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[math-fun] AGM3 (was Re: Iterated averaging of triples)
by Keith F. Lynch 19 Sep '20

19 Sep '20

M F Hasler <mhasler(a)dsi972.fr> wrote: > Brad Klee <bradklee(a)gmail.com> wrote: >> a = (a+b+c)/3 >> b = [(a*b)^(1/2)+(b*c)^(1/2)+(c*d)^(1/2)]/3 >> c = (a*b*c)^(1/3) >> Is this the best generalization of the agm to three (or more) variables? >> Does it even converge? >> I checked that: agm3[1,2,3] = 1.8932121023002878019777... converges > I think the most natural generalisation is to take as "update" for > the k-th variable the k-th root of the k-th symmetric polynomial in > the variables, normalized by the number of terms. > For 3 that would mean > a = (a+b+c)/3 > b = [( a*b + b*c + c*a )/3 ]^(1/2) > c = (a*b*c)^(1/3) > It gives AGM(1,2,3) = 1.9099262335408153... Let's compromise by taking the AGM of Klee's 1.8932121023002878019777 and Hasler's 1.9099262335408153, which is about 1.9015599859028125. :-) How about a fully recursive definition with no visible addition, multiplication, powers, or roots: a = agm(agm(a,(b,c)) b = agm(agm(b,(a,c)) c = agm(agm(c,(a,b)) (As I assume Klee and Hasler did, I'm not replacing any of a,b, and c until all three assignments are done.) With 1,2,3 that converges to 1.9091504422214686... For four terms: a = agm(agm(a,b),agm(c,d)) b = agm(agm(a,c),agm(b,d)) c = agm(agm(a,d),agm(b,c)) then use my above agm3 on those three. (I thought they might immediately be equal, but no.) For 1,2,3,4 that converges to 2.3587709092584177... Is there a natural way to extend this approach to 5 or more terms? Of course it's required that it give the same answer if the terms are permuted.

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Re: [math-fun] Iterated averaging of triples
by Dan Asimov 18 Sep '20

18 Sep '20

Arclength along an ellipse is notorfiously hard to quantify. But if the ellipse is rescaled to be a circle, points at equal angles apart seem easier to deal with. —Dan Tom Karzes wrote: ----- This suggests the following question: For any three distinct, non-collinear points, does their exist an ellipse such that (1) the points all lie on the ellipse and (2) the distances between the points, along the ellipse, are equal? -----

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Re: [math-fun] More iterations of triples
by Dan Asimov 18 Sep '20

18 Sep '20

Of course, a polynomial with the vertices as its set of roots does not contain any information about the original polygon's cyclic order (or more accurately, dihedral order). I'm curious how the limiting ellipses differ for the same set of initial vertices with varying order. A priori, the ellipses might just represent the correlation coefficient of the original set of points. (Specifically, via a level curve of the best-approximating 2-dimensional normal distribution of having the initial vertices as sample points.) —Dan Henry Baker wrote: ----- 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. ... ... -----

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[math-fun] Re Iterated averaging of triples
by Neil Sloane 18 Sep '20

18 Sep '20

This reminds me of a paper Elwyn Berlekamp wrote in (I think) the American Mathematical Monthly a long time ago.

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[math-fun] Iterated averaging of triples
by Henry Baker 18 Sep '20

18 Sep '20

Suppose we have a set of 3 distinct numbers, {a,b,c}. Then averaging them in pairs preserves distinctness: {(a+b)/2,(b+c)/2,(c+a)/2}. Proof: assume not, e.g., (a+b)/2=(b+c)/2 => a=c. Contradiction. QED. So we can go backwards: {a,b,c} => {a+b-c,b+c-a,c+a-b} So, {(a+b)/2,(b+c)/2,(c+a)/2} => {(a+b+b+c-c-a)/2=b,(b+c+c+a-a-b)/2=c,(c+a+a+b-b-c)/2=a} Note also that 'averaging' preserves the sum of the triple (hence its average): (a+b)/2+(b+c)/2+(c+a)/2=a+b+c. Thus, 'going forward' and 'going backwards' preserves this sum. OK, so now let's iterate this averaging process N times. I contend that this process makes the numbers more and more 'average'; i.e., all three converge towards the same number: (a+b+c)/3. Yet, we can still recover our original 3 numbers by running the process backwards N times. Even better, we can go backwards *further* than our original 3 numbers. Curiously, if we start with {-1,0,1}, we get {-2^k,0,2^k}, for all k. Note now that a,b,c need not be 'numbers'; they could be vectors of numbers, so we can play this game in higher dimensions. E.g., we can choose triples of (x,y) pairs to represent triangles, so does iterating them converge towards equilateral triangles?

