July 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Geometry question
by Dan Asimov 24 Jul '11

24 Jul '11

It's an amazing fact about 4-space that there exist orthogonal bases like the one Jakob describes, tilted so that each vector is perpendicular to 2 of the standard basis vectors, and at 45 degrees to the other 2. --Dan Jakob wrote: << What about orienting a tesseract with edge length sqrt(2) in the following manner. Let one vertex sit in the origin, and let (1,0,1,0), (-1,0,1,0), (0,1,0,1), (0,-1,0,1) be the coordinates of the four vertices that are (edge-)adjacent to the origin. Since the corresponding vectors are orthogonal and have the same length sqrt(2), this indeed gives rise to a regular tesseract. Now, consider the projection (x,y,z,w) -> (x,y). The vertex set of the tesseract is mapped to the set {-1,0,1} x {-1,0,1}, and the projection of each edge has direction (1,0) or (0,1). Thus the projection of the 1D-skeleton of the tesseract is a 2x2 grid. >> David wrote: << Can a regular tessertact can be oriented in R^4 so that the orthogonal projection of its 1D edges onto a 2D plane have the shape of a 2 x 2 grid? >> Sometimes the brain has a mind of its own.

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[math-fun] The Ramanujan Notebooks, the real thing
by Simon Plouffe 23 Jul '11

23 Jul '11

Hello, I have found these documents here : http://www.imsc.res.in/~rao/ramanujan/NotebookFirst.htm These are the real notebooks, !, I tought this might be of interest to some of the members. Bonne lecture, Simon Plouffe

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[math-fun] Hadamard conjecture
by Dan Asimov 23 Jul '11

23 Jul '11

P.S. Hadamard showed that if there exists an orthogonal basis for n-space (n >= 2) all of whose n^2 coordinates are +-1 (i.e., a basis lying on the corners of the 2^n n-cube centered at the origin), then n must be either 2 or a multiple of 4. The (still unresolved) Hadamard conjecture is that for each of these values of n, there always exists such an orthonormal basis. Amazing that such a simple statement has remained unresolved for over 100 years. Or a very long time, anyhow. --Dan I wrote: << (1 1 1 1) (1 -1 -1 1) (1 1 -1 -1) (1 -1 1 -1) >> Sometimes the brain has a mind of its own.

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Re: [math-fun] Geometry question
by Dan Asimov 22 Jul '11

22 Jul '11

Actually, while I'm fostering topic drift, maybe it's also worth mentioning the even more symmetrical (with respect to the standard basis) orthogonal basis for 4-space: (1 1 1 1) (1 -1 -1 1) (1 1 -1 -1) (1 -1 1 -1). Each axis in the direction of one of these vectors makes the angle of arccos(1/4) with each standard axis. Which is how I should have couched the claim below: in terms of new and old axes. (The angles I said are 45 degrees are actually either 45 or 135 degrees.) As someone recently said: I should ensure brain is engaged before mouth is opened. --Dan I wrote: << << It's an amazing fact about 4-space that there exist orthogonal bases like the one Jakob describes, tilted so that each vector is perpendicular to 2 of the standard basis vectors, and at 45 degrees to the other 2. Jakob wrote: << What about orienting a tesseract with edge length sqrt(2) in the following manner. Let one vertex sit in the origin, and let (1,0,1,0), (-1,0,1,0), (0,1,0,1), (0,-1,0,1) be the coordinates of the four vertices that are (edge-)adjacent to the origin. Since the corresponding vectors are orthogonal and have the same length sqrt(2), this indeed gives rise to a regular tesseract. Now, consider the projection (x,y,z,w) -> (x,y). The vertex set of the tesseract is mapped to the set {-1,0,1} x {-1,0,1}, and the projection of each edge has direction (1,0) or (0,1). Thus the projection of the 1D-skeleton of the tesseract is a 2x2 grid. >> >> Sometimes the brain has a mind of its own.

