Clayton W. Dodge. ..... JIGGER to FIFTH {hence the name of the language},
LITER- MAGNUM and. BLOTTO. Commands refer ..... ВҐi j = 1, ..., 7, a
comparison of column vectors for p and p reveals .... The last two plans went
quite some way in.
VOLUME7
FALL 1983 CONTENTS
NUMBER 9
Integration: Why You Can and Wh 557
Little Known Computer Languag Another Day at the Races On the Discrete Lyapunov and Riccati Matrix Equ The Fuzzy Plane Christopher Roesmer . Sum of Powers of 2 Hao-Nhien Qui V n Analysis of Monopoly
A curios Ratio of K-Stars Michael Eisenstei
.
...............
..
..........
Puzzle Section Joseph D.E Konhauser . . . . . . . . . . . . . . . . . . . . . . . . . 601 Problem Department Clayton W. Dodge.. . . . . . . . . . . . . . . . . . . . . . . 609
INTEGRATION:
WHY YOU CAN AND WHY YOU CAN'T
by ?tick W a n d a Cotonado S-tott U n i v m U y
PI MU EPSILON JOURNAL THE OFFICIAL PUBLICATION OF THE HONORARY MATHEMATICAL FRATERNITY
A t most c o l l e g e s and u n i v e r s i t i e s , a l a r g e p a r t of t h e second
semester of c a l c u l u s i s devoted t o t h e arcane s u b j e c t commonly known a s " techniques of i n t e g r a t i o n " .
David Ballew, Editor
form expression f o r v a r i a b l e x.
ASSOCIATE EDITORS Roger Opp
-
where
Joseph Konhauser
f(x) i s a s p e c i f i c f u n c t i o n o f t h e
Typically, t h e following methods a r e discussed:
'forward' s u b s t i t u t i o n s
- 'backward' s u b s t i t u t i o n s
Clayton Dodge
The b a s i c problem i s t o f i n d a closed-
x = g(u) u = h(x)
- i n t e g r a t i o n by p a r t s
OFFICERS OF THE FRATERNITY
t h e use o f exponentials and logarithms
- trigonometric s u b s t i t u t i o n s
President: E. Maurice Beesley, University of Nevada
- i n v e r s e hyperbolic t r i g f u n c t i o n s - powers of s i n e s and c o s i n e s
President-Elect: Milton Cox, Miami University
-
Secretary-Treasurer: R.A. Good, University of Maryland
i n t e g r a l s with q u a d r a t i c s
- p a r t i a l f r a c t i o n decompositions.
Past-President: Richard V. Andree, University of Oklahoma
There a r e two l o g i c a l r e a c t i o n s t o t h i s s u b j e c t :
COUNCILORS David Barnette, University of California, Davis
a)
There is t o o much m a t e r i a l h e r e .
b)
There i s not enough m a t e r i a l here. For anyone who has taken o r t a u g h t t h i s course, ( a ) h a r d l y needs
Robert Eslinger, Hendrix College Eileen L. Poiani, Saint Peter's College A.D. Stewart, Prairie View A&M University
explanation.
F i r s t l y , t h e mastery o f a l l t h e s e techniques r e q u i r e s
q u i t e a b i t of r o t e memorization of b a s i c i n t e g r a l s , e s p e c i a l l y f o r t h e average s t u d e n t . 'menu'
. books for review, problems for solution and solutions to problems, should be 'ectfy to the special editors found in this issue under the various sections. Editorial -"-ec including manuscripts (two copies) and news items should be mailed to THE
3 -E
Pi MU EPSILON JOURNAL. South Dakota School of Mines and Technology.
For manuscripts, authors are requested to identify themselves as to their
are undergraduates or graduates, and the college or university they are position if they are faculty members or in a non-academic profession. published at the South Dakota School of Mines and Technology One volume consists of five years (10 issues) beginning with the tarting in 1949. For rates, see inside back cover.
Secondly, when faced with a n i n t e g r a t i o n problem, t h e
of p o s s i b l e techniques t o t r y t o apply i s l a r g e enough t o make
t h e d e c i s i o n process f a i r l y complicated.
F i n a l l y , with e x t e n s i v e t a b l e s
and ( l a t e l y ) computer programs which i n t e g r a t e a l l f u n c t i o n s encountered i n t h i s course, t h e motivation t o delve i n t o t h i s s u b j e c t with one's s l e e v e s r o l l e d up i s n a t u r a l l y diminished, and t h i s i s made worse by t h e amount t h e r e seem t o be t o know. Have you e v e r heard ( b ) from a s t u d e n t of t h i s s u b j e c t ?
Well,
now you have, and l e t me e x p l a i n why.
A f t e r a good s o l i d course on t h e
techniques o f i n t e g r a t i o n , i n c l u d i n g a thorough d i s c u s s i o n o f t h e t o p i c s l i s t e d above, I could w e l l come away with t h e following broad c l a s s i -
m a t e r i a l devoted t o i n t e g r a t i o n techniques, t h i s is t h e a t r u e theo-
f i c a t i o n of integrals: i) ii)
I n my view, it is n o t u n f a i r t o say t h a t , even given t h e m-iss of rem i n t h i s course; t h e o t h e r t o p i c s covered a r e r e a l l y j u s t methods t o
The i n t e g r a l s which I can f i n d .
use a s t h e occasion a r i s e s .
The i n t e g r a l s which I cannot f i n d .
e v e r s t a t e d e x p l i c i t l y , and o f t e n t h e d e t a i l s o f t h e process o f p a r t i a l
Most o f t h e i n t e g r a l s encountered i n t h e course a r e o f t y p e ( i )
f r a c t i o n s (which i s t h e proof o f t h i s theorem) is given much more weight
A s t u d e n t , i n f a c t , may
a r e o f t y p e ( i ) , f o r t h e a p p r o p r i a t e choice of "I"; s i n c e he ( o r s h e )
s t a n d a b l e , s i n c e c a r r y i n g out t h e p a r t i a l f r a c t i o n decomposition i s a complicated and cumbersome t a s k , even i n f a i r l y simple s i t u a t i o n s , and
knows i n h i s g u t t h a t he c a n ' t p o s s i b l y s o l v e a l l i n t e g r a t i o n problems, t h e a p p r o p r i a t e choice f o r "I", and
r e q u i r e s some a t t e n t i o n .
