October 2020 - mathfun@mailman.xmission.com

Re: [math-fun] binary roulette wheels
by Dan Asimov 03 Oct '20

03 Oct '20

Allan and Andy have pointed out the ridiculousness of my original suggestion. But maybe a slight change could make it possible: Could there be a continuous function on the disk F : D^2 —> R such that for every continuous function g : [a, b] —> [0, 1] on an interval, there is a curve C = C_g C : [a, b] —> D^2 such that F(C(t)) = g(t) for all a ≤ t ≤ b ??? —Dan ----- Or even simpler than that, only 1/L constant functions, out of uncountably many, are possible. Andy On Sat, Oct 3, 2020 at 1:29 PM Allan Wechsler <acwacw(a)gmail.com> wrote: > I am feeling strongly that Dan's universal circular function is impossible. > For any L, let N be ceil(1/L). Consider just the family of N+1 functions > f[i](x) = ix, i ranging from 0 to N. It seems obvious to me that R/Z isn't > big enough to fit them all. > > On Sat, Oct 3, 2020 at 12:24 PM Dan Asimov <dasimov(a)earthlink.net> wrote: > > > I wonder if analogous continuous objects can exist. > > > > For instance, can there be a continuous function on the circle R/Z > > > > f : R/Z —> R > > > > and some fixed L > 0, such that f contains all continuous functions > > > > g : [0, L] —> R > > > > as represented on an arc [c, c+L] ⊂ R/Z of the circle, via > > > > g(x) = f(x+c) > > > > (addition modulo 1) ??? > > > > Or something along these lines, maybe for a restricted class of > > functions f and g ? > > -----

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Re: [math-fun] binary roulette wheels
by Dan Asimov 03 Oct '20

03 Oct '20

Jim Propp wrote: ----- I believe there is a sense in which the Riemann zeta function on the critical line exhibits the sort of behavior Dan is looking for. If I recall correctly, for any continuous real-valued function f on any finite interval, and for any positive epsilon, there’s an interval on the critical line such that the zeta function differs from f by less than epsilon everywhere on the interval. ----- Aha, maybe I heard of something like that once, too. Googling led to this article: <https://en.wikipedia.org/wiki/Zeta_function_universality>. It is apparently a 2-dimensional approximation in the critical strip (not line), which I guess allows for more wiggle room. —Dan

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Re: [math-fun] binary roulette wheels
by Dan Asimov 03 Oct '20

03 Oct '20

I wonder if analogous continuous objects can exist. For instance, can there be a continuous function on the circle R/Z f : R/Z —> R and some fixed L > 0, such that f contains all continuous functions g : [0, L] —> R as represented on an arc [c, c+L] ⊂ R/Z of the circle, via g(x) = f(x+c) (addition modulo 1) ??? Or something along these lines, maybe for a restricted class of functions f and g ? —Dan Adam Goucher wrote: ----- de Bruijn sequences Veit Elser wrote: ----- Is there a name for cyclic sequences (necklaces) of length 2^n that contain all the integers 0, … , 2^n-1 expressed in binary in the 2^n subsequences of length n? For example, for n=4 the sequence 0000100110101111 contains 0, 1, 2, 4, 9, 3, 6, 13, 10, 5, 11, 7, 15, 14, 12, 8. ----- -----

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[math-fun] Complete quaternion soln: AX=XB
by Henry Baker 03 Oct '20

