August 2008 - mathfun@mailman.xmission.com

Re: [math-fun] rolling a ball along a fractal
by James Propp 05 Aug '08

05 Aug '08

>This seems easy - since the limit of the length is infinite, you could >only get a fixed point as you go from level n to n+1 if there's some >very nice (rational) relationship between the size of the ball and the >length of the level n snowflake. Here's another way to state Josh's point: "How can you roll a ball along a Koch curve when you can't even roll a *hoop* along a Koch curve?" (If we mark a point on the hoop, and roll it along a polygonal approximation to the curve starting with the marked point touching the plane, then when the hoop has finished its journey, the angle between the marked point, the center of the hoop, and the new point of contact between the hoop and the plane will be L/R mod 2 pi, where L is the length of the polygonal approximation and R is the radius of the hoop. But there's no reason to expect L/R modulo 2 pi to have any kind of good limiting behavior as we consider a sequence of polygonal approximations with L going to infinity. Indeed, by continuity, we can construct polygonal approximations that will make L/R mod 2 pi equal to any angle you like.) Jim Propp

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Re: [math-fun] rolling a ball along a fractal
by James Propp 05 Aug '08

05 Aug '08

I realized last night that what I called a "no-slip" condition should properly be called a "no-spin" condition. Think of a billiard ball that's been hit off-center: it travels in a straight line with a spin. This point of contact with the table does not trace out a great circle, but some other sort of curve (can anyone tell me what family of curves on a sphere arises in this situation?). This is *not* the sort of thing we want when we're rolling a ball over a fractal curve. I now believe that if a nowhere-differentiable curve is sufficiently close to a straight line, there's a well-defined way to roll a ball along it. Specifically, use a variant of the Koch construction with free parameters epsilon_1, epsilon_2, epsilon_3, ..., where at the nth stage of the construction, all existing segments get little equilateral bumps on them, where the ratio of the size of the bump to the size of the segment is epsilon_n. I think one can show that if the epsilons go to 0 sufficiently fast, there's a well-defined limiting behavior if one rolls a ball along a polygonal approximation to the curve of mesh delta and sends delta to 0. And I'm fairly sure that this modified Koch construction gives a nowhere-differentiable curve, though not 100% positive. However, I'm fairly sure such a curve would have fractal dimension 1, and in fact have finite length. Indeed, I find Josh's argument >This seems easy - since the limit of the length is infinite, you could >only get a fixed point as you go from level n to n+1 if there's some >very nice (rational) relationship between the size of the ball and the >length of the level n snowflake. > >In other words, I think the answer to all your questions are "no, it >doesn't make sense to talk about rolling a ball on a fractal". quite compelling. At first I found Rich's argument >The difference between rolling the ball along one edge of the >polygonal path, and the 4 smaller edges of the next iteration, >should be O(edge^2). Even after multiplying by the number of >edges, this is O(edge). This sum converges. So we'd expect >the sphere-rolling-along-snowflake-path to converge. quite compelling too, which put me in the position of holding two contradictory beliefs! But I think what Rich's argument misses is the non-commutativity of the situation. Imagine one ball whose point of contact with the plane travels from (0,0) to (epsilon,epsilon) to (2*epsilon,0) and on to (N,0), and a second ball that rolls directly from (0,0) to (N,0). Even if epsilon is minuscule, the diference in orientation of the two balls can be sizeable when they arrive at (N,0), provided N is big enough, since the ball that goes from (0,0) to (2*epsilon,0) directly is already oriented differently from the ball that goes via (epsilon,epsilon), and this tiny difference gets magnified as the balls continue to roll. My current guess (I won't call it a belief yet) is that the class of rectifiable curves coincides with the class of curves that one can roll a ball along in a well-defined way. What do you all think? Jim Propp

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Re: [math-fun] ilg (integral log gamma) from polylog
by Dan Asimov 05 Aug '08

