August 2011 - mathfun@mailman.xmission.com

[math-fun] Infinite product expansion for 1
by Adam P. Goucher 10 Aug '11

10 Aug '11

Consider the infinite product: (24/16) * (48/64) * (120/144) * ... * ((p^2 - 1)/(16*[p/4]^2)) * ... In Sloane notation, this is: Product (n >= 1) of (A061066(n+1) / (2*A024698(n+1)^2)) (Perhaps it could be used to set a new record for computing the most decimal digits of 1? : ) Sincerely, Adam P. Goucher

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[math-fun] flowsnake bathroom tile
by Fred lunnon 10 Aug '11

10 Aug '11

[ Take 3 --- with apologies to Rich! ] After GoogleDocs rehashed their interface again --- which it admittedly needed --- I can't upload there any more. So I've attached a GIF graphic, which I hope will sneak under the 40Kb limit set by the math-fun server. An efficient algorithm (combinatorial, rather than generic polygon-filling) to fill the original snowflake curve required some thought on my part, although in the end it isn't very difficult. The natural tiles are rhombs, rather than RWG's hexagons [hexflo.png]. However, I haven't yet figured out a neat way to use them directly, rather than split into two triangles. The two regions are simply connected, unsurprisingly though not immediately obviously --- take the ends of a finite segment of curve and pull? Rather grotty Maple code & larger pics available on request. Fred Lunnon

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Re: [math-fun] new(?) equilateral dodecahedron
by Bill Gosper 10 Aug '11

10 Aug '11

> APGoucher>Bill Gosper,>>>Another example, where all faces are identical, is the endo-dodecahedron.>>(The Symmetries of Things, p. 328)>> Is this four tetrahedra clumped around one <http://gosper.org/endodo.png>? No, it has five non-convex equilateral pentagonal faces and pyritohedral symmetry. Specifically, it is the shape of the 'holes' left behind when regular dodecahedra are packed together in a FCC lattice. This reminds me of something only moderately well-known: Regular dodecahedra can lock together in an infinite airtight sheet. > There's always the rhombic dodecahedron, no? Again, not pentagonal. Although I was reading somewhere about the regular dodecahedron as being the limit of a set of pyritohedral dodecahedra with vertices in Q^3, where the points are determined by ratios of successive Fibonacci numbers. The rhombic dodecahedron is indeed the degenerate case when this ratio is zero. >>Which maths camp was this?>> http://www.mathcamp.org/currentstudents/visitingspeakers.php,> whereat Julian failed to show CNWH S. Silver's marvelous> 1x1013783 > breeder<http://pentadecathlon.com/lifeNews/2011/05/quadratic_population_growth_fr.h…>> because it was "just a stunt". Grr, I bet CNWH would have loved it. Agreed. Has the same fate befallen my pi calculator? Fortunately, no, we showed him at G4G9. His traveling biographer wanted *my* reaction to it. > Sincerely, > Adam P. Goucher Also my mentioning BAMO while answering your IMO question was a non-sequitur-- they're unconnected, but for O'Dorney's participation in both.

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[math-fun] Electrocuted figures
by Adam P. Goucher 09 Aug '11

09 Aug '11

To determine the area of most plane figures, one needs to place a ruler precisely on the plane. For example, for a circle, the easiest method is by ensuring it passes directly through the centre. For a square, it should be parallel to one of the sides, or pass through two vertices, et cetera. However, suppose the person required to determine the area is being electrocuted, so they cannot control precisely (to within epsilon) the position and orientation of the ruler. It is a remarkable fact that the area of any annulus can still be determined whilst being electrocuted. Indeed, as long as the ruler is dropped such that it intersects the inner circle of the annulus, the following formula works: Area = pi*a*b, where a and b are the distances from an intersection of the ruler with the inner circle to the intersection points of the ruler with the outer circle. (The reason this works is as a result of the power of a point theorem.) Are there any other *interesting* 'electrocuted figures'? Sincerely, Adam P. Goucher

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Re: [math-fun] new(?) equilateral dodecahedron
by Bill Gosper 09 Aug '11

