March 2009 - mathfun@mailman.xmission.com

[math-fun] R.I.P., John W. Wrench Jr.
by Michael Beeler 30 Mar '09

30 Mar '09

A nice obit for John Wrench appeared in the Boston Globe, reprinted from the Washington Post: http://www.boston.com/news/education/higher/articles/2009/03/27/john_wrench… He was 97. He did lots of good math, and for a while, with Shanks, held the record for digits of pi.

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[math-fun] Simpson's rule and Gaussian Quadrature
by Eugene Salamin 30 Mar '09

30 Mar '09

It seems that the weights in Gaussian Quadrature over the interval [-1,+1] are proportional to sqrt(1 - x^2). This means that means the weights do not become constant inside the interval as the number of evaluation points increases. The weights must vary inversely with the density of roots of the Legendre polynomial. On the other hand, the weights vary smoothly, rather than jumping around as in Simpson's rule. -- Gene

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[math-fun] Puzzle
by Dan Asimov 28 Mar '09

28 Mar '09

Find the missing term ___ in this sequence: 102, 102342, 1031345242, ___, 1042, 10467842, --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Correction Re: Villarceau circles
by Dan Asimov 28 Mar '09

28 Mar '09

No, I'm the one who's missing something -- it's dumb of me to spontaneously add comments I haven't thought out! Of course you are right -- there's no problem in giving a topological example of (round, but not congruent) k-spheres foliating an open set of R^(k+p), for any p >= 1. (It's a perfectly good foliation that you described, of the highest differentiability class, even -- real analytic.) It's a fascinating fact about foliations, by the way, that the 3-sphere admits many C^oo codimension-1 foliations (i.e., by surfaces), but does not admit any real analytic one. --Dan ------------ Allan wrote: << I think I must be missing something. Take a family of parallel, concentric 2-spheres whose radii span the open interval (1,2). Say they are in the abc-hyperplane. Their union is a 3-dimensional set. Now cross that set with an open 3-ball in the def-hyperplane. The resulting set is open, and 6-dimensional. I don't think any of the 2-spheres intersect any other. Doesn't that answer your question constructively, and in the affirmative? Maybe I'm missing one of the requirements for foliation. On Fri, Mar 27, 2009 at 10:51 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > Thanks to Fred for pointing out that I couldn't have meant 2-spheres in R^5 > (that case is already excluded, as mentioned). > > What I meant to write is 2-spheres in R^6. (In fact, even just > topologically: > I don't know if topological 2-spheres can foliate a non-empty open set of > R^6.) . . . >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Correction Re: Villarceau circles
by Dan Asimov 28 Mar '09

28 Mar '09

Thanks to Fred for pointing out that I couldn't have meant 2-spheres in R^5 (that case is already excluded, as mentioned). What I meant to write is 2-spheres in R^6. (In fact, even just topologically: I don't know if topological 2-spheres can foliate a non-empty open set of R^6.) --Dan I wrote: << On 3/28/09, Dan Asimov <dasimov(a)earthlink.net> wrote: > P.S. For completeness I'll add that the above readily implies that there do exist > non-empty open sets in R^(2k+1+p), for any p >= 0, that are foliated by congruent > round k-spheres for k = 0,1,3,7. And conversely, for any k not in {0,1,3,7}, there > does not exist a non-empty open set in R^(2k+1-p), for p >= 0 and 2k+1-p >= 0, > that is foliated by congruent round k-spheres. > > But this leaves unresolved infinitely many cases of k-spheres in n-space. First open case: > Can congruent round 2-spheres foliate a non-empty open set in R^5 ??? ^^^ >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Villarceau circles
by Dan Asimov 28 Mar '09