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[math-fun] Dandelin spheres, ellipses & projections of circles
by Henry Baker 17 Sep '20

17 Sep '20

Let's consider the version of Dandelin spheres internally tangent to a circular cylinder instead of a cone. The usual ellipse has a semi-minor axis the same as the radius of the cylinder. Is there a version where an ellipse is created whose semi-MAJOR axis the same as the radius of the cylinder? I'm trying to understand the meaning of the foci of the ellipse formed by the projection of a arbitrarily oriented circle in 3D orthographically projected onto the XY plane.

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Re: [math-fun] Tammes problem for the square torus (Dan Asimov)
by Steve Witham 13 Sep '20

13 Sep '20

Until yesterday I looked upon the terms "lim sup" and "lim inf" with suspicion and had no use for the concepts, but yesterday I had a need, so when I looked it up it made sense. With the phi-based method of placing points on a square torus, as N, the number of points increases, there is a lim sup *and* a lim inf of the ratio (minimum distance between lattice points) / (1/sqrt(N)) (the "figure of merit" to maximize). For reference, with a square lattice the ratio is 1. With an equilateral triangle lattice, it's sqrt(2/sqrt(3)), about 1.07457 -- that has to be the best case of any method. The best ratios of the phi method do gradually approach-- that is, their limit superior is-- 1.07457. You might think this method gets more reliable with more points, but no, there's also a limit inferior! It's about 0.8137. As I try larger N's (up to about 2 million now), the record worst case keeps approaching that, so phi is no panacea. I could look at one of the worst solutions and find the dimensions of the cells, but I don't see right away how to find the exact pessimal solution. I haven't found a match to 0.8137 among expressions of the form sqrt(a/sqrt(b)). > From: Yoav Kallus <colonelmustard(a)gmail.com> > Date: 9/11/20, 5:01 PM > > I asked a question similar to Steve's "peppering problem" for the circle on > MathOverflow a few years ago. My assumption was also that k phi (mod 1), k > = 1, 2, ... would be the way to maximize lim inf [N d(N)], where d(N) is > the minimum separation between the first N points. This turned out to be > wrong, and you can do substantially better (0.721 vs 0.618 for the golden > ratio process). > > https://mathoverflow.net/questions/275158/sequential-addition-of-points-on-… Awesome! Thank you Colonel Mustard, you are definitely in the study. --Steve

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Re: [math-fun] Tammes problem for the square torus
by Dan Asimov 11 Sep '20

