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[math-fun] pi^2/6 packing problem
by Richard Schroeppel 18 May '05

18 May '05

Dan asks > Hmm, what about other rectangles also with area pi^2/6 ? > For example the pi/sqrt(6) x pi/sqrt(6) square. No, you need one edge of the rectangle to be at least 1.5, if the squares are placed othogonally. It looks even worse if they're rotated 45 degrees. But pi/2 x pi/3 looks like a possibility. Rich

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RE: [math-fun] Re: Packing circles into a finite rectangle
by dasimov＠earthlink.net 18 May '05

18 May '05

Hmm, what about other rectangles also with area pi^2/6 ? For example the pi/sqrt(6) x pi/sqrt(6) square. Is there any one of these where such a packing is known to be possible or known to be impossible? And I have to add, of course: What about the potentially easier question of packing the 1/n x 1/n squares into a flat *torus* of area pi^2/6 ? (For that matter, what about other open sets of area pi^2/6 ? Conjecture: Given an open set O that for each n >=1 can contain disjoint open squares of sides 1, 1/2, 1/3,...,1/n , then O can contain disjoint squares of all sides 1/n, n >= 1 simultaneously. --Dan --------------------------------------------------------------------- Jim Propp writes: << Since the sum of the reciprocals of the squares of the positive integers is pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, etc can be packed into a rectangle of size 1 by pi^2/6. A picture of such a packing appears at http://www.pisquaredoversix.force9.co.uk/Tiling.htm I know of no proof that such a packing is possible or impossible. A program which I wrote has recently packed the first one million such squares into such a rectangle. . . . >>

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Re: [math-fun] Packing circles into a finite rectangle
by dasimov＠earthlink.net 18 May '05

18 May '05

Man, those are cool animations! --Dan ---------------------------------------- Erich wrote: << [Rich wrote]: > This suggests an interesting, though difficult, contour plot to make > in the 1st quadrant of the plane: F(x,y) = the largest number of diameter-1 > circles that can be packed in an x*y rectangle. i've managed to find all F(x,y) <= 8. you can see them animated here: >>

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RE: [math-fun] Packing circles into a finite rectangle
by Schroeppel, Richard 18 May '05

18 May '05

This suggest an interesting, though difficult, contour plot to make in the 1st quadrant of the plane: F(x,y) = the largest number of diameter-1 circles that can be packed in an x*y rectangle. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com on behalf of Erich Friedman Sent: Wed 5/18/2005 11:38 AM To: math-fun Subject: Re: [math-fun] Packing circles into a finite rectangle > 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? yes, they will fit inside a 1 x (11/6) rectangle for example. put each circle inside a square, and then pack the squares like this: http://www.stetson.edu/~efriedma/harmonic/ erich _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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

18 May '05

Dan Asimov asked: > 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? The limit does not exist. Since S(x) = x - S(x^2), the limit would have to be 1/2 if it existed. But I'll show that there are values of x arbitrarily close to 1 for which S(x) > 0.5008. Rewrite the definition as S(x) = SUM (x^4^k - x^(2*4^k)) k>=0 For 0 < x < 1, the summands are all positive, so any partial sum is a lower bound for S(x). In particular, calculation shows that, for x=199/200, the sum with k going from 0 to 5 is 0.50088158499... > 0.5008, so S(199/200) > 0.5008. Next note that S(x) = x - x^2 + S(x^4) > S(x^4). So S((199/200)^(1/4)) > S(199/200) > 0.5008, S((199/200)^(1/16)) > S((199/200)^(1/4)) > 0.5008, S((199/200)^(1/64)) > S((199/200)^(1/16)) > 0.5008, etc. Since the arguments of S approach 1, the limit can't be 1/2, so it doesn't exist. Dean Hickerson dean(a)math.ucdavis.edu

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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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