03 Oct '20

For the record, here's the complete solution. Let A,B,X be quaternions; A',B',X' are their conjugates. Here's some formulae we may need: S(A) = (A+A')/2 V(A) = (A-A')/2 A = S(A)+V(A) A' = S(A)-V(A) N(A) = AA'=A'A=|A|^2 N(V(A)) = N(A)-S(A)^2 V(A)^2 = -N(V(A)) A is a *unit* quaternion when N(A)=1. A is a 'pure' vector when S(A)=0. A is a 'pure unit vector' when N(A)=1 and S(A)=0. X=0 is the trivial solution of AX=XB. For X to have a nontrivial solution, N(AX)=N(A)N(X)=N(X)N(B)=N(XB) => N(A)=N(B). Also, since S(B) = S(1/X A X) = S(1/X (S(A)+V(A)) X) = S(1/X S(A) X + 1/X V(A) X) = S(S(A) X/X + 1/X V(A) X) = S(S(A) + 1/X V(A) X) = S(S(A)) = S(A) (Because A => (1/X V(A) X) merely rotates a pure vector into another pure vector w/o a scalar part.) So also for X to have a nontrivial solution, S(A)=S(B). Note also that N(A)=S(A)^2+N(V(A))=S(B)^2+N(V(B))=N(B), hence N(V(A))=N(V(B)). Let X = (N(V(A)) - V(A) V(B)) = |V(A)||V(B)| (1 - (V(A)/|V(A)|) (V(B)/|V(B)|)) (Because |V(A)|=|V(B)|, N(V(A))=|V(A)||V(B)|.) So this X is a real multiple of the X we found in the previous email when A,B were pure unit vectors. Indeed, V(A)/|V(A)|, V(B)/|V(B)| ARE pure unit vectors. Checking: AX = XB (S(A)+V(A)) X = X (S(B)+V(B)) S(A) X + V(A) X = S(B) X + X V(B) S(A) X + V(A) X = S(A) X + X V(B) V(A) X = X V(B) V(A) (N(V(A))-V(A)V(B)) = (N(V(A))-V(A)V(B)) V(B) V(A)N(V(A)) - V(A)^2 V(B) = N(V(A))V(B) - V(A)V(B)^2 N(V(A))V(A) - (-N(V(A))) V(B) = N(V(A)) V(B) - V(A) (-N(V(B))) N(V(A)) V(A) + N(V(A)) V(B) = N(V(A)) V(B) + N(V(B)) V(A) N(V(A)) (V(A)+V(B)) = N(V(A)) V(B) + N(V(A)) V(A) N(V(A)) (V(A)+V(B)) = N(V(A)) (V(B)+V(A)) V(A)+V(B) = V(B)+V(A) Check!

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Re: [math-fun] Trivial soln for AX=XB in quaternions
by Henry Baker 03 Oct '20

03 Oct '20

Hi Neil: Yes, a "pure" quaternion A has S(A)=(A+A')/2=0, while a "unit" quaternion A has AA'=N(A)=1 and a pure unit quaternion A has AA=-1. Proof: AA = (AxA-A.A) = 0-(AA') = -N(A) = -1. If AX=XB, then clearly *any* real multiple (including 0) of X will suffice: A(nX)=(nX)B n(AX)=n(XB) AX=XB In particular, n=-1 works, as does n=|X|. So once we find a solution X, then we can have a *unit* (i.e., 'normalize') solution X/|X|. At 07:58 PM 10/1/2020, Neil Bickford wrote: >Hi Henry, > >Just to clarify - in this context, does unit vector quaternion refer to >a quaternion > >xi + yj + zk > >such that > >x^2 + y^2 + z^2 = 1 > >(i.e. a quaternion that is unit and only has a vector part?) > >I haven't been able to find this in the literature yet - my best guess so far >is that because there are an infinite number of solutions (described below), >robotics applications seem to focus on the problem of finding a value of X >that minimizes a metric of error (e.g. least squares) in a system of >equations > >A_1 X = X B_1, >A_2 X = X B_2, >... > >It looks like many applications also consider the case where X, A, B, ... >are >general matrices, which could include translation (e.g. solutions involving >dual quaternions), as well as rotations that aren't odd multiples of pi/2 >(i.e. A, B not necessarily unit vector quaternions). > >For the infinite number of solutions, suppose X is a quaternion that >satisfies AX = XB, where A and B are nonzero quaternions. Then for any real >u, v, (defining the power of a quaternion in an appropriate way), > >Y = A^u X B^v is another solution to AY = YB, > >since > >A (A^u X B^v) B^-1 >= A^u A X B^-1 B^v >= A^u X B^v >= Y > >so AY = YB. > >In other words, if a unit quaternion solution to AX=XB exists, and A and B >are also unit quaternions (not necessarily unit vector quaternions) >(and assuming I haven't made any mistakes!), then there can be up to a >2-dimensional space of unit quaternion solutions. > >Best regards, > >--Neil Bickford

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[math-fun] a simple problem, but ?
by Simon Plouffe 02 Oct '20

