CALCULUS
Michael Sþimh
CALCULUS
Thini Edition
Copynght
© 1967, 1980, 1994 byMichaelSpivak
All nghts reserved
Libraryof Congnus
CatalogCardNumber80-82517
Publish or Perish, Inc.
1572 West Gray, #377
Houston, Texas 77019
in the UnitedStatesof America
Manufactured
ISBN 0-914098-89-6
Dedicated to the Memory of Y. P.
PREFACE
FRANCIS
BA00N
PREFACE TO THE FIRST EDITION
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 to expect,
rather
than avoid, the strengthening
of insight with logic. In addition to developing the students' intuition about the beautiful concepts of analysis, it is surely
to persuade them that precision and rigor are neither deterrents
to intuition, nor ends in themselves, but the natural medium in which to formulate
and think about mathematical questions.
equally important
which, in a sense, the entire book
This goal implies a view of mathematics
well
matter
particular
topics may be developed, the
how
attempts to defend. No
of
will
realized
only
succeeds
goals this book
be
if it
as a whole. For this reason, it
would be of little value merely to list the topics covered, or to mention pedagogical
practices and other innovations. Even the cursory glance customarily bestowed on
new calculus texts will probably tell more than any such extended advertisement,
and teachers with strong feelings about particular aspects of calculus will know just
where to look to see if this book fulfills their requirements.
A few features do require explicit comment, however. Of the twenty-nine chapchapters are optional, and the three chapters comters in the book, two (starred)
prising Part V have been included only for the benefit of those students who might
of the real numbers. Moreover, the
want to examine on their own a construction
and
contain
optional material.
appendices to Chapters 3
11 also
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
Since the most exciting concepts of calculus do not appear
until Part III, it should be pointed out that Parts I and II will probably require
the entire book covers a one-year
less time than their length suggests-although
chapters
covered
the
be
meant
at any uniform rate. A rather
not
to
course,
are
natural dividing point does occur between Parts II and III, so it is possible to
reach differentiation and integration even more quickly by treating Part II very
briefly, perhaps returning later for a more detailed treatment. This arrangement
of most calculus courses, but I feel
corresponds to the traditional organization
who have seen a
that it will only diminish the value of the book for students
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
but not overly simple, exercises which develop basic
range from straightforward,
techniques and test understanding
of concepts, to problems of considerable diffiand,
comparable
culty
interest. There are about 625 problems in all.
I hope, of
Those which emphasize manipulations usually contain many example, numbered
"topics."
vii
viii
Preface
with small
Roman numerals, while small letters are used to label interrelated parts
other
in
problems. Some indication of relative difficulty is provided by a system of
starring and double starring, but there are so many criteria for judgingdifficulty,
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 consulted, that the best of students will probably have to attempt only
those which especially interest them; from the less difficult problems it should be
easy to select a portion which will keep a good class busy, but not frustrated. The
answer section contains solutions to about half the examples from an assortment
of problems that should provided a good test of technical competence.
A 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 Reading list, to which the
problems often refer, and a glossary of symbols.
I am grateful for the opportunity to mention the many people to whom I owe my
thanks. JaneBjorkgren performed prodigious feats of typing that compensated for
Richard Serkey helped collect the material
my fitful production of the manuscript.
which provides historical sidelights in the problems, and Richard Weiss supplied
the answers appearing in the back of the book. I am especially grateful to my
friends Michael Freeman, JayGoldman, 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. Needles 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 success. This is always safeafter all, the class is unlikely to rise up in a body and protest publicly-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
February1967
MIC HAE L SPIV AK
PREFACE TO THE SECOND EDITION
I have often been told that the title of this book should really be something like "An
Introduction to Analysis," because the book is usually used in courses where the
students have already learned the mechanical aspects of calculus-such
courses are
standard in Europe, and they are becoming inore common in the United States.
After thirteen years it seems too late to change the title, but other changes, in
addition to the correction of numerous misprints and mistakes, seemed called for.
There are now separate Appendices for many topics that were previously slighted:
polar coordinates, uniform continuity, parameterized curves, Riemann sums, and
the use of integrals for evaluating lengths, volumes and surface areas. A few topics,
like manipulations with power series, have been discussedmore thoroughly in the
text, and there are also more problems on these topics, while other topics, like
Newton's method and the trapezoid rule and Simpson's rule, have been developed
in the problems. There are in all about 160 new problems, many of which are
intermediate in difficulty between the few routine problems at the beginning of
each chapter and the more difficult ones that occur later.
Most of the new problems are the work of Ted Shiflin. Frederick Gordon
pointed out several serious mistakes in the original problems, and supplied some
non-trivial
corrections,
as well as the neat proof of Theorem 12-2, which took
two Lemmas and two pages in the first edition. JosephLipman also told me
of this proof, together with the similar trick for the proof of the last theorem in
the Appendix to Chapter 11, which went unproved in the first edition. Roy O.
Davies told me the trick for Problem 11-66, which previously was proved only in
Problem 20-8 [21-8in the third edition], and Marina Ratner suggested several
interesting problems, especially ones on uniform continuity and infinite series. To
all these people
go my thanks, and the hope that in the process of fashioning the
contributions
weren't too badly botched.
edition
their
new
MICHAEL
ix
SPIVAK
PREFACE TO THE THIRD EDITION
The most significant change in this third edition is the inclusion of a new (starred)
Chapter 17 on planetary motion, in which calculus is employed for a substantial
physics problem.
In preparation for this, the old Appendix to Chapter 4 has been replaced by
three Appendices: the first two cover vectors and conic sections, while polar coordinates are now deferred until the third Appendix, which also discusses the polar
coordinate equations of the conic sections. Moreover, the Appendix to Chapter 12
has been extended to treat vector operations on vector-valued curves.
of old material.
Another large change is merely a rearrangement
"The Cosmopolitan Integral," previously a second Appendix to Chapter 13, is now an
ChapAppendix to the chapter on "Integration in Elementary Terms" (previously
which
chapter
used
those
that
problems from
ter 18, now Chapter 19); moreover,
the material from that Appendix now appear as problems in the newly placed
Appendix.
A few other changes and renumbering of Problems result from corrections, and
elimination of incorrect problems.
I was both startled and somewhat dismayed when I realized that after allowing 13 years to elapse between the first and second editions of the book, I have
allowed another 14 years to elapse before this third edition. During this time I
list of corrections, but no longer have
seem to have accumulated
a not-so-short
the original communications, and therefore cannot properly thank the various individuals involved (whoby now have probably lost interest anyway). I have had
time to make only a few changes to the Suggested Reading, which after all these
years probably requires a complete revision; this will have to wait until the next
edition, which I hope to make in a more timely fashion.
MICHAEL
x
SPIVAK
CONTENTS
PREFACE
PART I
vi
Prologne
l Basic Properties of Numbers 3
2 Numbers of Various Sorts 21
PART II
Foundations
3 Functions 39
Appendix.OrderedPairs 54
4 Graphs 56
Aþþendix1. Vectors75
Appendix2. The Cone Sections 80
Appendix3. Polar Coordinates 84
5 Limits 90
6 Continuous Functions 113
7 Three Hard Theorems 120
8 Least Upper Bounds 131
Appendix.UnformContinuity 142
PART III
Derivatives
and
Integrals
9 Derivatives 147
10 Differentiation 166
11 Significance of the Derivative 185
Appendix.Convexipand Concavig 216
12 Inverse Functions 227
of Curves 241
Appendir.ParametricRepresentation
Integrals
250
13
Appendix. RiemannSums 279
14 The Fundamental Theorem of Calculus 282
xi
xii
Contents
15 The Trigonometric Functions 300
x is Irrational 321
*17
Planetary Motion 327
18 The Logarithm and Exponential Functions
19 Integration in Elementary Terms 359
Integral 397
Appenda. The Cosmopolitan
*16
PART IV
Infinite Sequences
336
and Infinite Series
20 Approximation by Polynomial Functions 405
e is Transcendental 435
22 Infinite Sequences 445
23 Infinite Series 464
24 Uniform Convergence and Power Series 491
25 Complex Numbers 517
26 Complex Functions 532
27 Complex Power Series $46
*21
PART V
Epilogue
28 Fields 571
29 Construction of the Real Numbers 578
30 Uniqueness of the Real Numbers 591
Suggested
Reading 599
Answers(toselected problems)609
Glossaryof Symbols 655
Index 659
CALCULUS
PART
PROLOGUE
To be consdaus ihat
ym are ignorant is a grea¢ step
to knowledge.
HNþMB
DISRHO
BASIC PROPERTIES OF NUMBERS
CHAPTER
The title of this chapter expresses in a few words the mathematical knowledge
to read this book. In fact, this short chapter is simply an explanation of
what is meant by the
and
properties of numbers," all of which-addition
multiplication,
subtraction
and division, solutions of equations and inequalities,
already familiar to us. Neverfactoring and other algebraic manipulations-are
required
"basic
Despite the familiarity of the subject, the
theless, this chapter is not a review.
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. Some may even seem too
obvious to mention, but a surprising number of diverse and important facts turn
out to be consequences of the ones we shall emphasize.
Of the twelve properties which we shall study in this chapter, the first nine are
concerned with the fundamental operations of addition and multiplication. For
this operation is performed on a pair
the moment we consider only addition:
of numbers-the
sum a + b exists for any two given numbers a and b (which
may possibly be the same number, of course). It might seem reasonable to regard
addition
as an operation which can be performed on several numbers at once, and
consider the sum a1 +
, an as a basic concept.
It is
+ an of n numbers a1,
of
convenient,
only,
and
consider
addition
of
pairs
numbers
however,
to
more
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 two different ways. One can first add b and c,
obtaining b + c, and then add a to this number, obtaining a + (b + c); or one can
first add a and b, and then add the sum a + b to c, obtaining (a + b) + c. Of
course, the two compound sums obtained are equal, and this fact is the very first
property we shall list:
(Pl)
If a, b, and c are any numbers,
a+(b+c)=
then
(a+b)+c.
of this property clearly renders a separate concept of the sum of
three numbers superfluous; we simply agree that a + b + c denotes the number
a + (b+ c)
(a + b) + c. Addition of four numbers requires similar, though slightly
The symbol a + b + c + d is defined to mean
more involved, considerations.
The statement
=
or
or
or
or
3
(1) ((a + b) + c) + d,
(2) (a+(b+c))+d,
(3) a + ((b+ c) + d),
(4) a + (b + (c + d)),
(5) (a + b) + (c + d).
4 Prologue
since these numbers are all equal. Fortunately, this
This definition is unambiguous
need
not be listed separately, since it follows from the property P1 already
fact
listed. For example, we know from P1 that
(a+ b) +
and it follows immediately that
c
a +
=
c),
(b +
(1)and (2)are
equal. The equality of (2)and (3)
this
P1,
is a direct consequence
may not be apparent at first sight
role
of b in P1, and d the role of c). The equalities
(onemust let b + c play the
(3) (4) (5) are also simple to prove.
It is probably obvious that an appeal to P1 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.
although
of
=
=
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 inadequate to prove the equality
of all possible sums of an arbitrary fmite collection of numbers ai,
a.. This
about
would
who
the
like to worry
fact may be taken for granted, but for those
approach is outlined in
proof (andit is worth worrying about once) a reasonable
Problem 24. Henceforth, we shall usually make a tacit appeal to the results of this
problem and write sums ai +
+ as with a blithe disregard for the arrangement
.
-
of parentheses.
The number
(P2)
-
.
.
,
-
0 has one property so important that we list it next:
If a is any number,
then
a+0=0+a=a.
An important role is also played by 0 in the third property of our list:
(P3) For every number
a, there
a+
is a number
(-a)
(-a)
=
such that
-a
+a
=
0.
Property P2 ought to represent a distinguishing characteristic of the number 0,
it is comforting to note that we are already in a position to prove this. Indeed,
if a number x satisfies
and
a+x=a
for any one number a, then x = 0 (andconsequently this equation also holds for all
numbers a). The proof of this assertion involves nothing more than subtracting a
from both sides of the equation, in other words, adding
to both sides; as the
shows,
all
proof
three
properties
P1-P3 must be used to justify
following detailed
this operation.
---a
If
then
hence
hence
hence
a+ x
(-a) + (a + x)
((-a) + a) + x
0+ x
x
=
a,
=
(-a)
=
0;
=
0;
=
0.
+ a
=
0;
1. BasicPropertiesof Numbers5
As we have just hinted, it is convenient to regard subtraction as an operation
derived from addition: we consider a b to be an abbreviation for a + (-b). It
is then possible to find the solution of certain simple equations by a series of steps
justifiedby Pl, P2, or P3) similar to the ones just presented for the equation
(each
a. For example:
a + x
-
=
If
then
hence
hence
hence
x+3=5,
(x + 3) + (-3)
x +
(3 + (-3))
5 + (-3);
=
5 3
2;
2.
=
x+ 0
x
-
=
=
=
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 by indicating so explicitly the reliance on properties Pl, P2, and
P3 (orany of the further pmperties we shall list).
Only one other property of addition remains to be listed. When considering the
sums of three numbers a, b, and c, only two sums were mentioned: (a + b) + c
and a + (b + c). Actually, several other arrangements
are obtained if the order of
a, b, and c is changed. That these sums are all equal depends on
then
(P4) If a and b are any numbers,
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
well
example,
this property: usually
behaved.
For
subtraction
have
not
does
so
might
when
passing
ask
b ¢ b a. In
b does equal b a, and it
we
a
a
just
is amusing to discoverhow powerless we are if we rely only on properties P1-P4
Algebra of the most elementary variety shows that
to justifyour manipulations.
when
=
=
b b a only
b. Nevertheless, it is impossible to derive this fact
a
a
from properties P1-P4; it is instructive to examine the elementary algebra carefully and determine which step(s) cannot be justifiedby P1-P4. We will indeed
be able to justifyall steps in detail when a few more properties are listed. Oddly
enough, however, the crucial property involves multiplication.
The basic properties of multiplication
are fortunately so similar to those for adwill
needed;
dition that little comment
be
both the meaning and the consequences
should be clear. (Asin elementary algebra, the product of a and b will be denoted
by a b, or simply ab.)
-
-
-
-
-
-
(P5) If a, b, and c are any numbers, then
a
(P6)
-
(b
-
c)
=
If a is any number, then
a.1=1-a=a.
(a
-
b) c.
·
-
6 Prologue
Moreover, 1 ¢ 0.
(The assertion that 1 ¢ 0 may seem a strange fact to list, but we have to
list it, because there is no way it could possibly be proved on the basis of the
other properties listed-these properties would all hold if there were only one
number, namely, 0.)
(P7) For every number
a¯1 such that
number
¢ 0, there is a
a
a·a-1=a
-a=1.
(PS) If a and b are any numbers,
then
a-b=b-a.
One detail which deserves emphasis is the appearance of the condition a ¢ 0
in P7. This condition is quite necessary; since 0-b 0 for all numbers b, there is no
number 0¯* satisfying 0 -0¯I
1. This restriction has an important consequence
for division. Justas subtraction was defined in terms of addition, so division is
defined in terms of multiplication:
The symbol a/b means a b¯1. Since 0¯I is
meaningless, a/0 is also meaninglessaivision
by 0 is always undefined.
Property P7 has two important consequences.
If a b
a c, it does not
and
necessarily follow that b = c; for if a
then
0,
both a b
a c are 0, no matter
what b and c are. However, if a ¢ 0, then b
c; this can be deduced from P7 as
follows:
=
=
-
=
-
=
-
-
-
=
If
a¯*
then
a b=a-canda¢0,
(a b) a-1 (a 4
-
=
-
-
-
-1
(a¯1 4 b
hence
hence
hence
-
=
-
1 b
b
(It may happen that a
b
=
(a b)
=
·
a
then
a
hence
hence
hence
(a¯I
·
0, then either a
=
-
-1
-
·
1 c;
c.
