9
This book
the
is
to be returned on or before
last
date stamped below.
NAME
CALCULUS
Michael Spivak
BRANDEIS UNIVERSITY
WORLD STUDENT SERIES
EDITION
A complete and unabridged reprint a/the
American textbook,
may be
this
World Student
sold only in those countries
consigned by Addison-Wesley or
distributors. It
to
which
the
it
may not
to
its
United States of America or
is
which
it
is
authorized trade
be re-exportedfrom the country
has been consigned, and
The publisher
original
Series edition
pleased
to
its
it
may
not be sold in
possessions.
acknowledge the assistance
of Wtadistaw Finns, who designed the text and cover,
and F. W. Taylor, who produced the illustrations.
Copyright
reserved.
©
1967 by W. A. Benjamin, Inc. All rights
of this publication may be reproduced,
Ma part
stared in a retrieval system, or transmitted, in any form
or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written
permission of the publisher. Original edition published
in the United States of America, Published simultaneously
in
Canada. Philippines copyright 1967.
Library of Congress Catalog Card
Number: 67-20770
Dedicated
to the
Memory ofY.
P.
PREFACE
/ hold
every
man
a debtor
to his profession,
from
the
doe seeke
which as men of course
to receive
countenance and profit,
endeavour
so ought they
of duty
themselves by
way of amends,
to
be a help
to
and
ornament thereunto.
FRANCIS BACON
Every aspect of this book was influenced by the desire to present calculus not
merely as a prelude to but as the first real encounter with mathematics.
Since the foundations of analysis provided the arena in which modern modes
of mathematical thinking developed, calculus ought to be the place in which
than avoid, the strengthening of insight with logic. In addition to developing the students' intuition about the beautiful concepts of
analysis, it is surely equally important to persuade them that precision and
rigor are neither deterrents to intuition, nor ends in themselves, but the
to expect, rather
natural
medium
in
which
to
formulate and think about mathematical
questions.
This goal implies a view of mathematics which, in a sense, the entire book
No matter how well particular topics may be developed,
the goals of this book will be realized only if it succeeds as a whole. For this
attempts to defend.
reason,
it
would be of
little
tion pedagogical practices
customarily bestowed on
value merely to
calculus texts will probably
particular aspects of calculus will
men-
the topics covered, or to
and other innovations. Even
new
the cursory glance
tell
more than any
the teacher with strong feelings about
such extended advertisement, and
fulfills his
list
know just where
to look to see
if
this
book
requirements.
A few features do require explicit comment, however. Of the twenty-nine
chapters in the book, two (starred) chapters are optional, and the three chapters comprising Part V have been included only for the benefit of those
students who might want to examine on their own a construction of the real
numbers. Moreover, the appendices to Chapters 3 and
1 1
also contain optional
material.
The
order of the remaining chapters
the purpose of the book
is
is
intentionally quite inflexible, since
to present calculus as the evolution of one idea, not
most exciting concepts of calculus do not
and II will probably
although the entire book covers
require less time than their length suggests
a one-year course, the chapters are not meant to be covered at any uniform
as a collection of "topics." Since the
appear
rate.
A
until Part III, I should point out that Parts I
—
rather natural dividing point does occur between Parts II
and
III, so
quickly by
it is possible to reach differentiation and integration even more
detailed
more
a
treating Part II very briefly, perhaps returning later for
organization
traditional
treatment. This arrangement corresponds to the
of most calculus courses, but
I feel
that
it
will only
diminish the value of the
book for students who have seen a small amount of calculus previously, and
for bright students with a reasonable background.
The problems have been designed with this particular audience in mind.
They range from
straightforward, but not overly simple, exercises-
which
develop basic techniques and test understanding of concepts, to problems of
considerable difficulty and, I hope, of comparable interest. There are about
625 problems in
many
vii
all.
Those which emphasize manipulations usually contain
examples, numbered with small
Roman
numerals, while small
letters
are used to label interrelated parts in other problems. Some indication of
is provided by a system of starring and double starring, but
relative difficulty
there are so
many
criteria for
judging
difficulty,
and
so
many
hints
have been
provided, especially for harder problems, that this guide is not completely
reliable. Many problems are so difficult, especially if the hints are not con-
probably have to attempt only those which
difficult problems it should be easy to
from
the
less
them;
especially interest
class busy, but not frustrated. The
good
will
keep
a
which
portion
select a
answer section contains solutions to about half the examples from an assortment of problems that should provide a good test of technical competence. A
sulted, that the best of students will
separate answer book contains the solutions of the other parts of these problems, and of all the other problems as well. Finally, there is a Suggested Read-
ing
list,
to
which the problems often
refer,
and an index of symbols.
am grateful for the opportunity to mention the many people to whom I
owe my thanks. Mrs. Jane Bjorkgren performed prodigious feats of typing^
that compensated for my fitful production of the manuscript. Mr. Richard
I
Serkey helped collect the material which provides historical sidelights in the
problems, and Mr. Richard Weiss supplied the answers appearing in the
back of the book. I am especially grateful to my friends Michael Freeman,
Jay Goldman, Anthony Phillips, and Robert Wells for the care with which
they read, and the relentlessness with which they criticized, a preliminary
version of the book. Needless to say, they are not responsible for the deficiencies which remain, especially since I sometimes rejected suggestions which
would have made the book appear suitable for a larger group of students.
I must express my admiration for the editors and staff of W. A. Benjamin, Inc.,
who were always eager to increase the appeal of the book, while recognizing
the audience for which it was intended.
The inadequacies which preliminary editions always involve were gallantly
endured by a rugged group of freshmen in the honors mathematics course at
Brandeis University during the academic year 1965-1966. About half of this
course was devoted to algebra and topology, while the other half covered
calculus, with the preliminary edition as the text. It
is
almost obligatory in
such circumstances to report that the preliminary version was a gratifying
after all, the class is unlikely to rise up in a body
success. This is always safe
and
protest publically
—
—but the students themselves,
it
seems to me, deserve
the right to assign credit for the thoroughness with which they absorbed
an impressive amount of mathematics. I am content to hope that some
other students will be able to use the book to such good purpose, and with
such enthusiasm.
Waltham, Massachusetts
February 1967
MICHAEL SPIVAK
PREFACE TO THE
WORLD STUDENT SERIES EDITION
In American universities, "calculus" courses are so varied in content and
viewpoint that only tradition dictates that the same title be given to all. Some
of these courses cover only the most rudimentary techniques, sedulously shun-
ning all theory. At the other extreme, so-called "honors" courses concentrate
on the foundations, and emphasize the theoretical importance of the basic
processes of calculus, which have often already been seen by the students in
their high school classes. This book was written for just such courses. In the
United
States, they are so rare that
I
felt
constrained to begin
preface with a lengthy defense, or apology for the fact that
for the very type of course
and
I would
usually called
like to
which
in
it
Great Britain and Europe
was
is
my
original
really written
quite standard,
"An introduction to Real analysis". From that long prologue
salvage only one thought, and once again express
my
feeling
that the foundations of analysis provide one of the most beautiful instances of
the welding together of mathematical intuition
and mathematical
rigor,
and
that ignoring either one only weakens, rather than strengthens, the other.
A
few comments about the contents and arrangement should be made,
who is learning the material on his own. Of the twenty-
especially for the student
nine chapters in the book, two (starred) chapters are "optional", and the three
V
might be thought of in the same way. Moreover,
and 1 1 also contain "optional" material. Aside
from this, the order of the remaining chapters is intentionally quite inflexible,
since the purpose of the book is to present calculus as the evolution of one idea,
not as a collection of "topics". Since the most exciting concepts of calculus
do not appear until Part III, I should point out that Parts I and II will probably
require less time than their length suggests, and although the entire book
covers a one-year course, the chapters are not meant to be covered at any
uniform rate. A rather natural dividing point does occur between Parts II and
III, so it is possible to reach differentiation and integration even more quickly by
reading Part II very briefly, perhaps returning later for more detailed study. But
I personally feel that the route set forth here, though it requires perseverance
at the beginning, will be the most rewarding.
The problems, without which any mathematics text is incomplete, range from
straightforward (but not overly simple) exercises which develop basic techniques and test understanding of concepts, to problems of considerable difficulty and, I hope, of comparable interest. There are about 625 problems in all.
Those which emphasize manipulations usually contain many examples,
chapters comprising Part
the appendices to Chapters 3
numbered with small Roman numerals, while small
interrelated parts in other problems.
Some
letters are
used to label
indication of relative difficulty
is
provided by a system of starring and double starring, but there are so many
criteria forjudging difficulty, and so many hints have been provided, especially
for
harder problems, that
are so difficult, especially
will
this
if
is
not completely reliable.
Many problems
probably have to attempt only those which especially interest them. From
it should be easy to select a portion which will be
the less difficult problems
ix
guide
the hints are not consulted, that the best of students
*
Preface
to the
World Student
Series edition
challenging, but not frustrating.
The Answer
Section contains solutions to
about half the examples from an assortment of problems that should provide a
good test of technical competence. A separate answer book, Supplement to
Calculus, contains the solutions of the other parts of these
the other problems as well. Finally, there
is
problems, and of all
a Suggested Reading
list,
to
which
and a Glossary of Symbols.
Once again I will take the opportunity to mention the many people to whom
feats of typing
I owe my thanks. Mrs. Jane Bjorkgren performed prodigious
Mr. Richard
manuscript.
the
of
production
fitful
my
for
that compensated
in the
sidelights
historical
provided
which
material
the
collect
helped
Serkey
problems, and Mr. Richard Weiss supplied the answers appearing in the back
of the book. I am especially grateful to my friends Michael Freeman, Jay
Goldman, Anthony Phillips, and Robert Wells for the care with which they
read, and the relentlessness with which they criticized, a preliminary version
the problems often refer,
of the book.
editions always involve were gallantly
in the honors mathematics course at
freshmen
of
group
rugged
a
by
endured
Brandeis University during the academic year 1965-1966. It is almost obligatory in such circumstances to report that the preliminary version was a
The inadequacies which preliminary
gratifying success. This
is
always safe— after
—
all,
the class
is
unlikely to rise
but the students themselves,
seerns to
up
me,
in a body and protest publicly
deserve the right to assign credit for the thoroughness with which they absorbed
an impressive amount of mathematics. I am content to hope that some other
students will be able to use the book to such good purpose, and with such
it
enthusiasm.
Berkeley, California
MICHAEL SPIVAK
CONTENTS
PART
PART
I
II
Prologue
1
Basic Properties of
2
Numbers
Numbers
of Various Sorts
Foundations
3
Functions
37
Appendix. Ordered Pairs
Graphs
5
Limits
6
Continuous Functions 93
Three Hard Theorems 100
1 1
Least Upper Bounds
8
III
52
4
7
PART
3
21
Derivatives
9
54
72
and
Integrals
Derivatives
125
144
10
Differentiation
11
Significance of the Derivative
Appendix. Convexity and Concavity
12
214
Integrals
14
The Fundamental Theorem of Calculus
The Trigonometric Functions 256
*16
17
18
IV
191
200
13
15
PART
Inverse Functions
163
7r
*20
21
Irrational
22
Infinite Series
Uniform
Complex
Complex
Complex
25
26
xi
283
Sequences and Infinite Series
Approximation by Polynomial Functions
362
e is Transcendental
372
Infinite Sequences
23
24
277
The Logarithm and Exponential Functions
Integration in Elementary Terms 302
Infinite
19
is
240
333
388
Convergence and Power Series
Numbers 433
Functions 448
Power Series 462
412
xii
Contents
PART V
Epilogue
487
27
Fields
28
Construction of the Real
29
Uniqueness of the Real Numbers
Suggested Reading
513
523
Answers {to selected problems)
Glossary of Symbols
Index
575
571
Numbers
494
507
PART
PROLOGUE
To
you
to
be conscious that
are ignorant
is
a great step
knowledge.
BENJAMIN DISRAELI
CHAPTER
NUMBERS
BASIC PROPERTIES OF
I
The
title
of this chapter expresses in a few
words the mathematical knowledge
required to read this book. In fact, this short chapter is simply an explanation
addition
of what is meant by the "basic properties of numbers/' all of which
and multiplication, subtraction and division, solutions of equations and
—
and
factoring
inequalities,
manipulations— are
algebraic
other
already
not a review. Despite the familiarity
of the subject, the survey we are about to undertake will probably seem quite
novel; it does not aim to present an extended review of old material, but to
condense this knowledge into a few simple and obvious properties of numbers.
familiar to us. Nevertheless, this chapter
is
too obvious to mention, but a surprising number of
turn out to be consequences of the ones we shall
Some may even seem
diverse and important
facts
emphasize.
Of the twelve properties which we shall study in this chapter, the first nine
are concerned with the fundamental operations of addition and multiplicaperformed
tion. For the moment we consider only addition: this operation is
—the sum
a + b exists for any two given numbers a and b
number, of course). It might seem reasonsame
(which may possibly be the
which can be performed on several
operation
an
as
addition
regard
able to
+ a n of n numbers
numbers at once, and consider the sum a\ +
a% as a basic concept. It is more convenient, however, to consider
au
addition of pairs of numbers only, and to define other sums in terms of sums
of this type. For the sum of three numbers a, b, and c, this may be done in twodifferent ways. One can first add b and c, obtaining b + c, and then add a to
then
this number, obtaining a + (b + c); or one can first add a and b, and
compound
two
the
course,
Of
c.
b)
(a
obtaining
+
+
add the sum a + b to c,
sums obtained are equal, and this fact is the very first property we shall list:
on a pair of numbers
•
•
•
•
,
(PI)
If «, b,
and
c
are
a
The statement
of three
•
+
any numbers, then
(b
+
=
c)
(a
+
b)
+
c.
of this property clearly renders a separate concept of the
numbers superfluous; we simply agree that
=
+
+
a
+ +
b
c
sum
denotes the
Addition of four numbers requires
b
considerations. The symbol a
involved,
similar, though slightly more
mean
defined
to
d is
c
number
a
+
(b
+
c)
}
(a
b)
c.
+ +
+
(1)
3
or
(2)
or
(3)
or
(4)
or
(5)
+ b)+c)+d,
+ (b + e)) + d
a + {(b + c) + d),
+ d)),
a + (b +
(a + b) +
+ d).
((a
(a
9
(c
(c
This definition
unambiguous
is
since these
numbers are
all
equal. Fortunately,
need not be listed separately, since it follows from the property PI
already listed. For example, we know from PI that
this fact
(a
+
b)
+c=a+
(b
+ c),
equality of (2)
and
(3) is a direct consequence of PI, although this may not be apparent at
sight (one must let b
c play the role of b in PI, and d the role of c).
first
and
it
and
follows immediately that (1)
(2) are equal.
The
+
equalities (3)
=
(4)
=
The
(5) are also simple to prove.
probably obvious that an appeal to PI will also suffice to prove the
equality of the 14 possible ways of summing five numbers, but it may not be so
clear how we can reasonably arrange a proof that this is so without actually
listing these 14 sums. Such a procedure is feasible, but would soon cease to be
if we considered collections of six, seven, or more numbers; it would be totally
It is
inadequate to prove the equality of
possible
all
sums of an arbitrary
finite
be taken for granted, but
an
for
to worry about the proof (and it is worth worrying
about once) a reasonable approach is outlined in Problem 23. Henceforth, we
numbers ai,
those who would like
collection of
shall usually
+
ax
•
•
.
,
may
If a
0 has one property so important that
is
we
list it
next:
any number, then
important role
(P3)
This fact
tacit appeal to the results of this problem and write sums
with a blithe disregard for the arrangement of parentheses.
+
a
An
.
.
make a
+ an
The number
(P2)
.
is
0
also played
For every number
a+
a,
=
0
+
by 0
there
a.
in the third property of
number —a such
a
is
= (-
(-a)
=
a
+a =
fl )
our
list:
that
0.
Property P2 ought to represent a distinguishing characteristic of the number
and it is comforting to note that we are already in a position to prove
0,
this.
Indeed,
if
a number x
satisfies
a
+
x
=
a
any one number a, then x = 0 (and consequently this equation also holds
numbers a). The proof of this assertion involves nothing more than
subtracting a from both sides of the equation, in other words, adding — a to
both sides; as the following detailed proof shows, all three properties P1-P3
for
for all
must be used
to justify this operation.
If
then
hence
hence
hence
+x=
+ + x) =
(( — a) + a) + x =
0 + # =
a
(—a)
(a
x
=
<z,
{—a)
0;
0;
0.
+a =
0;
Basic Properties of Numbers
As we have just hinted,
derived from addition:
it is
we
5
convenient to regard subtraction as an operation
— b).
— b to be an abbreviation for a
(
+
consider a
then possible to find the solution of certain simple equations by a series
of steps (each justified by PI, P2, orP3) similar to the ones just presented for
x = a. For example:
the equation a
It is
+
=
(* + 3) + (-3) =
x + (3 + (-3)) =
x + 0 =
x —
+
x
If
then
hence
hence
hence
3
5,
+ (-3);
5-3 = 2:
5
2;
2.
Naturally, such elaborate solutions are of interest only until you become convinced that they can always be supplied. In practice, it is usually just a waste
of time to solve an equation
ties
Only one other property
c
so explicitly the reliance
and
a
+
the order of
(P4)
shall
of addition remains to be listed.
on proper-
list).
a, b,
and
When
considering
+
c,
+
(b
a, b,
If a
we
b)
only two sums were mentioned: (a
if
obtained
are
arrangements
c). Actually, several other
on
equal
depends
all
sums
are
these
That
changed.
and c is
the sums of three numbers
+
by indicating
PI, P2, and P3 (or any of the further properties
and
any numbers, then
b are
+
a
b
—
b
+ a.
The statement of P4 is meant to emphasize that although the operation of
addition of pairs of numbers might conceivably depend on the order of the
two numbers, in fact it does not. It is helpful to remember that not all operations are so well behaved.
erty: usually a
—
—
b 9± b
—
For example, subtraction does not have this propa. In passing we might ask just when a — b does
amusing to discover how powerless we are if we rely
only on properties P1-P4 to justify our manipulations. Algebra of the most
elementary variety shows that a — b — b — a only when a = b. Nevertheless,
it is impossible to derive this fact from properties P1-P4; it is instructive to
examine the elementary algebra carefully and determine which step(s)
equal
b
cannot be
detail
a,
and
it is
justified
when
We
by P1-P4.
indeed be able to justify all steps in
listed. Oddly enough, however, the
will
a few more properties are
crucial property involves multiplication.
basic properties of multiplication are fortunately so similar to those for
addition that little comment will be needed; both the meaning and the conse-
The
quences should be
will
(P5)
If a, b }
*
and
b>
c
or simply ab.)
are
a
(P6)
If a
is
elementary algebra, the product of a and
clear. (As in
be denoted by a
any numbers, then
'
(b
'
c)
=
(a
-
b)
'
any number, then
a
'
1
=
1
•
a
=
a.
c.
b
1^0.
Moreover,
1^0 may seem a strange fact to
(The assertion that
list it,
because there is'no
the other properties listed
way
—these properties would
only one number, namely,
a
(P8)
If a
and
*
detail
there
is
no
— a~
a~ l
l
=
b
'
=
means
a
*
b~~
l
.
if
there were
a.
'
the appearance of the condition
is
quite necessary; since 0 b = 0 for all numbers
-1 =
satisfying 0 0
1. This restriction has an impor-
1
•
is
•
in terms of
was defined
defined in terms of multiplication: the symbol a/b
is
CT" 1 is
Since
hold
all
1.
tant consequence for division. Just as subtraction
addition, so division
to
a number a~ l such that
is
a
•
b
which deserves emphasis
number CP
we have
any numbers, then
b are
a 7* 0 in P7. This condition
b,
there
=^ 0,
a
One
but
0.)
For every number a
(P7)
list,
could possibly be proved on the basis of
it
meaningless, a/0
also meaningless
is
—division by 0
always undefined.
is
Property P7 has two important consequences. If a
=
necessarily follow that b
matter what
from P7 as
b
and
c are.
for
c;
if
However,
In
a
l
then
a~~
hence
hence
hence
(a~ l
(a
•
•
if
a
b
•
then both a
0,
^
—
=
—
=
=
b
•
b)
•
a)
*
b
1
•
b
b
a consequence of P7 that
then
0,
—
b
b
c;
=
a
and
this
*
c, it
a
'
c
does not
are
no
0,
can be deduced
if
a
•
a
*
and
c
a~ x
(a~ l
*
1
(a
*
•
'
a)
a
0,
c)\
•
c;
c;
c.
b
=
0,
then either a
=
0 or £
=
0.
fact,
may happen
excluded
-1
then
a
hence
hence
hence
(a~ l
that a
when we
~
=
a)
b —
1-^ =
=
a
if
(It
—
follows:
If
It is also
a
=
(a
•
*
'
6
6)
•
•
/;
and
0
6
say "either a
0 and a
?± 0,
0;
0;
0;
0.
= 0 are both true;
= 0 or b = 0"; in
this possibility
is
not
mathematics "or"
is
always used in the sense of "one or the other, or both.")
This latter consequence of P7 is constantly used in the solution of equations.
Suppose, for example, that a number x is known to satisfy
~ 1)(* - 2) - 0.
— = 0 or x — 2 ~
(*
Then
On
it
follows that either x
1
the basis of the eight properties listed so far
0;
hence x
it is still
—
1
or x
=
2.
possible to prove
Basic Properties of Numbers
which combines the operations of addi-
Listing the next property,
very
little.
tion
and multiplication,
(P9)
and
If a, b,
will alter this situation drastically.
are any numbers, then
c
(b
+
c)
=
(Notice that the equation
(b
+
c)
a
•
a
—
b
+
b
'
=
a
As an example of the usefulness of P9 we
—
7
a
b
will
•
+
a
•
c.
c
•
a
is
also true,
by P8.)
now determine just when
-
a
b
a:
a
—
b
b)
+
b
If
then
-
(a
hence
hence
Consequently
—
=
=
=
=
—
a
a
a
(1
+a
+ 1)
and therefore
a
— a,
- a) + b =
b + b — a;
{b + b - a) +
b
(b
£
(1
*
+
b
+
(b
a
=
b
-
+
a);
b.
1),
b.
0 = 0 which we have
is the justification of the assertion a
already made, and even used in a proof on page 6 (can you find where?).
This fact was not listed as one of the basic properties, even though no proof
A second use of P9
mentioned. With P1-P8 alone a proof was not
possible, since the number 0 appears only in P2 and P3, which concern addiP9 the
tion, while the assertion in question involves multiplication. With
have
We
obvious:
not
perhaps
though
proof is simple,
was offered when
was
it
first
a
as
we have already
both
A
sides) that d
0
'
0
+
a
=
'
'
•
will establish the
b).
To
0)
this,
may
help explain the somewhat
is positive. To begin
+
(-&)
Consequently, adding
(a
(
— a)
•
b
=
*
=
a'b = [(-a) +a]'b
= 0-b
- (a
0.
•
b) to
both
sides) that
note that
•
to
0)
note that
immediately (by adding
(-*)
•
two negative numbers
=
— (a-b). Now
+
easily acceptable assertion that
more
(-a) 'b
It follows
(0
0;
0.
we
prove
a
a
consequences of P9
mysterious rule that the product of
— (a
=
=
0
noted, this immediately implies (by adding (a
series of further
with,
'
+
•
[-(a
6) to
*
&)]
both
= (-a) (-/>) + (-a)
= (-«)•[(-*) + *]
= (-tf)'O
= 0.
sides,
•
we obtain
•
b
(-a)
•
The fact that the product of two negative numbers is positive is thus a consequence of P1-P9. In other words, if we want PI to P9 to be true, the rule for the
product of two negative numbers
is forced upon us.
various consequences of P9 examined so far, although interesting and
important, do not really indicate the significance of P9; after all we could have
The
each of these properties separately. Actually, P9 is the justification for
all algebraic manipulations. For example, although we have shown
to solve the equation
listed
almost
how
-
(x
-
l)(x
=
2)
0,
we can hardly
expect to be presented with an equation in this form.
more likely to be confronted with the equation
x2
The
2
"factorization" x
-
2>x
+
-
3x
+
2
2
=
(x
-
=
We
are
0.
1)(*
-
2)
is
really a triple use
ofP9:
(*
-
1)
•
(*
-
2)
-
x2
+
+ (-1) (* - 2)
(-2) + (-1) x + (-1)
x[(-2) + (-l)] + 2
x2
-
3x
=
x
•
=
=
=
x
•
(*
+
x
2)
x
•
•
•
+
•
(-2)
2.
A
final illustration of the importance of P9 is the fact that this property is
actually used every time one multiplies arabic numerals. For example, the
calculation
13
X24
52
26
312
is
a concise arrangement for the following equations:
13
•
24
=
=
=
13
•
(2
13 -2
26
•
•
10
10
+ 4)
+ 13
+
52.
10
•
•
4
(Note that moving 26 to the left in the above calculation
26 10.) The multiplication 13 4 = 52 uses P9 also:
•
•
13-4 =
=
•
(1
10
+
3)
4
+
-4-10 +
12
1
•
10
•
-4
3
•
4
= 4-10 + 1-10 +
= (4 + 1) 10 + 2
= 5-10 + 2
= 52.
•
2
is
the
same
as writing
9
Basic Properties of Numbers
P1-P9 have descriptive names which are not essential to
remember, but which are often convenient for reference. We will take this
opportunity to list properties P1-P9 together and indicate the names by
which they are commonly designated.
The
properties
(PI)
(P2)
(Associative law for addition)
a
(Existence of an additive
a
+
+
+
(b
=
0
=
c)
+
0
{a
+
=
a
b)
+
c.
a.
identity)
(P3)
(Existence of additive inverses)
a
+
+
(P4)
(Commutative law for addition)
a
(P5)
(Associative law for multiplica-
a
-
(b
a
-
1
a
*
<T l
a
*
a
•
b
•
+
= (-a)
(-a)
+
=
b
c)
=
=
a
0.
a.
(a
b)
'
'
c.
tion)
(P6)
(Existence of a multiplicative
=
a
1
=
1^0.
a;
identity)
(P7)
(Existence of multiplicative
=
d~ l
'
=
a
1,
tor a
0.
inverses)
(P8)
(Commutative law
for multi-
=
b
b
'
a.
plication)
(P9)
(Distributive law)
(b
+
c)
—
a
•
b
+a
•
c.
The three basic properties of numbers which remain to be listed are concerned with inequalities. Although inequalities occur rarely in elementary
mathematics, they play a prominent role in calculus. The two notions of
inequality, a < b (a is less than b) and a > b (a is greater than b), are intimately
two
related: a < b means the same as b > a (thus 1 < 3 and 3 > 1 are merely
ways of writing the same
assertion).
The numbers
a satisfying a
>
0 are
0 are called negative.
is possible to reverse
it
terms
of
in
defined
<,
be
thus
can
While positivity
the procedure: a < b can be defined to mean that b — a is positive. In fact,
called positive, while those
it is
P
9
numbers
a satisfying a
convenient to consider the collection of
as the basic concept,
(P10)
and
all
<
positive
numbers, denoted by
state all properties in terms of P:
(Trichotomy law) For every number
a,
one and only one of the
following holds:
(i)
a
=
(ii)
a
is
(iii)
—a
0,
in the collection P,
is
in the collection P.
(PI 1)
(Closure under addition) If a and
(PI 2)
(Closure under multiplication) If a
in P.
b
are in P, then a
and
b
+
b is in P.
are in P, then a
b
is
These three properties should be complemented with the following
defini-
tions:
>
<
>
<
a
a
a
a
Note, in particular, that a
b
if
a
b
if
b
b
if
a
b
if
a
>
0
—
>
>
<
b is in
•
b
or a
b
or a
and only
if
P;
a;
if
a
= b\
= b*
in P.
is
however elementary they may
seem, are consequences of P10-P12. For example, if a and b are any two
numbers, then precisely one of the following holds:
All the familiar facts about inequalities,
(i)
a
(ii)
a
—
=
b
— (a —
(iii)
Using the definitions
0,
b is in
the collection P,
=
b)
made,
just
—
b
a
in the collection P.
is
follows that precisely one of the following
it
holds:
A
more
slightly
If a
<
b,
<
b,
then a
a
so that b
+
a
and
£
<
c
+
£
and
c
so
c
if
(iii)
b
=
>
>
a
<
b
b,
b,
a.
from the following manipulations.
in P, then surely (b
is
c.
+
c)
Similarly, suppose a
— a is in P,
— b is in P,
— a — (c —
b
This shows that
a
a
interesting fact results
—
<
c
(i)
(ii)
and
b
<
c,
+
(b
then a
<
b)
—
c.
<
a)
—
{a
b
and
is
+
<0 is
b
<
c.
in P; thus,
if
Then
in P.
(The two inequalities a < b
< b < c, which has
are usually written in the abbreviated form a
the third inequality a
<
c
almost built
in.)
somewhat less obvious: If a < 0 and b < 0, then
ab > 0. The only difficulty presented by the proof is the unraveling of definitions. The symbol a < 0 means, by definition, 0 > a, which means 0 — a —
— a is in P. Similarly — b is in P, and consequently, by PI 2, (—a)( — b) — ab
is in P. Thus ab > 0.
The fact that ab > 0 if a > 0, b > 0 and also if a < 0, b < 0 has one
special consequence: a 2 > 0 if a ^ 0. Thus squares of nonzer> numbers are
The
*
There
following assertion
is
one
is
slightly perplexing feature of the
we know
usefulness of the symbol
some values of
a
and
b,
is
is
< The
.
statements
1
that
bound
and
1
1
in the second. This sort of thing
>
+ <3
+ <2
1
are both true, even though
symbols
<
to
could be replaced by
occur when
revealed by a statement like
<
<
in the
first,
and by
=
used with specific numbers; the
Theorem 1 here equality holds for
is
while inequality holds for other values.
—
Basic Properties of Numbers
11
have
always positive, and in particular we have proved a result which might
properties:
of
list
1
our
> 0
seemed sufficiently elementary to be included in
(since
1
The
=
l
2
).
-a >
fact that
0
if
<
a
0
the basis of a concept which will play an
For any number a, we define the
is
role in this book.
extremely important
absolute value \a\ of a as follows:
III
~
\
Note that
\a\
0
when
a
a
a
always positive, except
is
>
<
a,
-a,
0.
=
For example, we have
0.
= + V2 - V3, and
— 3j = 3,
= 7, Jl + V2 |7J
7
V !^! = VlO — V2 — 1. In general, the most straightforward
1
|1
|
any
+ V2 -
approach
to
problem involving absolute values requires treating several cases sepaapproach
since absolute values are defined by cases to begin with. This
rately,
may
THEOREM
1
For
values,
be used to prove the following very important fact about absolute
all
numbers
a
and
we have
b,
+b
\a
PROOF
We
<
!
+
\a\
1*1-
will consider 4 cases:
In case
(1)
we
also
0)
a
a
>o,
>o,
b
(2)
(3)
a
<
o,
b
(4)
a
<o,
b
have a
+
b
+
b\
=
a
\
>
a
0,
b
>
<
>
<
0;
0;
0;
0.
and the theorem
+b-
+
\a\
is
obvious; in fact,
\b\,
so that in this case equality holds.
In case
we have
(4)
\a
In case
(2),
+
a
b\=
when
a
>
+ b < 0, and again equality holds:
+
-(a + b) = -a + (-*) =
\a\
<
0 and b
\a
+
we must prove
0,
b\
<
a
-
a
which
hand,
is
if
certainly true since b
a
+
b
<
0,
+b
b
is
<a —
< -b,
which
is
certainly true since a
b <
—a <
is
+
-b
is
positive
a
—
b,
a,
and
b
positive.
that
—a —
i.e.,
If a
>
0,
then
we
b,
negative and
we must prove
that
b.
This case may therefore be divided into two subcases.
must prove that
i.e.,
\b\.
—a
is
negative.
On
the other
may
Finally, note that case (3)
applying case
Although
(2)
this
with a and
method
b
be disposed of with no additional work, by
interchanged. |
of treating absolute values (separate consideration
sometimes the only approach available, there are often
simpler methods which may be used. In fact, it is possible to give a much
shorter proof of Theorem 1 this proof is motivated by the observation that
of various cases)
is
;
=
\a\
(Here, and throughout the book,
symbol
is
when
defined only
(\a
+ b\)
2
>
x
=
(a
V7
2
.
denotes the
0.)
+
b)
2
square root of x; this
positive
We may now
observe that
- a + lab + b
< a 2 + 2\a\ \b\ + b 2
= \a\ 2 + 2\a\ \b\ + \b\
= (W + H)
2
2
•
•
2
2
.
+
+
2
we can conclude that \a
b\ < \a\
\b\ because x
< y 2 implies
provided
that
and
x
< y,
y are both non-negative; a proof of this fact is left
to the reader (Problem 5).
From
this
x
One final observation may be made about the theorem we have just
proved a close examination of either proof offered shows that
:
+
\a
a and b have the same sign
one of the two is 0, while
if
(i.e.,
\a
if
a
and
We
b are
b\
+
=
\a\
+
\b\
are both positive or both negative), or
b\<
\a\
+
if
\b\
of opposite signs.
this chapter with a subtle point, neglected until now,
whose inclusion is required in a conscientious survey of the properties of
numbers. After stating property P9, we proved that a — b = b — a implies
a = b. The proof began by establishing that
will
conclude
a'
(1
+
1)
=
A
-
(1
+1),
from which we concluded that a = b. This result is obtained from the equation
1. Division by 0
(1 + 1) = b
(1 + 1) by dividing both sides by 1
should be avoided scrupulously, and it must therefore be admitted that the
validity of the argument depends on knowing that 1 + 1^0. Problem 24
is designed to convince you that this fact cannot possibly be proved from
properties P1-P9 alone Once P10, Pll, and PI 2 are available, however, the
proof is very simple: We have already seen that 1 > 0; it follows that 1 + 1 >
0, and in particular 1 + 1^0.
This last demonstration has perhaps only strengthened your feeling that it
is absurd to bother proving such obvious facts, but an honest assessment of our
a
•
+
'
!
present situation will help justify serious consideration of such details. In
.
Basic Properties of Numbers
13
numbers are familiar objects, and that
well-known properties of
PI -PI 2 are merely
numbers. It would be difficult, however, to justify this assumption. Although
one learns how to "work with" numbers in school, just what numbers are,
remains rather vague. A great deal of this book is devoted to elucidating the
concept of numbers, and by the end of the book we will have become quite
well acquainted with them. But it will be necessary to work with numbers
throughout the book. It is therefore reasonable to admit frankly that we do
not yet thoroughly understand numbers; we may still say that, in whatever
way numbers are finally defined, they should certainly have properties
this
we have assumed
chapter
that
explicit statements of obvious,
P1-P12.
Most
of this chapter has been an attempt to present convincing evidence
which we should assume in order to
deduce other familiar properties of numbers. Some of the problems (which
indicate the derivation of other facts about numbers from PI -PI 2) are offered
as further evidence. It is still a crucial question whether PI -PI 2 actually
account for all properties of numbers. As a matter of fact, we shall soon see
that they do not. In the next chapter the deficiencies of properties PI -PI 2 will
that PI -PI 2 are indeed basic properties
but the proper means for correcting these deficiencies is
not so easily discovered. The. crucial additional basic property of numbers
which we are seeking is profound and subtle, quite unlike PI -PI 2. The dis-
become quite
clear,
covery of this crucial property will require all the work of Part II of this book.
In the remainder of Part I we will begin to see why some additional property
is required; in order to investigate this we will have to consider a little more
carefully
what we mean by "numbers."
PROBLEMS
1.
Prove the following:
(i)
If ax
(ii)
x2
(iii)
If x 2
(iv)
(v)
(vi)
=
a for
-y =
2
=
y
(x
2
,
some number a
-
then x
=
^
0,
then x
—
1
+ y).
y){x
y or x
=
—y.
— y){x + xy + y
~
n'
x n -y n = (x - y)(x
+ xy n ~ + y n ).
+ x n~ y +
(There is a particularly
x + y = (x + y)(x — xy + y
x
3
—y =
3
2
2
(x
).
2
1
3
2
z
•
l
2
•
2
).
and it will show you how
way to do this, using
n
n
torization for x + y whenever n is odd.)
(iv),
2.
What
is
wrong with the following "proof? Let
-y
y(x - y),
xy
(x
+y)(x - y)
x
+y
y>
2
l.
2
,
x
=
y.
easy
to find a fac-
Then
14
Prologue
3.
Prove the following:
a
ad
c
be
~\-
(iii)
=
(ab)~ L
a~ l b~ l
5* 0.
if 6,
>
if a, b
,
(To do
0.
this
you must remember the
defining property of (ab)~~ l .)
(iv)
(vi)
*
d
If b 9
0,
=
then -
_
a
Find
numbers x
all
for
if
ad
=
5
(vi)
(vii)
(viii)
(ix)
0.
(When
is
a product of two numbers posi-
- 2x + 2 > 0.
+ x + > 2.
* - * + 10 > 16.
> 0.
x + x +
(* - 7r)Cr + 5)(* - 3) >
(x - \ 2){x - V2) > 0.
x2
x2
1
2
2
1
(xi)
2*
(xii)
*
(xiii)
-
1
<
+
0.
8.
<
3*
4.
1
+
x
5.
when
2
/
(x)
Also determine
2
tive?)
(v)
be.
which
- x < 3 - 2x.
- x < 8.
5 - x < -2.
(x — \){x - 3) >
4
(ii)
(iv)
and only
a
(i)
(iii)
if
b
b
4.
-
d
b
1
—
>
0.
X
Prove the following:
(i)
If a
(ii)
If a
<
<
b
b,
and
c
then
< d, then a
-b < -a.
+
c
<
b
+
d.
Basic Properties of Numbers
6.
(iii)
If a
(iv)
If a
(v)
If a
(vi)
If a
<
<
<
>
<
<
<
(vii)
If 0
(viii)
If 0
.(ix)
If 0
(x)
If a, £
Prove that
and
and
and
b
b
b
>
>
<
c
e
c
then a
0,
then ac
0,
then ac
then a 1 > a.
2
a < 1, then f2
<
a.
b
and 0 <
c
<
b,
then a 2
<
a
<
<
>
0 and a 2
<
0
<
then
a
if
<
a
b,
/
b
be.
2
,
d,
+
Va6 <
<
b.
bd.
(viii).)
<
then a
a
<
then
b.
(Use
(ix),
backwards.)
b
b.
+
£)/2 holds for a, b > 0, without
generalization of this fact occurs in
(a
A
2-20.
= y. Hint: First explain
y and n is odd, then x
suffices to consider only the case x, y > 0; then show that
Prove that
why it
x < y and
if
n
x
>
y
=
11
x are both impossible,
= y n and is even, then x = or x = —v.
Although the basic properties of inequalities were stated in terms of the
collection P of all positive numbers, and < was defined in terms of P,
this procedure can be reversed. Suppose that P10-P12 are replaced by
(b)
*8.
d.
be.
b\ (Use
—
the additional assumption a
(a)
<
>
—
b
< Vab < ~^r— <
Notice that the inequality
Problem
<
c
1,
a
*7.
—
d,
15
Prove that
(P'10)
if
n
x
;?
For any numbers a and
b
and only one, of the
one,
follow-
ing holds:
(P'll)
«
a
(ii)
a
(iii)
b
(P'13)
Show
that
b,
a.
<
a,
b,
and
<
if
c,
it
<
a
P10-P12 can then be deduced
signs.
\V2
(ii)
\(\a
(iii)
|(|fl
(iv)
(v)
7
+ V3 - V 5 +
+ - + +H ~ +b+
b\
|*|
|(|V2
\b\)\.
\a
6|
-_2xy
<
b
and
b
<
c,
then
if
c,
then a
b,
a
<
b
+c<b+
and
0
<
c,
e.
then
be.
as theorems.
Express each of the following with at least one
(i)
a
e.
For any numbers <2, b, and c,
For any numbers a, b, and
ac
9.
b,
For any numbers
a
(P'12)
=
<
<
c\)\.
+/|.
+ V3|-|V5-V7|)|.
less
pair of absolute value
16
Prologue
10.
Express each of the following without absolute value signs, treating
when
various cases separately
(i)
+ b\-
\a
necessary.
\b\.
(ii)
11.
|*|
(iv)
a
Find
(i)
(ii)
12.
- |^|.
- \{a -
(iii)
numbers *
all
-
3|
+
4|
|*
I*
(iii)
|*
(iv)
|*
(v)
|*
(vi)
|*
(vii)
I*
(viii)
|*
\a\)\.
-
=
<
<
3|
8.
8.
2.
+
+
+
lj
1|
lj
II-
I*
•
|*
1|
which
for
|*
-
2\
|*
+
1|
|*+
lj
>
<
<
+ =
+ 2| =
II
1.
2.
1.
0.
3.
Prove the following:
(i)
=
|*|
=
—
\xy\
(ii)
•
>
\y\.
if
x
9* 0.
(The best way
to
do
this is to
remember what
\x\
-1
is.)
X
if/?*
(iii)
0.
y
Civ)
\x
(v)
\x\
— y\ <
+
—
< \x — y\.
|#|
jv
|
I
-
\y\
(Give a very short proof.)
(A very short proof
is
possible, if
you write
things in the right way.)
(vi)
|(|#|
—
|jy|)|
<
— y\. (Why
\x
does this follow immediately from
(v)?)
(vii)
13.
+
+ +
+
z\ < \x\
|*
y
\y\
prove your statement.
\z\.
when
Indicate
equality holds,
and
The maximum of two numbers x and y is denoted by max(^, j). Thus
max( — 1, 3) = max (3, 3) = 3 and max( — 1, — 4) = max (—4, —1)
= — 1. The minimum of x and y is denoted by min(#, y). Prove that
,
N
min(*, y)
=
Derive a formula for max(x,
max(#,
y, z)
x
+y +
\y
—
x\
x
+y —
\y
—
x\
y, z)
and min(# y
= max
?
(at,
}
z), using, for
max(jy, z)).
example,
b
Basic Properties of Numbers
14.
(a)
Prove that
many
=
\a\
(c)
*15.
(a)
not to become confused by too
is
\
<
>
prove the statement for a
cases. First
obvious for a
(b)
—a\. (The trick
17
Why
0.
is
it
then
0?)
Prove that
—b<a<b
follows that
—
<
\a\
if
<
a
and only
,
if
<
\a\
In particular,
b.
it
\a\.
Use this fact to give a new proof that \a + b\ < \a\ + \b\,
Use Problem 1 and Problem 7 to prove that if x and j are not both
0,
then
x1
x
+ xy
4
z
+ xy + y ^ 0,
+ x y + xy + y*
2
2
z
2
9* 0.
For every number x 9^ 0, each of these expressions is positive for some
number y and also for some negative y (namely, y — ±x); it
therefore seems reasonable that the ^ signs can be replaced by >
signs. This maneuver is valid, but we are not yet in a position to prove
this (see Problem 7-9). Parts fb) and (d) of this problem provide a
positive
direct demonstration that the
(b)
Using the
>
signs hold.
fact that
+
+y = (* + y) > 0,
show that the assumption x + xy + y < 0 leads to a contradiction.
x2
2
2xy
2
2
(c)
Show
similarly that
4x
3x 2
**(d)
*16.
Show
that
Show
(a)
(b)
**(c)
+
+
6xy
5xy
+
+
Ay
3y
2
x2
z
z
x
+y
+y
2
only
z
only
when
when
(x
x
x
when
to find out
b
+ yY =
^5
17.
(a)
_|_
\
now be
the proof
Suppose that
-b
if
b
able to
is
0.
= 0,
= 0or# =
0 or y
0or;y
(x
+ yY =
5
-
Hint:
x*
—y.
+ y\
From
the assumption
be able to derive the equation
0.
This implies that
(x
+ yY =
2
+
satisfy the
make a good
when
(x
+ y) n =
guess as to
contained in Problem 11-41.
—
Ac
V
2
2
both
>
4
2
should
+y
z
2
—
=
+y
h
2
z
n
then
0,
2
2
then
0.
+ x y + x y + xy* + >
z
0,
0,
*4
+ yY = x + y you should
xy
x + 2x y + 2xy + y = 0,
xy
xy =
+ yy
xn
>
>
and y are not both
(x
You
2
x
if
Use Problem 15
Find out when
2
2
that
+ y) —
=
(x + y)
(x
x and y are not both
if
2
2
>
0.
-
4c
Show
that the
numbers
-b - Vb 2
'
equation x 2
2
+
bx
+c
=
0.
4c
Suppose that
(b)
satisfying x
—
2
<
Ac
Show
0.
+c=
bx
+
then x 2
+y
xy
>
2
numbers x
that there are no
2
0; in fact, x
+ bx + >
+ bx + c =
0 for all
c
"Complete the square," i.e., write x 2
Use this fact to give another proof that
(c)
if
x and
(x
+
Hint:
x.
£/2)
+
2
?
are not both 0,
jy
0.
For which numbers a is it true that x 2 + axy -{- y 2 > 0 whenever
x and y are not both 0?
Find the smallest possible value of x 2 + bx + c and of ax 2 + bx + c,
(d)
(e)
for a
18.
b
+
2
(Use the trick in part
0.
>
(b).)
numbers a, elementary as it may seem, is
nevertheless the fundamental idea upon which most important in-
The
fact that a 2
0 for
all
equalities are ultimately based.
is
The great-granddaddy
of
all
inequalities
the Schwartz inequality:
+
xiyi
The
2
2
2
2
common
—their reliance on the
Prove the Schwartz inequality by
(xi
(b)
+ x V yi + y
/
2
2
.
three proofs of the Schwartz inequality outlined below have only
one thing in
(a)
< V*!
x2y 2
2
+
x 2 2 )(yi 2
Prove that
+^
)
=
if x\
=
2
2
Xy
Oiji
+
first
0 for
all a.
proving that
+
x 2y 2 ) 2
and x 2 = X^ 2
x
>
fact that a 2
for
Oij 2
—
xoji)
2
.
some number
X,
then
equality holds in the Schwartz inequality. Prove the same thing
yi
—
=
V2
there
is
suppose that yi and y 2 are not both
no number X such that x\ = \y and x 2 = \y 2
<
=
(x>-i
-
\ 9 (yi 2
x
x
y + (Xv - x y
~ 2X(x iyi +
+y*
2
0,
if
and that
Then
2
2
x2y 2)
)
+
(x,
2
+x
2
2
).
Using Problem 17, complete the proof of the Schwartz inequality.
Prove the Schwartz inequality by using 2xy < x 2 + y 2 (how is this
derived?) with
=
x
first
(d)
.
x
o
(c)
Now
0.
for
i
—
Xi
V
Xl
2
+X
and then
1
=
y
2
2
for
yi
and x 2
=
=
y
=
2
Xy 2
0 or
when
;
Vy *+y
x
z
—
*
t
2.
Deduce, from each of these three
when
yi
there
is
proofs, that equality holds only
a
number X such
that Xi
—
Xyi
.
In our later work, three facts about inequalities will be crucial. Although
proofs will be supplied at the appropriate point in the text, a personal assault
on these problems is infinitely more enlightening than a perusal of a completely worked-out proof. The statements of these propositions involve some
weird numbers, but their basic message is very simple: if x is close enough to x Q
and y is close enough to y Q then x + y will be close to * 0 + ^o, and xy will be
,
,
19
Basic Properties of Numbers
close to x 0 y 0
and
,
/y will be close to l/y 0
1
these propositions
the
is
fifth letter
The symbol "e" which appears
.
in
and
of the Greek alphabet ("epsilon"),
could just as well be replaced by a less intimidating Roman letter; however,
tradition has made the use of £ almost sacrosanct in the contexts to which
these theorems apply.
19.
Prove that
if
-
\x
<
Xo\
-
and
-y <
0
\y
->
\
then
*20.
Prove that
\x
-
then
+y) -
(x 0
\(x
-
-
(x 0
y)
+y
-y
<
<
0 )\
0 )\
e,
6.
if
min
*o|'<
\xy
\(x
—
* 0 .)'o|
^_i^y l)
<
and
\y
- ,„| <
S.
(The notation "min" was defined in Problem 13, but the formula provided by that problem is irrelevant at the moment; the first inequality
in the hypothesis just
-
\x
means that
e
x0
'
< —r~
\
and
-
\x
-
x0
\
<
1
,
at
one point in the argument you
need the
will
first
inequality,
and at
another point you will need the second. One more word of advice: since
the hypotheses only provide information about x — x 0 and y — y 0 it is
almost a foregone conclusion that the proof will depend upon writing
,
xy
*21.
—
x 0yo in a
Prove that
if
way
0
y0
0
—y
x 0 and y
0 .)
and
!JV
then y
—
that involves x
—
/
—
<mm^—
z\yo\
f\yo\
•
,
,
JVo|
2
—
\
j>
and
1
1
y
yo
|
<
e.
I
*22.
i
Replace the question marks in the following statement by expressions
involving
If
yQ
^
0
s,
# 0 and y 0 so
,
9^
the conclusion will be true:
and
\y
then y
that,
0
—
Vol
<
?
and
\x
and
X
Xq
\y
y<>
|
<
e.
—
xQ
\
<
?
This problem
trivial in
is
the sense that
its
from Problems
solution follows
20 and 21 with almost no work at all (notice that x/y = x \/y). The
crucial point is not to become confused; decide which of the two
problems should be used first, and don't panic if your answer looks
•
unlikely.
This problem shows that the actual placement of parentheses in a
is
irrelevant.
The
proofs involve "mathematical induction";
sum
you are
problem, it can
if
not familiar with such proofs, but still want to tackle this
be saved until after Chapter 2, where proofs by induction are explained.
Let us agree, for definiteness, that a x
a n will denote
+
+
ai
Thus
ai
denotes
+
(a
+
_
+
+ n + (a n _ + a n )))
a
a
denotes
+ +
a + (a + a
and a + a +
+ (a + (a + # etc.
(a 2
*
'
'
2
z
2
2
3
),
a3
2
x
+a
4
4 )),
3
Prove that
(a)
+
(ai
•
'
+
•
+ a k+i
ak )
Hint: Use induction on
Prove that
(b)
+
(ai
(c)
).
x
x
2
•
•
•
•
•
if
>
n
k)
ai
+
•
+
*
a k+ x
k.
(a k+1
+
'
'
•
+ an
)
=
+
ai
.
.
,
+
*
•
Hint: Use part (a) to give a proof by induction on
Let s(a h
a k ) be some sum formed from ai,
.
.
then
fc 3
+a +
=
an
.
k.
.
.
.
ak
,
Show
.
that
s(a h
.
.
.
,
ak)
+
=
Hint: There must be two sums
s'(a h
•
.
•
.
•
+
.
,
ak
at )
.
and s ff {a i+h
.
.
.
,
ak )
such that
s(a u
.
.
.
,
ak )
=
s'(a u
.
.
.
,
ai)
+ s"(ai+ h
.
.
.
,
a*).
Suppose that we interpret "number" to mean either 0 or 1, and
and to be the operations defined by the following two tables.
•
+
0
1
0
1
0
0
1
0
0
0
1
1
0
1
0
1
Check that properties P1-P9
all
hold, even though
1
+ 1=0.
+
CHAPTER
*i
NUMBERS OF VARIOUS SORTS
In Chapter
1
we used
word "number" very
the
with the basic properties of numbers.
loosely, despite
now
It will
our concern
be necessary to distinguish
carefully various kinds of numbers.
The
simplest
numbers are the " counting numbers"
...
1, 2, 3,
.
significance of this collection of numbers is emphasized by
symbol N (for natural numbers). A brief glance at PI -PI 2 will show that
our basic properties of "numbers" do not apply to N for example, P2 and
P3 do not make sense for N. From this point of view the system N has many
deficiencies. Nevertheless, N is sufficiently important to deserve several comments before we consider larger collections of numbers.
The most basic property of N is the principle of "mathematical induction."
The fundamental
its
—
Suppose P(x) means that the property
P
holds for the
principle of mathematical induction states that P(x)
number
is
Then
x.
true for
the
natural
all
numbers x provided that
(1)
P(l)
(2)
Whenever
Note that condition
assumption that P(k)
if
is
(2)
true.
P(k)
merely
is
condition (1) also holds. In
number
P(k)
A
is
is
P(h
+
asserts the truth of
fact, if
+
P{k
is
true,
then
1).
Now,
since P(2)
P(\)
=
true (using (2) in the special case k
will eventually
true for all
1) is true.
1)
under the
true; this suffices to ensure the truth of P(x) for all x,
is
true (by using (2) in the special case k
that P(3)
true,
=
it
follows that P(2)
is
true
numbers
is
follows
that each
2). It is clear
be reached by a series of steps of this
it
sort, so
that
k.
favorite illustration of the reasoning
behind mathematical induction
envisions an infinite line of people,
person
number
1,
person
number
2,
person
number
3,
.
.
.
.
If each person has been instructed to tell any secret he hears to the person
behind him (the one with the next largest number) and a secret is told to
person number 1, then clearly every person will eventually learn the secret.
If P(x) is the assertion that person number x will learn the secret, then the
instructions given (to
tell all
secrets learned to the next person) assures that
and telling the secret to person number 1 makes (1) true.
The following example is a less facetious use of mathematical induction.
There is a useful and striking formula which expresses the sum of the first n
numbers in a simple way:
condition
(2) is true,
f
21
22
Prologue
1+
To
prove
that for
this
formula, note
some integer
k
•
•
first
+„
•
that
+
__-
n(n
=
1)
clearly true for n
it is
=
1.
Now assume
we have
2
Then
k(k
=
+
!)
+
1)
+*+
+2£ +
1
2
2
*2
=
+
3£
+
2
2
(A
+
+
!)(*
+ 2)
the principle of induction this proves
This particular example illustrates a
phenomenon that frequently occurs, especially in connection with formulas
like the one just proved. Although the proof by induction is often quite
straightforward, the method by which the formula was discovered remains a
so the
formula
is
the formula for
also true for k
all
1
By
.
natural numbers
mystery. Problems 4 and 5 indicate
n.
how some
formulas of
this
type
may
be
derived.
The principle of mathematical induction may be formulated in an equivway without speaking of "properties" of a number, a term which is
sufficiently vague to be eschewed in a mathematical discussion. A more precise
formulation states that if A is any collection (or "set"— a synonymous mathealent
matical term) of natural numbers and
(1)
1
is
(2)
k
+
in A,
is
1
in
A whenever
h
is
in A,
then A is the set of all natural numbers. It should be clear that this formulation
adequately replaces the less formal one given previously—we just consider the
the
set A of natural numbers x which satisfy P(x). For example, suppose A is
set of
natural numbers n for which
1+
Our
is
•
•
•
it is
+
„
true that
_+
n(n
=
1)
+
1
previous proof of this formula showed that A contains 1 and that k
that the
if h is. It follows that A is the set of all natural numbers, i.e.,
,
in A,
formula holds for all natural numbers n.
There is yet another rigorous formulation of the principle of mathematical
induction, which looks quite different. If
A
is
any
collection of natural
num-
!
Numbers of Various
bers,
it is
tempting to say that
statement can
A must have
a smallest member. Actually, this
to be true in a rather subtle
fail
23
Sorts
A
way.
particularly important
numbers is the collection A that contains no natural numbers at
all, the "empty collection" or "null set,"* denoted by 0. The null set 0 is a
in fact, it has no
collection of natural numbers that has no smallest member
set of natural
—
members
This
at all.
set of natural
the only possible exception, however;
A
numbers, then
known
statement,
is
as the "well-ordering principle."
principle of induction as follows. Suppose that the set
Let
B
be the
Clearly
1
numbers
1 were
set of natural
in
is
B
(because
such that
n
.
not in A. This shows that
every number n
is
in B,
i.e.,
number
n.
Thus
^4
natural
is
a nonnull
1,
can be proved from the
A has no least member.
.
.
.
,
.
,
k
if
=
+
in B, then k
is
the numbers
1,
.
.
.
+
.
as smallest
1
.
,
A
not in
is
1
.
k
+
1
are
in B. It follows that
1
is
,
n are not in
which completes the
0,
A.
n are all not in
k are not in A, surely k
member). Moreover, if 1,
(otherwise k + 1 would be the smallest member of A), so 1,
all
A
A would have
in A, then
if
.
if
has a least member. This "intuitively obvious"
A
for
any
proof.
prove the principle of induction from the well-ordering
It is also possible to
may
principle (Problem 9). Either principle
be considered as a basic assump-
tion about the natural numbers.
There is still another form of induction which should be mentioned. It
sometimes happens that in order to prove P(k + 1) we must assume not only
P(k), but also P(l) for
all
natural numbers
A
"principle of complete induction": If
<
/
k.
In
this case
we
rely
on the
a set of natural numbers and
is
in A,
(1)
1
is
(2)
k
+
1
A
in
is
if 1,
.
.
.
,
k are in A,
A is the set of all natural numbers.
Although the principle of complete induction may appear much stronger
than the ordinary principle of induction, it is actually a consequence of that
principle. The proof of this fact is left to the reader, with a hint (Problem 10).
Applications will be found in Problems 6, 16, 19, and 20.
Closely related to proofs by induction are "recursive definitions." For
then
example, the number
n
natural numbers
than or equal to
less
!
(read "w factorial")
n\
=
1
•
2
•
.
is
defined as the product of all the
n:
.
•
.
{n
-
1)
-
a.
This can be expressed more precisely as follows:
(1)
1!
(2)
n\
=
=
1,
n
-
(h
-
1)!.
This form of the definition exhibits the relationship between
*
Although
it
may
many
contexts.
satisfying Isome property P; often
A might
be
and
(n
—
1)
not strike you as a collection, in the ordinary sense of the word, the null set
arises quite naturally in
that
nl
0—in fact often
We
frequently consider the set A, consisting of
that
P is
satisfied
always
false
by showing that
we have no guarantee
one proves that
P is
by
any
all
number,
A —
0.
x
so
:
an
in
way
explicit
that
is
ideally suited for proofs
by induction. Problem 21
reviews a definition already familiar to you, which
may
be expressed more
succinctly as a recursive definition; as this problem shows, the recursive
definition is really necessary for a rigorous proof of some of the basic properties
of the definition.
One
tion
definition
which we
which
may
not be familiar involves some convenient nota-
will constantly
be using. Instead of writing
+
a\
we
employ the Greek
will usually
•
•
+ an
•
letter
2
,
(capital sigma, for
"sum") and
write
71
11
In other words,
i
/=1,2,
.
.
.
^=
n.
,
of the
numbers obtained by
letting
Thus
1
Notice that the
sum
a { denotes the
i
letter
i
+2+
=
+n
•
•
n(n
+
1)
2
really has nothing to
do with the number denoted by
n
^
i,
and can be replaced by any convenient symbol (except
n,
of course!):
i=i
n(n
3
=
=
+
V
To
define
^
1)
1
L
n=
1)
1
Hi
3
+
n
= Hi
~
+V
2
1
a precisely really requires a recursive definition
{
^
(1)
ai
=
ai
=
n
n
^
(2)
°1>
—1
^
ai
+ an
.
But only purveyors of mathematical austerity would
such precision. In practice,
used,
it
of this
too strongly
on
symbolism are
of modifications
necessary to add any words of explanation.
all sorts
and no one ever considers
insist
Numbers of Various
25
Sorts
The symbol
t
for
example,
is
an obvious way of writing
a\
more
or
+
+
a2
+a +
az
i
may be
set
.
+
'
an,
n
1
=5
i
numbers which we discovered at the beginremedied by extending this system to the
partially
.
-2, -1,
,
Z
denoted by
is
PI -PI 2, only P7
A
*
integers
.
This
=
deficiencies of the natural
ning of this chapter
set of
*
precisely,
3
The
+
#6
b
(from
0, 1, 2,
...
.
German "Zahl," number). Of
properties
Z.
fails for
numbers is obtained by taking quotients m/n of
These numbers are called rational numbers, and the
set of all rational numbers is denoted by
(for "quotients''). In this system of
numbers all of PI -PI 2 are true. It is tempting to conclude that the "properties
of numbers," which we studied in some detail in Chapter 1, refer to just one
set of numbers, namely, Q. There is, however, a still larger collection of
numbers to which properties PI -PI 2 apply the set of all real numbers,
denoted by R. The real numbers include not only the rational numbers, but
other numbers as well (the irrational numbers) which can be represented
still
larger system of
^
integers (with n
0).
Q
—
by
infinite decimals;
proof that
tt is
w and
irrational
V2 are both examples of irrational numbers.
is
not easy
The
Part III to a proof of this fact.
and was known
quite simple,
irrationality of
V2, on
The
Chapter 16 of
shall devote all of
the other hand,
is
Pythagorean theorem
of length 1 has a hypotenuse
to the Greeks. (Since the
shows that an
isosceles right triangle,
of length
it is
,
—we
with sides
,
not surprising that the Greeks should have investigated this
The proof depends on a few observations about the natural numEvery natural number n can be written either in the form 2k for some
integer k, or else in the form 2k + 1 for some integer k (this "obvious" fact has
a simple proof by induction (Problem 7)). Those natural numbers of the form
2k are called even; those of the form 2k + 1 are called odd. Note that even
numbers have even squares, and odd numbers have odd squares:
question.)
bers.
(2k)
(2k
In particular
even;
if
n
2
is
it
+
l)
2
2
=
-
4k 2
=
2
4k 2
+
4k
•
(2k 2 ),
+
1
follows that the converse
odd, then n
is
odd.
=
2
must
The proof
•
(2k 2
+
2k)
+
also hold: if n 2
that
is
1.
is
even, then n
irrational
is
now
is
quite
simple. Suppose that
numbers p and
We
V 2 were
rational; that
suppose there were natural
such that
q
^
can assume that p and
?
=
have no
2.
common
divisors could be divided out to begin with).
P
This shows that p 2
for
is,
is
2
k.
Now we
common
have
= 2q\
even, and consequently
some natural number
divisor (since all
p must be even; that
is,
p
=
2k
Then
so
2k 2
=
2
q
.
and q
2
This shows that q is even, and consequently that q is even. Thus both/?
divisor. This
common
no
have
and
that
fact
the
q
contradicting
p
are even,
contradiction completes the proof.
important to understand precisely what this proof shows. We have
2
demonstrated that there is no rational number x such that x = 2. This asserNote,
often expressed more briefly by saying that V2 is irrational.
It is
tion
is
number
however, that the use of the symbol Vl implies the existence of some
such a
that
proved
not
have
We
2.
is
(necessarily irrational) whose square
is
proof
a
present,
at
that,
confidently
assert
can
we
exists and
number
Any
impossible for us.
proof at
this stage
would have
to be based
on P1-P12
have mentioned); since PI -PI 2 are also true for
there is a rational number whose
Q, the exact same argument would show that
(the only properties of
R we
(Note that the reverse argument will
cannot
not work our proof that there is no rational number whose square is 2
our
because
is
square
2,
be used to show that there is no real number whose
every
that
fact
the
of
property
special
Q,
proof used not only PI -PI 2 but also a
square
2,
is
and
this
we know
is false.
—
can be written p/% for integers p and q.)
could,
This particular deficiency in our list of properties of the real numbers
existence
the
asserts
which
property
new
adding
a
by
corrected
of course, be
is, however,
of square roots of positive numbers. Resorting to such a measure
would still
neither aesthetically pleasing nor mathematically satisfactory; we
number
in Q,
know that every number has an nth root if n is odd, and that every positive
not
number has an nth root if n is even. Even if we assumed this, we could
though
(even
=0
x
1
+
prove the existence of a number x satisfying x +
the solution
there does happen to be one), since we do not know how to write
solution
the
that
known
it
is
(in
fact,
roots
of
nth
of the equation in terms
wish to
not
do
certainly
we
course,
And,
of
form).
this
in
cannot be written
x
number
that all equations have solutions, since this is false (no real
not
b
assume
satisfies x
2
+
1
-
0,
for
example). In
fact, this direction
of investigation
is
R
distinguishing
not a fruitful one. The most useful hints about the property
elucidating this
of
necessity
from Q, the most compelling evidence for the
—
Numbers of Various
27
Sorts
come from the study of numbers alone. In order to study the
numbers in a more profound way, we must study more
numbers. At this point we must begin with the foundations of
property, do not
properties of the real
than the real
fundamental concept on which calculus
calculus, in particular the
is
based
functions.
PROBLEMS
1.
Prove the following formulas by induction.
(i)
1
(ii)
l
2
+
•
•
•
3
+
•
•
•
+
n(n
l)(2n+
2
_
=
n3
=
(1
+
3
+
5
+
+
32
+
5
+n
1)
6
+
Find a formula
2
(i)
+
•
+
•
•
n)\
for
( 2i
-
*)
(2i
—
I)
=
1
+
•
2n
-
V-
+
(2n
-
(
n
(ii)
£=
i
Hint:
What do
In and
3.
If
0
<
k)
2
1
2
l
k
2
+
<
22
+
•
*
these expressions have to
+
32
+
+
•
•
(2n)
W
W W=
(
2
*
l)
2
.
)
1.
do with
1
+2+3+
*
•
+
•
?
the "binomial coefficient"
w,
k\(n-k)\
(^) =
(a)
=
2
l
^
is
defined by
k\
(This becomes a special case of the
define 0!
=
formula
first
if
we
1.)
Prove that
("rK:,K>
(The proof does not require an induction argument.)
"Pascal's triangle"
following configuration,
rise to the
This relation gives
—a number not on one of the
the two
numbers above
number
in the nth row.
it;
the binomial coefficient
1
1
13
4
6
1
1
1
2
1
5
10
sides
1
3
1
4
10
1
5
1
is
(^j
known
sum
the
is
as
of
the kth
numbers
(b) Notice that all the
Use part
bers.
prove by induction that
(a) to
number. (Your entire proof by induction
up in a glance by Pascal's triangle.)
(c)
num-
in Pascal's triangle are natural
Give another proof that
always a natural
a sense, be
will, in
summed
number by showing
a natural
is
is
(^j
that
(j^j
1,
(d)
the
is
(^j
.
.
.
,
of sets of exactly k integers each chosen
number
from
n.
Prove the "binomial theorem":
and
If a
b
any numbers and
are
rcis
a natural number, then
(a
+
b)
n
=
a
n
+
(j)
Q a»~W +••+(„",)
a^b +
+ *"
"SO"a""*.
(e)
Prove that
®2C)-G)
< ii
4.
(a)
+ •••*(;)=»•
»s<- K")=(o)-(0
i
Prove by induction on
1
+r+
r
2
that
+
•
+
•
•
r
=
n
—-
—
1
(if r
1
Derive
this
5.
=
n+1
r
sum certainly presents no problem),
+ r n multiplying
by setting S = 1 + r +
equation by r, and solving the two equations for S.
ifr
(b)
r
1
evaluating the
1,
this result
The formula
for
2
+
(k
+
l
*
•
•
,
+
•
2
may
—
3k*
n
be derived as follows.
We
with the formula
Writing
+
(n +
(n
this
_ p =
33
-
l)
3
3
-
3
formula for k
23
l)
l)
23
n*
1
3
.
12
= 3-2
2
= 3'n
= 3[1
2
2
—
kz
=
1,
.
.
+3 +
+ 3-2 +
+
+
.
i
3-n
•
•
•
+
+
.
+
,
n
3k
+
l.
and adding, we obtain
\
1
1
n 2]
+
3[1
+
•
•
•
+ n] +
n.
begin
Numbers of Various
n
Thus we can
if
we already know
found in a similar way). Use
l
(i)
(ii)
l
3
+
•
•
•
4
+
•
•
•
method
this
£ (which could have been
to find
3
.
+ -!-+•••+ »(« +
1-22-3
1
3
UVJ
£
+n
+ n\
-*-
(hi)
29
'*
k2
T
find
Sorts
2«
5
P-2
2
2
'
2
1)
-3
2
»
2
(rc
+
+
l
,
l) 2
71
Use the method of Problem
in the
5 to
show
Y P can always
be written
form
^—
+ An' + En*+
p
(The
that
first
1
+ Cn^ +
2
•
•
•
.
\
10 such expressions are
n
V
k
=
ih 2
+
^
+i/z 2
+
W
+
in
n
y F =
3
k=l
n
\
k3
=
in'
6
=
i*
+
|« 2
n
J=
:
/c
6
+ *« + A« -
8
+
4
5
tVrc
2
1
n
=1
:
tF = |«
fix
T F=
:
V
+V
+ ftll - A" + A"
4
4
+f
7
rc
- A« 5 +
l«
3
2
- A"
—1
n
y f=
^«>o
+ £n« +
f
8
rc
-
-jV
6
+
- A"
2
n
Notice that the coefficients in the second column are always i, and that
decrease
after the third column the powers of n with nonzero coefficients
30
Prologue
by 2 until n 2 or n is reached. The coefficients in all but the first two
columns seem to be rather haphazard, but there actually is some sort of
pattern; finding
Problem 26-16
may
it
for the
complete
story.)
7.
Prove that every natural number
8.
Prove that
A
a set
either even or odd.
is
numbers contains
of natural
w0
and contains
+
k
1
natural numbers > no.
Prove the principle of mathematical induction from the well-ordering
whenever
9.
if
See
test.
be regarded as a super-perspicacity
it
contains
A
then
k,
contains
all
principle.
10.
Prove the principle of complete induction from the ordinary principle of
1 whenever it coninduction. Hint: If A contains 1 and A contains n
+
tains
1,
.
.
.
n,
,
consider the set
B of all
k such that
1,
.
.
.
,
k are all
in A.
11.
(a)
If a
(b)
If a
and
rational
is
What
if
is
and
a
+
a
b is irrational, is
b necessarily irrational?
are both irrational?
b
and
rational
b
irrational,
is
ab necessarily irrational?
is
(Careful!)
there a
number
(c)
Is
(d)
Are there two
2
a such that a
is
numbers whose sum
irrational
but a 4 is rational?
and product are both
irrational,
rational?
12.
(a)
Prove that
V3, V5, and
Ve are irrational.
example, use the fact that every integer
or 3n
(b)
*13.
2.
Why
doesn't this proof
and
Prove that
(b)
V2 + V3 irrational.
V6 - V2 - V3 irrational.
(a)
Prove that
(b)
treat
V3,
for
for
+
1
and m
is
a
and
b.
V4?
is
is
if
x
=p
+
(a)
Prove also that (p Prove that if m and
In) 2 1 (m + n) 2
(m
+
where p and
V
+
=
a
m
=
'
q)
a
>
2;
q are rational,
for
-
are natural
72
£
numbers and
Prove the same
2
(c)
Prove that
if
results
m/n <
m'/n' with m/n
< m
V
f
V
m
2
/n 2
<
2,
then
2
^
2
with
/n'
a
show, moreover, that
2
(b)
some rational
Vy.
+ 2n) _ 2<2 _m
n
(m + n)
(m
*16.
To
are irrational.
natural number, then x m
15.
work
Hint:
of the form 3n or 3n
Prove that
(a)
14.
+
is
all
inequality signs reversed.
2,
then there
<
V
is
another rational number
2.
It seems likely that
n is irrational whenever the natural number n is
not the square of another natural number. Although the method of
Problem 1 2 may actually be used to treat any particular case, it is not
Numbers of Various
advance that
clear in
always work, and a proof for the general
A natural number p is called a
will
it
31
Sorts
case requires .ome extra information.
prime number if it is impossible to write p =
a and b unless one of these is p, and the other 1
;
ab for natural
for
numbers
convenience
we
also
few prime numbers are 2,
= ab, with a and b
3, 5, 7, 11, 13, 17, 19. If n > 1 is not a prime, then n
both < n; if either a or 6 is not a prime it can be factored similarly; continuing in this way proves that we can write n as a product of primes.
agree that
1
=
For example, 28
(a)
4
•
=
7
2
2
•
first
7.
•
argument into a rigorous proof by complete induction.
(To be sure, any reasonable mathematician would accept the
informal argument, but this is partly because it would be obvious
Turn
to
A
The
a prime number.
is not
this
him how
to state
rigorously.)
it
fundamental theorem about
states that this factorization
is
which we
integers,
will
not prove here,
unique, except for the order of the factors.
Thus, for example, 28 can never be written as a product of primes one
is 3, nor can it be written in a way that involves 2 only once
of which
(now you should appreciate why
(b)
Using
this fact,
natural
is
not allowed as a prime).
is
irrational unless n
= m
is
irrational unless n
—
V'n
prove that
number
1
2
some
for
m.
(c)
Prove more generally that
(d)
No
numbers should
discussion of prime
fail
nr.
to allude to Euclid's
many of them. Prove
many prime numbers pi, p pz,
beautiful proof that there are infinitely
there cannot be only finitely
p n by considering pi p 2
-
*17.
(a)
Prove that
if
x
for
some
integer.
(b)
first
18.
6
1
integers a n ^ u
is
this
is
•
pn
•
•
+
•
that
.
•
,
I.
V 2 + ^2
powers of
this
.
this trivial if h
>
The Fibonacci sequence
.
aQ
,
+a =
Q
then x
,
is
0,
irrational unless x
is
an
is
irrational. Hint: Start
by working out the
number.
+
0
=
=
1,
an
=
a n -i
> —1,
If h
h)
n
>
+
1
then
nh.
?
ai, a 2
ai
ai
.
•
a generalization of Problem 16?)
Prove Bernoulli's inequality:
Why
.
.
+ a n _ lX n - +
(1
19.
.
,
x satisfies
n
(Why
Prove that
'
2
,
a3
,
.
.
.
is
defined as follows:
1,
+
for n
>
3.
was discovered by
This sequence, which begins 1, 1. 2, 3, 5, 8,
Fibonacci (circa 1175-1250), in connection with a problem about
.
.
.
,
32
Prologue
assumed that an initial pair of rabbits gave birth to
one new pair of rabbits per month, and that after two months each new
pair behaved similarly. The number a n of pairs born in the nth month is
# n __i + a M _2, because a pair of rabbits is born for each pair born the
previous month, and moreover each pair born two months ago now
gives birth to another pair. The number of interesting results about this
sequence is truly amazing there is even a Fibonacci Association which
rabbits. Fibonacci
—
publishes a journal, The Fibonacci Quarterly. Prove that
V5
One way
of deriving this astonishing formula
is
presented in Problem
23-8.
20.
'
The
an
result in
>
0,
Problem
1-6 has
an important generalization:
If ai,
.
.
.
,
then
vai
'
.
.
+
ai
<
an
•
.
.
.
.
+
qn
n
(a)
Why
aj.
happens
to
ai
and
If a»
aj are
= a n ? Suppose not all a { are equal,
a^)/2 what
both replaced by (a {
+
mean" A n — (a\ +
"geometric mean Gn = Vai
the "arithmetic
What happens
Why
=
this true if a x
is
^
say
to the
•
+
*
•
a n )/ n
?
an
?
5 5
'
.
.
'
.
does repeating this process enough times eventually prove
G n < A n ? (This is another place where it is a good exercise to
provide a formal proof by induction, as well as informal reason.)
that
The
reasoning in the previous proof is closely related to another interest-
ing proof.
(b)
Using the
k,
(c)
that
fact that
Gn < A n
For a general
G n < A n when n — 2, prove, by induction on
= 2k
m
m
2 > n. Apply part (b) to the 2 numbers
for n
n, let
a\,
.
.
.
.
,
an
,
An
.
,
.
An
,
2 m — n times
to
21.
The
prove that
following
is
Gn < A n
.
a recursive definition of a n
a1
=
a,
Prove, by induction, that
a
n+m
(a
n
=
a
m =
)
a
n
•
a
m
,
:
Numbers of Various
Sorts
33
(Don't try to be fancy: use either induction on n or induction on m, not
22.
both at once.)
Suppose we know properties PI and P4 for the natural numbers, but
that multiplication has never' been mentioned. Then the following can
be used as a recursive definition of multiplication:
+
(a
23.
In
'
(b
a
'
1
a
-
b
+
=
=
b
a
'
+
b
=
=
b9
a-
b
+
b.
order suggested!):
a
c
•
on
(use induction
a),
a,
b
'
chapter
this
—
c)
-b
(in the
Prove the following
a
1
1)
a (you just finished proving the case b
we began with
the natural
=
1).
numbers and gradually
built
A
completely rigorous discussion of this process
requires a little book in itself (see Part V). No one has ever figured out
how to get to the real numbers without going through this process, but
up
if
to the real
we do
be
numbers.
accept the real numbers as given, then the natural numbers can
1
1,1
1, etc. The
the real numbers of the form 1,1
whole point of
matical
(a)
A
way
set
A
+
+
defined as
problem
this
is
to
show that there
is
+
a rigorous mathe-
of saying "etc."
numbers
of real
(1)
1
is
(2)
k
+
is
called inductive
if
in A,
is
1
in
A whenever
k
in A.
is
Prove that
R
(i)
(ii)
(iii)
(iv)
(v)
(b)
A
is
real
(i)
set of positive real
set of positive real
number
inductive
(ii)
inductive.
numbers is inductive.
numbers unequal to \ is inductive.
set of positive real numbers unequal to 5 is not inductive.
If A and B are inductive, then the set C of real numbers which
are in both A and B is also inductive.
The
The
The
n will be called a natural
number
if
n
is
in every
set.
Prove that 1
Prove that k
is
+
a natural number.
1
is
a natural
number
if
£
is
a natural number.
PART
FOUNDATIONS
The statement
is
so frequently
made
that the differential calculus deals with
continuous magnitude, and yet
an explanation of this continuity
nowhere given ;
is
even the most rigorous expositions
of the
differential calculus do not base
their proofs
upon continuity but,
with more or
less consciousness
of the fact,
they either appeal to geometric notions
or those suggested by geometry,
or depend upon theorems which are never
established in a purely arithmetic manner.
Among
these, for
example,
belongs the above-mentioned theorem,
and a more
careful investigation
me
convinced
that this theorem, or
any one equivalent
in
some way as a
to it,
can be regarded
sufficient basis
for infinitesimal analysis.
It then only
remained
origin in the elements
to
discover
its true-
of arithmetic
and thus
at the
to secure
a real definition of
same time
the essence of continuity.
I succeeded Nov. 24, 1858, and
a few days afterward I communicated
the results
of my meditations
to
my
dear friend
Durege with whom I had a long
and
lively discussion.
RICHARD DEDEKIND
>
FUNCTIONS
CHAPTER
that of a
Undoubtedly the most important concept in all of mathematics is
turn out
functions
mathematics
modern
of
branch
function—in almost every
not
probably
therefore
will
It
investigation.
of
the central objects
to be
surprise
of great generality.
to learn that the concept of a function is one
we will be able to
will be a relief to learn that, for the present,
you
Perhaps it
restrict our attention
even this small class
attention for quite
our
of functions will exhibit sufficient variety to engage
For the moment
definition.
proper
some time. We will not even begin with a
to functions of a very special kind;
and
will
at length,
a provisional definition will enable us to discuss functions
mathematicians.
by
understood
as
functions,
illustrate the intuitive notion of
modern mathethe
of
advantages
the
discuss
Later, we will consider and
matical definition. Let us therefore begin with the following:
PROVISIONAL DEFINITION
A function is a rule which
real
assigns, to
each of certain real numbers, some other
number.
and amplify this
following examples of functions are meant to illustrate
clarification.
such
some
definition, which, admittedly, requires
The
Example
1
The
which
rule
assigns to
each number the square of that
number.
Example 2
The
rule
which
number
y the
number
<?
assigns to each
y*
+ 3y +
r+
number
5
1
Example 3
The
rule
which
assigns to each
+
3c
+
2
-
1
c
-1
the
number
x satisfying
-17 <
^
1,
5
The rule which assigns to each number
2
x < 7r/3 the number x
Example 5 The rule which assigns to each number a
irrational, and the number 1 if a is rational.
Example 6 The rule which assigns
Example 4
.
to 2 the
number
to 17 the
5,
number
—
7T
37
the
number 0
it
a
is
— the number 28,
to
17
36
to
— the number 28,
,
7T
and
V2
b
any y ^2,
to
for a, b in
17,
7r
2
/17, or 36/tt, the
number
16
if
is
of the form a
+
Q
number / 3 + x.
we are really
describing infinitely many different functions, one for each number x.)
Example 8 The rule which assigns to each number z the number of 7's in
the decimal expansion of z, if this number is finite, and — tt if there are
Example 7
The
rule
which
assigns to each
(This rule depends, of course,
many
infinitely
One
7's in
number
2
on what the number x
the decimal expansion of
the
is,
so
z.
thing should be abundantly clear from these examples
—a function
is
numbers to certain other numbers, not just a rule which
can be expressed by an algebraic formula, or even by one uniform condition
which applies to every number; nor is it necessarily a rule which you, or anybody else, can actually apply in practice (no one knows, for example, what
any rule that assigns
tt). Moreover, the rule may neglect some numbers and it
not even be clear to which numbers the function applies (try to determine, for example, whether the function in Example 6 applies to tt). The set of
rule 8 associates to
may
numbers
to which a function does apply is called the domain of the function.
Before saying anything else about functions we badly need some notation.
Since throughout this book we shall frequently be talking about functions
(indeed we shall hardly ever talk about anything else) we need a convenient
way of naming functions, and
of referring to functions in general. The standard
denote a function by a letter. For obvious reasons the letter "/"
a favorite, thereby making "g" and
other obvious candidates, but any
practice
is
to
is
any reasonable symbol, for that matter) will do, not excluding
and "y," although these letters are usually reserved for indicating numbers.
letter (or
If/
by
is
a function, then the
/W—this symbol
Naturally,
is
number which / associates
read "/ of x" and
is
to a
number
x
often called the value
"/," leading to the
domain
and
we can
symbol
*(/)).
Note that the symbol
/(*)
makes sense only
of /; for other x the symbol f(x) is not defined.
If the functions defined in Examples 1-8 are denoted by
/, g h,
9
then
f(x)
=
x2
2
h(c)
=
rewrite their definitions as follows:
for all x.
y
(3)
at x.
we denote a function by x, some other letter must be chosen to
number (a perfectly legitimate, though perverse, choice would be
for x in the
(1)
denoted
off
if
denote the
y,
is
2
c'
+
1
— T
1
for aI1 '
*
1,
-1.
r, s, 0,
ax
,
Functions
r(x)
(4)
s{x)
(5)
=
x2
=
for all x
— 17 <
such that
0,
x irrational
1,
* rational.
x
<
39
tt/3.
7
x
=
2
'
x
=
17
17
'
*
17
x
—
5,
36
—
7T
7T
flW
(6)
28
=
<
'28,
36
=
7T
x
16,
(7)
ax (t) =
(8)
y(x)
_
-
t
|
z
+
x
^
17,
2,
—j
or
17
for all
—
and *
,
=
+
a
for a, b in
Q.
7T
numbers
t.
exactly «7's appear in the decimal expansion of x
n,
infinitely
I
many 7 's appear
in the
decimal expansion of x.
These definitions illustrate the common procedure adopted for defining a
function /—indicating what f(x) is for every number x in the domain of /.
(Notice that this is exactly the same as indicating f(a) for every number a, or
abbreviations are tolerated.
f{b) for every number b, etc.) In practice, certain
Definition (1) could be written simply
f(x)
(1)
the qualifying phrase "for
(4)
all
It is
r( x )
Of
x" being understood.
the only possible abbreviation
(4)
= x\
=
course, for definition
is
-17 <
x\
x
<
tt/3.
usually understood that a definition such as
kix)
=
-
+
X
x
x
can be shortened
———
.
x
*
0,
1
1
to
*W = - + -^-7;
—
1
a:
in other words, unless the domain
consist of all
numbers for which
is explicitly restricted
the definition
makes any
further,
it
is
understood
to
sense at all.
should have little difficulty checking the following assertions about the
functions defined above:
You
/(*+
1)
g(x)
r(x
+
1)
+2x+ 1;
x + 3x + 5 =
h(x)
if -17
r{x) + 2x +
=/(*)
=
=
if
3
1
0;
<
x
<| -
1;
=
+ y)
s(x
s(x) if y is rational;
a z (x) =
x
[f(x)
If the expression f(s(a)) looks
that s(a)
is
number
a
like
a matter of fact, f(s(a))
=
+
1];
unreasonable to you, then you are forgetting
any other number,
s(a) for all a.
sions than f(s(a)) are, after a
so that f(s(a))
more
exposure, no
first
makes
sense.
As
complicated expres-
Why? Even more
difficult to unravel.
The
expression
formidable as
it
appears,
may
be evaluated quite easily with a
little
patience:
f(r(s(0(az(y(i))))))
=
=
=
=
f(r(s(6(aM))))
/(rW9(3))))
/(rW16)))
/(r(l))
-/(I)
=
The
first
1.
few problems at the end of this chapter give further practice manipu-
lating this symbolism.
The
function defined in (1)
a rather special example of an extremely
function / is a
is
A
important class of functions, the polynomial functions.
polynomial function
=
/(*)
a nx
n
+
if
there are real
a n „ix
n~ l
+
•
•
•
numbers
+
a 2x
2
a0,
+
.
aiX
.
+
.
,
a0
,
a n such that
for all x
0).
usually tacitly assumed that a n
of/;
degree
the
called
The highest power of x with a nonzero coefficient is
6
137* 4 — tt
for example, the polynomial function / defined by /(*) = 5x
(when
is
f{x)
written in this form
it is
+
has degree
6.
functions defined in (2) and (3) belong to a somewhat larger class of
functions, the rational functions; these are the functions of the form p/q
where p and q are polynomial functions (and q is not the function which is
The
rational functions are themselves quite special examples of an
even larger class of functions, very thoroughly studied in calculus, which are
simpler than many of the functions first mentioned in this chapter. The
always
0).
The
following are examples of this kind of function:
ro\
(9)
f(
\
fix)
x
=
±
x2
;
(10)
(11)
fix)
fix)
+ * sin2
—*
+ x sm x
;
x sin x
=
=
s'mix
z
2
).
sin(sin(^ 2 )).
41
Functions
„,
v
,
(12)
=
s
/(*)
n
.
«^
sin 2 (sin(sin 2 (* sin 2 * 2 )))
(
\
By what
built
you may
criterion,
a monstrosity
—+ +
fx
.
sin
•
sin(# sin #)\
up from a few simple
)•
sin x
/
impelled to ask, can such functions, especially
feel
The answer is that they can be
means of combining
(9)— (12) we need to start with
be considered simple?
like (12),
-
x
functions using a few simple
functions. In order to construct the functions
=
and the "sine function" sin,
x. The following are some
is
whose
of the important ways in which functions may be combined to produce new
the "identity function"
which
for
7,
I(x)
x,
often written simply sin
value sin(#) at x
functions.
If /
the
and g are any two functions, we can define a new function /
of / and g, by the equation
+ g, called
sum
(f
+ g)(x)
=f(x)+g(x).
Note that according to the conventions we have adopted, the domain of / + g
makes sense, i.e., the set of all x in both
consists of all x for which "/(*) +
domain /and domain g. If A and B are any two sets, then A C\ B (read "A intersect B" or "the intersection of A and Z?") denotes the set of x in both A and B;
this notation allows us to write domain (/ + g) = domain / Pi domain g.
we
In a similar vein,
define the product /
•
g
and the quotient -
(or f/g)
g
of / and g by
(/•*)(*)=/(*)•*(*)
and
Moreover,
if
g
is
a function and
c is
(c
a number,
g)(x)
=
c
'
we
define a
new
function
c
g by
•
g(x).
This becomes a special case of the notation / g if we agree that the symbol c
should also represent the function / defined by fix) — c; such a function,
which has the same value for all numbers x, is called a constant function.
•
The domain of / g is domain / C\ domain g, and the domain of c g is
simply the domain of g. On the other hand, the domain of f/g is rather com*
*
plicated
symbol
—
general,
{x:
xz
may
it
\x: g(x)
+
{x:
3
.
{x: x*
+
3
<
1
1
}
=
\x:
have been written using y
guess that {x
than
denotes the set of
denotes the set of
11}
this notation are
less
domain g C\ {x: g(x) ^0}, the
be written domain /
0} denoting the set of numbers x such that g(x) ^ 0. In
.}
.
<
consequently
just as well
^
<
x such that ".
numbers
.
."
is
true.
3
x such that x
<
8,
Thus
and
Either of these symbols could
everywhere instead of x. Variations of
x
common, but hardly
2}
.
require any discussion.
Any one can
8} denotes the set of positive numbers whose cube is
could be expressed more formally as {x: x > 0 and x* < 8}.
>
8; it
all
all
0:
*3
<
Incidentally, this set
is
slightly less transparent,
equal to the
contains just the four numbers
x
{x:
=
=
or x
1
3 or ^
set
<
0
{x:
but very standard. The
=
2,
1,
=
2 or ^
and
3,
<
x
2}.
One
variation
set {1, 3, 2, 4}, for
4;
it
is
example,
can also be denoted by
4).
Certain facts about the sum, product, and quotient of functions are obvious
facts about sums, products, and quotients of numbers. For
consequences of
example,
very easy to prove that
it is
(f
The proof is
(
+ k).
g
=f
which demonstrates that two
—the two functions must be shown to have the same domain,
and the same value
+ g) + h
)
characteristic of almost every proof
functions are equal
(f
+ g + h = /+
any number
at
+
(g
+
domain. For example,
in the
to
prove that
note that unraveling the definition of the
h),
two
sides gives
Kf
=
=
+ g)+h](x)
+ g)(x) + h(x)
+ *(*)] + *W
(f
[/(*)
and
U+(g +
/(*)
+
+
h(x)
and
=/(*)
h)](x)
=
and the equality of
+ g(x)] +
[/(*)
+
(g
k)(x)
UW + Kx)l
f(x)
+
[g(x)
+
h(x)]
is
a fact
about numbers. In this proof the equality of the two domains was not explicitly
mentioned because this is obvious, as soon as we begin to write down these
domain of (/ + g) + h and of / + (g + h) is clearly domain
/ H domain g C\ domain h. We naturally write / + g + h for (/ + g) +
h = f + (g + h), precisely as we did for numbers.
h = f (g h), and this function is
It is just as easy to prove that (/ g)
denoted by / g h. The equations / + g = g + / and f'g=g '/should also
present no difficulty.
in
Using the operations +,
/ we can now express the function / defined
(9) by
equations; the
•
•
•
*
•
•
•,
_
+
I
/
It
•
I
We
require yet another
way
If / and g are any two
position of / and g, by
+ I sin sin
+ / sin sin
I
sin
•
'
•
•
express function (10) this way.
of combining functions. This combination, the
should be clear, however, that
composition of two functions,
•
is
we cannot
by
functions,
far the
we
(/•*)(*)
most important.
define a
new
function /
° g,
the
com-
=/(*«);
domain of fog is {x: x is in domain g and g(x) is in domain /}. The
symbol u/o g » is often read "/ circle g" Compared to the phrase "the comthere is
position of / and g" this has the advantage of brevity, of course, but
the
another advantage of far greater import: there
is
much less chance of confusing
Functions
fog
and
43
must not be confused, since they are not usually
/ and g chosen at random will illustrate this point
f = I 1 and g = sin, for example). Lest you become too apprehensive
with g
o /,
these
equal; in fact, almost any
(try
-
about the operation of composition,
is
let
us hasten to point out that composition
associative:
(fog)
(and the proof is a
triviality)
;
o
= fo
h
(g oh)
this function is
denoted by /
°
g
We can now
° h.
write the functions (10), (11), (12) as
(10)
(11)
(12)
/ = sine (/•/),
/ - sin o sine (/•/),
/ = (sin sin) o sin o
•
One fact
(sin
•
sin)
°
(/
[(sin
sin)
-
o
(/
•
•
/)])
has probably already become clear. Although this method of writing
it is hardly short or con-
functions reveals their "structure" very clearly,
2
/ such that f(x) = sin(x ) for all
2
x unfortunately seems to be "the function / such that f(x) = sin(x ) for
all x." The need for abbreviating this clumsy description has been clear for
two hundred years, but no reasonable abbreviation has received universal
acclaim. At present the strongest contender for this honor is something like
venient.
The
shortest
name
for the function
(read "x goes to sin(# 2 )" or just
among
x
—
u
x arrow sin(* 2 )"), but
sin(* 2 )
writers of calculus textbooks. In this
amount
popular
of
is
book we
and speak of "the function
ellipsis,
f(x)
it is
=
sin(x 2 )."
Even more
2
the quite drastic abbreviation: "the function sin(* )."
For the sake
speaking, con-
we will never use this description, which, strictly
number and a function, but it is so convenient that you
of precision
fuses a
hardly popular
will tolerate a certain
end up adopting
it
for personal use.
As with any convention,
will
probably
utility
is
the
reasonable so long as the slight logical
deficiencies cause no confusion. On occasion, confusion will arise unless a
3"
is an
*
more precise description is used. For example, "the function x
motivating factor, and
this criterion
is
+
ambiguous phrase;
x
—> x +
z
t
it
could
i.e.,
}
mean
either
the function / such that f(x)
=
x
+
/
=
x
+
t
3
for all
x
z
for all
t.
or
t—>x
+
t
3
,
i.e.,
the function
/ such that
/(/)
many
important concepts associated with functions, calculus has a notation which contains the "x —>" built in.
By now we have made a sufficiently extensive investigation of functions to
warrant reconsidering our definition. We have defined a function as a "rule,"
but it is hardly clear what this means. If we ask "What happens if you break
As we
shall see,
this rule?" it is
however, for
not easy to say whether this question
is
merely facetious or
A more substantial objection
actually profound.
is
to the use of the
word "rule"
that
f(x)
=
x*
f(x)
=
x2
and
are certainly
for
different rules, if
determining fix)
;
+
-
x2
fix)
=
x2
+
3
-
3(*
we mean
we want
by a
rule
nevertheless,
fix)
3x
+
1)
the actual instructions given
and
+
3*
+
3
- 30 +
1)
same function. For this reason, a function is sometimes defined
an "association" between numbers; unfortunately the word "association"
to define the
as
escapes the objections raised against "rule" only because
it
is
even more
vague.
There is, of course, a satisfactory way of defining functions, or we should
never have gone to the trouble of criticizing our original definition. But a
satisfactory definition
can never be constructed by finding synonyms for
The definition which mathematicians
English words which are troublesome.
have finally accepted for "function" is a beautiful example of the means by
which intuitive ideas have been incorporated into rigorous mathematics. The
correct question to ask about a function is not "What is a rule?" or "What is
an association?" but "What does one have to know about a function in
order to know all about it?" The answer to the last question is easy for each
number x one needs to know the number fix) we can imagine a table which
—
;
would display
/(*)=*»:
all
the information one could
X
about the function
fix)
1
1
-1
1
2
4
-2
4
V2
V2
It is
desire
2
2
7T
TT
— TT
TT
2
2
not even necessary to arrange the numbers in a table (which would
if we wanted to list all of them). Instead of a two
actually be impossible
column array we can consider various
(1, 1),
(-1,
1), (2, 4),
pairs of
(-2,
numbers
2
4), (t, tt ),
(V2,
2),
:
Functions
45
set.* To find /(I) we simply take the second
number of the pair whose first member is 1 to find f(w) we take the second
number of the pair whose first member is 7r. We seem to be saying that a func-
simply collected together into a
;
tion
be defined as a collection of pairs of numbers. For example,
given the following collection (which contains just 5 pairs)
might
as well
we were
if
/
=
then /(l)
=
{(1,7), (3,7), (5,3), (4,8), (8,4)},
=
7,/(3)
=
7,/(5)
/=
{(1,7),
If
8,/(8) = 4 and 1, 3,
consider the collection
=
3,/(4)
only numbers in the domain of/.
we
(3, 7), (2, 5),
4, 5, 8
are the
(1,8), (8, 4)},
decide whether
5, /(8) = 4; but it is impossible to
to be any
defined
be
cannot
function
words,
a
other
In
8.
7 or/(l)
/(l)
old collection of pairs of numbers; we must rule out the possibility which
then /(3)
=
7,
f{2)
arose in this case.
DEFINITION
A
=
=
=
We
are therefore led to the following definition.
a collection of pairs of numbers with the following property:
words, the
if (a, b) and {a, c) are both in the collection, then b = c; in other
element.
first
same
the
with
pairs
different
two
contain
not
collection must
function
is
This is our first full-fledged definition, and illustrates the format we shall
always use to define significant new concepts. These definitions are so important (at least as important as theorems) that it is essential to know when one
motivating
is actually at hand, and to distinguish them from comments,
remarks, and casual explanations. They will be preceded by the word
DEFINITION, contain the term being defined in boldface letters, and constitute a paragraph unto themselves.
There is one more definition (actually defining two things at once) which
can now be made rigorously:
If /
DEFINITION
is
a function, the domain of /
such that
(a, b) is
in /. If a
of a function that there
This unique
b is
is,
is
is
in the
the set of
domain
in fact, a unique
all
of/,
it
a for
which there is some b
from the definition
follows
number
b
such that
(a, b) is
in/.
denoted by /(a).
we have reached our goal: the important thing about
its
a function / is that a number fix) is determined for each number x in
intuitive
an
where
point
the
reached
also
domain. You may feel that we have
definition has been replaced by an abstraction with which the mind can
With
this definition
hardly grapple.
*
The
Two
consolations
may
be offered.
First,
pairs occurring here are often called "ordered pairs," to
(2, 4) is
not the same pair as
in terms of ordered pairs,
and an appendix
to this
(4, 2). It is
only
fair to
warn
that
although a function
emphasize
we are going
that, for
example,
to define functions
another undefined term. Ordered pairs can be defined, however,
chapter has been provided for skeptics.
has been defined as a collection of pairs, there
is
nothing to stop you from
thinking of a function as a rule.
Second, neither the intuitive nor the formal
definition indicates the best way of thinking about functions. The best way
to
is
draw
pictures; but this requires a chapter all
by
itself.
PROBLEMS
1.
Let/M =
is
(0
f(f( x )) (f° r which x does this
m
/©.
(hi)
f(cx).
(iv)
f(x+y).
(v)
/M+/G0.
(vi)
(vii)
=
Let g(x)
x 2 and
let
,
__
x\
0,
|
(1,
(ii)
For which
For which
(iii)
What
(v)
3.
_y
h(y)
jy is
w
s
g(w)
is
x rational
a irrational.
y?
g(y)?
< w?
= ^(s)?
g(g(e))
is
Find the domain of the functions defined by the following formulas.
f{x)
=
(ii)
/(*)
=
(iii)
fix)
=
(i)
V
- x\
- Vi -
]
(iv)
/(*)
(v)
/(*)
Let
x\
-J- + -L-.
—
* — 2
x
4.
<
<
is A(jy)
is
For which
For which
(iv)
sense?).
x?
fa(
(i)
make
For which numbers c is there a number x such that /(ex) = f(x).
Hint: There are a lot more than you might think at first glance.
For which numbers c is it true that f(cx) = f(x) for two different
numbers
2.
+*) What
1/(1
1
= Vj - Vl -
x
=
P(x)
2
a:
,
let
*2
+ Vx - 1.
+ V* -~2.
2
=
2*,
and
let s(x)
=
sin a.
Find each of the
following. In each case your answer should be a number.
(J"*) GO.
(i)
(ii)
,)(,).
(5.P.j)(i)
(iii)
+
(jo *)(/).
(iv)
5.
Express each of the following functions in terms of
+
-
,
,
and
o
(for
example, the answer to
(i) is
Po
s).
P, s, using only
In each case your
S,
Functions
47
answer should be a function.
(i)
f(x)
(ii)
f(x)
(iii)
/(*)
(iv)
/(*)
=
=
=
=
(sin x)
(
v)
f(t)
2 ain *.
sin 2*.
sin x 2
.
sin 2 x
(remember that
sin 2 x
is
an
abbreviation
for
2
).
=
2 2 \ (Note: a
bc
m
this convention
means a
more simply as a hc .)
always
b c
(vi)
because (a ) can be written
2- 2 ).
f{u) = sin (2*
(vii)
f(y)
;
is
adopted
+
(viii)
22iin,
=
=
sin(sin(sin(2
2
+
sin2a
')))•
+
sin (a 2 )
2 sin(a2+sina)
.
flexible, occupy a favored
most investigations of functions. The following two problems illustrate
their flexibility, and guide you through a derivation of their most important
Polynomial functions, because they are simple, yet
role in
elementary properties.
6.
(a)
If x h
fi
.
.
.
,
product of
is
x n are distinct numbers, find a polynomial function
of degree n
—
which
1
-
(x
all
is
and 0
at
1
^
for ;
x<)
i,
is
at Xj for;
0 at xj
if ;
^
i.
Hint:
The
(This product
usually denoted by
n
n
;
(*
-
*i)>
=1
the symbol II (capital pi) playing the
2
(b)
find a
where
polynomial function / of degree n — 1 such that /(*»•) =
a n are given numbers. (You should use the
ah
.
.
.
,
(a). The formula you will obtain is called the
"Lagrange interpolation formula.' )
Prove that for any polynomial function /, and any number a, there
=
is a polynomial function g, and a number b> such that /(*)
into
(x
a)
(x
fl^W + b for all *. (The idea is simply to divide
example,
/(*) by long division, until a constant remainder is left. For
functions
/<
from part
5
7.
(a)
role for products that
plays for sums.)
Now
Gi,
same
the calculation
x2
x
-
l)x
+x
x3
-2
-3x
3
+
1
—x 2
x2
—3x
x2
—x
^2x
-2x
+
+2
1
-1
48
Foundations
-
shows that x s
proof
is
+
3*
(
+x-
l)(x 2
by induction on the degree
possible
2)
-
1.
A
formal
of/.)
(c)
(d)
Show
n roots, i.e., there are at
most
that for each n there
n roots. If h
is
and if n is odd
For which numbers a, b
y
If
A
is
any
=
0.
one with only one root.
find
c,
and d
will the function
cx
satisfy /(/(*))
a with f{a)
is
ax
(a)
numbers
n
a polynomial function of degree n with
even find a polynomial function of degree n with no
roots,
9.
-
x
Prove that if f(a) = 0, then /(*) = (* - a)g(x) for some polynomial function g. (The converse is obvious.)
Prove that if/ is a polynomial function of degree w, then /has at most
(b)
8.
=
1
=
* for
set of real
all
x
+d
+b
?
numbers, define a function
CA (x) =
1
as follows:
A
* in
,
CA
# not in A.
0,
Find expressions for CA n B and C^ wB and Cr_ a in terms of CA and
CB (The symbol A C\ B was defined in this chapter, but the other
two may be new to you. They can be defined as follows:
,
.
A VJ B —
R —A =
(b)
{x:
x
is
in
A
or x
{x:
x
is
in
R
but *
Suppose /is a function such that f(x)
there is a set A such that / = CA
in B},
is
=
not in A}.)
is
0 or
1
for
each
x.
Prove that
.
(c)
10.
(a)
(b)
*(c)
Show that / = / if and only if / = CA for some set A.
For which functions / is there a function g such that / = g 2 ? Hint:
You can certainly answer this question if "function" is replaced by
"number. 55
For which functions / is there a function g such that / = 1 Ig ?
For which functions b and c can we find a function x such that
2
OW) +
2
for all
*(d)
numbers
t
b(t)x(t)
+
c(t)
=
0
?
What
conditions must the functions a
a function x such that
a{t)x(t)
+b{t)
=
and
b satisfy if there is to
be
0
numbers ^ ? How many such functions x will there be?
Suppose that His a function and y is a number such that H(H(y))
for all
11.
(a)
y.
What
is
H(H(H(-
(H(y)
80 times
•
•
•)
?
=
49
Functions
(b)
(c)
Same
Same
replaced by 81.
question
if
80
question
if
H(H{y))
is
=
H(y).
H
such that H(H(x)) = H(x) for all numbers x,
*(d) Find a function
and such that H{\) = 36, H(2) = tt/3, 7/(13) = 47, //(36) = 36,
//(tt/3)
are
-
=
tt/3, 7/(47)
many
47. (Don't try to "solve" for H(x); there
=
functions //with H{H(x))
The
H(x).
extra conditions
H are supposed to suggest a way of finding a suitable H.)
on
Find a function //such that H(H(x)) = //(#) for all #, and such that
H{\) = 7, //(17) = 18.
A function / is even if f(x) = f(—x) and odd if f(x) = —/(—*). For
example, / is even if f(x) = x 2 or /(*) = \x\ or /(*) = cos x, while / is
*(e)
12.
=
=
odd
if f(x)
(a)
Determine whether /
x or /(#)
sin
+g
is
even, odd, or not necessarily either, in
and g even or odd.
(Your answers can most conveniently be displayed in a 2 X 2
the four cases obtained by choosing /even or odd,
table.)
(c)
Do
Do
(d)
Prove that every even function / can be written f(x)
(b)
same
the same
the
infinitely
*13.
(a)
/
for
f
many
•
g.
° g.
E is
Prove that
first,
even and
this
—
g{\x\), f° r
functions g.
R can be written/ = E
Prove that any function /with domain
where
(b)
for
way
by "solving"
0
is
odd.
of writing /
for
E
4- 0,
is
unique.
and 0 you
will
(If
you
do pan;
try to
(b)
probably find the solution
to part (a).)
14.
any function, define a new function |/| by |/|(*) = /(*)!• ^ /and
g are functions, define two new functions, max(/, g) and min(/, g), by
If /is
|
max (/,£)(*) = max (/(*), g(x)),
min (/,$)(*) - min(/(*), £(*)).
Find an expression
max(/, g) and min(/, g) in terms of
0) + min(/, 0). This particular way of
useful; the functions max(/, 0) and min(/, 0) are
for
|.
|
15.
(a)
Show
/ = max(/,
that
writing
/
fairly
is
and negative parts of /.
nonnegative if f(x) > 0 for all x. Prove that
any function / can be written / = g — h, where g and h are nonnegative, in infinitely many ways. (The "standard way" is g —
max(/, 0) and h = — min(/, 0).) Hint: Any number can certainly
be written as the difference of two nonnegative numbers in infinitely
called the positive
(b)
A
function
many
*16.
Suppose /
(a)
/
is
called
ways.
satisfies
f(x
Prove that/fo
+ y)
-
f(x)
+ f(y)
and
for all x
/(*,)
+
•
•
•
y.
+/(*«).
50
Foundations
(b)
point we're not trying to say anything about fix)
Hint: First figure out what c must be. Now prove
=
that f{x)
all
If fix)
and
satisfies
fix)
(a)
(b)
=
rational
=
0 for
these
such that fix)
when x is a natural number, then when x is an
when x is the reciprocal of an integer and, finally, for
f(x
y) = fix)
fiy) for all x and
suppose that /
fiy) for all x and y.
properties, but that fix) is not always 0. Prove that
all x,
•
y)
two
then /
~
fix)
+
satisfies
+
Now
•
all x, as follows:
Prove that/(l)
Prove that fix)
=
=
>
1.
x
x
if
is
rational.
(c)
Prove that fix)
0 if x > 0. (This part is tricky, but if you have
been paying attention to the philosophical remarks accompanying
(d)
Prove that fix)
Prove that fix)
the problems in the last two chapters, you will
(e)
> fiy) if x > y.
= x for all x. Hint: Use
two numbers there
*18.
is
know what
to do.)
the fact that between
any
a rational number.
what conditions must /, g, h, and k satisfy in order that fix)giy)
and y?
Prove that there do not exist functions / and g with either of the
Precisely
=
*19.
cx for all rational
x.
also fix
x for
c
cx, first
integer, then
y,
—
is
(at this
for irrational x).
*17.
some number
Prove that there
numbers x
hix)kiy) for all x
(a)
following properties:
(i)
fix)
(ii)
fix
+ giy) = xy for all x and y.
+ y) = gix) — y for all x and y.
Hint: Try to get some information about / or g by choosing particular values of x and y,
(b) Find functions / and g such that fix
y) = gixy) for all x and y.
+
*20.
(a)
Find a function /, other than a constant function, such that fiy)
/Ml < \y-*V
(b)
21.
Suppose that fiy) - fix) < iy - x) 2 for all x and y. (Why does this
imply that \fiy) - f{x)\ < iy - x) 2 ?) Prove that / is a constant
function. Hint: Divide the interval [x,y] into n equal pieces.
Prove or give a counterexample
(a)
(b)
—
\
+ h) =fo g +fo h
+ h)of = gof+ k of.
foig
ig
.
for
each of the following assertions:
Functions
22.
(a)
Suppose g
=
h
Prove that
° /.
if
=
/(#)
/(j),
then £(*)
=
51
£(>»).
(b) Conversely, suppose that / and g are two functions such that g{x) —
g(y) whenever f(x) = f(y). Prove that g = h ° / for some function
k.
definition will not
23.
Suppose that f ° g
(a)
if
x
^ y,
(b) every
*24.
*25.
*26.
27.
when z is of the form z = f{x) (these
and use the hypotheses to show that your
Hint: Just try to define h(z)
are the only z that matter)
=
I,
run into trouble.
where I(x) = x. Prove that
then g{x) ^ g(y);
b can be written
number
b
=
f(a) for
some number
a.
Suppose g is a function with the property that g(x) 5* g(y) if x 9^ y.
Prove that there is a function / such that fog = I.
(b) Suppose that / is a function such that every number b can be
written b — f(a) for some number a. Prove that there is a function
(a)
g such that fog = I.
Find a function / such that g °f = / for some g, but such that there is no
function h with / ° h = /.
Suppose f ° g = I and h°f = I. Prove that g = h. Hint: Use the fact
that composition is associative.
(a) Suppose f(x) = x
1. Are there any functions g such that f ° g —
+
(b)
Suppose /
is
a constant function. For which functions g does f
°
g —
g°f?
(c)
Suppose that / o g
=
identity function, f(x)
28.
(a)
Let
F
be the
+
o
=
/
for all functions g.
*
Show
that
/
is
the
x.
set of all functions
and
except P7 hold
using
g
whose domain
is
R. Prove that,
as defined in this chapter, all of properties
for F, provided 0
and
1
P1-P9
are interpreted as constant
functions,
(b)
*(c)
(d)
Show
Show
P7 does not hold.
P10-P12 cannot hold. In other words, show that there is
no collection P of functions in F, such that P10-P12 hold for P. (It
is sufficient, and will simplify things, to consider only functions
which are 0 except at two points x 0 and xi.)
that
that
Suppose we define / < g to mean that f(x)
P'lO-P'H (in Problem 1-8) now hold?
of
(e)
If /
<
g, is h
of < h
o
g
? Is
/
o
h
<g
o
h
?
<
g(x) for all
x.
Which
52
Foundations
APPENDIX. ORDERED PAIRS
Not only
well,
it is
in the definition of a function, but in other parts of the book as
necessary to use the notion of an ordered pair of objects. A definition
we have never even stated explicitly what properties
supposed to have. The one property which we will require
formally that the ordered pair (a, b) should be determined by a and b,
has not yet been given, and
an ordered pair
states
and the order
is
which they are given:
in
if (a, b)
—
then a
d),
(c,
—
c
and
—
b
d.
Ordered pairs may be treated most conveniently by simply introducing
as an undefined term and adopting the basic property as an axiom
since this property is the only significant fact about ordered pairs, there is not
much point worrying about what an ordered pair "really" is. Those who find
this treatment satisfactory need read no further.
The rest of this short appendix is for the benefit of those readers who will
{a, b)
*
uncomfortable unless ordered pairs are somehow defined so that this
becomes a theorem. There is no point in restricting our attention to ordered pairs of numbers; it is just as reasonable, and just as important,
feel
basic property
have available the notion of an ordered pair of any two mathematical
This means that our definition ought to involve only concepts common to all branches of mathematics. The one common concept which pervades
all areas of mathematics is that of a set, and ordered pairs (like everything
else in mathematics) can be defined in this context; an ordered pair will turn
to
objects.
out to be a
The
set of
a rather special
sort.
containing the two elements a and b, is an obvious first
choice, but will not do as a definition for (a, b), because there is no way of
{a, b},
set
determining from
which of a or
{a, b\
more promising candidate
is
b is
meant
the rather startling
be the
to
first
element.
A
set:
{{a},{a,b}}.
This
set
set {a},
as
it
has two members, both of which are themselves
containing the single
may
we
seem,
for this choice
are going to define
and
PROOF
<0, b)
If (a, b)
The
=
(c,
d)
hypothesis
the other
(a, b)
to
is
be
sets;
one member
is
the
the set {a,b}. Shocking
this set.
The
justification
there really isn't anything else worth saying.
DEFINITION
i
a,
given by the theorem immediately following the definition
is
the definition works,
THEOREM
member
9
then a
means
=
c
and
=
b
{{a}, {a,b}).
=
d.
that
{{«}, {a,b}}
=
{{c}, {c,d}\
—
{
53
Functions
Now
only
the
a
=
{{a}, {a, b}} contains just
two members,
c.
We
and only the proof that
7.
b
=
a.
b
In this case,
one member, namely,
\a\, which implies that d
{a}.
Case
not in
=
=
=
and
a
the
is
d remains.
{a, b\
=
{{a}, \a,d\\,
It is
\a\, so the set
The same must be
=
a
—
convenient to distinguish 2 cases.
{a}, {a, b]
{
true of
a\
{
,
a,
{
really has only
}
d
\
\ ,
so
{
a,
d
}
=
b.
In this case, b is in one member of {a}, {a, b} but not in the
but
therefore be true that b is in one member of {{a}, {a, d\
must
It
the other. This can happen only if ^ is in a, d\ but b is not in {a}; thus
2. b 9* a.
other.
b
{a, b};
therefore have
{{a}, \a,b\\
Case
and
{a}
common element of these two members of {{a}, {a, b) }. Similarly, c is
unique common member of both members of {{c}, {c, d\\. Therefore
a or b
)
{
}
{
=
d,
but
b 9^ a; so b
=
d.
|
,
CHAPTER
GRAPHS
I
Mention the
real numbers to a mathematician and the image of a straight
probably form in his mind, quite involuntarily. And most likely he
neither banish nor too eagerly embrace this mental picture of the real
line will
will
numbers. "Geometric intuition" will allow him to interpret statements about
numbers in terms of this picture, and may even suggest methods of proving
them. Although the properties of the real numbers which were studied in
Part I are not greatly illuminated by a geometric picture, such an interpretation will be a great aid in Part II.
You
oil
-i
figure
are probably already familiar with the conventional
sidering the straight line as a picture of the real numbers,
2
3
to
each real number a point on a
a point which
trarily,
i
The
<
label 0,
b,
to
1,
but to the
left
of 0,
To do
and a point
point twice as far to the right
from 0
a
we
line>
is
labeled
this
then the point corresponding to a
(Figure 1)
of associating
we
pick, arbi-
which we label 1.
the point the same distance
to the right,
2,
labeled —1, etc.
is
method of con-
i.e.,
lies to
With
the
left
this
arrangement,
if
of the point corre-
sponding to b. We can also draw rational numbers, such as ^> in the obvious
way. It is usually taken for granted that the irrational numbers also somehow
fit into this scheme, so that every real number can be drawn as a point on the
line.
We
will not
make
too.
much
fuss
about justifying
this
assumption, since
method of "drawing" numbers is intended solely as a method of picturing
certain abstract ideas, and our proofs will never rely on these pictures (althis
though we
Because
will frequently use
this
a picture to suggest or help explain a proof).
geometric picture plays such a prominent, albeit inessential
geometric terminology
is
frequently employed
when speaking
of
role,
numbers
number is sometimes called a point, and R is often called the real line.
The number \a — b\ has a simple interpretation in terms of this geometric
thus a
picture:
it is
the distance between a
has a as one end point and
~
a
figure
~
2
£
—
.
.
\
a
a
v
a
+
£
b as
and
b,
the length of the line segment which
the other. This means, to choose an example
whose frequent occurrence justifies special consideration, that the set of
numbers x which satisfy \x — a\ < £ may be pictured as the collection of
points whose distance from a is less than s. This set of points is the "interval"
from a — £ to a + £, which may also be described as the points corresponding
to numbers x with a — e < x < a + e (Figure 2).
g ets Q £ num :)ers which correspond to intervals arise so frequently that it is
desirable to have special names for them. The set {x: a < x < b) is denoted
by (a, b) and called the open interval from a to b. This notation naturally
creates some ambiguity, since (a, b) is also used to denote a pair of numbers,
j
but in context
it is
always clear (or can easily be
talking about a pair or an interval.
54
Note that
if
a
made clear) whether one is
> b, then (a, b) = 0, the set
Graphs
55
with no elements; in practice, however,
it is almost always assumed (explicitly
one has been careful, and implicitly otherwise), that whenever an interval
(a, b) is mentioned, the number a is less than b.
The set {x: a < x < b] is denoted by [a, b] and is called the closed interval
from atob. This symbol is usually reserved for the case a < b, but it is some-
if
=
The
and
(<z, b)
no reasonably accurate picture could ever
indicate the difference between the two intervals, various conventions have
been adopted. Figure 3 also shows certain "infinite" intervals. The set
{x: x > a} is denoted by (a, oo), while the set {x: x > a} is denoted by [a, oo);
the sets ( — oo,
and (— oo, a] are defined similarly. At this point a standard
warning must be issued: the symbols °o and — oo, though usually read
"infinity" and "minus infinity," are purely suggestive; there is no number
" oc " which satisfies oo > a for all numbers a. While the symbols oo and — oo
will appear in many contexts, it is always necessary to define these uses in
ways that refer only to numbers. The set R of all real numbers is also considered to be an "interval," and is sometimes denoted by (— oo, oo).
Of even greater interest to us than the method of drawing numbers is a
method of drawing pairs of numbers. This procedure, probably also familiar
to you, requires a "coordinate system," two straight lines intersecting at right
angles. To distinguish these straight lines, we call one the horizontal axis, and
one the vertical axis. (More prosaic terminology, such as the "first" and
"second" axes, is probably preferable from a logical point of view, but most
people hold their books, or at least their blackboards, in the same way, so that
"horizontal" and "vertical" are more descriptive.) Each of the two axes could
be labeled with real numbers, but we can also label points on the horizontal
axis with pairs (a, 0) and points on the vertical axis with pairs (0, b), so that
the intersection of the two axes, the "origin" of the coordinate system, is
labeled (0, 0). Any point (a, b) can now be drawn as in Figure 4, lying at the
vertex of the rectangle whose other three vertices are labeled (0, 0), (a, 0),
and (0, b). The numbers a and b are called the first and second coordinates,
times used for a
are
[a, b]
shown
6,
also.
determined in
respectively, of the point
Our
real concern, let us recall,
function
is
called the
all
is
just a collection of pairs
drawing each of the pairs
is
usual pictures for the intervals
in Figure 3; since
graph of
this
way.
method of drawing functions. Since a
of numbers, we can draw a function by
a
in the function.
The drawing obtained
in this
way
the function. In other words, the graph of / contains
the points corresponding to pairs (*,/(#)). Since most functions contain
many pairs, drawing the graph promises to be a laborious under-
infinitely
taking, but, in fact,
Not
f(x)
=
surprisingly,
c,
tion f(x)
from
it
The
lines
many functions have
have the simplest graphs.
=
c is
graphs which are quite easy to draw.
the simplest functions of
It is
all,
the constant functions
easy to see that the graph of the func-
a straight line parallel to the horizontal axis, at distance
(Figure 5).
functions f(x)
through
~
(0, 0), as in
cx also
Figure
have particularly simple graphs
6.
—
c
straight
A proof of this fact is indicated in Figure 7:
56
Foundations
Let x be some number not equal to 0, and let L be the straight line which
passes through the origin 0, corresponding to (0, 0), and through the point
A, corresponding to (x, cx). A point A\ with first coordinate y, will lie on L
when
the triangle A'B'O
similar to the triangle
is
AB =
OB
A'B'
OB'
ABO,
thus
when
c;
precisely the condition that A' correspond to the pair (j, cy),
A' lies on the graph of /. The argument has implicitly assumed that c
this
is
the other cases are treated easily enough.
ratio of the sides of the triangles
appearing in the proof,
c,
is
>
but
0,
which measures the
called the slope of the
have slope c.
This demonstration has neither been labeled nor treated as a formal proof.
Indeed, a rigorous demonstration would necessitate a digression which we are
not at all prepared to follow. The rigorous proof of any statement connecting
geometric and algebraic concepts would first require a real proof (or a pre-
straight line,
FIGURE
The number
that
i.e.,
7
and a
line parallel to this line
is
also said to
on a straight line correspond in
this, it would be necessary to
from
Aside
an exact way to the real numbers.
to develop the properties of
intend
we
precisely
as
as
geometry
develop plane
real numbers. Now the detailed development of plane geometry is a beautiful
cisely stated assumption) that the points
length d
(a,
—
b
shall
it is by no means a prerequisite for the study of calculus. We
use geometric pictures only as an aid to intuition; for our purposes (and for
most of mathematics) it is perfectly satisfactory to define the plane to be the
subject, but
b)
length
c
—
a
set of all pairs of real
8
between
(a, b)
among others,
and
(c,
d) to
the collections
f(x)
=
cx
+
d
~/~
/
/
—
/
/
/
/
r
s
/
/
/
/
/ f( x
)
-
cx
motivation for
9
cY
+
(b-
d)\
this definition
is
been built into our geometry.*
Reverting once more to our informal geometric picture, it is not hard to
see (Figure 9) that the graph of the function f(x) = cx + d is a straight line
with slope c y passing through the point (0, d). For this reason, the functions
d are called linear functions. Simple as they are, linear functions
/(*) = cx
occur frequently, and you should feel comfortable working with them. The
+
a typical problem whose solution should not cause any trouble.
Given two distinct points {a, b) and (c, d), find the linear function / whose
following
FIGURE
.
not clear, Figure 8 should serve as
adequate explanation— with this definition the Pythagorean theorem has
If the
(0, d)
}
}
be
V( a -
//
to define straight lines as certain collections
To
(*, cx):xa real number
geometry
of
structure
the
all
with
geometry
provide this artificially constructed
studied in high school, one more definition is required. If (a, b) and (c, d) are
two points in the plane, i.e., pairs of real numbers, we define the distance
of pairs, including,
FIGURE
numbers, and
*
is
might object to this definition on the grounds that nonnegative numbers
at the moment
are not yet known to have square roots. This objection is really unanswerable
—the definition will just have to be accepted with reservations, until this little point is settled.
The
fastidious reader
57
Graphs
graph goes through
f(c)
=
d. If
/
(a, b)
and
(c,
d).
aa
ac
therefore
a =
,
This amounts to saying that f(a)
= ax
be of the form f(x)
to
is
-
(rf
d
.
c
-
£)/(<;
—
—
a)
+ =
+0=
d
c
then
=
b
and
we must have
rf;
-
/3
=
—
—
b
d
a
c
and
b
f3,
b,
(3
a
+
b
[(«/
-
—
—
b
b)/(c
-
a)]a, so
N
a
a formula most easily remembered by using the "point-slope form" (see
Problem
5).
Of course,
this solution
is
possible only
account only for the straight
if
a
^
c;
the graphs of linear functions
which are not parallel to the vertical axis.
The vertical straight lines are not the graph of any function at all; in fact, the
graph of a function can never contain even two distinct points on the same
vertical line. This conclusion is immediate from the definition of a function
two points on the same vertical line correspond to pairs of the form (a, b) and
(a, c) and, by definition, a function cannot contain (a, b) and (a, c) if b 9^ c.
Conversely, if a set of points in the plane has the property that no two points
lie on the same vertical line, then it is surely the graph of a function. Thus the
first
lines
some
the graph of a function whose
is
have no points on them at all.
2
is perhaps the function f(x) = x
we draw some of the pairs in /, i.e., some of the pairs of the form (x, x 2 ), we
vertical lines
After the linear functions the simplest
If
and the last 2 are; notice
domain is not all of R, since
2 sets in Figure 10 are not graphs of functions
that the fourth
obtain a picture like Figure
1 1
• (2,
•I*, f)
(-1, !)•
•d,
9 (h
•<*.*)
,
(0, 0)
FIGURE
11
i)
.
1)
4)
!
x 2)
lie along a curve
one shown in Figure 12; this curve is known as a parabola.
Since a graph is just a drawing on paper, made (in this case) with printer's
ink, the question "Is this what the graph really looks like?" is hard to phrase
in any sensible manner. No drawing is ever really correct since the line has
thickness. Nevertheless, there are some questions which one can ask: for
example, how can you be sure that the graph does not look like one of the
drawings in Figure 13? It is easy to see, and even to prove, that the graph
cannot look like (a); for if 0 < x < y, then x 2 < y 2 so the graph should be
It is
not hard to convince yourself that
all
the pairs
(x,
like the
,
higher at y than at x, which is not the case in (a).
by drawing a very accurate graph, first plotting
It is also
many
easy to
pairs
(x,
see,
simply
x 2 ), that the
graph cannot have a large "jump" as in (b) or a "corner" as in (c). In order
to prove these assertions, however, we first need to say, in a mathematical
way, what it means for a function not to have a "jump" or "corner"; these
ideas already involve some of the fundamental concepts of calculus. Eventually
we will be able to define them rigorously, but meanwhile you may amuse
yourself by attempting to define these concepts, and then examining your
may be compared with the ones
they compare favorably, you are cer-
definitions critically. Later these definitions
mathematicians have agreed upon.
If
tainly to be congratulated
The
called
functions f(x)
power
=
xn
,
for various natural
numbers
are sometimes
functions. Their graphs are most easily compared as in Figure
by drawing several at once.
The power functions are only special cases of polynomial functions, introduced in the previous chapter. Two particular polynomial functions are
14,
\
\
\
;
\
\
\
\
\
\
FIGURE
14
Graphs
graphed in Figure 15, while Figure 16
graph of the polynomial function
/(*)
in the case a n
>
=
a nx
+
n
is
an_ xxn
meant
~l
+
•
59
to give a general idea of the
•
•
+a
0,
0.
In general, the graph of / will have at most n — 1 "peaks" or "valleys" (a
is a point like (*,/(*)) in Figure 16, while a "valley" is a point like
"peak"
(y> f(y))-
power
The number
of peaks
and valleys may actually be much smaller (the
have at most one valley). Although these
functions, for example,
make, we will not even contemplate giving proofs until
Part III (once the powerful methods of Part III are available, the proofs will
be very easy).
assertions are easy to
Figure 17 illustrates the graphs of several rational functions. The rational
functions exhibit even greater variety than the polynomial functions, but their
behavior will also be easy to analyze once
tool of Part III.
we can
use the derivative, the basic
Many interesting graphs can be constructed by "piecing together" the
graphs of functions already studied. The graph in Figure 18 is made up
entirely of straight lines.
and
is
The
function / with this graph
a linear function on each interval
— \l(n +
1)].
(The number 0
is
[l/(w
+
satisfies
1), l/n]
not in the domain of /.)
Of
+
and
[
— 1/n,
course, one
can
write out an explicit formula for f(x), when x is in [\/{n
this is a
1), 1/w]
good exercise in the use of linear functions, and will also convince you that a
picture
is
worth a thousand words.
n
even
(a)
FIGURE
16
n
odd
(b)
;
60
Foundations
FIGURE
18
Graphs
—
.
1-1
.
=
"N. f(x)
/-360
61
sin *
\
/360
\
/720
--1
It is actually possible to define, in a
exhibits this
same property
In Chapter
sine function.
much
simpler way, a function which
of oscillating infinitely often near 0, by using the
15 we will discuss this function in detail, and
radian measure in particular; for the time being
it
will
be easiest to use degree
measurements for angles. The graph of the sine function is shown in Figure 19
(the scale on the horizontal axis has been altered so that the graph will be
clearer; radian measure has, besides important mathematical properties, the
additional advantage that such changes are unnecessary).
Now consider the function f(x) = sin 1/x. The graph of / is shown in
Figure 20. (To draw this graph it helps to first observe that
fix)
=0
for x
=
-^—> —!— —180 360 540
f(x)
=
for x
=
—
90
fix)
1
= —
for x
1
=
)
>
90
+
Notice that
FIGURE
20
when x
is
*
*
360 90
—i ——
^
270 270
j
+
9
720
j
+
large, so that 1/x
*
360 270
is
+
small, f(x)
)
720
is
•
*
•
/
also small;
when
x
62
Foundations
f(x)
\
=
-v
sin
|
/
\
/
%
A
u
\
/
\
/
/
\
\
/
FIGURE
/
/
/
/ /
/
\
1
\
/
\
21
"large negative, 55 that
is, when [*| is large for negative x, again f(x) is close
although /(*) < 0.
An interesting modification of this function is f(x) = x sin 1 fx. The graph
of this function is sketched in Figure 21 Since sin 1 /x oscillates infinitely often
near 0 between 1 and —1, the function f(x) = x sin 1/x oscillates infinitely
is
to 0,
.
often between x
FIGURE
22
and —x. The behavior of the graph
for x large or large
nega-
63
Graphs
tive
is
larger
harder to analyze. Since sin 1/x is getting close to 0. while x is getting
and larger, there seems to be no telling what the product will do. It is
possible to decide, but this
III.
The graph
=
of f(x)
another question that
is
is
best deferred to Part
x 2 sin 1/x has also been illustrated (Figure 22).
For these infinitely oscillating functions, it is clear that the graph cannot
hope to be really " accurate." The best we can do is to show part of it, and
leave out the part near 0 (which
much
to find
The graphs
the interesting part). Actually,
is
simpler functions whose graph cannot be "accurately
v
I
x2
x
,
<
1
,
,
f
x2
\
,
can only be distinguished by some convention similar
Our
last
x
<
1
to that used for
open
intervals (Figure 23).
example
is
a function whose graph
spectacularly nondrawable:
is
0,
x irrational
1,
x rational.
x rational
x irrational
1,
0,
FIGURE
drawn.
of
£f
and closed
easy
it is
55
24
The graph of/ must
contain infinitely
many
points
on the horizontal
axis
and
but it must
not contain either of these lines entirely. Figure 24 shows the usual textbook
picture of the graph. To distinguish the two parts of the graph, the dots are
placed closer together on the line corresponding to irrational x. (There is
also infinitely
many
points
on a
line parallel to the horizontal axis,
actually a mathematical reason behind this convention, but
some sophisticated
ideas, introduced in
20-6*
Problems
and
it
depends on
20-7.)
The peculiarities exhibited by some functions are so engrossing that it is
easy to forget some of the simplest, and most important, subsets of the plane,
which are not the graphs of functions. The most important example of all is
the circle.
all
A circle
the points
(#,
y)
and radius r > 0 contains, by definition,
whose distance from (a, b) is equal to r. The circle thus
with center
(a, b)
consists (Figure 25) of all points (x, y) with
V( x ~
a)
2
+
(y
-
b)
2
=
r
or
(x
-
a)
2
+
(y
-
b)
2
=
r
2
.
The
circle
copy,
is
with center
and radius
(0, 0)
called the unit
often regarded as a sort of standard
1,
circle.
A close relative of the circle is the ellipse. This is defined as the set of points,
the sum of whose distances from two fixed points
fixed points are the same,
we
points are taken to be (—c, 0)
to
obtain a
and
(c,
a constant.
is
circle.)
(When
the two
convenience, the two
for
If,
and the sum of the distances is taken
is on the ellipse if
0),
be 2a (the factor 2 simplifies some algebra), then (x,y)
and only
if
V( x -
{-c)) 2
+y + V( x 2
+y =
2
2
c)
2a
or
V(*
+
c)
+y =
2
2
- V( x -
2a
c)
2
+y
2
or
+
x2
lex
+
c
+y =
2
2
a 2)
=
V x - +y
-4a V x -
+
aA
=
-
4a 2
4a
c)
or
4{cx
-
2
c)
+x -
2
2
+y
2
lex
+
c
2
+y
2
2
or
c
2
x2
-
lexa 2
a 2 (x 2
-
+
lex
c
2
+y
2
)
or
2
(c
-
-
a 2 )x 2
a 2y
2
-
a
2
[c
2
-
a 2)
or
This
is
usually written simply
where
a
2
—
c
c
2
>
0).
A
2
(since
we must
picture of an ellipse
intersects the horizontal axis
when
y
=
clearly choose a
is
€, so that
*2
—=
a2
(0,
FIGURE
26
-b)
1
1,
shown
x
= +
±a,
>
c,
it
in Figure 26.
follows that
The
ellipse
Graphs
and
it
intersects the vertical axis
The hyperbola
is
when
=
x
0, so
65
that
defined analogously, except that
we
require the
difference
two distances to be constant. Choosing the points (— c, 0) and (c, 0)
once again, and the constant difference as 2a, we obtain, as the condition that
(x, y) be on the hyperbola,
of the
V(x + c y + y - V( x 2
which may be simplified
a
b
= Vc —
2
a
2
,
2
y2
we must
then (x,y)
is
a
picture
is
shown
2
—
=
c
z!
=
but
it
c
if
>
a,
so that a 2
and only
—
c
2
<
0. If
if
i
in Figure 27. It contains
two different orders. The hyperbola
= ± a,
±2a,
1
on the hyperbola
ence between the distances of (x,y) from
so that x
+f =
2
clearly choose
The
2
to
-+
In this case, however,
c)
(— c,
two pieces, because the differ0) and (e, 0) may be taken in
intersects the horizontal axis
never intersects the vertical
when y =
0,
axis.
It is interesting to compare (Figure 28) the hyperbola with a = b = V2
and the graph of the function f(x) = 1/x. The drawings look quite similar,
and the two sets are actually identical, except for a rotation through an angle
of 45° (Problem 21).
Clearly no rotation of the plane will change circles or ellipses into the
graphs of functions. Nevertheless, the study of these important geometric
figures can often be reduced to the study of functions. Ellipses, for example, are
FIGURE
28
made up
of the graphs of two functions,
Vl -
./(*)
=
g (x )
= -b Vl -
b
-a<x<a
(**/**),
and
Of
course, there are
many
(x
2
/a 2 ),
-a<x<a.
other pairs of functions with this same property.
For example, we can take
b
=
fix)
Vl -
(x*/a*),
-b Vl -
(x 2 /a 2 ),
0 <
-a <
x
x
<
<
a
0
and
-b Vl -
=
g(x)
b
We
Vl -
(x
2
/a 2 ),
2
(x /a 2 ),
0<x<a
-a<x<0.
could also choose
=
f( x )
— (x /a ),
Vl — (x /a
2
°
I
b
2
2
x rational,
2
),
x irrational,
—a <
—a <
x
x
<
<
a
a
and
^
=
|
—b
{ b
But
all
V
y/\
\
—
—
(x
(x
2
2
/a
/a 2 ),
x rational,
2
x irrational,
),
—a <
—a <
x
x
<
<
a
a.
these other pairs necessarily involve unreasonable functions
jump around.
A
proof, or even a precise statement of this fact,
is
which
too difficult
Although you have probably already begun to make a distinction
between those functions with reasonable graphs, and those with unreasonable
graphs, you may find it very difficult to state a reasonable definition of reasonat present.
A mathematical definition of this concept is by no means
and a great deal of this book may be viewed as successive attempts
to impose more and more conditions that a ''reasonable
function must
satisfy. As we define some of these conditions, we will take time out to ask if we
have really succeeded in isolating the functions which deserve to be called
able functions.
easy,
5
reasonable.
The answer,
'
unfortunately, will always be "no," or at best, a
qualified "yes."
PROBLEMS
1.
Indicate on a straight line the set of
tions.
you
Also
name each
will also
(i)
|*
(ii)
\x
(hi)
j*
(iv)
|*
2
set,
need the KJ
- 3| < 1.
- 3| < 1.
— a\ < e.
- 1| <
all
x satisfying the following condi-
using the notation for intervals (in some cases
sign).
67
Graphs
—
(vi)
+
1
<
a (give
>
2.
an answer
in terms of a, distinguishing various
x2
cases).
2.
+
+
(vii)
x2
(viii)
(x
Draw
the set of
1
1)(*
-
1)(*
all
-
2)
points
most cases your picture
(*,
will
>
0.
;y)
satisfying the following conditions. (In
be a sizable portion of a plane, not just a
line or curve.)
>
x
(i)
(vi)
>y+
y < x
2
y < x
|*-j|<1.
|*+j|<1.
(vii)
x
+y
*
+y
.
(iv)
.
(v)
(vm)
(ix)
(*
<
an
integer.
is
an
integer.
2
1)
+
Draw
the set of
«
M+
\y\
k|
H
(v)
(vi)
2)
2
y
<
1.
all
points
(x, y)
satisfying the following conditions.
-
2
2
2
2
2
(vii)
(viii)
.
Draw
the set of
x=y
(iii)
x
(iv)
x
Hint:
(a)
-
= 1= i.
1^-11 = 1^-11.
11-^ = 1^-11.
* + ^ = 0.
xy = 0.
- 2* + y = 4.
# = y
(iii)
(i)
(y
< x\
x
(iv)
5.
2
is
(x)
(ii)
4.
b.
2
(iii)
3.
y.
+a
x
(ii)
=
=
all
points
(x,
y) satisfying the following conditions:
2
.
\y\.
sin y.
You
Show
already
know
the answers
when
that the straight line through
of the function f(x)
= m(x —
a)
+
x
{a, b)
b.
and y are interchanged.
with slope
This formula,
m
is
the graph
known
as the
"point-slope form"
more convenient than the equivalent
far
is
expression fix) = mx
(b — ma)
point-slope form that the slope
+
a
/(*)
=
c
When are the
(c)
it is
immediately clear from the
m, and that the value of
is
/
at
is b.
For a ^ c, show that the straight
graph of the function
(b)
;
—
line
through
(x-a)+
and
(a, b)
(c,
d)
is
the
b.
a
graphs of fix)
= mx
+
+
and g(x) = m'x
b
b'
parallel
straight lines?
(a)
For any numbers A, B, and C, with
A and B
line
is
all
(x,
not both
+C=
+
By
y) satisfying Ax
(possibly a vertical one). Hint: First decide
the set of
0
when
show that
a vertical straight
described.
(b)
Show
(a)
can be described as the set of all (x,y) satisfying Ax
Prove that the graphs of the functions
conversely that every straight
are perpendicular
if
line,
including vertical ones,
By
+C=
0.
mn — —1, by computing the squares of the
(Why
lengths of the sides of the triangle in Figure 29.
where the
+
= mx + b,
= nx + c,
fix)
g(x)
case,
0,
a straight line
is
good
lines intersect at the origin, as
is
this special
as the general
case?)
(b)
Prove that the two straight
lines consisting of all
(x,y) satisfying
the conditions
+ By + C = 0,
A'x + B'y + C = 0,
if and only if A A' + BB' =
Ax
are perpendicular
(a)
+yi)
(b)
0.
Prove, using Problem 1-18, that
2
+
(x 2
+y
2
2)
<
vV +
X2 *
+ VyS+yS.