11 Sep '20

A related interesting question is the same question except for the hexagonal torus T^2_H = C / Z[w] where w = exp(2πi/3). The geometry of this torus is more compatible with triangular arrangements. The lattices Z^2 and Z[w] are the two maximally symmetric lattices in R^2, (The maximally symmetric lattices in R^3 are L = Z^3, L = fcc, L = bcc. So the related torus questions are naturally for (T^3)_L = R^3/L for one of these lattices L.) —Dan Steve Witham wrote: ----- > From: Allan Wechsler <acwacw(a)gmail.com> > Date: 9/11/20, 4:05 PM > > I would think that for big enough N = a^2 + b^2, the moderately > obvious "Pythagorean" solution would *not* be optimal, and that you > could do better with big patches of points arranged in a triangular > lattice. With this nearly-golden-ratio method, for N < 23, the nearly-square lattices hold their own matching 1/sqrt(n). As N increases, there appear patches of packings where the min distance is > 1/sqrt(N). These do tend to have nearly-equilateral triangular lattices; most of them are not Pythagorean so far. The first one appears at N=23. I only checked up to N=200 (easy to calculate but information overload for me). The best & second best so far are at N=146 and 151. min N distance 1/sqrt(N) ratio sum(sq) lattice type 2 0.707107 0.707107 1.000000 1,1 square 3 0.471405 < 0.577350 0.816497 4 0.500000 0.500000 1.000000 ~triangular 5 0.447214 0.447214 1.000000 1,2 ~square 6 0.372678 < 0.408248 0.912871 7 0.319438 < 0.377964 0.845154 ~triangular 8 0.353553 0.353553 1.000000 2,2 ~triangular 9 0.333333 0.333333 1.000000 10 0.316228 0.316228 1.000000 1,3 ~square 11 0.287480 < 0.301511 0.953463 12 0.250000 < 0.288675 0.866025 13 0.277350 0.277350 1.000000 2,3 ~square 14 0.225877 < 0.267261 0.845154 15 0.240370 < 0.258199 0.930949 16 0.225347 < 0.250000 0.901388 17 0.212091 < 0.242536 0.874475 1,4 18 0.235702 0.235702 1.000000 3,3 19 0.189766 < 0.229416 0.827170 20 0.223607 0.223607 1.000000 2,4 21 0.202031 < 0.218218 0.925820 22 0.203279 < 0.213201 0.953463 (Only those with min d > 1/sqrt(N) shown below.) 23 0.217391 >>> 0.208514 1.042572 ~diamonds 53 0.143694 >>> 0.137361 1.046107 2,7 ~triangular 56 0.135996 >>> 0.133631 1.017700 58 0.134659 >>> 0.131306 1.025536 3,7 61 0.132168 >>> 0.128037 1.032266 5,6 66 0.124943 >>> 0.123091 1.015038 71 0.120338 >>> 0.118678 1.013987 120 0.093169 >>> 0.091287 1.020621 133 0.090538 >>> 0.086711 1.044139 ~diamonds 138 0.087558 >>> 0.085126 1.028577 141 0.085401 >>> 0.084215 1.014085 143 0.084497 >>> 0.083624 1.010435 146 0.087714 >>> 0.082761 1.059853 5,11 ~triangular 151 0.086093 >>> 0.081379 1.057925 ~triangular 154 0.083157 >>> 0.080582 1.031957 156 0.083333 >>> 0.080064 1.040833 159 0.082723 >>> 0.079305 1.043097 161 0.080745 >>> 0.078811 1.024544 162 0.079051 >>> 0.078567 1.006154 9,9 164 0.079502 >>> 0.078087 1.018128 8,10 166 0.078313 >>> 0.077615 1.008996 167 0.079890 >>> 0.077382 1.032409 172 0.076471 >>> 0.076249 1.002903 175 0.077723 >>> 0.075593 1.028175 183 0.074325 >>> 0.073922 1.005450 -----

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Re: [math-fun] Tammes problem for the square torus
by Steve Witham 11 Sep '20