--
c l a s s of f u n c t i o n s which a r e a l l "of type ( i ) "
about.
is seen
However, I t h i n k i t i s a mistake n o t t o r i s e
above t h e f r a y and d r i v e t h e p o i n t home t h a t h e r e i s a l a r g e and common
t h a t t h e s u b j e c t i s much t o o complicated f o r mere m o r t a l s t o t h i n k I f a n i n t e g r a l of t y p e ( i i )
This i s under-
than t h e simple and obviously powerful statement i t s e l f .
never s e e an i n t e g r a l of type ( i i ) , and may conclude t h a t a l l i n t e g r a l s
t h e conclusion is t h a t he is
However, it i s h a r d l y
t h i s would be t h e f o c a l p o i n t o f t h i s c o u r s e .
Statement ( b ) i s one r e a c t i o n t o t h e e x i s t e n c e of t h e second c l a s s . ( o r should be, by t h e end o f t h e semester).
This being t h e c a s e , one would t h i n k t h a t
I can i n t e g r a t e them!
I f you g r a n t t h a t t h i s i s t h e 'only' theorem of t h i s type, then i n t h i s course, it is u s u a l l y
i n one o f t h e " s e t up b u t do n o t e v a l u a t e t h e i n t e g r a l which computes
..."
your mind should n a t u r a l l y t u r n t o t h e following:
can o t h e r i n t e g r a l s
problems on a n exam; when going over t h e q u e s t i o n s on t h e next day, t h e
be brought t o t h i s form by c l e v e r s u b s t i t u t i o n s , and can t h i s theorem t h e r e f o r e achieve a wider scope o f a p p l i c a t i o n ? The well-known answer
t e a c h e r nay make a remark t o t h e e f f e c t t h a t "we c a n ' t f i n d t h i s i n t e -
t o t h i s question is:
gral
..." and
t h e s u b j e c t i s embarrassingly dropped.
Sometimes, i f you g e t lucky.
Generally, no
E
attempt i s made t o e x p l a i n why some i n t e g r a l s can be found and some
x
.
Integrate
/+ +
2
x dc
.
c a n ' t , and we're back t o r e a c t i o n ( b ) (on a s l i g h t l y d i f f e r e n t l e v e l ) : There is something missing h e r e .
Solution.
(1
I n t h i s a r t i c l e I ' d l i k e t o d i s c u s s why t h e r e a r e i n t e g r a l s o f t y p e s ( i ) and ( i i ) , and t r y t o e x p l a i n t h e f a s c i n a t i n g r e l a t i o n s h i p between t h i s a p p a r e n t l y a n a l y t i c s u b j e c t and t h e much more geometric sub-
dc =
(
(1
- t2P
.
2t
Substitute x = .
-
Then 1
+ x2
= (( ,1
-+
Jand t2)
t2)
s o t h e above i n t e g r a l transforms t o
2
+
4t
2
t 2t
4
at,
j e c t o f a l g e b r a i c plane curves. Let me begin by s t a t i n g a theorem.
T h e 0 i ~ . Let R f t ) be a r a t i o n a l f u n c t i o n of t h e v a r i a b l e t , i - e . , t) i s t h e r a t i o of two polynomials.
Then
and t h e theorem a p p l i e s . This seemed p r e t t y lucky.
What i f I t r y
case I ' m s t u c k f o r a c l e v e r s u b s t i t u t i o n .
What i s going on here?
In
o r d e r t o f i x our a t t e n t i o n on a c e r t a i n g e n e r a l c l a s s of f u n c t i o n s , c o n s i d e r t h e following.
Ve6inLtion. l u r s e , a c t u a l l y f i n d i n g a closed-form expression f o r it i n r i n g polynomials and s o l v i n g l i n e a r e q u a t i o n s , and is a task i n i t s e l f
--
A f u n c t i o n y = y ( x ) i s a l g e b r a i c a l l y dependent on
.1-
i f t h e r e is a polynomial f f x 1 ,x 2 ) i n two v a r i a b l e s , such t h a t f f x , y f x ) )
i s identically
but I won't address t h e s e problems here.)
Examptu
.
y = 'fx f f f x1,x 2 ) = x 1
-
2
x2)
t h e o t h e r po , g i v i n g t h e e x p l i c i t parametrization above.
+
The i n t e g r a l s f l l
x
2
&,
The importance o f a r a t i o n a l p a r a m e t r i z a t i o n f o r t h e curve C is
j\^
+
X
3
&, e t c . , a r e examples o f i n t e -
g r a l s which involve f u n c t i o n s of x which a r e a l g e b r a i c a l l y dependent on
x , and t h i s i s t h e c l a s s o f f u n c t i o n s which I want t o focus on.
Our
g e n e r a l problem can be formulated a s follows.
The GenvwJL P/iob-ton
demonstrated by t h e following.
Thexifiem. Let R(x 1' x2 ) be a r a t i o n a l f u n c t i o n of two v a r i a b l e s and l e t y = y ( x ) be a l g e b r a i c a l l y dependent on x , with f ( x , y ( x ) ) i d e n t i c r a t i o n a l l y parametrized.
04 In^eg/io-tlon o(1 Atge.blcu.c Funatiov^.
Proof.
Let R ( x l , x ) be a r a t i o n a l f u n c t i o n o f two v a r i a b l e s . 2
y ( x ) ) dx be found?
with l u c k .
sometimes.
r a t i o n a l integrand.