-
=
It is also a consequence of P7 that if a b
if
.
=
·
a)
b
-
=
0 or b
=
0. In fact,
0 and a ¢ 0,
0;
0;
0;
=
1 b
b = 0.
=
-
0 and b 0 are both true; this possibility is not excluded
0 or b 0"; in mathematics
is always used in the
a
sense of
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
when we say
"either
=
=
=
"or"
=
"one
(x
Then it follows that either x
-
1
=
-
1) (x
2)
-
0 or x
-
2
=
=
0.
0; hence x
=
1 or x
=
2.
1. Basic 14opertiesof Numbers 7
On the basis of the eight properties listed so far it is still possible to prove very
little. Listing the next property, which combines the operations of addition and
multiplication,
will alter this situation drastically.
(P9) If a, b, and c are any numbers,
then
a-(b+c)=a-b+a-c.
(Notice that the equation (b + c) a
b a + c a is also true, by P8.)
=
-
-
-
As an example of the usefulness of P9 we will now determine just when a
b
-
-
b
=
a:
If
then
(a
hence
hence
Consequently
a-b=b-a,
b) + b
--
a
a + a
a
-
(1 + 1)
and therefore
a
a) + b
b + (b
=
(b
=
b + b a;
(b + b a) + a
b (1 + 1),
=
=
=
-
=
a);
-
--
=
-
b + b.
-
b.
0 which we have
A second use of P9 is the justificationof the assertion a 0
made,
already
and even used in a proof on page 6 (canyou find where?). This
fact was not listed as one of the basic properties, even though no proof was offered
when it was first mentioned.
With P1-P8 alone a proof was not possible, since the
number 0 appears only in P2 and P3, which concern addition, while the assertion
With P9 the proof is simple, though perhaps
in question involves multiplication.
obvious:
We have
not
-
=
a-0+a-0=a-(0+0)
=a.0;
-(a
as we have already noted,
sides) that a 0
0.
-
this immediately implies
(byadding
-
0) to both
=
A series of further consequences of P9 may help explain the somewhat mysterious rule that the product of two negative numbers is positive. To begin with,
b). To
we will establish the more easily acceptable assertion that (-a) b
-(a
=
-
-
prove this, note that
.b+a.b=
-b
[(-a)+a]
(-a)
=
It follows immediately
Now note that
(byadding
(-a) (-b)
-
+
[-(a
0.
-(a
--(a
-
b)]
b) to both sides) that
=
(-a) (-b)
-
(-a)
(-a) 0
-
=
=
+ (-a)
[(-b) + b]
-
0.
(-a) b
-
-
b
=
-
b).
8 Prologue
Consequently, adding (a b) to both sides, we obtain
-
(-a)
-
(-b)
=
a
b.
-
The fact that the product of two negative numbers is positive is thus a consequence
of two
of P1-P9. In other words, if we want P1 to P9 to be true, the rule fortheproduct
negativenumbers isforced
upon us.
The various consequences of P9 examined so far, although interesting and important, do not really indicate the significance of P9; after all, we could have listed
each of these properties separately. Actually, P9 is the justificationfor almost all
algebraic manipulations.
For example, although we have shown how to solve the
equation
(x 1)(x 2)
-
=
-
0,
we can hardly expect to be presented with an equation in this form. We are more
likely to be confronted with the equation
x2-3x+2=0.
The
x2
"factorization"
-
3x + 2
(x 1) (x 2)
--
·
-
(x
=
1)(x
-
-
2) is really a triple use of P9:
(x 2) + (-1) (x 2)
=
x
=
x x + x (-2) + (-1) x +
x2
+ x [(---2)
+ (-1)] + 2
=
-
-
-
-
-
-
.
(-1) (-2)
-
=x2-3x+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
×24
26
is a concise arrangement
for the following equations:
13-24=
13-(2-10+4)
=13-2·10+13·4
=26-10+52.
(Note that moving 26 to the left in the above calculation is the same as writing
26 10.) The multiplication
13 4 52 uses P9 also:
-
=
-
13-4=(1-10+3)-4
=1-10-4+3-4
=4-10+12
=4-10+1-10+2
=
(4 + 1) 10 + 2
-
=5.10+2
=
52.
1. BasicPropertiesofXumbers9
The properties 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.
(P1)
(Associativelaw for addition)
a
(P2)
(Existence of an additive
identity}
a + 0
(Existence of additive inverses)
a +
(b +
+
c)
(a + b) +
=
0+ a
=
=
c.
a.
.
(P3)
(P4) (Commutative law for addition)
(PS) (Associativelaw for multiplica-
(-a)
a + b
a
-
a
-
(b
=
a
0.
b + a.
=
·
(-a) +
=
c)
(a
=
b) c.
-
-
tion)
(P6) (Existence of a multiplicative
identity)
(P7) (Existence of multiplicative
1
1 a
=
a a¯1
·
=
a¯I
-
·
1 ¢ 0.
a;
-
a
=
1, for a ¢ 0.
inverses)
(PS) (Commutative law for multiplication)
a
-
(P9) (Distributive law)
a
-
b
=
(b +
b a.
-
c)
=
a
-
b + a c.
-
The three basic properties of numbers which remain to be listed are concerned
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 related: a < b
means the same as b > a (thus1 < 3 and 3 > 1 are merely two ways of writing
the same assertion). The numbers a satisfying a > 0 are called positive, while
those numbers a satisfying a < 0 are called negative.
While positivity can thus
of
<,
possible
the
it
is
procedure: a < b can be
be defined in terms
to reverse
defined to mean that b a is positive. In fact, it is convenient to consider the
collection of all positive numbers, denoted by P, as the basic concept, and state
all properties in terms of P:
with
-
(P10) (Trichotomy law) For every number
following holds:
(i)
(ii)
(iii)
(P11)
(P12)
a, one and only one of the
0,
a is in the collection P,
is in the collection P.
a
=
-a
(Closure under addition) If a and b are in P, then a + b is in P.
(Closure under multiplication) If a and b are in P, then a b is
in P.
-
10 Prologue
These three properties should be complemented
a>b
asb
following definitions:
a-bisinP;
b>a;
if
if
if
if
a>b
a<b
with the
a>bora=b;
a<bora=b.*
Note, in particular, that a > 0 if and only if a is in P.
All the familiar facts about inequalities, however elementary they may seem, are
of P1¾P12.
For example,
precisely one of the following holds:
consequences
(i)
if a and b are any two numbers,
then
a-b=0,
(ii) a
(iii)
-
-(a
b is in the collection P,
b)
b a is in the collection
=
-
-
P.
Using the definitions just made, it follows that precisely one of the followingholds:
(i)
a
=
(ii) a
>
(iii)b >
b,
b,
a.
A slightly more interesting fact results from the following manipulations.
b, so that b a is in P, then surely (b + c) (a + c) is in P; thus, if a
a
<
-
then a + c
<
-
b + c. Similarly, suppose a
b
-
and
c
-
so
c
-
<
b and b
<
<
If
b,
c. Then
is in P,
is
b in P,
a
a
=
(c
-
b) +
(b
-
a)
is in P.
This shows that if a < b and b < c, then a < c. (The two inequalities a < b and
b < c are usually written in the abbreviated form a < b < c, which has the third
inequality a < c almost built in.)
The following assertion is somewhat less obvious: If a < 0 and b < 0, then
ab > 0. The only difficulty presented by the proof is the unraveling of definitions.
is in P.
The symbol a < 0 means, by definition, O > a, which means 0 a
ab is in P. Thus
is in P, and consequently, by P12, (-a)(-b)
Similarly
ab > 0.
The fact that ab > 0 if a > 0, b > 0 and also if a < 0, b < 0 has one
special consequence: a2 > 0 if a ¢ 0. Thus squares of nonzero numbers are
always positive, and in particular we have proved a result which might have seemed
sufficiently elementary to be included in our list of properties: 1 > 0
(since1 12).
-a
-
=
-b
=
=
*
There is one slightly perplexing feature of the symbols
1+153
1+152
>
and 5. The statements
by < in the first, and by = in the
are both true, even though we know that 5 could be replaced
second.
This sort of thing is bound to occur when 5 is used with specific numbers; the usefulness
of the symbol is revealed by a statement
values of
like Theorem 1-here equality holds for
some
and b, while
inequality holds for
other
values.
a
1. BasicPropertiesofNumbers11
which will play an
The fact that
> 0 if a < 0 is the basis of a concept
extremely important role in this book. For any number a, we define the absolute
value |al of a as follows:
-a
0
a 5 0.
a,
a
-a,
Note that |a| is always positive, except when a
0. For example, we have |
=
-
3|
=
3,|7|=7,]l+Ã-h|=1+Ã-Ë,and|1+Ã-EÕ|=EÕ-Ä-1.
straightforward
In general, the most
approach to any problem involving absolute
several
treating
cases separately, since absolute values are defined
by cases to begin with. This approach may be used to prove the following very
important fact about absolute values.
values requires
THEOREM
1
For all numbers
a and b, we
have
|a+bl
PROOF
la|+ |b|.
5
We will consider 4 cases:
a 2 0,
a 2 0,
a 5 0,
a 5 0,
(1)
(2)
(3)
(4)
In case
(1)we
0;
b i 0;
b à 0;
b 5 0.
b
also have a + by 0, and the theorem is obvious;
|a+b|=a+b=
|a|+|b|,
so that in this case equality holds.
In case (4)we have a + b 5 0, and again equality
-(a+b)=
(2),when
a
>
holds:
-a+(-b)=
|a+b|=
In case
in fact,
|a
0 and b
<
+
|bl.
0, we must prove that
|a+b|5a-b.
This case may therefore be divided into two subcases.
prove that
If a + b 2 0, then we must
a+bsa-b,
-b,
i.e.,
b5
-b
which
is certainly true since b 5 0 and hence
a + b 5 0, we must prove that
à 0. On the other hand, if
-a-bsa-b,
i.e.,
which
-a
y a,
is certainly true since a 0 and hence
5 0.
with
that
of
Finally, note
case (3)may be disposed
no additional work, by apply-
ing case
-a
(2)with
a and b interchanged.
12 Prologue
consideration
ofvarAlthough this method of treating absolute values (separate
sometimes
often
the only approach available, there are
ious cases) is
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
la|
Ñ.
=
(Here, and throughout the book,
denotes the positive
square root of x; this
is defined only when x > 0.) We may now observe that
2
2
+ 2ab + b2
(|a + b|)2
a2
+ 2|a| |b| + b2
5
|a|2 + 2|a| b + 1b|2
symbol
_
_
-
=
.
(|a[ + |b|)2.
=
From this we can conclude that |a + b| 5 |a| + |b| because x2 < y2 implies x < y,
provided that x and y are both nonnegative; a proof of this fact is left to the reader
(Problem 5).
One final observation may be made about the theorem we have just proved: a
close examination of either proof ofTered shows that
la+b|= |a|+|b|
if a and b have the same sign
the two is 0, while
(i.e.,are
both positive or both negative), or if one of
la+bl<lal+|bi
if a and b are of opposite signs.
We will conclude 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
-
=
-
=
a-(1+1)=b-(1+1),
b. This result is obtained from the equation
from which we concluded that a
b (1 + 1) by dividing both sides by 1 + 1. Division by 0 should
a (1 + 1)
be avoided scrupulously, and it must therefore be admitted that the validity of the
argument depends on knowing that 1+1 ¢ 0. Problem 25 is designed to convince
you that this fact cannot possibly be proved from properties P1-P9 alone! Once
Pl0, Pll, and P12 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
present situation will help justifyserious consideration of such details. In this
chapter we have assumed that numbers are familiar objects, and that P1-Pl2 are
merely explicit statements of obvious, well-known properties of numbers. It would
be difficult, however, to justifythis assumption. Although one learns how to
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
=
-
=
-
"work
l. BasicPropertiesofNumbers 13
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
numbers; we may still
to admit frankly that we do not yet thoroughly understand
whatever
numbers
that,
they
should certainly have
in
defined,
finally
way
are
say
properties P1-P12.
Most of this chapter has been an attempt to present convincing evidence that
P1-Pl2 are indeed basic properties 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 P1-P12) are offered as further evidence. It is still a crucial question whether P1-P12 actuallÿ 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 Pl-P12 will become quite clear, 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 P1-P12. The discovery 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
I.
Prove the following:
(i)
If ax
x2
=
a
2
for some number
¢ 0,
a
then x
1.
=
y)(X
‡ y).
y2, then x
y or x
y)(x2 + xy + y2L
x2 y*
(x
(iv)
y)(In-1
n-2y
+Xyn-2
n-I
x"
y"
*
(x
(v)
X + y2). (There is a particularly easy way to
(vi) x3 + y3 (X # y)(X
do this, using (iv),and it will show you how to find a factorization for
x" + y" whenever n is odd.)
(ii)
(iii) If x2
_
-y.
-
=
-
=
-
-
-
=
2.
=
=
-
What is wrong with the following
"proof"?
X2
x2
(x +
y)(x
_
(ii)
a
-
ac
=
-,
b
a
-
b
bc
y) = y(x
x+y=y,
2y= y,
+
--
c
d
=
if b, c ¢ 0.
ad + bc
bd
,
if b, d
¢ 0.
2
_
-
Prove the following:
(i)
Xy,
y2
2
3.
=
Let x
=
1.
--
y),
=
y. Then
14 Prologue
a¯Ib¯1,
if a, b ¢ 0. (To do this you must remember
defining property of (ab)¯14
(iii) (ab)
a
b
(iv)
--
4.
c
d
ac
if b, d
db
=
a
(v)
(vi)
-
-
=
--
-
--,
ad
c
=
bc
If b, d ¢ 0, then
a
b
b
a
Find all numbers
¢ 0.
if b, e, d ¢ 0.
-,
d
b
if and only if ad
=
=
bc. Also determine when
for which
x
4-x<3-2x.
2
x < 8
(ii) 5 x2
<
(iii) 5
1)
(x 3)
(iv) (x
(i)
--
.
-2.
--
(v)
>
-
-
0. (When is a product of two numbers
positive?)
x2-2x+2>0,
(vi) x2+x+1>2.
x2
x + 10 > 16.
x2 + x + 1 > 0.
(viii)
(ix) (x ar) (x + 5) (x 3)
(vii)
--
--
-
(x) (x 6)(x
--
(xi) 22 <
Ã)
-
>
>
0.
0.
8.
(xii)x + 32 < 4.
1
(xiii)x
1
+
-
1 x
1
x
(xiv)x + 1 > 0.
>
0.
-
-
5.
Prove the following:
(i)
If a
If a
<
b and c
<
d, then a + c
b, then
b + d.
-a.
<
-
6.
<
-b
(ii)
(iii) If a < b and c > d, then a c < b d.
(iv) If a < b and c > 0, then ac < bc.
(v) If a < b and c a2< 0, then ac > bc.
(vi) If a > 1, then > a.
(vii) If0<a<1,thena2<a.
(viii)If 0 < a < b and 0 < c < d, then ac < bd.
(ix) If 0 5 a < b, then a2 < b2. (Use (viii).)
(x) If a, b > 0 and a2 < b2, then a < b. (Use (ix),backwards.)
(a) Prove that if 0 5 x < y, then x" < y", n 1, 2, 3,
(b) Prove that if x < y and n is odd, then x" < y".
(c) Prove that if x" y" and n is odd, then x y.
<
-
=
=
(d) Prove that
if x"
the
=
....
=
y" and n
is even, then x
--y.
=
y or x
=
1. BasicPropertiesof Numbers 15
7.
Prove that if 0
<
a
<
b, then
a<Ñ<a+b<b.
2
< (a + b)/2 holds for all a, b
Notice that the inequality 5
of
alization
this fact occurs in Problem 2-22.
