August 2011 - mathfun@mailman.xmission.com

Re: [math-fun] fractal optimizer
by Veit Elser 18 Aug '11

18 Aug '11

I've posted images of the fractal that solves this problem here: http://milou.msc.cornell.edu/images/ The sequence shows a greek cross being iteratively refined, with successive crosses rotated by 45 degrees. Each one is connected, and it looks like there is no pinch-off in the limit either. Each one of these shapes tiles the plane with square lattice translations, and in two ways. The two sets of translations (square lattices) are related by a rotation that approaches 45 degrees in the limit. Here's how you use these shapes to solve the optimization problem, an optimal matching. Center the shape on any point of one of the square lattices; it will contain exactly one point of the other lattice because it is a fundamental domain of that lattice. This also works the other way, and the correspondence of points of the two lattices is symmetric because the shape is a fundamental domain of both lattices and has central symmetry. There are many fractal shapes that serve as fundamental domains to both lattices, but this one is special because it has the smallest second moment (quantization error). Veit On Jul 31, 2011, at 2:14 PM, Veit Elser wrote: > The Mandelbrot posts reminded me of something: > > Are fractal sets ever the solution of an optimization problem? > > Here's one instance I discovered a couple of years ago: > > Take an infinite square grid of red points and an identical grid of blue points. > Now take the blue grid and rotate it by 45 degrees with respect to the red grid. > The problem is to form a perfect matching of red and blue points such that > the matching minimizes the sum of the squared distances between pairs. > > If the infinite value of the thing being minimized bothers you, do this: Use a > continued fraction approximation for sqrt(2) to approximate the blue grid > so that it becomes commensurate with the red grid (at periods that increase > with the order of the continued fraction approximation). Now solve the > optimization problem on just one (finite) period. To get the fractal, translate all the > matched (red, blue) pairs so the red point is at the origin, and observe how the set > of blue points evolves with order of approximation. > > Veit > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] mathworld.wolfram.com/TrigonometryAngles
by Bill Gosper 17 Aug '11

17 Aug '11

JBuddenhagen> On Mon, Aug 15, 2011 at 10:04 PM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote: > [...]> Simplify[Developer`TrigToRadicals[Sin[\[Pi]/25]]]> (after the death threat)> produces this marvelously simple quintic surd>> -((I (-(-1 - Sqrt[5] - I Sqrt[10 - 2 Sqrt[5]])^( 6/5) + (-1 - Sqrt[5] + I> Sqrt[10 - 2 Sqrt[5]])^(6/5)))/> (8 2^(2/5)))>> And ToRadicals doesn't.-(> --rwg> _______________________________________________> I guess this must be related to x=4*(sin(Pi/25))^2 being a root of the marvelously simple quintic x^5 - 10*x^4 + 35*x^3 - 50*x^2 + 25*x - 3 + t where t is the golden ratio. ----------- How is that marvelously simple, but not x^5-5*x^2-3? Whose roots are several times less marvelous: Out[1124]= (1/(10^(2/5)))((-1)^((2 #1)/5) (25-5 Sqrt[5]-Sqrt[30 (25-11 Sqrt[5])])^(1/5)+ (-(-1)^(3/5))^#1 (25-5 Sqrt[5]+Sqrt[30 (25-11 Sqrt[5])])^(1/5)- (-(-1)^(1/5))^#1 (-25-5 Sqrt[5]+Sqrt[30 (25+11 Sqrt[5])])^(1/5)+ (-1)^((4 #1)/5) (25+5 Sqrt[5]+Sqrt[30 (25+11 Sqrt[5])])^(1/5))& In[1125]:= Chop@N[#^5-5*#^2-3,69]&/@%/@Range@5 Out[1125]= {0,0,0,0,0} --rwg

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[math-fun] mathworld.wolfram.com/TrigonometryAngles
by Bill Gosper 16 Aug '11