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Re: [math-fun] Dupin cyclides --- another rhinoceros' pancreas?
by Dan Asimov 22 Jul '11

22 Jul '11

Sorry, missed that embedded space. Here's a better URL, anyway: < http://jalape.no/math/mathgal.htm >. Click on the middle image in the second column to get an enlarged version of a Dupin cyclide with two conical singularities. Click instead on its ",description" and you get an explanation of how the image was obtained. ----- But come to think of it, should there be more singularities lurking in the complex version? --Dan P.S. In any case one way to regularize the calculations is by using a conformal transformation of space to make all 3 "generating" 2-spheres be of equal radius with. Then I think there are only two essentially distinct cases to consider. But I regret I don't have time at the moment to think this out carefully. -------------------------------------------------------- Allan wrote: << Repair URL, please? I wrote: << I think the complex singularities can be brought into the real domain by an appropriate transformation, yielding a surface like this: [bad URL was here]. >> >> Sometimes the brain has a mind of its own.

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[math-fun] Graphic displays of money
by Henry Baker 22 Jul '11

22 Jul '11

Cool graphic displays of various large amounts of money: http://www.wtfnoway.com/

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[math-fun] Dupin cyclides --- another rhinoceros' pancreas?
by Fred lunnon 22 Jul '11

22 Jul '11

I have read in a number of places --- though now I come to check up, I'm having difficulty tracking down a definitive source --- that the quartic surface known as the "cyclide of Dupin" has 4 nodes: conical singularities at which the gradient vector vanishes. Furthermore, it is averred that one pair of these may may be real or may be complex; the other pair is always complex. Like the trusting innocent I am, I have for years cheerfully accepted this scenario, without ever properly understanding why it might --- or might not --- obtain. The first pair can easily be seen in action in spindle or horn (real pair) versus ring (conjugate complex pair) cyclides --- at least, they could easily be seen before some brainstorm overtook the two pages at http://en.wikipedia.org/wiki/Dupin_cyclide http://mathworld.wolfram.com/Cyclide.html which I was sure used to show informative plots of the various forms! Pinning down the second pair is more problematic. In the special "parabolic" case --- an infinite cubic "full twist" surface --- there are indeed 4 complex nodes: for example differentiating y(x^2 + y^2 + z^2 - 1) - 2 x z = 0 wrt x,y,z and solving, we find solutions at [x, y, z] = [i, 1, i], [-i, 1, -i], [-i, -1, i], [i, -1, -i], where as usual i^2 = -1. However, in the general quartic case, Maple is quite insistent that there is only a single pair of solutions; and in consideration of the contradictions in which my earlier credulity was beginning to embroil me, I am beginning to think that Maple might just be right about this. In the case of the torus in standard postion with central and tube radii p and q, we (at least, Maple and me) find (x^2 + y^2 + z^2 - p^2 - q^2)^2 - 4 p^2(q^2 - z^2) = 0 has just the two nodes [x, y, z] = [0, 0, +sqrt(q^2-p^2)] , [0, 0, -sqrt(q^2-p^2)] . Inverting this with centre x = e say, we get a general shape with again just two nodes [x, y, z] = [e(e^2-1+q^2-p^2), 0, (+/-)sqrt(q^2-p^2)] / (e^2+q^2-p^2) unless e^2 = p^2 - q^2 (parabolic case). Can anyone cast light on my confusion? Fred Lunnon

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[math-fun] Geometry question
by David Wilson 22 Jul '11

22 Jul '11

Can a regular tessertact can be oriented in R^4 so that the orthogonal projection of its 1D edges onto a 2D plane have the shape of a 2 x 2 grid? +-----------+------------+ | | | | | | +-----------+------------+ | | | | | | +-----------+------------+

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[math-fun] Fwd: Veit's disks, again
by Bill Gosper 21 Jul '11