*8.
>
0. A gener-
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 P1&P12 are replaced by
For any numbers a and b one, and only one, of the
following holds:
(i) a = b,
(P'10)
(ii)
a
b
(iii)
<
b,
<
a.
(P'11)
For any numbers
(P'12)
For any numbers
<
a
if a
<
b and b
b, and c, if a
<
b, then
<
b and 0
a, b, and c,
<
c, then
<
c, then
c.
a,
a+c<b+c.
For any numbers
ac < bc.
(P'13)
a, b, and c,
if a
Show that P1AP12 can then be deduced as theorems.
9.
Express each of the following with at least one less pair of absolute
value
signs.
(ii) |(|a+b| |a|- |b|)|.
(iii) |(|a+b|+|c|-la+b+c|)|.
(iv) |x2 24 + y2
-
-
(V)
10.
I(lÃ
Äl IE
+
-
-
Ë\)\.
Express each of the following without
when necessary.
cases separately
(i) |a+b|--|b|.
(ii) (|x|
1)|.
-
(iii) |x! |x2
(iv) a |(a |a|)|.
-
--
11.
-
Find all numbers x for which
(i) |x
(ii) |x
(iii)
-
-
3|
3|
=
<
8.
8.
x+4|<2.
(iv) |x
(v) |x
-
-
1| + |x 2|
1| + |x + 1|
-
>
<
l.
2.
absolute
value signs, treating various
16 Prologue
(vi) lx-li+lx+li<1.
(vii) [x 1 |x +
(viii)fx-1|·|x+2
12.
1
-
-
0.
=
=3.
Prove the following:
(i) |xy|= [x| |y|.
-
1
(ii)
1
=
-
ixi
x
|x[¯I
(iii)
-=
what
is.)
]x|
|y
(iv) x
(v) |x
if x ¢ 0. (The best way to do this is to remember
-,
x
-,ify¢0.
y
y| 5
-
-
|x|+ |y|. (Give a very short proof.)
y [x y|. (A very short proof is possible, if you
write things in
--
y
the right way.)
(vi) [(|x| |yl)| 5
-
(vii)|x +
y + z| 5
y[. (Why does this follow immediately from (v)?)
x
xi + ly!+ ]zi. Indicate when equality holds, and prove
-
your statement.
13.
of two numbers
x and y is denoted by max(x,
=
max(3, 3) = 3 and max(-1,
max(-4,
of x and y is denoted by min(x, y). Prove that
The maximum
-4)
max(-1,
The
3)
=
minimum
Thus
y).
-1)
-1.
=
x+y+!y-x|
max(x,
y)
=
min(x,
y)
=
,
2
x+y-|y-x|
2
Derive a formula for max(x, y, z) and min(x, y, z), using, for example
max(x,
14.
(a) Prove that |a
=
y, z)
=
max(x,
max(y,
z)).
to become confused by too many
for a > 0. Why is it then obvious for
|-al. (The trick is not
First prove the'statement
0?)
a 5
5 a 5 b if and only if la 5 b. In particular, it follows
(b) Prove that
cases.
-b
-|al
that
5 a 5
|a|.
(c) Use this fact to give a new proof that |a + b| 5 |a + bl.
*15.
Prove that if x and y are not both 0, then
x24y
X4
X3
+X2
p
2>0,
2
3
4
Hint: Use Problem 1.
*16.
(a) Show that
(x + y)2
(x + y)3
_
2
2
X3
3
only when x
only when x
=
=
0 or y
0 or y
=
=
0,
0 or x
--y.
=
1. BasicPropertiesof Numbers 17
(b) Using the
fact that
x2
show that
+ 2xy + y2
4x2 + 6xy + 4y2
>
y)2
(x +
=
à 0,
0 unless x and y are both 0.
(x + y)4
X4
4 y4
Hint: From the assumption (x + y)S
derive the equation x3+2x2y+2xy2+y3
you
2
2
0, if xy ¢ 0. This implies that (x + y)3
(c) Use part (b)to find out when5
(d) Find out when (x + y)5
x5+ya
should be able to
=
5.
You should now be able to make a good guess as to when (x + y)"
2x2
2(x
the smallest possible value of
square," i.e., write 2x2 3x + 4 =
(b) Find the smallest possible value of
(c) Find the smallest possible value of
(a) Find
3x + 4. Hint: "Complete the
3/4)2 + ?
x2
3x + 2y2 + 4y + 2.
x2 + 4xy + 5y2
6y + 7.
_
18.
b2
(a) Suppose that
x" + y";
in Problem 11-57.
the proof is contained
17.
=
=
-
-b
+
-
-
-
-4x
-
4c à 0. Show that the numbers
/b2
-
-b
4c
2
-
'
/b2
-
4c
2
both satisfy the equation x2 + bx + c 0.
Suppose that b2 4c < 0. Show that there are no numbers x satisfying
(b) x2
0; in fact, x2 + bx + c > 0 for all x. Hint: Complete the
+ bx + c
=
-
=
square.
Use
(c) x2
this fact to give another
2 >
(d) For which
proof that if x and y are not both 0, then
0.
numbers
x2 +
a is it true that
axy + y2
>
0 whenever
x and
y are not both 0?
x2
(e) Find the smallest possible value of + bx + c and of ax2 + bx + c, for
a
19.
>
0.
The fact that a2 à 0 for all numbers a, elementary
as it may seem, is
nevertheless
the fundamental idea upon which most important inequaliof all inequalities is the
ties are ultimately based. The great-granddaddy
Schwarzinequalig:
x171 +
X272
Il2
+
X22
2
+ 722
(A more general form occurs in Problem 2-21.) The three proofs of the
Schwarz inequality outlined below have only one thing in common-their
reliance on the fact that a2 à 0 for all a.
that if xi
Ayi and x2
ly2 for some number 1, then equality
holds in the Schwarz inequality. Prove the same thing if yr
0.
y2
Now suppose that yi and y2 are not both 0, and that there is no number
(a) Prove
=
=
=
=
18 Prologue
I such that xi
0
Ayr and x2
=
xi)2 #
12(yi2 + y22)
(lyt
<
=
=
(Ày2
-
-
Ay2. Then
X2)
X272)
2A(x1yi +
-
(112E X22L
+
Using Problem 18, complete the proof of the Schwarz inequality.
(b) Prove the Schwarz inequality by using 2xy 5 x2+y2 (howis this derived?)
with
x
x,
=
y
,
y,
=
°
»2
71 +
first for i 1 and then for i 2.
(c) Prove the Schwarz inequality by first proving that
=
=
(x12
)(71
X2
72
)
(XIfl
=
+X272)2
+
(X172
-X271)2
from each of these three proofs, that equality holds only when
ly1 and
0 or when there is a number A such that xi
y2
(d) Deduce,
yi
x2
=
=
=
Ay2.
-
In our later work, three facts about inequalities will be crucial. Although proofs
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 xo, and y is close enough to yo,
then x+ y will be close to xo+ yo, and xy will be close to xoyo, and 1/y will be close
which appears in these propositions is the fifth letter of the
to 1/yo. The symbol
alphabet
and could just as well be replaced by a less intimidating
Greek
("epsilon"),
Roman letter; however, tradition has made the use of e almost sacrosanct in the
will
"s"
contexts
20.
to which these theorems apply.
Prove that if
|x
xo|
-
<
and
|y
(xo+
yo)
-
yo|
<
,
then
I(x +
y)
-
I<
e,
I(x-y)-(zo-yo)l<E.
*21.
Prove that if
|x-xo]<min
then
|xy
--
xoyo|
(The notation
€
(
,1
2(|yo|+ 1)
<
and
€
ly-yo[< 2(\xo
was
defined in Problem 13, but the formula provided by
the first inequality in the hypothesis
that problem is irrelevant at the moment;
just means that
-
,
s.
"min"
|x
+ 1)
xol
8
<
2(|yo|+ 1)
and
|x
-
xo|
<
1;
1. Basic Propertiesof Numbers 19
at one point in the argument you will need the first inequality, and at another point you will need the second. One more word of advice: since the
hypotheses only provide information about x xo and y yo, it is almost a
foregone conclusion that the proof will depend upon writing xy xoyo in a
yo.)
xo and y
way that involves x
-
-
-
-
*22.
-
Prove that if yo ¢ 0 and
.
ly
then y
*23.
yol
-
mm
<
¢ 0 and
-,
[yo| elyo
2
1
1
y
yo
2
,
2
Replace the question marks in the following statement
ing e, xo, and yo so that the conclusion will be true:
by expressions involv-
If yo ¢ 0 and
|y
then y
--
yol
<
?
and
x
xo
y
yo
|x
-
xal
<
?
¢ 0 and
This problem is trivial in the sense that its solution followsfrom Problems 21
and 22 with almost no work at all (noticethat x/y = x 1/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.
-
*24.
This problem shows that the actual placement of parentheses in a sum is
induction"; if you are not fairrelevant. The proofs involve
miliar with such proofs, but still want to tackle this problem, it can be saved
until after Chapter 2, where proofs by induction are explained.
Let us agree, for definiteness, that ai +
+ a, will denote
"mathematical
ai +
(a2 + (as +
-
-
-
+
(an-2 ‡ (an-1 ‡
an)))
-
).
Thus ai + a2 + a3 denotes al + (a2 + as), and ai + a2 + as + a4 denotes
ai + (a2+ (a3+ a4)), etc.
(a) Prove that
(a) ‡···#ak)
·-*ak+l
#ak+l
=
al ‡
Hint: Use induction on k.
(b) Prove that if n > k, then
(ai+---+ag)+(ak+1
Hint: Use part
(a)to
"'
n)
1
""
give a proof by induction on k.
n·
20 Prolo.gue
(c) Let s(ai,
...,
ak)
be some sum formed from
s(ai,...,ak)=Ul
'
ai,
ak.
Show that
k•
Hint: There must be two sums s'(ai,...,ai)
that
and s"(aisi,...,ak)
such
l)#S"(G¡gi,...,ak)-
s(ai,...,ak)=S(Ul,---,
25.
...,
"number"
Suppose that we interpret
to mean either 0 or 1, and + and
be the operations defined by the following two tables.
-01
+01
000
001
1
10
101
Check that properties P1-P9 all hold, even though I + 1
=
0.
-
to
NUMBERS OF VARIOUS SORTS
CHAPTER
"number"
In Chapter 1 we used the word
very loosely, despite our concern with
of
numbers.
will
the basic properties
It
now be necessary to distinguish carefully
various kinds of numbers.
numbers"
The simplest numbers are the
"counting
1, 2, 3,
....
The fundamental significance of this collection of numbers is emphasized by its
numbers).
A brief glance at P1-P12 will show that our
N (fornatural
of
do not apply to N-for example, P2 and P3 do not
basic properties
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.
induction."
The most basic property of N is the principle of
Suppose P(x) means that the property P holds for the number x. Then the principle of mathematical
induction states that P(x) is true for all natural numbers x
symbol
"numbers"
"mathematical
provided that
(1)P(1)
is true.
(2)Whenever P(k) is true, P(k + 1) is true.
asserts the truth of P(k+1) under the assumption
that P(k) is true; this sufBces to ensure the truth of P(x) for all x, if condition
(1)also holds. In fact, if P(1) is true, then it follows that P(2) is true (byusing
(2)in the special case k 1). Now, since P(2) is true it follows that P(3) is true
(using(2)in the special case k 2). It is clear that each number will eventually be
reached by a series of steps of this sort, so that P(k) is true for all numbers k.
induction envisions
A favorite illustration of the reasoning behind mathematical
of
people,
an infinite line
Note that condition
(2)merely
=
=
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
(theone 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 (totell all
secrets learned to the next person) assures that condition
(2)is true, and telling
the secret to person number 1 makes (1)true. The following example is a less
him
induction. There is a useful and striking formula
which expresses the sum of the first n numbers in a simple way:
facetious use of mathematical
21
22 Prologue
n(n+1)
1+---+n=
.
2
To prove this formula, note first that it is clearly true for n
for some natural number k we have
1+---+k
k(k + 1)
2
=
=
1. Now assume that
.
Then
k(k + 1)
+k+1
2
1+---+k+(k+1)=
k(k+1)+2k+2
2
k2+3k+2
¯
2
=
(k + 1)(k+ 2)
'
2
so the formula is also true for k + 1. By the principle of induction this proves
the formula for all natural numbers n. This particular example illustrates a phethat frequently occurs, especially in connection with formulas like the
nomenon
one just proved. Although the proof by induction is often quite straightforward,
the method by which the formula was discovered remains a mystery. Problems 5
and 6 indicate how some formulas of this type may be derived.
The principle of mathematical
induction may be formulated in an equivalent
of a number,
way without speaking of
a term which is sufficiently
discussion. A more precise formulation
vague to be eschewed in a mathematical
mathematical
collection
term) of
states that if A is any
synonymous
(or
"properties"
"set"-a
natural
numbers
and
(1)
(2)
1 is in A,
k + 1 is in A whenever
then A is the set of all natural numbers.
adequately replaces the less formal one
set A of natural numbers x which satisfy
of natural numbers n for which it is true
1+---+n=
k is in A,
It should be clear that this formulation
given previously-we just consider the
P (x). For example, suppose A is the set
that
n(n + 1)
.
2
Our previous proof of this formula showed that A contains 1, and that k + 1 is
in A, if k is. It follows that A is the set of all natural numbers, i.e., that the 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 numbers, it
2. Nurnbers
of'VariousSorts 23
is tempting to say that A must have a smallest member. Actually, this statement
can fail to be true in a rather subtle way. A particularly important set of natural
numbers is the collection A that contains no natural numbers at all, the
collection" or
set,"* denoted by Ø. The null set 0 is a collection of natural
numbers that has no smallest member-in
fact, it has no members
at all. This
is the only possible exception, however; if A is a nonnull set of natural numbers,
then A has a least member. This
obvious" statement, known as the
principle," can be proved from the principle of induction as follows.
Suppose that the set A has no least member. Let B be the set of natural numbers
if 1 were in A, then
n such that 1,
n are all not in A. Clearly 1 is in B (because
A would have 1 as smallest member). Moreover, if 1,
k are not in A, surely
would
smallest
member
not
of A), so 1,
k + 1 is
k+ 1
in A (otherwise
be the
k + 1 are all not in A. This shows that if k is in B, then k + 1 is in B. It follows
that every number n is in B, i.e., the numbers 1,
n are not in A for any natural
number n. Thus A
which
completes
the
Ø,
proof.
It is also possible to prove the principle of induction from the well-ordering
principle (Problem 10). Either principle may be considered as a basic assumption
"empty
"null
"intuitively
"well-ordering
.
.
.
,
...,
.
.
.
.
.
.
,
,
=
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 I < k. In this case we rely on the "principle
of
complete induction": If A is a set of natural numbers and
(1)
(2)
1 is in A,
k + 1 is in A if 1,
.
.
.
,
k are in A,
then 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.
reader,
with
of
the
this
The proof
fact is left to
a hint (Problem 11). Applications
will be found in Problems 7, 17, 20 and 22.
definitions." For example,
Closely related to proofs by induction are
number
the
n! (read factorial") is defined as the product of all the natural
numbers less than or equal to n:
"recursive
"n
n!=1-2-...-(n-1)-n.
This can be expressed more precisely as follows:
(1)
1!
(2)
n!=n-(n-1)!.
=
1
This form of the definition exhibits the relationship
between n! and (n
-
1)! in an
*Although it may not strike you as a collection, in the ordinary sense of the word, the null set arises
quite naturally in many contexts. We frequently consider the set A, consisting of all x satisfying some
property P; often we have no guarantee that P is satisfied by may number, so that A might be 6-in
fact often one proves that P is always false by showing that A = 0.
