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n=1 Consequently n ! 0, which forces {since f ( ) is convex{ that xn ! 1. Thus yn & 0.
P
Since tan( )= ! 1 as ! 0, summation (3a) can be restated as 1 f 0 (xn ) < 1. Letting
n=1
sn be the absolutevalue of the slope of the line joining point Pn with Pn+1 , the convexity of f
implies that sn f 0 (xn ) . Consequently,
1
X
n=1 Printed: February 14, 1995 sn < 1: (3b) Filename is Article/18Billiards/billiards.ams.tex 3 Billiards inside a cusp Computing the slope sn . We have used the fact, when the cueball hits the upper cushion f , that
the angles of incidence and re ection are equal. But somewhere our argument had better use that
these angles are equal when the ball bounces up o the oor! Here it is:
tan( ) = yn + yn+1 :
n The upshot is that xn+1 = ytan(y n )
xn+1 xn
n + n+1
1 As a consequence, xn tan( 1 ) :
yn + yn+1 tan( 1 ) yn + yn+1 :
yn yn+1
So our summation condition mutates one last time, to become
note
sn === yn yn+1
xn+1 xn 1
X yn yn+1
y + yn+1
n=1 n < 1; with yn & 0. (3c) But this cannot besuch a sum as this last one must always be in nite. Its M th partial sum is PM 1
X yn yn+1
=
y + yn+1
n=M n 1
X yn yn+1
y + yM
n=M M 1
1
yM nl!1 yn+1 = :
im
yM + yM
2
Since the partial sums fPM g1=1 do not go to zero, conditions (3c,b,a) were all impossible, as was
M
gure 2. Any cueball shot out the cusp must turn around.
Post mortem re ection. This proof can be readily shown to a secondsemester calculus class
and gives a nontraditional and curious use for a seriesdivergence test. All that is used about the
upper cushion y = f (x) of the table is that f is an eventuallyconvex di erentiable function which
is asymptotic to the xaxis.
However, the argument is unsatisfactory from the point of view of understanding \why" the
cueball had to turn around. One test of the strength of a method of argument is whether it can be
used on related questions. Suppose we remove the convexity condition and allow the uppercushion
to have wiggles.
Can cueballs wander monotonically out the cusp for the
(4a)
table determined by, say, f (x) = 3 + sin(px) (x + 1)2 ?
While one could possibly use a seriesdivergence argument to exhibit a speci c cueball which fails
to escape, such an approach might require real delicacy to make a substantial general assertion.
Yet another natural question for which the seriesdivergence approach looks illadapted focuses
on a stronger sense in which cueballs might fail to escape.
Do cueballs return arbitrarily close (in both
(4b)
position and direction) to where they started?
By the way, a cueball which in nitelyoften returns arbitrarily near to its initial state is called
recurrent . Having developed more powerful tools, we will come back to recurrence later.
Math. Intelligencer, vol.17 no.1, (1995), 8{16. 4
J.L. King
Philosophy. Answering questions such as (4a,b) for an individual cueball may be di cult. Yet nearby cueballs have nearby trajectories {for a while{ and so it may be pro table to make assertions
about collections of cueballs. This suggests nding a useful measure on the space of cueballsa
measure which is preserved under the action of \rolling" and \bouncing o the cushion". It turns
out {this is wellknown to those who study dynamical systems but is not a commonplace among
mathematicians in general{ that the \billiard ow" on any billiard table has a natural invariant
volume. The theme of this article is the tool of an invariant measure hidden inside a problem which,
on the surface, has no mention of measures. Along the way we will encounter a few elementary
but usefu...
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This note was uploaded on 01/26/2012 for the course MTG 6401 taught by Professor Staff during the Fall '09 term at University of Florida.
 Fall '09
 Staff

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