- mathfun@mailman.xmission.com - mailman.xmission.com

Re: [math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

----- Okay, that makes things easier than T an arbitrary rotation around the same axis (it is possible to generalize though). ----- What does the word "axis" mean here? —Dan

3 3

[math-fun] theorem about normal matrices
by Jean Gallier 21 Sep '20

21 Sep '20

Hello everybody, There is a spectral decomposition theorem about real normal matrices (AA^T = A^T A) which says that there is an orthogonal matrix P and a block diagonal matrix D such that A = P^ D P, where the blocks of D are either a scalar (real), or a 2 x 2 matrix of the form a -b b a with b > 0. So indeed this amounts to finding an orthogonal basis and in some pairwise orthogonal planes, A behaves like a rotation composed with a positive scaling. In the special case of an orthogonal matrix, the scalars are +1 or -1 and the 2 x 2 matrices are 2D rotation matrices cos theta -sin theta sin theta cos theta 0 < theta < pi. The corresponding vectors in P give you a plane. The angle theta correspond to the two eigenvalues cos theta + i sin theta and cos theta - i sin theta. If det(A) = 1, then the number of -1 is even, and you can group them together in 2 x 2 matrices that correspond to 2D rotations of angle pi. I don’t know how old the theorem is. It know that it is proven in Gantmaker (Theory of matrices), Berger (Geometry I), and I have a version in my recent book https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/d… <https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/d…> Please email me if you want a pdf. Best, — Jean

1 0

Re: [math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

Fred, thanks for your note. (How are you doing, by the way?) Yes, it would have been better to write 0 ≤ j ≤ n initially, but isn't 1 ≤ j ≤ n+1 redeemed by "(indices are modulo n+1)"? As for the later reference to R^n, isn't that actually R^n because it's the orthogonal complement in R^(N+1) of the line in the (1,1,...,1) direction? —Dan ----- To Dan --- Multiple notational confusion alert! << For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). >> should read 0 <= j <= n ; also elsewhere, eg. O-P only for for n even! Then half-way thru', |R(n+1) mysteriously transmogrifies into |R(n) ... That theorem is not "Chasles' theorem"; however it might well be due to Chasles, or (more likely) Euler. Fred On 9/21/20, Dan Asimov <dasimov(a)earthlink.net> wrote: > I worked this out last fall and found it surprising: > > Let {e_0,...,e_n} be the standard basis of Euclidean n-space R^(n+1) > (e_j has a 1 as its jth coordinate and 0's elsewhere). > > Define the linear mapping T : R^(n+1) —> R^(n+1) as follows > by its effect on basis vectors: > > For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). > > It's easy to check that T is orientation-preserving only for n even, > and that T is always an isometry of R^(n+1). > > It's also clear that the vector (1,1,...,1) of all 1's is a fixed point > of T. This means that its orthogonal complement (1,1,...,1)^⊥, namely > > (1,1,...,1)^⊥ = {(x_0,...,x_n) in R^(n+1) | Sum x_j = 0} > > (which is a copy of R^n), is carried to itself by T. > > A nice theorem states that any rotation T of R^n for n even has n/2 > mutually orthogonal 2-planes that are each taken into themselves > by a rotation by some angle. (I don't know the name of this theorem; > can anyone tell me its name or who first discovered it?) > > So for n even, say n = 2k, our T as defined above, when restricted > to (1,1,...,1)^⊥, is actually a collection of rotations by angles > that we'll call theta_j, 1 ≤ j ≤ n/2. > > Question: What are these angles theta_j ? > > Suggestion: Try to guess before doing a calculation. > > —Dan -----

1 0

[math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

I worked this out last fall and found it surprising: Let {e_0,...,e_n} be the standard basis of Euclidean n-space R^(n+1) (e_j has a 1 as its jth coordinate and 0's elsewhere). Define the linear mapping T : R^(n+1) —> R^(n+1) as follows by its effect on basis vectors: For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). It's easy to check that T is orientation-preserving only for n even, and that T is always an isometry of R^(n+1). It's also clear that the vector (1,1,...,1) of all 1's is a fixed point of T. This means that its orthogonal complement (1,1,...,1)^⊥, namely (1,1,...,1)^⊥ = {(x_0,...,x_n) in R^(n+1) | Sum x_j = 0} (which is a copy of R^n), is carried to itself by T. A nice theorem states that any rotation T of R^n for n even has n/2 mutually orthogonal 2-planes that are each taken into themselves by a rotation by some angle. (I don't know the name of this theorem; can anyone tell me its name or who first discovered it?) So for n even, say n = 2k, our T as defined above, when restricted to (1,1,...,1)^⊥, is actually a collection of rotations by angles that we'll call theta_j, 1 ≤ j ≤ n/2. Question: What are these angles theta_j ? Suggestion: Try to guess before doing a calculation. —Dan

3 2

Re: [math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

----- I don't exactly know what you're asking for, but maybe you are trying to make a jump off Berlekamp's complex transformations by restricting to reals? I think the answer to your questions just falls out of basic representation theory of a finite cyclic group. ----- I thought it was clear I was asking for the angles theta_j. ----- In some choice of coordinates, your T commutes with a cyclic permutation of basis vectors, call that transformation T'. T' is a regular representation of the generator of a cyclic group, so we know its eigenbasis, which is shared with T. ----- The transformation T is *defined* as a cyclic permutation of (orthonormal) basis vectors. I don't know anything about representation theory, and didn't knowingly use it here. —Dan