02 Oct '20

Here is a simple problem I saw in one math marathon at the University of Orsay for youngsters, it is similar to the PUTNAM. If we have a*b + 7*a*c + 15*c = a*b*c then find all a,b,c that are primes that satisfies the equation. a,b,c are all >=2. By using brute force I can find 3 solution with small primes and 49 others with non-primes, 52 solutions in all. I am stuck with this problem, I do not see how to resolve this without using brute force. Normally, that problem should be feasible with simple means, I don't see how to find the 52 solutions and from that , the 3 triplets of primes. hint, a, b c are relatively small one of the solution is 2, 29, 2 for a, b, c, see below for the complete set (spoiler). here is the original paper : https://www.imo.universite-paris-saclay.fr/marathon/sept20.pdf Best regards, Simon Plouffe 1, 23, 23 1, 24, 12 1, 33, 3 1, 44, 2 2, 15, 30 2, 29, 2 3, 13, 13 3, 14, 7 3, 15, 5 3, 16, 4 3, 18, 3 3, 24, 2 4, 11, 44 5, 11, 11 5, 12, 6 5, 15, 3 5, 20, 2 6, 10, 20 6, 19, 2 8, 9, 72 9, 9, 27 9, 13, 3 10, 9, 18 10, 17, 2 12, 9, 12 12, 11, 4 15, 9, 9 15, 10, 5 15, 12, 3 15, 16, 2 16, 8, 128 17, 8, 68 18, 8, 48 19, 8, 38 20, 8, 32 21, 8, 28 21, 9, 7 23, 8, 23 25, 8, 20 27, 8, 18 30, 8, 16 30, 9, 6 30, 10, 4 30, 15, 2 35, 8, 14 39, 8, 13 45, 8, 12 45, 11, 3 55, 8, 11 75, 8, 10 75, 9, 5 135, 8, 9

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[math-fun] Trivial soln for AX=XB in quaternions
by Henry Baker 02 Oct '20

02 Oct '20

I assume that this solution is known, but I've not seen it before -- e.g., in Wikipedia, etc. I am mystified why not. Find quaternion X s.t. AX=XB, A,B unit vector quaternions. A,B are unit vector quaternions => AA = BB = -1. Consider X=(1-AB). A(1-AB)=(1-AB)B A-AAB = B-ABB A-(-1)B = B-A(-1) A+B = B+A (Check!) -X = AB-1 is also a solution: A(AB-1)=(AB-1)B AAB-A = ABB-B -B-A = -A-B (Check!) Example: A = J, B = (cos(2t) + I sin(2t)) J, X = 1-AB = 2*cos(t)*(cos(t) - I sin(t)). Note that X is actually a square root, as is typical for quaternion rotations. If we want to normalize X, simply divide by 2*cos(t). If X isn't normalized, then denote X' as the conjugate of X, and compute B = X'AX/(X'X); since X'X=|X|^2, we never need to compute |1-AB|. ----- The above is closely related to the following hack: To compute the square root of a complex number z = r*(cos(t)+i*sin(t)): Let sz = |z|+z = r*(1+cos(t)+i*sin(t)). Then sz^2 = 2*(1+cos(t))*r^2*(cos(t)+i*sin(t)) = 2*(1+cos(t))*r*z Thus sqrt(z) = sz/|sz| = sz/sqrt(2*r*(1+cos(t))). Note that we have utilized trig only to show that this hack works; we never have to calculate trig functions to compute the square root. This hack works because the points (0,|z|,z,|z|+z) form a rhombus, so |z|+z has 1/2 the angle of z. Computer graphics people also utilize this hack for quaternion square roots, so how come they don't compute (1-AB) for these rotations, as above?

