March 2012 - mathfun@mailman.xmission.com

[math-fun] Wikipedia glitch?
by Fred lunnon 26 Mar '12

26 Mar '12

Folks, it's hardly maths and it certainly isn't fun, I know; but ... I've come across a curious bug at http://en.wikipedia.org/wiki/Lagrange's_four-square_theorem in the section Proof using the Hurwitz integers where (under my browsers) one displayed formula terminates in the proud claim " (1/4)^2 + (1/2)^2 + (1/2)^2 + (1/2)^2 = 15/16 < 1 . " Why the fuss --- surely all I need to do is click the edit button to call up the source text, and correct the TeX. The latter certainly looks to be current --- it shows the same typo "consequense" (sic) on the following line --- but there the corresponding formula already reads correctly " ... =\frac{13}{16}<1. " Can somebody please check that this is not some individual glitch affecting only my ancient Mac? I have attempted to access the relevant bug search & report pages, but rapidly concluded that negotiating that particular jungle would prove a task for an explorer more heroically committed than myself ... It is a somewhat alarming prospect that numerical constants on Wikipedia pages may be subject to elementary errors which are impossible to correct! Fred Lunnon

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[math-fun] Asperger syndrome, Autism, & math talent
by Warren Smith 26 Mar '12

26 Mar '12

Supposedly, Asperger syndrome (which allegedly no longer "exists," having been merged into the "Autism spectrum"), is linked to "mathematical talent." paper Ioan James: Autism and Mathematical Talent, online here http://www.scribd.com/doc/48195975/autism-and-math-talent Mathematicians and physicists identified by him or by others as having Asperger or Autism or signs thereof include A.Einstein, I.Newton, P.A.M.Dirac, G.Perelman, S.Ramanujan, P.Erdos, Ronald Fisher, GH Hardy, Alan Turing, A Weil, Norbert Weiner, Robert Ammann (a post office clerk who made big contributions to tiling theory), WR Hamilton, and R. Borcherds. -- Warren D. Smith http://RangeVoting.org

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[math-fun] symmetrized or quaternionized sunflowers
by Warren Smith 26 Mar '12

26 Mar '12

The sunflower as originally defined has no symmetry. But if we take the set-union of it with the same thing rotated by j/k of a turn, for each j=1,2,...,k-1, then the resulting point set is (a) k-fold symmetric (b) still obeys my lattice approx theorem (c) still obeys theorems about decrease rate of max gaps and min gaps [if sunflower based on a quadratic irrational angle] and hence "no holes" and "no high density" theorems. Surely some plants like to have symmetry, no? -- Re 4D etc sunflower ideas with quaternions... the combing of 3 kinds of hairs on the surface x^2+y^2+z^2+t^2=1 arises as follows: view the 4 coordinates as a quaternion q with |q|=1, and multiply point q by i, j, and k (on left) to obtain the 3 direction arrows (tangent to the sphere at q), and mutually perpendicular, that are the three colors of "hair." So I suppose some sort of 4D sunflower analogue, point set parameterized by 3 integer parameters a,b,c, would be exp(i*a*g1+j*b*g2+k*c*g3) * (a^2+b^2+c^2)^(3/8) using quaternion exponentiation where g1,g2,g3 are three suitable quadratic irrationals. More generally still the argument of the exp() would be g1*a+g2*b+g3*c where now g1,g2,g3 are suitable irrational pure-imaginary quaternion constants such exp()=1 is ONLY possible if a=b=c=0. One can similarly make an 8D sunflower analogue using octonions.

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[math-fun] Four-dimensional sunflowers?
by Adam P. Goucher 26 Mar '12

26 Mar '12

It is convenient to consider sunflowers on the Argand plane. The two enantiomorphs of the normal sunflower are represented by: z_n = n^(1/2) * exp(2*pi*i*n/phi) and z_n = n^(1/2) * exp(-2*pi*i*n/phi) This suggests the possibility of a generalisation to Hamilton's quaternions. Indeed, we get a set of SO(3) sunflowers, by replacing 'i' with any of the square-roots of -1 in the above formula. They are all congruent to the normal sunflower, and therefore all two-dimensional. What a shame. However, this idea is not completely disastrous. If we use 2 parameters, instead of just one, we can create something like this: z_(l,m) = (l+m)^(1/2) * exp((2*pi/phi) * (i*l + j*m)) If we define exp(w) = 1 + w + w^2/2! + w^3/3! + w^4/4! + ..., then the non-commutativity of multiplication should pose no problems. I have no idea what this beast looks like, though. Sincerely, Adam P. Goucher

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[math-fun] sunflowers have no holes (answering Propp)
by Warren Smith 26 Mar '12

26 Mar '12

sure, one can prove sunflowers contain no large vacant disks because of theorems already mentioned/cited (Ravenstein cited/mentioned) on Weyl sequence max gaps always being small. Similarly, sunflowers contain no disks with point density greater than some constant... at least for those based on quadratic-irrational angles and for disks containing at least 2 points.

