February 2006 - mathfun@mailman.xmission.com

RE: [math-fun] Problem Puzzler 1999
by Eric Angelini 27 Feb '06

27 Feb '06

First pangrammatic self-describing french crossword, by Frédéric Martin, into a 22x22 grid : http://www.cetteadressecomportecinquantesignes.com/22x22.htm Both an English and a German version can be seen there: http://carremagique.site.voila.fr/ Best, É.

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Re: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
by dasimov＠earthlink.net 25 Feb '06

25 Feb '06

(*) (a_1 + a_2 + . . . )^2 = a_1^2 + a_2^2 + . . . is indeed a fascinating equation to ponder. Assume hereon that a_i -> 0 as i -> oo. Note that since sum (a_i) may not be absolutely convergent (or perhaps *must* not?), the LHS may (must?) depend on the order in which its terms are summed. Maybe the intelligentest order is to do (**) sum {n = 2 to oo} of (sum {i+j=n} of a_i*a_j). The truth of (*) <=> the cross terms (sum {i <>j} a_i*a_j) = 0. Of course, if the cross terms are NOT absolutely convergent, then there are always some (inf'ly many) orderings of them that sum to 0. (Or any other chosen sum.) So unless we make some assumption about the ordering of (a_1 + a_2 + . . .)^2, (*) is not that interesting; I propose (**). What would be really cool to see would be a *geometrical* proof of sum (1/n^2) = pi^2 / 6. That is, some object of volume pi^2 / 6 that can be partitioned into pieces of volume 1/n^2. P.S. Also note that 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 . . . . = integral {0 to 1} (1+x^2) / (1+x^4) dx --Dan

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[math-fun] where can I buy a flexing hyperboloid?
by Thane Plambeck 25 Feb '06

25 Feb '06

This is from Courant and Robbins, "What is Mathematics", pg 214 * * * One of the most remarkable properties of the hyperboloid is that although it contains two families of intersecting straight lines, these lines do not make the surface rigid. If a model of the surface is constructed from wire rods, free to rotate at each intersection, then the whole figure may be continuously deformed into a variety of shapes **** So, I'd like to play with such a model. Can I buy one somewhere? Does anything come to mind? I doubt I have patience to make one. I'm also a bit unclear on what "free to rotate at each intersection" means in the quoted passage. Can you fold it into nonhyperboloids? Google was unhelpful. Here's a photo from the book http://static.flickr.com/37/104048548_2327835121_o.gif -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
by Robert Baillie 25 Feb '06

25 Feb '06

HAKMEM 239 has this interesting question: "119 (Schroeppel): Can someone square some series for Pi to give the series Pi^2/6 = 1 + 1/2^2 + 1/3^2 + ... ?" Would you settle for squaring a series for Pi/Sqrt(8) to get a series for Pi^2/8? Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 --++ ... Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9+2 + 1/11^2 + ... Also, notice that you can square the first series by adding the squares of the individual terms! I first noticed this by comparing exercises 6a and 6b on page 398 in Wilfred Kaplan's "Advanced Calculus". The series for Pi/Sqrt(8) comes from the Fourier series for f1(x) = 0 for -Pi <= x < 0, f(x) = 1 for 0 <= x <= Pi. The series is f[x_] := (1/2) + (2/Pi) Sum[ Sin[(2n-1)x]/(2n-1), {n, 1, Infinity}] f[Pi/4] gives the series for Pi/Sqrt[8]. The series for Pi^2/8 comes from the Fourier series for g1(x)= Abs[x], -Pi <= x <= Pi. g[x_] := (Pi/2) - (4/Pi) Sum[ Cos[(2n-1)x]/((2n-1)^2), {n, 1, Infinity}] g[0] gives the series for Pi^2/8.

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[math-fun] A generalization of the Gamma function reflection formula
by Eugene Salamin 24 Feb '06

24 Feb '06

It is known that 1/(z! (-z)!) = sin(pi z)/(pi z). Does there exist a similar simplification formula for 1/(z! (wz)! (wwz)! ...), w an n-th root of 1, and the product is over all n roots? When n is even, we can use the reflection formula, but what happens when n is odd? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Tiling a square
by Kerry Mitchell 24 Feb '06

24 Feb '06

Hi all, I've found out, through sheer playing around, that a 15x15 square can be tiled exactly using 4 5x5 squares, 9 3x3 squares, and 11 2x2 squares. No big deal. What I'd like to do is determine whether or not a certain combination of smaller squares will tile a larger square, without actually finding the tiling. For example, will 2 5x5s, 11 3x3s and 19 2x2s also tile a 15x15? I've already figured out that the areas need to match. :-) Any pointers or pearls of wisdom? Thanks, Kerry Mitchell -- lkmitch(a)gmail.com www.fractalus.com/kerry

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[math-fun] 7-color torus animation ("Szilassi 42-hedron")
by R. William Gosper 22 Feb '06

22 Feb '06

Those of you with PCMacsyma 2.4 might enjoy www.tweedledum.com/rwg/tor7.mfe . Click on the last image, then the rocket icon. To tumble and zoom the moving image, click off the rocket, click the camera mount icon, Continuous Draw, then reclick the rocket. Control the speed with the movie camera icon. If you get sporadic gray frames, try reducing the section height or notebook width. If you wish to abuse it, the code is all there in the notebook. --rwg PS, rwg>>PS, I like Klarner's distinction: perfect packing -> tiling. JPropp>Can someone say what Klarner's distinction is? (I have used the >two phrases interchangeably in my own work.) Sorry, probably everybody but me had it right all along. I was misusing packing to mean tiling.

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[math-fun] perfect packings versus tilings
by James Propp 21 Feb '06

21 Feb '06

>PS, I like Klarner's distinction: perfect packing -> tiling. Can someone say what Klarner's distinction is? (I have used the two phrases interchangeably in my own work.) Jim Propp

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Re: [math-fun] AB=B.AA; Gosper's packing problem
by dasimov＠earthlink.net 21 Feb '06

21 Feb '06

Rich writes: << Gosper's packing puzzle is to fit as many 4x4x1 bricks as possible into a 7x7x7 cube. I'm assuming an "orthogonal, registered" fit, with each brick oriented parallel to a cube face, and placed at integral subcube boundaries. The answer is 18 bricks. [Here is a packing of 18.] My proof is the same as Michael Reid's: Consider the (orthogonal) lines going through the center of the 7x7x7 cube. Take the union, remove the center cell. Resulting collection of subcubes has 18 cells. Any 4x4x1 brick must occupy at least one of these cells. Note that the simple volume constraint gives an upper bound of 7^3/4^2 = 343/16 = 21.4375. >> Cute proof. As usual I'm interested in the torus version of this cube puzzle. QUESTION: Suppose we identify the opposite faces of this 7x7x7 cube, to get a 7x7x7 three-torus T. How many registered* 4x4x1 bricks will fit ? (Of course from the cube solution & the volume constraint, the answer lies in {18, 19, 20, 21}. And the proof of Michael & Rich for the cube no longer works for the torus.) --Dan ____________________________________________ * To register your bricks, just send me a check for $50.

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P.S. Re: [math-fun] Fractals and (real) snowflakes
by dasimov＠earthlink.net 20 Feb '06

20 Feb '06

3. I also wonder: What is this "fractal analysis"; Anyway? I'd expect someone trying to find a "fingerprint" of Pollock's word to use Fourier analysis: taking Fourier transforms of his known works and comparing these with the ones in question. But "fractal analysis" of paintings ???? Sounds kind of loosey-goosey to me. --dan

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