2 1

[math-fun] Two Teaching Points
by Brad Klee 21 Sep '20

21 Sep '20

No, Eigenvalues[H] is not broken, but buyer beware: https://arxiv.org/pdf/1809.08178.pdf S.M. Blinder's Mathematica solution of the Ammonia problem appears to suffer from "junk in, junk out", despite him having a canonical business address on the title page. One, the references seem to be out of date, and with too old data. Two, why choose a quartic potential? While "proof" checking, I compared quartic and sextic potentials. The sextic worked much better for extracting the barrier height from ground state splitting / first excitation. Generally I am interested in Blinder's demonstrations at: https://demonstrations.wolfram.com/author.html?author=S.+M.+Blinder Unfortunately, his tutorial on Ammonia appears to me, not right enough. Would anyone second this opinion, or is there dissent from my dissent? --Brad

1 0

[math-fun] Torus tiling puzzle
by Dan Asimov 21 Sep '20

21 Sep '20

It's known that a certain torus T^2 (namely, the hexagonal torus obtained by identifying opposite edges of a regular hexagon) can be tiled by regular hexagons so that each hexagon shares an edge with each other one. Puzzle: For which N ≥ 1 does there exist a tiling of some torus (either the square or hexagonal one) by N regular polygons, any two of which share an edge? —Dan

1 0

[math-fun] More iterations of triples
by Henry Baker 19 Sep '20

19 Sep '20

Here's one rationally reversible method to 'equilateralize' a triangle in the complex plane. This method is inspired by Gosper's continued fraction root extraction hack (perhaps in HAKMEM ??). Consider the cubic polynomial p(z) having the triangle vertices as roots. Translate the triangle centroid to the origin, then flip the polynomial end-for-end: z^3*p(1/z). (Of course, we need to remember this sequence of centroid translations to enable reversing.) Rinse & repeat. We are 'hollowing out' the internal coefficients of the polynomial, leaving only the z^3 and the z^0 terms in the limit. The result seems to converge to a 2-cycle of equilateral triangles, where the limit equilateral triangle inverts alternately inside and outside the unit circle. Below are 8 cycles starting from a 9-12-15 triangle; this iteration seems to converge pretty quickly. Q: The final result seems to remember two numbers, a 'size' and a 'phase' from the original triangle. I wonder what these quantities might be? BTW: This method is effectively Gosper's root extraction hack, because we can solve the limit equation z^3-C by inspection, and then reverse all of our steps. Note that unlike Gosper, we have produced *all three* of the roots of the original cubic equation. 3 2 2 (%o2) z - 9.0 %i z - 12.0 z + 108.0 %i z (%i3) forward(%); 3 2 (%o3) z + (0.12733403582719 - 0.045711275026344 %i) z - 0.002086406743941 %i - 0.0024868282402529 (%i4) forward(%); 3 2 (%o4) z + (0.23777047608189 - 1.844515505700051 %i) z + 208.5099028352554 %i - 222.3534715094485 (%i5) forward(%); 3 2 (%o5) z + (- 0.0031990745810353 %i - 0.0020077204240425) z - 0.0022424181810333 %i - 0.002388162593124 (%i6) forward(%); 3 2 (%o6) z + (0.0013849128634252 %i + 4.345783781893431E-4) z + 208.9519514637108 %i - 222.5322250905755 (%i7) forward(%); 3 2 (%o7) z + (- 3.3426031773526269E-10 %i - 2.2762130066155846E-9) z - 0.0022424186214955 %i - 0.0023881586265698 (%i8) forward(%); 3 2 (%o8) z + (4.8202318997299898E-16 - 2.4021268196176432E-16 %i) z + 208.9519514635722 %i - 222.5322250907546 (%i9) forward(%); 3 2 (%o9) z + (3.1212293864605986E-34 - 5.3805053878963621E-35 %i) z - 0.0022424186214955 %i - 0.0023881586265698 (%i10) forward(%); 3 2 (%o10) z + (4.6722777796490018E-66 - 9.0752194922195901E-66 %i) z + 208.9519514635722 %i - 222.5322250907546

1 1

Re: [math-fun] Geometric mean of values of sin(x) between x=0 and x=π
by Dan Asimov 19 Sep '20

19 Sep '20

Very nice! Even if there may exist a more intuitive proof, this is very short and very clear, and it avoids actually doing any nasty definite integration. —Dan ----- On 19/09/2020 22:19, Dan Asimov wrote: > Wolfram alpha tells me that to 100 digits, the geometric mean > of the values of sin(x) for 0 ≤ x ≤ π is exactly 1/2. > > Can someone prove this? Even better, is there an intuitive > reason that it ought to be true? The following things are clearly equal; call them all M. - GM of sin x on (0,π). - GM of sin x on (0,π/2). - GM of cos x on (0,π/2). Second times third equals M^2 but also equals GM of 1/2 sin 2x on (0,π/2) equals GM of 1/2 sin x on (0,π) equals M/2. So M^2 = M/2 and so (modulo checking that M isn't zero) M = 1/2. This is a very short proof but I'm not sure it counts as an intuitive reason. Maybe there's some geometrical way to think about sin 2x = 2 sin x cos x that would bridge the gap? -----

1 0

[math-fun] Geometric mean of values of sin(x) between x=0 and x=π
by Dan Asimov 19 Sep '20

19 Sep '20

Wolfram alpha tells me that to 100 digits, the geometric mean of the values of sin(x) for 0 ≤ x ≤ π is exactly 1/2. Can someone prove this? Even better, is there an intuitive reason that it ought to be true? —Dan

2 1