But i t d o e s n ' t have anything t o do If y = y ( x )
is a l g e b r a i c a l l y dependent on x , then t h e r e i s t h i s polynomial f ( x 1 , x 2 )
= 0.
x ) = 0 d e f i n e s a so1' 2 c a l l e d " a l g e b r a i c curve" i n t h e ( x , x )-plane, and ( x , y ( x ) ) always 1 2 l i e s on t h i s curve. The p r o p e r t i e s o f t h i s curve should t h e r e f o r e be Now t h e equation f ( x
important i n studying y ( x )
Ve6imAt.on.
.
C e n t r a l f o r us i s t h e following property.
Let f ( x
x ) be a polynomial i n two v a r i a b l e s . The 1' 2 curve C = { ( x , x ) \ f ( x 1 , x 2 ) = 01 i s r a t i o n a l l y parametrized i f t h e r e 1 2 a r e r a t i o n a l f u n c t i o n s x = x l ( t ) , x2 = x ( t ) , such t h a t f ( x l ( t ) , x 2 ( t ) ) 1 2 i s i d e n t i c a l l y zero a s a f u n c t i o n o f t . I n t h i s case t h e p o i n t ( x ( t ) , x f t ) ) w i l l l i e on t h e curve C f o r 1 2 a l l values o f t . L e t ' s look a t any e a s y example.
E
m
2 Let f ( x , x ) = x1 1 2
.
unit circle.
+
x22
-
1, s o t h a t t h e curve C i s t h e
t2)
.
(Check t h i s ! )
p o i n t P = (-1,O) i s on C .
Let L
an equation f o r Lt i s x2 = t ( x l
t
+
This is not magic.
be t h e l i n e through P with slope t ;
1).
For any
The above p r o p o s i t i o n seems t o be c o n s t r u c t i v e , t o o ; t h e only question is:
t , t h i s l i n e Lt w i l l inCall
I n p a r t i c u l a r , t h e immediate
Which curves C can be r a t i o n a l l y parametrized, and how?
? + bx2 Â c, s o t h a t t h e degree o f f i s one and C i s a I f f(x ,x ) = a 1 2 l i n e , t h e n c l e a r l y C may be r a t i o n a l l y parametrized; xl = b t + z l , x2
-
at
+
x (x ) = 2 1
an, where ( z l , ~ i) s any p o i n t on C.
-
a (-)x
-
In t h i s c a s e y ( x ) =
b (Ñ i s a l i n e a r f u n c t i o n of x and any r a t i o n a l expres-
s i o n i n x and y can be immediately reduced t o a r a t i o n a l f u n c t i o n o f x a l o n e , s o t h e above process i s not t o o e n l i g h t e n i n g . F o r t u n a t e l y , t h e r e i s one o t h e r l a r g e c l a s s o f curves which can be parametrized.
T ' t ~ p ~ i L b L o n Any . conic C ( i . e . ,
defined by f ( x 1' x 2 ) = 0
where
f ( x 1 , x 2 ) is of degree 2 ) can be r a t i o n a l l y parametrized. Let me p r e s e n t two p r o o f s of t h i s s t a t e m e n t , one a l g e b r a i c
and one geometric i n s p i r i t .
Note t h a t t h e
t e r s e c t t h e c i r c l e C i n two p o i n t s , one of which i s , o f course, P .
We can now apply t h e theorem.
h i t c h is i n parametrizing t h e curve C.
Proof.
7 .
(1 - t") Then C is r a t i o n a l l y parametrized by x l ( t ) = -.-, (1 + t 2 )
2t (1
+
= x l ( t ) , x2 = x 2 ( t ) be t h e p a r a m e t r i z a t i o n of C.
One g e t s / R ( x ( t ) , X 2 ( t ) ) ( 2 ) d t , which has a
into the integral.
L e t ' s t r y t o t h i n k about t h i s s y s t e m a t i c a l l y .
such t h a t f ( x , y )
Let x
x = x ( t ) , y = x ( t ) i n t h i s c a s e ; make t h i s s u b s t i t u t i o n 1 2
Note t h a t
Let y = y ( x ) be a l g e b r a i c a l l y dependent on x .
The answer i s again:
Assume t h a t t h e curve C = { ( x x ) \ f ( x 1 , x 2 ) = 0 ) caq be 1' 2 Then h ( x , y ( x ) ) & can be found.
a l l y zero.
The f i r s t s t e p o f t h e a l g e b r a i c proof i s
t o change c o o r d i n a t e s from ( x , x ) t o ( x , y ) s o t h a t f ( x , x ) becomes 1 2 1 2 2 2 x g(x, y ) = 2 J- 2 - 1 t h e " standard form" f o r a c o n i c . This i s a a b
a
l i n e a r change of c o o r d i n a t e s , s o t h a t i f we can parametrize g ( x , y ) = 0 by r a t i o n a l f u n c t i o n s , we w i l l be a b l e t o t r a n s p o r t t h i s p a r a m e t r i z a t i o n
to ffx1,x2). The second step is to explicitly parametrize the standard conic g(x, y) = 0. Here is one way.
2
2 2
z = a k2 F a 2t2 3
Y = = 2ab2t
b k a t
As long as we're here... Parametrizing conics has been fun for millenia. Let our parametrization of the circle
b ka t
=
2
1?-I/ = 2t 2
1 +t2'
1+t
A more geometric proof is afforded by following the hint of the circle example. Pick any point P on the conic C. Parametrize the lines through P by their slopes: if P = (xo,yo), let Lt be the line y - y = 0 t(x x ) through P with slope t. Now intersect L with the conic C; t 0 one will get two points, one of which is P, the other is P = (x(t), t y(t)); it is not hard to see that x(t) and y(t) are rational parametrization~of the conic C.
-
Q.E.D. Note that in the above argument, one might want to use a vertical line sometimes where the slope "is infinity". This leads naturally into some elementary concepts of projective geometry, which I do not wish to
v
clearing denominators, we see that
.
