August 2008 - mathfun@mailman.xmission.com

Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 17 Aug '08

17 Aug '08

Rich wrote: << Quoting Dan Asimov <dasimov(a)earthlink.net>: > on a theorem of Cauchy: > > Theorem: Given two convex curves A,B > in the plane with B lying inside A, > then length(B) < length(A). Is it fair to give this one to Cauchy? Archimedes must have assumed something like it to estimate the circumference of a circle as being between the in- & circumscribed polygons. Was Cauchy just bringing the rigor up to modern standards, or is some non-obvious insight required? >> I don't know the history of this theorem -- and hadn't realized Archimedes knew it. (How did he show this in the rigor of his day?) (Cauchy's proof goes something like this: First, show that if C is a convex curve in the plane, and you average the lengths of its projections onto lines of all directions, then that average is just half the length of the curve. (This is proved by first proving it for an interval, then approximating a convex curve by a convex polygon.) Now the above Theorem follows readily. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 15 Aug '08

15 Aug '08

I wrote: << this axiom looks logically equivalent to it. >> because there is, I think, a short chain of reasoning showing the equivalence of the two statements. But any two true statements are logically equivalent. Can "logically equivalent" be defined so it's not a trivial statement? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 15 Aug '08

15 Aug '08

Actually, there is a very relevant 1980 paper in Mathematics Magazine, by Scott Brodie, about Archimedes' axioms. Here is the Math Reviews review of it, by H.S.M. Coxeter: << Archimedes' treatise on the sphere and cylinder uses the axioms of Euclid along with five new ones. These include a statement to the effect that if two curves in the plane both intersect a line at the same two points so that each curve, together with the line, bounds a convex set, and if one of these convex sets is a proper subset of the other, then the curve bounding the larger set is the longer of the two. The author uses calculus to prove this statement in modern terms. >> As Rich suggested, this is indeed extremely close to Cauchy's theorem; in fact this axiom looks logically equivalent to it. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by James Propp 14 Aug '08

14 Aug '08

Dan Asimov writes: >This "puzzle" was evidently harder and/or >of less interest than I thought. Speaking for myself, I thought it was quite interesting; I just didn't see how to get started with it! >Theorem: Given two convex curves A,B >in the plane with B lying inside A, >then length(B) < length(A). Several posters have remarked that this theorem was known to Archimedes, and some have asked the question of how he proved it. The answer is, he didn't prove it; he assumed it as an axiomatic property of arc-length. I mean, what else could he have done? Lacking the notion of a limit, I don't think you can really give a satisfactory definition of length, let alone prove theorems about it. I once heard my teacher Andrew Gleason speak very approvingly of Archimedes' approach to arc-length. The monotonicity property of arc-length for convex curves that Dan cites is simple enough to be intuitive, and powerful enough to enable one to prove things about arc-length of various curves of interest (e.g., it enables one to bound the lengths of curves above and below by lengths of polygonal paths). >... >Thus for t close to 2, the curvature of >H(S^1,t) must be close to 1/r = 1/2, >and hence positive. So this curve is convex. >... A nice argument. Wish I'd thought of that. >It's easy to see there exist such H's >if the C^2 condition is dropped. Details, please? >(This observation was made in collaboration >with Joseph Gerver.) Please say hi to Joe for me. (He and Jamie Simpson and I collaborated on my very first published paper, way back when.) Jim

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 14 Aug '08

14 Aug '08

Joshua wrote: << On Thu, Aug 14, 2008 at 12:40 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > a) For a point outside the circle, its two tangent segments > have lengths that total more than the circle's arclength > between them. This amounts to showing that > > tan(theta) > theta > > for 0 < theta < pi/2. I believe Archimedes was capable > of verifying this inequality even without calculus and modern > terminology for trig functions. I think that's the hard part -- how do you show that? How did Archimedes do it? >> Consider the picture of the circle with one external point and one of its two tangent segments to the circle. My guess is that Archimedes considered a very small arc of the circle, and its radial projection onto the tangent segment. The very small segment is almost straight, whereas its projection onto the tangent segment is exactly straight but with endpoints clearly farther apart (from the radial projection). I can easily imagine Archimedes thinking in terms of the limit, where the small slightly rounded arcs became infinitesimal but straight. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 14 Aug '08

14 Aug '08

Rich wrote: << . . .a theorem of Cauchy: > > Theorem: Given two convex curves A,B > in the plane with B lying inside A, > then length(B) < length(A). Is it fair to give this one to Cauchy? Archimedes must have assumed something like it to estimate the circumference of a circle as being between the in- & circumscribed polygons. Was Cauchy just bringing the rigor up to modern standards, or is some non-obvious insight required? >> Actually, it seems easy to verify this in-betweenness for the circle directly. a) For a point outside the circle, its two tangent segments have lengths that total more than the circle's arclength between them. This amounts to showing that tan(theta) > theta for 0 < theta < pi/2. I believe Archimedes was capable of verifying this inequality even without calculus and modern terminology for trig functions. b) For a side of the inscribed polygon, that must be shorter than the circle's arc between the side's endpoints. a) and b) together prove the in-betweenness of the circle's length w.r.t. any in- & circumscribed polygons. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 14 Aug '08