Prove that
V(*8 -*i) 2
+
(yz
~yi) 2 < ^(x 2
~
x,Y
+
Interpret this inequality geometrically
inequality").
When
does
strict
+
(y 2
y/(x 9
(it
is
-
yi
y
-xt)*+(y -yt)\
9
called the "triangle
inequality hold?
Sketch the graphs of the following functions, plotting enough points to
get a good idea of the general appearance. (Part of the
make
a reasonable decision
problem is to
"enough"; the queries posed
thought will often be more valuable
how many
below are meant to show that a little
than hundreds of individual points.)
is
.
Graphs
f(x )
(i)
=
x
(What happens
-\
does the graph
Why
does
f(x)
=
f(x)
=
(iv) /(*)
=
(ii)
for x
0,
and
for large x?
Where
relation to the graph of the identify function?
lie in
only positive x at
suffice to consider
it
near
69
first?)
-
x
X
(iii)
10.
*
2
*2
i.
+
#2
--
-
2
Describe the general features of the graph of /
/
/
/
(i)
(ii)
(iii)
is
if
even.
is
odd.
is
nonnegative.
= f{x
a) for all x (a function with this property
periodic, with period a).
+
(iv) f(x)
11.
Graph the
way to do
is
called
for m = 1, 2, 3, 4. (There is an easy
functions fix) =
using Figure 14. Be sure to remember, however, that
this,
of x when m is even you should also note
difference between the graph when m is
important
that there will be an
means the positive mth root
even and when
12.
m
;
odd.)
is
Graph fix) = |*| and fix) = x\
Graph fix) = |sin*| and fix) =
(a)
(b)
(There is an important
which we cannot yet even describe
you can discover what it is; part (a) is meant to
sin 2 *.
difference between the graphs,
rigorously. See
if
be a clue.)
13.
14.
Describe the graph of g in terms of the graph of /
(i)
gix)
(ii)
gix)
(iii)
gix)
(iv)
gix)
=
=
=
=
(v)
gix)
=/(lA).
(vi)
gix) =fi\x\).
(vii)
gix)
(viii)
£(*)
(ix)
£(*)
(x)
£(*)
Draw
fix)
fix
+
+
c.
c).
cfix)
(
D
easy to
make
ngu sh t he
cases
(It is
i st i
i
a mistake here.)
c
=
0, c
>
0, c
<
0.)
ficx).
= l/M
= max(/, 0).
= min(/, 0).
= max(/, 1).
I
the graph of fix)
Problem
if
1-17.
=
ax 2
+
bx
+
c.
Hint: Use the methods of
15.
The symbol
=
[2]
denotes the largest integer which
[x]
=
2 and [-0.9]
functions
(they are
= -1. Draw
[-1.2]
is
Thus
[2.1]
=
the graph of the following
quite interesting, and several will reappear
all
frequently in other problems).
16.
(iv) f{x)
=
=
=
=
(v)
/(*)
=
[I]-
(vi) fix)
=
4r
(0
/(*)
(ii)
/(*)
(iii)
fix)
Graph
(i)
[•*]•
x
-
[v].
VX -
fx].
M + Vx -
[
X ].
the following functions.
/(*)
=
where
{*},
{*}
is
defined to be the distance from x to the
nearest integer.
(ii)
f(x)
(iii)
fix)
(iv) fix)
(v)
Many
fix)
=
=
=
{2*}.
+
{x}
%{2x).
{4*}.
+i\2x\ +i\4x\.
{x}
functions
may
be described in terms of the decimal expansion of a
number. Although we will not be in a position to describe infinite decimals
rigorously until Chapter 22, your intuitive notion of infinite decimals should
suffice to carry you through the following problem, and others which occur
before Chapter 22. There is one ambiguity about infinite decimals which
must be eliminated: Every decimal ending in a string of 9's is equal to another
ending in a string of 0's (e.g., 1.23999
= 1.24000
.). We will always
use the one ending in 9's
.
*17.
.
.
.
.
Describe as best you can the graphs of the following functions (a comis usually out of the question).
plete picture
(0
fix)
(ii)
fix)
(iii)
fix)
—
=
=
the 2nd
the
number
(iv) fix)
—
finite,
(
v)
number in the decimal expansion of x.
number in the decimal expansion of x.
number of 7 s in the decimal expansion
the 1st
0
?
is finite,
the
if
and
1
of x
if
this
and 0 otherwise.
number
of 7's in the decimal expansion of x
is
otherwise.
= the number obtained by replacing all digits in the decimal
expansion of x which come after the first 7 (if any) by 0.
/(*)
(vi) fix)
=
0
if 1 first
if 1
never appears in the decimal expansion of
appears in the
72th
place.
x,
and
n
71
Graphs
*18.
Let
x irrational
x
=
- rational in lowest terms.
q
(A number p/q
factor,
and
q
sprinkle points
bers with q
19.
(a)
=
is
>
in lowest
0).
Draw
terms Hp and q are integers with no common
the graph of / as well as you can (don't
randomly on the paper; consider
2,
=
=
then those with q
first
the rational
num-
3, etc.).
2
x 2 are the ones of the form (x, x ).
points on the graph of f(x)
Prove that each such point is equidistant from the point (0, i) and
The
= -|. (See Figure 30.)
Given a point P = (a, fi) and a horizontal line L, the graph of
= Y) show that the set of all points (*, y) equidistant from P and
g( x )
ax 2 + bx + c.
L is the graph of a function of the form f(x)
Show that the square of the distance from (c, d) to (x, mx) is
the graph of g(x)
(b)
*20.
(a)
x 2 (m 2
+
1)
+
to find the
Using Problem 1-17
that the distance from
\cm
(b)
*21.
(a)
x(-2md -
(c,
2c)
+
minimum
d) to the
2
.
of these numbers,
graph of /(*)
- d\/Vm +
2
+c
d2
= mx
show
is
1.
Find the distance from (c, d) to the graph of /(*) = mx +
(Reduce this case to part (a).)
Using Problem 20, show that the numbers x' and / indicated
Figure 31 are given by
1
b.
in
1
i
(b)
Show
that the set of
same
as the set of all
all (x,y)
(x,
with (x'/V2)*
y) with xy
=
1
- (vV^) =
2
1 is
the
CHAPTER
LIMITS
The concept
of a limit
one in
difficult
we
all
is
surely the most important,
of calculus.
The
and probably the most
goal of this chapter
is
the definition of
once more, going to begin with a provisional definition;
what we shall define is not the word "limit" but the notion of a function
limits,
but
approaching a
PROVISIONAL DEFINITION
The
are,
limit.
function / approaches the limit / near a, if we can make fix) as close as
/ by requiring that x be sufficiently close to, but unequal to, a.
we
like to
Of the
six functions
graphed
Notice that although g(a)
it is still
true that g
is
in
Figure
1,
only the
not defined, and h{a)
and h approach
/
near
a.
is
first
three approach
/
at a.
wrong way,"
because we explicitly
defined "the
This
is
ruled out, in our definition, the necessity of ever considering the value of the
function at a
but unequal
—
it is
to a.
only necessary that f(x) should be close to / for x close to a,
are simply not interested in the value of f(a). or even in
We
the question of whether f(a)
FIGURE
is
defined.
1
One convenient way of picturing the assertion that / approaches / near a is
provided by a method of drawing functions that was not mentioned in
Chapter 4. In this method, we draw two straight lines, each representing R,
and arrows from a point x in one, to f(x) in the other. Figure 2 illustrates such
a picture for two different functions.
72
Limits
73
consider a function / whose drawing looks like Figure 3. Suppose we
ask that /(*) be close to /, say within the open interval B which has been
drawn in Figure 3. This can be guaranteed if we consider only the numbers x
Now
in the interval
A
interval which
will
3. (In this diagram we have chosen the largest
work; any smaller interval containing a could have been
of Figure
chosen instead.) If we choose a smaller interval B' (Figure 4) we will, usually,
have to choose a smaller A\ but no matter how small we choose the open
interval B, there is always supposed to be some open interval A which works.
A similar pictorial interpretation is possible in terms of the graph of/, but
in this case the interval
B
must be drawn on the
vertical axis,
B when x is
and the
A means
set
A
that the
on the horizontal axis. The fact that f(x)
part of the graph lying over A is contained in the region which is bounded
by the horizontal lines through the end points of B; compare Figure 5(a),
where a valid interval A has been chosen, with Figure 5(b), where A is too
is
in
in
large.
In order to apply our definition to a particular function, let us consider
= x sin 1/x (Figure 6). Despite the erratic behavior of this function near
0 it is clear, at least intuitively, that / approaches 0 near 0. and it is certainly to be hoped that our definition will allow us to reach the same con-
/(*)
clusion. In the case
we must
ask
that x be
get x sin
if
we
we can
are considering, both a
=
get /(*)
/x within
^
—
1
-
succinctly,
\x
sin l/x\
sin -
x
0.
To
be
.11
< —
<
10
more
/
of the definition are 0, so
specific,
<
<
x sin x
to1,
Now
if
suppose
means we want
of 0. This
—
or,
^
sufficiently close to 0, but
1
and
x sin 1/x as close to 0 as desired
>
10
this
is
easy. Since
for all x
^
0,
we require
we wish to
74
Foundations
we have
s f(x)
=
* sin -
x sin
<
1*1,
*
for all x
0.
X
This means that
4
/
FIGURE
/
x sin 1/x
is
\
\
\
is
if |*|
within
1
y^
< xo an d
^
0,
then
—
take any positive
number
£
we can make
|*
<
sin 1/*|
y^-; in
other words,
within yo of 0, but ^ 0. There
is just as easy to guarantee that
is
nothing special about the number y^; it
— 0| < j-q-q simply require that |*|
\f(x)
|*| < £, and * ^ 0.
For the function /(*)
approaches 0 near 0. If,
6
x
of 0 provided that x
< y^,
\f(x)
—
but x
0|
<
£
^
0.
In
fact, if
we
simply by requiring
that
=
x 2 sin 1/x (Figure 7)
for
it
seems even clearer that /
example, we want
x £ sin *
then we certainly need only require that
that |* 2 < yJ~o and consequently
<
|*|
1
io
< y^ and
*
9± 0,
since this implies
|
FIGURE
7
(We could do even
better,
and allow
|*|
< 1 /Vio and
*
5^ 0,
but there
particular virtue in being as economical as possible.) In general,
f(x)
=
y/\x\ sin
ensure that
<
we need only
FIGURE
e,
require that
8
|*|
<
£
and
*
0,
if £
>
is
no
0, to
.
Limits
FIGURE
75
9
provided that £ < 1 If we are given an £ which is greater than 1 (it might be,
even though it is "small" e's which are of interest), then it does not suffice to
.
require that
I
<
but
e,
it
certainly suffices to require that
a third example^consider the function /(*)
As
iin
|*|
In order to
make
|*|
if £
<
or that
1,
<
sin l/x\
<
|*|
1
<
and x
£
5*
£
2
0
=
V|*|
we can
require that
and
^
if £
>
x
1
<
1
and x
5* 0.
0,
(the algebra
Finally, let us consider the function f(x)
|*|
sin \/x (Figure 8)-
=
sin 1/x
is left
to you).
(Figure 9). For this
/approaches 0 near 0. This amounts to saying that it is
£ > 0 that we can get \f(x) — 0| < £ by choosing
x sufficiently small, and ^ 0. To show this we simply have to find one £ > 0
for which the condition \f(x) - 0| < £ cannot be guaranteed, no matter how
function
it is
false that
not true for every
small
we
require
number
|*|
to be. In fact, £
= % will
do:
it is
impossible to ensure that
|/(x)| < i no matter how small we require |*| to be; for if A is any interval containing 0, there is some number * = 1/(90 + 360rc) which is in this interval,
and for this x we have f(x) = 1
This same argument can be used (Figure 10) to show that / does not
approach any number near 0. To show this we must again find, for any particular number /, some number £ > 0 so that \f(x) — /| < £ is not true, no
matter how small x is required to be. The choice £ = i works for any number
be, we cannot ensure that
/; that is, no matter how small we require |*| to
_
some
and
/|
#1
also
=
<
1
The
/(90
+
reason
is,
that for any interval
A
containing 0 there
360«) in this interval, so that
some x 2 - 1/(270
+
360m)
/(**)
in this interval, so that
= -I-
is
76
Foundations
But the interval from
total length
is
only
1
- £ to / + * cannot
/
-
|1
i
x,
I
0,
/(*)
no matter what
x rational
x irrational
/|
< iand
The phenomenon
we consider
also
=
exhibited by /(*)
the function
its
-
|-1
/[
<
no matter what a
sin 1/x
many
near 0 can occur in
x irrational
/(*)
FIGURE
since
1,
/ is.
ways. If
then,
-1 and
contain both
we cannot have
so
;
-(?:
x rational,
/ does not approach any number / near a. In fact
< \ n o matter how close we bring x to a, because
m any interval around a there are numbers x with /(*) = 0, and also numbers
x with f(x) = 1, so that we would need
|0 - l\ < \ and also |1 - l\ <
An amusing variation on this behavior is presented by the function \.
shown
we cannot make
11
is,
-
\f{x)
l\
^
in Figure
1 1
=
f(x)
0,
I
c_±J
m
c
f(X)
=
The behavior of this function is "opposite" to that
of g(x)
approaches 0 at 0 but does not approach any number
at a, if a
you should have no difficulty convincing
c
S
—
e
)
As a contrast
.
a
pathological,
FIGURE
* ^tional
# irrational.
*'
|
If/(*)
12
=
c,
we
=
sin 1/x;
*
0.
it
By now
yourself that this is true.
to the functions considered so far,
which have been quite
will now examine some of the simplest
functions.
then /approaches
c
near
a,
for every
-
number
a.
In
that I/O)
c\ < s one does not need to
restrict x to be near a at
tion is automatically satisfied (Figure
fact, to
all;
ensure
the condi-
12).
As a
slight variation, let
/ be the function shown
x
x
If a
1
^
/to = i,*> o
>
0,
then /approaches
1
near
certainly suffices to require that
\x
-
b
/(*)
= -1,
x < 0
so
.
-1
that/W =
that |/0)
you
may
-
1.
Similarly,
(-1)|
easily check,
The
FIGURE
13
function /(*)
ensure that |/0) -
The
<
function
a,
<
0,
a
<C
x
0
<
x,
a\
x
is
<
£
then / approaches
require that
|*
2
^r
easily dealt with. Clearly
/
we just have
requires a
we must
decide
-
k 2
a2
\
-1
£|
near
<
<
1
e
it
£.
b:
to ensure
Finally, as
0.
|*
-
a
\
<
more work. To show
to ensure that
<
1
approaches a near
to require that
little
how
—
a
/ does not approach any number near
=
/0) =
proaches a 2 near
if *
e it suffices to
0
0.
indeed, to ensure that \f(x)
a\ < a, since this implies
a:
or
<
>
in Figure 13:
a: to
£.
that
/ ap-
77
Limits
Factoring looks
most promising procedure: we want
like the
—
|*
a\
+
\x
'
a\
<
£.
Obviously the factor |*
a\ is the one that will cause trouble. On the other
hand, there is no need to make \x
a\ particularly small; as long as we know
some bound on the values of \x + a\ we will be in good shape. For example,
if \x
a\ < 1,000,000, then we will just need to require that |* — a\ <
e/1 ,000,000. Therefore, to begin with, let us require that \x — a\ < 1 (any
+
+
+
number other than
positive
ensure that x
As a matter of
would do
1
just as well)
not too large, and consequently that
is
presumably
;
+
\x
a\ is
this will
not too large.
Problem 1-12 shows that
fact,
-
\x\
\a\
<\x -
<
a\
1.
so
H< +
i
kl,
and consequently
+
\x
Now we
a\
<
+
\x\
<
\a\
+
2\a\
need only the additional requirement that
1.
—
\x
a\
<
z/{2\a\
+
1).
In other words,
if \x
-
< min
a\
(
l r
\
Naturally, min(l, e/(2\a\
same
Precisely the
a z near
if \x
In
a.
-
a\
1
+
+
1)) will just
sort of trick will
\
then
\x
2
-
<
2
a
\
s.
1/
+
be e/(2\a\
show that
if f(x)
1) for
=
small
e.
x s then / approaches
,
fact,
< min
( 1,
\
'
(1
+
—
~4
\a\r+
\a\(\
The proof
of this assertion will
from: If
—
\x
—
2\a\
a\
\x
<
2
then
1,
+
ax
+
a
|*|
2
\
<
<
<
\a\)
+
nz)
9
then
3
l^
a *\
<
e
'
\a\*J
show where the weird denominator comes
+ 1, and consequently
\a\
\x\
(1
rr,
+
+
+
+ \a\y+ \a\(l +
2
\a\
'
\a\*
\x\
\a\)
+
\a\\'
Therefore
_
fl 8|
=
\
x
< n
(1
= £.
_
.
fl
|
^2
+
ax
4-1\a\)
h 2 -Li hi
+
+
+
a2
\
+
+ iui^
\a\)
\a\
2
*
[(1
+
^ l)2
+
W(1
+
M)
+ W]
has now come to point out that of the many demonstrations about
which we have given, not one has been a real proof. The fault lies not
with our reasoning, but with our definition. If our provisional definition of
a function was open to criticism, our provisional definition of approaching a
The time
limits
even more vulnerable. This definition is simply not sufficiently precise
is hardly clear how one "makes" f(x) close to / (whatever "close" means) by "requiring" x to be sufficiently close to a (however
limit
is
to be used in proofs. It
:
78
Foundations
close "sufficiently" close
you
definition
may
is
supposed to be). Despite the criticisms of our
hope you do) that our arguments were
feel (I certainly
nevertheless quite convincing. In order to present any sort of argument at all,
practically forced to invent the real definition. It is possible to
we have been
arrive at this definition in several steps, each one clarifying
phrase which
some obscure
remains. Let us begin, once again, with the provisional
still
definition:
The function / approaches the limit / near a, if we can make/(*) as close
we like to / by requiring that x be sufficiently close to, but unequal to, a.
as
The very first change which we made in
/0)
close to
The
/
meant making
function
The
was
this definition
and
small,
making
to note that
and
similarly for x
a:
/ approaches the limit / near a, if we can make \f(x) - /|
like by requiring that \x - a\ be sufficiently small, and
a.
second,
small as
i\
we
as small as
x 9^
-
\f(x)
we
more crucial, change was
means making \f(x) -
like"
making
to note that
<
l\
any
£ for
£
>
-
|/(*)
"as
l\
0 that happens to
be given us
The function / approaches the limit / near a, if for every number £ > 0
we can make \f{x) — /| < e by requiring that \x - a\ be sufficiently
small,
and x
^
a.
There is a common pattern to
have given. For each number
with the property that
8 say,
For the function /(*)
=
>
if
x
x sin \Jx
we
the demonstrations about limits which
all
£
(
0
we found some
a
and
with a
«
\x
-
0, /
<
a\
=
other positive number,
<5,
then
0), the
—
\f(x)
number
<$
/|
<
e.
was just
number £; for f(x) = V\ x sin 1/x, it was £ if £ < 1, and it was 1 if
> 1; for/<» = x 2 it was the minimum of 1 and e/(2\a\ + 1). In general, it
may not be at all clear how to find the number given £, but it is the condition \x — a\ < 8 which expresses how small " sufficiently" small must be:
the
2
\
£
<5,
The
some
function / approaches the limit / near a,
5 > 0 such that, for all *, if \x - a\ < 6
<
£.
is
practically the definition
This
change, noting that "|*
"0
DEFINITION
<
The
is
\x
-
a\
<
-
a\
<
we
8
will adopt.
and x
^
for
>
0 there
is
then
-
l\
every £
^
a,
\f(x)
We will make only one
trivial
a" can just as well be expressed
5."
function / approaches the limit I near a
8 > 0 such that, for all *, if 0 < |*
some
if
and x
—
means
a\
<
:
8,
for every £
then f(x)
\
>
—
0 there
l\
<
£.
Limits
79
so important (everything we do from now on depends on it)
any further without knowing it is hopeless. If necessary
memorize it, like a poem! That, at least, is better than stating it incorrectly;
A good exercise in
if you do this you are doomed to give incorrect proofs.
This definition
is
that proceeding
giving correct proofs
approaching
tions
down
all
to
review every fact already demonstrated about funcThis requires writing
what you are proving, but not much more
work has been done already. When proving that / does not
the correct definition of
the algebraic
approach
If
is
limits, giving real proofs of each.
it is
/
be sure to negate the definition correctly:
at a,
not true that
for every £
>
0 there
then |/0)
-
/|
<
is
some
6
>
0 such that, for
all x, if
0
<
\x
-
a\
<
5,
e,
then
there
0
<
some £
is
\x
-
<
a\
>
d
0 such that for
but not
\f(x)
>
every 5
-
<
/|
0 there
is
some x which satisfies
£.
= sin 1/x does not approach 0 near 0, we
> 0 there is some x with 0 < |* — 0|
note that for every
36(k),
namely, an x of the form 1/(90
1 /x - 0| < i
Thus, to show that the function f(x)
consider £
<
8
= \ and
but not
where n
As an
limit,
is
|sin
<5
—
so large that 1/(90
+
+
360n)
<
5.
illustration of the use of the definition of a function
we have
reserved the function
shown
approaching a
example,
in Figure 14, a standard
but one of the most complicated:
fix)
(Recall that p/q
factor
and
q
>
is
0,
x irrational, 0
Vq,
x
= p/q
in lowest terms
<
<
x
1
in lowest terms, 0
if
<
x
<
1.
p and q are integers with no
0.)
x irrational
(0,
h
FIGURE
14
x
=£
in lowest
terms
common
80
Foundations
For any number
prove
that
be
/n
1
with 0
a,
consider any
this,
<
Notice that the
e.
< 1 the function / approaches 0 at a. To
> 0. Let n be a natural number so large
only numbers x for which \f(x) — 0| < 6 could
<
a
number
,
£
false are:
112131234
?
;
2
(If a
;
3
4
3
"
1
9
;
'
'
-
n
1
5' 5* s' 5 ;
4'*
n
*
'
'
n
rational, then a might be one of these numbers.) However many of
numbers there may be, there are, at any rate, only finitely many. There-
is
these
numbers, one
one p/q among these numbers.
fore, of all these
is
closest to a\ that
is,
—
\p/q
smallest for
a\ is
one of these numbers, then
consider only the values \p/q — a\ for p/q ?± a.) This closest distance may be
chosen as the 5. For if 0 < \x — a\ < 8, then x is not one of
(If a
happens
and
-
n
1
to be
1
—
0| < £ is true. This completes the proof. Note that our
which works for a given £ is completely adequate there
is no reason why we must give a formula for 8 in terms of £.
Armed with our definition, we are now prepared to prove our first theorem;
you have probably assumed the result all along, which is a very reasonable
thing to do. This theorem is really a test case for our definition: if the theorem
could not be proved, our definition would be useless.
therefore \f(x)
description of the
THEOREM
l
A
function cannot approach two different limits near
approaches
PROOF
—
<5
Since this
/
is
near
our
a,
m
and / approaches
theorem about
first
near
limits
then
a,
it
In other words,
a.
if
/
m.
/ ==
will certainly
be necessary to
translate the hypotheses according to the definition.
Since / approaches
ber 8 1
>
also
near
a,
we know
that for
any
£
-
/|
>
0 there
is
some num-
0 such that, for all x,
if
We
/
0<
|*
-
a|
<
8h
then
know, since / approaches m near
<
e.
that there
is
|/(*)
a,
some
82
>
0 such
that, for all x,
if
We
0
<
|*
—
a|
<
have had to use two numbers,
8 2)
8i
then |/0)
and
82
,
-
m\
<
since there
e.
no guarantee that
is
the 8 which works in one definition will work in the other. But, in fact,
now
£
>
0 there
\f(x)
-
/|
easy to conclude that for any
is
some
8
>
all X,
if
0
<
\x
-
we simply choose
a\
8
=
<
8,
then
min(6i, S 2 ).
<
6
and |/0)
-
it is
0 such that, for
m\
<
e;
~
Limits
To
complete the proof we just have to pick a particular
s
>
81
0 for which the
two conditions
-
I/O)
cannot both hold,
/
^
m, so that
there
is
a
-
|/
>
5
^
>
if /
m\
0
<
0,
This implies that for 0
-m =
\
-
|*
|/
<
-
and
6
<
m\
S
The proper choice is suggested by Figure 15. If
we can choose \l - m\/2 as our e. It follows that
m.
0 such that, for
if
|/
<
/|
a\
\x
all x,
<
-
b,
a\
then |/0)
"
and |/0)
-
<
-H
—j—
\l
<
/|
m|
<
^
"
2
we have
<5
- f(x) + fix) -
m\
<
\l
=
1/
- /0)l +
-
«l
2
2
-
l/to
m|,
a contradiction. |
The number
/
which / approaches near
a
is
denoted by lim fix) (read: the
x-*a
z—»o
approaches a). This definition is possible only because of
a: appro;
limit
mit of fix)
f(x) as *
ensures that lim }{x) never has to stand for two different
Theorem
"heorem 1, which ensui
numbers. The equation
lim/0) =
I
x-*a
has exactly the same meaning as the phrase
/ approaches
The
possibility
lim fix)
=
/
is
/
near
a.
remains that / does not approach / near a, for any /, so that
saying
false for every number 7. This is usually expressed by
still
x—*a
that "lim fix) does not exist."
x,
x—>a
letter
Notice that our new notation introduces an extra, utterly irrelevant
already
not
does
which
letter
which could be replaced by /, y, or any other
appear— the symbols
lim/O),
Hm/(0,
lim/0),
x-+a
t-*a
V-*a
has
denote precisely the same number, which depends on / and a, and
at
anything
denote
not
do
fact,
in
letters,
(these
or
f,
with
do
to
nothing
*,
y
notation,
lim/, but this
all). A more logical symbol would be something like
all
a
despite
its
brevity,
tried to use
it.
The
is
so infuriatingly rigid that almost
notation lim f(x)
is
much more
no one has
seriously
useful because a function
/ often has no simple name, even though it might be possible
by a simple formula involving x. Thus, the short symbol
+ sin x)
2
lim
to express /(*)
(.v
x-*a
could be paraphrased only by the awkward expression
lim
/,
=
where f(x)
x2
+
sin x.
a
Another advantage of the standard symbolism
is
illustrated
by the expres-
sions
lim x
+
J
+
t\
3
,
ar—>a
lim x
The
first
means the number which / approaches near
f(x)
x
+
t\
/(*)
should have
=
little difficulty
x
+
when
for all x;
means the number which / approaches near
the second
You
=
a
t\
for all
(especially
if
a
when
t.
you consult Theorem
2)
proving
that
lim x
+
t
+
t
3
=
a
+
3
=
x
+a
t\
x~*a
lim x
z
.
t->a
These examples
flexibility.
In
illustrate the
fact,
main advantage
the notation lim f(x)
is
of our notation, which
so flexible that there
is
is its
some danger
x—*a
what it really means. Here is a simple exercise in the use of this
notation, which will be important later: first interpret precisely, and then
of forgetting
prove the equality of the expressions
lim f(x)
and
lim f(a
x-*a
An
make
important part of
it
easy to find
of inequalities
this
many
+
h).
h-*0
chapter
is
the proof of a theorem which will
The proof depends upon certain properties
values, hardly surprising when one considers the
limits.
and absolute
Although these
have already been stated in Problems
importance they will be presented once
again, in the form of a lemma (a lemma is an auxiliary theorem, a result that
justifies its existence only by virtue of its prominent role in the proof of
another theorem) The lemma says, roughly, that if x is close to * 0 and y is
definition of limit.
1-19, 1-20,
and
facts
1-21, because of their
.
close to ^o, then x
,
+y
+
be close to x 0
yo, and xy will be close to x 0yo
and 1/y will be close to l/y 0 This intuitive statement is much easier to remember than the precise estimates of the lemma, and it is not unreasonable to read
will
3
.
the proof of
Theorem
2
first,
in order to see just
how
these estimates are used.
Limits
LEMMA
(1)
If
1*
-
<
*o|
|
and
- y0 <
\y
|>
\
then
\(x+y)(2)
<
+yo)\
(x 0
e.
If
'*
K min
* oi
"
0' 27i^f+T))
L
b ~ y° K
27M M)'
and
l
then
—
\xy
(3)
If y 0
^
proof
(i)
-j, 0
(2)
|
<
min^— —j-y
-
*o)
»
0 and
9±
|o
S.
0 and
|j
then y
<
Jfo^ol
+ j) -
Since
|*
—
Oo
*0
|
+^
<
0 )|
=
10
<
I*
-
+
+
*„|
-
(y
yo)\
- yo\ <
\y
\
+\=
e.
we have
1
—
Ul
|*o|
<
\x
—
*o|
<
1,
so that
W
<
1
+
|*o|-
Thus
|ry
-
* 0 )>o|
= 1*0 - yo) +yo(x < \x\ \y - yo\ + \yo\
•
<
(i
< ~
+ M)
+
2
(3)
We
=
~
2(W +
|jy|.
>
I*
-
*o|
+
M
1)
2(W + D
s.
2
have
W-M<
so
*o)|
i^ol/2.
In particular,
0,
jy
\y
- Jo| <
^
and
±<-L.
\y\
\yo\
Thus
i
y
_1
y<>
\y*-y\
M
*
M<
A
l>'o|
.
_L
|>o|
=
.
2
e
.
|
83
84
Foundations
THEOREM
=
lim /(*)
If
2
and lim
/
x—* a
= m
g(x)
then
9
x—*a
(1)
lim (f
(2)
lim"
+ g)(x) =l + m;
-l-m.
(f-g)(x)
x—+a
Moreover,
m ^
if
then
0,
lim (-)
(3)
PROOF
The
means
hypothesis
=
(x)
-•
m
\g/
x->a
>
that for every e
>
0 there are 5 h 8 2
0 such that, for
all x,
<\x < |* —
ifO
and
This means
>
52
<5i,
0
if
i(0
and
Now
0
<
let
\x
if
fi
a\
<5
a\
<
—
<z|
<
If 0
<
|#
2
+ m)\
To
If £
<
-
prove
>
This proves
£.
0
if
<
|*
-
l\
m\
< e,
< £.
number) that there are
then
\f(x)
-
/|
<5
then
\g(x)
—
m\
2,
<
a\
then 0
8,
<
<
<
"
lemma
H
<
Again
let 6
=
|/W So,
0
if
/|
<
|*
<
then
S l9
>
0
<
if
(l,
<
—
is
a 8
-
<
8,
|*
i
a\
-
|*
—
<z|
^^y)
0 there
i
a\
<
Si
((/
and
+ g)(x) —
~
>
,
I
°
£
< min
l\
<
£,
<
m|
(l,
2(|m|
|,
and
W
this
0 such that, for
i
/
\
then \g(x)
-
i
m\
—-
+
then
8,
and
<
m\
lemma.
all x,
-
|/(*)
then
Sa,
min(6i, § 2 ), If 0
if £
-
similarly, after consulting part (2) of the
by the lemma, \(f'g)(x)
Finally,
\x
this implies that
0 such that, for
- a\<
< min
-•
|
\
and
|,
(1).
>
a\
8h
< | and
/|
we proceed
(2)
0 there are 5 h 8 2
—
both
are true. But by part (1) of the
(I
\f(x)
\g(x)
-
|*
true, so
2
I/O)
then
then
2,
also a positive
is
<\x -
<
0
6i,
<5
all x,
= min(5i, ).
are both
<
5
—
e/2
(since, after all,
0 such that, for
<
<
a\
a\
<
-
m\
<
proves
(2).
all x,
mm
.
/\m\
I
^
£|m| 2 \
2/
\
1)/
to part (3) of the
But according
(l/g)(x) makes
lemma
85
^
0, so
that g(x)
first,
and second that
sense,
W
This proves
means,
this
Limits
<
e.
rn
(3). |
Using Theorem 2 we can prove,
™
trivially,
+ 7x =
"
x* +
5
x*
1
such facts as
+ la
« + r
b
az
2
without going through the laborious process of finding a
must begin with
=
7,
1
=
1,
lim *
=
a,
lim 7
5,
given an
6.
We
x-*a
lim
x-+a
If we want to find the 5, however, the
proof of Theorem 2 amounts to a prescription for doing this. Suppose, to take
a simpler example, that we want to find a 5 such that, for all x,
but these are easy to prove directly.
0
if
<
|*
-
5%
>
0 such that, for
and
Since
how
we have
to
do
5,
then
Theorem
Consulting the proof of
and
<
a\
\x
2
+x-
(a
2
+
a)
I
<
e.
we must
2(1),
we
<
8i,
then
<
<5
then
|*
-
a|
2
=
a2
and lim x -
see that
if
0
<
|*
-
if
0
<
\x
-
a\
a\
2,
already given proofs that lim
a:
2
I*
-
2
<z
|
<
<
•- mto
|-
1
( '2(HTi)/
Thus we can take
«
min(5 i?
find 8i
|
this:
8
first
all x,
<5
2)
= min C mm (
yi,
I
—-—
j
,--
y,
a,
we know
If a
the
0,
same method can be used
<
0
if
The proof of Theorem
find a 8
>
-
|*
<
0
-
|*
<5,
>
a 8
then
<
^2
v2
0 such that, for
all x,
6.
2(3) shows that the second condition will follow
0 such that, for
if
<
a\
to find
<
a\
if
we
all x,
then
8,
\x
2
-
a
2
\
< min
^£
Thus we can take
6
= min
2(W
+
1)
Naturally, these complicated expressions for 8 can be simplified considerably,
after they
One
have been derived.
technical detail in the proof of
In order for
Km f(x)
to be defined
it is,
Theorem 2 deserves some discussion.
as we know, not necessary for / to be
x—*a
defined at a nor
y
there must be
<
8;
necessary for / to be defined at
is it
some
>
8
0 such that/(*)
points x
all
^
a.
defined for x satisfying 0
is
However,
<
\x
—
a\
otherwise the clause
"if 0
<\x -
<
a\
8,
then
-
\f(x)
<
l\
s"
would make no sense at all, since the symbol f(x) would make no sense for
some #'s. If / and g are two functions for which the definition makes sense, it is
easy to see that the same is true for / + g and / g. But this is not so clear for
1/g, since 1/g is undefined for x with g(x) = 0. However, this fact gets
*
Theorem
established in the proof of
2(3).
There are times when we would like to speak of the limit which / approaches
at a, even though there is no 8 > 0 such that f(x) is defined for x satisfying
0 < \x — a\ < 8. For example, we want to distinguish the behavior of the two
functions shown in Figure 16, even though they are not defined for numbers
less than a. For the function of Figure 16(a) we write
=
lim f{x)
or
I
x—*a +
(The symbols on the
left
lim f(x)
=
/.
x la
are read: the limit of f(x) as x approaches a from
above). These "limits from above" are obviously closely related to ordinary
limits,
e
>
and the
0 there
is
definition
a 8
>
if
(The condition "0
x
>
very similar: lim f(x)
is
0 such that, for
0
<
<
x
x
—
a
a
<
<
8"
all
8,
/
means
that for every
a*,
then f(x)
\
is
=
-
l\
<
£.
equivalent to "0
<
\x
-
a\
<
8
and
a.")
"Limits from below"
(Figure
17)
are
defined similarly:
lim f(x)
—
I
87
Limits
=
(or lim f(x)
x
T
I)
means that
>
for every £
0 there
a 5
is
>
0 such that, for
a
all x,
if
0
<
-
a
<
x
8,
then
<
l\
S.
and below even if / is
Thus, for the function /
quite possible to consider limits from above
It is
defined for numbers both greater and
of Figure 13,
less
than
a.
we have
=
lim f(x)
It is
-
\f(x)
and
1
an easy exercise (Problem 27)
= — 1.
lim f(x)
show that lim
to
f(x) exists
and only
if
x—*a
if
lim f(x) and lim f{x) both exist and are equal.
x—>a +
x—*a~
Like the definitions of limits from above and below, which have been
smuggled into the text informally there are other modifications of the limit
concept which will be found useful. In Chapter 4 it was claimed that if x is
;
large, then sin \/x
is
This assertion
close to 0.
lim sin l/#
FIGURE
usually written
0.
18
The symbol lim
x—*
becomes
finity.
=
is
f(x)
read "the limit of f(x) as x approaches
is
infinite,"
and a
limit of the
form lim f(x)
often called a limit at in-
is
Figure 18 illustrates a general situation where lim f(x)
x—
*
lim f(x)
=
I
or "as x
<*>
»
means that
for
every
e
>
0 there
is
a
=
/.
Formally,
00
number
N such
that, for
all x,
if
The analogy with
x
> N,
then f(x)
\
l\
<
S.
the definition of ordinary limits should be clear: whereas
< \x — a\ < 5" expresses the fact
> N" expresses the fact that x is large.
the condition "0
condition "x
-
that x
is
close to a, the
We have spent so little time on limits from above and below, and at infinity,
because the general philosophy behind the definitions should be clear
if
you
understand the definition of ordinary limits (which are by far the most important) Many exercises on these definitions are provided in the Problems, which
.
also contain several other types of limits
which are occasionally
useful.
PROBLEMS
Find the following limits. (These limits all follow, after some algebraic
manipulations, from the various parts of Theorem 2 be sure you know
which ones are used in each case, but don't bother listing them.)
;
x2
(i)
lim
(ii)
lim
x*
x
x—>2
—
1
-8
-2
3
(iii)
lim
— 2
— v"
lim
x->y x — y
xn - y n
lim
y-+x x — y
X
x->3
xn
(iv)
(v)
,
(vi)
2.
:
+ h - Va
'
lim
In each of the following cases, find a 8 such that
# satisfying 0
<
\x
(i)
/(*)
=
(ii)
/(*)
= -;a =
*4
=
/
;
—
a\
a4
<
|/(.v)
—
l\
<
£ for all
8.
.
1,/
=
1.
x
+
/(*)
-
*4
(iv) /(*)
=
—+^
(iii)
sin 2 x
1
(v)
=
1, /
=
;
o,
2.
/
=
0.
= VH; = 0 / = 0.
= V*; = 1, / = 1.
f(x)
fl
(vi) /(*)
3.
=
-; a
5
fl
For each of the functions in Problem 4-15, decide
for
which numbers a
the limit lim fix) exists.
x—>a
*4.
5.
(a)
Do
(b)
Same problem,
the
same
for
each of the functions
if
we
in
Problem
4-17.
use infinite decimals ending in a string of 0's
instead of those ending in a string of 9's.
Suppose the functions / and g have the following property:
and
if
0<
|*
-
2|
<
sin 2
if
0
<
|*
-
2\
<
e
e
>
0 find a 8
>
<
\x
-
then
For each
ii)
for all s
all x,
if
0
2|
<
5,
2
,
(^j
+
6,
then \g(x)
then
|/(*)
-
<
4|
0 such that, for
|/(x)
+
-
2\
6.
all x,
—
6|
<
e.
<
6,
>
0
Limits
<
0
if
(ii)
-
\x
<
2|
1
(iii)
if
0
<
|*
-
2\
<
5,
then
(iv)
if
0
<
|*
-
2|
<
j,
then
_
/to
7.
Give an example of a function /
false:
If \f(x)
when
0
(a)
<
|*
- < e when
- a\ < 8/2.
i
lim f(x)
If
and lim
x—>a
e.
<
E.
which the following assertion
for
-
|.v
do not
g(x)
<
e.
2
<
0
l\
<
8|
1
£to
6.
-
then f(x)g(x)
<5,
89
<
a|
exist,
then
5,
-
|/(*)
<
+ g(x)]
can lim[/(x)
x—+a
/|
is
e/2
or
x—>a
lim f(x)g(x) exist?
x—>a
(b)
If
lim f(x) exists and lim [/(#)
x—+ a
(c)
If
x—
+ g(x)]
exists,
g(x) exist?
x—+ a
lim /to exists and lim g(x) does not
x—*a
must lim
a
exist,
can lim
x—+o
+ g(x)]
[f(x)
x—*a
exist?
(d)
If
lim f(x) exists and lim f(x)g(x)
x—
x—
o
exists,
does
follow that lim g(x)
it
a
i
—a
exists?
8.
=
Prove that lim f(x)
lim f(a
x—>a
+
(This
h).
mainly an exercise in under-
is
A—*0
standing what the terms mean.)
9.
(a)
=
Prove that lim /(*)
x— a
and only
/ if
if
lim[/(^)
-
x— a
>
=
/]
0.
(First see
why
the assertion is obvious; then provide a rigorous proof. In this
chapter most problems which ask for proofs should be treated in
same way.)
Prove that lim f(x)
the
=
(c)
lim f(x).
x- a ->0
Prove that lim /(*) = lim /(* 3 ).
(d)
Give an example where lim f(x 2 )
(b)
x—*a
x—>0
x—»0
exists,
x->0
10.
but lim f(x) does not.
x^O
Suppose there is a 8 > 0 such that f(x) = g(x) when 0 < \x — a\ < 8.
Prove that lim f(x) = lim g(x). In other words, lim f(x) depends only
x—
>
x—
a
a
*
on the values of /(x)
for x
near
a
—
x—+ a
this fact
is
often expressed
that limits are a "local property." (It will clearly help to use
other
11.
(a)
letter,
instead of
Suppose that
f(x)
in the definition of limits.)
8,
<
by saying
or some
8',
g(x) for
all x.
Prove that lim f(x)
x—ki
<
lim g(x),
x—*a
provided that these limits exist.
(b) How can the hypotheses be weakened?
(c)
If f(x)
<
g(x) for all
x,
does
it
necessarily follow that lim f(x)
<
x—*a
lim g(x)
?
x—*a
12.
Suppose that f(x)
<
g(x)
<
h{x)
and that lim
f(x)
that lim g(x) exists,
x—*a
picture!)
and that lim
x—>a
g(x)
=
lim f(x)
x—»a
=
lim h{x). Prove
x—>a
x-+a
=
lim h(x). (Draw a
x—>a
.
90
Foundations
*13.
(a)
Prove that
=
lim /(*)/*
if
and
/
b
^
b[f{bx)/bx\
=
(b)
What happens
(c)
Part (a) enables us to find lim (sin 2x)/x in terms of lim (sin x)/x.
(a)
Find this limit in another way.
Prove that if lim /(*) = /, then lim
if b
0?
x—>0
x—*Q
14.
(b)
Prove that
if
=
lim /(*)
and lim
/
i—»a
(a)
=
|/|.
x—*a
x—*a
15.
=
g(x)
m, then lim
= max(/, /«) and similarly for min.
Prove that lim \/x does not exist,
false for
show that lim \/x =
i.e.,
—
(b) Prove that lim \/{x
Prove that
if
=
lim f(x)
x-+a
Af such that
17.
+
for 0
1
<
\x
Prove that if f{x)
lim f(x) does not
/,
M
<
\f(x)\
Why
pictoriaily?) Hint:
I
/
is
*->0
number
every
max (/,$)(*)
x—»«
x—*a
x—>0
16.
bl.
x-+Q
=
Hint: Write f(bx)/x
=
then lim f(bx)/x
0,
x—*0
-
<
a\
=
/.
does not
1)
then there
if
<
0
does
a number
is
—
\x
exist.
a\
<
suffice to
it
d.
5
>
0 and a
(What does
prove that
/
—
number
mean
this
<
1
f(x)
<
8 ?
0 for irrational x
exist for
any
and
/(*)
=
1
for rational x,
then
a.
x—»o
*18.
Prove that if }{x) = x for rational *, and f(x)
then lim/W does not exist if a ^ 0.
= -x
x—»a
19.
(a)
Prove that
if
=
lim g(x)
0,
then lim g(x) sin
1
A
for irrational x,
=
0.
x-»0
z—>0
(b) Generalize this fact as follows
:
lim g (x)
If
= 0and|AW| <
M
for all
z-+ 0
a-,
then lim g(x)h(x)
you succeed
(b)
in
may make
=
0.
(Naturally
it is
unnecessary to do part
(a) if
doing part (b); actually the statement of part
that's one of the values of
easier than (a)
—
it
generalization.)
20.
Consider a function / with the following property:
which lim g(x) does not exist, then lim[/(x)
tion for
not
exist.
This
**21.
is
Prove that
this
happens
if
and only
if
if
g
is
+ g(x)]
lim f(x) does
any funcalso does
exist.
Hint:
actually very easy: the assumption that lim f(x) does not exist
leads to an immediate contradiction if you consider the right g.
g is replaced
This problem is the analogue of Problem 20 when /
+
by
considerably more complex, and the
/ g. In
analysis requires several steps (those in search of an especially challengthis case the situation is
•
ing problem can attempt an independent solution)
(a)
Suppose that lim f(x)
not
exist,
exists
and
is
^
0.
then lim f(x)g(x) also does not
x-+0
Prove that
exist.
if
lim g(x) does
91
Limits
(b)
Prove the same result
this sort of limit
(c)
Prove that
lim
if
(The precise definition of
<*>.
given in Problem 33.)
is
neither of these
if
=
\f{x)\
two conditions
holds, then there
is
a
function g such that lim g(x) does not exist, but lim f(x)g(x) does
Hint: Consider separately the following two cases: (1) For
> 0 we have |/(*)| > £ for all sufficiently small x. (2) For
exist.
some
e
every 6 > 0, there are arbitrarily small x with |/(*)|
second case, begin by choosing points x n with \x n
\
< £. In the
< 1/n and
\f(x n )\<l/n.
*22.
Suppose that A n is, for each natural number n, some finite set of numbers
[0, 1], and that A n and A m have no members in common if m 5* n.
in
Define / as follows:
X
=
f(x)
(
\
Prove that lim /(*)
=
ln
xin An
>
An
x not in
0,
0 for
all
a in
for
any
n.
[0, 1].
x-+a
23.
why
Explain
the following definitions of lim f(x)
=
are
/
all correct:
x—*a
For every
*24.
8
>
(i)
if
0
<
i*
(ii)
if
0
<
I*
<
<
(iii)
if
0
(iv)
if
0
I*
I*
0 there
-
Give examples
a\
a\
a\
a\
to
<
<
<
<
an
is
£
>
0 such that, for
e,
then
\f(x)
s,
then
\f{x)
e,
then
|/(at)
~
-
/|
l\
/|
e/10, then |/(*)
<
<
<
-
show that the following
all x,
5.
8.
56.
/|
<
«.
definitions of lim f(x)
=
I
5,
then
are
x—*a
not correct.
(a)
For
l/W (b)
For
0
25.
all
<
£
>
<
>
-
a\
all 8
I*
/I
0 there
£
>
0 such that
is
a 6
>
0 such that
in
Problem 4-15 indicate
and lim f(x) exist.
is
an
if
0
\x
—
a\
|/(*)
—
l\
<
<
e.
0 there
<
if
<
£,
then
8.
For each of the functions
for
which numbers
a
the one-sided limits lim f(x)
x—*a +
*26.
Do
same
x—*a~
27.
each of the functions in Problem 4-17.
(b) Also consider what happens if decimals ending in 0's are used
instead of decimals ending in 9's.
Prove that lim f(x) exists if lim /(*) = lim /(*).
28.
Prove that
(a)
the
for
x~*a +
x~*a
(i)
lim f(x)
(ii)
lim /(H)
x~»0
=
=
lim /(-*).
lim/(x).
x-*0+
x-*a~
=
lim f(x 2 )
(iii)
lim /(*).
x-»0+
x->0
(These equations, and others like them, are open to several interpretaThey might mean only that the two limits are equal if they both
tions.
exist;
is
or that
equal to
equal to
*29.
or that
<
Suppose that lim f(x)
<
only
if
—
<
a
m >
What
n.
one easy limit
is
•
•
X—>
•
=
•
•
do some algebra
0;
•
when m = n? When m > n? Hint: The
the limit
lim l/x k
this
some 8 > 0 such that f(x) < f(y) when8 and \y — a\ < 8. Is the converse true?
+ a 0 )/(b mx m +
+ b 0) exists if and
is
<
a\
+
is
suitable.)
lim /(*). (Draw a picture to illustrate
Prove that there
y and \x
n
Prove that lim (a n x
x— 00
ever x
and
and is
the other also exists
exists,
either limit exists, then the other exists
if
Decide for yourself which interpretations are
it.
assertion.)
*30.
a certain one of the limits
if
it;
so that this
is
the only
oo
information you need.
31.
Prove that lim f(l/x) = lim
x—*
x— +
Define " lim /(*) = /."
32.
X—>
(a)
— oo
Find lim
—
X—*
(b)
(a n x
n
+
•
+
•
Prove that lim
=
/( x)
We
m
+
•
*
•
+
b 0 ).
lim f(-x).
x—
oo
— oo
=
Prove that lim f(\/x)
lim f(x).
x— — oo
x—
33.
a 0 )/(b m x
00
X—+
(c)
/(*).
oo
»
define lim f(x)
= «
mean
to
that for
N
all
there
is
a 8
>
0 such
x—*a
that, for all x,
if
0
<
—
\x
<
a\
8,
then f(x)
>
N. (Draw an appropriate
picture!)
(a)
Show
that lim \/{x
-
3)
2
=
oo.
x-»3
(b)
Prove that
if
/(*)
>
>
£
0 for
all x,
and lim
g(x)
=
then
0,
X—Kl
34.
(a)
Define lim f(x)
\imf(x)/g(x)
=
oo.
lim f(x)
=
oo,
=
oo
;
x—*a +
least
and lim
f(x)
=
oo.
(Or
at
x—+ oo
x-+a~
convince yourself that you could write down the definitions if
How many other such symbols can you define?)
you had the energy.
(b)
Prove that lim 1/x
=
oo.
(c)
Prove that lim f(x)
=
oo
x->0 +
if
and only
if
lim f(\/x)
x~+
00
=
oo.
CONTINUOUS FUNCTIONS
CHAPTER
If /
is
an arbitrary function,
it is
not necessarily true that
lim/W
=/(*).
x-*a
In fact, there are many ways this can fail to be true. For example, / might not
even be defined at a, in which case the equation makes no sense (Figure 1).
Again, lim f(x) might not exist (Figure 2). Finally, as illustrated in Figure
x—*tt
3,
even
defined at a and lim f(x) exists, the limit might not equal f(a).
if / is
-A
(b)
(a)
FIGURE
(c)
2
We would like to regard all behavior of this type as abnormal and honor,
with some complimentary designation, functions which do not exhibit such
peculiarities. The term which has been adopted is "continuous." Intuitively,
a function /
oscillations.
is
continuous
Although
whether a function
worth cultivating)
is
it
this
if
the graph contains no breaks, jumps, or wild
continuous simply by looking at
is
you to decide
graph (a skill well
description will usually enable
its
easy to be fooled, and the precise definition
is
very
important.
DEFINITION
The
function /
is
continuous at a
if
lim/(*)
We
will
have no
difficulty finding
or are not, continuous at some
=
/(„).
many examples
number
a
of functions which are,
—every example involving
limits
provides an example about continuity, and Chapter 5 certainly provides
enough of
these.
function f(x) = sin 1/x is not continuous at 0, because it is not even
defined at 0, and the same is true of the function g(x) — x sin 1/x. On the
The
other hand,
93
if
we
are willing to extend the second of these functions, that
is, if
a
94
Foundations
we wish
to define a
new
G
function
by
^0
= (*sinlA,
G(x)
=
x
0,
then the choice of a = G(0) can be made in such a way that G will be continuous at 0 to do this we can (in fact, we must) define G(0) = 0 (Figure 4).
—
This sort of extension
is
not possible for
F{x)
then
F will
not be continuous at
we
/; if
sin 1/x,
x
a,
x
define
0
=
0,
no matter what
0,
a
is,
because lim f(x) does
x-*0
not
exist.
The
function
_
\
f(
J
x
|
is
not continuous at
a,
a
if
rational
x ratioi
i
rrational
x irrati
0,
\
since lim f(x) does not exist.
0,
However,
x—»a
=
lim f(x)
0
=
/(0), so
/
continuous at precisely one point,
is
0.
x—>0
The
functions /(*)
numbers
a,
=
=
g(x)
c,
=
and
x
are continuous at
all
since
=
lim f(x)
=
c
—
c
lim x
—
a
lim
x—*a
f(a),
x—>a
=
lim g(x)
x—*
x—>a
FIGURE
*,
2
—
lim A(*)
lim x 2
=
a
—
2
g(a),
=
h{a).
4
Finally, consider the function
}W
f(
In Chapter 5
\
—
~
we showed
\
0>
irration
x irrational
\
l/q,
x=p/qiin
that lim f(x)
=
lowest terms.
0 for
all a.
Since 0
=
}(a) only
when
x—*a
a
is
irrational, this function
is
continuous at a
if
a
is
irrational, but not
if
a
is
rational.
It is
even
easier to give
examples of continuity
if
theorems.
THEOREM
l
If /
and g are continuous
at
(1)
(2)
Moreover,
if
g(a)
^
0,
then
a,
/ + g i s continuous at a,
f g is continuous at a.
•
then
(3)
1/g
is
continuous at
a.
we prove two
simple
Continuous Functions
PROOF
Since / and g are continuous at
lim/O) =
By Theorem
2(1) of
a,
and
f(a)
+g
=
Starting with the functions /(*)
we can
Theorem
use
b„x
=
/(*)
+g{a) =
+ g)(x) = }{a)
which is just the assertion that /
(2) and (3) are left to you. |
at a, for every a,
(/
+
continuous at
e
and
=
*,
a.
proofs of parts
The
which are continuous
to conclude that a function
1
fen-i*"
/(*)
+
•
•
•
+b
+
0
c0
X
C
g(a).
is
-1
+
n
=
lim g(x)
5 this implies that
Chapter
lim (/
95
harder to get much further
prove
it will be easy to
than that. When we discuss the sine function in detail
funcA
meanwhile.
this fact
that sin is continuous at a for all a; let us assume
is
continuous at every point in
its
domain. But
it is
tion like
f{x)
sin 2 x
_
"
'
+
sin 27 x
x2
+
+
4x sin 2 x
x 4 sin x
2
domain. But we are
can now be proved continuous at every point in its
we
of a function like /(*) - sin(* );
still unable to prove the continuity
functions.
of continuous
obviously need a theorem about the composition
the definition of conabout
point
following
the
theorem,
Before stating this
Urn f(x) = f(a) according
tinuity is worth noting. If we translate the equation
to the definition of limits,
for
every
if
But in
this case,
0
e
<
we
obtain
>
0 there
|*
-
where the
a\
limit
0
is
<5
d,
>
|/(*)
a\
- f(a)\ <
e.-
phrase
f(a), the
<\x -
all x,
0 such that, for
then
<
d
be changed to the simpler condition
may
\x
since
if
x
=
a
it is
-
a\
<
certainly true that \f(x)
5,
-
<
f(a)\
e.
continuous at
o
continuous at a, and / is continuous at g(a), then / g is
a.)
at
not
g(a),
at
continuous
(Notice that / is required to be
2
Ug
is
PROOF
Let
£
THEOREM
is
<
>
0.
We
wish to find a
if |*
-a\<5,
S
>
0 such that for
then \(fog)(x)
i.e.,
|/UW)
-
(f
all x,
°
g)(a)\
<
2.
<
e,
a.
We
use continuity of / to estimate
first
how
close g(x)
that for
to g(a) in order
must be
for this inequality to hold. Since /is continuous atg(a), there
is
a
<5'
>
0 such
all y,
if \y
(1)
-
g(a)\
- f(g(a))\ <
<
8',
then
<
6',
then \f(g(x))
\f(y)
e.
In particular, this means that
We now
g(a)\
use continuity of g to estimate
—
the inequality \g{x)
any other
just like
-
if \g(x)
(2)
<
g(a)\
positive
8' to
- f(g(a))\ <
how close x must be
The number 8 is a
f
hold.
number; we can therefore take
definition of continuity of g at
We
a.
conclude that there
to a in order for
positive
8' as
a 8
is
e.
the
>
number
£ (!) in
the
0 such that,
for all x,
Combining
and
(2)
can
now
(3)
we
<
a\
8,
then \g(x)
-
g(a)\
<
- f(g(a))\ <
then \f(g(x))
8'.
S.
|
reconsider the function
m_
,/
x
~
We
<5,
see that for all x }
\x-a\<
if
We
-
if \x
(3)
I
\
x sin
1
0
x
/x,
x
0,
=
0.
have already noted that / is continuous at 0. A few applications of
1 and 2, together with the continuity of sin, show that / is also con-
Theorems
tinuous at
a,
for
a
^
0.
Functions
like f(x)
=
sin(* 2
+
sin(*
+
sin 2 (# 3 )))
should be equally easy for you to analyze.
The few theorems
functions at a
been related to continuity of
single point, but the concept of continuity doesn't begin to be
we
some
continuous on
all
points of
is
called
a
little differently;
all
on functions which are continu(a, b), then/
(a, b). Continuity on a closed interval must be defined
a function / is called continuous on [a, b] if
really interesting until
ous at
of this chapter have
focus our attention
interval. If /is continuous at x for all x in
(1)
/is continuous at x for
(2)
lim f(x)
=
f(a)
all
and lim
x in
f(x)
=
(a, b),
f(b).
£—>&~
x—*a +
Functions which are continuous on an interval are usually regarded as
might be specified as the first
condition which a "reasonable" function ought to satisfy. A continuous function is sometimes described, intuitively, as one whose graph can be drawn
without lifting your pencil from the paper. Consideration of the function
especially well behaved; indeed continuity
f (v
/W
\
_
|
x sin \/x,
x
i
0,
x
0
=
0
97
Continuous Functions
shows that
this description
many
that there are
is
a
little
too optimistic, but
nevertheless true
it is
which are continu-
results involving functions
important
ous on an interval. These theorems are generally much harder than the ones in
this chapter, but there is a simple theorem which forms a bridge between the
two kinds of results. The hypothesis of this theorem requires continuity at only
a single point, but the conclusion describes the behavior of the function on
some interval containing the point. Although this theorem is really a lemma
for later
THEOREM
3
arguments,
included here as a preview of things to come.
Suppose /is continuous at
that /(*)
is
PROOF
it is
a
>
0 for
number
8
>
a\
0 such that f(x)
<
0 for
>
if \x
>
0
we can
0.
-
A
<
Then
8.
a\
<
8,
then
take f(a) as the
8y
then
this last inequality implies f(x)
all
>
\f(x)
e.
Thus
)/(*)
number
a
is
if f(a)
x satisfying
<z,
there
is
\x
>
if s
- f(a)\ <
<
—
>
8
0,
a\
0 there
0 such
then there
<
is
8.
a 8
>
0
£.
8
>
0 so that for
all x,
~ f(a)\ < /(a),
0.
similar proof can be given in the case f(a)
can apply the
there
Similarly,
Since/ is continuous at
if\x-a\ <
and
0.
-
x satisfying
Consider the case f(a)
such that, for all x,
Since f(a)
and f(a) >
a,
\x
all
<
0; take £
=
—/(a).
Or one
case to the function —/. |
first
PROBLEMS
1.
For which of the following functions /
with domain
(i)
fix)
=
(ii)
f(x)
=
R
such that F(x)
=
is
there a continuous function
f(x) for all x in the
F
domain of/?
X
2.
3.
(iii)
f{x)
(iv)
fix)
=
-
0,
x irrational.
\/q, x
= p/q
rational in lowest terms.
At which points are the functions of Problems 4-1 5 and 4-17 continuous?
Show that
(a) Suppose that / is a function satisfying \f(x) < |*| for all x.
equal
must
0.)
that
(Notice
at
0.
/(0)
/ is continuous
at any
(b) Give an example of such a function / which is not continuous
|
a
^
0.
Suppose that g is continuous at 0 and g(0) = 0, and |/(*)| < \g(x)\.
Prove that / is continuous at 0.
Give an example of a function / such that / is continuous nowhere, but
(c)
4.
|
/| is
continuous everywhere.
98
Foundations
5.
For each number
any other
6.
Find a function / which
(a)
tinuous at
continuous at
but not at
a,
is
discontinuous at
1, £,.
Suppose that
.
but con-
.
.
other points,
all
/satisfies f(x
at 0. Prove that
8.
is
Find a function / which is discontinuous at
at 0, but continuous at all other points.
(b)
7.
which
find a function
0,
points.
/
+ y)
=
f(x)
+ f(y),
continuous at a for
is
1
J,
£-,
,
and that /
is
.
.
.
,
and
continuous
all a.
Suppose that / is continuous at a and f(a) = 0. Prove that if a 9* 0, then
a is nonzero in some open interval containing a.
/
(a) Suppose / is not continuous at a. Prove that for some number £ > 0
there are numbers x arbitrarily close to a with \f(x) — f(a) > 6.
+
9.
\
Illustrate graphically,
(b)
Conclude that
arbitrarily close to a with f(x)
10.
(a)
(b)
(c)
Prove that
is
continuous at
a,
then so
is
there are
there are
+
Prove that
/ and g are continuous, then
if
numbers x
numbers x
|/[
Prove that every continuous / can be written / = .E
even and continuous and 0 is odd and continuous.
min(/,
(d)
/
if
£ > 0 either
< f(a) — £ or
> f(a) + £.
some number
for
arbitrarily close to a with f(x)
so are
0, where
E
is
max(/, g) and
g).
Prove that every continuous / can be written /
=
g
—
h,
where g and
h are nonnegative and continuous.
11.
Prove Theorem
f(x)
*12.
(a)
=
1
by using Theorem 2 and continuity of the function
(3)
l/x.
Prove that
if /is
continuous at
/
and lim g(x)
=
L then lim f(g(x))
x—*a
/(/).
(You can go
sider the function
(b)
Show
that
ally true
if
right back to the definitions, but
G
with G(x)
continuity of / at
that lim f(g{x))
=
/ is
13.
and/(/)
(a) Prove that
is
if
/
is
=
g(x) for x
^
a,
it is
easier to con-
and G(a)
not assumed, then
it is
=
L)
not gener-
/(lim g{x)). Hint: Try /(*)
x—*a
x^l,
=
=
x—*a
=
0 for
x—
1.
continuous on
continuous on R, and which
[a,
b\ then there
satisfies g(x)
=
is
a function g which
f(x) for all x in
[a,
b\
Hint: Since you obviously have a great deal of choice, try making g
constant on (— <x> a] and [b, 00).
5
(b)
14.
(a)
Give an example to show that this assertion is false if [a, b] is replaced
by (a, b).
Suppose that g and h are continuous at a, and that g{a) = h (a).
Define f(x) to be g(x) if x > a and h(x) if x < a. Prove that / is continuous at
(b)
a.
continuous on [a, b] and h is continuous on [b c] and
Let f(x) be g(x) for x in [a, b] and h(x) for x in [b, c].
Show that / is continuous on [a, c]. (Thus, continuous functions can
be "pasted together.")
Suppose g
g(b)
—
is
h(b).
9
Continuous Functions
15.
(a)
Prove the following version of Theorem 3
Suppose that lim
nuity'':
=
f(x)
fia),
99
for " right-hand conti-
and f(a) >
0.
Then
there
is
a
x—*a +
number
6
>
Similarly,
fix)
(b)
<
0 for
>
0 such that fix)
if
fia)
all
<
0,
0 for
then there
x satisfying 0
<
x
all
is
—
a
x satisfying 0
number
a
<
8
< x — a < d.
> 0 such that
6.
Prove a version of Theorem 3 when lim fix)
=
fib).
x—>b~
16.
If lim fix)
x—*a
but
exists,
discontinuity at
=
sin
is
^ fia),
then /
is
0 and /(0)
=
said to
have a removable
a.
/x for #
^
does / have a removable
(a)
If fix)
(b)
^ 0, and /(0) = 1?
Suppose / has a removable discontinuity at a. Let gix) = /(*) for
= lim/(x). Prove that g is continuous at a.
*
a, and let
1
discontinuity at 0?
What
if fix)
=
x sin
1
1
,
fx for *
x—»o
(Don't work very hard; this
(c)
= 0 if
What is
Let /(*)
terms.
a:
is
quite easy.)
let fip/q) = l/q if p/q is in lowest
the function g defined by gix) = lim f(y)?
is
irrational,
and
y—*x
*(d) Let / be a function with the property that every point of discontinuity is a removable discontinuity. This means that lim fiy) exists for
y—*x
all x,
but /
numbers
is
**(e)
x.
be discontinuous at some (even infinitely many)
Define gix) = lim fiy). Prove that g is continuous. (This
may
not quite so easy as part
(b).)
and which
worth thinking about this
problem now, but mainly as a test of intuition; even if you suspect
the correct answer, you will almost certainly be unable to prove it
Is
there a function / which
is
discontinuous at every point,
has only removable discontinuities?
at the present time. See
(It is
Problem 21-24.)
CHAPTER
THREE HARD THEOREMS
This chapter
devoted to three theorems about continuous functions, and
The proofs of the three theorems themselves will
not be given until the next chapter, for reasons which are explained at the end
some of
is
their consequences.
of this chapter.
THEOREM
1
If / is continuous on
such that /(*) - 0.
and f(a)
[a, b]
<
0
< f(b),
then there
is
some x
in
[a, b]
means that the graph of a continuous function which
below the horizontal axis and ends above it must cross this axis at some
(Geometrically, this
starts
pointy as in Figure 1.)
THEOREM
2
If/
is
continuous on
some number
then /is bounded above on
b],
[a,
N such
that /(*)
(Geometrically, this theorem
<
N for
means
all
x in
[a,
[a,
b],
that
is,
there
is
b].
that the graph of /
lies
below some
line
parallel to the horizontal axis, as in Figure 2).
THEOREM
3
If /
is
continuous on
[a,
f(y) >/(*) for all x in
b],
[a,
then there
is
some number y
in
[a,
b]
such that
b] (Figure 3).
These three theorems differ markedly from the theorems of Chapter 6. The
hypotheses of those theorems always involved continuity at a single point,
Three
Hard Theorems
101
while the hypotheses of the present theorems require continuity on a whole
interval [a, £]— if continuity fails to hold at a single point, the conclusions
may fail. For example, let / be the function shown in Figure 4,
Then / is continuous
but there
FIGURE
0-<x<V2
V2 < x < 2.
/M
V2
is
4
single point
at every point of
no point x
V2
is
in [0, 2]
-
V2, and/(0) <
/
shown
the function
is
/(*)
0,
x
^
0
x
=
0.
<
0
f(2),
0; the discontinuity at the
sufficient to destroy the conclusion of
Similarly, suppose that
N
except
[0, 2]
such that /(*)
Theorem
1
in Figure 5,
but / is not bounded
> 0, we have f(£$) = 2N > N.
above on [0, 1], In fact for any number
This example also shows that the closed interval [a, b] in Theorem 2 cannot
be replaced by the open interval (a, b), for the function /is continuous on (0, 1),
Then /
is
continuous at every point of
[0, 1]
except
0,
N
•
but
not bounded there.
is
Finally, consider the function
/«
FIGURE
5
On
the interval
[0,
1]
conclusion of Theorem
"
2,
x
0,
x
is
even though /
Theorem
fix) for all x in [0,1]; in fact,
in Figure 6,
x\
the function /
not satisfy the conclusion of
/GO >
shown
it is
<
>
1
1.
bounded above,
is
so
/ does
not continuous on
3— there
[0, 1].
satisfy the
But / does
in [0, 1] such that
certainly not true that/(l) > /(*) for
is
no y
we cannot choosey = 1, nor can we choose 0 < y < 1 because
1
f{y) < f(x) if x is any number with y < x <
This example shows that Theorem 3 is considerably stronger than Theorem
function on
2. Theorem 3 is often paraphrased by saying that a continuous
all
x in
[0, 1] so
a closed interval "takes on
its
maximum
value" on that interval.
As a compensation for the stringency of the hypotheses of our three theorems,
the conclusions are of a totally different order than those of previous theorems.
They describe
the behavior of a function, not just near a point, but
on a whole
more
interval; such "global" properties of a function are always significantly
prove than "local" properties, and are correspondingly of much
greater power. To illustrate the usefulness of Theorems 1, 2, and 3, we will
soon deduce some important consequences, but it will help to first mention
difficult to
FIGURE
6
some simple generalizations of
THEOREM
4
If /
is
continuous on
such that fix)
=
c.
[a, b]
and
these theorems.
f{a)
<
c
<
fib),
then there
is
some x
in
[a, b]
PROOF
Let g
there
THEOREM
5
If / i s
=
is
f
Then g
c.
some x
in
=
/(at)
continuous, and g(a)
is
b]
[a,
continuous on
such that
PROOF
—
such that g{x)
[a, b]
'
and
/(a)
>
c
=
0.
>
fib),
<
But
0
gib). By Theorem 1,
means that /(*) = c. |
<
this
then there
some x
is
in
[a, b]
c.
The function — / is continuous on [a, b] and —f(a) < —c < —f(b). By
Theorem 4 there is some x in [a, b] such that —f(x) = — c, which means
that f(x)
=
c.
|
Theorems 4 and 5 together show that / takes on any value between f{a) and
We can do even better than this: if c and d are in [a, b], then / takes on
any value between f(c) and fid). The proof is simple; if, for example, c < d,
then just apply Theorems 4 and 5 to the interval [c, d]. Summarizing, if a
continuous function on an interval takes on two values, it takes on every value
f(b).
in between; this slight generalization of
Theorem
1
is
often called the Inter-
mediate Value Theorem.
THEOREM
6
PROOF
If / is continuous on [a, b], then / is bounded below on
some number TV such that f(x) > N for all x in [a, b].
The
function
— / is continuous on
so
[a, b],
M
such that -/(*) <
for all x in [a, b]. But
= — M. |
x in [a, b\ so we can let
[a,
b\ that
is,
there
is
M
by Theorem 2 there is a number
this means that /(*)" > -AT for all
N
Theorems 2 and 6 together show that a continuous function / on [a, b] is
bounded on [a, b], that is, there is a number N such that \f{x) < N for all x in
[a, b]. In fact, since Theorem 2 ensures the existence of a number Ni such that
f{x) < Ni for all x in [a, 6], and Theorem 6 ensures the existence of a number
\
N
THEOREM
7
2
such that fix)
If /is
>
continuous on
for all x in
[a,
N
2
for all x in
[a, b],
[a, b],
then there
is
we can
take
somejy in
N=
[a, b]
max(|#i|, \N 2 \).
<
such that/(j)
/W
b].
(A continuous function on a closed interval takes on
its
minimum
value on
that interval.)
PROOF
The function —/is continuous on
such that
all
x in
—fiy) >
b\
—fix) for
8
[a, b]
x in
by Theorem 3 there is some;
which means that f(y) <
the trivial consequences of Theorems
in
[a, b]
fix) for
1, 2,
and
3,
interesting things.
Every positive number has a square
is
\
[a, b],
|
Now that we have derived
we can begin proving a few
THEOREM
all
some number x such that x 2 =
root. In other words, if
a.
a >
0,
then there
Hard Theorems
Three
PROOF
103
Consider the function /(*) = * 2 which is certainly continuous. Notice that
the statement of the theorem can be expressed in terms of/: "the number a
,
The proof
has a square root" means that / takes on the value a.
about / will be an easy consequence of Theorem 4.
There
number b > 0 such that f(b) > a (as illustrated in
> 1 we can take b = a, while if a < 1 we can take
< f(b) Theorem 4 applied to [0, b] implies that for
we have /(*) = a, i.e., x = a. |
obviously a
is
Figure 7); in fact, if a
b = 1. Since /(0) < a
some x
of this fact
y
2
(in [Q, b]),
same argument can be used to prove that a positive number has
any natural number n. If n happens to be odd, one can do
better: every number has an nth root. To prove this we just note that if the posin =
n =
—a (since n is
tive number a has the nth root *. i.e., if x
a, then ( — x)
odd), so —a has the nth root —x. The assertion, that for odd n any number a
Precisely the
an nth
root, for
has an nth root,
equivalent to the statement that the equation
is
xn
has a root
if
n
- a =
odd. Expressed in this
is
0
way
the result
is
susceptible of great
generalization.
THEOREM
9
If n
is
odd, then any equation
xn
+
an
^x n
~l
+
+
a0
-
0
has a root.
PROOF
We
obviously want to consider the function
fix)
=
x
n
+
a n ^ 1x
n~ 1
+
we would like to prove that / is sometimes positive and sometimes negative.
The intuitive idea is that for large \x\, the function is very much like g(x) = x n
and, since n
is
large negative
The proper
fix)
=
xn
odd, this function
x.
A
algebra
little
is
positive for large positive x
is all
we need
analysis of the function
+
an
to make
and negative
this intuitive
/ depends on writing
^x n ~
Note that
<
+
[
+
•
•
Igoj
n
l*
Consequently,
if
we choose
(*)
then
|**|
>
|*|
|*|
x satisfying
>
1,
2n|a n _i|,
5
2w|a 0
and
\a n -k\
|gn-fc|
|**|
|*|
<
2n\a n - k
5
2^
\
|
:
l
for
idea work.
1_
xn
2n
In
2
n terms
In other words,
which implies that
Therefore,
if
we choose an
2
>
so that f(xi)
0.
*i
Now
<
an x in
On
it
then
xn/
if
<
x2
0
satisfies (*),
then x 2 n
<
0
and
*2 /
X2
0.
1
to the interval [x 2 x{]
,
such that /(*)
9 disposes of the
frustrating to leave the
Some equations,
At
like x
—what more
is
2
first
—
1
that there
is
problem of odd degree equations so happily that
problem of even degree equations comsight, however, the problem seems insuperable.
= 0, have a solution, and some, like x 2 + 1 =0,
Theorem
pletely undiscussed.
we conclude
«= 0.
|
would be
do not
satisfies (*),
xi
the other hand,
applying Theorem
[x2 , xt ]
0 which
\
\
so that f(x 2 )
>
there to say? If we are willing to consider a
more general
question, however, something interesting can be said. Instead of trying to solve
the equation
xn
let
+
a n -ix n
-1
+
+a «
•
•
0
;
0,
us ask about the possibility of solving the equations
xn
+ an^x^ +
1
•
•
•
+a
0
=
c
numbers c. This amounts to allowing the constant term a 0 to
information which can be given concerning the solution of these
for all possible
vary.
The
equations depends on a fact which is illustrated in Figure 8.
^1
71
an-ix
a 0 with n even,
The graph of the function /(*) = x n
contains, at least the way we have drawn it, a lowest point. In other words,
there is a number y such that f(y) < f(x) for all numbers x the function /
+
+
•
•
•
+
,
—
on a minimum value, not just on each closed
but on the whole
line. (Notice that this is false if n is odd.) The proof depends on Theorem 7, but
a tricky application will be required. We can apply Theorem 7 to any interval
takes
interval,
Three
Hard Theorems
105
and obtain a point y 0 such that f(y 0 ) is the minimum value of/ on [a, 6];
but if [a, ft] happens to be the interval shown in Figure 8, for example, then
the point y 0 will not be the place where / has its minimum value for the whole
line. In the next theorem the entire point of the proof is to choose an interval
THEOREM
io
[a,
ft],
[a>
ft]
If n
is
in
<
As
in the
that this cannot happen.
xn
=
even and f(x)
such that f(y)
PROOF
way
such a
+
<2
Theorem
proof of
all
+
+
<z
x with
2«|<2 n_i|
even, x n
provided that
such that
b
n
|^|
>
>
0 for
2
~
>
Af.
<
xn
?)-/<->.
x
\
Now consider the number /(0). Let > 0 be a number
we have (Figure 9)
and also b > M. Then, if * >
2/(0)
x
number y
so
all x,
ft
ft,
xn
if
a
2n|fl 0 |),
,
x
fix)
Similarly,
is
> M, we have
|x|
2
is
then there
9, if
!<!+— +
Since n
o,
f(x) for all x.
M — max(l,
then for
n_1
n -i*
b
n
>|>y >/(0).
then
—ft,
Summarizing:
if
>
^
ft
or *
<
-ft,
Now
apply Theorem 7 to the function
clude that there is a number y such that
if
(1)
In particular, f(y)
(2)
THEOREM
11
<
if
Combining
(1)
and
Theorem
10
now
/(0).
x
<
(2)
-ft
<
<
ft,
/on
/(0).
the interval
then /GO
[
>
ft,
then /(*)
see that f(y)
<
>
/(0)
>
/(>).
f(x) for all x. |
Consider the equation
Xn
ft,
< /(at).
allows us to prove the following result.
(*)
—
Thus
-ft or x
we
x
>
then /(*)
+
dn^x"- 1
+
+ ao =
c,
ft].
We
con-
106
Foundations
and suppose n is even. Then there is a number m such that
for c > m and has no solution for c < m.
PROOF
Let /(*)
=
+ an-ix*- +
xn
+
1
(*) has a solution
a 0 (Figure 10).
According to Theorem 1 0 there is a number y such that f(y) < f(x) for all
Let m — /(>')• If c < m, then the equation (*) obviously has no solution,
since the left side always has a value > m. If c — m, then (*) has y as a solution. Finally, suppose c > m. Let b be a number such that b > y and f(b) > c.
x.
Then f(y) = m <
ber x in
[y, b]
c
<
f(b)
Consequently, by
.
such that f(x)
=
c,
so x
is
Theorem
a solution of
4,
there
is
some num-
(*). |
These consequences of Theorems 1,2, and 3 are the only ones we will derive
(these theorems will play a fundamental role in everything we do later,
however). Only one task remains to prove Theorems 1, 2, and 3. Unfortunately, we cannot hope to do this
on the basis of our present knowledge
about the real numbers (namely, PI -PI 2) a proof is impossible. There are
several ways of convincing ourselves that this gloomy conclusion is actually the
case. For example, the proof of Theorem 8 relies only on the proof of Theorem
if we could prove Theorem 1, then the proof of Theorem 8 would be com1
plete, and we would have a proof that every positive number has a square root.
As pointed out in Part I, it is impossible to prove this on the basis of PI -PI 2.
now
—
—
;
Again, suppose we consider the function
FIGURE
1
10
x
2
-
2
If there were no number x with x = 2, then / would be continuous, since the
denominator would never = 0. But /is not bounded on [0, 2]. So Theorem 2
depends essentially on the existence of numbers other than rational numbers,
and therefore on some property of the reals other than PI -PI 2.
Despite our inability to prove Theorems 1, 2, and 3, they are certainly
results which we want to be true. If the pictures we have been drawing have
any connection with the mathematics we are doing, if our notion of continuous
function corresponds to any degree with our intuitive notion, Theorems 1, 2,
and 3 have got to be true. Since a proof of any of these theorems must require
some new property of R which has so far been overlooked, our present diffi2
culties suggest
proof of
One
FIGURE
11
that
is,
way
1,
to discover that property: let us try to construct a
for
the smallest x in
[a, b]
is
to locate the first point
such that /(*)
=
0.
To
where f(x)
find this point,
first
=
0,
con-
A which contains
In Figure
all numbers x in [a, b] such that / is negative on
such a point, while x' is not. The set A itself is indicated
Since / is negative at <2, and positive at b, the set A contains
1 1
,
x
is
by a heavy line.
some points greater than
(We
example, and see what goes wrong.
idea which seems promising
sider the set
[a, x].
a
Theorem
a,
while
all
points sufficiently close to
are here using the continuity of /on
[a, b] 9
as well as
b
are not in A.
Problem
6-15.)
Hard Theorems
Three
fix)
< 0
for all x in this interval
107
Now suppose a is the smallest number which is greater than all members of
< a < b. We claim that /(a) = 0, and to prove this we only have
A; clearly a
< 0 and /(a) >
Then, by Theorem
0.
to eliminate the possibilities f(a)
Suppose
first
that /(a)
<
0.
6-3, /(*)
would be
less
than
0 for all x in a small interval containing a, in particular for some numbers
bigger than a (Figure 12); but this contradicts the fact that a is bigger than
A would
all
every
also contain
/(a)
these points
On
FIGURE
member of A,
<
0
since the larger
numbers would
also be in A. Consequently,
is false.
the other hand, suppose /(a)
>
Again applying Theorem
0.
6-3,
we
would be positive for all a: in a small interval containing a, in
particular for some numbers smaller than a (Figure 13). This means that these
smaller numbers are all not in A. Consequently, one could have chosen an
even smaller a which would be greater than all members of A. Once again we
have a contradiction; f(a) > 0 is also false. Hence f(a) = 0 and, we are
see that f(x)
12
tempted to say, Q.E.D.
We know, however, that something must be wrong, since no new properties
or R were ever used, and it does not require much scrutiny to find the dubious
than all
point. It is clear that we can choose a number a which is greater
clear
that
=
so
not
is
but
it
b),
members of A (for example, we can choose a
x
numbers
> 0
all
of
consists
A
suppose
we can choose a smallest one. In fact,
V
2 did not exist, there would not be a least
number
number greater than all the members of A for any y > V2 we chose, we could
such that x 2
<
2. If
the
;
f{x)
>
0 for
all
in this interval
x
always choose a
Now
that
still
we have
smaller one.
tional property of the real
and use
FIGURE
13
it.
That
is
almost obvious what addineed. All we must do is say it properly
discovered the fallacy,
numbers we
it is
the business of the next chapter.
PROBLEMS
1.
For each of the following functions, decide which are bounded above or
below on the indicated interval, and which take on their maximum or
minimum value. (Notice that/ might have these properties even if/ is not
continuous, and even
(i)
/(*)
(ii)
f(x)
(iii)
f(x)
(iv)
/(*)
=
=
=
=
x*
x2
x2
if
the interval
is
not a closed interval.)
on (-1, 1).
on (-1, 1).
on R.
on [0, 00).
x
x
>
a
on {-a
-
1,
a
+
a
sary to consider several possibilities for
a.)
1).
(It will
be neces-
=
sf \
(via) f(x)
/•x
r,
(ix)
f
=
v
/(*)
xf
x )\
1,
x irrational
-i/q,
x
2.
/(*)
>
#
on
.
* irrational
0,
sin 2 (cos
on
[x]
*
(i)
/(*)
(ii)
/(*)
+ V] +
rn
(iv) /(*)
*i79
*5
n
[0, a],
a 2)
on
[0, a
3
].
sin
4
2
If n
is
=
—
such
n
1.
some number x such that
+ x + sin
2
x
This problem
(a)
an integer
1,
1^3
+
x
find
5
1
(ii)
[0, 1].
*3
Prove that there
(i)
rrk
[0, a]:
- x + 3.
+ 5x + 2x +
x + x+ 1.
4* - 4* +
=
=
=
=
in
[0, 1].
on
m lowest terms
p/q
For each of the following polynomial functions /,
that f(x) = 0 for some x between n and n + 1
(hi) f{x)
3.
=
=
f(x)
(xii)
=
x rational
x,
f
rn
on
.
in lowest terms
f
(
(xi)
=
1
{
=
\
f(x )
# irrational
x
(
I
f
(
9
/£,
\
k
—
= n9
#
1
a continuation of Problem 3-7.
is
is
2
even,
and > 0,
find a polynomial function of degree n
with exactly k roots.
(b)
A
if
root a of the polynomial function / is said to have multiplicity m
m
= (x
a) g(x), where g is a polynomial function that does
—
f(x)
have a as a root. Let / be a polynomial function of degree n.
Suppose that /has k roots, counting multiplicities, i.e., suppose that
not
k
is
sum
the
of the multiplicities of
all
the roots.
Show
that n
—
k
is
even.
5,
5.
Suppose that / is continuous on [a, b] and that f(x) is always rational.
What can be said about / ?
Suppose that/is a continuous function on [—1, 1] such that* 2 + (/(*)) 2 = 1
for all x. (This mean s that ( *, f(x)) always lies on the unit circle.) Show
.
How many
for all x
i.
K
V
V
2
2
=
1 - x for all x, or else /(*) = 1 - x for
continuous functions / are there which satisfy (/(*)) 2
that either f{x)
7
all *.
=
x2
?
Suppose that / and g are continuous, that f 2 = g 2 and that /(#) ^ 0 for
all x. Prove that either f(x) =
g (x) for all x, or else /(*) = — g(x) for all #.
(a) Suppose that / is continuous, that f(x) = 0 only for x = a, and that
,
/(*)
>
0 for some x
>
show that x 2
+
xy
some x <
a as well as for
about f(x) for all x ^ a ?
*(b) Using only the fact that x 2
+y >
2
to consider various fixed y
+ xy + y ^
2
0
^
if
0,
x and
jy
so that
0
if
a.
What can
x and
be said
are not both 0,
(The trick
you can define a function.)
are not both
0.
is
;
*(c) Discuss similarly the sign of x
not both
10.
+
3
x 2y
+
+y
xy 2
109
Hard Theorems
Three
z
when
x and y are
0.
Suppose / and g are continuous on [a, b] and that f(a) < g(a), but
> g(b). Prove that f{x) = g(x) for some x in [a, b). (If your proof
f(b)
very short,
isn't
11.
12.
it's
not the right one.)
Suppose that /is a continuous function on [0, 1] and that /(*) is in [0, 1]
for each x (draw a picture). Prove that /(*) = x for some number x.
(a) Problem 1 1 shows that / intersects the diagonal of the square in
Figure 14 (solid line). Show that / must also intersect the other
(dashed) diagonal.
Prove the following more general fact: If g
and^(0) = 0,g(\) = 1 or^(0) = 1,^(1) =
(b)
some
13.
continuous on
is
then /(*)
0,
=
[0,
1]
g(x) for
x.
0 and let /(0) = 0. Is / continuous on
[
1] ? Show that / satisfies the conclusion of the Intermediate
Value Theorem on [— 1, 1]; in other words, if / takes on two values
=
Let /(*)
(a)
\jx for x
sin
^
— 1,
somewhere on [ — 1, 1], it also takes on every value in between.
Suppose that / satisfies the conclusion of the Intermediate Value
Theorem, and that / takes on each value only once. Prove that / is
*(b)
continuous.
Generalize to the case where /takes on each value only finitely
*(c)
many
times.
14.
If
/
of
a continuous function on
is
!/|
on
1], let
[0,
Prove that for any number c we have
< ||/||
*(b) Prove that \\f
g\\
(a)
\\f
(c)
*15.
+
g\\
(b)
Hint:
*\7.
</>
11/11
is
+
+
+ hi
- /|] <
\\k
\\cf\\
\\h
-
g\\
+
continuous and lim
\\g
-
cf)(x)/x
=
Give
\c\
•
||/||.
an
example where
=
lim
f\\.
n
=
0
<j)(x)/x
<f>
0)
*16.
value
n
.
n
Prove that if /us odd, then there is a number* such that x + (x) =
n
that y
such
number
Prove that if n is even, then there is a
y
000 <
(*,
*
Prove that
Suppose that
(a)
maximum
be the
||/||
1].
[0,
xn
+
</>(*)
Of which
for all
0.
+
x.
proofs does this
problem
test
your understanding?
Let / be any polynomial funcdon. Prove that there is some number y
such that 1/001 < |/(*)| for all x.
Suppose that / is a continuous function with f(x) > 0 for all x, and
lim f(x) = 0 = lim f(x). (Draw a picture.) Prove that there is some
X—
x—*
00
—
00
y such that f(y) > f(x) for all x.
(a) Suppose that / is continuous on [a, b],
number
*18.
and
let
x be
any number.
Prove that there is a point on the graph of / which is closest to (x, 0)
in other words there is some y in [a, b] such that the distance from
<
(x,
0) to (y, f(y))
[a,
b\ (See Figure 15.)
is
distance from
(x, 0)
to (z, f(z)) for all z in
(b)
Show
same
that this
replaced by
(c)
Show
(d)
In cases
assertion
that the assertion
and
(a)
(c),
is
"19.
(a)
true
if [a. b] is
replaced by
minimum
letg(x) be the
a point on the graph of
(e)
not necessarily true
is
b]
if [a,
is
throughout.
(a> b)
Prove that g(y)
/.
R throughout.
distance from
<
g(x)
+
\x
(*,
0) to
— y\,
and
conclude that g is continuous.
Prove that there are numbers x 0 and x x in [a, b] such that the distance
from Oo, 0) to Oi,/(*i)) is < the distance from (*</, 0) to (xi\f(xi))
for any *</, x\ in [a, b\
.Suppose that /is continuous on
and /(0) = /(l). Let n be any
some number x such that
[0, 1]
natural number. Prove that there
is
= f(x
1/n), as shown in Figure 16 for n = 4. Hint: Consider
the function g(x) = f(x) — f(x
1/n); what would be true if
g(x) ^ 0 for all x ?
+
f(x)
+
(b)
Suppose 0
number
satisfies
any
20.
(a)
<
<
a
,
but that a
not equal to
is
Why?
b).
{a,
In the
than f(a)
<
implies that f(x)
}
first
any natural
case all values close to f(a), but
are taken on
f(a) for x
<
a
and x >
somewhere
in (a, b); this
b.
Refine part (a) by proving that there is no continuous function /
which takes on each value either 0 times or 2 times, i.e., which
takes on exactly twice each value that it does take on. Hint: The
previous hint implies that / has either a
value (which must be taken on twice).
values close to the
maximum
maximum or a minimum
What can be said about
value?
Find a continuous function / which takes on every value exactly
3 times. More generally, find one which takes on every value exactly
n times, if n
(d)
/n for
x.
x in
all
(c)
1
Prove that there does not exist a continuous function / defined on
R which takes on every value exactly twice. Hint: If f(a) — f{b) for
a < b, then either f(x) > f(a) for all x in (a, b) or f(x) < f{a) for
slightly larger
(b)
1
Find a function / which is continuous on [0, 1] and which
/(0) = /(l), but which does not satisfy f(x) = f(x + a) for
n.
Prove that
is
if
odd.
n
is
even, then there
is
every value exactly n times. Hint:
example,
all
x in two
f(x)
<
0 for
To
treat the case n
all
f{x 2 )
=
—
4.
for
/(* 8 ) = f(x A ). Then either /(*) > 0 for
of the three intervals (x h # 2 ), (*2, # 8 ), (#3, ^4), or else
let f(x{)
=
no continuous / which takes on
x in two of these three intervals.
LEAST UPPER BOUNDS
CHAPTER
This chapter reveals the most important property of the real numbers. Nevertheless, it is merely a sequel to Chapter 7; the path which must be followed
has already been indicated, and further discussion would be useless delay.
A
DEFINITION
set
A
numbers
of real
bounded above
is
>
x
Such a number x
is
if
there
a
number
x such that
for every a in A.
a
an upper bound
called
is
for A.
Obviously A is bounded above if and only if there is a number x which is an
upper bound for A (and in this case there will be lots of upper bounds for A)
we often say, as a concession to idiomatic English, that "^4 has an upper
;
bound" when we mean
that there
number which is an upper bound for A.
now been used in two ways
functions, and now in reference to sets. This
is
a
Notice that the term "bounded above" has
first,
in
Chapter
7, in
reference to
always be clear whether
the two definiMoreover,
of numbers or a function.
dual usage should cause no confusion, since
we
are talking about a set
tions are closely connected:
/
is
bounded above on
The
if
A
[a, b] if
entire collection
R
bounded above
will
the set {f(x): a
and only
if
the set
< x < b}, then the function
A is bounded above.
and the natural numbers N, are
bounded above. An example of a set which
of real numbers,
both examples of sets which are
is
is
it
not
is
A =
{x:
0
<
x
<
1}.
bounded above we need only name some upper bound for
A, which is easy enough; for example, 138 is an upper bound for A. and so are
phrase
2, l£, 1-^, and 1. Clearly, 1 is the least upper bound of A; although the
confusion
possible
avoid
any
in
order
to
just introduced is self-explanatory,
(in particular, to ensure that we all know what the superlative of "less"
To show
that
A
is
means), we define
DEFINITION
A number x is
a least
and
Ill
this explicitly.
upper bound
(1)
x
is
(2)
if
y
of
A
if
an upper bound of A,
an upper bound of A, then x
is
<
y.
The
use of the indefinite article "a" in this definition was merely a conces-
sion to
temporary ignorance. Now that we have made a precise definition, it is
if x andjv are both least upper bounds of A, then x = y. Indeed,
easily seen that
in this case
x
and y
it
<
<
y,
since y
is
an upper bound, and x
is
a least upper bound,
x,
since x
is
an upper bound, and y
is
a least upper bound;
=
follows that x
For
y.
The term supremum
of
reason we speak of the least upper bound of A.
synonymous and has one advantage. It abbrevi-
this
A
is
ates quite nicely to
A
sup
and
(pronounced "soup A")
saves us from the abbreviation
\\xbA
is nevertheless used by some authors).
There is a series of important definitions, analogous to those just given,
which can now be treated more briefly. A set A of real numbers is bounded
(which
below
if
there
is
a
number x such that
<
x
for every a in A.
a
Such a number x is called a lower bound
lower bound of A if
and
The
(1)
x
is
(2)
if
y
greatest lower
bound
for A.
A
number
x
is
the greatest
a lower bound of A,
is
of
a lower
A
is
bound
of A, then x
also called the
infimum
>
y.
of A, abbreviated
inf A;
some authors use the abbreviation
gib A.
One detail has been omitted from our discussion so far
—the question of which
have at least one, and hence exactly one, least upper bound or greatest
lower bound. We will consider only least upper bounds, since the question for
greatest lower bounds can then be answered easily (Problem 2).
HA is not bounded above, then A has no upper bound at all, so A certainly
cannot be expected to have a least upper bound. It is tempting to say that A
does have a least upper bound if it has some upper bound, but, like the principle
of mathematical induction, this assertion can fail to be true in a rather special
way. If A = 0, then A is bounded above. Indeed, any number x is an upper
sets
bound
for 0:
x
simply because there
there
is
surely
no
is
least
noy
>
y
in 0.
for every y in 0,
Since
upper bound
number is an upper bound for 0,
With this trivial exception however,
every
for 0.
113
Least Upper Bounds
our assertion
true^and very important,
is
warrant consideration of details.
enough
definitely important
to
We are finally ready to state the last property
numbers which we need.
of the real
(The
(PI 3)
A
upper bound property) If A is a set of real numbers,
and A is bounded above, then A has a least upper bound.
least
y£ 0,
may strike you as anticlimactic, but that is actually one of its
To complete our list of basic properties for the real numbers we require
Property PI 3
virtues.
no particularly abstruse proposition, but only a property so simple that we
might feel foolish for having overlooked it. Of course, the least upper bound
FIGURE
property
1
rational
not really so innocent as
is
numbers
|
< 2,
A and which is
satisfying x 2
for
all that; after all, it does not hold for the
For example, if A is the set of all rational numbers x
then there is no rational number y which is an upper bound
Q,.
an upper bound
PI 3
but
is.
we
than or equal to every other rational number which is
become clear only gradually how significant
less
for A. It will
are already in a position to demonstrate
the proofs which were omitted in Chapter
THEOREM
7-1
PROOF
If
/
is
continuous on
x in
[a,
Our
proof
and
[a, b]
=
such that f(x)
b]
f(a)
<
0
<
its
power, by supplying
7.
f(b),
then there
is
some number
0,
merely a rigorous version of the outline developed at the end of
[a, b] with f(x) = 0.
Define the set A, shown in Figure 1, as follows:
Chapter
Clearly
A =
{x: a
A ^
0,
all points
is
is
7—we will locate the smallest number x in
<
<
x
since a
is
continuous on
is
and /
A; in
x
<
and f(a)
[a, b]
>
a 8
in
<
x satisfying a
in fact, there
b,
is
negative on the interval
fact, there is
+
a
<
8; this
some
all
>
x]}.
0 such that
A
contains
Problem 6-15, since/
an upper b^und for A and,
follows from
0. Similarly, b is
0 such that
8
[a,
points x satisfying b
—
8
<
x
<
b
are
from Problem 6-15, since f(b) > 0.
From these remarks it follows that A has a least upper bound a and that
a < a < b. We now wish to show that f(a) = 0, by eliminating the possibilities f(a) < 0 and f(a) > 0.
Suppose first that f(a) < 0. By Theorem 6-3, there is a 8 > 0 such that
upper bounds
for
A;
this also follows
< 0 for a -~ 8 <
A which satisfies a —
f(x)
x
8
< a + 8 (Figure 2). Now there is some number x 0 in
< x 0 < a (because otherwise a would not be the least
upper bound of A). This means that / is negative on the whole interval
[a, xo]. But if X\ is a number between a and a + 5, then / is also negative on
the whole interval [x 0 x\]. Therefore / is negative on the interval [a, x{], so
x\ is in A. But this contradicts the fact that a is an upper bound for A; our
original assumption that f(a) < 0 must be false.
Suppose, on the other hand, that f(a) > 0. Then there is a number 8 > 0
,
such that f(x)
that there
FIGURE
3
is
negative on
>0foro; —
an x 0
[a,
in
A
8
<
x
satisfying
x 0 ], which
is
< a
a
—
+
8
8 (Figure 3).
Once again we know
<
means that / is
Thus the assumption
x0
<
impossible, since f(xo)
a; but this
>
0.
114
Foundations
f(a)
>
0 also leads to a contradiction, leaving f(a)
=
0 as the only possible
alternative.
The
result,
proofs of
which
Theorems
will play
2
and
3 of
Chapter 7 require a simple preliminary
as Theorem 6-3 played in the previ-
much the same role
ous proof.
THEOREM
l
If / is continuous at a, then there
above on the interval (a — 8, a
is
+
PROOF
Since lim f(x)
=
f(a), there
is,
number
a
8) (see
for every s
8
>
Figure
>
0,
0 such that
/
is
bounded
a 8
4).
>
0 such that, for all x,
x—*a
-
if |*
It is
only necessary to apply
do), for example, £
=
if
It
follows,
in
<
a\
this
8,
then
f(a)
1
.
e.
statement to some particular
e
(any one will
.
particular, that
+
- f(a)\ <
We conclude that there is a 8 > 0 such that, for all x,
\x-a\< 8, then |/(*) - f(a)\ < 1.
1
if
completes the proof: on the interval
above by
\f(x)
\x
— a\ < 5,
— 8, a +
(a
then f(x)
8)
— f(a) <
the function /is
1.
This
bounded
|
It should hardly be necessary to add that we can now also prove that / is
bounded below on some interval {a —
a + 5), and, finally, that / is bounded
on some open interval containing a.
A more significant point is the observation that if lim f(x) = f(a), then
<5,
x-*a +
1
1
1
a
FIGURE
4
—
5
a
a
+
5
.'
115
Least Upper Bounds
there
is
>
a 6
0 such that /
similar observation holds
bounded on the set \x\ a < x < a + 8\, and a
= f(b). Having made these observations
is
lim f(x)
if
x—
(and assuming that you
theorem.
THEOREM
7-2
If
/
continuous on
is
supply the proofs), we tackle our second major
will
then /
[a, b],
is
bounded above on
[a, b].
Let
A —
Clearly
A ^
both to the
i.e.,
<
x
b
and /
bounded above on
is
x]}.
[a,
is in A), and A is bounded above (by 6), so A has a least
Notice that we are here applying the term ''bounded above'
A which can be visualized as lying on the horizontal axis, and
0 (since a
upper bound
to /,
<
[x: a
a.
set
y
to the sets {f(y): a
<
<
y
x},
which can be visualized
as lying
on the
vertical axis (Figure 5).
Our first step is to prove that we actually have a = b. Suppose, instead, that
a < 6. By Theorem 1 there is 5 > 0 such that /is bounded on (a —
a + 8).
Since a is the least upper bound of A there is some 0 in A satisfying a — 8 <
x 0 < a. This means that / is bounded on [a, x 0 ]. But if xi is any number with
a < xi < a + 8, then / is also bounded on [x 0 xi]. Therefore /is bounded on
[a, #i], so xi is in A, contradicting the fact that a is an upper bound for A. This
contradiction shows that a — b. One detail should be mentioned: this demonstration implicitly assumed that a < a (so that / would be defined on some
interval (a — d, a + 5)); the possibility a = a can be ruled out similarly,
<5,
,v
,
using the existence of a
The proof
for every x
is
<
6
>
0 such that /is
not quite complete
by
bounded on
—we only know that
not necessarily that /is bounded on
[a,
\x: a
<
<
x
a
+
8}
bounded on [a, x]
b\ However, only one
/
is
small argument needs to be added.
There
A
#o in
[#o, b],
To
THEOREM
7-3
If /
is
is
a 8
so /
is
bounded on
We
[a, b],
continuous on
already
[a, b],
know
bounded. This
a >
x
[a,
<
b}
.
There
is
x 0 ] and also on
then there
is
a
number y
in
trick.
[a, b]
such that /(y)
>
[a, b].
that /
is
bounded on
{/(*):* in
is
<
|
prove the third important theorem we resort to a
f(x) for all x in
PROOF
0 such that / is bounded on {x: b — 8
b — 8 < x 0 < b. Thus / is bounded on
>
such that
set
is
f(x) for x in [a, b]
obviously not
it
Suppose instead that a
suffices to
^ f(y)
0,
b],
[a,
b]}
so
it
which means that the
jv
by
fix)
in
set
has a least upper bound a. Since
show that a
for all
a -
[a,
=
[a, b].
f(y) for
Then
some y
in
the function g
[a,
b\
denned
116
Foundations
continuous on
is
the other hand,
is
denominator of the right side
since the
[a, b],
a
the least upper
bound
of
{
f(x)
x in
:
never
is
[a 9 b]
}
;
this
0.
On
means
that
>
for every £
means
This, in turn,
means that g
this
theorem.
is
x in
[a, b]
with
a —
<
f(x)
e.
that
>
for every £
But
0 there
0 there
x in
is
not bounded on
is
with g(x)
[a y b]
>
1
/e.
contradicting the previous
[a, b],
|
At the beginning of this chapter the set of natural numbers N was given as
an example of an unbounded set. We are now going to prove that N is unbounded. After the difficult theorems proved in this chapter you may be
startled to find such an "obvious" theorem winding up our proceedings. If so,
you are, perhaps, allowing the geometrical picture of R to influence you too
strongly. "Look," you may say, "the real numbers look like
—
-
I
12
—
number
x
1
0
so every
1
I
•
*
—— +
i
1
•
3
is
i
x
n
n
between two integers
n,
1
+
n
1
ger)." Basing the argument on a geometric picture
(unless x
is itself
an
inte-
not a proof, however,
and even the geometric picture contains an assumption: that if you place
unit segments end-to-end
you
will eventually get a
given segment. This axiom, often omitted from a
etry,
is
segment larger than any
first
introduction to geom-
usually attributed (not quite justly) to Archimedes, and the corre-
is
sponding property
for
N
numbers, that
is
not bounded,
is
called the Archi-
consequence of PI -PI 2
(see reference [17] of the Suggested Reading), although it does hold for Q,,
of course. Once we have PI 3 however, there are no longer any problems.
median property of the reals. This property
THEOREM
2
PROOF
N
is
not a
not bounded above.
is
Suppose
N were
bound a
for
bounded above. Since
N ^
0,
there
would be a
least
upper
N. Then
a >
n
ct
>
n
is
in
N. But
for all n in
N.
for all n in
N,
Consequently,
since
+
72
1
is
in
N
if
n
a —
and
this
that
a
means that a
is
—
1 is
1
+
>
also
1
this
means
that
for all n in
N,
an upper bound
for
the least upper bound. |
n
N, contradicting the fact
\
Least Upper Bounds
117
There is a consequence of Theorem 2 (actually an equivalent formulation)
which we have very often assumed implicitly.
THEOREM
3
PROOF
For any
>
£
0 there
a natural
is
number
Suppose not; then 1/n > £ for all n
is an upper bound
means that 1/s
in
n with
1
/n
<
£.
< 1 /s for all n in N. But
N, contradicting Theorem 2. |
N. Thus n
for
this
A brief glance through Chapter 6 will show you that the result of Theorem 3
was used
in the discussion of
many
examples.
Of
course,
Theorem
3
was not
available at the time, but the examples were so important that in order to
give
them some cheating was
honesty
but
if
we can claim
tolerated.
As
partial justification for this dis-
that this result was never used in the proof of a theorem,
your faith has been shaken, a review of
all
the proofs given so far
order. Fortunately, such deception will not be necessary again.
stated every property of the real
no more
numbers that we
We
will ever need.
is
in
have now
Henceforth,
lies.
PROBLEMS
1.
Find the
upper bound and the greatest lower bound (if they exist)
sets. Also decide which sets have greatest and least eledecide when the least upper bound and greatest lower bound
least
of the following
ments (i.e.,
happens to belong
(i)
|-:rcinNj-
(ii)
(-: n in
to the set).
Z and
n
0 or x
=
^
0
•
\n
(iii)
{x:
x
(iv)
{x:
0
(v)
\x:
x2
(vi)
\x:
x2
(vii)
\x:
x
(viii)
2.
(a)
|-
+
=
<
+
+
<
1/n for some n in N}.
< V2 and
> 0}.
x +
x - 1 < 0}.
x
x
is
rational}.
1
0 and x 2
(»l)
n
:
+x-
n in
1
<
0}.
nJ-
Suppose A ^ 0 is bounded below. Let —A denote the set of all —x
x in A, Prove that — A ^ 0, that — A is bounded above, and that
for
—
(b)
sup ( — ^4)
is
the greatest lower
bound
of A.
A 7* 0 is bounded below, let B be the set of all lower bounds of A.
Show that B 7* 0, that B is bounded above, and that sup B is the
If
greatest lower
bound
of A.
118
Foundations
3.
Let/ be a continuous
The proof
(a)
function on
Theorem
of
[a> b]
showed that there
1
<
with }(a)
is
0
<
f(b).
a smallest x in
[a, 6]
with /(#) = 0. Is there necessarily a second smallest x in [a,
with f(x) = 0 ? Show that there is a largest * in [a, 6] with/(x) =
(Try to give an easy proof by considering a
new
6]
0.
function closely
related to /.)
depended upon consideration of A =
{x: a < x < b and / is negative on [a, x]}. Give another proof of
Theorem 1, which depends upon consideration of B = {x: a < x < b
and /(*) < 0}. Which point x in [a, b] with /(*) = 0 will this proof
locate? Give an example where the sets A and B are not the same.
The proof
(b)
4.
Theorem
of
1
Suppose that /is continuous on
>
also that /(*o)
and
0 for
some x 0
[a, b]
and that f(a)
=
/(£)
=
0.
Suppose
b\ Prove that there are numbers c
such that /(c) = f(d) = 0, but/(*) > 0
previous problem can be used to good
in
[a,
a<c<x <d<b
with
0
for all x in
The
Hint:
d).
(c,
advantage.
5.
x
<
<
k
consider
/
x
<
<
r
x
>
Hint: Let
y.
+
1.
<
y.
Suppose x
(b)
y.
Prove that there is an integer £ such that
be the largest integer satisfying / < x, and
1.
/
Prove that there
Hint: If
1
/n
<
y
-
is
x,
number
a rational
then ny
—
nx
(d)
from
1.
r
such that
Why
(Query:
(b)
and
(c).
A set A of real numbers is said to be dense if every open interval contains
a point of A. For example, Problem 5 shows that the
bers
(a)
and the
set of irrational
Prove that
if
/
is
(c)
If
we assume
f(x)
>
continuous and /(#)
instead that f(x)
g(x) for all
x.
set of rational
num-
numbers are each dense.
=
0 for all
dense set A, then f(x) = 0 for all x.
(b) Prove that if / and g are continuous and f{x)
dense set A, then f(x) = g(x) for all x.
7.
>
have parts (a) and (b) been postponed until this problem set?)
Suppose that r < s are rational numbers. Prove that there is an
irrational number between r and s. Hint: As a start, you know that
there is an irrational number between 0 and 1.
Suppose that x < y. Prove that there is an irrational number between
x and y. Hint: It is unnecessary to do any more work; this follows
(c)
*6.
—
Suppose that y
(a)
Can > be
>
—
numbers x
g{x) for all x in a
g(x) for all x in A,
replaced by
>
in a
show that
throughout?
Prove that if / is continuous and f(x + y) = f(x) + f(y) for all x and y,
then there is a number c such that f(x) = cx for all x. (This conclusion
can be demonstrated simply by combining the results of two previous
problems.) Point of information: There do exist noncontinuous functions/
prove this
satisfying /(#
y) = f(x)
f(y) for all x andy, but we cannot
now; in fact, this simple question involves ideas that are usually never
+
+
Least Upper Bounds
mentioned
any undergraduate
in
course.
119
The Suggested Reading
con-
tains references.
*8.
Suppose that /
(Figure
(a)
Prove that lim f(x) and lim f(x) both
-
o
i—
problem
(b)
(c)
whenever a
<
b
exist.
Hint:
Why
is
this
in this chapter?
Prove that / never has a removable discontinuity (this terminology
comes from Problem 6-16).
Prove that if / satisfies the conclusions of the Intermediate Value
If/is a
is
continuous.
bounded function on
=
[0, 1], let |||/|||
sup
{|/(*)|:
x in
[0, 1]}.
Problem 7-14.
number
x
can
be written uniquely in
every
that
Prove
a
0.
Suppose
>
the form x = ka + x where k is an integer, and 0 < x' < a.
is a sequence of positive numbers with
(a) Suppose that a h a 2 a z
fln+i < a n /2. Prove that for any £ > 0 there is some n with a n < S.
Prove analogues of the properties of
10.
f{b)
x—>a +
Theorem, then /
*9.
<
a function such that f(a)
is
6).
||
||
in
f
,
11.
,
(b)
Suppose
P
is
,
.
.
.
a regular polygon inscribed inside a
inscribed regular polygon with twice as
many
circle. If
sides,
P
f
is
the
show that the
f
and the area of P is less
than half the difference between the area of the circle and the area
difference between the area of the circle
P (use Figure 7).
Prove that there is a regular polygon P inscribed in a circle with
area as close as desired to the area of the circle. In order to do part
(c) you will need part (a). This was clear to the Greeks, who used
part (a) as the basis for their entire treatment of proportion and
of
(c)
area.
By calculating
the areas of polygons, this
of exhaustion") allows computations of
Archimedes used
it
to
show that
tt
<
to
ir
method ("the method
any desired accuracy;
< V- But it has far
greater theoretical importance:
*(d)
12.
two regular polygons with the same
number of sides have the same ratio as the square of their sides,
prove that the areas of two circles have the same ratios as the square
of their radii. Hint: Deduce a contradiction from the assumption
that the ratio of the areas is greater, or less, than the ratio of the
Using the
square of the radii by inscribing appropriate polygons.
A and B are two nonempty sets of numbers such that x
Suppose that
for all x in
(a)
(b)
13.
fact that the areas of
A and
all
Prove that sup
Prove that sup
<
y
y in B,
A <
A <
y for all y in B.
inf B.
A and B be two nonempty sets of numbers which are bounded above,
let A + B denote the set of all numbers x + y with x in A and y in B,
Prove that sup (A + B) = sup A + sup B. Hint: The inequality
Let
and
+
sup (A
B <
sup
+
(A
+
B)
A —
sup
FIGURE
<
B)
+
suffices to
£ for all £
>
0;
—
i?
+
+
<
y
e/2.
—
H
1"
1
8
A
prove that sup
sup B < sup
prove that sup A
in 5 with
begin by choosing x in A and
it
s/2 and sup
Why? To
easy.
is
B)
<
x
+ sup B
A
sup
sup (A
I
H
jj
14.
Consider a sequence of closed intervals h
(a)
that there
Show
(b)
is
<
and
a point x which
Suppose that a n
<z
£ n+ i
n+1
that this conclusion
is
<
[a i,
in every In
b\\,h
~
[a 2, £2],
•
•
•
.
(Figure 8). Prove
.
we
false if
is
=
6 n for all n
consider open intervals
instead of closed intervals.
The simple result of Problem
It
1
4(a)
is
called the "Nested Interval
may be used to give alternative proofs of Theorems 1 and
2.
Theorem.
5 '
The appropriate
reasoning, outlined in the next two problems, illustrates a general method,
called a "bisection
*15.
Suppose /
f((a
+
is
b) /2)
argument."
continuous on
= 0,
[a, b]
and
f(a)
<
0
<
Then
f(b).
or / has different signs at the end points of
+
[a, (a
either
+ b)/2],
b)/2 b]. Why? If
/ nas different signs at the end points of [(a
b)/2) y£ 0, let Ix be one of the two intervals on which / changes
f((a
or
}
+
Now bisect h. Either / is 0 at the midpoint, or / changes sign on
one of the two intervals. Let I2 be such an interval. Continue in this
way, to define In for each n (unless / is 0 at some midpoint). Use the
Nested Interval Theorem to find a point x where f(x) = 0.
Suppose / were continuous on [a, b], but not bounded on [a, b\ Then
/would be unbounded on either [a, (a + b)/2] or [(a + b)/2, b]. Why?
Let Ii be one of these intervals on which / is unbounded. Proceed as in
Problem 15 to obtain a contradiction.
(a) Let A — {x: x < a}. Prove the following (they are all easy):
sign.
*16.
17.
If x
(i)
(b)
A
in
(iii)
(iv)
If x
is
A and
y
<
x,
in A, then there
Suppose, conversely, that
{x:
*18.
is
A ?*0.
A * R.
(ii)
x
<
number
finitely
then y
is
is
in A.
some number
A
satisfies
x' in
A
such that x
Prove that
(i)-(iv).
<
x''.
A —
sup A}.
x
is
called
an almost upper bound
many numbers
y in
A
with y
>
x.
An
for
A
if
there are only
almost lower
bound
is
defined similarly.
(a)
all almost upper bounds and almost lower bounds of the sets
Problem 1.
Suppose that A is a bounded infinite set. Prove that the set B of all
almost upper bounds of A is nonempty, and bounded below.
Find
in
(b)
_
-
Least Upper Bounds
from part
B
number
121
(c)
It follows
(d)
and denoted by lim A or lim sup A. Find lim A
for each set A in Problem 1.
Define lim A, and find it for all A in Problem 1.
(b) that inf
exists; this
called- the
is
limit superior of A,
*19.
If
A
is
a bounded
prove
infinite set
(c)
A < lim A.
lim A < sup A.
If lim A < sup A,
(d)
The analogues
lim
(a)
(b)
,
A
then
contains a largest element.
of parts (b)
and
lim
(c) for
— _—
——
~j
.
—
L
-shadow points
FIGURE
*20.
9
Let / be a continuous function on R. A point x is called a shadow point
if there is a number y > x with f(y) > f(x). The rationale for this
terminology is indicated in Figure 9; the parallel lines are the rays of
of /
the sun rising in the east (you are facing north). Suppose that
of
(a, b)
(a)
are
For x in
shadow
(a, b),
points, but that a
prove that f(x)
<
and
f(b).
b
are not
Hint: Let
all
points
shadow
points.
A —
x
{y:
<
y
<
b
and f{x) < f(y) If sup A were less than b, then sup A would be a
shadow point. Use this fact to obtain a contradiction to the fact that
b is not a shadow point.
}
(b)
Now
.
prove that f(a)
<
f(b).
(This
is
a simple consequence of
continuity.)
(c)
Finally, using
the fact that a
'
f(a)
is
not a shadow point, prove that
=f(b).
is known as the Rising Sun Lemma. Aside from serving
good illustration of the use of least upper bounds, it is instrumental in proving several beautiful theorems that do not appear in
this book; see page 371.
This result
as a
PART
DERIVATIVES
AND
INTEGRALS
In 1604, at the height of
his scientific career, Galileo argued
that for a rectilinear motion
in
which speed increases proportionally
to
distance covered,
law of motion should be
the
just that (x
=
ct
2
)
which he had discovered
in the investigation offalling
Between 1695 and 1700
bodies.
not a single one
of the monthly issues
of Leipzig's Acta Eruditorum was published
without articles of Leibniz,
the Bernoulli brothers
or the
Marquis
de I'Hopital treating,
with notation only slightly different from
that which
the
we
use today,
most varied problems of
differential calculus, integral calculus
and
the calculus
Thus
of variations.
in the space
of almost precisely
one century
infinitesimal calculus or,
we now call
The Calculus,
as
it
in English,
the calculating tool
par
excellence,
had been forged;
and
nearly three centuries of
constant use have not completely dulled
this
incomparable instrument.
NICHOLAS BOURBAKI
CHAPTER
DERIVATIVES
The derivative of a
function
Together with the
derives
its
is
integral,
particular flavor.
the
it
first
of the
two major concepts of this section.
from which calculus
constitutes the source
While
it is
true that the concept of a function
is
fundamental, that you cannot do anything without limits or continuity, and
that least upper bounds are essential, everything we have done until now has
been preparation if adequate, this section will be easier than the preceding
ones
— for
—
the really exciting ideas to come, the powerful concepts that are
truly characteristic of calculus.
Perhaps (some would say "certainly") the interest of the ideas to be introduced in this section stems from the intimate connection between the mathematical concepts and certain physical ideas. Many definitions, and even some
theorems,
may
be described in terms of physical problems, often in a revealing
the demands of physics were the original inspiration for these
way. In fact,
fundamental ideas of calculus, and we shall frequently mention the physical
interpretations. But we shall always first define the ideas in precise mathematical form, and discuss their significance in terms of mathematical problems.
The collection of all functions exhibits such diversity that there is almost no
hope of discovering any
interesting general properties pertaining to
all.
Because continuous functions form such a restricted class, we might expect to
find some nontrivial theorems pertaining to them, and the sudden abundance
of theorems after Chapter 6 shows that this expectation is justified. But the
most interesting and most powerful results about functions will be obtained
only when we restrict our attention even further, to functions which have even
greater claim to be called "reasonable," which are even better behaved than
most continuous functions.
(a)
FIGURE
(b)
(c)
1
Figure 1 illustrates certain types of misbehavior which continuous functions
can display. The graphs of these functions are "bent" at (0, 0), unlike the
graph of Figure 2, where it is possible to draw a "tangent line" at each point.
The quotation marks have been used to avoid the suggestion that we have
125
126
Derivatives and Integrals
defined "bent" or "tangent line/ although we are suggesting that the graph
might be "bent" at a point where a "tangent line" cannot be drawn. You
have probably already noticed that a tangent line cannot be defined as a line
which intersects the graph only once such a definition would be both too
restrictive and too permissive. With such a definition, the straight line shown
in Figure 3 would not be a tangent line to the graph in that picture, while the
parabola would have two tangent lines at each point (Figure 4), and the three
functions in Figure 5 would have more than one tangent line at the points
where they are "bent."
5
—
FIGURE
3
A
more promising approach
to the definition of a tangent line
with "secant lines," and use the notion of limits.
points (a y f(a))
whose slope
and
(a
+
h,f(a
+
If h
^
h)) determine, as in
0,
might
start
then the two distinct
Figure
6,
a straight line
is
f(a
+
-/(«)
h)
h
a
+
>f(a
/
FIGURE
h /(<
[a
y
+
h)
)
+
-
h))
f(a)
H
6
As Figure 7 illustrates, the "tangent line" at (a, f(a)) seems to be the limit, in
some sense, of these "secant lines," as h approaches 0. We have never before
talked about a "limit" of lines, but
we
can talk
about the limit of their slopes:
127
Derivatives
the slope of the tangent line through
f{a
\ir*
+ h)
DEFINITION
are ready for a definition,
The
function /
and some comments.
differentiable at a
is
~~ f{a)
h
h-^o
We
f(a)) should be
(a,
if
exists.
h
h-^o
In this case the limit
a.
(We
is
also say that /
denoted by /'(a) and
is
differentiable
if
/
called the derivative of / at
is
differentiable at a for every a
is
domain of /.)
in the
an addendum; we define the
tangent line to the graph of /at (a, f(a)) to be the line through (a, f(a)) with
slope f{a). This means that the tangent line at (a, J{a)) is defined only if / is
The
first
comment on our
differentiable at
definition
is
really
a.
refers to notation. The symbol f(a) is certainly
reminiscent of functional notation. In fact, for any function/, we denote by/'
the function whose domain is the set of all numbers a such that / is differ-
The second comment
entiable at
a,
and whose value
at such a
h^o
(To be very
for
precise: /'
which lim
[f(a
is
number
is
h
the collections of
+ h) - f(a)]/h
a
pairs
all
exists.)
The
function /'
is
called the
derivative of /.
comment, somewhat longer than the previous two, refers to the
physical interpretation of the derivative. Consider a particle which is moving
along a straight line (Figure 8(a)) on which we have chosen an "origin" point 0,
Our
and a
third
direction in
which distances from 0
shall
be written as positive numbers,
of points in the other direction being written as negative
the distance from
0
numbers. Let
denote the distance of the particle from 0, at time
s(t) has been chosen purposely; since a distance
s{t)
suggestive notation
FIGURE
8(a)
0
line
along which particle
is
moving
t.
The
s(t)
is
128
Derivatives and Integrals
determined
for
each number
with a certain function
from 0, on the
s.
/,
the physical situation automatically supplies us
of s indicates the distance of the particle
The graph
vertical axis, in terms of the time, indicated
on the horizontal
axis (Figure 8(b)).
"distance"
The
quotient
s(a
+ h) —
s{a)
h
has a natural physical interpretation.
It is the "average velocity" of the partiduring the time interval from a to a
h. For any particular a, this average
speed depends on k, of course. On the other hand, the limit
+
cle
FIGURE
8(b)
lim
—+^ ~ W
h->0
h
J
depends only on a (as well as the particular function s) and there are important
physical reasons for considering this limit.
would like to speak of the
We
"velocity of the particle at time
a"
but the usual definition of velocity is
really a definition of average velocity; the only reasonable definition of
"velocity at time a" (so-called "instantaneous velocity") is the limit
Rm
s (a
+
define
Notice that
-
s(a)
h
h->0
Thus we
h)
the (instantaneous) velocity of the particle at a to be s'(a).
could easily be negative; the absolute value \s\a)\
times called the (instantaneous) speed.
s'(a)
is
some-
It is important to realize that instantaneous velocity is a theoretical concept,
an abstraction which does not correspond precisely to any observable quantity. While it would not be fair to say that instantaneous velocity has nothing
to do with average velocity, remember that s\t) is not
s(t
+
h)
-
s(t)
h
any particular
k, but merely the limit of these average velocities as h
Thus, when velocities are measured in physics, what a physicist
really measures is an average velocity over some (very small) time interval;
such a procedure cannot be expected to give an exact answer, but this is
for
approaches
really
The
no
0.
defect, because physical
velocity of a particle
is
measurements can never be exact anyway.
often called the "rate of change of
its
position."
This notion of the derivative, as a rate of change, applies to any other physical
situation in which some quantity varies with time. For example, the "rate of
change of mass" of a growing object means the derivative of the function m,
where m(t)
is the mass at time /.
In order to become familiar with the basic definitions of this chapter, we
will spend quite some time examining the derivatives of particular functions.
Before proving the important theoretical results of Chapter 11,
we want
to
Derivatives
129
have a good idea of what the derivative of a function looks like. The next
chapter is devoted exclusively to one aspect of this problem calculating the
derivative of complicated functions. In this chapter we will emphasize the
concepts, rather than the calculations, by considering a few simple examples.
—
Simplest of all
is
=
a constant function, f(x)
In this case
c.
=
h
k
a-»o
Thus /
is
differentiable at a for every
0.
h
fc-*o
h-*Q
number
and
a,
f'(a)
that the tangent line to the graph of / always has slope
always coincides with the graph.
0,
—
0.
This means
so the tangent line
Constant functions are not the only ones whose graphs coincide with their
this happens for any linear function f(x) = cx + d. Indeed
tangent lines
—
f'(a)
=
lim
+ h)
f(a
'
h-
= hm
-f(a)
h
o
—+
c(a
+
h)
d
lim
the slope of the tangent line
A
— =
h->o
h
is c,
the
same
d]
as the slope of the
=
+
h)
=
lim
=
lim
slope 2a
= hm
f(a
•
x2
graph of/.
Here
.
— f(a)
(a
+ h) -
a
a2
+
h2
2
+
2ah
a— o
Some
+
c\
refreshing difference occurs for f(x)
f'(a)
[ca
ch
..
=
-
h
h->o
2
-
a2
h
=
lim {2a
=
2a.
+
h)
of the tangent lines to the graph of / are
shown
in Figure 9. In this pic-
ture each tangent line appears to intersect the graph only once,
fairly easily: Since the tangent line
can be checked
it is the graph of the function
FIGURE
=
=
g(x)
9
Now,
if
2a(x
2ax
— a) +
—a
or
In other words,
(a,
a 2)
is
x
2
so
(x
or
x
=
2ax
2ax
=
a
2
)
and
this fact
has slope 2a,
a2
.
—
—
a)
(a,
2
the graph of / and g intersect at a point
x2
through
2
—
(x,
/(*))
a2
+
a2
=
0
=
0;
a.
the only point of intersection.
=
0, g(x)), then
130
Derivatives
and
Integrals
The
=
function /(#)
x2
happens
to be quite special in this regard; usually
more than once. Consider,
a tangent line will intersect the graph
=
the function f(x)
x3
.
for
example,
In this case
lim*
/>(,)-
h
h->0
+
lim^
=
-
A)3
a5
h
h-^o
a
3
+
=
lim
=
lim
=
lim 3a 2
=
3a 2
+
3a 2 h
+
3a 2h
3ah 2
3ah 2
+
+h 3
a
3
h3
h->0
Thus the tangent
is
of /
and g
graph of / at
has got to be x
is
easily solved
=
a,
—
=
+
=
3a x
— a) +
- 2a
.
This means
.
(*,
=
3a 2x
-
2a 3
-
3a 2 x
+
2a 3
=
f{x))
a) is
We
a factor of the
-
ax
—
2a 2 also has x
a)(x
+
-
g(x))
when
solution of the equation
left side;
the other factor
obtain
a)(x
2
—
0.
we remember that one
—
-
-
a3
3
(x
(x
j{x)
2
x3
if
so that (x
happens that x 2
3a 2 (x
x3
can then be found by dividing.
It so
2
a 3 ) has slope 3a
(a,
intersect at the point
or
This equation
2
the graph of
g(x)
The graphs
+h
3ah
.
line to the
that the tangent line
+
ax
a)(x
+
2a 2 )
—
=
0.
a as a factor;
2a)
=
we obtain
finally
0.
*»
Thus, as illustrated in Figure 10, the tangent line through (a, a 3 ) also intersects the graph at the point { — 2a, —8a 3 ). These two points are always dis//slope 3 a 2
-2a
1
Vfa
-^H
tinct,
except
when
a
=
0.
We
3
a
)
have already found the derivative of sufficiently many functions to
illustrate the classical, and still very popular, notation for derivatives. For a
given function /, the derivative /' is often denoted by
df(x)
dx
For example, the symbol
dx 2
FIGURE
10
dx
Derivatives
denotes the derivative of the function /(*)
parts of the expression
=
131
x 2 Needless to say, the separate
.
df(x)
dx
9
are not supposed to have any sort of independent existence— the d s are not
numbers, they cannot be canceled, and the entire expression is not the quotient
two other numbers "#(*)" and "dx" This notation is due to Leibniz
(generally considered an independent co-discoverer of calculus, along with
Newton), and is affectionately referred to as Leibnizian notation.* Although
of
the notation df(x)/dx seems very complicated, in concrete cases it may be
2
shorter; after all, the symbol dx /dx is actually more concise than the phrase
*V
"the derivative of the function /(*) =
The following formulas state in standard Leibnizian notation
mation that we have found
all
the infor-
so far:
dc
=
o,
=
a
=
2x,
—=
3* 2
dx
d(ax
+
b)
9
dx
dx 2
dx
dx z
.
dx
Although the meaning of these formulas is clear enough, attempts at literal
interpretation are hindered by the reasonable stricture that an equation should
not contain a function on one side and a number on the other. For example, if
be true, then either df(x)/dx must denote /'(x), rather
than/', or else 2x must denote, not a number, but the function whose value at
x is 2x. It is really impossible to assert that one or the other of these alternatives
/' and sometimes means
is intended; in practice df(x)/dx sometimes means
the third equation
is
to
denote either a number or a function. Because of this
are reluctant to denote f{a) by
authors
ambiguity, most
/'(*),
while 2x
may
—— W,
df(x)
dx
instead /'(a)
is
usually denoted
by the barbaric, but unambiguous, symbol
dx
*
Leibniz was led to this
symbol by
his intuitive notion of the derivative,
+ h) - f(x)]/h,
which he considered
but the "value" of this quotient when h
was denoted by dx and the
is an "infinitely small" number. This "infinitely small" quantity
dx) - /(*) by df(x). Although this point
corresponding "infinitely small" difference f(x
view is impossible to reconcile with properties (P1)-(P13) of the real numbers, some people
to be, not the limit of quotients [f(x
+
of
find this notion of the derivative congenial.
In addition to these
difficulties,
more ambiguity. Although
Leibnizian notation
the notation dx 2 /dx
is
associated with one
absolutely standard, the
is
notation df(x)/dx is often replaced by df/dx. This, of course, is in conformity
with the practice of confusing a function with its value at x. So strong is this
tendency that functions are often indicated by a phrase like the following:
We
"consider the function^ = x ."
will sometimes follow classical practice to
the extent of using y as the name of a function, but we will nevertheless carefully distinguish between the function and its values— thus we will always say
2
"consider the function (defined by) y(x) = x 2 ."
ambiguities of Leibnizian notation, it is used almost
exclusively in older mathematical writing, and is still used very frequently
something
like
Despite the
The
today.
around
many
staunchest opponents of Leibnizian notation admit that it will be
some time, while its most ardent admirers would say that it
for quite
be around forever, and a good thing too In any case, Leibnizian notation
cannot be ignored completely.
will
!
The
policy adopted in this book
the text, but to include
is
to disallow
Leibnizian notation within
in the Problems; several chapters contain a
few
(immediately recognizable) problems which are expressly designed to illustrate the vagaries of Leibnizian notation. Trusting that these problems will
provide ample practice in this notation, we return to our basic task of examining some simple examples of derivatives.
The few
it
functions examined so far have
been differentiable.
all
To
fully
appreciate the significance of the derivative it is equally important to know
some examples of functions which are not differentiable. The obvious candidates are the three functions
Figure
Now
1
;
\h\/h
if
first discussed in this chapter, and illustrated in
they turn out to be differentiable at 0 something has clearly gone
=
1
for h
>
0,
\h\/h
f(h)
- /(0)
Jim
h-+0
In
= -
and
1
for h
does not
<
0.
This shows that
exist.
h
fact,
lim
m
-f(Q) =
i
h
and
lim
Kh) -/(0)
h
(These two limits are sometimes called the right-hand derivative and the
left-hand derivative, respectively, of / at 0.)
Derivatives
If a 7* Q,
then
does
/''(a)
=
= -1
1
/'O)
The proof of this
fact
a linear function).
is left
In fact,
exist.
f(x)
to
you
The graphs
133
if
x
if
x
>
<
0,
0.
easy if you remember the derivative of
and of /' are shown in Figure 11.
(it is
of /
For the function
x
x2
=
fix)
i
-°
>
We
a similar difficulty arises in connection with f(0).
-=
- 7(0)
/(«
lim
0
h
=
\
Therefore,
<
A
A,
have
^-
A
1,
/(0)
=
>
0.
0,
but
Thus
/'(0) does not exist;
fix)
exists for x
The graphs
of /
^
0
and
—
/
it is
/' are
shown
Even worse things happen
f(h)
-
not differentiable at
is
0.
Once
again, however,
easy to see that
/(0)
in
for /(*)
_
)
2x,
x
1,
*
<
>
0
0.
Figure 12.
=
Vjlj. For this function
Vh
1
A
VA
h
>
0
h
<
0.
In this case the right-hand limit
1
lim
lim
h
does not
exist; instead
And, what's more,
1/Va becomes
h-+o+
_
Vh
arbitrarily large as h approaches 0.
— /V ~k becomes arbitrarily large in absolute value,
negative (Figure 13).
1
but
.
134
Derivatives and Integrals
The
better
=
function f(x)
behaved than
~
f(h)
/(*)
\/x, although not differentiable at 0,
this.
The
<Th
/(0)
_
h llz
=
h
simply becomes arbitrarily large as h goes to
= —
/(#)
0.
at 0. Geometrically this
is
at least a little
1
m
{<rh y
Sometimes one says that /has
means that the graph of/ has
parallel to the vertical axis (Figure 14).
Of course,
y/x has the same geometric property, but one would say that / has
a derivative of "negative infinity" at
FIGURE
1
=
h
an "infinite" derivative
a "tangent line" which
is
quotient
Remember
0.
supposed to be an improvement over
mere continuity. This idea is supported by the many examples of functions
which are continuous, but not differentiable; however, one important point
14
that differentiability
is
remains to be noted:
THEOREM
1
If
/
is
differentiable at
PROOF
lim /(a
+
a,
then /
h)
is
=
f(a)
•
continuous at
lim
f(a
+
= yim f{a
As we pointed out
in
Chapter
5,
f(a)
+ h) ~ f(a)
h
a-»o
f(a)
—
L
h
h-+o
=
=
h)
a.
•
lim h
h->0
0
0.
the equation lim f(a
+
h)
— f(a) =
0
is
h->0
equivalent to lim f(x)
It is
=
f(a)
;
thus /
very important to remember
remember
that the converse
is
continuous at
is
Theorem
not true.
1,
a. |
and
just as important to
A differentiable function is continuous,
but a continuous function need not be differentiable (keep in mind the func= |*|, and you will never forget which statement is true and which
tion f(x)
false)
The continuous
functions
examined
points with at most one exception, but
so far
it is
have been differentiable
functions which are not differentiable at several points, even an infinite
(Figure 15). Actually, one can do
FIGURE
15
at all
easy to give examples of continuous
much worse than
this.
There
is
number
a function
Derivatives
(d)
(c)
FIGURE
16
which
continuous everywhere
is
135
and
differ entiable
nowhere! Unfortunately, the defini-
Chapter 23, and I have been
unable to persuade the artist to draw it (consider carefully what the graph
should look like and you will sympathize with his point of view). It is possible
to draw some rough approximations to the graph, however; several successively better approximations are shown in Figure 16.
Although such spectacular examples of nondifFerentiability must be postponed, we can, with a little ingenuity, find a continuous function which is not
differentiable at infinitely many points, all of which are in [0, 1]. One such function of this function will be inaccessible to us until
tion
is
illustrated in Figure 17.
precisely;
it is
The
reader
r,
.
/(*)
=
[x sin -5
\
I 0,
This particular function /
ability.
is
given the problem of defining
it
a straight line version of the function
Indeed, for h
^
0
is itself
x
^
0
x
=
0.
x
quite sensitive to the question of differenti-
we have
,
/z
1
•
0
sin
A
=
1
.
sin -•
h
h
Long ago we proved
that lim sin
at 0. Geometrically,
one can
that the secant line through
between
—1 and
1,
1
/h does not exist, so f
see that a tangent line
and
(0, 0)
no matter how small we require
-i
0,
not differentiable
cannot
f{h)) in Figure 18
(h,
x sin
FIGURE
is
exist, by noting
can have any slope
h to be.
x
?±
0
x
=
0
18
This finding represents something of a triumph; although continuous, the
somehow quite unreasonable, and we can now enunciate one
mathematically undesirable feature of this function it is not differentiable
function /seems
—
one should not become too enthusiastic about the criterion
of differentiability. For example, the function
at 0. Nevertheless,
,
N
g(.x)
=
\
x 2 sin
x
j
differentiable at 0; in fact ^'(0)
=
=
x
0,
«•
0
X
{
0
0:
1
U2 sin r
•
= hm
lim
h-^o
h
h-+o
=
lim A sin -
=
0.
h
h-+o
The tangent
line to the
graph of g
h
at (0, 0)
is
therefore the horizontal axis
(Figure 19).
This example suggests that we should seek even more restrictive conditions
on a function than mere differentiability. We can actually use the derivative to
:
Derivatives
FIGURE
137
19
formulate such conditions
if
we introduce another
set of definitions,
the last of
this chapter.
For any function
we
/,
by taking the
obtain,
(whose domain
may
differentiability
can be applied to the function
function
at
The
a.
whose domain
(/')',
function
derivative of /.
a,
If
(/')' is
f
t
(a) exists,
of a particle
acceleration at time
new function /'
The notion of
t.
or course, yielding another
usually written simply /"
f
and the number/'^)
/',
consists of all points a such that/'
is
then /
called the
In physics the second derivative
tion at time
derivative, a
be considerably smaller than that of /).
is
and
is
is
differentiate
called the
second derivative off at
a.
particularly important. If s(t)
is
moving along a
second
said to be 2-times differentiable at
straight line, then s"{t)
is
is
the posi-
called the
Acceleration plays a special role in physics, because, as
on a particle is the product of its
you can feel the second derivative
stated in Newton's laws of motion, the force
mass and
its
acceleration. Consequently
when you
sit in an accelerating car.
There is no reason to stop at the second derivative we can define /'" =
(/")'» /"" = (/'")', etc. This notation rapidly becomes unwieldy, so the fol-
—
lowing abbreviation
is
usually adopted
=
/(*+D
(it is
really a recursive definition)
(/<*>)'.
Thus
/
(l
>
=
=
/',
= (/y,
fW =/"' = (/")',
/(« = fx" = {f » y>
/<«
f>
etc.
various functions f {h \ for k
derivatives of /.
The
Usually,
have f
for
{k)
we
>
resort to the notation
2,
f
defined for smaller k also. In
k)
are sometimes called higher-order
only for k
fact,
>
4,
but
it is
convenient to
a reasonable definition can be
made
(0
f \ namely,
/CO,
=
/.
Leibnizian notation for higher-order derivatives should also be mentioned.
138
Derivatives and Integrals
The
is
natural Leibnizian symbol for /"(*), namely,
abbreviated to
W
Similar notation
The
=
to
-
..
following example illustrates the notation
how
and
—
x2
.
Then,
=
=
/'"W =
/<*>(*) =
/'(*)
/"(*)
Figure 20 shows the function
-
whose graph
2
is
shown
/,
in
2x,
2,
0,
if
0,
>
*
3.
together with
various derivatives.
its
by the following function,
Figure 21(a):
/(*)
It is
one
we have already checked,
as
A rather more illuminating example is presented
/"(*)
also shows, in
various higher-order derivatives are related to the
original function. Let /(*)
2x
more frequently
used for / (n) (*)
is
very simple case,
'f'(x)
or
'
2
x2
=
x>0
,
-x\
x
<
0.
easy to see that
(c)
/'(«)
2a
/'(«)
-2a
AO) =
lim
if a > 0,
ifa<0.
Moreover,
/<*>(*) ** 0, k
>
lim
3
-/(0)
f{h)
f(h)
Now
(d)
FIGURE
h-*0 +
20
and
lim
^
=
lim
h
Z*' _
0)
so
/'(0)-lim^ =
*->o
This information can
all
be summarized as follows:
f{x)
It follows that /"(0)
0.
h
=
2|*|.
does not exist! Existence of the second derivative
is
thus a
:
139
Derivatives
rather strong criterion for a function to
some
tion like / reveals
Even
satisfy.
