April 2016 - mathfun@mailman.xmission.com

[math-fun] Distributive law without +? (6th Grade)
by Warren D Smith 27 Apr '16

27 Apr '16

> You can pick a or c: 6*(9*1) or 6*(9+0) why not b too? 6*(7+2)=6*(2+7) Not that I know much about educating children, but I suspect that demanding detailed educational standards about content is a waste of time. More effective might be to demand that the teachers actually know their topic. E.g. I had an elementary school teacher try to teach us that a/b + c/d = (a+c)/(b+d). Some parent, perhaps mine, later stepped in to try to correct her. So, maybe: don't test the children, test the teachers. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Prince Rupert's problem, cubes in cubes, Kay R.P.DV. Schulz
by Warren D Smith 27 Apr '16

27 Apr '16

I stumbled upon https://www.kitp.ucsb.edu/sites/default/files/preprints/2013/13-142.pdf which is a 123-page manuscript / note-collection by Kay R. Pechenick DeVicci Shultz, partly typed and partly handwritten, mostly dating to 1996 or before. It seems she knew most everything I knew about the cube problem and more. Here is a summary. She defines f(m,n) = the edgelength of the largest m-dimensional cube (or m-cube) which fits into an n-cube of unit side. f(m,n) = MAX_Q 1 / MAX(j=1..n) SUM(i=1..m) |Q_{i j}| where Q is an nXn orthogonal matrix, for the Rupert problem m=n-1, and the outermost max is over all such orthogonal matrices Q. I'd been calling this "Huber's magic formula" thinking it was due to Greg Huber, but actually it seems K.R.P.DV.S. knew it earlier. She then gives f(m,n)<f(m,n+1) because: add column of 0s f(m,n)<f(m-1,n) because: delete row f(k,m)*f(m,n)<=f(k,n) because: put k-cube in m-cube and m-cube in n-cube f(m,m+1)>1 (which implies the first two inequalities can use < not merely <=) because: she considers infinitesimal rotations and solves a linear program thus proving existence of an infinitesimal rotation that does better than the identity matrix f(k*m,k*n)>=f(m,n) because: kronecker matrix product f(m1+m2+...+mk, n1+n2+...+nk) >= min( f(m1,n1), f(m2,n2), ... f(mk,nk) ) f(m,n) <= sqrt(n/m) because of: diameter She claims as a major result: f(m,n) = sqrt(n/m) if and only if m divides n. Any orthogonal matrix Q accomplishing this contains only +1, -1, and 0 as entries, each row containing exactly n/m nonzero entries, and her example arises from a Kronecker product of mXm identity matrices with an all-1's vector with (n/m) entries. Finally, she claims in the handwritten part of the manuscript via an immense case analysis to have found the exact optimum to establish the value of f(3,4). She notes on p.54 that many authors assume without proof that the problem of finding f(n-1,n) and Prince Rupert's problem are equivalent, but she is herself carefully does not claim that, and I agree with her caution. But... The only reason the two might fail to be equivalent is that perhaps the optimum Rupert way of passing an n-cube A thru another B, happens NEVER during the motion, to have an (n-1)-cube cross-section of A, which lies entirely inside B. But if so, then there must exist some "tilted" cross-section of A which does lie entirely inside B, and this tilted thing is topologically the same as an (n-1)-cube, and indeed a linear transformation of it which as a map is necessarily expansive or length-preserving for every point-pair. Therefore, it seems to me, this tilted cross-section must include as a SUBSET, a set geometrically congruent to the untilted cross-section. If so, that proves the two problems really ARE equivalent -- any Rupert-motion must involve a moment when an (n-1)-cube cross-section of A lies inside B; and conversely the largest such inscribing yields a Rupert-motion. Meanwhile... I have been working on the Prince Rupert problem but not for cubes, rather for other regular polytopes, especially simplexes and octahedra. My current claim about that (which unfortunately differs from what I used to think) is the n-octahedral Rupert problem actually has the SAME expansion factor as the n-cube Rupert problem (!). For the n-simplex Rupert problem I have a formula involving orthogonal matrix, and my computer by applying that formula to random orthogonal matrices has rediscovered the known results for n=2 and n=3. (Also for the cube problem, for n=2, 3, and 4.) But it so far has failed to find a simplex solution for n=4, which failure I find surprising since I used to think it was clear a solution existed for all n. I think Kay R.P.DV.Schultz's idea about linear program and infinitesimal rotations is an excellent insight which could be applied to any other polytope besides n-cube to determine whether any slight rotation exists that would work. I.e. it will rapidly decide Rupertability for any convex polytope, provided the Rupert is achievable via an infinitesimally tiny rotation angle. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] The mathematician’s 90th-birthday party
by Bill Gosper 27 Apr '16

