# Tag Info

221

This question was cross-posted on Math Stack Exchange. Here is a copy of my answer for it there. This is it.  The perfectly centered billiards break.  Behold. Setup This break was computed in Mathematica using a numerical differential equations model. Here are a few details of the model: All balls are assumed to be perfectly elastic and ...

46

The idea to study what we call irrational rotation of a torus indeed belongs to Nicole Oresme, at least he clearly understood the density of the trajectories (which is not the same as ergodicity or equidistribution! One thing is to say that a trajectory visits every interval infinitely many times, and another thing to say HOW OFTEN. Oresme has absolutely no ...

45

If I understand correctly, there is already a counterexample on the torus: On the $xy$-plane $\mathbb{R}^2$, let $X$ be the vector field $$X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\partial\ }{\partial y}.$$ Now let $T^2=\mathbb{R}^2/\Lambda$ where $\Lambda$ is the lattice generated by $(2\pi,0)$ and $(0,2\pi)$. Since $X$ is invariant ...

42

Not a definitive answer, but a close upper bound. The same paper that had the first published computer-generated image of the Mandelbrot set also includes an image of the Julia set (no idea if it's the first one too). The paper is from precisely the same year Julia died, 1978! The conference was held on June 5-9, at the State University of New York, ...

40

Not an answer, but way too long for a comment: According to Ilyashenko ("Centennial history of Hilbert's 16th problem," http://www.ams.org/journals/bull/2002-39-03/S0273-0979-02-00946-1/S0273-0979-02-00946-1.pdf), the claimed result of Petrovski and Landis was disproved by Ilyashenko and Novikov (pg. 303). A citation to this disproof is not given, ...

36

Any surface of revolution in $3$-space with poles will have this property. The reason is that, in this case, any geodesic either goes through a pole (i.e., a point where the axis of revolution meets the surface) and is a profile curve that lies in a plane or else, because of the Clairaut integral, it avoids that pole by some positive distance. Thus, no ...

35

I can't comment on the Lion hypotheses. I'm pretty sure the SHLT is nothing more than the fact that: A linear endormophism of a $k$-dimensional vector space factors through a $(k-1)$-dimensional vector space iff it has nontrivial kernel iff its determinant is $0$. Stated without all the indices, this is completely obvious to any mathematician. So I ...

33

It is very unlikely that Gaston Julia saw a computer-generated image of a julia set. The first images were obtained at the beginning of the eighties. Note also that this would have been far less interesting and accurate images than the ones that were drawn by hand at the end of the nineteenth century. Credit: Fricke and Klein, 1897, hosted by Centre ...

32

Due to chaotic behaviour of the Solar System, it is not possible to precisely predict the evolution of the Solar System over 5 Gyr and the question of its long-term stability can only be answered in a statistical sense. For example, in http://www.nature.com/nature/journal/v459/n7248/full/nature08096.html (Existence of collisional trajectories of Mercury, ...