"smooth looking" func-
a
when examined with
irregularity
the second derivative.
This suggests that the irregular behavior of the function
sm
~>
0,
I,
might
^'(O)
by the second
also be revealed
=
but
0,
we
know
do not
x
7*
0
x
=
0
derivative.
g'(a) for
any
a
At the moment we know that
^
0, so it is
hopeless to begin
computing g"(0). We will return to this question at the end of the next chapter,
after
we have
perfected the technique of finding derivatives.
PROBLEMS
1.
(a)
Prove, working directly from the definition, that
= ~\/a\
(b)
2\x\
2.
(a)
for a
*
f(a)
Prove that the tangent line to the graph of / at
intersect the graph of /, except at {a, I /a).
Prove that
if
/(*)
- 1A
2
,
then f'(a)
which
point,
(*)
=
2,
Prove that if/W
you obtain
x > 0
A, then
1
= ~2/a*
(a,
for a
\/a) does not
/at
*
0.
(a,
= 1/2^, for a > 0. (The expression
- f(a)]/h will require some algebraic face
^x, then/' (a)
for [f(a
+
h)
For each natural number
Si'(x)
to
but the answer should suggest the right
lifting,
4.
lies
=
=
1/a 2 ) intersects / at one other
on the opposite side of the vertical axis.
(b) Prove that the tangent line
3.
if /(*)
0.
let
n,
£n (*) =
trick.)
xw
.
Remembering
that
and &'(*) = 3a: 2 conjecture a formula for
1,
n
Prove your conjecture. (The expression (x + A) may be expanded
=
=
&'(*)
2x
,
y
Sn '(x).
by the binomial theorem.)
(c)
/W
=
5.
Find/
6%
Prove, starting from the definition (and drawing a picture to illustrate)
7.
= /(*) + c, then g'(x) = /'(*);
= c/W, then £'(*) = </'(*).
Suppose that f(x) — #
(a)
if
if
[*].
g(x)
(b) if g(x)
3
.
(a)
(b)
(c)
What is /'(9), /'(25), /'(36)?
What is/'(3 ),/'(5 ),/'(6 2 )?
What isf(a 2 ),/'(* )?
2
2
2
you do not find this problem silly, you are missing a very important
2
point: f(x ) means the derivative of / at the number which we happen
If
2
to be calling x
;
it is
not the derivative at x of the
Just to drive the point
(d)
For /(*)
=
function g(x)
=
home:
x\ compare
f'(x
2
)
and
g'(x)
where g(x)
=
f{x
2
).
f{x
2
).
140
Derivatives and Integrals
8.
(a)
(b)
= fix +
why
+
=
a)
is
and
differentia ble
periodic, with period a
Prove that
f(x) for all x).
/'
(i.e.,
also periodic.
is
Find fix) and also fix + 3) in the following cases. Be very methodical,
or you will surely slip up somewhere. Consult the answers (after you do
the problem, naturally).
fx) =
+
f(x +
fx
(ii)
(iii)
Find f(x)
not
11.
a picture to illustrate
should be true, also.
this
Suppose that /
(i)
10.
Prove (starting from the definition) that
this. To do this prob-
c).
Draw
•
f(x
9.
+
f(x
c).
lem you must write out the definitions of g'{x) and f(x + c) correctly. The purpose of Problem 7 was to convince you that although
this problem is easy, it is not an utter triviality, and there is something to prove: you cannot simply put prime marks into the equation
To emphasize this point:
gi x ) — f( x + o).
Prove that if g(x) = f(cx), then g'{x) = c f(cx). Try to see pictorially
(c)
=
Suppose g{x)
g'(x)
(a)
fix)
= ix + 5)\
= git + x), and
Prove that Galileo was wrong:
form
f
and
sit)
s
=
proportional to
is
ct
-
if /(*)
+
git
x).
The answers
will
fact will
=
(i)
s"(t)
(ii)
[/(/)]*
if
a body
falls
then
cannot be a function of the
s,
s
a distance
s(t)
in
t
sec-
2
.
Prove that the following
first
(c)
3)
3)\
= x\
be the same.
onds,
(b)
if
+
ix
3)
about
facts are true
show why we switched from
a (the acceleration
=
is
s if sit)
c to
=
(a/2)t 2 (the
a/2):
constant).
lasit).
If s is measured in feet, the value of a is 32. How many seconds do
you have to get out of the way of a chandelier which falls from a 400foot ceiling? If you don't make it, how fast will the chandelier be
going when it hits you? Where was the chandelier when it was mov-
ing with half that speed?
12.
Imagine a road on which the speed limit
In other words, there
is
specified at every single point.
is
a certain function
miles from the beginning of the road
driving along this road; car
^4's
is
L such
£(*).
that the speed limit x
Two
position at time
t
is
cars,
a(t),
A and B are
and car B's is
y
bit).
(a)
What
equation expresses the fact that car
speed limit? (The answer
(b)
Suppose that
at time
t
A
is ^4's
Suppose
B
not a' it)
=
always goes at the speed
position at time
the speed limit at
(c)
is
all
t
—
1.
A
always travels at the
Lit).)
limit,
Show
and that
that
B
is
2?'s
position
also going at
times.
always stays a constant distance behind A. Under what
conditions will B still always travel at the speed limit?
141
Derivatives
13.
Suppose
=
that/(<z)
g(a)
and that the left-hand derivative of/ at a equals
< a, and
the right-hand derivative of g at a. Define h{x) = fix) for x
h(x) = g(x) for x > a. Prove that h is differentiate at a.
14.
Let fix)
—
x2
if
x
is
rational,
and fix) = 0 if x is irrational. Prove that /is
by this function. Just write out the
differentiable at 0. (Don't be scared
definition of /'(0).)
*15.
(a)
Let / be a function such that |/(*)| < x 2 for all x. Prove that / is
differentiable at 0. (If you have done Problem 14 you should be able
do
to
(b)
this.)
This result can be generalized
x2
if
is
replaced by
\g(x)\>
where g has
what property?
>
16.
Let a
17.
Let 0 < 0 < 1. Prove that
/ is not differentiable at 0.
*18.
Let /(*)
1.
prove
this for irrational a.
= -0.00
Suppose that
=
f(a)
and that f(a) =
a
,
prove that /
>
is
0
\x\
differentiable at 0,
and/(0)
=
then
0,
.
+ h)
.
=
0tf n+ ia n+2
.
=
g(a)
If a
...
n.aia 2 a 5
— f(a)]/h
is
for h rational,
the decimal
and
also for
....
h(a), that f(x)
<
g(x)
<
h(x) for all x,
Prove that g is differentiable at a, and that
f(a) = g'(a) = h'{a). (Begin with the definition of g'(a).)
Show that the conclusion does not follow if we omit the hypothesis
f(a)
20.
Why?
consider [f(a
a,
h
(b)
\x\
/satisfies |/(*)|
0 for irrational x, and \/q for x — p/q in lowest terms. Prove
not differentiable at a for any a. Hint: It obviously suffices to
is
(a)
if
=
that /
expansion of
19.
<
If / satisfies \f(x)\
=
=
g(a)
ti(a).
h(a).
Let / be any polynomial function; we will see in the next chapter that
/ is differentiable. The tangent line to / at {a, f(a)) is the graph oig(x) =
— a) + f(a). Thus f(x) — g(x) is the polynomial function d(x) =
— f(a)(x — a) — f(a). We have already seen that if fix) = x then
= (x - a) and iff(x) = x\ then dix) = (* - a) (x + 2a).
f(a)(x
2
f(x)
d(x)
,
2
2
,
when/(x)
=
show that it is divisible by (x — a) 2
be some evidence that d(x) is always
x 4 and
(a)
Find
(b)
There certainly seems to
divisible by (x — a) 2 Figure 22 provides an
,
.
argument:
graph at
two points; the tangent line intersects the graph only once near the
55
point, so the intersection should be a "double intersection. To give
.
intuitive
usually, lines parallel to the tangent line will intersect the
a rigorous proof,
first
d{x)
x
—
note that
=
a
/(*)
x
-}{a)
—
_ f(a)
a
Now answer the following questions. Why is fix) — fia) divisible by
(x — a)? Why is there a polynomial function h such that h(x) =
a ? Why is lim k(x) = 0 ? Why is h{a) = 0 ?
dix) fix - a) for x
Why
does this solve the problem?
142
Derivatives and Integrals
21.
(a)
Show
= Km
that f(a)
[/(*)
- /(*)]/(* -
a).
(Nothing deep here.)
x—*a
(b)
Show
that derivatives are a "local property": iff(x) = g{x) for all x
interval containing a, then f'(a) = g'(a). (This means
some open
in
that in computing/' (a),
Of
*22.
(a)
you can ignore f(x)
course you can't ignore /(*) for
Suppose that /
differentiable at
is
/(,)-
2A
Remember an
old algebraic trick
the
same quantity
is
more
added
=
/(*
Iim
h,k-^o +
Prove that
if
/
is
confusion, let g(x)
thing g
*24.
Draw
is.)
Prove that
if
/
is
to
+ *)-/(«-*),
h + k
even, then /'(*)
=
f(—x); find
=
—/'(—*). (In order to minimize
g'(x)
=
odd, then /'(*)
/'(—#).
Problems 23 and 24 say that /' is even
What can therefore be said about f {k) ?
26.
Find /"(*)
(ii)
(iii)
(iv)
27.
and
then
remember what other
a picture!
25.
(i)
—
a number is not changed
and then subtracted from it.
generally, that
f{x)
*23.
a.
Prove that
x.
Hint:
**(b) Prove,
^
any particular x
+
lim
fc-o
if
for
such x at once!)
all
if
/
Once
is
again,
draw a
odd, and odd
if
/
picture.
even.
is
if
= x*.
= xK
f(x) =
fix + 3) =
f(x)
f{x)
-
If
x n and 0
,
<
£
<
n,
prove that
-"CD*"'
*28.
(a)
(b)
Find /'(*) if/(*) Analyze / similarly
|*|».
Find /"(*). Does /"'(*)
if /(*)
=
x 4 for x
>
0
exist for all * ?
and /(a:) =
—x
4
for *
Let fix) = * for * > 0 and let /(*) = 0 for x < 0. Prove that
exists (and find a formula for it), but that (n) (0) does not exist.
/
30.
Interpret the following specimens of Leibnizian notation; each
restatement of some fact occuring in a previous problem.
—=
dxn
(i)
dx
nx n \
<
0.
/^"^
*29.
n
is
a
—
dy
.....
(ul)
dz
.
,
—IT+
v)
c]
d[f(x)
=
dx
dy
dx*
df{x)
ST
,
c
=
it
z
=
3a 4
f
dx
(vi)
_
y
.
dx
(Vll)
,
....
(viu)
—7—
—
rf/(ex)
(1X)
^T =
C
#00
"^~
+c
,
CHAPTER
DIFFERENTIATION
The
process of rinding the derivative of a function
the previous chapter you
may have
called differentiation.
is
the impression that this process
is
From
usually
laborious, requires recourse to the definition of the derivative, and depends
upon successfully recognizing some limit. It is true that such a procedure is
often the only possible approach— if you forget the definition of the derivative
you are
likely to
entiate a large
be
Nevertheless, in this chapter we will learn to differof functions, without the necessity of even recalling the
lost.
number
definition. A few theorems will provide a mechanical process for differentiating
a large class of functions, which are formed from a few simple functions by the
process of addition, multiplication, division, and composition. This description
should suggest what theorems will be proved. We will first find the derivative
and then prove theorems about the sum, products,
quotients, and compositions of differentiable functions. The first theorem is
of a few simple functions,
merely a formal recognition of a computation carried out in the previous
chapter.
THEOREM
l
/
If
is
=
a constant function, f(x)
f(a)
PROOF
—
c,
0
then
for all
numbers
/'(*)
a.
lim
h
h-^o
The second theorem
is
also a special case of a
computation
in the last
chapter.
THEOREM
2
If
/
PROOF
is
the identity function, f(x)
f'(a)
f'{a)
-
1
:
=
x,
then
for all
numbers
lim
tAjZ
h-+o
h
a
lim
+
—
M
a
h
h-+o
lim h
h
a.
=
l.|
h-+0
The derivative of the sum
sum of the derivatives.
the
144
of two functions
is
just
what one would hope-
+
145
Differentiation
THEOREM
3
If /
and g are
differentiable at
(/
PROOF
(/
+ gY(a) =
+g
then /
a,
^+
+
lim (/
*)-(/ + g )(g)
h
]im
^±
± g(g +
a-o
=
lim
=
lim
r
fa
/(°
= /'W
The formula
it is
A>
~ t/W + g(g)j
h
h-*o
wish, but
+
— /(g)
A)
£(a
^
+ ^)-/W +
h
+ A) —
*(«
+
THEOREM
4
If /
if
+ *'(«)•
for the derivative of
a product
not as simple as one might
and the proof requires only a
is
nevertheless pleasantly symmetric,
the
and g are
—
trick, which we have found useful before
a number
same quantity is added to and subtracted from it.
differentiable at
(/"*)'(*)
PROOF
(f-g)'(a)
=lim
=
*)-*(«)
h
h->o
simple algebraic
changed
and
+ *)'(*) = /'(«)+*'(«).
fc-+0
=
also differentiable at a,
is
lim
(f'g)(*
/(<2
+
then / g
a,
•
is
also differentiable at
is
not
and
a,
=f(a)'g(a) +f(a)-g'(a).
-
h)
+ %(* +
(/•*)(«)
- f(a)g(a)
A)
/i->0
lim r
f{a
+ h)[g{a +
h)
-
[/(a
g(a)l
+
A)
- /( a )]g(a)
|
-
lim /(a
=
/(«)•*'(«)
+ h)
lim
+ lim /(« + *)-/(«)
g(*
THEOREM
5
If g(x)
=
cf(x)
Theorem
and / is
If h{x)
=
c,
so that
g
(-)
diflPerentiable at a,
=
=
=
=
=
then g
is
A)
differentiable at
=c •/'(*).
h-f, then by
g'(a)
+
= /(a)-)
4 simplifies considerably:
*'(«)
PROOF
^
+ /'(«)•*«.
(Notice that we have used Theorem 9-1 to conclude that lim /(a
In one special case
.
Theorem
4,
(A •/)'(«)
A (a)
f(a)+k'(a)
c-f'(a)
+
c-f{a).\
0-f(a)
f(a)
a,
and
I
146
Derivatives
and
Integrals
(—f)'{a) = —f(a), and consequently
Notice, in particular, that
—
(/
g)'{a)
= (f+l-g]Y(a)=f'(a)-g'(a).
To
demonstrate what we have already achieved,
derivative of
THEOREM
€
If f(x )
=
some more
x n for some natural
number
=
f(a)
PROOF
we
compute the
will
special functions.
na
n~
n,
then
l
for all a,
The proof will be by induction on n. For n — 1 this is simply Theorem
assume that the theorem is true for n y so that if f(x) = x n then
2.
Now
,
=
=
Let g(x)
* n+1 If I(x)
=
.
x,
a.
the equation x n+1
=
—
g(x)
£=/•/.
thus
n
•
x can be written
for all x;
I(x)
f(x)
x
from Theorem 4 that
It follows
g'(a)
na*' 1
for all
f(a)
= U-D'(a) = f(a) 7( ) + /(«) I'(a)
= rca w_1 a + a n \
= «fl n + a n
= 0 + l)a n for all a.
•
fl
'
'
,
This
is
+
precisely the case
1
which we wished
Putting together the theorems proved so far
to prove. |
we can now
find /' for
/ of the
form
/(*)
We
a nx
n
+
a n ^ix
n ~' 1
+
-
•
+
•
+
a 2x 2
a xx
+a
0.
obtain
=
/'(*)
We
=
rc^**- 1
+
(n
- lK^i*"- +
2
•
•
+
•
2a 2 *
+
ax
.
can also find /":
/"(*)
=
»(„
- IK*"" +
2
(«
-
\){n
-
2)a n ^x
n
^+
:
•
•
+
2a,.
easily. Each differentiation reduces the highest
and eliminates one more a,-. It is a good idea to work out the
(4)
and perhaps / (5) until the pattern becomes quite clear.
/
This process can be continued
power of x by
1
derivatives /"',
The
for k
,
,
,
last interesting derivative is
>
n
Clearly, the next step in our
fig. It is
/<»>(*)
=
n\a n
/<*>(*)
=
0.
;
we have
program
is
to find the derivative of a quotient
quite a bit simpler, and, because of
to find the derivative of 1/g.
Theorem
4,
obviously sufficient
147
Differentiation
THEOREM
7
If g is differentiable at a,
and
*
g(a)
0,
0w
PROOF
we even
Before
then 1/g
is
differentiable at a,
and
wi?
write
©<«+*>-(>
we must be
(Vg)0*
tions.
Since g
that g
there
does
sure that this expression
+ A)
is
is
is
is,
by hypothesis,
continuous at
some
make
8
>
a.
h.
it is
necessary to check that
This requires only two observa-
from Theorem 9-1
from Theorem 6-3 that
differentiable at a, irfollows
Since g(a)
0 such that g(a
sense for small
ii„V
—
makes sense
defined for sufficiently small
+
enough
h)
h,
V_,
lim
=
lim
iA 0, it
^
follows
0for|A|
lim
Therefore (\/g) {a
+ h)
+ A)]
~[g(a
+
-g(a)}
h)
1
,
A
ft-o
_
6.
«<i±*L_«<5>
a-o
_ Um
<
and we can write
-[,(«
a-o
g(fl)g(a
+ *)-,(«)]
.
+ h)
lim
___L__
fc-»o
g(a)
A
'
g(a
+ «)
[gwr
(Notice that
we have used
continuity of g at a once again.) |
The general formula for the derivative of a quotient is now easy to derive.
Though not particularly appealing, it is important, and must simply be
memorized (I always use the incantation: "bottom times derivative of
minus top times derivative of bottom, over bottom squared.")
theorem
8
Iff and g are differentiable at a and g(a)
7* 0,
then f/g
and
{a)
kg)
=
sw?
is
top,
differentiable at a y
148
Derivatives and Integrals
PROOF
Since f/g
=/•
we have
(1/g)
/(gKzg^M)
2
teWl
/'(«)
=
i
*(«)
[*(«)?
We
•
r
lf
r
can now differentiate a few more functions. For example,
(
^~
v
/W
,
fM
if
/(*)
.
1
then
*
t
=
/W
(
v
-
1)(2*)
{x*+\)-x(2x)
= - ~ = (~1)*~
then /'(*)
1,
+1)(2^)
_
.
(*
,
,
X2
1
-*
2
+
i)
'
2
2
.
2
.
x2
x
Notice that the
f(x)
example can be generalized:
last
= x~ n =
—
for
>
xn
then
if
some natural number
n,
n~ l
—
nx
/'W = -fsr- =
thus
Theorem
interpret f(x)
6 actually holds both for positive
=
then Theorem
because
it is
6
mean
*° to
is
/(x)
=
true for n
not clear
how
=
1,
and
/'(*)
and negative
=
0
'
*
_1
to
integers. If
mean/'(*)
we
=
0,
word "interpret" is necessary
should be defined and, in any case, 0 (T is
0 also. (The
1
0°
•
meaningless.)
Further progress in differentiation requires the knowledge of the derivatives
of certain special functions to be studied later. One of these is the sine function.
For the moment we shall divulge, and use, the following information, without
proof:
sin' (a)
cos' (a)
— cos a
— — sin
for all a,
a
This information allows us to differentiate
for all a.
many other functions. For example,
if
f(x)
=
x sin
x,
then
/'(*)
/"(*)
= x cos x
= — * si n
= —x sin
+
x
x
sin *,
+
+
cos x
+
2 cos x;
cos x
Differentiation
149
if
g(x)
=
sin 2 x
=
=
=
=
=
sin x cos
sin x
•
sin
then
g"(x)
a:
2 sin ^ cos
+
cos x sin #
x,
— sin x) +
— sin *];
2[(sin #)(
2[cos 2 ^
cos x cos x]
2
if
A(#)
=
cos 2 x
—
cos x
cos
'
a:,
then
h'(x)
/?"(*)
= (cos #)( — sin x) + (—
= — 2 sin ^ cos
= — 2[cos 2 # — sin 2 x\
sin x) cos #
Notice that
+
g'{x)
h'(x)
=
0,
+
= 1. As we would
have g"{x)
The examples above involved only products of two functions. A function
involving triple products can be handled by Theorem 4 also; in fact it can be
handled in two ways. Remember that / g h is an abbreviation for
hardly surprising,
we
expect,
+ h)(x) = sin
+ h"(x) = 0.
since
also
•
(/*)•*
Choosing the
first
(f'g'
=
=
=
(/
•
g)'(x)
h{x)
•
a:
cos 2 x
•
f'(g'h).
of these, for example,
h)'(x)
The
or
2
we have
+
(/
•
g)(x)h'(x)
[f(x)g(x) +f(x)g'(x)]h(x) +f{x)g(x)h'{x)
f(x)g(x)h(x) +f(x)g'(x)h(x) +f(x)g(x)h'(x).
choice of / (g h) would, of course, have given the same result, with a
different intermediate step. The final answer is completely symmetric and
•
easily
•
remembered:
(/ g h)' is the sum of the three terms obtained by differentiating each
of/, g, and h and multiplying by the other two.
*
'
For example,
if
f(x)
=
x z sin x cos
x,
then
f(x)
=
3x 2 sin x cos x
+x
3
cos x cos x
+x
z
(sin
x)(—
sin x).
Products of more than 3 functions can be handled similarly. For example, you
should have
(f-g-h-
little difficulty
*)'(*)
You might even
=
deriving the formula
+ f{x)g'{x)h{x)k{x)
+ f(x)g{x)h'{x)k{x) + f(x)g(x)h(x)k'(x).
f'(x)g(x)h(x)k(x)
try to prove (by induction) the general formula:
n
(/x-...
•/»)'(*)
= J/ito
•
•
•
•
-/.-iW/.-'W/i+iW
•
•
•
•
•/.(*)•
Differentiating the most interesting functions obviously requires a formula
°
gYW in terms of /' and g'. To ensure that / ° g be differentiate at a,
one reasonable hypothesis would seem to be that g be differentiate at a. Since
the behavior oif°g near a depends on the behavior of/ near g(a) (not near a),
it also seems reasonable to assume that / is differentiate at g(a). Indeed we
shall prove that if g is differentiate at a and / is differentiate at g(a), then
f° r (/
/og
differentiate at
is
a,
and
=
(fog)' (a)
g'{a).
f'(g(a))
is called the Chain Rule, presumably because
a composition of functions might be called a "chain" of functions. Notice that
(f°g)' is practically the product off and g', but not quite: /' must be evalu-
This extremely important formula
ated at g(a) and g' at
a.
Before attempting to prove this theorem
we
will try
a
few applications. Suppose
=
f(x)
Let
temporarily, use
us,
.
S to denote the ("squaring")
/ =
Therefore
sin x 2
sin
function S(x)
=
x2
.
Then
o S.
we have
/'(*)
Quite a different result
is
=
=
S'(x)
S in'(S(x))
cos x 2
obtained
if
f(x)
=
•
2x.
sin 2 x.
In this case
/
= S o sin,
=
=
^(sin
so
/'(*)
Notice that this agrees
/
=
sin
•
x)
2 sin x
(as it should)
m
sin'(*)
•
cos
x.
with the result obtained by writing
and using the product formula.
sin
Although we have invented a
special symbol, S, to
name
the "squaring"
do problems like this without
bothering to write down special symbols for functions, and without even
bothering to write down the particular composition which / is one soon
becomes accustomed to taking / apart in one's head. The following differif you find it
entiations may be used as practice for such mental gymnastics
necessary to work a few out on paper, by all means do so, but try to develop
the knack of writing /' immediately after seeing the definition of /; problems of
this sort are so simple that, if you just remember the Chain Rule, there is no
function,
it
does not take
much
practice to
—
—
thought necessary.
if f(x)
f(x)
=
=
sin x 3
sin 3 x
then
/'(*)
/'(*)
=
=
cos x z
•
3 sin 2 x
3x 2
•
cos x
Differentiation
fix)
—
sin
«4—
—
x
fix)
fix)
fix)
A
=
=
=
sin (sin x)
sin(* 3
(*
3
151
+
+
=
=
=
f{x)
3* 2 )
fix)
3* 2) 53
fix)
cos (sin x)
cos(* 3
53(* 3
•
+
+ 3*
cos x
3* 2)
)
(3a:
•
52
2
•
2
(3*
+ 6x)
+ 6x).
2
function like
=
/(*)
which
is
sin 2 x 2
=
* 2] 2
[sin
,
the composition of three functions,
= S o sin o S,
/
can also be differentiated by the Chain Rule. It is only necessary to remember
that a triple composition f°g°h means (/ o g) o h or / © ig o h). Thus if
=
fix)
we can
The
sin 2 x 2
write
/ =
(S
f=
So
o
sin) o S,
(sin
o^).
by applying the Chain Rule
whether the two expressions lead to ec-ually
simple calculations; As a matter of fact, as any experienced tlifferentiator
knows, it is much better to use the second:
derivative of either expression can be found
twice; the only doubtful point
We can now write down fix)
first
is
f
= So
in
one
function to be differentiated
is
o^).
(sin
we must put sin * 2
Thus we begin by writing
Inside the parentheses
function, sin
° S.
fix)
=
To begin with, note that the
formula for fix) begins
swoop.
fell
S, so the
2 sin x 2
,
the value at x of the second
BllifSI
•
(the parentheses weren't really necessary, after
this
much
involves a composition of two functions,
handle.
We
We
must now multiply
—
S at x; this part is easy
which we already know how
©
it
to
obtain, for the final answer,
fix)
The
all).
of the answer by the derivative of sin
following example
is
=
2 sin x 2
handled
f{x)
•
cos x 2
similarly.
=
sin (sin
•
2x.
Suppose
x 2 ).
Without even bothering to write down / as a composition g o h © k of three
we can see that the left-most one will be sin, so our expression for
functions,
fix) begins
/'(*)= cos(
)
—
:
we must put
Inside the parenthesis
sin (sin x 2 )
(what you get from
the value of h
by deleting the
o
k(x); this
is
simply
sin x
2
So our expression for
first sin).
f(x) begins
=
fix)
cos(sin x 2 )
•
We
can now forget about the first sin in sin (sin x 2 ) we have to multiply what
we have so far by the derivative of the function whose value at x is sin x 2
which is again a problem we already know how to solve:
—
;
f{x)
=
cos (sin x 2 )
*
cos x 2
2x.
•
some other functions which are the comsome other triple compositions. You can proba"see" that the answers are correct if not, try writing out / as a
Finally, here are the derivatives of
position of sin
bly just
and S
as well as
y
—
composition
if
fx)
fx)
fx)
fx)
=
—
=
=
/(*)
=
sin ((sin x) 2 )
[sin (sin x)]
f(x)
=
=
=
=
f(x)
—
then /'(*)
2
f(x)
sin (sin (sin x))
fix)
2
sin (# sin x)
cos ((sin x) 2 )
2 sin (sin x)
2 sin x
*
cos (sin (sin x))
2 sin (a* sin x)
sin (sin (x
sin x))
cos (sin (x
•
2
cos x
•
cos x
cos (sin *)
*
cos x
*
cos(x sin x)
'
•
2
•
cos (sin x)
'
[sin *
+ x cos *]
sin x))
cos (a; 2 sin x)
[2x sin x
•
+
The rule for treating compositions of four (or even more) functions
always (mentally) put in parentheses starting from the right,
fo(go(hok))
and
start
smaller
x 2 cos
is
x].
easy
9
reducing the calculation to the derivative of a composition of a
number
of functions:
f(g(h(k( x ))))-z
For example,
if
fx) =
2
sin (sin 2 (*))
[f
= So
= So
sin o
So
sin
(sin o (5 o sin))]
then
=
fix)
2 sin(sin 2 x)
cos(sin 2 x)
*
*
2 sin *
•
cos
if
/(*)
=
sin((sin x 2 ) 2 )
[/
= sin o ^ o sin o
= sin © iS o (sin
.S*
o »S))]
then
fix)
=
2
cos((sin
a:
2
)
'•
)
2 sin a 2
cos x 2
•
2a:;
if
/(a:)
=
2
sin (sin(sin
[fill
a:))
in yourself,
if
necessary]
then
/'(*)
=
2 sin (sin (sin x))
•
cos (sin (sin x))
'
cos (sin x)
•
cos
x.
Differentiation
153
With these examples as reference, you require only one thing to become a
master differentiator practice. You can be safely turned loose on the exercises
—
and it is now high time that we proved the Chain Rule.
some of the tricks one
some of the difficulties encountered. We begin, of course,
at the end of the chapter,
The
might
following argument, while not a proof, indicates
try, as
well as
with the definition
+ *>-(/•*)(«>
(/'*)(«
(/.,)'(.)
h
-
lim
+ h))
f(s(a
Somewhere in here we would
it in by fiat:
-f(g(a))
h
h-+o
like the expression for g'(a).
One approach
is
to
put
\im
f(s(a
+ A))
h
h-+o
h-^o
This does not look bad, and
lim
(/°g>( g
+ *) ~
*-o
+ h)) - f(g(a))
g(a + h) — g(a)
-f^(a)) _
f^(a
lim
it
looks even better
if
we
g(a
+ h) - g{a)
h
write
(/«*)(«)
h
= im f(s(a)
+
l
W ++ h)-—
g(a
h-^o
h)
g(a)])
- f(g(a))
.
g(a)
^
g(a
+
h)
h
a—+o
The second limit is the factorg'(tf) which we want. Ifwelet^(a
(to be precise we should write k (ti)) then the first limit is
- g(a)
+ h) —
=
k
y
Hm /kM±*)
/i-»o
-/(*(«))
'
A;
It looks as if this limit should be f(g(a)), since continuity of g at a implies that
k goes to 0 as A does. In fact, one can, and we soon will, make this sort of reasoning precise. There is already a problem, however, which you will have
noticed if you are the kind of person who does not divide blindly. Even for
A^Owe might have g(a + h) — g(a) = 0, making the division and multiplication by g(a + h) — g(a) meaningless. True, we only care about small h,
but g(a + h) — g{a) could be 0 for arbitrarily small h. The easiest way this
can happen
for g to be a constant function, g{x) =
Then g(a + h) —
is
g(a)
=
so the
0 for
all h.
c.
In
this case, / ©
g
is
also a constant function, (/
°
g) (x)
=
f(c),
Chain Rule does indeed hold:
(fogY (a) =
However, there are
0
= f(g(a))-g'(a).
also nonconstant functions
for arbitrarily small h.
For example,
if
a
II
=
sin -i
£
0,
for
which g (a
(a)
the function g might be
#
5^
0
*
=
0.
*
0,
+ h) —
=
0
154
Derivatives and Integrals
=
In this case, ^'(O)
we must have
(f
we showed in Chapter 9.
0, as
=
g)'(0)
°
Chain Rule is correct,
and this is not exactly
If the
0 for any differentiable
/,
A
proof of the Chain Rule can be found by considering such recalcitrant functions separately, but it is easier simply to abandon this approach, and
obvious.
use a trick.
THEOREM
9
(THE CHAIN RULE)
If g
is
differentiable at
able at
and / is
a,
differentiable at g(a), then
/
°
g
differenti-
is
and
a,
(/•*)'(«) =/'(*(«))•*'(«)•
PROOF
0
Define a function
+ h)) - f(g(a))
g(a + h) — g{a)
f(g{a
(
4>{h)
as follows.
=
I
/(*(«)),
g(a
+
h)
—
g{a)
is
+
h)
-g(a)
g(a
+
h)
-g(a) =
if
should be intuitively clear that
It
also small, so
if
0
continuous at
is
+
g(a
—
h)
^0
ilg(a
g{a)
is
0.
When
0:
h
is
small,
not zero, then 0(A) will
be close to f(g(a)); and if it is zero, then <f>(h) actually equals f(g(a)), which
is even better. Since the continuity of
is the crux of the whole proof we will
provide a careful translation of this intuitive argument.
We know that / is differentiable at g(a). This means that
<j>
M±i^)) =/W)
]im
Thus,
if £
(1)
Now
g
is
>
if
0 there
0
<
\k\
some number
is
<
.
k
o
<5',
f(g(a)
then
differentiable at
a,
>
5'
0 such that, for
+k)- f(g(a))
all k,
-/'(*(«))
hence continuous at
a, so
-
g(a)\
there
is
<
£.
>
0 such
a 8
that, for all h,
Consider
(2)
if \h\
now any
h with
4>{h)
it
-
<
8,
\h\
<
8.
If k
—
+ h)) - f(g(a))
g(<i + A) —
f(g(a
follows from (2) that
\k\
<
+
then \g(a
8',
g(a
the other hand,
if
g(a
+
h)
=
<
We
0,
f(g(a))
.
-f(g(a))\
<
(1) that
then
S.
have therefore proved that
lim 0(A) =f'(g(a)),
A->0
g(a) 5* 0, then
6.
surely true that
\<t>(h)
8'.
A
and hence from
+ h) - g(a)
—
<
_ f(g(a)+k) -
\Mh) -f(g(a))\
On
h)
<t>(h)
=
f'(g(a))
9
so
it is
Differentiation
so
continuous at
is
4>
f(g(a
even
if g(a
+ A) —
{fog y (a)
=
lim
0.
The
rest of the
+ A) ~ f(g(a))
g(#)
M
= /ok*)
=
proof
=
(A)
easy. If h ?± 0, then
is
g(g
.
+
-
A)
155
we have
g(g)
0 (because in that case both sides are 0). Therefore
+ h))-f(i(a))
w«
•
_ Um
^
Hm
.
ii*
+
h)
— gM
i
many
Now that we can differentiate so
another look at the function
x 2 sin
-j
functions so easily
x
5*
0
x
=
0.
we can
take
/(*)
.
0,
In Chapter 9 we showed that f(0) = 0, working straight from the definition
(the only possible way). For x ^ 0 we can use the methods of this chapter.
We
have
f(x)
=
+
2x sin x
x 2 cos -
•
x
(—
\
^-Y
x 2/
Thus
2x sin -
=
/'(*)
—
cos -j
a:
0,
As
this
it is
formula reveals, the
first
derivative /'
not even continuous there. If
N
/(*)
=
(
we
5*
0
x
=
0.
indeed badly behaved at 0
is
consider instead
x 3 sin
-j
x
0
*
|
I
x
a:
*
0,
=
0,
then
, r/
/
\
(*)
=
(
\
3# 2 sin x
—
at
cos -j
I 0,
In
this case /'
sion
sion
3a:
2
is
continuous at
0,
jf
^
0
^
=
0.
X
but /"(0) does not exist (because the expresis differentiable at 0 but the expres-
sin 1/x defines a function which
— x cos
1/x does not).
As you may
improvement.
suspect, increasing the
power of x yet again produces another
If
x 4 sin -j
/« =
x
0
x
.
0,
*
=
0,
then
1
4# 3
1
x 2 cos
sin
->
0,
It is
x
5*
0
*
=
0.
easy to compute, right from the definition, that (/')'(O)
easy to find for x
„,
/
W
.
=
=
0,
and /"(*)
is
?± 0:
t
\2x 2 sin *
I
0,
(
—
—
Ax cos x
—
2x cos -
x
sin
a:
0
a:
*
=
0.
In this case, the second derivative /" is not continuous at 0. By now you may
have guessed the pattern, which two of the problems asks you to establish: if
1
/(*)
-
i
1
x
10,
then/'(0),
.
.
.
,/
(n)
(0) exist,
0
x
x
but
f
n)
is
=
0,
not continuous at 0;
if
/(*)
k
a:
=
0,
(n)
then/'(0),
entiable at
0,
.
n)
,/ (0) exist, and ^discontinuous at 0, but f is not differThese examples may suggest that "reasonable" functions can be
.
0.
.
characterized by the possession of higher-order derivatives
hard we
try to
sufficiently
fortunately,
mask
the infinite oscillation of f(x)
=
sin
1
—no matter how
/x, a derivative of
high order seems able to reveal the underlying irregularity.
we
Un-
much worse things can happen.
calculations, we will bring this chapter to
will see later that
After all these involved
a close
with a minor remark. It is often tempting, and seems more elegant, to write
some of the theorems in this chapter as equations about functions, rather than
about their values. Thus Theorem
3
might be written
(/+£)'=/'
Theorem 4 might be
written as
(/•£)'
and Theorem
+ g',
-/•*'+/•*,
9 often appears in the
form
= <f'g)-g'.
may be false, because
(fog)'
Strictly speaking, these equations
left-hand side might have a larger
the functions on the
domain than those on the right. Nevertheless, this is hardly worth worrying about. If / and g are differentiable everywhere in their domains, then these equations, and others like them, are true,
and this is the only case any one cares about.
Differentiation
157
PROBLEMS
1.
As a warm up exercise, find/'fV) for each of the following/. (Don't worryabout the domain of for /'; just get a formula for f(x) that gives the right
answer when it makes sense.)
(i)
/(*)
(ii)
f{x)
(iii)
f{x)
(iv)
f(x)
=
=
=
—
(v)
/W
=sin( -^i).
(vi)
/to
(vii)
/(a:)
(viii)
f(x)
+x
+ sin x
2
sin(*
).
sin x
2
.
sin (cos x).
sin (sin x).
C
sin (cos x)
X
=
=
+
sin(*
sin x).
sin (cos (sin x)).
Find f(x) for each of the following functions /. (It took the author 20
minutes to compute the derivatives for the answer section, and it should
not take you much longer. Although rapid calculation is not the goal of
mathematics, if you hope to treat theoretical applications of the Chain
Rule with aplomb, these concrete applications should be child's play
mathematicians like to pretend that they can't even add, but most of
them can when they have to.)
+ l)\x +
+ sin x).
((# + sin x)
(i)
/to
sin((*
(ii)
/(*)
sin (#
(iii)
/(*)
sin
(iv)
/CO =
sin
/(*)
sin(x sin x)
(v)
3
2
2)).
2
2
).
(~^—\
\COS X 3 /
+
sin (sin x 2 ).
31 '.
(vi)
/(*)
(cos*)
(vii)
sin 2 x sin x 2 sin 2 x 2
(xiv)
/to
/to
/to
/to
/to
/to
/to
/to
(XV)
/to
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(x
+
sin 5 #) 6
.
sin (sin (sin (sin (sin x)))).
+ l)
(((^ + ^) + ^ + ^)
sin(# + sin
+ sin x )).
sin((sin 7 x 1
7
3
4
).
5
.
2
(a;
2
2
sin (6 cos (6 sin (6 cos 6x))).
1
+
sin
x
1
(xvi)
.
sin 3 (sin 2 (sin x)).
/to
x
+
sin x
158
Derivatives and Integrals
(xvii)
/(*)
(xviii) /(*)
3.
Find the derivatives of the functions tan, cotan, sec, and cosec. (You
don't have to memorize these formulas, although they will be needed
once in a while; if you express your answers in the right way, they will be
simple and somewhat symmetrical.)
4.
For each of the following functions /, find f(f(x))
(i)
/(*)
(ii)
f(x)
=
—+j-.
—
=
=
sin x.
1
(hi) f(x)
(iv) f{x)
5.
/(*)
(ii)
fix)
(iii)
fix)
(iv) fix)
6.
Find
=
(/•/)'(*))
x
x\
17.
For each of the following functions
(i)
(not
/,
find f(f(x)).
-•
X
=
=
=
/' in
x\
17.
17*.
terms of
g' if
(0
fix)
=g(x+g(a)).
(ii)
fix)
=g(x-g(a)).
(iii)
fix)
=gix + gix)).
=g(x)(x-a).
fix) = g(a)(x-a).
/(* + 3) = gix>).
(iv) fix)
(v)
(vi)
7.
(a)
A
circular object
but
it is
known
is
that
increasing in size in
when
the radius
some unspecified manner,
the rate of change of the
is 6,
Find the rate of change of the area when the radius is 6.
(If r(0 and A(t) represent the radius and the area at time t, then the
functions r and A satisfy A = rr 2 a straightforward use of the Chain
radius
is 4.
;
(b)
Rule is called for.)
Suppose that we are now informed that the circular object we have
been watching is really the cross section of a spherical object. Find
the rate of change of the volume when the radius is 6. (You will clearly
need to know a formula for the volume of a sphere; in case you have
forgotten, the volume is ^tt times the cube of the radius.)
Differentiation
(c)
Now
159
suppose that the rate of change of the area of the circular
is 5 when the radius is 3. Find the rate of change of the
cross section
8.
volume when the radius is 3. You should be able to do this problem
in two ways: first, by using the formulas for the area and volume in
terms of the radius; and then by expressing the volume in terms of
the area (to use this method you will need Problem 9-3).
Let /(*) = x 2 sin 1 fx for x ^ 0, and let /(0) = 0. Suppose also that h and
k are two functions such that
h'{x)
A(0)
= sin
= 3
2
(sin(*
+
*'(*)
1))
k(0)
= /(*
= 0.
+
1)
Find
(/oA)'(0).
(i)
9.
(ii)
(W)'(O).
(iii)
a'O 2 ), where
Find
a(x)
=
h(x 2 ). Exercise great care.
/'(0) if
\
f(x)
=
sin ->
| g(x)
x
9*
0
x
=
0,
x
\
I 0,
and
*(0)
10.
11.
12.
=
s'(0)
-
0.
Using the derivative of /(*) = 1/x, as found in Problem 9-1, find
(l/g)'(x) by the Chain Rule.
= V} - x 2
(a) Using Problem 9-3, find /'(*) for -1 < x < 1, if fix)
2
(b) Prove that the tangent line to the graph of / at (a, y/\ — a ) interelementary
sects the graph only at that point (and thus show that the
geometry definition of the tangent line coincides with ours).
Prove similarly that the tangent lines to an ellipse or hyperbola intersect
.
these sets only once.
13.
If /
If /
+g
•
is
differentiable at a, are
g and / are
differentiable at a
14.
(a)
(b)
(c)
/ and g necessarily differentiable at a ?
what conditions on / imply that g is
differentiable at a,
?
Prove that if / is differentiable at a, then
provided that f(a) ^ 0.
Give a counterexample if f(a) = 0.
|/| is
also differentiable at a,
Prove that if / and g are differentiable at a, then the functions
max(/, g) and min(/, g) are differentiable at a, provided that
f(a)
*
g(a).
Give a counterexample if f(a) = g(a).
{n)
Suppose that f n) (a) and g (a) exist. Prove
(d)
15.
Leibniz's formula:
160
Derivatives and Integrals
*16.
Prove that if frigid)) and £ (n) 0) both exist, then (/ o gY n \a) exists. A
experimentation should convince you that it is unwise to seek a
formula for (f ° g) (n) (a). In order to prove that (f°g) (n) (a) exists you
little
have to devise a reasonable assertion about (f°g) {n) (a)
which can be proved by induction. Try something like: "(/°£) n (<z)
exists and is a sum of terms each of which is a product of terms of the
form
~
n
(a) If f(x) = a n x + a n -\x n l +
+ a 0 find a function g such that
r
g = /. Find another.
will therefore
17.
•
•
•
,
(b) If
\
r,
find a function
(c)
Is
b
b
+—+
2
=
fix)
g with
•
•
•
x*
=
g'
+ x—mm
b
3
.
x2
i
f.
there a function
=
/(*)
+
a n x«
-
•
+a +-+
•
•
Q
•
•
a:
such that /'(*)
18.
Show
(a)
that there
=
=
=
=
f(x)
(b) /'(#)
19.
is
no x
x, if
f(x)
(a)
The number
if
y
n
0 for exactly k
=
a
—
is
is
When
(x
is
even.
x, if n
—
k
is
odd.
double root of the polynomial function /
some polynomial function £. Prove that
if and only if a is a root of both / and /'.
= ax 2 + bx + c {a ^ 0) have a double root? What
fl)
does fix)
w
numbers
*.
called a
2
^*)
a double root of /
a
numbers
0 for exactly one
/'(*)
fix)
1
odd.
(c )
if
—
is
(d)
(b)
= 1 fx ?
a polynomial function / of degree n such that
0 for precisely n
0 for
+^
#
for
does the condition say geometrically?
20.
/
If
is
differentiable at a, let d(x)
Problem
*21.
£i,
(a)
.
.
,
If #i,
bn
(b)
is
a companion to Problem 3-6. Let
...
,
*22.
f(xi).
=
a\,
x n are distinct numbers, prove that there
=
di
and
Prove that there
/(*,•)
a) - /(a). Find
another solution for
- f'(a)(x -
.
.
.
,
an
and
be given numbers.
function / of degree In
and
f(x)
19, this gives
9-20.
This problem
.
=
In connection with Problem
(a).
0;
and
is
—
/'(#»)
is
a polynomial
such that f(x 3) = fix 3) = 0 for /
b{. Hint: Remember Problem 19.
1,
=
a polynomial function / of degree In
f'ixi)
=
for all
—
1
5^
i,
with
i.
Suppose that a and b are two consecutive roots of a polynomial function
but that a and b are not double roots, so that we can write fix) =
5^ 0.
0 and
(* — a) ix — b)g{x) where ^(a)
/,
161
Differentiation
(a)
Prove that g{a) and g(b) have the same
(b)
sign.
(Remember
that a
and
consecutive roots.)
b are
is some number x with a < x < b and /'(*) = 0.
draw a picture to illustrate this fact.) Hint: Compare the sign
of f{a) and f{b).
Now prove the same fact, even if a and b are multiple roots. Hint:
m
n
If f(a) = (x - a) (x - £) £(*) where
^ 0 and g(£) ^ 0, con— m~ — n~
=
Prove that there
(Also
(c)
sider the polynomial function h(x)
a)
f'(x)/(x
l
{x
l
b)
.
This theorem was proved by the French mathematician Rolle, in connection with the problem of approximating roots of polynomials, but the
result was not originally stated in terms of derivatives. In fact, Rolle was
one of the mathematicians who never accepted the new notions of calcuThis was not such a pigheaded attitude, in view of the fact that for
one hundred years no one could define limits in terms that did not verge
on the mystic, but on the whole history has been particularly kind to
lus.
Rolle; his
name
has become attached to a
much more
appear in the next chapter, which forms the
general result, to
basis for the
most important
theoretical results of calculus.
23.
*24.
25.
Suppose that f(x) = xg(x) for some function g which is continuous at 0.
Prove that / is differentiable at 0, and find /'(0) in terms of g.
Suppose / is differentiable at 0, and that /(0) = 0. Prove that f(x) =
xg(x) for some function g which is continuous at 0. Hint: What happens
if you try to write g(x) = f(x)/x ?
n
If /W = x~ for n in N, prove that
(n
+ k-
1)!
*
(*-D!
*26.
Prove that
it
differentiable
27.
What
is
*(b) f(x)
Let/W =
f
n)
l/(*
l/(*
2
- a) n
- 1)
(n)
=
^
and
0.
=
f(x)g(x)
where / and g are
Hint: Differentiate.
?
is
=
0.
not continuous at
0.
0,
^
(n)
is
let /(0)
Prove that/' (0),
will encounter the
.
.
.
,
(You
Problem 15.)
and let /(0) = 0. Prove that /'(0),
continuous at 0, and that / (n) is not differentiable
basic difficulty as that in
= x 2n+l sin \/x Hx
(0) exist, that /
at 0.
^(0)
?
and that / (n)
Let f(x)
/
=
x 2n sin 1/xifx
(0) exist,
same
*29.
/(0)
/<*>(*) if
(a) fix)
*28.
impossible to write x
is
and
0,
.
.
.
,
162
Derivatives
and Integrals
30.
In Leibnizian notation the Chain Rule ought to read:
_
df(g(x))
dg{x)
df{y)
dx
dy
v~q(x)
dx
Instead, one usually finds the following statement: "Let y
z
=
f(y).
=
g(x)
and
Then
dz
= dz^dy"
dx
dy
dx
Notice that the z in dz/dx denotes the composite function f ° g, while the
z in dz/dy denotes the function /; it is also understood that dz/dy will be
"an expression involving
y" and
that in the final answer g(x)
formula; then compare with Problem
(i)
z
(ii)
z
(iii)
z
(iv)
z
—
=
—
=
sin y,
sin y,
cos
sin
u,
v,
must be
In each of the following cases, find dz/dx by using this
substituted for
y
y
u
v
=
=
=
—
x
+x
cos
2
.
x.
sin x.
cos
u,
u
=
sin x.
1.
CHAPTER
SIGNIFICANCE OF THE DERIVATIVE
| I
One aim in
this
chapter
derivative of a function.
is
to justify the time
As we
we have
spent learning to find the
about/' tells us a
shall see, knowing just a little
about/. Extracting information about /from information about/' requires
some difficult work, however, and we shall begin with the one theorem which
lot
is
really easy.
This theorem
is
concerned with the
value of a function on an
maximum
Although we have used this term informally
while to be precise, and also more general.
interval.
DEFINITION
in
Chapter
7, it is
worth-
of/.
Let / be a function and A a set of numbers contained in the domain
A point x in A is a maximum point for / on A, if
/(*)
The number /(*)
say that / "has
itself is
its
>
for
f(y)
called the
maximum
every y in A.
maximum
value on
A
value of/ on
A
(and
we
also
at x").
different x
Notice that the maximum value of / on A could be /(*) for several
different maximum
several
have
can
function
a
words,
/
other
in
(Figure 1);
value. Usually we
points on A, although it can have at most one maximum
interval [a, b]\ if / is continushall be interested in the case where A is a closed
have a maximum value
ous, then Theorem 7-3 guarantees that / does indeed
on
b\
[a,
The
definition of a
definition
on A
We
is
1
on A will be left
minimum on A at x,
of /
the following: / has a
are
to you.
if
(One
-/ has
a
possible
maximum
'
at x.)
a theorem which does not even depend upon the
upper bounds.
now ready
existence of least
THEOREM
minimum
for
Let / be any function defined on
(a, b).
If x
is
a
maximum
(or a
then
=
/'(*)
minimum)
0.
point for / on (a, b), and / is differentiate at *,
continuity, of / at
(Notice that we do not assume differentiability, or even
other points.)
PROOF
the simple
Consider the case where /has a maximum at x. (Figure 2 illustrates
the left of
to
points
idea behind the whole argument— secants drawn through
right of
the
to
points
through
(x, /(*)) have slopes > 0, and secants drawn
follows.
as
proceeds
argument
this
(x)) have slopes < 0.) Analytically,
(x[f
163
164
Derivatives and Integrals
If h is
+h
any number such that x
>f(x
/(*)
/ has a maximum on
since
Thus,
x
+
if
h
>
0
(a, b)
in
is
at
then
(a, b),
+ h),
This means that
*.
/(*
+ A) " /«
/(*
+ *)
<
0.
we have
-f(x)
h
and consequently
FIGURE
2
<
lim
On
the other hand,
<
if
we have
0,
/(*
o.
+ *)
-/(*)
h
>
~ nU>
so
—
lim
By
hypothesis, /
is
differentiable at
fact equal to /'(*). This
FIGURE
The
<
0
it
follows that /'(*)
=
case where / has a
0.
so these
#,
two
limits
must be equal,
in
means that
f(x)
3
from which
>
minimum
and
f(x)
>
0,
0.
at x
is left
to
you
(give a one-line proof). |
we cannot replace {a, b) by [a, b] in the statement of
we add to the hypothesis the condition that x is in (a, b)).
Notice (Figure 3) that
the theorem (unless
Since f(x) depends only on the values of / near
to get a stronger version of
Theorem
1
.
We
x> it is
almost obvious
how
begin with a definition which
is
illustrated in Figure 4.
DEFINITION
Let / be a function, and A a
A point x in A is a local
of /.
there
AO
theorem
2
If /
is
is
(*
some
-
8,
8
+
x
defined on
>
0 such that
You
should see
numbers contained
in the
domain
8).
(a, b)
and has a
differentiable at x, then /'(*)
PROOF
set of
maximum [minimum] point for / on A if
x is a maximum [minimum] point for / on
why
this
is
=
local
maximum
(or
minimum)
at x,
0.
an easy application of Theorem
1
.
|
and / is
165
Significance of the Derivative
The
converse of
be 0 even
if
x
is
Theorem
2
is
definitely not true
maximum
not a local
—
minimum
or
it is
possible for /'(#) to
point for
/.
The
simplest
example is provided by the function /(*) = * 3 in this case f(0) = 0, but /has
no local maximum or minimum anywhere.
Probably the most widespread misconceptions about calculus are concerned
with the behavior of a function /near x when f(x) = 0. The point made in the
previous paragraph is so quickly forgotten by those who want the world to
;
be simpler than
true
it is,
—the condition
minimum
point of
/.
we
that
=
/'(*)
will repeat
Precisely for this
adopted to describe numbers x which
DEFINITION
A
critical
point of a function /
is
The
/(*) itself
is
the converse of
satisfy the
condition /'(*)
=
the ones which must be considered
2
is
not
=
0.
0.
called a critical value of
critical values of /, together
Theorem
a local maximum or
reason, special terminology has been
is
a number x such that
f(x)
The number
it:
0 does not imply that x
/.
with a few other numbers, turn out to be
in order to find the maximum and mini-
/. To the uninitiated, finding the maximum and
value of a function represents one of the most intriguing aspects of
calculus, and there is no denying that problems of this sort are fun (until you
mum
of a given function
minimum
have done your
first
Let us consider
on a closed
hundred or so).
the problem of finding the maximum or minimum of/
[a, b]. (Then, if / is continuous, we can at least be sure
first
interval
maximum and minimum value exist.) In order to locate
and minimum of / three kinds of points must be considered:
that a
(2)
The critical points of / in
The end points a and b.
(3)
Points x in
(1)
*2
a local
minimum
FIGURE
4
point
maximum
point
such that /
maximum
b\
is
not differentiable at
x.
point for / on [a, b], then x must be
in one of the three classes listed above: for if x is not in the second or third
group, then x is in (a, b) and / is differentiable at x; consequently /'(*) = 0,
by Theorem 1, and this means that x is in the first group.
If
a local
[a, b]
[a,
the
x
is
a
maximum
If there are
minimum
point or a
minimum
many points in these three categories, finding the maximum and
may still be a hopeless proposition, but when there are only a
of /
few critical points, and only a few points where / is not differentiable, the
procedure is fairly straightforward: one simply finds /(*) for each x satisfying
/'(*) = 0, and/(;t) for each x such that / is not differentiable at x and, finally,
and the
f(a) and f(b). The biggest of these will be the maximum value of /,
smallest will be the
minimum.
A
simple example follows.
Suppose we wish to find the
maximum and minimum
/(*)
on the
[—1,
interval
To
2].
so /'(*)
=
when 3*
0
-
=
1
=
that
0,
= VT73
x
— Vl/3
The numbers Vl/3 and
candidates for the location of the
The
third group
is
-
1,
when
is,
-
or
both
lie
VT/3.
in
[
— 1,
maximum and
-
2],
the
so the
first
minimum
group of
is
V173.
contains the end points of the interval
-1,2.
(2)
is
x
we have
3*'
Via
(i)
The second group
—
xz
begin with,
f(x)
2
=
value of the function
empty, since /
is
differentiate everywhere.
The
final step
compute
to
- VT/3 = i VT73 - VI73 = -I VT73,
=
V173)
- (- VT73) = -iVT/3 + VI73 = f VT73,
(/(- Vl/3)
/(-i) = 0,
/(2) = 6.
/(VT/3) = (VT/3)»
3
Clearly the
mum
minimum
value
is 6,
value
is
occurring at
This sort of procedure,
— f Vl/3,
occurring at
VT/3, and
the maxi-
2.
if feasible,
will
maximum and
always locate the
minimum
we
value of a continuous function on a closed interval. If the function
are dealing with is not continuous, however, or if we are seeking the maxi-
mum or minimum on an open interval or the whole line, then we cannot even
be sure beforehand that the
maximum and minimum values exist, so all the
may say nothing. Nevertheless, a
information obtained by this procedure
ingenuity will often reveal the nature of things. In Chapter 7
little
just such a
problem when we showed that
f( x )
has a
minimum value on
must occur
at
f(x)
n
+ a n ^x n - +
1
•
is
•
nx
l
we
solved
even, then the function
•
+a
0
the whole line. This proves that the
some number x
= n~
=
0
=x*
if
minimum
value
satisfying
+
(n
-
\)a n ^x
n
^+
•
•
•
+a
Q.
If we can solve this equation, and compare the values of f(x) for such x, we
can actually find the minimum of /. One more example may be helpful. Sup-
pose
we wish
to find the
maximum and minimum,
if
they
exist,
of the function
167
Significance of the Derivative
on the open
(-1,
interval
We
1).
/'(*)
have
2x
=
-
(1
so/'(x)
=
0 only for x
* 2) 2
= O.We can see immediately that for x close to
1
or
—1
not have a maxithe values of /(*) become arbitrarily large, so /certainly does
minimum at 0.
a
has
that
/
mum. This observation also makes it easy to show
with
b,
and
a
numbers
be
just note (Figure 5) that there will
We
-1 <
such that /(*)
>
a
<
0
and
0
<
b
<
1,
x
<
a
and
b
<
x
<
1.
/(0) for
—1 <
minimum of / on [a, b] is the minimum of / on all of
only place where
(-1,1). Now on |>, b] the minimum occurs either at 0 (the
minimum
/' =
or at a or b, and a and £ have already been ruled out, so the
This means that the
0),
value is/(0)
=
1.
_
did not draw the graphs of f(x)
graph
2
the
draw
to
cheating
z =
not
is
* ), but it
1/(1
x and /(*)
x
prove anything.
(Figure 6) as long as you do not rely solely on your picture to
sketching the
of
method
a
discuss
to
going
now
As a matter of fact, we are
in discussused
be
to
information
enough
gives
really
that
graph of a function
local maxima
even
locate
to
able
be
will
we
fact
in
minima—
and
ing maxima
In solving these problems
we purposely
involves consideration of the sign of/'(x),
and minima. This method
on some deep theorems.
and
relies
which have been proved so far, always
This is true even
yield information about/' in terms of information about /.
to determine
used
be
sometimes
can
theorem
of Theorem 1, although this
minima.
and
maxima
of
location
the
namely,
certain information about /,
that
/'(*) is not
emphasized
we
introduced,
first
was
When the derivative
numbers as h
for any particular h, but only a limit of these
The theorems about
[f( x
+ h)
derivatives
f(x)]/h
becomes painfully relevant when one tries to extract
and most frusinformation about / from information about /'. The simplest
the following
by
afforded
is
trating illustration of the difficulties encountered
approaches
0; this fact
x must*/ be a constant function? It is impossible
conviction is strengthened
to imagine how / could be anything else, and this
velocity of a particle is
the
if
interpretation—
by considering the physical
it is difficult
Nevertheless
still!
standing
be
must
particle
surely the
always
question: If /'(*)
=
0 for
all
y
0,
even to begin a proof that only the constant functions
x. The hypothesis /'(*) = 0 only means that
lim
h-+o
and
it is
not at
all
obvious
/(*
+ *)-/(*)
=
satisfy
fix)
=
0 for
all
0>
h
how one can use the information about the limit to
derive information about the function.
—
168
Derivatives and Integrals
The
/ is a constant function if /'(*) = 0 for all x, and many other
same sort, can all be derived from a fundamental theorem, called
the Mean Value Theorem, which states much stronger results. Figure 7 makes
it plausible that if/ is differentiable on [a, b], then there is some x in (a, b) such
fact that
facts of the
that
Kb) ~f(a)
/'(*)
this means that some tangent line is parallel to the line between
and (b,f(b)). The Mean Value Theorem asserts that this is true
some x in (a, b) such that /'(*), the instantaneous rate of change of/ at
Geometrically
(a,f(a))
there
FIGURE
is
7
"mean" change of / on [a, b], this average
change being [f(b) - f(a)]/[b - a]. (For example, if you travel 60 miles in
one hour, then at some time you must have been traveling exactly 60 miles per
hour.) This theorem is one of the most important theoretical tools of calculus
probably the deepest result about derivatives. From this statement you might
x, is
exactly equal to the average or
—
FIGURE
conclude that the proof is difficult, but there you would be wrong the hard
theorems in this book have occurred long ago, in Chapter 7. It is true that if
you try to prove the Mean Value Theorem yourself you will probably fail,
but this is neither evidence that the theorem is hard, nor something to be
8
ashamed
of.
The
proof of the theorem was an achievement, but today
first
can supply a proof which
is
we
quite simple. It helps to begin with a very special
case.
THEOREM
3 (ROLLE'S
THEOREM)
If /
is
there
PROOF
continuous on
is
a
It follows
mum
[a, b]
number x in
and
(a, b)
differentiable
such that/'(*)
from the continuity of / on
value on
[a, b]
=
on
(a, b),
and
f(a)
=
f(b),
then
0.
that / has a
maximum and
a mini-
[a, b].
Suppose first that the maximum value occurs at a point x in (a, b). Then
f(x) = 0 by Theorem 1, and we are done (Figure 8).
Suppose next that the minimum value of / occurs at some point x in (a, b).
Then, again,
/'(*)
=
0 by
Theorem
1
(Figure
9).
maximum and minimum values both occur at the end
= f(b), the maximum and minimum values of / are equal,
Finally, suppose the
points. Since f(a)
FIGURE
9
so / is a constant function (Figure 10), and for a constant function
choose any x in (a, b). |
Notice that we really needed the hypothesis that / is differentiable everywhere on (a, b) in order to apply Theorem 1 Without this assumption the
theorem is false (Figure 11).
You may wonder why a special name should be attached to a theorem as
easily proved as Rolle's Theorem. The reason is, that although Rolle's
Theorem is a special case of the Mean Value Theorem, it also yields a simple
proof of the Mean Value Theorem. In order to prove the Mean Value
.
FIGURE
10
we can
Significance of the Derivative
169
apply Rolle's Theorem to the function which gives the length
shown in Figure 12; this is the difference between /(#),
and the height at x of the line L between (a, f(a)) and (b, f{b)). Since L is the
graph of
Theorem we
will
of the vertical segment
we want
FIGURE
to look at
11
As
THEOREM
4
(THE
MEAN VALUE
it
If /
THEOREM)
turns out, the constant f(a)
is
continuous on
j in
[a, b]
irrelevant.
is
and differentiate on
/'« =
/(*)
6
proof
is
continuous on
=
and
[a, b]
is
a
number
-/(«)
-
.
a
differentiable
on
(a, b),
and
/(*),
AW=/w
_[/_M]
(4
_
a)
/(*))
=
Consequently,
some x
(a, /(*))'
in
/(«)•
we may apply
(a, b)
Rolle's
Theorem
=
-
to h
and conclude that there
is
such that
o
=
*'(*)
/'(*)
-/(«)
/(*)
b
FIGURE
then there
Let
Clearly, A
'(*,
(a, b),
such that
—
a
so that
12
/'(*)
=
fib)
b
Notice that the
Mean Value Theorem
corollary
l
If /
is
is
I
a
still fits
into the pattern exhibited
by
—information
about / yields information about /'. This
so strong, however, that we can now go in the other direction.
previous theorems
information
-f(a)
—
defined on an interval and /'(*)
=
0 for
all
x in the interval, then /is
constant on the interval.
PROOF
Let a and
b
be any two points in the interval with a
9^ b.
Then
there
is
some x
170
Derivatives
and
Integrals
in (a> b) such that
m
=
But f(x)
FIGURE
0 for
_ /(*)- /GO
o
and consequently
/(#)
the same,
is
a
b
x in the interval, so
all
13
interval
&L=M.
—
=
=
/(£).
/
i.e.,
Naturally, Corollary
1
is
Thus the value
of
/
at
constant on the interval.
any two points
in the
|
does not hold for functions defined on two or more
intervals (Figure 13).
COROLLARY
2
and g are defined on the same interval, and /'(*) = g'(x)
is some number c such that / = g + c.
If /
for all x in the
interval, then there
PROOF
For
there
1,
The statement
A
is
a
of the next corollary requires
Figure
illustrated in
DEFINITION
we have (/ — g)'(x) = /'(*)
number c such that / — g = c. |
x in the interval
all
Corollary
—
g'(x)
—
0 so,
by
some terminology, which
is
14.
/ is increasing on an interval if f(a) < f(b) whenever a and b
numbers in the interval with a < b. The function / is decreasing
on an interval if f(a) > f(b) for all a and b in the interval with a < b. (We
often say simply that / is increasing or decreasing, in which case the interval
is understood to be the domain of /.)
function
are two
COROLLARY
3
If /'(*)
/'(*)
PROOF
<
>
0 for
0 for
x in an interval, then / is increasing on the interval;
x in the interval, then / is decreasing on the interval.
all
all
Consider the case where /'(*)
with a
<
b.
Then
there
is
>
Let a and
0.
some x
in
(a, b)
b
But f(x)
>
0 for
all
x in
(a, b),
b
b
—
a
>
0
it
The proof when
~f(a)
—
follows that f(b)
/'(#)
<
0 for
be two points in the interval
with
a
so
fW
Since
—
b
all
> Q
a
>
*
f(a).
is left
if
to you. |
171
Significance of the Derivative
are true
Notice that although the converses of Corollary 1 and Corollary 2
is increasing, it is
If/
true.
not
is
3
Corollary
of
converse
(and obvious), the
hold for some x
easy to see that /'(*) > 0 for all *, but the equality sign might
(consider
/W
=
x 3 ).
graph ot
Corollary 3 provides enough information to get a good idea of the
the
more,
once
Consider,
a function with a minimal amount of point plotting.
function /(*)
=
x*
-
x.
We
have
=
/'(*)
3x 2
—
1.
-Vl/3, and
have already noted that /'(*) = 0 for x = Vl/3 and x =
x. Note that
other
all
for
is also possible to determine the sign of /'(*)
We
it
3x 2
—
>
1
0 precisely
when
>
3x 2
1,
x*>h
x
—
thus 3* 2
(a)
1
<
> VI73
0 precisely
x
or
when
- VT73 <
an increasing function
< - VT73;
Thus / is increasing for x < Vl73, and once again increasing
V
x
<
Via
1/3, decreasing
for x
>
V
1/3.
- VV3
between
Combining
this
and
information
with the following facts
(1)
/(- Vi/3) = f
/(Vl/3) =
(2) f(x)
=
0 for x
VTT^
-f Vl/3,
= -1,
0, 1,
gets large,
(3) /(*) gets large as x
(b)
14
as x gets large
negative,
a decreasing function
FIGURE
and large negative
it
is
to the
possible to sketch a pretty respectable approximation
graph
(Figure 15).
on which / increases and decreases
to examinejhe sign off. For
bothering
even
could have been found without
By
the way, notice that the intervals
example, since/'
is
continuous, and vanishes only at
- Vl/3
and Vl/3, we
Vl/3). Since
know that/' always has the same sign on the interval (-Vl/3,
y(_VT73) > /(VT73), it follows that / decreases on this interval. Similarly,
/'
always has the same signon
(VlA
00)
and
/(*)
is
large for large
x,
so
/
If/' is conmust be increasing on (Vl/3, 00). Another point worth noting:
points
critical
adjacent
two
tinuous, then the sign of/' on the interval between
this
in
x
one
any
for
of
/'(*)
sign
the
by finding
can be determined simply
interval.
Our
sketch of the graph of /(*)
=
x*
-^contains
- Vl/3
allow us to say with confidence that
VT/3 a local minimum point. In fact,
we
is
sufficient
a local
information to
maximum
point,
and
can give a general scheme for
:
172
Derivatives and Integrals
deciding whether a
critical
point, or neither (Figure
if/'
(1)
>
0 in
some
1
point
if/'
<
(3) if/'
is
x,
then x
is
a local
minimum
is
no point
point, a local
of x
and
is
/'
maximum
left
of x
minimum
<
0 in
some
interval
some
interval
point.
and
minimum
has the same sign in some interval to the
interval to the right, then x
(There
left
a local
0 in some interval to the
to the right of
maximum
a local
interval to the
to the right of x, then x
(2)
is
6)
/'
>
0 in
point.
it
has in some
maximum
nor a local
left
neither a local
of x as
point.
in
memorizing these
rules
—you can always draw the
pic-
tures yourself.)
The polynomial
functions can
all
be analyzed in this way, and
graph of such functions.
possible to describe the general form of the
fig URE
16
it is
To
even
begin,
Significance of the Derivative
we need
a result already mentioned in Problem 3-7:
f{x)
=
+
n
a nx
a n -ix
n~ l
+
•
•
•
173
If
+
fl 0 ,
there are at most n numbers x such that
an algebraic theorem, calculus can be used
to give an easy proof. Notice that if x\ and x 2 are roots of / (Figure 17), so that
/Oi) = f(x 2 ) = 0, then by Rolle's Theorem there is a number x between
then / has at most w "roots,"
fx) =
0.
Although
this
X\
and x 2 such that f(x) =
xi
<
#i
x2
<
'
'
<
'
i.e.,
really
is
This means that
0.
—
x k then /' has at least k
,
and x 2 one between x 2 and # 3
,
now
etc. It is
,
1
if
/ has k
different roots
different roots:
one between
easy to prove by induction that
a polynomial function
fx) =
has at most n roots
that
it is
true for
:
n,
anx
n
+
a n -ix
The statement
n~ l
+
'
•
"
+
=
surely true for n
is
flo
1
,
and
if
we assume
then the polynomial
=
g(x)
n+1
£ n+ i*
could not have more than n
+
+
b nx
+
n
roots, since if
'
1
it
'
+
did, g'
*0
would have more than
n roots.
With
information
this
f(x)
The
n
—
a
=
anx
+
n
tf
n -i*
n_1
+
'
'
+
'
flo.
derivative, being a polynomial function of degree n
1
point
if
not hard to describe the graph of
it is
roots.
is
and
Therefore / has at most n
negative on
1
maximum
not necessarily a local
b are
—
critical points.
or
minimum
—
1,
has at most
Of course,
point, but at
a critical
any
rate,
adjacent critical points of /, then /' will remain either positive or
since /'
(a, b),
creasing or decreasing on
continuous; consequently, / will be either inThus / has at most n regions of decrease or
is
(a, b).
increase.
As a
specific
example, consider the function
-
f( x )
=
xA
f(x)
=
4* 3
2x\
Since
the critical points of / are
— 1,
-
4x
0,
=
and
/(-I)
/(0)
/(I)
The behavior
Axix
1,
-
l)ix
+
1),
and
= "I,
= 0,
= "I.
on the intervals between the critical points can be determined by one of the methods mentioned before. In particular, we could determine the sign of /' on these intervals simply by examining the formula for fix).
On the other hand, from the three critical values alone we can see (Figure 18)
that / increases on ( — 1, 0) and decreases on (0, 1). To determine the sign of
—1) and (1, oo) we can compute
f on
of /
3
ft-!) = 4-(-2) -4-(-2) = -24,
f(2)
and conclude that /
These conclusions
large negative
is
=
4
23
•
-
4
2
•
=
24,
decreasing on (-qo,
also follow
from the
—1) and
fact that /(*)
is
increasing on
large for large x
(1,
<*>).
and
for
x.
We
can already produce a good sketch of the graph; two other pieces of
information provide the finishing touches (Figure 19). First, it is easy to deter-
mine
/(— x),
f(x)
=
that f(x)
so the
=
x3
—
0 for x
graph
is
=
0,
;
symmetric with
second,
clear that
/
even, f(x)
is
is
odd, /(*)
=
The function
respect to the vertical axis.
already sketched in Figure 15,
x,
it is
= —/(—*), and
is
consequently symmetric with respect to the origin. Half the work of graph
sketching may be saved by noticing these things in the beginning.
Several problems in this and succeeding chapters ask you to sketch the
graphs of functions. In each case you should determine
the critical points of/,
(1)
(2)
the value of / at the critical points,
(3)
the sign of /' in the regions between critical points
(if
this
is
not
already clear),
numbers x such that f(x) = 0 (if possible),
the behavior of f(x) as x becomes large or large negative
the
(4)
(5)
Finally, bear in
even,
may
mind
(if
that a quick check, to see whether the function
possible).
is
odd or
save a lot of work.
if performed with care, will usually reveal the basic
shape of the graph, but sometimes there are special features which require a
little more thought. It is impossible to anticipate all of these, but one piece of
This sort of analysis,
is not defined at certain points (for
a rational function whose denominator vanishes at some
points), then the behavior of /near these points should be determined.
information
is
often very important. If /
example,
/
is
if
For example, consider the function
=
/(*)
ff(x)
AV2,
(-1, -1)
FIGURE
19
(1,
-1)
0)
=
-
2*
x
-
x*
-
+
1
2x 2
2
Significance of the Derivative
which
is
not defined at
/'(*)
We have
(x - l)(2s -
175
1.
=
- (x» - I)
+ 2)
2x
2)
2
(*
*(«
(*
- 2)
- I)
2
Thus
the critical points of / are 0,
(1)
2.
Moreover,
=
=
(2) /(0)
/(2)
-2,
2.
Because / is not denned on the whole interval (0, 2), the sign of/' must be
determined separately on the intervals (0, 1) and (1,2), as well as on the
points in
intervals (- oo, 0) and (2, oo). We can do this by picking particular
/'. Either
for
formula
the
at
hard
staring
by
simply
or
intervals,
each of these
way we
find that
(3) /'(*)
/'(*)
f
(x).
/'(*)
>
<
<
>
0
if
0
if
0
0
if
1
0
if
x
<
<
<
x
x
<
<
2
<
x.
0,
1,
2,
we must determine the behavior of /(*) as x becomes large or large
give us
negative, as well as when x approaches 1 (this information will also
decreases).
and
increases
which
on
regions
the
/
determine
another way to
To examine the behavior as x becomes large we write
Finally,
*2
clearly /(*)
close to x
of / near
close to x
is
-
1
1
is
-2* +
x — 1
-
1
2
=
x
X
—
.
1
+x,
1
I
i
(and slightly larger)
(but slightly smaller)
when
x
is
1
when
x
is
large,
large negative.
and fx)
is
The behavior
also easy to determine; since
lim
(x
2
-2x +
2)
= 1^0,
the fraction
x2
-
2x
+
x
-
1
2
becomes large as x approaches 1 from above and large negative as x approaches
1 from below.
one
All this information may seem a bit overwhelming; but there is only
account
can
you
that
way that it can be pieced together (Figure 20); be sure
for each feature of the graph.
When this sketch has been completed, we might note that it looks like the
176
Derivatives
and Integrals
FIGURE
20
graph of an odd function shoved over
x2
-
2x
x
-
+
2
1
unit,
(x
1
and the expression
-
l)
x
-
+
2
1
1
shows that this is indeed the case. However, this is one of those special features
which should be investigated only after you have used the other information
to get a good idea of the appearance of the graph.
Although the location of local maxima and minima of a function is always
revealed by a detailed sketch of its graph, it is usually unnecessary to do so
much work. There is a popular test for local maxima and minima which
depends on the behavior of the function only at its critical points.
theorem
5
Suppose /'(a)
=
PROOF
By
>
maximum
0. If f'{a)
then / has a local
0,
at
then /has a local
minimum
at a;
if /"(a)
<
0,
a.
definition,
/» = umtk±A^m.
h
h-^o
Since f(a)
=
0, this
can be written
/>)=i im
fc->o
Suppose now that f
small
h.
Therefore:
f
(a)
>
0.
Then f(a
^.
h
+
h)/h must be positive for sufficiently
177
Significance of the Derivative
f(a
and
f'(a
+
+
must be positive for sufficiently small h > 0
must be negative for sufficiently small h < 0.
h)
h)
This means (Corollary
a
\/{x)
= -x*
and /is decreasing
minimum
at
3) that
some
in
/
is
increasing in
some
interval to the right of
interval to the left of a. Consequently, / has a local
a.
<
The proof
for the case f'ia)
Theorem
may be applied to the function fx) =
0
is
similar. |
(a)
5
been considered.
We
/"(*)
- Vl/3
critical points,
- Vl/3
Consequently,
mum
/(*)
-
*B
*,
which has already
3x 2
-
1
6*.
and Vl/3, we have
is
a local
maximum
0,
point and
Vl/3
is
a local mini-
point.
Although Theorem
5 will be
the second derivative of
many
found quite useful for polynomial functions,
is so complicated that it is easier to
functions
a critical point of / it
may happen that f'ia) = 0. In this case, Theorem 5 provides no information:
minimum point, or
it is possible that a is a local maximum point, a local
functions
neither, as shown (Figure 21) by the
consider the sign of the
(c)
FIGURE
=
=
/"(- Vl/3) = -6 VT/3 <
/"(Vl/3) = 6 Vl/3 > 0.
(b)
-
have
/'(*)
At the
*3
first
/«
21
in
derivative.
= ~x\
f{x)
Moreover,
= x\
if
fix)
a
=
is
x*;
= /"(0) = 0, but 0 is a local maximum point for the first, a
point for the second, and neither a local maximum nor minipoint for the third. This point will be pursued further in Part IV.
It is interesting to note that Theorem 5 automatically proves a partial coneach case/'(0)
minimum
local
mum
verse of
THEOREM
6
Suppose }"ia)
local
PROOF
itself.
exists. If /
maximum
at
a,
has a local
then f'ia)
<
minimum
at
a,
then f'ia)
>
0; if /
has a
0.
a. If f'ia) < 0, then / would also have a
/would be constant in some interval
Thus
local maximum at a, by Theorem
=
Thus we must have f'ia) > 0.
contradiction.
a
that
0,
containing a, so
f'ia)
Suppose / has a
local
minimum
at
5.
The
case of a local
maximum
(This partial converse to
and <
fix)
=
signs
x*
and
is
handled
Theorem
cannot be replaced by
/(*)
The remainder
= -*
5
is
similarly. |
the best
> and <,
as
we can hope
for:
the
>
shown by the functions
4
.)
of this chapter deals, not with graph sketching, or
and minima, but with three consequences
of the
maxima
Mean Value Theorem. The
178
Derivatives
and Integrals
first is
a simple, but very beautiful, theorem which plays an important role in
1 5, and which also sheds light on many examples which have occurred
Chapter
in previous chapters.
theorem
7
Suppose that / is continuous at
containing a, except perhaps
a,
and that f'(x)
=
for x
exists for all
x in some interval
Suppose, moreover, that lim f'(x)
a.
x—>a
exists.
Then f(a)
and
also exists,
f{a)
PROOF
By
lim /'(*).
definition,
=
f{d)
For
sufficiently small h
on
differentiable
h
=
<
0).
By
the
(a,
a
>
lim
+ h)
f{a
-f(a)
+
h] and
0 the function / will be continuous on [a, a
(a similar assertion holds for sufficiently small
+ h)
Mean Value Theorem
there
a
is
number a h
in
(a,
a
+ h)
such
that
fia
+ k) -/W
=/W
h
Now a h
lim /'(*)
approaches a as h approaches
exists, it follows
=
/'(«)
(It is
+ A) ~ /(a)
/(a
lim
a good idea to supply a rigorous
we have
Even
treated
if
/
is
0,
because
somewhat
=
e-5
lim /'(a*)
^0
argument
an everywhere differentiable function,
„
22
x
fix)
=
(
x 2 sin
7,
in
(«,
a
+ h)
;
since
=
lim /'(*).
for this final step,
which
->
it is still
possible for /' to
if
x
9*
0
a:
=
0.
*
^
I 0,
According to Theorem
is
informally.) |
be discontinuous. This happens, for example,
FIGURE
ah
that
however, the graph of /' can never exhibit a discon-
shown in Figure 22. Problem 39 outlines the proof of another
beautiful theorem which gives further information about the function /'.
The next theorem, a generalization of the Mean Value Theorem, is of
tinuity of the type
interest
THEOREM
8
(THE CAUCHY
MEAN
VALUE THEOREM)
If /
mainly because of
its
and g are continuous on
number
x in
(a, b)
applications.
[a, b]
and
differentiable
on
(a, b),
such that
[fib)
~fia)]g'ix)
=
[gib)
-
g(a)]f(x).
then there
is
a
Significance of the Derivative
(If gib) 9* g(a),
and
^
g'(x)
179
equation can be written
0, this
fib) -fja) _fjx)
g\x)
g{b)-g{a)
Notice that
if
= x for all x, then g'(x) = 1, and we obtain the Mean
On the other hand, applying the Mean Value Theorem to/
g{x)
Value Theorem.
and g separately, we
find that there are x
and
M -M _
y in
(<z,
b)
with
/'(*).
g(b)~g(a)
g'iyY
but there is no guarantee that the x and y found in this way will be equal.
These remarks may suggest that the Cauchy Mean Value Theorem will be
quite difficult to prove, but actually the simplest of tricks suffices.)
PROOF
Let
-
h(x) =f{x)[g{b)
Then
h
is
continuous on
It follows
-
g(x)[f(b) -f(a)].
b\ differentiate on
[a,
from Rolle's Theorem that h\x)
=
(a, b),
=
=f(a)g(b) -g(a)f(b)
h(a)
means
g(a)}
and
h(b).
0 for
some x
in
(a, b),
which
that
0
=
f(x)[g(b)
-
-
g(a)]
The Cauchy Mean Value Theorem
theorem which
g'(x)[f(b)
-
(a)]. |
the basic tool needed to prove a
is
facilitates evaluation of limits of the form
M
Km-
x^a g{x)
when
lim f(x)
In
this case,
=
0
and
lim gix)
=
0.
5-3 is of no use. Every derivative is a limit of this form,
derivatives frequently requires a great deal of work. If some
Theorem
and computing
derivatives are
known, however, many
limits of this
form can now be evalu-
ated easily.
theorem
9 (L'HOPITAL'S
RULE)
Suppose that
lim f(x)
=
x—>a
and suppose
0
and
lim g(x)
also that lim
/'
1
'(*)
g '(*)
exists.
—
hm J
x-^agix)
Theorem
7
0,
is
Then lim /(*)/$(*)
x->a
x->a
(Notice that
=
x—*a
= hm
—
x^agix)
a special case.)
•
exists,
and
180
Derivatives and Integrals
PROOF
The
hypothesis that lim f'{x)/g'{x) exists contains two implicit assumptions:
(1)
there
(2)
is
an interval
—
{a
—
a
8,
+
8)
such that /'(*) and g'(x) exist for
+
8)
except, perhaps, for x
in this interval g'(x)
^
0 with, once again, the possible exception of
x in (a
all
=
x
8,
a
=
a,
a.
On the other hand, / and g are not even assumed to be defined at a. If we
define f(a) = g(a) = 0 (changing the previous values of f(a) and g(a), if
necessary), then /and g are continuous at a. If a < x < a
8, then the Mean
+
Value Theorem and the Gauchy
the interval
Mean Value Theorem
Mean Value Theorem to g,
would be some xi in (a, x) with
applying the Cauchy Mean Value Theorem
applying the
there
number ax
in
(a,
apply to / and g on
— 8 < x < a). First
we see that g(x) ^ 0, for if g(x) = 0
g'(x{) = 0, contradicting (2). Now
(and a similar statement holds for a
[a, x]
to
/ and
g,
we
see that there
is
a
such that
x)
[/(*)
-
f
-
0]g (ax )
[g(x)
-
0]f(ax )
or
/(*)
Now a x
_
f'(ax )
approaches a as x approaches
lim f(y)/g'(y)
exists, it follows
y—>a
lim
(Once again, the reader
argument.)
M
is
a,
because
ax
is
in
(a, x)
;
since
that
lim
f^
lim
m.
invited to supply the details of this part of the
|
PROBLEMS
1.
For each of the following functions, find the maximum and minimum
by finding the points in the interval
where the derivative is 0, and comparing the values at these points with
the values at the end points.
values on the indicated intervals,
(i)
(ii)
/(*)
f{x)
(hi) f(x)
(iv) f(x)
(vi)
=
=
x*
*
5
x*
-
+ *+
3* 4
1
=
/W =
-
8*3
Sx
+
—
Xb
+X+
AT
—
+
1
6* 2
1
on [-2,
2].
on [-1,
1].
on [- J,
J].
on [-i,
1].
1
on
1
[0, 5].
181
Significance of the Derivative
Now sketch
the graph of each of the functions in Problem
maximum and minimum
local
1
,
and
find all
points.
n
(a)
<
If ai
•
<
•
•
an
find the
,
minimum
value of f(x)
= ^
i
—
(x
ai)
2
.
=i
n
*(b)
Now
minimum
find the
= ^
value of f{x)
t
lem where calculus won't help at
on the
all:
the function /is linear, so that the
en's
—
\x
This
is
a prob-
=i
intervals
between the
minimum clearly occurs at one
and these are precisely the points where / is not differHowever, the answer is easy to find if you consider how
changes as you pass from one such interval to another.
of the
ai,
entiable.
f(x)
*(c)
Let a
>
Show
0.
/<*)
that the
is
(2
+
+
a)/{\
(—
+
+
|*|
l
+
(i)
=
/
x,
x
\
5,
x
-3,
*
9,
*
{
/
/•
• •
(m)\
f
r/
/(*)
x/
/W
v
.
x
c/
\
\
(iv)7 y
/(*)
=
=
=
W
v
*|
(0, a),
all local
maximum and
mini-
//
\
JW
__
-
0,
(
1/?,
'
=
=
5
7
9.
a:
=
/
*
x.
x rational
0,
x irrational.
1,'
*
0,
otherwise.
=
i
in lowest terms.
\
(
9
3
(
1/n for some n in
N
{
I
|
A straight line
then back to
5* 3, 5, 7,
* irrational
f
I
,
*
7,
I
(n)
-
points.
fix)
/..\
|*
(The derivative can be found on each of the
and (a, oo) separately.)
a).
<x, 0),
For each of the following functions, find
mum
value of
1
=
i
intervals
maximum
the decimal expansion of x contains a 5
1,
if
^
otherwise
drawn from the point
is
(1, b)>
(0, a)
to the horizontal axis,
as in Figure 23. Prove that the total length
is
and
shortest
must bring a function
where (*, 0) is the point
on the horizontal axis. The dashed line in Figure 23 suggests an alternative geometric proof; in either case the problem can be solved without
when
the angles
a and
/3
are equal. (Naturally you
into the picture: express the length in terms of*,
actually finding the point
Prove that of
greatest area.
all
(*, 0).)
rectangles with given perimeter, the square has the
7.
among
Find,
all
volume
right circular cylinders of fixed
V, the
smallest surface area (counting the areas of the faces at top
as in Figure 24).
8.
one with
and bottom,
,
Figure 25 shows the graph of the
derivative of/.
Find
all local
maximum
and minimum points of/.
FIGURE
*9.
25
n~
n
l
+
Suppose that / is a polynomial function, f(x) = x + a n -ix
—
values
critical
corresponding
and
critical
points
with
4,
3,
1,
2,
1,
+ 0o,
6, 1, 2, 4, 3. Sketch the graph of /, distinguishing the cases n even and n
•
odd.
*10.
(a)
~
•
•
on the basis of this information.
Does there exist a polynomial function with the above properties,
except that 3
11.
is
not a
critical point?
Describe the graph of a rational function
to the text's description of the
12.
(a)
•
n l
n
+
Suppose that the polynomial function f(x) = x + a n -ix
=
+ a 0 has critical points -1, 1,2, 3, and/"(-l) 0,/"(l) > 0,
/"(2) < 0, /"(3) = 0. Sketch the graph of/ as accurately as possible
•
(b)
•
(in
very general terms, similar
graph of a polynomial function).
Prove that two polynomial functions of degree
m and
n,
respectively,
most max(m, n) points,
For each m and n exhibit two polynomial functions of degree m and n
which intersect max(m, n) times.
n ~~ l
n
+
Suppose that the polynomial function f(x) = x + a n -\x
critical
all
for
0
points
and
^
critical
/"(#)
exactly
k
has
a
+ 0
intersect in at
(b)
*13.
(a)
•
•
•
points
(b)
(c)
x.
•
•
•
+
points.
(d)
Show
—
that n
k
odd.
is
For each n, show that there is a polynomial function / of degree n
with k critical points if n — k is odd.
n" 1
n
+
Suppose that the polynomial function f(x) = x + a n ^\X
Let
k%
=
n,
maximum
a 0 has ki local
Show
that k 2
—
kY
+
1
if
n
even,
k h k 2 be three integers with k 2
ki if n
is
odd, and ki
+
k2
<
n.
and k 2 local minimum
and k 2 = £i if n is odd.
= k\ + 1 if n is even, and
points
is
Show
that there
is
a polynomial
function / of degree n, with k\ local maximum points and k 2 local
< a kl +k and try
minimum points. Hint: Pick a\ < a 2 <
•
•
•
t
kl+k2
f{x)
=
n
~
ad
'
(1
+
* 2 )* for an appropriate
number
/.
Significance of the Derivative
14.
Prove that
(a)
M(b -
for
x in
all
Prove that if/'(*) < m for all x in [a,
Formulate a similar theorem when
(a)
Suppose that/'(*)
/(*)
>
g(x) for x
> g'(x)
> a and
Show by an example
(b)
=
the hypothesis f(a)
16.
M
>
/'(*)
Find
(a)
all
>
then f{b)
[a, b],
f{a)
+
-
a).
a).
(c)
(b)
15.
if
183
functions
= sin x.
= x\
/'"(*) = x +
/ such
for all
fix)
a:,
<
<
then /(£)
b],
|/(*)|
<
M
and that f(a)
g(x) for x
that these conclusions
<
/(*)
+
for all x in
=
g(a).
[>, b].
Show that
a.
do not follow without
g(a).
that
f(x)
(b) /"(*)
(c)
17.
x\
true that a weight dropped from rest will fall s(t) = 16/ 2
seconds, this experimental fact does not mention the behavior
of weights which are thrown upwards or downwards. On the other hand
Although
feet after
it is
t
= 32 is always true and has just enough ambiguity to
account for the behavior of a weight released from any height, with any
initial velocity. For simplicity let us agree to measure heights upwards
from ground level; in this case velocities are positive for rising bodies and
the law s"(t)
negative for falling bodies, and
s"(t)
(a)
(b)
all
bodies
fall
according to the law
= -32.
Show that s is of the form s(t) = —l6t 2 + at-\-/3.
By setting t = 0 in the formula for s, and then in the formula
for /,
show that s(t) = — 16t 2 + v 0 t + s 0 where s 0 is the height from which
the body is released at time 0, and v 0 is the velocity with which it is
,
released.
(c)
A
weight
ground
is
level.
maximum
it
thrown upwards with velocity v feet per second, at
How high will it go? ("How high" means "what is the
height for
achieves
its
all
What is its velocity at the moment
What is its acceleration at that
ground again? What will its velocity
times.")
greatest height?
moment? When will it hit the
be when it hits the ground again?
18.
A
cannon
ball
is
shot from the ground with velocity v at
(Figure 26) so that
horizontal
the law
(a)
Show
v
cos a.
= — 16t
+
(v sin
v
2
distance
Its
a)t while
9
an angle a
v sin
a and a
above the ground obeys
horizontal velocity remains
s(t)
its
cos a.
cannon ball is a parabola (find the position
and show that these points lie on a parabola).
Find the angle a which will maximize the horizontal distance
traveled by the cannon ball before striking the ground.
that the path of the
at each time
(b)
has a vertical component of velocity
component
s(t)
constantly
it
/,
*19.
(a)
Give an example of a function /
for
which lim
X—
exists,
f(x)
but
so
lim /'(*) does not
x—»
Prove that
(b)
exist.
°o
lim /(*) and lim /'(*) both
if
x—» oo
Prove that
(c)
a;—*
exists,
0.
=
0
X— «
>
=
=
00
then lim /"(*)
X— W
00
also lim f(x)
X—»
then \imf(x)
exist,
oo
lim /(*) exists and lim /"(*)
if
X—>
and
x—*
0.
oo
(d) Generalize this result.
20.
Suppose that / and g are two differ en tiable functions which satisfy
— fg = 0. Prove that if a and b are adjacent zeros of/, and g(a) and
(Natug(b) are not both 0, then g(x) = 0 for some * between a and
same result holds with / and g interchanged; thus, the zeros of
each other.) Hint: Derive a contradiction from the
separate
/ and g
assumption that g(x) ^ 0 for all x between a and b: if a number is not 0,
rally the
there
21.
stant
22.
is
a natural thing to do with
Suppose that
-
\f(x)
by considering
Prove that
/O0|
/'.
it.
< 0 ~
n
3>)
for n
>
1.
Compare with Problem
Prove that /
is
con-
3-20.
if
ai
—+—
an
ao
12
-r
*
'
n
'
+—7
;
n
_
"*
u
=
0
>
1
then
a0
some #
for
23.
+ aix +
•
•
•
+
an*
n
in [0, 1].
m never has two
Prove that the polynomial function fm (x) = x* - 3x
roots in [0, 1], no matter what m may be. (This is an easy consequence of
Rolle's Theorem. It is instructive, after giving an analytic proof, to graph
and /i, and consider where the graph of fm lies in relation to them.)
+
f0
24.
*25.
Suppose that / is continuous and differentiable on [0, 1], that /(*) is in
* 1 for all x in [0, 1]. Show that there is
[0, 1] for each x, and that/'OO
exactly one number x in [0, 1] such that /(*) = x. (Half of this problem
has been done already, in Problem 7-11.)
Prove that if/ is a twice differentiable function with /(0) = 0and/(l) = 1
an d
/'(0)
=
f{\)
=
0,
then
|/"(jc)|
>
4 for some x in
[0,
1].
In more
A
particle which travels a unit distance in a unit time,
picturesque terms:
and starts and ends with velocity 0, has at some time an acceleration > 4.
Hint: Prove that either f"(x)
—4
26.
for
some x
>
4 for some x in
Suppose that / is a function such that /'(*)
= 0. Prove that f(xy) = f(x) + /(>) for
/(l)
*27.
/(*) when^W
Suppose that /
[0, J],
or else /"(*)
<
in [J, 1].
=
=
f(xy).
satisfies
/"(*) +f'(x)g(x)
-/W
=
1/x for
all x,
0
y
>
all
0.
x > 0 and
Hint: Find
185
Significance of the Derivative
for
some function g. Prove that if / is 0 at two points, then /
between them. Hint: Use Theorem 6.
is
0 on the
interval
28
Suppose that / is rc-times differentia ble and that
{n)
x. Prove that
f (x) = 0 for some x.
f(x)
=
0 for n
+
1
different
29.
Prove that
< V66 -
i
(without computing
30.
V66
<i
8
to 2 decimal places!).
Prove the following slight generalization of the Mean Value Theorem:
If/ is continuous and differentiable on (a, b) and lim
f(y) and lim
y
exist,
then there
is
some x
in
b
-
31.
y—*b~
a+
f(y)
—
lim f(y)
a
(Your proof should begin: "This is a
Value Theorem because
.")
.
*
such that
(a, b)
lim f(y)
—
trivial
consequence of the
Mean
.
.
Prove that the conclusion of the Cauchy
written in the form
fib)
- fjfl)
g(b)
-g(a)
Mean Value Theorem
can be
f(x)
=
g'{x)
under the additional assumptions that g(b)
never simultaneously 0 on (a, b).
^
g(a)
and that f(x) and
g'(x) are
*32.
Prove that if/ and g are continuous on [a, b] and differentiable on
g'(x) ^ 0 for x in (a, b), then there is some x in (a, b) with
and
/'(*)
g'M
Hint: Multiply out
33.
What
is
first,
_
m -~
gW
to see
what
—
*
2
34.
limit
is
actually —4.)
Find the following
x
(i)
lim
*-*o
(ii)
lim
+0
this really says.
wrong with the following use of FHopital's Rule:
x
x -2
-— = hm
hm — +
z-i x - 3x + 2
x~>i
(The
fia)
g(x)
tan x
cos 2 x
x*
—
1
limits:
3x*
2x
+1 = hm —
6*
=
-
3
2
,
3.
(a, b),
1
186
Derivatives and Integrals
35.
Find
/'(O)
if
x
=
x
10,
and 5 (0) = ^(0) - 0 and ^"(O) =
36.
0
a:
i
=
fix)
0,
17.
Prove the following forms of l'Hopital's Rule (none requiring any essen-
new
tially
(a)
reasoning).
=
lim f{x)
If
lim
f(x)/g(x)
(b)
=
=
(c)
—
>
<z
=
£—
/,
then lim
x—+a +
from below).
and lim f(x)/g
0,
f
(x)
=
oo,
then lim
x—*a
x—>a
— oo
(and similarly for
oo
or
s
+ or x —» - )-
if
—>
x
a
is
replaced by
<3
=
lim /(*)
If
=
x—*a
f(x)/g(x)
=
f'(x)/g'(x)
x—*a +
lim g(x)
x—*a
*
and lim
0,
(and similarly for limits
I
lim fix)
If
=
g(x)
x—*a +
x—*a +
-
lim
x—
»
f(x)/g(x)
=
(and
I
=
and lim f(x)/g'(x)
0
x— »
flO
similarly
then lim
x—» 00
— oo).
for
/,
oo
Consider
Hint:
lim
x-»0 +
f(l/x)/g(l/x).
(d)
X—>
lim
£(»
=0
and lim
x—» «
oo
=
fix)/gix)
37.
=
lim /(*)
If
x—*
f'(x)/g'(x)
= »,
then lim
X—
oo
«0
oo,
another form of l'Hopital's Rule which requires more than
algebraic manipulations: If lim f(x) = lim g(x) = oo, and lim
There
is
X—
>
f(x)/g
,
(x)
=
X—*
oo
=
then lim f(x)/g(x)
X—
/,
I.
Prove
X—> 00
oo
this as follows.
oo
(a)
For every
£
>
0 there
m
is
a
-
number
<
a such that
for x
s
>
a.
*'(*)
Apply the Cauchy Mean Value Theorem
to /and
g on
[a, x]
to
show
that
fix)
g(x)
~ fia)
- g(a)
(Why can we assume
(b)
Now
g(x)
<
—
£
for x
g(a) 9^ 0
>
a.
?)
write
_
f(x)
gix)
f{x)
-
gix)
-gia)
f(x)
fja)
~
gja)
- fia)
gix)
— fia) ^
0 for large x
fix)
(why can we assume that fix)
g(x)
that
/(*)
*(*)
-
/
<
2e
for sufficiently large x.
?)
and conclude
.
187
Significance of the Derivative
38.
To complete
to prove a
many
the orgy of variations on l'Hopital's Rule, use Problem 37
few more cases of the following general statement (there are so
possibilities that
you should
select just
a few,
if
any, that interest
you):
If lim /(*)
=
f(*)/g(x)
=
*39.
(a)
(b)
or
co
in
(a, b).
oc
(
[
+ or a~ or oo or ] can be a or a
and ( ) can be / or oo or — oo
,
minimum
Hint: Consider the
in
then lim
),
qo
,
and
{
}
of /on
[a, b];
why must
it
be
b)?
(a,
<c < f'(b),
then /'(*) - c for some x in (a, 3).
Darboux's Theorem.) Hint: Cook up an
appropriate function to which part (b) may be applied.
Prove that
if
(This result
*40.
-
{and lim f(x)/g'(x) =
{
Suppose that / is differentiable on [a, b]. Prove that if the minimum
of/ on
is at a, then /'(a) > 0, and if it is at b, then
f(b) < 0.
(One half of the proof of Theorem 1 will go through.)
Suppose that /'(a) < 0 and fib) > 0. Show that /'(*) = 0 for some x
somewhere
(c)
Here
).
(
can be 0 or
=
lim g{x)
f(a)
is
known
as
It is easy to find a function
/ such that |/| is differentiable but / is not.
For example, we can choose /(*) = 1 for x rational and
f(x) = - 1 for
x irrational. In this example /is not even continuous, nor is this a mere
coincidence: Prove that if |/| is differentiable at a, and/ is continuous at
then /is also differentiable at a. Hint: It suffices to consider only a with
f(a) = 0. Why? In this case, what must \f\'(a) be?
a,
*41.
(a)
Let y
^
when
x
n
0 and let n be even. Prove that x n + n = (* +
only
y
y)
= 0. Hint: If x 0 n + y n = (* + y) n apply Rolle's Theorem
n
to f(x) = x n +y« - (x +
on [0, * 0
y)
Prove that if y ^ 0 and n is odd, then x n + n = (x
+ y) n only if
y
x = 0 or x = —y.
0
,
],
(b)
**42
**43.
Use the method of Problem 41 to prove that if n is even and
f(x)
then every tangent line to / intersects
/ only once.
Prove even more generally that
intersects
*44.
is
-
0
increasing on
Prove that
*45.
it is
essential, as
Use
and f
is
increasing. Prove that the function g(x)
+
Hint: Obviously you should look at g'(x).
positive by applying the Mean Value Theorem to
/ on
(it will help to remember that the hypothesis
shown by the function /(*) = 1 -f
if
n
>
1,
then
<
x
for
-
1
(notice that equality holds for x
=
0).
x)
=
oo).
(0,
derivatives to prove that
(1
,
increasing, then every tangent line
the right interval
is
xn
/ only once.
Suppose that/(0)
f(x)/x
;
if / is
=
n
>
1
+
nx
<
0
and 0 <
x.
/(0)
=
0
188
Derivatives and Integrals
46.
and
x* sin 2 l/x for x 9* 0,
a local
(a)
Prove that 0
(b)
Prove that /'(0)
is
=
minimum
= 0.
let /(0)
=
point for
0 (Figure 27).
/.
/"(0)
This function thus provides another example to show that Theorem 6
cannot be improved. It also illustrates a subtlety about maxima and
minima that often goes unnoticed: a function may not be increasing in
any interval to the right of a local minimum point, nor decreasing in any
M
FIGURE
Let/0) =
interval to the
27
*47.
(a)
left.
f(a) > 0 and /' is continuous at
in some interval containing a.
Prove that
The next two
If g(x)
(b)
if
parts of this
=
1
=
0 (see
problem show that continuity of/'
increasing
is
essential.
1
by showing that in any interval there are
interval containing 0,
points x with /'(*)
The behavior
is
/*,
=
< a <
Suppose 0
/(0)
then /
show that there are numbers x arbitrarily close
and also with g'(x) = — 1.
2
1. Let /(*) = ax + x sin l/x for x 9* 0, and let
Figure 28). Show that / is not increasing in any open
x 2 sin
to 0 with g'(x)
(c)
a,
of /for
>
and
0
a >
1,
also points x with
which
is
much more
<
/'(at)
0.
difficult to analyze, is
discussed in the next problem.
FIGURE
*48.
Let f(x)
28
=
ax
+x
the sign of /'(*)
1
/x
is
< -1
for
2
sin l/x for x
when a >
1 it is
any numbers x
if
^00 < -1
0,
and
let/(0)
=
close to 0. It
-
is
a
In order to find
2x sin l/x - cos
more convenient
0.
if
little
cos y for y * 0; we want to
the most
for large y. This question is quite delicate;
to consider the function g(y)
know
*
necessary to decide
=
2(sin y)/y
—
}
189
Significance of the Derivative
is - cos y, which does reach the value -1, but
happens only when sin y = 0, and it is not at all clear whether
g
itself can have values < — 1 The obvious approach to this problem is to
significant part of g(y)
this
.
minimum
find the local
values of
=
solve the equation g'(j)
(a)
Show
that
=
f
if
g {y)
it is
impossible to
more ingenuity
is
required.
then
0,
=
sin y ^
=
s'my
cos y
Unfortunately,
g.
0 explicitly, so
~2~/
and conclude that
g(y)
(b)
Now show
that
if
=
0,
sin 2 y
=
g\y)
(—2y
then
Ay 2
+ y*
4
and conclude that
1*001
(c)
(d)
A
is
+>
Va + y
2
2
+
4
+/
2
fact that (2
> 1, show that
y )/VA
not increasing in any interval around 0.
Using the
/
=
Using the
fact that lim (2
then /
is
increasing in
function
/
is
+y
some
increasing at a
2
interval
if
+
)/V~A
4
>/
=
around
there
is
1,
show
>
f(a)
if
a
<
x
<
a
+
5
fix)
<
fid)
if
a
-
8
<
x
<
a.
does
not
a =
that
if
1,
then
a >
1,
0.
some number
fix)
if
5
>
0 such that
and
Notice that
a
+
8); for
this
mean
that /
is
—
8,
increasing at
0,
increasing in the interval (a
example, the function shown in Figure 28
is
but
is
(a)
Suppose that /is continuous on [0, 1] and that /is increasing at a for
[0, 1]. Prove that /is increasing on [0, 1]. (First convince
yourself that there is something to be proved.) Hint: For 0 < b < 1,
prove that the minimum of / on [b, 1] must be at b.
Prove part (a) without the assumption that / is continuous, by considering for each b in [0, 1] the set Sh = {x: fiy) > fib) for ally in
not an increasing function in any open interval containing
0.
every a in
(b)
[b, x]
(This part of the problem is not necessary for the other parts.)
Hint: Prove that Sb = {x: b < x < 1 } by considering sup Sb
If/ is increasing at a and /is differentiable at a, prove that f'ia) > 0
.
.
(c)
(this
is
easy).
190
Derivatives and Integrals
>
If f(a)
(d)
0,
prove that /is increasing at a (go right back to the
defi-
nition of f(a)).
(e)
(f)
Use parts (a) and (d) to show, without using the Mean Value
Theorem, that if / is continuous on [0, 1] and f'(a) > 0 for all a in
[0, 1], then / is increasing on [0, 1],
Suppose that /is continuous on [0, l]and/'(a) = 0 for all a in (0, 1).
Apply part (e) to the function g(x) = f(x) + ex to show that
/(l) -/(0) > -e. Similarly, show that /(l) - /(0) < £ by considering h(x) = ex - /(*). Conclude that/(0) = /(l).
This particular proof that a function with zero derivative must be constant has many points in common with a proof of H. A. Schwartz, which
may be the
to think
it
first
rigorous proof ever given. Its discoverer, at least, seemed
was. See his exuberant letter in reference [41] of the Suggested
Reading.
**50.
a constant function, then every point is a local maximum point
is quite possible for this to happen even if / is not a constant function: for example, if f(x) = 0 for x < 0 and /(*) = 1 for x > 0. Prove,
however, that if / is continuous on [a, b] and every point of [a, b] is a
If
/
is
for/. It
local
maximum
mum
of / occurs at x 0 consider the set {x in
point, then
/
is
a constant function. Hint: If the mini-
,
[#<>,
b]\
f(y)
—
f(x 0 ) for
all
yin[x Q ,x]\.
**51.
(a)
A
point x
f(x)
>
an ordinary
maximum point for / on A
^ x (compare with the definition
point). A local strict maximum point
way. Find all local strict maximum points
a
strict
if
A
maximum
with y
of
is
called
f(y) for all y in
defined in the obvious
is
of
the function
0,
x irrational
~j
x
— -
in lowest terms.
seems quite unlikely that a function can have a local strict maximum at every point (although the above example might give one
It
pause for thought). Prove
.(b)
this as follows.
Suppose that every point is a local strict maximum point for /. Let
x\ be any number and choose a\ < Xi < bi with b\ — ai < 1 such
that f(xi) > f{x) for all x in [a h b{\. Let x 2 5* *i be any point in
(a i, bi) and choose a x < a 2 < x 2 < b 2 < bi with b 2 — a 2 < \ such
that f(x 2 ) > f{x) for all x in [a 2 b 2 \ Continue in this way, and
,
use the Nested
contradiction.
Interval
Theorem (Problem
8-14)
to obtain
a
Significance of the Derivative
191
CONVEXITY AND CONCAVITY
APPENDIX.
Although the graph of a function can be sketched quite accurately on the
by the derivative, some subtle aspects of the
graph are revealed only by examining the second derivative. These details
were purposely omitted previously because graph sketching is complicated
enough without worrying about them, and the additional information obtained is often not worth the effort. Also, correct proofs of the relevant facts
are sufficiently difficult to be placed in an appendix. Despite these discouraging remarks, the information presented here is well worth assimilating,
because the notions of convexity and concavity are far more important than
as mere aids to graph sketching. Moreover, the proofs have a pleasantly
geometric flavor not often found in calculus theorems. Indeed, the basic
basis of the information provided
definition
DEFINITION
1
A
geometric in nature (see Figure
is
function /
line
is
convex on an
segment joining
(a>
interval, if for all a
and
f(a))
(b,
The geometric condition appearing in this
way that is sometimes more useful in
and
f(a))
(b,
f(b))
=
m^m
—
definition
above the graph of / at x
b
a)
g(x)
>
by
.
f(x), that
is,
if
+f(a)>f(x)
a
fib)
b
2
if
- fja)
(x-a)>
-a
fib)
DEFINITION
-
(x
—
b
We
can be expressed in an
straight line between
The
proofs.
- a)+ /w
{x
a
b
lies
the interval, the
b in
the graph of the function g defined
is
g(x)
This line
and
above the graph of /.
f{b)) lies
analytic
(a,
1).
-fja)
—
>
fix)
x
a
- f(a)
f(x)
-fja)
—
_
a
therefore have an equivalent definition of convexity.
A
function /
a
<
x
<
b
is
convex on an
fix)
If the
interval
if
for a, x,
and
b in
the interval with
we have
word "over"
- fja)
in Definition
1
<
is
fjb )
- fja)
.
replaced by "under" or, equivalently,
192
Derivatives and Integrals
if
the inequality in Definition 2
/(*)
(*,
is
replaced by
-/(«)
/(*))
/W
>
b
"/(*)
—
a
is
not hard to see
that the concave functions are precisely the ones of the form
— /, where / is
we obtain
the definition of a concave function (Figure 2). It
convex. For this reason, the next three theorems about convex functions have
immediate corollaries about concave functions, so simple that we will not even
FIGURE
bother to state them.
2
Two
Figure 3 shows some tangent lines of a convex function.
things
seem
to
be true:
The graph
(1)
point
(a,
tangent
If a
(2)
<
of /
lies
above the tangent
f(a)) itself (this point
As a matter of fact
THEOREM
l
PROOF
line at (a, f(a)) except at the
called the point of contact of the
line).
b,
then the slope of the tangent line at
slope of the tangent line at
FIGURE
is
(b,
f{b))\ that
these observations are true,
is,
(a,
/'
f(a))
is
is
than the
less
increasing.
and the proofs are not
difficult.
3
Let / be convex. If / is differentiable at a, then the graph of / lies above the
tangent line through (a, /(a)), except at (a, f(a)) itself. If a < b and /is differentiable at a and b, then f(a) < f'(b).
If 0
<
hi
<
A 2 then, as Figure 4 indicates,
,
fja
+
hQ -f(a)
(1)
<
f(a
+
h 2)
- f(a)
hi
hi
A nonpictorial proof can be derived immediately from Definition 2
a
<
a
+ hi
<
a
+h
2
.
Inequality (1) shows that the values of
/(*
+ *)
-/(*)
applied to
193
Significance of the Derivative
1
FIGURE
1
1
a
a
+ hi
a -f h 2
4
decrease as h
—
>
0+
.
Consequently,
/(„)<MiHW
forA>0
h
(in fact/'(tf)
that for h
is
>
the greatest lower
bound of all
0 the secant line through
(a,
and
larger slope than the tangent line, which implies that
above the tangent
means
these numbers). But this
f(a))
+ h,
{a +
(a
line (an analytic translation of this
+ h))
f(a + h))
f(a
argument
is
has
lies
easily
supplied).
For negative h there
is
f(a
a similar situation (Figure
+
-f(a)
h,)
>
f(a
+
h 2)
h2
<
hi
<
0,
then
- f(a)
hi
hi
This shows that the slope of the tangent line
f(°
5): if
+
»-M
is
greater than
{orh<0
h
(in fact
f(a)
is
the least upper
above the tangent
lies
\
1
a
+
h2
FIGURE
a
5
+
line
1
hi
a
if
h
bound
<
0.
of
all
these numbers), so that f(a
This proves the
first
+ h)
part of the theorem.
194
Derivatives and Integrals
FIGURE
Now
6
suppose that a
.
v
ff(
/
<
{a)
<
Then,
b.
we have already
f(a+(b-a)) -f(a)
-
b
_
as
fW
b
—
seen (Figure 6),
since b
—
sincefl
-, <0
a
>
0
a
-f(a\
-
a
and
-»))-/(»)
yW
=
/(g)
a
Combining
figure
Theorem
7
We
1
if /' is
=
- /(a)
/(*)
b
a
b
we
these inequalities,
<
obtain
/'(£). I
has two converses. Here the proofs will be a
begin with a
Rolle's
-/(*)
—
Theorem
lemma
that plays the
plays in the proof of the
increasing, then the graph of / lies
same
little
more
role in the next
Mean Value Theorem.
below any
difficult.
theorem that
It states
that
secant line which happens to
be horizontal.
LEMMA
Suppose /
/(*)
PROOF
<
is
/(«)
differentiable
=
/(*) for a
and
<
x
/' is increasing. If a
<
<
b
and
f(a)
=
f(b),
then
b.
Suppose first that f(x) > f(a) = f(b) for some x in (a, b). Then the maximum
/ on [a, b] occurs at some point x 0 in (a, b) with f(x 0) > f(a) and, of course,
f(x 0) = 0 (Figure 7). On the other hand, applying the Mean Value Theorem
to the interval [a, x 0], we find that there is x\ with a < x\ < #o and
of
f{xi)
= /(*)-/(«) >
Xo
—
a
0)
195
Significance of the Derivative
contradicting the fact that /'
<
for a
x
for x in
<
b,
and
(if it
=
[xi, x]
some x
f(a) for
in
(a, b).
<
<
x and/(*i)
<
f(a).
we conclude
that there
is
x 2 with x\
a
x\
X
=
the other hand, f(x)
0, since
contradicts the hypothesis that /'
We now attack the general
we used
that
THEOREM
2
PROOF
If
/
is
Let a
and
<
b.
is
case
in the proof of the
differentiable
f(b)
<
x2
<
not constant on
is
(Figure 8)
Mean Value Theorem
x
and
0.
maximum
occurs at
x.
Again
this
increasing. |
by the same
sort of algebraic
machinations
Mean Value Theorem.
then /
is
convex.
(x
-
Define g by
fib)
=
/(*)
b
It is
Applying the
a local
/' is increasing,
g(x)
easy to see that g'
ing the
—
also impossible
so there
[a, x])
- /to)
>
— X\
/to
/'(*2)
On
is
We know that / is
were, /' would not be increasing on
some #1 with
to
f(a)
f(a)
(a, b).
Suppose f(x)
[a, x]
=
only remains to prove that f(x)
it
<
increasing. This proves that f(x)
is
lemma
to
is
a).
moreover, g(a)
also increasing;
=
g(b)
=
f(a).
Apply-
g we conclude that
<
g(x)
In other words,
- f(a)
— a
if
<
a
<
x
/to
b,
-
f{a)
if
<
a
x
<
b.
then
Kb)
b
- f(a)
(x-a)< f(a)
—a
or
/(*)
Hence /
THEOREM
3
If /
is
is
"/«
.
Kb)
- f(a)
convex. |
differentiable
and the graph of /
/ is convex.
lies
above each tangent
line except at
the point of contact, then
PROOF
Let a
<
at
/(a)),
from Figure 9 that if (b, f(b)) lies above the tangent line
above the tangent line at (b, fib)), then the slope
of the tangent line at (£, fib)) must be larger than the slope of the tangent line
at (a,f(a)). The following argument just says this with equations.
Since the tangent line at (a, f(a)) is the graph of the function
(a,
b. It is
and
clear
(a,
f(a)) lies
g(x)
=f(a)(x-a)+f(a),
and
since
(b,
f(b)) lies
above the tangent
Similarly, since the tangent line at
(b,
h(x) =f'(b)(x
and
(a,
f(a)) lies
above the tangent
from
a)
f(b))
is
-
+f(b),
b)
+f(a).
the graph of
we have
line at (b, f(b)),
f( a )>f'(bKa-b)+f(b).
(2)
It follows
we have
-
f(b)> f{a){b
(1)
line,
and (2) that f'(a) < f'(b).
from Theorem 2 that / is convex.
(1)
It
now
If
a function / has a reasonable second derivative, the information given in
which / is convex or
follows
|
these theorems can be used to discover the regions in
concave. Consider, for example, the function
1
=
/(*)
+
1
For
x2
this function,
-2x
=
/'(*)
+
(1
Thus
/'(*)
=
=
0 only for x
0,
and
>
<
/'(*)
/'(*)
/(0)
x2) 2
=
0
if
x
0
if
x
1,
while
<
>
0.
0,
Moreover,
FIGURE
/(*)
>
f(x)
—> 0
/
even.
is
0
for all x,
as #
—
>
co
or
-
<x>
10
The graph
of / therefore looks something like Figure 10.
rod =
(1
+x
2
2
)
{-2)
+ 2x [2(1 +
(1 + x Y
•
2
2{2>x 2
(i
-
1)
+x y'
2
x 2)
We now compute
;
2* 1
It is
Significance of the Derivative
197
=
0 only
not hard to determine the sign of /"(*). Note
= Vl/3
when
x
same
sign
-
or
Vl/3. Since /"
on each of the
that /"(*)
first
clearly continuous,
is
it
must keep the
sets
(-«,_! ^T/3),
(- Vl/3, Vl/3),
(Vl/3,
Since
we
oo).
compute, for example, that
easily
*>o,
/"(-I) =
= -2 <
= i>
/"(0)
/"(I)
we conclude
/"
on (_
>
> 0 on (- oo _-^VT/3^and
< Oon (- Vl/3, Vl/3).
0 means
5
/
_ VT73)
00)
o,
that
/"
Since /"
0,
is
increasing,
and (VT73,
follows
it
oo),
from Theorem 2 that/ is convex
while on
<»),
(VlA
(-
V
1/3,
V 1/3) /
is
con-
cave (Figure 11).
/
is
convex
/
is
concave
- Vl/3
FIGURE
to the right,
(V
left,
1
/3,
f)
since
/
is
- Vl73
number
(a,
are inflection points of f(x)
= 0
= x\
does
not
ensure that a
then /"(0)
=
0,
but /
lies
below the part of the graph
and above the part of the graph
oo),
(-Vl/3, Vl/3);
concave on
the tangent line to the graph of /at
/(*)
convex
the tangent line
since /is convex on (Vl/3,
crosses the graph. In general, a
f"(a)
is
11
Notice that at
to the
/
Vl/3
is
is
a
is
f(a)) crosses the graph; thus
=
an
1/(1
+*
2
).
/, it is
right of
a.
This example
Vl/3 and
inflection point of /; for
example,
convex, so the tangent line at
an
procedure which
may
if
(0, 0) cer-
inflection point of
necessary that /" should have different signs to the
illustrates the
if
Note that the condition
tainly doesn't cross the graph of /. In order for a to be
function
thus the tangent line
called an inflection point of /
left
a
and
be used to analyze any
function/. After the graph has been sketched, using the information provided
\
198
Derivatives and Integrals
by
the zeros of /" are
computed and the sign of /" is determined on the
between consecutive zeros. On intervals where /" > 0 the function is
convex; on intervals where /" < 0 the function is concave. Knowledge of the
/',
intervals
and concavity of / can often prevent absurd misinter/. Several functions, which can be analyzed in
way, are given in the problems, which also contain some theoretical
regions of convexity
pretation of other data about
this
questions.
There
is
one further
fact
which
cannot
I
resist
mentioning here.
We
have
seen that convex and concave functions have the property that every tangent
line intersects the graph just once, and a few drawings will probably convince
you that no other functions have
only proof
omit
know
I
this property.
many
contains so
This
indeed the case, but the
is
delicate points that I feel
it is
wiser to
it.
PROBLEMS
1.
Sketch, indicating regions of convexity
the functions in
inflection,
Problem
and concavity and points of
11-1
(consider
double
as
(iv)
starred)
2.
Figure 25 in Chapter 11 shows the graph of /'. Sketch the graph of/.
3.
Find two convex functions / and g such that f(x) = g(x) if and only if x is
an integer.
Show that / is convex on an interval if and only if for all x and y in the
interval we have
4.
f(tx
5.
*6.
+
-
(1
t)y)
<
tf(x)
+
(1
t)f(y), for
0
<
t
<
1.
(This is just a restatement of the definition, but a useful one.)
Prove that if / and g are convex and / is increasing, then g o f is convex.
Let / be a twice-differentiable function with the following properties:
f(x)
> 0 for x > 0, and / is
= 0 for some x > 0 (so
/"(#)
is
not positive for
f(x)
=
l/(x
with/'(*o)
+
<
1),
0.
decreasing,
is
by
all x;
and
/'(0)
=
that in reasonable cases
an example
inflection point at
hypothesis in this theorem
is
given by f(x)
=
0.
/
1/(1
Prove that
have an
x 2 )). Every
will
+
shown byf(x) = 1 — * 2 which
which is not decreasing; and by
essential, as
f(x)
=
x2
,
,
which does not satisfy /'(0) = 0. Hint: Choose x 0 > 0
cannot have f'(y) < f(x 0) for all y > * 0 Why not?
for some xi > * 0 Consider /' on [0, Xi].
We
.
So f(xi) > f'(xo)
(a) Prove that if / is convex, then f([x + y]/2) < [/(*) +
f(y)]/2.
(b) Suppose that /satisfies this condition. Show that f(kx + (1 — k)y) <
kf(x) + (1 — k)f(y) whenever k is a rational number, between 0 and
n
Hint: Part (a) is the special case n = 1. Use
1, of the form m/2
induction, employing part (a) at each step.
(c) Suppose that / satisfies the condition in part (a) and / is continuous.
.
*7.
-
.
Show
that
/
is
convex.
199
Significance of the Derivative
*8.
Let
pi,
.
.
.
,
p n be
numbers with
positive
i
^=
=
pi
1.
i
n
(a)
For any numbers
xi,
.
.
.
,
x n show that
i
smallest
and the
^=
largest
n-1
n-1
(b)
Show
same
the
between the
piXi lies
i
for (1/t)
^
piXi,
where
t
=
^
i=i
t=i
n
(c)
Prove Jensen's
inequality:
If/ is convex, then/
n
^
A**) <
^
Hint: Use Problem
n
show that (1/0
^
t
*9.
(a)
by
is
/_'(#).
and
in the
—
1
domain
of /
t.
if
(Part (b)
xu
.
.
.
is
,
needed to
* n are.)
i
/,
the right-hand derivative, lim [f(a
+
h)
—
if
/
is
convex. Check this assertion, and also show that
/_' are increasing,
Show that
if
is
tinuous.) Hint:
close to
**11.
is
—
and that f-'(a) < f+(a),
then
convex,
f+{a) = fJ{a) if and only if /+' is con/
tinuous at a. (Thus / is differentiate precisely when /+' is con-
/+'
**10.
noting that p n
denoted by /+'(«), and the left-hand derivative is denoted
The proof of Theorem 1 actually shows that /+' and /_'
always exist
**(b)
=
For any function
f(a)]/h,
4,
-
pif(xi).
i=i
i=i
<z,
and
[f(b)
/+'(£)
— f(a)]/(b —
is less
than
a)
is
close to /_'(«) for b
<
a
this quotient.
Prove that a convex function must be continuous.
Prove that if / is convex on R, then either / is decreasing, or else / is
increasing, or else there is a number c such that / is decreasing on
(-oo t] and increasing on [c, oo ). This problem requires a lot of careful
;
bookkeeping, and
is
perhaps best postponed until after the next chapter,
where a similar theorem is proved in detail. An important step for this
problem is to establish that if f(a) < f(b) for some a < b, then / is
increasing on [b, cc), while / is decreasing on (— oo, a ] if f(a) > f(b).
12
CHAPTER
INVERSE FUNCTIONS
I
We now
have at our disposal quite powerful methods for investigating funcis an adequate supply of functions to which these methods
be applied. We have studied various ways of forming new functions from
tions;
may
what we lack
— addition,
old
alone,
and composition
multiplication, division,
we can produce only
— but
using these
the rational functions (even the sine function,
although frequently used for examples, has never been defined). In the next
few chapters we will begin to construct new functions in quite sophisticated
ways, but there is one important method which will practically double the
usefulness of
If
we
any other method we
recall that a function
is
discover.
a collection of pairs of numbers,
upon the bright idea of simply reversing
we
we might hit
Thus from the function
the pairs.
all
/=
{(1,2),
(3, 4), (5, 9), (13, 8)},
=
{(2,1),
(4, 3), (9, 5),
obtain
g
While /(l)
-
2
and
/(3)
-
we have
4,
=
^(2)
(8,13)}.
1
=
and g(4)
3.
Unfortunately, this bright idea does not always work. If
/=
{(1,2),
(3, 4), (5, 9), (13, 4)},
then the collection
(4,13)}
{(2, 1), (4, 3), (9, 5),
is
not a function at
where the trouble
sort of thing that
since
all,
lies:
contains both
it
=
/(3)
/(13), even
can go wrong, and
it is
and
(4, 3)
(4, 13).
It is clear
though 3 ^ 13. This is the only
worthwhile giving a name to the
functions for which this does not happen.
DEFINITION
A
function
The
/
is
one-one (read "one-to-one")
identity function /
is
if f(a)
^ f(b)
obviously one-one, and so
is
whenever a
^
b.
the following modifi-
cation:
x
The
function f(x)
=
x2
is
x
5,
x
3, 5
=
=
5
3.
not one-one, since /( —
g{x)
(and leave g undefined for x
200
Ix,
3,
<
0)
=
,
x\
x
then g
>
is
1)
=
/(l), but
if
we
define
0
one-one, because g
is
increasing
Inverse Functions
(since g'(x)
=
2x
>
0, for
>
x
-
/(*)
then /
is
one-one. If n
is
is
easily generalized: If n
one-one (since
xn
>
x
,
0,
odd, one can do better: the function
=
f{x)
is
This observation
0).
number and
a natural
is
201
x
n
for all x
n~ l
= nx
> 0, for all x ^ 0).
easy to decide from the graph of /whether /is one-one: the
fib) for a ^ b means that no horizontal line intersects the
/'(*)
It is particularly
condition f{a) ^
graph of / twice (Figure
1).
a function which
a one one function
FIGURE
If
in
we
any
not one one
is
(b)
(a)
1
reverse all the pairs in (a not necessarily one-one function)
some
case,
collection of pairs. It
is
/ we obtain,
popular to abstain from
this
pro-
—
cedure unless / is one-one, but there is no particular reason to do so instead
of a definition with restrictive conditions we obtain a definition and a theorem.
For any function /, the inverse of/, denoted by/- 1
(a, b) for which the pair (b, a) is in /.
DEFINITION
THEOREM
l
PROOF
/
1
is
a function
Suppose
first
and
that /
(c,
a)
implies that b
=
(b, a)
if
and only
is
/,
Since /
is
and
(*,/(*))
=
so a
Thus f~
l
is
-1
Conversely, suppose that/
pairs ib,fib))
_1
is
one-one. Let
are in
c.
/
if
=
,
is
the set of all pairs
one-one.
and (a,
and a =
(a, b)
fib)
be two pairs in f~\ Then
f(c); since / is one-one this
c)
a function.
is
ic,
a function.
a function this implies that
The graphs off and /_1
b
=
c.
=
/(c), then / contains the
and (/(*),<;) are in / l
Thus / is one-one. |
If fib)
fib)), so (/(*), b)
.
are so closely related that it is possible to use the
_1
_1
consists of all
Since the graph of /
graph of / to visualize the graph of /
1
from the
graph
the
off'
obtains
one
graph
of
/,
with (b, a) in the
.
pairs
(a, b)
202
Derivatives and Integrals
graph of / by interchanging the horizontal and
shown
FIGURE
2(a)
a
"<
rs
n
_
-i
C.
0)
oq "o
tu
.3" oq
d)
/
/
2
n
C «
n
5. °
/
/
/
This procedure
it is
C
C
3
3
C/q_
ST
F
I(x)
=
is
awkward with books and impossible with blackboards,
-1
is another way of constructing the graph of/
The
fortunate that there
and
which is
(a, b)
x,
we merely
.
are reflections of each other through the graph of
-1
called the diagonal (Figure 4). To obtain the graph of/
(b, a)
reflect the
graph of / through
this line (Figure 5).
Reflecting through the diagonal twice will clearly leave us right back where
we started; this means that (/_1 ) _1 = /, which is also clear from the definition.
In conjunction with
FIGURE
crq
6>
dia S°»aI
points
/
!-i
oq
/
so
/
Q
v
^
o
•
gj
crq
-
(c,
0-
f
a.
to
^
/
graph
vertical axis. If/ has the
in Figure 2(a),
4
Theorem
1,
this
equation has a significant consequence:
)~ l is
is also one-one (since
a one-one function, then the function /_1
a function).
if
/
is
l
{f~~
There are a few other simple manipulations with inverse functions of which
203
Inverse Functions
you should be aware. Since
/
Thus j~\b)
/ /
/
/
/
/
is
in
precisely
/
when
(b, a)
is
in
/ \
it
follows that
=
b
1/
{a, b)
r
/
f(x)
=
ber
is,
x
much
J
,
a
as
~
number a such that f(a) = b; for example, if
unique number a such that a 3 = b, and this num-
the (unique)
is
1
then/" ^)
by
The
s/
3
means the same
f(a)
is
the
definition, \fb.
fact that
f
1
(x) is
the
number y such
that f(y)
=
x can be restated in a
more compact form:
_1
/(/
W) =
*>
domain
for a11 * in the
of
Moreover,
FIGURE
Z"
1
(/(*))
= ^
f° r a ^ *
^
n
tlie
domain of /;
5
from the previous equation upon replacing / by f~\ These two
important equations can be written
this follows
r
i
o/
=
/
(except that the right side will have a bigger
not
all
is
of R).
Since
tions,
-1
domain if the domain of /or/
standard functions will be defined as the inverses of other funcquite important that we be able to tell which functions are one-one.
many
it is
already hinted which class of functions are most easily dealt with—
is
increasing and decreasing functions are obviously one-one. Moreover, if/
-1
-1
decreasis
then/
decreasing,
if/
is
and
increasing,
is also
increasing, then/
We have
FIGURE
ing (the proof
6
is left
to you). In addition,
/
is
increasing
decreasing, a very useful fact to remember.
It is certainly not true that every one-one function
decreasing.
Figure
One example
and only
if
— / is
either increasing or
has already been mentioned, and
is
now graphed in
6:
X
FIGURE
is
if
3,
x
5,
I,
x
3, 5
=
=
5
3.
are
Figure 7 shows that there are even continuous one-one functions which
you
neither increasing nor decreasing. But if you try drawing a few pictures
an interval
will soon agree that every one-one continuous function defined on
the
delight;
bureaucrat's
is either increasing or decreasing. This theorem is a
to
helps
It
organization.
proof uses no new ideas, but it does require careful
7
begin with two lemmas, the second being a corollary of the
LEMMA
1
PROOF
first.
Let / be a continuous one-one function defined on an interval, and let a,
and c be points of the interval with a < c < b. If f(a) < f(b), then f(a)
(as in Figure 8).
f(c) < f(b); and if }{a) > f(b), then f{a) > f(c) > f(b)
Consider
first
the case where f(a)
<
f(b).
Since
c
^
a, b
we have /(c) ^
b,
<
/(*),
204
Derivatives and Integrals
^
h
(a)
FIGURE
and
(b)
8
f(c) 7*f(b), so
either
H
FIGURE
h
f(c)
< f(a)
or
f(a ) <f(c)
or
f{b) <f(c),
<f(b)
and
it is only necessary to rule out the first and third possibilities.
Suppose that f(c) < f(a) were true (Figure 9). Apply the Intermediate
Value Theorem to the interval [c, b]\ since f(c) < f(a) < f(b) there is some x
in [c, b] such that f(x) = f(a). But this contradicts the fact that / is one-one,
9
since a
^
Thus
x.
<
f{a)
is
Similarly, suppose that f{c)
>
f(c)
impossible.
f(b) were true (Figure 10). Apply the Intermediate Value Theorem to the interval [a, c] since f(a) < f(b) < f(c) there is
some x in [a, c] such that f(x) = f(b), again contradicting the fact that / is
:
one-one.
FIGURE
Thus
f(c)
> f(b)
is
also false, leaving only the possibilitv f(a)
<
f(b) <f{c).
10
The case where f(a) > f(b) may be handled in precisely the same way.
one can be elegant and apply the first case to — /. |
LEMMA
2
Or
Let / be a continuous one-one function defined on an interval, let a and b be
two points of the interval, and suppose that f(a) < f(b). If c is a point of the
interval with c < a, then f(c) < f(a); and similarly, if b < c, then f(b) < f(c).
(Analogous statements could be made if f(a) > f(b), but we will not need
them.)
PROOF
a
fif)
<
<f{a)<f{b).
fQ>).
case
Finally,
2
<
c
f(c)
The
theorem
c < a. Suppose first that f(c) > f(b). Then Lemma 1,
< b, would imply that f{a) > f(b), which is false. Hence
But now Lemma 1 may be applied to c < a < b to conclude that
Consider the case
applied to
If
/
is
b
we
<
c is
handled
similarly. |
are ready for the theorem.
continuous and one-one on an interval, then /
is
either increasing or
decreasing on that interval.
PROOF
Let a and
b
be two numbers in the interval with a
<
b.
Since f(a)
^ /(£),
205
Inverse Functions
<
either f(a)
>
f(b) or f(a)
fib).
We
and show that / is increasing.
with
c and d be two points in the interval
fib),
Let
follows immediately from
b, then it
other possibility leads to six cases:
equals a or
The
f{d).
(1)
c
< d <
a
<
b
(2)
c
<
a
<
d
<
b
(3)
c
<
a
<
b
< d
(4)
a
<c <d <
(5)
a
<
c
<
b
< d
(6) a
<
b
<
c
< d
—
—
—
1
1
1
c
a
1
d
b
1
1
—
1
a
first
then apply
b
I—
a
c
b
1
1
1
1
1
d
1
bed
!
1
1-
1
cases can be treated easily, using the
Lemma
apply
Lemma
1
to
c
one (or both) of c, d
1 and 2 that /(c) <
——
d
c
a
case (1)
d. If
1
\
1
b
Each of these
<
<
——
a
c
c
Lemmas
f(a)
1
1
\
on the case
will first concentrate
2 to
<
d
c
<
<
a
a to
<
lemmas. For example, in
/(c) < f(a), and
< }{d). The other
conclude that
b to
conclude that
f(c)
an easy exercise.
prove similarly that /is decreasing. Or one can apply
can
one
If /(a) > f{b),
method). |
the result just proved to -/ (clearly the preferable
cases are
left
to you, as
with continuous
Henceforth we shall be concerned almost exclusively
If/ is such
interval.
an
on
defined
are
which
increasing or decreasing functions
1
will be
of/
domain
the
what
precisely
quite
is possible to say
a function,
it
like.
closed interval
Suppose first that /is a continuous increasing function on the
Value Theorem, / takes on every value
[a, b\ Then, by the Intermediate
closed interval
between f{a) and fib). Therefore, the domain of/" is the
b\ then the
[a,
on
Similarly, if / is continuous and decreasing
1
[f(a),f(b)l
1
of f'
domain
is [f(b)J(a)\
b) the analysis
a continuous increasing function on an open interval (a,
point c in
some
choose
us
let
with,
becomes a bit more difficult. To begin
possiOne
on
by
taken
are
values
/
>f(c)
(a, b). We will first decide which
1 1). In this case / takes
(Figure
values
large
arbitrarily
bility is that / takes on
the other hand,
on all values >f(c) by the Intermediate Value Theorem. If, on
c < x < b} is
=
A
then
values,
{/(*):
large
arbitrarily
on
/ does not take
suppose y is
bounded above, so A has a least upper bound a (Figure 12). Now
(because
any number with /(c) < y < a. Then / takes on some value /(*) > y
Theorem,
Value
Intermediate
/
a is the least upper bound of A). By the
itself;
a
value
the
on
take
cannot
that
/
on the value y. Notice
If /
is
,
actually takes
for
if
a =
which
is
/(*) for a
impossible.
<
x
<
b
and we choose
t
with x
<
t
<
b,
then
/(/)
>
a,
206
Derivatives
and
Integrals
same arguments work
Precisely the
on
all
values
than f(c) or there
less
values between
for values less
number
a
is
<
($
than
f(c)
either
:
/ takes
such that / takes on
f(c)
all
and f(c), but not fi itself.
This entire argument can be repeated if / is decreasing, and if the domain
off is R or (a, <x>) or (— oo, a). Summarizing: if / is a continuous increasing,
or decreasing, function whose domain is an interval having one of the forms
/3
(-oc,
{a, b),
or R,
oo),
b), {a,
then the domain of f' is also an interval which has one of these four forms.
Now that we have completed this preliminary analysis of continuous one1
one functions, it is possible to begin asking which important properties of a
one-one function are inherited by its inverse. For continuity there is no
problem.
THEOREM
3
PROOF
If/
is
continuous and one-one on an interval, then/ -1
We know by Theorem 2
well assume that
/
that /
by applying the usual
We must show that
also continuous.
either increasing or decreasing.
is
we can then
increasing, since
is
is
We might as
take care of the other case
— /.
trick of considering
lim /-*(*) =/-'(*)
for
each
in the
b in
the
domain of/" Such a number b is of the form /(a) for some a
For any s > 0, we want to find a 5 > 0 such that, for all x,
1
.
domain
of/.
if
-
f(a)
8
<
Figure 13 suggests the
you
see the
x
<
way
8,
-
then a
FIGURE
+
a
+
e.
—
<
e
<
a
+
a
Z,
-
+
£);
account of the information contained in
this picture.
choice of
Consequently,
a
<
and f(a) - f(a - e). Figure 13 conand what follows is simply a verbal
8
fifl
a
(x)
we let 8 be the smaller of f(a
e)
f(a)
tains the entire proof that this 8 works,
Our
£
l
follows that
+
a
< f~
of finding 8 (remember that by looking sideways
/(*-e) </(*) </(*
—
£
graph of f~ l ): since
a
it
+
f{a)
ensures that
-
e)
<
f(a)
-
8
and
+
f(a)
8
<
f(a
+
if
/(a)
s
-
5
<
*
<
e)
<
at
</(
+
f{a)
5,
then
13
f(a
Since /
is
/~
increasing,
x
is
r U(a -
-
also increasing,
i
£))
</-*(*)
fl
+
e).
and we obtain
</"
1
(/(«
+
e)),
e).
Inverse Functions
207
i.e.,
a
which
is
precisely
Having
-
e
<
<
what we want.
+
a
e,
|
successfully investigated continuity of
f~\
tackle differentiability. Again, a picture indicates just
/
true. Figure 14
/
through
shows the graph of a one-one function / with a tangent line
If this entire picture is reflected through the diagonal,
(a, f(a)).