27 Apr '16

On 2016-04-25 07:59, Dan Asimov wrote: > http://www.nytimes.com/2016/04/25/opinion/the-mathematicians-90th-birthday-… > —Dan When Neil Bickford was barely 12, I admonished him not to get too comfortable in the child prodigy role, since most of the great mathematicians burnt out at 19 and spent the rest of their lives mining their old notebooks. "Oh, no! Paul Erdös was proving theorems in his 80s!" (Had his parents even heard of Paul Erdös?) "Paul Erdös is a myth we tell children, like Santa Claus and the Easter Bunny, to keep them from becoming discouraged. What actually happens at 19 is your nose fills up with itchy hairs and you can no longer think clearly." --rwg Apropos the "right" answer: at MIT, many of my fellow freshmen were from New York State, and were traumatized by losing credit for this question on the Regents Exam: As you approach the source of a sound, the *pitch* a) increases b) decreases c) stays the same The answer was c. They were testing whether you knew the difference between pitch and intensity.

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Re: [math-fun] IQ
by Warren D Smith 26 Apr '16

26 Apr '16

> As for IQ, I dismiss the whole concept.? IQ is the score on an IQ test, and > nothing more. > > ? --? Gene --there is much to be said for that stance, but ultimately it is that of a proverbial ostrich burying head in sand. The nature of human intellect is important, and therefore should be studied, and therefore one has to try to do it right and infer whatever conclusions one can from the evidence. Unfortunately, I looked into that area quite a lot once (related to me discovering "mathematical definition of intelligence" which turned out to be essentially a rediscovery of same by Marcus Hutter a few years earlier) and it was rather blatantly obvious that this entire subfield of psychology had been done AWESOMELY badly by a bunch of incredibly poor researchers, who nevertheless are lauded as "greats." And there also are other things about human intellect worth researching too, such as memory; and again, the amount of ultra-poor-quality work about that has to be seen to be believed. Which all is a pity. I have also had some experience with other parts of psychology/psychiatry and maybe it is just bad luck, but every time I look into some subarea of their literature, the experience just blows my mind at how incredibly awful they have been and how many thousands of people have been hurt as a consequence.

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Re: [math-fun] Distributive law without +? (6th Grade)
by Hilarie Orman 26 Apr '16

26 Apr '16

You can pick a or c: 6*(9*1) or 6*(9+0) I mean, what do you tell a 6th grader who is struggling with order of operations and by carefully following the rules, comes down to 6(9) ? Did you previously prove "parens around singletons can be thrown away"? How? Common Core the standard says that students should understand and be able to use the distributive law. The textbook implementing Common Core may be idiotic, but the same idiocy existed before Common Core. I don't think anyone knows very much about the psychology of math. There's no easy pathway to the circuits and programming of it. This makes it difficult to teach, in general. So much can go wrong if the student gets off track. Maybe square brackets are a useful crutch for most people, I don't know. Auto close of parens wouldn't have been invented if nesting wasn't so confusing. Hilarie > From: Bill Gosper <billgosper(a)gmail.com> > The simplification 6(9) = 6*9 is an example of (choose one) > a) associativity b) commutativity c) distributivity. > "Answer": c. Has anybody ever seen this usage of "the distributive law"? > This is from the same Common Core idiots who outlaw "improper" fractions. > They also insist that parens enclosing parens be changed to square brackets. > --rwg

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[math-fun] intersections of convex polyhedra, solution of Propp's simul-pierce problem
by Warren D Smith 25 Apr '16