05 Aug '08

RWG writes: << . . . Despite many stratagems, I can't get the sum to converge. >> Shouldn't "many stratagems" read "much stratajewelry" ? --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] Theta function halfangles
by rwg＠sdf.lonestar.org 05 Aug '08

05 Aug '08

> From: rwg(a)sdf.lonestar.org [mailto:rwg@sdf.lonestar.org] > Sent: Wed 7/9/2008 9:22 AM > To: math-fun(a)mailman.xmission.com > Cc: "rwg(a)sdf.lonestar.orgrwg"@sdf.lonestar.org > Subject: Theta function halfangles > > (I promised Cheny Xu, proprietor of Jade Palace Restaurant, www.jadepalace.us > that I'd solve a problem with the help of some fish he fed me last night.) > > It's not obvious how to extract a tractable halfangle formula from the many > double angle formulae in W&W. After you eliminate the three undesired > species, the resultant polynomial is too hairy. Actually, it would > probably simplify if Macsyma or Mma only knew how, but I can't even get Mma > 6.0 to EllipticTheta[1, Pi, q] -> 0 ! (Hey guys, need a programmer?) > > Fortunately, I (re?)conjectured a family of theta determinant identities, > http://www.tweedledum.com/rwg/thet.PNG (too tall for IE 7), > arxiv.org/pdf/math/0703470, the first few (including the one used here) proved by > Rich (and Hilarie?). These are nice when actually typeset: > > x > theta (-, q) = > 1 2 > > 3 3 3 > - theta (0, q) theta (x, q) + theta (0, q) theta (x, q) - theta (0, q) theta (x, q) > 4 4 3 3 2 2 1/4 > (-----------------------------------------------------------------------------------) , > 2 > > x > theta (-, q) = > 2 2 > > 3 3 3 > - theta (0, q) theta (x, q) + theta (0, q) theta (x, q) + theta (0, q) theta (x, q) > 4 4 3 3 2 2 1/4 > (-----------------------------------------------------------------------------------) , > 2 > > x > theta (-, q) = > 3 2 > > 3 3 3 > theta (0, q) theta (x, q) + theta (0, q) theta (x, q) + theta (0, q) theta (x, q) > 4 4 3 3 2 2 1/4 > (---------------------------------------------------------------------------------) , > 2 > > x > theta (-, q) = > 4 2 > > 3 3 3 > theta (0, q) theta (x, q) + theta (0, q) theta (x, q) - theta (0, q) theta (x, q) > 4 4 3 3 2 2 1/4 > (---------------------------------------------------------------------------------) . > 2 > > E.g., > pi pi > theta (--, q) = theta (--, q) = > 1 4 2 4 > > 2 2 > theta (0, q) theta (0, q) (theta (0, q) - theta (0, q)) > 3 4 3 4 1/4 > (-------------------------------------------------------) , > 2 > > pi pi > theta (--, q) = theta (--, q) = > 3 4 4 4 > > 2 2 > theta (0, q) theta (0, q) (theta (0, q) + theta (0, q)) > 3 4 4 3 1/4 > (-------------------------------------------------------) . > 2 But, duh, %pi %pi 4 theta (---, q) = theta (---, q) = theta (0, q ) 4 4 3 4 4 > Finally, this must be in Tannery & Molk, but I don't recall seeing a Lambert type > series come out as a pure theta instead of a log derivative theta: > > inf 2 > ==== j 2 j - 1 theta (0, q) - 1 > \ (- 1) q 4 > > --------------- = ----------------. > / 2 j - 1 4 > ==== q + 1 > j = 1 There seems to be a lot of these. theta (0, q) 2 log(-------------------) inf 1/4 ==== n n 2 theta (0, q) q \ (- 1) q 3 > ---------- = ------------------------. / n 2 ==== n (q + 1) n = 1 > --rwg >

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Re: [math-fun] rolling a ball along a fractal
by James Propp 04 Aug '08

