August 2008 - mathfun@mailman.xmission.com

[math-fun] Limit question
by Dan Asimov 30 Aug '08

30 Aug '08

The expression f(n) = sqrt(1 + sqrt(2 + . . . + sqrt(n)...)) approaches about 1.7579 . Rather quickly, too: For n >= 20, the first 16 digits after the decimal point remain the same. Is there a closed form for lim n->oo f(n) ? --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] an almost-uniform random variable?
by James Propp 30 Aug '08

30 Aug '08

Or I could have written: << And here's a fun fact related to Archimedes' insight about the sphere and the cylinder: If X, Y, Z are independent Gaussians of mean 0 and variance 1, |X|/(X^2+Y^2+Z^2) is uniform on [0,1]. >> Another fun result Oded told me is that if X and Y are uniform on [0,1], so is min(X,Y)/max(X,Y) (nice, not-too-hard puzzle: give a pictorial proof). >From this, it's easy to deduce that if X and Y are uniform on [0,a], for any a, then min(X,Y)/max(X,Y) is uniform on [0,1]. Oded says he came up with this as a way you can generate uniform random real numbers in [0,1] if you're writing code in a language with a built-in routine that generates random integers in some range of the form [0,a] but you forget what a is. Jim Propp

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Re: [math-fun] an almost-uniform random variable?
by James Propp 30 Aug '08

30 Aug '08

Dan Asimov is right; I should have written << And here's a fun fact related to Archimedes' insight about the sphere and the cylinder: If X, Y, Z are independent Gaussians of mean 0 and variance 1, X/(X^2+Y^2+Z^2) is uniform on [-1,1]. >> To elaborate on what Dan wrote: Note that if X, Y, and Z are independent Gaussians, then (X,Y,Z) is spherically symmetric in distribution, so if you map to the unit sphere by dividing by sqrt(X^2+Y^2+Z^2) you get a uniform point on the unit sphere. Now proceed as Archimedes did. Jim Propp

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Re: [math-fun] an almost-uniform random variable?
by Dan Asimov 29 Aug '08

29 Aug '08

In fact, for any R^2, assume that X^2 + Y^2 + Z^2 = R^2. Then by symmetry the distribution of X/R^2 is the same of the x-coordinate on [a sphere of radius 1 with the uniform distribution]. Since the area of this sphere lying between a and b for -1 <= a <= b <= 1 is proportional to b-a, it follows that the original X/(X^2+Y^2+Z^2) has a uniform distribution on [-1,1]. --Dan << << And here's a fun fact related to Archimedes' insight about the sphere and the cylinder: If X, Y, Z are independent Gaussians of mean 0 and variance 1, X/(X^2+Y^2+Z^2) is uniform on [0,1]. >> This doesn't seem possible, since there's 1/2 a chance that X, and thus X/(X^2+Y^2+Z^2), is < 0. >> _____________________________________________________________________ "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] an almost-uniform random variable?
by Dan Asimov 29 Aug '08

29 Aug '08

<< And here's a fun fact related to Archimedes' insight about the sphere and the cylinder: If X, Y, Z are independent Gaussians of mean 0 and variance 1, X/(X^2+Y^2+Z^2) is uniform on [0,1]. >> This doesn't seem possible, since there's 1/2 a chance that X, and thus X/(X^2+Y^2+Z^2), is < 0. --Dan _____________________________________________________________________ "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] DedekindEta.pdf document available on the net
by Simon Plouffe 29 Aug '08

29 Aug '08

Hello everybody, I sent a file to the math-fun list, I goofed, and realized the goof once I clicked <send> . Here is where the file is now, http://www.lacim.uqam.ca:16080/~plouffe/articles/DedekindEta.pdf This is about a document which was hard to find or get about the dedekind eta function. Simon Plouffe

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Re: [math-fun] Skolem-like puzzle sequence
by Eric Angelini 28 Aug '08

28 Aug '08

Hello Math-Fun & SeqFans, Dan Hoey found this (finite) sequence -- do you think it is worth adding it to the OEIS ? I would say yes ! S = 4, 5, 6, 7, 8, 41, 51, 63, 72, 83, 200 (there are 4 digits between the two 4's 5 digits between the two 5's 6 digits between the two 6's 7 digits between the two 7's 8 digits between the two 8's 1 digit between the two 1's ... 0 digit between the two 0's) Best, É. --- [Dan Hoey]: > The best is 456784151637283200, which can be chunked into > 4+5+6+7+8+41+51+63+72+83+200 = 540. > > (...) There are 10216 such 18-digit strings, or 9060 such > integers if we prohibit leading zeroes. --- [Eric A., a week ago]: We are looking for an 18-digit integer (like 946131483695200285) where we have "d" digits between two d's (here: one digit between two 1's, zero digit between the two 0's, nine digits between the two 9's, etc.) Those 18 digits form thus 9 pairs of digits -- the 9 pairs being different one from another. Now cut this integer into chunks so to make a finite monotonically increasing sequence (like this one, for instance, among others): 9,46,131,483,695,200285 ... and we sum the terms: 9+46+131+483+695+200285 = 201649 Question: Find the integer which, properly chunked, will give the smallest possible sum. (...)

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Re: [math-fun] AGM for Gamma(n/12)
by Dan Asimov 28 Aug '08

28 Aug '08

P.S. Given p in [0,1], experiment seems to show that for a,b > 0, (a,b) -> (p*a + (1-p)*b, a^p b^(1-p)) always converges. Does this have a name? --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] AGM for Gamma(n/12)
by Dan Asimov 28 Aug '08

28 Aug '08

I haven't read Borwein & Borwein, and I don't know much about the AGM beyond its definition. (Or one def. at least: for a,b > 0 define F(a,b) := ((a+b)/2, sqrt(ab)) and iterate; the limit is then (AGM(a,b), AGM(a,b)) if memory serves.)* Is much known about when AGM(a,b) is rational, algebraic, or transcendental? --Dan ___________________________________ * Btw, is there a natural continuum of "means" with the AM & GM at the extremes, and the AGM halfway between them? _____________________________________________________________________ "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] rolling a ball along a fractal
by James Propp 27 Aug '08

27 Aug '08

Bill Thurston writes: >For any circle on the sphere, just look at the cone tangent to the >sphere along that circle. If you roll the sphere along the circle, >it's the same as rolling the cone on this circle. The cone develops >into the plane, so you see that the circle unrolls to a circle in the >plane whose radius is its distance to the apex of the cone. This >geometric picture is better than the algebraic formula, but of course >the formula is immediate. The net angle is given by ratio of arc >lengths, i.e. 2 pi * (radius of circle in space)/(distance to apex of >cone). Thanks! >Also, once I did (and recommend) the very easy computer experiment of >drawing computer-generated Brownian curves equivalent to Brownian >rolling of a sphere. One way to do it is take a Brownian path in the >plane, then transform by using the same X-coordinates but use the Y- >coordinate as a slope. As the slope of what line? I gather that this is a description of some rational map from R^2 to R^3, but I don't know which. >When you see the picture it becomes intuitively >clear. Is this written up somewhere (with the picture included)? Jim Propp

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