2 2 2 In other y 2 = 1 means that (v2 u2)' + ( 2 u ~ =) ~ fu + v ) 2 2 2 2 words, (v u , Zuv, u + u ) is a Pythagorean triple. Moreover, it is
and x
2
-
+
-
an elementary theorem from number theory that all Pythagorean triples come this way. This very geometric approach to number theory was pioneered by the Greek Diophantus, and has been refined into some amazing
discuss at this time. As promised by our theorem, a proposition about parametrizing curves should give us a nice application to integrals. Here's the re-
results relating the geometry of solutions to equations and the number theory which naturally arises. But back to integration. Recall the following magic trick for integrating an expression involving sin8 and cos6 : make the substitu-
sult for conics restated for this purpose:
Coto&khg.
Note that if t is a rational number, then x and y will both be rational numbers also. So what? Well, write t = u --, with u and v integers. Then,
For any numbers a, b and a, the integral
tion 6 = Zarctanft).
Why does this work? A little trigonometry and
differentiation formulas (including the dreaded half-angle formulas) will produce can be found (where R(x ,x ) is a rational expression in two variables). 1 2
=b2 + bx + a,
Proof. If y 2 x; ffx,y) = y 2 - ax
-
bx
-
then y is algebraically dependent on a is identically zero. Since f(x,y) has
degree 2, the curve f(x ,x ) = 0 defines a conic, and therefore may be 1 2 rationally parametrized. Now the theorem applies. Q.E.D. In our course on techniques of integration, a lot of time is spent developing methods for handling integrals involving
+
bx
+
c, but -- I
and so this substitution replaces the trigonometric integrand with a rational integrand, and now we use the theorem. From our vantage point, this amazing and ad hoc substitution, which at first glance works "because it works", is seen as exactly substituting the rational parametrization of the circle which we've become quite familiar with for the trigonometric parametrization x = cos 9, y = sin 6. Hence we have the following (without any magic!):
the general result above is very rarely brought out into the open think it should be.
CokoUMy.
If R(x1 ,x2 ) is a rational expression in two variables,
then / ~ ~ ~ sin6)d8 ~ 6 ,can be found.
Recall the hyperbolic functions sinhCx) and cosh(x), so called 2 = 1; because they give a parametrization of the hyperbola x2 - x2 cosh2 (x) - sinh2 fx) = 1 for any x , so (cosh(x) , sinhfx)) always lies on the unit hyperbola. We now know that the unit hyperbola can
3 you might try to prove that y2 - x
also be rationally parametrized by
the integrals involving the square root of a cubic polynomial i n x are
-
1 = 0 can't be rationally para-
- -3 metrized. 1 One corollary of our discussion, then, is that141 t x d x can't be expected to be found with our present techniques. In general,
classically called elliptic (they arise in computing various quantities associated to an ellipse, e-g., arclength, etc.) and can't be solved in closed form using elementary functions. Now we know why: behind the whole problem lies an unparametrizable curve!
Our main theorem now yields the following immediately. CcJh~Uafl-y. If
R(x
ables, then /~(coshfx)
1
,X ) is any rational expression in two vari 2
, sinh(x))dx
can be found.
J
(Using the chain rule it is easy to see that dc = above substitutions for cosh(d and sinhfx).)
--%using the 1 - t
This just about exhausts the applications of the existence of
The problem of parametrizing curves actually led to the invention of topology.
Assume {ffx,y) = 0) is parametrized. This gives a nice
continuous function from
{t-space} to {solutions to f(x,y)
a real number; after all, we went "backward" to rational t's for a
-- why not go "forward" to complex t's?
rational parametrizations for conics to the theory of integration. Can we proceed to higher degree curves? Well, there curves which are
number-theoretic application
not conics, but which can still be rationally parametrized:
(which, again, we saw earlier was not unreasonable).
q
E x . y = sfi satisfies f(x,y) q metrized by x = t , y = if. Hence, CcJh~-!.hh~. fi(x,giq)dx
= yq
-
sf E 0.
This is para-
can be found, where R(x
2 The lemniscate ffz,y) = fx this!) has a rational parametrization
Recall that {complex t-space} is a 2-sphere, if you add the point at
-
So the above para-
metrization can be viewed as a nice continuous function from the 2-sphere to complex- solutions (x,y) to ffx,y) = 0.
Therefore, intuitively, these
complex solutions better look pretty much like a sphere. However, in x ) is any 1' 2
rational expression in two variables.
E ~ mp&. -
= 01, send-
There's nothing in all of the above ing a typical t to fxft),y(t)). discussion which says that t can't be a complex number instead of just
lots of examples, this solutions set doesn't look anything like a sphere. + x3 made up, topologically,
For example, the complex solutions to y2 = 1
+
y2)2
-
2 (x
-
y2~: 0 (draw
a torus. - So there seems to be a real topological obstruction here to parametrizing this curve, and the attempt to understand this phenomenon led to the development of modern topology. It turns out that the general curve of degree at least 3 (i.e.,
>. 3) cannot be rationally parametrized; however, there are special curves which can be, as the examples above illustrate. The ffx,y) has degree
To find this, one intersects the lemniscate C with a circle C centered t In fact, C O C t consist of g a n d one other point Pt, which has the above coordinates. at ft, -t) of radius 42t, so that 0_ = (0, 0) is on C .