25 Apr '16

If P and Q are convex polyhedra (or nondegenerate convex objects described peicewise by a finite set of algebraic surfaces of bounded degree) in 3D, then I claim that either P intersect boundary(Q) or Q intersect boundary(P) must contain at least one component that is a topological disk. That solves Propp's problem -- showing "P pierces Q" and "Q pierces P" are not simultaneously possible, because the disk will cut either P or Q, as a consequence of whatever the Jordan-curve-theorem analogue is for well-behaved surfaces and sets in 3D. Proof of claim (modulo the fact, which is probably highly apparent, that I am not a topologist): Walk round the boundary of a component of P intersect boundary(Q) . This curve must close. So you get a circle-like curve on a sphere-like-surface. So each component of P intersect boundary(Q) must be a disc with holes in it. If it has no holes, done. If it has holes, then consider one of them as part of Q intersect boundary(P) and then redo the argument starting from there recursively. In this way we keep descending the "tree of recursion" to smaller & smaller topologically-circular loops. Since for polyhedra said tree must be finite, we eventually must stop, and which point we get a circle with no holes and proof is done. For nonpolyhedral convex objects, perhaps this tree could be infinite and the argument will never hit bottom. (Can anybody construct two convex surfaces with that behavior??) However, for convex bodies described by algebraic surfaces of bounded polynomial degree, with finite number of pieces, the argument always will hit bottom. (I have implicitly used well known facts about algebraic surfaces and curves...) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Jim Propp's Rupert-ish questions
by Warren D Smith 25 Apr '16

25 Apr '16

Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected. Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.) Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P? (One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.) Perhaps Pechenik and Shultz already consider such questions? Jim Propp -- 1. Pechenik & Shultz are the same person. She kept changing her name. The worst instance I saw of that was this Frenchwoman named "Yvonne" who proved some good results in general relativity. Her last name(s) kept changing plus there also was a combinatorial explosion of different hyphenation possibilities which she also eagerly employed at random. Finally it dawned on me: ignore all her names except for Yvonne, there was only one Yvonne in France doing general relativity during 1900-2000. 2. It is not possible for P to pierce Q and Q simultaneously to pierce P, (both convex 3D objects) if P\Q and Q\P each have the topology of a solid torus. Because then Q must "drill a hole thru" P and the two outer surfaces of this hole, which are topological disks, each must cut Q disconnecting it. So any example (if one exists) must have a more complex topology. More generally the intersection of Q with boundary(P) is a collection of connected components and it is not permitted for any of them to be a topological disk, because that disk would cut Q. Anyhow, my competence in such topological arguments is nil, and here I am inherently depending on some kind of Jordan-curve-like theorem for 3D convex objects, which I presume is already known. Some real topologists should do better... 3. I can sketch a proof there are M equal cubes in R^3 such that cube(j) pierces cube(j+1 mod M) for all j. For some modulus M>=3, whose value I do not know. You keep applying appropriate rotation matrices one at a time until the product of all M of them equals identity matrix. We know each step of this process is always possible with a certain amount of "free play" freedom in each rotation matrix, and since the process cannot continue forever without two rotation matrices getting arbitrarily near one another, near enough to be within the free play zone, there must be a "cycle", which proves my claim for some M>=3. This argument works not only for cubes, but actually anything "Rupertable (or piercable) with a positive radius of free play" in any finite dimension.

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[math-fun] Ruperting the cube, octahedron, & simplex, computer results dims<7, conjecture
by Warren D Smith 25 Apr '16