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[math-fun] sunflowers in 3D impossible?
by Warren Smith 25 Mar '12

25 Mar '12

>The 2D sunflower constitutes an expanding *ball* (disk) in R^2, not a circle. > >So the 3D analogy would be an expanding ball in 3D. There's no reason that >a sunflower-like arrangement can't be improvised on an expanding ball in 3D. > >For example, the spherical component could be defined by saying that >any 3 consecutive points s_n, s_(n+1), s_(n+2) of the 3D sunflower >sequence satisfy that their unit vectors > > s_n/||s_n||, s_(n+1)/||s_(n+1), s_(n+2)/s_(n+2) > >are congruent as a triple on the unit sphere, for all n. >While at the same time their lengths ||s_n|| slowly grow larger >exponentially, say ||s_n|| := e^(n*(1+eps)), for very small eps > 0. >Dan Asimov --My impossibility argument (such as it was -- as I said it is not as satisfactory as one would want) focused on the spherical boundary of the expanding large ball. Each point on said boundary feels locally (where "locally" means within a distance large enough compared to 1 but small enough compared to radius of ball) that it is a member of a periodic "finite set of lattices" point arrangement, to within epsilon [note this is NOT true of the whole sunflower ball, only little pieces of it near boundary], and all boundary points "feel the same" in some statistical sense. All these feelings, for all boundary points at once, are mutually incompatible with each other, I tried to argue. I think we should also demand that an N-point sunflower in D dimensions have radius of order N^(1/D), min point separation of order 1, and a simple formula for the point locations. Your (Asimov's) 3D proposal seems to fail the radius desideratum. The existence of a "spiral" in 2D which "covers the plane at distance=1," is something that seems pretty unique to 2D. One can in 3D make curves or surfaces that do it, but they unlike the 2D spiral can make no claim to being "nearly rotationally symmetric." What is this vague "near rotationally symmetric" notion, you ask. Well, one possible answer could perhaps be that if a 2D spiral is scaled and rotated, you get the spiral back again (or something very closely approximating it) if you look only within a narrow-width annulus. The scaling factor has to be chosen as a continuous function of the rotation angle which is 1 for small angles. So a spiral is not continuum-rotationally symmetric, but it can be regarded as such to a high degree of approximation if "rescaling is used to correct matters" and if you only look within a narrow width annulus. But I do not think this is true of any 1D curve or 2D surface that tries to cover the whole of 3D. So I'm putting forward here a vague idea that the spiral as so-characterized, is a uniquely 2D notion. It would be better to formalize all this non-vaguely... -- Warren D. Smith http://RangeVoting.org

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Re: [math-fun] math-fun Digest, Vol 109, Issue 36
by Dan Asimov 24 Mar '12

24 Mar '12

Right, if surface means the d-dimensional sphere S^d. ((( A famous problem was how many differently colored hairs can be orthogonal at each point of S^d, if each color of hair is continuous and of positive length over all of S^d. The curious answer (1962) is that if 2^k is the largest power of 2 dividing d+1, then the maximum number of colors is max(d) = 2^(k mod 4) + 8*floor(k/4) - 1. ))) --Dan << --yah, but really, we want d different color hairs, and the green hairs are combed, the red hairs are combed in a direction always orthogonal to the green hairs etc. For that more powerful kind of combing, you can do it for a surface that is 1,3, or 7 dimensional using complex, quaternion, octonion tricks in the d+1 dimensional space... >> ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] math-fun Digest, Vol 109, Issue 36
by Warren Smith 24 Mar '12

24 Mar '12

> Actually, it's true that combings are possible > for all odd-dimensional spheres. (Or in the language used > below, for a sphere in any even dimension: > > S^(2n-1) := {x in R^(2n) | ||x|| = 1}. > > That can be readily seen because the above definition is equivalent > to > > S^(2n-1) := {z in C^n | ||z|| = 1} > > and so there is a free SO(2) action on S^(2n-1) via > > e^it z := (e^it * z_1, e^it * z_2,...,e^it * z_n), > > which means that S^(2n-1) is fibred by consistently oriented > circles -- hence there is a "combing". > > --Dan --yah, but really, we want d different color hairs, and the green hairs are combed, the red hairs are combed in a direction always orthogonal to the green hairs etc. For that more powerful kind of combing, you can do it for a surface that is 1,3, or 7 dimensional using complex, quaternion, octonion tricks in the d+1 dimensional space...

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Re: [math-fun] big sunflowers
by Henry Baker 24 Mar '12

24 Mar '12

I saw a recent TED Talk about Phyllotaxis, but when I Google searched for it, I couldn't find it. According to this talk, the Fibonacci sequence comes quite naturally in phyllotaxis, because when a branch on a plant splits off, Kirchoff's current law says that the water flow in the branches equals the water flow in the trunk. Since branches tend to split off one at a time, you get flow_{i-2}=flow_{i-1}+flow_{i}, etc. The sizes of the branches is organized to support this amount of flow. At 09:24 PM 3/23/2012, Stuart Anderson wrote: >A reference for some of that recent (and not so recent work - von Iterson) >http://www.scribd.com/doc/39484505/Phyllotaxis > >> with real sunflowers evidently one needs to understand the dynamics of >> plant buds; 'primordia' which has been the subject of much recent research

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Re: [math-fun] 3D sunflowers are impossible?
by Dan Asimov 24 Mar '12

24 Mar '12

The 2D sunflower constitutes an expanding *ball* (disk) in R^2, not a circle. So the 3D analogy would be an expanding ball in 3D. There's no reason that a sunflower-like arrangement can't be improvised on an expanding ball in 3D. For example, the spherical component could be defined by saying that any 3 consecutive points s_n, s_(n+1), s_(n+2) of the 3D sunflower sequence satisfy that their unit vectors s_n/||s_n||, s_(n+1)/||s_(n+1), s_(n+2)/s_(n+2) are congruent as a triple on the unit sphere, for all n. While at the same time their lengths ||s_n|| slowly grow larger exponentially, say ||s_n|| := e^(n*(1+eps)), for very small eps > 0. --Dan << For a lattice, there are 3 distinguished basis directions, call them the red, blue and green directions. Were these directions to vary smoothly as we moved around on the surface of a big sphere, then either we would violate the hairy ball theorem, or we would have to have some points on the big sphere at which the red lattice direction was "radially outward." >> ________________________________________________________________________________________ It goes without saying that .

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