16 Aug '11

>From http://mathworld.wolfram.com/TrigonometryAnglesPi11.html : [...] hence [image: sin(pi/11)], can be expressed in terms of radicals (of complex numbers). The explicit expression is quite complicated, but can be generated in *Mathematica <http://www.wolfram.com/products/mathematica/>*using Developer`TrigToRadicals<http://reference.wolfram.com/mathematica/Developer/ref/TrigToRadicals.html> [Sin <http://reference.wolfram.com/mathematica/ref/Sin.html>[Pi<http://reference.wolfram.com/mathematica/ref/Pi.html> /11]]. However, Developer`TrigToRadicals[Sin[\[Pi]/11]] Developer`TrigToRadicals::obs: Developer`TrigToRadicals has been superseded by ToRadicals, and is now obsolete. It will not be included in future versions of Mathematica. >> and then disgorges 13 pages(!) of 5th roots. Trying In[1058]:= ToRadicals[Sin[Pi/11]] Out[1058]= (-(1/2))*(-1)^(9/22)*(-1 + (-1)^(2/11)) Foo! I wanted 5th roots. Just far fewer. Like this: Out[1097]= -(1/5) + (1/(5 2^(2/5)))11^(1/5) E^((2 I \[Pi] #1)/ 5) ((-89 - 25 Sqrt[5] - 5 I Sqrt[2 (205 - 89 Sqrt[5])])^(1/5) - (-1)^( 3/5) (-89 + 25 Sqrt[5] - 5 I Sqrt[2 (205 + 89 Sqrt[5])])^(1/5) E^((2 I \[Pi] #1)/5) + (-1)^( 2/5) (-89 + 25 Sqrt[5] + 5 I Sqrt[2 (205 + 89 Sqrt[5])])^(1/5) E^((4 I \[Pi] #1)/5) + (-89 - 25 Sqrt[5] + 5 I Sqrt[2 (205 - 89 Sqrt[5])])^(1/5) E^((6 I \[Pi] #1)/5)) & In[1099]:= Rationalize[ArcCos[Chop[(%1097 /@ Range[5])/2.]]/\[Pi]] Out[1099]= {2/11, 10/11, 6/11, 8/11, 4/11} In[1100]:= Rationalize[ArcCos[Chop[(%1097 /@ Range[5])/-2.]]/\[Pi]] Out[1100]= {9/11, 1/11, 5/11, 3/11, 7/11} In[1101]:= Rationalize[ArcSin[Chop[(%1097 /@ Range[5])/2.]]/\[Pi]] Out[1101]= {7/22, -9/22, -1/22, -5/22, 3/22} I.e., the pi/11 article could give all the sins and coses with just %1097, and all the cscs and secs with a similar expression. One wonders how much smaller could be the expression for Sin[pi/23], whose article says it takes .5GB! This may have discouraged trying pi/25, which http://mathworld.wolfram.com/TrigonometryAngles.html skips, yet Simplify[Developer`TrigToRadicals[Sin[\[Pi]/25]]] (after the death threat) produces this marvelously simple quintic surd -((I (-(-1 - Sqrt[5] - I Sqrt[10 - 2 Sqrt[5]])^( 6/5) + (-1 - Sqrt[5] + I Sqrt[10 - 2 Sqrt[5]])^(6/5)))/ (8 2^(2/5))) And ToRadicals doesn't.-( --rwg

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Re: [math-fun] Real analytic surface with uncountable base for its topology (correction)
by Dan Asimov 16 Aug '11

16 Aug '11

Correction: This << (Of course as a real manifold S is 4-dimensional.) should have read << (Of course as a real manifold S* is 4-dimensional.) --Dan Sometimes the brain has a mind of its own.

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[math-fun] Real analytic surface with uncountable base for its topology
by Dan Asimov 16 Aug '11

16 Aug '11

I find the following to be amazing: On one hand, every Riemann surface -- i.e., 1-dimensional complex analytic manifold -- has a countable base for its topology. (Due to Tibor Rado.) On the other hand, there exist real analytic surfaces that have an uncountable base for their topology. (The standard example is the Pruefer manifold, an astonishingly simple construction that is almost always described in the opaquest possible terms. First described in print by Rado.) But given such a real analytic surface S with an uncountable base for its topology, it can always be extended to a 2-dimensional complex analytic manifold S* which must then have an uncountable base for its topology. (Of course as a real manifold S is 4-dimensional.) CONCLUSION: Complex analytic manifolds must have a countable base if they are 1-dimensional, but can have an uncountable base if their dimension is >= 2. (Of course all manifolds mentioned are assumed to be connected.) --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] 3D fractal/sponge models via "3D printing" technology
by Bill Gosper 15 Aug '11