21 Jul '11

rwg>How about your semisecret fourteen disk solution, > which has no symmetry at all? > (http://milou.msc.cornell.edu/images/>>>> seems to > be [intermittently] down.)>>>> --rwg > > Whoa, what about twelve? http://gosper.org/HTMLFiles/12disks.gif > I thought this was the "six dimes and three quarters around > three nickels" I Mobius-transformed to make Twubblesome Twelve, but it > seems to > have been Mobius transformed already! It appears to retain bilateral > symmetry, > and the transformation should be described by a single real parameter. If > I > can reconstruct my "six dimes" formulae, it should become a Calculus I > problem > with an exact algebraic solution. > --rwg For (unbeknownst to me) a stupid technical reason, this discussion shrank off-list to just Veit, David Cantrell, and me. And then took some surprising turns. Using commercial software, David has improved upon Veit's 10, (11, slightly?), 12, 13, and 14 packings, which I thought had been proved optimal! See http://www2.stetson.edu/~efriedma/cirRcir/ In "14", the extremely close pair that Veit reported is no longer the closest. The new closest differ substantially more, but are probably (uncomfortably) swappable, even in aluminum, and might need to be artificially equalized to make a decent physical puzzle. And my proposed Calculus I problem for "12" is obviated--David finds the optimal Möbius transformation is the identity. He sent the polynomial for the exact Sum(radii), which factors into nasty cubics. David's packings go all the way to "32", with "31" perhaps the most amazing, for lack of the D(ihedral)_5 symmetry that "32" almost has. Talk about a sucker bet-- who would believe the "31" concentric shell packing was suboptimal? <Insert Baskin-Robbins joke here.> So the "14" puzzle is somewhat ruined, and the next totally asymmetric packing is "25" (! Almost inconceivable.) Are we out of metagrobological luck? Probably not--many of David's bilaterally symmetric packings have about the same number of different sized pieces as Veit's "14"! Veit adds: VE>I should explain why my symmetries were off. I never doubted the 12 packing had D_3 symmetry. My code requires a target value for the radius sum (it's a constraint solver). My procedure was to converge on the optimum by bracketing the optimal radius sum on a sequence of shrinking intervals. Clearly I terminated my sequence too early for the 12 packing in that the imperfect convergence didn't escape your notice! >I don't have a good excuse for missing the obvious 10 packing! >Veit --rwg

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[math-fun] the volume of the debt, us and france, or the volume of a pile of euros or american quarter.
by Simon Plouffe 20 Jul '11

20 Jul '11

Hello, I was talking to a person the other day, I could clearly see that the person could not understand what enomous amount is represented by the debt of countries like US or France. So I made this exercise , let's say we pile up quarters of the total amount of the debt , what volume would it make. For the US : about 7 times the volume of 1 tower of the former WTC. For France : in euro, would represent approximately the volume of the 2 principal pyramids of Gizeh, Kheops and the other, not the 3rd one. For the calculation i simply used the volume taken by 1 coin , either euro : http://en.wikipedia.org/wiki/Euro_coins or american quarters : http://en.wikipedia.org/wiki/Quarter_%28United_States_coin%29 as if they piled in a square grid, this is the best approximation of volume I could find because the volume of loose coins is hard to measure precisely. That problems is of course related to the packing of spheres, some people here are experts in this area, does someone has an idea what a pile of loose euros or american quarter will occupy ? My approximation is not very good but it gives an idea of the volume, also, for those remembered in the sept. 11 eleven, please tell me what would be a standard volume of reference, something that most people would recognize as a large volume, like the size in cubic meters of the queen mary II, but that one is not easily found. I took the egyptian pyramids because it has to do with french history and Napoleon and it is well known. The debt of usa as everybody knows now is 14300 billion dollars and the french debt is about 5000 billion dollars, I converted to euro for the computation and of course the volume of the 2 pyramids. found here : http://en.wikipedia.org/wiki/Egyptian_pyramids I sent my computations to the french journal 'Le Monde' and had no reaction at all as well as the Canard EnchaÃ®nÃ©. Well, I my opinion : we are doomed, we are toasted. Best regards. Simon Plouffe

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