September 2020 - mathfun@mailman.xmission.com

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

11 Sep '20

Good point. —Dan Allan Wechsler wrote: ----- 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. -----

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

11 Sep '20

I think that for an approximate incremental algorithm, probably Steve's idea is optimal (in some precise sense). (Years ago, I used something like that to get geodesic segments S(L) of length L on the n-torus T^n = R^n/Z^n that were intended to minimize the radius of the largest open ball B(r), r=r(L), in the complement T^n - S(L).) For the exact question, I think that for N = a^2 + b^2 points in T^2, where gcd(a,b) = 1, you can tile T^2 by N squares (usually tilted). Then the vertices of the tiling (or equally the centers of the tiles) form an optimal arrangement. Giving the max-min distance as sqrt(1/N). (I think this is closely related to finite factor rings of the Gaussian integers.) Checking N = 5, N = 10, N = 13, this looks an awful lot like exactly what Steve got for numbers of the form N = a^2 + b^2, gcd(a,b) = 1. I'm not sure about exact solutions on T^3. But another method for approximating the exact answer is to consider the function M : (T^3)^N —> R on arrangements of N points on T^3 (or any-dimensional torus), that gives the max-min distance, and try to climb the gradient. One problem is that this function isn't always differentiable. Another is that, a priori, there might be local maxes that aren't global maxes, though I don't know if this can occur. —Dan Steve Witham wrote: ----- You can get a pretty nice packing on a square torus using the golden ratio. For the unit square, pick one of these spacings (Yes, they are > 1.) h = floor(N phi) / N or h = ceil(N phi) / N, put the points at (k h (mod 1), k / N) for k = 1 to N, and pick the better of the two packings. They are uniform, which makes it < O(N) to measure the closest spacing. > But what would seem to be the best arrangements > for 1 <= N <= 10 on T^2 and T^3 ??? Here are my numbers. The sequence is not monotonic, which means you can sometimes do better by using the next spacing and leaving out a point! Using the best of `floor()` or `ceil()` (on a unit square): N min distance 2 0.7071067811865476 3 0.4714045207910316 4 0.5 5 0.44721359549995765 6 0.3726779962499649 7 0.3194382824999699 8 0.3535533905932738 9 0.3333333333333333 10 0.3162277660168369 11 0.2874797872880344 12 0.25 13 0.27735009811261385 14 0.22587697572631218 I don't know an analogous method for the cube 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 > From: Dan Asimov <dasimov(a)earthlink.net> > Date: 9/10/20, 6:26 PM > > The Tammes problem is the well-known problem to find the arrangement > of N points on the unit sphere S^2 that maximizes the minimum distance > between any pair of them. > > What about the same question for N points on the square torus > > T^2 = R^2/Z^2 > > or the cubical torus > > T^3 = R^3/Z^3 > > ??? -----

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

11 Sep '20

You can get a pretty nice packing on a square torus using the golden ratio. For the unit square, pick one of these spacings (Yes, they are > 1.) h = floor(N phi) / N or h = ceil(N phi) / N, put the points at (k h (mod 1), k / N) for k = 1 to N, and pick the better of the two packings. They are uniform, which makes it < O(N) to measure the closest spacing. > But what would seem to be the best arrangements > for 1 <= N <= 10 on T^2 and T^3 ??? Here are my numbers. The sequence is not monotonic, which means you can sometimes do better by using the next spacing and leaving out a point! Using the best of `floor()` or `ceil()` (on a unit square): N min distance 2 0.7071067811865476 3 0.4714045207910316 4 0.5 5 0.44721359549995765 6 0.3726779962499649 7 0.3194382824999699 8 0.3535533905932738 9 0.3333333333333333 10 0.3162277660168369 11 0.2874797872880344 12 0.25 13 0.27735009811261385 14 0.22587697572631218 I don't know an analogous method for the cube 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 > From: Dan Asimov <dasimov(a)earthlink.net> > Date: 9/10/20, 6:26 PM > > The Tammes problem is the well-known problem to find the arrangement > of N points on the unit sphere S^2 that maximizes the minimum distance > between any pair of them. > > What about the same question for N points on the square torus > > T^2 = R^2/Z^2 > > or the cubical torus > > T^3 = R^3/Z^3 > > ???

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[math-fun] stochastic analysis
by Bernie Cosell 11 Sep '20

11 Sep '20

https://www.theguardian.com/science/2020/sep/10/uk-mathematician-martin-hai… /Bernie\ Bernie Cosell bernie(a)fantasyfarm.com -- Too many people; too few sheep --

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