15 Aug '11

Neil's Thing-o-matic is higher res and less drooly than George's Cupcake. E.g., http://www.youtube.com/watch?v=pvT5ySMnO8U, which Neil has been withholding for lack of editing. --rwg > On 8/14/2011 1:05 PM, Michael Kleber wrote: >> ...You need to be somewhat cautious of the plastic home 3D printers... > > > > Yes, the inexpensive home-use 3D printers are limited in many ways, but > there are still many cool things you can make with them. You can see a > variety of mathematical objects I've made on my Makerbot here: > > http://www.georgehart.com/rp/makerbot/makerbot.html > > George > http://momath.org/ > http://georgehart.com/

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[math-fun] 3D fractal/sponge models via "3D printing" technology
by Henry Baker 14 Aug '11

14 Aug '11

FYI -- Metal 3D printers are very expensive, but plastic 3D printers can be built at home by hackers (see Maker Faire) for not a lot of money. I recently heard a talk about this technique, which enables the direct production of a metal part using 3D printing ("additive manufacturing") technology. One German company can even build _titanium_ parts using this technique. The important thing about this technology (in my opinion) is that it allows the production of parts that could never be manufactured _any other way_. For example, if you want to save weight in a part, you can "drill out" portions of the metal that contribute to the weight, but not to the strength. But this process is quite limited, because there are many places you can't get a straight drill hole, and many places where the drill holes would be far too small to actually drill. You could, however, build something like a bicycle crank as a lacy, fractal structure with microscopic holes _everywhere_, which would weigh 1/2, 1/3, or less than a standard crank, but without giving up any strength. Human bones are similar in structure: if they were _solid_, bones would weigh far more than they do for the same strength. So far, the only thing missing from this picture is the incorporation of very strong fibers -- e.g., carbon fibers or even carbon nanotubes -- which could provide even more strength with less weight. Ideally, one would like to incorporate both types of technologies at the same time -- i.e., a kind of "weaving" machine hooked up to a 3D printing machine. In any case, get ready for a true revolution in strength/weight ratios in new materials & manufacturing technology. http://en.wikipedia.org/wiki/Direct_metal_laser_sintering

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Re: [math-fun] Non-isomorphic finite simple groups of the same order
by N. J. A. Sloane 12 Aug '11

12 Aug '11

Andy, Yes, I agree! I was wrong!

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Re: [math-fun] The Real 3D Mandelbrot Set
by Tom Karzes 12 Aug '11

12 Aug '11

Me neither. There is, however, a very natural 4-dimensional set that combines both the standard Mandelbrot set and all of its corresponding Julia sets into a single entity. Let one plane determine c0, and an orthogonal plane determine c1. The product is 4-dimensional, and the iteration is: z(0) = c0 z(i) = z(i-1)^2 + c1 The planar slice obtained by setting c0 to 0 and varying c1 is the Mandelbrot set. The planar slice obtained by setting c1 to some constant and varying c0 is the Julia set corresponding to c1. Unfortunately, having a 2-dimensional retina makes it difficult for me to visualize such a set. Being three dimensional can be very frustrating. Tom > I cannot understand the basis for the assumption that there is > a "real" 3D Mandelbrot set. > > --Dan > > Sometimes the brain has a mind of its own. > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >

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Re: [math-fun] Non-isomorphic finite simple groups of the same order
by N. J. A. Sloane 12 Aug '11

12 Aug '11

>On 7/17/2011 8:06 PM, Dan Asimov wrote: >> According to this website:< http://www.madore.org/~david/math/simplegroups.html>, >> the smallest order for which there are more than one isomorphism class of >> finite simple groups is 2160. >That's 20160. One must also say NON-CYCLIC simple groups. See A119648 in the OEIS. Neil

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