November 2002 - mathfun@mailman.xmission.com

Re: [math-fun] A problem from the files of David Klarner
by Mike Stay 13 Nov '02

13 Nov '02

There's a simple extension of the trivial case you mentioned: Let E,F be the midpoints of AB, CD respectively. Then you can cut the rectangle AEFD lengthwise. I suspect there's a way to cut it where you turn one of the pieces over... -- Mike Stay staym(a)datawest.net ----- Original Message ----- From: Thane Plambeck Sent: 11/13/2002 13:54:32 To: math-fun(a)mailman.xmission.com Subject: [math-fun] A problem from the files of David Klarner > Draw a square ABCD with center M. Cut out the triangle T= BMC. > Rotate and reattach T to an adjacent side of whats left of the square > by lining up side BC with either CD or AB. > > Now you've got a six-sided figure. Suppose you want to cut this > figure into two pieces that can reassembled into the original square shape. > > Of course, one way to do it is to lop of the original triangle and put it back > where it came from. > > Find another solution. > > * * * * * > > In Klarner's files there are some notes that credit this highly addictive little > problem to some unknown person in Eindhoven. Does anyone know anything > more about it? > > For more on Klarner's files (he died in 1999) > > http://www.qxmail.com/mathematics/klarner > > > > Thane Plambeck > 650 321 4884 office > 650 323 4928 fax > http://www.qxmail.com/home.htm >

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[math-fun] A problem from the files of David Klarner
by Thane Plambeck 13 Nov '02

13 Nov '02

Draw a square ABCD with center M. Cut out the triangle T= BMC. Rotate and reattach T to an adjacent side of whats left of the square by lining up side BC with either CD or AB. Now you've got a six-sided figure. Suppose you want to cut this figure into two pieces that can reassembled into the original square shape. Of course, one way to do it is to lop of the original triangle and put it back where it came from. Find another solution. * * * * * In Klarner's files there are some notes that credit this highly addictive little problem to some unknown person in Eindhoven. Does anyone know anything more about it? For more on Klarner's files (he died in 1999) http://www.qxmail.com/mathematics/klarner Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm

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[math-fun] Re: seven
by hoey+l＠aic.nrl.navy.mil 13 Nov '02

13 Nov '02

Jim Propp wrote: > My mistake; as Dan Asimov pointed out, the sequence has period 5. That was Dan Hoey, not that it matters. > Here are the true facts I should have asked about (neither of which > involves the number seven): > 1) The map (x,y) |-> (y,(y+1)/x) is of order 5; > 2) The map (x,y,z) |-> (y,z,(y+z+1)/x) is of order 8. Also the map (x,y) |-> (y,y/x) is of order 6. Is there any c other than 0 or 1 for which (x,y) |-> (y,(y+c)/x) has finite order? Dan Hoey(a)AIC.NRL.Navy.Mil

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[math-fun] Re: seven
by James Propp 13 Nov '02

13 Nov '02

I wrote: >Does the group of order 168 shed any light on the curious fact that >the map > (x,y) |-> (y,(y+1)/x) >(from C(x,y) to itself) is of order 7? > >(Another way of stating this fact is that the sequence x,y,(y+1)/x, >((y+1)/x)+1)/y,... satisfying a(n+1) = (a(n)+1)/a(n-1) is periodic >with period 7. This sequence is usually attributed to someone named >Lyness, but I don't know why Lyness studied it.) My mistake; as Dan Asimov pointed out, the sequence has period 5. Here are the true facts I should have asked about (neither of which involves the number seven): 1) The map (x,y) |-> (y,(y+1)/x) is of order 5; 2) The map (x,y,z) |-> (y,z,(y+z+1)/x) is of order 8. I'd like to know if there's a nice picture to explain what's going on with these recurrences. (But I don't expect it to involve the simple group of order 168!) Jim Propp

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[math-fun] Review 0976.91008
by Richard Guy 13 Nov '02

13 Nov '02

In your review of Fraenkel & Rahat, Infinite cyclic impartial games, Theor. Comput. Sci. 252, No.1-2, 13--22(2001), the reviewer has failed to understand the import of the Sprague-Grundy theory and the late Cedric Smith's infinite extension thereof. In the present context, in the comparatively uninteresting case of play with a single token on a graph with a single component, it would indeed suffice to classify the positions as N-positions, O-positions or P-positions. But as soon as there are two or more N-positions we cannot decide the outcome. We need the simple but powerful theorem that the nim-value (Sprague-Grundy value) of the sum (i.e., disjunctive compound) of games is the nim-sum of their nim-values. Every impartial game is equivalent to playing Nim -- but we need to know how many beans there are in each heap. Mit bestem Gr"usse, Richard K. Guy.

