January 2007 - mathfun@mailman.xmission.com

Re: [math-fun] Ultraviolet catastrophe & Maxwell's demon
by Daniel Asimov 23 Jan '07

23 Jan '07

Henry wrote: << I attended a talk last night by an MIT professor who is working with "photonic crystals", which have the property that they can selectively allow photons of certain wavelength/energy to propagate, while disallowing other photons. . . . This is not a perpetual motion machine of the first type, because energy is conserved, but isn't it a perpetual motion machine of the second type? >> I don't know from "PMM's of the nth type" -- but wouldn't a photon moving through empty space without hitting anything be a kind of PMM ? Hmm, I suppose that since force fields, though attenuated, extend (AFAIK) indefinitely through space, any photon would realistically experience "friction" with various force fields, so would gradually slow down . . . --Dan

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[math-fun] Ultraviolet catastrophe & Maxwell's demon
by Henry Baker 23 Jan '07

23 Jan '07

(I apologize if this is more physics than math.) I attended a talk last night by an MIT professor who is working with "photonic crystals", which have the property that they can selectively allow photons of certain wavelength/energy to propagate, while disallowing other photons. These devices are non-dissipative, so that the non-selected photons simply bounce back. One type of crystal is made of a conductor with small holes, and the size of the holes determines which wavelengths can get through. (Think Faraday cage.) This professor suggests that one can put such a photonic crystal "filter" in front of a "blackbody", and thereby allow only the most energetic photons to leave. _Presumably_, the photons that bounce back into the blackbody exchange energy with the blackbody to reconstitute the classic blackbody radiation spectrum, so that the only photons leaving the blackbody are those with high energies. This is not a perpetual motion machine of the first type, because energy is conserved, but isn't it a perpetual motion machine of the second type? Why doesn't this idea violate the non-existence of Maxwell's demon?

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[math-fun] Puzzle: Can you ID this sequence?
by Daniel Asimov 21 Jan '07

21 Jan '07

Here are the first 20 terms of an infinite sequence of nonnegative integers: 0,1,0,0,1,0,2,0,1,1,2,2,1,0,0,1,0,1,0,2,... Can you ID this (somewhat novel) sequence? (Hint: It's based on pi/4, and each term is the floor of a certain finite expression in closed form.) --Dan

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[math-fun] some pi trivia
by R. William Gosper 21 Jan '07

21 Jan '07

(With some prodding from Jason H) The Newton iteration for e^(ix)=-1 is (c184) %i*%e^-(%i*x)+x+%i - %i x (d184) %i %e + x + %i which indeed converges quadratically (c185) subst(3.b0,x,%) (d185) 1.0007503399554542729b-2 %i + 3.1411200080598672221b0 (c186) subst(%,x,d184) (d186) 3.1415974073206700026b0 - 5.0129702705366286305b-5 %i (c187) subst(%,x,d184) (d187) 3.141592653828090399b0 - 1.2451741388209226812b-9 %i (c188) subst(%,x,d184) (d188) 3.1415926535897932388b0 - 7.4708305947829289906b-19 %i (c189) subst(%,x,d184) (d189) 3.1415926535897932385b0 (interesting that it got imagpart exact. Macsyma doesn't, e.g., force 3.1... + .69b-105*%i to real.) But note that the convergence is *cubic* if we just subst realparts: (c190) subst(realpart(d185),x,d184) (d190) 1.1169689640015603699b-7 %i + 3.1415926535721955587b0 (c191) subst(realpart(%),x,d184) (d191) 3.1415926535897932385b0 In fact, this has become Salamin's x+sin(x) iteration, which is simply Newton's method for cot(x/2). But Jason objected to using e, so we cheat: x 1 (d245) (%i - -) (----------- + 1) + x n %i x n (---- + 1) n (c253) (bfprecision : 2,subst([x = 3.0b0,n = 2^4],d245)) (d253) 3.0b-1 %i + 3.1b0 (c254) (bfprecision : 4,subst([x = realpart(%),n = 2^8],d245)) (d254) 2.455b-2 %i + 3.14b0 (c255) (bfprecision : 8,subst([x = realpart(%),n = 2^16],d245)) (d255) 1.6573659b-4 %i + 3.1415924b0 (c256) (bfprecision : 16,subst([x = realpart(%),n = 2^32],d245)) (d256) 3.141592653589794b0 - 1.244747560102321b-9 %i (c257) (bfprecision : 32,subst([x = realpart(%),n = 2^64],d245)) (d257) 3.1415926535897932384626433832795b0 - 1.8571745299313218459235921249323b-17 %i I carefully wrote a recursive (1+epsilon)^2^n - 1 function. Why is it needless? --rwg

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[math-fun] Prime magic squares
by Richard Guy 20 Jan '07

20 Jan '07

Apologies if this is a repetition of earlier messages, which I've deleted, but I've just chanced upon Allan William Johnson, Problem 872, Prime magic squares, J. Recreational Math, 12:3(1980-81) 224 which gives 1255 as the smallest possible magic sum for a 5 x 5 symmetrical prime magic square 191 401 11 389 263 431 419 23 233 149 41 53 251 449 461 353 269 479 83 71 239 113 491 101 311 and 630 for a 6 x 6 such square 131 199 17 37 163 83 29 19 157 167 61 197 151 103 139 109 97 31 179 113 101 71 107 59 13 149 43 53 191 181 127 47 173 193 11 79 P.S. I've unsubscribed from the seqfan network -- if this was (also) on that, perhaps someone who's on both will forward it. Best to all, R.

