puzzle actually has a valid solution. It does. --Dan << In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2. >> ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] Quaternion meta-commutation
by Fred lunnon 29 Mar '12

29 Mar '12

On 3/29/12, Henry Baker <hbaker1(a)pipeline.com> wrote: > Isn't Q' P' always equivalent to (P Q)' ?? Am I missing something ?? Notation malfunction! It's an understandable misunderstanding, so my reply is addressed to the list. P' does not denote conjugate of P here; it is simply another quaternion with the same norm. In the case that some given Lipschitzian X is primitive, its factorisation is unique modulo unit migration, once the order in which the norms of its prime factors occur is specified. So there are essentially just two factorisations of X = P Q = Q' P'; however, their components (as the examples illustrate) may fail to correspond in any obvious fashion. Incidentally, my conjecture about the closure of factorisations under meta- commutation turns out (inconsiderately) to fail for p = 3; so the proviso "odd (prime) norms" requires amendment to "prime norms > 4". Fred Lunnon > At 06:42 AM 3/28/2012, you wrote: >> Just in case anyone on math-fun is interested, I'm forwarding my response >>to an enquiry from Derek Smith :--- >> >> I've been doing some thinking about the matter raised in Conway & Smith >>(2003) sect.5.4; for what it's worth, my conclusions are the following. >> >> At first sight it might appear that the transformation from X = P Q to >>X = Q' P', with P,P' primes of norm p and Q,Q' primes of norm q, >>might prove amenable to some simple combinatorial manipulation of >> components. >> >> For example if X has i- and j-components equal, then (modulo unit >>migration) set P',Q' equal to P,Q with i- and j-components >> interchanged. >>Again, if p = q, then interchange the factors with P',Q' equal to Q,P >>so that the factorisation (unique modulo unit migration) remains unaltered. >> >> However consider a more general example: X has norm 323 = 17*19. >>Meta-commutation may change the component bag of neither factor: >> [3,2,2,0]*[1,3,3,0] = [1,3,3,0]*[3,2,2,0] = [-9,11,11,0] (commutative); >> of both: [3,-2,0,2]*[3,3,-1,0] = [4,1,1,1]*[4,0,0,1] = [15,5,3,8]; >> of norm-17 only: [3,2,2,0]*[3,3,1,0] = [0,3,3,-1]*[4,0,0,1] = >> [1,15,9,-4]; >> of norm-19 only: [3,-2,0,-2]*[0,1,3,3] = [4,1,1,1]*[3,2,2,0] = >> [8,9,13,3]; >>above X = s + p i + q j + r k is denoted [s,p,q,r] . >> >> Further numerical exploration soon suggests this conjecture: >>Given primes P,Q with unequal odd norms p,q, denote by S,T >>closure sets under inclusion of Y' in S and Z' in T, where >>Z' Y' = Y Z under meta-commutation, for Y,Z already in S,T; >>initially S = {P}, T = {Q}. Then (modulo unit migration) S,T >>finally comprehend _all_ Lipschitz integers with norm p,q resp. >>[By Jacobi's theorem, these sets have sizes p+1, q+1 resp.] >> >> It follows that any general meta-commutation algorithm may be expected >>to incorporate a mechanism enumerating representations of a rational >>integer as a sum of four squares, rendering any component-diddling >>solution highly improbable. >> >> The following refactorisation algorithm is almost certainly essentially >>optimal in general: set Q' = GCD_L(q, X) and P' = GCD_R(p, X), where >>GCD denotes the Euclidean algorithm applied appropriately, on left or >> right, >>to Lipschitz integers with at least one norm odd. >> >>Fred Lunnon > >

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[math-fun] Quaternion meta-commutation
by Fred lunnon 28 Mar '12

28 Mar '12

Just in case anyone on math-fun is interested, I'm forwarding my response to an enquiry from Derek Smith :--- I've been doing some thinking about the matter raised in Conway & Smith (2003) sect.5.4; for what it's worth, my conclusions are the following. At first sight it might appear that the transformation from X = P Q to X = Q' P', with P,P' primes of norm p and Q,Q' primes of norm q, might prove amenable to some simple combinatorial manipulation of components. For example if X has i- and j-components equal, then (modulo unit migration) set P',Q' equal to P,Q with i- and j-components interchanged. Again, if p = q, then interchange the factors with P',Q' equal to Q,P so that the factorisation (unique modulo unit migration) remains unaltered. However consider a more general example: X has norm 323 = 17*19. Meta-commutation may change the component bag of neither factor: [3,2,2,0]*[1,3,3,0] = [1,3,3,0]*[3,2,2,0] = [-9,11,11,0] (commutative); of both: [3,-2,0,2]*[3,3,-1,0] = [4,1,1,1]*[4,0,0,1] = [15,5,3,8]; of norm-17 only: [3,2,2,0]*[3,3,1,0] = [0,3,3,-1]*[4,0,0,1] = [1,15,9,-4]; of norm-19 only: [3,-2,0,-2]*[0,1,3,3] = [4,1,1,1]*[3,2,2,0] = [8,9,13,3]; above X = s + p i + q j + r k is denoted [s,p,q,r] . Further numerical exploration soon suggests this conjecture: Given primes P,Q with unequal odd norms p,q, denote by S,T closure sets under inclusion of Y' in S and Z' in T, where Z' Y' = Y Z under meta-commutation, for Y,Z already in S,T; initially S = {P}, T = {Q}. Then (modulo unit migration) S,T finally comprehend _all_ Lipschitz integers with norm p,q resp. [By Jacobi's theorem, these sets have sizes p+1, q+1 resp.] It follows that any general meta-commutation algorithm may be expected to incorporate a mechanism enumerating representations of a rational integer as a sum of four squares, rendering any component-diddling solution highly improbable. The following refactorisation algorithm is almost certainly essentially optimal in general: set Q' = GCD_L(q, X) and P' = GCD_R(p, X), where GCD denotes the Euclidean algorithm applied appropriately, on left or right, to Lipschitz integers with at least one norm odd. Fred Lunnon

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[math-fun] pdfs for some sunflowers
by Warren Smith 27 Mar '12

27 Mar '12

I used Speciner's ps code to produce these viewable pdf files http://dl.dropbox.com/u/3507527/Sunflower2colorRt2m1.pdf http://dl.dropbox.com/u/3507527/Sunflower2colorPhi.pdf http://dl.dropbox.com/u/3507527/Sunflower1colorPhi.pdf there is for some reason a very freaky mental effect when I view these.

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[math-fun] The sequence that gets densest in the circle the fastest
by Dan Asimov 26 Mar '12

26 Mar '12

A propos the sunflower: Let a circle have unit circumference: C = R/Z, and let tau be (sqrt(5)-1)/2. For any sequence S = {x_n in C | n=1,2,3,...}, define M_n(S) as the length of the largest arc of C in the complement of {x_1,...,x_n}. Claim (?): The sequence in C S_tau := {x_n := n*tau (mod Z) | n=1,2,3,...} is the (unique ?) sequence S whose maximum gap M_n(S), as a function of n, decreases *in some sense* faster than that of any other sequence in C. Of course, there are sequences T for which M_n(T) = 1/n (the minimum possible and for n > 1 always < M_n(S_tau)) for an unbounded set of values of n. SO: Can someone state the claim rigorously and correctly? --Dan ________________________________________________________________________________________ It goes without saying that .

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