28 Mar '09

I accidentally (re-)discovered the circles of Villarceau c. 1992 when trying to foliate an open subset of 3-space by congruent circles. But in fact Villarceau discovered these in 1848. One way to generate them is to start with the Hopf fibration of the 3-sphere S^3 := {(z,w) in C^2 | |z|^2 + |w|^2 = 1}. (C = the complex plane.) Its fibres are circles defined as the equivalence classes of (z,w) ~ (gz,gw) for any (z,w) in S^3 and where g is any complex number of length 1. (These are all great circles of the 3-sphere.) Then project S^3 - {(0,i)} stereographically onto R^3 via S((a+bi, c+di)) := (1/(1-d))(a,b,c). Since S is conformal, it carries the Hopf fibres to circles (except that the incomplete fibre through (0,i) goes to the z-axis). Now, any Hopf fibre in S^3 will lie on a unique torus of the form T_s := S_r x S_s in C^2 with 0 <= s <= 1, where S_t denotes the circle of radius t about the origin in C. Here r^2 + s^2 = 1 by the definition of S^3. (Actually, note that the "tori" T_1 and T_0 are degenerate, and are in fact great circles of S^3.) It is left as an exercise for the reader, to show that for any Hopf fibre F, the size of its image circle S(F) in R^3 depends only on the s for which F lies on T_s. Thus for each s, 0 <= s < 1, we can modify the stereographic projection (S | T_s) with a post-scaling, (S' | T_s) := f(s)*(S | T_s), so that S' takes each Hopf fibre to a circle of radius 1. It is straightforward to verify that S' : S^3 - T_0 -> R^3 is a diffeomorphism onto the open solid torus U of revolution of all points in R^3 that lie at a distance < 1 from the unit circle S^2 in the xy-plane. Then the images of all the Hopf fibres of S^3 - T_1 are unit circles in the image of S', foliating the open solid torus U by congruent circles. Incidentally, the analogous procedure works for all four Hopf fibrations (based on the normed real division algebras, i,e., on R, C, H, O): S^(2k+1) fibred by great S^k's for k = 0,1,3,7 . . . yielding a foliation of an open set of R^1, R^3, R^7, R^15 by S^0's, S^1's, S^3's, S^7's, respectively. (I am told that Bill Thurston knew of these foliations by the mid-1970s.) Interestingly, there are no other values of k for which a non-empty open set in R^(2k+1) can be foliated by congruent round k-spheres. But for k = 1,3,7 there are infinitely many different ways to foliate an open set in R^(2k+1) with (round) unit k-spheres as the leaves. An amusing fact about these is that there is an upper bound M_k to the volume of such a foliated open set U in R^(2k+1). (One needs to assume the leaf space of the foliation is connected, which for k=1,3,7 is the same as assuming that U is connected (i.e., except for k = 0). It's easy to check that for k = 0 this upper bound is 4. I've conjectured that the Hopf fibration method also gives the largest-volume such foliated open set in the other three cases, k = 1,3,7 but have not so far proven it. The volume in the Hopf cases is always A(k)*V(k+1) where A(k) is the k-dimensional content of the (surface of the) unit sphere in R^(k+1), and V(k+1) is the (k+1)-dimensional content of the unit ball in R^(k+1). So we know M_0 = 4 in R^1, and conjecturally M_1 = 2 pi^2 in R^3, M_3 = pi^4 in R^7, M_7 = pi^8/72 in R^15. --Dan P.S. For completeness I'll add that the above readily implies that there do exist non-empty open sets in R^(2k+1+p), for any p >= 0, that are foliated by congruent round k-spheres for k = 0,1,3,7. And conversely, for any k not in {0,1,3,7}, there does not exist a non-empty open set in R^(2k+1-p), for p >= 0 and 2k+1-p >= 0, that is foliated by congruent round k-spheres. But this leaves unresolved infinitely many cases of k-spheres in n-space. First open case: Can congruent round 2-spheres foliate a non-empty open set in R^5 ??? _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Villarceau circles
by Fred lunnon 27 Mar '09