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Re: [math-fun] Trivial soln for AX=XB in quaternions
by Dan Asimov 02 Oct '20

02 Oct '20

This seems like a very limited question, since if A^2 = -1, B^2 = -1 that means both A and B lie on the sphere S^2 = {a+bi+cj+dk ∊ H | a=0, b^2+c^2+d^2 = 1} of the quaternions H. And what is the general solution, in any case, of AX = XB ??? —Dan Henry Baker wrote: ----- I assume that this solution is known, but I've not seen it before -- e.g., in Wikipedia, etc. I am mystified why not. Find quaternion X s.t. AX=XB, A,B unit vector quaternions. A,B are unit vector quaternions => AA = BB = -1. Consider X=(1-AB). A(1-AB)=(1-AB)B A-AAB = B-ABB A-(-1)B = B-A(-1) A+B = B+A (Check!) -X = AB-1 is also a solution: A(AB-1)=(AB-1)B AAB-A = ABB-B -B-A = -A-B (Check!) Example: A = J, B = (cos(2t) + I sin(2t)) J, X = 1-AB = 2*cos(t)*(cos(t) - I sin(t)). Note that X is actually a square root, as is typical for quaternion rotations. If we want to normalize X, simply divide by 2*cos(t). If X isn't normalized, then denote X' as the conjugate of X, and compute B = X'AX/(X'X); since X'X=|X|^2, we never need to compute |1-AB|. -----

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Re: [math-fun] Three-dimensional flat torus composed of seven cubes each touching the other six?
by Dan Asimov 01 Oct '20

01 Oct '20

Yes, that's exactly (one way that) I solved this, and agrees with what Fred Lunnon wrote — and also Allan Wechsler had another solution that works in Hilbert space. Note that the Voronoi regions* of the integer points in R^3 are unit cubes. The way I came to this originally was from realizing that if we define the "3D plus sign" as the union of the 7 Voronoi cubes Plus = V(0) ∪ V(e_1) ∪ V(-e_1) ∪ V(e_2) ∪ V(-e_2) ∪ V(e_3) ∪ V(-e_3) (where {e_1,e_2,e_3} is the standard basis of R^3), then appropriate translates of Plus will tessellate R^3. One lattice of translations that works is generated by the vectors {(2,-1,0), (0,2,-1), (-1,0,2)}. I think the Voronoi region of this lattice is a (non-uniform) truncated octahedron, but I'm not sure yet. —Dan ————— * For a discrete point set X in R^n, the Voronoi region V(x) of x ∊ X is V(x) = {v ∊ R^n | ‖x-v‖ ≤ ‖v-y‖ for all y ∊ X} — the points of R^n at least as close to x as to any other y ∊ X. Scott Huddleston wrote: ----- Yes. Let L = { (a_1, a_2, a_3) | a_1 + 2a_2 + 3a_3 = 0 mod 7 } This also generalizes to d dimensions, using L = { (a_1, a_2, ..., a_d) | a_1 + 2a_2 + ... + d*a_d = 0 mod 2d+1 } I wonder what the Voronoi regions look like. On Tue, Sep 22, 2020 at 11:01 AM Dan Asimov <dasimov(a)earthlink.net> wrote: > Let L be any discrete subgroup of R^3 isomorphic to Z^3. > > The the quotient R^3/L is a "flat 3-torus", and it carries > a well-defined metric. Its metric can be be described as the > result of identifying opposite faces of a parallelepiped > (the bounded intersection of three slabs* in R^3) in the > obvious way. > > Of course the cubical 3-torus R^3/Z^3 is one example, but > a flat 3-torus can have surprising properties. Note that > the cubical 3-torus can be tessellated by 8 cubes each > touching three of the other seven along two common square > faces. > > > Puzzle: > ------- > Does there exist a flat 3-torus that can be tessellated by > seven cubes each touching the other six along a common face? > > —Dan > > > ————— > * A "slab" in R^3 is the closed region between two parallel planes. -----

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[math-fun] Three-dimensional flat torus composed of seven cubes each touching the other six?
by Dan Asimov 01 Oct '20

01 Oct '20

Let L be any discrete subgroup of R^3 isomorphic to Z^3. The the quotient R^3/L is a "flat 3-torus", and it carries a well-defined metric. Its metric can be be described as the result of identifying opposite faces of a parallelepiped (the bounded intersection of three slabs* in R^3) in the obvious way. Of course the cubical 3-torus R^3/Z^3 is one example, but a flat 3-torus can have surprising properties. Note that the cubical 3-torus can be tessellated by 8 cubes each touching three of the other seven along two common square faces. Puzzle: ------- Does there exist a flat 3-torus that can be tessellated by seven cubes each touching the other six along a common face? —Dan ————— * A "slab" in R^3 is the closed region between two parallel planes.

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