Unbounded Orbits for
the Penrose Kite

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.