The above example looks like I'm just showing off -- maybe that's right. Finding parametrizations for plane curves is not easy, and in fact most curves u(x,y) = 0) EOJ ke rationally parametrized! One 3 example is y2 - x - 1 = 0, which defines the algebraic function y =
general problem of the existence of rational parametrizations of plane curves ultimately led to the flowering: of the field of algebraic geometry, and is quite complicated. Have we then simply substituted one field of ignorance for another? No, not really. I think we have isolated the essential problem, which is one of parametrization, not integration, and along the way elucidated many of the standard results of integration theory, all in terms of one basic idea. This kind of overview can only benefit any student of this subject, can put into its proper perspective the more mundane aspects
o f t h e techniques of i n t e g r a t i o n , and hopefully motivate both s t u d e n t and t e a c h e r with a broader p i c t u r e of t h e f i e l d , One l a s t h i g h l y b e n e f i c i a l s i d e e f f e c t t o t h i s approach i s t h a t , on t h e horizon o f t h i s s u b j e c t , which seems t o some, a t first glance,
LITTLE KNOWN COMPUTER LANGUAGES
t o be a "dead end" mathematically, we s e e t h e following t o p i c s r i s i n g
Au-tfiohUnknown
t a n t a l i z i n g l y o u t o f t h e mist:
-
t h e theory of conics number t h e o r y , and diophantine equations t h e s e programming languages a r e w e l l known and There a r e
- complex v a r i a b l e s
{more o r l e s s } w e l l loved throughout t h e computer i n d u s t r y .
- higher analysis
numerous o t h e r languages, however, t h a t a r e l e s s w e l l known y e t s t i l l
-
have a r d e n t devotees.
a l g e b r a i c geometry.
This i s a l a r g e p a r t of modern mathematics!
Do a l l hard problems ( l i k e why I c a n ' t i n t e g r a t e e v e r y t h i n g ) l e a d t o such unexpected, d i v e r s e areas?
--
PASCAL, FORTRAN, COBOL
topology
I d o n ' t know, b u t even one example is an occasion f o r celebra-
I n f a c t , t h e s e l i t t l e known languages g e n e r a l l y
have t h e most f a n a t i c admirers. t h e s e obscure languages
-- and
For t h o s e who wish t o know more about why they a r e obscure
--
we p r e s e n t t h e
following c a t a l o g :
t i o n by a l o v e r of mathematics. --SIMPLE--SIMPLE
i s a n acronym f o r Sheer I d i o t s Monopurpose Programming L i n g u i s t i c Environment. This language, developed a t t h e Hanover College
f o r Technological M i s f i t s , was designed t o make it impossible t o w r i t e code with e r r o r s i n it. BEGIN, END and STOP.
The s t a t e m e n t s a r e . t h e r e f o r e , confined t o
No m a t t e r how you a r r a n g e t h e s t a t e m e n t s , you c a n ' t
make a syntax e r r o r . --SLOBOL---
REGIONAL MEETINGS
SLOBOL is b e s t known f o r t h e speed, o r lack o f it.
Although
many compilers allow you t o t a k e a c o f f e e break while they compile,
Many hegi0na.e me&tin.g~ of, the. ftathKIttOticat
COBOL compilers allow you t o t r a v e l t o Bolivia t o pick t h e c o f f e e .
A~soc^aAtonof, Am£A(.co heghave A & A A ~ O ~ & ~ O J Lunde~g~fldua-te papekb. Id -ftoo oh mohe coU.eg~iand at iea&t one l o c a t chapter. h e l p ~ ~ o n b oohh pwuXCA.paXe -in such undvi.ghaduoJts. A ~ A A ~ (A.nawAJCUi. O ~ A , M p LA avaJUiabte up to $ 5 0 . Wsuite t o :
Three o r f o u r programmers a r e known t o have d i e d o f boredom s i t t i n g a t
Dr. Richard Good Department o f Mathematics U n i v e r s i t y o f Maryland College Park, Maryland 20742
t h e i r t e r m i n a l s while w a i t i n g f o r a SLOBOL program t o compile.
Weary
SLOBOL programmers t r y t o r e t u r n t o a r e l a t e d {but i n f i n i t e l y f a s t e r } language, COCAINE. --VALGOL---
From its modest beginnings i n Southern C a l i f o r n i a ' s San
Fernando Valley, VALGOL i s enjoying a dramatic surge o f p o p u l a r i t y across t h e industry. VALGOL commands include REALLY- LIKE, WELL AND Y'NOW. assigned with t h e =LIKE and "TOTALLY o p e r a t o r s .
Variables a r e
Other o p e r a t o r s include
t h e " C a l i f o r n i a Booleans," FERSURE and NOWAY. R e p e t i t i o n s o f code a r e
handled i n FERSURE loops.
Here i s a sample VALGOL program.
BLOTTO.
Commands r e f e r t o i n g r e d i e n t s such a s CHABLIS- CHARDONNAY-
CABERNET- G I N i VERNOUTHi VODKA, SCOTCH and WHATEVERSAROUND-
LIKEiY'NOW {INEAN} START IF
The many v e r s i o n s o f t h e FIFTH language r e f l e c t t h e s o p h i s t i c a t i o n - a n d
A = LIKE BITCHEN AND B = LIKE TUBULAR AND C = LIKE GRODY**N4
financial s t a t u s of its users.
Commands i n t h e ELITE d i a l e c t include
{FERSURE}**Z
VSOR and LAFITE, while commands i n t h e GUTTER d i a l e c t i n c l u d e HOOTCH
THEN FOR I = LIKE 1 TO OH MAYBE 100 DO WAH + {DITTY**Z} BARF{ I } = TOTALLY GROSS{OUT}
and RIPPLE.
who end up using t h i s language.
SURE
--c---
LIKE BAG THIS PROGRAM REALLY LIKE TOTALLY {Y'NOW} VALGOL c h a r a c t e r i z e d by i t s unfriendly e r r o r messages.
The l a t t e r is a f a v o r i t e o f f r u s t r a t e d FORTH programmers
This language was named f o r t h e grade received by i t s c r e a t o r when
he submitted it a s a c l a s s p r o j e c t i n a graduate programming c l a s s .
i s b e s t described a s a "low l e v e l " programming language. For example,
C-
In f a c t , t h e
language g e n e r a l l y r e q u i r e s more C- statements than machine-code s t a t e -
when t h e u s e r makes a syntax e r r o r , t h e i n t e r p r e t e r d i s p l a y s t h e message
ments t o e x e r c i s e a given t a s k .