09 Aug '11

APGoucher>Bill Gosper, >Another example, where all faces are identical, is the endo-dodecahedron. >(The Symmetries of Things, p. 328) Is this four tetrahedra clumped around one <http://gosper.org/endodo.png>? There's always the rhombic dodecahedron, no? >Which maths camp was this? http://www.mathcamp.org/currentstudents/visitingspeakers.php, whereat Julian failed to show CNWH S. Silver's marvelous 1x1013783 breeder<http://pentadecathlon.com/lifeNews/2011/05/quadratic_population_growth_fr.h…> because it was "just a stunt". Grr, I bet CNWH would have loved it. >Was Julian at IMO 2011, also? No, Evan O'dorney<http://mathcircle.berkeley.edu/pictures2/BAMO09/bamo09_evan_julian.JPG>won the B(ay)A(rea)MO again this year. >Sincerely, > Adam P. Goucher --rwg

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Re: [math-fun] The Real 3D Mandelbrot Set (Tom Karzes)
by Stuart Anderson 08 Aug '11

08 Aug '11

Melinda Green has developed some software for projecting a 4D set to a 2D plane, with holographic like animation http://www.superliminal.com/fractals/ Here's some renderings of a 4D Mandelbrot set, http://www.superliminal.com/fractals/bgram/ZrZiIn.htm

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[math-fun] new(?) equilateral dodecahedron
by Bill Gosper 08 Aug '11

08 Aug '11

Julian flew back from math camp today (on one hr's sleep) and, generalizing a question of mine, found a dodecahedron with (planar) equilateral pentagonal faces, but not the usual ones. Before peeking <http://gosper.org/dodohedron.png>, you might try to find one. After peeking, please speak up if you've seen it before. Tnx, --rwg

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Re: [math-fun] The Real 3D Mandelbrot Set
by Dan Asimov 08 Aug '11

08 Aug '11

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.

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[math-fun] The Real 3D Mandelbrot Set
by Ray Tayek 08 Aug '11

08 Aug '11

https://christopherolah.wordpress.com/2011/08/08/the-real-3d-mandelbrot-set/ Perhaps more than any other area of serious mathematics, fractals and more specifically the Mandelbrot set, have attracted a great deal of public interest. Continuing this pattern, the <http://www.skytopia.com/project/fractal/mandelbulb.html>Mandelbulb set caused a great deal of excitement. As the Mandelbulb website it says, however, theres good reason to believe that it isnt the real McCoy. And so the question is left hanging, what is the real 3D Mandelbrot Set? In this essay, I will present the fractal that I believe deserves this position and why this is the case. <http://christopherolah.files.wordpress.com/2011/08/zrxc800001_doover_fixed_…> [] What is the Mandelbrot Set? Before we can try and generalize the Mandelbrot set, we need to understand what it is. Answering this requires a substantial amount of mathematical context and I will move somewhat quickly through the basics. If you have trouble following, look up complex numbers or ask me in the comments. Some Basic Complex Analysis The real numbers (the numbers on the number line, 0, 2, -½, ) are extremely useful things that we use in our day to day lives, but in a certain sense, they are missing something. In particular, not all polynomials have real roots, for example, x² = -1 has no real solution for x. If we introduce a number i such that i²=-1, however, it turns out that all polynomials have roots of the form a+bi where a and b are real numbers. While it took mathematicians a surprising amount of time to realize it, the complex numbers have a simple geometric interpretation. Ask yourself, where is i relative to the real number line? It isnt a positive number, since positive numbers times positive numbers are positive numbers and isnt a negative number since negative numbers times negative numbers are postive numbers and it certainly isnt 0! So it isnt on the number line. Then it must be beside the number line! Once one thinks about it this way, the complex plan interpretation is obvious: ... --- co-chair http://ocjug.org/

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[math-fun] Rational Parametric Polys vs Conic Sections vs (Un)?Solvable Quintics
by James Cloos 06 Aug '11

06 Aug '11

AIUI, in order to form conic setion curves using rational parametric polynomials such that the (Cartesian) coordinates of the loci, the radii and the coefficients are all in Q, the polymonials have to be at least quintic. Otherwise, at least one of the loci, radii or coefficients are radicals. Presuming that is true, is there any relation between that and the fact that quintics are the lowest degree polys which are not always solvable in terms of radicals (given coefficients in Q)? -JimC -- James Cloos <cloos(a)jhcloos.com> OpenPGP: 1024D/ED7DAEA6

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