-1
shows the graph of/
V
,
is
only reasonable to
result ought to be
it is
what
and the tangent
line
U through (f(a),
the reciprocal of the slope of L. In other words,
it
a).
The
L
it
slope of
appears that
/(«))
f(a)
This formula can
directly, for each
equally well be written in a
b in
the
domain
of
f
way which
expresses (/ *)'(*)
1
:
i
(r )'w =
i
Unlike the argument for continuity, this pictorial "proof becomes somewhat involved when formulated analytically. There is another approach
which might be
it is
tried.
Since
we know
that
tempting to prove the desired formula by applying the Chain Rule:
(rrw
-
W)
_1
is differentiable, since the Chain
is not a proof that /
Rule cannot be applied unless f~ is already known to be differentiable. But
l
differentiable,
this argument does show what (/""*) 'M wil1 have to be if f~ is
information.
preliminary
and it can also be used to obtain some important
Unfortunately, this
l
1
THEOREM
4
/ is a continuous one-one function defined on an interval and/'Cf" ^))
1
is not differentiable at a.
then
If
f
PROOF
We
have
If/""
1
aw*
differentiable at a, the
hence
which
is
absurd. |
Chain Rule would imply that
=
0,
208
Derivatives and Integrals
A
Theorem 4
simple example to which
=
Since /'(0)
0
and
=
0
applies
f~\0), the function
f~
l
=
#3
is
the function f(x)
is
not differentiable at 0
.
(Figure 15).
Having decided where an
now ready
inverse function cannot be differentiable,
for the rigorous proof that in all other cases the derivative
by the formula which we have already "derived"
Notice that the following argument uses
in
continuity of
we
is
are
given
two different ways.
which we have
already proved.
THEOREM
5
Let / be a continuous one-one function defined on an interval, and suppose
l
is differentiable at f"\b), with derivative
f(f~~ (b)) ^ 0. Then f~ is
l
that /
differentiable at
b,
and
i
f(r\b))
PROOF
Let
b
=
f(a).
Then
lim
r\b +
=
]
f{x)
=
h)
- r\b)
f~\b
lim
+ h) -
a
x*
Now
every
number
b
+
h in the
b
for a
domain of /
can be written in the form
+ h - f(a + k)
unique k (we should really write
plicity).
1
k(h),
but
we
will stick
with k for sim-
Then
lim
}-\b
+h)-a
h->Q
=
lim
a-*o
=
We
f(a
lim
;.-.o
+ k)) - a
+ k) — b
f'Kfia
f(a
are clearly on the right track
!
+
It is
k)
- f{a)
not hard to get an explicit expression
for k; since
b
(b)
FIGURE
+ h = f(a + k)
we have
15
f~\b
+ h) =a + k
or
k=n(b+h)
Now
by Theorem
3,
_1
the function /
is
-/-'(*).
continuous at
b.
This means that k
Inverse Functions
approaches 0 as h approaches
f(a
lim
209
Since
0.
+ k)-f(a) = fl{a) = f{nm ^
0>
K
k->0
this implies that
The work we have done on
here
inverse functions will be
an immediate dividend. For
is
=
/„(#)
odd,
n
x
n
amply repaid
later,
but
let
for all x;
for n even, let
=x n
fn (x)
Then fn
is
x>
,
0.
a continuous one-one function, whose inverse function
= Vx =
gn(x)
By Theorem
5
we
have, for x
^
is
0,
1
/»'(/« "'to)
1
n{fn~~
1
1
- I
1
Thus,
if /(a:)
then/'(*)
rational
ber;
=
=
xa
,
is
an integer or the reciprocal of a natural number,
now easy to check that this formula is true if a is any
a
=
and
It
number: Let
a
is
m/n, where
m
is
n
=
an
integer,
and
if
f( x )
then,
= xm
'
1
(j '*)",
by the Chain Rule,
f(x)
= mix
_ ^
.
1
'")™- 1
x
•
i
•
j™**- 1
[(m/n)-(l/n)l + ((l/n)-ll
= ™ _y(m/n)-l_
n
is
a natural
num-
Although we now have a formula
when
for /'(*)
~
f(x)
—
x a and a
is
rational,
a
the treatment of the function f(x)
x for irrational a will have to be
saved for later at the moment we do not even know the meaning of a symbol
2
like x^
Actually, inverse functions will be involved crucially in the definition
—
.
of x a for irrational
Indeed, in the next few chapters several important func-
a.
tions will be defined in terms of their inverse functions.
PROBLEMS
1
.
Find /
1
for
(i)
f{x)
(ii)
f(X )
each of the following
- x*+
= (X -
1.
1)1
x rational
x,
(iii)
f(x)
(iv)
—x,
f{x)
x irrational.
-
\
x\
Ix,
=
+
x
fix)
(vii)
/(0.<zia 2 tf3
<
0.
7* ai 9
x
=
=
an
.
.
a i9
i
.
=
an
,
1,
.
.
.
,
n
-
1
.
[x].
•
•
x
x
x
ai,
(vi)
/.
•)
=
0.a 2 aia 3
....
(Decimal representation
is
be-
ing used.)
(viii)
fix)
—
= -—
1
2.
(ii)
(iii)
(iv)
4.
,
-1 <
/
/
/
/
is
is
is
is
<
x
1.
X
Describe the graph of/"" 1
(i)
3.
—
and
and
decreasing and
decreasing and
when
increasing
always positive.
increasing
always negative.
always positive.
always negative.
Complete the proof of Theorem
Prove that
if
/
2.
increasing, then so
is
is
f~\ and
similarly for decreasing
functions.
/ and g are increasing,
5.
If
6.
(a)
Prove that
(/
(b)
7.
^
0
Find £ -1
Show
if
/and g
find
/
Or / g
•
?
Or/og?
then/o^ is also one-one. Find
The answer is not f~ o g _1
are one-one,
.
=
—
ex
_1
+g?
terms of f~~ l and g -1 Hint:
in terms of
that fix)
/
is
in this case.
-f-
d
f~
is
l
if
g{x)
=
one-one
1
if
l
.
+/(*)•
and only
if
ad
—
be
^
0,
and
:
211
Inverse Functions
8.
On
(i)
which
f{x)
(ii)
f(x)
(iii)
f(x)
(iv) /(*)
9.
intervals
will the following functions
[a, b]
be one-one?
= x*- 3x
= x + x.
= (1 + **)"'.
2
.
h
^±L
=
+
X2
1
Suppose that / is a one-one function and that Z" 1 has a derivative which
nowhere 0. Prove that / is differentiate. Hint: There is a one-step
is
proof.
*10.
(a)
/ such that [f(x)] b +
f(x) + x = 0 for all x. Hint: Show that / can be expressed as an
And the
inverse function. The easiest way to do this is to find
easiest way to do this is to set x = f~ iy).
Find /' in terms of /, using an appropriate theorem of this chapter.
Find /' in another way, by simply differentiating the equation
Prove that there
a differentiable function
is
l
(b)
(c)
defining
The
/.
function in Problem 10
by the
often said to be defined implicitly
is
v
y +
however. As the next problem shows, an equation does not usually define a
function implicitly on the whole line, and in some regions more than one func5
equation
tion
may
11.
(a)
+
=
x
0.
The
situation for this equation
be defined implicitly.
What
are the two differentiable functions / which are defined
on
implicitly
satisfy x
2
+
(
— 1,
[f(x)]
2
1)
=
1
by the equation x 2 + y 2 = 1, i.e., which
for all x in (-1, 1)? Notice that there are
no solutions defined outside [— 1,
(b)
*(c)
quite special,
is
It will
1].
+
2
functions / satisfy x 2
[fix)] = -1?
differentiable functions / satisfy [fix)]'3
Which
Which
help to
first
draw
—
3/(#)
the graph of the function g(x)
— x? Hint:
= x — 3x.
z
In general, determining on what intervals a differentiable function
defined implicitly by a particular equation
may
be a delicate
affair,
and
is
is
we
assume that / is such
a differentiable solution, however, then a formula for f(x) can be derived,
best discussed in the context of
advanced
calculus. If
exactly as in Problem 10(c), by differentiating both sides of the equation
defining / (a process
12.
(a)
Apply
this
known
method
2
y
to the
equation
[f(x)]
2
+
x2
=
1.
Notice that
only to be expected, since there
more than one function defined implicitly by the equation
your answer
is
as "implicit differentiation")
+
x2
-
1.
will involve fix)
;
this is
212
Derivatives and Integrals
(b)
(c)
13.
But check that your answer works
in Problem 11(a).
Apply this same method to [/(*)] 3
Leibnizian notation
ation. Because y
is
both of the functions / found
for
—
3/(x)
=
x.
particularly convenient for implicit differenti-
is
so consistently used as
an abbreviation
for /(*), the
equation in x and y which defines / implicitly will automatically stand
for the equation which / is supposed to satisfy. How would the following
computation be written
our notation?
in
+ y* + xy =
+ ax +, +
y*
ax
dy
14.
As long
4
=
0,
dx
-y
_
dx
1,
4y
3
+
3y
+x
2
as Leibnizian notation has entered the picture, the Leibnizian
notation for derivatives of inverse functions should be mentioned. If
_1
dy/dx denotes the derivative of /, then the derivative of /
is denoted by
dx/dy. Write out Theorem 5 in this notation. The resulting equation will
show you another reason why Leibnizian notation has such a large
following. It will also explain at which point {f Y is to be calculated
l
when
What
using the dx/dy notation.
the significance of the following
is
computation?
y
=
=
dx 1 !"
_
x
y
x 1/n
,
dy
dx
dx
dx
ny
n— l
dy
15.
Suppose that /
is
a differentiate one-one function with a nowhere zero
xf~\x) - F(f~ l (x)). Prove that
G'{x) —
(Disregarding details, this problem tells us a very interesting fact: if we know a function whose derivative is /, then we also
know one whose derivative is f~ l But how could anyone ever guess the
function G? Two different ways are outlined in Problems 14-10 and
and
derivative
that
/ =
F'.
=
Let G{x)
.
18-13.)
16.
Suppose h
Find
is
a function such that
17.
sin 2 (sin(#
+
1))
and
h(0)
=
3.
l
(i)
(h~~ )'(?>).
(ii)
(/?- 1 ) (3),
/
*18.
=
h! (x)
where
Find a formula
Prove that
if
/
for
(/
=
h{x
+
1).
(/^"M.
_1
(fc)
(3(x)
(*)) exists,
and
is
nonzero, then
U~
ik
l
)
\x )
exists.
213
Inverse Functions
19.
(a)
Prove that an increasing and a decreasing function intersect at most
once.
(b)
Find two continuous increasing functions / and g such that f(x)
g(x) precisely
(c)
*20.
(a)
x
an
geometrically?)
1
Prove that
f(x)
=
if
x for
/
is
all x.
One
possibility
is
f(x)
= — x.
an increasing function such that / = /~"\ then
Hint: Although the geometric interpretation will
be immediately convincing, the simplest proof (about 2 lines)
rule out the possibilities f(x) < x and f(x) > x.
*21.
Which
functions have the property that the graph
function
22.
A
when
function /
more
A
=
integer.
Give several examples of continuous / such that / = f~ andf(x) = x
for exactly one x. Hint: Try decreasing /, and remember the geometric interpretation.
(c)
is
Find a continuous increasing function / and a continuous decreasing
function g, defined on R, which do not intersect at all.
If / is a continuous function on R and / = f~\ prove that there is at
l
least one x such that f(x) = x. (What does the condition / =
f~
mean
(b)
when
precise
through the graph of
reflected
is
nondecreasing
we should
is
to
the graph of a
— / (the "antidiagonal")
?
<
f(y) whenever x < y. (To be
that the domain of / be an interval.)
if
stipulate
nonincreasing function
is still
is
f(x)
defined similarly. Caution:
Some
writers
use "increasing" instead of "nondecreasing," and "strictly increasing"
for
(a)
our "increasing."
Prove that
on some
is
(b)
*23.
(a)
/ is nondecreasing, but not increasing, then / is constant
(Beware of unintentional puns: "not increasing"
not the same as "nonincreasing.")
Prove that
for all
(c)
if
interval.
if
/
is
Prove that
if /'(*)
Suppose that f(x)
there
and nondecreasing, then f{x) > 0
differentiate
.v.
is
>
>
0 for
0 for
all x,
all x,
then /
is
nondecreasing.
and that /is decreasing. Prove that
a continuous decreasing function g such that 0
<
g(x)
<
f{x)
for all x.
(b)
Show
that
we can even arrange
that g will satisfy lim g(x)/f(x)
=
0.
CHAPTER
^/
I
INTEGRALS
The derivative
does not display
its full
strength until allied with the "integral,"
topic may seem to be a comdo not appear even once! The
study of integrals does require a long preparation, but once this preliminary
work has been completed, integrals will be an invaluable tool for creating new
functions, and the derivative will reappear in Chapter 14, more powerful than
the second
main concept
plete digression
—
of Part III.
At
first this
in this chapter derivatives
ever.
Although ultimately
i
to
be denned in a quite complicated way, the integral
that of area. By now it should come
formalizes a simple, intuitive concept
as
no
—
surprise to learn that the definition of
great difficulties
—"area"
certainly
is
an
intuitive concept
can present
no exception.
In elementary geometry, formulas are derived for the areas of many plane
figures, but a little reflection shows that an acceptable definition of area is
seldom given. The area of a region is sometimes defined as the number of
1, which fit in the region. But this definition is
hopelessly inadequate for any but the simplest regions. For example, a circle
of radius 1 supposedly has as area the irrational number 7r, but it is not at all
squares, with sides of length
^
(a,
_
0)
0)
what u w squares" means. Even if we consider a circle of radius 1 /vV,
which supposedly has area 1 it is hard to say in what way a unit square fits in
this circle, since it does not seem possible to divide the unit square into pieces
which can be arranged to form a circle.
In this chapter we will only try to define the area of some very special
those which are bounded by the horizontal axis, the vertiregions (Figure 1)
cal lines through (a, 0) and (6, 0), and the graph of a function /such that
f(x) > 0 for all x in [a, b]. It is convenient to indicate this region by R(f, a, b).
Notice that these regions include rectangles and triangles, as well as many
clear
FIGURE
1
,
—
other important geometric figures.
area of R(f, a, b) will be
b\ Actually, the integral will be defined even for
functions / which do not satisfy the condition f(x) > 0 for all x in [a, b). If / is
the function graphed in Figure 2, the integral will represent the difference of
The number which we
called the integral off
on
will eventually assign as the
[a,
the area of the lightly shaded region
FIGURE
2
interval
[a, b]
to,
a
214
is
h,
=
t2>
to
t 3,
<
t\
t±
with
< h <
ts
<
ti
of the heavily shaded
indicated in Figure
has been divided into four subintervals
by means of numbers
i
and the area
region (the "algebraic area" of R(f, a, b)).
The idea behind the prospective definition
=
b
3.
The
Integrals
(the
numbering of the
will
equal the
maximum
value Mi;
m and
value of / be
s
subscripts begins with 0 so that the largest subscript
number
On the first interval
£
= m
x
(h
-
of subintervals).
h] the function /has the minimum value mi and the
ti] let the minimum
similarly, on the ith interval
[h,
let
f0
maximum
the
)'+
rn 2 (t2
-
The sum
value be Mi.
+m
h)
-
(h
z
+m
t%)
4
(*4
-
represents the total area of rectangles lying inside the region #(/,
the
FIGURE
215
h)
a, b),
while
sum
3
S = Mi(h -
to)
M
+
2 (t 2
~
M (h - h)
+
h)
+
z
M,(U -
h)
represents the total area of rectangles containing the region R(f, a, b). The
guiding principle of our attempt to define the area A of R(f, a, b) is the obser-
vation that
A
should
satisfy
< A
s
A <
and
S,
and that this should be true, no matter how the interval [a, b] is subdivided. It is to
be hoped that these requirements will determine A. The following definitions
begin to formalize, and eliminate some of the implicit assumptions in, this
discussion.
DEFINITION
Let a
[a, b\,
The
<
b.
A partition of the interval
one of which
points in a partition can be
a
we
DEFINITION
shall
<
to
ti
<
•
a finite collection of points in
is b.
numbered
t Q,
.
<
<
tn
•
•
fw
-l
.
.
=
5
tn
so that
b\
always assume that such a numbering has been assigned.
Suppose /
[a, b].
=
[a, b] is
and one of which
is a,
is
bounded on
[a, b]
and
P =
{* 0 ,
•
•
•
,
*n} is
a partition of
Let
m =
Mi =
{
The lower sum
inf {/(*):
sup {/(*):
fe-i
<
<
x
x
of / for P, denoted by L(f, P),
<
<
ti},
h}.
defined as
is
n
Uj, P) =
The upper sum
J
rmiu
-
ti-i).
of / for P, denoted by U(f, P),
is
defined as
n
U(f, P)
=
J
Miih
The lower and upper sums correspond
-
to the
sums
s
and S in the previous
216
Derivatives and Integrals
example; they are supposed to represent the total areas of rectangles lying
below and above the graph of /. Notice, however, that despite the geometric
motivation, these sums have been defined precisely without any appeal to a
concept of "area."
Two
comment. The requirement that / be
rrii and Mi be defined. Note,
also, that it was necessary to define the numbers m and Mi as inf s and sup's,
rather than as minima and maxima, since / was not assumed continuous.
One thing is clear about lower and upper sums: If P is any partition, then
details of the definition deserve
bounded on
[a, b] is
essential in order that all the
{
<
L(f, P)
£/(/, P),
because
n
Uf,
=
P)
J=
i
m&t l
n
U(f,
P)= T
=
i
and
for
each
i
we have
mi(ti
On the other hand,
FIG URE
any
4
two
-
Mi<M
l
partitions of
-
< Mi(k -
something
[a, b],
then
less
it
jf
t
_i).
obvious ought to be true:
If
Pi and Ps are
should be the case that
<
L(f, Pi)
P
£/(/,
2 ),
because L(f, Pi) should be < area P(/, a, b), and U(f, P 2 ) should be > area
b). This remark proves nothing (since the "area of P(/, a, b)" has not
even been defined yet), but it does indicate that if there is to be any hope of
P(/>
defining the area of P(/,
first.
<z,
The proof which we
b), a proof that L(f, Pi) < U(f, P2 ) should come
are about to give depends upon a lemma which
concerns the behavior of lower and upper sums when more points are included
in a partition. In Figure 4 the partition P contains the points in black, and Q
contains both the points in black and the points in grey.
FIGURE
5
that the rectangles
region R(f,
lemma
If
Q
a, b)
contains
P
drawn
for the partition
than those
(i.e., if all
for the original partition P.
points of
L(/, P)
P)
£/(/,
PROOF
Consider
first
The
P
picture indicates
are a better approximation to the
Q
To
be precise:
are also in Q), then
< L(/,
> U(f
9
Q),
Q).
the special case (Figure 5) in which
Q
contains just one
point than P:
P =
Q =
{^o,
•
•
.
{*o>
•
•
•
,
tn
},
•
,
•
•
,
t n },
where
a
=
t0
< h <
-
•
•
<
<
m
<
ffc
<
•
•
•
<
/„
=
b.
more
217
Integrals
Let
m! =
m" =
inf {/(#):
tk
-i
<
#
<
*
<
inf {/(*): w
<
u}
9
fe}.
Then
^
n
= £
L(/, P)
?
To
=
-
fe-i),
i
Jb-1
n
t=l
i=k+l
prove that
<
P)
Q)
m k {tk -
Now
x
<
<
the set {/(#):
u
and
,
<
a:
than or equal
-
m'iu
tk
tk
+
-i)
contains
]
some smaller
possibly
first set is less
<
m"(th
u).
numbers
the
all
—
that
in {/(*):
ones, so the greatest lower
the greatest lower
to
show
therefore suffices to
it
mk <
bound
-
tk
bound
-i
<
of the
of the second; thus
m!
Similarly,
m k < m"
Therefore,
= m k (u —
m k {tk —
This proves, in
>
U(f, P)
-i)
+m
k {tk
—
<
u)
m'(u
special case, that L(f, P)
this
U(f, Q)
tk
is
similar,
and
<
you
left to
is
- k_
t
x)
L(f, Q).
as
+ mn
(t k
—
u).
The proof
that
an easy, but valuable,
exercise.
The
now be deduced quite easily. The partition Q can be
by adding one point at a time; in other words, there is a
general case can
P
obtained from
sequence of partitions
P = Ph P
such that
Pj+
contains just one
i
=
L(f, P)
L(f, Pi)
2,
.
.
.
Pa = Q
,
more point than Pjt Then
< US,
Pi)
<
U(f, ft)
>
•
•
•
< US, Pa) =
L(f, Q),
and
U(f, P)
=
U(f,
PO >
The theorem we wish
THEOREM
i
Let
P and P
x
[a, b].
2
to
prove
be partitions of [a,
b],
is
There
•
•
>
U(f, P.)
=
£/(/, Q). I
a simple consequence of this lemma.
and
be a function which
let /
is
bounded on
Then
L(f, Pi)
PROOF
•
is
a partition
in both Pi
and
P
P which
2 ).
US,
9
2 ).
and
lemma,
contains both Pi
According
Pi)
< U(f P
< US,
to the
P)
<
u(f, P)
<
P
2
(let
u(f,
P consist of all points
p ).
2
|
n
from Theorem 1 that any upper sum £/(/, P') is an upper bound
lower sums L(f, P). Consequently, any upper sum £/(/, P ) is
greater than or equal to the least upper bound of all lower sums:
It follows
f
for the set of all
?):?a
sup {£(/,
partition of
for every
P
set of all
upper sums of /. Consequently,
f
.
means
This, in turn,
that sup {L(f, P)}
sup {£(/,/>)}
It is clear that
sum
of
/
<
[a, b]}
<
U(f, P'),
is
a lower bound for the
inf {£/(/, P)}.
both of these numbers are between the lower sum and upper
for all partitions:
£(/,/*)< sup \L(f, P)} < £/(/, />'),
P') <inf ([/(/,?)) < U(f,P'),
L(f
y
for all partitions P'.
It
may
well
happen that
sup{L(/,P)}
in this case, this
/ for
is
all partitions,
area of R(f,
a, b).
the <m/y
=
inf {£/(/,
P};
number between the lower sum and upper sum of
number is consequently an ideal candidate for the
and
this
On
the other hand,
sup [L{f,P)\
if
<
inf \U(f,P)}>
then every number x between sup {£(/, P)
p')<
L{f,
<
x
and
}
inf {U(f, P)} will satisfy
u(f,
for all partitions P'.
It is
The
not at
all
clear just
when such an embarrassment
of riches will occur.
following two examples, although not as interesting as
many which
will
soon appear, show that both phenomena are possible.
Suppose first that f(x) = c for
any partition of [a, b\ then
is
all
x in
rrii
— Mi =
[a, b]
(Figure 6). If P
=
c,
so
n
L(f, P)
=
<*<
~
c(h
-
^=
= <b ~
a ).
-
a).
l
i
n
(/(/,
P)
=
J=
t
In
this case, all
=
c{b
lower sums and upper sums are equal, and
sup
Now
<,_,)
1
P)}
-
inf {U(f, P)}
=
c(b
consider (Figure 7) the function / defined by
v
*
'
_
~
I
0,
x irrational
I
1,
x rational.
-
a).
{^o,
.
•
•
,
tn \
Integrals
If
P =
{fo,
•
•
•
>
-
0,
Af,=
1,
rrii
any
tn] is
219
partition, then
since there
an
is
irrational
number
in
[ff-_i,
/»•],
and
since there
number
a rational
is
in
Therefore,
n
a
=
t0
h
t%
tn-l
tn
L(/, P)
=b
=
('«•
1
=
-
**-0
=
°»
-
fe-i)
=
b
1
n
FIGURE
£/(/,
7
P)
=
i
Thus, in
The
this case
(U
'
1
-
a.
i
certainly no/ true that sup {£(/, P)
it is
upon which
principle
J=
insufficient information to
the definition of area
was
=
}
inf
{
U(f, P)
}.
to be based provides
determine a specific area for R(f,
6)—any num-
a,
—
a seems equally good. On the other hand, the region
ber between 0 and b
refuse to assign it any area at
R(J, a, b) is so weird that we might with justice
In
all.
fact,
we can maintain, more
*inf
sup {L(f,P)}
the region R(f,
whenever
generally, that
P)l
{£/(/,
too unreasonable to deserve having an area. As our
"unreasonable" suggests, we are about to cloak our
a, b) is
appeal to the word
ignorance in terminology.
DEFINITION
A function / which
sup
{
L(f, P)
:
Pa
bounded on
partition of
[a, b]
integrable on
[a, b] is
=
}
common number
In this case, this
is
is
is
inf
{
U(f, P)
:
[a, b] if
P a partition of [a,
called the integral of
/on
[a, b]
b]
.
and
denoted by
/.'/•
(The symbol J is called an
"sum;" the numbers a and
The
in
If
integral
j* f
is
integral sign
an elongated s, for
and upper limits of integration.)
and was
originally
b are called the lower
also called the
area of R(J
9
a, b)
when /(*) >
0 for
all
x
[a, b].
/
is
integrable, then according to this definition,
L(f, P)
<
b
J
f^ U(f>
b
Moreover,
j f
is
the unique
^)
for a11 Partitions
number with
P
of
[a,
b].
this property.
solve, the problem discussed
(nor do we know how
integrable
are
functions
which
before: we do not know
present we know only
to find the integral of/ on [a, b] when / is integrable). At
This definition merely pinpoints,
and does not
220
Derivatives and Integrals
two examples:
(1)
if
f(x)
=
then /is integrable on
c,
[a, b]
and
J* f
—
c
•
(b
—
a).
(Notice that this integral assigns the expected area to a rectangle.)
(2)
if
/(*)
=
(
1,
I
Several
Even
ther.
* irratlonal
x rational,
^'
more examples
will
for these examples,
then y
not integrable on
[s
[a, b].
be given before discussing these problems furhowever, it helps to have the following simple
criterion for integrability stated explicitly.
THEOREM
2
If /
S
>
is
bounded on
0 there
is
j>, b],
then / is integrable on
P of [a, b] such that
U(f, P)
PROOF
Suppose
that for every s
first
[a, b] if
and only
if
for every
a partition
>
-
0 there
-
U(f, P)
<
L(f, P)
is
s.
a partition
<
L(f, P)
P
with
e.
Since
vai\U<J,F)\ < U(f,P),
supjL(/,P')} >L{f,P),
it
follows that
inf {£/(/, P')}
Since this
is
true for
all £
>
-
sup {£(/,/>')}
0, it follows
<£
that
sup {L(f,P')} =inf{t/(/,P')};
by
definition, then,
lar. If
/
is
/
is
integrable.
The proof of the converse
=
sup \L{f,P)}
This means that for each e
>
P
-wS{U{f,T)).
0 there are partitions P',
U(f, P")
Let
assertion is simi-
integrable, then
-
L(f, P')
<
P"
with
e.
be a partition which contains both P' and P". Then, according to the
lemma,
U(J, P)
L(f, P)
<
>
U(f, P"),
L(f, P');
consequently,
U(f, P)
-
L(f, P)
<
U(J, P")
-
L(f, P')
<
s.
|
Although the mechanics of the proof take up a little space, it should be clear
Theorem 2 amounts to nothing more than a restatement of the definition
that
Integrals
of integrability. Nevertheless,
221
a very convenient restatement because there
it is
no mention of sup's and inf s, which are often difficult to work with. The
next example illustrates this point, and also serves as a good introduction to
the type of reasoning which the complicated definition of the integral necessi-
is
tates,
even in very simple situations.
Let / be defined on
by
[0, 2]
:-\.
Suppose
P=
{*o,
•
.
.
,
tn
a partition of
is
]
tj-!
(see
Figure
8).
<
<
1
t
[0, 2]
with
}
Then
mi
= Mi =
mj
—
0
if
^ /,
i
but
and
0
Mj =
1.
Since
t=J+l
we have
-
U(f, P)
This certainly shows that /
L(f P)
9
tj
-
h-x.
integrable: to obtain a partition
is
-
U(f, P)
it is
=
L(f, P)
<
P
with
e,
only necessary to choose a partition with
<
t,-i
Moreover,
it is
/
is
and
tj
tj
—
tj-i
<
s.
clear that
<
L(J, P)
Since
<
1
<
0
integrable, there
is
U(f, P)
for all partitions P.
only one number between
all
lower and upper
sums, namely, the integral of /, so
f/ =
o.
Jo
Although the discontinuity of / was responsible for the difficulties in this
example, even worse problems arise for very simple continuous functions. For
example, let /(#) = x, and for simplicity consider an interval [0, b], where
is a partition of [0, b\ then (Figure 9)
tn
b > 0. If P = {t^
.
.
.
)
,
mi
—
and
Af»-
=
U
and
therefore
n
L(f, P)
=
2=
=
ti-iCh
-
l
t
to(h
-
t 0)
+
h(t 2
-
+
h)
•
•
•
+
*n _i(*„
-
f„_i),
n
U(f, P)
=
^=
t
U(U
-
Neither of these formulas
siderably for partitions
case, the length
U
ti-x)
1
—
is
Pn =
particularly appealing, but both simplify con{^ 0 ,
•
.
t
,
n
into
)
of each subinterval
=
to
is
equal subintervals. In this
b/n so
f
o,
h = —
ft
>
n
<2
2b
—
=
>
etc;
n
in general,
ti
ib
=
n
Then
n
Uf, Pn) =
^
-
/,-l(<<
/<-l)
-m
n-l
J
Remembering
—o
the formula
1+
...
+A
_*C*±2>,
2
this
can be written
=
"
n
b
1
.
'
_l.
2
Integrals
223
Similarly,
n
U(f, Pn)
-
=
^
=
V+
n(n
=
-
•
1)
2
+
*
=
^
*
1
.
*
n
2
rz
h
l.
2
2
very large, both L(f, Pn ) and U(f, Pn ) are close to b /2, and this remark
that
first
Notice
makes it easy to show that / is integrable.
If *
is
Pn ) -
U(f,
Pn
This shows that there are partitions
desired.
By Theorem
/
2 the function
be found with only a
little
L(f,
/>„)=-•
L(f,
with
is
£/(/,
Pn ) -
Pn < | <
)
U(f, Pn)
as small as
may now
Jq f
that
It is clear, first of all,
work.
Pn )
L(f,
integrable. Moreover,
for all n.
/2 lies between certain special upper and
lower sums, but we have just seen that U(f, P n ) - Uf, Pn) can be made as
small as desired, so there is only one number with this property. Since the
This inequality shows only that
b
2
we can conclude
integral certainly has this property,
that
jT'-r
equation assigns area b 2 /2 to a right triangle with base and
(Figure 10). Using more involved calculations, or appealing to
Notice that
altitude b
Theorem
this
4, it
can be shown that
2
2
The
function f(x)
(Figure 11),
if
P=
mi
=
=
{t 0 ,
x
2
.
/(<*_!)
presents even greater difficulties.
.
.
=
,
tn }
is
fe_i)
2
a partition of
and
Choosing, once again, a partition P»
so that
=
Mi =
{fo,
.
-
[0,
f(ti)
.
,
In this case
b\ then
=
k
2
.
**} into «
equal parts,
the lower and upper sums
become
n
Uf, P.)
£<•-<>'-;•n2
Z-/
i
=
n
l
n-l
n
^
U(f, Pn)
k
L4
2
*
-
fa-i)
+
1)(2*
{U
n2
n
Recalling the formula
1"
from Problem
+••*+* —
2-1, these
sums can be written
+
1)
as
=V7(»-D(»)(2«-1),
rr 6
f/(/,P»)
=-'^»(«+l)(2«
w
It is
s
+
l).
6
not too hard to show that
L(f,
and that U(f, Pn )
sufficiently large.
-
Pn < y <
U(f,
)
Pn ),
can be made as small as desired, by choosing n
The same sort of reasoning as before then shows that
L(f,
Pn )
J
Jo
3
—
This calculation already represents a nontrivial result the area of the region
hounded by a parabola is not usually derived in elementary geometry. Neverthe
theless, the result was known to Archimedes, who derived it in essentially
we
chapter
same way. The only superiority we can claim is that in the next
will discover a
much
simpler
way
to arrive at this result.
225
Integrals
Some
of our investigations can be summarized as follows:
f =
j
c
'
—
(b
/
=
^2
£2
-—
/
3
=
*!
a)
if
f(x)
=
c
if
/(*)
=
* for
if
/(*)
=
for all *,
all *,
3
This
~^
3
for all x
3
already reveals that the notation j* f suffers from the lack of a con-
list
venient notation for
naming
functions defined
by formulas. For
alternative notation,* analogous to the notation lim /(#),
J*
means
f(x) dx
same
precisely the
is
this
reason an
also useful:
j* f.
as
Thus
rb
dx
=
x ax
=
c
c
6
(b
•
2
—
a),
a2
—
2
2
= -
x 2 dx
3
3
Notice that, as in the notation lim /(*), the symbol x can be replaced by any
x—*a
or
6,
of course)
= p(t)
dt
=
other letter (except
*
/,
/;/(«)
<**
= /;/«
<*»
=
*.
has no meaning in isolation, any more than the symbol
has any meaning, except in the context lim /(*). In the equation
The symbol
=
x 2 dx
°—i
3
3
the
<?nf*re
may
symbol x 2 dx
be regarded as an abbreviation
the function
/ such
that /(*)
=
x 2 for
&
*
The
notation
J /M
dx
is
actually the older,
x—>
and was
for
many
for:
all x.
years the only, symbol for
the integral. Leibniz used this symbol because he considered the integral to be the sum
of infinitely many rectangles with height /(*) and "infinitely small" width dx.
(denoted by
J)
Later writers used x 0
by
Axi.
The
,
integral
.
.
.
,
xn to denote the points of a partition, and abbreviated Xi
was defined as the
limit as
Ax { approaches
0 of the
sums
^=
i
(analogous to lower and upper sums).
f(xi) to /(*),
and Axi
to dx, delights
—
n
The
fact that the limit
many
people.
is
/(*»•)
Ax {
i
obtained by changing
2
to /,
226
Derivatives
and
Integrals
This notation for the integral
examples
may
we have made
rb
/
x
+y)dx=
/
Ja
fx
(y
+
+^
(1
(3)
(fa
Ja
Ja
£
fx
xi
dX
=
t
+
0 j'
f(f (,+,)*)*- £[x
=
of
f*
x dx
(x
-
a)
t(x
-
a).
dX
dx
(i+4f-?-«(*-4
=
The computations
£
a2
~^
la
(1
a2
---+
+
(1
)
=
dy
6.*
-~--+y(b-a).
fx
=
(4)
ydx =
/
Ja
dZ
and
5
k2
rb
xdx+
jjdy + ^
t)dy=
Theorems
use of
rb
(
(2)
as flexible as the notation lim fix). Several
x—>a
aid in the interpretation of various types of formulas which
frequently appear;
(1)
is
{d
-c)+ <l-l]dx
C
(~- ^(b-a)
+
^
(d-c)
f\dx
and j* x 2 dx may suggest that evaluating
integrals is generally difficult or impossible. As a matter of fact, the integrals
of most functions are impossible to determine exactly (although they may be com-
puted
less,
to
as
any degree of accuracy desired by calculating lower and upper sums). Nevertheshall see in the next chapter, the integral of many functions can be
we
computed very easily.
Even though most integrals cannot be computed exactly, it is important
at least to know when a function / is integrable on [a, b\ Although it is possible
to say precisely which functions are integrable, the criterion for integrability
is a little too difficult to be stated here, and we will have to settle for partial
results. The most useful result will be stated now, but proved at the end of the
next chapter.
THEOREM
3
If
/
is
continuous on
(Notice that
it is
[a, b],
then /
integrable on
is
[a, b].
unnecessary to assume that / is bounded on
[a,
b\ because of
continuity.)
* Lest
chaos overtake the reader
when he
reads other books, equation (1) requires an impor-
tant qualification. This equation interprets
/
y dx to
mean
the integral of the function / such
f&
that each value /(*)
is
the
number^. But
classical notation often uses y for_y(x), so
might mean the integral of some arbitrary
function y.
/
y dx
227
Integrals
Although
this
theorem
of integrals in this book,
will provide all the information necessary for the use
more
it is
have a somewhat larger supply
satisfying to
of integrable functions. Several problems treat this question in detail. It will
help to know the following three theorems, which show that / is integrable on
[a, b], if it is integrable on [a, c] and [c, b]; that / + g is integrable if /andg are;
and that c / is integrable if / is integrable and c is any number.
As a simple application of these theorems, recall that if / is 0 except at one
point, where its value is 1 then / is integrable. Multiplying this function by c, it
follows that the same is true if the value of / at the exceptional point is c.
Adding such a function to an integrable function, we see that the value of an
integrable function may be changed arbitrarily at one point without destroying integrability. By breaking up the interval into many subintervals, we see
that the value can be changed at finitely many points.
•
,
The
proofs of these theorems usually use the alternative criterion for inteTheorem 2; as some of our previous demonstrations illustrate, the
grability in
obscure the point of the proof. It is a
good idea to attempt proofs of your own, consulting those given here as a
last resort, or as a check. This will probably clarify the proofs, and will certainly give good practice in the techniques used in some of the problems.
details of the
THEOREM
4
argument often conspire
Let a
<
[c, b].
Conversely,
[a, b].
c
<
b.
Finally,
If /
if
is
integrable
if /is
/
on
b\ then /
[a,
integrable on
integrable on
is
to
integrable
is
and on
[a, c]
b],
[c,
on
and on
[a, c]
then /is integrable on
then
[a, b],
/>-/:/+/>
PROOF
Suppose/ is integrable on
of
>
b].If£
[a,
there
is
P =
a partition
{t
.
Q,
.
.
,
t
n}
such that
[a, b]
-
U(f, P)
We
might as well assume that
partition which contains t 0
.
,
L(f, Q)
<
Now
P
U(f, P)
=
{/ 0 ,
a partition of
y
0,
[c,
.
b]
.
.
L(f, P)
.
is
tj}
,
=
c
.
<
,
tn
£(/, P)
some
for
tj
and
<
e.
then
c;
(Otherwise,
/.
Q
Q be
let
the
contains P, so £/(/, Q)
-
e.)
a partition of
[a, c]
and P"
=
{*,-,
.
.
.
,
*»} is
(Figure 12). Since
a/,
p)
U(J, P)
= us,
= U(f,
n
P')
+
+
uf,
n,
U(J, P"),
we have
p"
P'
n
a
i
i
i
i
i
n=
r
i
i
c
i
i
i
m
/„
[U(f, P')
=
b
-
Uf,
P')]
+
[u(f, p")
Since each of the terms in brackets
shows that /
is
integrable on
[a, c]
Uf,n
is
and
Uf, P")} =
u(f, p)
nonnegative, each
[c,
b].
Note
Uf, p") < f*f <
is less
also that
< p<u(f,p'),
u(f, P"),
- Uf,
than
p)
e.
<
«•
This
228
Derivatives and Integrals
so that
Since this
any
true for
is
b
<
L{f,P)
f+
fo
fb°f<U(f,P).
P, this proves that
/:/+/;/-/.*/•
Now
suppose that /
partition P' of
integrable on
is
and a
[a, c]
partition
U(f, P')
U{f, P")
If
P is
the partition of
[a,
b]
=
=
L(f, P)
of
+
+
U(f,P')
If £
>
there
0,
is
a
e.
I
6/2,
e/2.
the points of P'
all
L(f, P')
[c, b].
such that
b]
[c,
- L(f, P') <
- L(f, P") <
containing
U(f,P)
and on
[a, c]
P"
and P", then
L(f, P"),
U(f,P");
consequently,
-
U(f, P)
=
L(f, P)
Theorem
4
[U(f, P')
the basis for
is
was defined only
Ja f
for a
/;/=0
With
even
-
L(f, P')]
a
<
<
c
b
is
[U(f, P")
-
L(f, />")]
some minor notational conventions. The
<
b.
and
We now
add the
///--/;/
these definitions, the equation
if
+
fcf+
<
integral
definitions
if«>*.
= f*f
f*f
not true (the proof of this assertion
holds for
is
all a,
5
If /
and g are integrable on
[a, b],
then /
+g
is
integrable on
[a, b]
//(/+*)-/:/+/>
PROOF
Let
P =
{*<>,
.
•
-
,
U
be any partition of
mi
=
m/ =
m/ =
r
and
define
M
iy
M/,
inf {(/
inf \g(x):
M" similarly.
it is
t^i
<
t^ <
x
It is
*
x
<
<
+
m/',
> m/
+
m/'.
Mi < Mi
+
true (Problem 7) that
rrii
<
x
Let
<
ti\,
fe},
t-),
not necessarily true that
— m/
rrii
but
+ g)(x):
inf j/0):
[>, £].
Similarly,
Mi".
b
a rather tedious
case-by-case check).
THEOREM
c,
and
Integrals
229
Therefore,
L(f,P)+L(g, P)<L(J
+ g,P)
and
+ g,
U(f
< U(f,P)
P)
+
U(g, P).
Thus,
Uf,
+
P)
P)
Ug,
<
Uj + g,P)<
U(J
+ g,P)<
Since / and g are integrable, there are partitions
P')
U(f, P') U(g, P") - Ug, P")
US
If
P
contains both P'
s/2.
+
Ug,
U(g, P).
P" with
P',
<
<
+
8/2,
and P", then
+
U(f, P)
U(g, P)
and consequently
-
[Uf, P)
+ g,P) -L(f + g,P)<
U(f
+ g integrable on b\
US P) + Lis, P) ^ L (f + s, P)
This proves that /
(1)
U(f, P)
[a,
is
P)]
<
«,
S.
Moreover,
</;(/+,)
< U(f +
and
Since
US,
desired,
it
Uf,
P)
~
P)
+
Uf,
Ug,
P)
<
b
J +f g< Wf,
\
+
is
on
[a,
integrable
b]
- Ug,
-
u(g, P)
/:</+*>
proof
+
U{g,P);
+
u(g, P).
P) can both be
made
as small as
follows that
can also be made as small as desired;
If /
P)
a
P) and U{g, p)
u(f, p)
6
< U(f,P)
also
(2)
THEOREM
g,P)
on
it
[Uf, P)
+
Ug,
P)]
therefore follows from (1)
and
(2)
that
-/:/+/>•
[a, b],
then for any number
much
easier
c,
the function cf
is
integrable
and
The proof (which
good idea
is
than that of Theorem
to treat separately the cases c
>
0 and
c
<
5)
0.
is left
to you. It
Why?
is
a
|
theorem that / g is
(Theorem 6 is just a special case of the more general
hard to prove
quite
is
result
this
but
are,
integrable on [a, b], if / and g
(see Problem 32).)
complicated definition, a tew
In this chapter we have acquired only one
theorem with no proof at all.
simple theorems with intricate proofs, and one
•
230
Derivatives and Integrals
This
is not because integrals constitute a more difficult topic than derivatives,
but because powerful tools developed in previous chapters have been allowed
remain dormant. The most significant discovery of calculus is the fact that
and the derivative are intimately related—once we learn the con-
to
the integral
nection, the integral will
become
The connection between
derivatives
but the preparations which
We
first
many
THEOREM
7
we
as useful as the derivative,
and
and
as easy to use.
integrals deserves a separate chapter,
make in this chapter may serve as a hint.
concerning integrals, which plays a role in
will
state a simple inequality
important theorems.
Suppose /
on
integrable
is
m <
and that
[a, b]
M
<
f(x)
for all x in
b].
[a,
Then
i(b
PROOF
It is clear
-
b
<
a)
f<M(b -
Ja
a).
that
m(b
-a)<
and
L(f, P)
for every partition P. Since
J* f
=
< M(b -
U(f, P)
sup {£(/, P)\
=
a)
inf {£/(/, P)}, the desired
inequality follows immediately. |
on
Suppose now that /
[a, b] by
integrable on
is
[a, b].
We can
define a
new
function
F
c+h)-F(c)
= /;/ =
f{x)
c
c
+
dt.
Theorem 4.) We have seen that /may be integrable even if
not continuous, and the Problems give examples of integrable functions
(This depends on
h
it is
FIGURE
/;/(*)
13
which are quite pathological. The behavior of
F is
therefore a very pleasant
surprise.
THEOREM
8
If /
integrable on
is
[a, b]
and
F is
defined on
f(x)
F
then
PROOF
is
Suppose c
on [a, b];
continuous on
is
in
let
[a, b\.
[a,
Since /
>
0,
by
/;/,
b].
is
integrable on
M be a number such that
< M
[a, b] it is,
for all x in
|/(*)|
If h
=
[a, b]
[a,
by
definition,
b\
then (Figure 13)
f(,
+ a)-f
( ,)
=
+
/;
v-/;/ =
Since
—M < f(x)
<
M
for all x,
+
/; V.
bounded
/
—
t
1
a
^J-
F
f
FIGURE
it
14
follows from
Theorem
7 that
-Af 'A< j
C+h
f<Mh;
c
in other words,
(1)
— Af
•
h
<
F(c
+
h)
-
<
AT-
A.
If h
<
0,
a similar inequality can be derived. Note that
Applying Theorem 7
to the interval
MA <
— 1,
multiplying by
Mh <
and
Inequalities (1)
if £
>
0,
+ h) -
\h\
words
F
< e/M.
is
F is
true this
M
-
\h\.
F(c)\
<
e,
This proves that
continuous at
+
c.
=
h)
F(c);
\
=
Figure 14 compares / and F(x)
that
< -Mh.
F{c)
<
F(c)\
+ h) -
lim F(c
in other
we have
we have
\F{c
provided that
-Mh-
-
h)
we obtain
of length —h,
the inequalities,
all
+
F{c
c],
can be combined:
(2)
\F{c
Therefore,
f <
f +h
Jc
which reverses
(2)
+ h,
0
e
for various functions /;
J* f
always better behaved than
In the next chapter
/.
we
it
appears
will see
how
is.
PROBLEMS
1.
Prove that
J*
x z dx
=
4
by considering
b /4,
partitions into n equal sub-
n
intervals, using the
formula for
^
=
i
i
z
which was found
in
Problem
2-5.
i
This problem requires only a straightforward imitation of calculations
in the text, but you should write it up as a formal proof to make certain
argument are
that all the fine points of the
2.
*3.
Prove, similarly, that
(a)
Using Problem
4
J*
2-6,
x dx
=
b /5.
n
show
Y
that
*
to l/(p
(b)
4.
+
=
=
k p /n p+1 can be
b
p+l
/{p
+
made
n large
enough,
1).
Decide which of the following functions are integrable on
calculate the integral
/
•
\
(0
when you
X,
0 <
/W = U-2, 1<
r,
\
(
can.
X
<
X
<2.
1
as close
i
by choosing
1) as desired,
Prove that j* x* dx
clear.
b
[0, 2],
and
233
Integrals
/(*)
(iii)
/(*)
/•
/W
//
\
(1V)
(v)
-
x
=
+
*
=
*
0,
1
2,
* rational
[*],
# of the form a
* not of this form.
=
/
5.
<
*
*
=
0 or *
shown
the function
is
<
0
for rational #
b
and
b
1
[i]
0,
(vii)
+
1,
0,
1
f{x)
1
2.
* irrational.
f{x)
(vi)
<
<
x
*
[x].
+
f
(
\
<
<
0
x,
(ii)
>
1.
in Figure 15.
Find the areas of the regions bounded by
=
(i)
the graphs of fix)
(ii)
the graphs of fix)
x2
x2
=
= —
and
(iii)
(iv)
(v)
2.
= — x 2 and
x 2 and g(x)
through (-1, 0) and (1, 0).
the graphs of fix) — x 2 and g(#)
the graphs of /(*) = x 2 and
the graphs of /(*) = a: 2 and g(x)
+
=
=
=
the vertical lines
—x
- x and hix) = 2.
- 2x + 4 and the
2
1
.
2
1
x2
vertical
axis.
= V*,
the graph of fix)
(vi)
the horizontal axis,
2
through
(2,
0). (Don't try to find
and the
vertical line
you should
|q
see a
way
of guessing the answer, using only integrals that you already know
how to evaluate. The questions that this example should suggest are
considered in Problem 16.)
6.
Find
I!Uc f^siy)dy)dx
d
b
in terms of
and
f
j f
a vengeance;
7.
(This problem
crucial that
Prove, using the notation of
m.:/
8.
it is
g.
(a)
+
Which
mi"
=
inf {/(a)
is
an exercise
in notation, with
you recognize a constant when
Theorem
+ g(x
2 ):
5,
it
appears.)
that
<
*i, *2
<
**}
<m
functions have the property that every lower
f.
sum
equals
every upper sum?
(b)
Which
some
(c)
functions have the property that
(other) lower
Which continuous
are equal?
some upper sum equals
sum?
functions have the property that all lower
sums
234
Derivatives and Integrals
Which
*(d)
integrable functions have the property that
mind
are equal? (Bear in
=
that one such function
lower sums
all
is
=
f(x)
0 for x
/q for x — p/q in lowest terms.) Hint: You will
need the notion of a dense set, introduced in Problem 8-6, as well as
irrational, f(x)
Problem
the results of
9.
Ha<b<c<d and /
on
10.
Prove that
(a)
in
[a,
warn
/
if
>
f* f
Prove that
(b)
21.
integrable on
is
d\ prove that /
[a,
integrable
is
(Don't work hard.)
[b, c].
then
11.
1
if
integrable on
/and g
b\ then
that
is
and
[a, b]
f(x)
>
0 for
x in
all
[a, b],
0.
are integrable on
>
Ja f
g>
you work hard on part
if
[a,
now
(By
and f(x) >
b~\
g(x) for all x
should be unnecessary to
it
you are wasting time.)
(b)
Prove that
=
J*
I! X*)
f(*
-
<) d*.
(The geometric interpretation should make this very plausible.) Hint:
Every partition P =
of [a, b] gives rise to a partition
tn
0
{
P'
*12.
=
{to
+
c,
•
•
,
,
.
.
Ji
Hint:
P=
of
can
This
{t 0 ,
.
[b, ab],
.
.
,
and
tn
I
Ji
be
}
+
t
written
b
c,
+
and conversely.
c],
of
[1
,
-dt
fab
=
I
-dt.
Ji
t
t
j"\/tdt = j*
a] gives rise to
h
\/tdt.
a partition
Every partition
P =
r
{bt 0 ,
.
.
.
,
bt n
\
conversely.
Prove that
b
jl
c
f(t) dt
(Notice that Problem 12
=
e
£
f(ct) dt.
a special case.)
is
Suppose that /is bounded on
in
[a, b]
and that /is continuous
with the exception of # 0 in (a, b). Prove that /
Hint: Imitate one of the examples in the text.
[a, b]
[a, b].
15.
+
rb\
l
~dt
14.
\
,
of [a
Prove that
fa
*13.
.
+ c]
tn
Suppose that / is nondecreasing on [a, b]. Notice that /
bounded on [a, b], because f(a) < f(x) < f(b) for x in
(a)
If
P=
{fo,
...,*»}
is
a partition of
-
=
d for
[a, b],
is
what
at each point
integrable
is
on
automatically
[a,
is
b\
L(f, P)
and
U(f, P)?
(b)
Suppose that k
each
i.
Prove that U(f, P)
-
L(f, P)
=
(c)
8[f(b) -/(«)].
Prove that / is integrable.
(d)
Give an example of a nondecreasing function on
discontinuous at infinitely
many
points.
[0, 1]
which
is
235
Integrals
might be of
It
compare
interest to
problem with the following
this
extract from Newton's Principia.*
LEMMA
II
AacE,
terminated by the right lines
there be inscribed
any number of parallelograms
If in any figure
acE,
prehended under equal bases
&c,
bLcm, cMdn, &c,
AB, BC, CD, &c, and
Aa
parallel to one side
and
their
which
say, that the ultimate ratios
number
aKbl,
be augmented in infinitum,
to
the inscribed figure
AKbLcMdD,
AalbmcndoE, and curvilinear figure AabcdE,
circumscribed figure
to
Bb, Cc, Dd,
the sides,
of the figure; and the parallelograms
are completed: then if the breadth of those parallelograms
be supposed to be diminished,
/
Aa, AE, and the curve
Ab, Be, Cd, &c, com-
the
will have
one another, are ratios of equality.
For the difference of the inscribed and circumscribed figures is the
sum of the parallelograms Kl, Lm, Mn, Do, that is (from the equality
of all their bases), the rectangle under one of their bases Kb and the
sum of their altitudes Aa, that is, the rectangle ABla. But this recits breadth AB is supposed diminished in infinitum,
than any given space. And therefore (by Lem. 1) the
figures inscribed and circumscribed become ultimately equal one to
the other; and much more will the intermediate curvilinear figure be
tangle, because
becomes
less
ultimately equal to either. Q.E.D.
*16.
Suppose that /is increasing. Figure 16 suggests that
fr = rw -ama)(a)
If P =
f~
{h,
.
.
.
{t n )\.
L{f~\ P)
17.
Now
(c)
Find j*
A
P =
V* dx
s
defined on
{t Q ,
.
.
Prove that
if
2
Newton's
P'
=
[f~Kt Q ),
.
.
.
.
P) =
f
U(f,
bf~Kb)
-
af~Ka).
.
/
is
a
<
[a, b] is
*„}
,
<
of
b.
may
is
a parti-
constant on each
0
called a step function
[a, b]
such that
s is
if
there
be arbitrary).
integrable on
-
[a, b],
then for any
Principia,
si
and
s
>
0 there
is
a
also a step function
2
2
*
[a, b], let
< f with f
si < e,
J*
f*
—
^ > / with
s
< e.
Ja f
Suppose that for all s > 0 there are step functions si < f and
> / such that
s ^
Si < e. Prove that / is integrable.
step function
(b)
+
for 0
(the values of s at h
(a)
a partition of
prove the formula stated above.
function
tion
is
)
Prove that, as suggested in Figure 17,
l
(b)
tn
,
/;;;:/.
J*
A
2
f*
Revision of Mott's Translation, by Florian Cajori. University of Cali-
fornia Press, Berkeley, California, 1946.
(c)
Find a function / which
=
j f
18.
Prove that
some
L(f, P) for
if
/
<f
tinuous function g
P
partition
on
integrabie
is
not a step function, but which
is
f
[a, b].
>
then for any £
[a, b],
with
of
—
satisfies
g <
0 there
is
a con-
Hint: First get a step
£.
function with this property, and then a continuous one.
A
picture will
si
+
help immensely.
19.
(a)
Show that
(b) Prove,
(c)
Use part
such that
integrabie on
is
=
J* f
f.
ble to choose x to be in
21.
are step functions on
The purpose
s 2 is also.
Si
Ja
+
Ja
s2 .
Prove that there
[a, b].
that
in
always possi-
not
it is
number x
a
is
(a, b).
of this problem
P=
{/ 0 ,
<
•
.
—
b
sup {/(*):
choose
(c)
=
s2 )
to
is
show that
many
/
if
points in
is
integrabie
on
[a, b],
[a, b].
tn
be a partition of [0, 1] with U(f, P) Prove that for some i we have Mi — mi < 1.
Prove that there are numbers a\ and b\ with a < a\ < b\ < b and
Let
L(f, P)
(b)
+
(si
Show by example
then / must be continuous at
(a)
then
[a, b],
that
5,
5.
Suppose that /
[a, b]
s2
(and Problem 17) to give an alternative proof of
(b)
Theorem
20.
and
if Si
without using Theorem
<
=
ai
[ai, bi]
•
,
]
<
bi}
a.
*
/J
—
inf {/(#): ai
from part
<
<
=
x
(a) unless
i
<
bi)
or
1
two cases a very simple device solves the problem.)
Prove that there are numbers a 2 and b 2 with a\ < a 2
<
sup {/(*): a 2
(d) Continue in
x
<
b2
]
-
inf \f(x): a 2
<
x
<
b2
}
n;
(You can
and in these
<
b2
<
1.
<
and
bi
i.
this way to find a sequence of intervals In = [a n b n ]
such that sup {/(*): a: in 7n - inf \f(x): x in In < 1/n. Apply the
Nested Intervals Theorem (Problem 8-14) to find a point x at which
,
)
}
/
(e)
22.
is
continuous.
Prove that /
Recall, from
(a)
is
continuous at infinitely
Problem
f*f >
10, that
>
Give an example where f(x)
b\ and f* f = 0.
(b) Suppose f(x) > 0 for
0
many
>
0 for
if /(*)
all x,
points in
0 for
and f(x)
all
>
[a, b],
x in
0 for
[a,
b\
some x
in
[a,
and f(x 0 ) >
0.
all
x in
lower sum L(f, P) which
(c)
Suppose /
is
integrabie on
is
b
Prove that
>
j f
0.
Hint:
[«, b]
f >
Prove that
and / is continuous
0.
Hint:
at * 0 in
[a, b]
It suffices to find
one
positive.
[a, b]
You
and
will
f(x)
>
0 for
all
x in
[a, b].
need Problem 21; indeed that
was one reason for including Problem 21.
Integrals
*23.
(a)
Suppose that /
is
continuous on
and
[a, b]
=
tinuous functions ^ on [a, b]. Prove that /
is an obvious g to choose.)
(b)
Suppose /
is
continuous on
tinuous functions g on
=
g(a)
g(b)
=
which
[a, b]
=
Prove that /
0.
and that
[a, b]
f* fg
0.
=
0 for
(This
=
J fg
satisfy the
is
all
237
con-
easy; there
0 for those conextra conditions
(This innocent looking fact
0.
is
an important lemma in the calculus of variations; see the Suggested
Reading for references.) Hint: Derive a contradiction from the
assumption f(x Q ) > 0 or f(x 0 ) < 0; the g you pick will depend on
the behavior of / near x 0
24.
*25.
—
Let f(x)
(a)
Compute
(bj
Find inf
=
Let f(x)
that /
is
27.
L(f, P) for
{£/(/,
0 for irrational
you must
.
and
all
P):Pa
integrable on
clearly 0;
*26.
x for x rational
=
f(x)
0 for x irrational.
P
partitions
of
partition of
[0,
and \/q
x
x,
[0, 1]
if
and that
figure out
how
to
[0, 1].
1]}.
—
p/q in lowest terms. Show
= 0. (Every lower sum
JQ f
make upper sums small.)
is
Find two functions / and g which are integrable, but whose composition
g o f is not. Hint: Problem 25 is relevant.
for all x in [a, b], then
(a) Prove that if m < f(x) <
M
f*f(x) dx
for
(b)
some number
Prove that
if
/
is
=
-
(b
a) ii
fx
< M.
continuous on
[a, b],
jjl
with
m <
then
b
dx= (b- a)M)
jJ{x)
some
for
J in
[a, b];
This fact
essential.
is
and show by an example that continuity is
called the (First) Mean Value Theorem for
Integrals.
28.
(a)
Suppose that / is continuous on [a,
nonnegative on [a, b]. Prove that
'/(*)*(*) dx
\l
for
some
J in
[a, b].
=
/(*)
This result
is
b]
and that g
is
integrable and
\lg(x)dx
called the
Second
(or Generalized)
Mean Value Theorem for Integrals. Hint: The
< f(x) <
implies that mg(x) < f(x)g(x) < Mg(x),
M
inequality
since g
is
nonnegative.
(b)
The
Show
First
that the hypothesis for g
is
essential.
and Second Mean Value Theorems
not very essential way) in Chapter 19.
The
(Tryg(*)
=
# on
[— 1,
1].)
be used (in a
problem, which
for integrals will
result of the next
I have cribbed from several older calculus texts, is called the Third Mean
Value Theorem for Integrals. To be quite frank, I haven't the slightest idea
what it is good for.
238
Derivatives
and
Integrals
*29.
Suppose that g is increasing on [a, b] and that / is integrable on
[a, b]. Prove that there is a number £ in [a, b] such that
(a)
p(t)g(t)
+ g(b)
g{a) ///(<) dt
f*f(t)
dt.
= a and for £ = b?
on g is essential. (Try g(a) = g(b) = 0.)
Let / be a bounded function on [a, b] and let P be a partition of [a, b].
Let Af and rrii have their usual meanings, and let M/ and m/ have the
corresponding meanings for the function |/|.
(b)
*30.
=
dt
Hint:
What
Show
that the hypothesis
the value of the right side for £
is
t
(a)
(b)
(c)
(d)
Prove that
31.
max(|Af<[,
m/ =
min(|Afi|,
\rrii\),
|iw,-|).
Prove that Af/ — m/ < Mi —
Hint: You will have to consider
two separate cases, because that's how max and min are defined.
Prove that if / is integrable on [a, b], then so is |/|.
Prove that if / and g are integrable on [a, b], then so are max(/, g)
and min(/,
(e)
M/ =
g).
Prove that / is integrable on [a, b] if and only if its "positive part"
max(/, 0) and its "negative part" min(/, 0) are integrable on [a, b].
Prove that
/
if
is
integrable on
[a, b],
|/.Vw*|<
then
Jl i/w
dt.
i
Hint: This follows easily from a certain string of inequalities; Problem
1-14 is relevant.
*32.
and f{x) > 0 for all x in [a, b].
and m/ denote the appropriate
and m/' similarly for g, and define Mi
Suppose / and g are integrable on
P be a partition of [a, b]. Let
Let
and inf s for /,
and mi similarly for
sup's
(a)
Prove that
(b)
Show
M
define
M/
fg.
<
{
M"
[a, b]
M-M"
and
m > m/m"{
that
n
-
U(fg, P)
L(fg,
P)<2
i
(c)
Using the
for x in
U(fg, P)
=
[M/M/' - m/m/'^U -
/ and g are bounded,
show that
fact that
[a, b],
-
M
{
so that |/(*)|,
<
M
fc_0
}.
L(fg, P)
n
<
/,•_,).
l
n
\
[Mi'
-
m/](ti
- t^) + T
t=i
(d)
Prove that /g
(e)
Now
is
[Mi"
- m/ ](ti f
t=i
integrable.
eliminate the restriction that fix)
>
0 for x in
[a, b].
239
Integrals
33.
Suppose that / and g are integrable on
[a,
b].
The
Cauchy -Schwartz
inequality states that
(/»(/>>
(a)
Prove the Gauchy-Schwartz inequality by imitating one of the
(b)
If equality holds,
proofs of the Schwartz inequality in
is it
Problem
necessarily true that f
(c)
/ and g are continuous?
Show that the Schwartz inequality
Schwartz inequality.
(d)
Prove
=
1-18.
\g for some X ?
What
if
that^
replaced by a
*34.
Suppose that /
is
1
2
/)
and
<
2
(J* f
Is
)-
continuous and lim /(*)
—
f
1
lim -
The
> some
in
and
1
are
condition lim f(x)
x — 00
N. This means that
comparison to N, then
—
a.
Prove that
oo
X
f{t) dt
/
=
*
t
this result true if 0
b ?
X
Hint:
a special case of the Cauchy-
is
f
a
=
a.
implies that fit)
is
close to a for
N+M
/
Ma/ (N
/(f) dt is close to Afa. If
+ M)
is
close to a.
M
is
large
THE FUNDAMENTAL THEOREM
OF CALCULUS
CHAPTER
From
the
the hints given in the previous chapter you
first
fa f
theorem of
continuous;
is
original function
derivative
THEOREM 1 (THE FIRST
FUNDAMENTAL THEOREM
is
/
it
is
only
is
we
fitting that
if
may
/
is
have already guessed
integrable, then F(x)
=
what happens when the
ask
F is
differentiate (and
its
especially simple).
[a, b],
continuous at
c
in
F on
and define
OF CALCULUS)
is
that
continuous. It turns out that
Let / be integrable on
Iff
We know
this chapter.
*w
= Ij.
then
F is
[a, b],
[a, b]
by
differentiate at
c,
and
F'{c) =/(,).
(If c
=
a or
then F'(c)
b,
mean
understood to
is
the right- or left-hand deriva-
tive of F.)
PROOF
We will
assume that
in
c is
supplied by the reader.
{a, b)
By
;
the easy modifications for
F'(c)=l^
first
that h
>
0.
mh and
M
h
may
h
+
h)-F(c) =
+k
f.
f;
as follows (Figure 1):
mh =
h =
inf {/(*):
M
It follows
a or b
Then
F(c
Define
=
+ h) - F^.
F{c
A-+0
Suppose
c
definition,
sup
<
<
c
\f(x): c
x
<
c
x
<
c
+
+
h],
h\.
from Theorem 13-7 that
h<
[c + h
f<M
h
Jc
-h.
Therefore
^
F(c
+
ti)
—
F(c)
n
If h
<
0,
only a few details of the argument have to be changed. Let
mh =
h =
M
240
inf {/(*):
c
sup {/(*):
c
+
+
h
h
<
<
x
x
<
<
c],
c\.
be
The Fundamental Theorem of Calculus
241
A
nh
c-i-h
l
FIGURE
1
Then
C
mh {-h) < f +h
-
c
f<M
h
-(-h).
Since
F {c+h)-F(c) =
+h
f= -f;+ j,
f:
this yields
mh
Since h
<
0,
•
h
>
F{c
+ h) -
F{c)
> Mh
'
h.
dividing by h reverses the inequality again, yielding the
same
result as before:
mh <
F{c
+
-
h)
F{c)
;
<
h
This inequality
/
is
is
continuous at
true for
c,
this
h
.
any integrable function, continuous or
not. Since
however,
lim
and
M
mh =
lim
M
h
—
f(c),
proves that
F'(c)
=
F(c
lim
+
h)-F{c)
_
t
k
h-yO
Although Theorem 1 deals only with the function obtained by varying the
upper limit of integration, a simple trick shows what happens when the lower
limit is varied. If G is defined by
CM
-
/„*/,
then
= /:/-/:/•
242
Derivatives
and
Integrals
Consequently,
if
/
is
continuous at
=
G\c)
The minus
Theorem 1
sign appearing here
defined even for x
<
very fortunate, and allows us to extend
is
if c
<
a
= /;/
we can
In this case
a.
™
so
-/(,).
where the function
to the situation
*w
is
then
c,
= -
write
/;/,
we have
=
F'(c)
-(-/(c))
=
/(,),
exactly as before.
F at c is ensured by continuity
most interesting when / is con-
Notice that in either case, differentiability of
of
/
Theorem
at c alone. Nevertheless,
tinuous at
[a, b]
points in
all
[a, b].
In
is
1
this case
F is
differentiable at all points in
and
F'=f.
In general,
it is
extremely
difficult to
decide whether a given function /
Theorem
derivative of some other function; for this reason
Theorem (Problem
tain properties
problem
at all
11-7
is
the
and Darboux's
11-39) are particularly interesting, since they reveal cer-
which / must have.
—according
to
If
/
Theorem
is
1,
/
continuous, however, there
is
the derivative of
some
is
no
function,
namely, the function
*w
Theorem
1
=
/."/•
has a simple corollary which frequently reduces computations
of integrals to a triviality.
COROLLARY
If
/
is
continuous on
[a, b]
/.*/
proof
=
and /
r
for
g
some function
g,
then
= *(*)-*«.
Let
F{X)
Then F' = / =
g'
on
[a, b].
c
can be evaluated
0
so c
=
/;/.
Consequently, there
F=
The number
=
=
g
easily:
F{a)
=
+c
note that
g{a)
+
c,
—g(a); thus
F(x)
=
g(x)
is
-
g(a).
a number
c
such that
243
The Fundamental Theorem of Calculus
This
is
=
f*f
The proof of
for
T(b)
-
g{b)
to
know
/.*/
=
gib)
=
J* f
g(a). |
?
make
the corollary seem
that
-
The point,
g(a)
of course,
is
that one might
happen
quite different function g with this property. For example,
x3
= -
g(x)
then g'{x)
=
is it
what good
example, g(x)
know a
to
Thus
b.
this corollary tends, at first sight, to
useless: after all,
if g is,
—
true, in particular, for x
—
we
f(x) so
and
if
= x\
f(x)
computing lower and upper
obtain, without ever
sums:
—
dx
One can
x n + l /(n
powers similarly;
treat other
+
=
then g'(x)
1),
x
rx
For any natural number
interval containing 0, but
n,
if
n
3
n
dx
=
and
b are
is
b~~
=
f.
formula
n
is
a natural number and g(x)
=
n
1
+
1
= a: n is not bounded on any
both positive or both negative, then
the function /(*)
a
-n dx
this
+
n
b
Naturally
if
so
n+1
-72+1
only true for n
-n
+
1
— 1 We do not know a simple expression
.
for
a
X
The problem of computing this integral is discussed later, but it provides a
good opportunity to warn against a serious error. The conclusion of Corollary 1
is
often confused with the definition of integrals
—
—many students think that
f
where g is a function whose derivative is /." This
"definition" is not only wrong it is useless. One reason is that a function /
may be integrable without being the derivative of another function. For examis
defined as: "g(b)
g(a),
—
=
=
then / is integrable, but / cannot be a
is also another reason that is much more imporderivative (why not?).
tant: If / is continuous, then we know that / = g' for some function g; but we
know this only because of Theorem 1. The function /(*) = \/x provides an excelple, if /(*)
0 for x
and
There
9* 1
/(I)
1,
=
244
1
Derivatives and Integrals
lent illustration:
>
x
if
0,
The
g'(x) 9
f
=
8(x)
and we know
=
then f(x)
x
1
dt >
h
i
no simpler function g with
of
where
this property.
Theorem 1 is so useful that it is frequently called the Second Fundamental Theorem of Calculus. In this book, that name is reserved for
a somewhat stronger result (which in practice, however, is not much more
useful). As we have just mentioned, a function / might be of the form g' even if
/
corollary to
not continuous.
is
/
If
integrable, then
is
it is still
&% - g(a).
/„7 =
The proof, however, must be entirely different
we must return to the definition of integrals.
THEOREM 2 (THE SECOND FUNDAMENTAL THEOREM OF CALCULUS)
If
/
integrable on
is
[a, b]
=
and /
/„7 =
PROOF
Let
P =
there
is
{t
0i
.
.
.
a point
X{
,
tn
g,
1,
so
then
gW - g{a).
/J
such that
-
g{ti^)
g{k)
—we cannot use Theorem
some function
g' for
be any partition of
}
in
true that
=
=
By the Mean Value Theorem
[a, b].
g'(xt)(ti
f(Xi)(ti
- t^)
- k-l).
If
=
Mi =
inf {/(#):
U-\
sup {/(#): U-\
rrii
<
<
<
<
x
x
k},
t{],
then clearly
that
rriiiti
-
ti_t)
<
f( Xi)(ti
rriiiti
-
U-i)
<
g(ti)
-
M
t
)
<
-
Mi(ti
M
t
),
is,
Adding
i
=
ti-i)
<
these equations for
1,
—
.
g(U-i)
.
.
,
n
< Mi(U —
we
obtain
n
n
T
%
nn(fi
-
gib)
-
g(a)
<
2
Miih
-
<,_,)
i=l
1
so that
Uf, P) <
for every partition P.
£ut
this
means
gib)
We
gib)
-
-
g(a)
<
U(f, P)
that
gia)
=
p. |
have already used the corollary to Theorem 1 (or, equivalently,
2) to find the integrals of a few elementary functions:
Theorem
b
x
n
dx
=
fon+l
n+\
n n+l
n+
-' 72
7*
-1.
(a
and
b
both positive or
both negative
if
n
<
0.)
The Fundamental Theorem of Calculus
245
As we pointed out in Chapter 1 3, this integral does not always represent the
area bounded by the graph of the function, the horizontal axis, and the vertical
lines through (a, 0) and (b, 0). For example, if a < 0 < b, then
does not represent the area of the region shown in Figure
instead
2,
which
is
given
by
Similar care must be exercised in finding the areas of regions which are
bounded by the graphs of more than one function
—a problem which may
frequently involve considerable ingenuity in any case. Suppose, to take a sim-
example first, that we wish to find the area of the region, shown
between the graphs of the functions
ple
=
f(x)
x2
and
g(x)
=
in Figure 3,
x3
on the interval [0, I]. If 0 < x < 1, then 0 < x z < x 2 so that the graph of g
lies below that of /. The area of the region of interest to us is therefore
,
area R(f,
which
0, 1)
-
area R(g,
0, 1),
is
£x
2
dx
-
\l
x 3 dx
=
i
-
i
=
iV
This area could have been expressed as
/:</-„).
If g (x)
/ and
<
fix) for all x in
g, even if f
in Figure 4. If
[a, b],
i? 2 ,
[a, b] ,
then this integral always gives the area bounded by
negative.
The
easiest
a number such that f
+c
and g
and g are sometimes
c is
then the region
bounded by /
c
to see this
bounded by / and g, has the same area
and g + c. Consequently,
R =
x
area
R =
2
=
is
shown
are nonnegative on
Rh
+c
area
way
+
as the region
+ c) — j* (g + c)
- (g + c)]
L(f +
J* (f
b
fa
c)
-/;(/-*)•
This observation
is
useful in the following problem:
region bounded by the graphs of
f(x)
=
x3
—
x
and
g(x)
=
x2
.
Find the area of the
246
Derivatives and Integrals
The
necessity
first
/ and g
intersect
determine
to
is
region more precisely.
this
x3
z
—
—
or
x
or
#(# 2
or
*
=
x
the interval
+
(0, [1
f(x)
-
([1
x2
,
x — x = 0,
— x — 1) = 0,
+ V5 - V5
n
1
1
=
0
j
5
V5]/2,
2
we have
0)
—
V5]/2) we have * 2 > x 3
the graphs (Figure
=
g(x) only
5),
if
x
xs
—
x
>
x2
and on the
interval
These assertions are apparent from
but they can also be checked easily, as follows. Since
=
0,
[1
+
a:.
V5]/2
does not change sign on the intervals
it is
of
2
2
On
The graphs
when
S
or
—
([1
Vs]/2, the function / - g
V5]/2, 0) and [0, [1 + V5]/2);
[1
therefore only necessary to observe, for example, that
v
(-i)
- (-*) - (-*) = i>
P _ _ 12 = -1 < 0,
s
2
o,
1
to conclude that
FIGURE
f-g>0
f-g<0
4
The area
on
on(0,
of the region in question
Q
/l
-V5
"
*
~
([1
is
[1
V5]/2,
+
0),
V5]/2).
thus
^+/
0
2
^ 2
(*
3
-
*)1
^«
one of the major problems involved in finding the
areas of a region may be the exact determination of the region. There are,
however, more substantial problems of a logical nature we have thus far
defined the areas of some very special regions only, which do not even include
some of the regions whose areas have just been computed! We have simply
As
this
example
reveals,
—
assumed that area made sense for these regions, and that certain reasonable
properties of "area" do hold. These remarks are not meant to suggest that you
should regard exercising ingenuity to compute areas as beneath you, but are
meant to indicate that a better approach to the definition of area is available,
although its proper place is somewhere in advanced calculus. The desire to
define area was the motivation, both in this book and historically, for the
definition of the integral, but the integral does not really provide the best
method of defining areas, although it is frequently the proper tool for computing
them.
be discouraging to learn that integrals are not suitable for the very
purpose for which they were invented, but we will soon see how essential they
are for other purposes. The most important use of integrals has already been
It
may
The Fundamental Theorem of Calculus
1
FIGURE
247
+ V5
5
emphasized:
if
/
is
continuous, the integral provides a function y such that
/(*)
This equation
=/«.
is
the simplest example of a "differential equation" (an equation
for a function y
which involves derivatives of y). The Fundamental Theorem
of Calculus says that this differential equation has a solution,
if /is continuous.
In succeeding chapters, and in various problems, we will solve more complicated equations, but the solution almost always depends somehow on the
integral; in order to solve a differential equation it is necessary to construct a
new
function, and the integral is one of the best ways of doing this.
Since the differentiable functions provided by the Fundamental Theorem
of Calculus will play such a prominent role in later work,
to realize that these functions
to yield
still
more
functions,
may
be combined,
it is
very important
like less esoteric functions,
whose derivatives can be found by the Chain Rule.
Suppose, for example, that
Although the notation tends
to disguise the fact
somewhat, / is the composition
of the functions
C(x)
In
fact, /(*)
=
=
xs
and
F(C(x)); in other words, f
= Fo
Rule,
fix)
=
=
F'(C(x))
F'(x*)
1
+
•
•
CM
3x 2
sin 2
xs
C. Therefore,
by the Chain
248
Derivatives
and Integrals
If
/
is
defined, instead, as
fa
=
i
dt >
+ sin r*
/
i
Jx* 1
2
•
/
then
If /
is
defined as the
reverse
1
=
w
/'(*)
7
3* 2
•
+
1
.
sin 2 x 3
composition,
\Ja
+
1
sin 2
/
1
then
f(x)
Similarly,
=
C(F(x))
~
3
(/
F'W
•
+ sin *)
2
1
+ sin
1
1
2
x
if
M
x
/*sin
1
A/
T
+ sin
:
1
r°
2
£
i
J sin
x 1
+
sin 2
£
X
=
sin
( [
\Ja
+ sin
1
dt\
2
/
f
then
/'(#)
g'(x)
h'{x)
—
=
+
sin 2 (sin x)
1
+
sin 2 (sin x)
=
=
•
cos ( [*
Via
The formidable appearing
/W
also a composition
/'« =
;
~~
-
+
1
cos
x,
COS X,
dt)
i
sin 2
•
/
f
1
+ sin
+sin't
d<
)
•
*
1
ia
in fact,
F'(F(*))
+
1
/ =
•
F ° F.
sin 2
*
Therefore
F'M
1
1
0
1
a
As
2
function
fUal
is
*
1
1
+
sin
^0A
2
1
1
+ sin
sir
2
x
)
the
these examples reveal, the expression occurring above (or below)
249
The Fundamental Theorem of Calculus
which
integral sign indicates the function
written as a composition.
As a
appear on the
will
example, consider the
final
j"
dt
1 + sin*
r(Iaa 1+sin^
f\J
t
/W
=
J/ a
)
J
fl
ft
1
|
1
1
+
.
,
sin
=
g(x)
dt,
2
t
(/a
/
i
+\imt
/al +
when /
right
dt
is
compositions
triple
)
sinM
J
\
/
.
J°
1
+ sm
.
.
dt,
2
t
which can be written
f =
FoFo C
and
g
= F° F°
may
Omitting the intermediate steps (which you
insecure),
F.
supply,
f(x ) =
feel
still
+
1
sin 2
)
1
1
g'W =
+
sm
1
A
2
/
Ua
,'
3* 2
1
1
\Ja
1
you
if
we obtain
1
+
:
sin 2
Via
1
+
sin 2
/
*
J
1
1
1
+ sin
2
#
Like the simpler differentiations of Chapter 10, these manipulations should
become much easier after the practice provided by some of the problems, and,
problems of Chapter 10, these differentiations are simply a test of your
understanding of the Chain Rule, in the somewhat unfamiliar context provided by the Fundamental Theorem of Calculus.
The powerful uses to which the integral will be put in the following chapters
like the
all depend on the Fundamental Theorem of Calculus, yet the proof of that
theorem was quite easy it seems that all the real work went into the definition
of the integral. Actually, this is not quite true. In order to apply Theorem 1 to
a continuous function we need the one theorem whose proof has not yet been
given: If /is continuous on [a, b\ then /is integrable on [a, b\ The proof which
will be given depends upon a trick which is not very illuminating, but which
at least serves to fill this gap in our knowledge.
If / is any bounded function on [a> b], then
—
sup{L(/,P)}
will
both
exist,
even
integral of /on
will
if
[a, b]
and
inf
{
U(f, P)
}
/ is not integrable. These numbers are called the lower
and the upper integral of /on [a, b], respectively, and
be denoted by
L/;/
The lower and upper
integrals both
and
u/>.
have several properties which the integral
250
Derivatives and Integrals
<
possesses. In particular, if a
l /:/= l /:/+ l
and
if
m <
fix)
<
M
m{b
The
<
c
proofs of these facts are
precisely
We
-
M(Z>
a).
exercise, since they are quite similar to
an
left as
The
the corresponding proofs for integrals.
corollary of the results for
then
[a, b],
£/ < U £/ <
< L
a)
u /:/= u /:/+ u />
/>
for all x in
-
then
b,
results for integrals are actually
upper and lower integrals, because /
is
a
integrable
when
prove that a continuous function / is integrable by showing that this
equality always holds for continuous functions. It is actually easier to show
will
that
for all x in
[a, b];
the trick
most of the proof of Theorem
to note that
is
1
didn't
even depend on the fact that / was integrable
THEOREM
13-3
PROOF
If
/
is
continuous on
Define functions
b\ then /
[a,
U on
L and
integrable on
is
(a, b).
>
If h
b\
by
[a, b]
LOO = L J*f and
Let x be in
[a,
= U j* f.
U(x)
0 and
mh =
h =
sup
<
<
x
inf {/(*):
M
{/(*): x
<
<
t
t
x
x
+ h}>
+ h),
then
mh
< L
-h
x+h
Jx
U
f <
<
M
k
•
h,
so
mh
•
h
<
L(x
+
-
h)
< U(x
L(x)
+
h)
—
U(x)
<
M
<
M
h
'h
or
772 a
<
<
;
:
lih<0
h-
h
h
and
mh =
M
h
=
inf {/(f):
*
sup
x
{/(*):
+
+
h
h
<
<
t
t
<
<
*},
x\,
one obtains the same inequality, precisely as in the proof of Theorem
Since / is continuous at x we have
}
lim
A— 0
mh =
lim
A->0
M
h
—
/(#),
1
—
The Fundamental Theorem
and
this
of Calculus
251
proves that
L'(x)
This means that there
-
=
U'(x)
a
number
=
L(x)
is
U(x)
such that
c
+
for x in {a, b).
fix)
for all x in
c
6].
Since
=
U(a)
the
number
c
must equal
=
0,
0, so
U(x)
=
for all x in
L(x)
[a, b].
In particular,
U/ V=
1Kb)
U
and
this
means that /
is
=
integrable on
= L./ V,
a
L{b)
[a, b]. |
PROBLEMS
1.
Find the derivatives of each of the following functions
(i)
F(x)
=
sin 3
1
dt.
J*
(ii)
(iii)
dv)
(vi)
F(*)
F(*)
fw
F(x)
=
—+
^
f*
J 15 \Js
+
1
+
2
t
1
sin 2
/
J*
1
+
r
+
sm
Ja
1
+
r
+
sm 2
= sm
.
sm
2
(vii)
F
,
where F(x)
—
-
/
F
1
where F(x)
=
^
1
/(*)?
(Caution:
it
of/.
(Find (F' 1 )'^) in terms of
1
,
/
For each of the following /,
/
t
F-'(x).)
-
A.
VI
Vl -
Jo
2.
dt) dy.
7
/
s * n3
Ji
(viii)
t
£
(^J
l
2
,
&
1
r
+
shT 6
1
- [*(
=
1
1
if
F(x)
2
t
=
J* f,
might happen that F'(x)
continuous at
(i)
fix)
(ii)
/(*)
(iii)
f{x)
=
=
=
0
if
*
o
if
*
0
if
a:
at
<
<
*
1,
/(*)
1,
/(*)
1,
/(*)
=
=
=
1
if
x
1
if
x
1
if
*
>
>
=
1.
1.
1.
which points x
=
/(*),
even
is
F'(x)
if
/
is
=
not
252
Derivatives
and
Integrals
FIGURE
(vi)
=
fa) =
f(x) =
(vii)
/
(viii)
f(x)
Find
(f
(iv)
(
3.
6
/(*)
v)
0
if
a-
0
if
x
0
if
a
is
<
<
i
=
1
if
y(0)
if
(i)
/(a)
=
(ii)
/(*)
=
J*
^
x
+
1
=
0, /(*)
shown
1
0
*
if
in lowest terms.
a:
in Figure 6.
=
/n for some n in N, /(a)
sin (sin
sin (sin
=
=
0, /(*)
the function
is
= \/q if x = p/q
> 0.
1/[1 A] if * > 0.
irrational, /(*)
0 otherwise.
0
a?/.
(Don't try to evaluate / explicitly.)
=
(The answer
you should
4.
Find F'(*)
5.
perform an obvious manipulation on the integral before trying to find F
Prove that if / is continuous, then
if
F(*)
x
j0
**6.
Ku)(x-u)du = /; (/; /(o *)
x
*8.
is
mrt */(*);
f(u)( X
- uYdu =
2
2
/;(/;
f
.)
du.
Hint: Differentiate both sides, making use of Problem
Use Problem 5 to prove that
jo
7.
dt.
f*xf(t)
4.
(f /(o ^)^)^
2
+
sin 2 x. (This problem is
Find a function / such that /'"(*) = 1/Vl
"find.")
word
the
misinterpret
supposed to be easy; don't
= /(*) for all x.
}{x
if
a)
period
a,
with
periodic,
is
function
A
/
+
(a)
If /
is
periodic with period a
/o°/
=
+
-//
and integrable on
7 ^
all *.
[0, a],
show that
The Fundamental Theorem of Calculus
(b)
Find a function /such that /is not periodic, but /'
periodic g for which
is.
Hint: Choose a
=
can be guaranteed that f(x)
it
253
J*
g
not
is
periodic.
(c)
9.
Suppose that /' is periodic with period
if and only if f{a) = /(0).
Find
J*
Vxdx, by simply
a.
Prove that /
guessing a function / with /'(*)
using the Second Fundamental
Theorem
of Calculus.
is
=
periodic
V**,
and
Then check with
Problem 13-16.
10.
Use the Fundamental Theorem of Calculus and Problem 13-16
to
derive the result stated in Problem 12-15.
11.
12.
13.
(a)
Find the derivatives of F(x)
(b)
Now
give a
new proof
=
1/t dt
J*
Problem
for
and G(x) = j**
1/t
dt.
13-12.
Use the Fundamental Theorem of Calculus and Darboux's Theorem
(Problem 11-39) to give another proof of the Intermediate Value
Theorem.
Prove that if h is continuous, / and g are differentiable, and
=
then F'(x)
h(g(x))
-
m
=
g'(x)
-
hit) dt,
/;;?
h(f(x))
•
f(x). Hint:
Try
to reduce this
two cases you can already handle, with a constant either as the
lower or the upper limit of integration.
(a) Suppose G' = g and F' = /. Prove that if the function y satisfies the
to the
*14.
differential equation
(*)
g(y(*))
then there
(**)
(b)
(c)
is
*/W = /(*)
a
G(y(x))
number
=
c
+
F{x)
for all x in
some
interval,
such that
for all x in this interval.
c
Show, conversely, that ii y satisfies (**), then y
Find what condition y must satisfy if
y
(In this case g(t)
=
w
>(\
1
+
1
=
+
is
a solution of (*).
*2
l+y(x)
t
and
/(/)
=
1
+
2
t
.)
Then
"solve" the
resulting equations to find all possible solutions y (no solution will
have
as its domain),
R
(d)
Find what condition y must
/(*)
=
1
(An appeal
to
satisfy if
+
Problem 12-10
—
S[y(x)Y
will
satisfying the resulting equation.)
show that there
are functions
254
Derivatives
and
Integrals
(e)
Find
all
functions y satisfying
=
y(x)y'(x)
-x.
Find the solution^ satisfying y(0)
*15.
The
limit lim
denoted by
/, if it exists, is
/
= — 1.
-/°
called
x r dx>
Determine
(b)
Use Problem 13-12
f
What can you
Suppose that
0
(d)
*16.
<
<
g(x)
Explain
(b)
/(*) dx),
say about
/(*)
>
J*
0 for x
1/(1
+
1/x dx
/
>
fix) for all x
why
it
is
jj f
Explain
+
J_
why
f,
aa
1/x dx does not
show that
to
>
and
0,
then
exist.
Hint:
?
"
Jo /
and that
0
JJ g
exists.
Prove that
if
also exists.
x 2 ) dx exists. Hint: Split this integral
up
at
1
/is denned in the obvious way, as ^lim
But another kind of improper integral
(a)
/
Ja
< —1
if r
The improper integral j"^
way:
(or
an "improper integral."
(a)
(c)
/
/
Ja
J"^ f
is
denned
in a
nonobvious
provided these improper integrals both
+
1/(1
f.
exist.
x 2 ) dx exists.
J"^
Explain why jf^ x dx does not
exist.
(But notice that lim
j N
xdx
does exist.)
(c)
Prove that iff"
/exists, then lim
[* f
exists
and equals f"m
1
Show moreover,
equal
Can you
/.
J"^
(If you
facts?
that lim
you
can't,
f and lim
f***
f both
[
exist
f.
and
state a reasonable generalization of these
will
have a miserable time trying to do these
special cases!)
*17.
'
There is another kind of 'improper integral"
bounded, but the function is unbounded:
(a)
If a
>
a
0,
find lim [
1
/Vxdx. This
even though the function fix)
matter how we define /(0).
^ ^
r
— <
(b)
Find
(c)
Use Problem 13-12
Ar
if
1
to
r
<
limit
= 1/V*
is
is
in
which the interval
denoted by
P
not bounded on
is
1/Vxdx
[0, a],
9
no
0.
show that
x~ dx does not make
l
J*
sense,
even
as a limit.
(d)
Invent a reasonable definition of
for
-1 <
r
<
0.
ja
\x\
r
dx for a
<
0 and compute
it
255
The Fundamental Theorem of Calculus
(e)
Invent a reasonable definition of
and show that the
of two limits,
j
(1
+
(1
i
-
x)~
112
How
dx exist?
-1 <
X 2 )' 112 for
It is possible, finally, to
<
x
(1
—
x 2)
limits exists.
does
+
(1
1/2
dx,
Hint:
x)~ 112
a
as
Why
compare
sum
does
with
0?
combine the two
possible extensions of the notion
of the integral.
(a)
If /(*)
= \/Vx
for
0
<
x
<
1
and
f(x)
=
\/x 2 for x
>
find
1,
f(x) dx (after deciding what this should mean).
(b)
JJ
Show
—
1
<
that
r
<
JJ
x T dx
0 and
other case at
qo
;
r
never makes
< —
for
r
1
.
sense.
(Distinguish
the
In one case things go wrong at
= —
1
things go
wrong
cases
0, in
at both places.)
the
THE TRIGONOMETRIC FUNCTIONS
CHAPTER
The definitions of the functions sin and cos are considerably more subtle than
one might suspect. For this reason, this chapter begins with some informal and
intuitive definitions, which should not be scrutinized too carefully, as they
shall soon be replaced by the formal definitions which we really intend to use.
In elementary geometry an angle is simply the union of two half-lines with
a
common
figure
More
as pairs
initial
point (Figure
1).
1
useful for trigonometry are "directed angles,"
(/i, / 2 )
of half-lines with the
same
which
may be regarded
initial point, visualized as in
Figure
2.
ti
FIGURE
If for
h we always choose
the positive half of the horizontal axis, a directed
described completely by the second half-line (Figure 3).
Since each half-line intersects the unit circle precisely once, a directed angle
described, even more simply, by a point on the unit circle (Figure 4), that is,
angle
is
2
is
by a point
256
(x,
y)
with x 2
+y
2
=
1
;
257
The Trigonometric Functions
The
and cosine of a directed angle can now be defined
2
is determined by a point (*, y) with x
of the angle is defined as y, and the cosine as x.
sine
(Figure 5)
the sine
:
a directed angle
as follows
+y
=
2
1
Despite the aura of precision surrounding the previous paragraph, we are
not yet finished with the definitions of sin and cos. Indeed, we have barely
begun.
we
What we have
want to define
is
defined
the sine
is
and cos x
sin x
for
and cosine
each number
what
of a directed angle;
The
x.
usual procedure for
depends on associating an angle to every number. The oldest
"measure angles in degrees." An angle "all the way around" is
associated to 360, an angle "half-way around" is associated to 180, an angle
"a quarter way around" to 90, etc. (Figure 6). The angle associated, in this
doing
this
method
to
is
manner,
to the
number
x,
is
called "the angle of x degrees."
The angle
of
ambiguity is puralso an angle of
is
degrees
angle
of
90
posely extended further, so that an
which
we will denote
function,
define
now
a
360 + 90 degrees, etc. One can
0 degrees
by
is
the
same
as the angle of
360 degrees, and
this
sin°, as follows:
=
sm°(#)
sine of the angle of x degrees.
There are two difficulties with this approach. Although it may be clear
what we mean by an angle of 90 or 45 degrees, it is not quite clear what an
angle of V2 degrees is, for example. Even if this difficulty could be circumvented, it is unlikely that this system, depending as it does on the arbitrary
choice of 360, will lead to elegant results it would be sheer luck if the func0
tion sin had mathematically pleasing properties.
"Radian measure" appears to offer a remedy for both these defects. Given
any number x, choose a point P on the unit circle such that x is the length of
the arc of the circle beginning at (1, 0) and running counterclockwise to P
(Figure 7). The directed angle determined by P is called "the angle of x
radians." Since the length of the whole circle is 2tt, the angle of x radians and
—
the angle of
2ir
+
x radians are identical.
A
function sin r can
now be
defined
as follows:
=
sin r (Ar)
sine of the angle of x radians.
This same method can easily be adopted to define
have sin° 360 = sin r 2w, we can define
.
sin
x
=
,
sin
r
2irx
—— =
.
sin
r
shall
soon drop the superscript
r
in sin
only function which will interest us; before
we want
to
•
180
360
We
—ttx—
sin°; since
r
we
since sin
,
r
(and not
sin°)
is
the
do, a few words of warning are
advisable.
The
expressions sin° x
and
sin
r
x are sometimes written
sin x°
sin x radians,
but
this
notation
is
quite misleading; a
number
x
is
simply a number
—
it
does
*
258
Derivatives
and
:
Integrals
not carry a banner indicating that it is "in degrees" or "in radians." If the
the notation "sin x" is in doubt one usually asks:
meaning of
p
"Is x in degrees or radians?"
length
but what one means
is:
"Do you mean
Even
for
£
or
sin°'
'sin
r
'?"
mathematicians, addicted to precision, these remarks might be
it not for the fact that failure to take them into account will
dispensable, were
lead to incorrect answers to certain problems (an example
is
given in Problem
18).
FIGURE
Although the function sin r is the function which we wish to denote simply
by sin (and use exclusively henceforth), there is a difficulty involved even in
the definition of sin r Our proposed definition depends on the concept of
length of a curve. Although we could digress, and define length, at the present
time it is much easier to reformulate the definition in terms of areas, which we
can treat by means of the integral. (A treatment in terms of length is outlined
7
.
Problems 32 to 34.)
Suppose that x is the length of the arc of the unit circle from (1 0) to P; this
arc thus contains x/2tt of the total length 2w of the circumference of the unit
circle. Let S denote the "sector" shown in Figure 8; S is bounded by the unit
in
,
/
/
Yvlength
\
x
11
area
circle, the
horizontal axis,
S should be x/2t times the
thus S should have area
and the
half-line
x
•
With
and
circle
7T
=
x
—
2
now
—more
begins.
The
(This definition
is
x
<
through
the
sum
=
2
•
jl %
it
as the area of the unit
a semicircle (Figure
Vl -
9).
x 2 dx.
it
will
be necessary to
first
define sin x
and
cos x only
tt.)
The second
of the sector
definition identifies
not offered simply as an embellishment; to define the
trigonometric functions
<
first
precisely, as twice the area of
w
for 0
P which
these remarks as background, the rigorous definition of the functions
cos
DEFINITION
9
and P. The area of
which we expect to be t;
(0, 0)
We can therefore define cos x and sin x as the coordinates of the point
determines a sector of area x/2.
8
sin
FIGURE
circle,
^
2tt
FIGURE
through
area inside the unit
definition is meant
bounded by the unit
to describe, for
circle,
—1 <
x
<
the horizontal axis,
1,
the area A(x)
and the
half-line
y/l — x 2 ). If 0 < x < 1, this area can be expressed (Figure 10) as
of the area of a triangle and the area of a region under the unit circle
(x,
The Trigonometric Functions
-
This same formula happens to work for
(x,
VI -
x
2
<
x
0 also. In this case (Figure
term
11), the
Vl -
x
is
<
1
259
x2
and represents the area of the
negative,
X
triangle
which must be subtracted
from the term
FIGURE
Vl -
2
t
dt.
10
DEFINITION
Notice that
if
—1 <
Fundamental Theorem
=
A'(x)
x
<
-2x
-
FIGURE
-x
1
(1
-
V\ -
X2
+
2
2
2
1
- Vi -
X2
X 2)
- 2x
- Vi Vl - X
- 2x - 2(1 - x
2 Vl — x
2
1
11
differentiate at x and (using the
is
+ vT
2Vl
2
_
"
A
then
1,
of Calculus),
x2
2
2
2
)
2
-1
2
Vl -
x2
Notice also (Figure 12) that on the interval [-1,
1]
the function
A
decreases
from
1
a{-\) =
to A{\)
FIGURE
12
=
fact that
For 0
P =
0.
its
<
x
DEFINITION
If
0
<
x
<
+ ^ VT^
f
7T
2
t
dt
This follows directly from the definition of A, and also from the
derivative
<
ir
13).
7r,
is
we wish
(cos x, sin x)
x/2 (Figure
0
negative on
(
— 1,
1).
and sin x as the coordinates of a point
which determines a sector whose area is
to define cos x
on the unit
circle
In other words:
then cos x
is
the unique
A (cos
x)
number
—
-;
and
sin x
= Vl —
2
(cos x)
.
in
[
— 1,
1]
such that
260
Derivatives
and
Integrals
This definition actually requires a few words of justification. In order to
know
that there
ij
a
number^
continuous, and that
A
=
satisfying A(y)
x/2,
we
use the fact that
A
is
on the values 0 and w/2. This tacit appeal to the
Intermediate Value Theorem is crucial, if we want to make our preliminary
definition precise. Having made, and justified, our definition, we can now
takes
proceed quite rapidly.
THEOREM
1
If 0
<
x
<
7T,
then
cos'(*)
sin'(*)
PROOF
If
B —
2A, then the definition
in other words, cos
is
= — sin
= cos x.
A (cos
x)
—
x/2 can be written
B (cos
x)
=
x;
= -
have already computed that
1
Vt -
2
(cos x, sin x)
We
just the inverse of B.
A'(x)
P ~
x,
x2
from which we conclude that
= -
B\x)
Consequently,
cos'M = (B-yix)
FIGURE
13
1
B'{B-\x))
1
1
= = —
V
-
1
(cos x) 2
sin x.
Since
sin x
we
= Vl —
(cos #) 2 ,
also obtain
.
/ /
\
sin (x)
=
_
—2
1
2
cos x
•
.
Vl -
f
•
cos (x)
=^
(cos x) 2
cos x sin x
sin x
=
The information contained
in
cos
^. I
Theorem
1
can be used to sketch the graphs
The Trigonometric Functions
of sin
and
cos
on the
interval
cos'(x)
Since
[0, ir].
= —
=
<
sin x
the function cos decreases from cos 0
quently, cos y
0 for a unique y in
261
=
1
[0, w].
<
0
0,
x
<
x,
to cos t = — 1 (Figure 14). ConseTo find y, we note that the definition
of cos,
A (cos
=
x)
->
2
means that
SO
y
It
is
2
we can
Now we
2
dt.
t
so sin increases
The
Vi -
j*
? dt,
have
=
'
cos ^
I
[t/2,
=
dt
t*
also write
sin'W
on
~
£ Vl
easy to see that
Vi so
=
7t]
on
to sin
[0, ir/2]
=
7r
values of sin x
from
<
=
sin 0
<
X
<
7T,
w/2
=
1,
7T/2
0,
0 to sin
and then decreases
0 (Figure 15).
and
cos x for x not in
t] are most easily defined by a
[0,
two-step piecing together process:
(1)
If
7T
<
x
<
2tt,
then
sin
a*
COS X
— — sin(27r — x),
= cos(27r — x).
Figure 16 shows the graphs of sin and cos on
(2)
If x
=
2irk
+
x' for
[0, 2ir].
some integer k and some x
f
}
sin
x
cos
=
=
sin
[0,
2w], then
x\
sin
cos x\
Figure 17 shows the graphs of sin and cos,
Having extended the functions
in
and
now
defined on
cos to R,
all
of R.
we must now check
that
the basic properties of these functions continue to hold. In most cases this
easy.
For example,
it is
clear that the equation
sin 2 x
+
cos 2 x
—
1
is
—
262
—
Derivatives and Integrals
(a)
FIGURE
17
holds for
not hard to prove that
all x. It is also
sin' (a:)
cos' (a:)
x
if
is
not a multiple of
7r.
= cos x,
= — sin
For example,
sin x
= —
sin(27r
if ir
—
x,
<
x
<
2ir,
then
x),
so
sin'O)
If x
is
= - sin'(27r — C0S(27T — x)
= COS X.
x)
•
(-1)
a multiple of t we resort to a trick; it is only necessary to apply
11-7 to conclude that the same formulas are true in this case also.
Theorem
Jt
——
i
1
I
1
1
7T
1
TAP'
1-
1
3tT
IT
.
'
A;
\
Ay
/esc
1
t
i
\ cot
'
\
|
i
1
A
FIGURE
18
IT
'
i
(-
2x
Ai
ft
1!
The Trigonometric Functions
The
other standard trigonometric functions present no difficulty at
263
all.
We
define
sec x
cos x
sin x
+
kw
x
t±
X
9* k-K.
tt/2,
tan x
cos x
1
esc x
sin x
cos x
cot x
The graphs
are sketched in Figure 18. It
is
a good idea to convince yourself
that the general features of these graphs can be predicted from the derivatives
of these functions, which are listed in the next theorem (there is no need to
of the theorem, since the results
memorize the statement
whenever needed.)
THEOREM
2
If x ?± kir
+ 7r/2,
then
=
=
sec'(#)
tan' (^)
If x
^
kir,
sec x tan x,
sec 2
x.
then
= —
= —
csc'(Y)
cot'(Y)
PROOF
can be rederived
esc x cot x,
esc 2 x,
Left to you (a straightforward computation).
|
The inverses of the trigonometric functions are also easily differentiated.
The trigonometric functions are not one-one, so it is first necessary to restrict
them to suitable intervals; the largest possible length obtainable is t, and the
intervals usually chosen are (Figure 19)
[-tt/2, tt/2] for
sin,
for cos,
[0, 7r]
(-tt/2, tt/2) for tan.
(The inverses of the other trigonometric functions are so rarely used that they
not even be discussed here.)
will
The
inverse of the function
f{x)
=
sin x,
—tt/2
<
x
<
ir/2
denoted by arcsin (Figure 20); the domain of arcsin is [ — 1, 1]. The
-1
has been avoided because arcsin is not the inverse of sin
-1
has
(which is not one-one), but of the restricted function /; the notation sin
_1
the additional disadvantage that sin (*) might be construed as 1/sin x.
is
notation sin
The
inverse of the function
g(x)
—
COS
X,
0
<
X
<
7T
264
Derivatives
and
Integrals
denoted by arccos (Figure 21); the domain of arccos
is
The
is
[
— 1,
11.
inverse of the function
h(x)
=
tan
—w/2 <
x,
x
< t/2
denoted by arctan (Figure 22) arctan is one of the simplest examples of a
which is bounded even though it is one-one on all of R.
is
;
differentiable function
The
derivatives of the inverse trigonometric functions are surprisingly sim-
and do not involve trigonometric functions at all. Finding the derivatives
a simple matter, but to express them in a suitable form we will have to sim-
ple,
is
plify expressions like
cos(arcsin x),
sec (arctan x).
FIGURE
22
A
picture
little
is
the best
way
to
remember the
correct simplifications.
example, Figure 23 shows a directed angle whose sine
is
x
thus an angle of (arcsin x) radians; consequently cos(arcsin x)
theorem
FIGURE
3
If
will not resort to
-1 <
x
<
1,
.
then
23
arcsin '(x)
arccos
Moreover, for
all
x
f
(x)
1
—
Vi -
x2
we have
l
arctan' (x)
1
PROOF
is
is
the length of
x 2 However, in the proof of the next theorem
such pictures.
the other side, namely,
we
For
—the angle shown
+
x*
arcsin '(#)
i
i
sin' (arcsin x)
1
cos (arcsin x)
The Trigonometric Functions
265
Now
+
2
[sin(arcsin x)]
that
2
[cos(arcsin x)]
=
1,
is,
x2
+
=
1;
= Vl —
x2
[cos(arcsin x)] 2
therefore,
cos(arcsin x)
(The
positive square root
cos(arcsin x)
>
0.)
is
.
to be taken because arcsin x
This proves the
is
in
(— 7r/2,
ir/2), so
formula.
first
The second formula has already been established (in the proof of
Theorem 1). It is also possible to imitate the proof for the first formula, a
valuable exercise if that proof presented any difficulties. The third formula
is
proved
as follows.
arctan'M = (/T 1 )'^)
"
_J
h'(h~ l {x))
_
1
tan '(arc tan x)
1
-
sec 2 (arctan x)
Dividing both sides of the identity
a
+
cos 2 a
tan 2 a
+
1
sin
2
by cos a
2
=
1
yields
=
sec 2
a.
It follows that
[tan (arc tan x)]
2
+
1
=
sec 2 (arc tan x),
or
x2
+
1
which proves the third formula.
The
=
sec 2 (arctan
|
traditional proof of the formula
the one given here)
is
sin'(Ar)
=
cos x (quite different
outlined in Problem 26. This proof depends
from
upon
first
establishing the limit
sin h
—
hm—
h->o
=
1,
h
and the "addition formula"
sin(#
+ y) =
sin x cos y
+ cos x sin y.
Both of these formulas can be derived easily now that the derivative of sin and
cos are known. The first is just the special case sin'(O) = cos 0. The second
depends on a beautiful characterization of the functions sin and cos. In order
to derive this result we need a lemma whose proof involves a clever trick; a
more straightforward proof
will
be supplied in Part IV.
266
Derivatives
and
Integrals
LEMMA
Suppose / has a second derivative everywhere and that
f"+f=0,
Then / =
PROOF
AO) =
0,
AO) =
o.
0.
Multiplying both sides of the
equation by
first
=
/'/"+//'
/' yields
0.
Thus
+f
2
[(f)
so (/')'
+/
2
is
2
=
]'
From
a constant function.
that the constant
is
+ //')
2(/'/"
=
/(0)
=
0,
and
0
/'(0)
=
0
it
follows
0; thus
[/'MP
+ [/MP =
0
for all
This implies that
/M =
THEOREM
4
If
/ has a second
0
for all *. |
derivative everywhere
and
r+f-o,
/(0)
/'(0)
=
=
a,
A,
then
(In particular,
/'(0)
proof
=
0,
if
then /
/(0)
=
=
=
sin
+
a
cos.
0 and /'(0)
=
1,
then /
/
b
'
=
sin;
if
cos.)
Let
g(x )
~
/M
b sin
—
x
a cos
x.
Then
£'M = /'M — 6 cos * + a sin
£"M = /"(*) + * sin * + a cos *.
a:,
Consequently,
g'(0)
=
=
=
o,
x
—
a cos
g(0)
o,
o,
which shows that
0
THEOREM
5
If
=
g(x)
=
f(x)
—
b sin
at,
for all #. |
* and y are any two numbers, then
sin(#
cos(*
+ y)
+ y)
—
=
j>
+
cos x sin
>>,
cos ^
—
sin # sin
jy.
sin x cos
cos
a-
/(0)
=
1
and
The Trigonometric Functions
PROOF
267
For any particular number y we can define a function / by
=
/(*)
sin(x
+y).
Then
= cos(* + y),
= - sinO +y).
f(x)
/"(*)
Consequently,
r +/ =
/(0)
/'(0)
It follows
sin y,
cos y.
from Theorem 4 that
/
that
o,
=
=
=
(cos y)
-
+
sin
(sin y)
•
cos;
is,
sin (a:
Since any
+ y)
=
sin x
cos
+
for all
sin j cos
number^ could have been chosen
formula for all x and y.
The second formula is proved similarly.
to begin with, this proves the first
|
As a conclusion to this chapter, and as a prelude to Chapter 17, we will
mention an alternative approach to the definition of the function sin. Since
arcsin'C*)
it
follows from the
=
1
for
f
Vl -
—1 <
*
<
1,
x2
Second Fundamental Theorem of Calculus that
arcsin x
=
—
arcsin x
arcsin 0
=
/
Jo
This equation could have been taken as the
,
VI -
definition
- dt.
2
t
of arcsin. It
would follow
immediately that
arcsin (x)
=
.
VI-
t
2
-1
the function sin could then be defined as (arcsin)
derivative of an inverse function would show that
sin^*)
which could be defined
as cos
x.
= Vl —
and the formula
for the
sin 2 x,
Eventually, one could
show that
,4
(cos x)
=
end of the development the definition with which
While much of this presentation would proceed more rapidly, the
definition would be utterly unmotivated; the reasonableness of the definitions
would be known to the author, but not to the student, for whom it was
intended! Nevertheless, as we shall see in Chapter 17, an approach of this sort
x/2, recovering at the very
we
is
started.
sometimes very reasonable indeed.
PROBLEMS
Differentiate each of the following functions.
(0
/(*)
(ii)
f{x)
(in) f(x)
(iv) f(x)
=
=
=
arctan(arctan(arctan x)).
arcsin (arctan (arccos x)).
arctan (tan x arctan
1
=
arcsin
lim
(ii)
lim
(hi)
lim
-
x
+*
3
/6
sin x
-
*
+*
3
/6
cos
-
1
+x
2
/2
+x
2
/2
cos
lim
(v)
lim
(vi)
a:
a*
-
lim (1
*-»o V*
-
2
1
—
arctan x
—
at
,
1,
At
in
at
4.
+
3
a:
/3
J-Y
sin
(a)
a:
sin x/
I
(b)
by PHopitaPs Rule,
limits
sin x
X
(iv)
Y
Wl + W
(
Find the following
(i)
x).
x*0
a:
=
0.
Find /'(0).
Find /"(0).
this point, you will almost certainly have to use PHopitaPs Rule, but
Chapter 23 we will be able to find / (0) for all k, with almost no work
(fc)
all.
Graph
the following functions.
(a) f(x)
=
=
sin 2x.
sin(* 2 ).
(A pretty respectable sketch of this graph can be
obtained using only a picture of the graph of sin. Indeed, pure
thought is your only hope in this problem, because determining the
(b) f(x)
sign of the derivative f'(x)
=
cos(x 2 )
2x is no easier than determining the behavior of /directly. The formula for f'(x) does indicate
one important fact, however /'(0) = 0, which must be true since /
is even, and which should be clear in your graph.)
•
—
(c)
f(x)
=
sin x
+
sin 2*. (It will
the graphs of g(x)
=
sin
probably be instructive
x and h(x)
=
sin 2a: carefully
to first
draw
on the same
:
269
The Trigonometric Functions
and guess what the sum will look like. You
many
critical points / has on [0, 2ir] by concan easily find out how
sidering the derivative of/. You can then determine the nature of
from 0
set of axes,
to 2w,
by finding out the sign of /
these critical points
at
each point; your
sketch will probably suggest the answer.)
(d) f(x)
=
tan x
—
determine the behavior of /in (— tt/2, tt/2);
x. (First
(kw
in the intervals
—
tt/2, kir
(e)
f{x)
=
sin x
—
(The material
x.
+
the graph of
7r/2)
moved up
exactly the same, except
a certain amount.
in the
Appendix
to
/
will look
Why?)
Chapter
11 will
this function.)
be particularly helpful for
sin x
I
=
x
1,
0.
determine approximately where the
are located. Notice that / is even and continuous at 0; also
you
(Part (d) should enable
zeros of /'
to
consider the size of / for large
5.
Prove the addition formula
6.
(a)
From
for cos.
and
the addition formula for sin
cos 2x, sin 3x,
(b)
x.)
and
cos derive formulas for sin 2x,
cos 3*.
to find the following values of the trigonometric
Use these formulas
by geometric arguments
functions (usually deduced
in
elementary
trigonometry)
7T
sm - =
.
7T
cos -
4
7T
tan 4
=
-
_
sin
'
4
2
1,
1
2
6
7T
cos 6
7.
(a)
Show
A
that
able a and
proof.
You
sin (a:
+
+
b cos x for suitB) can be written as a sin x
in this chapter provides a one-line
(One of the theorems
should also be able to figure out what a and b are.)
b, find numbers A and B such that a sin x
(b)
Conversely, given a and
(c)
Use part
(a)
Prove that
+
8.
b.
'
2
b cos
= A
x
(b) to
sin(*
+
graph
,
tan(*
B)
f{x)
for all x.
=
tan x
+ y) =
x
1
+
V3
—
sin x
+
cos
x.
+ tan v
tan x tan y
x, y, and x
y are not of the form
addition formulas for sin and cos.)
provided that
kir
+ tt/2.
(Use the
.
Derivatives
and
Integrals
(b)
Prove that
+
arc tan x
arctan ( *
=
arc tan y
\1
^
—
Y
ry/
indicating any necessary restrictions on # andjy. Hint: Replace #
9.
arctan # and y by arctan
Prove that
arcsin
+
a
=
arcsin
by
in part (a)
jy
—
arcsin (aV'l
2
/3
+
/3
\/l
—a
2
),
a and /5.
any numbers, then
indicating any restrictions on
10.
Prove that
if
m and
mx
sin
w are
sin nx
sin rax cos nx
cos
11.
Prove that
cos
if
rax
m and
f*
/-*
/
—
^[cos (m
n)x
12.
cos (m
+ sin
+
w)#
+ cos
+
i[cos (m
sin
mx
sin nx dx
—
(m
(m
0,3
m
m =
m
TT,
772
^'
|
[TT,
cos
mx
cos w#
,
=
(
{
(
J*
—
n)x
^-[sin
+
—
—
n)x],
n)x],
»)*].
n are natural numbers, then
/*
•/-*
—
—
=
sin
m#
=
cos
—
n
n,
n
n,
0.
These relations are particularly important in the theory of Fourier
series. Although this topic will receive serious attention only in the Suggested Reading, the next problem provides a hint as to their importance.
(a) If / is integrable on [ — ir, x], show that the minimum value of
Jl w
occurs
(/W "
a cos nx ^ 2 ^*
when
a
=
- \* fix) cos nx
and the minimum value
dx,
of
when
a
=
1
7T
r*
/
fix) sin n#
./
(In each case, bring a outside the integral sign, obtaining a quadratic expression in a,)
(b)
Define
an
= —
bn
=
f' f{x) cos nx dx,
- [* fix)
sin n# i*,
n
=
0, 1, 2,
«
=
1, 2, 3,
271
The Trigonometric Functions
Show
that
and
if Ci
are any numbers, then
di
N
-
J
^+ ^
n—
+
C0S nX
Cn
dn
Sin
</*
fMf
1
N
N
= /'
[/(*)]*
-
dx
2t
Va
+
ftp
+ M.) + * (f +
B <„
n=l
£
+
2
<„
</»
2
)
n=l
AT
=
£
-*
[/(*)]« dx
+
£
n=
+
an *
*„
2
)
1
2
+ ((7-2-^) +
n=
thus showing that the
bi
=
functions
=
s n (x)
integral
first
d^ In other words,
among
"-
a"
)2+( ^-
W2)'
1
smallest
is
when
cn
=
f<
and
"linear combinations" of the
all
c n (x)
-
cos nx for
a n cos nx
+
b n sin
and
sin nx
(c
t
ir
1
<
n
<
N, the
particular function
N
=
g( x)
13.
14.
~+
^
nx
has the "closest fit" to / on [— 7r, 7r].
Find a formula for sin * + sin y and for cos x + cos > Hint: First find a
formula for sin {a + b) + sin (a - 6). What good does that do?
2
(a) Starting from the formula for cos 2x, derive formulas for sin x and
cos 2 x in terms of cos 2x.
(b)
Prove that
+
\
2^2
x
-
<
x
for 0
(c)
15.
cos *
/l
cos
v
<
Use part
16.
If
x
=
=
x
1
=
\
O
and
sin 2 x
and cos(arctan
x)
trigonometric functions. Hint: y
sinjy/cosjy
.
Sm
tt/2.
(a) to find
Find sin(arctan
.
and
o
sin^y/v'l
—
sin 2
tan u/2, express sin u
=
expressions
as
x)
cos 2 x
ja
not
arctan x means that x
involving
=
tan ^
=
}'.
and
cos u in terms of
x.
(Use Problem 15;
the answers should be very simple expressions.)
17.
(a)
Prove that sin(x
+
the graphs of sin
(b)
18.
W
(a)
What
is
arcsin(cos x)
ri
Find
1
+
dt.
2
t
=
cos
cos as
and
\
/
Jo
tt/2)
and
Hint:
if
x.
along we have been drawing
were the case.)
(All
this
arccos(sin x)?
The answer
is
not 45.
272
Derivatives and Integrals
Find
(b)
Jo
19.
Find lim x
X—*
20.
cos°(*)
same
functions
=
sin
0
and cos 0 by sin°(» =
0
(sin )' and (cos°)'
cos(7r*/180). Find
and
sin(7r*/180)
in terms of these
functions.
Find lim
(b)
—
X
00
Define
(a)
sin
dt.
+
1
Sm
-
and lim x
X
x-+0
sin° -•
x
x->*>
21.
Prove that every point on the unit circle is of the form (cos
at least one (and hence for infinitely many) numbers 6.
22.
(a)
0, sin 6)
for
is the maximum possible length of an interval on which
one-one, and that such an interval must be of the form
(2kir - t/2, 2kw
tt/2).
Prove that w
sin
is
+
Suppose we
(b)
or
is
23.
Let f(x)
1
=
=
let g(x)
sin x for x in (2/for
—
t/2, 2kir
+ tt/2). What
)'?
<
sec x for 0
<
x
t.
_1
Find the domain of /
and sketch
its
graph.
24.
Prove that |sin x — sin jy| < |^ —
for all numbers x and y. Hint: The
same statement, with < replaced by < is a very straightforward consequence of a well-known theorem; simple supplementary considerations
then allow < to be improved to <.
It is an excellent test of intuition to predict the value of
,
'25.
lim
/>)
X*
dx.
Continuous functions should be most accessible to intuition, but once
you get the right idea for a proof the limit can easily be established for
any integrable /.
(a)
Show
(b)
Show
d
lim (
that
=
\x dx
sin
0,
computing
by
the
integral
explicitly.
that
if s is
a step function on
b
13-17), then lim [ s(x) sin \x dx
a
[a, b]
=
(terminology from Problem
0.
\_>oo J
(c)
Finally, use
for
This
B=
0
FIGURE
24
(1,
0)
26.
any function / which
result, like
Fourier
Problem 13-17
series; it
Problem
is
known
1 1
,
is
b
to
show that lim f
X->oo J a
integrable on
f(x) sin \x dx
=
0
[a, b].
plays an important role in the theory of
as the
Riemann-Lebesgue Lemma.
This problem outlines the classical approach to the trigonometric functions.
The shaded
sector in Figure 24 has area x/2.
.
273
The Trigonometric Functions
(a)
By
and
OCB
prove that
if
0
<
x
<
then
xx
< <
sin
sin x
2
2
(b)
OAB
considering the triangles
7r/4,
2 cos *
Conclude that
sin x
<
cos x
<
1,
x
and prove that
hm
sin x
(c)
Use
—
1
Using parts
(b)
sin'(Y), starting
cos x
x
x->0
*27.
1
this limit to find
hm
(d)
=
X
x->0
and (c), and the addition formula
from the definition of the derivative.
for sin, find
Yet another development of the trigonometric functions was briefly
mentioned in the text starting with inverse functions defined by inte-
—
grals. It
for all x.
is
this function is
convenient to begin with arctan, since
To do
defined
problem, pretend that you have never heard of the
this
trigonometric functions.
(a)
Let a{x)
=
f*
+
(1
2
t
)" 1
Prove that a
dt.
and that lim a(x) and lim a{x) both
X—
each other.
X
oo
If
we
is
exist,
odd and increasing,
and are negatives of
— »~
define
=
7r
2 lim a(x) } then
aT
1
is
defined on
(-x/2,t/2).
(b)
Show
(c)
For x
(cO'M =
that
1
+
l
[<x~ (x)]*.
^ 7r/2 or — 7r/2, define tan x = aT^*')Then define cos x = /VI + tan x, for x not of the form kir + x/2
or kir
7r/2, and cos {kir ± tt/2) = 0. Prove first that cos'(^) =
— tan x cos x, and then that cos"(.x:) = — cos x for all x.
=
kir
+
x'
with
x'
2
1
*28.
If
we
tions,
are willing to assume that certain differential equations have solu-
another approach to the trigonometric functions
pose, in particular, that there
and which
(a)
satisfies
Prove thatjy 0 2
ory(O) ^
(b)
yo"
+
+y =
2
is
possible.
is
Sup-
not always 0
0.
0
(joO
is
some function y 0 which
constant,
and conclude that
either ;> 0 (0)
^
0
+s—
+ byo'.
0 and s(0)
=
0
0.
Prove that there
and /(O) =
is
1.
is
a function
Hint:
Try
s
s satisfying
s"
of the form ay 0
:
274
Derivatives
and Integrals
If
we
define sin
=
=
and cos
s
trigonometric functions become
/, then almost all facts about
one point which
7r. This is most
done using an exercise from the Appendix to Chapter 1 1
requires work, however
easily
There
trivial.
—producing
the
is
number
(d)
Use Problem 6 of the Appendix to Chapter 1 1 to prove that cos x
cannot be positive for all x > 0. It follows that there is a smallest
#o > 0 with cos # 0 = 0, and we can define t = 2x 0
Prove that sin ir/2 = 1. (Since sin 2
cos 2 = 1, we have sin x/2 =
(e)
Find cos
(c)
.
+
±1; the problem
7r,
why
to decide
is
and
sin w, cos 27r,
sin 7r/2
is
positive.)
sin 2ir. (Naturally
sin'
(f)
29.
(a)
=
and
cos
= —
cos'
Prove that cos and
After
all
the
use any
we know
that
sin.)
sin are periodic
work involved
concerting to find that sin
(b)
may
you
addition formulas, since these can be derived once
with period
2ir.
in the definition of sin,
is
would be
it
dis-
actually a rational function. Prove that
it isn't. (There is a simple property of sin which a rational function
cannot possible have.)
Prove that sin isn't even defined implicitly by an algebraic equation;
that
is,
there do not exist rational functions /0 ,
(sin x)
n
+ /„_!(*) (sin x)
n~ l
+
•
•
.
.
+ /„(*) =
•
.
,
0
fn -i such that
for all x.
Hint: Prove that f0 = 0, so that sin x can be factored out. The
remaining factor is 0 except perhaps at multiples of 2ir. But this
implies that it is 0 for all x. (Why?) You are now set up for a proof
by induction.
*30.
Suppose that 0i and 0 2
satisfy
+ £i0i
0i"
and that g 2 >
(a)
Show
Show
0,
gi.
that
0l"02
(b)
=
that
if
—
0i(#)
02"01
>
(g2
—
gl)0102
and 0 2 (*) > 0
0
,,
I*
—
02
[0i
- 0
2
"0i]
=
0.
for all x in (a, b),
>
then
0,
and conclude that
[0i'W0 2
(c)
Show
W
-
0i'(a)0 2 (<2 )][0 1 (£)0 2 '(£)
(d)
Show
0i
>
we cannot have
that in this case
Consider the sign of
<f>i(a)
and
that the equations <pi(a)
0,
0 <
2
0 or 0!
(You should be able
-
<
to
0,
do
02
this
0i(*)0 2 'M]
0i(#)
=
>
0i(£)
o.
=
0.
Hint:
<t>i(b).
=
>
0i(£)
0,
=
or 0i
0 are also impossible
<
0,
0 <
2
0 on
with almost no extra work.)
if
(a, b).
275
The Trigonometric Functions
The net result of this problem may be stated as follows:
consecutive zeros of <f>i, then
</> 2
if a and b are
must have a zero somewhere between
and b. This result, in a slightly more general form, is known as the
Sturm Comparison Theorem. As a particular example, any solution
a
of the differential equation
+
y"
must have zeros on the
+
(x
-
\)y
0
positive horizontal axis
which are within w of
each other.
31.
Prove that
+
-
cos x
^
x/2
if sin
+
0,
cos 2x
then
+
•
•
+
•
sin(n
cos nx
+ ^)x
2 sin
At
A
this stage of the
game,
problem
this
is
:
a mathematical
idiot's delight.
proof by induction depends on establishing the formula
—
sin(n
+
£)*
.
f-
(
cos («
+
,
1)#
sinO
=
+
£)*
2 sin -
2 sin 2
2
This will require repeated use of addition formulas, in just the right
A more reasonable derivation of the
formula will occur in Problem 26-13. Like two other results in this probplaces, but very little thinking.
lem set,
and we
this
equation
also
make
is
very important in the study of Fourier
use of
it
in
series,
Problems 18-28 and 22-17.
The remaining problems
in this chapter form a brief introduction to the
and show how trigonometric functions can be defined in
topic of arc length,
terms of arc length.
*32.
Let /be a continuous function on
tion of
b],
[<z,
[a,
b\
If
P =
{^o,
.
•
.
?
is
tn ]
a parti-
define
n
i
The number
=l
t(f P) represents the length of a polygonal curve inscribed
9
graph of / (see Figure 25). We define the length of / on [a, b] to
be the least upper bound of all t(f, P) for all partitions P (provided that
in the
the set of
(a)
If /
is
all
such
If
/
is
[a, b]
P)
is
bounded above).
a linear function on
tance from
(b)
-l(f,
(a,
[a, b],
prove that the length of / is the dis-
f(a)) to (b, f(b)).
not linear, prove that there
such that /(/,
\bj\b)). (You will
is
a partition
P =
P) is greater than the distance from
need Problem 4-8.)
{a,
(a,
t9
b]
of
f(a)) to
Derivatives and Integrals
1
1
1
1
1
:
a
FIGURE
=
t0
ti
t2
(Or, in
straight
Suppose that
(a)
b
=
and f(b) = d,
than the length of any other.
conventional but hopelessly muddled terminology: "A
line is the shortest distance between two points. 55
)
all
functions /on [a y b] with f(a)
the length of the linear function
*33.
=
25
Conclude that of
(c)
t4
t3
/' is
bounded on
c
is less
[a, b].
UP
any
is
partition of
[a, b]
show that
L(Vi
+
cm p)
Why
Now
(c)
sup {L(Vl
is
show that sup
integrable on
any two
[a, b].
*34.
Let
t{f,
P)?
/(at)
= Vl -
/on
[x,
[a, b] is
{/(/,
{U( V\
P)}? (This
+
Vl +
f*
P ) < ^7(Vl
r
both P' and P",
x 2 for
p).
(fy,
-1 <
x
<
1.
how
+
is
easy.)
(f)\ P)\, thereby
(f)\
(f)
Vl +
if
(/')«
P and P" are
P ). If P conf
if
;/
2
,
does /(/, P')
compare
to
Define £(x) to be the length of
1].
(a)
Show
(b)
Show
(This
that
is
actually
Define
w
as
and define
sin'(*)
an improper
integral, as defined in
Problem 14-17.)
that
£'(*)
(c)
inf
+
Hint: It suffices to show that
partitions, then
tains the points of
<
{/(/, P)\
proving that the length off on
is
< u(Vi
t{L P)
Mean Value Theorem.
+ {f)\ p)\ < sup
Hint: Use the
(b)
<
=
=
1
for
r
VI -
£( — 1). For 0 < x <
= Vl — cos
for 0 < x < ir.
sin x
cos x
2
x.
-1 <
*
<
1.
t, define cos *
Prove that
by
«£(cos x)
cos'W =
=
x,
—sin x and
"CHAPTER
I
TT
IRRATIONAL
IS
This short chapter, diverging from the main stream of the book,
demonstrate that we
matics. This entire chapter
tional.
Like
many steps
many
in
4
is
is
included to
do some sophisticated mathe-
are already in a position to
devoted to an elementary proof that ir is irraproofs of deep theorems, the motivation for
'elementary'
5
our proof cannot be supplied; nevertheless,
it is still
quite possi-
ble to follow the proof step-by-step.
Two
observations must be
made
The
before the proof.
first
concerns the
function
/»(*)
- xY
*»(!
=
'
.
nl
which clearly
satisfies
0<
fn (x) <
—
<
for 0
x
<
1.
nl
An important property of the function fn is revealed by considering the expresby actually multiplying out x n {\ —
appearing will be n and the highest power will be
in the form
sion obtained
x)
n
.
2n.
The lowest power of x
Thus fn can be written
2n
u{x)
=
hX
i
where the numbers a are
integers. It
/n <*>(0)
=0
if
is
k
CiXi '
=n
clear
<
from
n or k
>
this expression that
2n.
Moreover,
fn
(n)
(x)
=
— [n!
cn
+
terms involving
x]
nl
fn
(n+1)
(x)
=—[(« +
1)! c n +i
nl
A
277
(2W)
W
=~[(2n)\c 2n l
nl
+
terms involving
x]
278
Derivatives
and
Integrals
This means that
/n
/n
/«<
=
-
(n)
(0)
(W+1)
(0)
=
2n)
(0)
C n)
+
(»
l), n+1 ,
-
(2»)(2«
where the numbers on the right are
(fc)
/n
The
(0) is
«...-(«
1)
all integers.
an
+
1)* 2 «,
Thus
integer for all £,
relation
=
/»(*)
/n(l
"
*)
implies that
A W
= (-i)Yn
tt)
(ft)
-*);
(i
therefore,
/n
The
proof that
number, and
s
>
x
(fc)
an
(l) is also
integer for all k.
irrational requires
is
one further observation:
we will have
if
a
is
then for sufficiently large n
0,
n
a
n\
To prove
this,
notice that
>
n
if
then
2a,
a n+l
(n
Now
let w 0
+
a
+
n
1)!
a
1
may
(n 0
(n 0 +l)
+1)!
2
(no+2)
(n 0
(n 0
+
+
a
is
a
I
2
>
2a.
n
r!
Then, whatever value
I
<
2)!
<
k)\
n0
(n 0 )!
(n.+ i)
J
fl
A;
•
have, the succeeding values satisfy
a
If
-
—
a
1
<
n!
be any natural number with n 0
(no)
n
fl
-
*
2
(» 0
1
a
~7
2*
n°
(n 0 )
110
so large that
(«o)!e
<
2*,
—TTT
+ 1)!
then
<
!
J
2
2
a
n0
(n 0 )!
any
0
x
(n 0
for the
THEOREM
l
is
the desired result.
one theorem in
The number x
Having made
PROOF
we
these observations,
are ready
this chapter.
x2
irrational. (Notice that the irration-
is
ality of x 2 implies the irrationality of x, for if 7r
would
279
+ k)\
irrational; in fact,
is
Irrational
(no+k)
a
which
is
were
rational, then
7r
2
certainly
be.)
Suppose x 2 were
rational, so that
0
T2 =
b
for
some
positive integers a
g(x)
(i)
=
-
*"[*•'»/»(*)
and
b.
Let
T 2n 2/«"to
r«»-yw w(*)
+
+
(-l)7n (2n) (*)].
Notice that each of the factors
n
b TT
is
an
integer. Since
2n ~ 2k
fn
(k)
=
(0)
b
n
2
(TT )
n -k
i-Jfe
=
£
n
^
and fn {k) (^) are
integers, this
shows that
G(0) and G(l) are integers.
Differentiating
G"(*)
(2)
The
last
G
-
n
n
b [T* fn "(X )
- T 2n ~ fn ^(x) +
2
term, (-l) n/n (2n+2) (*),
(3)
Now
twice yields
G"(x)
+
H(x)
—
ir
2
is
G(x)
b
n
•
+
•
Thus, adding
zero.
=
•
ir
(-l) n/n (2n+2) (*)].
(1)
and
(2) gives
= xV/n (*).
2n+2
fn {x)
let
—
G'(x) sin irx
xG(x) cos
irx.
Then
#'(*)
= xG'O) cos x# + G"0) sin
= \G"{x) + x 2 G(*)] sin irx
= ir a nfn (x) sin xx, by (3).
irx
-
xG'O)
cos irx
+
ir
2
G(x) sin xx
2
By
the Second
Fundamental Theorem of Calculus,
x 2 f*a nfn (x)simrxdx =
=
-
-
H(0)
G'{\) sinx
x[G(l)
+
- xG(l) cosx -
G'(0) sin 0
G(0)].
Thus
x
JQ
a n fn (x) sin
xx
</x
is
an
integer.
+ xG(0) cos
—
On
the other hand, 0
0
< fn (x) <
< Ta n fn (x)
l/n\ for 0
<
sin irx
<
x
<
so
1,
n
ita
for 0
<
x
<
1.
nl
Consequently,
0
<
7r
a
f
Jo
n
fn (x)
sin
wx dx
<
nl
This reasoning was completely independent of the value of n.
enough, then
0
<
7r
a
/
Jo
n
fn (x)
sin irx dx
<
—-
<
Now
if
n
is
large
1.
n\
But this is absurd, because the integral is an integer, and there is no integer
between 0 and 1. Thus our original assumption must have been incorrect:
7r
2
irrational. |
is
This proof
way
is
admittedly mysterious; perhaps most mysterious of
—
we have proved
all is
the
irrational
it almost looks as if
without ever mentioning a definition of it. A close reexamination of the proof
will show that precisely one property of x is essential
that
7r
enters the proof
sin(7r)
The proof really depends on
=
0.
the properties of the function
number x with sin x
properties of sin are required, namely,
irrationality of the smallest positive
sin'
cos'
sin(0)
cos(0)
Even
it
=
sin,
0.
and proves the
In
fact,
very few
= cos,
= — sin,
= 0,
= 1.
could be shortened; as far as the proof is concerned, cos might
be defined as sin'. The properties of sin required in the proof may
this list
just as well
then be written
sin"
+
sin
sin(0)
sin'(0)
=
=
=
0,
0,
1.
this is not really very surprising at all, since, as we have seen in the
previous chapter, these properties characterize the function sin completely.
Of course,
PROBLEMS
1.
(a)
Prove that the areas of triangles
by the equation
OAB and OAC in Figure
1
are related
7r is
OAC =
area
1
-
- Vl -
jl
=
Hint: Solve the equations *y
(b)
Pm be the regular polygon of m
A m is the area of Pm show that
+/ =
1,
for y.
sides inscribed in the unit circle.
V2 - 2 Vl -
= ~
Ai m
281
OAS) 2
2(area OAS), x 2
Let
If
16 (area
Irrational
(2^ m /m) 2
.
This result allows one to obtain (more and more complicated) expressions for A 2 % starting with A 4 = 2, and thus to compute t as accurately as desired (according to
ods will appear in Chapter 19, a slight
a very interesting expression for
(a)
Using the
Although better methvariant of this approach yields
Problem
8-11).
7r:
fact that
area(0A8)
= OB,
area((MC)
show that
(b)
Show
if
am
is
the distance from
0
to
one side of
Pm
,
then
that
2
A2
(c)
Using the
«4
'
'
OL S
«
fact that
am =
cos
1
and the formula cos x
etc.
t
+
7T
—
m
COS X
(Problem 15-14), prove that
Together with part
"infinite product"
to
be precise,
(b), this
shows that
2/tt
can be written as an
this equation means that the product of the first
n faccan be made as close to 2/tt as desired, by choosing n sufficiently
large. This product was discovered by Francois Viete in
1579, and is
only one of many fascinating expressions for ir, some of which are
mentioned later.
tors
THE LOGARITHM AND
EXPONENTIAL FUNCTIONS
CHAPTER
for a preliminary
In Chapter 15 the integral provided a rigorous formulation
integral plays a more
definition of the functions sin and cos. In this chapter the
definition presents
essential role. For certain functions even a preliminary
difficulties.
For example, consider the function
=
/(*)
assumed to be defined for all x and to have an inverse function,
10,"
positive x, which is the "logarithm to the base
This function
defined for
10-.
is
f
l
=
(x)
logio x.
definition for
usually defined only for rational x, while the
rational x
for
definition
the
of
review
brief
irrational x is quietly ignored.
principle
important
an
recall
also
but
will not only explain this omission,
In algebra, 10*
is
A
x
behind the definition of 10
turns
The symbol 10" is first defined for natural numbers n. This notation
numbers,
large
out to be extremely convenient, especially for multiplying very
.
because
x
by the desire to
extension of the definition of 10 to rational x is motivated
the customary
us
upon
preserve this equation; this requirement actually forces
The
the equation
we want
definition. Since
10°
to
be
true,
we must
=
define 10°
lCT n
to
be true, we must define lO"
10
1/n
.
.
71
10
•
•
.
=
10 n
•
1
•
10
n
1/n
•
.
m
to
.
10°
=
;
=
io 1 /
the equation
we want
1
we want
n +"'* +1/n
=
10 1
the equation
=
10
n times
1/n
be true, we must define 10
io
=
10 n
=
l/10 n since
n times
to
0+n
since
;
=
1/n
10
•
.
io
=
1/n
vT<); and since
we want
"- +1/n
m/n
=
n+
=
10
m t imes
times
be true, we must define
io 1 /
the equation
10 mln
= (V 10) m
.
halt. We have
Unfortunately, at this point the program comes to a dead
to ensure that
as
so
defined
been guided by the principle that 10* should be
algebraic way
simple
any
suggest
10*+v = 10*10*; but this principle does not
of defining 10* for irrational
283
x.
For
this
reason
we
will try
some more
sophisti-
284
Derivatives
and
Integrals
cated ways of finding a function
(*)
/ such
+ y) - /(*)
/(*
that
for all x
/GO
'
and
y.
Of course, we are interested in a function which is not always zero, so we might
add the condition /(l) ^ 0. If we add the more specific condition /(l) = 10,
then
(*) will
=
imply that /(*)
10* for rational *,
and 10 x could be
defined as
f{x) for other x; in general /(*) will equal [/(l)] x for rational x.
One way to
more
difficult
find such a function
problem: find a
f(*
suggested
is
+ y) = /(*)
we
for all
/GO
'
=
/(I)
if
try to solve
an apparently
function / such that
differentiable
x and
jy,
io.
—
Assuming that such a function exists, we can try to find/7 knowing the derivative of / might provide a clue to the definition of / itself. Now
h
h~+o
=
lim
h
h-*o
= fix) hm
-
n
h-+o
The answer
thus depends
on
/(0)
for the
moment assume
Even
a
= hm
;
and denote
this limit exists,
fix) =<x-fix)
if
tive of /
If
in a
The
x
we examine
new light:
a.