GAG HE WITH A SPOON!
t o COBOL-
--LAIDBACK---Historically, VALGOL i s a d e r i v a t i v e o f LAIDBACK, which was
--LITHP---This otherwise unremarkable language is d i s t i n g u i s h e d by t h e
developed a t t h e {now defunct} Marin County Center f o r T ' a i Chi, Mellow-
absence of an "S" i n i t s c h a r a c t e r s e t .
n e s s and Computer Programming, a s an a l t e r n a t i v e t o t h e more i n t e n s e
s u b s t i t u t e "TH".
In t h i s r e s p e c t , it i s very s i m i l a r
Programmers and u s e r s must
LITHP is s a i d t o be u s e f u l i n p r o t h e t h i n g l i t h t t h .
atmosphere i n nearby S i l i c o n Valley. The Center was i d e a l f o r programmers who l i k e d t o soak i n h o t t u b s while
--DOGOÑÑDevelop
they worked.
DOG0 h e r a l d s a new e r a o f c o m p u t e r - l i t e r a t e p e t s .
Unfortunately, few programmers could s u r v i v e t h e r e f o r
long, s i n c e t h e Center outlawed p i z z a and RC Cola i n f a v o r o f bean curd
Many mourn t h e demise of LAIDBACK because of its r e p u t a t i o n a s a g e n t l e and nonthreatening language.
For example, LAIDBACK responded t o syntax
e r r o r s with t h e message: SORRY NAN- I CAN'T DEAL BEHIND THAT. SARTRE---Named a f t e r t h e l a t e e x i s t e n t i a l philosopher, SARTRE is an
extremely u n s t m c t u r e d language. they j u s t a r e . functions.
SIT, STAY, HEEL and ROLL OVER.
DOG0 commands i n r l u d e
An innovative f e a t u r e o f DOG0 is " p ~ p p y
graphics," a small cocker s p a n i e l t h a t o c c a s i o n a l l y l e a v e s a d e p o s i t a s
and P e r r i e r .
--
a t t h e Massachusetts I n s t i t u t e o f Obedience Training,
Statements i n SARTRE have no purpose,
Thus, SARTRE programs a r e l e f t t o d e f i n e t h e i r own
he t r a v e l s a c r o s s t h e screen. language d e s i g n ~ dt o run on small DEC machines with
--FOCUSALL---a minimal memory.
I t s only supported d i s t r i b u t i o n is paper t a p e , f o r
loading i n from an ASR-33 t e l e t y p e .
This t a k e s 20 minutes, a f t e r which
t h e user is g r e e t e d with t h e message: CONGRATULATIONS!
YOU HAVE JUST LOADED FOCUSALL!
SARTRE programmers tend t o be b o r i n g and depressed and a r e The i n t e r p r e t e r is then ready t o accept any v a l i d command.
no fun a t p a r t i e s .
Th- only
v a l i d command i s : --FIFTH---FIFTH
i s a p r e c i s i o n mathematical language i n which t h e d a t a
types r e f p r t o q u a n t i t y .
The d a t a types range from CCi DUNCE- SHOT and
JIGGER t o FIFTH {hence t h e name of t h e language}, LITER- MAGNUM and
LOAD FOCUSALL which causes t h e system t o once again load t h e tape.
interpret^^- from paper
The power o f t h e language comes f r o m t h e f a c t t h a t preceding a command w i t h a statement l i n e causes it t o be s t o r e d a s a program l i n e f o r l a t e r execution a s i n t h e following example:
100 LOAD FOCUSALL 110 LOAD FOCUSALL 150 LOAD FOCUSALL
ANOTHER DAY AT THE RACES
The pronunciation o f t h e name is much more f l e x i b l e than t h e language itself.
You pronounce it according t o your mood.
I n p r e p a r i n g a p r e s e n t a t i o n o f "A Day a t t h e Races" by William Tomcsanyi
(Pi Mu. Epsilon J o w m Z , Spring 1982) f o r o u r c h a p t e r o f P i Mu
Actually, t h e name came f r o m a combination o f DEC FOCAL, a PDPB DELIGHT, and t h e h a b i t we
Epsilon, West V i r g i n i a Alpha, I came a c r o s s a s i m p l i f i c a t i o n o f t h e pro-
Optics Lab t y p e s i n days o f yore had o f r e f e r r i n g t o a l e a d engineer
cedure which reduces t h e c a l c u l a t i o n s i n M r . Tomcsanyi's a r t i c l e .
a s "Focus Man" {should b e "Focus Person"".}.
Somebody would t h e n chime
o u t : "and h e ' s gonna focus every chance he gets!" -PINBOL---PINBOL program run.
is b e s t known f o r t h e chance involved i n making its Three t r i e s a t running a r e allowed, a f t e r which t h e
message "GAIIE OVER-
INSERT QUARTER AND TRY AGAIN" is displayed.
Some allowable PINBOL i n s t r u c t i o n s and t h e i r meanings a r e : LEFT FLIPPER RIGHT FLIPPER SHOOT
I1l o g i c a l L e f t Shi f t I l l o g i c a l Right S h i f t Try to Run
PINBOL is known t o be extremely a d d i c t i v e .
Those who a r e f l u e n t PINBOL
The impetus f o r t h i s s i m p l i f i c a t i o n was M r . Tomcsanyi's attempt t o s o l v e a n 8 x 8 matrix r e p r e s e n t i n g every h o r s e i n a n eight- horse r a c e , which i f s u c c e s s f u l would have shown t h a t it i s p o s s i b l e t o b e t on every h o r s e i n a given r a c e and come o u t ahead no m a t t e r which one wins. H i s example produced n e g a t i v e r e s u l t s , i n d i c a t i n g no s o l u t i o n , b u t he s t a t e d , "There probably does e x i s t some combination o f odds t h a t would somehow y i e l d p o s i t i v e r e s u l t s t o t h e e i g h t equations." I n f a c t , t h i s conjecture is not true. two important r e l a t i o n s a r e used: I)
E r r o r messages include:
"COIIPUTUS INTERRUPTUS-" A c l o s e l y r e l a t e d language i s NOONER-
users.