24 Prologue
explicit way that is
ideally suited for proofs by induction. Problem 23
reviews a
which
definition
familiar to you,
may be expressed more succinctly as a recursive definition;as this problem shows, the recursive definitionis really necessary
for a rigorous proof of some of the basic properties of the definition.
already
One definition which may not
which we will constantly be using.
be familiar involves some convenient notation
Instead of writing
al +
we will usually
+as,
·
employ the Greek letter E
(capitalsigma,
for
"sum")
and write
a¿.
i=1
In other words,
a¿ denotes the sum of the numbers
obtained
by letting
i=1
i
=
1, 2,
n.
...,
Thus
"
i
=
1+2+---+n
n(n +
=
'
2
i=1
Notice that the letter i really has nothing
1)
to do with the number
denoted by
i,
i=l
and can
be replaced
by any convenient
symbol
"
=
'
=
n
=
i (i + 1)
2
j=1
'
j(j+1)
°
2
n=1
To define
'
2
Í
of course!):
n(n + 1)
.
21
j=1
(exceptn,
a¿ precisely really requires a recursive
definition:
i=l
1
(1)
i
a¿
=
a¿
=
ai,
i=1
n-1
n
(2)
i=1
L a¿ + a..
i=l
austerity would insist too strongly on such
But only purveyors of mathematical
precision. In practice, all sorts of modifications of this symbolism are used, and
no one ever considers it necessary to add any words of explanation. The symbol
of VmiousSorts 25
2. Numbers
i=1
i¢4
for example, is an obvious way of writing
ai + a2 + as + as + a6
·
n.
or more precisely,
3
n
La¿+
i=I
The deficiencies of the natural numbers
of this chapter may be partially remedied
integers
---2,
a
.
i=5
which we
discovered at the beginning
by extending this system to the set of
-1,
0, 1, 2,
...,
....
This set is denoted by Z (from
German "Zahl," number). Of properties Pl-Pl2,
P7 fails for Z.
A still larger system of numbers is obtained by taking quotients m/n of integers
and the set of all
(withn 0). These numbers are called rational numbers,
rational numbers is denoted by Q (for
In this system of numbers all
of P1-Pl2 are true. It is tempting to conclude that the
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 P1-Pl2 apply-the
denoted by R. The real numbers
set of all real numbers,
only
rational
the
numbers,
but other numbers as well (theirrational
include not
numbers)
which can be represented
by infinite decimals; x and å are both
examples of irrational numbers.
The proof that x is irrational is not easy-we
shall devote all of Chapter 16 of Part III to a proof of this fact. The irrationality
of Å, on the other hand, is quite simple, and was known to the Greeks. (Since the
Pythagorean theorem shows that an isosceles right triangle, with sides of length 1,
has a hypotenuse of length Å, it is not surprising that the Greeks should have
investigated this question.) The proof depends on a few observations about the
natural numbers.
Every 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
fact has a simple proof by induction (Problem 8)). 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:
only
-y
"quotients").
"properties
"obvious"
(2k)2 4k2
=
(2k + 1)2
=
2 (2kl),
4k2 + 4k + 1 2 (2k + 2k) + 1.
=
-
=
-
In particular it follows that the converse must also hold: if n2 is even, then n is even;
if n2 is odd, then n is odd. The proof that à is irrational is now quite simple.
Suppose that à were rational; that is, suppose there were natural numbers p
26 Prologue
and q such that
all common divisors
We can assume that p and q have no common divisor(since
could be divided out to begin with). Now we have
p2
2q2.
=
This shows that p2 is even, and consequently
some natural number k. Then
p2
=
4k2
__
p must be even; that is, p
=
2k for
2
so
2k2
=
q2.
This shows that q2 is even, and consequently that q is even. Thus both p and q
the fact that p and q have no common divisor. This
are even, contradicting
contradiction completes the proof.
It is important to understand precisely what this proof shows. We have demonnumber x such that x2
strated that there is no rational
2. This assertion is often
expressed more briefly by saying that à is irrational. Note, however, that the
irrational)
use of the symbol å implies the existence of some number (necessarily
whose square is 2. We have not proved that such a number exists and we can asfor us. Any proof at this stage
sert confidently that, at present, a proof is impossible
would have to be based on P1-P12 (theonly properties of R we have mentioned);
since P1-P12 are also true for
Q the exact same argument would show that there
rational
number whose square is 2, and this we know is false. (Note that the
is a
proof that there is no rational number whose
reverse argument will not workeur
used
show
that
there is no real number whose square is 2,
square is 2 cannot be
to
because our proof used not only P1-P12 but also a special property of Q, the fact
that every number in Q can be written p/q for integers p and q.)
This particular deficiencyin our list of properties of the real numbers could,
of course, be corrected by adding a new property which asserts the existence of
square roots of positive numbers. Resorting to such a measure is, however, neither
aesthetically pleasing nor mathematically satisfactory; we would still not know that
every number has an nth root if n is odd, and that every positive number has an
nth root if n is even. Even if we assumed this, we could not prove the existence of
x5 + + 1 0
x
a number x satisfying
(eventhough there does happen to be one),
of nth
since we do not know how to write the solution of the equation in ter
solution
written
that
the
this
in
form). And,
roots (infact, it is known
cannot be
of course, we certainly do not wish to assume that all equations have solutions,
since this is false
(noreal number x satisfies x2 + 1 0, for example). In fact,
this direction of investigation is not a fruitful one. The most useful hints about the
property distinguishing R from Q, the most compelling evidence for the necessity
of elucidating this property, do not come from the study of numbers alone. In
order to study the properties of the real numbers in a more profound way, we
=
=
=
2. Numbersof VmiousSorts 27
must study more than the real numbers.
foundations of calculus, in particular the
At this point we must begin with the
fundamental concept on which calculus
is based-fimetions.
PROBLEMS
1.
Prove the following formulas by induction.
2.
n(n+1)(2n+1)
12+--·+n
(i)
(ii)
=
6
(1+---+n)2.
13+--·+n3=
Find a formula for
(2i-1)=1+3+5+---+(2n-1).
(i)
i-1
(2i-1)2--12+32+52+---+(2n-1)2
(ii)
i=1
Hint: What do these expressions have to do with 1 + 2 + 3 +
12 + 22 + 32 +
+ (2n)2?
-
3. If 0
sk
)(
-
5 n, the
k!(n
=
-
-
+ 2n and
-
"binomial
coefficient"
is defined by
n(n-1)---(n-k+l)
n!
-4
,
k!
k)!
--
-
if k
0, n
1. (This becomes a special case of the first formula if we
define 0! 1.)
=
=
(a) Prove that
n+1
kl
(The proof does not require an induction argument.)
This relation gives rise to the following configuration, known as "Pascal's triangle"-a
number not on one of the sides is the sum of the two
numbers
above
is the (k + 1)st number
it; the binomial coefficient
in the (n + 1)st row.
1
1
1
1
2
1
1
1
3
3
14641
1
10 10
5
5
1
28 Prologue
(b) Notice that all the numbers in Pascal's triangle are natural
is always a natural
part (a)to prove by induction that
by induction will, in a sense, be summed
triangle.)
Pascal's
entire proof
another
(c) Give
,
(Your
up in a glance
number
by showing
by
that
n.
(d) Prove the
natural
"binomial
If a and b are any numbers
theorem":
a"¯Ib+
a"¯2b2+···+
ab"¯1+b"
1
n
(e) Prove that
(i)
+
=
(-1)
(ii)
+
-
+
+
(iv)
+
+
+
l even
/
/
2".
=
-
(iii)
-
-
-
-
-
-
-
±
=
0.
2"¯1
=
-
-
=
2"¯
/
(a) Prove that
CX,
n+m
k
Hint: Apply the binomial theorem to (1 + x)"(1 + x)m
(b) Prove
and n
number, then
(a+b)"=a"+
4.
number.
Use
of sets of exactly k integers each chosen from 1,
is the number
...
is a natural
proof that
numbers.
that
is a
2. Numbersof VanousSorts 29
5.
on n that
(a) Prove by induction
1-r"+1
l+r+r2+---+r"=
1
r
---
1, evaluating the sum certainly presents no problem).
if r ¢ 1 (ifr
multiplying this equation
result
by setting S 1+r +-·
(b) Derive this
by r, and solving the two equations for S.
=
-+r",
=
6.
The formula for 12 +
formula
-
-
+ n2 may be derived as follows. We begin with the
-
(k+1)3-k3=3k2+3k+1.
Writing this formula for k
=
1,
...
,
n and adding, we obtain
23-13=3-12+3-1+1
33-23=3.22+3-2+1
(n+1)3-n3=3-n2+3-n+1
=3[12+-·-+n2
-1
(n+1)
k2 if we already know
Thus we can find
(whichcould
k
have been
k=1
k=1
found in a similar way). Use this method to fmd
(i)
13 4
14 4
(ii)
1
(iii) 1.2
--
(iv)
*7.
.
.
.
.
.
3
.
4
1
1
+
+
n(n+1)
2-3
2n+1
5
3
+
+ n2(n
+
12 22 22 32
+ Î)2
Use the
+
-
--
-
·
.
·
method
·
·
-
-
-
of Problem
kP
6 to show that
can always be written
i=i
the form
n
p+1
+ An? + Bn*¯' + Cn*¯2 +
p + 1
(The first 10 such expressions are
-
-
·
.
in
30 Prologue
k=
+
n
n
k=I
+n2+n
k2=n
k=1
k3
_
4
3
2
5
4
3
k=1
k4
k-I
ks
=
n6
+
k6
=
n'
+
n6 #
k'
=
n°
+
ni +
k"
=
n°
+
n° +
k
=
5
n4
+
-
n2
k=1
n5
3
_
n
k=1
n6
-
2
n4
k=1
n'
-
n° +
n
3
-
n
k-1
nl°+
n°
+
(n"
ni
+
þ©
k-1
klo
_
11
+
k=1
-
-
4
n6
In'
+
NS
-
-
n2
n'
-I-
gn.
and that after
Notice that the coefficients in the second column are always
of
with
coefficients
column
the third
the powers
decrease by 2 until
n
nonzero
n2 or
n is reached. The coefEcients in all but the first two columns seem to
rather
haphazard, but there actually is some sort of pattern; finding it may
be
be regarded as a super-perspicacity
test. See Problem 27-17 for the complete
(,
story)
8.
9.
Prove that every
natural
number
is either even or odd.
Prove that if a set A of natural numbers contains no and contains
it contains k, then A contains all natural numbers > no.
k+ 1
whenever
10.
Prove the principle of mathematical
induction from the well-ordering
prin-
ciple.
11.
Prove the principle of complete induction from the ordinary principle of
induction. Hint: If A contains 1 and A contains n + 1 whenever it contains
k are all in A.
1,
n, consider the set B of all k such that 1,
...,
12.
...,
is rational and b is irrational, is a + b necessarily irrational? What
if a and b are both irrational?
(b) If a is rational and b is irrational, is ab necessarily a4irrational? (Careful!)
(c) Is there a number a such that a2 is irrational, but is rational?
(a) If a
2. Numbersof VanousSorts 31
whose sum and product
there two irrational numbers
tional?
(d) Are
13.
(a) Prove that d, Ã,
and à are irrational. Hint: To treat ß, for example, use the fact that every integer is of the form 3n or 3n + 1 or 3n + 2.
Why doesn't this proof work for Ã?
(b) Prove that
14.
W are
and
+
Ä
Ä
is irrational.
is irrational.
(b) Ä +
(a) Prove that if x p + g
=
number,
(b) Prove
16.
irrationai.
Prove that
(a) Ä
15.
are both ra-
then xm = a +
also that (p
-
where p and q are rational, and m
b
for some rational a and b.
)m
=
a
-
bg.
that if m and n are natural numbers
2n)2/(m
+ n)2 > 2; show, moreover, that
(m+
(a) Prove
(m+ 2n)2
(m+n)2
is a natural
and
m2/n2
<
2, then
m2
-2
<
2---. n2
same results with all inequality signs reversed.
then there is another rational
that if m/n <
(b) Prove the
(c) Prove
with m/n
*17.
m'/n'
<
<
n,
n.
number
m'/n'
the natural number n is not
is irrational whenever
It seems likely that
the square of another natural number. Although the method of Problem 13
may actually be used to treat any particular case, it is not clear in advance
that it will always work, and a proof for the general case requires some extra
information. A natural number p is called a prime number
if it is impossible to write p
ab for natural numbers a and b unless one of these is p,
and the other 1; for convenience we also agree that 1 is not a prime number.
The first few prime numbers
are 2, 3, 5, 7, 11, 13, 17, 19. If n > 1 is not a
with
and
ab,
then
b both < n; if either a or b is not a prime it
prime,
a
n
similarly;
continuing in this way proves that we can write n
can be factored
product
of
primes.
example, 28 4 7 2 27.
For
as a
=
=
=
-
=
-
this argument into a rigorous proof by complete induction. (To
would accept the informal argube sure, any reasonable mathematician
ment, but this is partly because it would be obvious to her how to state
(a) Turn
it rigorously.)
A fundamental theorem about integers, which we will not prove here, states
that this factorization is unique, except for the order of the factors. Thus,
for example, 28 can never be written as a product of primes one of which
is 3, nor can it be written in a way that involves 2 only once (now
you should
appreciate why 1 is not allowed as a prime).
32 Prolqgue
is irrational unless n
Using this fact, prove that
(b) number
=
m2 for som€ naturaÏ
m.
(c) Prove more generally that 6
(d) No discussion of prime numbers
is irrational unless n
=
m*.
should fail to allude to Euclid's beautiful
proof that there are infmitely many of them. Prove that there cannot be
only finitely many prime numbers pt, P2, P3,
Pn by considering
---,
pl ·P2·---·Pn+1.
*18.
(a) Prove that
if x satisfies
a,_ix"¯I
x" +
+
-
-
+ ao
-
0,
=
for some integers a._i,
, ao, then x is irrationalunless x is an integer.
(Why is this a generalization of Problem 17?)
(b) Prove that à E à is irrational
...
-
(c) Prove that Ã
powers
-
W is irrational.
+
Hint: Start by working out the first 6
of this number.
-1,
19.
Prove Bernoulli's inequality: If h
(1 + h)"
Why is this trivial if h
20.
>
then
>
1 + nh.
>
0?
The Fibonacci sequence al, a29 63,
·
-
·
is defined as follows:
ai=1,
az=1,
a,
This sequence,
=
a._i
for n
+ a._2
which
3.
>
begins 1, 1, 2, 3, 5, 8,
, was discovered by Fibonacci
connection
with
a problem about rabbits. Fibonacci
(circa1175-1250), in
assumed that an initial pair of rabbits
gave birth to one new pair of rabbits
month, and that after two months each new pair behaved similarly. The
per
number a, of pairs born in the nth month is an-1 + an-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
is even a Fiinteresting results about this sequence is truly amazing-there
Prove
bonacci Association which publishes a journal, TheFibonacciQuarterly.
..
that
"
(l+ß
¯
2
.
"
1-Ã
2
One way of deriving this astonishing formula is presented in Problem 24-15.
21.
The Schwarz inequality (Problem 1-19) actually has a more general form:
x, y¿ 5
i=l
x,
\
i=1
2
2
\
i=1
of VariousSorts 33
2. Numbers
Give three proofs of this, analogous
22.
to the three proofs in Problem 1-19.
The result in Problem 1-7 has an important generalization: If ai,
then the
"arithmetic
.
..
,
a.
>
0,
mean"
A,=ai+---+a.
n
and
"geometric
mean"
G,
=
Jai
...a.
satisfy
G, 5 A..
ai < A.. Then some at satisfies a¿ > A,; for convenience,
A.. Let ãi
A, and let ã2
ã1. Show that
ai + a2
(a) Suppose that
say a2
>
=
=
äiã2
>
--
aia2.
this process enough times eventually prove that G, 5
place where it is a good exercise to provide a formal
proof by induction, as well as an informal reason.) When does equality
hold in the formula G, 5 An?
