May 2005 - mathfun@mailman.xmission.com

Re: [math-fun] Re: A puzzle I read about
by dasimov＠earthlink.net 19 May '05

19 May '05

Mike Speciner writes << Well, how about 0 1 k k+1 2k 2k+1 ... Then (for x < 1) the alternating sum is just (1-x)/(1-x^k) which is 1/(1+x+x^2+ ... x^(k-1)) As x->1, this -> 1/k. --ms >> This puts the kibosh to what I thought was a proof that if the alternating series sum k=0 to oo of (-1)^k x^(n_k) converges, then it converges to 1/2 (as Gareth asked). [The "proof" just looked at the terms of the Cesaro summation, which for a convergent series also converges. That is, the averages of the first n terms of the given series for n = 1,2,3,.... The sequence of these averages' limits as x -> oo approaches 1/2. (But I see from Mike counterexample that one can't argue this implies the conclusion I wanted.)] --Dan

3 3

Re: [math-fun] Re: A puzzle I read about
by dasimov＠earthlink.net 19 May '05

19 May '05

Note: I was the one with the failed "proof", not Gareth. --Dan

1 0

RE: [math-fun] Re: A puzzle I read about
by dasimov＠earthlink.net 19 May '05

19 May '05

This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1. --Dan --------------------------------------------------------------------------------- << 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. >>

5 4

Re: [math-fun] Re: A puzzle I read about
by Eugene Salamin 19 May '05

19 May '05

I plotted Dan's function S(x) = x - x^2 + x^4 - x^8 + x^16 ... for x near 1. To get a nice scaling, I did S(1-exp(-x)). It looks periodic: sinusoidal oscillations, 0.500 +- 0.003, period 1.39 ?=? 2 log(2). Maple bombs at x=37.5, so the plots end there. Then I tried f(1-exp(-x)) with f(x) = 1 - x + x^2! - x^3! + x^4! ... Again oscillating, but with increasing amplitude and period. Amplitude was 0.5 +- 0.3 when Maple bombed, and period was 5.6. Is there a theory for such functions that explains their values near x = 1 or elsewhere on the natural boundary, in analogy to the theory for modular functions? Gene __________________________________ Do you Yahoo!? Yahoo! Mail - Find what you need with new enhanced search. http://info.mail.yahoo.com/mail_250

1 0

Re: [math-fun] Random polyominoes
by Michael D Beeler 19 May '05

19 May '05

Black and white tiles in the shower at a Palo Alto Hyatt got me working on random polyominoes. I think it's a great topic for high school algebra class. Paraphrasing from a short, unpublished paper I wrote in 1988: b = number of tiles in body of figure (= order of polyomino) n = number of tiles in neighborhood to define figure r = number of distinct orientations p = probability of a tile being figure color q = 1-p = probability of a tile being background color P = probability of finding a given figure in any orientation and "centered" on a specified tile = r p^b q^n A convenient set to analyze is the polyominoes through order 5. Some have identical b, n and r values, and therefore have the same chance of occurring for all p. For example, the T and Z tetrominoes have b=4, n=8, and r=4. Because of cases like this, the 21 polyominoes through order 5 are described by only 16 probability equations. For each of the 16 classes, it is not hard to find the p and P where the chance of occurrence peaks, and where its curve touches or crosses the curves of the other classes. It's easy to generalize the method to several colors, or to non-connected figures. Some problems, most of which I don't know the answer to, are: 1. If we look only for black polyominoes on a white background, the P curve for most polyominoes is asymmetric. For polyominoes with b=n, it is symmetric. An easy example is the 4X4 16-omino. Is there a lower-order polyomino whose P curve is symmetric? Are there any other 16-ominoes with a symmetric P curve? What is the smallest irregular (define: r=8) polyomino with a symmetrical curve? 2. Is there any polyiamond or polyhex whose b, n and r values are the same as those of some polyomino? 3. The point at p=1/2 where the P curves for four classes of polyomino intersect is difficult to graph, because of the relative values of the curves. What scaling of the axes would show the area around this point clearly? (Members of those 4 classes are: straight tetromino; L pentomino; T, U & Z pentominoes; X pentomino.) 4. For polyominoes of any given order, there is a minimum and a maximum n. Consider a histogram of the n values for a given b. Is this histogram always unimodal, for every b? If so, what is the most common n, as a function of b, as b increases without bound? If not, what is the smallest order of polyomino with a multimodal histogram? 5. What is the smallest order polycube whose P curve is symmetric? What are the two smallest polycubes whose P curves intersect? 6. It is easy to show that p of the peak = b / (b+n). Using this, it is easy to show that the peak likelihood is P_peak = r (b^b n^n) / (b+n)^(b+n). The symmetry of this formula in b and n leads to the following conclusion. Consider two polyominoes, the b of one equal to the n of the other, and vice versa. The symmetry of the P_peak equation tells us that the likelihood curves of the two shapes have the same maximum value. Can you find any such pairs of polyominoes? Better yet, can you find a way of generating an infinite family of such pairs? One idea is pairs of rectangles, but this does not work to make an infinite family. With some effort, you can show there are only 5 integer rectangle pairs where the area of one is the perimeter of the other, and vice versa: 1 X 34 and 7 X 10 1 X 38 and 6 X 13 1 X 54 and 5 X 22 2 X 10 and 4 X 6 2 X 13 and 3 X 10 (The related 3-dimensional problem of surface-volume pairs of bricks is much harder. Rich Schroeppel and Dean Hickerson worked some theory on this. My computer search found there are 246 pairs of different bricks, plus 10 bricks whose S = V.)

1 0

[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

1 0

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. . . . >>

1 0

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: >>

1 0

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

2 1

[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

1 0