Then
for all x.
the inverse function/
derivative of f~ l
1
by
could be computed, this approach seems self-defeating.
has been expressed in terms of / again.
is
dx
is
—
1
logio, the
1
1
vfiTKx))
"x
the only one which
/
x
n
1
f
J
x
l
~
dt
=
-
is
even
dx examined previously, the integral
we cannot
logio x
deriva-
whole situation appears
evaluate. Since logio
should have
-
The
about as simple as one could ask! And, what
more interesting, of all the integrals
/
it
logio
1
=
logio x.
1
=
0
we
285
The Logarithm and Exponential Functions
This suggests that
a
is
we
unknown. One way of evading
this difficulty
kg*and hope
t~ l
define logi 0 x as (\/a)
is
The
dt.
difficulty
is
that
to define
-dt,
I
that this integral will be the logarithm to some base, which might be
later. In any case, the function defined in this way is surely more
determined
reasonable, from a mathematical point of view, than logi 0
on the important
of logio depends
role of the
number
.
The
usefulness
10 in arabic notation
(and thus ultimately on the fact that we have ten fingers), while the function
log provides a notation for an extremely simple integral which cannot be
evaluated in terms of any functions already
DEFINITION
>
If x
known
to us.
then
0,
dt.
The graph
and
<
0
if
x
of log
<
1,
is
shown
in Figure
<
then log x
x
(
<
0,
a
number log
not bounded on
l
If x,
y
>
Notice
Now
>
1,
then log x
>
0,
1/t
is
l
{
Jx
t
-dt<o.
t
x cannot be defined in this way, because f(t)
=
log
=
,v
0,
for the notation "log"
first
comes from the following theorem.
then
log(*y)
PROOF
x
1
The justification
theorem
if
[x, 1].
area
FIGURE
Notice that
since, by our conventions,
-dt= -
Jl
For x
0,
1.
that log'(*)
choose a
=
log x
0
=
+
log y.
by the Fundamental Theorem
and let
1/x,
number y >
/(*)
=
log(*y).
of Calculus.
286
Derivatives and Integrals
Then
=
/'(*)
—
1
= -
1
=
log'foO -y
•
y
x
xy
Thus
/'
=
log'.
This means that there
—
f(x)
that
+
log x
number
a
is
for all x
c
such that
c
>
0,
is,
=
log(ry)
The number
c
+c
log #
for all *
>
0.
—
can be evaluated by noting that when x
=
—
log(l -y)
log
1
+
1
we obtain
c
c.
Thus
=
\og(xy)
Since this
COROLLARY
1
If n
is
is
true for
all
y
>
COROLLARY
2
a natural number and x
>
0,
then
n
=
)
is
log
proved. |
x.
Left to you (use induction). |
If x y y
>
0,
then
log0^ =
PROOF
for all x.
the theorem
0,
log(#
PROOF
+ log
log x
log x
-
logy.
This follows from the equations
log x
Theorem 1
The function
=
log
=
m
(^ yj
lo g
+
lo S J- 1
provides some important information about the graph of log.
log
is
=
clearly increasing, but since \og'(x)
1/x, the derivative
becomes large, and log consequently grows more and
not immediately clear whether log is bounded or unbounded
becomes very small
as x
more slowly. It is
on R. Observe, however, that
log(2
it
follows that log
is,
log
therefore log
takes
on
is
w
)
=
for a natural
log
Therefore
R
important function has a special
come
clear.
bounded above.
1
-
not bounded below on
all values.
(and log 2
n log 2
in fact, not
=
number
log 2 n
(0, 1).
= -n
>
n,
0);
Similarly,
log 2;
Since log
is
continuous,
it
actually
-1
This
domain of the function log
name, whose appropriateness will soon beis
the
.
The Logarithm and Exponential Functions
DEFINITION
The "exponential
function," exp,
287
defined as log
is
The graph of exp is shown in Figure 2. Since log x is defined only for x >
we always have exp(x) > 0. The derivative of the function exp is easy
0,
to
determine.
THEOREM
2
For
all
numbers
x,
=
exp'(x)
PROOF
exp'(*)
exp(#).
1
= Gog-yw
log'flog-K*))
1
1
log \x)
=
A
THEOREM
3
If x
-1
= expO).
<>)
second important property of exp
is
|
an easy consequence of Theorem
1
and y are any two numbers, then
exp (a:
PROOF
log
Let x
f
—
exp(#)
and
/ =
+ y)
=
exp(x)
exp(j).
•
exp (y), so that
= log
= log/.
x
y
Then
x
+y
=
log x'
+
logy'
=
log(x'y').
This means that
cxp(x
+ y) =
= expO)
x'y'
exp(jv). |
•
This theorem, and the discussion at the beginning of
FIGURE
that exp(l)
is
particularly important. There
is,
2
this
number.
DEFINITION
e
This definition
is
=
exp(l).
equivalent to the equation
1
=
log
e
=
f
/
Ji
e
1
t
this chapter, suggest
in fact, a special
dt.
symbol
for
288
Derivatives and Integrals
As
illustrated in Figure 3,
[2
i
-
dt
<
since
1,
1
—
(2
•
an upper sum
1) is
fpr
J
= l/*on
/(/)
[1,2],
and
/
-
7i
t
</*
>
since
1,
sum
£
-
(2
•
for /(/)
+£
1)
=
1/t
on
-dt
<
•
(4
-
2)
=
1
is
a lower
[1, 4].
Thus
FIGURE
3
-dt
\
Ji
<
t
Ji
t
<
e
-dt,
Ji
t
which shows that
2
In Chapter
that
1
9
we
<
4.
much better approximations for e, and also prove
much easier than the proof that tt is irrational !)
will find
irrational (the proof is
e is
As we remarked
at the beginning of the chapter, the equation
+ y) =
exp(#
expO)
exp(j)
implies that
exp(*)
=
=
[exp(l)]*
for all rational x.
e*,
Since exp is defined for all x and expO) =
for rational x, it is consistent with
our earlier use of the exponential notation to define ex as exp(Y) for all x.
DEFINITION
For any number
x,
e
The terminology
£
x
not yet defined a x
,
a
the attempt. If x
is rational,
for
e,
but there
If a
(If a
last expression is
>
=
0,
then, for
e this
any
is
now be
irrational)
clear.
exponent
x.
We have
We have
a reasonable principle to guide us in
then
a
DEFINITION
exp(#).
an arbitrary (even
e
But the
=
'exponential function" should
succeeded in defining
if
x
x
=
(yoga)*
defined for
real
=
^loga^
all x, so
number
we can
use
it
to define a
x
.
x,
definition clearly agrees with the previous one.)
The requirement
a
>
0
is
necessary, in order that log a be defined. This
is
289
The Logarithm and Exponential Functions
we would not even expect
not unduly restrictive since, for example,
\
x
sense,
be denned. (Of course, for certain rational *, the symbol a will make
according to the old definition; for example,
to
=
1
Our
definition of a
x
=
=
(-1)1/3
/(a)
=
_1.)
to ensure that
was designed
(f)v
e*v
and
for all x
y.
equation turns out to be true when e is replaced by any
number a > 0. The proof is a moderately involved unraveling of terminology
x
At the same time we will prove the other important properties of a
As we would hope,
FIGURE
v^T
i
(-1)1/2
4
THEOREM
4
If a
>
this
then
0,
(a
(1)
b
c
)
=
a bc
for all
b, c.
h
(Notice that a will automatically be positive, so
a1
(2)
=
and
a
a
x+v
=
a
x
•
av
(a
b c
will
)
be defined);
for all *, y.
x
old one for
(Notice that (2) implies that this definition of a agrees with the
rational x.)
^
PROOF
^J>y
_
(Each of the steps in
or the fact that exp
a
(2)
a
l
=
x+y
e
lloga
=
_
^cloga*
e
^clog
(e
=
log
=
e
=
l
=
1.
c (&
log a)
Suppose
^c6Ioga
_
^
depends upon our
last definition,
.)
a
=
=
a,
^loga-h/loga
=
on whether
>
a
_
-1
lo *
(x+y)loga
of the function depends
f(x)
g
this string of equalities
Figure 4 shows the graphs of /(*)
x
=
blo «°)
1.
^sloga
.
*
^ loga
a
i.e.,
=
fl
*
.
x
>
0.
|
a.
0<l,<z = l,or<z>l.
x log a
so
^l/loga
a* for several different
In this case log a
if
then
=
The behavior
If a
=
1,
then
Thus,
< 7,
< y log a,
< ^ loga
< av
,
.
so
Thus the function /(*) - a* is increasing. On the other hand, if 0 < a < 1,x
is
a
=
function
the
that
/(*)
shows
that log a < 0, the same sort of reasoning
Since
one-one.
=
is
a*
then/(*)
and
a
0
*
1,
decreasing. In either case, if a >
x
on every
exp takes on every positive value it is also easy to see that a takes
FIGURE
5
positive value.
Thus
and takes on all
by loga (Figure
the inverse function
values. If
5).
/W
=
a
x
,
is
numbers,
the function usually denoted
defined for
then f~
l
is
all
positive
Just as a* can be expressed in terms of exp, so log
a can be expressed in terms
of log. Indeed,
if
=
=
=
y
X
then
log*
so
or
_
y
loga X,
aV
=
eV\oga^
y log
log *
a,
loga
In other words,
log X
loga X
loga
The
=
derivatives of f(x)
/to
=
«* Iog °,
a1
and
£(*)
=
so /'(*)
,(,)-|28f,
is
also easy to differentiate, if
it
follows
=
that the
There
The
g(*)
that, by definition,
w
•
Fa'W
[*'(*)
L
+
log
log^to
+
first
A(*)
£to
no point in remembering this formula
in any specific case that arises; it does
is
a*,
hix)
you remember
•
= £to
it
•
from the Chain Rule that
/'to
is
log a
like
/to =
behind
<?^ a =
log a
* log a
more complicated function
There
loga x are both easy to find:
sog'W=_J_.
log a
A
=
J
—simply apply the principle
help, however, to
factor in the derivative will be g(x) Hx)
remember
.
one special case of the above formula which is worth remembering.
= xa was previously defined only for rational a. We can now
function f(x)
define
and
result
is
find the derivative of the function f(x)
just
=
x a for any
number
a;
the
what we would expect:
f(x)
=
xa
=
a
/'(*)
=
e*
10**,
so
?.,«l og x
=
°L.
xa
.
ax a- K
x
Algebraic manipulations with the exponential functions will become second
nature after a little practice—just remember that all the rules which ought to
work actually do. The
2 and 3:
basic properties of exp are
still
those stated in
Theorems
The Logarithm and Exponential Functions
=
exp'(*)
exp(*
-
+ y)
291
exp(tf),
exp(*)
•
exp(jy).
each of these properties comes close to characterizing the function exp.
Naturally, exp is not the only function / satisfying /' = /, for if / = cf , then
= /(*); these functions are the only ones with this property,
f(x) =
In
fact,
however.
THEOREM
5
If
/
differentiable
is
and
/'(*)
then there
is
number
a
c
for all *,
/(*)
such that
/(*)
proof
=
=
for all *.
ce*
Let
*w
(This
is
permissible, since
Therefore there
The second
function exp
is
**
a number
^
—
-
0 for
all x.)
c
such that
=
M.
=
c
Then
for all x. |
basic property of exp requires a
is
clearly
more involved
not the only function / which
f(*+y)
discussion.
The
satisfies
-/W-/W-
any function of the form f(x) = a* also satisfies this equathere are infinitely
tion. But the true story is much more complex than this
without
impossible,
is
it
but
many other functions which satisfy this property,
one
even
is
there
that
prove
to
appealing to more advanced mathematics,
In
fact, /(*)
=
0 or
—
the
function other than those already mentioned! It is for this reason that
which
functions
many
infinitely
are
/
10*
there
difficult:
is
so
definition of
satisfy
=/W7W
f(*+y)
=
/(i)
but which are
not
continuous function
the function /(*)
/
=
?
io,
10*!
One
thing
is
true
however—any
satisfying
f(x+y)
must be of the form /(*) =
a
x
or /(*)
=/W'/W
=
0.
(Problem 27 indicates the way to
.
.
292
Derivatives
and
Integrals
prove
and
this,
few words to say about discontinuous functions with
also has a
this property.)
In addition to the two basic properties stated in Theorems
2 and 3, the
function exp has one further property which is very
important—exp "grows
than any polynomial." In other words,
faster
THEOREM
6
For any natural number
n,
lim
PROOF
The proof
consists of several steps.
Step 2.
>
e*
x for
To
prove
this
=
<
1
this
is
an upper sum
>
is
=
1/t
log x
on
<
clear for x
for all
clearly true, since log x
for f(t)
e
=
x
oo
(this
may
be con-
0).
statement (which
x
If x
oo
and consequently lim
all x,
sidered to be the case n
—=
[1, x],
<
x
>
0. If x
so log x
0)
it
suffices to
show that
0.
>
<
1,
x
then (Figure 6) x
-
1
<
1,
and lim
-
1 is
=
oo
x.
x
Step 2.
lim
it—
To
oo
prove
e
—
=
oo.
X
this,
note that
e* 12
e*
By Step
FIGURE
6
this
1,
e
x/2
the expression in parentheses
x
e /x
shows that lim
lim
Step 3,
•
—=
x
=
is
greater than
e* 12
oo.
oo
11
Note that
e_
The
_
(f
ln
y
i
xln
(e
\
n
expression in parentheses becomes arbitrarily large,
power certainly becomes
It is
now
tion: f{x)
=
possible to
e-v*\ x
^
by Step
2, so
the wth
arbitrarily large. |
examine carefully the following very interesting func0. We have
A*)
X3
;
The Logarithm and Exponential Functions
293
Therefore,
<
>
/'(*)
/'(*)
FIGURE
is
large, so
— 1 fx
e
llxl
with
for x
<
>
2
is
0,
0,
for positive x.
close to 0, so e~
lfxi
is
Moreover,
close to
1
if \x\ is
(Figure 7).
7
The behavior of/ near
so
for
0
and increasing
so /is decreasing for negative x
large, then x 2
x
0
large, so e~ llx2
is
e's
and
5's,
0
=
is
more
1/
interesting. If x
1/xi
(e
)
is
small. This
small, then 1/x 2 is large,
argument, suitably stated
is
shows that
lim e~ l '
xl
=
0.
x—>0
Therefore,
if
we
define
/w
1
then the function / is continuous (Figure
FIGURE
at 0:
=
*
o,
8).
In
fact,
8
Indeed
/'(0)
=
lim
*->o
—
h
1
=
h
lim
0,
/ is actually differentiable
Wc
already
know
that
— =
lim
the
it is all
more
e
(x>)
=
oo,
X
Z-+00
this
;
true that
lim
and
oo
X
»
x—*
means that
—=
lim
0.
Thus
We
can
now compute
e-U**
x
9*
0
0,
x
=
0.
that
h
*->o
2
=
A»
lim
A-0
an argument similar
A
r
= hm
2^-^
—
to the
one above shows that /"(0)
4
A
= hm —77=
x
0,
This argument can be continued. In
(Problem 29) that
(0)
=
0 for
The
\
=
function /
sin 1/x,
(
\
9
=
—2*-;
=
0.
0,
Thus
0.
using induction
fact,
every k.
Lx
FIGURE
4
lim
is
x
x
it
can be shown
extremely flat at 0,
^
=
0
0
The Logarithm and Exponential Functions
and approaches 0 so quickly that it can mask many
example (Figure 9), suppose that
295
irregularities of other
functions. For
X
5*
0
x
=
0.
X
fix)
0,
can be shown (Problem 30) that for this function it is also true that
= 0 for all k. This example shows, perhaps more strikingly than any
other, just how bad a function can be, and still be infinitely differentiable. In
Part IV we will investigate even more restrictive conditions on a function,
which will finally rule out behavior of this sort.
It
/< fc >(0)
PROBLEMS
1.
Differentiate each of the following functions
denotes a m
(i)
/(*)
(ii)
f(x)
(hi)
f(x)
(iv)
f(x)
(v)
f(x )
(vi)
f(x)
2.
(viii)
f(x)
f{x)
(x)
f{x)
=
=
=
+
log(l
(sin *)
(a) fix)
(
c ) fix)
(d) fix)
(e) fix)
=
=
=
=
....
(II)
+
+
.
[/
=
=
\~| log (sine*)
x
-7— II
(arcsin *) lo « 3 .
(log xY°**.
x*.
of the following functions.
e
x+1
e*
e
(
+ e^))e^ +
(log(3
.
+e
^ _
=
^+
x
x
.
e~\
\
(Compare the graphs with the graphs
I
1/exp.)
-x
x
'—_4 =
2X
=
1
2
+
limits
by
e~
e
*
1
1
x
e
-
1
-
x
-
* 2 /2
-
\
-
x
-
x 2 /2
-
9^
+
1
l'Hopital's Rule.
hm
hm
+ e^ *))).
log(l
sin (sinX)
1
Find the following
(I)
log(l
= eUo'**)= sin x*' **™'.
= logc) sin x.
Graph each
(b) fix)
a b * always
).
arcsin
(ix)
(remember that
-
x 3 /6
of exp
and
)
296
Derivatives
and Integrals
log(l
lim
(iv)
+
-
x)
X
log(l
(v)
lim
(vi)
lim
(cos x, sin x)
4.
The
+
jfl
x
+
x 2 /2
*
+
* 2/2
2
-
log0+^)^^ + ^/2-^/3
functions
sinh
g
=
a:
x
-
e~
2
cosh x
=
tanh #
=
e*
e
x
+C
+
=1
e~x
e
2x
+
1
1
are called the hyperbolic sine, hyperbolic cosine,
and hyperbolic
tangent, respectively (but usually read sinch, cosh, and tanch). There
are many analogies between these functions and their ordinary trigo-
One analogy
nometric counterparts.
shown
that the region
is
illustrated in Figure 10; a
in Figure 10(b) really has area x/2
is
proof
best deferred
when we will develop methods of computing
Other analogies are discussed in the following three problems,
but the deepest analogies must wait until Chapter 26. If you have not
already done Problem 2, graph the functions sinh, cosh, and tanh.
until the next chapter,
integrals.
5.
Prove that
(a)
cosh 2
(b)
tanh 2
(b)
FIGURE
=1.
sinh 2
(d)
+ 1/cosh =
sinh(# + y) = sinh x cosh y + cosh x sinh y.
cosh(* + y) = cosh x cosh y + sinh x sinh y.
(e)
sinh'
cosh'
=
=
cosh.
(f)
(g)
tanh '
-
53?
10
(c)
6.
—
2
1.
sinh.
The
functions sinh and tanh are one-one; their inverses, denoted by arg
and arg tanh (the "argument" of the hyperbolic sine and tangent),
are defined on R and ( — 1, 1), respectively. If cosh is restricted to [0, «>
it has an inverse, denoted by arg cosh, which is defined on [1, oo).
Prove, using the information in Problem 5, that
sinh
(a) sinh (cosh
-1
(b) cosh(sinh(c)
1
x)
Ar)
(sinh-TM =
=
=
Vx
V
1
1
2
-
+*
1
2
.
.
!
The Logarithm and Exponential Functions
(d)
(e)
7.
(a)
(b)
1
(cosh-yM =
2
-
1.
1
(tadT^'M =
for
< L
1*1
1
1
1
Find an explicit formula for sintT cosh" and tanh" (by solving
1
the equation y = sinlT" x for x in terms of y, etc.).
Find
,
,
A Vl +
H X
/
;a
dX9
:
2
v> _
—^— dx
for fl,^
</x
>
or
1
K
a,
1,
i
b
f
for
W,
2
|*|
Compare your answer
—
writing
1
8.
>
for x
,
Vx
297
-
x2
=
<1.
for the third integral
with that obtained
by
-[—
+ —}
+
211 x
x]
1
Find
(a)
lim ax for 0
X—
(b)
<
a
<
1.
(Remember
the definition!)
00
lim
(log *)«
(log *)»
(c)
lim
(d)
lim *(log x) n Hint: *(log
(
(e)
.
lim
n
x)
-i)^iogiy
—
=
x*.
x-»0+
9.
10.
Graph f(x) =
(a)
Find the
x* for x > 0. (Use Problem 8(e).)
minimum value of /(*) = ex /x n for
> e n /n n for > «.
at
that /(*)
(b) Using the expression /'(*)
n+1 for x > n
n+1 /(n
e
l)
+
+
lim f(x)
X—
=
e*(x
1,
-
and
>
0,
and conclude
n)/x n+1 prove that /'(*) >
thus obtain another proof that
,
co.
00
11.
Graph /(*) = *7* n
12.
(a)
Find lim
-
log(l
+ y)/y.
l/-»0
would be
(b)
=
x
silly.)
Find lim x log(l
+
1
A).
(You can use l'Hopital's Rule, but that
298
Derivatives
and
Integrals
Prove that
(c)
=
^
lim
x—*
Prove that
(d)
e
=
a
+ /*)*
x
(1 + a/x)
lim
x—» 00
part
with just a
(c)
Prove that log
*(e)
13.
Graph
14.
If a
/(*)
bank
=
+
(1
=
b
1
(1
00
little
algebraic fiddling.)
-
lim x(b 1/x
\/x)
x
derive this from
(It is possible to
.
for *
>
1).
(Use Problem 12(c).)
annum, then an initial investment I
If the bank compounds the interest
0.
gives a percent interest per
+
yields 1(1
a/100) after
year.
1
(counts the accrued interest as part of the capital for computing interest
the next year), then the initial investment grows to 1(1
a/100) n after
+
n years.
Now suppose that interest is given twice a year. The final amount
after n years
is,
although interest
is
+
—
a/100) 2n but merely 1(1
a/ 200) 2n
awarded twice as often, the interest must be halved
not 1(1
alas,
+
y
in each calculation, since the interest
is
+ a/100)
larger than 1(1
is a/2 per half year. This amount
but not that much larger. Suppose that the
71
,
bank now compounds the interest continuously, i.e., the bank considers
what the investment would yield when compounding k times a year, and
then takes the least upper bound of all these numbers. How much will an
investment of 1 dollar yield after 1 year?
Let/(*) = log |*| for x ^ 0. Prove that f(x) = 1/x for x ^ 0.
Prove that if /' = cf for some number c, then f(x) — ke cz for some
initial
15.
*16.
number
*17.
k.
A radioactive substance diminishes at a
present (since
total
rate proportional to the
amount
atoms have equal probability of disintegrating, the
disintegration is proportional to the number of atoms remaining).
all
is the amount at time t, this means that A'(t) = cA(t) for some
(which represents the probability that an atom will disintegrate).
If A(t)
(a)
Find A(t)
(b)
Show
amount A 0 = A(0) present at time 0.
number r (the "half-life" of the radioactive
in terms of the
that there
is
a
ment) with the property that A(t
*18.
c
+
r)
=
ele-
A(t)/2.
Newton's law of cooling states that an object cools at a rate proportional
to the difference of its
temperature and the temperature of the surroundT(t) of the object at time t, in terms
of its temperature T 0 at time 0, assuming that the temperature of the
surrounding medium is kept at a constant, M. Hint: To solve the differential equation expressing Newton's law, remember that
= (T — M)'.
ing
medium. Find the temperature
V
*19.
Prove that
*20.
Find
21.
(a)
all
if
= j* f(t)
/(*)
functions
/
dt,
then
/ -
satisfying /'(0
=
f(t)
+
+
-*
0.
+
J*
/(/) dt.
Prove that
1
+*+
Tx
Ti
+
Hint: Use induction on
'
n,
'
'
^
e*
and compare
for *
>
0.
derivatives.
.
299
The Logarithm and Exponential Functions
(b)
Give a new proof that lim
e
x
=
/x n
<*>
X-*oo
Give yet another proof of this fact, using the appropriate form of PHopital's Rule. (See Problem 11-38.)
A point P is moving along a line segment AB of length 10 7 while another
point Q moves along an infinite ray (Figure 11). The velocity of P is
always equal to the distance from P to B (in other words, if P[t) is the
7
position of P at time t, then P'{i) = 10 - P(t)), while Q moves with
7
=
traveled by Q after time t is
distance
10
The
constant velocity Q'(t)
.
defined to be the Napierian logarithm of the distance from
P to B at time
/.
Thus
10 7 *
= Nap
-
log[10 7
P(t)l
This was the definition of logarithms given by Napier (1550-1617) in his
publication of 1614, Mirifici logarithmonum canonis description (A Descrip-
work which was done before
10 7 was chosen because
Napier's tables (intended for astronomical and navigational calculations), listed the logarithms of sines of angles, for which the best possible
available tables extended to seven decimal places, and Napier wanted to
Law
Wonderful
tion of the
of Logarithms)
the use of exponents was invented!
;
The number
avoid fractions. Prove that
Nap
log x
-
10 7 log
10 7
Hint: Use the same trick as in Problem 18 to solve the equation for P
(a) Sketch the graph of f(x) = (log x)/x (paying particular attention
.
to the behavior
(b)
(c)
Which
is
=
y
x
is
number y
e
0
<
<
1,
or x
=
x;
but
if
x
if
y
5*
x
**(f)
=
e,
then the only number y satisfying
then there is precisely one
if x < e, then y > e,
moreover,
y
(Interpret these statements in terms of the
1,
x
x satisfying xy
=
x
5* e,
\
if x > e, then y < e.
graph in part (a) !)
Prove that if x and y are natural numbers and x y
=
—
=
Show
4, or x
4, y
y
that the set of all pairs
and a
straight line
or x
(e)
?
>
and
(d)
oo).
ir
Prove that
xy
near 0 and
larger, e* or
2,
which
=
=
y
x
>
then x
=
y
2,
(x,
y)
with xy
=
y
x
consists of a
intersect; find the intersection
curve
and draw a
rough sketch.
ix) =
For 1 < x < e let g{x) be the unique number > e with x°
x Prove that
consider
g(x)
g is differentiate. (It is a good idea to
.
separate functions,
f
x
(x)
0<x<e
=
x
x
300
Derivatives
and
Integrals
and write g
and /2
in terms of fi
If
.
you do
part properly you
this
should be able to show that
rt4.JfflL.lzi.)
*25.
X2
log g(x)
1
/
This problem uses the material from the Appendix to Chapter
Prove that exp
(a)
convex and log
is
is
11.
concave.
n
Prove that
(b)
£=
if
t
'
=
t
and
1
all pi
>
0,
then
.
.
•
.
+
< p lZl
z n *>»
•
+
•
/> n
zn
.
(Use Problem 8 from the Appendix to Chapter 11.)
Deduce another proof that Gn < A n (Problem 2-20).
(c)
26.
p
i
Suppose / satisfies /' = / and f(x + y) = f(x)f(y) for all x and y. Prove
— exp or / = 0.
Prove that if / is continuous and f(x + y) = f(x)f(y) for all x and jy, then
that /
*27.
either
/
=
=
0 or /(*)
[/(l)] z for all x. Hint:
Show
that /(*)
=
[/(l)]*
and then use Problem 8-6. This problem is closely related
to Problem 8-7, and the information mentioned at the end of Problem
8-7 can be used to show that there are discontinuous functions / satisfy-
for rational
ing /(*
*28.
=
f(xy)
f(e) log
*29.
+ y) = f(x)f(y).
Prove that
if
x for
Prove that
/ is a continuous function defined on the positive reals, and
for all positive x and y, then / = 0 or f(x) =
+ f(y)
f(x)
if
>
=
x
all
f(x)
=
Hint: Consider g(x)
0.
e~ 1
^ for x ^
and
0,
/(0)
/(<?*).
=
0,
then
/< fc) (0)
=
0 for
all k.
*30.
31.
Prove that
0 for all £.
(a)
if f(x)
Prove that
(*)
anx
n
if
+
=
^^
a
is
1
2
sin 1/x for ^
(**) an y
0,
and /(0)
=
then f k \0)
0,
=
a root of the equation
fln-i*""
+
1
•
+ aix + a =
•
=
then the function y (x)
{n)
^
0
satisfies
+ a n ^y^ +
•
•
•
0,
the differential equation
+
+
-
0.
a is a double root of (*), then y(x) = ^ax also satisfies
Remember that if a is a double root of a polynomial
equation f(x) = 0, then f'(a) = 0.
Prove that if a is a root of (*) of order r, then y(x) = x k e ax is a solu-
*(b) Prove that
if
(**). Hint:
*(c)
tion for 0
If (*)
<
k
<
r
-
has « real numbers as roots (counting multiplicities), part
gives n solutions y l9
(d)
1.
Prove that in
satisfies (**).
this
.
.
.
,
(c)
y n of (**).
case the function Ciyi
+
•
*
+ c ny n
also
301
The Logarithm and Exponential Functions
It is a theorem that in this case these are the only solutions of (**).
Problem 16 and the next two problems prove special cases of this
theorem, and the general case is considered in Problem 19-17.
In Chapter 26 we will see what to do when (*) does not have n real
numbers as roots.
*32.
Suppose that /
/ =
(a)
Show that/ 2 -
(b)
Suppose that
constant
**(c)
/"
-/=
0 and /(0)
=
/'(0)
=
0.
Prove that
0 as follows.
=
either f(x)
(/')
ce
x
2
^
f{x)
=
0.
0 for
all
or else /(#)
x in
=
some
interval
ce~x for all x in
(a, b).
Show
that
for
some
{a, b) )
c.
If/(*o)
^
0 for x 0
0
<
a
that
Why? Use
*33.
satisfies
<
this
> 0, say, then there would be a number a such
and f(a) = 0, while fix) ^ 0 for a < x < x 0
fact and part (b) to deduce a contradiction.
x0
.
that if / satisfies /" - / = 0, then f(x) = ae x + be~ x for some
and b. (First figure out what a and b should be in terms of /(0)
and f(0), and then use Problem 32.)
(b) Show also that / = a sinh + b cosh for some (other) a and b.
Find all functions / satisfying
Show
(a)
a
34.
(a) /<«>
(b) /<*>
*35.
= /(--i).
= f n ~*\
This problem, a companion to Problem 15-28, outlines a treatment of
the exponential function starting from the assumption that the differential
(a)
equation
f'=f has
a nonzero solution.
Suppose there is a function / 9^ 0 with /' = /. Prove that f(x) ^ 0
each x by considering the function g(x) = f(x 0 + x)f(x 0 — x),
where f(x 0 ) 5* 0.
and /(0) = 1
Show that there is a function / satisfying
For this / show that f(x + y) = f(x) f(y) by considering the func-
for
(b)
(c)
f~f
•
= f(x+y)/f(x).
Prove that / is one-one and that
tion g(x)
(d)
**36.
A
function
/
is
said to
grow
faster
(/~~ *)'(*)
than g
if
=
l/x.
lim f(x)/g(x)
X—
—
00.
For
00
example, exp grows faster than any polynomial function. Suppose that
are continuous functions. Show that there is a continuous
gi, g2, gi,
37.
.
.
•
function / which grows faster than each
Prove that logi 0 2 is irrational.
gi.
CHAPTER
W7
IS
INTEGRATION
|
ELEMENTARY TERMS
IN
Every computation of a derivative yields, according to the Second Fundamental Theorem of Calculus, a formula about integrals. For example,
if
F(x)
—
—
*(log x)
F
then
x,
=
f
(x)
log x;
consequently,
J*\ogxdx =
Formulas of
F{b)
-
F(a)
=
this sort are simplified
[a{\og a)
-
F(b)
-
we adopt
if
a\
0
<
a, b.
the notation
F{a).
write
J*
This evaluation of
J*
—
x(log x)
x
|\
depended on the lucky guess that log
log x dx
— f is called
=
log x dx
derivative of the function F(x)
satisfying F'
-
b
considerably
=
F(x) T
We may then
-
b(log b)
=
x(\og x)
—
x.
Of course,
a primitive of /.
is
the
In general, a function
F
a continuous function /
always has a primitive, namely,
*w
we
=
/;/>
which can be written in
terms of familiar functions like sin, log, etc. A function which can be written
in this way is called an elementary function. To be precise,* an elementary
but in
this
chapter
will try to find a primitive
function is one which can be obtained by addition, multiplication, division,
and composition from the rational functions, the trigonometric functions and
their inverses,
and the functions log and exp.
should be stated at the very outset that elementary primitives usually
cannot be found. For example, there is no elementary function F such that
It
F'(x)
=
e~ x *
for all x
not merely a report on the present state of mathematical ignorance; it
theorem that no such function exists) And, what is even worse,
you will have no way of knowing whether or not an elementary primitive can
be found (you will just have to hope that the problems for this chapter contain
(this
is
*
a
is
(difficult)
.
The definition which we will give is precise, but not really accurate, or at least not quite
standard. Usually the elementary functions are defined to include "algebraic" functions,
that
is,
functions g satisfying an equation
(g(x)) n +fn-l(x)(g(*)) n
where the /»•
302
-l
+
•
•
'
+/oW =
0,
are rational functions. But for our purposes these functions can be ignored.
—
Integration in Elementary
Terms
303
no misprints). Because the search for elementary primitives is so uncertain,
finding one is often peculiarly satisfying. If we observe that the function
F(x)
— +*
log(l
=
x arctan x
2
)
satisfies
F
(just
f
(x)
=
arctan x
how we would ever be led to such an observation is quite another matter),
so that
*
arctan x ax
—
log(l
x arctan x
+
x 2)
f.
then
we may
feel
that
we have
This chapter consists of
"really" evaluated j* arctan x dx.
little
more than methods
primitives of given elementary functions (a process
tion"), together with
for finding
known simply
elementary
as "integra-
some notation, abbreviations, and conventions designed
elementary functions can
to facilitate this procedure. This preoccupation with
be justified by three considerations:
(1)
(2)
(3)
Integration
is
know about
it.
a standard topic in calculus,
Every once in a while you might actually need to evaluate an integral,
under conditions which do not allow you to consult any of the
standard integral tables (for example, you might take a (physics)
course in which you are expected to be able to integrate).
The most useful "methods" of integration are actually very important theorems (that apply to all functions, not just elementary ones).
Naturally, the last reason
how
and everyone should
is
the crucial one.
Even
if
you intend
to forget
(and you probably will forget some details the first time
through), you must never forget the basic methods.
These basic methods are theorems which allow us to express primitives of
one function in terms of primitives of other functions. To begin integrating we
will therefore need a list of primitives for some functions; such a list can be
to integrate
obtained simply by differentiating various well-known functions. The list
given below makes use of a standard symbol which requires some explanation.
The symbol
//or
means "a primitive of/"
of /."
The symbol jf
or,
more
will often
most useful in formulas
Jf(x)dx
precisely, "the collection of all primitives
be used in stating theorems, while //(*) dx is
like the following:
This "equation" means that the function F(x) = * 4 /4 satisfies F'(x) = # 3 It
cannot be interpreted literally because the right side is a number, not a function, but in this one context we will allow such discrepancies; our aim is to
.
make
any
the integration process as mechanical as possible,
possible device.
and we
will resort to
Another feature of the equation deserves mention. Most
people write
/
4
emphasize that the primitives of f(x) = x 3 are precisely the functions of the
form F{x) = x z + Cfor some number C. Although it is possible (Problem 1 1) to
obtain contradictions if this point is disregarded, in practice such difficulties
do not arise, and concern for this constant is merely an annoyance.
There is one important convention accompanying this notation: the letter
appearing on the right side of the equation should match with the letter
to
appearing after "d" on the
left side
—thus
4
u z du
/
—
—)
4
***
f
j =
-,
jtxdx
/
A
function in jf(x) dx,
integral" of
while
/,
J*
i.e.,
tx dt
=
xt 2
—2
•
a primitive of
f(x) dx
is
called,
/, is
often called
by way of
an "indefinite
contrast, a "definite
integral." This suggestive notation works out quite well in practice, but
it is
important not to be led astray. At the risk of boring you, the following fact
emphasized once again: the integral
where Fis an
Ja
f(x) dx
is
not
you do not
defined as "F(b)
—
is
F(a),
find this statement repe-
of/"
time to reread Chapter 13).
We can verify the formulas in the following short table of indefinite integrals
simply by differentiating the functions indicated on the right side.
indefinite integral
(if
titious, it is
/
/
J
adx
=
x n dx
ax
=
n
-dx =
+
log x
,
n 9*
—1
1
- dx
is
often written
J
— for convenience; similar
abbreviations are used in the last two examples of this
table.)
/
e
x
dx
=
e
x
Integration in Elementary
J
sin x dx
= —
cos x dx
—
J
y sec
2
y sec
a:
305
cos x
sin x
=
x dx
Terms
tan #
tan x dx
—
sec #
/*
/
J
1
+
=
arctan x
*2
/
==
/
VI -
./
Two
arcsm *
*2
general formulas of the same nature are consequences of theorems
about differentiation:
=
SUM + g(*)] dx
jc
•
f{x) dx
//(*) dx
=
+
fg(x) dx,
c ff(x) dx.
These equations should be interpreted as meaning that a primitive of / + g
can be obtained by adding a primitive of / to a primitive of g, while a primitive
of c f can be obtained by multiplying a primitive of / by c.
-
Notice the consequences of these formulas for definite integrals: If / and g
are continuous, then
j*
[/«
+ g(x)] dx
c
•
=
f(x) dx
J"
=
+
f{x) dx
c
'
J*
f"
g(x) dx,
f(x) dx.
These follow from the previous formulas, since each definite integral may be
written as the difference of the values at a and b of a corresponding primitive.
Continuity is required in order to know that these primitives exist. (Of course,
the formulas are also true when / and g are merely integrable, but recall how
much more difficult the
The product formula
which
THEOREM
1
(INTEGRATION BY PARTS)
If /'
proofs are in this case.)
for the derivative yields a
more
interesting theorem,
be written in several different ways.
will
f
and g are continuous, then
ffg'=fg-/ fg,
f
f(x)g'(x) dx
pMg'(x)
dx
=
=
f(x)g(x)
- / f{x)g(x) dx,
- J'{x)g{x) dx.
b
f{x)g{x)
I
j
(Notice that in the second equation f{x)g{x) denotes the function
f
g.)
306
Derivatives and Integrals
PROOF
The formula
<Jgy=f'g+fg'
can be written
fg'
=
(fg)'
- fg.
Thus
jfg'
and fg can be chosen
the
first
as
= / (fg)' ~
Jf'g,
one of the functions denoted by
f(fg)'.
This proves
formula.
is merely a restatement of the first, and the third
formula follows immediately from either of the first two. |
The second fomula
As the following examples illustrate, integration by parts is useful when the
function to be integrated can be considered as a product of a function /, whose
derivative is simpler than /, and another function which is obviously of the
form
g'.
There are two
The
first is
written
U
U
fg'
fg
special tricks
f
g
which often work with integration by parts.
1, which can always be
to consider the function g' to be the factor
in.
The second
and then
trick
is
by parts to find fh in terms of fh again,
simple example is the calculation
to use integration
solve for fh.
A
Integration in Elementary
Terms
307
or
1,
/
A more complicated
J
e* sin
f
calculation
=
x dx
e
x
•
(
—
(log *) 2
2
—
J
cos x
= — e* cos
e*
(
•
—
cos x) dx
f
g
= — e*
.
often required:
is
cos x)
f
g'
—
=
.
- log xdx
x
+
I
+
x
g
cos x dx
e*
X
i
u
v'
•
(sin x)
\e"
u
—
J
v
^(sin x) dx];
u
f
V
therefore,
2
j
e
x
sin
=
^(sin #
—
cos #)
xdx —
**(sin *
—
cos x)
a:
dx
or
e
x
sin
upon recognizing that a function is of
you can already integrate, the greater your
chances for success. It is frequently reasonable to do a preliminary integration
before tackling the main problem. For example, we can use parts to integrate
Since integration by parts depends
the form
g',
the
more
functions
j
= j
(log x) 2 dx
/ log x dx = x(log
integration by parts) we have
if
we
recall that
x)
—
(log x) (log x) dx
x
I
I
f
g'
(this
formula was
itself
derived by
;
j
(log *)(log x) dx
=
(log *)[*(log
*)-*]-/
(l/*)|>(log
i
A
i
i
i
i
f
i'
f
g
f
g
=
=
=
=
(log *)[x(log x)
(log x)[x(log x)
(log *)[*(log *)
x(log *) 2
The most important method
The use of this method
Rule.
integrating
-
at)
-
*] <fc
- x]- [log x - 1]
f
— x] — log xdx + 1 dx
J
J
- *] - [*(log x) - x] + x
2*(log *)
of integration
+
2x.
is
a consequence of the Chain
more ingenuity than
by parts, and even the explanation of the method is more difficult.
requires considerably
308
Derivatives
end
Integrals
We
will therefore
nite integrals
theorem
2
If /
(THE SUBSTITUTION FORMULA)
this method in stages, stating the theorem for defiand saving the treatment of indefinite integrals for later.
develop
first,
f
and g are continuous, then
O(«0«fcPROOF
If
F is
a primitive of /, then the
/.*/(*«)*'(*>*•
-
F{g(b))
left side is
F(g(a)).
On
the other
hand,
(Fog)'= (F'*g)-g' = (fog)-g',
soFo^isa
primitive of (fog)
-
(Fo g)(b)
The
g'
•
is
facilitated
=
=
j*
by the appearance of the
g'(x) for g(x)
5
—
.
a;
cos
=
depend upon recognizing
sm 6
a;
a
tan
-
,
a;
5
,
sin x
is
g(x)
tan x dx
/(«)
/
sinx]
Uh
J
f(u)du
sin
6
a
6
J
/
J cob
a
U
a;
write
.
a;
=
*) for f(u)
=
cos X
=
1
=
cos
where
g'(x) y
*=
=
-
/
log(cos a)
-
/(jrMk'M
/"cos 6
sin
we
if
/
Ja
'6
= -
(sin x)
£
—
= — p
can then be written /(cos
= -
will be the factor
can be written as
which
dx can be treated similarly
ya
this case the factor
x,
6
b
f
/
cos x dx^j
f(g(fiW(x)dx-
/
fan
In
=
=
!>(*)'
,
•sin 6
tan
5
factor cos
lf(u)
J*
F(g(a)). |
For example, the integration of
.
(sin *)
tf*
a:
7a
factor 1/cos x
g
is
=
b
[ sin B5
integration of
f
•
remaining expression,
u\ Thus
sin x; the
f(g(*)), for f(u)
/
The
-
F(g(b))
of the form (fog)
sin 5 x cos x dx (
(£(.'.'))
=
(Fog)(a )
right side
simplest uses of the substitution formula
that a given function
is
and the
cos x; the remaining
1/u.
Hence
/(«)
log(cos
*
£).
Integration in Elementary
Terms
309
Finally, to find
dx,
X lOg X
a
notice that 1/x
f(u)
=
=
=
where g{x)
g'(x)
log
and that 1/log x =
x,
f(g(x)) for
Thus
1/w.
f
a
'g(x)
=
log x~
M
=
-
X log X
/•»(»
/*
=
f(g(*))g'(x) dx
/
=
Jo
/(«)&
7<7(o)
log 6
=1
1
=
log(log 6)
-
log(log a).
U
log a
Fortunately, these uses of the substitution formula can be shortened conThe intermediate steps, which involve writing
siderably.
p(g(x))g'(x)dx
can
easily be eliminated
= /*>/(«>
by noticing the
du,
following:
To go from
the
left
side
to the right side,
u for p(x)
,
,/\
du for g (x) dx
(and change the limits of integration);
,
(
.
substitute
_
,
(
the substitutions can be performed directly on the original function (accounting for the name of this theorem). For example,
f
.
sm 5
L
and
/"sin 6
-i
6
x cos x
afx
u for sin a:
r
for cos x
substitute
L
I
,
=
u* du,
/
J sin
tfxrj
a
similarly
f
b
—
sin x
_
f
X^T^r
Usually
we
abbreviate this
,
w for cos
.
bStltUtC
1
a:
rcoB&i
_
du.
^for-sin,^ J
J cos
a
U
method even more, and say simply:
"Let
u
du
=
=
g(x)
g'(x) dx."
Thus
let
-
du
7a X 10
log X
In
this
integrals,
if
=
=
log x
rioz b
=
- dx
/
./log a
i
-du.
x
we are usually interested in primitives rather than definite
certainly
we can find
f(x) dx for all a and b, then we can
chapter
but
w
(fx
J*
find J7(*) dx.
For example, since
b
f
it
=
.
sin 5 x cos x dx
sin 6 £
sin 6 a
6
6
follows that
/
sin 6 *
=
sin 5 x cos *
Similarly,
y tan x dx
/^lcb^
log X
It is quite
first
uneconomical
— —
=
log x cos
l0g(l0g
to obtain primitives
#,
^
from the substitution formula by
finding definite integrals. Instead, the two steps can be combined, to
yield the following procedure:
(1)
Let
u
du
(after this
=
—
g(x),
g\x) dx;
manipulation only the
letter u
should appear,
letter x).
(2)
(3)
Find a primitive (as an expression involving
Substitute g(x) back for u.
u).
Thus, to find
J
sin 5 x cos x dx,
(1) let
—
—
u
du
so that
we
sin x,
cos x dx,
obtain
j u*du;
(2)
evaluate
f *
u du
J
(3)
remember
to substitute sin x
f
y
•
5
sin 5
.*•
=
u6
V
back
cos
for u, so that
j
=
xdx
Similarly, if
u
=
log
du
—
- dx,
x
#,
Sm6 *
—
-
not the
Integration in Elementary
Terms
311
then
f
I
—
fl-
1
becomes
dx
/
x log X
J
J
=
du
log
u,
u
so that
/ x log #
To
dx
=
log (log x).
evaluate
/ +*
1
2
dx,
let
a
du
the factor 2 which has just
=
=
1
+ x\
2x dx;
popped up causes no problem—the
integral
becomes
so
I
be combined with integration by parts to yield a formula
already mentioned:
(This result
may
j
1
•
arctan x dx
—
=
x arctan x
* arctan *
—
Jy+
2
a:
— i log(l +
x 2 ).)
These applications of the substitution formula* illustrate the most straightforward and least interesting types once the suitable factor g'(x) is recognized, the whole problem may even become simple enough to do mentally.
The following three problems require only the information provided by the
—
short table of indefinite integrals at the beginning of the chapter and, of
course, the right substitution (the third problem has been disguised a little by
some algebraic chicanery).
*
The
substitution formula
f
is
often written in the form
f(u)du=
This formula cannot be taken
f
f(g(x))g'(x) dx,
u=g(x).
literally (after all, //(«) du
should
mean
a primitive of /
and
not equal). HowIf(gM)g'(x) dx should mean a primitive of (Jog)- g'; these are certainly
developed.
ever, it may be regarded as a symbolic summary of the procedure which we have
well:
particularly
reads
formula
fudging,
the
little
and
a
If we use Leibnitz's notation,
du
f
f{u)du=
j m^dx,
j sec
j
2
x tan 5 x dx,
x
(cos x)e
dx,
dx.
/
you have not succeeded in finding the right substitutions, you should be
them from the answers, which are (tan 6 x)/6, e silix and arcsin e x
At first you may find these problems too hard to do in your head, but at least
when g is of the very simple form g(x) — ax + b you should not have to waste
time writing out the substitution. The following integrations should all be
clear. (The only worrisome detail is the proper positioning of the constant
should the answer to the second be e Zx /3 or 3e 3x ? I always take care of these
problems as follows. Clearly Je 3x dx = e Zx (something). Now if I differentiate
F(x) = e Zx I get F'(x) = 3e* x so "something" must be i, to cancel the 3.)
If
able to guess
.
,
•
,
,
/x +
/
I
I
log(*
3
dx
=
cos 4x dx
=
—3
3),
>
4x
sin
4
+
sin (2x
1)
cos (2*.+ 1)
dx
2
arctan 2x
dx
/ + 4*
1
More
+
2
interesting uses of the substitution formula occur
g'(x) does not appear.
happens. Consider
when
the factor
There are two main types of substitutions where
this
first
1
+
e
x
dx.
f -e
i
The prominent appearance
of the expression
?
suggests the simplifying
substitution
u
du
Although the expression
J
1
e*
=
=
x
e
e
,
x
dx.
dx does not appear,
-?
J
1
-
e*
it
e*
can always be put
in:
Integration in Elementary
We
Terms
313
therefore obtain
+u
1
/ —
1
u
u
1
which can be evaluated by the algebraic
[LLa.lto^
J
—
1
u
trick
-l— + \ du=
f
u
.
du,
J
u
1
+ io g «,
-2iog(i -«)
u
so that
+
e
/ -
e
1
1
There
is
x
= -2
dx
x
-
log(l
+
*»)
= -2
log
an alternative and preferable way of handling
does not require multiplying and dividing by
u
=
x
—
log
dx
=
- du,
,
we
If
.
x
x
e
e
~
log(l
this
«*)
+
x.
problem, which
write
u,
then
f
y
Most
1
-
"
^
lb
_
immediately becomes
e*
substitution problems are
expressing x in terms of
hard to see why
in terms of x
u,
this trick
is
f
J
and dx
much
\
one
if
-
*
u
-
afo.
u
resorts to this trick of
in terms of du, instead of vice versa. It
all
(as
substitution
u
=
= g~\u)
= (g-V(u)
x
g(x),
dx
du
to the integral
/
we
On
//M(f )'WA.
!
the other hand,
if
we apply
the straightforward substitution
u
du
to the
same
=
=
g{x)
g'(x) dx
integral,
/ /(*(*))
we
/(*(*)) dx,
obtain
(i)
J
f(g(x))
•
g'(x) dx,
obtain
(2)
f /(«)
J
The
integrals (1)
and
(2)
is
not
long as the function expressing u
x under consideration): If we apply the
always works
one-one for
easier
j
•
*
,.
y
g (g H«)
are identical, since
_1
(|r
du.
)'(")
=
As another concrete example, consider
dx.
/
In
we
this case
by one
letter.
will
+
Ve>
1
go whole hog and replace the entire expression
Thus we choose the
-
*2
V> +
=
=
u
u2
+
e*
1,
=
log(« 2
dx
=
—
-
w2
The
integral then
-
1),
1
becomes
-
u2
u
J
1
1,
*
=
1
vV +
substitution
J
1
3
Thus
/
A
:
+
is
In
this case, instead of
we
will replace x
we
1)
3/2
~
2[f
+
1)
1/2
by
Vl -
x2
dx.
replacing a complicated expression by a simpler one,
sin w,
because
VI —
=
sin 2 u
x
dx
=
=
sin u,
=
cos
arcsin x, but
which helps us find the expression
let
[u
=
u.
This really means
the expression for x
it is
to be substituted for dx.
arcsin x]
cos u du;
then the integral becomes
j Vl —
The
.
second main type of substitution
illustrates the
are using the substitution u
in terms of u
+
the integral
f
that
\ {f
3
Another example, which
that can occur,
=
1
= j
sin 2 u cos u du
cos 2 w du.
evaluation of this integral depends on the equation
cos 2 u
=
1
+
cos 2u
2
(see the discussion of trigonometric functions
j cos
2
u du
=
~+
J
cos 2u
below) so that
_
</w
2
=
a
2
+
.
sin 2w
4
Thus,
Integration in Elementary
Terms
315
and
//Vl-x*dX
^-+
arcsin x
=
arcsin x
=
sin (2 arcsin x)
-
.
1
.
2
ar^
=
.
—
/
N
h - sin (arcsin x)
2
V
1
2
*
/
N
cosfarcsm x)
2
2
Substitution and integration by parts are the only fundamental methods
which you have to learn; with their aid primitives can be found for
a large number of functions. Nevertheless, as some of our examples reveal,
success often depends upon some additional tricks. The most important are
below. Using these you should be able to integrate all the functions in
Problems 1 to 7 (a few other interesting tricks are explained in some of the
remaining problems).
listed
TRIGONOMETRIC FUNCTIONS
1.
Since
+ cos
sin 2 x
2
=
x
1
and
cos 2x
we
=
cos 2 x
—
sin 2 x,
obtain
cos 2x
cos 2x
—
—
—
cos 2 x
—
(1
sin
— cos
x) — sin
2
(1
2
x)
2
x
=
=
2 cos 2 x
1
—
—
2 sin 2
1,
*,
or
sin
2
x
=
1
—
=
1
+
cos 2x
2
cos 2 x
cos 2x
2
These formulas
may
be used to integrate
J
j
if
n
is
sin
71
cos
n
x dx,
x dx,
even. Substituting
(1
-
cos 2x)
(1
or
for sin 2 x or cos 2 x yields a
sum
+
cos 2x)
2
2
of terms involving lower powers of cos. For
example,
j sin*xdx
=
j^
-^os2*
y
dx
=
l
-dx- j cos
j
2xdx
+
cos 2
2^^
and
If n
is
odd,
=
/z
2k
+
J
sin
2xdx =
cos 2
y
dx.
J
1,
n
then
=
x dx
J
—
sin *(1
cos 2 x) k dx;
the latter expression, multiplied out, involves terms of the form sin* cos* x,
of which can be integrated easily. The integral for cos" x is treated simi-
all
larly.
An
integral
sin
j
n
x cos m x dx
is handled the same way if n or m is odd. If n and
formulas for sin 2 x and cos 2 x.
A final important trigonometric integral is
/——
= f
dx
cos x
sec x dx
=
m
are both even, use the
+ tan x).
log(sec x
J
Although there are several ways of "deriving" this result, by means of the
methods already at our disposal (Problem 10), it is simplest to check this
formula by differentiating the right side, and to memorize it.
2.
REDUCTION FORMULAS
Integration by parts yields (Problem 17)
/sin n
— — - sin n_1
x dx
x cos x
+
/cos n
1
f
J
x dx
(x
2
+
\)
n
dx
=
=
In
+
- cos n_1 x sin x
n
—-
*
2 (x
f
n
n
2
+
l)"'
1
^
-
In
n~ 2
x dx,
J
n
+
sin
f
cos
n-2
x
dx,
J
1
f
2 J
(x
2
+
l)
71
"" 1
*,
'
and many similar formulas. The first two, used repeatedly, give a different
method for evaluating primitives of sin n or cos w The third is very important
.
for integrating a large general class of functions,
which
will
complete our
discussion.
3.
RATIONAL FUNCTIONS
Consider a rational function p/q where
= anx n + an -ix n ~ +
+ 0
m
m' +
=
q(x)
b mx + b m^x
+ bo.
We might as well assume that a n = b m — 1. Moreover, we can assume that
n < m, for otherwise we may express
as a polynomial function plus a
p(x)
l
•
1
•
•
•
•
<2
•
,
J
Integration in Elementary
which
rational function
integration of
first
follows
317
of this form by dividing (for example:
is
u+l +
-
u
The
Terms
1
an arbitrary rational function depends on two facts; the
from the "Fundamental Theorem of Algebra" (see Chapter 25,
Theorem 2 and Problem 25-3), but the second will not be proved in this book.
THEOREM
Every polynomial function
=X m +
q(x)
brn^X**-
+
1
•
•
•
+
bo
can be written as a product
q( x )
= (x-
ai)
•-..'(*-
r
'
a*) r *(* 2
+ fa +
7i)'«
*
(*
(where
r\
+
+ rk +
2(jx
+
+
-
si)
2
+ ft* + 70
-«
m).
(In this expression, identical factors have been collected together, so that all
x
—
en
+ fax +
and x 2
may
7»
be assumed
distinct.
Moreover, we assume
means that
that each quadratic factor cannot be factored further. This
ft
since otherwise
2
a:
+ ft* +
we can
=
7,-
[*
- 47 <
2
0,
;
factor
~
(
)]
into linear factors).
THEOREM
If n
< m and
=
+ an__i* n-~ +
+a
+ b m_ lXm + — + b 0
^0) =
= (x- ai) r •-..•(*- at) r *(*
1
^(jf)
•
•
•
0j
l
*
»
2
+ ft* + 7i)*
•
(*
2
+ ft* + yi)>i,
then p(x)/q(x) can be written in the form
q(x)
L(*
-
«i)
+
(at
+
^i
I"
L(*
2
(*
t
a*)
i-y
+
is
known
so complicated that
r
~
...
.
+
J
a*) r
'
2
J
bi,*i x
,
'
71)
L(*
.(*
This expression,
ai)
(*
+
fax
-
+ ft* +
(*
2
+
7 ,)
as the "partial fraction
it is
+ £Mi
+ 7i)
ft*
1
Sl
+
J
(*
2
+ fax +
70
s
'
decomposition" ofp(x)/q(x),
simpler to examine the following example, which
318
Derivatives
and
Integrals
theorem,
2x 7
+
how
such an expression and shows
illustrates
to find
According to the
it.
possible to write
it is
+ 13x + 20x + 15* + 16* + 7x + 10
(x + x + 1) (* + 2x + 2)0 - l)
b
a
cx + d
ex + /
7
+
+ * + „ + 2 + 7-T-;
(x +
x (* - l)
+T-7T
1)
4
5
8* 6
2
3
2
2
2
2
7T~n
2
1
o
2
.
2
2a:
+k
—
gx
1
H
+*+
2
(*
at
I)
2
To find the numbers a, b, c, d, e, f, g, and h, write the right side as a polynomial
2
the
over the common denominator (x 2 + x + l) 2 (x + 2x + 3)(* — 1)
2
;
numerator becomes
«(*
+
(cx
1)(*
+
2
2x
+ d)(x-
+ x + I)
+ x+\) +(ex+f)(x-\) (x + + 2)(x + x + l)
+ (gx + h)(x- l) (x + 2x + 2).
+
2){x 2
2
2
l) (x
+x+
l)
+
2
2
£(*
+
2x
+
2
2
2)(*
2
2
2
2
2x
2
Actually multiplying
coefficients are
this
+
the coefficients of 2x 7
coefficient of x 8
out
is
8x 6
a,
+
we obtain
0)
we obtain
(!)
combinations of
.
.
.
+
13x 5
a polynomial of degree
h.
,
2
20x 4
whose
8,
Equating these
coefficients
+
+
15x*
+
16x 2
Ix
unknowns
8 equations in the eight
+
a,
with
10 (the
.
.
.
h.
3
After heroic calculations these can be solved to give
a
<
=
=
1,
b
=
2,
c
0,
/=
0,
£
=
=
1,
d
=
3,
0,
A
=
1.
Ix
+
4
Thus
2x 7
+
5x«
/
(*
2
+ 13x + 20* + 21x + 16x +
+ x + 1) + 2x + 2)(x - l)
4
5
3
2
V
J
2
r
-
j jx^T)
dx
+
+
i^c^^^^
i
We
it
+
,
*
r
example by starting with the
into one fraction.)
I
,
obtained
partial fraction decomposition
will illustrate all the difficulties
in integrating rational functions.
first
two
integrals are simple:
1
jX -
'
dx
=
log(x
-
l)
third integration depends
x2
+x+
1
-
1),
1
2
(*
The
3
and
are already in a position to find each of the integrals appearing in the
above expression; the calculations
The
+
/^ + ^ + 2-
may actually be feasible.
(In simpler cases the requisite calculations
this particular
converting
f
/c^H-i + i>»^
2
x
"2
- 1
on " completing the square
=
(*
+
i)
2
+
f
5
which
arise
Terms
319
— f instead of f we could not take the square root,
but in
Integration in Elementary
(If
we had
obtained
this case
our original quadratic factor could have been factored into linear
factors.)
We
The
can now write
substitution
changes
—+ i
=
x
u
du
=
-j^dx,
this integral to
i f
3
J
(
W
yl
+ l)
d
2
2
which can be computed using the third reduction formula given above.
Finally, to evaluate
/
we
(*
x + 3
+ 2x +
•
dx
2)
write
+ 3 dx-'-f 2X + 2 dx+
^
2 J x> + 2x + 3
x* + 2x + 2
J
The
first
it
du
The second
integral
simply 2 arctan
J
(x
J
+
l)
2
dx.
+
1
on the right side has been purposely constructed
by using the substitution
u
(x>
f
integral
can evaluate
f
2
*
[
is
2
-*
+
3
+
2x
+
(x
on the
+
*
right,
1). If
1
2)»
=
=
+ 2x + 3,
+ 2) dx.
x2
(2x
which
is
just the difference of the other two,
the original integral were
2*
/•
2 J
we
so that
(x>
+
+
2x
+ 2Y
^
2
2
*
^
f
[(x
J
+
l)
2
+
l]
n
'
on the right would still be evaluated by the same substitution.
would be evaluated by means of a reduction formula.
This example has probably convinced you that integration of rational func-
the
first
integral
The second
tions
is
integral
a theoretical curiosity only, especially since it
you can even begin. This
factorization of q{x) before
is
is
necessary to find the
only partly true. We
have already seen that simple rational functions sometimes
integration
1
+ * dx;
j
J - e>
:
1
arise, as in
the
another important example
-
is
the integral
i_ A
-p
1
=
ilog(*
-
1)
-
llog(*+l).
Z
z
1
if a problem has heen reduced to the integration of a rational functhen certain that an elementary primitive exists, even when the difficulty or impossibility of finding the factors of the denominator may preclude
Moreover,
tion,
it is
writing this primitive explicitly.
PROBLEMS
1.
This problem contains some integrals which require little more than
algebraic manipulation, and consequently test your ability to discover
algebraic tricks, rather than your understanding of the integration
processes. Nevertheless, any one of these tricks might be an important
preliminary step in an honest integration problem. Moreover, you want
to
have some
feel for
which
integrals are easy, so that
the end of an integration process
resort to
(v)
/
it,
will
tan 2 x
is
in sight.
you can see when
section, if you
The answer
only reveal what algebra you should have used.
dx.
(Trigonometric integrals are always very touchy,
because there are so many trigonometric identities
that
an easy problem can
easily look hard.)
Integration in Elementary
2.
The
Terms
321
following integrations involve simple substitutions, most of which
you should be able
(i)
J
J
do
in
your head.
e* sin e* dx.
xl
(ii)
to
xe~
/log
dx.
x
(In the text this
dx.
was done by
parts.)
x
r \
(iv)
(v)
,
,
x
f
/
J
j
^
e**e
j
(viii)
J
(ix)
J
+
x
1
dx.
~^dx.
xVl -
x 2 dx.
log (cos x) tan x dx.
* log X
7
3.
x
dx
2e
x dx
f
(vii)
+
2
e *
Integration by parts.
x 2e* dx.
(i)
J
(iii)
j
(iv)
y
(v)
j (log*)
e
ax sin
bx dx.
sin
si
#
^MA
l
(vi)
(vii)
J
/
dx
scc 3 xdx. (This
is
a tricky and important integral that often
If you do not succeed in evaluating it, be
comes up.
sure to consult the answers.)
(viii)
j cos (log x)
dx.
(ix)
j V# log x dx.
(x)
j *(log x)
The
x
=
2
dx.
following integrations can
=
sin u, x
cos
all
be done with substitutions of the form
To do some of these you will need to remember
u, etc.
that
/ sec x dx
=
log (sec x
as well as the following formula,
+
tan x)
which can
also
be checked by
differ-
entiation:
J esc x dx
= —
log (esc x
+
cot x).
In addition, at this point the derivatives of
dx
f
(i)
/
v
the trigonometric func-
(You already know this integral, but use the sub— sin u anyway, just to see how it works
x
~~
*
all
handy.
tions should be kept
stitution x
out.)
(ii)
f
^X
V +x
dx
(iii)
,
J
/
(iv)
2
(vi)
/
(x)
The
is
no
=
sec 2
u,
you want
to use
the
u.)
1
—
(The answer will be a certain inverse function that
was gi ven short shrift in the text.)
1
dx
J
x2
,
xV\ + X
x Vl - x
2
s
J
-
dx
f -
//
V
(ix)
tan
.
v x2
/.- Vl
J
(vii)
1
=
dx
x
w
+
—
Vx -
r
(Since tan 2 u
substitution x
2
1
2
dx.
(Yqu wiU need
I
1
—
Vl +
x 2 dx.
(
tQ
remember the methods
and cos.)
for
integrating powers of sin
J
x 2 dx.
|V?=T*.
following integrations involve substitutions of various types. There
substitute for cleverness, but there
is
a general rule to follow: substi-
an expression which appears frequently or prominently; if two
different troublesome expressions appear, try to express them both in
terms of some new expression. And don't forget that it usually helps to
tute for
Integration in Elementary
Terms
323
express x directly in terms of u, to find out the proper expression to substitute for dx.
r
a)
+ Vx +
1
(
(H)
—
—
J
dx
/*
.....
1
Vx+ Vx
J
/dx
(The substitution
V +^
w
= ^
leads to
an
ing yet anot her substitution; this
1
integral requir-
is
all right,
but
both substitutions can be done at once.)
« /it
tan #
dx
(vi)
(viii)
made
1
+
+
4*
(vii)J
V
(Another place where one substitution can be
|
f
X
J 2
to
do the work of two.)
1
1
y *V»</*.
/>/ —^=# dx.
— Vx
1
(In this case
substitution,
*(x)
two successive substitutions work out
two obvious candidates for the first
and either will work.)
best; there are
1
[J*Z±.Ldx.
v
J
*
+
*2
1
previous problem provided gratis a haphazard selection of rational
functions to be integrated. Here is a more systematic selection.
The
#
y
W
( ii)
'
(iv)
f
J
(v)
+x —
2
A
—
x
*±i
+
3x
-
x 33 4. 7^.2
7x*
_
-
c„
5x
+
(*-
i)\x
+
2* 2
+x+
1
+
+ 4 *.
Jx^Ti
AT
,
1
3x 2
-
/ (* + 3)(* /•
,
3
J x*
r
v
^ + 7,-1—
f
(i)
1
5
,
-
1)
dx.
l) 2
*.
324
Derivatives
and
Integrals
(V
°
dx
2
(vii)
( ix )
J*
-
+ 2x >+l
3x + 3* +
+ 2x + Ix +
J x*
3
1
2
y (x + x —
+ l)
—
1
2x_
=
2
*
2
3
«. / , 2+ ; +1) 3^
(
*7.
(No holds barred.) The following integrations involve
methods of the previous problems.
Potpourri.
»
w
r*i
(iii)
(iv)
arctan x
/
/"
the
.
ax.
+
1
all
*2
—
arctan x
;
J
log
Vl +
dx.
y * log y/\ + *
2
(v)
(vi)
(vii)
y arcsin
(viii)
(ix)
*
f
J
+
1
•am x
dx.
sin x
x cos*
.
J>
^"
dx.
__
Vtan
/dx
*
8.
If you
6+1
at
—
sin X
_
.
fa
m
COS^ X
a:
dx.
+
(To factor x 6
1, first
factor y*
+
1,
using Problem
14.)
have done Problem
1
7-7, the integrals
(ii)
look very familiar. In general, the substitution x
—
for integrals involving
integrals involving
V# +
2
1
.
1
,
while x
Try
—
—
Problem 4 will
cosh u often works
is
the thing to try for
and
sinh u
(iii)
these substitutions
in
on the other
inte-
Integration in Elementary
grals in
Problem
(The method
4.
is
325
Terms
not really recommended;
it is
easier
to stick with trigonometric substitutions.)
*9.
The
world's sneakiest substitution
x
—
t
tan
—
is
x
=
dx
=
'
undoubtedly
2 arctan L
2
2
+
1
As we found
this substitution leads to the expressions
Problem 15-16,
in
™
2t
X
dt.
2
t
cosx =
= TT?'
1
-
2
t
TT7
2
This substitution thus transforms any integral which involves only sin
and
cos,
combined by addition, multiplication, and
division, into the
integral of a rational function. Find
f
(j)
J
(ii)
+
1
—
f
J
(iii)
1
(Compare your answer with Problem
*L
—
f
/
(i n this
sin
2
(
*10.
)
f
J 3
(There
+
b cos
x dx.
(An
it is
better to let
x
also another
is
problem
——
+
t
=
tan
Why?)
x.
exercise to convince
(A
way
to
do
this,
using
15 . 7> )
should be used only as a
*
Wv
case
sin 2 x
J a sin x
(iv)
l(viii).)
sin x
you that
this substitution
last resort.)
last resort.)
5 sin x
Derive the formula for J sec x dx in the following two ways:
(a)
By writing
cos x
1
cos x
_
~
1
—
1
["
2 Ll
sin
2
x
cos X
COS X
+ sin x
—
1
~1
sin *J
an expression obviously inspired by partial fraction decompositions.
Be sure to note that f cos x/{\ - sin x) dx = - log(l - sin at); the
very important. And remember that % log a
there on, keep doing algebra, and trust to luck.
minus sign
From
is
=
log
V a.
t = tan x/2. Once again, quite a bit of
required to put the answer in the desired form; the
expression tan x/2 can be attacked by using Problem 15-8, or both
answers can be expressed in terms of t. There is another expression
By using
(b)
the substitution
manipulation
for
is
which
sec x dx,
/
using Problem 15-8,
r
sec x dx
I
—
log
is less cumbersome than
we obtain
f
l+ tan
log (sec x
+
tan x);
t\ =
log(tang + ?)}
I
(
J
This
last expression was actually the one first discovered, and was
due, not to any mathematician's cleverness, but to a curious histori-
In
accident:
cal
amounted
Wright computed nautical tables that
1599
When the first tables for the
logarithms of tangents were produced, the correspondence between
the two tables was immediately noticed (but remained unexplained
to definite integrals of sec.
until the invention of calculus).
The
derivation of je x sin x dx given in the text seems to prove that the
only primitive of f(x) = e x sin x is F(x) = e x (sin x - cos x)/2, whereas
F(x) = x (sin x — cos x)/2
C is also a primitive for any number C.
Where does C come from? (What is the meaning of the equation
+
<?
xdx =
je x sin
(a)
Find / arcsin x
e
dx,
x
sin x
—
e
x
cos x
— J>
sin x dx?)
using the same trick that worked for log and
arctan.
*(b) Generalize this trick:
(a)
(b)
Find $f~ l (x) dx in terms of ff(x) dx. Compare
with Problems 12-15 and 14-10.
Find / sin 4 x dx in two different ways: first using the reduction
formula, and then using the formula for sin 2 x.
Combine your answers
to
obtain an impressive trigonometric
identity.
Express / log (log x) dx in terms of (log x)~ l
J
dx.
(Neither
is
expressable
in terms of elementary functions.)
Express jx 2e~x2 dx in terms of je~ x2 dx.
Prove that the function f(x) = e x /(e bx
primitive. (Do not try to find it!)
Prove the reduction formulas in the
dx
f
J
(1
+
and work on the
tion x
=
tan
u.)
X
r
n
2
)
J
(1
last integral.
+
+ ex +
text.
dx
*2)n-l
1)
has an elementary
For the third one write
x 2 dx
f
J
(1
+
(Another possibility
is
x *)n
to use the substitu-
:
Integration in Elementary
18.
*19.
Find a reduction formula
Terms
327
for
n
(a)
fx e°dx.
(b)
/(log x) n dx.
Prove that
/—2
Vt A
L
,
\dt
=
cosh x sinh x
x
-
2
2
i
(See Problem 17-4 for the significance of this computation.)
20.
Prove that
h
ja
h
=
f{x) dx
ja
f(a
+
-
b
x) dx.
21.
(A geometric interpretation makes this clear, but it is also a good exerhandling of limits of integration during a substitution.)
2
Prove that the area of a circle of radius r is wr (Naturally you must
22.
remember that w is defined as the area
Use induction and integration by parts
cise in the
.
of the unit circle.)
to generalize
Problem
14-6:
/;(/;(... (/>„)„,)...)„„.
23.
If /'is
continuous on
Lebesgue
Lemma
b\ use integration by parts to prove theRiemann-
[a,
for /:
6
lim f
This result
is
/(/) sin (X*) dt
=
0.
Problem 15-25, but it can be used to
the same way that the RiemannProblem 1 5-25 from the special case in
just a special case of
much
prove the general case (in
Lebesgue Lemma was derived in
which /
24.
is a step function).
Prove the following version of integration by parts for improper integrals
P
u'{x)v{x) dx
=
u(x)v(x)
Ja
The
first
—
|o
symbol on the right
side
lim u(x)v(x)
X—*
*25.
f
00
f
;
u(x)v'{x) dx.
fl
means, of course,
—
u(a)v(a).
96
One of the most important functions in
T(x)
= jj
analysis
e-H*- 1
is
the
gamma function,
dt.
Prove that the improper integral T(x) is defined if x > 0.
(b) Use integration by parts (more precisely, the improper integral
version in the previous problem) to prove that
(a)
I\*
(c)
Show
that T(l)
natural numbers
=
n.
1,
+
1)
=
xY{x).
and conclude that
I»
=
(n
-
1)! for all
328
Derivatives
and
Integrals
The gamma function thus provides a simple example of a continuous
function which 'interpolates" the values of n for natural numbers x.
Of course there are infinitely many continuous functions / with
'
!
(n — 1 )
there are even infinitely many continuous funcwith
f(x
1) = xf(x) for all x > 0. However, the gamma
/
function has the important additional property that log of is
convex, a condition which expresses the extreme smoothness of
=
f(n)
;
!
+
tions
A beautiful theorem due to Harald Bohr and Johannes
this function,
Molierup
=
/(l)
+
and f(x
1
T
states that
1)
is
the only function
=
xf(x). See the Suggested Reading for a
/ with
log
°
/ convex,
reference.
*26.
(a)
Use the reduction formula
/v/2
sm n x
/
—
n
=
dx
for
Now show
/2
r^
\
w
n
x dx to show that
2
sin
/
Jo
(b)
sin
J
n~2
x dx.
Jo
that
2
sin^,^ =
-
o
.i.^
3
T/2
,n
sin 2 "
.
,
x <&
7
5
+
In
x
1
3 5
-•-•-•-
=
2
o
4
2
1
-
2»
1
,
In
6
and conclude that
72
7r_2244
2
(c)
3
1
Show
3
<
This
5
5
sin
1
2n
7
-
2«
2n+1
result,
can be
1
+
<
sin
and
x
2w
*
2n
1
+
sin2n *
/o"
1
^
^Si^n+i ~
^*
2
2
4
1
3
3
made
271
<
x
sin
271
"1
x
for 0
U
<
x
<
tt/2.
2n
2n
6
5*5*7
'
2n
-
1* 2n
as close to ir/2 as desired,
7T
known
=
2
Show
is
l/2n, starting with the inequalities
which shows that the products
infinite product,
(d)
6
that the quotient of the two integrals in this expression
between
0
6
2
2
as Wallis
4
4
6
r3*3*5*5
5
is
+
1
usually written as
an
product:
6
'7*
also that the products
1
\Tn
1
•
2
•
3
•
4
5
;
•
6
.
2n
•
.
.
•
(2n
-
1)
can be made as close to Vx as desired. (This
problem and in Problem 26-18.)
fact
is
used in the next
Terms
Integration in Elementary
27.
It is
an astonishing
computed
where ordinary
in cases
no elementary formula
J
improper integrals
fact that
for
can often be
f(x) dx cannot.
integrals
Ja
but we can
e~ x * dx,
ja
f(x) dx
Jq
329
There
is
find the value of
e~ x dx precisely! There are many ways of evaluating this integral,
J
but most require some advanced techniques; the following method
involves a fair amount of work, but no facts that you do not already
*
know.
(a)
Show
that
,24
i
3
o
1
i
(1
—+
In
=
(1
—7'
2n
5
1
7T
1
3
2n
-
2
2
4
2n
-
dx
+
x 2)
71
2
(This can be done using reduction formulas, or by appropriate
substitutions,
(b)
combined with the previous problem.)
Prove, using the derivative, that
1
-
x2
<
xi
<
~
e~
(c)
e~ x *
for 0
<
x
for 0
<
x.
1
1
+
<
1.
x2
Integrate the nth powers of these inequalities from 0 tol
to oo
,
respectively.
Then
use the substitution y
3
and from 0
V nx to show that
'
In
4
Vn
=
+
2n
5
1
\fn
<
e~y* dy
j
<
/
e~y" dy
q
<-Vn2
2
(d)
Now
(a)
2n
2n
-
3
2
use Problem 26(c) to show that
I
*28.
3
4
rv
Use integration by parts
f
b
'
sin
.v
Ja
,
f
dx
/
=
X
and conclude that
gate the limit as
Jj
to
dy
-
T-
show that
—+
cos a
b
cos £
a
0
(sin
*)/*
P cos x dx,
/
Ja
exists.
+
a -» 0 and the right
b
f
.
X2
(Use the
left
side to investi-
side for the limit as b
—
>
co
.)
330
Derivatives
and
/
Integrals
(b)
Use Problem 15-31
to
•
show that
sin (n
+ \)t dt
=
7T
i:
for
(c)
any natural number
n.
Prove that
lim
sin(X
+
\)t
dt
=
0.
sin
(d)
Hint: The term in brackets is bounded by Problem 15-2(vi); the
Riemann-Lebesgue Lemma then applies,
Use the substitution u = (X + %)t and part (b) to show that
sin x
~
~
X
Jo
*29.
(a)
(b)
Use the
Find
=
x to
t
ro) =
x Jo
substitution u
X
_
~
7T
2
show that
r e-« *dx.
l
PART
INFINITE
SEQUENCES
AND
INFINITE
SERIES
One of the most remarkable
algebraic analysis
m
m(m —
+-x+
1-2
1
of
series
is the following
I)
2
X'
1
m(m -
1)
(
m-
2)
„.
1-2-3
m(m -
+
1)
•
•
-
[m
-
(n
1-2
1)]
n
+
m
when
which
as
is
a positive whole number
sum of the
the
is
is
series,
then finite, can be expressed,
known, by
When m
is
( 1
+
m
x)
.
not an integer,
the series goes on to infinity,
and
it
will
converge or diverge according
as the quantities
m
and % have
this or that value.
In this case, one writes the same equality
+x) m =
(1
1
+™x
—
A jT2~
m(rn
+
.
.
.
It is
-
1)
+ "'
etc
assumed that
the numerical equality will
whenever the
this
x
always occur
series is convergent, but
has never yet been proved.
NIELS HENRIK ABEL
'
APPROXIMATION BY
POLYNOMIAL FUNCTIONS
CHAPTER
There
at
all.
one sense in which the "elementary functions" are not elementary
If p is a polynomial function,
is
p{x)
=
<2
0
+
+
aix
*
+
'
•
then p(x) can be computed easily for any number
At
functions like sin, log, or exp.
a nx
x.
n
,
This
present, to find log x
not at
is
=
the error involved in accepting such a
for log x
is
true for
1/tdt approxi-
J*
mately, we must compute some upper or lower sums, and make
sum
all
certain that
not too great.
Com-
log" (a:) would be even more difficult: we would have to computing e
pute log a for many values of a until we found a number a such that log a is
x
approximately x then a would be approximately e
In this chapter we will obtain important theoretical results which reduce
the computation of/(*), for many functions /, to the evaluation of polynomial
—
x
1
—
.
finding polynomial functions which are
appropriate,
close approximations to/. In order to guess a polynomial which is
thoroughly.
more
themselves
it is useful to first examine polynomial functions
The method depends on
functions.
Suppose that
p(x)
It
is
at 0.
To
a0
+
+
aix
'
•
+ anxn
•
.
our purposes very important, to note that the coeffican be expressed in terms of the value of p and its various derivatives
interesting,
cients en
=
and
for
begin with, note that
p(0)
=
a,.
Differentiating the original expression for p(x) yields
=
p'(x)
ax
+
2a 2x
+
'
•
•
+
na n x
n '\
Therefore,
P'(0)
Differentiating again
V'(jr)
=
we
2a 2
=/> (1) (0) =ai.
obtain
+
3
•
2
•
a zx
+
•
•
•
+
n(n
-
1)
a nx
•
n "\
Therefore,
2
^"(0) =/,< >(0)
=
2a+
|
In general,
we
will
have
/><*>(0)
agree to define 0! =
formula holds for k = 0 also.
If
wc
=
1,
kla k
and
or
ak
= ~jr'
recall the notation
(0)
/>
=
p, then this
333
I
334
Infinite Sequences
and
Infinite Series
If we had begun with a
(*-«),"
p(x)
=
a0
function
p
that
was written
"polynomial in
as a
+ ai(x -*)+'••+ a n (* -
a)
n
,
then a similar argument would show that
—
=P
flt
Suppose now that /
(a)
a function (not necessarily a polynomial) such that
is
/<»(*),
all exist.
ik)
.
.
,/»>(«)
.
Let
=
a*
——
<
0
>
k
<n,
"
k\
and define
The polynomial
Pn.a
(Strictly speaking,
Pn.aj, to indicate the
will
be
useful.)
called the
is
we should
Taylor polynomial of degree n for / at
dependence on
/;
The Taylor polynomial has been
Pn/
k)
a.
more complicated expression, like
at times this more precise notation
use an even
(a)
= f k) (a)
for 0
<
defined so that
<
k
n;
it is clearly the only polynomial of degree < n with this property.
Although the coefficients of Pn a j seem to depend upon / in a fairly complicated way, the most important elementary functions have extremely simple
Taylor polynomials. Consider first the function sin. We have
in fact,
,
sin(O)
sin'(0)
sin"(0)
sin"'(0)
sin (4) (0)
From
this
= 0,
= cos 0 = 1,
= - sin 0 = 0,
= - cos 0 = -1,
= sin 0 = 0.
point on, the derivatives repeat in a cycle of
4.
The numbers
sin ( *>(0)
are
11—
0, 1, 0,
>
0,
31
5
,
0,
11—
>
7!
!
0,
•
•
•
Therefore the Taylor polynomial ftn+i.o of degree 2n
Oct
X3
,
3!
(Of course,
P
2 »+i,o
=
iV^.o).
X5
X1
5!
7!
•
•
9!
+
1
for sin at 0
—I— i
(2n
+
1)1
is
—
,
335
Approximation by Polynomial Functions
The Taylor polynomial P 2n ,o
are
left
of degree In for cos at 0
is
(the computations
to you)
x2
,
2!
x4
x6
4!
6!
.
mxm
,
,
x 2n
(2n)l
The Taylor polynomial for exp is especially easy to compute.
=
=
exp(O)
1
Taylor polynomial of degree n at 0
for all k, the
u3
y
^.o(x)
The Taylor polynomial
since log
0.
The standard
•
*
choice
1
log'W =
•
+^r-
must be computed at some point a
for log
not even defined at
is
is
yi4
1+- + - + - + -+
=
Since exp (fc) (0)
—>
log'(l)
=
is
a
=
1.
0,
Then
1;
X
1
log"W =
X2
—2
log"'W =
X3
in general
(-l)*- 1 ^ --1)!
log<*>(*)
1
**
Therefore the Taylor polynomial of degree n for log at
Pn.l(x)
=
(X
~
—
1)
+
2
It is often
this case
more convenient
"
n
to consider the function f(x)
—
log(l
+ x).
In
We have
a
=
0.
/<*>(*)
=
log (A) (l
/<*>(0)
=
log<*>(l)
we can choose
+
+
3
1 is
+ *),
so
=
(-l)*-»(*
-
0!.
Therefore the Taylor polynomial of degree n for / at 0
-*-- +
/»„..(*)
2
— -+
4
3
•••
+
is
(-1)"-V
/»
There is one other elementary function whose Taylor polynomial
arctam The computations of the derivatives begin
tant
is
impor-
—
arctan'O)
=
—
-
1
arctan"(*)
=
+x
~ 2*
+#
+*
=
(1
=
1;
=
0;
2
2
)
2
x
arctan"(0)
>
/
(1
arctan"'(*)
arctan'(O)
>
2
——
2
)
*
(-2)
-f
(1
—
+2*
+*
2
-2(1
4
)
+*5_v2*
>
m = — 2.9
„„
,
arctan'"(0)
336
Infinite Sequences
and
Infinite Series
It is clear that this brute force computation will never do. However, the
Taylor polynomials of arctan will be easy to find after we have examined the
properties of Taylor polynomials more closely
although the Taylor poly-
—
a as
/,
Pn ,a,/
was simply defined so as to have the same first n derivatives at
the connection between / and Pn a j will actually turn out to be much
nomial
,
deeper.
One line of evidence for a closer connection between / and the Taylor polynomials for / may be uncovered by examining the Taylor polynomial of
degree 1, which is
*..(*)-/(«)+/(«)(*-«).
Notice that
/(*)
Now, by
FIGURE
- PUx)
=
-f(a)
/(*)
_
we have
the definition of f(a)
1
In other words, as x approaches a the difference /(*) — Pi, a (x) not only
small, but actually becomes small even compared to x — a. Figure 1
becomes
illustrates the
graph of f(x)
—
e
x
and
of
Pi.oM =/(0) +f(P)x
which is the Taylor polynomial of degree
the graph of
*..(*)
which
is
= /(0) + /'(0)
+
1
=l+x,
for / at 0.
^
x*
=
—
also
shows
1+ x + £
0. As x approaches 0, the
seems
small
to
getting
even faster than the
be
i\o(*)
the Taylor polynomial of degree 2 for / at
difference f(x)
The diagram
Approximation by Polynomial Functions
difference f(x)
we
are
—
Pi,o(x).
now prepared
As
it
in general
i
im
x—ki
=
For f(x)
e
x
and
a
=
lim
On
0 this
is
not very precise, but
We
have just noted that
stands, this assertion
it
to give
a definite meaning.
^M
—
x
means
=
337
o.
a
that
/w-A,w
g
=
lim
--i-,
=
Q-
the other hand, an easy double application of PHopital's Rule shows
that
eX
i
Z
im
z->o
"
1
a:
*
=
2
±
j
o.
— Pi.oW becomes small compared to x, as x approaches
become small compared to x 2 For ft.oW the situation is quite
Thus, although /(#)
does
0, it
not
.
different; the extra
term x 2 /2 provides just the right compensation:
e
x
—
1
—
x
2
X
a?->0
e
=
lim
—
if f{a)
Suppose that /
is
—
x
1
=
0.
2
and f'{a)
Pn>a
exist,
then
0;
also true.
a function for which
is
/'M,
all exist.
—
lim
x— o
=
in fact, the analogous assertion for
l
x
1
2X
e
lim
THEOREM
—
x-+0
=
This result holds in general
x
lim
2
.
.
,/->(«)
.
Let
ak
=
f
{k)
—
(a)
0
'
7i
<
A
'
'
<
n,
k\
and
define
^n
t
aW =
a0
+ ai(x —
a)
+
'
+ an {x —
Then
lim^j ~
*->a
(x
—
^-W
a)
n
=
0.
a)
n
.
338
Infinite Sequences
and
Infinite Series
PROOF
Writing out
Pn ,a(x)
we
explicitly,
obtain
h
- pu*) =
n
(x - a)
zoo
i
-
(x
It will help to introduce the
new
<w)
-
/
a)
(«)
n
,
n\
functions
n-l
Q(x)
now we must
=
Z^tp
-
(*
=
*W
and
«)'
(
*
~
prove that
lim
/W - QW
_/
<n)
(«)
.
n\
g(x)
Notice that
= /<*>(«),
Q(«(a )
£<»(*)
=
Q(*)]
=
-
n!(*
k
<
-
n
1,
*)"-*/(» -A)!.
Thus
-
lim [/(*)
-
/(«)
Q(«)
=
0,
X—>fl
lim
-
[/'(*)
lim [/ (B
- 2)
= /'(*) - Q'OO =
Q'(*)]
W
-
= /<"-
Q<»- 2) (*)]
J)
o,
- Q (n-
(a)
2)
(a)
=
=
0.
0,
2
and
limgW =
lim*'(*)
x—<*
x—*a
We may therefore
Q is a
/(*)
=/
x-^a
and
this last limit
•
•
-
is
(n
" 1)
(x
Q(«)
(
"- 2)
(*)
-
(n - 1}
_
L/
-
1
times to obtain
M - Q^W
1, its (n
-
.
l)st derivative is
a constant;
Thus
(^).
-
= lim£
x—*a
polynomial of degree n
in fact,
•
apply THopitaPs Rule n
lim
Since
=
a)
n
f n) (a)/nl
x-~*a
by
n\(x
definition of
f
n)
-
a)
(a). |
One simple consequence of Theorem 1 allows us to perfect the test for local
maxima and minima which was developed in Chapter 11. If a is a critical
local minipoint of /, then, according to Theorem 11-5, the function / has a
mum at a if f"{a) >
0,
and a
conclusion was possible, but
further information;
and
0. If/" (a) = 0 no
7
" (a) might give
of/
sign
that
the
conceivable
local
it is
if f"(a)
maximum at a if/" (a) <
-
0,
(4)
then the sign of/ (a) might be
signifi-
.
Approximation by Polynomial Functions
Even more
cant.
generally,
=/»
/'(«)
(*)
we can
f n) (a) *
The
•
•
=0,
(a)
•
0.
can be guessed by examining the functions
situation in this case
- (* - «)•
= -(*-«)",
f(x)
*(*)
which
when
ask what happens
=
339
Notice (Figure 2) that if « is odd, then a is neither a local
nor a local minimum point for / or g. On the other hand, if n is
even, then /, with a positive nth derivative, has a local minimum at a, while g,
satisfy (*).
maximum
with a negative wth derivative, has a local
maximum
at
a.
Of
functions
all
about the simplest available; nevertheless they indicate
the general situation exactly. In fact, the whole point of the next proof is that
any function satisfying (*) looks very much like one of these functions, in a
satisfying (*), these are
sense that
is
made
precise
by Theorem
1
Suppose that
=
f(a)
f
(1)
If n is
(2)
If n
is
(3)
If n
is
There
is
n)
•
(a)
*
>
<
0,
even and f (n) (a)
even and f (n) (a)
=
•
then / has a local minimum at a.
then / has a local maximum at a.
odd, then / has neither a local maximum or minimum at
clearly
no
0,
of generality in assuming that f(a)
loss
the hypotheses nor the conclusion are affected
Then, since the
Pn>a
of /
—
n
first
1
if
/
is
derivatives of / at a are 0,
=
a.
0, since
neither
replaced by / — f(a).
the Taylor polynomial
is
PnM) = f(a)
(X
1
(n)
f
Thus, Theorem
0
=
Consequently,
if
lim
x
is
(x
Suppose now that
—
a)
(x
•
_
a)
n
n\
states that
1
/M - />„.„(*) _ _
J
1:
—
a)
n
—a
*-»a
sufficiently close to a,
/(*)
n
-«)+•
!
(a)
(#
f(x)/(x
0,
0.
n
—
r
L(*
f n)
/«
—
-
fl)
a)
n
has the same sign as
7f
(n)
0 for
all
(a)
n!
even. In this case (x
has the same sign as f
J
then
a)*
is
(
nl
(n)
—
a)
n
>
x
a.
Since
(a)/n\ for x sufficiently close to a, it
340
Infinite
Sequences and Infinite Series
same
follows that f(x) itself has the
a.
lif
n)
>
{a)
for x close to a. Consequently,
for the case
Now
f
(n)
<
(a)
suppose that n
sufficiently close to a,
is
-
a)
n
>
-
0 for x
signs for x
different
mum nor a local
for x sufficiently close to
>
0
/ has a
=
f(a)
minimum
local
at
a.
A
similar proof
0.
The same argument
odd.
/(*)
(x
\
as before shows that
if xis
then
(x
But
n
f (a)/n
means that
0, this
f(x)
works
sign as
—
a)
>
u
u
the same sign.
always has
n
and
a
-
(x
> a and x < a. This
minimum at a. \
n
< 0 for x < a. Therefore f(x) has
proves that / has neither a local maxi-
a)
Although Theorem 2 will settle the question of local maxima and minima
about any function which arises in practice, it does have some theoretical limitations, because f (k) (a) may be 0 for all k. This happens (Figure
for just
3(a)) for the function
K
)
=
x
\ 0,
0,
which has a minimum at 0, and also for the negative of this function (Figure
3(b)), which has a maximum at 0. Moreover (Figure 3(c)), if
-
/(*)
-e~
{
then f
(k)
(0)
maximum
(c)
FIGURE
The
3
=
0 for all
k,
lix
\
x
=
<
0
0,
but / has neither a local
minimum nor
a local
at 0.
conclusion of
Theorem
is
1
concept of "order of equality."
at a
x
Jo,
often expressed in terms of an important
Two functions / and g are
equal up to order n
if
lhn^-y
—
(x
x~+a
a)
n
=
0.
Theorem 1 says that the Taylor polynomial
The Taylor polynomial might very well have
In the language of this definition,
Pn.aj equals /
up
to
order n at
a.
been designed to make this fact true, because there is at most one polynomial
< n with this property. This assertion is a consequence of the following elementary theorem.
of degree
THEOREM
3
PROOF
Let P and Q be two polynomials in (x — a), of degree <
P and Q are equal up to order n at a. Then P = Q.
Let
R =P—
Q. Since
R is a polynomial of degree <
n,
and suppose that
n, it is
only necessary to
341
Approximation by Polynomial Functions
prove that
if
R(x)
=
+
b0
+
-
b n (x
a)"
satisfies
R(x)
—
x^a (x
then
R —
Now
0.
RW
lim
/
x-* a (x
For
=
2
—
=
.
surely imply that
0
<
for 0
=
x—>a
=
+
[b 0
<
n.
=
0;
on the other hand,
-«)+••• +
bi(x
-
<0
n
]
x—*a
=
b0
lim
i
ay
0 this condition reads simply lim R(x)
x —*a
\imR(x)
Thus
R
the hypotheses on
n
a)
6„.
0 and
/?(*)
=
4i(*
=
*i
+*
-«)+•
Therefore,
2
(*
-
a)
+
•
+
•
b n (x
-
a)
n~ l
and
X—>a
Thus
#i
=
*(*)
Continuing in
COROLLARY
Let / be
(x
PROOF
—
—
a
this
=
b 2 (x
-
a
y+
•
way, we find that
a, and suppose that P is a polynomial in
which equals / up to order n at a. Then P = Pn 0
times differentiable at
rc-
a) of
at
0 and
degree
<
n,
.
,
P and Pn a both equal / up to order n at a, it is easy to see that P equals
Pnta up to order n at a. Consequently, P = Pn a by the Theorem. |
Since
,
,
At
first
sight this corollary appears to
hypotheses;
it
have unnecessarily complicated
might seem that the existence of the polynomial
matically imply that /is sufficiently differentiable for
this is
not
so.
For example (Figure
s(
J
If P(x)
=
0,
is
undefined.
_
{
* n+ \
10,
,
P would
to exist.
auto-
But in fact
suppose that
x irrational
x rational.
P is certainly a polynomial of degree < n which equals / up
On the other hand, f(a) does not exist for any a 7± 0, so /"(0)
then
to order n at 0.
\
4),
Pn a
342
Infinite Sequences
and
Infinite Series
V
+1
x irrational
,
x rational
0,
FIGURE
4
When
useful
/ does have
method
that our
first
however, the corollary may provide a
Taylor polynomial of /. In particular, remember
find the Taylor polynomial for arctan ended in failure.
n derivatives at a,
for finding the
attempt
to
The equation
=
arctan x
/
Jo
suggests a promising
1
by
1
+
t
2
,
1
+
dt
2
t
method of finding a polynomial
close to arctan
—divide
to obtain a polynomial plus a remainder:
This formula, which can be checked easily by multiplying both sides by
shows that
=
f
/
x
+
1
.
•
fx
+ (-1)V»*+
•
/
J°
JO
3
y 2n+I
+(- )"^xr +
2n +
+
1
5
(-i) n+1
1
fx
/
Jo
1
+
t
2
,
f2n+2
1
1
f
+
dt
t
2
2n+2
hrrt*1 +
t
2
According to our corollary, the polynomial which appears here will be the
Taylor polynomial of degree In + 1 for arctan at 0, provided that
^2n+2
Jo
lim
1
—+
dt
2
2
t'
X 2n+1
X-+Q
=
0.
Since
^2w+2
rx
7o
this is clearly true.
2n
+
1
i
+
rx
2
y0
/
2/z
+
3
Thus we have found that the Taylor polynomial of degree
for arctan at 0 is
X
=*--X + ---•+
(-1)"
3
ft«+uW
3
h
5
~2n+l
2n
+
1
:
!
343
Approximation by Polynomial Functions
By
is
now
the way,
that
w
we have
*)
and
=
is,
by
find arctan a) (0)
-... +( -ir
2n
I
+
.
i
definition,
+ arctan-(O), +
arctan-(O)
it
find arctan a) (0) for all k: Since
*-- +
since this polynomial
we can
discovered the Taylor polynomials of arctan,
work backwards and
possible to
2
*+
+
(2/2
!
by simply equating the
+
^
1)
coefficients of x k in these
two
polynomials
arctan a) (0)
=
.
....
0
if
k
is
even,
k\
arctan » +1) ( 0)
(
(2/+
= (^}Y
arctan <« +l (0)
>
or
21+1
1)!
=
.
(2 /)|.
A much more interesting fact emerges if we go back to the original equation
—
y3
-
*
y5
T+?-
-
«
y 2n-fl
+
•
27+T
+ - ,r+I
fx
t
2n+2
(
/.
and remember the estimate
^2n f 2
o
When
<
|#|
as small as
<
|*|
as
we
1,
we
we can
1
this expression
1
is
dt
+
at
2n
+
most 1/(2h
3
+
3),
and we can make
use
The most important theorems about Taylor polynomials extend
like.
this
simply by choosing n large enough. In other words, for
the Taylor polynomials for arctan to compute arctan x as accurately
like
this
isolated result to other functions, and the Taylor polynomials will soon play
quite a new role. The theorems proved so far have always examined the
behavior of the Taylor polynomial Pn a for fixed n, as x approaches a. Henceforth we will compare Taylor polynomials P nia for fixed x, and different n. In
,
anticipation of the coming theorem
If
/
a function for which
is
we introduce some new notation.
exists, we define the remainder term
P n a (x)
,
Rn a{x) by
t
/(*)
- Pn ,a(x)
= /GO
+ Pn.a&
+/'(«)(*
-*)+•••+ ^-7^ (* ~ *) n + RnA*).
n
I
We would like to have an expression for R n>a (x) whose size is easy to estimate.
There
is
such an expression, involving an integral, just as in the case for arctan.
to guess this expression is to begin with the case n = 0:
One way
/(*)
«/(n)
+RM-
344
Infinite Sequences
and
Infinite Series
The Fundamental Theorem
of Calculus enables us to write
/w =/oo + j'mdt,
so that
AuM
=
A similar expression for Ri, a {x)
x
f(t)dt.
ja
can be derived from
this
formula using inte-
gration by parts in a rather tricky way: Let
=
"(0
(notice that x represents
=
v'(t)
1);
and
f(t)
some
—
v(t)
number
fixed
t
—
x
in the expression for v(t), so
then
X
(
f (t)dt= P/'M-l
dt
i
I
u(t) v'(t)
=
-
u(t)v(t)\*
r
f
{t){t-x)
u'(t)
Since v(x)
=
dt.
v(t)
we obtain
0,
/w
=
+ j'mdt
- «(«>(«) +
/(«)
= /(«)
= f(a)
+f'(a)(x
-
*
-a) +
/"«(*
f'
/; /"W(*
o
t) dt.
Thus
RU*) = fa* f"(t)(x It is
v(t)
=
hard
t.
it is
any motivation for choosing v(t) = t — x, rather than
happens to be the choice which works out, the sort of thing one
to give
It just
might discover
ever,
now
after sufficiently
v'(t)
—
many
similar but futile manipulations.
easy to guess the formula
u (t)
then
t) dt.
(x
—
t),
=
f"(t)
(
and
v(t)
so
x
r f"(tKx Ja
How-
R 2ta (x). If
~
* - ^\
=
for
1)
dt
=
-
u(t)v(t)\
\a
r
Ja
-/'(«)('-«>'
2
'(0
+
•
~(x-t)
—
r^
Ja
{x
dt
-tyd,
2
This shows that
./«>(/)
(x
2
You
should
now have
little difficulty
-
2
O ^-
giving a rigorous proof, by induction,
1
345
Approximation by Polynomial Functions
that
if
f
(n+1)
continuous on
is
=
*»..(*)
then
[a, x],
/
^—t^
"
(*
0"
dt.
form of the remainder, it is
expressions for /?„,«(*):
important
possible (Problem 13) to derive two other
remainder,
the Cauchy form of the
From
formula, which
this
R n aix) =
called the integral
is
-
-
(*
t)
n
(x
-
for
a)
some
t
in (a, *),
n\
and the Lagrange form
of the remainder,
f(n+l)(A
Rn,a(*)
=
7
i
A
(*
-
SOme
f° r
we
will derive all three
One
virtue of this proof
In the proof of the next theorem (Taylor's Theorem)
forms of the remainder in an entirely different way.
in
1
Cauchy and Lagrange forms of
(aside from its cleverness) is the fact that the
the remainder will be proved without assuming the extra hypothesis that
(n+1)
direct generalis continuous. In this way Taylor's Theorem appears as a
/
ization of the
is
=
0,
and which
suggest a strategy for proving Taylor's
Theorem.
Mean Value Theorem,
to
which
it
reduces for n
the crucial tool used in the proof.
may
These remarks
R n a (a) =
Since
0,
we might
try to apply the
Mean Value Theorem
to the
,
expression
x
On
is
—
second thought, however,
not at
Indeed,
clear
all
if
we
this
how / (n+1) (f)
=
which
is
is
—
a
idea does not look very promising, since it
ever going to be involved in the answer.
take the most straightforward route,
of the equation which defines
/'(*)
x
a
f(a) +f"(a)(x
useless.
R n>a we
,
-*)+•••+
The proper
and
differentiate
both sides
obtain
(*
application of the
-
a)*" 1
+ /?„«'(*),
Mean Value Theorem
has a
This proof
lot in common with the integration by parts proof outlined above.
was fixed.
which
number
denoted
a
x
which
in
involved the derivative of a function
This
THEOREM
4
(TAYLOR'S theorem)
is
just
how
x will be treated in the following proof.
(n+1)
are defined on
Suppose that /',..., /
by
/(*)
=
f( a )
+ f(a )(x _«)+...+
[a,
x\ and that
(x
R n a (x)
is
denned
,
- aY + R n ,a(x )-
346
Infinite
Sequences and Infinite Series
Then
/(»+i>
(1)
ft)
Rn,aW =
-
(x
n
/)
~
(*
some
a)
for
for
some
in
/
(a, x).
,
(2)
Moreover,
(3)
(If
a:
<
f
Rnta (x) =
if /<
n+1)
a,
(t)
-
(*
;/
J
;
\n
-+-
is
integrable
n+1
a)
/
in
(a, x).
(n
+
1
t
!
1
on
then
[a, at],
/•<»+!) (V)
=
/?„..(*)
(n+1)
/
I
A.
./aft!
then the hypothesis should state that /is
)
-times differentiable
on [x, a]; the number / in (1) and (2) will then be in (x, a)> while
remain true as stated, provided that / (n+1) is integrable on [x, a].)
PROOF
For each number
/(*)
=
fit)
t
in
[a,
we have
x]
+f(t)(x
Let us denote the number
-
+
t)
R Htt (x)
•
by
•
+
•
^
S(t); the
-
(x
function
+
0"
(3) will
HA*)-
defined on
6" is
[a, x],
and we have
(*) fix)
=
fit)
-
+f'(t)(x
+
0
'
•
+
•
(*
-
0
n
+
J(0
for all
We
will
now
t
in
[a, *].
differentiate both sides of this equation,
function whose value at
t
is
/(#), equals the function
which asserts that the
whose value at t is
{n)
f(0
(In
+
f
+ ^—y
(t)
•
•
(*
-
ty
+ sit).
common parlance we are considering both sides of (*)
Just to
make
sure that the letter x causes
git)
=
for all
/(*)
"as a function of /.")
no confusion, notice that
t,
then
g'it)
and
=
0
for all
t;
if
f
(k)
(t)
then
g'it)
=l-Mk{x-Jf
{k)
w_
ty-\-\)
(t)
r
x
_
t
\k-i
+
L_w
f
_i_
Z
(k+i)
(x
(t)
\ll rx
_
_
t)
k
rt*
if
347
Approximation by Polynomial Functions
Applying these formulas
0
= f'{i)
+
[-f(t)
+
In
this beautiful
•
each term of
to
+
•
-
(x
obtain
+ [=£@ {x-t)+^(x-
0]
+
•
we
(*),
C*
0- +
-
1
^
<*
-
0-]
02
]
+ S'(t).
formula practically eveything in sight cancels out, and we
obtain
= -
S>( t )
I
-ifi
X
(
-
t
)\
n\
Now we can apply the Mean Value Theorem to
is
some
/
in
(a, x)
™ ~—
s(a)
at
Remember
=
means
(x
-
S on
[a,
x]
:
there
o-.
n!
<2
that
S(t)
this
- -
S'(t)
the function
^
such that
=
*„,,(*);
in particular that
= Rn x (x) =
= R„,«(x).
S(x)
,
J(a)
0,
Thus
o-i M W
x
—
/
.
<n+1)
W
a
or
Rn,*(*)
this
the
is
To
Cauchy form
= f—-^-(x-tnx--a);
of the remainder.
derive the Lagrange form
to the functions
S and g(t) =
(x
we apply
—
t)
n+ l
:
the
/
-
n\
g(*)-g(a)
g'(t)
+
*...(*)
(x
w
.
K
S'(t)
Thus
-
a)
n+1
or
is
(n+l)
S(a)
S(x)
which
Cauchy Mean Value Theorem
is some t in {a, x) such that
there
the Lagrange form.
= /
(n+1)
(0
(n+1)!
_. t)n
348
Infinite
Sequences and Infinite Series
Finally,
if
is
-
S(x)
on
integrable
fx
=
S(a)
fx f(*+i)(A
= -
S'(t)
/
then
[a, x],
Ja
^
I
/
Ja
(x
-
t)
n
dt
7l\
or
/(n+i)// f\
Cx /(n+i)
fx
~
7
=
Rn,a(*)
/
:
Ja
-
(x
t)
n
dt.
|
n\
Although the Lagrange and Cauchy forms of the remainder are more than
Problem 22-16), the integral form of the
remainder will usually be quite adequate. If this form is applied to the functheoretical curiosities (see, e.g.,
tions sin, cos,
and exp, with
a
=
0,
Taylor's
Theorem
yields the following
formulas:
sin *
=
*
-
—+—3!
5!
•
•
+
•
(-l) n
-
v
2
cos x
=
1
-
X
X*
+ --•••+
2!
**
To
1
(-1)"
k
;
4!
^.n
X 2n
(2«
+ l)!
1
fi cos (»+DM
+
—
/
(x
-
t)
2n
dt,
(2n)
Jo
/*x
-+...+ ?L +/
+* +
—
.»(2n+2)
Jo
(2n)\
y.2
=
+ 1)!
+
{In
'
.L
(*_,)»
evaluate any of these integrals explicitly would be supreme foolishness
the answer of course will be exactly the difference of the
other terms on the right side
!
To
and worthwhile.
The first two integrals are
sin
estimate these integrals,
left
side
however,
and
is
especially easy. Since
(2n+2)
|
W
|
<
!
for
aU
t}
we have
f oTn?
(2n +
Jo
~
(x
l)\
t)2n+1 dt
* ?^4lTT
f
{2n + l)\\Jo
(x
I
~
Since
(
x
-ty n+i
dt
= -(x
o
2rc
we conclude
2n+2
t=sx
\
2n
+
+
2
2
<-o
that
S in (2n+2) (Y)
Jo
/;
Similarly,
t)
(2«
we can show
fx
<
+ l)!
(2«
+ 2)!
that
coS
(2n+1)
(/)
{X
(2,)!
-Vndt
(2n
+
l)!
<)2
" +1
all
the
both easy
dt
—
Approximation by Polynomial Functions
proved in Chapter 16)
interesting, because (as
These estimates are particularly
any £ > 0 we can make
349
for
xn
1
n\
<
8
by choosing n large enough (how large n must be will depend on x). This
enables us to compute sin x to any degree of accuracy desired simply by
evaluating the proper Taylor polynomial P n ,o(x). For example, suppose we
wish to compute sin 2 with an error of
less
than
10~~ 4
Since
.
22n+2
-
sin 2
we can
P
use
+
/5 2n+i,o(2)
2 n+i,o(2)
where
R,
\R\
—--
<
-
>
as our answer, provided that
22n+2
A
it
number
<
+
(2n
.
can be found by a straightforward search
n with this property
n
obviously helps to have a table of values for n and 2 (see page 356) In this
!
case
it
happens that
—
n
sin 2
=
is
i>n,o(2)
+R
3!
5!
even easier to calculate
sin
To
sin
= P 2n +i,o(l)
1
obtain an error
than
less
.
5 works, so that
where
It
1(T 4
2)\
e
7!
\R\
11!
9!
< 10~
4
.
approximately, since
1
+
R,
where
1
<
(2n
we need only
_1
(2rc
\R\
+
<
find
an
n
+
2)\
such that
£?
2)!
only a brief glance at a table of factorials. (Moreover, the
P 2n +i,o(l) will be easier to handle.)
For very small x the estimates will be even easier. For example,
and
this requires
individual terms of
1
.
sin
-
10
P2n+M
+ (l^)
\10/
'
+
,
*'
'
lnt
where|i?l
1
'
1
.
<
10
2"
+2 (2/2
+
2)!
< 10~ 10 we can clearly take n = 4 (and we could even get away
These methods are actually used to compute tables of sin and
cos. A high-speed computer can compute P 2 n+i,o(*) for many different x in
almost no time at all.
Estimating the remainder for e* is only slightly harder. For simplicity
assume that x > 0 (the estimates for x < 0 are obtained in Problem 9). On
To obtain
with n
=
\R\
3).
.
350
Infinite Sequences
and
Infinite Series
the interval
maximum
the
[0, x]
value of
-{x-tYdt<^-
Since
we
since
is
(»_,)-
that
*
e
(«
<
x
pX Y n
(n
x n+l
Once
+
(n
!)!
x
,
< R follows
= 4, then
(The inequality 0
if
n
1)
!
sufficiently large.
will
make
x. If
<
0
How
things
more
<
then
x
1,
n
x
x
•••+=l+x + -+
+
2!
nl
In particular,
+
1)!
again, the estimates are easier for small
2
e
+X
4xx n+l
<
which can be made as small as desired by choosing n
large n must be will depend on x (and the factor 4 X
difficult).
increasing, so
is
we have
4,
+
exp
A
n\ Jo
know
already
l
rx
pX
fx ep t
/
Jo nl
e
where 0
/?,9
<R <
+
(n
1)!
immediately from the integral form for R.)
0<R< h
<
h
so
^
= ^1=1
+
1+1+1 + 1 + ^
2!
65
= Ya
5
where 0
< R < ^~
10
4!
3!
+r
-l + '-l + R,
'
24
which shows that
<
2
e
<
3.
(This allows us to improve our estimate of
R
slightly:
3 x n+i
*
By taking
«
=
7 you can
compute that the
*
=
2.718
.
x
first
3 decimals for
e
are
.
.
(you should check that n = 7 does give this degree of accuracy, but it would
be cruel to insist that you actually do the computations)
The function arctan is also important but, as you may recall, an expression
for arctan a) (*) is hopelessly complicated, so that the integral form of the
remainder is useless. On the other hand, our derivation of the Taylor polynomial for arctan automatically provided a formula for the remainder:
arctan x
=
x
x*
--+'•'
+
3
,
,
—+—
(~l)V> n+1
2n
1
p(-l)"+^+
,
h
/
Jo
TT~^
+t
1
2
2
dL
351
Approximation by Polynomial Functions
As we have already estimated,
dt
+
1
+
2n
!
t*
j
3
For the moment we will consider only numbers x with \x\ < 1. In this case,
the remainder term can clearly be made as small as desired by choosing n
sufficiently large. In particular,
arctan
With
1
=
1-
this estimate
-
1
+ --
3
5
1
it is
--
—
-+ ±~\}J- +
In +
n
(
where
R,
2n
for
to compute t. Fortunately, there are some
surmount these difficulties. Since
smaller
puting
shows
7r is
how
3
the remainder
less
w's will
work
for only
to express arctan
this
will usually
for arctan
clever tricks
should allow us
which enable us
somewhat smaller
#'s.
The
comProblem 5
trick for
in terms of arctan x for smaller x;
1
to
<
l**+x..(*)|
much
+
have to be so
To obtain a remainder
3)/2. This is really a shame,
4
example, we must take n > (10 =
polynomial
Taylor
1
the
tt/4, so
because arctan
,
make
easy to find an n which will
than any preassigned number; on the other hand, n
large as to make computations hopelessly long.
<10~ 4
1
<
\R\
1
can be done in a convenient way.
The Taylor polynomial for the function f(x) = log (a:
handled in the same manner as the Taylor polynomial
+
1) at a
=
1
is
best
Although
for arctan.
the integral form of the remainder for / is not hard to write down, it is difficult
to estimate. On the other hand, we obtain a simple formula if we begin with
the equation
'
\+t
=
- +
1
t
~
t*
•
*
+
'
(-l)"- 1 *"-
1
+
}
^
*
1
+
t
this implies that
1
^
y2
1
,+ '
)
=
/.r^'*-'-i
jr
+
xn
3
+( - 1) ""v
i--"
fx
for all x
> -1.
If
x
>
0,
Jo
and there
is
n
then
fx
/
t
t
+
y n ^~l
fx
fn
-=—dt <
1
n
t
dt
=
—
n
Jo
+
\
when — 1 < x < 0 (Problem
remainder term can be made as small as desired by
a slightly more complicated estimate
10). For this function the
choosing n sufficiently large, provided that
— <
1
x
<
1
352
Infinite Sequences
and
Infinite Series
The behavior
of the remainder terms for arc tan
when
quite another matter
>
|*|
—+—
-
2n
=
\og(x
+
1) is
n
for arctan,
3
-^7
+
<
l*».o(*)|
f(x)
2n+3
Ui
<
|#2n+i,o0)|
and
In this case, the estimates
1.
(*>0)for/,
1
m /m become large as m
becomes large. This predicament is unavoidable, and is not just a deficiency
of our estimates. It is easy to get estimates in the other direction which show
that the remainders actually do remain large. To obtain such an estimate for
when
are of no use, because
arctan, note that
+
t
<
2
+x
1
the bounds x
1
in [0, x] (or in
if t is
1
>
\x\
[x,
<
2
<
0] if x
2x\
then
0),
>
if |*|
1,
so
1
*2n+2
fx
\
1
Similarly,
if
x
>
Jo
0,
1
+
>
~
dt
t
2
then for
t
±
fX
2n
\l
2x \J
in [0, x]
+ < +x
1
+2
dt
4n
+
6
we have
<
1
t
t
2
2x,
if
>
x
1,
so
fx
t
/
Jo
t
n
+
i
dt
>
fx
-
/
r dt =
2n
2x Jo
1
+
2
These estimates show that if \x\ > 1, then the remainder terms become large
as n becomes large. In other words, for \x\ > 1, the Taylor polynomials for
arctan and / are of no use whatsoever in computing arctan x and log(x + 1). This is
no tragedy, because the values of these functions can be found for any x once
they are
known
for all x with
|#|
<
1.
This same situation occurs in a spectacular
We have already seen that f
that the Taylor polynomial
(k)
(0)
Pn
,
0
=
for
for the function
0 for every natural
/
number k. This means
is
Pn.oM =/(0)+/'(0)*+
7f"(0)
-^*
+
Z
!
=
way
2
in)
•
•
+ Jf-^x»
72
(0)
I
0.
In other words, the remainder term R n ,o(x) always equals f(x), and the Taylor
polynomial is useless for computing /(*), except for x — 0. Eventually we will
be able to offer some explanation for the behavior of this function, which is
such a disconcerting illustration of the limitations of Taylor's Theorem.
The word "compute 55 has been used so often in connection with our estimates for the remainder term, that the significance of Taylor's Theorem might
<
353
Approximation by Polynomial Functions
be misconstrued.
It is true
tional aid (despite
that Taylor's
its
THEOREM
5
PROOF
it
has
these will be developed
will illustrate
some ways
The first illustration
who have waded through the proof,
Theorem may be
impressive to those
an almost ideal computa-
consequences. Most of
two proofs
in succeeding chapters, but
7r is
is
ignominious failure in the previous example), but
equally important theoretical
Taylor's
Theorem
used.
in
which
be particularly
in Chapter 16, that
will
irrational.
e is irrational.
We know
e
=
el
that, for
=
1
any
n,
+h+h+
+ h + Rn
"
•
'
Suppose that e were rational, say e =
Choose n > b and also n > 3. Then
a/b,
where 0
*• <
>
where a and
ottoi"
b are positive integers.
so
—=
n\
b
Every term
+
+ n\ + —
2!
'
•
•
+ n\Rn+—
n\
than n \R n is an integer (the
Consequently, n\R n must be an integer
in this equation other
integer because n
>
b).
0
< Rn <
0
<n\R n
left
side
also.
is
an
But
3
(n+l)\
so
<-^-<\<\,
4
n
which
is
impossible for an integer.
The second
illustration
proved in Chapter 15:
is
+
1
|
merely a straightforward demonstration of a fact
If
/"+/=0,
f(0)
/'(0)
then /
=
0.
To
this,
observe
f* =
=
/
/(5) =
(/")'
prove
(4)
etc.
(/
(/
3
)'
first
=
=
=
that
=
f
k)
exists for
-/'>
=("/')'
(4))/
o,
o,
^
=-/"=/,
every
k\ in fact
354
Infinite Sequences
and
Infinite Series
This shows, not only that
ones: /,
Theorem
f
all
k)
exist,
=
-/, -/'. Since /(0)
/',
any
states, for
is
—
f
/
is
(x
-
continuous (since f {n+2)
(n+1)
|/
0.
Now
Taylor's
<
(0l
(we can add the phrase "and
M
all
*)" dt.
nl
M such that
number
a
are
0, all /<*>(())
fx f(n+D(A
=
Jo
there
=
that
n,
fix)
Each function f {nJrl)
but also that there are at most 4 different
/'(0)
for 0
<
exists), so for
<
*
and
x,
a//
any particular x
tz
n" because there are only four different
f
k)
).
Thus
\
Since this
is
true for every
by choosing n
M\x\ n+l
< Ml ['<*—
\f(x)\
Jo
n
and
rc,
(»+
nl
can be
shows that |/(*)
since x /n
sufficiently large, this
!
|
1)1
made
<
as small as desired
s for
any
£
>
0; conse-
quently, f(x) = 0.
The other uses to which Taylor's
Theorem will be put in succeeding chapcomputational considerations which have concerned us for much of this chapter. If the remainder term R n a (x ) can be made
as small as desired by choosing n sufficiently large, then f(x) can be computed
to any degree of accuracy desired by using the polynomials Pn ,a( x )- As we
require greater and greater accuracy we must add on more and more terms.
ters are closely related to the
,
If we are willing to add up infinitely many terms (in theory at least!), then
we ought to be able to ignore the remainder completely. There should be
4
infinite
sums"
like
sm
x
=
xz
,
=
X2
,
1
arctan x
+
*)
7!
X4
X6
4!
6!
,
*
'
'
,
xs
,
=
x
——
=
x
Y2
YZ
v4
2!
3!
4!
^5
^7
5
7
h
- X^ + X4 - ^ +
z
3
4
infinite
>
•
•
,
•
•
•
if |*|
•
•
•
if
We are almost completely prepared for this step.
necessary definitions.
•
,
H
—we have never even defined an
•
•
8!
=\+x + -+-+-+
3
log(l
5!
h
2!
e*
x7
h
3!
COS x
x5
x
<
-1<
1,
X
<
1.
Only one obstacle remains
sum. Chapters 21 and 22 contain the
Approximation by Polynomial Functions
355
PROBLEMS
1.
(of the indicated degree,
Find the Taylor polynomials
and
at the indi-
cated point) for the following functions.
(i)
/(*)
=
e
(ii)
/(*)
=
e*
(iii)
sin;
degree
(iv)
cos; degree 2n, at t.
(v)
exp; degree
(vi)
log; degree n, at 2.
(vii)
/(*)
=
x6
(viii)
/(*)
=
xh
(ix)
f(x)
=
~^-
/(*)
=
e
degree
';
x
in
at 0.
3,
degree
3, at 0.
7T
(x)
2.
2n, at -•
at
n,
1.
+ x + x;
+ x + x;
3
degree
4,
at 0.
z
degree
4,
at
2
7^t~;
x ~t~ -V
degree 2n
;
degree
n,
+
1,
1.
at 0.
at 0.
Write each of the following polynomials in x as a polynomial in (x — 3).
(It is only necessary to compute the Taylor polynomial at 3, of the same
degree as the original polynomial. Why?)
3.
-
(i)
x2
-
Ax
(ii)
x4
-
I2x*
(iii)
x\
(iv)
ax 2
+
bx
9.
+
+
44a:
2
+
2x
+
1.
c.
notation) which equals each of the following
numbers to within the specified accuracy. To minimize needless comn
putation, consult the tables for 2 and n on the next page.
Write down a sum (using
2
\
(i)
sin 1; error
< 1(T 17
(ii)
sin 2; error
<
1(T 12
(iii)
sin J; error
<
10~ 20
(iv)
e;
(v)
e
error
2
;
error
<
<
10
-4
.
10" 5
.
.
.
.
—
Z
71
•
1
2
1
2
4
2
3
8
6
4
16
24
120
720
5
32
6
64
7
128
5,040
8
40,320
9
256
512
362,880
10
1,024
3,628,800
11
2,048
39,916,800
12
4,096
479,001,600
13
8,192
6,227,020,800
14
16,384
87,178,291,200
15
32,768
1,307,674,368,000
16
65,536
20,922,789,888,000
17
131,072
355,687,428,096,000
18
262,144
6,402,373,705,728,000
19
524,288
121,645,100,408,832,000
20
1,048,576
2,432,902,008,176,640,000
This problem
is similar to the previous one, except that the errors
are so small that the tables cannot be used. You will have to
thinking, and in some cases it may be necessary to consult the
demanded
do a
little
n
Chapter 16, that x /n can be made small by choosing n large
the proof actually provides a method for finding the appropriate n. In
the previous problem it was possible to find rather short sums; in fact, it
proof, in
!
was possible to find the smallest n which makes the estimate of the
remainder given by Taylor's Theorem less than the desired error. But in
this problem finding any specific sum is a moral victory (provided you
can demonstrate that the sum works).
< 10- (1 ° 10)
10" 1>00 °.
(i)
sin 1; error
(ii)
e;
error
<
< 10~ 20
10~ 30
(iii)
sin 10; error
(iv)
e
(v)
arctan toS error
(a)
10
;
error
<
.
.
.
<
10~ (1 ° 10)
.
Prove, using Problem 15-8, that
7T
- =
4
1
arctan -
+
2
*
- = 4A arctan
1
4
5
1
arctan ->
3
arctan
—
1
239
Approximation by Polynomial Functions
Show
(b)
that
=
ir
3.14159
.
.
man
(Every young
.
.
357
should verify a
few decimals of ir for himself, but the purpose of this exercise is not
to set you off on an immense calculation. If the second expression in
part (a)
used, the
is
remarkably
decimals for
first 5
can be computed with
ir
work.)
little
For every number a, and every nonnegative integer
"binomial coefficient"
6.
\n)
a
If
n
>
Show
a.
define the
n\
not an integer, then
is
we
n,
never
is
(^j
and
0,
alternates in sign for
that the Taylor polynomial of degree n for f(x)
=
(1
+ x)«
n
at 0
P ,oW =
is
x k and that the Cauchy and Lagrange forms
^
rt
k
,
(j^J
=0
of the remainder are the following:
Cauchy form:
_
,
N
RnAx) =
a(a
—
—
1)
(a
•
•
•
"
a(a
—
—
"
n
1)
•
i
•
n
—
n)
*(*
n
- 0 m/A
(l +
'( 1
+ ')-'(rr-;)"
,
(a
•
—
n)
!
+ •>(,+ J** +
-("
n
.
.
tY
1
!
1
(rd)"'
'
in
[0
-
or
[*•
0I
-
Lagrange form:
_
a(a
Ai.oW =
.
.
'=
(n
—
+
1)
*
•
.
1N
(a
—
+
^
(w+1)!
J
,
* n+1 d
n)
„,., A
* n+1 0
a_n_1
'
'
+
,
N
0°
„.
n
1
in [°' *1 or
&
Estimates for these remainder terms are rather difficult to handle, and
are postponed to Problem 22-16.
7.
Suppose that
a of
g
(i)
/ and
(a)/i\.
ai
and
bi
are the coefficients in the Taylor polynomials at
g, respectively.
Find the
In other words,
coefficients
a
+ g.
(0
f
fg.
(iii) /'•
(iv)
h{x)
=
(v)
k(x)
=
j'f{t)dt.
J*
f(t) dt.
=
f
{i)
(a)/i\
and
bi
=
of the Taylor polynomials at a of the
following functions, in terms of the afs
(ii)
ai
and
b/s.
358
Infinite
Sequences and Infinite Series
*8.
(a)
An
+
1
at 0
P
is
the Taylor polynomial of degree 2n
+
1
=
P(x)
R(x)/x 2n+1
=
0.
this
(b)
Find f
(c)
In general,
gat
+
imply about lim
ik)
(0) for all
if f(x)
where
R(x),
^(Ar
2
)A
4n+1
Km
if
x
<
What
does
?
A:.
=
g(x n ), find
f
k)
terms of the derivatives of
(0) in
then
0,
x
- (x -
t)
n
dt
7o «
Prove that
if
-1 <
x
f.o
I
(n
<
+
0,
dt
(\
t
show that
(a) to
l)\
+x)(n+
Show that if \g (x)\ < M\x - a\ n
M\x - a\ n+1 /(n + 1) for \x - a\ <
Use part
+
then
f
(b)
for sin at 0,
0.
Prove that
(a)
of degree
v 4n
y 10
then sin x
Hint: If
*11.
sin(# 2 )
is
y6
10.
=
Prove that the Taylor polynomial of f(x)
for
\x
-
a\
<
8,
then
n
=
0,
then
\g(x)\
<
8.
r
if
1)
lim g (x)/(x
—
a)
x—*a
lim g(x)/(x
(c)
(d)
-
a)
n+l
-
0.
n aJ (x),theng'(x) = f'(x) - Pn -i,aAx).
Give an inductive proof of Theorem 1, without using PHopitaPs
Show that iig{x) =f(x)
-P
,
Rule.
12.
Deduce Theorem
13.
14.
Theorem, with any form
be necessary to assume one
as a corollary of Taylor's
1
of the remainder. (The catch
is
that
it
will
more derivative than in the hypotheses for Theorem 1.)
Deduce the Cauchy and Lagrange forms of the remainder from the
integral form, using Problems 13-27 and 13-28. There will be the same
catch as in Problem 12.
(a)
Prove that
if
f
f"(a)
f
(a) exists,
- Um-f(a
then
+ h)+ ^ ~
h)
-
W
Hint: Use the Taylor polynomial of degree 2 with x
with x
—
a
—
h.
=
a
+
h and
Approximation by Polynomial Functions
(b)
exists,
15.
=
Let /(*)
>
x 2 for x
and -x 2
0,
for x
<
Show
0.
359
that
even though /"(0) does not.
Use the Taylor polynomial Pi, a ,/, together with the remainder, to prove
a weak form of Theorem 2 of the Appendix to Chapter 11: If /" > 0,
then the graph of / always
above the tangent
lies
line of /,
except at the
point of contact.
*16.
!
*17.
Problem 17-32 presented a rather complicated proof that / — 0 if
/" - / = 0 and /(0) = /'(0) = 0. Give another proof, using Taylor's
Theorem. (This problem is really a preliminary skirmish before doing
battle with the general case in Problem 1 7, and is meant to convince you
that Taylor's Theorem is a good tool for tackling such problems, even
though tricks work out more neatly for special cases.)
Consider a function / which satisfies the differential equation
B
/<»>
for certain
numbers
ao,
.
.
f
.
aj f«\
a n -\. Several special cases
,
have already
received detailed treatment, either in the text or in other problems; in
we have found all functions satisfying /' = /, or /" + / = 0,
f"—f= 0. The trick in Problem 17-31 enables us to find many solu-
particular,
or
whether these are the only
which will be supplied by this
some (necessarily sketchy) remarks
tions for such equations, but doesn't say
solutions. This requires a uniqueness result,
problem. At the end you will
about the general solution.
(a)
find
Derive the following formula for / (n+1)
(let
us agree that
will
beO):
n-1
(b)
Deduce a formula
The formula
for
in part (b)
y
is
(n
+ 2)
.
not going to be used;
it
was inserted only
to
convince you that a general formula for f n+k) is out of the question. On
the other hand, as part (c) shows, it is not very hard to obtain estimates
on the
(c)
size of
(n+A:)
(*).
Let
N
this
means that
=
max(l,
|<2
0
|,
.
.
.
,
\a n -i\).
Then
|<zy_i
n-1
=
£
bffi$\
where |V| <
2#
2
.
+
tf
n _i<z;|
< 2N
2
;
/
360
Infinite Sequences
and
Infinite Series
Show
that
n-l
fn+2)
= £
^yCfl
where
<
|£ y 2|
4
^
3j
j=0
more
and,
generally,
= ^
fn+k)
j
(d)
bj
Conclude from part
a
number
k
f
Now suppose
M
<
that /(0)
-
<
2k
Nk+1
9
2W* +1
•
=
/'(0)
M
<
"
O+
is
/<»>
(
/i »(0)
=/
(i)
2
T
=
=f
•
=
0.
Show
that
1)!
n + k+1
'
1)!
solutions of the differential equation
«y/«>,
<; <
0
(0) for
•
+
A:
+ k+
(n
for all £.
•
\2Nx\
•
and conclude that / = 0.
Show that if /1 and f2 are both
and
\b*\
any particular number x there
(c) that, for
I
(f)
where
\
M such that
!/<«+*>(*)
(e)
j
=0
-
n
1,
then
x
-/
2.
In other words, the solutions of this differential equation are deter-
mined by the "initial conditions" (the values f j) (0) for 0 <j <
n — 1). This means that we can find all solutions once we can find
enough solutions to obtain any given set of initial conditions. If the
equation
ah
has n distinct roots
/(*)
is
.
.
.
a ny then any
,
= de™ +
•
function of the form
+ c n ea x
«
•
a solution, and
/'(0)
= a+
= OtxCi +
/n-i(Q)
=
/(0)
•
•
'
•
+
+c
+
n>
•
'
-
.
a^n,
+ a/-Vn
.
As a matter of fact, every solution is of this form, because we can
obtain any set of numbers on the left side by choosing the c's
361
Approximation by Polynomial Functions
properly, but
we
18.
(a)
will
not try to prove
this last assertion.
(It is
a
which you can easily check for n = 2 or 3.)
These remarks are also true if some of the roots are multiple roots,
and even in the more general situation considered in Chapter 26.
Let/0) = * 4 sin 1/x 2 for x * 0, and/(0) = 0. Show that / = 0 up
to order 2 at 0, even though /"(0) does not exist.
purely algebraic
This example
fact,
is
more complex, but
slightly
also slightly
impressive, than the example in the text, because both f(a)
f\a)
exist for a
^
0.
Thus,
for
each number a there
is
another
more
and
num-
ber m(a) such that
(*) /(*)
=
/(«)
+ mix -
= f"(a) for a
defined in this way
namely, m(a)
function
(b)
m
a)
- a y + Ra
"^>.(x
and m(0)
0,
is
+
=
0.
(
x)
Notice that the
not continuous.
Let / be a differentiable function. Suppose that there
is
a function
m
such that (*) holds for all numbers a, and that m is continuous.
Prove that /"(a) exists for all a and equals m(a). Hint: Write (*) for
h.
x = a
h, a = a, and also for x = a, a — a
+
+
*(c)
Suppose that there
f(x)
is
a continuous function
=f(a)+f'ia)(x-a)
+ f'ia)
2
ix
m
- ay +
such that
mja)
(x
3!
- ay
+ Ra(x),
where lim
Ra(x)
(x
Does
it
follow that f"'(a)
=
m(a)?
- ay
=
0.
*
CHAPTER
AlV
TRANSCENDENTAL
IS
e
The irrationality of e was so easy to prove that in this optional chapter we will
attempt a more difficult feat, and prove that the number e is not merely
irrational, but actually much worse. Just how a number might be even worse
than irrational is suggested by a slight rewording of definitions. A number x
irrational
is
b 7* 0.
This
if it is
is
and
—
solution of
=
0, b
+
aQ
-
2
—
in this light, the irra-
is
it
not the
0,
the solution of the equation
=
0,
how
of one higher degree. Problem 2-17 shows
numbers x which
satisfy
a nx
where the
is
Viewed
0.
an equation
x2
that
with
0
V2 just barely manages to be irrational —although V2
aix
it is
—
a
b,
any equation
does not seem to be such a terrible deficiency; rather,
tionality of
appears that
=
except for a
b,
a/b for any integers a and
as saying that x does not satisfy
bx
for integers a
—
not possible to write x
same
the
all ai
en
=
is
+
a n -ix
and
a0
n~ l
^
+
•
•
produce
ao
=
many
irrational
•
+
0,
0 (this condition rules out the possibility
A number which satisfies an
called an algebraic
encountered
n
are integers
0).
to
higher-degree equations
number, and
4 4
algebraic" equation of this sort
practically every
number we have ever
(ir and e
defined in terms of solutions of algebraic equations
are the great exceptions in our limited mathematical experience). All roots,
such as
V2,
^3,
<Tl,
are clearly algebraic numbers, and even complicated combinations, like
^3 + V5 +
</\
+ V2 +
</l
are algebraic (although we will not try to prove this) Numbers which cannot
be obtained by the process of solving algebraic equations are called transcendental; the main result of this chapter states that e is a number of this
.
anomalous
sort.
The proof
that
retically possible
this proof,
e is
is
well within our grasp,
19. Nevertheless,
and was theo-
with the inclusion of
we can justifiably classify ourselves as something more than novices
mathematics; while many irrationality proofs depend
in the study of higher
362
transcendental
even before Chapter
;
363
e Is Transcendental
only on elementary properties of numbers, the proof that a number is transcendental usually involves some really high-powered mathematics. Even the
dates connected with the transcendence of
proof that
we
that
e is
are impressively recent
e
transcendental, due to Hermite, dates from 1873.
will give
—the
first
The proof
a simplification, due to Hilbert.
is
Before tackling the proof
a good idea to
itself, it is
which depends on an idea used even
map
in the proof that e
out the strategy,
is
Two
irrational.
features of the expression
e
were important
=
for the proof that e
1
+ h +R "
---
+h+h+
1
irrational:
is
+^+
•
•
On
number
the one hand, the
+-.
•
can be written as a fraction p/q with q < n (so that n (/>/ q) is an integer)
on the other hand, 0 < R n < 3/(w + 1) (so n\R n is not an integer). These
two facts show that e can be approximated particularly well by rational
numbers. Of course, every number x can be approximated arbitrarily closely
by rational numbers if £ > 0 there is a rational number r with \x — r\ < e;
the catch, however, is that it may be necessary to allow a very large denomi!
!
!
—
nator for
r,
as large as
1
/s perhaps. For
e
we
are assured that this
not the
is
case: there is a fraction p/q within 3/(n + 1) of e, whose denominator q is at
most n\. If you look carefully at the proof that e is irrational, you will see that
only this fact about e is ever used. The number e is by no means unique in this
respect: generally speaking, the better a number can be approximated by
!
rational numbers, the worse
Problem
in
3).
(some evidence for
it is
The proof
that
extension of this idea: not only
e
is
but any
e,
this assertion
is
presented
transcendental depends on a natural
finite
number
of powers
e,
e
2
,
.
.
.
,
e
can be simultaneously approximated especially well by rational numbers.
In our proof we will begin by assuming that e is algebraic, so that
n
,
ane
(*)
for
some
integers a 0
find
certain
€i,
.
.
.
,
€n
,
.
integers
.
n
+
.
,
M,
'
'
+ aie +
'
a0
=
0,
a0
^
0
a n In order to reach a contradiction
.
Mh
.
.
.
M
,
such that
e
L
9
=
+
Mi
e'i
'
M
M +
M
2
M +
M
n
we
will
then
and certain "small" numbers
n
e2
€n
364
Infinite Sequences
and
Infinite Series
how
Just
e's must be will appear when these expressions are subassumed equation (*). After multiplying through by
we
small the
M
stituted into the
obtain
[a 0
The
M + aiMi +
term
first
+
in brackets
an
is
an
M +
n]
Oitfi
integer,
and we
We
be a nonzero integer.
will necessarily
+
+
0.
choose the M's so that
will
manage
will also
=
en a n]
it
to find e's so small
that
+
|ei<2i
*
*
+
*
this will lead to the desired contradiction
number
As a
en an
—the sum of a nonzero integer and a
^ cannot be zero
very reasonable and
basic strategy this
all
is
you
will
function! (This function was introduced in
PROOF
e
is
quite straightforward.
that the Af's
and
need to know about the
Problem 18-25.)
e's
are
gamma
transcendental.
Suppose there were integers
ane
(*)
Define numbers
M,
Mh
n
a0
+
...
xp
k
.
.
,
.
a n ^e
M
,
x*- l [{x
M
~
yP-l
l
an
,
n'
and
n
unspecified
later.
number p
^
with a 0
,
+
x
ei,
.
[(x
0,
such that
+a =
•
0
.
.
,
-!)•••(*(p - 1)!
en
0.
as follows.
n)] p e-*
dx,
-!)•••(*
(P - 1)!
»)]'
dx,
[(*-!)•(*(p
The
way
part of the proof will be the
defined. In order to read the proof
l
^;
of absolute value less than
The remarkable
THEOREM
<
\
n)] p e~*
dx.
-
1)!
represents a prime
number* which we
will
choose
Despite the forbidding aspect of these three expressions, with a
work they
will
appear
much more
reasonable.
We
-
n)l
concentrate on
M
little
first.
If
the expression in brackets,
is
actually multiplied out,
[(*
-
we
obtain a polynomial
xn
*
The term "prime number" was
numbers
will be
1)
+
•
•
•
•
+
nl
defined in Problem 2-16.
used in the proof, although
number which does not
(*
•
•
divide the integer
it is
a,
An
important fact about prime
this book: If p is a prime
not proved in
and which does not divide the integer
b,
then p also does not divide ab. The Suggested Reading mentions references for this theorem
(which is crucial in proving that the factorization of an integer into primes is unique). We
will also use the result of
is
Problem
2-1 6(d), that there are infinitely
asked to determine at precisely which points
this
information
is
many
primes
required.
—the reader
"
e Is
with integer
When
coefficients.
power
raised to the pth
Transcendental
this
365
becomes an even
more complicated polynomial
***+••• ±
Thus
M can be
(n\) p
.
form
written in the
np
m = Li
y
Ca
-
(p
a=0
where the
— c„
1
-
are certain integers,
f
Jo
l)!
and Co = ±{n\) v But
.
*^ _x
/
—
k\.
Jo
Thus
(,-!+«)!