Why do- s repeating
A,? (This is another
The reasoning
in this proof is related to another
interesting proof.
fact that G, 5 A, when n 2, prove, by induction on k, that
2k
G, 5 A. for n
2m > n. Apply part (b)to the 2m numbers
general
let
For
n,
(c) a
(b) Using the
=
=
A.,...,
al,...,a.,
A,
to prove that G. 5 a..
23.
The following is a recursive definition of a":
a1=a,
an*1
=
a"
-
a.
Prove, by induction, that
(Don't try to be fancy:
an+m
=
a" am
(a")"'
=
an.
use either
·
induction on n or induction on
m, not
both
at once.)
24.
know properties P1 and P4 for the natural numbers,
but that
mentioned.
the
Then
following can be used
has never been
multiplication:
as a recursive definition of
Suppose
we
multiplication
1·b=b,
(a+1)-b=a-b+b.
34 Prologue
Pove the following (inthe order suggested!):
a
·
(b +
c)
=
a-1=a,
b a
a b
·
25.
=
-
a
-
b+ a c
-
(useinduction
(youjust finished
on a),
proving the case b
=
1).
In this chapter we began with the natural numbers and gradually built up to
the real numbers. A completely rigorous discussion of this process requires
a little book in itself (seePart V). No one has ever figured out how to get to
the real numbers without going through this process, but if we do accept the
real numbers as given, then the natural numbers can be dejinedas the real
numbers of the form 1, 1+ 1, 1+ 1+ 1, etc. The whole point of this problem
is to show that there is a rigorous mathematical way of saying
"etc."
(a) A set
A of
real
numbers
is called inductive
(1)
1 is in A,
(2)
k + 1 is in A whenever
if
k is in A.
Prove that
(i)
(ii) The
(iii) The
set of positive real numbers
set of positive real numbers
set of positive real numbers
is inductive.
j
to is inductive.
to 5 is not inductive.
(iv) The
real numbers which
of
then
the
and
A
B
set
inductive,
C
are
(v) If
and
also
inductive.
B is
are in both A
will be called a natural
number
real
number
if n is in every inductive
A
n
(b)
2
FIGURE
R is inductive.
unequal
unequal
Set.
1
(i)
(ii)
26.
Prove that 1 is a natural number.
Prove that k + 1 is a natural number
if k is a natural
number.
There is a puzzle consisting of three spindles, with n concentric rings of
decreasing diameter stacked on the first (Figure 1). A ring at the top of a
stack may be moved from one spindle to another spindle, provided that it
is not placed on top of a smaller ring. For example, if the smallest ring is
ring is moved to spindle 3, then
moved to spindle 2 and the next-smallest
the smallest ring may be moved to spindle 3 also, on top of the next-smallest.
Prove that the entire stack of n rings can be moved onto spindle 3 in 2" 1
1 moves.
moves, and that this cannot be done in fewer than 2"
--
-
*27.
TradiUniversity B. Once boasted 17 tenured professors of mathematics.
tion prescribed that at their weekly luncheon meeting, faithfully attended by
all 17, any members who had discovered an error in their published work
should make an announcement of this fact, and promptly resign. Such an anhad never actually been made, because no professor was aware
nouncement
of any errors in her or his work. This is not to say that no errors existed,
however. In fact, over the years, in the work of every member of the department at least one error had been found, by some other member of the
2. Numbersof VariousSorts 35
This error had been mentioned to all other members of the
department, but the actual author of the error had been kept ignorant of the
fact, to forestall any resignations.
One fateful year, the department was augmented by a visitor from another
university, one Prof. X, who had come with hopes of being offered a
permanent position at the end of the academic year. Naturally, he was apprised, by
various members of the department, of the published errors which had been
discovered. When the hoped-for appointment failed to materialize,
Prof. X
obtained his revenge at the last luncheon of the year. "I have enjoyed my visit
here very much," he said, "but I feel that there is one thing that I have to tell
you. At least one of you has published an incorrect result, which has been
discovered by others in the department." What happened the next year?
department.
**28.
After figuring out, or looking up, the answer to Problem 27, consider the following: Each member of the department already knew what Prof. X asserted,
so how could his saying it change anything?
PART
FOUNDATIONS
mado
The statement is se frequently
that the djerential calculus deals with
continuens magnidade, and yet
explanation of this antinuity is
nowhere given;
an
even the mori rigorens expositions
of the differmtialcalculus do not base
their proofsuþen continuity but,
with more er less consciousnessof the fact
they either appeal to geometric notwns
_geometry,
those IN,g,gtsted by
or deþenduþan theoremswhich are never
es¢ablished in a purelyarithmetic manner.
Among these,forexample,
beleggsthe abow-mentioned theorem,
or
and
a more careful invest(gation
corwinced me that this theorem,or
regarded
any one equivalent to it, can be
in some Rey as a sigicient basis
forinfinitesimalanalysis.
It then only ranained to discoverits true
or(gin in the elementsof arithmetic
and that at the
same time
to secure a real de(inition of
the essence of continuity.
I succeededNov. 2-( 1858, and
a fewdays afinward I commamrated
the results
of my meditations to my dear riend
f
Durà.ge with whom I had a long
and livelydiscassion.
RICHARD DEDEK[ND
FUNCTIONS
CHAPTER
Undoubtedly the most important concept in all of mathematics
is that of a
modern
mathematics
of
function-in almost every branch
functions turn out to
be the central objects of investigation. It will therefore probably not surprise you
to learn that the concept of a function is one of great generality. Perhaps it will
be a relief to learn that, for the present, we will be able to restrict our attention to
functions of a very special kind; even this small class of functions will exhibit sufficient variety to engage our attention for quite some time. We will not even begin
with a
proper definition. For the moment a provisional definition will enable us to
discuss functions at length, and will illustrate the intuitive notion of functions, as
understood by mathematicians.
Later, we will consider and discuss the advantages
of the modern mathematical
definition. Let us therefore begin with the following:
PILOVISIONAL DEFINITION
A function is a rule which assigns, to each of certain real numbers,
some other real
number.
The following examples of functions are meant to illustrate and amplify this definition,
which,
admittedly,
Example1 The
Example 2
requires some such clarification.
rule which assigns to each number
The rule which assigns to each number
the square of that number.
y the number
y3+3y+5
y2 + 1
-1
Example3
The rule which assigns to each number
c
¢ 1,
the number
c3+3c+5
c2
-
1
-17
Example4
the number
The rule which assigns to each number x satisfying
Example 5 The rule which assigns to each number
irrational, and the number 1 if a is rational.
Example 6 The rule which assigns
to
2 the
number
to 17 the number
39
Ex 5 x/3
x2.
5,
36
--,
a the number
0 if a is
40
Foundations
2
to
to
and to any y
¢ 2, 17, x2/17,
or
-
17
36
--
the
number
28,
the number
28,
36/x, the number
16 if y is of the form a +
bÃ
fora,binQ.
Example7 The rule which assigns to each number t the number t3 + x. (This
rule depends, of course, on what the number x is, so we are really describing
infinitely many different functions, one for each number x.)
Example8 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
if there are infinitely many
expansion
of
7's in the decimal
z.
-x
One thing should be abundantly
clear
from these examples-a
function is any
rule that assigns 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 (noone knows, for example, what rule 8 associates
to x). Moreover, the rule may neglect some numbers and it may not even be clear
to which numbers the function applies
to determine, for example, whether the
function in Example 6 applies to x). The set of numbers to which a function does
(try
apply is called the domainof 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
anything
need
of
about
naming
talk
else)
convenient
hardly ever
funcwe
a
way
tions, and of referring to functions in general. The standard practice is to denote
a function by a letter. For obvious reasons the letter f" is a favorite, thereby
making
and
other obvious candidates, but any letter (orany reasonable
symbol, for that matter) will do, not excluding
although these letters
and
reserved
numbers.
usually
then the number
for
indicating
If
is
function,
are
f a
"
"h"
"g"
"x"
which
x" and
"y",
"
associates to a number x
is often called the value of
is denoted by f (x)-this symbol is read f of
f at x. Naturally, if we denote a function by x,
other
chosen
must
letter
be
to denote the number (a perfectly legitimate,
some
would
though perverse, choice
leading to the symbol x(f)). Note that the
be
symbol f (x) makes sense only for x in the domain of f; for other x the symbol
f (x) is not defined.
If the functions defined in Examples 1-8 are denoted by f, g, h, r, s, 0, ex,
and y, then we can rewrite their definitions as follows:
f
"f,"
(1) f (x)
=
x2
for all x.
y3+3y+5
(2)
g(y)
=
(3)
h(c)
=
y2
1
c3 +
3c + 5
1
c
-
for all y.
-1.
-j
for all c
1,
3. Functions 41
(4)
r(x)
=
(5)
s(x)
=
x2
--17
5 x 5 x/3.
for all x such that
0, x irrational
1, x rational.
x=2
5,
36
17
x
=
28,
x
=
28,
x
=
16,
x¢2
---,
2
(6) 0 (x)
(7)
ax
(8)
y(x)
=
(t)
=
=
36
t3 + x
-
17,"-,or
17
for all numbers
x
,andx=a+bÃfora,binQ.
t.
7's appear in the decimal expansion of x
infinitely many 7's appear in the decimal expansion of x.
exactly
n,
I
-x,
n
These definitions illustrate the common procedure adopted for defining a function f-indicating what f (x) is for every number x in the domain of f. (Notice
that this is exactly the same as indicating f(a) for every number a, or f(b) for evb, etc.) In practice, certain abbreviations
ery number
are tolerated. Definition (1)
could be written simple
(1) f (x)
x2
=
"for
all x" being understood.
the qualifying phrase
the only possible abbreviation is
(4)
It is usually understood
r(x)
be shortened
-Î7 5
X
$
(4)
3.
X
that a definition such as
k(x)=-+
can
x2,
=
Of course, for definition
1
1
x
x-1
x¢0,1
,
to
k(x)=-+
1
x
1
x-1
;
in other words, unless thedomainis explicitly restricted further,
it is understood to consist of
which
makes
all.
all numbers
thedefinition
any sense at
for
should
checking
the following assertions about the funclittle
have
difficulty
You
tions defined above:
f (x +
1)
=
f (x) +
2x + 1;
g(x)=h(x)ifx3+3x+5=0;
-17
r (x +
1)
=
r
(x) + 2x + 1 if
5 xs
-
-
3
1;
42 Foundations
s(x + y)
=
if y is rational;
s(x)
36
(x2
0-=0-•
17
'
x
ax(x)=x·[f(x)+1];
If the expression f(s(a)) looks unreasonable to you, then you are forgetting that
s(a) is a number like any other number, so that f(s(a)) makes sense. As a matter
of fact, f(s(a))
s(a) for all a. Why? Even more complicated expressions than
after
The expression
a first exposure, no more difficult to unravet
f(s(a)) are,
=
f (r(s(9(a3(7(g)))))),
formidable as it appears,
be evaluated quite easily with a little patience:
may
f (r(s(0(as(y(y))))))
=
f(r(s(Ð(as(0)))))
=
f(r(s(0(3))))
=
f(r(s(16)))
f(r(l))
f(1)
=
=
=
1.
The first few problems at the end of this chapter give further practice manipulating
this symbolism.
is a rather special example of an extremely imporof
functions, the polynomial functions. A function f is a polynomial
tant class
such that
function if there are real numbers ao,
, as
The fimetion defined in
(1)
...
f(x)=anx"+a.
ixn-i+···+a2x2‡aix+a0,
foralÌx
(whenf (x) is written
in this form it is usually tacitly assumed that a, ¢ 0). The
coefficient is called the degree of f; for
highest power
a nonzero
5x6 + 137x4 x has
example, the polynomial function f defined by f(x)
degree 6.
The 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 (andq is not the function which is always 0). 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 following are examples of this kind
of function:
of x with
=
x+x2+xsin2x
(9) f (x)
=
(10) f (x)
(11) f (x)
=
=
x sin x + x
sin(x2).
sin(sin(x2g
sin2 x
-
3. Functions 43
(12) f (x)
=
sin2(sin(sin2(x
+ sin(x sin x)
sin2 x2
(x
x+smx
By what criterion, you may feel impelled to ask, can such functions, especially a
like (12),be considered simple? The answer is that they can be built
simple functions using a few simple means of combining functions.
from
few
up
a
In order to construct the functions (9)-(12)
we need to start with the
"sine
function" I, for which I(x)
function" sin, whose value sin(x) at
x, and the
often
written
simple
sin
x is
x. The following are some of the important ways in
which functions may be combined to produce new functions.
If f and g are any two functions, we can define a new function f + g, called
the sum of f and g, by the equation
monstrosity
"identity
=
(f + g)(x)
g(x).
f (x) +
=
Note that according to the conventions we have adopted, the domain of f + g
for which f(x) + g(x)" makes sense, i.e., the set of all x in both
domain f and domain g. If A and B are any two sets, then A nB (read"A
intersect B" or
intersection of A and B") denotes the set of x in both A
and B; this notation allows us to write domain( + g)
domain fn domain g.
f
"
consists of all x
"the
=
In a similar vein, we define the product
f
and g
f
-
g and the quotient
-
-g)(x)=
(orf/g) of
g
by
(f
f
g(x)
f(x)
and
f
(
-
g
(x)
=
f (x)
Moreover, if g is a function and c is a number,
(c
-
g)(x)
=
.
g(x)
we
define a
new
function c g by
·
c g(x).
-
This becomes a special case of the notation f g if we agree that the symbol c
should also represent
the function f defined by f(x)
c; such a function, which
value
all
numbers x, is called a constant
has the same
for
function.
-
=
The domain of f g is domain fn 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 complicated-it
may be written domain fn domaing n {x: g(x) ¢ 0}, the symbol {x: g(x) ¢ 0}
denoting the set of numbers x such that g(x) ¢ 0. In general, {x :
denotes
is true. Thus {x: x3 + 3 < 11} denotes the set of
the set of all x such that
all numbers x such that x3 < 8, and consequently
{x: x3 + 3 < 11) (x : x < 2).
of
symbols
could
well
Either
these
have been written using y everywhere
just as
instead of x. Variations of this notation are common, but hardly require any
discussion. Any one can guess that {x > 0 : x3 < 8} denotes the set of positive
numbers whose cube is less than 8; it could be expressed more formally as {x :
x3 < 8}. Incidentally, this set is equal to the set
x > 0 and
{x : 0 < x < 2}. One
-
-
...)
"
"
...
=
44 Foundations
is slightly less transparent, but very standard.
contains just the four numbers 1, 2, 3, and 4;
variation
example,
it
The set
{1,3, 2, 4}, for
can also
be denoted by
{x:x=lorx=3orx=2orx=4}.
Certain facts about the sum, product, and quotient of functions are obvious conFor example,
sequences of facts about sums, products, and quotients of numbers.
that
it is very easy to prove
f+(g+h).
(f+g)+h=
The proof is characteristic of almost every proof which demonstrates that two
functions are equal-the
two functions must be shown to have the same domain,
and the same value at any number in the domain. For example, to
prove that
(f + g) + h f + (g + h), note that unraveling the definition of the two sides gives
=
[(f+g)+h](x)=(f+g)(x)+h(x)
[f(x)+g(x)]+h(x)
=
and
[f+(g+h)](x)= f(x)+(g+h)(x)
f(x) + [g(x)+h(x)],
=
and the equality of
+ h(x)] is a fact about
f(x) + g(x)] + h(x) and f (x)+
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 equations;
[
the domain of
[g(x)
g) + h and of f + (g + h) is clearly
domain h. We naturally write f + g + h for (f + g) + h
as we did for numbers.