32

It is known (though I don't have a reference except for the statement at http://ocw.mit.edu/courses/mathematics/18-s34-problem-solving-seminar-fall-2007/assignments/roots.pdf, #2) that the only real polynomials with nonzero constant term whose roots equal their coefficients are $x^2+x-2$, $x^3+x^2-x-1$, and (approximately) $$x^3 + .56519772x^2 - 1.... 28 Edit: I've updated this answer to reflect the helpful comments made by Andres Koropecki and Ian Morris. As the other answers mentioned, the first crucial distinction you must make is that some properties refer to a topological dynamical system (X,T), while others refer to a measure-preserving dynamical system (X,T,\mu). Thus there are two different ... 28 A possibly new example: a ball bouncing on a parabola (= graph of a quadratic polynomial) is never chaotic. The associated dynamical system has an extra invariant that makes it "integrable" (if I have the terminology correctly). The general orbit is an elliptic curve \cal E equipped with a point P such that going from one bounce to the next corresponds ... 28 the proof goes back to Nicole Oresme in his paper De commensurabilitate vel incommensurabilitate motuum celi [On the Commensurability or Incommensurability of the Motions of the Heavens], dated around 1360, see Nicole Oresme and the commensurability or incommensurability of celestial motions (contains an annotated English translation of Oresme's Latin text) ... 27 Post-critically pre-periodic quadratic polynomials, i.e. those for which the orbit of the critical point 0 is pre-periodic, are well-known to be dense in the boundary of the Mandelbrot set. (This is essentially a normality argument.) Each of these is determined by an algebraic equation. This answers your first question (the question in the title). EDIT. ... 26 The first question is false as stated. By Artin's encoding, geodesics on SL_{2}(\mathbb{R})/SL_{2}(\mathbb{Z}) corresponding to continued fractions, and the geodesic flow corresponds to the shift. It's easy to find one fraction where you'll see any given prefix (hence dense), but you won't be equidistributed (say think about larger and larger blocks ... 26 These so-called "whispering gallery modes" are familiar from studies of microcavity lasers; they can trap the light indefinitely, only limited by diffraction; this web site by Jens Nöckel nicely summarizes the issues; an efficient way to untrap the trajectory is to introduce flattened portions in the boundary (in 2D this would be a stadium rather than a ... 25 Donnay and Pugh proved that every embedded surface S\subset R^3 can be C^0-perturbed so that the new metric has ergodic geodesic flow, see here. In particular, the new metric will have dense geodesics (moreover, "generic" geodesics will be dense in the unit tangent bundle). 24 I have run some simple simulations of this system in Python, and it looks nothing like a nice integrable system once you zoom in on the orbits. Certainly, the orbits do not look like closed curves about the the fixed point, as suggested by the plot in the question. I'll give some numerical evidence of the chaotic behaviour of the map here, rather than actual ... 24 Over {\bf C}, An easy counterexample to question 3 is f(x) = x^2, g(x) = cx^2 where c is a nontrivial cube root of unity. Then f(f(x)) = g(g(x)) = x^4 but f and g do not commute. There are similar examples for higher iterates. [Added later] A more exotic construction yields further examples, some defined over {\bf Q}, such as the degree-4 ... 24 Douglas Zare's shortest path idea seems to me very well-suited for this. Intuitively, we can view the circles as being rings, and the reflected ray like a rope going through the rings. We pull to obtain the shortest rope (considering the rings fixed, and other suitable idealizations). The picture below shows how a path connecting (0,0) with C(m,n) may ... 24 Here is a partial answer: in \mathbb R^3 there are cycles of length 2. To show this, it is enough to find non-trivial systems of unit vectors x_i, y_j (i,j=1,\dots,d+1) such that (x_i,y_j)=t for some t>0 and all i\ne j. If d=3, we can take the normalized copies of (1,0,4),(1,0,-4),(-1,4,0),(-1,-4,0) and (-2,0,-1),(-2,0,1),(2,-1,0),(2,1,... 24 The quantity R_n is asymptotic to {4\over \pi}(n\log n)^2, see "Cover times for Brownian motion and random walks in two dimensions" by Dembo, Peres, Rosen and Zeitouni. This was previously conjectured by Aldous. 24 There are several puzzling things about the question: Firstly of course \theta must be irrational, and it is intended for \{ x\} to denote the Bernoulli polynomial x-[x]-1/2 rather than the more usual fractional part. Secondly, where is the result of Hardy from? I did find this statement in the Cambridge ICM paper of Hardy and Littlewood where they ... 23 As Stefan Kohl has pointed out it is not possible to give good lower bounds for T(N) without making progress on the Collatz problem itself. However it is not difficult to understand what the truth is, and the argument below will yield a rigorous upper bound. Instead of the map C(n) as defined, it is better to consider C_0(n) = n/2 if n is even, ... 23 For the unordered version, the solutions for degree 3 are$$ \eqalign{x^3&\cr x^3& {}+x^2-2\,x\cr x^3&{}+x^2-x-1\cr x^3&{}+ \left( r^2/2-1 \right) x^2+rx-r-r^2+2 \ \text{where}\ r^3 - 2 r + 2 = 0\cr} $$The solutions for degree 4 are$$ \eqalign{ x^4\cr x^4&{}+x^3-x^2-x\cr x^4&{}+x^3-2\; x^2 \cr x^4&{}+ \left( \frac{r^2}{2}-1 \...

23

There is a related iterated function system with two functions, $f_0(x) = 1+zx$ $f_1(x) = -1+zx.$ $X_z$ is the unique nonempty compact fixed set of this iterated function system. It is sometimes called a generalized dragon set with parameter $z$, and particular values of $z$ can produce some well-known fractals called dragons. A relevant result on ...

23

As remarked by Joe Silverman, a natural way to look at this question is by phrasing it in terms of the map $f(x):=\frac{1}{2}x(x+1)$ on the $2$-adic integers $\mathbb{Z}_2$. We are then asking about the behaviour of the orbit $f^n(2)$ with respect to the partition of $\mathbb{Z}_2$ into two clopen sets $U_1:=2\mathbb{Z}_2$ and $U_2:=1+2\mathbb{Z}_2$. Most ...

22

Let $C_r(x,y)$ or $C(x,y)$ be the circle of radius $r$ about the lattice point $(x,y)$. Suppose we choose a sequence of circles to hit, and ask for the piecewise linear path of shortest length from the origin hitting each of the circles along the path. If this doesn't go inside a circle, then by the least action principle, the angle of incidence will equal ...

22

I'm sorry that my PhD thesis wasn't published indeed and is hard to find (up to my knowledge it is only available at university of Paris 7). As I have a scanned copy of it, I asked people at my software company Riskdata to include it in the list of posted papers, althogh it is not really related to math finance, which I now study. It should be available in ...

22

Ilyashenko explaines very well the strategy and the main error of Petrovski-Landis in this lecture: http://www.mathnet.ru/php/seminars.phtml?option_lang=rus&presentid=8316 Watch from 50m.

Only top voted, non community-wiki answers of a minimum length are eligible