„.£
Now,
for
a =
we
0
c<>
.
obtain the term
We will now consider only primes p > n; then this term is an integer which is
not divisible by p. On the other hand, if a > 0, then
C*
(P
1
= C°(P
7—
which
by p.
«
Now
divisible
+
M+
~
a
1J!
(/>
M
by p. Therefore
We
consider M^.
= ,
«
^
have
-!)••(,-
r-
(p-i)\
f x»- ~A
l
[{x
1)
(x
•
•
2)
•
an integer which
itself is
)k
-
-
n)]^- (I
•
is
A
not divisible
dx
- A>
^
(p-i)>.
This can be transformed into an expression looking very
much
like
M by the
substitution
u
du
The
changed
limits of integration are
"+
/•-(«
*
io
ky-*[{u
=
=
x
—
k,
dx.
to 0
and
°o
,
and
+ k-\) •••«••(« + *- n)]V ^
" ~~
(>- D!
There is one very significant difference between this expression and that for M.
The term in brackets contains the factor u in the kth place. Thus the pth power
contains the factor u p This means that the entire expression
.
(u
+
k) p ^[(u
+ *-!)••(« + *-
n)] p
1
366
Infinite Sequences
and
Infinite Series
is
a polynomial with integer coefficients,
has degree at
every term of which
least/?.
Thus
np
np
U(p-\)\
Jo
a=l
where the
a —
1
Da
Now
it is
term in the sum
divisible by p.
is
M
divisible
is
clear that
k
e
M
_
_
—
+
k
M
ek
—
k
,
M + a\M\ +
*
'
+
'
an
M
+
n]
I,
.
.
n.
,
.
M we obtain
Substituting into (*) and multiplying by
[a 0
summation begins with
by p. Thus each
k is an
are certain integers. Notice that the
which
integer
(p-i)\
a=
in this case every
;
(p-\+a)\
L
[aid
+
*
*
*
+ an
en ]
=
0.
In addition to requiring that p > n let us also stipulate that p > \ao\. This
and a 0 are not divisible by p, so a Q
is also not divisible
M
M
means that both
by p. Since each
M
divisible
k is
clqM
is
by
not divisible
p.
+
by
+
a\M\
In particular
follows that
it
p,
•
+
*
•
an
M
n
a nonzero integer.
it is
In order to obtain a contradiction to the assumed equation
prove that
e is
transcendental,
it is
|ai€i
+
(*),
and thereby
only necessary to show that
•
•
+ an
•
€n
\
can be made as small as desired, by choosing p large enough; it is clearly suffishow that each |€*| can be made as small as desired. This requires
nothing more than some simple estimates; for the remainder of the argument
remember that n is a certain fixed number (the degree of the assumed polycient to
nomial equation
(*)).
To
n
\e k
\
<
e
k
begin with,
\x
p~ l
[{x
be the
n,
then
n)] p \e~
x
(/,-!)!
np
-
1[
\[(x
-
1)
•
maximum
of
\(x
—
1)
•
•
•
•
(x
-
<
A
~.
(P
e
~
r-
/
(x ~~ n)\ for
•
e~x dx
D!
n p -i
n
Ap
(P
~ D!
cVA p
(p-\)\
x
1)!
-
D!
(p
--i p
n
e np
n)] p \e~
dx.
-
(P
A
<
dx
f
n
let
k
-!)••(*-
Q
Now
<
if 1
e
n
{nAY
{p-\)\
x in
[0, «].
Then
e Is
—
But n and A are fixed; thus (nA) p /(p
by making p sufficiently large. |
Transcendental
367
can be made as small as desired
1)1
w is irrational, deserves some philosophic
argument seems quite "advanced"—after all,
we use integrals, and integrals from 0 to co at that. Actually, as many mathematicians have observed, integrals can be eliminated from the argument completely; the only integrals essential to the proof are of the form
This proof,
like the
At
afterthoughts.
proof that
the
first sight,
xke~ x dx,
fj
and these
for integral k,
Thus M,
where
for
Ca
can be replaced by k whenever they occur.
integrals
!
example, could have been defined
are the coefficients of the polynomial
[(*
If this idea
proof that
is
-
1)
•
•
(*
*
n)Y.
transcendental, depending only on the fact that
=
+i
+i
+i
+
3!
1!
2!
1
Unfortunately, this "elementary" proof
one
-
developed consistently, one obtains a "completely elementary"
e is
,
original
initially as
....
harder to understand than the
just to eliminate
is
—the whole structure of the proof must be hidden
is by no means peculiar to this specific
theorem "elementary" arguments are frequently more difficult than
"advanced" ones. Our proof that ir is irrational is a case in point. You probably remember nothing about this proof except that it involves quite a few com-
a few integral signs! This situation
—
more advanced, but much more conwhich is of great
historical, as well as intrinsic, interest. One of the classical problems of Greek
mathematics was to construct, with compass and straightedge alone, a square
whose area is that of a circle of radius 1 This requires the construction of a
plicated functions.
There
is
actually a
ceptual proof, which shows that
7r
is
transcendental, a fact
.
line
segment whose length
of length
7r is
is
constructible.
VV, which can
be accomplished
The Greeks were
totally
if
a line segment
unable to decide whether
such a line segment could be constructed, and even the full resources of
modern mathematics were unable to settle this question until 1882. In that
year Lindemann proved that w is transcendental; since the length of any seg-
ment
that can be constructed with straightedge
in terms of
line
+,
•,
— —, and
,
segment of length
The proof that ir is
7r
V
,
and
is
and compass can be written
therefore algebraic, this proves that a
cannot be constructed.
transcendental requires a sizable
amount of mathematics
368
Infinite Sequences
and
Infinite Series
which
not
too advanced to be reached in this book. Nevertheless, the proof
is
much more
proof for
difficult
than the proof that
practically the
7r is
certainly surprise you.
same
as the proof for
The proof
that
is
transcendental. In fact, the
e is
e.
This
last
statement should
transcendental seems to depend so
e is
thoroughly on particular properties of e that it is almost inconceivable how
any modifications could ever be used for 7r; after all, what does e have to do
with 7r? Just wait and see!
PROBLEMS
1.
(a)
Prove that
if
(b)
Prove that
if
a > 0 is algebraic, then
is algebraic.
a is algebraic and r is rational, then a + r and ar
are
algebraic.
Part (b) can actually be strengthened considerably: the sum, product,
and quotient
of algebraic
numbers
some
cult for us to prove here, but
2.
Prove that
degree
(a)
a be an
diffi-
by actually findneed equations of
are algebraic,
(You
satisfy.
will
number which
algebraic
is
not rational. Suppose that
a
=
anx
+ a n ^x n ~ +
n
l
•
•
•
+a =
0
0,
and that no polynomial function of lower degree has this property.
Show that fip/q) 5* 0 for any rational number p/q. Hint: Use Problem 3-7 (b).
Now show that \f(p/q)\ > \/q n for all rational numbers p/q with
q > 0. Hint: Write f(p/q) as a fraction over the common denominator
q\
M
= sup {|/'(*)
Let
to prove that if p/q
a
\
\a
*4.
too
the polynomial equation
/(*)
(c)
is
can be examined:
4.)
Let
satisfies
(b)
algebraic. This fact
special cases
V2 + V3 and V2(l + V3)
ing algebraic equations which they
*3.
is
I
is
- p/q\ > \/Mq n
— p/q\ > c/q n for
.
-
\x
:
a\
<
1}.
Use the Mean Value Theorem
a rational
that for
(It follows
all rational
\a — p/q\ < 1, then
max(l, \/M) we have
number with
c
=
p/q.)
Let
a = 0.110010000000000000000001000
where the Ts occur
that
a
is
in the n
\
place, for each
transcendental. (For each
equation of degree
n,
n.
Use Problem
show that
a
is
3 to prove
not the root of an
n.)
Although Problem 4 mentions only one specific transcendental number,
should be clear that one can easily construct infinitely many other numbers
c/q n for
a
any c and n. Such numbers were
considered by Liouville (1809-1882), and the inequality in Problem 3 is
which do not
first
— p/q\ >
it
satisfy \a
)
e Is
often called Liouville's inequality.
structed in this
way happens
None
Transcendental
numbers con-
of the transcendental
to be particularly interesting, but
369
for a long time
numbers were the only ones known. This situation
was changed quite radically by the work of Cantor (1845-1918), who showed,
without exhibiting a single transcendental number, that most numbers are
transcendental. The next two problems provide an introduction to the ideas
that allow us to make sense of such statements. The basic definition with which
we must work is the following: A set A is called countable if its elements can
Liouville's transcendental
be arranged in a sequence
01, 02, 03, 04,
The
set is
obvious example (in
N, the
more
fact,
set of natural
•
•
•
or less the Platonic ideal of) a countable
numbers; clearly the
set of
even natural numbers
is
also countable:
2, 4, 6, 8,
set of all integers (positive,
more surprising to learn that Z, the
negative, and 0) is also countable, but seeing is
It is
a
little
0, 1,
-1,
2,
-2,
3,
believing:
-3, ...
.
next two problems, which outline the basic features of countable sets, are
really a series of examples to show that (1) a lot more sets are countable than
one might think and (2) nevertheless, some sets are not countable.
The
*5.
(a)
Show
or x
(b)
is
Show
if A and B are countable, then so is A VJ B = {x: x is in
B}. Hint: Use the same trick that worked for Z.
that the set of positive rational numbers is countable. (This
that
A
in
is
really quite startling; use the following look-and-see proof:
\
\
\
\
/ /f /
\
t
t
(c)
Show
Ah A2
,
.
same
.
.
is
also countable.
Prove that the
triple
(f)
\
i
i
I
•
•
'
—
(/,
is
countable. (This
is
as part (b).)
are each countable, prove that
A
(e)
\
that the set of all pairs (m, n) of integers
practically the
(d) If
\
X
\J
A
2
\J
A3
KJ
'
'
'
(Again use the same trick as in part
set of all triples
(/,
m, n) of integers
m, n) can be described by a pair
Prove that the set of all rc-tuples (a h
you have done part (e), you can do
a 2}
(/,
(b).)
countable. (A
m) and a number
...,*„)
this,
is
n.)
are countable. (If
using induction.)
370
Infinite Sequences
and
Infinite Series
Prove that the
polynomial functions of degree n is
shows that the set of all polynomial functions of
degree n can be arranged in a sequence, and each of these functions
has at most n roots.)
(g)
set of all roots of
countable. (Part
Now use
(h)
bers
*6.
Since so
is
parts (d)
is
and
(g) to
prove that the
set of all algebraic
num-
countable.
many sets turn out to be countable,
numbers between 0 and
set of all real
there
(f)
no way of
(decimal notation
suppose such a
listing all these real
1
is
it is
important to note that the
In other words,
not countable.
numbers
in a sequence
being used on the right). To prove that
were possible and consider the decimal
is
list
this
is
so,
2
3
4
0.5i 5 2 5 3 5 4
1
where
n
=
n
=
=
n
Show
and
an
list,
thus obtaining a contradiction.
an
5 if a n
Problems 5 and 6 can be
bers
n
5
cannot possibly be in the
is
summed up
6
if
an
as follows.
5.
The
that this
number
set of algebraic
num-
countable. If the set of transcendental numbers were also countable,
numbers would be countable, by Problem 5(a), and
numbers between 0 and 1 would be countable.
But this is false. Thus, the set of algebraic numbers is countable and the set of
transcendental numbers is not ("there are more transcendental numbers than
algebraic numbers")- The remaining two problems illustrate further how
important it can be to distinguish between sets which are countable and sets
which are not.
then the
set of all real
consequently the
*7.
set of real
Let / be a nondecreasing function on
lim f(x) and lim f(x) both exist.
x—*a +
(a)
Recall (Problem 8-8) that
x-*a~
For any
in
(b)
[0, 1].
[0, 1]
e > 0 prove that there are only finitely many numbers a
with lim /(*) — lim /(#) > e. Hint: There are, in fact, at
most [/(l) - /(0)]/e of them.
Prove that the set of points at which /
Hint: If lim f(x)
x~*a +
number
—
lim /(*)
>
0,
then
is
discontinuous
it is
>
is
countable.
1/n for some natural
x—*a~
n.
This problem shows that a nondecreasing function is automatically
continuous at most points. For differentiability the situation is more
difficult to
analyze and also more interesting.
A nondecreasing func-
371
e Is Transcendental
tion can
able, but
at
to be differentiable at a set of points
fail
most points
[33]
(in
Problem
is
not count-
a different sense of the word "most"). Reference
of the Suggested
Lemma
Rising Sun
which
true that nondecreasing functions are differentiable
it is still
9 of the
Reading
gives a beautiful proof, using the
of Problem 8-20. For those
Appendix
Chapter
to
11,
it is
who have done
possible to provide
at least one application to differentiability of the ideas already
developed in
(a)
this
problem
set: If
except at those points where
its
tinuous; but the function /+'
is
/
only a countable
bers a x
/ on
and
(a x , b x )
bx
differentiable
is
disconis
/
is
.
set
a constant function. Suppose
now
dropped. Prove that / takes on
of values. Hint: For each x choose rational num-
such that a x
<
x
is
<
Then every value f(x)
interval (a x , b x ).
(d)
is
increasing, so a convex function
that the hypothesis of continuity
(c)
convex, then /
automatically differentiable except at a countable set of points.
Problem 11-50 showed that if every point is a local maximum point
for a continuous function /, then
(b)
is
right-hand derivative /+'
How many such
bx
is
is a maximum point for
maximum value of / on some
and x
the
intervals are there?
Now
deduce Problem 11-50 as a corollary of part (a).
Suppose that every point of / is either a local maximum or a local
minimum point for /. Prove that / takes on only a countable set of
values. (This is a minor variant of part (a).)
Now suppose that / is continuous, and that every point is either a
local
maximum
or a local
minimum
point for /.
stant function. (Although this statement
Problem 11-50, the only proof
I
know
is
Show
that /
is
a con-
only a minor variant of
uses part (c),
very different ideas than the ones in Chapter 11.)
and thus
relies
on
CHAPTER
*m I
The
INFINITE SEQUENCES
idea of an infinite sequence
is
so natural a concept that
dispense with a definition altogether.
One
it is
tempting to
frequently writes simply "an infinite
sequence
#2} #3,
<Zl)
(35,
.
.
.
the three dots indicating that the numbers
<z»
continue to the right "forever."
A rigorous definition of an infinite sequence
is
not hard to formulate, however;
the important point about an
ber,,
ft,
there
is
functions are
DEFINITION
<Z4 3
An
infinite
From
by a
a real
meant
number
infinite
sequence
is
num-
that for each natural
a n This sort of correspondence
.
is
precisely
what
to formalize.
sequence of
numbers
real
a function whose domain
is
is
N.
the point of view of this definition, a sequence should be designated
single letter like a,
and particular values by
a(l), a(2),
fl
(3),
but the subscript notation
tfij
is
#2,
a Si
.
.
.
itself is usually denoted by a
{(-l) n }, and {\/n} denote the sequences a, ft
almost always used instead, and the sequence
symbol
like {a n
}.
Thus
{n},
and 7 defined by
fin
= ("l) n
Tn
= -
,
1
n
A
sequence, like any function, can be graphed (Figure 1) but the graph is
usually rather unrevealing, since most of the function cannot be fit on the page.
'Hill
(c)
(a)
FIGURE
372
1
(b)
5
373
Infinite Sequences
05
=
=
03
01
H
=
04
06
:
76 7s
0
FIGURE
=
02
2
A
more convenient representation of a sequence is obtained by simply
as, ... on a line (Figure 2). This sort of picture
ai,
55
shows where the sequence 'is going.
The sequence {a n "goes out to
55
infinity,
the sequence {fi n "jumps back and forth between —1 and 1," and
labeling the points
4
}
\
the sequence {y n j "converges to 0. Of the three phrases in quotation marks,
is the crucial concept associated with sequences, and will be defined
5
the last
precisely (the definition
a N+ t
/
FIGURE
A
DEFINITION
illustrated in Figure 3).
is
/+
a.y
6
3
sequence
{
an
}
converges
to
I
(in
is
a natural
number
n
if
N such
>
/) if
that, for all natural
N, then
\a n
—
l\
for every g
<
numbers
said to
converge
if it
converges to
I
for
/
or has the limit
some
/,
0
n,
s.
In addition to the terminology introduced in this definition,
say that the sequence {a n } approaches
>
»o
n-~*
there
=
symbols lim a n
and
to
/.
we sometimes
A sequence
diverge
if it
{a n
}
is
does not
converge.
To show
that the sequence
following. If e
n
>
N we
>
0,
there
is
=
-
N
converges to
0, it suffices to
N such that
1/N <
observe the
e.
Then,
if
-
0
- Vn -
0
so
e,
Yn
<
e.
limit
lim
will
}
11
< - <
n
is
yn
have
Tn
The
{
a natural number
Vn +
probably seem reasonable after a
practically the
same
as
1
little
for large n) ,
reflection
(it
just says that
+
1
but a mathematical proof might not
« »
374
Infinite Sequences
and
Infinite Series
be so obvious.
To
Vn +
f
estimate
—
(Vn
/-
Vrc=
— Vn we
1
+
can use an algebraic
-V~n ){Vn +7= +
+Vn
\
1
trick:
Vn)
.
Vn +
1
+ —n
Vn + + Vn
n
1
1
1
+ + Vrc
Vn + — Vra by
Vrc
It is also possible to
Theorem
estimate
1
applying the
1
Vx on the interval
=
to the function /(*)
Vn +
1
[n,
n
+
Mean Value
1].
We obtain
+
1)
- Vn
1
for
>
2Vx
some x
in
(n, n
1
2
may
Either of these estimates
proof
is left
The
V^'
be used to prove the above
but valuable exercise.
limit; the detailed
to you, as a simple
limit
3rc
,™
4/2
+ =
+ 63
3
+
7n 2
3
-
8n
3
1
4
z
should also seem reasonable, because the terms involving n are the most
important when n is large. If you remember the proof of Theorem 7-9 you will
be able to guess the trick that translates this idea into a proof dividing top
—
and bottom by
n z yields
+
+ 63
3
+
In 2
An z
-
8n
3rc
nz
n
1
A
4
63
8
"~
+
n
T
,
n3
2
Using this expression, the proof of the above limit
one uses the following facts:
If lim a n and lim b n both exist, then
not
difficult, especially if
n—* »
n— »
»
lim
n—
(a n
+
bn)
=
lim
if
lim
71—
6n
lim a n
+
(a n
'
b n)
=
lim a n
n—» w
•
lim
»
bn
^
lim a n /b n
=
lim a w /lim
0,
0 for
b ni
n—
then
^
lim
n—>
n— «
cc
n—> ao
moreover,
is
all
72
greater than
00
£ ft .
some
iV,
and
375
Infinite Sequences
(If we wanted to be utterly precise, the third statement would have to be even
more complicated. As it stands, we are considering the limit of the sequence
= {a n /b n }, where the numbers c n might not even be defined for certain
\c n
n < N. This doesn't really matter
we could define c n any way we liked for
such n because the limit of a sequence is not changed if we change the
sequence at a finite number of points.)
Although these facts are very useful, we will not bother stating them as a
]
—
—
theorem
—you
should have no difficulty proving these results for yourself,
=
because the definition of lim a n
/
so similar to previous definitions of
is
n—+ oo
X—
The
=
lim f(x)
limits, especially
/.
00
=
between the definitions of lim a n
similarity
n —*
actually closer than
mere analogy;
it is
I
and lim
x—
oo
possible to define the
f(x)
=
I
is
oo
first
in terms of
the second. If/ is the function whose graph (Figure 4) consists of line segments
FIGURE
4
joining the points in the graph of the sequence {a n }, so that
=
f(x)
-
(fln+i
—
a n )(x
+
n)
an
x
<
lim f{x)
—
n
,
<
n
+
1,
then
lim a n
This observation
0
<
a
<
1.
is
—
and only
if
I
prove
this
frequently very useful. For example, suppose that
we
—
lim a x
x.
<
0,
0.
lim
e
x—
oo
=
large
=
note that
X—»
since log a
so that x log a
Notice that
we
is
xlo * a
oo
o,
negative and large in absolute value for
actually have
lim a n
-
0
if \a\
<
1;
n—* oo
for if a
<
0
1.
Then
lim a n
To
if
we can
write
lim a n
=
lim (~\) n \a\ n
=
0.
376
Infinite Sequences
and
Infinite Series
The behavior
becomes
of the logarithm function also shows that
becomes
arbitrarily large as n
=
lim a n
This assertion
large.
oo,
>
a
if
>
a
1,
then a n
often written
is
1,
n—> oo
and
sometimes even said that
it is
{a
71
approaches
}
<x>
We also write equations
.
like
—a n = —
lim
n—
and say that
lim a
n—*
n
— an
{
does not
co
5
00
approaches — oo. Notice, however, that
even in this extended sense.
\
if
< — 1,
a
then
exist,
oo
more important to
on a line
(Figure 3). There is another connection between limits of functions and limits
of sequences which is related to this picture. This connection is somewhat less
obvious, but considerably more interesting, than the one previously mentioned
Despite
this
connection with a familiar concept,
it is
picture convergence in terms of the picture of a sequence as points
—instead of defining limits of sequences in
terms of limits of functions,
it is
possible to reverse the procedure.
THEOREM
1
Let / be a function defined in an open interval containing
at c itself, with
-
lim /(*)
c,
except perhaps
/.
x—*c
Suppose that
{a n
a sequence such that
is
}
each a n
each a n
(1)
(2)
lim a n
(3)
n—*
Then
the sequence
{
f(a n )
}
is
in the
^
c,
=
domain of /,
c.
oo
satisfies
=
lim f(a n )
I.
n—»oo
Conversely,
tions,
PROOF
if this is
Suppose
first
true for every sequence
=
then lim f(x)
{
an
satisfying the
}
above condi-
/.
that lim f(x)
=
/.
Then
for every e
>
0 there
is
a 5
>
0 such
x—*c
that, for all #,
if
If the
0
sequence {a n }
number
N such
<
|*
satisfies
c\
<
d,
lim a n
then
=
c,
\f(x)
-
l\
choice of
8, this
n
>
N, then
\a n
—
<
6,
means that
!/(*»)
-
1\
<
then (Figure
that,
if
By our
-
c\
<
8.
s.
3) there
is
a natural
377
Infinite Sequences
showing that
lim f(a n )
n—
=
1.
*>
=/
Suppose, conversely, that lim f(a n )
for
every sequence
{a n
with
}
n—* *>
=
lim a n
n—
c.
=
If lim f(x)
were
/
not true, there
would be some
£
>
0 such
x—>c
<»
>
that for *z#ry 5
>
0 there
<
0
an # with
is
-
|*
c\
<
but
8
\f(x)
-
/|
>
s.
In particular, for each n there would be a number x n such that
<
0
<r|
<
-
but
\f(x n )
-
l\
>
e.
the sequence {# n } clearly converges to c but, since |/(# n ) — l\ > s for
the sequence |/(^ n )} does not converge to /. This contradicts the
Now
all
-
\x n
n,
=
hypothesis, so lim f(x)
Theorem
1
provides
the sequences {a n }
must be
/
true. |
many examples
and
{b n
}
-.in
—
bn
of convergent sequences. For example,
defined by
(.3+1)
cos
clearly converge to sin(13) and cos(sin(l)), respectively. It is important,
however, to have some criteria guaranteeing convergence of sequences which
are not obviously of this sort. There is one important criterion which is very
easy to prove, but which
the basis for
is
all
other results. This criterion
is
stated
|
ai
FIGURE
ai
03
a*
a
which therefore apply also to
sequences: a sequence {a n is increasing if a n+ i > a n for all rc, nondecreasing
if tfn+
> o-n for all n, and bounded above if there is a number
such that
for all n; there are similar definitions for sequences which are dean <
creasing, nonincr easing, and bounded below.
in terms of concepts defined for functions,
5
\
M
i
M
THEOREM
2
If {a n
is
}
statement
PROOF
The
set
A
nondecreasing and bounded above, then
is
true
if
{a n
is
}
numbers a n
consisting of all
has a least upper bound a.
if £
>
bound
0, there
of A.
is
some
Then
if
n
an
We
>
This proves that lim a n
is,
}
converges (a similar
by assumption, bounded above,
claim that lim a n
cin satisfying
>
[a n
nonincreasing and bounded below).
= a
a —
aN
—
< a — aN <
<
£,
since
(Figure
a
N we have
aN,
so
=
|
ot.
a
an
£.
is
5).
In
so
A
fact,
the least upper
378
Infinite Sequences
and
Infinite Series
The
hypothesis that {a n }
2: if {a n
{a n
\
is
clearly diverges.
}
should be
sequence
j
is
bounded above
clearly essential in
is
not bounded above, then (whether or not {a n
}
consideration,
first
Theorem
nondecreasing)
might appear that there
it
trouble deciding whether or not a given nondecreasing
little
an
Upon
is
}
bounded above, and consequently whether or not
is
verges. In the next chapter such sequences
j
an
}
con-
very naturally and, as
will arise
we
shall see, deciding whether or not they converge is hardly a trivial matter.
For the present, you might try to decide whether or not the following (obviously increasing) sequence is bounded above:
+* + *,
+
l, i
+*+
i
Although Theorem 2 treats only a very special class of sequences, it is more
might appear at first, because it is alwayc possible to extract from
an arbitrary sequence \a n another sequence which is either nonincreasing or
else nondecreasing. To be precise, let us define a subsequence of the sequence
{^n} to be a sequence of the form
useful than
\
\graph ofa
R
%d
where the
2
and
nj are
1234
1
1
1
6 are peak points
1
1
56
789
1
1
1
1
10
a na
•
•
j
<
Hi
<
72
3
<
*
*
*
h
11
Then
every sequence contains a subsequence which
or nonincreasing. It
is
although the proof
worth recording as a lemma.
6
LEMMA
Any
sequence
{
an
}
is
either nondecreasing
become quite befuddled trying to prove
very short if you think of the right idea;
this
possible to
assertion,
FIGURE
2i
natural numbers with
ni
H
an
is
contains a subsequence which
is
it is
either nondecreasing or
nonincreasing.
PROOF
number
Call a natural
all
m >
Case
n3
<
'
n (Figure
•
a "peak point" of the sequence
{a n
\
<
if a m
a n for
6).
The sequence has
1.
'
n
infinitely
many peak
are the peak points, then a ni
>
points.
a n2
>
In this case,
a n}
>
•
*
,
if Wi
so {a Uk
<
}
ft 2
is
<
the
desired (nonincreasing) subsequence.
Case
2.
The sequence has
greater than
all
such that a n2
greater than
>
only finitely
is
peak points)
way we
many peak
m
is
COROLLARY (THE
In this case,
not a peak point, there
let n\
be
is
obtain the desired (nondecreasing) subsequence. |
If we assume that our original sequence
an extra corollary along the way.
BOLZANO- WEIERSTRASS THEOREM)
points.
some n 2 > «i
not a peak point (it is greater than m, and hence
there is some n 3 > n 2 such that a nt > a nr Con-
points. Since
a nr Since n 2
all
tinuing in this
peak
{a n
\
is
bounded, we can pick up
Every bounded sequence has a convergent subsequence.
Infinite Sequences
379
Without some additional assumptions this is as far as we can go: it is easy
having many, evenly infinitely many, subsequences
converging to different numbers (see Problem 2). There is a reasonable
assumption to add, which yields a necessary and sufficient condition for
convergence of any sequence. Although this condition will not be crucial for
to construct sequences
our work,
it
does simplify
many proofs. Moreover,
mental role in more advanced
worth stating now.
If
this
condition plays a funda-
for this reason alone
and
investigations,
it is
a sequence converges, so that the individual terms are eventually all close
same number, then the difference of any two such individual terms
to the
To be
should be very small.
precise,
=
lim a n
if
some
for
/
then for any
/,
n—* oo
£ > 0 there is an
and m > N, then
\a n
This
N such that
\a n
—
—
am
final inequality,
can be used
<
\
\a n
\a n
—
am
l\
<
\
-
£,
<
l\
+
—
\l
> N; now
e/2 for n
if
both n
>
N
6
<-£ + -=£•
,
an
2
2
which eliminates mention of the limit /,
Cauchy condition) which is clearly
to formulate a condition (the
necessary for convergence of a sequence.
A sequence
DEFINITION
number
{a n
is
\
a
Cauchy sequence
N such that,
is
>
for every £
if
\a n
usually written lim
\a
m, n
>
0 there
is
a natural
n,
N, then
if
(This condition
m and
for all
—
m
am
—
<
\
an
£.
=
\
0.)
The beauty of the Cauchy condition is that it is also sufficient to ensure
convergence of a sequence. After all our preliminary work, there is very little
left to do in order to prove this.
THEOREM
3
PROOF
A
sequence
\a n
}
converges
if
and only
a
if it is
Cauchy sequence.
have already shown that {a n is a Cauchy sequence if it converges. The
proof of the converse assertion contains only one tricky feature showing that
every Cauchy sequence {a n is bounded. If we take £ = 1 in the definition of
We
}
:
\
a
Cauchy sequence we
find that there
\a
In particular,
this
means
\a
Thus
{a m
:
m > N}
the whole sequence
is
is
m
m
—
an
\
<
1
is
some
iV such that
for m, n
>
for all
m >
N.
that
—
a N +i\
<
1
N.
bounded; since there are only
bounded.
finitely
many
other
ah
a
380
Infinite Sequences
and
»
.
Infinite Series
The
Lemma
corollary to the
thus implies that some subsequence of {a n
}
converges.
Only one point remains, whose proof will be left to you: if a subsequence of
a Cauchy sequence converges, then the Cauchy sequence itself converges. |
PROBLEMS
1.
Verify each of the following limits.
lim
(i)
n-^» n
+
3
1.
4
'
+
lim V'
(iii)
=
-J-
-
1
+
1
=
0.
You
Hint:
should at least be able
n—* »
vV +
to prove that lim
vV =
-
1
0.
n—»
=
lim
(iv)
n—
for
<
A:
n—*
lim
(vii)
lim
=
1,
V« =
1.
2
V
lim
n—*
1)
*
•
•
k\ for k
<
n,
in particular,
>
a
0.
n
+
n
+
b
—
n
1
=
max{a,
b).
oo
=
lim
(ix)
= »(«—
«j
(vi)
(viii)
Hint: «!
n/2.
lim
(v)
0.
oo
n—
» oo
where a(n)
0,
is
The fact that each prime
how small a{n) must be.
Hint:
of
the
number
of primes which divide
n.
72
is
>
2 gives a very simple estimate
i
n v^ X
n-»oo
(a)
What can
is
an
Find
all
an
(b)
—
1
,
limits
(c)
.
.
p
+
1
be said about the sequence
{a n
}
if it
converges and each
integer?
convergent subsequences of the sequence 1, —1, 1, — 1, 1,
(There are infinitely many, although there are only two
.
.
which such subsequences can have.)
convergent subsequences of the sequence 1, 2, 1, 2, 3, 1, 2,
... (There are infinitely many limits which
Find
all
3, 4,
1, 2, 3, 4, 5,
.
such subsequences can have.)
Infinite Sequences
(d)
Consider the sequence
11
2i_23.i_23_Al
2) 3) 3> 4j 4? 4) 5> 5? 5j 5j 6j
3.
4.
381
•
•
•
•
(a)
For which numbers a is there a subsequence converging to a?
Prove that if a subsequence of a Cauchy sequence converges, then
(b)
so does the original Cauchy sequence,
Prove that any subsequence of a convergent sequence converges.
<
<
(a)
Prove that
(b)
Prove that the sequence
if
0
V2,
a
Vl
V2,
< Via <
then a
2,
Vl
y/ 2 V2,
.
.
2.
.
converges.
(c)
Find the
limit.
Hint: Notice that
if
=
lim a n
/,
then lim
n—
n—>
* oo
V2/, by Theorem
1.
=
Identify the function /(*)
lim (lim (cos n
n—* oo
many
tioned
\irx)
2k
)
.
(It
has been men-
oo
times in this book.)
Many impressive looking limits can
person
Vla n =
op
who makes them
be evaluated easily (especially by the
up), because they are really lower or upper
sums
in disguise. With this remark as hint, evaluate each of the following.
(Warning: the list contains one red herring which can be evaluated by
elementary considerations.)
(l)
r
hm
Ve+Ve* +
-
•
•
:
;
+
n
,..v
(n)
r
hm
n—
(iii)
(v)
+ V? +
m
+
•
---
(^fry2+
"
lim (
+
+
^>
oTT2T»
+
'
4-
-
limits like lim
n— oo
V«
+
"
'
"
7
Although
ife"
Yl
^(^TT +
!l .
(vi)
V~e
oo
•
•
•
+
(^r^)'
"
-
)•
and lim a" can be evaluated using
facts
n— »
*
about the behavior of the logarithm and exponential functions, this
approach is vaguely dissatisfying, because integral roots and powers can
be defined without using the exponential function. Some of the standard
"elementary" arguments for such limits are outlined here; the basic
tooks are inequalities derived from the binomial theorem, notably
(1
+
h)
n
>
1
+
nh>
for h
>
0;
382
Infinite
Sequences and Infinite Series
and, for part
{1
(a)
(c),
n
+ hr > l+nh + »±^ h >> ±Z^) h
71—>
h
>
oo
if
>
a
by
setting a
1,
by
setting
<
1.
1,
rh>0.
—
1
+
h,
=
1
+
where
OO
0.
(b)
Prove that lim a n
(c)
Prove that lim
n—*
estimating
(d)
=
Prove that lim a n
{o
*,
=
0
if
=
1
<
0
if
<
a
>
a
1.
h
and
oo
h.
Va =
Prove that lim
<
0
if
1
a
n—
> oo
(e)
=
Prove that lim
1.
n—* oo
8.
(a)
(b)
Prove that a convergent sequence is always bounded.
Suppose that lim a n = 0, and that each a n > 0. Prove that the
n~>
of
9.
(a)
all
numbers
a n actually has a
maximum member.
Prove that
1
n
(b)
<
+
+
log (n
—
1 )
show
=
+
1
+
-
-
+
=
'
is
lim ( 1
a
refractory;
known
it is
Suppose that /
j
decreasing,
is
/(i)
+
•
•
+
•
choose /
—
=
log
as Euler's
number, has proved
* </(2) +
and show that
rc!
<
<
follows that
=
lim
n^oo
<z
n
>
0.
log n)
/
known whether y is
increasing on [1, oo). Show
<
it
n
+•••+/(«- ix
Now
and that each
number
not even
is
n,
??
n-^« \
This number,
log
H
'
'
3
that the sequence \a n
It follows that there
(b)
1
<
n
2
(a)
log n
1
If
an
10.
set
oo
n
e
-;
to
be quite
rational.
that
•
•
•
+/(»).
383
Infinite Sequences
Vn
This result shows that
approximately
is
\
the ratio of these two quantities
conclude that n
An
close to (n/e)
is
\
because
But we cannot
in this sense; in fact, this
false.
is
n\
is
theorem which provides the
Problem 26-18.
which x > 0 the symbol
The standard (and
tables.
n
n/e, in the sense that
for large n.
1
very desirable, even for concrete computacannot be calculated easily even with logarithm
estimate for n\
tions,
close to
is
difficult)
right information will be found in
*11.
This problem investigates for
'
x
makes
sense. In other words,
when does
b(x)
=
zX
if
'
we
define ai{x)
=
0n
x,
+iM =
x anix \
lim a n (x) exist?
n—
oo
Prove that if b{x) exists, then x b(x) = b(x).
(The situation is similar to that in Problem 4.)
According to part (a), if b{x) exists, then x can be written in the form
lle
lly
Hint: Consider the
for some y. Conclude that 0 < x < e
y
(a)
(b)
.
graph of f(y)
(log y)/y, as in
<
Suppose, conversely, that 0
(c)
since {a n (x)
\
also that b(x)
is
<
x
Problem 17-24.
<
(e)
b'(x) in
(f)
terms of
(Of course,
Prove that
if
.
Show
that each a n {x)
r
b {x)
and
1/e
<
e\
),
find a formula for
Problem 17-24(/).
by differentiating the equation x h{x) =
similar to
is
method does not prove
this
=
(0, e
Hint: This part
b{x).
lim a n
n—
lle
e).
Derive a formula for
b(x).
e
clearly increasing, this proves that b{x) exists (and
Find b{Vl) and b(e^).
Prove that b is differentiate on
(d)
*12.
=
that b
is
differentiate.)
then
/,
oo
lim
n—*
•
•
•
+
=
/.
n
Hint: This problem
Problem
Suppose that a n
+
(gi
oo
is
very similar to
(in fact
it is
a special case of)
13-34.
*13.
^
0 for each n
and that lim
a n +i/a n
=
L Prove that
n—
* oo
V an
—
Hint: This requires the same sort of argument that works
n /—
a = 1, for a > 0.
in Problem 12, together with the fact that lim
lim
n—
I.
oo
V
n—
>
14.
(a)
Suppose that {a n
Prove that lim a n
\
n—
(b)
a convergent sequence of points
is
in [0, 1].
* oo
{a n
}
of points all in (0, 1) such that
oo
Suppose that /
all
also in [0, 1].
Find a convergent sequence
lim a n is not in (0, 1).
n—>
15.
is
is
oo
continuous and that the sequence
*,/«,/(/«),/(/(/(*))),
•
•
•
384
Infinite
Sequences and Infinite Series
Prove that / is a "fixed point'* for /, i.e., /(/) = /. Hint:
have occurred already.
Suppose that / is continuous on [0, 1] and that 0 < f(x) < 1 for all
x in [0, 1]. Problem 7-1 i shows that / has a fixed point (in the
converges to
Two
16.
(a)
/.
special cases
terminology of Problem 15). If /
ment can be made: For any x
is
increasing,
a
in [0, 1], the
*, /(*), /(/(*)),
much
stronger state-
sequence
•
•
is necessarily a fixed point, by Problem 1 5) Prove
by examining the behavior of the sequence for f(x) > x
< x, or by looking at Figure 7. A diagram of this sort is
has a limit (which
FIGURE
7
.
this assertion,
and
f(x)
used in Littlewood's Mathematician's Miscellany to preach the value
of drawing pictures: "For the professional the only proof needed is
Figure]."
[this
*(b)
Suppose that / and g are two continuous functions on [0, 1], with
0 < f(x) < 1 and 0 < g(x) < 1 for all x in [0, 1], which satisfy
fog = g o f. Suppose, moreover, that / is increasing. Show that
/ and g have a common fixed point; in other words, there is a number / such that /(/) = / = g(l). Hint: Begin by choosing a fixed
point for
***(c)
g.
Does the conclusion of part
/
is
(b) hold
without the assumption that
increasing?
Problem 15 is really much more valuable than Problem 15
might suggest, and some of the most important "fixed point theorems" depend
upon looking at sequences of the form from x, /(*), /(/(*)), .... A special,
but representative, case of one such theorem is treated in Problem 18 (for
which the next problem is preparation).
The
17.
trick in
(a)
Use Problem 2-4
to
show
that
if c t± 1,
+
c
n
then
m
c
=
1
(b)
Suppose that
\c\
<
1.
- c n+1
-c
Prove that
lim c m
+
+
c
n
=
0.
Suppose that {* n is a sequence with \x n
is a Cauchy sequence.
c < 1. Prove that {x n
Suppose that / is a function on [a, b] such that
(c)
}
—
x n +i\
}
*18.
\f(x )
(*)
where
(a)
(b)
c
<
1.
— /GO ^
I
x
c
~~
\
(Such a function
is
jvl>
f°r
a ^ * ancl y in [a
called a contraction.)
Prove that / is continuous.
Prove that / has at most one fixed point.
>
<
c
n
>
where
Infinite Sequences
(c)
385
considering the sequence
By
*,/(*),/(/(*)),
for
in
any x in [a, b], prove that / does have a fixed point. (This result,
a more general setting, is known as the "contraction lemma.")
Let {x n
}
be a sequence which
—
yn
(a)
bounded, and
is
SUp {x n * n +l, *n+2,
•
,
Prove that the sequence {y n
converges.
\
•
let
•}•
The
limit lim y n
is
denoted
by Urn * n or lim sup x n and called the limit superior, or upper
,
n—
n—» so
*>
limit, of the sequence {x n \.
(b)
Find lim x n
each of the following:
for
n—>
(c)
(ii)
Xn
= (-i)"--
(iii)
Xn
=
(iv)
*n
=
vV
Define lim x n (or lim inf x n ) and prove that
n—
n— *
oo
*
lim x n
<
lim # n
(d)
Prove that lim x n
this case
=
lim x n
(e)
and only
exists if
=
lim x n
lim * n
n—
n—* oo
n—> oo
.
n—> «o
n—* oo
if
=
lim x n
.
00
A
Recall the definition, in Problem 8-18, of lim
A. Prove that
the
if
numbers
lim x n and that in
x n are distinct,
for a
n->
where
(a)
^4
=
{x n
Suppose that
:
n
{
n in
}
is
for all n.
=
set
lim 4,
oo
N}.
nonincreasing, with b n
ai
+
aibi
+
m <
bounded
then lim x n
•
•
•
+
an
+
<z»& n
<
>
n,
and that
may
not look
0 for each
M
Prove that
<
bitn
This result
is
known
•
•
as Abel's
<
6iM.
Lemma. Although
it
very interesting, it is the basis for several very interesting results
which will appear in later problem sets; in addition, the proof
386
Infinite Sequences
and
Infinite Series
depends on a rather tricky way of writing
+
a\bi
•
*
•
+
anb n
.
Deduce that
(b)
bm <
(a trivial
+
akbk
4
+ an b n <
*
*
consequence of part
bk
M
only because
(a), inserted
it is
needed
later).
*21.
The
Theorem
Bolzano-Weierstrass
quite differently than in the text
of limit points.
there
(a)
is
A
\x
—
<
a\
£
but x
limit points of the following sets.
(i)
j-L-
tfinNj.
(ii)
(--
+—
{(-1)*
(iv)
Z.
(v)
Q.
and
n
:
m
m
in
n)
A
if
for every £
>
0
9* a.
.
J
+^J:rcinN).
[l
a limit point of A
many points a of A
lim A is the largest
Prove that x
also proved,
statement uses the notion
a limit point of the set
is
with
all
(hi)
(c)
point x
classical
Find
\n
(b)
A
a point a in
and
usually stated,
is
—the
is
if
and only
if
for every £
—
are infinitely
satisfying
Prove that
limit point of A,
\x
a\
<
>
0 there
£.
and lim A the
smallest.
The
usual form of the Bolzano-Weierstrass Theorem states that if A
an infinite set of numbers contained in a closed interval [a, b], then
some point of [a, b] is a limit point of A. Prove this in two ways:
Using the form already proved in the text. Hint: Since A is infinite,
is
(d)
there are distinct
(e)
*22.
numbers x h # 2 # 3
,
.
in A.
.
Use the Bolzano-Weierstrass Theorem to prove that if / is continuous
on [a, b], then / is bounded above on [a, b]. Hint: If / is not bounded
above, then there are points x n in
**23.
.
,
Using the Nested Intervals Theorem. Hint: If [a, b] is divided into
two intervals, at least one must contain infinitely many points of A.
(a)
Let {a n
}
h
f
Suppose that 0
<
with f(x n )
f
h
>
n.
be the sequence
i,
gers j
[a, b]
j
is
<
a
f f
,
<
,
b
n such that aj
i, t>
<
is
1.
,
Let
in
%
,
•
.
•
•
be the number of inte(Thus N(2;
f) = 2, and
A^(rz; a, 6)
[a, b].
-g-,
Infinite Sequences
h f) =
387
Prove that
3.)
lim
n-+oo
(b)
A
sequence
tributed in
{a n
of
}
ft
numbers
in
[0, 1]
is
called
uniformly
dis-
[0, 1] if
N(n;
a, b)
lim
=
b
-
a
n
and b with 0 < a < b < 1 Prove that if s is a step function
defined on [0, 1], and {a n is uniformly distributed in [0, 1], then
for all a
.
}
(c)
Prove that if \a n
[0, 1], then
)
is
uniformly distributed in
[0, 1]
and /is integrable
on
(a)
Let / be a function defined on
a in
[0, 1].
points a in
set of
such that lim f(y)
exists for all
For any s > 0 prove that there are only finitely many
- f(a)\ > £. Hint: Show that the
[0, 1] with |lim f(y)
such points cannot have a limit point
lim f(y) could not
(b)
[0, 1]
x,
by showing that
exist.
Prove that, in the terminology of Problem 20-5, the
set of points
where / is discontinuous is countable. This finally answers the question of Problem 6-16: If / has only removable discontinuities, then
/ is continuous except at a countable set of points, and in particular,
/ cannot be discontinuous everywhere.
CHAPTER
*m*^
INFINITE SERIES
Infinite sequences
were introduced
+
a.\
in this chapter. This
of infinitely
+
a2
+
az
•
•
•
as yet completely undefined.
is
sum
What can be
5 '
=
sn
infinite
with the specific
not an entirely straightforward matter, for the
is
many numbers
defined are the " partial sums
and the
in the previous chapter
"sums"
intention of considering their
+
a\
+
•
•
sum must presumably be
an
,
defined in terms of these partial
sums. Fortunately, the mechanism for formulating this definition has already
been developed in the previous chapter.
puting the infinite
represent closer
This
sum
+
a\
a2
•
3
•
•
,
is
to
be any hope of comsums s n should
the partial
and closer approximations as n is chosen larger and larger.
amounts to little more than a sloppy definition of limits:
last assertion
the "infinite
sum"
a\
+
a2
will necessarily leave the
sequence
{jn
}
may
+
a$
with a n
= (—
which lim
71—
yields the
sn
'
of
*
1,
ought
'
many
have a
-1,
be lim
sn
.
This approach
sequences undefined, since the
limit.
-1,
to
.
For example, the sequence
.
.
new sequence
si
=
ai
^2
=
ai
S3
=
a\
Si
=
a\
does not
+
"sum"
easily fail to
1,
for
If there
+a +
=
1,
+ a = 0,
+a +a =
+ a + az +
exist.
2
2
2
z
1,
~
0,
Although there happen
to
be some clever
00
extensions of the definition suggested here (see Problems 8
and 23-11)
it
seems unavoidable that some sequences will have no sum. For this reason, an
acceptable definition of the sum of a sequence should contain, as an essential
component, terminology which distinguishes sequences for which sums can be
defined from less fortunate sequences.
388
Infinite Series
The sequence
DEFINITION
{a n
}
summable
is
—
sn
In
this case,
lim
»
if
ai
the sequence
+
'
'
+
'
an
{s n
}
389
converges, where
.
denoted by
s n is
n—
^
n=
and
is
an
(or, less
formally, a\
called the
sum
of the sequence {a n
a2
+
as
+
*
'
*)
precise expressions; indeed the
An
infinite
\
in this definition
The terminology introduced
day language.
+
1
title
sum
of this chapter
£
n=
an
is
is
is
usually replaced by less
derived from such every-
usually called an
infinite series,
the
l
word "series" emphasizing the connection with the infinite sequence \a n \.
The statement that [a n is, or is not, summable is conventionally replaced by
\
the statement that the series
£=
71
a n does, or does not, converge. This terminol1
oo
ogy
is
somewhat
(so it can't
peculiar, because at best the symbol
"converge"), and
it
£
a n denotes a
doesn't denote anything at
summable. Nevertheless, this informal language
unlikely to yield to attacks on logical grounds.
is
all
number
unless \a n
}
is
convenient, standard, and
direct
Certain elementary arithmetical operations on infinite series are
if
{a
that
show
n \ and
consequences of the definition. It is a simple exercise to
{
bn
j
are summable, then
n
=
1
X
C
'
°n
examples of
As yet these equations are not very interesting, since we have no
terms are
the
which
summable sequences (except for the trivial examples in
some
sequence,
summable
eventually all 0). Before we actually exhibit a
general conditions for summability will be recorded.
can be
There is one necessary and sufficient condition for summability which
if the sequence
only
and
if
summable
is
{a
sequence
The
n
stated immediately.
\
only if
converges, which happens, according to Theorem 21-3, if and
original
the
terms of
s n * 0; this condition can be rephrased in
lim sm
\s n
\
m,n—»*>
-
390
Infinite Sequences
and
Infinite Series
sequence as follows.
THE CAUCHY CRITERION
The sequence
{a n
summable
is
}
lim
m,n—
a n +i
if
and only
if
+••+«*=•().
0°
Although the Cauchy criterion is of theoretical importance, it is not very
useful for deciding the summability of any particular sequence. However, one
simple consequence of the Cauchy criterion provides a necessary condition for
summability which is too important not to be mentioned explicitly.
THE VANISHING CONDITION
If \a n
}
summable, then
is
=
lim a n
0.
This condition follows from the Cauchy criterion by taking
also be proved directly as follows. If lim s n = I, then
n—
=
lim a n
n— »
lim
n—*
»
—
(s n
=
Jn -i)
lim
n
Unfortunately, this condition
n
+
sn
—
lim
»
*
far
is
from
sufficient.
is not summable; in
numbers \/n shows that the sequence \sn
}
>
>
\
(2 terms,
each
The method
never
>
can
jn _i
For example, lim
fact the following
1
jn
=
oo
grouping of
not bounded:
is
>
*
(4 terms,
each > i)
i)
it
0.
but the sequence {\/n}
the
;
n—*
n—
0,
1
to
oo
= 1-1 =
m =
h
(8 terms,
each > ft)
of proof used in this example, a clever trick which one might
need for some more standard methods for attacking these
problems. These methods shall be developed soon (one of them will give an
see, reveals the
alternate proof that
n=
1/n does not converge) but
^
it
will
be necessary to
first
1
procure a few examples of convergent series.
The most important of all infinite series are the "geometric series"
|j
=0
=
r«
+r+r +
2
1
r
3
+
•
•
•
.
n
Only the
cases
approach 0
if
\r\
\r\
<
1
>
1.
These
are interesting, since the individual terms do not
sn
can be managed because the partial sums
series
=
1
+ +
r
•
•
•
+
r
n
can be evaluated in simple terms. The two equations
sn
=
rsn
=
1
+r+ +
r + r +
r
2
2
+ n
+ rn +
r
r
n+1
Infinite Series
391
lead to
-
s n {\
r)
=
*n
=
1
-
r»
+1
or
1
- r n+1
-r
1
by
(division
lim
r
n
=
1
—
r is
0, since
\r\
valid since
<
Xr
1. It
=
n
we
are not considering the case
r
=
1).
Now
follows that
lim
——
1
—
i
+
n-*<*
1
r
1
[r|
,
—
<
1.
r
n=0
In particular,
n=0
n=l
that
is,
i
an
infinite
FIGURE
+
i
+
•
•
-
•
1,
sum which can always be remembered from the picture in Figure
1
1
from which
Special as they are, geometric series are standard examples
derived.
will
be
summability
for
important tests
with each a n > 0; such
then
sequences are called nonnegative. If \a n is a nonnegative sequence,
with
combined
is clearly nondecreasing. This remark,
the sequence {s n
summability
Theorem 21-2, provides a simple-minded test for
shall consider only sequences \a n
For a while we
\
\
\
:
A nonnegative sequence
THE BOUNDEDNESS CRITERION
sums
sn is
{a n
}
is
summable
if
and only
if
the set of partial
bounded.
not very helpful— deciding whether or not the set of
On the other hand, if some
all s n is bounded is just what we are unable to do.
this criterion can be
comparison,
for
available
already
are
convergent series
By
itself, this
criterion
is
used to obtain a result whose simplicity belies
almost all other tests).
THEOREM
1
(THE COMPARISON TEST)
its
importance
Suppose that
0
<
an
<
bn
for all n.
(it is
the basis for
—
392
Infinite Sequences
and
Infinite Series
Then
^
if
b n converges, so
does
n=l
PROOF
^
an
.
n=l
If
= ai +
= bi+
sn
tn
•
'
•
•
+ an
+b n
•
,
,
then
<
0
sn
<
for all
tn
n.
00
Now
{/ n
}
is
bounded, since
^
n=
Therefore
£ n converges.
{s n
}
is
bounded;
l
00
consequently, by the boundedness criterion
^
n=
Quite frequently the comparison
# n converges.
1
can be used to analyze very com-
test
plicated looking series in which most of the complication
is
irrelevant.
For
example,
2
+
sin
2
3
(rc
+
n
+
1)
n2
converges because
o
2
<
±
+
sin3 (*
2n
+
< A,
'
2n
n2
and
00
00
X
n
is
a convergent (geometric)
=
~
2n
3
y
n
tion really
means
we would
expect the series
1
1
+
sin 2 n 3
1
term of the
series
is
practically
1
/2
n
.
This
last asser-
that
Hm
n-oo
( n
\2
-
1
and this formulation shows us how
number c > 1, then it is certainly
0
1
series. Similarly,
n L(2
—
to converge, since the nth
X r*
n=
1
* 2 "-l +
L_y±
W/
+sin 2
to verify
2*
=
1
,
our conjecture.
If
true that
<C
sin*»'
h
f° r lmge emUgh ^
we choose any
393
Infinite Series
This shows that
+
converges for some N, and of course
sin
2
n
z
this implies
convergence of the original
series.
the comparison test backwards. Consider, for
It is also possible to use
exam-
ple, the series
n+
Sn + r
n=
1
2
1
We
1
would expect
/n for large
n.
this series to diverge, since (n
To
prove
<
0
this,
<
-
c
n2
n
This shows that the
Y
series
+
choose any number
+
for large
with 0
enough
1) is practically
<
c
<
1.
Then
n.
1
+
(n
+
l)/(n 2
c
\)/{n 2
+
1) diverges,
because
if it
con-
00
verged, then by the comparison test
know
is
\/n would also converge, and this
£
n=
we
1
false.
The comparison
analyzed
other important
test yields
tests
when we
Choosing the geometric
series as catalysts.
^=
series
n
vergent series par
we
excellence,
use previously
the con "
0
obtain the most important of
all
tests for
summability.
THEOREM
2
(THE ratio TEST)
Let a n
>
0 for
all n,
and suppose that
lim
——
n— «
<3
=
r.
n
00
Then
a n converges
£
n=
<
if r
1.
On
the other hand,
if r
>
1,
then the terms
1
so
an
do not approach
0, so
£
n
to
PROOF
compute lim
Suppose
first
a n +i/a n
that
r
<
1
.
=
<2
n diverges.
(Notice that
it is
therefore essential
1
and not lim
a n /a n +\ 0
Choose any number
hm
—^
== r
j
<
with
1
r
<
s
<
1
.
The hypothesis
394
Infinite Sequences
and
Infinite Series
implies that there
some
is
N such
^
<
that
> N.
for n
s
an
This can be written
<
tfn+i
>
for n
sa n
N.
Thus
Y
Since
fc
0^
=
V
ajv
=0
fc
=
j*
converges, the comparison test shows that
0
00
Y
00
tf
n
= Y
fltf+fc
converges. This implies the convergence of the series
^=
n
<z
n,
which has only
1
00
finitely
The
many
case
r
Y
terms not included in
>
1
is
even
easier. If
1
# rt
n=
iV
<
j
.
<
a
number
N such
do not approach
0, so {a n
}
r,
then there
is
that
a
-^>s
for
n>N,
an
which means that
aN+k
>
tfj\rJ*
>
—
k
<zn
This shows that the individual terms of {# n
summable.
}
0, 1,
...
.
is
not
|
oo
As a simple application of the
ratio test, consider the series
<z
n
=
we obtain
1
an+i
= (n+l) =
»!
fln
1
0+1)!
\
n\
Thus
Iim
^
=
0,
= _J
»
+
r
1/n
^
n=
1
!.
Letting
Infinite Series
which shows that the
l/»! converges. If
£
series
n=
395
consider instead the
we
1
so
series
^
r
n
/rc!,
where
r is
some
number, then
fixed positive
-J+
= Km
lim
.R
so
^
r
n
/w
=
0,
1
converges. It follows that
!
r
lim
=
0,
n->« n!
was based on the
a result already proved in Chapter 16 (the proof given there
the series
consider
we
if
Finally,
same ideas as those used in the ratio test).
^
nr
n
we have
lim
n _>„
since lim (n
7i
—
verges,
+
l)/»
=
nr
1.
=
n
lim
=
r
r,
n
n->«=
This proves that
0
if
<
r
<
1,
then
rcr*
J
n=
con-
1
and consequently
lim nr n
=
0.
n— »
>
(This result clearly holds for
-
1
<
r
<
0, also.) It is
a useful exercise to proan intermediary.
using the ratio test as
vide a direct proof of this limit, without
as a
Although the ratio test will be of the utmost theoretical importance,
of the
drawback
One
disappointing.
found
be
practical tool it will frequently
determine, and
fact that lim a +Ja may be quite difficult to
ratio test
is
the
may not even
regularity,
is
is
A more
exist.
will
if
serious deficiency,
which appears with maddening
the fact that the limit might equal
precisely the one
example,
n
n
an
show that
=
£
which
is
it
(1/n) 2 converges, even
/
i
The
case lim a n+ i/a n
might not be summable
might be. In fact, our very next
inconclusive: \a n
but then again
1.
1
(for
\
though
y
=
test
1
396
Infinite Sequences
and
Infinite Series
This
test
of the comparison
THEOREM
3
(THE integral TEST)
method
provides a quite different
divergence of infinite series
Suppose that /
—
but the
test,
it is
chosen for comparison
series
and decreasing on
positive
is
determining convergence or
an immediate consequence
for
like the ratio test,
oo),
[1,
quite novel.
is
and that
=
f(n)
a n for
00
Then
all n.
a n converges if
^
and only
if
the limit
/,•/->.»///
exists.
PROOF
The
existence of lim
/
/
is
equivalent to convergence of the series
/,'/+/;/+/,'/+••.
Now,
/
since
we have
decreasing
is
f{n
(Figure 2)
+ 1)<
/ <
f{n).
oo
The
first
double inequality shows that the
half of this
^=
series
n
compared
to the series
converges
if
lim
/
^
/,
proving that
n=l
^
^
a n +i
may
be
^=
a n)
l
a n +i (and hence
n=l
n
exists.
as
The second
half of the inequality shows that the series
f may be
^
n=
1
00
FIGURE
2
compared
to
00
Y
the series
A
a ni proving that lim
f
f must
Y
an
settles
the
exist
if
converges. |
Only one example using
the integral test will be given here, but
many
question of convergence for infinitely
vergence of
^
n=
\/n p
is
series at once. If p
equivalent, by the integral
>
it
0,
the con-
to the existence of
test,
1
r
Now
L
(P
dx.
1
1
-
A*-i
1)
log A,
+
A
This shows that lim
A—
» oo
f
^
1
1
/x p dx exists
1
p
-
P
1
p
if
p >
1,
but notify
=
1.
< l.Thus Y
J^,
l/n p
1
Infinite Series
>
converges precisely for p
1.
^ V*
In particular,
n=
The
tests
397
diverges.
l
considered so far apply only to nonnegative sequences, but nonmay be handled in precisely the same way. In fact, since
positive sequences
an
I
n=l
= -(
I
n=
«,),
considerations about nonpositive sequences can be reduced to questions
involving nonnegative sequences. Sequences which contain both positive and
all
negative terms are quite another story.
^
If
n=
an
is
a sequence with both positive and negative terms, one can con-
1
sider instead the sequence
£
\a n
all
\,
whose terms are nonnegative. Cheer-
of
fully ignoring the possibility that we may have thrown away all the interesting
information about the original sequence, we proceed to eulogize those
sequences which are converted by this procedure into convergent sequences.
00
oo
DEFINITION
The
series
vergent.
Y
(In
summable
if
an
absolutely convergent
is
more formal language, the sequence
the sequence
\
Although we have no right
is
\a n \\
THEOREM
4
is
\a n
\
to expect this definition to
The
is
con-
\
absolutely
be of any
interest, it
following theorem shows that the
at least not entirely useless.
Every absolutely convergent series is convergent. Moreover, a series is absothe
lutely convergent if and only if the series formed from its positive terms and
series formed from its negative terms both converge.
00
PROOF
is
\a n
summable.)
turns out to be exceedingly important.
definition
Y
the series
if
If
^
n=
\a n
Cauchy
converges, then, by the
criterion,
\
l
lim
|a n+ i|
+
*
*
*
+
\
a m\
=
0.
m,n—K*>
Since
|fl
it
n +l
+
'
'
'
+
am
\
+
<
•
"
•
follows that
lim
a n +\
+
m,n—> »
00
which shows that
^
a n converges.
'
'
'
+
am
=
0,
+
km|,
398
Infinite Sequences
and
Infinite Series
To
prove the second part of the theorem,
An,
if
an
o,
if
an
let
>
<
<
>
if O-n
if
so that
an~
^
n=
an+
^
n=
is
is
dn
o
o,
0
o,
the series formed from the positive terms of
1
an
^
n
=
,
and
1
the series formed from the negative terms.
l
^
If
and
n= 1
^
a n ~ both converge, then
n=l
OO
00
n
=
n=
1
00
n=
1
00
n=
1
l
00
also converges, so
^
n=
<z
converges absolutely.
n
l
00
On
the other hand,
^
if
n=
|<z
n
|
converges, then, as
we have
just
shown,
1
oo
^=
n
a n also converges. Therefore
1
t
*»
+
-
id * + s m
)
and
both converge.
|
It follows from Theorem 4 that every convergent series with positive terms
can be used to obtain infinitely many other convergent series, simply by putting in minus signs at random. Not every convergent series can be obtained in
this way, however
there are series which are convergent but not absolutely
convergent (such series are called conditionally convergent). In order to
prove this statement we need a test for convergence which applies specifi-
—
cally to series with positive
THEOREM
5
(LEIBNIZ'S
THEOREM)
and negative terms.
Suppose that
a\
>
a-i
>
az
>
•
and that
lim a n
=
0.
>
0,
H
«
399
Infinite Series
Then
the series
00
£—
71
(-l) n+1
=
n
<Z
—
ai
+
02
<Z
—
3
+
<Z4
<Z
-
5
*
'
*
1
converges.
PROOF
between the partial sums which we
illustrates relationships
Figure 3
will
establish:
(1)
S2
(2)
Si
(3)
Sk
<
>
<
Si
S3
<
>
k
is
*S
'
•
'
'
'
,
,
—H—
/ is
odd.
—
I
•SlO -^12
3
prove the
first
two
inequalities, observe that
(1)
(2)
To
•
even and
H—t—
56
To
S5
Sq
if
si
H
FIGURE
<
>
=
>
=
>
S 2n +2
S2n+S
+
S2n
S2n,
^n+l
"
<
S2n
#2n
Since fl2n+l
>
—
#2n+3
#2n + 2
+
first
S2n-\ ~~
>
fl
«2n+3-
that
2n
>
since a 2n
i-2n-i,
^2nJ
Since #2n+2
^2n+lj
prove the third inequality, notice
-
«2n+l
0.
and
This proves only a special case of (3), but in conjunction with (1)
that
general case is easy: if k is even and / is odd, choose n such
2n>
k
Sk
<
and
In
-
>
1
(2)
the
/;
then
which proves
Now,
S2n
<
<
S2n-\
^>
(3).
the sequence
bounded above (by
{s 2n
s t for
}
converges, because
any odd
a = sup
/).
is
it
nondecreasing and
Let
{s 2n
}
= hm
2n .
jr
n— »
*
Similarly, let
(S
=
inf {^2n+l}
— nm
S2n+1-
n—
* 00
If follows
from
(3) that
*2n+l
a <
-
0; since
^2n
=
«2n+l
and
UlH
fl
n
=
0
n—
»
it is
actually the case that
a =
0.
This proves that a
= 0 =
lim sn
,
|
is
400
Infinite Sequences
and
Infinite Series
The standard example
Theorem
5
is
-* + *-* + *
i
which
derived from
the series
,
convergent, but not absolutely convergent (since
is
^=
71
converge). If the
sum of this series
denoted by
is
\/n does not
1
the following manipulations
x,
lead to quite a paradoxical result:
*
= i -i + i-i + i -£+
= l- i- i + |- i- i +
(the pattern here
=
(i
-
i)
-i+
•
i-T o- TV + + -i r-iV+--1
1
one positive term followed by two negative ones)
is
-
(i
•
•
i)
-i+
(*
- A) -
A+
*- i + *- i + A- A + A- A+ --id - * + *- * + *- * + + -*+•)
=
so x
=
^
0:
x
*/2, implying that *
>
that x
sum
the partial
s2
=
On
(t
A
- A) -
+
.
.
-
the other hand,
it is
easy to see that
equals ^, and the proof of Leibniz's
Theorem shows
0.
sz.
This contradiction depends on a step which takes for granted that operations valid for finite sums necessarily have analogues for infinite sums. It is
true that the sequence
{b n
contains
{a n
=
}
1,
1,
— i,
-h h -h -h h - TV,
-i,
numbers
the
all
=
}
-tV,
•
•
•
in the sequence
h —h
t?
y
?
—h
— A> A> — A>
i>
*
'
"
•
is a rearrangement of {<z n
in the following precise sense: each
where / is a certain function which "permutes" the natural numbers, that is, every natural number m is f(n) for precisely one n. In our
example
In
fact, \b n
bn
—
/(2m
)
}
<2/(n)
+
1)
= 3m
+
1
(the terms
... go
1,
into the 1st, 4th, 7th,
.
.
.
places),
= 3m
/(4m)
/(4m
+
2)
= 3m
—
(the terms
9th,
+
2
.
.
— A,
-A,
(the terms
5th, 8th,
•
•
go into the 3rd, 6th,
places),
.
.
.
.
... go
into the 2nd,
places).
00
SB
Nevertheless, there
is
no reason
to
assume that
^
n=
these
sums
are,
by
definition, lim b\
n—
<*>
+
•
•
+
b„ should equal
bn
^=
n
l
and lim
n—*
ai
+
*
•
an
:
1
+
an,
oo
so the particular order of the terms can quite conceivably matter.
The
series
Infinite Series
401
00
2*
n
=
is
not special in this regard; indeed,
behavior
its
typical of
is
1
which are not absolutely convergent— the following
series
convergent
result (really
how bad
of a grand counterexample than a theorem) shows
more
conditionally
series are.
00
THEOREM
6
a n converges, but does not converge absolutely, then for any
^
If
n=
number a
l
00
there
a rearrangement \b n
is
of {a n
]
\
such that
^
n=
=
bn
a.
1
00
PROOF
Let
pn denote the
^
n=
y
series of
denote the
one of these
least
positive terms of {a n
formed from the
series
}
and
let
l
to converge, for
negative terms.
It
follows
does not converge. As a matter of fact, both must fail
one had bounded partial sums, and the other had un-
series
if
V
partial sums, then the original series
bounded
n=
Now let a be any number.
Assume,
an
would
also
have un-
1
assumption that
partial sums, contradicting the
bounded
from Theorem 4 that at
for simplicity, that
it
converges.
a >
0 (the proof for
Pn
not convergent,
oo
a <
0 will be a simple modification). Since the series
£
n=
there
is
a
number
N such that
n =
We
will
choose
is
1
JVi to
1
be the smallest
N with
this property.
This means that
JVi-l
(1)
1
n=
but
(2)
<
>
^p n
n=
Then
Pn
a,
l
a.
l
if
N^
Si
- £
n=
j»„,
1
we have
This relation, which
is
clear
^1
from Figure
—
<
4,
— ^
follows immediately
/> n
=
pNi>
from equation
402
Infinite Sequences
and
Infinite Series
0
+
pi
FIGURE
To
the
4
sum
T\ which
for
Si
we now add on just enough
than
is less
a.
negative terms to obtain a
we choose
In other words,
new sum
the smallest integer
M
i
which
Mi
T =Si+
J=
q n <Oi.
x
71
As
before,
we have
a —
We now
larger
1
continue
procedure indefinitely, obtaining sums alternately
this
and smaller than
< —qM v
Ti
a, each time choosing the smallest
N
k
or
•
•
M
possible.
k
The sequence
•
Pu
•
•
•
,pN
qi,
s9
.
.
.
,
qM„pN l+
\,
.
.
.
,pN
t,
•
an
The partial sums of this rearrangement increase
S u then decrease to T h then increase to S 2 then decrease to T 2 etc. To
complete the proof we simply note that \Sk — a\ and Tk — a\ are less than
or equal to pN or —qM respectively, and that these terms, being members
is
a rearrangement of
}
j
.
to
,
,
\
k,
k
of the original sequence {a n
must decrease
},
to 0, since
^=
n
a n converges. |
1
Together with Theorem 6, the next theorem establishes conclusively the
between conditionally convergent and absolutely convergent
distinction
series.
theorem
7
n=
and
a n converges absolutely,
^
If
{b n
}
is
any rearrangement of
{a n
},
then
l
00
^
n
b n also
converges (absolutely), and
=1
n=l
n=l
PROOF
Let us denote the partial sums of
by
{a n
}
by
sn ,
and the
partial
sums of
tn .
00
Suppose that
£
>
0.
Since
a n converges, there
^
n=
|
l
^
an
—
<
j-at
|
e.
is
some
N such that
{b n
}
403
Infinite Series
oe
Moreover, since
^=
n
\a n
converges,
we can
f
n=
N so
that
- (W +
knl
•
•
+ M) <
•
s,
1
so that
i.e.,
Now choose M so large that each of a u
Then whenever m > M, the difference
At,
also choose
\
1
.
.
.
aw
,
Thus,
if
—
m
m > M,
SN
\
<
•
t
•
aa-
,
-
m
appear among b h
s N is
the
sum
.
.
.
,
b
M
.
of certain a u where
Consequently,
are definitely excluded.
\t
•
+
|flA'+l|
|flJV+2|
+
|flAT +3|
+
'
*
*
•
then
|
£
an
—
/
^an -sN + U-sN
=
m
\
I
I
n-l
n=l
06
<
^
|
n=
<
Since this
is
true for every £
>
s
0,
tf
—
n
it
\tm
—
Sn\
J
+
s.
Y
the series
„
prove that
+
JA
l
bn
converges to
£
an
To
.
n-
=1
converges absolutely, notice that the above argument can be
applied to the positive terms of
shows that the
£
<2
n
and the negative terms separately;
formed from the positive and negative terms of
series
this
£=
n
bn
1
00
each converge, so
^
n=
6n
converges absolutely.
|
1
of this chapter have been concerned with summability of
reason
sequences, but not with the actual sums. Generally speaking, there is no
presume that a given infinite sum can be "evaluated' in any simpler terms.
The theorems
5
to
However, many simple expressions can be equated to infinite sums by using
which
Taylor's Theorem. Chapter 19 provides many examples of functions for
n
where lim
n—>
R n ,a{*) =
0-
This
is
precisely equivalent to
oo
n-*
oo
Z—/
i
!
•
404
Infinite
•
Sequences and Infinite Series
which means,
in turn, that
/(*)
=0
i
As particular examples we have
*3
x
+-_
3!
5!
sin x
X2
cos X
2!
e
x
=
1
+
b
X1
rL
+
X*
—_
+4!
-f
.
.
.
.
.
.
X
+-+-+
U + 2! 3! 4!
X
+ _ fL +
3
2
X
3
a:
3
a:
4
5
a:
arctan x
log(l
.
T
X2
+ *)
xz
2
.
--+-+
4
#4
a:
>
1; in
addition,
when
<
5
-1 <
,
X
<
1.
5
+
(Notice that the series for arctan * and log(l
\x\
M
.
7
= — 1,
#
do not even converge
+ x) becomes
x)
for
the series for log(l
-i-i-i
-i
which does not converge.)
Some
pretty impressive results are obtained with particular values of x:
0
=
=
TZ
7T
3!
5!
5
TT
7
7T
+
'
*
'
,
7!
+ h + + + ---'
h h
E-,_i + i_i + ...
4
3
5
7
1,1
^ =
log2
l-- + ---+....
e
1
f
More
1
,
,
,
developments may be anticipated if we compare the series
and cos x a little more carefully. The series for cos x is just the one
we would have obtained if we had enthusiastically differentiated both sides
significant
for sin x
of the equation
sin
x
=
x*
.
xb
x
•
3!
*
*
5!
we have never proved anything about the
if we differentiate both sides of the
formula for cos x formally (i.e., without justification) we obtain the formula
cos'(*) = — sin x, and if we differentiate the formula for e x we obtain
exp'(*) = exp(*). In the next chapter we shall see that such term-by-term
term-by-term, ignoring the fact that
derivatives of infinite sums. Likewise,
differentiation of infinite
sums
is
indeed valid in certain important cases.
Infinite Series
405
PROBLEMS
1.
Decide whether each of the following infinite series is convergent or
divergent. The tools which you will need are Leibniz's Theorem and
the comparison, ratio, and integral tests. A few examples have been
picked with malice aforethought; two series which look quite similar
may require different tests (and then again, they may not). The hint
below indicates which tests may be used.
eo
nd
sin
<"
i-.
-* + *-*+
GO
1
(iii)
1
(iv)
£(n
(v)
logn
n
=i
Y—
(The summation begins with
\
~
1
y </n}
+
1
—
1
00
(viii)
n=
1
(ix)
Z-/ log n
n=2
eo
S
=2
(log n)"'
00
(xi)
(xii)
S Gog nY
(-!)»—1_.
Y
Zv
n
(log n)»
=2
eo
(xiii)
n3
eo
(xiy
+
1
n
=
2 simply to
avoid the meaningless term obtained for n
(vii)
n
-
:
1
(x)
•
1)
~2 ~^n<1
n
*
-i+f - i+*-i+t-*+
n
(vi)
'
=
1.)
406
Infinite Sequences
and
Infinite Series
V-L-.
(xv)
1-1 n log n
n-2
00
^
XVi )
/
"71
\~2
n—2
00
1
(xvii)
_,n
2
(log n)
oo
(xvii
°
s?*
=
n
n
l
=
l
Hint: Use the comparison
test for (i), (ii), (v), (vi), (ix), (x), (xi), (xiii),
(xiv), (xvii); the ratio test for (vii), (xviii), (xix), (xx); the integral test
for (viii), (xv), (xvi).
The
next two problems examine, with hints, some infinite series that require
more
delicate analysis than those in
*2.
(a)
you have
If
Problem
1.
examples
successfully solved
(xix)
and
(xx)
from
00
Problem
1, it
should be clear that
and diverges
for a
>
For a
e.
=
e
n
^
n=
a n\/n
n
<
converges for a
e
1
the ratio test
show that
fails;
00
^=
71
e
n
n\/n
n
actually diverges,
by using Problem 21-10.
1
00
(b)
Decide when
21-10
when
n
n
n /a n\
^
n=
converges, again resorting to Problem
l
the ratio test
fails.
00
00
*3.
Problem
1
presented the two series
^
(log n)~
k
and
which the
first
^
(log n)~
n
of
,
n=2
n=2
diverges while the second converges.
The
series
00
1
(log n) los
n=
which
lies
between these two,
n
2
is
analyzed in parts
(a)
and
(b).
00
(a)
Show
that
JJ
e v /yv
dy exists,
by considering the series
£
{e/n)
n
.
407
Infinite Series
Show
(b)
that
logn
(log n)
;
converges, by using the integral
substitution
Show
(c)
and part
Hint: Use an appropriate
test.
(a),
that
V*
Z/(log
1
n)
l0 « (l0 «
n)
by using the integral test. Hint: Use the same substitution
and show directly that the resulting integral diverges.
diverges,
as in part (b),
4.
Let {a n
(a)
be a sequence of integers with 0
}
a n 10~ n exists (and
y
{a n
<
with 0
}
<
an
1). (This,
usually denote by G.aia2a 3 ai
number which we
(b) Suppose that 0
between 0 and
lies
<
x
an
<
<
1
Prove that there
.
a n \0~
£
and
9
=
n
is
.
.
<
Prove that
9.
of course,
the
is
.)
.
a sequence of integers
Hint: For example,
x.
n-l
ai = [10*] (where [y] denotes the greatest integer which
(c) Show that if \a n } is repeating, i.e., is of the form a h a 2
is
<
.
.
,
y).
.
,
CO
ai, 02,
•
•
•
ak, ai, a 2 ,
,
.
.
.
,
then
(and find
it).
repeating,
The same
i.e., if
a n \0
^
n=
n
is
a rational
result naturally holds if \a n
the sequence {a N+k }
number
l
eventually
is
}
repeating for some N.
is
00
Prove that
(d)
x
if
=
a n \0'
£
n=
n
rational, then
is
{a n
eventually
is
\
1
repeating. (Just look at the process of finding the decimal expansion
of p/q
5.
—dividing
Suppose that
\a n
)
proof of Leibniz's
|
^—
n
6.
p by long
q into
satisfies
the hypothesis of Leibniz's Theorem.
Theorem
(-l) n+1«»
-
division.)
Use the
to obtain the following estimate:
[«i
~
at
+
•
•
± a N ]\< a N
•
.
l
Prove the following theorem (another kind of "comparison
test"). If
00
an
,
bn
>
0 and lim a n /b n
00
if
^
converges.
=
c
^
0,
then
T
a n converges
if
and only
408
Infinite
Sequences and Infinite Series
7.
Prove that
an
>
0 and lim
if r
>
1.
if
and diverges
This result
known
is
which the ratio test
test shows that the
i
Va n =
r,
S
then
(The proof is very similar
as the "root test." It
while the root
fails,
<
if r
1,
to that of the ratio test.)
easy to construct series for
is
test
a n converges
works. For example, the root
series
+
+
2
(i)
+
2
(i)
3
(i)
+
3
(i)
+
•
•
•
converges, even, though the ratios of successive terms do not approach a
limit.
Most examples
are of this rather artificial nature, but the root test
nevertheless quite an important theoretical tool,
and if the ratio test
works the root test will also (by Problem 21-13). It is possible to elimirnate limits from the root test; a simple modification of the proof shows
is
00
that
^
n=
a n converges
there
if
is
some s <
1
such that
all
but
finitely
many
l
00
^/a n are
<
s,
and that
^
n-
This result
is
known
a n diverges
if infinitely
many
>
are
as the "delicate root test" (there
ratio test). It follows, using the notation of
is
a similar delicate
Problem 21-19, that
if
<
lim
1
and diverges
if
8.
possible
A
sequence \a n )
lim
=
is
<2i
+
*
*
•
"
+
•
•
•
+ f* =
/,
if
/
n
oo
*
sk
no conclusion
summable, with Cesaro sum
called Cesaro
is
n—
(where
:
lim
is
sequence
1
1
n—> oo
n—* *
if
>
lim
an
^
n=
converges
1.
1
+
Problem 21-12 shows that a summable
a k ).
sum equal
automatically Cesaro summable, with
Cesaro sum. Find a sequence which
is
not
to
its
summable, but which
is
Cesaro summable.
9.
This problem outlines an alternative proof of Theorem 7 which does not
rely
on the Cauchy
(a)
Suppose that a n
{a n
(b)
\,
and
let
criterion.
> 0 for each
= a\ +
sn
•
Show
that for each n there
Show
that
^
an
<
£
bn
is
•
n.
Let
+
an
\b n
be a rearrangement of
}
and
some m with
^n
t
= bi
< m
n
t
.
+
*
'
+
bn
.
409
Infinite Series
00
00
Show
(c)
that
=
an
^
^
bn
.
n=l
n=l
00
Now
(d)
replace the condition a n
>
0 by the hypothesis that
n
Theorem
converges absolutely, using the second part of
10.
Prove that
(a)
a n converges absolutely,
^
if
n=
and
{b n
an
1
4.
any subse-
is
}
^=
1
00
quence of
{a n
then
},
£
n=
converges (absolutely).
6n
1
00
Show
(b)
that this
n=
Prove that
*(c)
1
a n converges absolutely, then
^=
if
a n does not converge absolutely.
^
false if
is
n
l
oo
n=
+
= Ol
^
+
«3
^5
+
'
')
'
+
(«a
+ ^4 +
«6
Prove that
Problem
JQ
*13.
18-28
(sin x)/x\
00
00
Y
if
an
is
absolutely convergent, then
^
|
shows
that
j^" (sin x) fx dx
an
|
<
"=1
n=l
*12.
'
'
l
00
11.
+
£=
»
converges.
\a n
\.
1
Prove
that
dx diverges.
|
Find a continuous function / with /(*)
exists,
but lim f(x) does not
>
0 for
all x,
such that
J"
f(x) dx
exist.
X-+00
*14.
1 /x for 0 < a: < 1, and let /(0) = 0. Recall the definifrom Problem 15-32. Show that the set of all /(/, P) for
P a partition of [0, 1] is not bounded (thus / has "infinite length").
Hint: Try partitions of the form
Let f(x)
=
x sin
tion of t(f, P)
15.
Let / be the function shown in Figure
the shaded region in Figure
*16.
In this problem
we
5.
Find
J*
/,
and
also the area of
5.
will establish the
"binomial
5
series'
00
(i
+ *)« = £(«>*, W<i,
fc-0
for
any
steps,
a,
by showing that lim
Rn ,o(x) =
°-
The P roof
is
in several
and uses the Cauchy and Lagrange forms as found in Problem
1
9-6.
410
Infinite Sequences
and
Infinite Series
so
(a)
Use the
ratio test to
verge for
+
(1
(b)
<
|r|
a
r) ). It
Suppose
show
(this is
1
that the series
^ (^j
not to say that
it
that 0
<
x
<
1.
Show
k
does indeed con-
necessarily converges to
follows in particular that lim (
n—>oo \
first
r
a
) r
TL
that lim
n
=
0 for
/?„,<>(*)
=
Lagrange's form of the remainder, noticing that
(c)
+
1
>
<
1.
+
(1
by using
~
a n~
0,
n—» oo
for n
\r\
J
l
t)
<
1
a.
Now suppose that — 1 < x < 0; the number * in Cauchy's form of the
remainder satisfies — 1 < x < t < 0. Show that
+'0*"
1
|
<
M = max(l,
where
\x\M,
+ x)*-
(1
1
),
and
x
-
1
+
t
/
Using Cauchy's form of the remainder, and the
show that lim
/? n(0 (Ar)
=
fact that
0.
n—
» oo
*17.
(a)
Suppose that the partial sums of the sequence
and that \b n is a decreasing sequence with lim
)
{a n
bn
\
=
are bounded,
0.
Prove that
n—* oo
00
a rt 6„ converges. Hint:
^
n=
Use Abel's
Lemma
(Problem 21-20) to
1
(b)
check the Cauchy criterion.
Derive Leibniz's Theorem from
(c)
Prove, using Problem 15-31, that the series
this result.
^=
n
if*
is
(cos nx)/n converges
l
not of the form 2kir for any integer £ (in which case
it
clearly
diverges)
OO
*18.
Suppose \a n
verges,
then
]
is
decreasing and lim a n
2 n <2 2 »
^
n—
also
=
converges
0.
(the
Prove that
if
\
a»
con "
"Cauchy Condensation
l
OO
Theorem"). Notice that the divergence of
1/n
is
a special case, for
if
l
00
00
£
^
n=
1/n converged, then
n= 1
may
serve as a hint.
^
n=l
2 n (l/2
n
)
would
also converge; this
remark
411
Infinite Series
*19.
Prove that
(a)
n=
Prove that
(b)
an 2
£
if
^
n=
£
if
and
l
<z
n
2
6n
2
converge, then
^ nA
converges, then
fl
Suppose
{a n
}
>
decreasing and each a n
is
1
converges.
n=l
n=l
*20.
0 n £ n converges.
£
n=
l
Prove that
0.
if
^=
n
=
verges, then lim na n
be sure to use the
down
Hint: Write
0.
fact that \a n
]
the
Cauchy
a n con1
and
criterion
decreasing.
is
00
*21.
If
0.
a n converges, then the partial
;
sums
sn
are bounded,
and lim
an
=
=1
n
tempting to conjecture that boundedness of the partial sums,
It is
00
=
together with the condition lim a n
"
This
is
will
have
true,
not
As a
ingenuity.
but
n
a counterexample requires a
finding
hint, notice that
some
subsequence of the partial
you must somehow allow
to converge;
\
0, implies convergence of
this to
an
.
=1
little
sums
happen, without
letting the sequence itself converge.
oo
*22.
The divergence
of
remarkable
fact:
\/n
^
n=
is
a particular consequence of the following
1
Any
positive rational
number
x can be written as a
numbers of the form \/n. Such a representation can
finite
always be found by the straightforward method of continually adding
on the next number in the sequence which is not too large. For example,
sum
of distinct
the calculation
»-i=U
<
tt
f
shows that
Notice that the numerators 23, 13, 11, 1 are decreasing. Prove that
every positive rational x can be written in such a way by showing that
the numerators in this sort of calculation must always decrease, so that
eventually the remainder has numerator 1. Hint: You just have to check
that
l(n
if
+
p/q
<
1)
less
is,
\/n, but
than
p.
p/q
>
\/(n
+
1),
then the numerator of p/q
-
CHAPTER
g^+J
23
The
way
UNIFORM CONVERGENCE AND
POWER SERIES
considerations at the end of the previous chapter suggest an entirely
of looking at infinite series.
infinite
sums
Our
new
attention will shift from particular
to equations like
=
e*
1
X
x^
1!
2!
+ — + f. +
.
.
.
which concern sums of quantities that depend on x. In other words, we are
interested in functions defined by equations of the form
/(*)
(in the
will
=AW
previous example
+/tW +/aW +
n~
fn {x) = x
l
—
/(n
we
•
In such a situation {/„}
1)!).
be some sequence of functions; for each x
•
obtain a sequence of
num-
bers {/»(*)}, and /(#) is the sum of this sequence. In order to analyze such
functions it will certainly be necessary to remember that each sum
/iW
by
is,
+ /*(*)
definition, the limit of the
we
then
define a
we can
new sequence
•
•
•
sequence
/iW, /iW +M*)>
If
+
+/•(*)
AW
+Mx) + /.(*), ....
of functions
Jn
= fi
express this fact
more
/(*)
+
'
n
{
'
'
+ /»,
succinctly
=
lim
by
}
by writing
sn (x).
n—*to
For some time we
shall therefore concentrate
/(*)
=
on functions defined
as limits,
lim /„(*),
rather than on functions defined as infinite sums.
The
total
body of
results
about such functions can be summed up very easily: nothing one would hope
to be true actually is
instead we have a splendid collection of counterexamples. The first of these shows that even if each fn is continuous, the function / may not be Contrary to what you may expect, the functions fn will be
very simple. Figure 1 shows the graphs of the functions
—
!
/»(*)
=
,
j 1,
412
0
x
<
>
x
1.
<
1
413
Uniform Convergence and Power Series
These functions are
=
continuous, but the function f(x)
all
lim fn (x)
is
not
n—» oo
continuous; in fact,
hm/n (*) =
0,
0
<
1}
x
>\
|
Another example of
this
same phenomenon
<
*
1
m
illustrated in Figure 2; the
is
functions fn are defined by
-1,
x
< -
n
FIGURE
AW
2
=
-
< nx,
1
-
<
x
<
1
n
n
i,
n
In this case,
-1, and
to
if
if
x
< 0, then fn (x) is eventually (i.e., for large enough n) equal
> 0, then fn (x) is eventually 1, while n (0) = 0 for all n. Thus
x
lim/n
so,
W
once again, the function f(x)
=
=
[-1,
*<0
(0,
x
U,
^>0;
lim fn (x)
=
is
0
not continuous.
n—> oo
By rounding
FIGURE
off the corners in the previous
examples
differ entiable functions
for
produce a sequence of
3
f(x)
= lim/n 0)
n—*
is
not continuous.
One
\jn \
it is
even possible to
which the function
such sequence
is
easy to define
so
explicitly:
-1,
/
n
3
-
1,
<x.
n
2-
These functions are differentiable (Figure
A
1
1-
lim/n (*) =
still
have
< 0
= 0
*>o.
-1,
x
*
Continuity and differentiability are, moreover, not the only properties for
which problems arise. Another difficulty is illustrated by the sequence {/n }
shown
in Figure 4;
on the
interval [0, \/n] the
triangle of altitude n y while
4
but we
0,
U,
\ A
l
FIGURE
3),
defined explicitly as follows:
fn (x) = 0
for x
>
graph of fn forms an
\/n. These functions
isosceles
may
be
414
Infinite Sequences
and
Infinite Series
2n 2x,
AW
=
{
2n
0
<
-
<
~
x
—
2n
!<*<!
- 2n%
2n
n
1
<
0,
1.
Because this sequence varies so erratically near 0, our primitive mathematical instincts might suggest that lim fn (x) does not always exist. Nevern—»
does exist for
theless, this limit
continuous. In fact,
=
moreover, /n (0)
if
x
>
0,
0 for all
all x,
oo
and the function f(x)
then /„(*)
0,
lim /„(*)
n—
certainly
so lim
other words, f(x)
=
lim fn {x)
n—
>
=
0 for
On
all x.
=
dx
\q fn(x)
5
fn (x) =
0;
=
In
0.
oo
the other hand, the integral
oo
quickly reveals the strange behavior of this sequence;
FIGURE
even
oo
have lim /n (0)
n—
is
oo
n—
we
so that
n,
eventually
is
=
we have
-J",
but
Thus,
lim
n—^oo
lim/n (*)
C fn(x) dx ^ r n—
u
u
•/
» oo
This particular sequence of functions behaves in a way that we really never
imagined when we first considered functions defined by limits. Although it is
true that
f(x)
=
lim fn (x)
for
n—* oo
each x in
[0, 1],
the graphs of the functions fn do not "approach" the graph of / in the sense of
—
5, we draw a strip around /of total width 2e
and below), then the graphs of fn do not lie completely within this strip, no matter how large an n we choose. Of course, for
each x there is some N such that the point (x, fn (x)) lies in this strip for n > N;
this assertion just amounts to the fact that lim fn (x) = f(x). But it is necessary
lying close to
it
if,
as in Figure
(allowing a width of e above
71—>
to choose larger
and larger
A^'s as
x
is
oo
chosen closer and closer to
and no one
0,
N will work for all x at once.
FIGURE
6
The same
situation actually occurs,
though
less blatantly, for
each of the
other examples given previously. Figure 6 illustrates this point for the sequence
0
X
<
>
x
<
1
1.
A strip of total width 2s has been drawn around the graph oif{x) =
lim /„(*).
n—*
If e
<
this strip consists of
two
pieces,
«j
which contain no points with second
Uniform Convergence and Power Series
coordinate equal to
of each
N such that
since each function
-5-;
within this
fn fails to lie
strip.
takes
fn
Once
on the value
415
the graph
-g-,
again, for each point x there
is
fjx)) lies in the strip for n > N; but it is not possible to
pick one TV which works for all x at once.
It is easy to check that precisely the same situation occurs for each of the
other examples. In each case we have a function /, and a sequence of functions
{fn }, all defined on some set A, such that
some
(x,
— Km fn {x)
f(x)
for all x in A.
This means that
FIGURE
7
for all e
then
>
0,
and
-/»(*)
\f(x)
for all x in A, there
<
I
But in each case different N's must be chosen
it is
is
some
N
such that
if
n
>
N
9
e.
for different
x% and in each case
not true that
>
for all £
0 there
l/W - AMI <
Although
is
some
N such
that for all x in A,
if
n
>
N, then
«.
from the
this condition differs
first
only by a minor displacement
has a totally different significance. If a
sequence {/„} satisfies this second condition, then the graphs of/n eventually
lie close to the graph of /, as illustrated in Figure 7. This condition turns out to
of the phrase "for
all
x in A,"
it
be just the one which makes the study of limit functions
DEFINITION
Let \fn
which
if
]
is
feasible.
be a sequence of functions defined on A, and
let / be a function
on A. Then / is called the uniform limit of {/„ on A
such that for all x in A,
0 there is some
also defined
>
for every £
}
N
ifn> N,
We also say that {fn
}
then
|/(*)
-/«(*)!
<
8-
converges uniformly to /on A, or that
fn
approaches
/ uniformly on A.
As a contrast
to this definition,
f(x)
=
if
we know only
lim fn (x)
71—>
for
that
each x in A,
00
we say that {/M converges pointwise to /on A, Clearly, uniform convergence implies pointwise convergence (but not conversely!).
Evidence for the usefulness of uniform convergence is not at all difficult to
amass. Integrals represent a particularly easy topic; Figure 7 makes it almost
then
}
obvious that
made
if
{fn \ converges uniformly to
/,
then the integral of fn can be
as close to the integral of /as desired. Expressed
more
precisely,
we have
the following theorem.
THEOREM
1
a sequence of functions which are integrable on [a, b],
converges uniformly on [a, b] to a function /which is integrable
Suppose that {/„}
and
that {/n }
is
—
416
Infinite
Sequences and Infinite Series
on
PROOF
Let
[a, b].
e
>
Then
0.
There
is
N such that for all n
some
~ /ntol <
l/to
Thus,
if
N we have
for all * in
S
[a, b].
N we have
>
n
>
f*f(x) dx
-
|
b
(x)
fa fn
dx\
= /; [f(x) - /„(,)] dx
* //l/W -/»(*)!<&
\
I
< f\dx
= £(A - a).
Since this
is
true for
any
£
>
0, it
follows that
/0V=lim/>|
The treatment of continuity is only a little more difficult, involving an
"s/3-argument," a three-step estimate of \f(x) - f(x
If {fn
is a
ti)\.
sequence of continuous functions which converges uniformly to /, then there
is some n such that
+
Moreover, since fn
is
-/nto| <
(1)
l/to
(2)
|/(*-M) ~/n(*
J
+ A)|
<|.
continuous, for sufficiently small h
(3)
l/nto
-/n(*
\
+ *)|
we have
<|-
from (1), (2), and (3) that |/(*) - f(x
h)\ < e. In order to
however, we must restrict the size of \h\ in a way that cannot be
predicted until n has already been chosen; it is therefore quite essential that
there be some fixed n which makes (2) true, no matter how small \h\ may be
+
It will follow
obtain
it is
THEOREM
2
(3),
precisely at this point that uniform convergence enters the proof.
Suppose that {/n
and that {/n
}
}
is
a sequence of functions which are continuous on
converges uniformly on
[a, b]
to /.
Then /is
[a, b] 9
also continuous
on
[«,*]•
PROOF
For each x in
[a, b]
with x in
b)
fications.
(a,
;
we* must prove that /
the cases x
=
a
and x
is
=
continuous at
x.
We will deal only
b require the usual
simple modi-
Uniform Convergence and Power
Let e
>
0.
Since {/„} converges uniformly to / on
[a, b],
Series
there
is
417
some n
such that
\f{y)
In particular, for
Now /„
is
all
h such that x
(1)
|/(*)
(2)
|/(*
continuous, so there
(3)
Thus,
|/(*
-/.Ml <|
if |A|
<
for all
+h
-/B (*)|
+ A)
is
in
is
8
|/nW -/„(*
[a, b],
[a,*].
we have
<|.
-/„(*
some
>in
>
+ A)
|
<|-
0 such that for
+ A)|
|A|
<
S
we have
<|-
then
5,
+ A) -/Ml
+ A) — /„(* + h) + /„(* + h) - /„(*) + /»(*) - /(at)
~ /(*)|
|/(* + A) - /»(* + A) + l/»(* + h) - /»(*)! + |/»to
s
s
s
< + +
I
3
=
<
|/(*
|
I
,
,
3
This proves that /
FIGURE
is
continuous at
*. |
8
After the two noteworthy successes provided by Theorem 1 and Theorem 2,
the situation for differentiability turns out to be very disappointing. If each
not necesfn is differentiate, and if }/n } converges uniformly to /, it is still
there is a
that
shows
Figure
8
sarily true that / is differentiate. For example,
sequence of differentiable functions {/n
}
which converges uniformly
to the
418
Infinite Sequences
and
Infinite Series
function f(x)
=
Even
|*|.
if
/
is
differentiable,
this
is
not at
all
surprising
if
we
may
reflect that
a smooth function can be approxi-
oscillating functions.
=
fn(x)
For example (Figure
=
fn(x)
n—
x
=
9), if
-sin(rc 2#),
n
then {fn } converges uniformly to the function /(*)
and lim
not be true that
= Hm /.'(*);
/'(*)
mated by very rapidly
it
=
0,
but
2
n cos(n x),
2
n cos(w *) does not always exist (for example,
it
does not exist
if
oo
0).
n
FIGURE
9
Despite such examples, the Fundamental Theorem of Calculus practically
guarantees that some sort of theorem about derivatives will be a consequence
of Theorem 1; the crucial hypothesis is that
{//} converge uniformly (to
some continuous function).
theorem
3
Suppose that {fn }
is a sequence of functions which are differentiable on
[a,b\ and that {/n } converges (pointwise) to /. Suppose, moreover, that
{fn} converges uniformly on [a, b] to some continuous function g. Then / is
differentiable
and
/'(*)
-
lim /„'(*).
n—
PROOF
Applying Theorem
1
to the interval
=
oo
[a, x],
lim [/„(*)
= fix) -
we
see that for
- /„(*)]
f(a).
each x we have
419
Uniform Convergence and Power Series
Since 0p
continuous,
is
=
follows that f{x)
=
g(x)
lim fn (x) for
x—*
the interval
[a, b].
x in
all
*>
|
that the basic facts about uniform limits have been established,
Now
clear
it
how
it is
to treat functions defined as infinite sums,
/« =/iW +/2W + /»(*) +
*
'
*
•
This equation means that
/(*)
=
+
lim/i(jr)
n—> oo
•
•
•
+/„W;
our previous theorems apply when the new sequence
/i,/i+/2,A
+/«+/*
•
we
converges uniformly to /. Since this is the only case
ested in, we single it out with a definition.
DEFINITION
The
£
series
n=
is
shall ever
be
inter-
(more formally: the sequence {fn
/„ converges uniformly
\
1
uniformly summable)
to
/ on A,
if
the sequence
+/»+/•
•
•
•
converges uniformly to / on A.
We
series;
convergent
can now apply each of Theorems 1, 2, and 3 to uniformly
corollary.
common
in
one
stated
be
may
results
the
oo
COROLLARY
Let
^
(1)
(2)
fn converge uniformly
to
/on
[a, b].
each fn is continuous on [a, b], then / is continuous on
If / and each fn is integrable on [a, b], then
If
[a,
b\
/:/=!/>
n=
l
00
Moreover,
if
uniformly on
T
/n converges (pointwise)
to
[a, b]
to
/ on
fix)
- ^
ft],
some continuous function, then
00
(3)
[a,
for
aU * in
[a
'
bl
and
£
fn converges
420
Infinite Sequences
and
Infinite Series
PROOF
If
(1)
each/n
is
continuous, then so
uniform limit of the sequence f h
continuous by Theorem 2.
(2)
Since
/i,
/i
+/
+/ +/
/i
2,
2
from Theorem
follows
is
+f
/i
.
3}
each /i
•
•
+ /2
fi
+ /n, and /is the
•
+/s,
.
.
.
so /is
,
converges uniformly to
.
.
2,
+
/,
it
that
1
///.to /.'</, + ••+/.)
-«-(/>+••
+/;/.)
00
Each function
(3)
'
•
;
+
/i
+ /»',
•
•
+ /n
•
and //,
with derivative
differentiable,
is
+/
/i'
2 ',
/
x
'
+ /,' + /,',
.
.
con-
•
verges uniformly to a continuous function, by hypothesis. It follows
from Theorem 3 that
/'(*)
=
lim [//(*)+
n—
•
•
•
+/„'(*)]
oo
00
= 2 /.'w.
=
n
At the moment
to predict
uniformly.
i
1
this corollary
is
not very useful, since
it
seems quite
difficult
when the sequence /i, /i + / /i + /2 + /3
will converge
The most important condition which ensures such uniform con2
,
,
.
.
.
is provided by the following theorem; the proof is almost a triviality
because of the cleverness with which the very simple hypotheses have been
vergence
chosen.
THEOREM
4
(THE WEIERSTRASS M-TEST)
Let {fn be a sequence of functions defined on A, and suppose that
sequence of numbers such that
\
M
<
\fn(x)\
{M n
}
is
a
for all x in A.
n
CO
Suppose moreover that
M
T
«
=
n
converges.
Then
for
each x in
£
A
the series
i
oo
00
fn(x) converges (in fact,
it
converges absolutely), and
n~l
uniformly on
A
fn converges
^
n==
1
to the function
00
/(*)
=
I=
n
PROOF
For each x in
A
the series
^
/.(*).
3
|/nW| converges, by the comparison
test;
Uniform Convergence and Power Series
consequently
^=
n
/„(*) converges (absolutely).
Moreover, for
all
x in
421
^ we have
1
00
i/w -
[/iw
+
•
•
•
+/-WH =
I
<
n
2 /»w
f
I
1/nWI
= 7?+l
00
<
fM
=
i(2*i
M
n.
00
00
Y
Since
I
M
n converges, the
desired,
by choosing
FIGURE
10
number
Y
Af n can be made as small as
TV sufficiently large. |
following sequence {/n } illustrates a simple application of the Weierinteger (the
strass M-test. Let {*} denote the distance from x to the nearest
define
Now
Figure
=
in
10).
illustrated
is
{x}
graph of /(*)
The
(a)
/.W -iJ:
drawings
functions f x and /2 are shown in Figure 11 (but to make the
n
n
has been
functions
simpler, 10 has been replaced by 2 ). This sequence of
clearly
applies:
defined so that the Weierstrass M-test automatically
The
/.to
=
1
|/«Ml<^
for all*,
(b)
FIGURE
*
00
11
and
^ V10
n
converges.
Thus
£
since each /«
fn converges uniformly;
continuous, the corollary implies that the function
is
!
422
Infinite
Sequences and Infinite Series
is
also continuous. Figure 12
fi
+
*
*
•
+
/n
As
.
shows the graph of the first few partial sums
graphs become harder and harder to
n increases, the
so
draw, and the
infinite
sum
^
n—
fn
shown by the
quite undrawable, as
is
follow-
1
ing theorem (included mainly as an interesting sidelight, to be skipped
if
you
find the going too rough).
THEOREM
5
The
function
/w =
tro"»
—
n
l
continuous everywhere and differentiable nowhere
is
PROOF
{10 "* }
We have just shown that /is continuous; this is the only part of the proof which
We will prove that / is not differentiable at a, for
uses uniform convergence.
any
{h m
<z,
by the straightforward method of exhibiting a particular sequence
for which
approaching 0
}
f(a
lim
+
hm)
m-K»
does not
< a<
0
exist. It
~ f{a)
hm
obviously suffices to consider only those numbers a satisfying
1.
Suppose that the decimal expansion of a
=
a
Let h m
=
for these
is
....
= -10~ m if am =
0.<Zi<22<23tf4
10~ m if a m ^ 4 or 9, but let k m
two exceptions will appear soon). Then
f(a
+ h m - f(a)
\lO n (a
)
+ h m )} -
±
infinite series
i<r-*[{io*(a
really a finite
is
n
{10 a}
±10~ m
10
This
4 or 9 (the reason
+
sum, because
h m )}
if
n
-
>
{I0
n
a}l
n
m, then 10 h m
is
an
integer, so
{10
On
the other hand, for n
n
10 a
10 n (a
(in
hm
n
+
hm)
=
=
(a
+
h m )\
-
< m we can
integer
integer
+
+
n
\10 a]
9).
0.
write
0.a n+ ia n+2 a n +3
.
.
•
0.a n+ ia n +2a n +3
.
.
.
order for the second equation to be true
~ — 10 -m when a m =
-
Now
it
is
am
(a m
•
•
•
.
±
•
1)
•
essential that
suppose that
«m
.
2
•
•
•
we choose
423
Uniform Convergence and Power Series
Then we
also
have
O.0 n +i<Z n + 2 0n+3
+
(flm
-
.
.
1)
•
=
•
^i
m = n + 1 the second equation is true
= 4). This means that
±10*-™,
{10"(fl + A»)| - {10"*} =
(in the special case
hm
•
because
we chose
-10" m when a m
and exactly the same equation can be derived when 0.a n+ ia n -^2a n+ 3
Thus, for n < m we have
m- n [{\0*(a + k m)\ - {10M] = ±1\0
.
.
.
h
>
In other words,
+ k m - f(a)
f(a
)
hm
is
/!+/.+/•
to
sum of m - 1 numbers, each of which is ± 1. Now adding +1 or
The sum of m
a number changes it from odd to even, and vice versa.
the
±1
numbers each
is
is
therefore
an
even integer if
m
is
odd, and an odd
—
1
-
1
integer if
m
even. Consequently the sequence of ratios
f(a
+ hm
)
-f(a)
hm
cannot possibly converge, since
nately
odd and even.
a sequence of integers which are alter-
it is
|
Weierstrass M-test is an
In addition to its role in the previous theorem, the
behaved. We will give
well
very
are
ideal tool for analyzing functions which
special attention to functions of the
form
/l+/.+/l+/4
a n (x
/(*)
which can
also
fn (x)
=
a n {x
on powers of
simplicity,
(*
we
a)
n
,
be described by the equation
/to
for
-
sum, of functions which depend only
centered at a. For the sake of
is called a power series
series centered at 0,
power
on
concentrate
usually
- a) n
- a),
will
=
.
Such an
infinite
eo
/(*)
=
£
a n x\
n=0
One
especially important
group of power
n =0
series are those of
the form
424
Infinite Sequences
and
Infinite Series
where /
is
called the
some function which has derivatives of all orders at a; this series
Taylor series for / at a. Of course, it is not necessarily true that
tea (*-«)•;
/W-
=0
71
this
equation holds only
when
is
R n Ax —
the remainder terms satisfy lim
n—+
We already know that a power series ^
)
0.
as
a n x n does not necessarily converge
n=0
for all x.
For example the power
series
X3
converges only for
|#|
<
#2
,
#3
*4
3
4
2
—1 <
converges only for
X7
x
for x
<
=
1
.
0,
.
^
xh
for
any x
.
It is even possible to produce a power
For example, the power series
series
n
+
l)!(*
n+1
=
)
+
(n
nix
unbounded
^
0; indeed, the ratios
(n
are
,
5
i\x
does not converge for x
,
while the power series
1,
#
which converges only
Xb
t
9* 0.
If
\)x
a power series
^
a nx
n
does converge for
n=0
00
some
for
theorem
6
|*|
#0^0
<
however, then a great deal can be said about the
series
^
a nx
n
|# 0 |.
Suppose that the
series
/Oo)
converges,
and
let
<z
=
^
^n^0
be any number with 0
n
<
a
<
\x 0 \.
Then on [ — a,
a]
the
series
00
/(*)
=
£=
n
0
converges uniformly (and absolutely). Moreover, the same
is
true for the
i
Uniform Convergence and Power Series
425
series
na n x "
«=
/
Finally,
and
differentiable
is
= J
n=
/'(*)
for all x with
<
|x|
nawX
^
1
1
|* 0 |-
00
PROOF
Since
£=
71
tf
n* 0
n
converges, the terms
is
*
in
is
[
— a,
|<z«*
n
|
=
<
=
n
|
then
a],
=
0.
Hence they
are surely
\x
\On\
•
i^l
\
<
M
<
|
for all n.
so
\a\,
|«n|
*
|*0]
\
\
(this
Xq
the clever step)
is
I
0
< Af
\a/x 0
n
n
'
\a n
\
<
|*|
|x 0
•
\a n
I
But
approach
some number Af such that
|a n *o
if
n
0
bounded: there
Now
n*0
tf
I
so the (geometric) series
1,
n
converges. Choosing Af
=0
\a/x 0
n
as the
number Af n
in the Weierstrass Af-test,
\
CO
it
follows that
^
a nx
n
converges uniformly on [-a,
a].
n=0
op
To prove
the
same
assertion for g(x)
= ^
n=
<
._
<
Since \a/x 0
\
<
1,
«W
W
Af
\a
•
u
.
1
n_1 notice that
wa n ^
l
n-l|
|»
„
±r
|
a
n «!-
the series
Af
=
M\
A
j
n
—a
h„4
converges
(this fact
was proved
in
Chapter 22 as an application of the
ratio
426
Infinite Sequences
and
Infinite Series
test).
Another appeal
to the Weierstrass M-test proves that
^
na n x
n
1
con-
n-l
[— a,
verges uniformly on
a).
Finally, our corollary proves,
=
/'(*)
= £
g(x)
n
Since
that g
first
=
na n x
n~ l
Theorem
\x\
6,
<
|x 0 |.
for x in
[-a,
a].
1
we could have chosen any number
for all x with
continuous, and then that
is
<
with 0
a
<
a
\x 0 \,
this result
holds
I
applied to the infinite series
=
sin x
x7
3!
5!
7!
X4
X*
2!
4!
6!
X
x^
*9
.
•
•
•
,
,
9!
X8
,
1
=
eX
x^
X2
—
cos X
x*
x
x^
+v+v+
1
—
•
•
'
,
8!
x^
+
Ti
4i
-'->
+
which are expected. Each of these converges for any
hence the conclusions of Theorem 6 apply for any x:
yields precisely the results
#o,
—+—=
= - — + —- —-+•••
= + — + — +*•' =
=
sin'(*)
„
cos'(x)
,
-
1
5x i
3x 2
2x
4x*
6x>
21
41
61
2x
exp'(*)
3>x
=
1
-
= -
sin x,
2
-
+
log(l
~
^3
x
x) the situation
is
only slightly
+ x* -
—+
x«
•
'
7
5
also converges for
x2
^7
j£\>
1
3
arctan'M
x,
the series
arctan x
1, it
cos
exp(x).
For the functions arctan and f(x)
=
•
1
more complicated. Since
converges for x 0
•
•
<
\x\
+
•
1,
=
•
and
—J—
+x
1
In this case, the series happens to converge for x
formula for the derivative is not correct for x =
= —1
1
for
|x|
2
or x
also.
<
1.
However, the
= — 1;
indeed the
series
- x + x* - * +
= — 1 Notice that this does not contradict Theorem
= 1 and x
which proves that the derivative
diverges for x
6,
\x\
<
\xo\.
6
2
1
•
•
•
.
is
given by the expected formula only for
.
Uniform Convergence and Power
Series
427
Since the series
+ ,)^-| + |-|+|
log(l
=
converges for x 0
1
In
=
+*
1
1, it also
log'(l
+
x)
converges for
=
-
1
+x 2
x
<
|#|
*3
and
1,
+
•
"
•
does not converge for x
this case, the original series
=
differentiated series does not converge for x
for
= —1
power
its
power series
where the derivative
derivative, at the points
;
<
1.
moreover, the
1
All the considerations which apply to a
apply to
\x\
will automatically
is
represented by a
series. If
00
=
/(*)
I=
71
converges for
all
(—R,R), then Theorem
x in some interval
6 implies that
n
na n x ~i
= y
f{x)
a n x»
0
n-l
for all * in
(—R
9
Theorem
R). Applying
6 once again
we
find that
00
=
/"(*)
J
n=
»(»
- IK*"-2
,
2
and proceeding by induction we
find that
oo
/<*>(*)
=
J
n(n
-
1)
Thus, a function defined by a power
(
— R,R)
is
•
•
(«
•
series
-
k
+
IK**"*.
which converges
in
some interval
automatically infinitely differentiable in that interval.
Moreover,
the previous equation implies that
/<*>(0)
-
k\a k3
so that
/<*>(0)
A!
In other words, a convergent power
of the function
On
this
Taylor
which
it
series centered at
happy note we could
series.
A
0
is
always
the
Taylor series at 0
defines.
easily
end our study of power
careful assessment of our situation will reveal
and
some un-
series
explained facts however.
The Taylor
Taylor
and exp are as satisfactory as we could
and can be differentiated term-by-term for all
series of sin, cos,
they converge for
all #,
series of the function f(x)
=
log(l
+
x) is slightly less pleasing,
desire;
x.
The
because
428
Infinite Sequences
and
Infinite Series
—1 <
converges only for
it
x
<
but
1,
a necessary conse-
this deficiency is
quence of the basic nature of power series. If the Taylor series for / converged
for any x 0 with |# 0 > 1, then it would converge on the interval ( — 1* 0
|*o|), and
on this interval the function which it defines would be differentiable, and thus
continuous. But this is impossible, since it is unbounded on the interval
( — 1, 1), where it equals log(l + x).
The Taylor series for arctan is more difficult to comprehend there seems
[,
I
—
be no possible excuse for the refusal of
to
This mysterious behavior
=
f(x)
+
1/(1
is
exemplified even
= 7-7— 2 =
/(*)
+x
1
\x\
>
more
~
1
i
;
*2
4
+* -
*6
+
x*
of /
~
all.
'
question
always dangerous, since
is
unsympathetic answer:
are
!
In
it
happens because
it
>
1.
'
the next best
is
is
given by
'
•
Why? What unseen
obstacle prevents the Taylor series from extending past
this sort of
\x\
by the function
strikingly
The Taylor polynomial
the Taylor series does not converge at
1
converge when
x 2 ), an infinitely differentiable function which
thing to a polynomial function.
If
this series to
1
and —1? Asking
to settle for an
we may have
happens
—
that's the
way
things
does happen to be an explanation, but this explanation
this case there
is about real
can be answered intelligently only when placed in a broader
context. It will therefore be necessary to devote two chapters to quite new
material before completing our discussion of Taylor series in Chapter 26.
is
impossible to give at the present time; although the question
numbers,
it
PROBLEMS
1
For each of the following sequences {/n }, determine the pointwise limit
(if it exists) on the indicated interval, and decide whether {fn
of {fn
}
\
converges uniformly to this function.
(i)
/nW = Vx,
(ii)
fn(x)
x
o,
(iii)
/«(*)
(iv) f(x)
e
on
[a, b],
and on R.
c
j
x
=
e
e
(v)
n
z
—n
=
<
—
x
—nx*
5
-x*
j
f(x)
on R.
n
2-
Find the Taylor
(i)
fix)
/(*)
=
each of the following functions.
1
=
x
(ii)
series at 0 for
—
a 9^0.
a
log(*
-
a),
a*
0.
429
Uniform Convergence and Power Series
(iii)
1
=
/(*)
-
V —
(
3.
v)
=
/(*)
arcsin
Find each of the following
(ii)
_
i
-
^!
(iii)
i
v
If
FT
=
/W
power
5.
In
_
xs
—+—
3-2
(1VJ
^
+
.
.
—+
•
•
0
•
-
+* -* +
3
2
*
?
*
Problem
(See
10.)
[
=
/(0)
1,
find
/
(A)
(0).
Hint: Find the
series for /.
problem we deduce the binomial
this
<
*
Hint: Differentiate.
1.
series (1
+ x) a =
without
1
all
fact established in part (a) of that
problem
—the
series
,
=0
work of Problem 22-16, although we
the
*n
^
n
|*|
•
1
'
4
and
1
W <
for
•
L±...
5
^
for *
*)A
What is
5-4
J
,
Hint:
.
§
4-3
i
sums.
infinite
3-2^4-3
(sin
(Use Problem 19-6.)
.
y«4
V>3
+
*3
*)- 1/2
x.
y.2
4.
-
(1
x
1
will use
a
x n does
^
n=0
converge for
(a)
Prove that
(b)
Now
f(x)
<
|*|
(1
1.
+ *)/'(*) =
af(x) for
<
[x|
1.
show that any function / satisfying part (a) is of the form
= c(l + x) a for some constant c, and use this fact to establish
the binomial
series.
Hint: Consider g(x)
=
f(x)/(l
+ x) a
.
CO
6.
(a)
Suppose that
/(*)
=
anX " converS es for
£
a11
* in
some
interval
n=0
(-R,R) and
<z
(b)
n
=
(If
0.
that /(*)
=
0 for
all
x in (-/?, R). Prove that each
you remember the formula
=
Suppose we know only that /(*»)
lim x n = 0. Prove again that each a n
71—
for a n this
0 for
=
is
easy.)
some sequence
0.
{xn } with
Hint: First show that
00
f(0)
=
ao
=
0;
then that /'(0)
=
ai
=
0, etc.
1/xt
sin 1 /x for * 5* 0, then / cannot
This result shows that if f(x) = e~
It also shows that a function
series.
possibly be written as a power
be
0 for x < 0 but nonzero for
cannot
series
power
defined by a
the motion of a particle
0
describe
cannot
series
power
thus
a
x >
—
which has remained
at rest until time 0,
and then begins
to
move
430
Infinite Sequences
and
Infinite Series
= £
n
a n x and g(x) = V b n x n converge for all x
n=0
n=0
in some interval containing 0 and that f(t m ) = £(/m ) for some
Suppose that /(*)
(c)
=
sequence
{tm } converging to 0. Show that a n
b n for each n. In
particular, a function can have only one representation as a power series
centered at 0.
00
7.
Prove that
if
=
/W
^
n=
a n*
n
an even function, then
is
<z
n
=
0 for n odd,
0
*8.
and if / is an odd function, then a n = 0 for n even.
Recall that the Fibonacci sequence {a n \ is defined by ai
tf
n+ i
=
an
(a)
Show
(b)
Let
+
=
=
a2
1,
a n -i.
<
2.
B_l
=
that a n+1 /a n
00
/to =
2
n=
Use the
(c)
*n*
1
+*+
if
<
|4
•
•
•
.
|*|
<
1/2.
=
2
+x-
1
Hint: This equation can be written /(*) - xf(x) - x 2f(x) = 1.
Use the partial fraction decomposition for I/O 2
x - 1), and the
power series for l/(x - a), to obtain another power series for /.
+
It follows from Problem 6 that the two power
must be the same. Use this fact to show that
an
9.
+
1/2, then
a:
(e)
3* 3
ratio test to prove that /(*) converges if
Prove that
fix)
(d)
+
2* 2
1
Show that
-1 < x <
—
series
obtained for /
~
r
V5
power series for /(*) = log(l — x) converges only
and that the power series for g (x) = log[(l + x)/(l -
the
for
1,
x)]
converges only for
a:
in
(
— 1,
1).
*
*10.
Suppose that
£
00
a n converges.
We know that
the series f(x)
= Y
n =o
a nx n
n =o
must converge uniformly on [-a, a] for 0 < a < 1, but it may not
converge uniformly on [- 1, 1]; in fact, it may not even converge at the
point -1 (for example, if f(x) = log(l
x)). However, a beautiful
+
theorem of Abel shows that the
series does
converge uniformly on
[0, 1].
00
Consequently, /
is
continuous on
[0, 1]
and, in particular,
^
an
=
Uniform Convergence and Power
Km 7
<
e,
n
a nx
.
then
Prove Abel's Theorem by noticing that if \a m
\a
mx
m
+
•
'
+ 0n*
n
<
|
+
Lemma
by Abel's
e,
431
Series
•
•
•
+ an
\
(Problem
21-20).
11.
A sequence
{a n
}
is
called
Abel summable if
hm £
exists;
Problem
necessarily Abel summable. Find
which is not summable. Hint:
but
a sequence which is Abel summable,
find one which does not conyou
until
Look over the list of Taylor series
is continuous at 1.
represents
it
function
the
verge at 1, even though
10 shows that a
12.
summable sequence
is
Show
3 we assumed only that {/» converges pointwise to/.
converges
actually
that
ensure
hypotheses
n
{/
remaining
that the
uniformly to /.
necessarily conis a sequence of bounded (not
(a) Suppose that |/n
to / on [a 9 b].
uniformly
tinuous) functions on [a, b] which converge
In
Theorem
}
}
13.
J
Prove that / is bounded on [a, b\
converge
Find a sequence of continuous functions on [a, b] which
b].
[a,
on
function
unbounded
pointwise to an
pointwise
Suppose that/is differentiate. Prove that the function/ is the
know
already
we
(Since
functions.
continuous
of
sequence
limit of a
(b)
*14.
15.
16.
example
examples of discontinuous derivatives, this provides another
continuous.)
not
where the pointwise limit of continuous functions is
to the
Find a sequence of integrable functions {/„} which converges
irrathe
on
and
rationals
0
the
(nonintegrable) function / that is 1 on
points.
few
at
a
except
tionals. Hint: Each fn will be 0
the integral;
This problem outlines a completely different approach to
learned
integrals
about
facts
any
use
to
unfair
consequently, it is
previously.
(a)
Let
s
some
be a step function on
partition
where
Si
is
{* 0>
[a,
...,'»}
b\ so that
of
constant on
Define
[a, b].
the (constant) value of
s is
on
s
s as
(/ t-_i, ft)
-
£
f*
for
i=i
(U-i, k).
Show
that this
definition does not depend on the partition {t 0
b] if it is the uniform
(b) A function / is called a regulated function on [a,
that in this
limit of a sequence of step functions {* n on [a, b\ Show
m,
n
for
that
> we have
such
case there is, for every e > 0, some
,
}
\s n
(d)
(e)
(x)
-
Sm(x)\
<
S for all
x in
[a,
.
.
N
N
(c)
.
b\
Show that the sequence of numbers f* s n
{
}
will
be a Gauchy sequence.
on [a, b]
Suppose that {*„} is another sequence of step functions
there is
every
0
£
>
for
that
Show
to
uniformly
/.
which converges
[a, b\.
in
x
an N such that for n > N we have \sn (x) - t n (x)\ < £ for
&
we can
Conclude that lim f s n - lim /* t n This means that
.
h
define
*(f)
[*
f
to be lim
f
sn for
any sequence of
The only remaining
converging uniformly to
/.
functions are regulated?
Here
is
step functions
question
is:
{j- n
}
Which
a partial answer.
Prove that a continuous function is regulated. Hint: To find a step
s on [a, b] with \f(x) — s(x)\ < 6 for all x in [a, b\ consider
all y for which there is such a step function on [a, y].
function
Find a sequence {fn approaching / uniformly on [0, 1] for which
lim (length of fn on [0, 1]) ^ length of /on [0, 1]. (Length is defined in
\
n—*
oo
Problem 15-32, but the simplest example
length of whose graphs will be obvious.)
will involve functions the
COMPLEX NUMBERS
CHAPTER
With the exception
of the last few paragraphs of the previous chapter, this
book has presented unremitting propaganda for the real numbers. Neverthenot every polynomial
less, the real numbers do have a great deficiency
is the fact that no
example
function has a root. The simplest and most notable
2
that long ago
severe
is
so
deficiency
=0.
This
number x can satisfy x + 1
—
mathematicians felt the need to "invent" a number i with the property that
was quite mysterious:
{2
i = o. For a long time the status of the "number" i
2
1 = 0, it is nonsensical to say "let
since there is no number x satisfying x
2
Nevertheless, admission of the
1 = 0."
the number satisfying i
i be
+
+
+
simplify greatly
i to the family of numbers seemed to
bi
algebraic computations, especially when "complex numbers" a
computation
arithmetical
(for a and b in R) were allowed, and all the laws of
enumerated in Chapter 1 were assumed to be valid. For example, every
"imaginary" number
+
many
quadratic equation
+
ax 2
can be solved formally
x
= -b
+c=
bx
0
^
{a
0)
to give
+ Vb 2
Aac
2
-b - Vb
"
=
x
or
2a
2a
Ifb 2
—
Aac
>
0,
Aac
these formulas give correct solutions;
are allowed the formulas seem
make
to
sense in
when complex numbers
all cases.
For example, the
equation
+ x+\ =
x2
0
has no real root, since
x2
But the formula
+x+
1
=
(x
for the roots of
x
+
+f
2
i)
>
0,
for all x.
a quadratic equation suggest the "solutions"
=
and
x
=
;
2
2
if
we understand
these
to
mean
,
V3.
i
2
is
433
•
(-1)
-
V?>
V-l
•
= VI
i,
then
numbers would be
1
It
V3
2
.
and
—
1
V3
—
2
2
.
i.
not hard to check that these, as yet purely formal, numbers do indeed
434
Infinite Sequences
and
Infinite Series
equation
satisfy the
+*+
*2
It is
1
=
0.
even possible to "solve" quadratic equations whose coefficients are themcomplex numbers. For example, the equation
selves
+x+ + -
x2
1
0
i
ought to have the solutions
=
*
~
-1 ± Vl -
—
where the symbol
square
—3 —
is
4(1
+
0 = -1 ±
V-3 -
4i
2
'
means a complex number a
4i
+
fit
whose
In order to have
4i.
a2 - P2
+ #) =
2
(a
+
= -3 -
2aPi
4i
we need
a 2 - 0 2 = -3,
2a0 = -4.
These two equations can
two possible solutions:
be solved for real a and
easily
a =
1
,
0.-2
Thus the two "square
no reasonable way
and which
real
x
>
to decide
—
0, in
roots" of
—
4z ;
which case
and
in fact, there are
a — —
/3
—3 —
/3;
=
4i are
2.
—
1
2i
and —1
+
which one of these should be called
the conventional usage of
V*
2 i. There
V—3 —
is
4*',
V# makes sense only for
denotes the (real) nonnegative root. For this
reason, the solution
X
must be understood
^
With
as
this
—
— _+ r
as
1
,
V-3 -
-1 ±
where
we
r is
4i
2
an abbreviation
understanding
you can
—
=
for:
one of the square roots of
—3 —
4i.
arrive at the solutions
+ l-2i =
*
- -l
*
= — 1 — 21
easily check, these
+
2»
=
1
numbers do provide formal
equation
x2
.
+x+ + =
1
i
0.
solutions for the
Complex Numbers
For cubic equations complex numbers are equally
useful.
435
Every cubic
equation
ax z
with real coefficients
divide ax z
+
bx 2
+
+
bx 2
a, b, c,
cx
+
+ cx + d =
and
d by x
d,
0
(a 5* 0)
we know, a real root a, and if we
obtain a second-degree polynomial
has, as
—a
we
whose roots are the other roots of ax z + bx 2 + cx + d = 0; the roots of this
second-degree polynomial may be complex numbers. Thus a cubic equation
will have either three real roots or one real root and 2 complex roots. The
existence of the real root is guaranteed by our theorem that every odd degree
equation has a real root, but it is not really necessary to appeal to this theorem
(which
is
of
no use
at all
if
the coefficients are complex); in the case of a cubic
we can, with sufficient cleverness, actually find a formula for all the
The following derivation is presented not only as an interesting illustra-
equation
roots.
tion of the ingenuity of early mathematicians, but as further evidence for the
importance of complex numbers (whatever they
To
most general cubic equation,
only equations of the form
solve the
x*
+
bx 2
it
+ cx + d -
may
be).
obviously suffices to consider
0.
2
even possible to eliminate the term involving x by a
let
we
If
forward manipulation.
It is
,
x
=
y
—
fairly straight-
b
-»
then
so
0
=
xz
+
bx 2
+ cx + d
The right-hand side now contains no term with^ 2 If we can solve the equation
for y we can find x; this shows that it suffices to consider in the first place only
.
equations of the form
xz
+ px +
q
-
0.
z
In the special case p = 0 we obtain the equation x = — q. We shall see later
root,
in fact it has three, so
on that every complex number does have a cube
on the other hand,
case
solutions.
The
three
0,
that this equation has
p ^
436
Infinite Sequences
and
Infinite Series
requires quite an ingenious step. Let
P_
3w
Then
0
=
+px +
*3
w°
3w 2p
—
+p(w
= (w -
q
+—
3wp 2
2>w
p
9w
+
27q(w*)
q
2
z
—
bw
VF
s
27 w*
2
+
p
~
3w
V q
This equation can be written
27(> 3 ) 2
which is a quadratic equation
Thus
in
w
z
-p =
3
(!!).
= -27? ± V(27)V
2
2
Remember
•
^
4
•
Hp*
+ 27
* 4
+
r,
where
r is
a square root of
4
+
•
27
can therefore write
^9
this
+
27
that this really means:
2
We
~
0,
equation means that
square root of q 2 /4
are substituted into
+p
3
if is
'A
97'
27
7
some cube root of
/27. This
— q/2 +
r is
some
when
these
where
r,
but
allows six possibilities for w,
(*), yielding
3/"
= ^Zl ±
*
2
Jt^I
M
p
3
27
^
it
_„
2
" V
27
4
turns out that only 3 different values for x will be obtained
surprising feature of this solution arises
all
when we
!
An
even more
consider a cubic equation
of whose roots are real; the formula derived above
may
still
involve
com-
plex numbers in an essential way. For example, the roots of
xz
are 4,
-2
+
-
15*
-
VJ, and -2 - V3. On
4
=
0
the other hand, the formula derived
Complex Numbers
above (with p
= — 15,
x
=
= — 4)
q
one solution
gives as
-15
+ V4 -
</l
125
3
+
= ^2
^2 + V4 -
'
125
15
+
11/
437
3
+
</i
•
11/
Now,
(2
so
+
+3 2 +3 2 +
8 + 122 — 6 —
2 + Hi,
+ 11 2 + /.Thus, for one solution of the equation
=
=
=
3
/)
one of the cube roots of 2
23
2
•
•
/
/
•
3
2
/
2
is
£
we obtain
x
=
+ +
2
i
+
6
3/
=
+ +6
+
2
=
/
t
3i
3t
45/
+
36
6
3*
-
90
+ +
2
=
-
6
15
9
4(0.
found if the other cube roots of 2 + 11/ are
The
known. The fact that even one of these real roots is obtained from an expression which depends on complex numbers is impressive enough to suggest that
the use of complex numbers cannot be entirely nonsense. As a matter of fact,
the formulas for the solutions of the quadratic and cubic equations can be
other roots can also be
interpreted entirely in terms of real numbers.
Suppose we agree,
writing the real
for the
number
+ bi) +
(a + bi)
(a
to write all
i
2
as 0
that
+
+
d)h
{ad
+
c)
-
(ac
bd)
(b
+
+
If.
as a
The
+
bi,
laws
bc)i.
like
(1
+
2f)
•
(3
+
= 1+7/
If)
for the two equations
be regarded as simply an abbreviation
=
3 - 2
1-1+2.3 =
1
The
/
= — 1 show
+
{a
complex numbers
number
and the
Oi
relation
+ di) =
+ di) -
(c
(c
'
may
+
a as a
and the
of ordinary arithmetic
Thus, an equation
moment,
•
•
1
2
solution of the quadratic equation ax
1,
7.
+
bx
+
c
=
could be paraphrased as follows:
u2
_
v2
UV
=
0,
If
'
(i.e., if
=
(ii
b
2
+
_
vi)
4aCj
2
«
b
2
-
4*;),
0 with real coefficients
438
Infinite
Sequences and Infinite Series
+
-b
«[(
•H
It is
+c=
0,
2a
then
writing
u
+
-b
u
not very hard to check this assertion about real numbers without
down
a single
C£
z',"
but the complications of the statement
itself
should
convince you that equations about complex numbers are worthwhile as
abbreviations for pairs of equations about real numbers. (If you are
convinced, try paraphrasing the solution of the cubic equation.) If
intend to use complex numbers consistently, however,
some reasonable
sary to present
One
it is
still
we
not
really
going to be neces-
definition.
been implicit in this whole discussion. All mathematical
complex number a + bi are determined completely by the real
numbers a and b; any mathematical object with this same property may
reasonably be used to define a complex number. The obvious candidate is the
ordered pair (a, b) of real numbers; we shall accordingly define a complex
number to be a pair of real numbers, and likewise define what addition and
multiplication of complex numbers is to mean.
possibility has
properties of a
DEFINITION
complex number is an ordered pair of real numbers; if z = (a, b) is a
complex number, then a is called the real part of z, and b is called the
imaginary part of z. The set of all complex numbers is denoted by C. If
(a, b) and (c, d) are two complex numbers we define
A
b)
+
(a, b)
(The
+
and
while the
+
-
(c,
d)
(c,
d)
=
—
(a
+
(a
'
cy b
—
c
+
b
d)
d,
•
a
+
d
•
b
•
c).
appearing on the left side are new symbols being defined,
appearing on the right side are the familiar addition and
*
and
multiplication for real numbers.)
When complex numbers were first introduced, it was understood that real
numbers were, in particular, complex numbers; if our definition is taken
seriously this
all.
is
(a,
0)
(a,
this
+
0)
•
is
—
a real number is not a pair of real numbers, after
only a minor annoyance, however. Notice that
not true
This difficulty
(*,
0)
(b, 0)
=
=
{a
+
(a
b
b,
-
0
0
+
0)
0,
a
=
•
0
(a
+
+
shows that the complex numbers of the form
same with respect
to addition
and multiplication
b } 0),
0
b)
{a
•
b,
0);
0) behave precisely the
complex numbers as real
(a,
of
=
Complex Numbers
numbers do with
their
own
addition and multiplication. For this reason
will adopt the convention that
a
+
bi
(a, 0) will
be denoted simply by
now
notation for complex numbers can
definition
=
i
(a, b)
You may feel
if
we
familiar
one more
it is
(a,
0)
(a,
0)
+
a
+
+
= -1
(the last equality sign
(0, b)
(*,
0)
•
(0, 1)
bi.
was merely an elaborate device for defining
form a + bi" That is essentially
a firmly established prejudice of modern mathematics that new
as "expressions of the
must be defined
is
=
=
0)
that our definition
complex numbers
theless, it
be recovered
The
(0, 1).
Notice that i 2 = (0, 1) (0, 1) = (-1,
depends on our convention). Moreover
objects
a.
made.
is
DEFINITION
correct;
439
as
something
specific,
not as "expressions." Neverwere sincerely worried
interesting to note that mathematicians
about using complex numbers until the modern definition was proposed.
Moreover, the precise definition emphasizes one important point. Our aim in
introducing complex numbers was to avoid the necessity of paraphrasing
statements about complex numbers in terms of their real and imaginary parts.
This means that we wish to work with complex numbers in the same way that
we worked with rational or real numbers. For example, the solution of the
= w — p/3w, so we want to know that
w was found by solving a quadratic equation,
cubic equation required writing x
1
/w makes
sense.
Moreover,
2
which requires numerous other algebraic manipulations. In short, we are
likely to use, at some time or other, any manipulations performed on real
numbers. We certainly do not want to stop each time and justify every step.
Fortunately this is not necessary. Since all algebraic manipulations performed
on real numbers can be justified by the properties listed in Chapter 1, it is only
necessary to check that these properties are also true for complex numbers.
In most cases this
is
quite easy,
and
these facts will not be listed as formal
theorems. For example, the proof of PI,
[(*, b)
+
(c,
d)}
+
=
(e,f)
(a, b)
+
[(c,
d)
+
(ej)]
requires only the application of the definition of addition for complex
bers.
The
left
side
becomes
([a
and the
right side
+
c]
+
e,
[b
+ d]+f)
3
becomes
(a
+
[c
+ elb +
[d+f]);
num-
is true for real numbers. It is a good idea to
check P2-P7 and P9. Notice that the complex numbers playing the role of 0
and 1 in P2 and P6 are (0, 0) and (1, 0), respectively. It is not hard to figure
these two are equal because PI
out what
a
— (a,
b)
but the multiplicative inverse for
is,
^
little trickier: if {a, b)
+
then a 2
(0, 0),
b
^
2
This fact could have been guessed in two ways.
it is
The
=
b)-(x,y)
(*,
(a, b)
0
and
To
find
(x,
required in P8
is
y) with
(1,0)
only necessary to solve the equations
=
solutions are x
reason that
if
\/{a
+
a
Once
a/{a 2
bi)
+
ax
—
bx
+ ay
2
b ),
=
—
by
1,
0.
= -b/(a
y
means anything, then
1
1
+
+
a
bi
—
—
a
a
bi
2
+
a
a
bi
2
It is also possible to
).
should be true that
it
bi __
b
2
—
bi
+
b
2
the existence of inverses has actually been proved (after guessing the
by some method), it follows that this manipulation is really valid; it is
the easiest one to remember when the inverse of a complex number is actually
being sought it was precisely this trick which we used to evaluate
inverse
—
15
6
+
3*
=
~
_
~
15
6
#
6
+
90
3i
-
'
6
-
3i
3i
45i
"
36
+
9
Unlike P1-P9, the rules P10-P12 do not have analogues: it is easy to prove
that there is no set P of complex numbers such that P10-P12 are satisfied for all
complex numbers. In fact, if there were, then P would have to contain 1 (since
1
=
l
2
)
and
also
-1
(since
-1 =
2
z"
),
and
tHis
would contradict P10. The
absence of P10-P12 will not have disastrous consequences, but it does mean
that we cannot define z < w for complex z and w. Also, you may remember
1^0. Forthat for the real numbers, P10-P12 were used to prove that 1
to this
reduced
be
can
tunately, the corresponding fact for complex numbers
+
one: clearly
(1, 0)
+
(1, 0)
^
(0, 0).
Although we will usually write complex numbers in the form a + bi, it is
worth remembering that the set of all complex numbers C is just the collection
of all pairs of real numbers. Long ago this collection was identified with the
plane, and for this reason the plane is often called the "complex plane." The
horizontal axis, which consists of
all
points
(a,
0) for a in
R,
is
often called the
Complex Numbers
and the
real axis,
vertical axis
called the imaginary axis.
is
Two
441
important
definitions are also related to this geometric picture.
+
If z
=
x
of z
is
defined as
a complex
iy is
number
=
z
x
-
and the absolute value or modulus
+y
2
>
0, so
\z\
iy,
of z
-
x
that
V# + 7
2
2
is
defined unambiguously;
+y
it
2
.)
•
Geometrically, z
FIGURE
defined as
is
denotes the nonnegative real square root of x 2
z
then the conjugate z
real),
= Vx 2 +y\
\z\
(Notice that x 2
and y
(with x
1
distance from z to
simply the reflection of z in the real
is
0) (Figure
(0,
1).
while
axis,
\z\ is
the
Notice that the absolute value notation for
complex numbers is consistent with that for real numbers. The distance
between two complex numbers z and w can be defined quite easily as \z — w\.
The following theorem lists all the important properties of conjugates and
absolute values.
THEOREM
1
Let z and
w
(1)
z
(2)
z
=
=
(3)
z
be complex numbers. Then
z.
z
and only
if
number
if
z
real
is
(i.e., is
of the form a
+
0/,
for
some
real
a).
+w=
z
+ w.
-(z).
(4)
w = z w.
= (z)~\ if z 9*0.
= z-z.
\z\
\z-w\ = |*| H|* + H < 1*1 + M.
(5)
z
(6)
F
-
•
1
2
(7)
(8)
(9)
PROOF
•
Assertions (1) and (2) are obvious. Equations (3) and (5) may be checked
by straightforward calculations and (4) and (6) may then be proved by a trick:
0
1
Equations
The only
(7)
=
=
=
J =
and
difficult
0
(8)
z
z
+
(- *) = z_+ -z,
=
•
may
2
•
z"
1
so
so
,
-z =
=
-(*),
_1
(*)
.
proved by a straightforward calculation.
is (9). This inequality has, in fact,
but the proof will be repeated here, using
also be
part of the theorem
already occurred (Problem 4-8),
slightly different terminology.