04 Aug '08

Here's another way to clarify my question: Consider a level-zero snowflake curve (a path of length 1) and a level-one snowflake curve (a path consisting of four segments of length 1/3). If we roll a ball along the former, (with its south pole starting facing downward), the final orientation of the ball is NOT THE SAME as if we roll the ball along the latter (also with the south pole starting facing downward). (Actually, I suppose that if one chooses the radius cleverly, the orientation of the ball *could* be the same, but for generic radii, the ball will be in a different orientation.) If we replace the level-one curve by a level-two curve (consisting of sixteen segments of length 1/9), the final orientation of the ball could once again change. And so on. Is there any reason to think that this sequence of orientations of the ball should converge to some particular orientation in the configuration space of the ball (where we mod out by translation --- all we care about is rotation --- so we're really looking at the configuration space of a sphere)? And even if it does converge, is there any reason to think that a different sequence of polygonal approximations to the snowflake curve would give the same limit? Hopefully this is clear. So this I why I don't think the Pearse references are relevant. But thanks anyway to Mike for sending his message and "getting the ball rolling", as it were. :-) Jim Propp

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Re: [math-fun] rolling a ball along a fractal
by James Propp 04 Aug '08

04 Aug '08

Mike Stay writes: >A tube formula talks about the area within a certain distance of the >fractal; essentially you're rolling a circle or a ball over the >surface of the fractal. By defining the zeta function of a fractal >and looking at its zeros, you get the Hausdorff dimension as one of >the zeros and a lattice of other zeros. The tube formula is a sum >over terms involving those complex zeros of the fractal's zeta. So a tube formula is about the area of the Minkowski sum of the fractal and a disk (or, one dimension up, the volume of the Minkowski sum of the fractal and a ball). But that's very different from what I'm asking about, which is how a ball rotates if it rolls "without slipping" so that the point of contact with the plane follows the fractal. Now that I think about it, I'm not sure how to phrase the "without slipping" condition in a way that depends solely on rectifiability of the curve. It seems that the derivative is involved. So for a nowhere-differentiable curve like the Koch snowflake, I don't even know how to abstractly characterize the property of the path in the configuration space of a sphere that I'm looking for. But it still makes sense to approximate the snowflake curve by polygonal paths, and we can ask whether the path in sphere-configuration-space converges as the mesh of our polygonal approximation goes to zero, so we can duck the issue of what "no-slip" really means for smooth curves. A related issue is parallel transport of tangent planes. If you have a particle moving smoothly on the surface of the sphere, there's a natural way to carry a coordinate frame along with the particle. So you get a way to identify points in one tangent plane with points in another tangent plane. (But watch out: if you come back to where you started, your coordinate frame will have rotated, according to how much area on the sphere your path has enclosed.) Piecewise smooth motions are okay too. But what about a fractal curve on the sphere? It's not at all clear to me when and how parallel transport would work. Note that, as in my problem, we're trying to shadow one path with another, but this time we're going in the other direction (from the sphere to the plane instead of the reverse). I hope I've made it clear how my question is different from Pearse's. If not, I trust someone will ask me to clarify (or tell me what I'm missing). Jim Propp

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Re: [math-fun] rolling a ball along a fractal
by James Propp 04 Aug '08

04 Aug '08

Mike Stay wrote: >> Maybe there's a dichotomy here, and it makes sense to roll a ball along a >> fractal curve as long as its fractal dimension is sufficiently low, but not >> if the fractal dimension exceeds some cutoff? >> >> (This question has "Dan Asimov" written all over it > > Well, Erin Pearse wrote all over *about* the question: > http://arxiv.org/abs/math-ph/0412029 > http://arxiv.org/abs/math.DS/0605527 > http://www.math.cornell.edu/~erin/dissertation/dissertation.pdf Can someone who knows more geometry than I do explain the connection between my question and Pearse's work? (I've taken a look at the links Mike provided, but I don't know how things like tube formulas and complex dimensions are pertinent to questions about shadowing a path in the plane with a compatible path in the configuration space of a sphere.) Jim Propp

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[math-fun] rolling a ball along a fractal
by James Propp 04 Aug '08