SUB
o f t h e b e t s placed on each o f t h e horses i s c o n s t a n t
x1 + x , + . 11)
--GERITOL---This
The
throughout t h e problem:
programmers a r e known a s PINBOL WIZARDS. --FASTBOL---commonly known a s a QUICKIE.
language i s c h a r a c t e r i z e d by t h e h a b i t s o f i t s a r d e n t
I n s t r u c t i o n s f r e q u e n t l y f o r g e t t h e i r f u n c t i o n while executing
and conclude with t h e "I USED TO KNOW THATn c o n d i t i o n code.
Loops tend t o r e p e a t f r e q u e n t l y a t sporadic i n t e r v a l s , even when n o t i n f l a t e d .
To prove t h a t it i s n o t ,
. . +xg=S.
The sum o f t h e f r a c t i o n s o f t h e t o t a l p o t b e t on each horse
as computed from t h e odds i s g r e a t e r t h a n one. The second o f t h e s e occurs because t h e t r a c k and t h e s t a t e t a k e a p o r t i o n o f t h e t o t a l amount b e t , l e a v i n g l e s s t h a n 100% f o r t h e winner's pool. Typically, t h e winners' pool would be about 80% of t h e t o t a l , b u t t o be a s g e n e r a l as
Track's percentage
p o s s i b l e , l e t u s s a y t h a t t h e t o t a l b e t by everyone a t t h e t r a c k i s B and t h e winners' pool is US, where k < 1.00.
Winners ' pool
Pictorially,
t h e winners' pool would look l i k e t h i s : Now, t h e odds on each horse a r e computed based on t h e amount b e t I f t h e odds a r e a
on t h a t h o r s e compared t o t h e t o t a l winners' pool. t o 1, then a = (kB
-
h)/h, where h is t h e t o t a l amount b e t on t h a t h o r s e ,
or (a
+
KB / h .
1) =
I n an i d e a l r a c e , where t h e winners' pool were 100%
of t h e amount b e t , we would have ( a
+
1 ) = B / h, o r I/ ( a
t h e f r a c t i o n of t h e t o t a l which was bet on t h i s horse.
+
1 ) = h/ B ,
From 11, t h e sum of t h e 1 / ( a; showed t h a t t h i s number was
1/k
.
+
1 ) 's i s g r e a t e r than one.
We
Thus
Here, though, we
have 1 / (a + 1 ) = h /kB, and t h e sum of t h e e i g h t f r a c t i o n s i s g r e a t e r than one:
Since we have chosen P t o be p o s i t i v e and follows t h a t S must be negative.
k i s l e s s than one, it
Thus t h e r e i s never any way t o b e t on
every horse i n a r a c e and come out ahead no matter who wins, even i n an i d e a l s i t u a t i o n where t h e t r a c k takes nothing f o r i t s e l f . It i s a simple matter t o generalize t h i s argument f o r any s i z e race. But t h e s t e p s used i n t h i s argument do more than j u s t prove t h a t no s o l u t i o n e x i s t s involving every horse i n t h e race; some i n t e r e s t i n g Let us now go back t o t h e 8 x 8 system used by M r . Tomcsanyi,
i and P = t h e desired p r o f i t , a s before, t o t h e t o t a l d o l l a r amount t o be b e t on horse i, i n s t e a d
l e t t i n g ai = t h e odds on horse but changing x .
of t h e number of $2. b e t s .
a
1 -1
s i d e b e n e f i t s f a l l out along t h e way. 1)
I
Determine whether t h e r e e x i s t s a combination of b e t s y i e l d i n g
a p o s i t i v e s o l u t i o n f o r any number of horses l e s s than t h e t o t a l ; 2)
The revised matrix i s a s follows:
We can now
Estimate t h e t o t a l amount which must be b e t t o produce t h e
desired p r o f i t before t h e i n d i v i d u a l amounts a r e computed; and
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
Taken t o g e t h e r , these form a three- step process which i s a s gen-
-1
-1
-1
-1
-1
e r a l a s t h e matrix form and, once you g e t used t o it, a s simple t o com-
-1
-1
-1
-1
-1
-1
a2
-1
-1
-1
a3
-1
-1
-1
-1
-1
-1
a,, -1
a5
-1
-1
-1
-1
-1
-1
a6
-1
-1
-1
-1
-1
-1
-1
-1
a7
-1
-1
-1
-1
-1
-1
-1
-1
3)
pute a s t h e general form of t h e 3 x 3: 1)
writing each row equation would be
-
fai
or
+
(S - x . )
l)x
i
=
P
=
i
+
S)
(
Using t h e sum above a s g, plug it i n t o t h e equation
P =
(
-a1 - 1) S and
solve f o r S.
You may choose any value f o r
P, but t h e bigger P i s , t h e bigger S becomes. 3)
P,
Once you have chosen P a n d computedS, compute t h e amount t o
bet on each horse by
giving
With t h i s three- step system, it i s j u s t a s easy t o bet on four o r
P + S -. - a +I
f i v e horses a s on t h r e e , and s i n c e t h e c a l c u l a t i o n s a r e s h o r t e r , it can
i
a l l be done using l a t e r odds.
To f i n d 5 , we use t h i s formula t o s u b s t i t u t e f o r each xi
(P
Compute
+ S.
This i s e a s i l y solved f o r x;
x
2)
Since t h e amount l o s t i s the t o t a l
amount b e t minus t h e amount b e t on t h e winning horse, a simple means of
a2. x2.
You wish t o b e t on n h o r s e s i n t h e race.
I f t h i s sum i s l e s s than one, t h e r e e x i s t s a p o s i t i v e solution.
Notice t h a t each row i s j u s t an expression f o r t h e p r o f i t a s t h e amount won minus t h e amount l o s t .