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Re: [math-fun] seven
by Dan Hoey 13 Nov '02

13 Nov '02

James Propp <propp(a)math.wisc.edu> writes: > Does the group of order 168 shed any light on the curious fact that > the map > (x,y) |-> (y,(y+1)/x) > (from C(x,y) to itself) is of order 7? > (Another way of stating this fact is that the sequence x,y,(y+1)/x, > ((y+1)/x)+1)/y,... satisfying a(n+1) = (a(n)+1)/a(n-1) is periodic > with period 7. This sequence is usually attributed to someone named > Lyness, but I don't know why Lyness studied it.) > Jim Propp Are you sure? I get x, y, (y+1)/x, (x+y+1)/(xy), (x+1)/y, x, y, ... with period 5. Cute, the way it follows the same rule forwards as backwards. But of course the rule a(n)+1 = a(n+1)a(n-1) is symmetric. Dan

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[math-fun] misc admin
by Richard Schroeppel 13 Nov '02

13 Nov '02

There's no need to send Math-Fun postings to both the Arizona and the Xmission addresses. Going to Xmission directly will be a little faster. I've set the mailing-list's Reply-To option to send replies to the group. If you want to send a private reply to a poster, you'll need to edit your message. A reminder that we don't have any filters for de-Mime-ing messages; please try to stick to plain ascii text. Rich rcs(a)cs.arizona.edu

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[math-fun] seven
by James Propp 13 Nov '02

13 Nov '02

Does the group of order 168 shed any light on the curious fact that the map (x,y) |-> (y,(y+1)/x) (from C(x,y) to itself) is of order 7? (Another way of stating this fact is that the sequence x,y,(y+1)/x, ((y+1)/x)+1)/y,... satisfying a(n+1) = (a(n)+1)/a(n-1) is periodic with period 7. This sequence is usually attributed to someone named Lyness, but I don't know why Lyness studied it.) Jim Propp

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Re: [math-fun] Re: product integrals
by MCKAY＠vax2.concordia.ca 12 Nov '02

12 Nov '02

dddddddqOBFrom: IN%"math-fun(a)mailman.xmission.com" 12-NOV-2002 02:37:50.86 To: IN%"math-fun(a)mailman.xmission.com" CC: IN%"mcintosh(a)servidor.unam.mx" Subj: [math-fun] Re: product integrals Return-path: <math-fun-admin(a)mailman.xmission.com> Received: from clyde.concordia.ca (clyde.Concordia.CA [132.205.1.1]) by vax2.concordia.ca (PMDF V6.1 #40130) with ESMTP id <01KORJ7ZCWGC005UDP(a)vax2.concordia.ca> for mckay(a)vax2.concordia.ca (ORCPT mckay(a)vax2.concordia.ca) Tue, 12 Nov 2002 02:37:49 -0400 (EDT) Received: from mailman.xmission.com (mailman.xmission.com [198.60.22.29]) by clyde.concordia.ca (8.12.3/8.12.3) with ESMTP id gAC7cGbc030948 for <mckay(a)vax2.concordia.ca>; Tue, 12 Nov 2002 02:38:16 -0500 (EST) Received: from localhost ([127.0.0.1] helo=mailman.xmission.com) by mailman.xmission.com with esmtp (Exim 3.35 #1 (Debian)) id 18BVcY-0005hD-00; Tue, 12 Nov 2002 00:38:02 -0700 Received: from servidor.unam.mx ([132.248.10.1]) by mailman.xmission.com with esmtp (Exim 3.35 #1 (Debian)) id 18BVbx-0005g2-00 for <math-fun(a)mailman.xmission.com>; Tue, 12 Nov 2002 00:37:25 -0700 Received: from pine.servidores.unam.mx (pine.servidores.unam.mx [132.248.10.10]) by servidor.unam.mx (8.11.1/8.11.1) with SMTP id gAC7UcK11963; Tue, 12 Nov 2002 01:30:38 -0600 (CST) Received: by pine.servidores.unam.mx (sSMTP sendmail emulation); Tue, 12 Nov 2002 01:40:01 -0600 Date: Tue, 12 Nov 2002 01:40:01 -0600 From: mcintosh(a)servidor.unam.mx Subject: [math-fun] Re: product integrals Sender: math-fun-admin(a)mailman.xmission.com To: math-fun(a)mailman.xmission.com Cc: mcintosh(a)servidor.unam.mx Errors-to: math-fun-admin(a)mailman.xmission.com Reply-to: math-fun(a)mailman.xmission.com Message-id: <200211120730.gAC7UcK11963(a)servidor.unam.mx> Precedence: bulk X-BeenThere: math-fun(a)mailman.xmission.com X-Mailman-Version: 2.0.11 X-Scanned-By: MIMEDefang 2.23 (www . roaringpenguin . com / mimedefang) List-Post: <mailto:math-fun@mailman.xmission.com> List-Subscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=subscribe> List-Unsubscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=unsubscribe> List-Archive: <http://mailman.xmission.com/cgi-bin/mailman/private/math-fun/> List-Help: <mailto:math-fun-request@mailman.xmission.com?subject=help> List-Id: math-fun <math-fun.mailman.xmission.com> Original-recipient: rfc822;mckay(a)vax2.concordia.ca > [editorial comment: Gosper calls these Prodigals. I don't think > they have any non-obvious proerties, but it is strange there's no > "official" name. --Rich] On the contrary, they can be seen in all places and at all times (at least since we've had calculus). The reason that you don't hear more about them is that, after the initial excitement of discovery, they don't do much for you. In solving differential equations, they are equivalent to using Euler's method. There, their one great advantage (or two) is that by Gerschgorin's criterion, the solution matrix is invertible and solutions from initial conditions are thereby unique. #2 is that Huyghen's principle is valid, the solution after a while is a valid initial condition for continuing. Any attempt to multiply out the (1 + matrix epsilons) to get an integral formula makes the matrizant, which also arises from Picard's method. The problem is that those epsilons may not commute, in fact most assuredly will not, so terms cannot be just gathered up to form an exponential. Nevertheless the solution matrix for systems od linear differential equations has all the essential properties of an exponential, such as convergence, growth, and so on. Most of yesterday's postings cover this ground pretty well. One thing which I remember particularly well was "Feynman's Magnificent New Operator Calculus" wherein he had just doscovered time ordering. This was Cornell, 1949; one can look up the paper in Physical Review. The same ideas in field theory were called "Wick Ordering." I am kind of busy discussing all this right now in my complex variable class, under the rubrik "complex functions which satisfy differential equations." One of the nice techniques is the sum law for exponents, which changes form in the noncommutative case. Pursuing this in a certain direction leads to the Campbell-Hausdorff series. This is usually a bit complicated to derive, so I was interested in the remark that such a thing is guaranteed in matrix groups when the two matrices do not have a negative pair of eigenvalues. Can a reference to this be supplied? A good reference on the topic is an article by Wilhelm Magnus in Communications in Mathematical Physics, the Courant Institute house organ, late 50's. The formula for compounding rotations is a magnificent application of these ideas; what is not so well known is that if one makes pseudo-quaternions out of the real 2x2 matrices you get a similar formula with a lorentz metric. That is what you should use for the arc length of a rectangular hyperbola, and avoid elliptic integrals. But only if it is rectangular. - hvm _______________________________________________ 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: product integrals
by mcintosh＠servidor.unam.mx 12 Nov '02