(f +
domain fn domain gn
f + (g + h), precisely
=
It is just as easy to prove that (f g) h f (g h), and this function is denoted
by f g h. The equations f + g g + f and f g
g f should also present
-
-
=
-
-
-
=
-
=
-
-
difficulty.
no
Using the operations
+,
-,
/
we can now
express the function
f
defined in
(9)
by
f
[+I-I+I-sin·sin
=
.
I-sin+I-sm-sm
.
.
It should be clear, however, that we cannot express function (10)this way. We require yet another way of combining functiqns. This combination,
the composition
of two functions, is by far the most important.
If f and g are any two functions, we define a new function fo g, the cornposition of f and g, by
(f
the domain of
"
fo
g"
og)(x)
=
f(g(x));
fog is {x: x is in domain g and g(x) is in domain f}. The symbol
composition of f
f circle g." Compared to the phrase
is often read
"
"the
and g" this has the advantage of brevity, of course, but there is another advantage
of far greater import: there is much less chance of confusing fog with go f, and
3. Functions 45
these must not be confused, since they are not usually equal; in fact, almost any f
and g chosen at random will illustrate this point (tryf
sin, for
I I and g
apprehensive
about
operation
of
the
composition,
example). Lest you become too
let us hasten to point out that composition is associative:
=
og)oh=
(f
(andthe
(10) f
(11) f
(12) f
fo(goh)
proof is a triviality); this function is denoted by
fimetions (10),(11),(12)as
write the
sin o (I I),
sin o sin o (I
=
=
·
fogo
h. We can now
-
=
(sin
=
-
sin)
o
-
sin
I),
o
(sin
-
sin)
o
(I
-
[(sinsin)
(I
o
-
-
I)])
-
I + sin o (I sin)
I + sm
-
.
SIR
O
.
One fact has probably already become clear. Although this method
functions
shortest
"structure"
reveals
.
of writing
The
very clearly, it is hardly short or convenient.
sin(x2) for all
=
unfortunately
such
that
x
f
f(x)
sin(x2) for all x." The need for
=
such that
their
name for the function
be
seems to
"the
function f
f (x)
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
sin(x2)
->
x
"x
goes to sin(x2
m
just "X
Sin(X2)"),
but it is hardly popular among
of calculus textbooks. In this book we will tolerate a certain amount of
"the
ellipsis, and speak of
function f (x) = sin(x2)." Even more popular is the
"the
quite drastic abbreviation:
function sin(x2)." For the sake of precision we
which,
strictly speaking, confuses a number and
will never use this description,
(read
Or
arrOW
writers
a function, but it is so convenient that you will probably end up adopting it for
personal use. As with any convention, utility is the motivating factor, and this
criterion is reasonable so long as the slight logical deficiencies cause no confusion.
On occasion,
"the
example,
confusion
function
will arise unless a more
x + t3» is an ambiguous
x + t3, i.e., the
-+
x
precise description is used. For
phrase; it could mean either
function f such that f(x)
=
t3
x +
=
t3 for all t.
x+
for all x
or
t
4
t3, i.e., the function
x +
f
such that
f (t)
As we shall see, however, for many important concepts associated with functions,
4"
which contains the
calculus has a notation
built in.
By now we have made a sufficiently extensive investigation of functions to warbut it is
rant reconsidering
our definition. We have defined a function as a
what
rule?"
this
ask
this
clear
hardly
"What happens if you break
it
means. If we
is not easy to say whether this question is merely facetious or actually profound.
"x
"rule,"
46 Foundations
A more substantial
objection
to the use of the word
"rule"
is that
f(x)=x2
and
f(x)=x2+3x+3-3(x+1)
are certainly
dafwentrules, if by a rule we mean the actual instructions given for
determining f(x); nevertheless,
f (x)
we want
x2
=
and
f(x)=x2+3x+3-c(x+1)
to define the same function. For this reason, a function is sometimes defined as an
between numbers; unfortunately the word
escapes the
objections raised against
only because it is even more vague.
"association"
"association"
"rule"
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 English words which
are troublesome. The definition which mathematicians have finally accepted for
is a beautiful example of the means by which intuitive ideas have been
incorporated into rigorous mathematics.
The correct question to ask about a
rule?"
function is not "What is a
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 f (x);
we can imagine a table which would display all the information one could desire
x2·
about the function f(x)
"function"
=
x
f (x)
1
1
1
4
4
-1
2
-2
would actually
It is not even necessary to arrange the numbers in a table (which
wanted
all
of
of
column
them). Instead
be impossible if we
array we
to list
a two
of
numbers
consider
various
pairs
can
.
(1, 1), (-1, 1), (2, 4), (-2, 4), (x, x2),
(Ä, 2),
..
.
3. Functions 47
simply
collected
together into a set.* To find f (1) we simply take the second
number of the pair whose first member is 1; to find f (x) we take the second
number
of the pair whose first member is x. We seem to be saying that a function
might as well be defined as a collection of pairs of numbers.
For example, if we
contains just 5 pairs):
were given the following collection (which
f
=
{(1,7), (3, 7), (5,3), (4, 8), (8,4)},
then f (1) 7, f (3) 7, f (5)
only numbers in the domain of
=
=
f
=
3, f (4)
=
f.
=
8, f (8)
4 and 1, 3, 4, 5, 8 are the
=
If we consider the collection
{(1, 7), (3,7), (2,5), (1,8), (8, 4) },
7, f (2) 5, f (8) 4; but it is impossible to decide whether f (1) 7
or f(1) 8. In other words, a function cannot be defined to be any old collection
of pairs of numbers;
we must rule out the possibility which arose in this case. We
therefore
led
the
following definition.
to
are
then
f (3)
=
=
=
=
=
DEFINITION
A function is a collection of pairs of numbers with the following property: if
c; in other words, the
(a, b) and (a, c) are both in the collection, then b
collection must not contain two different pairs with the same first element.
=
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 is actually
remarks,
and casual
at hand, and to distinguish them from comments, motivating
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:
DEFINITION
If f is a function, the dornain of f is the set of all a for which there is some b
such that (a, b) is in f. If a is in the domain of f, it followsfrom the definition
of a function that there is, in fact, a unique number
b such that (a, b) is in f.
This unique b is denoted by f(a).
With this definition we have reached our goal: the important thing about a
function f is that a number f (x) is determined for each number x in its domain.
You may feel that we have also reached the point where an intuitive definition has
been replaced by an abstraction with which the mind can hardly grapple. Two
consolations
may be ofTered. First, although
a function has been defined as a
"ordered
pairs," to emphasize that, for example, (2, 4) is
here are often caRed
the
pair
not
same
as (4, 2). It is only fair to warn that we are going to define functions in terms of
ordered pairs, another undefined
term. Ordered pairs can be defined, however, and an appendix to
this chapter has been provided for skeptics.
*The pairs
occurring
48 Foundations
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 is to draw pictures; but this requires a
chapter all by itself
PROBLEMS
1.
Let f(x)
1/(1+ x). What is
=
(i) f (f (x)) (forwhich
x does this make sense?).
(ii) f
(iii) f (cx).
(iv) f (x + y).
.
(v) f (x) + f (y).
(vi) For which numbers
c is there a number x such that f(cx)
f(x).
Hint: There are a lot more than you might think at first glance.
(vii)For which numbers c is it true that f (cx) f (x) for two different
numbers x?
=
=
2.
Let g(x)
x2, and let
=
h(x)
0, x rational
1, x irrational.
=
For which y is h(y) < y?
(ii) For which y is h(y) 5 g(y)?
h(z)?
is g(h(z))
(iii) Whatwhich
g(w)
5 w?
w is
(iv) For
which e is g(g(s))
g(e)?
For
(v)
Find the domain of the functions defined by the following formulas.
(i)
-
=
3.
(i) f(x)=
(ii) f (x)
(iii) f (x)
=
=
1-x2.
1
1
(v) f(x)=
4.
V1
--
x2.
-
+
x-1
(iv) f (x) VI
=
-
1
.
x-2
X2
x2 #
_
¡
1-x+fx-2.
x2, let P(x)
sin x. Find each of the following.
Let S(x)
22, and let s(x)
In each case you answer should be a number.
=
(So P)(y).
(So s)(y).
(SoPo s)(t) +
(iii) s(t3).
=
=
(i)
(ii)
(so P)(t).
(iv)
5.
Express each of the following functions in terms of S, P, s, using only
and o (forexample, the answer to
+,
(i)is Po s). In each case your
.,
3. Functions 49
answer
should
(i) f (x)
(ii) f (x)
(iii) f (x)
(iv) f (x)
(v) f(t)
be a function.
2sinx
sin 22.
=
=
sin x2.
sin2
=
=
x
sin2
that
(remember
ab
for (sinx)2).
is an abbreviation
x
22. (Note:
always means aW; this convention
abe
written
because (a")ccan be
more simply as
=
(vi) f(u)=sin(2"+2").
sin(sin(sin(222
(vii) f(y)
(viii)f(a) 2662 + sag2)
=
=
is adopted
'Bþ
a2÷
+ 2sin
mg
Polynomial functions, because they are simple, yet flexible, occupy a favored
in most investigations of functions. The following two problems illustrate their
flexibility,and guide you through a derivation of their most important elementary
properties.
role
6.
distinct numbers, find a polynomial function f¿ of
is 1 at x¿ and 0 at x; for j ¢ i. Hint: the product of
degree n
all (x x;) for j ¢ i, is O at x; if j ¢ i. (This product is usually denoted
by
(a) If
xi,
...
x, are
1 which
,
-
-
(x
xj),
-
j=1
the symbol
n (capitalpi) playing
for products that E plays
for sums.)
(b) Now find a polynomial function f of degree n 1 such that f (xt) ai,
numbers.
where ai,
(You should use the functions
, an are given
will
obtain
is called the "Lagrange
from
The
formula
part (a).
you
f¿
interpolation formula.")
same role
t
/
=
-
...
7.
for any polynomial function f, and any number a, there is a
polynomial function g, and a number b, such that f (x) (x
for all x. (The idea is simply to divide (x a) into f (x) by long division,
until a constant remainder
is left. For example, the calculation
(a) Prove that
-a)g(x)+b
=
-
x2
x
-
_;
1)x3
-3x
+ 1
-x2
x3
x2
x2
-3
-2x
+ 1
-2x
+ 2
-
1
50
Foundations
shows that x-3
3x + 1
(x 1)(x2 + x
possible by induction on the degree of f.)
(b) Prove that if f(a) 0, then f(x) (x
=
-
-
2)
-
-
1. A formal proof is
-a)g(x)
=
for some polynomial
=
function g. (The converse is obvious.)
(c) Prove that if f is a polynomial function of degree n, then f has at most
n roots, i.e., there are at most n numbers a with f (a) 0.
(d) Show that for each n there is a polynomial function of degree n with
=
n roots. If n
roots, and if n
8.
For which numbers
is even find a polynomial function of degree n with no
is odd find one with only one root.
a, b, c, and
d will the function
f (x)
9.
ax + b
cx + d
=
satisfy
f (f (x))
(a) If
A is any set of real numbers,
x for all x?
=
Ca(x)
1
=
define a function Ca as follows:
'
0,
xinA
x not
in A.
Find expressions for Cana and Ca a and Ca._a, in terms of Ca and Ca.
(The symbol A nB was defined in this chapter, but the other two may
be new to you. They can be defined as follows:
AUB={x:xisinAorxisinB},
A
{x: x is in R but x is not in A}.)
R
(b) Suppose f
=
-
is a function such that
there is a set A such that f
Ca.
only
that
and
f2
Show
if
if f
f
(c)
f (x)
0 or 1 for each x. Prove that
=
=
=
10.
Ca for some set A.
=
is there a function g such that f g2? Hint: You
is replaced by
can certainly answer this question if
such
that f
which
there
1/g?
functions f is
a function g
(b) For
*(c)
For which functionsb and c can we find a function x such that
(a) For which
functions
f
=
"function"
"number."
=
(x(t))2+ b(t)x(t) +
*(d)
c(t)
=
0
for all numbers t?
What conditions must the functions a and b satisfy if there is to be a
function x such that
a(t)x(t)+b(t)=0
for all numbers
11.
(a) Suppose that
t? How many such functions x will there be?
H is a function and y is a number
What is
H(H(H(-·-(H(y)-·-)?
80 times
such that H(H(y))
=
y.
3. Functions 51
(b) Same question
if 80 is replaced by 81.
(c) Same question if H(H(y)) = H(y).
*(d)
Find a function H such that H(H(x))
H(x) for all numbers x, and
=
such that H(l)
H(2)
x|3,
H(13)
36,
47, H(36)
36, H(x|3)
n/3, H(47)
47. (Don't try to
for H(x); there are many functions H with H(H(x))
H(x). The extra conditions on H are supposed
of
finding a suitable H.)
to suggest a way
*(e)
Find a function H such that H(H(x)) = H(x) for all x, and such that
H(1)
18.
7, H(17)
=
=
=
=
=
"solve"
=
=
=
=
12.
A function f is even if f (x) f (-x) and odd if f (x)
f (-x). For
x2
example, f is even if f(x)
or f(x) = |x| or f(x)
cosx, while f is
odd if f(x)
sinx.
x or f(x)
=
=
=
-
=
=
=
+ g is even, odd, or not necessarily either, in the
four
f even or odd, and g even or odd. (Your
answers can most conveniently be displayed in a 2 x 2 table.)
Do
(b) the same for f g.
(c) Do the same for fo g.
(d) Prove that every even function f can be written f(x) g(|x|), for inwhether
obtained
cases
(a) Determine
f
by choosing
.
=
fmitely many functions g.
*13.
function f with domain R can be written f
E + O,
and
odd.
E is even
0 is
that
this
of
writing f is unique. (If you try to do part (b)first,
Prove
way
(b) "solving"
by
for E and O you will pobably find the solution to part (a).)
(a) Prove that
any
=
where
14.
If f is any function, define a new function |f| by |f|(x)
|f(x)|. If f
min(f,
functions, define two new functions, max(f, g) and
g), by
=
and g are
max(f,g)(x)=max(f(x),g(x)),
min(f,g)(x)=min(f(x),g(x)).
Find an expression
15.
for max(f, g) and min(f, g) in terms of | |.
max( f, 0) + min( f, 0). This particular way of writing
useful; the functions max(f,0)
and min(f,0)
is
fairly
are called the
f
negative
and
positive
parts of f.
called
nonnegative
A
function
is
if f(x) :> 0 for all x. Prove that any
f
(b)
function f can be written f = g h, where g and h are nonnegative,
way" is g
max( f, 0) and h
in infinitely many ways. (The
min( f, 0).) Hint: Any number can certainly be written as the difference
of two nonnegative
numbers in infmitely many ways.
(a) Show that f
=
-
"standard
=
=
-
*16.
Suppose f satisfies f (x + y)
=
(a) Prove that f(xi+--·+x.)
f (x) + f (y) for all x
=
and y.
f(xt)+---+ f(x.).
(b) Prove that there is some number c such that f (x) cx for all rational
numbers x (atthis point we're not trying to say anything about f (x) for
irrational x). Hint: First figure out what c must be. Now prove that
=
52
Foundations
cx, first when x is a natural
then when x is the reciprocal of an
f (x)
*17.
=
number,
then when x is an integer,
integer and, finally,for all rational x.
0 for all x, then f satisfies f (x + y)
f (x) + f (y) for all x and y,
all
and
y)
for
x
y. Now suppose that f satisfies
f(x
f(x) f(y)
always
these two properties, but that f (x) is not
0. Prove that f (x) x for
all x, as follows:
If f (x)
=
=
and also
=
-
-
=
(a) Prove that f (1)
1.
that
Prove
f (x) x if x is rational.