Find t h e individual b e t s with a minimum of calculations.
+ -a 2 + l + . . . + al+l
)=S. a&
i n I:
ii ) For any real n x n matrices R and S such that R =
> 0, -
ON THE DISCRETE LYAPUNOV AND R I C C A T I MATRIX EQUATIONS
by Aknh T. Tian Boe-tng I4LLutamj Ahplane Comprtni~ and Mahmoud E. Sawan WickLti~State U n i v m L t y
iii ) For any m x n matrix Y , n x m matrix 2 , n y n matrix W and
rnxm
matrix x , we have the following property ABSTRACT
I n t h i s note, the i n e q u a l i t i e s , which are s a t i s f i e d by the determinants o f the positive d e f i n i t e solutions of the discrete algebraic Rioaati and Lyapunov matrix equations, are presented. The r e s u l t s give lower bounds for t h e product o f the eigenvalues of the matrix solutions. Also for a discrete Lyapunov equation, we present an algorithm t o determine wider what conditions a positive diagonal solution will e x i s t . I f a l l the conditions care s a t i s f i e d , the algorithm also provides such a diagonal solution.
[w +
Z X - ~ Y I - ~=
-
w-lz[z
+ YW-~Z]-~YW-~
(3)
11. THE RICCATI EQUATION In this section, we derive a lower bound for the determinant of the discrete algebraic Riccati matrix equation
-
P = A ~ P A A~PB(I + ~PB)-'$PA
",B
where A, P, Q&
+Q
d x m, Q = QT > 0.
(4) Here we assume
1~ $ 1 2
\Q\
and the matrix A is stable, therefore the solution matrix P is positive definite. Thwhem 1.
The discrete algebraic Riccati and Lyapunov matrix equations have
w-I
The determinant of the positive definite matrix solu-
tion P of equation (4) satisfies the following inequality
been used widely in various areas of engineering system theory, particularly in control system theory. The techniques of solving these equations numerically are well-established [ 4 ] . Those techniques are mostly iterative algorithms which require making an initial guess of the solution. So if these initial guesses are chosen wisely, one can save a lot
IPI
where M =
of unnecessary computations. Therefore, to obtain precise estimates of the "sizes" of the solutions, we provide here lower bounds for the determinants of the matrix solutions of the two equations. Also for the discrete Lyapunov equations, we address the question of the existence of a positive diagonal solution for this equation and derive an algorithm to provide such a solution if all the conditions are satisfied. In the following, the notations xT, \.(x),
Proof.
Using (3) with W = P - I ,
-1
Z = B, x
= I
comes P = ~ ~ 1 p + - lB
Multiplying (6) by [ P -
~
+
+Q
A
i) For any n x n matrices L and H with L > 0
(6)
BB^] from the left yields
t v f x ) and 1x1 denote the transpose, eigenvalue, trace and determinant of the matrix x , respectively. Also for our derivation later, we will make use of the following results [I ,3].
T
and Y = B , (4) be-
Computing the traces of both sides of (7) and using (1) with LT [ P - + ~ $ 1 , H = A and rearranging terms, we have
1 =
Now using (2) with R = P and note that 1x1 = I/ \ x - 1 and tr(x)
= trfxT 1 ,
(8) becomes nlBf12'11/nlp12/n
-
Solving (9) for
1~1'~
Mlpll/n
-
nIQI
1/n
Lyapunov equation (10) exists if and only if A is a stable matrix, i.e., However, we are concerned here with the question of \\!A) < 1 [5] whether or not there exists a positive diagonal matrix solution P. In
.
1
-
(9
, we get the inequality
other words, given a real square stable matrix A, we pose the problem:
(5 1.
Lower Bound for the Solution of the Lyapunov Equation
Find the conditions on A such that a positive diagonal matrix P exists T where A PA - P < 0. Such a matrix has been used widely in the stability In the analysis, control theory and many of its applications [6,7].
Setting B = 0 in (4), the result is the discrete algebraic
following sections we present an algorithm to determine under what con-
Ill. THE LYAPUNOV EQUATION A.
B. Positive Diagonal Solution of the Lyapunov Equation It is well-known that the solution of the discrete algebraic
Lyapunov matrix equation
T
P=APA+Q
(10)
ditions such a matrix P will exist and if a11 the conditions are satisfied, the algorithm also provides the value of P. 1. Derivation of the algorithm
and we present the following theorem.
The following definitions are needed for our derivation.
T
Thtotem 2 .
a) x = ( r , x 2
The determinant of the positive definite matrix solu-
tionP of equation(10) satisfies the following inequality
--., xn)
b) P = diag fx,xy c) E = { e d ,
. .., xn)
llell = I 1
d) &E is an eigenvector corresponding
. .., -en) Proof.
1I
Multiplying (10) by P-I from the left and computing the
traces of both sides yield
With the above definitions, let
n = ~PCP'A~PA) + tr~p"'~)
(12
Then, using (1 ) with L = P and H = A
1
and noting that Ai ( A ) = \-(A T ), we
max eT (ATPA eeE = max ffx,e) ee-E =
n
have
~P(P-Q)2 n -
\\.(A)
\2
i=l
-
V)e
T
Substituting R = P - and
S =
- into (2) IQI'~~
leads to
-
We observe that a diagonal matrix P > 0 exists such that A PA P < 0 if and only if we can find a point XEX such that hfx) < 0. So the problem now is to determine whether or not such a point 2 exists. Before a search algorithm is derived, we need to point out some properties of h(x). Let
Rewriting (14), we get the inequality (11).
be an interior point of X and
To find x;
, let
then
xi = qkxk +
then it has been shown [2] that
(1
-
qkZk
So, we can compute x k+1 using the following formula
and h(x) 2.
2 xTg , for
all xeX c) Compute gk+l
The search procedure a) Start with xleX and compute h ( x l ) . If h f x l ) < 0 , stop; otherwise compute g , and choose x2^X such that xT g 2 1
