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Majoring in Mathematics (view source)
Revision as of 04:57, November 13, 2009
, 04:57, November 13, 2009rough draft of periodic orbits in billiards. i don't actually remember what the poincare recurrence theorem tells us here -- i'll work through it tomorrow and make sure i'm right
Another celebrated example of a fixed point theorem is known as Sharkovsky's theorem. In addition to fixed points, some functions have points for which repeating the function several times yields the same result: for example, a value of x for which cos(cos(cos(x))) = x is called a 3-periodic point of the function cosine. Sharkovsky's theorem is the surprising result that if a continuous function f has points of period 3, then it has points of all other periods! For example, consider the quadratic <math>f(x) = -\frac{5}{2} x^2 + \frac{7}{2} +1</math> has f(0) = 1, f(f(0)) = f(1) = 2, f(f(f(0))) = f(f(1))=f(2) = 0. So 0 is a point of period 3 of f(x). Sharkovsky's theorem implies that there are points of every other period. For example, there is some x such that f(f(f(f(f(f(f(f(x)))))))=x! Solving this explicitly would involve solving a polynomial of degree 14, an impossible task. But Sharkovsky's theorem guarantees that there is '''some''' value of x satisfying this.
Another celebrated example of a fixed point theorem is known as Sharkovsky's theorem. In addition to fixed points, some functions have points for which repeating the function several times yields the same result: for example, a value of x for which cos(cos(cos(x))) = x is called a 3-periodic point of the function cosine. Sharkovsky's theorem is the surprising result that if a continuous function f has points of period 3, then it has points of all other periods! For example, consider the quadratic <math>f(x) = -\frac{5}{2} x^2 + \frac{7}{2} +1</math> has f(0) = 1, f(f(0)) = f(1) = 2, f(f(f(0))) = f(f(1))=f(2) = 0. So 0 is a point of period 3 of f(x). Sharkovsky's theorem implies that there are points of every other period. For example, there is some x such that f(f(f(f(f(f(f(f(x)))))))=x! Solving this explicitly would involve solving a polynomial of degree 14, an impossible task. But Sharkovsky's theorem guarantees that there is '''some''' value of x satisfying this.
==== Mathematical Billiards ====
Imagine that you're playing a game of pool on a standard rectangular table, but with a twist: there's no friction, so after you hit the ball, it continues along its trajectory forever. Obviously if you hit the ball so that its path is perpendicular to one of the walls of the table, it will bounce straight back, then off the opposite wall, straight back again, etc.: the ball will follow a repeating pattern of bouncing between the two walls forever. Similarly, if you aim just right, you can hit the ball so that it moves in a diamond-shaped path forever, bouncing off the middle of the sides of the walls repeatedly. A trajectory of the cue ball is called "periodic" if it's like one of these, in that the ball eventually ends up back where it started, and moving in the same direction. A trajectory is said to have "period p" if it hits the wall exactly <math>p</math> times in the course of this repeating cycle. The two examples given above have periods of two and four, respectively. One question we might ask is whether there is an orbit of period <math>p</math> for every possible choice of p. Can you come up with a path on the standard billiards table that has period 3? If not, can you prove that one doesn't exist?
As is often the case is mathematics, we attempt to generalize these results on periodic orbits to other situations. What if the billiards table is not rectangular, but triangular? It's true, but extremely difficult to prove, that given any triangular billiards table, there exists a periodic orbit of '''some''' period. The existence of periodic orbits is an extremely subtle question. This can be generalized even further, to ask whether all polygons admit periodic orbits. What about polyhedra? Can you imagine some periodic billiards orbits on a cube-shaped "table" (no gravity! think billiards in space).
The orbits that aren't periodic turn out to be just as interesting as those that are. Imagine that you place the cue ball at random on your billiards table, and smack it in some random direction. Chances are it won't end up in a periodic orbit -- putting the ball in a periodic orbit requires a very careful shot. So the ball will bounce around forever without retracing its path. But will it ever come back to the point where it started, perhaps moving in a different direction? The answer to this question is "no", but the Poincare Recurrence Theorem gives a somewhat satisfactory replacement: given any distance (e.g., an eighth of an inch), and any angle (e.g. two degrees), after some amount of time, the ball will be within that distance of its starting point, and directed within that angle of its initial path! This is a very suprising fact.
While this problem seems extremely elementary, sophisticated mathematical techiques from a range of fields of geometry and analysis have been brought to bear on it. Even so, many easily-stated questions remain open.
# Periodic orbits in billiards in polyhedra. I think it should be possible to say some interesting things about this but with some real content. Possibly there's a better idea.
# Periodic orbits in billiards in polyhedra. I think it should be possible to say some interesting things about this but with some real content. Possibly there's a better idea.