12 Nov '02

> [editorial comment: Gosper calls these Prodigals. I don't think > they have any non-obvious proerties, but it is strange there's no > "official" name. --Rich] On the contrary, they can be seen in all places and at all times (at least since we've had calculus). The reason that you don't hear more about them is that, after the initial excitement of discovery, they don't do much for you. In solving differential equations, they are equivalent to using Euler's method. There, their one great advantage (or two) is that by Gerschgorin's criterion, the solution matrix is invertible and solutions from initial conditions are thereby unique. #2 is that Huyghen's principle is valid, the solution after a while is a valid initial condition for continuing. Any attempt to multiply out the (1 + matrix epsilons) to get an integral formula makes the matrizant, which also arises from Picard's method. The problem is that those epsilons may not commute, in fact most assuredly will not, so terms cannot be just gathered up to form an exponential. Nevertheless the solution matrix for systems od linear differential equations has all the essential properties of an exponential, such as convergence, growth, and so on. Most of yesterday's postings cover this ground pretty well. One thing which I remember particularly well was "Feynman's Magnificent New Operator Calculus" wherein he had just doscovered time ordering. This was Cornell, 1949; one can look up the paper in Physical Review. The same ideas in field theory were called "Wick Ordering." I am kind of busy discussing all this right now in my complex variable class, under the rubrik "complex functions which satisfy differential equations." One of the nice techniques is the sum law for exponents, which changes form in the noncommutative case. Pursuing this in a certain direction leads to the Campbell-Hausdorff series. This is usually a bit complicated to derive, so I was interested in the remark that such a thing is guaranteed in matrix groups when the two matrices do not have a negative pair of eigenvalues. Can a reference to this be supplied? A good reference on the topic is an article by Wilhelm Magnus in Communications in Mathematical Physics, the Courant Institute house organ, late 50's. The formula for compounding rotations is a magnificent application of these ideas; what is not so well known is that if one makes pseudo-quaternions out of the real 2x2 matrices you get a similar formula with a lorentz metric. That is what you should use for the arc length of a rectangular hyperbola, and avoid elliptic integrals. But only if it is rectangular. - hvm

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