27 Mar '09

I didn't know about these --- though perhaps I should've --- so maybe they're new to a few more people out there. http://en.wikipedia.org/wiki/Villarceau_circles http://mathworld.wolfram.com/VillarceauCircles.html are reasonably informative pages on the subject [the Mathworld page on "torus" currently has captions transposed for "spindle" and "horn" tori]. Imagine a torus in canonical position, generated by rotating about the z-axis a circle in the xz-plane with centre on the x-axis. Now tilt the xy-plane about (say) the y-axis, until it is tangential to the torus at their (two) furthest points of contact. Their intersection comprises two intersecting loops, each of which spirals around the torus tube once as it travels around the circumference; and rotating the plane about the z-axis generates two complete systems of such loops, each linked to all the others in its system. Well now, I came to study these loops as stereographic projections of Clifford parallels in elliptic space (see also Hopf fibration) --- see e.g. Berger, Cole & Levy sect 18.9 p 305 --- and spent some time wrestling with a conundrum: how can these spirals be correct, when (for various technical reasons) it's necessary that such projections be plane circles? And in the course of time it dawned on me that they're both spirals and circles! [Immediately following which, of course, I also discovered that everybody else knew this long ago.] Incidentally, once you know they're circles, its easy to see that their radius must be the same as that described by the generating circle around the origin. The web pages give rather sordidly computational proofs of these facts, along with an incomplete sketch of a more elegant argument (ascribed to Hirsch) using algebraic geometry; my own argument, though but one line in length, relies similarly on a theorem from geometric algebra. Can anybody supply an simple, elementary, synthetic proof that the intersection is (are?) circles? Also, Berger, Cole & Levy mentions in passing that there are Dupin cyclides (Moebius inversions of tori) with not only 4 but 6 systems of circles. I can't find anything out about these chimaeras --- indeed, I don't believe in them, but then, I didn't believe in Villarceau circles either! Fred Lunnon

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Re: [math-fun] Freeman Dyson integer problem
by Dan Asimov 27 Mar '09

27 Mar '09

The original problem has a nice little solution in the lowest possible base, base 3: (Find the smallest positive integer x such that x_3 = [K]d and (2x)_3 = d[K] where d = 1 or 2, [K] is some number in base 3 (without leading 0's), and adjacency just means concatenation.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Freeman Dyson integer problem
by N. J. A. Sloane 27 Mar '09

27 Mar '09

210526315789473684 appears twice in the OEIS, but only once in a relevant entry: %I A094676 %S A094676 210526315789473684,3103448275862068965517241379,410256, %T A094676 510204081632653061224489795918367346938775, %U A094676 6101694915254237288135593220338983050847457627118644067796,7101449275362318840579,8101265822784,91011235955056179775280898876404494382022471 %N A094676 Least n-transposable number. %C A094676 A k-transposable number,2<=k<=9, is one equal to k times the number whose digits are merely a cyclic permutation of its own leftmost digit to the right. %D A094676 H. Camous, Jouer Avec Les Maths, "Chassez le naturel", Section I, Problem 3 pp. 20;31-2, Les Editions D'Organisation, Paris 1984. %D A094676 L. A. Graham, Ingenious Mathematical Problems and Methods, "End At The Beginning", Problem 72 pp. 44;212-3 Dover NY 1959. %F A094676 n prepended to n*(10^m - n)/(10*n - 1), where m=A094224(n)-1. %e A094676 a(4)=410256=4*102564 %K A094676 fini,nonn %O A094676 2,1 %A A094676 Lekraj Beedassy (blekraj(AT)yahoo.com) Jun 07 2004 - though the definition doesn't look right! I will edit it. Anyway, could someone send me more terms of the sequence 210526315789473684, ... of numbers with the Freeman Dyson doubling property? Neil

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Re: [math-fun] Freeman Dyson integer problem (cycle)
by Eric Angelini 27 Mar '09

27 Mar '09

Yes Dave, you're right, of course -- thanks ! Best, É. -----Message d'origine----- De : math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Dave Blackston Envoyé : jeudi 26 mars 2009 20:45 À : math-fun Objet : Re: [math-fun] Freeman Dyson integer problem How about 11111111111? ;-) Actually, for mabcdef...jkl to be m times abc...klm, a would have to be 1, so there's no real way for your cycle to work unless all the values are 1.

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