(b)
that
(c) Prove
f (x) > 0 if x > 0. (This part is tricky, but if you have
the
been paying attention to the philosophical remarks accompanying
problems in the last two chapters, you will know what to do.)
(d) Prove that f (x) > f (y) if x > y.
(e) Prove that f (x) x for all x. Hint: Use the fact that between any two
numbers there is a rational number.
=
=
=
*18.
Precisely what conditions must
h(x)k(y) for all x and y?
*19.
(a) Prove that
g, h, and k satisfy
f,
in order that f(x)g(y)
=
there do not exist functions f and g with either of the following
properties:
(i) f(x) + g(y)
(ii) f(x) g(y)
-
xy for all x and y.
x + y for all x and y.
=
=
Hint: Try to get some information about f or g by choosing particular
values of x and y.
(b) Find functions f and g such that f(x + y) g(xy) for all x and y.
=
*20.
(a) Find
a
function f, other than a constant
f(x)|_<|y-x|.
function, such that |f (y)
-
-x)2
for all x and y. (Why does this
that f is a constant function.
?)
Prove
|f (y) f (x)| (y
the
from
Hint: Divide
interval
x to y into n equal pieces.
(b) Suppose that
f(y)
imply that
21.
f(x)
--
5 (y
-x)2
5
-
Prove or give a counterexample
for each of the following assertions:
(a) fo(g+h)=fog+foh.
(b) (g+h)of=gof+hof.
I
(c) fog
=-og,
f
1
(d) fog
22.
1
=
fo
1
-
.
g
then g(x)
g(y).
g(y)
that f and g are two functions such that g(x)
whenever
f (x) f (y). Prove that g hof for some function h. Hint:
h(z) when z is of the form z f(x) (these
define
try
to
are the only z
Just
that matter) and use the hypotheses to show that your definition will not
run into trouble.
(a) Suppose g ho f.
(b) Conversely, suppose
=
=
Prove that if f(x)
=
f(y),
=
=
=
=
3. Functions 53
23.
*24.
Suppose that
fog
=
I, where I(x)
x. Prove that
=
(a) if x ¢ y, then g(x) ¢ g(y);
(b) every number b can be written b
(a) Suppose g is a function with the
=
f (a) for some
number
a.
property that g(x) ¢ g(y) if x ¢ y.
Prove that there is a function
b can be written
(b) Suppose that f is a function
number
that
there
is a function g such that
b f (a) for some
a. Prove
f such that fog = I.
such that every number
=
fog=I.
*25.
*26.
Find a function
function h with
f such
foh
that gof
I.
=
I and hof
Suppose fog
associative.
is
=
=
=
I
for some g, but such that there is no
I. Prove that g
=
h. Hint: Use the fact that
composition
27.
(a) Suppose f (x) x + 1. Are there any functions g such that fog go f?
(b) Suppose f is a constant function. For which functions g does fog
go f?
(c) Suppose that f og gof for all functions g. Show that f is the identity
=
=
=
=
function, f (x)
=
x.
28. (a) Let F be the set of all functions whose domain is R. Prove that,
using +
as defined in this chapter, all of properties P1-P9 except P7 hold
for F, provided 0 and 1 are interpreted as constant functions.
(b) Show that P7 does not hold.
*(c)
Show that P1 Pl2 cannot hold. In other words, show that there is
no collection P of functions in F, such that P10-Pl2 hold for P. (It is
sufficient, and will simplify things, to consider only functions which are 0
and
-
except at two points xo and xi.)
(d) Suppose we define f < g to mean that
f(x) <
P'10-P'13 (inProblem 1-8) now hold?
Is fah<gah?
(e) If f<g,ishof<hog?
g(x) for all x. Which of
54
Foundations
APPENDIX.
ORDERED PAIRS
Not only in the definition of a function, but in other parts of the book as well,
it is necessary to use the notion of an ordered pair of objects. A definition has
not yet been given, and we have never even stated explicitly what properties an
ordered pair is supposed
to have. The one property which we will require states
formally that the ordered pair (a, b) should be determined by a and b, and the
order in which they are given:
if (a, b)
=
(c,d),
then a
=
c and
b
=
d.
Ordered pairs may be treated most conveniently by simply introducing (a, b)
this
term and adopting the basic property as an axiom-since
as an undefined
ordered
much
the
only
significant
about
there
point
is
fact
pairs,
is
not
property
worrying about what an ordered pair
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 feel
uncomfortable unless ordered pairs are somehow defined so that this basic property
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, to have available the
"really"
objects. This means that our
of an ordered pair of any two mathematical
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 (likeeverything else in mathematics) can be defined in this
context; an ordered pair will turn out to be a set of a rather special sort.
The set {a,b}, 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 determining
notion
from {a,b} which of a or b is meant to be the first element.
candidate is the rather startling set:
A more promising
{{a},{a,b} }.
members,
both of which are themselvessets; one member is the set
{a),containing the single member a, the other is the set {a,b}. Shocking as it may
seem, we are going to define (a, b) to be this set. The justificationfor this choice is
given by the theorem immediately following the definition-the definition works,
and there really isn't anything else worth saying.
This set has two
(a, b)
DEFINITION
THEOREM
1
PROOF
If (a, b)
=
(c,d),
then a
c and
=
b
=
=
{{a),{a,b} }.
d.
The hypothesis means that
{{a},{a,b)} {{c},{c,d}}.
=
{a) and {a,b}; and a is the only
{{a},{a,b} }. Similarly, c is the unique
Now {{a), {a,b} } contains just two members,
common element of these two members
of
3, Appendix.OrderedPairs 55
common member
of
both members
of
{{c), {c,d) ). Therefore
a
=
c.
We there-
fore have
{{a},{a,b}} {{a), {a,d)),
=
and only the proof that
b
=
d remains.
It is convenient
to distinguish 2 cases.
Case1. b = a. In this case, {a,b} = {a},so the set {{a},{a,b) }really
member, namely, {a}. The same must be true of ({a), {a,d) ), so
which implies that d = a = b.
has only one
{a,d} (a),
=
Case2. b ¢ a. In this case, b is in one member of {(a}, {a,b} } but not in the
other. It must therefore be true that b is in one member of {{a},{a,d) } but not
in the other. This can happen only if b is in {a,d), but b is not in {a};thus b a
=
orb=d,butb¢a;sob=d.
CHAPTER
-1
0
FIGURE
i
1
2
3
1
GRAPHS
Mention the real numbers to a mathematician and the image of a straight line will
probably form in her mind, quite involuntarily. And most likely she will neither
banish nor too eagerly embrace this mental picture of the real numbers. "Geometric intuition" will allow her 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 IL
You are probably already familiar with the conventional method of considering
the straight line as a picture of the real numbers, i.e., of associating to each real
number a point on a line. To do this (Figure 1) we pick, arbitrarily, a point which
which we label 1. The point twice as far to
We label0, and a point to the right,
the right is labeled 2, the point the same distance from 0 to 1, but to the left of 0,
if a < b, then the point corresponding
is labeled
etc. With this arrangement,
of
corresponding
the point
to b. We can also draw rational
to a lies to the left
numbers, such as
in the obvious way. It is usually taken for granted that the
numbers
also
somehow fit into this scheme, so that every real number
irrational
can be drawn as a point on the line. We will not make too much fuss about
numbers is intended
justifyingthis assumption, since this method of
solely as a method of picturing certain abstract ideas, and our proofs will never
rely on these pictures
we will frequently use a picture to suggest or help
(although
explain a proof). Because this geometric picture plays such a prominent, albeit
inessential role, geometric terminology is frequently employed when speaking of
numbers-thus
is sometimes called a point,and R is often called the
a number
real kne.
The number |a-b| has a simple interpretation in terms of this geometric picture:
it is the distance between a and b, the length of the line segment which has a as one
end point and b as the other. This means, to choose an example whose frequent
-1,
(,
"drawing"
that the set of numbers
x which satisfy
of
|x a| < e may be pictured as the collection points whose distance from a is
less than e. This set of points is the
from a e to a + e, which may also
numbers
be described as the points corresponding
to
x with a
e < x < a + e
(Figure 2).
occurrence
justifiesspecial
consideration,
-
"interval"
-
-
r
a FIGURE
2
e
o
,
a +
Sets of numbers 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, but in context it is
always clear (orcan easily be made clear) whether one is talking about a pair or
Ø, the set with no elements; in pracan interval. Note that if a > b, then (a, b)
=
56
4. Graphs 57
the open interval
.
the closed interval
(a, b)
[a,b]
a
the interval
(-
o,
a)
the interval
the interval
(-
o,
oo)
a]
The set {x: a Exs b} is denoted by [a,b] and is called the closed interval
from a to b. This symbol is usually reserved for the case a < b, but it is sometimes
used for a
b, also. The usual pictures for the intervals (a, b) and [a,b] are shown
in Figure 3; since no reasonably accurate picture could ever indicate the difference
between the two intervals, various conventions have been adopted. Figure 3 also
shows certain
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, a) and (-oo, a] are
similarly.
standard
warning must be issued: the symbols ao
point
this
defined
At
a
=
"infinite"
[a, m)
the interval
FIGURE
(a,
tice, however, it is almost always assumed (explicitly
if one has been careful, and
implicitly otherwise), that whenever an interval (a, b) is mentioned, the number a
is less than b.
3
"infinity"
-oo,
though usually read
and
"minus
suggestive;
infinity," are purely
which satisfies oo > a for all numbers a. While the
there is no number
symbols oo and
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
and is sometimes denoted by (-oo, oo).
and
"oo"
-oo
(0, b)
_
_
(a, b)
"interval,"
i
I
(1, 1)
1)
(-1,
Of even greater interest to us than the method of drawing numbers is a method
drawing pairs of numbers.
This procedure, probably also familiar to you, resystem,"
quires a
two straight lines intersecting at right angles. To
distinguish these straight lines, we call one the horizontalaxis, and one the vertical
axis. (More prosaic terminology, such as the
and
axes, is probably
of
view, but most people hold their books, or at
preferable from a logical point
and
least their blackboards, in the same way, so that
are
with
real numbers,
could
of
the
descriptive.)
be
labeled
Each
but
two axes
more
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
of the coordinate system, is labeled (0, 0). Any pair (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 firstand second
coordinates, respectively, of the point determined in this way.
of
"coordinate
(0, 0)
(a,o)
"first"
-1)
.
.
(-1,
-1)
(1,
"second"
"horizontal"
FIGURE
4
"vertical"
"origin"
f(x)
FIGURE
f(x)
-
i
Our real concern, let us recall, is a method of drawing functions. Since a function is just a collection of pairs of numbers, we can draw a function by drawing
each of the pairs in the function. The drawing obtained in this way is called the
graph of the function. In other words, the graph of f contains all the points corresponding
to pairs (x,f (x)). Since most functions contain infinitely many pairs,
drawing the graph promises to be a laborious undertaking, but, in fact, many
functions have graphs which are quite easy to draw.
5
--x
=
f(x) y
=
Not surprisingly, the simplest functions of all, the constant functions f(x)
c,
have the simplest graphs. It is easy to see that the graph of the function f (x) c
is a straight line parallel to the horizontal axis, at distance c from it (Figure 5).
=
=
The functions f (x)
lines
cx also have particularly simple graphsstraight
of
Figure
this
6. A proof
fact is indicated in Figure 7:
(0, 0), as in
=
FIGURE
6
through
58 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, correwith
sponding to (x, cx). A point A',
first coordinate y, will lie on L when the
triangle A'B'O is similar to the triangle ABO, thus when
L
(x,ex)
A'
--=-=c;
OB'
-(0,
FIGURE
o)
D'
7
E
(a, b)
'
length
FIGURE
this is precisely the condition that A' corresponds to the pair (y, cy), i.e., that A'
lies on the graph of f. The argument has implicitly assumed that c > 0, but the
other cases are treated easily enough. The number c, which measures the ratio of
the sides of the triangles appearing in the proof, is called the slope of the straight
line, and a line parallel to this line is also said to have slope c.
B
length d
-
a
a
-
b
This demonstration has neither been labeled nor treated as a formal pmof.
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 (ora precisely
stated assumption) that the points on a straight line correspond in an exact way
Aside from this, it would be necessary to develop plane
to the real numbers.
geometry as precisely as we intend to develop the properties of real numbers.
Now the detailed development of plane geometry is a beautiful subject, but it is by
no means a prerequisite for the study of calculus. We shall use geometric pictures
only as an aid to intuition; for our purposes (andfor most of mathematics)
it is
perfectly satisfactory to defmethe plane to be the set of all pairs of real numbers,
and to dafmestraight lines as certain collections of pairs, including, among others,
the collections {(x,cx) : x a real number}. To provide this artificially constructed
geometry with all the structure of geometry studied in high school, one more
defmition is required. If (a, b) and (c,d) are two points in the plane, i.e., pairs of
real numbers, we dfme the distance between (a, b) and (c, d) to be
(a
f(x)
-
cx
‡ d
/
(o,
OB
-
c)2 +
(b
-
d)2.
If the motivation for this definition is not clear, Figure 8 should serve as adequate
explanation-with
this definition the Pythagorean theorem has 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, passing through the point (0, d). For this reason, the functions
f (x) cx + d are called linear functions. Simple as they are, linear functions occur frequently, and you should feel comfortable working with them. The
following is a typical problem whose solution should not cause any trouble. Given
two distinct points (a, b) and (c,d), find the linear function f whose graph goes
through (a, b) and (c, d). This amounts to saying that f(a) b and f(c) d. If
=
=
FIGURE
9
=
*
=
numbers
The fastidious reader might object to this definition on the grounds that nonnegative
objection
really
unanswerable
the
moment-the
have
This
is
are not yet known to
square roots.
at
definition will just have to be accepted with reservations, until this little point is settled.
4. Graphs 59
e
f is to be of
the form
f (x)
=
ax +
ß, then
we must
have
ac+ß=d;
a)
therefore a
=
(d
-
b)/(c
f(x)=
-
d-b
c-a
a) and
x+b-
ß
=
b
-
[(d
d-b
c-a
-
b)/(c
d-b
a=
c-a
"point-slope
(b)
/
/
FIGURE
e
(c)
lo
(d)
-
a)]a,
so
(x-a)+b,
form" (seeProblem 6).
a formula most easily remembered by using the
Of course, this solution is possible only if a ¢ c; the graphs of linear functions
account only for the straight lines 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 defmition, a
function cannot contain (a, b) and (a, c) if b ¢ 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 fimction. Thus, the first two sets in Figure 10 are not
graphs of functions and the last two are; notice that the fourth is the graph of a
function whose domain is not all of R, since some vertical lines have no points on
them at all.
After the linear functions the simplest is perhaps the function f (x) = x2 gyy
draw some of the pairs in f, i.e., some of the pairs of the form (x,x2), we obtain
a picture like Figure 11.
e (2, 4)
FIGURE
11
60 Foundations
M
"'
It is not hard to convince yourself that all the pairs (x, x2) lie along a curve like
the one shown in Figure 12; this curve is known as a parabola.
Since a graph is just a drawing on paper, made (inthis 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.
there
Nevertheless,
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 x2 < y2, so the graph should be higher at y than at x, which is
not the case in (a) It is also easy to see, simply by drawing a very accurate graph,
first plotting many pairs (x, x2), that the graph cannot have a large
as in (b)
order
these
assertions,
however, we first need
to prove
or a
as in (c).In
to say, in a mathematical
way, what it means for a function not to have a
"corner";
already
these
ideas
involve some of the fundamental concepts of
or
calculus. Eventually we will be able to define them rigorously, but meanwhile
you
to define these concepts, and then examining
may amuse yourself by attempting
your definitions critically Later these definitions may be compared with the ones
mathematicians
have agreed upon. If they compare favorably, you are certainly
to be congratulated!
called
The functions f (x) x", for various natural numbers n, are sometimes
easily
compared
as in Figure 14, by
power functions. Their graphs are most
drawing several at once.