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[math-fun] Surveying a dig (or graphing on blank paper)
by Jason Holt 20 Jan '07

20 Jan '07

When we teach people how to use cartesian coordinates, we tell them to move x units along the x axis, y units up the y axis, and then take the intersection of the perpendiculars. The other day it occurred to me to wonder what would happen if instead, we drew a line from (x,0) to (0,y) and called the midpoint our destination. So if [a,b] are coordinates in Midpointesian coordinates (corresponding to the cartesian endpoints of the line (a,0) and (0,y)), and the point (x,y) is the corresponding point in Cartesian space, we could have a function f([a,b]) = (x,y). That function turns out quite trivial, but was nonetheless fun to try to visualize before I realized how straightforward it actually is, so I won't spoil it by giving it explicitly :) Then I realized it would actually be a rather useful thing to do. Say you're sketching out the graph of a function, but you don't have any graph paper. Use a compass and ruler to create X and Y axes, and mark at regular intervals. To plot a point (x,y) s.t. f^-1((x,y))=[a,b], draw a line from (a,0) to (0,y) and then use the compass to find the midpoint. Or say you're mapping out things you find at an archaeological dig. Create an X and Y axis at your site, and when you want to find the placement of an item, have a friend walk along each axis holding ends of a tape measure, with you holding it in the middle. When you read half the value seen by the friend holding the other end, they record their positions on the axes. (Or, in fact you could also record [a,b,r] for any line passing through the item, (a,0), (0,y) and showing a distance of r on the tape).

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[math-fun] Add and write sequence
by Eric Angelini 19 Jan '07

19 Jan '07

Hello Math-Fun and SeqFan, Start with 1, then a(n) is the smallest integer not used so far whose leftmost digits show the sum of a(n-1)'s digits: 1,10,11,2,20,21,3,30,31,4,40,41,5,50,51,6,60,61,7,70,71,8,80,81, 9,90,91,100,12,32,52,72,92,110,22,42,62,82,101,23,53,83,111,33, 63,93,120... This infinite seq. is certainly a rearrangement of the natural integers. Starting with my birthday year produces: 1951,16,7,70,71,8,80,81,9,90,91,10,1,11,2,20,21,... The present year gives: 2007,9,90,91,10,1,11,2,20,21,3,30,31,... Will the 1951-sequence and the 2007-sequence merge with the first one, sooner or later? This is the 11-sequence: 11,2,20,21,3,30,31,4,40,41,5,50,51,6,60,61,7,70,71,8,80,81,9,90, 91,10,1,12... 12 is there because 11 is forbidden; we could thus assign to 11 a "pseudo-loop" length -- the number of integers which separate the starting "11" from his expected second appearance. Here, this pseudo-loop has length 27. Could someone compute a few pseudo-loop lenghtes (say for n=1 to n=200)? This seq. obviously begins like this (n=1 to n=11): 1,1,1,1,1,1,1,1,1,2,27... Best, É.

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[math-fun] Chess -- mark legal squares
by Eric Angelini 19 Jan '07

19 Jan '07

Published in Die Schwalbe, Heft 222, december 2006 by Gerald Irsigler: How many squares can you mark at most on a chess board so that every possible position with pieces on those squares (including wK and bK) is legal? (e.g.: mark e1 g2 h1 isn't valid, since wKe1, wBh1 wKg2 isn't legal. e1 g3 h1 is valid though) Best, É.

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Re: [math-fun] two kinds of partitions
by Daniel Asimov 18 Jan '07

18 Jan '07

I wrote in response to Rich's post: << The only thing I notice worth mentioning is that (clearly) in counting the partitions of a multiset, the count depends only on the "shape" of the multiset, and such shapes are in one-to-one correspondence with the set of integer partitions. So this defines an interesting function from the set of integer partitions to the integers. >> Another potentially interesting function here goes from integers to multisets. Given any integer N, one has its set of partitions, P(N). For each partition p in P(N), let K(p) be the number of partitions of [a multiset having shape p]. Then the set M(N) := { K(p) : p is in P(N) } is a multiset (of integers) depending on N. Going yet farther, M(N) has a shape which is an integer partition, say part(N). This defines a function taking the integer N to the integer partiion part(N). --Dan

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[math-fun] crossword puzzle clue, 29 Dec 2006 New York Times
by Thane Plambeck 17 Jan '07

17 Jan '07

52 Across: Pioneer in the math of Sudoku (5 letter answer) I won't spoil it by giving the answer, which you can have by clicking this URL http://tinyurl.com/yqvedr My question is this: is there something interesting that Mr. X had to do with the "math of Sudoku"? I'm aware of general things, but is there something specific that comes to mind? -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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