14 Aug '08

This "puzzle" was evidently harder and/or of less interest than I thought. This may be in part because it depends on a theorem of Cauchy: ----- Theorem: Given two convex curves A,B in the plane with B lying inside A, then length(B) < length(A). ----- If such a mapping as H exists, then since it is C^2, the curvature of the closed curves H(S^1, t) for 1 <= t <= 2 must vary continuously, and in particular for parameter values t close enough to 2, the curvature of H(S^1, t) must remain everywhere close to the boundary component of A -- a circle of radius 2. Thus for t close to 2, the curvature of H(S^1,t) must be close to 1/r = 1/2, and hence positive. So this curve is convex. Then by Cauchy's theorem, the length of H(S^1,t) for t close enough to 2 must be strictly less than L(2) = 4pi, contradicting the condition on lengths. Thus there is no such H, QED. ----- It's easy to see there exist such H's if the C^2 condition is dropped. I find it surprising that such an innocuous- sounding condition as constant lengths of H(S^1,t) implies H must have less than C^2 differentiability. (This observation was made in collaboration with Joseph Gerver.) --Dan << So let me ask a simpler and straightforward version of the question: ----------------------------------------- Let A = { p in R^2 : 1 < ||p|| <= 2 }. PUZZLE: Prove there is no C^2 homeomorphism H : S^1 x (1,2] -> A such that if L(r) denotes the length of the curve H(S^1, r) in A, then L(r) is independent of r. ----------------------------------------- >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Josephus = Fractal
by Dan Asimov 11 Aug '08

11 Aug '08

I would say that this description could be written *far* more clearly. (And I don't see the problem as related to facility with English.) For example, are eliminated people included in the process of counting, or not? --Dan << The classical version of the Josephus problem counts off uneliminated people arranged in a circle and eliminates the person. The process continues until only one person is left. In this variant of the Josephus problem, two people are to be eliminated at each stage, but the two processes of elimination go in opposite directions. Suppose that there are people and every person is eliminated. The first process starts at the first person and eliminates the people at positions , , …. The second process starts with the person and eliminates the people at positions , , …. Suppose that the first process comes first and the second process second at each stage. Denote the position of the survivor by . The graph of the list is very interesting. >> -----Forwarded Message----- >From: Ed Pegg Jr <ed(a)mathpuzzle.com> >Sent: Aug 11, 2008 5:52 PM >To: math-fun <math-fun(a)mailman.xmission.com> >Subject: [math-fun] Josephus = Fractal > >Student Daisuke Minematsu and his classmates have noticed that the Josephus problem has fractal behavior. > >http://demonstrations.wolfram.com/TheJosephusProblemInBothDirections/ > >Here's a specific image: >http://demonstrations.wolfram.com/TheJosephusProblemInBothDirections/HTMLIm… > >Seems like a great result for a classic problem. It's new to me. Is it actually new? > >--Ed Pegg Jr > > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] CORRECTION Re: Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 11 Aug '08

11 Aug '08

Let me try that again. Let me ask a simpler and straightforward version of the question: ----------------------------------------- Let A = { p in R^2 : 1 < ||p|| <= 2 }. PUZZLE: Prove there is no C^2 homeomorphism H : S^1 x (1,2] -> A such that if L(r) denotes the length of the curve H(S^1, r) in A, then L(r) is independent of r. ----------------------------------------- (The important difference between the e-mail I sent a few minutes ago is changing a <= sign to a < sign, and changing [1,2] to (1,2].) Sorry about that. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 11 Aug '08

11 Aug '08

Time to fess up. The answer to the question below is "No." Furthermore, there are a number of irrelevant conditions. So let me ask a simpler and straightforward version of the question: ----------------------------------------- Let A = { p in R^2 : 1 <= ||p|| <= 2 }. PUZZLE: Prove there is no C^2 homeomorphism H : S^1 x [1,2] -> A such that if L(r) denotes the length of the curve H(S^1, r) in A, then L(r) is independent of r. ----------------------------------------- --Dan << Can you find a smooth (C^oo) function f : D -> [0,1] (where D is the closed unit disk in R^2) such that 1) f is onto. 2) The unit circle bd(D) is finv(1), and 3) For 0 < c <= 1, finv(c) is a smooth curve with length = 2pi. where finv(c) denotes the inverse image of c. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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