25 Apr '16

For the problem of "Ruperting" the n-dimensional cube, octahedron, and regular simplex (i.e. -- what is the max expansion factor E so an E-scaled cube can pass thru a hole drilled in a congruent-copy cube?) Here is a quick summary of my computer's findings so far (expansion factor is a lower bound found by computer; prob is based on Monte Carlo; it tries random rotation matrices then applies a crude optimizer to the more promising candidates): CUBE (& octahedron): n expansion factor theoretical opt prob random orthog matrix works 2 1.41421 sqrt(2) 0.99999 3 1.06066017 sqrt(9/8) 0.01295 4 1.00732 1.00743475688 1 / [ (2.5+-0.4)*10^8 ] 5 1.00074 1.000839446859 below 10^(-11) 6 1.0000627 ? below 10^(-11) SIMPLEX: n expansion factor theoretical opt prob random orthog matrix works 2 1.15470 2/sqrt(3) 0.99999 3 1.01461187 sqrt(6)/(1+sqrt(2))=(2-sqrt(2))*sqrt(3) 0.001282(3) 4 1.000715 ? below 10^(-11) 5 1.00002711 ? below 10^(-11) 6 1.0000006807 ? below 10^(-11) I have proven the cube & octahedron Emax are equal; this is a special case of the more general Rupert-duality theorem. I also have proofs E>1 in all dimensions n>1 for all 3 problems. My proofs yield E>1+const/9^n for cube & oct (same proof found by Greg Huber earlier), and E>1+const^(2^(n-1)) for the simplex. Both proofs involve a rotation matrix which is a small perturbation of the identity matrix. For the simplex, however, the best rotation probably is NOT a small perturbation of the identity, in which case my proof is being intentionally stupid. The best upper bound on E that I have is sqrt(n/(n-1)) so there is a large gap between lower & upper bounds, and the asymptotic behavior of E for large n remains very unknown. The empirical results for the cube do indeed suggest the 1+const/9^n lower bound is approximately tight (up to altering the value of 9 and perhaps some subexponential factor). For the simplex, empirical results suggest the falloff of E-1 is faster than any simple exponential, but not as fast as double exponential. Perhaps for example E-1 =approx= 10^(1-n) / (n-2)!. Kay R. Shultz (and perhaps others?) had introduced the function f(m,n) defined as the largest edgelength for an m-cube which will fit inside a unit n-cube, 0<m<=n. This function can be generalized to apply to any convex n-polytope P_n, not just the n-cube, as follows: f(m,n) is the supremum of the scale factors s for a rotated P_m, such that the cartesian set-product s*P_m x R^(n-m) (which is an "infinite prism with base P_m") can "pass through" a rotated P_n having scale=1. That is, such that there exist mXm and nXn orthogonal matrices Q1 and Q2, such that for each point of s*Q1 P_m, n real numbers x1,x2,...,xn exist (the first m of which are independent of the point) so that this point plus (x1,x2,...,xn), lies strictly inside Q2 P_n. Note, if Q1 and Q2 were somehow known then the maximum possible s and the (x1,x2,...,xn) could be deduced by solving a linear program. If Q1 and Q2 were both demanded to be infinitesimal perturbations of the identity matrix, Q=Ident+epsilon*AntiSymMatrix, then the question of whether s>1 would be answerable by solving a linear program in n dimensions. It can be argued using Hadamard's determinant bounds (or the cruder factorial determinant bound arising by expansion by minors) and Cramer's formula for matrix inverse, that for ANY class of convex polytopes describable using polynomial(n)-bounded integer coefficients (e.g. cubes & octahedra, regular simplex by going to n+1 dimensional representation, permutohedron), this latter linear program shows that if s>1, then s-1 can be made larger than 1/FactoriallyGrowingFunction(n). This immediately improves upon my crude lower bound on E for the n-simplex, which involved double-exponential falloff of E-1 for large n, to show factorial style falloff, specifically k^n/(c*n)! for some positive constants c, k. (I could work out explicit bounds on all these constants if I were less lazy.) And the empirical computer findings for the simplex suggest this sort of behavior actually is the truth. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Simple finite group problem
by Warren D Smith 25 Apr '16

25 Apr '16

I vaguely recall (perhaps wrongly) that the conjecture K.Lynch alluded to (every abelian group structure arises as a Z_n*) was proven as a theorem in Daniel Shanks' book "solved & unsolved problems in number theory." -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Simple finite group problem
by Keith F. Lynch 25 Apr '16

25 Apr '16

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > 4 14 16 26 Note that that's isomorphic to the other one of the two abstract groups of order 4, the one that isn't a cyclic group. Inspired by that, here's an absurdly bold conjecture with remarkably little evidence: *Every* finite abelian group is isomorphic to some multiplicative group mod N. Someone prove me wrong. After I get my program to find all such groups through at least N=1000 working, I'll check and see if it appears to be true. Of course whatever I find will constitute neither a proof or a disproof, but it may inspire insight. In support of that, is there any online database of small abstract abelian groups? I see little point in reinventing that wheel. Thanks.

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