It
is
clear that equality holds in (9)
that (9)
is
true
if
z
= \w
for
any
real
if
z
=
0'or
number X
w =
0. It is also
easy to see
(consider separately the cases
442
Infinite Sequences
and
Infinite Series
X
>
and X < 0). Suppose, on the other hand, that z ^ Xw
X, and that w ^ 0. Then, for all real numbers^ X,
0
for
any
real
number
0
(*)
<
\z
-
=
=
Xw\ 2
—
(z
(z
=
+
= XV|
+
zw
wz
+
Notice that wz
+
2
—
Xw)
Xw)
— X(wz + -zw)
2 X(wz + zw).
\z\
Xw)
Xw^
X 2 ww
•
(z
•
(z
real, since
is
+
zw = wz
+
zw = wz
zw = wz
+
zw.
Thus the right side of (*) is a quadratic equation in X with real coefficients
and no real solutions; its discriminant must therefore be negative. Thus
(wz
it
follows, since
wz
+
+
zw and
zw) 2
\w\
•
\z\
+
From
—
4\w\
2
\z\
-
2
<
0;
are real numbers,
<
zw)
2\w\
•
and
\w\
*
|z|
>
0,
that
\z\.
this inequality it follows that
\z
+
w\ 2
=
=
<
=
+ w) (z + w)
+ \w\ + (wf +
+ |w| + 2\w\
(kl + l^l)
(z
\z\
\z\
-
2
2
2
2
'
zw)
\z\
2
?
which implies that
\z
z
+
IV
proof of
The
then z
a
a
+
c
this
geometric fact
is left
-
w =
|w|. |
unnecessary
is
more
involved. If z
=
0 or
w =
0 (a one-line computational proof can be given, but even this
—the assertion has already been shown
restrict
our attention to
putting every nonzero complex
2
+
to you).
interpretation of multiplication
we may
FIGURE
\z\
The operations of addition and multiplication of complex numbers both
have important geometric interpretations. The picture for addition is very
simple (Figure 2). Two complex numbers z = (a, b) and w = (c, d) determine
a parallelogram having for two of its sides the line segment from (0, 0) to z,
and the line segment from (0, 0) to w; the vertex opposite (0, 0) is z + w
(a
c
+ w\<
For any complex number z
in this expression,
\z\
is
a positive real number, while
kl
is
from P1-P9), so
nonzero complex numbers. We begin by
to follow
number into a special form.
0 we can write
—
0,
=
M=
kl
1
'
443
Complex Numbers
so that z/\z\
—
ber a
x
is
+
with
some number
for
=
1
=
\a\
=
a
of 8 radians
number
a complex
iy
some
but 0
is
>
r
2
—
.
+
cos 6
i
sin 6
Thus every nonzero complex number z can be written
6.
=
if
0O
0.
i
sin 0)
The number
r is
unique
(it
equals
\z\),
+
then the others are do
2kw for k in
called an argument of z. Figure 3 shows z in
one
is
+
r(cos 6
number
0 and some
not unique;
+y
(cos 0, sin 0)
z
for
Now any complex num1
can be written in the form
of absolute value
x2
possibility,
any one of these numbers is
r and 6. (To find an argument
terms of
—
8 for z
x
+
iy
we may
note that the
equation
x
means
+
iy
x
y
If x ?± 0
=
6
?r/2
Now
z
~
=
\z\
+
i
sin 0)
sin
0.
we can therefore take 6 = arctan y/x, and
when y > 0 and 6 = 3?r/2 when y < 0.)
the product of two nonzero complex
z
-
(cos 6
\z\
|z|cos 0,
==
r(cos 0
w —
z
=
=
that
w =
=
=
+
rs(cos 6
i
rj[(cos 6 cos
+
™[cos (6
s(cos
sin 6) (cos
0 —
0)
+
<j>
<t>
+
+
+
sin (6
+
x
=
0
+
0,
we can
take
numbers
i
sm
i
sin 0),
6),
sin 0)
i
sin 6 sin 0)
i
if
+
z'( s
m
6 cos
cos 6 sin 0)]
0)].
is the product of the absolute values of
of any argument for each of the factors will be an
Thus, the absolute value of a product
the factors, while the
argument
sum
For a nonzero complex number
for the product.
z
=
r(cos 6
+
i
sin 6)
now an easy matter to prove by induction the following very important
formula (sometimes known as De Moivre's Theorem):
it is
z
n
=
n
|^|
(cos nd
+
i
sin nd), for
any argument
This formula describes z n so explicitly that
z
THEOREM
2
n
=
it
is
z.
easy to decide just
when
w:
Every nonzero complex number has exactly n complex nth roots. More preany complex number w ^ 0, and any natural number n, there are
n
precisely n different complex numbers z satisfying z = w.
cisely, fcr
proof
6 of
Let
w —
j-(cos
0
+
2
sin 0)
.
1
444
Infinite
Sequences and Infinite Series
for s
=
and some number
\w\
—
2-
satisfies
— w
zn
and only
ii
r
which happens
n
and only
if
From
the
first
equation
=
sin nd)
2
sin 0)
i
,r(cos
0
+
i
sin 0),
if
+
cos nd
+
r(cos 6
number
a complex
if
+
(cos nd
Then
0.
=
sin rc0
i
cos
+
0
2
sin 0.
follows that
it
= Vs,
r
where y/s denotes the positive real nth root of s. From the second equation
follows that for some integer k we have
Conversely,
z
=
r(cos 0
roots of w,
Now
we choose
if
+
and 6
r
it is
any integer
some integer
w.
nq
+
=
and some integer
q,
+
i
sin 0
k'
=
fc
\J
-
=
z
6k for k
it is
=
0,
then the
number
v'j (cos
+
A*,
between 0 and
= w
sin 0*)
z
k!
cos Ov
n
This shows that every z satisfying z
Moreover,
some
Bk for
To
can be written
Jc
cos 6k
Vr
=
=
determine the number of rcth
therefore only necessary to determine which such z are distinct.
sin 0) will satisfy z
i
h
for
n
it
+
.
.
,
n
—
1
In the course of proving
differ
by
Theorem
less
2,
—
1.
—
1
Then
sin 0^.
*
can be written
£
=
0,
easy to see that these numbers are
.
n
.
.
,
w
all different,
than 2w.
we have
.
since
any two
|
actually developed a
method
for finding the nth roots of a complex number. For example, to find the cube
roots of
(Figure 4) note that
i
cube roots of
1
i
cos
•
T
+
/ir
cos
'
T
'
L
and that
1
i
sin
-
7r/2
is
an argument
for
>
6J
2tt\1
2x\ .../*
+ t) + sin + t)J
Vg
.
.
1
V6
L
1
—
6
L
1
=
|i|
are therefore
,
cos
\6
4tt\
7/
,
.
.
4ir\~|
/tt
,
sin
\6
17 J
5tt
=
cos
T
=
cos
y+
3tt
+
.
.
1
.
'
.
sm
.
sin
5tt
T
T
3tt
i.
The
.
445
Complex Numbers
Since
7T
-
cos
.
sin
'
2
6
—
—
5tt
.
sm
'
2
6
—
—
3tt
.
sin
i
?
2
= -1,
are
VI +
- VI +
i
we cannot
—
i
,
,
2
In general,
1
2
z
2
z
the cube roots of
>
_
~
6
3tt
cos
= —
2
6
57T
cos
1
-
2
expect to obtain such simple results. For exa mpl e, to
V
125 and
cube roots of 2 + 1 li, note that |2 + 1 li\ = V2 2 + ll 2 =
that arctan Af is an argument for 2 + 1 h*. One of the cube roots of 2 + 1 li is
find the
therefore
•\VToe
T
V125 ^cos
(—3—J +
Previously
1
2
+
i|
=
can write
AM
/arctan
we noted
V2 +
2
this
l
.
.
/arctan
(—3—jj
=
VI
[cos
+
is
.
that 2
= VI, and
2
AMI
sm
z
(««|L*) +
,•
sin
(5=^)}
cube root of 2
also a
\
since arctan
is
+
Hi. Since
an argument of 2
+
i,
we
cube root as
2
+
=
i
Vs
(cos arctan
^
+
z
sin arctan ^)
These two cube roots are actually the same number, because
arctan
1
V" =
arctan -
2
3
(you can check this by using the formula in Problem 15-8), but this
is
hardly
the sort of thing one might notice!
The
fact that every
complex number has an nth root for all n is just a special
The number i was originally introduced in
case of a very important theorem.
order to provide a solution for the equation x 2
+
1
=0. The
Fundamental
Theorem of Algebra states the remarkable fact that this one addition automatically provides solutions for all other polynomial equations: every equation
zn
+
an-iz
71
"
has a complex root!
In the next chapter
1
+
we
•
•
•
+
shall give
mental Theorem of Algebra; the
a0
=
0
a 0,
.
.
.
,
a n -i in
C
an almost complete proof of the Fundagap left in the text can be filled in as an
slight
446
Infinite
Sequences and Infinite Series
The proof
exercise (Problem 25-5).
come up
concepts which
on
of the theorem will rely
quite naturally in a
more thorough
several
new
investigation of
complex numbers.
PROBLEMS
1.
2.
Find the absolute value and argument of each of the following.
+ AL
+ Ai)'\
(i)
3
(ii)
(3
(hi)
(1
(iv)
a/3
(v)
[3
+Q
5
.
+
+
Ai.
4z[.
Solve the following equations.
+
+
+
+ = 0.
+ = 0.
2ix -1=0.
x
ix " (1 + l)y = 3
= 4.
\(2 + i> +
(i)
x2
(ii)
x*
2
(hi)
(iv)
K
J
ix
1
x2
1
'
(
*>
(v)
3.
—
= -z.
= z-\
|z — a\ =
(i)
z
(ii)
t
(iv)
|z
(v)
5.
x2
x
—
Describe the set of
(hi)
4.
—
#3
—
|z|
<
|z
—
2
complex numbers z such that
all
—
—
0.
6|.
=
a)
+
|*
1
—
real part of z.
Prove that \z\ =
imaginary part is
Prove that
+
\z
6|
c.
and that the
(z
—
+
w\ 2
real part of z
(z
is
+
z)/2, while the
z)/2i.
-
\z
=
w| 2
2(|z|
H
+
2
and
2
),
interpret
this
statement geometrically.
6.
What is
the pictorial relation between z
line goes into the real axis
7.
(a)
Prove that
if
a 0,
.
.
.
Vi
and
•
z
V-f ?Hint:
under multiplication by V —
,
a n -i are real
and
a
+
+
fo"
/
Which
?
(for a
and
£ real)
- hi
n _iz
equation. (Thus the nonreal roots of such an equation always occur in pairs, and the number of such roots is even.)
2
n~ l
n
a 0 is divisible by z a n -iz
(b) Conclude that z
2
2
b ) (whose coefficients are real).
(a
2az
satisfies
n
the equation z
+
n_1
<z
+
•
•
•
a0
=
0,
then a
also satisfies this
+
+
+
*8.
(a)
Let
and
a
+
c
b
b
•
•
*
+
+
be an integer which
are integers
y/c, as a
we
is
not the square of another integer. If a
define the conjugate of a
— bVc Show
+
b
that the conjugate
Vc, denoted by
is
well defined
447
Complex Numbers
by showing that a number can be written
and b, in only one way.
(b)
+
bVc.
for integers a
Show that for all a and (3 of the form a + b Ve. we have a = a,
a = a if and only if a is an integer, o: + /3 = a + /3, —ol— —a,
-1
= a
if a
0.
and a -1 = (a)
a
-
•
(c)
a
Prove that
~j3,
if ao,
.
.
.
+
n
the equation z
and z =
a n _i are integers
,
a n _\z
n~ x
+
•
•
•
+
a0
=
a
+
then z
0,
satisfies
£
=
—
a
y/c
b
also satisfies this equation.
9.
the 4th roots of i; express the one having smallest argument
form that does not involve any trigonometric functions.
Prove that if co is an nth root of 1, then so is u k
A number co is called a primitive nth root of 1 if 1, a>, co 2
Find
all
in a
*10.
(a)
(b)
.
{
co
n_1
of
is
}
1
the set of
all
are there for n
nth roots of
=
3, 4, 5,
1.
How many
.
,
.
.
,
primitive nth roots
9?
n-l
(c)
Let
co
be an nth root of
1,
with
5* 1.
co
Prove that
V
k
*11.
(a)
.
.
.
0.
,
clear
is
if
reduce the general case to
Show
line
+^
•
•
•
also easy: the assertion
will
*12.
=
lie on one side of some straight line
Prove that if zi,
^ 0. Hint: This is obvious from
through 0, then zi
the geometric interpretation of addition, but an analytic proof is
+
(b)
co*
=0
further that Zi
through
Prove that
if \zi\
0, so
—
\z 2
-1
that
\
=
the line
~~
,
.
.
.
zC +
l
\z 3
\
,
the real axis,
is
zk
•
and z
1
l
all lie
on one
+z~
+z +z
•
•
l
k
2
z
are the vertices of an equilateral triangle. Hint:
that zi
is
real,
and
this
and a
trick
this one.
can be done with no
side of a straight
9* 0.
=
0,
then z h z 2 and z z
It will
,
help to assume
loss of generality.
Why?
CHAPTER
W
25
You
COMPLEX FUNCTIONS
probably not be surprised
will
deeper investigation of
to learn that a
complex numbers depends on the notion of functions. Until now a function
was (intuitively) a rule which assigned real numbers to certain other real
numbers. But there is no reason why this concept should not be extended; we
might just as well consider a rule which assigns complex numbers to certain
other complex numbers. A rigorous definition presents no problems (we will
not even accord
it
the
honors of a formal definition) a function
full
:
is
a collec-
complex numbers which does not contain two distinct pairs
with the same first element. Since we consider real numbers to be certain complex numbers, the old definition is really a special case of the new one. Nevertion of pairs of
we
theless,
sometimes resort to special terminology in order to clarify
is being considered. A function / is called
will
the context in which a function
real-valued iff(z) is a real number for all z in the domain of /, and complexto emphasize that it is not necessarily real-valued. Similarly, we will
valued
is defined on [a subset of] R in those
where the domain of / is [a subset of] R; in other cases we sometimes
mention that / is defined on [a subset of] C to emphasize that f{z) is defined
for complex z as well as real z.
Among the multitude of functions defined on C, certain ones are particularly important. Foremost among these are the functions of the form
usually state explicitly that a function /
cases
=
/(*)
where a 0)
.
.
.
,
a n are
a nz
n
+
an^z"- 1
+
+
•
•
a0
,
complex numbers. These functions are
called, as in
the real case, polynomial functions; they include the function f(z)
=
number
a
familiar function
Two
is
the "absolute value function" f(z)
and Im
Re(#
Im(*
The "conjugate
function"
(the
is
^
^
for all z in C.
produce new
rea j
=
z
=
ReW - am
(z).
R may
be combined in
complex-valued functions defined on
the function
,f(x
448
\z\
defined by
Familiar real-valued functions defined on
to
=
numbers are Re (the
"imaginary part function"), defined by
+ iy) — x,
+ iy) = y,
f(z)
ways
z (the
functions of particular importance for complex
"real part function")
is
—
some complex
and functions of the form f(z)
("constant functions"). Another important generalization of a
"identity function")
a for
+
iy)
=
e v sin (x
— y) +
ix
z
cos y.
C— an
many
example
.
449
Complex Functions
The formula
always
for this particular function illustrates a
possible.
Any complex-valued
/
=
function
u
+
decomposition which
/ can be written
is
form
in the
iv
—
some real-valued functions u and v simply define u(z) as the real part of
This decomposition is often very useful,
f(z), and v(z) as the imaginary part.
be inconvenient to describe a polywould
it
example,
but not always; for
for
nomial function in
this
way.
other function will play an important role in this chapter. Recall that
an argument of a nonzero complex number z is a (real) number 6 such that
One
z
There are
infinitely
=
|*|
(cos 0
many arguments
+
i
sin 6).
but just one which
for z,
satisfies
unique argument d(z), then 6 is a (real-valued)
0 < 6 < 2tt. If
on {z in C: z ^ 0}
function")
"argument
(the
function
" Graphs" of complex- valued functions defined on C, since they lie in
4-dimensional space, are presumably not very useful for visualization. The
alternative picture of a function mentioned in Chapter 4 can be used instead:
we draw two copies of C, and arrows from z in one copy, to f(z) in the other
we
(Figure
call this
1).
'•/(*)
FIGURE
The most common
pictorial representation of a
1
complex-valued function
is
produced by labeling a point in the plane with the value /(z), instead of with z
Figure
(which can be estimated from the position of the point in the picture).
Certain features of
2 shows this sort of picture for several different functions.
For example, the
the function are illustrated very clearly by such a "graph."
around
circles
0, the funcabsolute value function is constant on concentric
respectively,
lines,
horizontal
and
tions Re and Im are constant on the vertical
2
and the function /(z) = z wraps the
of radius
r
circle of radius r twice
around the
circle
2
.
functions in
Despite the problems involved in visualizing complex-valued
properties preimportant
of
analogues
define
to
possible
still
is
it
general,
cases these
viously defined for real-valued functions on R, and in some
the
example,
For
case.
properties may be easier to visualize in the complex
notion of limit can be defined as follows:
lim f(z)
=
rmmber
d
/
>
means that
for every (real)
0 such that, for
all z, if
0
<
number
\z
-
a\
<
£
>
d,
0 there
then
\f(z)
is
a
-
(real)
/|
<
S.
FIGURE
2
451
Complex Functions
Although the definition reads precisely as before, the interpretation is slightly
different. Since \z — w\ is the distance between the complex numbers z and w
the equation lim f(z) = / means that the values of f(z) can be made to lie
9
z—>a
inside
any given
circle
sufficiently small circle
using
trie
FIGURE
around /, provided that z is restricted to lie inside a
around a. This assertion is particularly easy to visualize
"two copy" picture of a function (Figure
3).
3
Certain facts about limits can be proved exactly as in the real case. In
particular,
c
=
c,
lim z
=
a,
lim
Z—Kl
+ g(z)]
lim [f(z)
=
lim /(*)
+
lim g(z),
z—*a
2—* a
z—>a
lim f(z)
g(z)
'
lim
-\- =
z~+a
g(z)
=
lim f(z)
•
lim g{z)
1
lim g(Z)
if
>
^
lim g(z)
y
0.
z-+a
upon which these results are based
this inequality holds for complex
and
is
+
numbers as well as for real numbers. These facts already provide quite a few
limits, but many more can be obtained from the following theorem.
The
essential property of absolute values
the inequality
THEOREM
l
<
w\
\z
+
\z\
\w\,
Let/(» = u(z) + iv{z) for real- valued functions u and v, and
for real numbers a and 0. Then lim f{z) = I if and only if
let /
= a
+
z—>a
lim u{z)
=
a,
=
0.
z-+a
lim v(z)
2—>0
proof
Suppose
first
that lim f(z)
if
The second
0
<
=
|*
/.
-
If e
>
0,
<
«,
then
fl|
there
is
|/(*)
6
>
-
0 such that, for
l\
inequality can be written
|[«(z)
-a] +
i[v(z)
-jS]|
<
e,
<
e.
all
iff
452
Infinite
Sequences and Infinite Series
or
-
[«(*)
Since u{z)
— a and v(z) —
+
a} 2
-
[v(z)
0]
2
<
e
2
.
are both real numbers, their squares are positive;
ft
this inequality therefore implies that
-
[u(z)
a] 2
<
£
2
and
[v{z)
-
p]
and
\v{z)
-
0\
<
2
£
2
,
which implies that
-
\u(z)
Since this
is
true for
>
all £
<
a\
£
= a and
Now suppose
=
/3.
>
0,
there
a\
<
->
lim v(z)
z—* a
z—> a
that these two equations hold. If £
that, for all z, if 0
<
\z
|w(z)
£.
follows that
0, it
lim u(z)
<
-
a\
<
—
a\
<
is
a 5
>
0 such
then
<5,
£
-
and
—
|z/(z)
£
which implies that
|/(z)
-
/|
=
<
-
-a\
£
£
,
In order to apply
the limit lim z
=
a,
=
I.
+
+
i[v{z)
\i\
•
-
\v(z)
0]\
-
jS|
2
2
This proves that lim f(z)
a]
\u(z)
|
Theorem 1 fruitfully, notice
we can conclude that
that since
we already know
z—>a
lim Re(z)
=
Re(a),
=
Im(fl).
2—*a
lim Im(z)
z—>a
A
limit like
lim sin (Re (z))
=
sin (Re (a))
z—»a
follows easily, using continuity of sin.
Many
applications of these principles
prove such limits as the following:
lim z
—
a
=
\a\,
}
z—>a
lim
|,?|
z->a
e y sin
lim
(x-fi3/)-*a
Now
+
x
+
ix
3
cos ^
=
e
b
sin a
+
ia
3
cos
fci
complex functions, the
continuous at a if lim f(z) —
that the notion of limit has been extended to
notion of continuity can also be extended: /
is
Complex Functions
and /is continuous
f(a),
if /is
continuous at a for
previous work on limits shows that
f(z)
/(')
/(*)
/(*
+
zy)
all a in
domain
the
of/.
453
The
the following functions are continuous:
all
= a n z n + a n ^z n ~ +
=
= 14
= ^ sin x +
cos
l
•
•
•
+
fl 0
,
3
z'*
jy.
Examples of discontinuous functions are easy
to produce,
and certain ones
come up very naturally. One particularly frustrating example is the "argument function" 0, which is discontinuous at all nonnegative real numbers
(see the "graph" in Figure 2). By suitably redefining 6 it is possible to change
f
the discontinuities; for example (Figure 4),
ment
of z with 7r/2
<
0'{z)
nonnegative real number
tinuities will
a.
<
57r/2,
if
8 (z) denotes the unique argu-
6' is
then
But, no matter
discontinuous at ai for every
how
6
is
redefined,
some discon-
always occur.
•2A»
7T
2
•2^tt .2f»
7T
2
TT
2
7T
2
7T
-2i7T
.2|7T
2
•2I7T
7T
.2£tt
2
•2i7T
7T
2
•2|tt
7T
2
T
T
IT
IT
TT
TT
TT
TT
TT
27r
2tt
2tt
2tt
2tt
2t
2ir
2t
2w
3ir
2
2
3tt
2
3tt
2
3tt
2
3tt
2
37T
2
FIGURE
The
=
0'{z)
4
discontinuity of 0 has important bearing on the problem of defining a
a function /such that (/(*)) 2 = zforall^r. For
real numbers the function y/ had as domain only the nonnegative real numbers. If complex numbers are allowed, then every number has two square roots
"square-root function," that
is,
0, which has only one). Although this situation may seem better, it is
some ways worse; since the square roots of z are complex numbers, there is
no clear criterion for selecting one root to be f(z)> in preference to the other.
(except
in
One way
to define
/
We set /(0) =
the following.
is
and
0,
for z 9^ 0
we
set
/W =VH(cos^+*sinM).
Clearly (f(z)) 2
ous.
=
but the function /
z,
As a matter of
2
(/(*))
=
z for
without any
all z.
is
discontinuous, since 6
it
discontinu-
is
the argument, given below,
fuss,
is
impossible to find a continuous / such that
Although you may be willing to accept this assertion
fact,
rather tricky.
is
We will actually show that a continuous square root function / cannot even
be defined on the set of all z with \z\ = 1. Suppose there were such a function /. We must have /(l) = 1 or /(l) = — 1, and we can therefore assume
that /(I) = —1, since —/would also be a continuous square root function.
In this case, since / does not have the value 1 at 1 it never takes on the value 1
This means that if we define g by
,
=
g{z)
for
8(f(z))
|*|
=
1,
then g will be continuous, since 6 is continuous at all w with
w = 1, which is not f(z) for any z. But the equation
(f(z)Y
=
\w\
—
1
except for
z
implies that
8(z)
=
20(f(z))
=
2g(z),
which would imply that 0 is continuous at z for all z with \z\ = 1 Since this is
false, no such function /can exist. Similar arguments show that it is impossible
to define continuous "r/th-root functions" for any n > 2.
For continuous complex functions there are important analogues of certain
theorems which describe the behavior of real-valued functions on closed intervals. A natural analogue of the interval [a, b] is the set of all complex numbers
.
z
=
+
x
iy
<
with a
<
x
closed rectangle, and
b
and
c
<
denoted by
is
y
<
d (Figure
[a, b]
X
[c,
5).
This
set
is
d].
a continuous complex-valued function whose domain is [a, b]
seems reasonable, and is indeed true, that / is bounded on [a, b]
If /is
then
it
That
is,
there
is
some
real
|/(<r)
|
<
number
M
M such that
for all z in
[a, b]
called a
X
[c,
X
X
[c,
d],
[c,
d].
d].
/ has a maximum and a minimum value on
[a, b] X [c, d], since there is no notion of order for complex numbers. If / is a
real-valued function, however, then this assertion does make sense, and is true.
In particular, if /is any complex-valued continuous function on [a, b] X [c, d\
then /| is also continuous, so there is some z 0 in [a, b] X [c, d] such that
It
does not
make
sense to say that
|
\f(zo)\
a similar statement
that
"/ attains
The
its
is
<
\f(z)\
for all z in
[a, b]
X
[c,
d];
true with the inequality reversed. It
maximum and minimum modulus on
various facts listed in the previous paragraph
will
is
[a, b]
sometimes said
X
[c,
d]."
not be proved here,
Complex Functions
455
although proofs are outlined in Problem 5. Assuming these facts, however, we
can now give a proof of the Fundamental Theorem of Algebra, which is
really quite surprising, since we have not yet said much to distinguish polynomial functions from other continuous functions.
THEOREM 2 (THE fundamental
THEOREM OF ALGEBRA)
Let a 0
.
,
.
.
,
<z
n _i
Then
be any complex numbers.
there
+
PROOF
is
a complex
number
, such that
=
a0
0.
Let
f(z)
Then /
is
|
/|
to
=
l/to
+
I*"
I
^z n ~ +
l
+
+a
•
•
the function
is
=
an
defined by
|/|
tfn-i^" 1
0.
+
*
•
'
+
a 0 \.
based on the observation that a point z 0 with f(z 0 ) = 0 would
point for |/|. To prove the theorem we will first show
does indeed have a smallest value on the whole complex plane. The proof
proof
is
clearly be a
that
zn
continuous, and so
l/l
Our
=
minimum
be almost identical to the proof, in Chapter 7, that a polynomial function
of even degree (with real coefficients) has a smallest value on all of R; both
will
depend on the
proofs
We
fact that
begin by writing, for z
if \z\
^
is
large, then \f(z)\
is
large.
0,
+
?)'
so that
+
<20
Let
M = max(l,
Then
for all z with
\z\
2w|<2 n _i|,
> M, we have
\z
k
\
.
.
>
.
\z\
,
2rc|<2 0 |).
and
2«|a n _A;|
SO
<2n-l
+
+
<
'
+
<1,
~ 2
which implies that
+
>
1
-
+
This means that
|/to
>
I
for
|*|
>
Af.
zn
>
456
Infinite
Sequences and Infinite Series
In particular,
if \z\
>
M and also
\z\
>
V2\f(0)\, then
I/Ml > l/WI-
Now
{z:
\z\
[a, b]
let
X
[a, b]
d]
[c,
be a closed rectangle (Figure
< max(M,
X
[c,
d]
is
|/(* 0 )|
(1)
V2\f(0)\)}, and suppose that the
attained at z 0 so that
if |*|
Combining
its
<
X
for * in
FIGURE
> max(Af,
(1)
minimum
minimum
<
and
(2)
|/(0)|.
^2|/(0)|), then
we
[c,
|/|
on
see that |/(z 0 )|
d].
Thus
\f(z)\
>
|/(0)|
< |/W|
>
\f(z 0 )\.
for all z, so that
value on the whole complex plane at z 0
|/|
attains
=
0.
.
6
To complete the proof of the theorem we now show that f(z Q )
convenient to introduce the function g defined by
=
g(z)
Then g
of
,
It follows, in particular, that |/(^ 0 )|
(2)
which contains
6)
is
occurs at
fit
+
a polynomial function of degree
0.
We
want
to
show that g(0) =
It is
^o).
n,
whose minimum absolute value
0.
Suppose instead that g(0) = a ^ 0. If m is the smallest positive power of*
which occurs in the expression for g, we can write
g{z)
where 0
9* 0.
such that
=
<x
+
/3z
Now, according
m
to
+
cm+l z
m+l
+
Theorem 24-2
•
•
•
there
+
is
cn z
n
,
a complex
number 7
Complex Functions
Then,
setting
dk
=
Cky k
we have
,
+ (3y mzm + dm+1 zm+1 +
m
m+1
\a - az + dm+1 z
+
=
=
\g(yz)\
\a
•
= a (l - zm
+ ^S±!
+ zm
= a (l -
a
L
This expression, so torturously arrived
first
that
at, will
<
a
If
we
choose, from
among
enable us to reach a quick
we
will
have
1.
z for which this inequality holds, some z which
and positive^ then
zm
\~ *+••]!<
<
z
I
Consequently,
1
This
all
I
chosen small enough,
if \z\ is
—-2- z +
is real
+ dn z»\
~zm + zm \ d^-z +
1
I
contradiction. Notice
•
-|
*+'•])!
|
\a\-
•
•
•
+---)|
|
=
457
is
if
0
<
1
=
zm
-
we have
-zm + zm [~*+
*
the desired contradiction
:
=
1
-
zm
+
<
1
-
zm
+ zn
=
1.
for
such a
\g(yz)\
<
+
+
']|<|1
'
\z
number
•
•
-]|
n
z
we have
\a\,
contradicting the fact that ol\ is the minimum of \g\ on the whole plane.
Hence, the original assumption must be incorrect, andg(O) = 0. This implies,
I
finally, that
f(z 0 )
=
0.
|
Even taking into account our omission of the proofs for the basic facts about
continuous complex functions, this proof verified a deep fact with surprisingly
little work. It is only natural to hope that other interesting developments will
arise if we pursue further the analogues of properties of real functions. The
next obvious step is to define derivatives: a function /is differentiable at a if
r Jf(a
- +
hm
*->o
z)
z
^
-f(a)
.
exists,
458
Infinite
Sequences and Infinite Series
in
which case the
limit
denoted by f'(a).
is
=
=
f(a)
f(a)
It
0
if /(*)
1
if /(^)
is
=
=
easy to prove that
c,
z,
+
gY(fl) =/'(«)
(f
(/•*)'(«)
=/W«) + /(«)*'(«),
=f(g(a))-g'{a);
(/«•«)'(«)
the proofs of
all
same
these formulas are exactly the
as before. It follows, in
z n then f'(z) - nz n ~\ These formulas only prove
the differentiability of rational functions however. Many other obvious
candidates are not differentiate. Suppose, for example, that
particular, that
if
f(z)
=
+
/(*
If /
is
,
=
iy)
iy
(i.e.,
f(z)
=
z).
to be differentiable at 0, then the limit
** +
lim
exist.
»>
*
(x+t W )->o
must
-
x
+
-AO),
ita
(x+iy)->o
iy
x
+
iy
Notice however, that
ify
=
0,
then^
x
+
=
1,
iy
and
x
if
=
0,
= -1
then
therefore this limit cannot possibly exist, since the quotient has both the values
1
and
—
1
for x
+
iy
arbitrarily close to 0.
where other differentiable funcand exp, you will see
tions are to come
that there is no hope at all of generalizing these definitions to complex numbers. At the moment the outlook is bleak, but all our problems will soon be
In view of this example,
it is
not at
all
clear
from. If you recall the definitions of sin
solved.
PROBLEMS
1.
(a)
For any real number y, define a (x) = x + iy (so that a is a complexvalued function defined on R). Show that a is continuous. (This
follows immediately
=
that P(y)
(b)
x
+
from a theorem
iy is
in this chapter.)
Show
similarly
continuous,
Let / be a continuous function denned on C. For fixed y, let g(x) =
iy) Show that g is a continuous function (defined on R) Show
f(x
iy) is continuous. Hint: Use part (a).
similarly that h{y) = f(x
+
.
.
+
2.
(a)
a continuous real-valued function defined on a
closed rectangle [a, b] X [c, d]. Prove that if/ takes on the values f(z)
Suppose that /
is
459
Complex Functions
for z and w in [a, b] X [c, d], then / also takes all values
between f{z) and f(w). Hint: Consider g(t) = /(£? + (1 - 0«0 for *
and f(w)
in
*(b) If
[0, 1].
/
is
[c, d],
a continuous complex-valued function defined on [a, b]
the assertion in part (a) no longer makes any sense, since
X
we
cannot talk of complex numbers between f(z) and f(w). We might
conjecture that / takes on all values on the line segment between /0)
and f(w), but even this is false. Find an example which shows this.
(a)
Prove that
if
a 0,
.
.
.
any complex numbers, then there are
zn (not necessarily distinct) such that
a n _ x are
,
complex numbers z u
.
.
.
,
n
+ a n ^z n - +
zn
1
•
+
•
•
a0
=
[I
t = i
0"
*<)•
n~
(b)
+
n
x
a0
+
a n -i are real, then z + a n _iz
Prove that if a 0)
quadratic
and
a
z
factors
linear
of
can be written as a product
factors z 2
az + £ all of whose coefficients are real. (Use Problem
.
.
.
'
'
'
,
+
+
24-7.)
this problem we
Such a polynomial
In
^2
_|_
.
.
.
_|_
will consider
is
called a
only polynomials with real coefficients.
of squares if it can be written as
sum
h
^ n 2 for polynomials
with real coefficients.
Prove that if / is a sum of squares, then f(x) > 0 for all x,
(b) Prove that if / and g are sums of squares, then so is / g.
(c) Suppose that/(*) > 0 for all x. Show that / is a sum of squares. Hint:
k
First write f(x) = x g(x), where g(x) 7* 0 for all x. Then £ must be
(a)
•
>
even (why?), and g(x)
(a)
Let
A
0 for
all x.
be a set of complex numbers.
Now
use Problem 3(b).
A number z is called, as in the real
point of the set A if for every (real) £ > 0, there is a
point a in A with \z — a\ < £ but z ^ a. Prove the two-dimensional
version of the Bolzano- Weierstrass Theorem: If ^4 is an infinite subset
Hint:
of [a, 6] X [c, d], then ^4 has a limit point in [a, b] X |>,
First divide [a, b] X [c, d] ijh half by a vertical line as in Figure 7(a).
case, a limit
Since A is infinite, at least one half contains infinitely many points of
A. Divide this in half by a horizontal line, as in Figure 7(b). Continue
in this way, alternately dividing by vertical and horizontal lines.
(The two-dimensional bisection argument outlined
standard that the
the
c
in this hint
is
so
Bolzano- Weierstrass" often serves to describe
in addition to the
theorem itself. See,
for
example,
"A Contribution to the Mathematical Theory of Big Game
Arner. Math. Monthly, 45 (1938), 446-447.)
Prove that a continuous (complex-valued) function on
is
(c)
title
method of proof,
H. Petard,
Hunting,"
(b)
1
bounded on
[a, b]
X
[c,
d].
[a, b]
a real-valued continuous function on [a, b]
then / takes on a maximum and minimum value on [a, b]
(You can use the same trick that works for Theorem 7-3.)
Prove that if/
is
X
[c,
d]
(Imitate Problem 21-22.)
X
X
[c,
d\
[c,
d].
460
Infinite
Sequences and Infinite Series
*6.
The proof
Theorem
of
2 cannot be considered to be completely ele-
mentary because the possibility of choosing y with y m = —a//3 depends
on Theorem 24-2, and thus on the trigonometric functions. It is therefore
of some interest to provide an elementary proof that there is a solution for
the equation z
(a)
(b)
a convex subset of the plane
(c)
(d)
—
c
=
0.
computation to show that solutions of z 2 — c = 0
can be found for any complex number c.
Explain why the solution of z n — c — 0 can be reduced to the case
Make an
where
(a)
n
n
explicit
is
odd.
n
Let z 0 be the point where the function f(z) = z — c has its minimum
absolute value. If z 0 ^ 0, show that the integer m in the proof of
Theorem 2 is equal to 1; since we can certainly find y with 7 1 —
— a/ /3, the remainder of the proof works for /. It therefore suffices to
show that the minimum absolute value of /does not occur at 0.
Suppose instead that /has its minimum absolute value at 0. Since n is
n
—c± 8 n i. Show
odd, the points ±5, ± 8i go under /into —c ± 5
,
that for small 6 at least one of these points has smaller absolute value
—
than
(b)
a
nonconvex subset
FIGURE
of the plane
7.
Let f(z)
c,
thereby obtaining a contradiction.
= 0 - z0 Wl •••(*-
**)"*.
k
8
(a)
Show
(b)
Let^O) =
that f{z)
=
-
(z
zi)
mi
••(*- z
k)
m"
£
m a (z -
za )~\
k
^
m a (z -
za )~ x
.
Show that if g(z) =
0,
then^i,
.
.
.
,
zk
a= 1
all lie on the same
Problem 24-11.
cannot
(c)
through
z.
Hint: Use
K contains the line segment joinFor any set A, there is a smallest
convex set containing it, which is called the convex hull of A (Figure
9) if a point P is not in the convex hull of A, then all of A is contained
A subset K of the plane
ing any two points in
;
FIGURE
side of a straight line
9
it
is
convex
(Figure
if
8).
Complex Functions
on one
side of
some
straight line through P.
prove that the roots of
f (z) =
0
lie
Using
461
this information,
within the convex hull of the set
(The definition of convexity in the Appendix to
a function / is convex
if the set of points lying above or on the graph of / is a convex set, and
if the graph contains no straight line segments. Further details about
convex sets will be found in reference [19] of the Suggested Reading.)
{zi,
.
.
Chapter
8.
*9.
.
,
1 1
zjt}.
is
related to the one given here
Prove that if / is differentiate at z, then /
Suppose that / — u + iv where u and v are
=
+
is
continuous at
z.
real- valued functions.
(c)
and h(x) = v(x + iy 0 ). Show that if
and
+ if$ for real a
ft then g' fa) — a and h! (x 0) = ft
f(x 0 + iyo)
k(y) = u(x 0 + iy) and l{y) =
that
suppose
hand,
On the other
v(x Q + iy). Show that l'(y 0 ) = a and k'(y 0 ) = -ft
Suppose that/(z) = 0 for all z. Show that / is a constant function.
(a)
Using the expression
find
f
(b)
Use
this result to find
(a)
For fixed y 0 \ctg(x)
u(x
iyo)
= a
(b)
10.
—
k)
(x) for all
L
arctan a) (0) for
all k.
COMPLEX POWER SERIES
CHAPTER
If
you have not already guessed where
going to come
from, the
intend to define functions by means of infinite
discussion of infinite
complex functions are
differentiable
of this chapter should give the secret away:
title
we
This will necessitate a
sequences of complex numbers, and sums of such
series.
sequences, but (as was the case with limits and continuity) the basic definitions
ai
are almost exactly the same as for real sequences and series.
#
•
An
sequence of complex numbers is, formally, a complex-valued
domain is N; the convenient subscript notation for sequences
of real numbers will also be used for sequences of complex numbers. A
sequence {a n of complex numbers is most conveniently pictured by labeling
as
•
#
infinite
function whose
•
%
•
a 13
•<2io
•«„
an
}
the points a n in the plane (Figure
FIGURE
The sequence shown
in Figure
1).
converges to
1
"convergence' of complex
5
0,
1
sequences being defined precisely as for real sequences: the sequence \a n
converges
to
/,
in
—
lim a n
n—»
if
for
every
e
>
0 there
is
a natural
if
n
>
/,
oo
N such that,
number
N, then
—
\a n
l\
<
sufficiently large n (Figure 2); expressed
eventually inside any circle
drawn around
Convergence of complex sequences
sequences, but can even be reduced to
2
THEOREM
1
is
/
will contain a n for all
colloquially, the sequence
is
/.
not only defined precisely for real
this familiar case.
Let
an
and
=
bn
+
= 0
+
for real b n
ic n
and
cn ,
let
I
Then lim
an
=
I if
and only
lim
PROOF
more
for all n,
S.
This condition means that any circle drawn around
FIGURE
)
symbols
The proof is left
as
consult the similar
The sum
an easy
iy
for real
0 and
y.
if
— 0
bn
and
lim
exercise. If there
Theorem
of a sequence \a n
1
is
=
cn
y.
any doubt
as to
\
is
defined, once again, as lim s n
n—
>
sn
462
how
to proceed,
of Chapter 25. |
=
a\
+
'
'
'
+
an
.
oo
,
where
1
1
Complex Power
Sequences
which
for
say that the infinite series
^=
n
THEOREM
2
summable;
this limit exists are
an
converges
alternatively,
limit exists,
if this
Series
463
we may
and diverges
1
new
unnecessary to develop any
otherwise. It
is
infinite series,
because of the following theorem.
convergence of
tests for
Let
an
Then
a n converges if
£
n=
—
+
bn
and only
if
£
n=
l
and
for real b n
ic n
and
b*
£
n=
1
cn
cn
.
both converge, and in
1
this case
an
2
n—
PROOF
This
00
00
00
=
+
£
an immediate consequence of Theorem
sums of {a n }. |
is
partial
There
is
^=
an
converges absolutely
if
the series
n
1
complex
series:
the series
^—
\a n
converges
(this is
a series of
\
l
numbers, and consequently one to which our earlier tests may be applied).
following theorem is not quite so easy as the preceding two.
The
3
applied to the sequence of
°o
real
THEOREM
1
also a notion of absolute convergence for
00
n
c n ).
n=
l
Let
an
=
bn
+
ic n
for real b n
^=
rt
cn .
»
00
00
Then
and
an
converges absolutely
if
and only
if
^
n=
l
bn
and
^
n
1
cn
both COU-
=l
verge absolutely.
00
PROOF
Suppose
n
=
6n
^
and
n=
1
cn
both converge absolutely,
2
and
I
I
that
00
£
n=
1
i.e.,
1
00
00
n=
00
£
that
first
|c n
|
both converge.
It follows
that
£
|£ n
|
+
\c n
converges.
\
n-
1
Now,
k| =
It follows
\b n
from the comparison
+
lCn\
test that
<
\bn\
^
n=
\a n
+ M»
converges (the numbers
\a n
\
\
1
00
and
\b n
\
+
\c n
\
are real
and nonnegative) Thus
.
£
a n converges absolutely.
464
Infinite Sequences
and
Infinite Series
Now
it is
suppose that
^
\a n
converges. Since
\
clear that
\b n
<
\
\a n
and
\
\c n
<
\
\a n
\.
00
Once
again, the comparison test shows that
00
^
n=
\b n
\
and
^
n=
1
\c n
converge. |
\
1
00
Two
consequences of Theorem 3 are particularly noteworthy.
^=
If
n
00
converges absolutely, then
00
^
and
bn
T^l
quently
^
and
bn
n=l
Theorem
^=
71
also
cn
converge absolutely; COnSe-
1
00
00
2.
an
1
00
converge, by
^
Theorem
^
22-4, so
a n converges
by
n=l
n—1
In other words, absolute convergence implies convergence.
Similar reasoning shows that any rearrangement of an absolutely convergent
series
has the same sum. These facts can also be proved directly, without
using the corresponding theorems for real numbers, by
analogue of the Cauchy criterion
With
power
(see
Problem
these preliminaries safely disposed
series, that
/0) =
^
first
we can now
of,
establishing
an
12).
consider
complex
functions of the form
is,
-
a n (z
a)
n
=
a0
+ ai(z —
a)
+
—
a 2 {z
a)
2
+
'
'
'
n=0
Here the numbers
a
and
a n are allowed to be complex,
interested in the behavior of
usually consider
power
series
/
for
complex
z.
centered at
0,
/w = I
n=
a n z"
As
and we are naturally
in the real case,
we
shall
0
in this case, if f(z 0 ) converges, then /(z) will also converge for \z\ < \z 0 The
proof of this fact will be similar to the proof of Theorem 23-6, but, for reasons
\.
that will soon become clear, we will not use all the paraphernalia of uniform
convergence and the Weierstrass Af-test, even though they have complex
analogues. Our next theorem consequently generalizes only a small part of
Theorem
THEOREM
4
23-6.
Suppose that
n
) an z 0
=
+ aiz +
a0
0
a^z 0 2
+
•
•
•
nt'o
converges for some z Q
^
0.
Then
if \z\
<
\z 0 \,
the two series
Series
465
As in the proof of Theorem 23-6, we will need only the fact that the
such that
numbers a n z 0 n is bounded: there is a number
set of
Complex Power
=
n
anz
^=
+
a0
aiz
+
2
a 2z
+
•
•
0
71
00
both converge absolutely.
PROOF
M
\a n z 0
We
n
M
<
\
for all n
then have
\a n z
n
=
\
\a n z 0
^
\
M
<
and, for z
n
0,
\na n z
Ti
n\a n z 0
l
\z\
M
Since the series
^= 0
n
\z/z Q
and
\
oo
n
n\z/z 0
converge, this shows that both
\
1
00
CO
^
71
£=
71
71
n
T~\
\z\
<z
n£
n
Y M n ^ n-1
and
=0
71
^
0,
but
this series certainly
converges for z
=
na n z
^
n=
=1
assumed that z
Theorem 4
converge absolutely (the argument for
n~ x
1
0 also). |
evidently restricts greatly the possibilities for the set
{z:
^=
71
For example, the shaded
set
A
anz
n
converges}.
0
in Figure 3 cannot be the set of all z
where
00
anz
^
n
converges,
since
it
contains z
y
but not the number
w
satisfying
n
\w\
<
\z\.
seems quite unlikely that the set of points where a power series converges
could be anything except the set of points inside a circle. If we allow "circles
of radius 0" (when the power series converges only at 0) and "circles of
radius oo" (when the power series converges at all points), then this assertion
is true (with one complication which we will soon mention) the proof requires
It
;
only Theorem 4 and a knack for good organization.
466
Infinite Sequences
and
Infinite Series
THEOREM
5
For any power
series
00
anz n
)
=
one of the following three
+ aiz +
a0
possibilities
a 2z
n
+
must be
+
a zz 2
•
•
•
true:
00
anz
^=
(1)
n
converges only for z
n
converges absolutely for
=
0.
0
n
00
a n^
^
(2)
z in C.
ail
n=0
There
(3)
R>
number
a
is
^ a^
0 such that
71
converges absolutely
if
n=0
and diverges if \z\ > R. (Notice that we do not mention w hat
|*| <
happens when \z\ — R.)
r
proof
Let
00
S —
{x in
V
R:
Suppose
first
£
that
<2
n
wn
converges for some
=
with
a;
#}.
=0
n
is
unbounded. Then
for
any complex number
z,
there
00
number x in S such that
a
is
|z|
<
x.
By definition of £,
this
means that
^
71
w
converges for some
with
=
\w\
x
>
n
wn
from Theorem 4 that
follows
It
\z\.
tf
=0
00
^
<z
nz
n
converges absolutely. Thus, in this case possibility
(2)
is
true.
71=0
Now suppose
R =
then
0,
that
^
tf
n£
let
R
be the least upper bound of
S. If
converges only for z
=
0, so possibility (1) is true.
Sup-
S is bounded, and
n
71=0
on the other hand, that
pose,
<
|z|
R, there
is
a
number
R >
x in
0.
Then
if
<
x.
£ with
\z\
z
is
a complex
Once
number with
means that
again, this
00
00
a nw
^
n
converges for some
w
with
<
\z\
\w\,
so that
n=0
71
^=
0 nz
n
converges
0
00
absolutely. Moreover,
if \z\
>
/?,
then
^=
n
not in
#nz
n
does not converge, since
|z|
is
0
S. |
The number R which
occurs in case (3)
is
called the radius of convergence
so
of
^
anz
n
.
In cases
(1)
and
(2)
it is
customary
to say that the radius of con-
n=0
vergence
is
is
0
and
<x>
?
called the circle of
respectively.
When
convergence of
^
0
< R <
a nz
n
.
If z
<x>,
is
the circle \z\
\z\
= R\
outside the circle, then,
Complex Power
of course,
anz
^
n
does not converge, but actually a
much
467
Series
stronger statement
n=0
can be made: the terms a n z n are not even bounded. To prove this, let w be
any number with \z\ > \w\ > R; if the terms a n z n were bounded, then the
proof of Theorem 4 would show that
anw
^
n
converges, which
is false.
Thus
71=0
00
(Figure 4), inside the circle of convergence the series
anz
^
n
converges in the
n=0
bounded
way (absolutely) and outside the circle the series diverges in the
way (the terms a n z n are not bounded).
What happens on the circle of convergence is a much more difficult question.
best possible
worst possible
We will not consider that question at all, except to mention that there are
power series which converge everywhere on the circle of convergence, power
series which converge nowhere on the circle of convergence, and power series
that do just about anything in between. (See Problem 5.)
Remember
circle of
convergence
that our goal in this chapter
produce differentiable funcproved for real power series
in Chapter 23, that a function defined by a power series can be differentiated
term-by-term inside the circle of convergence. At this point we can no longer
imitate the proof of Chapter 23, even if we were willing to introduce uniform
convergence, because no analogue of Theorem 23-3 seems available. Instead
we will use a direct argument (which could also have been used in Chapter 23).
Before beginning the proof, we notice that at least there is no problem about
the convergence of the series produced by term- by- term differentiation. If the
tions.
We
therefore
want
is
to
to generalize the result
oo
series
^=
n
a nz
n
has radius of convergence R, then
Theorem 4 immediately
0
implies that the series
na n z n
^
~
l
also converges for
\z\
<
R. Moreover,
n=l
\z\
>
R, so that the terms a n z n are unbounded, then the terms na n z
n~
l
if
are
OS
surely
unbounded,
so
^
na n z
n~ l
does not converge. This shows that the
n—1
00
radius of convergence of
theorem
6
If the
power
^
na n z
n ~~ x
has radius of convergence
<
R,
also exactly R.
series
f(z)
\z\
is
R >
0,
=
J
then /
anz
is
n
differentiable at z for all z with
and
na n z
n
468
Infinite Sequences
and
Infinite Series
PROOF
We will use another "s/3 argument.
for
55
The fact
that the theorem
is
clearly true
polynomial functions suggests writing
00
00
/(*
+ A)
-/(*)
-
(*)
^
na n z
n~ l
+
((*
=
£)
-
n
|
n=0
n=l
AT
00
+
((z
—
j
h)
n
—
z
+
n
)
-
A)"
) na n z
n
k
z")
\l;
n=
0
((£
+
+
-
A)"
z»)
—
N
y na n z
k
»
n=1
We will show that for any
made <
>
£
n=»l
each absolute value on the right side can be
This will
0,
N sufficiently large and h sufficiently small.
e/3 by choosing
clearly prove the theorem.
Only
the
first
term
in the right side of (*) will present
\z\ < \z 0
<
The expression {{z +
more convenient way if we remember that
begin with, choose some z 0 with
only h with
in a
h\
<
\z 0 \.
*1\.ZJ1
=
x n-l
\z
x
Applying
+
—
+
yv
\
+ x n-y
x n-2y
any
_|_
h)
-
n
.
.
difficulties.
we
R; henceforth
To
will consider
n
z )/h can be written
+ y«~\
.
this to
0+
n
h)
-
zn
+ h) n - z n
(z + h) -z'
_
(z
.
,
;
h
we obtain
(z
+ h) n -
zn
=
(z
+
z{z
+ hy~ +
l
z(z
+
h) n
~2
+
•
•
'
+
zn
~
l
.
Since
\(z
+
h)
n' 1
+
h)
n~ 2
+
•
•
•
+ zn
~l
\
<
n\zo\
n -\
we have
j(z
+
h)
n
- g)
<
n\a n
•
\
|z 0
n_1
.
|
00
But the
series
^
n\a n
'
\
ko|
n_1 converges, so
n= 1
^
n
=N+l
n\a n
\
if
N
is
sufficiently large,
then
Complex Power
469
Series
This means that
I
V
+1 hY -
On ((*
\l
n=
V
z")
+
^—
l
.
v
I
ccg
«•
—
+ *)" -
g")
2,
v
-2/
^
i
1
n = AT-fl
*">
h
n=0
0
~
A )"
Kg
an
-
n=
JV
((z
+ A)» -
i
+1
r
h
00
V
<
In short,
(1)
N
if
V
sufficiently large,
is
((*
an
l
+ h) n -
for all h
with
+
|*
kd— <
•
v
((*
an
l
+ hY - z n
<
S
)
<3
h
n=0
h\
f-
then
zn )
IT
n=0
1
«kl
\z<\.
CO
It is
easy to deal with the third term on the right side of (*)
:
Since
^=
71
converges,
follows that
it
if
N
is
sufficiently large,
na n z
n~ x
1
then
IV
(2)
y^na n z«-i n=
^
na n z
n
<3
1
Finally, choosing
an
hm
N such
that (1)
+
((Z
)
an
h)
n
~
and
(2) are true,
Zn)
=
£>
na n z
we
note that
n
J
n=l
n=0
AT
since the polynomial function
gO) =
^=
71
<z
n 2:
n
is
certainly differentiable.
0
Therefore
N
AT
<3
^
:
>I2
=
71
na n z n
~l
0
for sufficiently small h.
As we have already indicated,
Theorem
6 has
an obvious
(1), (2),
and
(3)
prove the theorem.
is
series
is its
Taylor
series at 0. It follows, in particular,
that/ is continuous inside
is continuous at z
the circle of convergence, since a function differentiable at z
(Problem 25-8).
by a power
and the power
corollary: a function represented
infinitely differentiable inside the circle of convergence,
series
|
and
Infinite Series
The continuity of a power series inside its circle of convergence helps explain
the behavior of certain Taylor series obtained for real functions, and gives the
to the questions raised at the end of Chapter 23. We have
already seen that the Taylor series for the function f(z) = 1/(1
z 2 ), namely,
promised answers
\
+
-
1
when
converges for real z only
vergence
J
points
i
\z\
<
and consequently has radius of con-
1,
no accident that the circle of convergence contains the two
and —i at which / is undefined. If this power series converged in a
1.
It is
circle of radius greater
than
which was continuous
in that circle, in particular at
would represent a function
i and
—i. But this is
impossible, since it equals 1/(1 + z 2 ) inside the unit circle, and 1/(1 + z 2 ) does
not approach a limit as z approaches i or —i from inside the unit circle.
The use of complex numbers also sheds some light on the strange behavior
then (Figure
1,
5) it
of the Taylor series for the function
X
=
/(*)
x
0,
0
=
0.
Although we have not yet defined e z for complex
that if y is real and unequal to 0, then
-l/fiy)«
/(«»
The
interesting fact about this expression
small.
Thus /
numbers,
z
=
so
_
presumably be true
becomes large
it
asjy
becomes
not even be continuous at 0 when defined for complex
hardly surprising that it is equal to its Taylor series only for
will
it is
0.
The method by which we will
complex z should by now be
for
*
sin
cos x
=
—
actually define
clear.
x^
*
X"*
1
3!
5!
2
4
x
e
x
-
1
cos z
x
,
—
cos
z, cos'(z)
•
•
•
,
•
+" + —+•*
•
•
2!
=
z
z
+ --
z
3
5
3!
5!
zz
z*
-++
2!
4!
1
-
1
+-z + -z
2
=
z and cos z)
4!
1!
sin' (z)
(as well as sin
define
sin z
exp(z)
z
l
1!
For complex z we therefore
e
For real x we know that
2!
Then
will
p \\y*
that
is
z, it
= — sin
z,
-
2!
and exp
f
(z)
=
exp(z) by
Theorem
6.
Complex Power Series
Moreover,
we
if
replace z by iz in the series for
e
z
>
something
471
particularly-
interesting happens:
e
=
lz
1
+
iz
*2
(this step
2!
wr»
e
iz
from the definitions
iz'
/
V
—
cos z
also
3!
/
5!
n z
power
— z) = —
=
series) that
sin z,
cos
z,
have
g
From
1
_|_ i s [
the
(i.e.,
sin(
we
'
fact that the series converges absolutely), so
cos (—z)
so
'
5!
z4
4!
can be justified by the
It is clear
4!
3!
21
V
'
r
-\
the equations for
e
iz
-iz
^
cos z
—
'
2
we can
and
si
n
Zt
derive the formulas
sm z =
2
l
=
cos z
2
The development of complex power
series thus places the
exponential function
at the very core of the development of the elementary funotions
—
it
reveals a
connection between the trigonometric and exponential functions which was
never imagined when these functions were first defined, and which could never
have been discovered without the use of complex numbers. As a by-product of
this relationship, we obtain a hitherto unsuspected connection between the
numbers e and 7r: if in the formula
e
we
take z
=
7r,
we
iz
—
CQS z
With
these remarks
And
we
n z
=
ir
-1.
will bring to a close
yet there are
still
have not been mentioned. Thus
centered at
I gj
obtain the remarkable result
e
functions.
_|_
our investigation of complex
several basic facts about
far,
power
series
we have seldom considered power
which
series
a,
00
f(z)
=
^=
n
except for a
—
0.
«»(*
-
«)".
0
This omission was adopted partly to simplify the exposition.
472
Infinite Sequences
and
Infinite Series
For power
this
centered at a there are obvious versions of all the theorems in
series
chapter (the proofs require only
modifications) there
trivial
:
is
number R
a
00
(possibly 0 or "oo") such that the series
^=
n
for z
with
—
\z
moreover, for
a\
all
<
—
\z
<R
a\
—
unbounded terms
R, and has
z with
a n (z
a)
n
converges absolutely
0
for z
with
\z
—
a\
>
R;
the function
00
f{z)
r\
=
-
a n {z
I
a)*
n=0
has derivative
a
f(z)
- |
n
It is less
tion as a
n -\
straightforward to investigate the possibility of representing a func-
power
centered at
FIGURE
na n (z-a)
=1
series
centered at
b, if it is
already written as a power series
a. If
00
6
=
f(z)
71
has radius of convergence
it is
R
y
and
a n (z
b is
=
numbers
bn
are necessarily
„)»
a point with
2=
n
-
0
true that f(z) can also be written as a
f(z)
(the
I=
power
»•(*-
b)
—
\b
a\
series
< R
(Figure 6), then
centered at
b,
n
0
f n) (b)/n !);
moreover,
this series
has radius of
convergence at least R — \b — a\ (it may be larger).
We will not prove the facts mentioned in the previous paragraph, and there
are several other important facts we shall not prove. If
00
f(z)
00
= 2 a ^z -
a) n
and
*w
converge in some circle around
/ g
1
*»(*
a, it is
certainly to be
converge in a circle around
a;
00
and ^(^) =
at
a,
= 2
then
it
hoped that f
moreover, ifg(a)
/g should have a power series representation. Finally,
f(z)
-
n=0
n=0
will also
= £
^
0,
+g
and
the function
if
00
««(*
-
and
g(z)
=
should be possible to write /
J
o
6 n (^
-
*)»,
g as a power
series
centered
6.
introducing any basic new
/ g presents a few problems, and
\/g presents even more problems. The possibility of changing a power series
centered at a into one centered at b becomes yet more involved, and the
treatment offog requires real skill. Rather than end this section with a tour
All of these facts could be proved
ideas.
The treatment off
+g
is
now without
easy, while
•
Complex Power
we
de force of computations,
will instead give a
473
Series
55
preview of " complex analysis,
one of the most beautiful branches of mathematics, where all these facts are
derived as straightforward consequences of some fundamental results.
Power series were introduced in this chapter in order to provide complex
functions which are differentiable. Since these functions are actually infinitely
differentiable, it is natural to suppose that we have therefore selected only a
very special collection of differentiable complex functions. The basic theorems
of complex analysis
show that
If a complex function
A, then
in
A
it is
is
not at
is
all true:
A
defined in some region
and
of the plane
is differentiable in
automatically infinitely differentiable in A. Moreover, fcr each point a
Taylor series for
the
this
at a will converge to
f
f
any
in
circle
A
contained in
(Figure 7).
These
among
facts are
the
be proved in complex analysis. It is
the methods used are
to
first
impossible to give any idea of the proofs themselves
—
quite different from anything in elementary calculus.
If these
are
facts
granted, however, then the facts mentioned before can be proved very easily.
Suppose, for example, that / and g are functions which can be written as
power
Then,
series.
we have shown, / and g
as
are differentiable
Appealing
differentiable.
to the results
they can be written as power
Similarly,
series.
= £
=
n
<
a\
centered at
b,
R}. Thus,
which
if b is
in A,
is
a)»
differentiable in the region
A =
power
series
—
This
possible to write / as a
it is
converge in the
will
-
a n (z
0
has radius of convergence R, then /
—
then
it
'
if
/(*)
\z\ \z
—
f + g, f g, 1/g and fog are also
from complex analysis, it follows that
follows from easy general theorems that
circle of radius
R —
\b
a\.
00
series
may
^= 0a n (z —
actually converge in a larger circle, because
a)
n
may
be
n
the series for a function differentiable in a larger region than A. For example,
suppose that f(z)
=
1/(1
+
z 2 ).
Then /
is
differentiable, except at
i
and
—
i,
00
where
Y
not defined. Thus f(z) can be written as a power series
it is
anz
n
n=0
with radius of convergence
and
ak
=
0
if
k
is
odd). It
1
is
(as
a matter of fact,
we know
that
am —
(~~l) n
also possible to write
00
f(z)
= 2
=
n
where the numbers
bn
b n {z
are necessarily
bn
the radius of convergence of this series:
to
i
or
—i
(Figure
8).
-
*)",
0
=
it is
f
(n)
(j)/n\.
Vl +
We can easily predict
2
(^-)
,
the distance from
\
474
Infinite Sequences
and
Infinite Series
As an added incentive
complex analysis
to investigate
mentioned, which
result will be
quite near the surface,
lies
found in any treatment of the subject.
For real z the values of sin z always lie between
z this
not at
is
all true.
In
=
z
fact, if
gi(iy)
sin
iy
e
=
large, then sin
is
iy is
—
-~i(iy)
g—y
=
1
and
and which
1
,
but for complex
then
for y real,
:
2
If y
iy,
one more
will be
further,
—
ey
:
T
l
also large in absolute value. This behavior of sin
is
which are defined and differentiable on the whole complex
plane (such functions are called entire). A result which comes quite early in
complex analysis is the following:
typical of functions
Liouville's Theorem:
The
only bounded entire functions are the constant functions.
As a simple application of
Liouville's
Theorem, consider a polynomial
function
f(z)
where
large
>
n
z,
1,
=
zn
+
ai z
n
~
so that /is not a constant.
so Liouville's
Theorem
tells
+
•
•
•
+a
H9
We already know that f(z)
us nothing interesting about
is
/.
large for
But con-
sider the function
g
&
=
7TT
/0)
were never 0, then g would be entire; since f{z) becomes large for large
would also be bounded, contradicting Liouville's Theorem.
Thus f(z) = 0 for some z, and we have proved the Fundamental Theorem of
If f(z)
z,
the function g
Algebra.
PROBLEMS
1.
Decide whether each of the following
converges absolutely.
n=
1
series converges,
and whether
it
'
Complex Power
Series
475
00
(i
v
;
yiog_«
L4
+
.
B
iog».
n
n
n=2
Use the
ratio test to
show that the radius of convergence of each
following power series
will
to 00 or to
a limit
is
<
approach a limit
>
1.
(In each
1 if \z\
<
1,
of the
case the ratios of successive terms
but for
\z\
>
1
the ratios will tend
1.)
n-\
» It
n
=
1
(iii)
00
(iv)
00
(v)
2-")^ n
^(« +
_
^ 2^
Use the root
n!
.
test
of the following
z
....
(»)
.
,
(Problem 22-7)
power
z*
z*
z*
to find the radius of convergence of each
series.
z6
z*
_
_
_
—
<?n\z n
>
.
(iv)
t**'
cv)
^ 2V
!
,
of
root test can always be used, in theory at least, to find the radius
convergence of a power series; in fact, a close analysis of the situation
The
476
Infinite
Sequences and Infinite Series
known
leads to a formula for the radius of convergence,
Hadamard
formula." Suppose
first
as the
"Cauchy-
numbers V|tf n
that the set of
is
|
bounded.
00
Use Problem 22-7
(a)
to
show
that
V|^ n z <
lim
if
|
1,
Y
then
n ~*~
n
anz
n
=0
converges.
00
show that
(b) Also
>
lim \Z\a n z
if
\
"-*
Y
then
1,
00
n
anz
n
has unbounded
=0
terms.
00
(c)
Parts (a)
and
(b)
show
that the radius of convergence of
_
1/ihn a/I^I (where "1/0" means " oo").
n—*
a nz
^
n
is
n=0
To complete
the formula,
oo
V/ |^ n =
define lim
oo
the set of
if
|
is
unbounded. Prove
0, so
that the radius of
^/\a n
all
\
n—+oo
00
anz
^
that in this case,
n
diverges for z
^
n=0
convergence
5.
may
0 (which
is
Consider the following three
be considered as "1/°°").
from Problem
series
2:
n
y^
n=
Prove that the
n
1
,
Vxn=
=1
l
converges everywhere on the unit circle; that
nowhere on the unit circle; and that the second
least one point on the unit circle and diverges for
first series
the third series converges
series
converges for at
on the unit
at least one point
6.
(a)
Prove that
e
z
e
w
=
e
zJrW
circle.
for all
complex z and w. (You
just
have
to
imitate the relevant parts of Problem 17-35.)
(b)
Show
=
7.
(a)
Prove
iv
e
(b)
8.
(a)
(b)
9.
(a)
cos z sin w and cos(z + w)
complex z and w.
that every complex number of absolute value 1 can be written
some real number y.
x
x
that \e + iy — e for real x and y.
that exp takes on every complex value except 0.
that sin takes on every complex value.
that a complex- valued function / = u + iv satisfies the
for
Prove
Prove
Prove
Show
+
that sin(z
cos z cos
w —
w)
=
sin z sin
sin z cos
w
w +
for all
\
equation
f n) + an-i/^ +
15
(*)
if
(b)
and only
Show
that
a n -iz
+
e
bx
n~ l
cos
a:
if
u
if
a
*
and
=
•
b
v
do. Hint:
+
+a
0
ci is
=
0,
•
•
•
+ aof =
Use Problem
0,
25-9.
n
a complex root of the equation z
then
are both solutions of (*).
/(*)
=
sin cx
and
/(*)
+
=
Complex Power
10.
(a)
Show
(b)
Given «/^0, show that
#
=
that exp
log
is
Series
477
+
with
one-one on C.
not
e
= w
z
if
and only
if
z
=
#
(here log denotes the real logarithm function),
\w\
iy
and j> an
argument of w.
*(c)
Show
that there does not exist a continuous function log defined for
nonzero complex numbers, such that exp (log (z)) = z for all z ^ 0.
(Show that log cannot even be defined continuously for \z\ — 1.)
Since there is no way to define a continuous logarithm function we
cannot speak of the logarithm of a complex number, but only of "a
logarithm for w," meaning one of the infinitely many numbers z with
e
z
=
w.
(d)
Find
all
logarithms for
(e)
Find
all
values of i\ that
i.
is,
of
e
iz
where z
is
a logarithm of
(The
i.
answers will be real numbers.)
(f)
Show
that
many
has infinitely
(1*)*
possible values, while
1***
has
only one.
11.
(a)
For real x show that we can choose log(*
(It will
(b)
The
+
^ — arctan
log (a;
+
i)
—
log(l
+
x 2)
+
log(x
—
i)
=
log(l
+
x 2)
—
—
arctan x
help to note that t/2
t
—
i
=
expression
-L-i(J—
1
+
2
2^ \a:
a:
?
and \og(x
i)
^
—
be
to
i)
xj,
arctan xj.
arctan \/x for x
^
0.)
U
+
z/
yields, formally,
^
/ +x
1
Use part
12.
(a)
(a) to
A sequence
lim
if
m,n—>
cn
\a
m
{a n
—
1
[log(*
,
2
check that
}
an
\
0.
]
log(^
+
0].
answer agrees with the usual one.
numbers
is
Suppose that a n
called a
=
bn
+
Cauchy sequence
ic n ,
where
bn
and
oo
is a Cauchy sequence if and only if
{a n
Cauchy sequences.
every Cauchy sequence of complex numbers converges.
are real. Prove that
\b n
this
of complex
=
-0 -
2i
and
{c n
}
}
are
(b)
Prove that
(c)
Give direct proofs, without using theorems about real series, that an
absolutely convergent series is convergent and that any rearrangement has the same sum. (It is permitted, and in fact advisable, to
use the proofs of the corresponding theorems for real series.)
>
478
Infinite
Sequences and Infinite Series
13.
Prove the formula
1
+
-
+
cos 8
+
cos 26
+
•
•
sin (n
=
cos nd
2
—
+
i)6
2 sin*
2
by writing cos nd = (^ + e~ )/2, summing the resulting geometric
and multiplying top and bottom by e~ i9l K
in6
n*
series,
14.
Let
{a n
be the Fibonacci sequence,
}
-
(a)
If r n
(b)
Show
^n+iAn, show that
that
=
r
lim
n—
>
=
r
Show
(c)
+
(1
rn
a\
1
+
and
r
=
r n +i
exists,
=
—
a2
1,
a n+2
—
an
+
a n+ \.
l/r„.
=
1
+
1/r.
Conclude that
oo
V5)/2.
^=
that
n
fl
n^
n
has radius of convergence 2/(1
+
\/S). (Using
1
oo
unproved theorems
the
in this chapter,
and the
fact that
^
n
-\/{z 2
+
z
-
+
-
z
1
15.
—
the smallest absolute value of the roots
/
0; since the roots are (-1 ± "V 5)/2, the radius
=
indeed equal to 2/(1
+
+
Vs)/2. Notice that
this
number
Vs).)
can be written as the power series 1 + z/2\ + z /3\
z
which is nonzero at 0, it follows that if g(z) = z/(e — 1) for
+
around 0.
circle
some
=
inside
1, then ^ is differentiate
z
0, and g(0)
there is a
that
chapter
this
in
theorem
It then follows from the unproved
Since
•
(**
•
power
- 1)A
2
•
series
VK
z
-
Lf
=
1
n
convergence;
bers z
=
it is 2tt,
Ikiri for
since this
which
are called the Bernoulli
(a)
Clearly
b0
=
g(0)
<?*
1.
-
+
z
-
e
e~
can even predict the radius of
the smallest absolute value of the
1
=
Now
_ _
The numbers
0.
bn
num-
appearing here
show that
z
+
z
z_
\
2
2^-1
1
e*+l
1
e
z
e~ z
We
is
numbers.*
=
z
-
n\
0
with nonzero radius of convergence.
*
n
is
of convergence should be (-!_
is
anz
l
from Problem 23-8, we could have predicted that
1)
the radius of convergence
of z 2
—
z
-
e
1
1
because b n = 0
Sometimes the numbers fl n = (-l)"-^ 2 « are called the Bernoulli numbers,
numbers b 2n alternate in sign, although we
n is odd and > 1 (see part ((a)) and because the
are also in use.
will not prove this. Other modifications of this nomenclature
if
Complex Power
Series
479
and deduce that
b
(b)
By
l
=
bn
=
0
if
is
72
finding the coefficient of z
n
odd and
>
n
1.
in the right side of the equation
z
show that
SO-
0
for n
>
1.
This formula allows us to compute any b k in terms of previous ones,
and shows that each is rational. Calculate two or three of the
following:
b2
*(c)
Part
(a)
=
= "Mi
bi
=
=
4V5
_i_
"
3 0-
shows that
b 2n
,
(In
(2n)
z
e>+\
z
2
e*
-
2
2n
!
1
+
e
2
'
2
-
^
2/2
n=U
Replace z by 2iz and show that
00
zcotz=
Show
*(d)
Show
n
*\n^<Ln„in
2 2
(_1)
.
•
that
tan z
*(e)
b<i
A(2T!
n=0
=
cot z
—
2 cot 2z.
that
00
tan z
=
V
^ 2n
Li
(2ft)
(-\)n-i2 2n (2 2n
-
\)z
2n
~\
!
(This series converges for
|*|
<
7r/2.)
numbers play an important role in a theorem which is
denote the
best introduced by some notational nonsense. Let us use D to
k
will mean
D
Then
/'.
denotes
that
so
f
operator,"
Df
"differentiation
The
f
(k)
Bernoulli
and e D f will mean
(n)
£= 0 / A!
(of course this series
makes no sense
in
71
general, but
Finally,
f( x
_|_
i)
let
it
A
will
make
sense
if
/is a polynomial function, for example).
denote the "difference operator" for which Af(x)
- f( x y Now Taylor's Theorem implies,
=
disregarding questions
!
480
Infinite Sequences
and
Infinite Series
of convergence, that
or
n=
we may
write this symbolically as
e
D
—
1,
—
A/ =
Even more symbolically
"identity operator."
A =
l
where 1 is the
can be written
1)/,
this
which suggests that
Thus we obviously ought
to
have
i.e.
The
(a)
beautiful thing about all this nonsense
Prove that (**)
which case the
fk\
(*) to
is
literally true if
infinite
sum
is
/
that
it
works
a polynomial function (in
By applying
really a finite sum). Hint:
find a formula for /<*>(*
formula in Problem
is
is
+
1)
-
/<*>(*);
15(b) to find the coefficient of
then use the
f
ij)
(x)
in the
right side of (**).
(b)
Deduce from
(**) that
00
(c)
g(0)
+
(d)
Show
•
•
•
that for
+ g(n)
Apply
any polynomial function g we have
= f^git)
this to g(x)
=
dt
x v to
+
££
show that
[*<*-»(!,
+
1)
-
g«-»(0)l
Complex Power
Using the
=
fact that bi
—h
481
Series
show that
ten instances of this formula were written out in Problem
pattern.
2-6, which offered as a challenge the discovery of the general
Berthe
but
suggestion,
This may now seem to be a preposterous
After
this
way!
precisely
in
discovered
noulli numbers were actually
The
first
writing out these 10 formulas, Bernoulli claims (in his posthumously
printed work Ars Conjectandi, 1713): " Whoever will examine the
be able to continue the table. He
then writes down the above formula, offering no proof at all, merely
noting that the coefficients b k (which he denoted simply by A, B,
series as to their regularity
C,
.
.
.)
satisfy the
55
may
equation in Problem 15(b).
tween these numbers and the coefficients
z/{f — 1) was discovered by Euler.
in
The
relation be-
the power
series for
16(c) can be generalized to the case where g is
the infinite sum must be replaced by a finite
function;
not a polynomial
In order to find an expression for the
term.
remainder
sum plus a
The formula
remainder,
(a)
The
Problem
in
it is
useful to introduce
Bernoulli polynomials
some new
functions.
are defined by
<p n
71
The
first
three are
<Pi(x)
Show
that
= bn
= b n if > 1,
(p n '(x) = n<p n -i(x),
*>n« = (-l)Vn(l -*)
*>n(0)
<Pn(X)
,
Tl
for,*
>
Hint: Prove the last equation by induction on
1.
«,
starting with n
-
482
Infinite
Sequences and Infinite Series
(b)
R N k (x)
Let
interval
[x,
be the remainder term in Taylor's formula for f k \ on the
x
+
so that
1],
N
/<*>(*
(*)
+
-
1)
^+
U+n)(
V ^—
fM( X) =
Prove that
AT
=
fW
[/U>(
b
~% i RN
+ 0 ~ /<4)W]
*
Xir*
-
N—
(c)
Hint: Imitate Problem 16(a). Notice the subscript
Use the integral form of the remainder to show that
4
(d)
*(*)
=0
Deduce the "Euler-Maclaurin Summation Formula":
+*(*
+
1)
+
•
•
+ g(x + n)
•
N
+l
=
k on R.
s(t) dt
J^"
V £ lgV°-»(x + n-l)-
+
g<*-*>(x)]
+ SK (x, n),
where
i
(e)
Let
$n(t)
^n
is
=o
n
be the periodic function, with period
=
<p n (t)
\//
<
for 0
continuous, since
even and odd
if
n
is
<
/
<p n
1.
(l)
odd.)
1,
(Part (a) implies that
=
<p n
Show
(0),
and
also that
which
if n
<p n is
tions
this).
become
SN (x,
n)
=
0,
satisfies
1,
even
then
if
n
is
that
Unlike the remainder in Taylor's Theorem, the remainder Sn(x,
does not satisfy lim
>
n)
usually
because the Bernoulli numbers and func-
large very rapidly (although the
first
few examples do not suggest
Nevertheless, important information can often be obtained from the
summation formula. The general
situation
is
best discussed within the context
of a specialized study ("asymptotic series"), but the next problem shows one
particularly important example.
5
!
Complex Power
*18.
N=
Use the Euler-Maclaurin Formula, with
(a)
n
1
logtdt--logn
/
(b)
Show
(c)
Explain
+
show that
[
n
Ji
—
^(0
dt.
that
why
show that
if
the improper integral 0
a =
g
(d)
2, to
483
Series
=
"
|t
2
fa(t)/2t dt exists,
and
then
\a/2»
+1 '
V"V
Jn
It
2
Problem 18-26(d) shows that
0z!)
V^r = lim
2
2 2"
(2n)
Use part
(c) to
show that
V/-
a 2n 2n+1 e- 2n 2 2n
=
7r
lim
7=>
a(2n)
and conclude that
(e)
Show
that the
o:
2w " H/
V~ 2B
Vn
= V2^.
maximum
value of
*J 4
i/:
2/
2
<
"
\(p*(x)\
for x in [0, 1]
is
£, so that
1
12w
Conclude that
W »+1/2<T»-1/I2n
The
final result of
Problem
/
that n
is
!
approximately
V
18,
<
„!
^+l/V" n+1/12n
<
.
a strong form of Stirling's Formula, shows
n+1/2<T n in the sense that this expression differs
2tt w
,
examfrom n by an amount which is small compared to n when n is large. For
<1%.
error
an
with
=
3628800,
10 we obtain 3598696 instead of
ple, for n
"
asymptotic
the
illustrates
Formula
A more general form of Stirling's
used in
nature of the summation formula. The same argument which was
have
we
iV
>
2
that
for
show
to
used
be
now
18
can
Problem
!
5
i
(
ni
\
=
v
bk
+
r
^
dt
484
Infinite Sequences
and
Infinite Series
Since
\j/
N
is
bounded, we can obtain estimates of the form
00
1/
UN
n1
N
is
large, the constant
will
make
MN
jMf)
N
tV!
dt
<
Mn
t
will also
be large; but for very large n the factor
the product very small. Thus, the expression
N
V^r> + l/2,-n
may
.
(
y
**
\
be a very bad approximation for n! when * is small, but for
large n (how
on N) it will be an extremely good one {how good depends on N).
large depends
There was a most ingenious Architect
who had contrived a new Method
for building Houses,
by beginning at the Roof, and working
downwards
to the
JONATHAN SWIFT
Foundation.
PART
EPILOGUE
CHAPTER tff
FIELDS
I
this book a conscientious attempt has been made to define all
important concepts, even terms like "function," for which an intuitive definition is often considered sufficient. But Q, and R, the two main protagonists of
this story, have only been named, never defined. What has never been defined
can never be analyzed thoroughly, and "properties" PI -PI 3 must be con-
Throughout
sidered assumptions, not theorems, about numbers. Nevertheless, the term
"axiom" has been purposely avoided, and in this chapter the logical status
more carefully.
Like Q, and R, the sets N and Z have also remained undefined. True, some
talk about all four was inserted in Chapter 2, but those rough descriptions are
far from a definition. To say, for example, that N consists of 1, 2, 3, etc.,
merely names some elements of N without identifying them (and the "etc." is
useless). The natural numbers can be defined, but the procedure is involved
and not quite pertinent to the rest of the book. The Suggested Reading list
of PI -PI 3 will be scrutinized
contains references to this problem, as well as to the other steps that are
one wishes to develop calculus from its basic logical starting point.
program would proceed with the definition
definition
in terms of Z. This program results
terms of N, and the
of
required
The
if
further development of this
of Z, in
Q
in a certain well-defined set
and properties PI -PI 2
as theorems.
struction of R, in terms of
+ and
program is the conconstruction which concerns us.
certain explicitly defined operations
Q,
The
•,
final step in this
Q,. It is this last
Assuming that Qhas been defined, and that PI -PI 2 have been proved for Q,
we shall ultimately define R and prove all of PI -PI 3 for R.
Our intention of proving PI -PI 3 means that we must define not only real
numbers, but also addition and multiplication of real numbers. Indeed, the
real numbers are of interest only as a set together with these operations: how
the real numbers behave with respect to addition and multiplication is crucial;
what the real numbers may actually be is quite irrelevant. This assertion can
be expressed in a meaningful mathematical way, by using the concept of a
"field," which includes as special cases the three important number systems
of this book. This extraordinarily important abstraction of modern mathematics incorporates the properties P1-P9 common to Q,, R, and C. A field is
a set F (of objects of any sort whatsoever), together with two "binary operations"
and defined on F (that is, two rules which associate to elements
a and b in F, other elements a
b and a
b in F) for which the following con-
+
•
+
•
ditions are satisfied:
{a
(2)
There
(i)
(ii)
487
+ b) +
(1)
a
is
c
=
a
+
(b
+
some element 0
+0
=
a
for all a,
c)
in
F such
b,
and
c
in F.
that
for all a in F,
for every a in F, there
is
some element b
in
F such that a
+b
=
0.
+b
=
(3)
a
(4)
(a
(5)
There
b)
•
a
(i)
•
+a
b
=
c
•
some element
1
=
=
(6)
a
*
b
(7)
<z
•
(b
b
+
•
F with
a in
=
b
and
in F.
c
that
1^0
and
=
a
0, there
a
is
some element
•
and
+a
b
•
6 in
F such
F
for all a,
c
6,
and
F
in
c
familiar examples of fields are, as already indicated, Q,, R,
The
b in
1.
for all a
a
c)
b,
F such
1 in
for all a in F,
a
•
b in F.
for all a,
c)
*
is
that a
and
for all a
(b
-
For every
(ii)
+
a
and C, with
+
and \ It is probably unnecessary
being the familiar operations of
to explain why these are fields, but the explanation is, at any rate, quite brief.
the rules 1, 3,
and
and are understood to mean the ordinary
When
the elements which play
4, 6, 7 are simply restatements of PI, P4, P5, P8, P9;
the role of 0 and 1 are the numbers 0 and 1 (which accounts for the choice of
and
•
+
+
•
the symbols 0, 1); and the number b in (2) or (5) is -a or aT\ respectively.
(For this reason, in an arbitrary field F we denote by —a the element such that
1
(_ fl ) = 0, and by a -1 the element such that a a" = 1, for a ^ 0.)
a
+
•
In addition to
R, and C, there are several other
Q,,
One example is
The operations + and
the collection
described easily.
Q
for a, b in
for real
numbers.
It is
•
will,
F
of
Y
fields
which can be
numbers
all
+ b V2
+ and
a
once again, be the usual
necessary to point out that these operations really do
produce new elements of F\\
{a
+ b V2) +
{a + b V2)
Conditions
for all real
a
_|_
&
satisfies
field are
a
+
0
in
(2)
+ b V2
=
0.
Similarly,
verification of (5ii)
obvious for
Fu
Fl
since these hold
numbers of_the form
=
0 + 0 V2_is in Fi
number
0
the
holds because
=
(-a) + ( — b) V2 in Fi
in Fi the number j3
numbers, they certainly hold
a
i;
is
(1), (3), (4), (6), (7) for
a =
a
is
c)
V2. Condition
and, for
The
+ d V2) = (a + + (b + d) V2, which in F
(c + d V2) = (ac + 2bd) + (be + ad) V 2, which
(c
*
1
=
1
for all real
+ 0 V2
is
in Fi, so (5i)
is
the only slightly difficult point. If a
a
+ b V2
is
satisfied.
+ b V2 ^
0,
then
1
it
is
therefore necessary to
—
7=
<z
+ 6V2
show that \/(a
=
1;
+ b Vl)
is
in Fi. This
is
true
because
1
aTTVl
_
"
(The division by a
a
(
fl
-
b
A
- bV2
V2)(
V2
is
fl
+
^
b
V~2)
a
,
a*-2b>
valid because the relation a
J^llL V2
a*
-
-2b*
6
=
0 could
489
Fields
=
be true only
if
hypothesis a
+ b V2
a
b
=
V2"
0 (since
irrational)
is
which
is
ruled out by the
0.)
The next example of a field, F
is considerably simpler in one respect: it con2
two elements, which we might as well denote by 0 and 1. The
and are described by the following tables.
operations
,
tains only
+
The
•
+
0
1
0
0
1
1
1
0
equation
may
Our
final
a in R,
and
a, b, c
—
0 or
also be written 1
+
(a, a)
+
(The
and
0
0
1
0
1
Notice that in
1.
—
=
this field 1
+
8 equations
=
1
0; this
1.
silly:
F
3
consists of all pairs
(0, a)
for
•
(a, a)
cation for R.)
0
may be proved by checking the
example of a field is rather
and are defined by
+
1
(l)-(7) are straightforward, case-by-case
verification of conditions
checks. For example, condition (1)
obtained by setting
0
-
=
=
(b, b)
(b, b)
(a
+
{a
*
b,
b,a
a
•
+
b),
b).
appearing on the right side are ordinary addition and multipliThe verification that Fz is a field is left to you as a simple
exercise.
A
for
detailed investigation of the properties of fields
our purposes,
fields
is
a study in
itself,
but
provide an ideal framework in which to discuss the
numbers in the most economical way. For example, the consequences of P1-P9 which were derived for "numbers" in Chapter 1 actually
hold for any field; in particular, they are true for the fields Q, R, and C.
Notice that certain common properties of Q, R, and C do not hold for all
properties of
fields.
fields,
a
=
b.
For example,
it is
For the
field
C
+ 1 = 0 to hold in some
— a does not necessarily imply that
+ 1^0 was derived from the explicit
possible for the equation 1
and consequently
—b
a
=
the assertion
description of C; for the fields
b
1
Qand R, however, this assertion was derived from
further properties which do not have analogues in the conditions for a
There
is
a
is
field
F
(with operations
+
and
•)
together with a
(the "positive" elements) with the following properties:
(8)
For
(i)
(ii)
(iii)
(9)
If a
(10) If a
all a in
a
a
F, one
=
0,
is
in P,
—a
and
and
is
b
and only one
of the following
in P.
are in P, then a
b are in P,
then a
+b
•
b
is
is
in P.
in P.
field.
An ordered field
certain subset P of F
a related concept which does use these properties.
is
true:
We have already seen that the field C cannot be made into an ordered field.
field F with only two elements, likewise cannot be made into an ordered
The
2,
field: in fact,
then
Fh
the field
be
made
It
applied to
(8),
+
=
1
consisting of all
into an ordered
positive real
into
condition
(9) implies that 1
numbers
an ordered
field;
0
is
—
=
1
numbers
+ b V2
a
P
field: let
be the
the description of
P
is
with
in
a, b
The
field
left to
you.
1
must be
in
P;
On the other hand,
+
set of all a
ordinary sense).
(in the
shows that
1,
in P, contradicting (8).
F
s
Q, certainly can
which are
can also be made
natural to introduce notation for an arbitrary ordered field which
is
R we
corresponds to that used for Q, and
a
a
a
a
>
<
<
>
define
:
—b
b
if
a
b
if
b
b
if
a
b
if
a
>
<
>
in P,
is
a,
b
or a
b
or a
Using these definitions we can reproduce,
=
=
for
b,
b.
an arbitrary ordered
field F,
the definitions of Chapter 7:
A set A of elements of Fis bounded above if there is some x in Fsuch that
x
>
a for all a in A.
ment x of F is a
x
<
Finally,
Any
least
y for every y in
it is
such x
an upper bound for A. An eleif x is an upper bound for A and
an upper bound for A.
is
called
upper bound
F which
possible to state
is
A
for
an analogue of property PI 3
for
R;
this leads to
the last abstraction of this chapter:
A
complete ordered field is an ordered field in which every nonempty
set which is bounded above has a least upper bound.
The
consideration of fields
may seem
to
have taken us
constructing the real numbers. However,
intelligible
will
means of formulating
this goal.
we
are
There are two questions which
be answered in the remaining two chapters:
1.
Is
there a complete ordered field?
2.
Is
there only one complete ordered field?
Our
field, containing N and Z as certain subsets.
be necessary to assume another fact about Q:
Let x be an element of Q, with x
in
assumed to be an
At one crucial point it
starting point for these considerations will be Q,,
ordered
will
from the goal of
provided with an
far
now
N
such that nx
>
>
0,
Then
for
any^
in Q, there
is
some
n
y.
This assumption, which asserts that the rational numbers have the Archimedian property of the reals, does not follow from the other properties of an
ordered field (for the example that demonstrates this conclusively see [17]).
The important
point for us
is
that
when
Q,
is
explicitly constructed, proper-
491
Fields
ties
if
PI -PI 2 appear as theorems, and so does this additional assumption;
really began from the beginning, no assumptions about Q, would be
we
necessary.
PROBLEMS
1.
Let
F
be the set
{0, 1, 2}
and define operations
+
following table. (The rule for constructing this table
and
is
•
on
F
as follows:
by the
add or
multiply in the usual way, and then subtract the highest possible multiple
of 3; thus 2
=
2
4
=
+
0
1
2
0
0
1
2
1
1
2
2
2
0
Show
that
F is
a
field,
+
3
1,
so 2
=
2
•
1.)
0
1
2
0
0
0
0
0
1
0
1
2
1
2
0
2
1
and prove that
it
cannot be
made
into
an ordered
field.
2.
Suppose now that we try to construct a field F having elements 0, 1, 2, 3
and defined as in the previous example, by adding
or multiplying in the usual way, and then subtracting the highest possible
with operations
multiple of
3.
Let
F=
4.
+
•
Show
{0, 1, a,
that
F will
not
be a
field.
0} and define operations
+ and
•
on
Fby
the follow-
ing tables.
+
0
1
a
0
0
0
1
a
0
1
1
0
/3
a
a
a
0
Show
4.
(a)
that
F is
a
OL
0
1
1
0
1
F
0
0
0
0
0
0
1
0
1
a
a
0
a
0
0
+
1 = 0. Show that a
field in which 1
a).
can also be written a =
Suppose that a
a = 0 for some a ^ 0. Show that
consequently £
£ = 0 for all b).
Let
a
1
a
1
field.
be a
a (this
(b)
0
+
+
—
+a
1
+
= 0
1
for all
= 0
(and
492
Epilogue
5.
(a)
Show
m
mn
n times
times
numbers m and
for all natural
(b)
we have
that in any field
Suppose that
n.
F we
in the field
have
1+^^+1
m
times
= o
times
some natural number m. Show that the smallest m with this
number (this prime number is called the
for
property must be a prime
characteristic of F)
6.
Let
F
(a)
Show
be any
with only finitely
field
must be
that there
m
(b)
many
distinct natural
numbers m and
n
with
n times
times
Conclude that there
elements.
some natural number
is
M^_^fi
=
k with
o.
k times
7.
Let
a, b, c,
and d be elements of a
field
F
with a
—
•
6
^
c
•
0.
Show
that the equations
8.
•
x
£
•
at
+
b2
(a)
(b)
b
•
b
=
Consider an equation x 2
(
In the
On
b
+ r)/2
field
F
2
is
A
a,
/3,
"square root" of a
is
an element
b of
1
+
=
1
0,
+
be incorrect.
b
b
2
x +
—4
=
c
•
•
c
0,
where
b
and
c
has a square root
are elements
r
in F.
Show
a solution of this equation,
of the text, both elements clearly have a square root.
the other hand,
the equation x 2
Let
F.
=
=
which case a has only one.
that
10.
j;
a.
of a field F. Suppose that
(b)
•
*
How many square roots does 0 have?
Suppose a ^ 0. Show that a has two square roots, unless
in
(a)
=
£
d y
can be solved for x and y.
Let a be an element of a field
.Pwith
9.
a
it is
+x+
What
easy to check that neither element satisfies
= 0. Thus some detail in part (a) must
1
is it?
F be a field and a an element of F which does not have a square root.
This problem shows how to construct a bigger field F\ containing F, in
which a does have a square root. (This construction has already been
carried through in a special case, namely, F = R and a = — 1 this
;
special case should guide you through this example.)
.
.
493
Fields
Let F' consist of
are
+
and
0
O
0, y)
(x,
y)
Prove that
Prove that
(a)
(b)
all
pairs
w)
jF',
(x,
Let
+
F
+
(#
•
0
0
0)
0)
we may agree
be the
and
u, v)
•
z,
(y,
w, y
•
(y,
= (x+y,
0)
+x
z
0 and
=
O,
•
is
a>)
a
field.
0),
0),
to abbreviate
=
•
(at,
0)
by
0) in F'
(a,
set of all four-tuples (w, x, y, z) of real
+
u, v)
t,
(w, x, y, z)
(w, x,y,z)
—
(sw
Show
that
—
F satisfies
the algebra will
plicative inverses
(b)
•
0)
tx
=
(s
+
w,
—
uy
—
vz, sx
+
sy
(a)
+ w),
y
+a
z
F
numbers. Define
by
•
(s,
,
(x
Find a square root of a
(c)
0
with the operations
(x,
so that
=
=
w)
(z }
and y in F. If the operations on
and O on F' as follows:
y) with x
(x,
define operations
It is
tz,
field,
except
(6).
ty
—
ux).
At times
quite ornate, but the existence of multi-
the only point requiring
any thought.
customary to denote
(0,
1,0,0) byi,
0,1,0) by;,
(0,
0,0,
(0,
Find
uw
conditions for a
all
become
is
+ x, u + y, v + z),
+ tw + uz — vy,
+ vx — sz + vw +
t
all 9
products of pairs
ticular that condition (6)
known
i,
is
as the quarternions.
1)
by
j,
and
k.
k.
The
show in par"skew field" F is
results will
definitely false. This
CHAPTER
28
CONSTRUCTION OF THE
REAL NUMBERS
of drudgery which this chapter necessarily contains is relieved by
one truly first-rate idea. In order to prove that a complete ordered field exists
we will have to explicitly describe one in detail; verifying conditions (1)— (10)
for an ordered field will be a straightforward ordeal, but the description of the
The mass
the elements in
field itself, of
is
it,
ingenious indeed.
raw material
the real
called
be
ultimately
will
which
it is
numbers. To the uninitiated this must seem utterly hopeless if only the
rational numbers are known, where are the others to come from? By now we
At our disposal
is
the set of rational numbers, and from this
necessary to produce the field
have had enough experience to
—
realize that the situation
hopeless as that casual consideration suggests.
The
may
not be quite so
strategy to be adopted in
our construction has already been used effectively for defining functions and
complex numbers. Instead of trying to determine the "real nature" of these
concepts, we settled for a definition that described enough about them to
determine their mathematical properties completely.
A
numbers requires
similar proposal for defining real
numbers
in
terms of rational numbers.
The
ought to be determined completely by the
suggests a strikingly simple
and quite
a description of real
observation, that a real
set of rational
numbers
less
attractive possibility: a real
number
than
it,
number
might (and in fact eventually will) be described as a collection of rational
numbers. In order to make this proposal effective, however, some means must
be found for describing "the set of rational numbers less than a real number'
without mentioning real numbers, which are still nothing more than heuristic
figments of our mathematical imagination.
If A is to be regarded as the set of rational numbers which are less than the
5
real
is
number
a, then
a rational
A ought
number
If x is in A andy
In addition to this
have the following property:
A should have
number x < a, the set A
property, the set
rational
to
satisfying y
<
x,
then y
is
in A.
a few others. Since there should be some
should not be empty. Likewise, since there
should be some rational number x > a, the set A should not be all of Q,.
Finally, if x < a, then there should be another rational number y with
x
<
y
<
a, so
A
should not contain a greatest member.
the real numbers as known, then it is not hard to
check (Problem 8-17) that a set A with these properties is indeed the set of
rational numbers less than some real number a. Since the real numbers are
presently in limbo, your proof, if you supply one, must be regarded only as an
unofficial comment on these proceedings. It will serve to convince you, how*If
we temporarily regard
we have not failed to notice any crucial property of the
appears to be no reason for hesitating any longer.
ever, that
494
set A.
There
Construction of the Real
DEFINITION
A
number
real
is
495
Numbers
a set a, of rational numbers, with the following four
properties:
number withjy <
If x
(2)
OL
(3)
Q.
There is no greatest element in a\
there is some y in a with y > x.
(4)
The
a and^
(1)
is
*
a *
in
is
a rational
set of all real
numbers
is
also in a.
should be clear that
a
is
if
x
is
then
in a,
denoted by R.
is
you of the philosophy behind our
example of a real number:
a =
It
thenjy
in other words,
Just to remind
explicit
x,
0.
\x in
Q: # < 0 or x 2
the real
<
number which
definition, here
is
an
2}
will eventually
V2,
be
known
is a
but it is not an entirely trivial exercise to show that a
number. The whole point of such an exercise is to prove this using only
facts about Q; the hard part will be checking condition (4), but this has already
appeared as a problem in a previous chapter (finding out which one is up to
you). Notice that condition (4), although quite bothersome here, is really
as
actually
real
essential in order to avoid ambiguity;
\
x in
without
Q: x <
1}
<
1}
it
both
and
{x in Q/. x
would be candidates for the "real number 1."
The shift from A to a in our definition indicates both a conceptual and a
notational concern. Henceforth, a real number is, by definition, a set of
number (a
number x has a
rational numbers. This means, in particular, that a rational
member
of
Q)
is
not
a real number; instead every rational
is a real number, namely, {y in Q: y < x). After
completing the construction of the real numbers, we can mentally throw away
will henceforth denote these special sets.
the elements of
and agree that
For the moment, however, it will be necessary to work at the same time with
natural counterpart which
Q
Q
rational numbers, real
real
numbers
(sets
numbers
(sets
of sets of rational
of rational numbers)
and even
numbers). Some confusion
inevitable, but proper notation should
keep
this to
numbers will be denoted by lower case Roman letters (x, y, z, a, b,
real numbers by lower case Greek letters (ex, /3, 7) capital Roman
;
(A, B, C) will
be used to denote
c)
and
letters
numbers.
devoted to the definition of +, •, and P
R, and a proof that with these structures R is indeed a complete ordered
The remainder
for
sets of real
sets of
perhaps
a minimum. Rational
is
field.
of this chapter
is
496
Epilogue
We
we
We first define a <
P
shall define
properties for P.
we
begin with the definition of P, and even here
shall actually
work backwards.
P; later,
when +,
and 0 are
shall
available,
a with 0 < a, and prove the necessary
with the definition of < is the
beginning
reason for
as the set of all
The
simplicity of this concept in our present setup:
Definition. If
in
a and p
a < P means that a is contained
an element of/3), but a ^ p.
are real numbers, then
every element of
a
also
P
(that
A
repetition of the definitions of <, >,
is,
is
>
would be
but
stultifying,
it is
be expressed more simply than < if a and
<
only if a is contained in p.
and
then
a
if
numbers,
are
real
P
P
If A is a bounded collection of real numbers, it is almost obvious that A
interesting to note that
<
now
can
;
should have a least upper bound. Each a in A is a collection of rational numbers; if these rational numbers are all put in one collection P, then P is presumably sup A. In the proof of the following theorem we check all the little
which have not been mentioned, not least of which is the assertion that
(We will not bother numbering theorems in this chapter,
P is
since they all add up to one big Theorem: There is a complete ordered field.)
details
a real number.
theorem
If
A
least
PROOF
a set of real numbers and A
upper bound.
is
Let P
=
{x:
x
is
in
some a
numbers; the proof that P
(1)
in A}.
is
a real
^
0 and
A
is
bounded above, then A has a
Then p is certainly a collection of rational
number requires checking four facts.
Suppose that x is in P and y < x. The first condition means that x is
a for some a in A. Since a is a real number, the assumption y < x
in
implies that y
(2)
Since
A ^
0,
is
in a. Therefore
there
is
some a
certainly true that y
in A. Since
some x in a. This means that x
(3)
it is
is
a
in P, so
is
p
is
in p.
a real number, there
^
is
0.
bounded above, there is some real number y such that
a < 7 for every a in A. Since 7 is a real number, there is some
rational number x which is not in 7. Now a < 7 means that a is
contained in 7, so it is also true that x is not in a for any a in A. This
Since
A
is
that x is not in p; so P ^ Q.
Suppose that x is in p. Then x is in a for some a in A. Since a does
not have a greatest member, there is some rational number y with
x < y and y in a. But this means that y is in p; thus P does not have
means
(4)
a greatest member.
is a real number. The proof that P is
bound of A is easier. If a is in A, then clearly a is contained in P; this means that a < P, so P is an upper bound for A. On the other
hand, if 7 is an upper bound for A, then a < 7 for every a in A; this means
These four observations prove that P
the least upper
Construction of the Real
that
a
contained in 7, for every
is
a in
A,
The
definition of
with a proof that
Definition. If
a
THEOREM
PROOF
If
=
'
'obvious'
5
definition
this surely implies that
thus
0
is
0
the least upper
is
con-
bound
it must be complemented
makes any sense at all.
are real numbers, then
=
x
{x:
y
+
z for some y in
are real numbers, then
Once again
;
both obvious and easy, but
is
a and 0
+0
a and 0
+
this
and
< y
tained in y. This, in turn, means that 0
of A. I
497
Numbers
a
+£
is
a and some
z in 0).
a real number.
four facts must be verified.
some x in a
/3. Then a: = jy + 2- for some)' in a
which means that w < y + z, and consequently,
w — y < z. This shows that w — y is in 0 (since ^ is in 0, and ^ is a
real number). Since w = y + (w — y), it follows that 2x> is in a
It is clear that a
0.
0, since a 5^ 0 and /3
0
Since a
and 0 5* Q, there are rational numbers a and b with a
not in a and 6 not in 0. Any x in a satisfies # < a (for if a < *, then
condition (1) for a real number would imply that a is in a); similarly
any y in 0 satisfies y < b. Thus x + < a + £ for any x in a and ^
in 0. This shows that a + 6 is not in o:
/3 ^ Q,.
0, so a
If x is in a
/3, then a: = y + ^ for y in a and z in 0. There are
/ in a and in 0 with _y < / and z < z then x < y + z and
Thus a
is in a
0 has no greatest member. |
_/ +
Suppose w < x
and some z in
(1)
+
for
/3,
+
+
(2)
(3)
+
+
+
(4)
f
f
f
;
+
+
By now you can see how tiresome this whole procedure is going to be. Every
we mention a new real number, we must prove that it is a real number;
this requires checking four conditions, and even when trivial they require
time
is really no help for this (except that it will be less
you check the four conditions for yourself) Fortunately, however, a
few points of interest will arise now and then, and some of our theorems will be
easy. In particular, two properties of
present no problems.
concentration. There
boring
if
.
+
THEOREM
PROOF
theorem
PROOF
and 7 are
If a, 0,
Since (x
+ y) +
member
of (a
If
a and 0
Left to
z
real
numbers, then (a
=
+
x
+ 0) + 7
(y
is
+
z) for all rational
also a
member
are real numbers, then
you (even
easier). |
+ 0) + 7
a
+0
of
a
= 0
+
= a
+
(0
numbers
x, y,
+ y)>
and
(0
+ a.
+ 7).
and
z,
every
vice versa. |
498
Epilogue
To
+
prove the other properties of
=
0
Definition.
Q: x <
{x in
we
PROOF
If
a
is
+0
a
a real number, then
define 0.
0}.
It is, thank goodness, obvious that 0
theorem is also simple.
THEOREM
first
=
a real number, and the following
is
a.
in a and y is in 0, then y < 0, so x + y < x. This implies that x + y is
Thus every member of a + 0 is also a member of a.
On the other hand, if x is in a, then there is a rational number y in a such
that y > x. Since x = y + (x — y), where y is in a, and x — y < 0 (so that
* - y is in 0), this shows that x is in a + 0. Thus every member of a is also a
If *
is
in a.
member
The
of
+ 0. |
a
—a
reasonable candidate for
{x in Q,:
(since
—x not
in
a means,
would seem
—x
is
set
not in a}
intuitively, that
certain cases this set will not even be a real
a
be the
to
—x >
a, so that x
< — a).
number. Although a
But in
number
real
does not have a greatest member, the set
Q—
—x is
{
x in
Q,:
x
is
a
not in
element x 0 when a is a real number of this kind, the set
not in a will have a greatest element —x 0 It is therefore necessary
may have
{x:
a —
a
;
.
}
to introduce a slight modification into the definition of
—a, which comes
equipped with a theorem.
Definition. If
—a =
THEOREM
PROOF
If
a
is
a
Q:
{x in
a real number, then
is
—*
is
a real number, then
(1)
it is
is
Since
in
is
not the least element of
Q—
a}.
a real number.
a ^
Q,,
is
jy
there
can assume that y
(since
—y
in
is
is
is
— a.
some
rational
number
—a
Since
a ^
—a ^
which
not the smallest rational
can always be replaced by any y
Thus
so
—a and y <
also true that
This shows that
(3)
—a
is
x. Then — y > —x. Since —x is not
not in a. Moreover, it is clear that — y is
not the smallest element of Q, - a, since —x is a smaller element.
Suppose that x
in a,
(2)
—x
not in a, but
1
>
y).
is
Q,
We
— a
in
—a.
not in a.
number in
Then —y is
9* 0.
0,
Q.
there
is
some x
in a.
Then
— * cannot possibly be in —a,
—
If
(4)
x
— a,
in
is
y
< —x
y
<
<
z
—.v.
element of
—a
—x
then
which
is
Q
is
number
number with
a rational
also not in a. Let z be a rational
Then
—
not in a, and there
is
a.
z
also not in a,
is
—z
So
is
in
— a.
and z
is
Since
499
Numbers
Construction of the Real
clearly not the smallest
—z >
shows that
x, this
does not have a greatest element. |
+ — a)
= 0 is not entirely straightforward. The diffiyou might presume, by the finicky details in the
definition of
a. Rather, at this point we require the Archimedian property of
Q, stated on page 490, which does not follow from PI -PI 2. This property
is needed to prove the following lemma, which plays a crucial role in the next
The proof
a
that
(
culties are not caused, as
—
theorem.
LEMMA
Let
a be
we may assume
over,
PROOF
Suppose
that z
first
that y
3
then
all in a,
number w
satisfies
nz for
.
.
some
w,
a
490. This contradicts the fact that
=
—
a.
z.
More-
.
number would be
every rational
w <
there are
x
numbers
in a. If the
is
—
not the smallest element of Q,
is
z 2z, 3z,
were
Then
a real number, and z a positive rational number.
(Figure 1) rational numbers x in a, and;? not in a, such thatjy
is
every rational
in a, since
by the additional assumption on page
a real number, so there is some k such
not in a. Clearly y — x = z.
r
Moreover, if y happens to be the smallest element of Q, — a, let x > x be
—
x
(V
x).
x\
and
an element of a, and replace
by
y by y
If z is not in a, there is a similar proof, based on the fact that the numbers
that x
—
kz
is
in
a and
y
=
(k
+
\)z
is
+
(
— n)z
cannot
be in a.
all fail to
|
x
FIGURE
THEOREM
If
a
is
a real number, then
a
PROOF
Suppose *
x
+y <
It is
—z >
a
0.
is
in
a and
y
+y
in 0.
0, so
x
little
more
According
is
is
can be written x
proves that z
is
in
a
+
(
0.
—a. Then — y is not in
Thus every member of a
a). |
a, so
+
go in the other direction.
lemma, there
(—y) =
+
(—a) =
in
difficult to
to the
+
with;/ not the smallest element of
tion
y
1
Q—
z.
is
some x
is
If
z
is
is
x.
in «,
Hence
in 0.
in 0, then
and some y not in a,
— x = —z. This equaand — y is in
«, this
in a,
a, such that;;
Since x
—y >
(—a)
—
500
Epilogue
Before proceeding with multiplication,
we
define the "positive elements"
and prove a basic property:
Definition.
P =
+
Notice that a
THEOREM
If
a
If
a
/? is
0}.
clearly in
P
if
a and 0
are.
a real number, then one and only one of the following conditions holds:
is
(ii)
a =
a is
in P,
(iii)
—a
is
(i)
PROOF
a >
{a in R:
0,
in P.
contains any positive rational number, then
numbers, so a contains 0 and a ^
no positive rational numbers, then one of two
tive rational
(1)
(2)
a
contains
certainly contains all nega-
a
i.e.,
negative rational numbers; then
all
is
in P. If
possibilities
a
contains
must hold:
a =
0.
some negative rational number x which is not in a; it can be
assumed that x is not the least element of Q, — a (since x could be
replaced by x/2 > x); then —a contains the positive rational number
there
— x,
is
we have
so, as
This shows that
just proved,
—a
is
in P.
must hold. If a = 0, it is clearly impossiMoreover, it is impossible that a > 0 and
(—a) > 0. |
would imply that 0 = a
at least one of (i)-(iii)
ble for condition
—a >
a
0,
(ii)
or
(iii)
0 both hold, since
to hold.
this
+
but is unequal to
a > 0 was defined to mean that a contains
This definition was fine for proving completeness, but now we have to show
that it is equivalent to the definition which would be made in terms of P.
Thus, we must show that a
0 > 0 is equivalent to a > 0. This is clearly a
Recall that
/5.
—
consequence of the next theorem.
theorem
PROOF
If a, ft
The
and y are
hypothesis
real
a > 0
from the definition of
a
+y
>
numbers and a
implies that
+
that
/3
0
+y
is
>
/3,
then
+y
a
contained in a;
is
contained in
a
it
> 0
+ 7-
follows immediately
+ y.
This shows that
+ y. We can easily rule out the possibility of equality, for
« + 7 = 0 + 7,
if
then
a =
which
is false.
(a
Thus
+ 7) + (-7) = (0 + 7) +
a+y >
+ y. |
ft
/3
Multiplication presents difficulties of
be defined as follows.
("7) =
its
own.
If a,
£ >
0,
then
a 0 can
•
a and 0
Definition. If
a
THEOREM
PROOF
•
j8
=
a and 0
If
As
usual,
(1)
<
z
{z:
are real numbers
0 or
are real
2:
=
a
•
0.
a and
tive x in
<
Since 0
(2)
(3)
a
<
a;
then
0,
0,
then
0 with
in
a 0
is
•
>
x,
0}
a real number.
in
>
0.
a 0. If u> <
Then z > 0,
•
positive ^ in 0.
then
is
ze;
=
so z
x
•
j>
automatically
for
some
posi-
Now
\z
z
we have
z,
0,
<
zt>/z
1,
)
x
so (w/z)
Thus
in a.
is
u> is
0.
•
a 0 ^
Clearly
•
0.
and
for all such positive *' and /.
Hence ^ >
is not in a
So
ft thus a j3 5^ Q,.
Suppose w is in a ft and w < 0. There is some * in a with x > 0
in £ with y > 0. Then z = xy is in a 0 and z >
and some
Now suppose > 0. Then w = xy for some positive x in a and some
positive y in ft Moreover, a contains some x > x; if z = x'.y, then
z > xy =
and z is in a ft Thus a 0 does not have a greatest
If *
j
>
is
not in a, and
/
y' for all
is
not in 0, then x
>
*' for all *' in a,
in 0.
•
*
(4)
0 >
is
z
in
>
0
four conditions.
w < z, where z
Suppose that a;
Suppose
in
•
a,
a,
xina and y
* j for some
numbers with
we much check
and
501
Numbers
Construction of the Real
•
•
zt>
r
•
•
element. |
Notice that
a
•
j8 is
clearly in
tion of all properties of P.
Definition. If
a
is
To
P
a and 0
if
complete the definition of
if
=
a,
a and 0
we
•
first
|a|
k
if
If
or,
0,
\a\.
a >0
a < 0.
if
if
-
-(|a|
•
I0|),
if
a = 0 or 0 = 0
a > 0, 0 > 0 or a <
a > 0, 0 < 0 or a <
0,
0,
0
0
<
>
0
0.
suspect, the proofs of the properties of multiplication usually
involve reduction to the case of positive numbers.
THEOREM
define
are real numbers, then
0,
As one might
This completes the verifica-
a real number, then
\a\
Definition. If
are.
and 7 are
real
numbers, then a
•
(0
•
7)
=
(a
•
0)
•
7-
502
Epilogue
PROOF
This
is
clear
if
a, 0,
7>
separate cases (and
THEOREM
PROOF
If
a and
This
(3
THEOREM
PROOF
If
is
a
Let
On
a,
if
=
Definition. 1
(It
0
>
The proof for
is
0. It is
•
/3
the general case requires considering
if
one uses the following theorem).
= 0
•
a.
cases are easily checked. |
}
1
a
=
1
•
a.
easy to see that every
the other hand, suppose x
member of a
in a. If x
is
<
•
1 is also
0, then x
is
a
member of a.
automatically in
a 1. If x > 0, then there is some rational number y in a such that
Then x = y (x/y), and #/y is in 1, so * is in a 1. This proves that a
ifa>0.
If a < 0, then, applying the result just proved, we have
•
a-1 =
Definition. If
a- 1 =
{
a
theorem
is
THEOREM
PROOF
a<
If
a
is
0,
then
when a =
a real number and
<
x in Q: x
a" =
1
>
or x
0,
= -(H) -
-(|a|-|l|)
obvious
is
member
if
0
of Q,
— (M"
a>
and 1/*
— a
0,
is
(1)
it
then x
>
<
0, so
not in a, and
in
(2)
(3)
(4)
a
x,
only
y.
a.
0. |
not in a, but 1/x
is
not the smallest
}
is
a>
in
0.
a"1
.
is
a real number.
Four conditions must be checked.
If
y
not in a. Since
is
1 /jy is
= a
).
and x
1/*
<
1
1
suffices to consider
Suppose y
x
then
1
a real number unequal to 0, then a"
Clearly
•
•
•
Finally, the
|
a real number.)
a real number, then
a >
a
and the other
0,
Q: x <
{x in
clear that 1
is
0.
simplified slightly
are real numbers, then
clear
is
is
<
0,
1/jy
then
>
1/x,
y is
it
in
a -1
.
If
j
follows that
clearly not the smallest element of Q,
—
>
0,
1 /jy is
a, so y
is
-1
Clearly
a" 1 ^
0.
a > 0, there is some positive rational number x in a. Then l/#
in a \ so a"
not
is
1
Suppose x is in a" 1 If x < 0, there is clearly some y in a:"" with
-1
contains some positive rationals. If x > 0, then
y > x because a
1 /x is not in a. Since 1 /x is not the smallest member of Q, — a, there
not in a, with y < 1/x. Choose a rational
is a rational number
number ^ with y < z < 1/x. Then 1/z is in a, and 1/z > *. Thus
a-1 does not contain a largest member. |
Since
.
jy
Construction of the Real
a" 1
In order to prove that
is
really the multiplicative inverse of a,
have another lemma, which
lemma.
to
LEMMA
PROOF
is
503
Numbers
it
helps
the multiplicative analogue of our
first
Let a be a real number with a > 0, and z a rational number with z >
Then there are rational numbers x in a, and y not in a, such that y/x =
Moreover, we can assume that y is not the least element of Q — a.
Suppose
first
that z
in a. Since z
is
+
*»=(!
l))
1
>
0 and
n
>
1
+ n(z -
z.
1),
numbers
follows that the
it
(*-
—
1.
z,
z 2 z\
.
,
.
.
h
cannot all be in a. So there is some k such that x = z is in a, and ^ =
=
to
be the least element of
happens
if
z. Moreover,
is not in a. Clearly y /x
y
x' and j byyx'/x.
by
replace
x
and
a,
of
element
an
a, let x' > x be
Q,
If z is not in a, there is a similar proof, based on the fact that the numbers
l/z k cannot all fail to be in a. |
number and a ^
THEOREM
If
a
PROOF
It
obviously suffices to consider only
a real
is
that x
or 1 Then
in
that xy
every
in 1.
is
member
1.
•
a>
in a,
0, in
and ^
-1
which case a
is
>
Suppose
0.
a positive rational
number
not in a, so \/y > x; consequently xy < 1, which means
Since all rational numbers x < 0 are also in 1, this shows that
\/y
.
a a" 1 =
then
number
a positive rational
is
0,
is
a
of
•
a""
1
in 1.
is
prove the converse assertion, let z be in 1. If z < 0, then clearly z is in
a or1 Suppose 0 < z < 1 According to the lemma, there are positive
rational numbers x in a, and y not in a, such that y/x = \/z\ and we can
To
•
.
.
assume that y is not the smallest element of Q, — a. But this means that
1
where * is in a, and l/> is in or 1 Consequently, z is in a a" |
z = x
•
•
.
We are almost done
The
we must consider many cases, but do not despair.
numbers are positive contains an interesting point, and the other
again
all
If a, 0,
PROOF
Assume
and y are
*
•
(y
a
•
03
+
0.
•
is
>
0.
A
a
•
(/3
+ y)
cases
is
also
•
•
+a
•
can
y.
Then both numbers in the equation
number in a (/? +
also in
•
7.
a 0
•
•
in 7. Since x
and z
a positive element of
number is
in a 0 + a
7, this
= a &
positive rational
z) for positive x in a, y in 0,
•
a
+ 7)
y
where * y
z,
element of
numbers, then
numbers <
rational
+
•
real
that a, ft
first
the form x
*
case
be taken care of very neatly.
theorem
all
Once
when all
the distributive law remains.
Only the proof of
!
.
a
+a
•
•
is
+
z)
of
=
and x z is a positive
Thus, every element of
0,
7-
(y
contain
7)
•
On
form
the other hand, a positive rational
xi
-
(*i/*0
'
+
y
y
<
x2
'
a
in
To
•
(/3
a
in
•
+a
j8
7
5
y, so (*i/* 2 )
xi
is
number
z for positive #i * 2 in a, y in ft and z in 7. If x x
+ 7).
Of
-jV is
-
y
+
in ft
x2
*
is
is
of the
x 2 then
,
Thus
=
+
x 2 [(xi/x 2 )y
same
course, the
complete the proof it
z
<
trick
works
z]
if
<
#2
xi.
necessary to consider the cases
whema,
ft
and 7
any one of the three equals 0, the proof is easy and the cases
are not all >
involving a < 0 can be derived immediately once all the possibilities for 0
and 7 have been accounted for. Thus we assume a > 0 and consider three
cases: ft 7 < 0, and 0 < 0, 7 > 0, and 0 > 0, 7 < 0. The first follows
immediately from the case already proved, and the third follows from the
second by interchanging (3 and 7. Therefore we concentrate on the case
0 < 0, 7 > 0. There are then two possibilities:
0. If
(1)
+ 7 >0.
0
Then
a 7 = «
•
(W
•
+ 7] +
= «
'
08
+ 7) + «
•
101,
so
«.
+ 7)
03
= -(a- |/3|)
= a (3 + a
*
(2)
0
a
•
+ 7 < 0. Then
= « (10 + 7l + 7)
101
•
=
ol
•
|0
+ a -7
•
7.
+ 7I +
<*
•
7,
so
a
•
(0
+ 7)
= -(«
•
10
+ 7l)
= -(a
101)
+ a- 7
=
<*-0
+ <*'7.|
This proof completes the work of the chapter. Although long and frequently
tedious, this chapter contains results sufficiently important to be read in detail
more than once !). For the first time we know
there is indeed a complete
in a vacuum
operating
been
that we have not
ordered field, the theorems of this book are not based on assumptions which
can never be realized. One interesting and horrid possibility remains: there
may be several complete ordered fields. If this is true, then the theorems of
calculus are unexpectedly rich in content, but the properties P1-P13 are
at least once (and preferably not
disappointingly incomplete.
—
The
last
chapter disposes of
this
possibility;
—
properties PI -PI 3 completely characterize the real numbers anything that
can be proved about real numbers can be proved on the basis of these properties alone.
PROBLEMS
There are only two problems in this set, but each asks for an entirely different
construction of the real numbers The detailed examination of another construction is recommended only for masochists, but the main idea behind these
other constructions is worth knowing. The real numbers constructed in this
!
Construction of the Real
505
Numbers
chapter might be called "the algebraist's real numbers," since they were
purposely defined so as to guarantee the least upper bound property, which
involves the ordering <, an algebraic notion. The real number system constructed in the next problem might be called "the analyst's real numbers,"
since they are devised so that Cauchy sequences will always converge.
1.
number ought to be the limit of some Cauchy sequence
we might try to define a. real number to be a Cauchy
sequence of rational numbers. Since two Cauchy sequences might converge to the same real number, however, this proposal requires some
Since every real
of rational numbers,
modifications.
(a)
Define two Cauchy sequences of rational numbers { a n } and { b n } to be
{b n }) if lim (a n - b n ) = 0. Prove that
equivalent (denoted by {a n }
~
\a n
)
~
if {a n
(b)
\
{a n
~
},
{b n
that [a n
and
}
Suppose that a
is
]
~ {b n
~ n
}
{b n
[c
\
the set of
all
if {b n
~
}
{a n
},
and that
{a n
}
~
{c n
}
\.
sequences equivalent to {a n }, and
/3 is
the set of all sequences equivalent to b n Prove that either a Pi /3 = 0
or a = 13. (If a C\ 0 ^ 0, then there is some [c n in both a and 0. Show
}
{
.
\
a and
that in this case
equivalent to
/8
both consist precisely of those sequences
{c n }.)
Part (b) shows that the collection of all Cauchy sequences can be split
up into disjoint sets, each set consisting of all sequences equivalent to
(c)
some fixed sequence. We define a real number to be such a collection,
and denote the set of all real numbers by R.
a
If a and /3 are real numbers, let \a n be a sequence in a, and {b n
)
)
a
sequence in 0. Define
+
j3
to
equivalent to the sequence {a n
be the collection of
+
bn
show that
on the particular sequences \a n and
Cauchy sequence and
also
}
Show
}.
that {a n
this definition
{£ n
}
Show that R is a field with
tive inverse
(e)
(f)
+
bn
\
is
a
is
a and
fi.
Check
well defined.
these operations; existence of a multiplica-
the only interesting point to check.
Define the positive real numbers P so that R will be an ordered field.
Prove that every Cauchy sequence of real numbers converges.
Remember that if { a n } is a sequence of real numbers, then each a n is
itself
2.
is
sequences
does not depend
chosen for
also that the analogous definition of multiplication
(d)
all
a collection of Cauchy sequences of rational numbers.
This problem outlines a construction of "the high-school student's real
numbers." We define a real number to be a pair (a, {b n }), where a is an
integer and \b n is a sequence of natural numbers from 0 to 9, with the
proviso that the sequence is not eventually 9; intuitively, this pair repre]
sents a
+
)
b n \0~
n
.
With
this definition,
a real number
is
crete object, but the difficulties involved in defining addition
a very con-
and multi-
plication are formidable (how do you add infinite decimals without
worrying about carrying digits infinitely far out?). A reasonable approach
is
outlined below; the trick
to use least
is
upper bounds right from the
start.
(a)
Define
have
(a,
{b n })
< dn
bn
<
(c,
but b s
=
bound
the least upper
{dn
d3
-
}) if
for
1
< c, or
< j < n.
a
if
a
=
Using
and
c
some
for
n
we
prove
this definition,
property.
k
(b)
Given a
=
a
+ ^
n-
bn
r
and
=
j8
0.
^ nl0
6 n for
=
(c,
a
(the least
ak =
a
+ £
n=
£ n lCT
>
1
all
Conversely, given a rational
~n
<
let
«
r'
<
A
n
£n'
=
0 for «
intuitively,
>
r
is
of the form
f
(a,
where
{b n }),
Now for a =
(a,
{b n })
\dn \) define
+0
=
sup {(a*
upper bound
exists
+
&)': A a natural number}
by part
(a)). If
multiplication
similarly, then the verification of all conditions for a field
forward
ak
decimal places after
number
denote th e real number
and
;
1
number obtained by changing
the rational
the kth to
{£„}), define
(a,
task,
not highly recommended.
is
is
defined
a straight-
Once more, however,
existence of multiplicative inverses will be the hardest.
^
CHAPTER
UNIQUENESS OF THE REAL NUMBERS
We shall now revert to the usual notation for real numbers, reserving boldface
fields which may turn up. Moreover, we will regard integers
numbers as special kinds of real numbers, and forget about the
specific way in which real numbers were defined. In this chapter we are
interested in only one question: are there any complete ordered fields other
than R? The answer to this question, if taken literally, is "yes." For example,
the field F z introduced in Chapter 25 is a complete ordered field, and it is certainly not R. This field is a "silly example because the pair {a, a) can be
symbols for other
and
rational
5
'
regarded as just another
name
+
(a, a)
(a, a)
•
for the real
number
(A, b)
=
{a
+
(6, b)
«
(a
•
b,
a
b9 a
•
+
a; the
operations
b),
b)>
are consistent with this renaming. This sort of example shows that any
intelligent consideration of the question requires some mathematical means
of discussing such renaming procedures.
If the elements of a field F are going to be used to
rename elements of R,
R there should correspond a "name" f(a)
then for each a in
in F.
The notation
functions. In order
f(a) suggests that renaming can be formulated in terms of
general than any
more
much
function
of
to do this we will need a concept
general notion
most
the
require
will
we
in
fact,
now;
until
which has occurred
of "function" used in mathematics.
A function,
in this general sense,
is
simply
a rule which assigns to some things, other things. To be formal, a function is a
collection of ordered pairs (of objects of any sort) which does not contain two
distinct pairs with the same first element. The domain of a function / is the
set A of all objects a such that (a, b) is in /for some b; this (unique) b is denoted
by /(a). Iff (a) is in the
to B. For example,
=
if f(x)
sin x for all x in
function from
if f(z)
if f(z)
z for
s in
e
z
R
to R;
all
R
it is
all a in
(and /
6
is
a function from \z in C: z
if
/ is
the collection of
to
is
is
called a function
from A
defined only for x in R), then /
is
from
R
[— 1,
to
a function from
a function from
C
is
a
1];
to C;
C to C; it is also a func-
to \z in C: z ?± 0};
from
R
is
z in C, then /
for all z in C, then /
C
A, then /
also a function
tion
from
507
=
=
B for
set
F
3.
all
pairs
^
0} to {x in R: 0
(a, (a,
a)) for a in
<
x
<
R, then /
2t};
is
a function
—
:
508
Epilogue
Suppose that F\ and F2 are two fields; we will denote the operations in Fi
by ©j O, etc. and the operations in F2 by +, •, etc. If F2 is going to be considered as a collection of new names for elements of Fu then there should be a
function from Fi to F2 with the following properties:
The
(1)
function / should be one-one, that
have fix) ^ f(y);
same name.
(2)
The
this
function / should be "onto," that
some x in Fi such that z
there should be
(3)
is,
if
x
^ y,
then
means that no two elements of
is,
=
for every
f(x)
;
element of F2 is used to name some element of
For all x and y in Fi we should have
F
we should
x
have the
element z in
F
2
means that every
this
Fi.
=f(x)+fiy),
fixOy) =fix)-f(y};
f(x (By)
means that the renaming procedure
this
is
consistent with the opera-
tions of the field.
If
we
are also considering Fi
and
F
2
as ordered fields,
we add one more
requirement:
(4)
A
lfx©y
then /(*)</(>).
function with these properties
This definition
DEFINITION
}
is
so important that
is
we
called
an isomorphism from Fi to
restate
it
If Fx and F2 are two fields, an isomorphism from Fi
from Fi to F2 with the following properties:
(1)
Ux^y,
(2)
If
(3)
If
F
2.
formally.
to
F
2 is
a function /
then fix) * f(y).
= fix) for some x in Fi.
2) then z
x and y are in Fi, then
z
is
in
F
f(x®y) =f(x)+f(y),
f(xOy) =f(x)-f(y).
If
Fx and
(4)
F
If
2
x
are ordered fields
€> y,
then /(*)
we
also require
< fiy).
The fields Fi and F2 are called isomorphic if there is an isomorphism
between them. Isomorphic fields may be regarded as essentially the same any
important property of one will automatically hold for the other. Therefore,
we can, and should, reformulate the question asked at the beginning of the
chapter; if F is a complete ordered field it is silly to expect F to equal R
rather, we would like to know if Fis isomorphic to R. In the following theorem,
and "positive elements" P; we
and
F will be a field, with operations
write a < b to mean that b
a is in P, and so forth.
—
—
+
Uniqueness of the Real Numbers
THEOREM
PROOF
F is
If
a complete ordered
Since two
F is
then
field,
isomorphic to R.
are defined to be isomorphic
fields
509
if
there
an isomorphism
to F which is an
is
between them, we must actually construct a function /from R
isomorphism.
We
begin by defining / on the integers as follows:
M
=
=
f(n)
0,
+
1
•
+1
•
•
>
for n
0,
n times
+
= -(1
f(n)
•
•
jn|
+ 1)
•
<
for n
0.
times
easy to check that
It is
f(m + n)
f(m-n)
=
=
f(m)
+ f(n),
f(m)-f(n),
m and n, and it is convenient to denote
rational numbers by
the
on
/
for all integers
define
f(m/n)
(notice that
so
n" 1 =
•
•
•
•
+1^0ifn>0,
makes sense because
definition
m
1+
ft
•
t"
1
.
It
is
The
any
definition of
real
number
is
/W
m/n =
then
n" 1
•
since
F is
an ordered
then ml
k/l,
+ u) =
fin)
=
/(ri)
/(ri
numbers
if
We
n.
=
nk so
9
field).
m
•
=
I
This
fc
•
n,
easy to check that
f(n
for all rational
m
= m/n =
by
/(n)
ri
'
and
n)
r 2,
+ /(r
•
and that
for arbitrary x
is
2 ),
/(n),
/(ri)
<
/(r 2 ) if ri
based on the
now
determined by the rational numbers
<
r2 .
familiar idea that
less
than
it.
For any
numbers
let A x be the subset of F
The set A x is certainly not empty, and it is also bounded above, for if
>/(r) for all/(r) in A x Since F
r 0 is a rational number with r 0 > x, then/(r 0 )
upper bound; we define /(*)
least
is a complete ordered field, the set A x has a
consisting of all f(r), for all rational
x in R,
r
<
x.
.
as
sup
Ax
.
have /(*) defined in two different ways, first for rational x, and
that these two
then for any x. Before proceeding further, it is necessary to show
rational number, we
definitions agree for rational x. In other words, if x is a
We now
want
to
show that
sup
Ax =
f{x),
depends
where /(x) here denotes m/n, for x = m/n. This is not automatic, but
required.
thus
is
digression
slight
on the completeness of F; a
Since
F is
complete, the elements
1
+1
+
n times
for natural
numbers
n
which is not bounded above; the proof is exactly the same as the
(Theorem 8-2). The consequences of this fact for R have exact
analogues in F: in particular, if a and b are elements of F with a < b then
form a
set
proof for
R
y
there
is
a rational number
such that
r
< /(r) <
a
Having made
we
return to the proof that the two definitions
a rational number with y < x, then we have
Thus every element of A x is < f(x). Consequently,
this observation,
of f(x) agree for rational
already seen that f(y)
x. If y is
< f(x).
sup
On
sup
there
But the condition /(r)
Ax
Ax
< /(*) means
this clearly contradicts the
\
A x < f(x).
would be a rational number
sup
set
A x < f(x).
we had
the other hand, suppose that
Then
the original assumption
b.
is false,
such that
r
<f(r) </(*).
that
<
r
x,
which means that f(r) is in the
A x < f{r). This shows that
condition sup
so
sup
Ax =
/(*).
thus have a certain well-defined function / from R to F. In order to
verify conditions (l)-(4) of the defini-
We
show that / is an isomorphism we must
We
tion.
If
will
begin with
(4).
x and y are real numbers with x
< y,
then clearly
Ax
is
contained in
Ay
.
Thus
/(*)
To
r
=
sup
Ax <
sup
Av =
f(y).
rule out the possibility of equality, notice that there are rational
and
s
numbers
with
x
We know that f(r) < f(s).
<
r
<
s
< y.
It follows that
/w</w</w</w.
This proves (4).
Condition (1) follows immediately from
or y
<
x; in
the
either case /(*)
first
case /(*)
< f(y), and
(4): If
x
^ y,
then either x
<
y
in the second case f(y) </(*); in
^ f(y).
prove (2), let a be an element of F, and let B be the set of all rational
numbers r with f{r) < a. The set B is not empty, and it is also bounded above,
because there is a rational number s with f(s) > a, so that/(j) >/(r) for r in
B, which implies that s > r. Let x be the least upper bound of B; we claim
To
Uniqueness of the Real Numbers
that f(x)
=
a.
In order to prove
this
it
/(*)
<
511
suffices to eliminate the alternatives
<h
«</(*).
In the
first
would be a rational number
case there
/(*)
But
x
=
this
means that x <
<
and that
r
This implies that
r
<
x.
Since x
<
a.
in B, which contradicts the fact that
would be a rational number r with
r is
sup B. In the second case there
a
f(r)
with
r
< fir) < fix).
= sup B, this means that r <
s for
some j in 2?.
Hence
<fis)<a,
fir)
again a contradiction. Thus f(x) = a, proving (2).
To check (3), let x and y be real numbers and suppose that fix
/(*)
+ fiy)-
Then
In the
first
this
would mean
+ y)
r
Then, using the
=
facts
f(r)
+> <
a contradiction.
Finally,
if a:
ri
+r
2,
sum
where x
checked about /for
=
The
number
r
< /(*
+ y).
such that
<m </(*)+/W-
could be written as the
r
+ /GO
/(*)
that
^
Therefore
or
case there would be a rational
fr.
But
5*
either
+ y)< /« + /GO
/(*
+ y)
fin
+
n)
other case
=
/(ri)
is
r.
of
two rational numbers
<
ti
and
rational
+/(ri)
handled
<
r2 .
numbers,
it
would follow that
> /(*) +/W,
similarly.
and^ are positive real numbers,
the
same
sort of reasoning
shows
that
/<*-j0
the general case
is
-/W-/W;
then a simple consequence.
|
This theorem brings to an end our investigation of the real numbers, and
any doubts about them: There is a complete ordered field and, up to
isomorphism, only one complete ordered field. It is an important part
of a mathematical education to follow a construction of the real numbers in
detail, but it is not necessary to refer ever again to this particular construction.
It is utterly irrelevant that a real number happens to be a collection of rational
numbers, and such a fact should never enter the proof of any important
resolves
theorem about the real numbers. Reasonable proofs should use only the fact
numbers are a complete ordered field, because this property of
the real numbers characterize them up to isomorphism, and any significant
mathematical property of the real numbers will be true for all isomorphic
fields. To be candid I should admit that this last assertion is just a prejudice
of the author, but it is one shared by almost all other mathematicians.
that the real
PROBLEMS
1.
(a)
(b)
2.
F
Let / be an isomorphism from Fx to
2.
Show that /(0) = 0 and /(I) = 1. (Here 0 and 1 on the left denote
elements in Fx, while 0 and 1 on the right denote elements of F2 .)
Show
Here
that
=
/(-a)
=
-/(a) and /(a- 1 )
f(a)~\ for a
*
0.
an opportunity to convince yourself that any significant property
is shared by any field isomorphic to it. The point of this problem
is to write out very formal proofs until you are certain that all statements
of this sort are obvious. Fi and F2 will be two fields which are isomorphic;
for simplicity we will denote the operations in both by
and •. Show that:
is
of a field
+
(a)
If the
in
F
2
equation x 2
with a 0y
.
.
equation x n
2
(c)
(d)
+
1
in
F
If
F
Fh
0 has a solution in
then
it
has a solution
,
<2
*
*
n _ x in
+ b n ^x
+1
•
•
xn
+
<z
n __i
Fi has a root in
~
l
+
•
•
+b
•
0
•
xn
Fu
— 0
~l
+
+a
•
•
•
0
— 0
then every polynomial
with
b 0,
.
.
.
in
,
F
2.
(summed m
times)
=
0 in
F
ly
then the same
is
true
2.
and
x
/(*)
3.
.
has a root in
If
=
polynomial equation x n
(b) If every
F
+1
.
<
F
2
are ordered fields (and the isomorphism
fiy) f° r x
<
y)
an d Fi
is
complete, then
F
2 is
/
satisfies
complete.
Let / be an isomorphism from Fx to F2 and g an isomorphism from F2 to
Fz Define the function g of from Fi to F3 by (g © /)(#) = g(f(x)). Show
that g o / is an isomorphism.
.
4.
Suppose that Fis a complete ordered
/ from
R
to F.
Show
that there
is
field, so
is
an isomorphism
F = R, this is Problem 3-17. Now if / and g are
isomorphisms from R to F consider g~ ° f.
Find an isomorphism from C to C other than the identity function.
to F. Hint:
In case
l
5.
that there
actually only one isomorphism from
R
two
SUGGESTED
READING
A man
ought
to
read
just as inclination leads him;
for what he reads as a task
will do
him
little
SAMUEL JOHNSON
good.
.
purpose of this bibliography is to guide the reader to other
sources, but the most important function it can serve is to indicate the variety
of mathematical reading available. Consequently, there is an attempt to
One
achieve diversity, but no pretense of being complete. The present plethora of
mathematics books would make such an undertaking almost hopeless in any
case, and since I have tried to encourage independent reading, the more
standard a text, the less likely it is to appear here. In some cases, this philosophy may seem to have been carried to extremes, as some entries in the list
cannot be read by a student just finishing a first course of calculus until several
years have elapsed. Nevertheless, there are many selections which can be read
now, and I can't believe that it hurts to have some idea of what lies ahead.
One of the most elementary unproved theorems mentioned in this book is
the fact that every natural number can be written as a product of primes in
only one way. A proof of this basic theorem will be found near the beginning of
almost any book on elementary number theory. Few books have won so
enthusiastic
[1]
An
an audience
as
Introduction to the Theory of
Numbers (third edition), by G. H. Hardy
Press, Oxford, 1960.
and E. M. Wright; Clarendon
The Pergamon
Press publishes a series, Popular Lectures in Mathematics,
with several worthwhile
[2]
A
Selection of
Problems
millan (Pergamon),
Finally, I will
[3]
mention a
Three Pearls
of
among them
titles,
in the
New
Theory of Numbers,
book which
little
Number
by W.
Sierpinski;
Mac-
York, 1964.
Theory,
I
hope
is still
in print:
by A. Khinchin; Graylock
Press,
Rochester, N.Y., 1952.
The
subject of irrational
analysis.
An
numbers
straddles the fields of
excellent introduction will be found
[4] Irrational
Numbers, by
I.
M.
number theory and
in
Niven; Wiley,
New
York, 1956.