04 Aug '08

Is there a sensible way to say what it means to roll a ball across a plane, without slipping, so that the point of contact traces out a snowflake curve? (Equivalently, the center of the ball should trace out a snowflake curve on a parallel plane.) It's clear how to roll a ball so that the point of contact traces out a polygonal path, and from this one can define what it means to roll a ball so that the point of contact traces out a specified rectifiable path (undoubtedly there's a classic theorem that says that the behavior of the ball converges to a limit as one looks at ever-finer polygonal approximations to the path), but it's unclear whether this limit theorem would apply to a non-rectifiable curve like the snowflake curve. Maybe there's a dichotomy here, and it makes sense to roll a ball along a fractal curve as long as its fractal dimension is sufficiently low, but not if the fractal dimension exceeds some cutoff? (This question has "Dan Asimov" written all over it, but others may reply as well. :-) ) Jim Propp

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[math-fun] Superfluous Multinomials
by wouter meeussen 04 Aug '08

04 Aug '08

dear funsters, when posting stuff like this, I presume it is generally known, or at least could be. Motivation: just fun and beauty. 1/. Intro consider the expansion of (a+b+c+d)^3 it is the sum over M(u,v,w,x) a^u b^v c^w d^x with positive or zero integers u+v+w+x=3 and M(u,v,w,x) is the multinomial u! v! w! x! /(u+v+w+x)! When we drop the multinomial coefficients, then we get the 'compositions' of 4 elements in groups of three. in Mma: Expand[(a+b+c+d)^3 /3! ]/. q_^n_ ->q^n n! gives a^3 + a^2*b + a*b^2 + b^3 + a^2*c + a*b*c + b^2*c + a*c^2 + b*c^2 + c^3 + a^2*d + a*b*d + b^2*d + a*c*d + b*c*d + c^2*d + a*d^2 + b*d^2 + c*d^2 + d^3 this is equivalent to Plus @@ Apply[Times,{a,b,c,d}^#& /@Compositions[3,4],{1}] 2/. Generalise replace the (a+b+c+d) above by Sum(j=1..n, t[ j ] ) , t stands for 'temporary', now replace t[j] with E^(2 Pi I j/n) and it's obvious that this sum equals zero. And that any power of zero is also zero. But now, dropping multinomials, we find: Table[ Plus@@ Apply[Times,(Exp[2 Pi I/n]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] and we find that it's also zero, except it's 1 when k equals n, that is, we take all n roots of (-1) and put it to the n-th (multinomial-less) power. 3/. an instructive typo : Exp[2 Pi I/k] instead of Exp[2 Pi I/n]: Table[ Plus@@ Apply[Times,(Exp[2 Pi I/k]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] equals Table[Ceiling[n/m],{n,9},{m,n}] alias A120885 That leads to this silly remark: Table[ Plus@@ (Exp[2 Pi I/k]^Range[n])^k ,{n,16},{k,n}] is integer, except for the following couples of {n,k} : {7, 5}, {8, 5}, {9, 7}, {10, 7}, {10, 8}, {11, 7}, {11, 8}, {11, 9}, {12, 5}, {12, 7}, {12, 8}, {12, 9}, {12, 10}, {13, 5}, {13, 8}, {13, 9}, {13, 10}, {13, 11}, {14, 8}, {14, 9}, {14, 10}, {14, 11}, {14, 12}, {15, 9}, {15, 10}, {15, 11}, {15, 12}, {15, 13}, {16, 7}, {16, 9}, {16, 10}, {16, 11}, {16, 12}, {16, 13}, {16, 14}, ... and, where's the system in *this* madness? Wouter.

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[math-fun] another universal CA?
by rcs＠xmission.com 01 Aug '08

01 Aug '08

2nd WORKSHOP ON REACHABILITY PROBLEMS, RP'08 (September 15-17, 2008, Liverpool, UK) http://www.csc.liv.ac.uk/~rp2008/ List of Accepted papers: - Maurice Margenstern and Yu Song. A universal cellular automaton on the ternary heptagrid.

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