11 Sep '20

> From: Allan Wechsler <acwacw(a)gmail.com> > Date: 9/11/20, 4:05 PM > > I would think that for big enough N = a^2 + b^2, the moderately > obvious "Pythagorean" solution would *not* be optimal, and that you > could do better with big patches of points arranged in a triangular > lattice. With this nearly-golden-ratio method, for N < 23, the nearly-square lattices hold their own matching 1/sqrt(n). As N increases, there appear patches of packings where the min distance is > 1/sqrt(N). These do tend to have nearly-equilateral triangular lattices; most of them are not Pythagorean so far. The first one appears at N=23. I only checked up to N=200 (easy to calculate but information overload for me). The best & second best so far are at N=146 and 151. min N distance 1/sqrt(N) ratio sum(sq) lattice type 2 0.707107 0.707107 1.000000 1,1 square 3 0.471405 < 0.577350 0.816497 4 0.500000 0.500000 1.000000 ~triangular 5 0.447214 0.447214 1.000000 1,2 ~square 6 0.372678 < 0.408248 0.912871 7 0.319438 < 0.377964 0.845154 ~triangular 8 0.353553 0.353553 1.000000 2,2 ~triangular 9 0.333333 0.333333 1.000000 10 0.316228 0.316228 1.000000 1,3 ~square 11 0.287480 < 0.301511 0.953463 12 0.250000 < 0.288675 0.866025 13 0.277350 0.277350 1.000000 2,3 ~square 14 0.225877 < 0.267261 0.845154 15 0.240370 < 0.258199 0.930949 16 0.225347 < 0.250000 0.901388 17 0.212091 < 0.242536 0.874475 1,4 18 0.235702 0.235702 1.000000 3,3 19 0.189766 < 0.229416 0.827170 20 0.223607 0.223607 1.000000 2,4 21 0.202031 < 0.218218 0.925820 22 0.203279 < 0.213201 0.953463 (Only those with min d > 1/sqrt(N) shown below.) 23 0.217391 >>> 0.208514 1.042572 ~diamonds 53 0.143694 >>> 0.137361 1.046107 2,7 ~triangular 56 0.135996 >>> 0.133631 1.017700 58 0.134659 >>> 0.131306 1.025536 3,7 61 0.132168 >>> 0.128037 1.032266 5,6 66 0.124943 >>> 0.123091 1.015038 71 0.120338 >>> 0.118678 1.013987 120 0.093169 >>> 0.091287 1.020621 133 0.090538 >>> 0.086711 1.044139 ~diamonds 138 0.087558 >>> 0.085126 1.028577 141 0.085401 >>> 0.084215 1.014085 143 0.084497 >>> 0.083624 1.010435 146 0.087714 >>> 0.082761 1.059853 5,11 ~triangular 151 0.086093 >>> 0.081379 1.057925 ~triangular 154 0.083157 >>> 0.080582 1.031957 156 0.083333 >>> 0.080064 1.040833 159 0.082723 >>> 0.079305 1.043097 161 0.080745 >>> 0.078811 1.024544 162 0.079051 >>> 0.078567 1.006154 9,9 164 0.079502 >>> 0.078087 1.018128 8,10 166 0.078313 >>> 0.077615 1.008996 167 0.079890 >>> 0.077382 1.032409 172 0.076471 >>> 0.076249 1.002903 175 0.077723 >>> 0.075593 1.028175 183 0.074325 >>> 0.073922 1.005450 --Steve

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Re: [math-fun] math-fun Digest, Vol 211, Issue 15
by Yoav Kallus 11 Sep '20

11 Sep '20

I asked a question similar to Steve's "peppering problem" for the circle on MathOverflow a few years ago. My assumption was also that k phi (mod 1), k = 1, 2, ... would be the way to maximize lim inf [N d(N)], where d(N) is the minimum separation between the first N points. This turned out to be wrong, and you can do substantially better (0.721 vs 0.618 for the golden ratio process). https://mathoverflow.net/questions/275158/sequential-addition-of-points-on-… On the 2-torus or the sphere for that matter I have no idea what to expect. Yoav On Fri, Sep 11, 2020 at 4:08 PM <math-fun-request(a)mailman.xmission.com> wrote: > > ---------- Forwarded message ---------- > From: Steve Witham <sw(a)tiac.net> > To: math-fun(a)mailman.xmission.com > Cc: > Bcc: > Date: Fri, 11 Sep 2020 14:23:10 -0400 > Subject: Re: [math-fun] Tammes problem for the square and cubical torus > [...] > But you *could* "pack" points on a unit interval: > k phi (mod 1) for k = 1 to N. > > ...Why would you want to do that!? you ask. Because you > can keep adding points to the interval incrementally, > and the neighbor distances (mod 1) are always within some > constant factor of 1/N. > > I have searched but not found a similarly simple way to add > points incrementally to the square torus that gets remotely > good packing. (I call this the peppering problem.) > > --Steve > >

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