Together with many historical notes, there are references to some fairly
elementary articles in journals. There is also a proof that ir is transcendental
(see also [53]) and, finally, a proof of the "Gelfond-Schneider theorem":
and
If a
b
are algebraic, with a
^
0 or
1,
and
b
irrational,
is
then a h
is
transcendental
All the books listed so far begin witn natural numbers, but whenever necessary take for granted the irrational numbers, not to mention the integers and
rational numbers. Several recent books present a construction of the rational
numbers from the natural numbers, but it is hard to believe that there can
be a more lucid treatment than the one
[5] Foundations of Analysis,
Incidentally, the original
[6]
to be
found in
by E. Landau; Chelsea,
German
New
York, 1951.
edition,
Grundlagen der Analysis (fourth edition),
by
E.
Landau; Chelsea,
New
York, 1965.
now available in paper back, together with a complete German-English
dictionary (of about 300 words) for the whole book an excellent way to begin
is
—
The basic idea for constructing the real
derived from Dedekind, whose contributions can be found in
reading mathematical German.
numbers
is
Essays on the Theory of Numbers,
[7]
by R. Dedekind; Dover,
New
York,
1963.
While many mathematicians are content to accept the natural numbers
as a natural starting point, numbers can be defined in terms of sets, the most
basic starting point of all. A charming exposition of set theory can be
found in a sophisticated little book called
Naive Set Theory, by P. R. Halmos;
[8]
Van
Nostrand, Princeton, N.J.,
1961.
Another very good introduction
Kamke; Dover, New York,
Theory of Sets, by E.
[9]
is
1950.
Perhaps it is necessary to assure some victims of the "new math" that set
theory does have some mathematical content (in fact, some very deep
theorems). Using these deep results, Kamke proves that there is a disconthose
tinuous function / such that f(x
y) = /(*)
f(y) for all x and y. For
+
+
who
first
enjoy reading the classics, the most important notions of
introduced by Cantor, whose work is reproduced in
set
theory were
[10] Contributions to the Founding of the Theory of Transfinite Numbers,
Cantor; Dover,
Inequalities,
New
which were treated
actually form a specialized
by G.
York, 1952.
field.
an elementary topic in Chapters 1 and 2,
good elementary introduction is provided
as
A
by
[11] Analytic Inequalities,
New
by N. Kazarinoff; Holt, Rinehart and Winston,
York, 1961.
mean is less than or equal to the
principle, can be found in the
different
a
each
based
arithmetic mean,
on
beginning of the more advanced book
Twelve
different proofs that the geometric
[12] Inequalities,
by
E.
Beckenbach and R. Bellman; Springer,
New
York,
1961.
The
classic
work on
inequalities
is
[13] Inequalities (second edition),
by G. H. Hardy,
G. Polya; Cambridge University Press,
New
J.
E. Littlewood,
of the authors of this triple collaboration has provided his
Each
and
York, 1952.
own
con-
tribution to the sparse literature about the nature of mathematical thinking,
written from a mathematician's point of view.
[14]
A
Mathematician's Apology, by G.
Press,
New
York, 1940.
My favorite
is
H. Hardy; Cambridge University
Suggested Reading
517
Littlewood's anecdotal selections are entitled
[15]
A
Mathematician's Miscellany,
Polya's contribution
is
pedagogy
by
E. Littlewood;
J.
at the highest level:
by G. Polya (Vol.
[16] Mathematics and Plausible Reasoning,
Analogy
in
Methuen, 1953.
I: Induction
and
Mathematics; Vol. II: Patterns of Plausible Inference); Princeton
University Press, Princeton, N.J., 1954.
which can be considered as background
a masterful mathematical work, but
preparation has been made. Probably
some
until
postponed
be
perhaps
should
the best modern work on "classical geometry" is
Geometry
is
the other
main
field
for calculus. Euclid's Elements
is
still
[17] Elementary Geometry from an Advanced Standpoint,
by E. Moise; Addison-
Wesley, Reading, Mass., 1963.
This beautiful book provides excellent historical perspectives and contains a
thorough discussion of the role of the "Archimedean axiom" in geometry; in
which the Archimedean
geometry books, all sorts of
addition, Chapter 28 describes an ordered field in
axiom does not hold. Speaking
of beautiful
fascinating things can be found in
[18] Introduction
to
Geometry,
by H.
S.
Coxeter; Wiley,
New
York, 1961.
Almost all treatments of geometry at least mention convexity, which forms
another specialized topic. I cannot imagine a better introduction to convexity,
or a better mathematical experience in general, than reading and working
through
[19] Convex Figures,
by
I.
New
and Winston,
M. Yaglom and W.
G. Boltyanskii; Holt, Rinehart
York, 1961.
This book contains a carefully arranged sequence of definitions and statements
of theorems, whose proofs are to be supplied by the reader (worked-out proofs
are supplied in the back of the book). Another geometry book has been
modeled on the same
principle:
[20] Combinatorial Geometry in the Plane,
Holt, Rinehart and Winston,
by H. Hadwiger and H. Debrunner;
New
York, 1964.
Along with these two out-of-the-ordinary books,
valuable
little
book, also of a specialized
[21] Counterexamples in Analysis,
Day, San
Many
by
B.
I
might mention an extremely
sort,
Gelbaum and
J.
Olmsted; Holden-
Francisco, 1964.
book come from more advanced topics in
few can be appreciated by someone who knows calculus.
of the examples in this
analysis, but quite a
Of
calculus books
A
[22]
mention only two, each something
I will
Course of Pure Mathematics (tenth edition),
bridge University Press,
New
of a classic:
by G. H. Hardy; Cam-
York, 1952.
by R. Courant; Vol. I, second edition,
Wiley (Interscience), New York, 1936.
[23] Differential and Integral Calculus,
1937; Vol. II,
Courant
is
Volume
I
first
edition,
especially strong on applications. In addition, the latter parts of
contain material usually found in advanced calculus, including
differential equations and Fourier series. An introduction to Fourier series
(requiring a little advanced calculus) will also be found in
An
[24]
Introduction to Fourier Series
New
jamin,
The second volume
tional material
and
by R. Seeley; W. A. Ben-
Integrals,
York, 1966.
on
of
Courant (advanced calculus
in earnest) contains addi-
differential equations, as well as
calculus of variations.
an introduction
to the
not mention any books devoted solely to the
calculus of variations, since they are (of necessity) quite difficult. Innumerable
I will
books on differential equations have been written, but none of the elementary
ones seems to generate much enthusiasm. There is a somewhat more advanced
book, however, which is universally admired:
Ordinary Differential Equations,
[25] Lectures on
Press,
by W. Hurewicz; M.I.T.
Cambridge, Mass., 1958.
bypass the more or less standard advanced calculus books (which can
be found by the reader himself) since nowadays there is a move-
I will
easily
ment
to revise the
linear algebra.
whole presentation of advanced calculus, basing it upon
of the first, and still one of the nicest, treatments of
One
advanced calculus using linear algebra
[26] Calculus of Vector Functions,
Prentice-Hall,
Englewood
is
by R. H. Crowell and R. E. Williamson;
Cliffs, N.J.,
1962.
Several recent books on advanced calculus attempt to acquaint undergraduates with very large areas of modern mathematics.
favorite, of
My
course,
is
[27] Calculus on Manifolds,
by M. Spivak; W. A. Benjamin, New York, 1965.
There are three other topics which are somewhat out of place in this
bibliography because they are rapidly becoming established as part of a
standard undergraduate curriculum. The purposeful study of fields and
related systems
is
the
domain
of "algebra."
The
first
undergraduate text on
modern algebra was
[28]
A
Survey
of
Modern Algebra (third edition), by G. BirkhofT and
MacLane; Macmillan, New York,
1941.
S.
519
Suggested Reading
One
of the strongest competitors
by
Topics in Algebra,
[29]
is
now
N. Herstein; Ginn
I.
(Blaisdell), Boston, Mass.,
1963.
is meant to be intermediate in
Modern
Algebra
and one of the great classics:
of
Herstein's book
[30]
difficulty
Modern Algebra, by B. L. van der Waerden; Ungar,
between
New
A
Survey
York, 1953.
the way, this book contains a proof of the partial fraction decomposition
By
of a rational function.
One
of the standard texts in
complex analysis
[31] Complex Analysis (second edition),
New
I
am
is
by L. V. Ahlfors; McGraw-Hill,
York, 1966.
personally looking forward to the publication of
[32] Lectures on the Theory of Functions of a Complex Variable,
Van
by G. W. Mackey
Nostrand, Princeton, N.J., 1967.
Finally, the study of topology, although never
mentioned before, has really
background of many discussions, since it is the natural generalization of the ideas about limits and continuity which play such a prominent role
in Part II of this book. There are now many elementary books on topology,
but I am dismayed by the prospect of reading and evaluating them all, so I
will merely call attention to the fact that they exist, and warn that some are
been
in the
not very good.
next few topics, ranging from elementary to very difficult, are included
have been alluded to in the text. The proof
that a nondecreasing function is differentiable at almost all points (and an
explanation of just what this means) receives a beautiful exposition in
The
in this bibliography because they
[33] Functional Analysis,
by
F. Riesz
and
B. Sz.-Nagy;
Ungar,
New
York,
1955.
book moves on to quite advanced
little book devoted entirely to
its properties, most of them proved by using the theorem of Bohr and Mollerup
which was mentioned in Problem 18-25:
(After this elementary beginning, the
material.)
[34]
The gamma
The
Gamma
function has an elegant
Function,
by E. Artin; Holt, Rinehart and Winston,
New
York, 1964.
The gamma
function
is
only one of several important improper integrals in
x%
mathematics. In particular, the calculation^ jj e~ dx (see Problem 18-27)
is
important in probability theory, where the "normal distribution function"
:
plays a fundamental role.
The
following book has already
become something
of a classic:
An Introduction to* Probability Theory and Its
by W. Feller; Wiley, New York, 1957.
[35]
Applications (second edition),
certain functions in elementary terms (among
one of the most esoteric subjects in mathematics. An
interesting discussion of the possibilities of integrating in elementary terms,
with an outline of the impossibility proofs, and references to the original
papers of Liouville, will be found in
The impossibility of integrating
them
=
f(x)
The
[36]
e~x *)
is
Variable (second edition),
Integration of Functions of a Single
G. H. Hardy; Cambridge University Press,
A complete
New
by
York, 1958.
presentation of the impossibility proofs will be found in
[37] Integration in Finite Terms,
by
J.
Ritt;
Columbia University
Press,
New
York, 1948.
Oddly enough, a
more
related but seemingly
difficult
problem has a much
xy — 0 is a
example) whose solutions cannot be expressed even in terms of
indefinite integrals of elementary functions. This fact is proved on page 43
of the (60-page) book:
neater solution. There are simple differential equations (y"
+
specific
An
[38]
Introduction to Differential Algebra,
by
I.
Kaplansky; Hermann, Paris,
1957.
To
read
(Just
this
book you
will
need to know quite a
bit of algebra,
however.
equally algebraic and polished treatments of Liouville's
recently,
theorems have been obtained, and presumably they will be published
soon.)
A few words should also be said in defense of the process of integrating in
elementary terms, which many mathematicians, look upon as an art (unlike
differentiation,
which
is
merely a
You
skill).
are probably already aware that
the process of integration can be expedited by tables of indefinite integrals.
For those who enjoy perusing tables there
is
a really beautiful collection, that
and a great deal
you should ever happen to need the value of the thirty-fourth
Bernoulli number, this is the place to look)
includes indefinite integrals, definite improper integrals,
more
besides
[39]
(if
Tables of Integrals, Series, and Products,
Ryzhik; Academic Press,
For the
[40]
thrifty,
there
is
New
by
I. S.
Gradschteyn and
I.
W.
York, 1965.
a paperback table of integrals:
Tables of Indefinite Integrals,
by G.
Petit Bois; Dover,
New
York, 1961.
521
Suggested Reading
The remaining
references are of a
three categories, of which the
referred to in
will
different sort.
The
letter of
They
fall
into
H. A. Schwartz
be found in
Ways of Thought of Great Mathematicians, by H. Meschkowski; HoldenDay, San Francisco, 1964.
[41]
Some
Problem 11-49
somewhat
first is historical.
historical remarks,
and an attempt
to incorporate
them
into the teaching
of calculus, will be found in
[42]
The Calculus: A Genetic Approach, by O. Toeplitz; University of
Chicago Press, Chicago, 111., 1963.
Most textbooks on the
An
history of mathematics are both superficial
admirable exception
[43]
An Introduction
and Winston,
and
dull.
is
to the
History of Mathematics,
New
York, 1964.
by H. Eves; Holt, Rinehart
Three good scholarly works are
[44] History of Analytic Geometry,
'
[45]
A
Press,
New
York,
by C. Boyer;
History of the Calculus, and Its Conceptual Development,
Dover,
[46]
by C. Boyer; Academic
1965,
New
York, 1959.
The Mathematics of Great Amateurs, by
J.
Coolidge; Dover,
New
York,
1963.
A
recent book presents philosophical ruminations on the history of matheis all the more impressive
matics by an author whose exceptional erudition
in a first-rate mathematician:
[47]
The Role of Mathematics
in the Rise of Science,
by
S.
Bochner; Princeton
University Press, Princeton, N.J., 1966.
Finally, extracts
[48]
A
Source
from original sources
Book
in
will
be found in
Mathematics (2 vols.), by D. Smith; Dover,
New
York,
1959.
large number of books listed here might easily give too hopeful a view of
the availability of historical information. Although the very early origins of
mathematics are receiving the most careful scrutiny, no one seems to care very
The
much about the origins of calculus. For example, it is almost impossible to find
out who first proved the Mean Value Theorem (according to the Encyhlopddie
der Mathematischen Wissenschaften, Volume II, it was O. Bonnet, whose name
Gauss-Bonnet
Theorem"). Similarly, nearly every history book tells us that Wallis proved
Wallis's formula by a "complicated method of interpolation," but almost
none bothers to mention what it was, even though it inspired Euler's investiga-
is
familiar to students of differential geometry from the
6
'
:
gamma
tions of the
function (a description
is
given in the answer book, along
with the solution to Problem 18-26).
The second category
in this final
group of books might be described as
"popularizations." The
editorials notwithstanding, there are
a surprisingly large
ones by real mathematicians:
[49]
[50]
New York Times
number of first-rate
What is Mathematics? (fourth edition), by R. Courant and H. Robbins;
Oxford University Press, New York, 1947.
Geometry and the Imagination, by D. Hilbert and S. Cohn-Vossen.
Chelsea,
New
York, 1956.
The Enjoyment of Mathematics, by H. Rademacher and O. Toeplitz;
Princeton University Press, Princeton, N.J., 1957.
[52] Famous Problems of Mathematics (second edition), by H. Tietze; Graylock Press, Rochester, N.Y., 1965.
[51]
One
of the
most renowned "popularizations"
is
especially concerned with the
teaching of mathematics:
[53] Elementary Mathematics from an Advanced Standpoint,
Arithmetic, Algebra, Analysis; vol. 2: Geometry)',
by
Dover,
F. Klein (vol.
New York,
1
1948.
Volume 1 contains a proof of the transcendence of x which, although not so
elementary as the one in [4], is a direct analogue of the proof that e is transcendental, replacing integrals with complex line integrals. It can be read as
soon as the basic facts about complex analysis are known.
The third category is the very opposite extreme original papers. The
difficulties encountered here are formidable, and I have only had the courage
to list one such paper, the source of the quotation for Part IV. It is not even
—
you do have a choice of foreign languages. The
French is in
in English, although
in the original
[54] Oeuvres
New
Competes
a" Abel;
article
Christiania. Johnson Reprint Corporation,
York, 1965.
appeared in a German translation in the Journal fiir die reine und
angewandte Mathematik (1) 1826. To compound the difficulties, these references
will usually be available only in university libraries. Yet the study of this
paper will probably be as valuable as any other reading mentioned here. The
reason is suggested by a remark of Abel himself, who attributed his profound
knowledge of mathematics to the fact that he read the masters, rather than
It first
the pupils.
ANSWERS
TO SELECTED
PROBLEMS
CHAPTER
1
1.
=
1
(i)
=
2
If x
(iii)
3.
One
,
.y
-
=
a' 1 (ax)
=
then 0
2
+y
or x
x
that
0,
2
l
(a
=
a)x
- j» -
1
•
=
x
x.
- (* + y), so either - y =
= —^ or x = j.
jy)
(iv).
—y =
by x
step requires dividing
0.
(iii)) = ac/bc.
)
(by
bc)(b~^)
=
=
bc){bd)+
(ad
+
+ fc)/(W)
= a^r + aT = <z/4 + c/d.
l
^(a"^" ) « (a a" ) (4 r ) - 1, so a- 6 - («*)
= (« ^X*
(a lb) I (c/d) = (a/bKc/d)'
=
= ( r )(c-1 <0 =
= M/(4<).
(i)
*/*
(ii)
(flrf
=
=
ab-i
=
l
(ae)(b^e~
(ac)(bc)^ (by
1
1
1
1
1
•
•
l
4.
(i)
*<-l.
(iii)
*
(v)
All
2*
7.
+
2
=
b
+
-
(*
l)
2
+
1.
(iii)
(xiii)
(a
(v)
+
(b
-
ac
>
c)
is
in P.
Thus,
d
>
a
+
c.
-c < -d; then (i) implies that a + (-*) < b + (-<*).
that is,
and -c are in />, so -c(b - a ) = ac - be is in P,
(ii),
fl )
be.
a
>
0 and a
<
so a 2
<
a.
(vii)
Using
(ix)
Substitute a for <;_and 4 for d in (viii).
(i)
(iii)
(
V)
(i)
(iii)
|a
(iii)
(v)
1,
4|
> -4 and 4 > 0;
< —4 and b < 0;
a + 2b if a > -4 and 4 < 0;
-a - 24 if a < -4 and 4 > 0.
a
if
x
-
a
if
<z
x2
-x (i)
(iv),
V2 + V3 - V5 + V7.
+ +>| - |a + *
V2 + V3 + V5 - V7.
—«
11.
-
2
a:
1
Using
(xi)
10.
V
< -
a:
W
'
fl<J
(i)
(ix)
9.
or
since
*,
*
*
-2 are the roots of x 2 - x - 6 = 0.
x > 3 or * < -2, since 3 and
* > 7T or -5 < x < 3.
x < 3.
x > 1 or 0 < x < 1.
- * are in P, so (4 - a) + (d - *) = (* + d)
& - « and
(vii)
5.
> Vl
-
1
•
•
fl
»
•
1
(v)
(hi))
1
1
(iii)
0
a:
(*
either *
is,
Replace > by -jy in
(vi)
2.
=
a" 1 a
if
*2
x
>
0;
if
*
<
0.
x = 11, -5.
-6 < x < -2.
No * (the distance from x to
plus the distance from x to
1
-1
is
at
least 2).
(vii)
12.
(i)
= 1, — 1.
.1,1111
and 1*1 w
= (W
since
= *V =
(M) 2 =
=
|xy|
that
proves
\y\.
|*|
are both > 0, this
(")) - I^T'I b y ®> = Ml\x\/\y\ = \x\
bl" = |*| Ir
x
M
WW
2
M
'
•
1
(iii)
525
1
(
•
!
'
526
Answers
to Selected
Problems
from
It follows
(v)
|*|
-
\y\
+y + z\ <
then \x + y\ =
\x
same sign
2
1.
that
Since
(i)
l
\z\
\x\
2
—
\y
—
(y
<
x)\
\y\
+
\y
1
•
+
•
•
•
+v+
one
(2
•
(2)
•
+
1
l)/6, the formula
true for
is
+
*(*
+
i)( 2 *
*)
+
[k(2k
= (A±ll
[{k
6
_
+
(*
l)(*
is
true for «
{k
+
Then
k.
+
1)
6
1)2
+ 6(* +
1)]
+ 2 )(2k + 3)]
+
2)(2[*
+
1]
+
1) ,
6
so the formula
+
true for k
is
1.
(0
(2i
-
1)
=
1
=
1
+3+5+
•
+2+3+
(2n)(2n + 1)
-
_
•
•
•
•
•
+ (2n - 1)
+ 2n - 2(1 +
n(n
2
=
4.
(a)
so
\z\.
\y\
=
£
x\,
0).
is
\y =
+
(k
\x\
6
2.
—
[y|,
(unless
=
2
\x
Suppose that the formula
i
=
|*|
If equality holds,
+ y\ + < + +
so # and y have the same sign. Moreover,
+
z must have the same sign as x + y, so x, y, and z must all have the
(vii)
chapter
(iv)
<\x- y\.
+
•
•
•
+ n)
1)
n\
Since
1
the formula
1
=
true for n
is
+r+
•
•
•
1.
+
-r
->
—r
2
1
l+r =
Suppose that
r
n
=
-
1
r
n+l
'
—
\
r
Then
1
+r+
*
•
•
+
r
n
+
r
n+1
=
1
=
1
=
— r n +l
n+l
r
1 — r
- r* +1 + r w+1 (l - 0
1 - r
- r w+1
\-
1
1
-
r
=
1.
Answers
to
527
Selected Problems
(b)
S =
+r+
r +
1
=
rS
•
•
•
•
•
•
+
+
r
r
n
+
n
r
n+1
.
Thus
-
J(l
r)
= J-
rS
=
1
-
r
1
-
+ 4k +
1,
r*
+1
,
so
From
(i)
+
(*
we
-
4
l)
=
k4
+
4k 9
6k 2
k
=
1,
.
.
.
,
n
obtain
n
(n
+
-
1)4
1
=
4
V
1
-
6
n
n
J P+
3
+
6
"("
+
^
A:
Y
4
*
+
so
n
+
(B
,_
1)
—r —«2 H
"
4
+
^-4^±i>- W
2
«2
3
=
2"
6
,
,
-*
4
From
(iii)
1
1
1
k
it
+
+
k{h
\
1)
we obtain
1
+
n
Li
1
+
k{k
l)
even or odd;
1 is either even or odd, in fact it is odd. Suppose n is either
=
1
1. In the first case n
then n can be written either as 2k or 2k
1) is
1 = 2{k
1
1 = 2A;
1 is odd; in the second case n
2A;
1 is either even or odd. (Admittedly, this looks
even. In either case, n
+
+
+
+
+
+
+
+
fishy,
Let
B
Then
but
set of all natural
be the
1
is
really correct.)
it is
in
B
y
and
/
+
1
is
numbers
in
B
if / is
/
such that n 0
in B, so
B
—
1
contains
+
/ is
all
in A.
natural
numbers, which means that A contains all natural numbers > n 0
b were rational, then b = (a + b) - a would be
(a) Yes, for if a
rational. If a and b are irrational, then a + b could be rational, for b
-
+
could be
(b)
If a
=
0,
r
—
some
a for
then ab
is
rational, for then b
rational
=
(ab)
number
a.
But if a * 0, then ab could not be
-1
would be rational.
a
rational.
(c)
Yes; for example, \^2.
(a)
V2
example,
(d) Yes; for
12.
— V2.
and
Since
2
(3„
Now
/?
+
+
9n 2
9n
follows that \ik 2
it
q
=
=
+ 1)2
+ 2)
(3„
suppose that
2
is
=
x
= 3( 3w 2 + 2w )
= 3(3rc + 4rc
+
+
2
Yin
4
divisible
by
3,
l
?
+
1)
1,
then £ must also be divisible by
3.
VI were rational, and let V3 = p/q where p and
have no common
must be. Thus, p
quently (3/0 2
+
+
6n
Then p 2 =
3# so /> is divisible by 3, so
3p' for some natural number p\ and conse-
factor.
=
3?
2
or 3(//)
,
=
2
2
?
2
2
,
Thus, 9
-
is
also divisible
by
3,
a
contradiction.
The same
+
+
+
+
(5n
(5 W
(5n
(Sn
2
l)
2
2)
2
3)
2
4)
work
proofs
=
=
=
=
+ lOn + = 5(5n + 2«) +
+ 2(k + 4 = 5(5rc + An) + 4,
+ 2>0n + 9 = 5(5« + 6n + 1) + 4,
+ AOn + 16 = 5(5n + + 3) +
2
25rc
2
2
2
and the corresponding equations
show that if k 2 is divisible by 5 or
V
for
this
a
is
4,
1,
2
2Sn 2
25rc
2
1
2
25rc
vi because the equations
and
for
because {An
+ 2)
2
1,
then k must be.
6,
by
The proof fails
(For precisely this reason
4.
Va
is irrational if
proof cannot be used to show that in general
m) 2 might
{an
that
not a perfect square we have no guarantee
—
+
not be a multiple of a for some m < a. Actually, this assertion
but the proof requires the information in Problem 16.)
(b)
+ m,
numbers of the form 6n
for
divisible
is
8/2
is
true,
Since
{In
+
l)
3
=
Sn*
+
\2n 2
+
+
6n
=
1
2{An s
+
6n 2
+
3n)
+
1,
follows that if k* is even, then k is even. If $2 — p/q where p and
had no common factors, then p 3 — 2q z so p z is divisible by 2, so p
must be. Thus, /? = 2p' for some natural number //, and conseit
#
,
quently
{2p')
z
=
2q\ or
4(//)
3
=
3
?
-
Thus, ?
is
also even, a contra-
diction.
The proof
(3„
(3„
+
+
1)3
2)
=
If n
18.
3
=
=
27n
1,
then
27w 3
3
for
+
+
(1
vi
27/2 2
54n
+
2
+
+
A)
n
is
similar, using the equations
+
36/z +
1
9rc
=
1
+
= 3(9« 3 + 9n 2 + 3n) + 1,
2
8 = 3(9rc + 18/z + 12n + 2)
3
Suppose that
nA.
(1
+
h)
n
>
+
1
2.
+
Then
(1
+
h
)n+i
=
=
(i
1
+ h){\ + A) n
+ (n + \)h +
>
nh 2
(1
>
+
1
A)(l
+
{n
+ nh),
+ l)h.
since
1
+h
>
0
»A.
Answers
For h
>
to Selected
529
Problems
the inequality follows directly from the binomial theorem,
n
are
h)
the other terms appearing in the expansion of (1
0,
+
since all
positive.
chapter
3
1.
0+
(i)
x
l)/0 + 2);
— 1 and x ^
1/(1
(v)
(*
Only
be
1,
since /(*)
0
(iii)
0.
(v)
-1,
(i)
{x:
(iii)
\x:
and
0,
=
rational, or y
(for x,
y
*
when
and
must
-1).
/(«:) implies that x
one x
true for at least
>
(i)
c
sense only
0).
if
=
makes
— 2.
+ cx) (for x * -l/c
+^ + 2)/(* + l)(y + 1)
(iii)
(vii)
the expression /(/(*))
>
=
cx,
this
0.
1
1.
—1 < <
and
x ^
at
1
1 }.
a:
^
2j.
(v)
4.
5.
11.
(i)
2 2y
(iii)
2 2sin<
(i)
Po
(iii)
S o S.
(v)
PoP.
(vii)
Sos°SoPoPoPos.
.
+
sin(2 f ).
S.
(a) y.
(b) H{y).
(c)
12.
H(y).
even
even
neither
odd
neither
odd
even
odd
even
even
odd
odd
odd
even
(b)
/ even
/ odd
g even
even
even
g odd
even
odd
(c)
(d)
odd
even
(a)
Let g(x)
=
/(*) for x
>
0
and define g
arbitrarily for x
<
0.
.
530
Answers
to Selected
Problems
21.
= 1 and let / be a function for which
Then/o(^ +
^fog + foh.
=
o
f]( X = (g + h)(f(x)) = *(/(*)) + *(/(*))
0
0
o
+ (A /) = [(5 /) + (* /)] W=
Let g(x)
(i)
/(l)
(ii)
°
(5
4
1.
2.
A)
W
A
V
•
/
Let g(x) = 2 and let/ be a function
Then l/(/o^) *fo(l/g).
/(£(*))
(i)
(2, 4).
(iii)
[2, 4].
(v)
(vii)
(-2, 2).
(-oo, 1]U[1,
(i)
All points below the
(iii)
All points below the graph of fix)
(v)
All points
(vii)
A
(ix)
All points inside the circle of radius
g) to/
(i)
(iii)
00).
= x.
= x2
between the graphs of f(x) = x
graph of f(x)
.
+
collection of straight lines parallel to the
The
The
(vii)
point
1
+y
2
—
=
*
(n,
—
1
= —x,
0) for integers n.
1
around
2x
+y =
2
(a, b).
(The important point about
remember the point slope form.)
The straight line through (a, b) and
so the equation follows
m' and
b
from part
b'
.
(c,
d)
m(x — a) + b = mx +
which goes through the
=
f(x)
a straight line with slope m,
When m =
—
since x 2
(1, 0),
is
ma)
point
(c)
f(x)
-\.
2
Simply observe that the graph of
(b
(b)
and
graph of fix)
(0, 0).
circle of radius
ix-\)
(a)
l//(2).
around (1, 2).
A square with vertices (1, 0), (0, 1), (-1, 0), and (0, -1).
The union of the graph of f(x) = x and of f(x) = 2 — x.
(v)
5.
^
which /(£)
for
intersecting the horizontal axis at the points
3.
^
)
W
t
/°£
(iv)
CHAPTER
/)
1- to - 77~rrr = (*M) =
7
(iii)
/(2)
yi)
+
[(£
h(x)
this exercise
has slope (d
is
—
simply to
b)/(c
—
a),
(a).
In that case, there
is
clearly
with fix) — g(x), while such a number x always
namely, x — (b — b)/(m — m').
no number x
exists if
m
m'
y
f
6.
(a)
If
B =
by
all
0
and A
points
(x,
then the set is the vertical straight line formed
with
x = — C/A. If B ^ 0, the set is the graph of
y)
5* 0,
/to = (~A/B)x
(b)
The
points
(x,
y)
+
(-C/A).
on the vertical
ones which satisfy
1
the graph of f(x)
=
•
+0
mx + b
*
•
jy
line with x
+ — a)
(
=
0.
— a are precisely the
The points (*, y) on
are precisely the ones which satisfy
i-m)x + 1 y+ {-b) = 0.
The graph of / is symmetric with respect to the vertical axis.
The graph of / is symmetric with respect to the origin. Equiva-
10.
(i)
(ii)
lently, the part of the
graph to the
left
of the vertical axis
is
Answers
(iii)
(iv)
19.
(a)
obtained by reflecting
first
through the horizontal
axis.
through the vertical
axis,
=
x*
=
(*
+
which
is
2
-+±
16
+ i)
2
,
y) satisfies this condition
(x,
- ay +
(*
-
(y
2
/3)
over.
x
the square of the distance from
The point
531
and then
The graph of / lies above or on the horizontal axis.
The graph of / repeats the part between 0 and a over and
The square of the distance from (#, x 2 ) to (0, \) is
2
(b)
Problems
to Selected
(x>
x 2) to the graph of g.
and only
if
= (y-y)
if
2
,
or
-
x2
lax
+
+y -
a2
2
2/3y
+£
2
=
y
-
2
+
a2
-
.2/3
2y)
-
\y
y2
/5
-y
-
2y
2
20
\
(3/
+
2yy
2
,
\
/
(3 ^ y, which is just the condition that
on L, then the solution is the vertical line
(This solution works only for
P
is
not on L.
P
If
is
through P.)
chapter
5
1.
= lim (x + 2x + 4) =
—2
x-»2
~ ^ = lim x n ~ + x n ~ +
lim —
y
2
lim
(ii)
=
(vi)
y
=
(V a
lim
It
is
If |*
\x*
+
ax
2
+
2
a x
—
+
a|
=
a4
<
a*\
1,
<
<
(*
-
then
(1
h
+ ^ n"
wy"" 1
-
.
+7^A +
h(V a +
+
h
1
Vfl)
Va)
by beginning with the equation
+ a* + a * + a
so
< +
+
+
+
+ \a\(l +
2
a)(* 3
1
|*|
+
+
\x\*
*)>
lim
possible to find 8
*4
+
k-+o
=
n~ 2
+
•
+ y n ~* =
x
A
(i)
•
+ y n' +
n~l
A-V^
Va +
lim
,
2.
2
l
(iv)
12.
X
x->2
\a\
\a\)
|*|
s
3
).
\a\,
2
•
2
|a|
2
|*|
\a\)
2
|a|
\a\
3
2
(l
+
\a\)
+
|a|
3
;
532
Answers
to
Selected Problems
we can choose
therefore
S
mm 0'
=
It
+
(I
+
\a\(l
+
\a\y
This shows that
\x
—
4
a
<
4
\
is
= min
1
(
—
By part
problem,
\x
—
1
(v)
Let 5
(i)
We
We
f-> ->
1,5
\2 4
=
£
need
<
0
<
1|
e/2.
2
,
-
i)>
—
<
£
1|
)
+
4
when
=
-> \2 2/
-
1/x)
According
-
2|
2|
——
-
8
<
since 0
i/(x)
|*
+
I
|(*
i)
<
2|
£
to parts
when
|1
and
(i)
A(ii)
1|
<
of this
happens when
this
< min
1|
1
(hi)
—
4
(a:
lemma.
1)
i)(H
|l/#
in
< min
1|
lemma,
(1) of the
e/2 and
+
'4(| a |^
lemma,
(3) of the
|#
(hi)
\a\)'
to use part (2) of the
|«+
+
2(W
By part
+
when
true
|,- a |< min^l,
(ii)
\a\)
when
£
2(| a
which
+
\a\*(l
and probably easier,
instructive,
is
+
\a\)>
|*|
<
•
2
<
= min
^
£
2
8.
\2 32/
<
implies that V|*|
-
e/2 and \g(x)
< min
=
(\>
2/
•
+ |,
(sin*
<
4|
~) =
e.
e/2, so
we need
5.
need
2
|£M-4|<min(H^- )>
so
we need
0
Let
=
/
<
\x
-
<
2|
=
[min(2, 8e)] 2
lim f(x) and define g(h)
=
/(a
+
5.
Then
h).
for every £
>
0
x—>a
there
is
a 5
Now,
if
0
<
>
0 such that, for
|A|
<
5,
then 0
all x, if
<
\(h
+
can also be written lim f(a
that
if
lim /(a
h-+Q
+ h)
—
+ h) =
/.
|*
-
a)
—
This inequality can be written
<
0
/|
-
<
<
a\
<
a\
£.
5,
=
then
sort of
— <
- <
I
\f(x)
+
h)
Thus, lim g(A)
The same
m, then lim f(x)
5,
so \f(a
l\
=
/,
which
argument shows
m. So either limit exists
x—*a
other does, and in this case they are equal.
e.
e.
if
the
Answers
9.
(a)
we can
Intuitively,
—
can get f(x)
/ as we like if and only if we
we like. The formal proof is so
of work to make it look like a proof at all.
get f(x) as close to
as close to 0 as
/
that it takes a bit
be very precise, suppose lim f(x)
trivial
To
533
Problems
to Selected
=
and
/
—
let g(x)
—
f(x)
I.
x-*a
Then
—
\x
<
—
\g(x)
tion
(b)
>
for all £
a\
<
0|
0 there
then
8,
a 8 > 0 such that, for all x, if 0 <
< s. This last inequality can be written
= 0. The argument in the other direc-
is
—
\f(x)
l\
so lim g(x)
£,
similarly uninteresting.
is
making x
Intuitively,
close to a
same as making x — a close
/, and let g(x) = f(x — a).
the
is
=
Suppose that lim f(x)
to 0. Formally:
x—> a,
Then for
<
<
+
so \f(y
8,
~
tion
<
l\
—
lim f(x)
<
<
l\
—
a)
Now,
£.
<
/|
0 such that, for
=
<
0
if
But
£.
So lim g(y)
£.
>
a 6
is
<
\y\
x
<
8,
|*
|*|
close to 0
is
For every
I.
have 0 <
<
|
we assume
if
>
l\
so |/0 3 )
8,
and only
if
£
-
then |/0)
3
other hand,
then for
|/(*
|/([\^]
(d)
Let
<
m\
-
3
)
=
/(a:)
>
£
all
-
)
a\
in the reverse direc-
there
0
a
is
<
£.
Then
-
l\
<
x3
if
<
0
if
>
<
|*|
£.
<
m|
1
or
£,
>
for x
if
0,
-
/.
8),
we
the
=
0
<
-
\f(x)
and
<
|*|
0
Thus lim
£.
= -
/(*)
0
if
we have
5 3,
<
m\
m,
x—>0
a 8 such that
is
if
On
min(l,
that lim f(x z ) exists, say lim /(* 3 )
0 there
Then
such that
0
Thus, lim f(x)
£.
Let
Formally:
is.
8
x—>0
3
—
\x
- a\
\(y + a)
<
inequality can be written
this last
The argument
I.
<
0
all x, if
then 0
8,
similar.
is
Intuitively,
0
0 there
-
then \f(x)
5,
\g(y)
(c)
>
all £
1
<
for x
<
<
<
|
=
then
5,
\Tx\
/(*)
0.
<
\x\
so
6,
m.
Then
lim f(x 2 )
2->0
=
15.
(a)
but lim /(*) does not
1,
The
=
function f(x)
1/* cannot approach a limit at
becomes
arbitrarily large near
be, there
is
<
+
(|/|
1/0 -
but
5,
£)).
This
a*
£)).
0
1)
In
fact,
|#|
<
0.
some x satisfying 0 <
= min(<5, \/{\l\ +
No matter what 5 >
x
(b)
exist.
5,
This x does not
may be,
>
+
there
£,
|/|
is
Problem
satisfy \\/x
some
—
9(b): lim 1/x
—
=
1)
=
min(l
-
/|
<
it
<
£.
—
\x
8, 1
£.
(It is also
1)
+
1
+
1/0 —
lim
<
l\
x satisfying 0
namely, x
does not satisfy \\/(x
possible to apply
since
0,
no matter what 8 > 0 may
butl/* > |/[ + £, namely,
if
1/
the
2-+1
latter exists, so this limit does
23.
(i)
This
is
not
exist,
because of part
the usual definition, simply calling the
and 8.
a minor modification
(a).)
numbers
8
and
£,
instead of £
(ii)
This
8
0
(in)
>
<
This
is
of
(i)
:
if
the condition
0,
then
it
applies to 8/2, so there
|*
-
<
£,
is
-
is
an
£
>
is
true for all
0 such that
<
8/2 < 8.
a similar modification: apply it to 8/5 to obtain
a|
then I/O)
l\
(i).
if
534
Answers
to Selected
Problems
(iv)
also a modification:
This
is
e/10
>
and
0,
it
same thing
says the
it
only the existence of some £
is
as
>
(i),
0 that
since
is
in
question.
27.
If
lim /(*)
=
5
28.
<
<5
a:
(i)
then for every
/,
>
£
0 there are 8 h 8 2
>
0
x—»a
such that, for
Let
=
lim /(*)
_
x—*a +
all x,
if
a
<
x
if
a
-
82
=
<
a or else
If
/
< a + Si,
< x < a,
min(Si 5 2 ).
=
\f{x)
\f(x)
—
|*
-
<
a|
£
all
-
£,
£.
—
so \f(x)
5i,
>
<
<
/|
l\
—
then either a
6,
a<#<0+5<<z +
lim /(*), then for
0 there
is
/|
a 5
>
<
/|
£ for
-
so |/(-*)
<
0
<
l\
<
x
-5 <
If
8.
x
Thus lim /(-*) =
e.
x
is
a
—
e.
0 such that
< -x <
<
0,
/.
Similarly,
then 0
if
and
close to 0
and has the same
exists
positive
if
and only
5,
lim /(*)
co-
x-^o-
then lim /O)
exists,
<
82
<
+
x-»0
I/O)
<
0
If
3
then
then
value. (Intuitively,
—a:
if
close to 0
is
and
negative.)
If
(ii)
/
=
lim f(x), then for
£
all
>
0 there
is
a 8
>
0 such, that
x->0 +
I/O)
<
-
<
/|
Thus lim
£.
(Intuitively,
If
(iii)
/
=
0
£ for
if
at
<
<
=
x
f(\x\)
So
8.
if
The
/.
\x\ is
>
£
all
<
|*|
reverse
close to 0, then
is
lim f(x), then for
0
<
close to 0
0 there
then
5,
|/(|*|)
direction
is
and
a 8
>
-
/|
similar.
is
positive.)
0 such that
x-*0 +
I/O) - l\ < £ for 0 < x < 8. If 0 < |*|
so |/0 2 ) - /| < £. Thus lim /0 2 ) = /.
< VS,
The
<
then 0
x2
<
reverse direction
5,
is
x—()
if
x
is
close to 0, then x
lim /0), then for every
£
>
0 there
similar. (Intuitively,
2
is
close to 0
and
positive.)
31.
If /
=
is
some
N such that |/0) —
N
< £ for x > N, and we can clearly assume that > 0. Now, if 0
1/N, then 1/x > N, so \f(l/x) - l\ < s. Thus lim /(l/*) =
reverse direction
chapter
CHAPTER
6
7
1
.
1.
+
FO) =
FO) =
(i)
Bounded above and below; minimum value
x
x
<
The
similar.
is
(iii)
(i)
<
/.
l\
2 for
0 for
all *.
all x.
0;
no
maximum
value.
(iii)
(v)
Bounded below but not above; minimum value 0.
Bounded above and below. It is understood that a >
-a a
+
1
<
a
+
2 for all * in
1). If
-1 <
(-a -
1,
a
a
<
+
\,
1),
then a
so a
+
—
1
(so that
< -a - 1, so /(*) =
2 is the maximum and
Answers
—\ <
to Selected
535
Problems
0, then /has the minimum value
minimum value 0. Since a + 2 >
only for [-1 - V5]/2 < a < [1 + V5]/2, when
{a + l)
a > — i the function / has a maximum value only for a <
[1 + \/5]/2 (the maximum value being a + 2).
Bounded above and below; maximum value 1 minimum value 0.
Bounded above and below; maximum value 1; minimum value
minimum
2
a
and
y
if
value. If
>
a
<
a
then / has the
0,
2
(vii)
;
(ix)
-1.
/ has a maximum and minimum value, since / is continuous.
n = -2, since /(-2) < 0 < /(-l).
n = -1, since /( — 1) = — 1 < 0 < /(0).
179
If /(*) = *
+ 163/(1 + * 2 + sin 2 a:), then / is continuous on
R and /(l) > 0, while /(-2) < 0, so f(x) = 0 for some * in
(xi)
2.
(i)
3.
(i)
5.
/
(iii)
(-2,
all
7.
1).
/ took on two different values, then / would take on
between, which would include irrational values.
constant, for
is
values in
if
= x;
= -x;
=
\x\;
f(x)
fa) = -|4
/(*)
(1)
(2) f(x)
(3)
(4)
10.
Apply Theorem
11.
If /(0)
<
=
/(l)
fix)
CHAPTER
8
1
.
=0
=
1
7(1),
* for
(i)
1
is
to
1
/
—
g.
choose x = 0 or 1. If /(0) > 0 = 7(0) and
then Problem 10 applied to / and I implies that
-
or /(l)
1,
some
the greatest element,
is
not in the
(iii)
1
is
(v)
Since
and the
and
the greatest element,
{x:
+x+ >
x2
greatest lower
bound
is
03
which
set.
=
0}
1
0
is
the least element,
R, there
is
no
least
upper bound or
greatest lower bound,
Since
(vii)
{x:
x
<
greatest lower
is
2.
(a)
0; neither
0
and * 2
bound
x - -
+
is [
1
1
belongs to the
<
V
0}
5]/2,
= ([-1 - Vs]/2,
and
the least
0),
the
upper bound
set.
A ^ 0, there is some x in A. Then — x is in — A, so —A ^ 0.
A is bounded above, there is some y such that y > x for all x in
A. Then -y < -x for all x in 4, so —y < z for all z in -^4, so
—4 is bounded below. Let a = sup (--4). Then a is an upper
Since
Since
bound
bound
for
Moreover,
for
-A,
bound
5.
(a)
If
x
/ is
+
1
—A,
so,
reversing the
argument just given,
—a
is
a lower
for A.
any lower bound for A, then — (3 is an upper bound
-/S > a, so 0 < -a. Thus -a is the greatest lower
if (3 is
so
for A.
the largest integer with
<
y.
So we can
let
£
/
=
<
/
x,
+
1 <
then /
1 > x } but /
(Proof that a largest such
+
1.
+
536
Answers
to Selected
Problems
integer
exists:
/
number
number
Since
<
of integers
ny,
+ V2
(c)
Choose
r
By part
(b),
10.
n.
—n <
<
I
Pick the largest.)
x.
-
(s
is
r)/2.
number r with x <
a rational
is
number
<
with x
s
r
<
s
<
y.
r < y, and
Apply part
0
>
<
(a)
(k
x
1
and
exists),
< x/a (the solution to Problem 5 shows that
= x — ka > 0. If x — ka = x > a, then
< x/a, contradicting the choice of k. So
+
l)a,
<
a.
+
k
so
f
let x'
1
Since any y in B satisfies y > x for
bound for A, so y > sup A.
all
x in A, any y in
Since x
<
sup
+
+
+
7
sup(A
9
1.
B
is
an upper
shows that sup A is a lower bound for B, so sup A < inf i?.
A and < sup B for every a: in A, and in
it follows
that x + y < sup A
sup B. Thus, sup ^4 + sup B is an upper bound
B so sup (^4
for A
B) < sup ^4
sup B. If # and y are chosen in
^4 and B, respectively, so that sup A — x < s/2 and sup B — y < e/2,
then sup A + sup B — (x + y) < B. Hence,
(b) Part (a)
CHAPTER
to
Let k be the largest integer
x
13.
there(c)
s.
such a k
12.
is some natural
There are consequently only a finite
not bounded above, there
<
with
there
there
fore a rational
<
/
is
x
some natural number n with 1/n < y — x.
there
is, by part (a), an integer k with nx <
1,
which means that x < k/n < y.
(d)
r
N
—n <
— x > 0,
ny — nx >
(b) Since y
k
Since
n with
+
+ B) > x + y >
A
sup
+ sup B —
£.
(a)
1
f {a) =
lim
=
lim
fc-+o
(b)
The tangent
line
*>-/(«> =
l im
/fr
+
a(a
+ A)
a2
g(x)
=
(a,
—
1/a)
1
-
(x
is
a2
If /(*)
=
g(x),
so #
=
a.
+a
a
then
1
the graph of
1
a)
a1
^ _^
+
2
1
l±A_i
—
= —
through
—
_
Answers
2.
to Selected
Problems
537
(a)
_1
r(a)
,
lim
=
lim
/(«
The tangent
-
(-2*A
Aa (a
line
A 2)
+ A)
2
fc-o
(b)
+ *)-/(«) m
through
g(x)
lim
1/a 2 )
(a,
(*
"
3
a3
a2
a)
+i
"
—+1
1=
the graph of
is
/7 2
= _2^
*2
;
a3
- ~»
/to «*(*), then
+ *)«
2
=
2
/7*
If
(,
'
a2
a3
or
2x 3
-
+a =
3a* 2
3
0,
or
=
0
So x
-
(*
=
a)(2* 2
a or x
-
ax
— —a/2;
side of the vertical axis
-
=
a 2)
(*
-
a)(2*
the point (—a/2, 4/a 2 )
from
=
lim
&+
lim
(Va
Vtf)(Va
A(V a
^
h
+ —A +
+A+
on the opposite
Vfl)
=
A
r
lim
2
=
Conjecture:
=
lim
S» (X
nx n ~\ Proof:
+ h) ~
S" (X)
A
h-+o
=
lim
+
h
h-+o
^-
=
lim 5
=
lim
20--'*'-
h-+Q
/-I
i-i
=
^
11
a:
"1
=
nx n ~\
—
A(VT+A + vi)
Va)
1
=
4.
~
lim
h~>0
-
+A
^°
~
M=
A
fc-»0
=
~
k)
lies
a).
1/a 2).
(a,
3.
f(a)
+ a)(* -
since lim A*" 1
=
0 for ;
>
1.
538
Answers
to Selected
Problems
5
#
6.
=
f(x)
0 for x not an integer, and /'(*)
is
not defined
if
x
is
an
integer.
(a)
g(x +
„, = hm
v
2
-
h)
gjx)
=
S (x)
=
h
a-*o
+
+
h)
-
c]
[f(x)
+
c]
=m
+ *)-/(*)
/(*
lim
[fx
lim
(b)
+
= hm g(*
v
g
(x)
-
h)
lim
cf(x
+
2
•
.
;
.
4
)
(25)
2
2
2
•
2
2
)
;
4
2
(36)
•
;
;
2
(d)
2
•
•
2
(c)
cfjx)
h
2
(b)
-
n
= 3 9 2 /'(25) = 3 (25) /'(36) = 3
2
/'(3 ) = /'(9) = 3 9 ;/'(5 ) = /'(25) = 3
/'(6 ) = /'(36) = 3
(36)
2
= 3a f'(x = 3(x 2 = 3*<.
/'(a 2 ) = 3(«
= 6*
but g(x) = * so
/'(* ) = 3x
(a) /'(9)
h)
*->o
h-^o
8.
=
h
h-*o
7.
g{x)
)
5
6
.
,
;
(a)
„
..
*
=
lim
g(x
+
/([*
+
h)
*]
—
g(x)
+ h + c)-
f(x
+ *)-/(« + '> =
f(x
+C
f(x
+ c)
).
A
ji-»o
(b)
,'(,)
=
lim
SiX
+
= hm c[f(cx
k)
=
-lim
+
fcx)]
k)
9.
g(x)
=
f(x
+
a),
=
= f
is
periodic, with period
so f{x)
+
-
~
*)
-
f(cx)]
K
k^O
then
g
9
..
lim
/{CX)
=c-f'(cx).
(Compare the manipulations
If
=
+ £*> - *">
k
h->o
(c)
lim /if*
Ch
/(CX
c
eix)
+ ch) -
h-*0
=
~
g'(x)
in this calculation with
g'(x)
= fix
+
+
= fix
a),
a) for all x,
Problem
by part
(a).
5-13.)
But
which means that
/'
a.
= 5x\ Now fx) = g(x + 5), so by Prob= Six + S)\
=
g\x
+5)
f{x)
- 3)\ as in part (i).
=
=
so
5ix
(ii)
3)
ix
fix)
fx)
(hi)
fx) = (* + 2) so /'(*) = 7(* + 2) as in part (i).
+ *), by Problem 8(a). If ft) =
If fx) = git + *), then fix) =
so fix) = £'(2*).
git + x), then fit) = g\t + *), by Problem 8(a),
k such that
number
no
=
is
=
there
2c/, and
then s\t)
ct
(a) If s[t)
=
all
=
for
'(
to
is,
2c/
[that
]
s
)
(i)
Ifg(x)
lem
=
x\ then
g'(x)
8(a),
5
,
6
7
,
5
10.
11.
2
,
2
t
(By the way, at this point
t.
we do not know any nonzero function / for
Answers
which
to
(b)
(c)
/'
proportional to
is
If s(t)
(ii)
[/(<)]>
The
in
After Chapter 17
/.
determine what the world would be
(i)
t
=
=
(a/2)t 2
(at)
chandelier
seconds
2a
=
=
\6t
=
s'(t)
2
=
t
5.
it
Problems
539
might be amusing
were correct.)
like if Galileo
=
so s"(t)
a.
2as(t).
16/ 2 feet in
or
,
at,
=
(a/2)t 2
•
falls s(t)
400
if
then
,
=
2
to Selected
seconds, so
/
it falls
400
feet
After 5 seconds the velocity will
(a)
16 = 80 feet per second. The speed was half
amount when 40 = s'(t) = 16*, or t = -f.
This is another way of writing the definition (see Problem 5-8).
(b)
This follows from Problem 5-10, applied to the functions a(h)
=
be /(5)
=
5a
5
•
this
21.
[f(a
26.
30.
/"(*)
(hi)
f
f
(x)
(ix)
(i)
(1
(iii)
(—
(vii)
(vii)
2.
=
=
means
means
means
means
means
(i)
(v)
10
h)
(i)
(iii)
CHAPTER
+
(i)
(iii)
-
+
f(a)]/h and
=
0(A)
[g(a
+
-
h)
6x.
4x 3
.
=
=
that f'(a)
that g'(a)
the
same
as
that g'(b)
2x)
*
if
71
a:
.
+
/(*)
c
(iii).
+x
cos(x
•
/'(a)
=
=
/(x)
if
+
= f(b + a) if g(x) = f(x
= cf{cb) if g(x) = f(cx).
that g'(b)
sin x)
n~l
na
a).
2
).
cos (cos x).
+ sin x)) (1 + cos x).
+ l) (x + 2))) [2(x + 1)(* + 2) + (x + l)
2(x + sinx)(l + cos *)
cos((x + sin x)
[2 sin((x + sin x)
2x.
(sin x + x cos x) + (cos(sin x )(cos
))
(cos(x sin
x
sin
+ (2x cos x sin x
(2 sin x cos x sin x sin x
+ (sin x cos sin x sin x
6(x + sin x) (l + 5 sin x cos x).
7x
cos x
7(sin x + l)
(7 sin x
cos(sin x + I)
x
sin
cos(x
+
sin
x
sin(x
cos(x +
+
[(2* +
))
(2x + 2x cos x ))].
2 sin x cos x)
sin x + sin x
(1 + sin x)(2x cos x
(cos(x
•
2
2
(cos((x
].
•
2
2
)]
)
'
2
2
(v)
(vii)
a:))
(xi)
(xiii)
2
2
2
2
2
2
2
)
)
2
2
2
).
5
7
•
a*
•
2
(ix)
=
g(a)]/h.
4
5
7
1
7
•
6
6
1
•
2
2
6
7
7
).
2
2
)
2
-
2
•
2
2
•
—
(XV)
(1
+
sin x)
L
cos x sin x sin' x
5
540
Answers
to Selected
Problems
4.
(i)
{x
(iii)
2x*.
(i)
-x
(iii)
17.
(i)
/'(*)
(iii)
fix)
(v)
7.
(a)
(b)
2Y
2
.
fix)
A'(t)
+
=
=
=
=
g'ix
gia).
=
F(0
the
is
First
it
2tt
=
4?rr(0 V(0
(c)
gia))
g(x))
=
2wr(ty(t). Since r\t)
that A'(t)
If
+
+
g'ix
4
•
=
6
•
4?r
4
•
method: Since
r(0
f
=
with
then V(t)
t,
=
r(t)
4, it
3
=
f
—
_
—
il'W
5
2wr(t)
5 for r{t)
when
r(/)
=
3.
6tt
Thus
=
To apply
when
30
/« = ^W 3/2 =
=
r{t)
we
the second method,
then, using
first
3.
note that
if
VA(t)\
Problem 9-3 and the Chain Rule,
fit)
%A(t) ll2A'(i)
(just as
we might have
Now
V(t)
=
=
/3, so
•
follows that
r'H)
follows
4.
= 47rr(/)
6 = 384tt when r(t) = 4.
A'(t) = 2Trr(t)r (t), and A'(t) =
volume at time
2
6 for that
when
48?r
4wr(t) 3
=
3
3
4[7rr(Q
3tt
2 3/ 2
]
1/2
4A(t) 312
37T
4*-[r(0»]"»
1/
2
'
guessed).
=
3,
Answers
to Selected
Problems
541
So
8.
(i)
(iii)
10.
=
=
(/• A)'(0)
=
a'(x 2 )
A
=
2
5
=
30.
A'(0)
=
/'(3)
•
=
sin 2 (sin(x
=
l = d£. dl =
(/•«)'(*)
d
(i)
dx
d
(iii)
±
^ *
.
11
1.
(i)
0
=
[
(cos ,)
•
(1
= (-
S in
w)
-
2*
-
=
/'(*)
are in
4
+
1))
•
2x 2
.
= /'g?w
+ 2x) =
w«
(cos(*
+x
2
))
•
(1
+
2x).
•
= (-
(cos x)
(cos x).
cos(sin *))
d/x
dfo
CHAPTER
2
sin (sin 1)
dx
dy
=
•
implies that
Q'w
30.
(t)
-g-)sin (sin 1).
2x 2
•
>
2
9 (sin
4
T u2r(t)A
.
3
•
f(h(0))
h'ix )
The Chain Rule
=
— 2,
3x 2
8 for x
=
2
=
and x
both of which
2];
= W;
maximum = -gy3 minimum = —11.
2
0 = /'(*) = 12x - 24x + 12x = 12x(x =
in
is
[—
only
which
0
of
and x
1,
2, i];
fi-h) = «,/(*) =H,/(0) =0;
maximum — yf minimum = 0.
0 = /'(x) =
/(-2) = 5,/(2) = -n,/(-i)
-,
(iii)
2
3
2x
+
1) for
x
=
0
,
(V)
+
x2
1
(x
forx
is
in
= -1
[-1,
(i)
(iii)
(x
+
+
l)
l)2x
1
2
(x
+ V2andx -
2
2x
+
l)
x2
2
-1 - V2, of which only -1
+ V2
\]\
= (1 + V2)/2;
maximum = (1 + V2)/2, minimum = 0.
— is a local maximum point, and 2 is a local minimum point,
0 is a local minimum point, and there are no local maximum
/(-I) =
2.
2
0,
f{\)
=
f,
/(-l
+ V2)
-g-
points.
(v)
- + V2
1
minimum
is
a local
point.
maximum
point,
and
— 1 — V2
is
a local
542
Answers
to Selected
Problems
3.
(a)
Notice that / actually has a
function of even degree.
minimum value,
The minimum
since
/ is a polynomial
occurs at a point x with
n
=
0
=
/'(*)
T
2
=
so *
4.
+
(ai
and
•
*
+
•
7 are local
(i)
3
(iii)
All irrational x
-
(*
at),
=1
1
a n )/n.
maximum points, and 5 and 9 are local minimum
points.
(v)
x
<
x
is
> 0 are local minimum points, and all irrational
maximum points,
maximum (minimum) point if the decimal expansion
0 are local
a local
contains (does not contain) a
5.
If /(*)
is
/(*)
The
positive
lim f(x)
=
=
Vx +
2
a2
oo,
and /
is
V(l -
Vx +
a
=
2
0.
Now,
(1
-
V(l -
This equation says that arctan
6.
is
.
since lim f(x)
X—*
a =
/'(#)
*)
2
x)
=
+
is
=
0
—
0°
minimum
when
Q
b2
arctan
also possible to notice that f{x)
shortest
2
minimum,
j3.
equal to the
sum
and the line segment from
when the two line segments lie along a
of the dashed line segment
This
+b
x) 2
differentiate everywhere, so the
_
*
2
is
+
function / clearly has a
occurs at a point * with f'{x)
It
5.
the total length of the path, then
of the lengths
(x,
0) to (1, b).
line (because of
Problem 4-8 (b), if a rigorous reason is required); a little plane geometry
shows that this happens when a — /5.
If at is the length of one side of a rectangle of perimeter P, then the length
of the other side is (P — 2x)/2, so the area is
= XJP
W
2
So the rectangle with greatest area occurs when x is the maximum point
=
/ on (0, P/2). Since A is continuous on [0, P/2], and A(0)
A(P/2) = 0, and A{x) > 0 for x in (0, P/2), the maximum exists.
Since A is differentiate on (0, P/2), the minimum point x satisfies
for
0
=
A'(x)
= 1^l2± 2
= PZ
2
so x
-
P/4.
AX
:
x
Answers
7.
Let
S(r)
Problems
543
V
be the surface area of the right circular cylinder of volume
with radius
Since
r.
V=
we have
to Selected
=
h
where h
>irr?h
the height,
is
F/xr 2 so
,
S(r)
=
2irr
2
+
2wrh
=
2wr 2
+
2V
—
r
We want the minimum point of S on
«)
(0,
this exists, since
;
lim £(r)
=
r->0
lim 5(r)
=
S is
Since
oo.
differentiable
on
oo),
(0,
minimum
the
point
r
cos x
is
r- »
satisfies
=
0
=
£'(r)
-
4?rr
—
2V
r
47rr
3
2
- 2V
or
3
r
=
2r
8.
14.
1
is
(a)
a local
We
maximum
point,
and
3
is
minimum
a local
point.
have
m —-
f(a)
=
f{x)
for
some x
for
some
in
(a, b)
a
b
> M,
so f(b)
(b)
We
-
a).
have
f(b)
b
-
so
(c)
> M(b -
f(a)
If |/(*)|
~
—
/(«)
<
/(«)
f(a)
^
=
a
<
iti(*
-
Af(A
-
a)
(
^
6)
a).
for all x in 0, b],
-
^ in
<
f(b)
then
<
-Af <
f(a)
/(*)
M(b -
+
< M,
so
«),
or
|/(*)
16.
(a) /(*)
= —
cos
a:
~ /Ml < M(b -
+ a for some number a
a).
(because fix)
= —
one such function, and any two such functions differ by a constant
function).
544
Answers
to Selected
Problems
(b)
=
fix)
x 4 /4
+
some numbers
(c)
(a)
some number
and
so f(x)
a,
=
x 5 /20
+
+
ax
b for
b.
+ * 3 /3 + a for some a, so /'(*) = x z /6 + x*/12 +
some a and b, so /(*) = * 4 /24 + x 5 /60 + ax 2 /2 + bx + c
for some numbers a, b, and £. Equivalently, and more simply,
fx) — x 4 /24 + x b /60 + ax 2 + bx + c for some numbers a, b, and c.
= -32, we have
Since
= -32* + a for some a, so
2
j(0 = — \(>t + a* + j8 for some a and
Clearly, s(0) = 0 + 0 + 0 and j'(0) = 0 + a. Thus, a: = v 0 and
0 = JoIn this case, j 0 = Oand^o = v, so s(t) = — 16/ 2 + ^. The maximum
/'(*)
+
ax
17,
=
a for
a
x 2 /2
b for
jff.
(b)
(c)
value of
when
occurs
s
maximum
value
=
0
— zr
~
=
At
that
moment
any
(as at
or
t
top).
v/\6
The
(it
~64
same
29.
=
= -I6t 2
s (t)
is
<
81,
fix)
fall
•
+
back down as
it
was
initially
=
to f(x)
V*
8
<
<
-
some x
for
7^
8
2
9
3* 2
lim
x-»l
+1^0.
0
is
—32
when
it
took to reach the
v
9,
so
1
<
2
8*
•
L'H6pitaPs Rule does not lead to the equation
because lim 3* 2
>
vt,
(^) +
Vx
a/66
:r-+l
t
2*
moving upward).
on
[64, 66]:
1
=
we have
2
33.
which
2
1
ground at time
then
velocity with
—
x
y/32, so the
L
hits the
takes as long to
velocity
V66 - ~
V64
=
66
64
<
=
clearly 0, but the acceleration
is
Apply the Mean Value Theorem
Since 64
*
EL.
64*
(»/16) = -32
(the
or
v,
32
The weight
0
=
V
the velocity
time).
= — 32/ +
s'(t)
is
+— =
1
—
3
—
„
6x
lim
x— 1 2
in [64, 66].
.
I
Answers
34.
545
Problems
to Selected
(i)
x
=
lim
tan x
x->o
——
1
lim
=
—
lim cos 2 x
sec 2 x
x-»o
1.
x-*o
(ii)
—
cos 2 x
1
lim
x->o
CHAPTER
12
1.
(i)
f-\x)
=
=
-
y
r
(iii)
(*
=
1
a:
-
(x
1)
1/3
(If
.
y
(If >
= rK*)>
=
*
±y
is
*
is
if
„
1
= fK*)*
/+
= /(» =
then *
1,
so
then
^ rati0nal
^
—y, y
(
y
=
2x
-)
,-/(*)-(
yw
since
cos x
at
lim
x->o
1/3
l)
/.
2 sin
..
=
2
irrational;
rational or irrational
rational
= —*
and
if
if
and only
if
is,
y
irrational, so
x
is
,
an
y
we have
=
/(*).)
(v)
Ix,
r
f~
f
2.
(i)
(iii)
4.
^
x
=
=
x
an,
(vii)
x
ai,
(b),
l
is
1
is
and f~ {x)
-1
decreasing and / (^)
l
increasing
a contradiction.
5.
—/
Clearly,
<
•
6.
(a)
is
is
If
/
+
g
The proof
is
increasing
is
.
is
is
,
n
Then f~
b.
< 0.
< 0.
not defined for x
not defined for x
or f~
l
(a)
l
(a)
>
*
f~
l
since
(b) }
f'Kb)- But
if
-1
/
is
f^a) >
=
«,
similar for decreasing
increasing, for
/,
or one can con-
=
< f(b) and g(a) < g(b),
+ g{b) = (/ + g)(b).
for example, if /(x) = g(x) = x.
if
f(a)
> 0 for all
if a < b, then
/(x)
if
increasing, for
(fo g)(x)
f(g(y))> so
If/(x)
.
instead.
(/o
*O0 = r'to,
=
=
so ,
<
>>,
since g
cx
+
+
6
d
_
ay
cy
=
(But
is
so f(g(a)) < f(g(b)).
=
f(g(y)), then g(x)
one-one.
(/o^)-i(x),thenx
= g-KHW).
/(>), then
ax
then
x.)
so that /(*(*))
g(y) y since / is one-one, so x
)-i = ^-i o /-i; for if j
(f og
7.
.
= fU~Kb)) < fiTKa)) =
(f+g)(a) = f{a) + g(a) < f(b)
f-gis not necessarily increasing;
/ g
fo g
2,
then
*
sider
.
=
ai.
-TW
l
.
i
-/-
1
Suppose / is increasing. Let a
one-one. So either fK*) <
f~
.
fl t-,
+
+
6
i
=
(f
*
g)(y)
=
546
Answers
to Selected
Problems
so
+
acxy
+
bd
<z<f(#
—
jy)
=
adx
+
bey
+ ady +
flcry
+
ftc*
bd,
or
If <2^ =^
ftc,
If
y
this implies that x
and y
for all *
= /-1 (V)> men
—
domain
in the
= /W?
*
—
bc(x
=
y
0.
jy).
(ButifW =
ft£,
then /(#)
=
f(y)
of /.)
so
#v +
= —
x
—
ft
+
>
so
f~
1
—
(x)
y
—— —
=
cx
8.
Those
(i)
[2,
(iii)
since
/
The formula
is
for x 9* a/c.
a
which are contained in (—
increasing on (— qo, 0] and [2,
[a,
/ is
ing on [0, 2].
Those intervals
since
14.
intervals
oo ),
—
[a,
ft]
oo, 0]
oo) }
or
[0, 2]
which are contained in (—<*>, 0] or [0,
— <», 0] and decreasing on [0, oo).
(
ft]
or
and decreas<*>)>
increasing on
for the derivative reads:
dx
=
J_
dy
dy
1x
understood that dx/dy means (f~ l )'(y) while dy/dx
an "expression involving
and in the final answer x must be replaced
(In this formula,
is
by
y,
it is
3
by means of the equation y = /(#).)
in Problem 14, when completed, shows that
The computation
dx" n
dx
nix
=
is.
16.
1
^rrw + r w
-
j
71
'
1
)
y (l/n)-l
-
ax) = x(ry(x) +
=
1
1
11 *1
F'irK*))
/(/->«)
•
-
(ro'w
cryw
(i)
A\A~
1
(3))
A'(0)
sin 2 (sin 1)
Answers
17.
to Selected
547
Problems
Since
(ri) '
w = /lrW
we have
i
crrw
CHAPTER
13
1.
If
Pn ={*„,...,<„}
= -/"(r w) (r )'w
is
i
•
[/'cnw)]
2
=
the partition with
ib/n,
then
n
n-1
b* f
and
(»-
*)
-
4
("
,
i) 3
,
- D' l
(»
similarly
n
3
=1
"«44 +
Clearly L(f,
Pn and
)
U(f,
Pn
+
2
4j"
can be made as close
)
-
to b 4 /4 as desired
by
Pn )
can be made as
£(/,
small as desired, by choosing n large enough. This shows that / is
integrable. Moreover, there is only one number a with L(/, Pn ) < a ^
choosing
/.
£/(/,
Pn
)
=
/4
will
6
4
sufficiently large, so £/(/, />n )
for all w; since
x 3 dx has this property, the proof that
he complete once
actually follows from the fact that L(f,
4
as close to 6 /4 as desired
were true that
U(f,
Pn
)
V(f, Pn)
b
4
/4
<
x 3 dx, then
as close as desired to b
>
x3
Jq
dx,
and
AO
4
Pn ) and
by choosing n
4
3
* dx
that £(/, P„) < £ /4 < £/(/,
a straightforward computation, but
we show
This can be done by
for all
JQ
it
£/(/, jPn )
it
can be made
sufficiently large. In fact, if it
would not be
possible to
make
/4 by choosing n large enough, since each
similarly
we cannot have
6
4
/4
>
j* x*
dx.
2.
We
have
30
ly L(f,
Pn ) and
/
(v)
=
is
£
5
Pn
can be made as close to
)
enough. As in Problem
large
loosing n
dx
U(f,
b
1,
this
/5 as desired
implies that
(i),
(iii),
and
b
A
not integrable.
(vii)
(For a rigorous proof that the functions in
integrable, see
Problem
15.
The
(vii)
are
values of the integrals, which are
from the geometric picture, can also be deduced rigorously
by using the ideas in the proof of Problem 1 5, together with known
clear
integrals.)
5.
(i)
(iii)
(v)
6.
jl (//
^^
g
dy
)
dx
=
f*
= //
(
^
g(y)
fo
g
^
)
dy /; /(*)
24.
27.
^
j
g(y) dy
Clearly L(f, P) > 0 for every partition P.
(b) Apply part (a) to / — g, and use the fact that
(a)
(a) 0.
(b)
i
(a)
Clearly
m{b
-
a)
<
L(f, P)
<
iS
thC constant)
dx
(here
10.
(herC
dx
dy
U(f, P)
< M(b -
a)
is
the constant).
Answers
P
for all partitions
m(b
of
-
[a, b].
a)
<
to
549
Selected Problems
Consequently,
j*
f(x) dx
< M(b -
a).
Thus
m <
m and
satisfies
(b)
< M.
fi
M be the minimum and maximum values of /on
Let
Since / is continuous, it takes on the values
quently the number fx of part (a).
31.
m and
Af,
[a, b].
and conse-
Since
-l/l</<l/l,
we have
so
b
(Problem 30 implies that
CHAPTER
14
1.
(i)
(sin 3
x 3)
3* 2
•
fx
(iii)
J
K
W
(v)
2.
1
+
t
1
+
t
/
Ja
(i)
All x 9*
(iii)
All x
(v)
All
(vii)
All
+
sin 2
2
+
sin 2
f
t
1.
^
x.
0.
11
(F is not differentiable at 0 because F(x)
~
(/ 1)/(0)
>
0 arbitrarily close to 0 with F(x)
=
=
/XrW)
=
* /;/(<)
dt,
+
1
»
1
=
sense.)
5* 1.
(0
F(*)
makes
-dt.
*
4.
\
dt.
2
but there are x
3
\f
.
I
1
/
Js
f
+ sin (sin 0))
=
sin(sin(/- 1 (0)))
1.
so
f'w = */« +
/;/(<)*.
=
=
0 for x
\.)
<
0,
550
Answers
to Selected
Problems
9.
We
can choose
=
/to
+
1
1
n
Then
£<l/n) + l
V^rf* =
-/(0) =
/(*)
i+1
n
CHAPTER
15
1.
(i)
1
+
(hi)
1
1
2.
(i)
0.
(hi)
0.
(v)
6.
(a)
+
+
(.
arctan 2 (arctan x)
(tan x arctan
1
arctan 2 x
+x
1
2
tan x \
sec 2 x arctan x
+
+
sin
a:
1
—
0.
sin 2a:
2x
cos 2x
sin
cos
3a:
3a:
=
=
=
=
=
=
=
+ x)
sin(#
x
—
sin
sin (2a:
+
a:)
cos
2
sin * cos #
2
=
x
=
2 cos
sin
a*
cos 2 x
+
(cos 2 x
3 sin
a:
cos 2 x
—
sin 3
>
(b) Since cos 7r/4
0
=
+
x)
=
=
=
=
=
cos
cos
3
cos
2
=
1,
x
4 cos
3
_
7T
7T
- =
cos 2
cos
7T
cos
- =
2
2 sin
a;.
2
sin x
— sin 2a: sin x
— sin x) cos = 2 sin * cos
— sin 2 cos x — 2 sin x cos
— 3 sin cos
x — 3 cos
cos x
2
a:
2
a:
2
a:
a:
a:
a:
a:.
*
-
=
„
2 cos 2
7T
-
—
It follows, since sin ir/A
= V2/2, and
7r/6 > 0 and
*
6.6
-
4 cos
>
0 and sin 2
+
=
1.
consequently tan 7r/4
^^.q^o
=
cos 3
.
1,
4
that sin 7r/4
=
2a:
2 sin # cos
sin 2 x) sin #
4
= V2/2.
Similarly, since cos
0
1
+
=
0 and
cos
cos tt/4
=
2
a:
cos 3 x
2
we have
—
a*
a:.
2a:
(cos 2
x
cos
—
cos x
2a:
2 sin
cos (2a:
2
"
71
3 cos ->
6
-
Answers
we have
sinx/6
8.
=
w/6
cos
\^3/2.
to Selected
since
follows,
It
551
w/6 >
that
sin
0,
= ^1 - (V3/2) = i
2
(a)
+ y) =
tan(*
—+—
sin (a:
jp
-
+
cos (a:
jy)
sin # cos
cos
From
part (a)
we have
+
arc tan y)
.
tan (arc tan x
jy
sin x sin
cos x cos
jy
cos # cos y
—
=
.
+
a:
tan x tan
<
+
tan (arc tan x) tan (arc tan y)
x
+y
1
—
xy
and arctan x
y,
+
}
arctan y
+
provided that arctan x
arctan
<
y
0
jy
>
and
The
first
<
xy
=
/ x
arctan
(
formula
—
n)x
cos (m
+
n)x
The
—
=
—
we must add
so that arctan x
1,
is
arctan y 5*
is always
7r
+
~\~
y \
-
xy)
)
(— 7r/2,
>
1,
which
is
so that arctan *
+
7r/2),
to the right side,
arctan y
<
and
if x,
—ir/2, then
we
derived by subtracting the second of the following
cos (mx
—
first:
nx)
=
mx cos(-nx)
mx sin nx,
cos
cos nx
+
sin
cos 01* cos nx
—
sin tw* sin nx.
cos
+
7r/2, this
7r.)
two equations from the
cos (m
arctan y lies in
> 0 and xy
(If x, y
1.
then
7r/2,
ry •>
must subtract
<
arctan x arctan
\1
whenever
tan(arctan y)
when arctan x + arctan^ = ±7r/2, which is equivaFrom this equation we can conclude that
arctan x
true
jy
jy
tan(arctan x)
—
j>
tan y
arctan
+
1
sin ^ sin y
cos * cos
provided that arctan
kir
7r/2. Since —7r/ 2
lent to xy
—
cos # cos y
1
=
+ cos # sin y
cos # cos y
.
the case except
jy
cos y
cos x sin
tan
(b)
a:
sin x cos j
1
10.
Problems
mx
other formulas are derived similarly.
—
sin
mx
sin(
— nx)
.
It follows
11.
sin
mx
from Problem 10 that
— \
sin nx dx
_
—
[cos(m
—
[sin (m
1
m ^
if
—
n)x
m —
+
cos(m
sm(m
n)w
|
2 IL
then
n,
+ «)tt]
m+
n
n)x] dx
n
J
sin(m — n)ir _
m—n
[
=
But
if
sin
m =
m*
The
(a)
m
+ n)T
+ n JJ
*l|
0.
then
n,
sin n*
</*
=
ir
=
=
14.
sin(m
~~
1
J*^
—
k{[ir
+ n)x dx
cos(m
cos(m
+ n)7r] —
[— x
—
cos(m
+
7T.
other formulas are proved similarly.
We
have
cos 2x
=
=
-
—
cos 2 x
sin 2 x
1-2 sin
2 cos 2 *
2
-
*
1.
So
=
sin 2 *
1
—
cos 2x
j
2
=
cos 2 *
(b)
1
+ cos 2x
These formulas follow from part
sin x/2 > 0 (since 0 < x < ir/2).
(a),
because cos x/2
>
0 and
(c)
sin 2 x dx
j
=
[
/
h
*
j
cos 2 x dx
/* 6
=
</x
—
arctan
lim arctan x
1
—
*
(b
—
—
sin 2a).
+ -4 (sin 2b —
sin 2a).
(sin
a)
—
(b
—
x
a)
t/4.
arctan 0
=
tt/2.
x—* •
1 9.
lim x sin x
x->«
20.
=
lim - sin x
x-*0+
=
2b
4
1
2
2
arctan 0
(a)
2
+ cos 2x
1
Ja
(b)
=
2
j
Ja
18.
dx
I
Ja
Ja
1
x
(a)(sinTW=^cos(^)=^cos^).
,
1 ,
.
.
-
v
Answers
sin°*
=
S^T
< b)
17
1.
Si
x-^o +
e **
ee
e
(iii)
(sin *)
•
'*
•
•
8in(sin:r)
sin
(v)
x
•
180
•
180
a:
a*
[(log(sin *))
cos(sin *)
•
•
cos x
+
ein
?r
"
„/180
18<)'
553
Problems
sin(ir*/180)
tt
™
——
=
= hm
*
(i)
=
i
Jim # sin x_>«
CHAPTER
sin(7T*/180)
to Selected
*,in *[(log(sin
*))
{(log(sin *))
sin
'
cos x
'
*
+
(cos */sin x)
sin(sin
•
a:)].
8in x
(cos x/sin x)
sin *}
•
+
(cos jc/sin
at)
•
sin *
8in
*].
(cos
[arcsin
(vii)
(^J^^fa
(arcsin
arcsin
3.
(i)
0.
*)
•
i
?
cos *
+ log *
4
sin-
a:
\sm */
*
at/
'
'
i
i
(iii)
(v)
5.
[logOog
sin
•
|
\sm
(log *) log *
—
sin x
+ log (sin **)
(ix)
(^)))
(a)
cosh 2 x
—
sinh 2 *
=
—
(
—
—
I
J
J
=
1.
(c)
sinh x coshjy
+ cosh
gx+y
e
a:
-x-v
ir + ~i
sinhjy
=
I
(
-r
1
J
e
-*+v
e*-v~\
V ex +y
e
-*^
~
^
e~x+v
^
^
e"*
y
-2
']
r + "Tj + L"T + "r + "T""~^J
=
= smh
!
2
(e)
^
Since
sinh x
=e
+ e"
(a:
+ y).
554
Answers
to Selected
Problems
we have
=
=
sinh' (at)
K }
cosh
x.
2
(g)
Since
.
=
tanh x
sinh x
1
cosh x
we have
—
(cosh x) 2
=
tanh'(*)
(sinh x) 2
by part
(a).
cosh 2 x
6.
(a)
If y
=
arg cosh
x
then *
x,
= cosh^ =
>
0
+
sinh 2
y/\
and
by part
;;
(a).
So
sinh (arg cosh x)
—
= V* 2 —
sinh y
1
>
since sinh y
(c)
1
=
(arg sinh)' (*)
sinh '(arg sinh (a:))
1
cosh (arg sinh(#))
1
VT+x~
bypart
(b)
*
2
(arg tanh)
(e)
7
l
W
7
tanh (arg
=
X.axih(x))
2
cosh (arg tanh(#)).
Now,
tanh 2
jy
+
—— =
-
1
by Problem
5(b),
cosh 2 j/
so
tanh 2 (arg tanh(*))
+
——
—
=
cosh 2 (arg tanh(#))
or
cosh 2 (arg tanh(*))
1
=
1
7.
(a)
If
y
=
arg sinh
x,
then
x
=
sinh
=
-
x2
1,
0 for y
>
0.
Answers
to Selected
Problems
555
I
so
-
ev
-
e™
€
=
-y
2x,
-1=0,
v
2xe
± Vax
2x
2
+
4
>
0
so
+ Vl +
ev
=
=
arg sinh x
x
x2
ey
since
or
y
=
+ Vl + a
log (a*
2
).
Similarly,
=
arg cosh x
= i
arg tanh x
(b)
6
f
/
J
a
=
—— 1
Vl +
=
dx
log(l
arg sinh
*2
P
,
8.
fl
.
2
f —-^—
r
;
<fe
12.
(a)
(b)
1),
ilog(l
arg sinh a
—
*).
by Problem
6(c)
+ Vl + b - log(a + Vl + a
+ V* - 1) - log(a + Va - 1).
=
2
2
).
)
2
2
log(£
[log(l
+
-
b)
-
log(l
A)
-
log(l
+
+ log(l -
a)
«)].
lim
x—
>
e
e
xlo * a
Since log a
.
<
0,
we have
'
oo
=
xlo * a
lim x log a
x—*
= -
oo,
oe
0.
jM^-lim£-0.
^
x-^
(e)
=
oo
so lim
(c)1
—
—
a)
2
lim a*
X—
6
+
1
- -
1
(a)
&
1
Vx -
2
~
= logO
Ja
+ V* —
\og(x
y-*
oo
lim
=
lim
e
* ey
zlogX
.
x->0+
lim log(l
+y)/y =
lim x log(l
x—
Now, lim
log x
a:
=
0 by part (d), so lim x*
+
l/x)
=
log'(l)
=
=
1.
x->0 +
x->0+
x->0+
1-
lim log(l
+ y)/y =
1.
y—*0 +
oo
(c)
«
=
exp(l)
=
>
(*)
=
+ l/x))
+ l/x))
exp(lim x log(l
X—
lim exp(xlog(l
= lim
x-+
(1
oo
+
l/x)
x
.
00
(The starred equality depends on the continuity of exp at 1, and
can be justified as follows. For every £ > 0 there is some 8 > 0 such
that \e - exp y\ < £ for \y — 1| < 8. Moreover, there is some N such
that|*log(l + l/x) - 1| < 8forx > N.So\e - *log(l + \/x)\ < 6
for x
>
N.
556
Answers
to Selected
Problems
(d)
e
=
a
(1
»
X—*
=
+ \/x) f =
+ /x) ax
+ a/y) v
x
[lim
lim
(1
lim
y —*
14.
After one year the
one dollar
will
(1
.
number
of dollars yielded
by an
initial
=
+
(1
a/100
=
x
x)
e
a ' 100
.
00
>
15.
Clearly /'(*)
16.
f{x) = (_!).!/(-*) = l/x
Lctg(x) = f(x)/e cx Then
1/x for x
0.
<
x
If
then f(x)
0,
so there
(a)
(b)
- ^/(^) -
\
some number
is
UA(t
=
16, there
.
=
A(t)/2
9
so
so <?<V
this r
= e ct /2 or
does work.
T
Newton's law
*
CT
—
k for
all
2)/c. It
= c(T - M)
is
- MY =
(i)
k
=
ke*'
(V^a
=
7(0)
1
(n)
,
7
x
+
(iii)
(*
(iv)
o*/P
= 7V
+ ^c/Vx
+*
=
( fl
3x
—
=
-
c(T
y
M).
some number
T(t)
and
ct
.
a;
-
M=
So T(t)
1'
10
+ aT
Vx =
)A 4 * == e'* x
/*)* = ,*io« («/&>_
+
A:
such that
ke ct ,
=
M+
7V
cf
.
1/3
).
1
- V* +
+ e~x
.
easy to check that
number
which can be written
is
£<?
2
= ^,sor = - (log
T'{t)
So by Problem 16 there
=
c *.
states that, for a certain (positive)
(T
a:.
k such that A{t)
then
4^ =
1.
log(-x),
cx
some
is
^
=
- A 0 So
+ r)
f(*)ce
such that g(x)
A:
According to Problem
Then k
18
=
.
.
aU
CHAPTER
investment of
be
lim
18.
ax
oo
X—>
17.
\/x)
oo
1
OX—* W
=
+
lira (1
X—*
1
c,
Answers
=
(v)
tan 2 x
(vi)
a+x
—
sec 2 x
1.
+
1/a
1
Va -x
2
——
Vl - (j/*)
— sin
—^ =
=
— sin —x
2
-
+
1
8a:
x
:
sin x
+
2
—+— =
+
x
4
(iii)
(log *) 2 /2. (Let u
c
2
2^.
=
We
(iii)
e
ar
ax
e
.
—
sec x tan
x.
+*+
2
1
e
X
(
-
l)
2
e*.)
=
log
x.)
\)
= V*.)
(Let u
=
f
sec 2 x
-(log(cos *)) 2 /2. (Let u = log(cos *).)
= x 2ez - J2a:^ </* = x 2e* - [2xex
J*V
(i)
/
=
(Let u
.
\ (Let a
(vii)
(ix)
3.
Vl -
(i)
AT
<?
=
1
V2* ~
- cos ex
(v)
sin #
cos 2 x
6
--
8*
1
1
2.
—
1
2
1
6*
(ix)
(x)
2
1
1
....
,
(viu)
557
Problems
=
i
(vii)
to Selected
x 2ex
-
+
2***
2e
-
x
fe dx]
x
.
have
,
,
1
sin 6a
afr
sin
=
6
£a:
a
J
=
/"
ex
/
e
ax
cos bx dx
a J
sin bx
b [e
ax
cos bx
b
[
e
/
ax
(-
sin
bx)dx
]
L
so
y
(v)
e
ax
(log x) z
"
0
^2
Using the
the text,
j
^
sin bx dx
dx=
+
- -
az
sin 6*
c>
yo
£2"
result /(log *) 2
<fe
=
+
2x]\og x
[*(log x)
2
-
2*(log *)
A:(log x) 3
-
2x(\og x) 2
*(log
a:)
3
=
*(log
(log
b
2
—
a:)
,
„
6>
*
2
ax
cos £a\
2a: (log
a*)
—
2A*(log x) 2
2
-
3*(log
a-)
[>(log x)
3
—
y ~ [x(log x) 2
+
=
a:
+
x)
+
2x from
we have
=
"
0
a2
(log x) 2 dx
2x log
2A*(log *)
2
+
+
2*]
&
2x log x
J
+
2*ftog *)
6a:
2[x log x
-
x]
-
2x
+
2[* log x
-
x]
-
2x
a:
+
log
+
a*
2x]
—
6a*.
(vii)
sec 3 x dx
J*
=
/(see 2
=
=
tan x sec #
=
(sec x) dx
a:)
tan x sec #
—
tan
—
—
sec x
a;
/ (tan
2
J sec x(sec #
3
dx
sec
#
J
+
—
#) (sec
x tan #)
<£*
1)
/ sec *
cfc,
so
=
3
J sec x dx
+ log(sec x +
%\tan~x sec *
//V log x dx
—
37 2
2a:
= —
a:
2
- /f * 3/2
—
log x
3
=
4.
(i)
=
Let x
=
sin w,
(iii)
=
Let #
/sec
—
sec w,
(v)
Let *
=
=
1
* 3/2
^
The
.
.
1
du
* 1/2
/
-
becomes
integral
=
=
u
arcsin
*.
7
f
.
The
.
=
sec a du
/
integral
+
,
.
log(sec w
becomes
>.
.
tan a)
7
+ Vx -
=
log(*
=
cos w du.
==
~
VI —
f
/
-
sec u tan w du.
H-l
sin u, dx
CQ S
sin w
=
<Z*
—=
Vsec
-
sin 2 w
u tan u du
2
log *
=
.
VI -
-
cos w du.
/cos u du=
-
log x
*)].
- dx
x
*
3 J
—
—
=
tan
sin 2 u
f
2
The
1).
— —
esc u du
becomes
integral
log(csc u
+ cot u)
J
__ log (i + ^IIi).
X
\X
(vii)
J
Let x
=
—
sin u, dx
(sin 3 u cos w)cos u du
=
The
2
sin 3 a cos u du
J
f
=
cos w du.
/
,
w
.
o
2
(sin u) (cos w
(1
-
x 2) 3 ' 2
Let x
2
J sec w sec udu
=
—
=
=
tan u dx
y
3
f sec w
=
sec 2 u du.
(sin a) (1
J
^
cos*
(1
3
(ix)
becomes
integral
=
—
/
The
-
v
=
.
u)du
—
cos 2 m)cos 2 u du
cos 3 a
,
1
cos 5 u
—
* 2) 5 ' 2
5*
integral
becomes
<fo
+ log (sec u + tan n)]
±[x VTT7 + log(* + Vl + *
by Problem 3 (vii)
^[tan w sec w
2
2
)].
Answers
5.
=
5
f 6u du
J^+^ =
=
=
f /
,
6
a
u2
~
u
-
u*
=
u 6 , dx
6w 5 du.
2u
du.
The
The
integral
becomes
+ Vx +
2 log(l
1).
becomes
integral
\
1
+
=
dx
1,
+ U )=2V, + l-
=
* 1/6 , x
=
x
1,
2 log(l
J (
-^r )du = 2u^3u^ + 6u-6log(u+l)
1
1
2^-3^ + 6^-6 log(\^ +
Let w
(v)
-
2u
Let u
(iii)
Vx +
=
Let u
(i)
559
Problems
to Selected
=
tan x} x
—
arc tan
a,
=
</*
1).
+
</w/(l
The
u 2).
integral
becomes
f
7
_
du
(1
«-2 \
+ «y
r/_i
1
112
Sj\2 +
+ « )(2 + u)
2
=
=
7 2
+
10 7
-
w)
d
=
Let u
(vii)
=
+
^log(2
=
2X x
,
2u
JL (
+«
- log(2
"
1
du
1 f
5
u
—
10
tan x)
+a
1
+
w 2)
—
f
7
5
+
log(l
+2
d
2
1
+a
- arctan a
5
^ log(l
-
=
(log a) /(log 2), dx
+
<fc/( M
tan 2 *)
log 2).
+ | *.
The
becomes
u*
1
f
log 2 7
+
r/
1
1
"
(«+l)«
1
-
\
g
*
log2 7 V
log 2 7
V
a
^+
=
1
[2-
log u
«
"
+
Let u
= V#,
x
—
u 2 , dx
r
J
—
+ * log 2 -
Vl vi
1
2u.
The
2w
u 22 2x
w
—
w
M
«/
1/
2 log( "
+
21og(2*
log 2
(ix)
2
integral
1)]
+
becomes
1)].
integral
Now
let u
~——~
2 cos y sin y cosy
/
.
dy
=
sin^, du
=
2
_
sin y
1
cosy
integral
becomes
2
(1
J
,
sin y
1
=
2
=
2 y* sin
j
(1
+
sinjy)sin^
</y
+^ —
cos 2y dy
1
</y
^=—
sin 2y
cosjy
+jy
substitution a
+y —
.
2 cos;;
+ arcsin u — u Vl — w
x + arcsin V* - V* V -
= — 2 Vl = -2 Vl The
The
dy.
^—— —sin fy)sin J-y dy
f
= —2
=
sin^cos.y
2
u2
1
= Vl -
*,
*
=
1
-
u 2 dx
,
x.
= —2k <fa leads
to
-2m 2 du
VT
and the
/— 2
sin
1
2
—
y
cos y
—
substitution u
</v
= — 2.
sin ^
—y —
.
sin y
then leads to
sin y cos
y
cosjy
= —2u — arcsin u — u Vl — u
= -2 Vl - x - arcsin Vl - x - Vl 2
x
V*.
These answers agree, since
arcsin
(check
x
6.
=
this
V* —
—
—
by comparing
arcsin
Vl —
x
their derivatives
and
0).
In these problems / will denote the original integral,
(i)
=
2 log(*
-
1)
x
+
1
(iii)
_4_
x
-iy~""
(*
-
(
(x
+ iy
2
1_
1)
J
"
(*
+
l) 2
their values for
1
Answers
to Selected
Problems
(v)
/=
=
f-^-dx T
+ f-^—dx
+
; ,* +
ilog(* + 1) + 4 arctan *.
1
2 J x2
1
2
(vii)
Now
f
x*
+ x +i
dx
J
+ i;
(X
FT-
I
/
4
3'
V3
2
-~ (&(• +
!))
so
/
=
log(*
+
1)
+ log(* + x +
2
1)
-
(x
arctan
(ix)
/=
2* +
(^+x
+d
y
2a: +
1
/"
j
*
2
_ [
j
_
1
16
1
(*
2
+* +
/*
1
/
Now
the substitution
V3
2
changes the second integral to
VI
16
'
9
2
f
J
__du_
'
(u
2
+
l)
2
*
2
1)
+
i)).
561
562
Answers
to
Selected Problems
Using the reduction formula,
*
[
L2(w
9
= _
2
+
+
1)
_
1
*2
+*+
2
1 f _*L- 1
2 7 w2
1J
V3
can be written
^
= -
+
l0
^
[
L
9
Q(* + *+
2
+
2
V3
The equation J>
J
sin x dx
=
e
z
sin
a:
—
e
x
A(x + -1\\
/ 2
arctan
9
.
+ 1 arctan J,
)
4
1))
4
1 1
1
2
log
9
1
this
— JV
cos #
,
W3 V
I
[
) J.
2//
sin x dx
means
= f sin # can be written Fix) = ^ sin x —
any function F with
x
x
e cos x - G(x) where G is another function with G'O) = e sin *. Of
but it is not necessarily true that
course, G = F + c for some number
that
F=
12.
G.
(a)
j arcsin x dx = J
=
13.
1
—
arcsin x dx
•
x arcsin x
—
x arcsin x
+ Vl —
x2
.
J
dx
.
(a)
/sin
sin 4 x dx
3
cos x
=
-—4— +
sin 3
=
sin
x cos *
3
,
/
sin # cos #
3 f
4
—
sin 2x
4
4
.
^_
sin
3a:
2*
4~~
(b) It follows that these
same value
for x
=
+
1
Yx
+
,
1
/*
2i
1A 1
.
,
J
3
x.
8
cos 4#
fa
/
2
,
4 L2
2
sin 2x
"8
1
_ f
4 J
1
4~
4
—
.
8
x
*
sin 2 x dx
3 sin x cos #
cos #
a"
/*
3
h -
sin 4#~
4#~l
8
J
sin 4#
32
two answers are the same, since they have the
0.
+
+
+
x
Answers
563
Problems
to Selected
(a)
17.
sin
n
— /(sin ^(sin*""
— — cos x sin 71-1
= — cos x sin ^
1
x dx
7
x) dx
+
+
x
1
a:
—
—
(n
(n
l)/(cos
1)/ (sin
a*)
(sin
n~ 2
71-2
a:)
x
—
f
sin
sin
(cos x) dx
n
a:)
ak,
so
/sin
71
= —
x dx
- cos * sin
71
"1
+
a:
n
n
7*
-2
a:
dx.
J
(b)
n
J cos x dx
—
—
=
/(cos x)(cos n_1
sin x cos
sin x cos
71-1
71
"
1
x
dx
a:)
+
+
x
—
—
(n
(n
~"
n 2
a*) sin x dx
1)/ sin #(cos
n~ 2
x — cos n x) dx,
l)J(cos
so
/cos n x dx =
- sin
a*
cos
71-1
a*
+
n
cos n
f
n
~2
x dx.
J
(c)
/dx
xy
(i
«
2
(i
f
y
fx
dx
[
" j
a*
)^ "
y
)-
[
y *
1
dx
2
(i
a-
/dx
+
(1
1
2
dx
+
(i
)
*
xt
(i
dx
a:
I"
~
)*" 1
2
A*
w
2
-
L2(l
«)(1 ~h
2
A"
71
-1
"
)
-
*
f
-
7 2(1
+
n)(l
w
2
a:
)
1
^J
so
dx
r
y
+*
(1
a:
1
n
2
-
2(»
)
1)
(a:
2
+
l)
71
'1
_
We
can also use the substitution x
changes the integral to
/sec 2
sec
u du
271
u
f
2
cos 271
2(n
-
=
tan
(2w
/*
3)
1
y
1)
+
2
(a:
=
a, ^a-
l)
71
"1
sec 2 w du,
which
u du
J
2n
-
cos 271
3
u sin a
2
1
In
2
'
(Vi~+V
2
-
2n - Z
)
1) (1
+
'
2n
Vr+T
x
2k
* 2) n_1
2n
2
—
-
3
4
271
f cos
u du
J
x
1
1
2{n
+—
~
-2
2w
-
(1
+
a*
2
71
)
dx
3
f
2 y
dx
f
2 J
2n
-
'
1
(1
+
2
a-
71
)
"1
564
Answers
to Selected
Problems
CHAPTER
19
1.
(i)
p 3i0
(
=
x)
+ ex + ex +
2
e
<P2n.iT/2M
=
(-\) n (x
-
-
—
1
(5e/3\)x\
2w
tt/2)
(2n)\
(v)
Pn,i(x)
=
+ e(x -
e
—
H
1)
•
f-
H
•
•
2!
2.
(vii)
P4,oW =
(ix)
A H ,,oW
If /is
Theorem
-12
(iii)
243
W
(iii)
1
-
1
+
*2
-
*4
that R„, a (x)
•
= 0.
= Pn a (x).
then /
n,
0, so f(x)
(
* +1)
from
It follows
,
.
-
(x
3)
Y
90(*
-
3)
17f° r2
"
+2
+
2
3)
3
+
-
15(*
3)
4
5
.
(27^ <10
SinCe
=0
(since
——
V
2
1 -*'
-
J""
L(2>(2i+
1
=
+ (-i)V.
•
•
2
t(2^X
t'
V
.
+ 2(* - 3) + (x - 3)
+ 405(x - 3) + 270(x -
(i)
+
-
=
n\
3
a polynomial function of degree
Taylor's
3
+x
at
1)!
=0
"
-
2re+2
(2«
<
+
1CT 20
^
for 2n
9 '° r
^
+2>
9
)
18,
2)
>
or n
8^
11
- (since
i
7.
(i)
(iii)
=0
a =
a =
21
i.
ai
(t
-
(i)
i
(iii)
lim
n— w
+
(n
+
+
=
(v)
chapter
\
i\
10~ 5 for
n
+
>
1
+
=
1 )
vV +
1
a =
i
+
+
/( w
-
<
1)
=
1
i
>
e for n
(Vn 2
lim
n—
+ > 1/e.
+ - V^) +
1
lim (V
1
0
=
0.
(Each of
proved in the same way that lim(V
71—»
proved in the
1) =
< Vn _+
2
=
n—* »
<
0.
Vw,
n-+
<
hese
/
+
??
«
t
wo
i
/?
limits
- Vn) =
can be
0 was
=
So lim
lim
*< lo *<*>/*
=
e
o
(
by
1.
n
and lim (Vn) 2 = lim
Clearly a(n)
t
7
n— »
text.)
Clearly lim (log a)/n
Theorem
(ix)
).
0.
oo
/
(vii)
11
/
a t-_iA for
— V w + l)=0 +
(v)
>
12, or n
!
\)cn.
f(t) dt;
n / (n
<
1)
log 2
so
Vn +
(Vn) 2 <
Vl (V n) =
2
1
2
by parts
°°
n,
and lim
(log 2 n)/n
=
0.
n
< Vl (Vn) 2
(v)
and
(vi).
,
»
Answers
4.
If
(a)
<
0
<
a
(b) Part (a)
<
then a 2
2,
"^2
the
5.
If
sequence
x
is
same
tional,
then
for all
1
ti\tx
is
such
a multiple of
=
so f(x)
0,
<
2,
2.
w
= V2/,
/
w
any
for
or
/
=
is
}
2.
for sufficiently large w, so
2k
so lim(lim(cos n\itx) )
n— 00 k—> 00
n,
not a multiple of
is
=
lim(cos n\irx) 2k
-
-
-
So the hint shows that
as {a n +i}.
=
2k
<
2.
denoted by {a n \, then the sequence {V/2a w
is
rational, then nlirx
(cos n\irx)
^2 V2
V2a <
by Theorem
so the sequence converges
If this
<
4,
shows that
V2 < ^2 V2 <
(c)
so a
<
2a
565
Problems
to Selected
n,
=
1.
so |cos
x
If
«!7ta:|
is
<
irra1,
so
0.
a— *
*
6.
j
(i)
(hi)
/
Jo
e
x
—
dx
—
e
(Use partitions of
\.
—— dx = log
1
+
into n equal parts.)
[0, 1]
2.
*
1
Jo
CHAPTER
22
1.
(1 +
a+*)
^
2
~"
2'
(i)
(Absolutely) convergent, since [(sin nd)/n 2
(iii)
Divergent, since the
(Leibnitz's
Theorem does not apply
<
\/n 2
sum \
2n terms have
first
\
.
+
"
*
decreasing in absolute value.)
Convergent, since
(v)
1
<
A
for sufficiently large n.
Convergent, since
(vii)
lim
(«
+
!);/(*
2
/Z
+
!)!
. Hm (n_±AV
n-*»
n-*
« \
/tt!
(ix)
Divergent, since l/(log
(xi)
Convergent, since l/(log
(xiii)
Divergent, since
«3
for large
enough
+
/
_
.
n
+
Q
1
>
rc)
n2
72
ft
n)
n
>
<
^ for «
>
9.
1
2n
1
n.
Divergent, since
(xv)
L
3
dx
a:
log x
=
log(log
AO
—
log(log 3)
*
+
1
In.
since the terms are not
—*
oo
as
N—
>
oo.
.
566
Answers
to Selected
Problems
=
(Notice that /(*)
m
~ [1 +
-
Convergent, since
(xix)
Convergent, since
2n+l n
(
<
2
1/n 2 for n
1)n " " 1
+
1
lim
n™
a n 10~
^
so
n=
<
only
n
2.
2
r
lim
=
2
-j
n
2 ^10" <
The
when
1.
i(*
1
=
=
1,
1
lies
+
i
1
number is denoted by 0.aifl2«3«4
(Actually, this
the sequence {a n
}
is
not eventually
The
is
integral
•
+
-
•
•
.
is
m
ttV]
•
0.)
- 1] + » - tVi +
+ •••)- Ui +
+
+ +
+
0
+ <nr +•••)- id + i + + +
tf*
8
n
area of the shaded region
i(i
lCT n
^
9
1
converges by the boundedness criterion, and
w - « + +a - +«A+ +
1.
>
l
between 0 and
23
-
N
n=
CHAPTER
oo), since
by Problem 17-12.
For each
we clearly have
(a)
=
=
(«
[3,
+ l)nU
+ l) n+1
2{jl
=
2 n «!/fl n
0
15.
iorx>e
0
<
2
=
4.
decreasing on
is
x]
t
l/rc (log n)
+
n™
lo
,
(x log
X)
(xvii)
lim
l/(x log x)
•
•
•)
•
•)
•
•)
•
•
1
1-1
(0
f(x)
=
lim
n-> «
(iii)
/nW =
*
0
°>
(
11,
"
<
°
X
{/n } does not converge uniformly to /.
n
/(*) = lim fn {x) = 0 (since lim x =
n—
oo
n—
>
<
oo
,
1
for
*
>
1).
The
se-
oo
quence {/„} does not converge uniformly to
we have /„(*) large for sufficiently large x.
/; in fact, for
any
n
Answers
(
v)
=
/(*)
!/»(*)
2.
(i)
-
1
I
=
lim fn (x)
<
1/n for
X
X
a2
az
and {fn
0,
Selected Problems
converges uniformly to
/,
567
since
all *.
-- --
a
}
to
.
(iii)
k
3.
=0
(i)
e~*
(iii)
If
yw
yZ
YA
y.'k
2
3-2
4-3
11
'
then
= log(l+*)
so for
(1
4.
<
|*|
1
we have
+
*)log(l
+
=
f(x)
(1
+ *)log(l + x) -
Since /(0) = 0,
x) - x for \x\ < 1.
some number
for
|*|<1,
c.
we have
c
=
+ x) + c
(1
1,
so f(x)
=
Since
sin *
=
—
>
n =o
0
we have
(-l) n x 2n
+
n=0
(notice that the right side
is
1
—
for x
1)!
0).
n
I(-l)
(STi)?
0,
chapter
24
1.
(i)
|3
(iii)
|(1
+ Ai\
+0
ment
(v)
|(|3
5
for
+
|
= 5; 0 =
= (|1 +
1
4t|)|
+
=
i,
A;
_
=
2n
odd.
arctan
5
*1)
=
1;
since tt/4
an argument
|5|
k
So
=
5; 6
=
0.
for (1
+
=
i)
arctan 1/1
b
is
57r/4.
is
an argu-
568
Answers
to
Selected Problems
2.
(i)
X
V-l -
— -i ±
-i ±_ Vb
+
(-1
(iii)
x2
+
2ix
s
-
2
x
(v)
x
2
+x+
2)(x 2
1
^
2
2
1
•
2
2
1).
All z
=
All z
on the perpendicular bisector of the
and
No
2
a-
=
+ ry|
(z + z)/2
(*
-
z)/2
\z
+
w\
2^w =
2
since
2
j>
since
[a-
1
*
2
=
=
+
with
=
—i.
solutions are
line
segment between
real.
b.
z,
+
iy
x
is
The
'*
(i)
(v)
5.
so the only solution
,
(iii)
a
4.
L.
or
z')
-
(-1 - Vl)i
^—
—
Vs)i
1
'
3.
i
— = (x +
- x - 2 = (* 2
4
<
—
2x
+
y
[(x
[(*
\z
2
1
—
+
2
+
+
+
|#
—
2x
<
=
x2
]
<
x2
>
2
+
<
—
1
that x
>
which
2^-,
and that
1,
is
impossible,
1.
(
(z
implies
x
2
jv
+ — y) =
+0- (* -
=
—
1
so
,
0 for x
z»
i»
w\ 2
z'jy
+
—
\x
iy\.
z»]/2 = *.
iy)yii = y.
w)(z
+
+
fi?)
(*
—
w) (z
—
w)
=
2zz
+
Geometrically, this says that the sum of the
squares of the diagonals of a parallelogram equal the sum of the squares
of the sides.
CHAPTER
26
1.
2(|z|
2
\w\ ).
Converges absolutely, since
(i)
|(1
+
n
i)
/n\\
< (V2) n /n\, and
00
V
(iii)
(V2)»/„! converges,
Converges, but not absolutely, since the real terms form the series
~i +
i
~i+
I"
—
*
and the imaginary terms form the
«•(*-*
(v)
'
'
•
series
+ *- + +•••).
Diverges, since the real terms form the series
log 3
log 4
log 5
log 7
log 8
log 9
3
4
5
7
8
9
Answers
2.
(i)
The
n+x /{n
\z\
/ n;
n-> •
\z\
<1 for |*| <
The limit
+
/n
1,
is
<1 for |*| <
The limit
1
is
(v)
Problems
569
limit
lim
(iii)
to Selected
l)
2
= hm
2
(n
+
IV,
n
/
>1
but
>1
but
for
for
=
,
f
n->« \
>
\z\
>
\z\
\z\
)
,
,
\z\
1.
1.
on+l| J(n+l)l
lim -
7
0 for
is
3.
(i)
The
2n
3*
|*|
for
<1
for
and
a/3
|*|
|*|
>
1.
< V2,
so
h *"Ej
V2
V
2 n+1
the series converges absolutely for
< y/l. But the series does not converge
r
/O
> V2, so the radius of convergence is
11^
(iii)
2\z\^~*
lim
*r+*
°o
fcC- W
f
|*|
but
1,
,
limits
lim
are
,
<
|*|
=
1
2 n \z\ nl
n--
.
•
i •
1
for
Since
lim
n—> *
>_M:
=
n
\
72
lim
!£!i5 <„„!£! =
n-> «
n—
72
<»
the series converges absolutely for all
vergence
(v)
absolutely
-v
The
is
2,
so the radius of con-
oo
limit
V 2 n* =
/
lim
n—
is
»,
71
0 for
|*|
<
1,
n!
(n
2 lim *
n—*
oo
but
oo for
|*|
>
1
" 1)!
oo
,
so the radius of convergence
is 1.
GLOSSARY
OF SYMBOLS
P
(— co, a)
(-•,*]
(- 00, 00)
9
\a\_U
Vx
12
max (*,;;)
16
min(x,y)
16
("epsilon")
e
N
M
19
lim/
81
lim f(x)
*
lim /(*)
25
Q
lim /(*)
25
x—>a~
R
25,
x
38, 45,
/
41
f
+g
lim
41
x—
41
f
.
.
.
42
a„\
,
42
h
f-g-h
fo g
41
.)
.
.
+g+
112
112
/'
121
CA
48
*?(*)
130
48
49
max(/, g)
m in(/,
f<g
(a, 6)
137
55
(a,
oo)
55
[a,
oo)
55
92
52
263
264
265
275
Z(x) 276
log 285
exp 287
287
e
288
e*
288
289
296
296
296
arg sinh
296
arg tanh
296
296
299
arg cosh
/"'
137
Nap
137
F(x)
log
*
>»
if
'
302
303
If(x)dx
201
rj>)
214
*(/,«,*)
P
* n,a
215
(a, 6)
t/(/,P)
54
263
arcsin
/<*>
/-»
51
the open interval
[a, b]
131
^
49
49
the pair
cot
tanh
48
l/l
92
cosh
dx
/"
R-A
263
sinh
*
47
263
esc
lo go
df(x)
a 6'
263
tan
a*
127
127
43
47
259, 261
sec
*<J,P)
121
f'(a)
n
«
A
4
sin
arctan
121
42
*-»/(*)
573
112
infi4
259
259, 261
arccos
112
glb4
lim A
lim
43
oo
*>
lim sup
42
f°g°h
Ft
i = i
oo
254
cos
92
/(*) ==
sup 4
Iub4
41
{at,
87
lim /(*)
— 00
lim /(*) =
41
f-g
{*:
87
X—
507
/
A(x)
41
AC\B
//<
86
lim /(*)
254
257
sin r
257
t 258
a
lim /(*)
495
27
G>
X
/
sin°
86
I,
Z
/(*)
86
x—*a +
24
254
'/(*)<**
78
81
23
»!
254
/
70
lim /(*)
23
0
249
55
("delta")
5
249
70
W
21
55
55
215
219
334
334
343
357
(:)
225
\a n )
303
327
372
Glossary of Symbols
lim a n
n— oo
C
373
-
>
lim a n
71—»
=
376
co
Ke~
Im
A A O
445
T>
382
lim x n
385
a
n—
> oo
lim sup x n
385
479
"70
AA
441
\z\
oo
7
^
438
A A A
441
z
11
AAA
A
4/V
<fn
481
V'n
40Z
T
448
4o/, 49/
44v
lim /(^)
4o
449
U
/,
dUI
4o/, 4vo
n—» oo
lim * n
n—
>
sin
lim inf x n
n—
>
385
cos
exp
a, b)
386
bn
00
an
389
Bn
Z)
i
458
470
470
oo
N(n;
y
/'(a)
385
oo
433, 439
Dk
470
478
479
479
479
—a^
fl"
P
>
<
>
<
|<x|
1
A QQ AClQ
4oo,
4Vo
488, 502
489
490, 496
490, 496
490, 496
490, 496
501
INDEX
I
I
!
1
AalbmcndoE, 235
Arctan, 264
Abel, Niels Henrik, 332, 430
addition formula
Abel summable, 431
Abel's lemma, 385
derivative
remainder term
Arg
Arg
Arg
Absolutely convergent, 397, 463
Absolutely summable, 397
Acceleration, 137
Acta Eruditorum, 124
Argument
Addition, 3
law for, 9
commutative law of, 9
of complex numbers, 438
geometric interpretation
Addition formula
Argument
function, 449
453
of the hyperbolic functions,
296
Arithmetic mean, 32
of,
442
Arrow, "x arrow
sin (a: 2 ),"
43
Ars conjectandi, 481
Associative law
270
270
for addition, 9
266
for multiplication, 9
Average
266
for sin, 265,
for tan,
sinh,
discontinuities of,
associative
for cos,
342
351
296
cosh,
296
tanh, 296
Argument, 443
for arctan,
for, 335,
for,
Area, 214, 219
1
of a complex number, 441
for arcsin,
270
for,
264
Taylor polynomials
Abel's Theorem, 431
Absolute value,
of,
269
velocity, 128
Axis
Additive identity, existence
of,
9
Additive inverses, existence
of,
9
Algebra, Fundamental
Theorem
horizontal, 55
imaginary, 441
real,
of,
317, 445, 455, 474
441
vertical, 55
Algebraic functions, 302
Algebraic number, 362
Bacon, Francis, vi
Algebraist's real numbers, 505
Bandes, D., handiwork
Almost lower bound, 120
Almost upper bound, 120
Analyst's real numbers, 505
Angle, 256
directed, 256
Antidiagonal, 213
Arabic numerals, multiplication
Arccos, 264
Archimedian property
for the rational
for the real
numbers, 490
numbers, 116
Arcsec, 272, 322
8
Bernoulli's inequality, 31
game
hunting, mathematical theory
459
Binary operation, 487
Binomial
coefficient, 27,
Binomial
series,
357
409, 429
Binomial theorem, 28
Bisection argument, 120, 459
Bolzano- Weierstrass Theorem, 378,
addition formula
Taylor
Bernoulli polynomials, 481
of,
Bohr, Harald, 328
Arcsin, 263
577
Bernoulli numbers, 478
of,
Archimedes, 116, 119, 224
derivative
"Bent" graphs, 125
Bernoulli, 124, 481
Big
Arc length, 275
of,
Basic properties of numbers, 3
of,
for,
264
series for,
270
386, 459
Bound
429
almost lower, 120
Bound
Complex function
(Cont.)
almost upper, 120
continuous, 453
greatest lower, 112
differentiable,
upper, 111, 490
least
graph
of,
457
449
lower, 112
limit
upper, 111, 490
nondifferentiable, 458
Bounded above,
Bounded below,
100, 111, 377, 490
449
of,
Taylor
series for,
469
Complex nth. root, 443
Complex numbers, 433, 438
112, 377
Bourbaki, Nicholas, 124
absolute value
addition
of,
441
of,
438
Cauchy, 239
Cauchy Condensation Theorem, 410
geometric interpretation
Cauchy criterion, 390
Cauchy form of the remainder,
imaginary part of, 438
logarithm of, 477
345,
modulus
347
Cauchy Mean Value Theorem, 178
Cauchy sequence, 379, 477
Cauchy sequences, equivalence of, 505
Cauchy-Hadamard formula, 476
Cauchy-Schwartz inequality, 239
Cesaro summable, 408
Chain Rule, 150 ff.
proof
geometric interpretation
154
of,
442
442
of,
of,
441
multiplication
of,
438
geometric interpretation
of,
real part of,
438
sequence
462
of,
Complex plane, 440
Complex power series, 464
Characteristic (of a field), 492
466
466
Complex sequences, 462-463
Circle, 63
Complex
of,
Change, rate
"/ circle
unit,
circle of
of,
128
of,
series,
of,
463
Complex-valued functions, 448
Composition of functions, 42
Concave function, 192
64
Circle of convergence, 466
Classical notation
Conditionally convergent
142-
143, 162, 212
for integrals,
convergence
radius of convergence
42
for derivatives, 130-131, 138,
442-
443
225
Cleio, 161
Closed interval, 55
Closed rectangle, 454
Closure under addition, 9
Closure under multiplication, 9
Commutative law
398
Constant function, 41
Construction of the real numbers, 494
Continuous at a, 93, 452
Continuous function, 93, 453
nowhere
differentiable, 135,
Continuous on (a,
Continuous on [a,
Contraction, 384
b),
96
b],
96
for addition, 9
Contraction lemma, 385
for multiplication, 9
Converge
Comparison test, 391, 407
Comparison Theorem, Sturm, 275
Complete induction, 23
Complete ordered field, 490, 509
Completing the square, 18, 318
Complex analysis, 473
series,
Conjugate, 441, 446
pointwise, 415
uniformly, 415, 419
Convergent sequence, 373, 462
Convergent series, 389, 463
absolutely, 397, 463
conditionally, 398
422
Index
Convex function, 191
Convex hull, 460
Convex subset of the plane, 460
Cooling, Newton's law
of,
Deritative, left-hand, 132
Leibniz ian notation
right-hand, 132
second, 137
55
Derivative of quotient, incantation for,
Coordinate system, 55
147
Diagonal, 202
55
of,
Difference operator, 479
"Corner," 58
DifTerentiable, 127, 457
Cos, 256, 259, 273-274, 276, 470
addition formula
derivative
inverse
Deriva-
"negative infinity," 134
298
second, 55
origin
for, see
notation for
tive, classical
Coordinate
first,
579
of,
of, see
for,
266
Differential equation, 247, 253, 273,
148, 260
275, 301. 359
Arccos
initial
conditions for, 360
Differentiation, 144
Taylor polynomials for, 334
remainder term for, 348
ff.
implicit, 211
Cosh, 296
Differentiation operator, 479
Cosine, hyperbolic, 296
Directed angle, 256
Cot, 263
Discontinuities of a nondecreasing
function, 370
Countable, 369
Counting numbers, 21
Discontinuity, removable, 99
Critical point, 165
Disraeli,
Critical value, 165
Distance, 56, 441
Csc, 263
Benjamin, 2
between two
shortest
Cubic equation, general
solution,
435
points,
276
Distributive law, 9
Diverge, 373, 463
Division, 6
Darboux's Theorem, 187, 242, 253
Decimal expansion, 70, 407
Division by zero, 6
Dedekind, Richard, 36
Domain, 38, 39, 45, 507
Double intersection, 141
Double root, 160
Defined implicitly, 211
Durege, 36
Decreasing function, 170
Definite integral, 304
DEFINITION, 45
Definition, recursive, 33
e9
Degree (of a polynomial), 40
Degree measurements, 61, 257-258
Delicate ratio test, 408
Delicate root test, 408
De Moivre's Theorem, 443
Dense, 118
Derivative, 125
ff.,
127
classical notation for,
130-131, 138,
142-143, 162, 212
a,
irrationality of,
353
relation with
368, 471
127
higher-order, 137
"infinite," 134
7r,
transcendentality
value
of,
of,
364
288, 350
Elementary function, 302
Ellipse, 64
Empty collection, 23
Entire function, 474
Epsilon, 19
Equal up
of/, 127
of/ at
287
to order n,
Equality, order
of,
340
340
Equations, differential,
equations
see Differential
Equivalent Cauchy sequences, 505
Euler, 481
number, 382
474
Eventually inside, 462
fixed point
Exhaustion, method
Exp, 287 ff., 470
of,
exponential, 287-288
119
graphs
93
of,
384
of,
54-63, 173, 449
hyperbolic, 296
of,
Taylor polynomials for, 335
remainder term for, 349-350
Expansion, decimal, 70, 407
increasing, 170
Extension of a function, 93
integrable, 219
identity, 41
imaginary part, 448
integral
219
of,
inverse, 201
Factorial, 23
ff.
linear, 56
Factorials, table of, 356
local
Factorization into primes, 31
Fibonacci, 31
local
local
Fibonacci Association, 32
logarithm, 285, 289
Fibonacci Quarterly, The, 32
Fibonacci sequence, 31, 430, 478
Field, 487
characteristic of, 492
complete ordered, 490, 509
ordered, 489
First coordinate, 55
First
240
maximum
maximum
point
of,
value
of, 1
minimum
value
for Inte-
grals, 237
Fixed point of a function, 384
Fourier series, 270, 272
Function, 37, 45
of,
163
63
163
most general definition
negative part
of,
of,
nonnegative, 49
43
for,
odd, 49, 174
periodic, 69, 140, 252
polynomial, 40
positive part of, 49
power, 58
507
product
absolute value, 448
argument, 449, 453
complex- valued, 448
composition of, 42
of,
quotient
of,
rational,
40
real part,
41
41
448
real-valued, 448
concave, 192
conjugate, 448
"reasonable," 96, 125, 156
constant, 41
regulated, 431
continuous, 93
strict
critical point of,
value
235
step,
ff.
convex, 191
critical
507
49
nondecreasing, 213
notation
Mean Value Theorem
to B,
maximum point of, 164
minimum point of, 164
strict maximum point of,
nonincreasing, 213
Fundamental Theorem of Calcu-
from A
ff.
even, 49, 174
extension
lus,
125
elementary, 203
entire,
Even function, 49, 174
Even number, 25
First
of,
differentiate, 127
Euler-Maclaurin Summation Formula,
482
Euler's
Function, derivative
of,
decreasing, 170
1
65
165
sum
maximum
of,
point
41
trigonometric, 256
value of at
x,
38
ff.
of,
190
190
Index
Fundamental Theorem of Algebra,
317,
Indefinite integrals, short table
445, 455, 474
of,
304-
305
Fundamental Theorem of Calculus
First,
581
Induction, mathematical, 21
240
complete, 23
Second, 244
Inductive set of real numbers, 33
Inequalities, 9
in
Galileo, 124, 140
Gamma
function, 327, 364
Gauchy-Schwartz, 239
geometric-arithmetic mean, 32
Jensen's, 199
"x goes to sin(x 2 )," 43
Graph sketching, 171-176
to,
Liouville's,
369
Schwartz, 18, 239
Graphs, 54-63, 72-73, 173, 449
Greatest lower bound, 112
Grow
490
field,
Bernoulli's, 31
Geometric mean, 32
Geometric series, 390
Global properties, 101
Goes
an ordered
Inequality
triangle, 68
Infimum, 112
faster than, 292, 301
"Infinite" derivative, 134
Infinite intervals, 55
Hadamard, 476
Half-life (of radioactive substance),
298
Infinite products, 282,
328
Infinite sequence, 372,
462
Infinite series, 389
Hermite, 363
Infinite
High-school student's real numbers, 505
Higher-order derivatives, 137
Infinitely
"Infinitely small," 131, 225
Hilbert, 363
Infinity, 55
Horizontal
axis,
55
sum, 354, 388
many
primes, 31
minus, 55
Hyperbola, 65
Inflection point, 197
Hyperbolic functions, 296
Initial conditions in differential
equa-
360
Instantaneous speed, 128
tions,
Identity
Instantaneous velocity, 128
additive, 9
Integer, 25
multiplicative, 9
Identity function, 41
Identity operator, 480
Imaginary axis, 441
Imaginary part of complex number,
438
Imaginary part function, 448
Integrable, 219
Integral, 219
classical notation for, 225,
definite,
First
Mean Value Theorem
Improper integral, 254, 329
Incantation for derivative of quotient,
237
indefinite,
304
short table
of,
304-305
Leibnizian notation
for, see
Integral,
classical notation for
lower, 249
147
Increasing at
for,
improper, 254-255, 329
Implicit differentiation, 211
Implicitly denned, 211
226
304
a,
189
Increasing function, 170
Second
Mean Value Theorem
for,
237
Mean Value Theorem for,
Increasing /sequence, 377
Third
Indefinite integral, 304
upper, 249
237
Integral form of the remainder, 345,
Leibnizian notation for derivatives, 131,
346
142-143, 162, 212
Integral sign, 219
Integral
test,
for higher-order derivatives, 138
396
Leibniz's formula, 159
Integration
219
by
305
parts,
Theorem, 398
Leibniz's
limits of,
Lemma,
ff.
substitution, 307
82
Length, 275
of rational functions, 316
by
Leibniz, 124, 131, 225
L'Hopital, Marquis de, 124
fT.
Interest (finance), 298
L'Hopital's Rule, 179, 186-187
Intermediate Value Theorem, 102, 109,
113, 253
Interpolation, Lagrange, 47
Limit, 72
Intersection of
sets,
at infinity, 87
of integration, 219
closed, 55
of a sequence, 373
55
uniqueness
open, 54
see also
Inverse
Lindemann, 367
additive, 9
Line, real, 54
multiplicative, 9
Inverse of a function, 201
Line, tangent,
ff.
Inverses of trigonometric functions,
see
Trigonometric functions
Tangent
see
line
Linear functions, 56
Liouville, 368
Liouville's inequality, 369
Irrational numbers, 25
fields,
80
of,
Limit point, 386, 459
Limit superior, 121, 385
Nested Intervals Theorem
Isomorphic
449
78,
"does not exist," 81
41
Interval, 54
infinite,
ff.,
from above, 86
from below, 86
Theorem, 474
Liouville' s
508
Local
Isomorphism, 508
point of function, 164
maximum
higher-order derivative
second derivative
Local
Jensen's inequality, 199
Johnson, Samuel, 514
minimum
see also
test for,
test for,
point of function, 164,
maximum point
Local
Local property, 89, 101, 142
Local strict maximum point, 190
Jump, 58
Log, 285
Klingenstein, K.,
trials
and
tribulations
of,
Taylor polynomials for, 335
remainder term for, 351
Logarithm
to the base 10,
Lagrange form of the remainder, 345,
347
Lagrange interpolation formula, 47
Large negative, 62
Least upper bound, 111 ff., 490
Least upper bound property, 113
Lebesgue, see Riemann-Lebesgue
Lemma
Left-hand derivative, 132
283
of a complex number, 477
Napierian, 299
Lower bound, 112
almost, 120
greatest, 112
Lower integral, 249
Lower limit of integration, 219
Lower sum, 215
Lowest terms, 71
339
176
Index
Newton's law of cooling, 298
Maclaurin, 482
Mass, rate of change
128
of,
Mathematical induction, 21
Maximum
point of a function, 163
local, 164, see also
Local
maximum
point
local strict, 190
strict,
Mean,
of
Newton's laws of motion, 137
Nondecreasing function, 213
Nondecreasing sequence, 377
Nondifferentiable complex functions,
458
Nonincreasing function, 213
Nonnegative function, 49
Nonnegative sequence, 391
Notational nonsense, 479
190
Maximum
Maximum
two numbers, 16
value of function, 163
Nowhere-difTerentiable continuous
arithmetic, 32
function, 422
geometric, 32
Mean Value Theorem,
nth root, 69, 443
168, 169
Cauchy, 178
existence
for integrals (First,
Second, and
1
complex, 433, 438
point of a function, local,
counting, 21
Minus
of
Local
minimum
point
two numbers, 16
infinity,
Modulus
even, 25
imaginary, 433
irrational, 25
55
natural, 25, 33
Mirifici logarithmonum canonis
description,
460
algebraic, 362
of function, 163
64, see also
Minimum
103, 443,
Number
of exhaustion, 119
Minimum
Minimum
of,
primitive, 447
Third), 237
Method
583
odd, 25
299
of a complex number, 441
prime, 31
Mollerup, Johannes, 328
rational, 25
Multiplication, 5
real, 25, 441,
495
transcendental, 362
of arabic numerals, 8
associative law for, 9
Numbers, basic properties
closure under, 9
Null
set,
of,
3
23
commutative law of, 9
of complex numbers, 438
geometric interpretation, 442-443
Multiplicative identity, existence
of,
Multiplicative inverses, existence
of, 9
9
Odd function, 49,
Odd number, 25
One-one
174
function, 200
Multiplicity (of a root), 108
Open
Napier, 299
"Or," 6
Order of equality, 340
Ordered field, 489
complete, 490
Ordered pair, 45, 52
Napierian Logarithm, 299
Natural numbers, 21, 33
Negative, large, 62
"Negative
infinity," derivative,
interval, 54
Origin (of a coordinate system), 55
134
Negative number, 9
Negative numbers, product of two, 7
Negative part of a function, 49
Nested Intervals Theorem, 120
Newton, 131, 235
Pair,
44
ordered, 45, 52
Parabola, 58
area under, 224
Partial sums, 388
258
Archimedes's approximation
Partition, 215
irrationality of,
Partial fraction decomposition, 317
7T,
relation to
"Peak," 59
Peak point, 378
transcendental ity
value
of,
of,
368
357
Period of a function, 69, 140, 252
Vieta's product for 2/x, 282
Periodic function, 69, 140, 252
Wallis's product for tt/2, 328
Perpendicularity of
119
All
Pascal's triangle, 27
e,
of,
279
68
lines,
Petard, H., 459
Pig, yellow, v,
Quaternions, 493
314
Quotient of functions, 41
Pigheaded, 161
Plane, 56
complex, 440
Rabbits, growth of population, 32
Point, 54
257-258
Point of contact, 192
Radian measure,
Point-slope form of equation of a line,
Radioactive decay, 298
Radius of convergence of complex
57, 68
power series, 466
Rate of change of mass, 128
Rate of change of position, 128
Polynomial function, 40
graph of, 58-59, 173
multiplicity of roots, 108, 160
specific
ff.,
see also
change
Positive elements of
field,
489
Positive part of a function, 49
Power
Power
Power
Power
of,
number, 9
functions, 58
series,
series
complex, 464
centered at
<z,
uniqueness
2,
table
of,
356
Prime number, 31
characteristic of a field, 492
many
of,
31
unique factorization
into, 31
Primitive, 302
Primitive
Principia,
nth. root,
447
235
Product, 5
of functions, 41
of
316
Real axis, 441
Real line, 54
Real number (formal definition), 495
Real numbers, 25
algebraist's, 505
505
Archimedian property
construction of, 494 ff.
of,
116, 490
high-school student's, 505
430
infinitely
of,
Rational numbers, 25
analysist's,
423, 471
series representation,
Powers of
408
integration
128
of,
an ordered
Positive element of R, 500
Positive
393
Rational functions, 40
functions
Position, rate of
test,
delicate,
Ratio
Polynomials, Bernoulli, 481
Polynomials, Taylor, 334
61,
two negative numbers, 7
Pythagorean theorem, 25, 56
inductive set of, 505
Real part of a complex number, 438
Real part function, 448
Real-valued function, 448
Rearrangement of a sequence, 400
"Reasonable" function, 66, 96, 125, 156
Rectangle, closed, 454
Recursive definition, 23
Reduction formulas, 316
Regulated function, 431
Remainder term for Taylor polynomials, 343
Removable
discontinuity, 99
585
Index
Riemann-Lebesgue lemma, 272, 327
Right-hand derivative, 132
Rising
Sun Lemma, 121
Theorem, 168
Root of a polynomial
function, 48
multiplicity
test,
power, 423, 471
Taylor, 424
108
of,
see also nth. roots
delicate,
convergent, 389
geometric, 390
double, 160
Root
conditionally convergent, 398
Fourier, 270, 272
Rolle, 161
Roile's
Series (Cont.)
408
408
22
empty, 23
Set,
Sets
intersection of, 41
notation
Shadow
Same
sign, 12
Sec, 263
inverse
Sign, 12
Sin, 41, 256, 259, 273-274, 276,
addition formula
of, see
line,
Arcsec
126
Second coordinate, 55
Second derivative, 137
Second derivative test for maxima and
minima, 176
Second Fundamental Theorem of Calculus, 244
Second Mean Value Theorem for integrals,
41-42
Sigma, 24
Schwartz, H. A., 18, 190, 239
Schwartz inequality, 18, 239
Secant
for,
point, 121
237
Sine, hyperbolic, 296
Sine function, 41
Sinh, 296
Sketching graphs, 171-176
Skew-field, 493
Slope of a straight
Sequence
470
266
derivative of, 148, 260, 272-273
inverse of, see Arcsin
Taylor polynomials for, 334
remainder term for, 348
for,
line,
56
Speed, instantaneous, 128
absolutely summable, 397
Cauchy, 379, 505
complex, 477
Square
434
102-103
root, 12,
existence
of,
pointwise, 415
Square root in a field, 492
Square root function, 453
Squaring the circle, 367
Step function, 235
Stirling's formula, 483
uniformly, 415
Straight line (analytic definition), 56
equivalence
of,
505
of complex numbers, 462
convergent, 373
decreasing, 377
shortest distance
divergent, 373
Fibonacci, 31, 430, 478
increasing, 377
infinite,
between two points,
276
slope
Strict
372
of,
56
maximum
point, 190
Sturm Comparison Theorem, 275
limit of, 373
Subsequence, 378
nondecreasing, 377
Substitution, world's sneakiest, 325
nonincreasing, 377
Substitution formula, 308
rearrangement
of,
400
summable, 389
Series
absolutely convergent, 397
Subtraction, 5
Sum
finite,
3-4
of functions, 41
Sum
Two-times
(Cont.)
infinite, 354,
of
of
an infinite sequence, 389
an infinite sequence of complex
numbers, 462
lower, 215
388
partial,
sigma notation
upper, 215
Sum
differentiable, 137
388
for,
24
limit, 415
Uniformly convergent sequence, 415
Uniformly convergent series, 419
Uniformly distributed sequence, 387
Uniformly summable, 419
Uniqueness
Uniform
of factorization into primes, 31
of squares, 459
of limits, 80
Summable, 389, 463
of
power
absolutely, 397
Unit
circle,
Cesaro, 408
Upper bound,
Abel, 431
111, 490
490
least, 111,
Supremum, 112
Symmetry
430
64
almost, 120
uniformly, 419
Swift, Jonathan,
series representations,
Upper integral, 249
Upper limit of integration, 219
Upper sum, 215
486
in graphs,
174
Tan, 263
"Valley," 59
Arctan
Taylor series for, 497
Tangent, hyperbolic, 296
inverse
Tangent
of, see
line, 125,
point of contact
"Tangent line,"
Tanh, 296
Value, absolute,
127
of,
average, 128
134
instantaneous, 128
Vertical axis, 55
Taylor polynomial, 334
remainder term
Taylor
series,
Absolute value
Velocity
192
vertical,
see
Value of / at x, 38
Vanishing condition, 390
of,
343, 345, 346
Viete, Francois, 282
424, 469
Taylor's Theorem, 345
Third
Mean Value Theorem
for inte-
grals, 237
Transcendental number, 362
Triangle inequality, 68
Trichotomy law,
Cos, Cot, Csc, Sec, Sin,
see
Bolzano- Weierstrass
Theorem
Weierstrass M-test, 420
Well-ordering principle, 23
9
Trigonometric functions, 256
Wallis's product, 328
Weierstrass,
see also
Wright, 326
Tan
of, 315-316
263 see also Arccos, Arcsec,
Arcsin, Arctan
integration
inverses
of,
Zahl, 25
Zero, division by, 6
GLOSSARY OF SYMBOLS
P
(—
.9
oo,
w _«
(-oo,
V*
(—
12
max (#,;>)
16
M
min(*, ^)
16
{*}
21
0
23
23
55
55
oo
\j
u/aV
55
oo )
5
L
70
'/
Km/(x)
81
oo
«!
lim/
81
25
R
25,495
lim /(*)
0
27
lim /(*)
/(*)
38, 45, 507
lim
+*
Pi
a
t
—
oo
42
an ]
,
42
A
^4
121
121
127
127
df(x)
131
47
CA
48
dx
48
R-A
48
49
l/l
max(/, £)
49
min (/,
49
5)
f<g
'»
/-»
51
the pair
(a, 6)
52
the open interval
55
(«,
[a>
)
55
oo)
55
00
214
215
215
&
Pn,a
/>»,«,/
Rn,a
for
219
/.'/(*)
*
302
303
)f(x) dx 303
327
r(x)
*(/,«,*)
(a, 6)
arccos
//
201
£/(/,/>)
54
[a, b]
263
F(x)
121
130
a 6'
92
137
112
dx
47
263
arc tan
df(x)
n
cot
f»
/'
*<
92
137
/'(a)
f°g°h
lim
43
oo
112
4
ihn ^
f-g-h 42
/«| 42
x-*f(x)
=
112
lim sup
43
263
arcsin
112
gib
41
.
+£ + A
n
=
sup
.}
.
263
esc
/"'
infil
.
263
tan
f"
41
.
sec
x=a
137
41
.
259, 261
264
265
t(f,P) 275
£(*) 276
log 285
exp 287
287
e
e 288
288
a*
289
log„
sinh 296
cosh 296
tanh 296
arg sinh 296
arg tanh 296
arg cosh 296
Nap log 299
A
lub4
-g
259, 261
sin
92
/(*)
lim /(#)
41
{a h
87
> oo
x—+a
41
cos
87
lim f(x)
41
f-g
/
a:
5
//*
{*:
a;
86
/(a:)
lim
4
254
257
257
258
A(x)
259
o
J.
Q,
/
f
sin r
X
41
254
254
7T
25
/
/
J
0
sin
86
lim/(;c)
Z
c
86
/(a:)
24
a<
254
oo J
a
lim
7
249
f(x) dx
78
("delta")
5
19
249
70
'
e ("epsilon")
N
fl )
a]
334
334
343
357
(:)
225
{a n
372
Glossary of Symbols
lim a n
lim a n
373
=
oo
Re
Im
385
17m sup xn
n-»
^
lim x„
*
^
385
a, b)
Um/(z)
y
Ll
an
389
i
Printed by Interprint (Malta)
Ltd
433,439
+
.
449
487,498
1
488, 502
_a
458
470
a-1
cos
470
p
470
>
<
>
<
386
*»
D
Z>*
478
479
479
479
487, 497
4g7) 501
0
sin
bn
n = -l
482
448
449
exp
N(n;
481
f>n
448
/'(«)
385
479
479
A
*-*«
385
lim inf x„
eD
441
\z\
382
lim x n
438
441
376
n—> oo
7
C
|«|
488> 498
488, 502
489
490, 496
490,496
490 496
490, 496
>
501
CDEF798765
SELECTED "WORLD STUDENT SERIES' EDITIONS
OF OUTSTANDING UNIVERSITY TEXTBOOKS
FOR STUDENTS OF MATHEMATICS
•CALCULUS
586 pp,
ilius
527pp. 61
ilius
By Michael Spivak, 8randeis University
•PROS ABILITY WITH STATISTICAL APPLICATIONS, Second
By Frederic Mosteller, Harvard
Edition
University,
Rourke. St. Stephen's School, Rome,
Thomas, Jr, Massachusetts Institute of Technology
Robert
E. K.
George
B.
•GENERAL STATISTICS, Second
Edition
By Audrey Haber, University of California at Los Angeles
Richard P. Runyon, C. W. Post College of Long island University
414 pp, 43 ilius
•AN INTRODUCTION TO LINEAR ANALYSIS
By Donald
L,
Kr eider, Dartmouth College,
Robert G. K utter,
Wayne
State University,
Donald R Ostberg, Indiana University,
Fred W. Perkins, Dartmouth College
792 pp 177 ilius
t
FUNDAMENTALS OF BEHAVIORAL STATISTICS, Second
Edition
By Richard P. Runyon, C W. Post College of Long Island University
Audrey Haber, University ol California at Los Angeles
A FIRST COURSE IN CALCULUS, Third Edition
A SECOND COURSE IN CALCULUS, Second Edition
ANALYSIS
500 pp, 28 ilius
305 pp, 116 ilius
460pp
I
LINEAR ALGEBRA, Second
Alt
351 pp, Hlus
400 pp, 59 illus
Edition
by Serge Lang, Columbia University
ELEMENTS OF DIFFERENTIAL EQUATIONS
270 pp, 65 iffus
ADVANCED CALCULUS
673 pp, 241
ilius
Both by Wilfred Kaplan, University of Michigan
Ask your
*
local bookseller for a complete
An Open
University Set
fist
of inexpensive World Student Series editions
Book
ADDISON-WESLEY PUBLISHING COMPANY
lONDGfo
READING, MASSACHUSETTS
AMSTERDAM
SINGAPORE
SYDNEY
TUK V U