Unbounded Orbits for
the Penrose Kite

Monograph guide
Billiard King homepage
The Penrose  kite  is the
kite K(A) with parameter A=Sqrt(5)-2.  
The other parameter of importance is
the parameter e=(1-A)/2.  In my
 paper ,  I prove the following result

Theorem   On K(A), the forwards orbit
of (e,1) is unbounded.  The backwards orbit 
of (e,1) is unbounded and also enters every 
neighborhood of the vertex (0,1).

This demo illustrates this theorem.
To activate the applet at right,
press the red square button.

Top half

The top half of the applet shows orbits of some octagons. The outer billiards map is entirely defined on each octagon. 0. The octagon O(0), magenta contains the point (e+1,1). 1. The octagon O(1), red, contains the point (e-A,1) 2. The octagon O(2), green contains the point (e+A^2,1) 3. the octagon O(3), orange, the point (e-A^3,1) 4. the octagon O(4), purple, contains the point (e+A^4,1) The pattern continues forever. The octagons converge to the point (e,1). This point is the bottom of the light blue marker. You can zoom in on this point by clicking on the applet. Let U(n) be the orbit of O(n) under the square of the outer billiards map. We draw U(0) and U(1) entirely. We draw the intersection of U(2) and U(3) with the lines y=-1 and y=1. We don't draw U(4), because it is quite large. You can rescale the picture to see the extent of the orbits. Click on the picture to rescale. Here are 2 observations. I. The diameter of U(n+1) is about 1/A times the diameter of U(n). II. The orbit of U(n) contains 2^n octagons, forming the nth level gaps of a self-similar Cantor set C. We don't quite prove these statements in our paper, though our methods would give a proof if we pursued it. However, we do prove that all the poiints in C' have unbounded orbits that also return densely to C'. Here C' is C minus the points in the boundaries of the octagons. That is, C' is C minus the gap points.

bottom half

The bottom half of the applet shows the arithmetic graphs of points in the octagons discussed above, as well as the arithmetic graph of the unbounded orbit. The offset value is e. The arithmetic graph of the unbounded orbit is the component containing (0,0). 0. The arithmetic graph of U(0) is trivial. 1. The arithmetic graph of U(1) is the component containing (-13,3) 2. The arithmetic graph of U(2) is the component containing (-55,13) 3. The arithmetic graph of U(3) is the component containing (-233,55) 4. The arithmetic graph of U(4) is the component containing (-987,233) The sequence 1,3,13,55,233,987... satisfies the Fibonacci-like recurrence relation a(n+2)=4a(n+1)+a(n). The successive quotients of this sequence converge to A. The continued fraction expansion for A is [4,4,4,4...]. If you use the arrow keys, you can rescale the picture to focus on any of the compact components we have described. If you want to see part of the unbounded orbit, as well as the global picture, you need to zoom out of the picture. One can see that each compact component is a kind of elaboration of the previous one. This suggests that the unbounded component has a kind of quasi-self similarity. If you zoom out, you will see that we have plotted the unbounded orbit U at two scales. We have plotted U itself in white. In magenta, we have plotted T(U), where T is dilation about the origin by 1/A. In our paper we prove that T(U) and U are with a uniformly bounded distance of each other. We then translate this result into a proof of the main theorem.