May 2005 - mathfun@mailman.xmission.com

Re: [math-fun] A puzzle I read about
by Eugene Salamin 18 May '05

18 May '05

The simple limit argument is justified because S(x) is monotonically increasing as x --> 1 from below along the real axis. This is clear by grouping the terms of the series as S(x) = (x - x^2) + (x^4 - x^8) + ... . Without some restriction on the path, the limit does not exist. The same grouping of terms shows that S(x) diverges to infinity when x --> w (w being a cube root of 1) along a line from the origin. Hence S(x) must also diverge to infinity along the line to exp(i t) with t of the form t = t[n] = (2 pi)/(3 2^n). Thus there exists a sequence of points x[n] --> 1, with arg(x[n]) = t[n] such that S(x[n]) --> infinity. Gene __________________________________ Do you Yahoo!? Yahoo! Mail - Helps protect you from nasty viruses. http://promotions.yahoo.com/new_mail

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[math-fun] Packing circles into a finite rectangle
by Robert Baillie 18 May '05

18 May '05

If you have circles of radii 1/1, 1/2, 1/3, ..., 1/n for all n >= 1, then their total area is finite, although the sum of their radii is infinite. Can one fit these circles into a finite-sized rectangle? Bob Baillie

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Re: [math-fun] A puzzle I read about
by Eugene Salamin 18 May '05

18 May '05

--- <dasimov(a)earthlink.net> wrote: > If you haven't seen this one before, it's worth a try: > > (Provenance to be provided with solution.) > > << > For positive x<1, consider the alternating sum > S(x) = x - x2 + x4 - x8 + x16 - x32 + - ... > Does S(x) approach a limit as x approaches 1 from below, and if so > what is this limit? > >> > > --Dan f(x) + f(x^2) = x. Let x --> 1. 2 limit = 1. limit = 1/2. Gene __________________________________ Yahoo! Mail Mobile Take Yahoo! Mail with you! Check email on your mobile phone. http://mobile.yahoo.com/learn/mail

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[math-fun] A puzzle I read about
by dasimov＠earthlink.net 18 May '05

18 May '05

If you haven't seen this one before, it's worth a try: (Provenance to be provided with solution.) << For positive x<1, consider the alternating sum S(x) = x - x2 + x4 - x8 + x16 - x32 + - ... Does S(x) approach a limit as x approaches 1 from below, and if so what is this limit? >> --Dan

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Re: [math-fun] hexagons, sequences, omino
by Richard Schroeppel 18 May '05

18 May '05

This is from Dylan Thurston. --Rich ----- Date: Wed, 18 May 2005 01:18:25 -0400 Reply-To: dpt(a)math.harvard.edu From: dpt(a)lotus.bostoncoop.net (Dylan Thurston) Subject: Re: [math-fun] hexagons, sequences, omino On Tue, May 17, 2005 at 08:50:52AM -0600, Schroeppel, Richard wrote: > Scott asks ... > > Is there a nice general formula or a good approximation for the number > > of nondecreasing sequences in N^k, for integers N and k ? > > For an exact formula: > The 1D case is already answered, a quotient of factorials. > Based on experiments done long long ago, at an Institute far far away, > the 2D case seems to be a product of something like the 1D answers. > Strict vs. non-strict vs. semi-strict perturbs the formulas slightly. The 2D case is actually a well-known problem, called "plane partitions"; it's equivalent to counting lozenge tilings of a hexagon. (A 'lozenge' is a 60-120-60-120 rhombus obtained by gluing two triangles.) You have an rectangular array of numbers, non-decreasing in the rows and columns. Above each square, make a stack of cubes with height given by the number in that square. Add in two walls and the floor of the room. Tilt the whole construction and balance it on one corner; looking at it from the corner, you get a lozenge tiling of a hexagon. (Recall that if you look at a cube from a corner you see a hexagon.) There are many different symmetry classes of such tilings. Many have nice counts. > This also worked for rectangles, MxN. But I couldn't find a similar > answer for the 3D case. I always figured there was a tie-in with linear > and planar partitions having a nice generating function, while cubic > partitions don't seem to have one. Other people have noticed this too; I haven't heard a convincing explanation why this should be true. Peace, Dylan

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[math-fun] Random polyominoes
by David Wilson 18 May '05

18 May '05

Suppose we tile the plane randomly with black and white unit square tiles in a grid pattern, black and white tiles occuring with equal probablity (sort of like the static you used to see on your old black and white TV). Consider the polyominoes formed by connecting like-colored tiles along their edges. Since every possible coloring of an nxn subsquare should occur, I'm thinking that every finite polyomino should occur in every orientation. What is the probability that a randomly chosen square is in a unomino (I find only one page on the web with that word!)? What is the probability that a randomly chosen polyomino is a unomino? What is the most frequent polyomino in both senses? I am thinking that if black squares occur with probability 1/2, the probability of an infinite polyomino is 0. Certainly if they occur with probability 1, the probability of an infinite polyomino is 1. What about choosing black squares with probability 3/4? Other probabilities? - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] Lattice percolation constant
by Richard Schroeppel 18 May '05

18 May '05

Michael Kleber sent this this morning. [Date: Tue, 17 May 2005 12:06:30 -0400] > Oof: I wrote all of > The standard big results here are: (1) there is a critical probability, > usually denoted p_c, such that if you declare squares open with > probability p < p_c, the probability of an infinite open component is > zero, while with p > p_c it is 1. In dimension two, it's even known > that p <= p_c is the probability zero condition; what happens at p_c > exactly is known for some other lattices, but I don't know the full story. ... but I neglected to mention that Robert Ziff's 1992 paper "Spanning probability in 2D percolation" (Phys. Rev. Lett. 69, 2670-2673 (1992)) calculated that p_c for the 2d square lattice is 0.5927460 +- .0000005 But his message included a special character for +-, and another for the dash between the page numbers in the reference. This seems to have poisoned it for ascii readers. I'd like to get an idea of what OS/browser/mailsoftware and other likely configuration stuff affected the reading of this message. Would those readers with a spare minute please send ME (not the list) a brief note saying which way they got the message, junk or intelligible, and what OS/&c/&c they are using. I'll try to figure out what's going on, and perhaps be able to do something about it. Rich rcs(a)cs.arizona.edu

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Re: [math-fun] Random polyominoes
by dasimov＠earthlink.net 18 May '05

18 May '05

Re percolation probabilities: If we're talking about about site percolation on the hexagonal lattice in the plane -- i.e., each site is surrounded by 6 others -- then the critical probability = 1/2. For the square lattice, bond percolation (i.e., a unit edge of the square tiling is present with a given probability) also has critical probability = 1/2 (of having an infinite component made of these edges). --Dan

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[math-fun] Questions about written numbers
by David Wilson 17 May '05

17 May '05

1. I had a professor who tried to teach us to write the digit 1 with the serif extended to the baseline, like a very narrow upper case lambda. Ostensibly this was easier to write and gave the digit 1 more body than a single stroke. Has anyone ever heard of this technique of writing the digit 1? 2. I was also at some point taught to add a short downstroke at the end of the vinculum in a radical sign, which ostensibly better demarcates the argument. I still do this. Has anyone ever heard of this technique? 3. Printed fonts in which the digits are all h-height and rest on the baseline seem to be fairly old, for example, I have a facsimile version of a 1611 King James Bible in which all of the digits are equal h-height. This seems to be fairly universal in modern fonts. However, there are printed fonts from the 19th and early 20th century, and probably earlier, in which 0, 1, and 2 are x-height; 3, 4, 5, 7 and 9 descend below the baseline, and 6 and 8 reach up to h-height. In my old book collection, I have several books in which the page numbers are printed using this sort of font, including a reprint of Vega's 1856 logarithmic and trigonometric tables, in which the preface uses equal-height digits, while the tables have a mix of equal-height and ascending and descending digits. I find the ascending and descending digits very beautiful and I wondered if anyone might have a font with such digits. If anyone is interested, I will scan an image of a page of Vega's tables for an example.

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RE: [math-fun] hexagons, sequences, omino
by dasimov＠earthlink.net 17 May '05

17 May '05

Rich asks: << Dan asks ... > Let T denote the torus obtained by identifying > each pair of opposite edges of a regular hexagon. How does this work? I can't get the edges to match up. >> The regular hexagon's 3 pairs of opposite edge are each identified to become 1 edge. The 6 edges you began with end up as 3 edges. The 6 vertices you began with end up as 2 vertices. The identified space is locally Euclidean, and has Euler char = V-E+F = 2-3+1 = 0. Since this surface is orientable, it must be a torus. (Equivalently, tile the plane with regular hexagons so that the origin is the center of some tile. Denote the group of translations that preserve the tiling as G. Then the quotient R^2/G is a metric space that is exactly the same as the result of identifying the hexagon as above: a certain flat torus.) --Dan

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