The power functions are only special cases of polynomial functions, introduced
in the previous chapter. Two particular polynomial functions are graphed in
.
"jump"
"corner"
"jump"
FIGURE
12
=
(a)
(c)
FIGURE
13
FIGURE
14
4. Graphs 61
f(x)
-
Figure 15, while Figure 16 is
polynomial function
x + x
f (x)
-1
-1
=
to give a general idea of the graph of the
meant
anx"
+
a,_ix"¯I
+
-··+
ao,
in the case a, > 0.
In general, the graph of f will have at most n 1
or
(a
is a point like (x,f (x)) in Figure 16, while a
is a point like (y, f (y)). The
number of peaks and valleys may actually be much smaller
(thepower functions,
for example, have at most one valley). Although these assertions are easy to make,
giving proofs until Part III (oncethe powerful methwe will not even contemplate
ods of Part III are available, the proofs will be very easy).
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 we can use the derivative, the basic tool of Part III.
together" the graphs of
Many interesting graphs can be constructed by
functions already studied. The graph in Figure 18 is made up entirely of straight
lines. The function f with this graph satisfies
"peaks"
"valleys"
"peak"
-
"valley"
(a)
-2
-
"piecing
a
__
3
f
1
=
(-1)n+1
=
(-1).÷1
=
1,
-1
f
I
-2--
-
E
f(x)
(b)
|x|
>
1,
-l/(n+
FIGURE
15
and is a linear function on each interval [l/(n+ 1), 1/n] and [-1/n,
1)].
(The number 0 is not in the domain of f.) Of course, one can write out an explicit
formula for f (x), when x is in [l/(n+ 1), 1/n]; this is a good exercise in the use
of linear functions, and will also convince you that a picture is worth a thousand
words.
1
I
I
I
I
I
I
I
I
i
I
FIGURE
16
I
n even
n odd
(a)
(b)
02
Poundahons
f(x)
f(x)
f(x)
=
1
1
f(x) p
-
i
(b)
(c)
(d)
FIGURE
17
i
FIGURE
18
-
4. Graphs 63
1-
-
f(x)
=
sin x
-360
720
360
---1
FIGURE
19
It is actually possible to define, in a much simpler way, a function which exhibits
this same property of oscillating infinitely often near 0, by using the sine function.
In Chapter 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 (thescale 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).
sinl/x.
The graph of
Now consider the function f(x)
observe
that
ure 20. To draw this graph it helps to first
f(x)=0
f(x)=1
f(x)=--1
1
1
1
180 360 540
1
1
1
forx=90 90 + 360 90 + 720
1
1
1
forx=270 270 + 360 270 + 720
forx=-
-
'
-,...,
'
'
'
'
is shown in Fig-
f
=
'
'
...
'
....
'
Notice that when x is large, so that 1/x is small, f (x) is also small; when x is
f (x)
FIGURE
20
-
sin 1
-
64 Foundations
f(x)
FIGURE
"large
=
x sin
1
-
21
negative,"
although
f (x)
that is, when
<
|x| is large for negative
x, again
f (x) is close
to 0,
0.
An interesting modification of this fimetion is f (x) x sin 1/x. The graph of
this function is sketched in Figure 21. Since sin 1/x oscillates infinitely often near 0
the function f (x) x sin 1/x oscillates infinitely often between
between 1 and
and
The behavior of the graph for x large or large negative is harder to
x
=
-1,
=
-x.
\
FIGURE
22
/
4. Graphs 65
j
2·-
Since sin 1/x is getting close to 0, while x is getting larger and larger, there
seems to be no telling what the product will do. It is possible to decide, but this is
x2 sin 1/x
another question that is best deferred to Part III. The graph of f(x)
has also been illustrated (Figure 22).
For these infinitely oscillating functions, it is clear that the graph cannot hope to
The best we can do is to show part of it, and leave out the
be really
the
is
interesting part). Actually, it is easy to find much simpler
part near 0 (which
whose
functions
graphs cannot be
drawn. The graphs of
analyze.
=
(a)
"accurate."
g
"accurately"
x2
f (x)
f
2,
'
741
2
and
g(x)
1
x>
23
=
x
>
1
FIGURE
(x,y)
(a, b)
irrational
0,
x
1,
x rational.
1, x rational
0, x irrational
¯
25
2,
41
'
CIOsed
f (x)
FIGURE
=
be distinguished by some convention similar to that used for open and
intervals (Figure 23).
Out last example is a function whose graph is spectacularly nondrawable:
can only
(b)
FIGURE
=
24
The graph of f must contain infinitely many points on the horizontal axis and
also infinitely many points on a line parallel to the horizontal axis, 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 actually a mathematical reason behind this convention, but it depends on some sophisticated ideas,
introduced in Problems 21-5 and 21-6.)
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.
A circle with center (a, b) and radius r > 0 contains, by definition, all the points
(X, y) Whose
distance from (a, b) is equal to r. The circle thus consists (Figure 25)
of all points
(x, y)
with
(x-a)2+Q-b2=r
or
(x
-
a)2 +
Q
-
2
2
66 Foundations
and radius 1, often regarded as a sort of standard copy,
The circle with center (0,0)
is called the unit cirde.
A close relative of the circle is the ellipse. This is defined as the set of points,
the sum of whose distances from two
points is a constant. (When the two
foci are the same, we obtain a circle.) If, for convenience, the focus points are
taken to be (-c, 0) and (c, 0), and the sum of the distances is taken to be 2a (the
factor 2 simplifies some algebra), then (x, y) is on the ellipse if and only if
"focus"
2
(x (-c))2
-
or
(x
+
2
c)2
-
=
2a
(x+c)2+y2=2a-d(x-c)2+y2
or
x2+2cx+c2+y2=4a2-4ad(x-c)2+y2+x2-2cx+c2+y2
or
4(cx
or
c2x2
-
-
a2)
2cxa2 # a4
or
(c2_
or
2
(x
2
_
2__a2
2
_
b
=
a2
-
c2
2
-
C2
x2
y2
a2
b2
0, so that
=
2
=1,
x=±a,
(a,0)
(-a, 0)
(0,
26
2
'
must clearly
of an ellipse is shown
.
FIGURE
2
4 y2
1.
=
(sincewe
0). A picture
horizontal axis when y
>
sects the
2
_
simply
-+-=1
42
+ y2
72
a2 + a2
-
where
a2
-
2_
x2
This is usually written
c)2
-4a
=
-
b)
choose
it follows that
in Figure 26. The ellipse intera
>
c,
4. Graphs 67
and
it intersects the vertical axis when x
=
0, so that
2
1,
=
y
=
±b.
The hyperbola is defined analogously, except that we require the dagence of
the two distances to be constant. Choosing the points (-c, 0) and (c,0) once
again, and the constant difTerence as 2a, we obtain, as the condition that (x, y)
be on the hyperbola,
(-a,0)
(a,o)
which may
FIGURE
27
2-
(x+c)2
(X-C)
+y2=±2a,
be simplified to
In this case, however, we must clearly choose c > a, so that a2
a2, then (x, y) is on the hyperbola if and only if
b
=
c2
-
<
0. If
Ác2
-
x
2
---=1
a2
2
y
b2
The picture is shown in Figure 27. It contains two pieces, because the difference
between the distances of (x, y) from (-c, 0) and (c,0) may be taken in two different orders. The hyperbola intersects the horizontal axis when y
0, so that
vertical
the
axis.
±a, but it never intersects
x
b
It is interesting to compare (Figure 28) the hyperbola with a
à 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 23).
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 made up of the
=
=
=
=
(a)
FIGURE
28
(b)
=
68 Foundations
graphs of two functions,
f (x)
=
g(x)
=
Ál
b
(x2/a2),
-
-a
5 x 5 a
and
(x2g2),
-b
1
-
-a
5 x 5 a.
Of course, there are many other pairs of functions with this same property For
example, we can take
f (x)
b
1
(x2/a2),
-
O < xs
=
-bÅl-(x2/a2),
a
-a5x50
and
Ál
-b
g(x)
=
1
b
-
(x2/a2),
-
(x2 a2,
O< x 5 a
-a
5 x 5 0.
We could also choose
f(x)
b /1
=
(x2/a2),
-
-b
1
--
X
rational,
(x2/a ),
x irrational,
(x2/a2),
x rational,
-
--
a
X
a
a 5 x 5 a
and
-b
g(x)
1
-
--
=
b Ál
-
(x2
2),
x
irradonal,
a 5 x 5 a
_<
--
G
X
G.
But all these other pairs necessarily involve unreasonable
functions which jump
of
this fact, is too difficult at present.
around. A proof, or even a precise statement
Although you have probably already begun to make a distinction between those
functions with reasonable graphs, and those with unreasonable
graphs, you may
reasonable
reasonable
of
definition
find it very difficult to state a
functions. A
mathematical
definition of this concept is by no means easy, and a great deal of this
book may be viewed as successive attempts to impose more and more conditions
that a
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 reasonable.
The answer, unfortunately, will always be
"yes."
or at best, a qualified
"reasonable"
"no,"
PROBLEMS
1.
Indicate on a straight line the set of all x satisfying the following conditions.
Also name each set, using the notation for intervals (insome cases you will
also need the U sign).
(i)
(ii)
(iii)
(iv)
|x--3|<1.
|x 3| 5 1.
|x a| < e.
|x2 1| <
-
-
-
.
4. Graphs 69
+1
1
1
5 a (give
an answer in terms of a, distinguishing various cases).
1 + x2
(vii)x2 + 1 > 2.
(viii)(x + 1)(x 1) (x 2) > 0.
(vi)
-
2.
-
There is a very useful way of describing the points of the closed interval
(wherewe
as usual, that a
assume,
<
b).
[a,b]
the interval [0,b], for b > 0. Prove that if x is in [0,b],
tb
1. What is the significance of the
then x
for some t with 0 sts
mid-point
of
the interval [0,b]?
number t? What is the
b],
then
that
if
is
in
(1 t)a + tb for some t with
x
x
[a,
(b) Now prove
also
expression
a).
be written as a + t(b
0 5 t 5 1. Hint: This
can
What is the midpoint of the interval [a, b]? What is the point 1/3 of the
(a) First consider
=
=
-
-
from a to b?
(c) Prove, conversely, that if 0 sts 1, then (1 t)a + tb is in [a,b].
(d) The points of the open interval (a, b) are those of the form (1 t)a + tb
for0<t<1.
way
-
-
3.
Draw the set of all points (x, y) satisfying the following conditions. (In most
cases your picture will be a sizable portion of a plane, not just a line or curve.)
(i)
>
y.
+
(ii) x a 2> y + b.
(iii) y < x 2.
(iv) y 5 x
x
(v) |x-y|<1.
(vi) |x+y|<1.
(vii)x +
y is an integer.
(viii)x +
y
1
(ix) (x2
(x)
4.
-
is an integer.
1)2 + (y
-
2)2
4
y
4
x
Draw the set of all points (x, y) satisfying the following conditions:
(i) |x|+|y|=1.
(ii) |xi |yl
-
(iii) |x
1|
-
=
|1 x|
x2+y2=0.
(iv)
(v)
(vi) xy
x2
(vii)x2
(viii)
=
-
=
-
--
=
1.
|y
|y
0.
2x + y2
y2.
1|.
1|.
-
-
=
4.
70 Foundations
5.
Draw the set of all points
(i)
x
(iv) x
the following conditions:
y2.
=
2
(ii)
(iii) x
satisfying
(x, y)
2
=
-
=
=
|y
1.
.
sin y.
Hint: You already know the answers when x and y are interchanged.
6.
(a) Show that the straight
function f (x) m(x
=
line through (a, b) with slope m is the graph of the
a) + b. This formula, known as the
"point-slope
-
form" is far more convenient than the equivalent expression f (x)
it is immediately clear from the point-slope form that the
mx + (b
slope is m, and that the value of f at a is b.
(b) For a y c, show that the straight line through (a, b) and (c, d) is the
graph of the function
=
-ma);
d-b(x-a)+b.
f(x)=
-a
c
are the graphs of
straight lines?
(c) When
7.
f(x)
=
mx + b and g(x)
=
m'x + b' parallel
nurnbers A, B, and C, with A and B not both 0, show that the
0 is a straight line (possibly
set of all (x, y) satisfying Ax + By + C
a
vertical one). Hint: First decide when a vertical straight line is described.
(b) Show conversely that every straight line, including vertical ones, can be
described as the set of all (x, y) satisfying Ax + By + C 0.
(a) For any
=
(1, m)
=
8.
(a) Prove that
the graphs of the functions
f(x)=mx+b,
g(x)=nx+c,
-1,
by computing the squares of the lengths
of the sides of the triangle in Figure 29. (Why is this special case, where
the lines intersect at the origin, as good as the general case?)
(b) Prove that the two straight lines consisting of all (x, y) satisfying the conare
gi,
FIGURE
29
perpendicular if mn
=
ditions
Ax+By+C=0,
A'x+B'y+C'=0,
are perpendicular
9.
(a) Prove, using
if and only if AA'+ BB'
=
0.
Problem 1-19, that
(x1+71)2+(X2+72)
Il2+X22
2+122
4. Graphs 71
(b) Prove that
(xs
-
xi)2
(ya yi)2
+
-
<
V(x2 Il)2
-
Ÿ
Q2
+
-
B)2
Á(xa x2)
-
Interpret this inequality geometrically (itis called the
ity"). When does strict inequality hold?
10.
"triangle
13
-
72)2
inequal-
Sketch the graphs of the following functions, plotting enough points to get
(Part of the problem is to make
a good idea of the general appearance.
the queries posed below are
decision how many is
a reasonable
meant to show that a little thought will often be more valuable than hundreds
of individual points.)
"enough";
(i) f (x)
=
1
x+
(What happens for x
-.
x
near
0, and for large x? Where
does the graph lie in relation to the graph of the identify function? Why
does it suffice to consider only positive x at first?)
11.
(ii) f (x)
=
(iii) f (x)
=
(iv) f (x)
=
1
x
-
-.
x
x2 + 1
4.
x
x
2
1
-
yx
.
Describe the general features of the graph of f if
is even.
is
(ii) f odd.
(iii) f is nonnegative.
(i) f
(iv) f (x)
odic,
12.
=
f (x +
a) for all x
with period
(afunction with
this property is called peri-
a.
Graph the fimetions f(x)
6 for m = 1, 2, 3, 4. (There is an easy way to
do this, using Figure 14. Be sure to remember, however, that 6 means the
mth root of x when m is even; you should also note that there will be
positive
an important difference between the graphs when m is even and when m is
=
odd.)
13.
(a) Graph f(x)
(b) Graph f(x)
=
|x| and f(x) x2.
|sinx| and f(x)
=
sin2x.
(There is an important differwhich
the
graphs,
we cannot yet even describe rigorously.
ence between
what
See if you can discover
it is; part (a)is meant to be a clue.)
14.
=
=
Describe the graph of g in terms of the graph of f if
72 Foundations
g(x)
(ii) g(x)
(iii) g(x)
(iv) g(x)
(v) g(x)=
(vi) g(x)
(i)
(vii)g(x)
(viii)g(x)
(ix) g(x)
(x)
g(x)
=
=
=
=
=
=
f (x) + c.
f (x + c). (It is easy
cf
(x)
(Distinguish the
f (cx).
cases c
=
0, c
>
0, c
<
0.)
f(1/x).
f (|x|).
If(x)l.
=
max(f,
=
min(
=
to make a mistake here.)
0).
f, 0).
max(f, 1).
ax2
15.
Draw the graph of f (x)
lem 1-18.
16.
Suppose that A and C are not both 0. Show that the set of all (x, y) satisfying
=
+ bx + c. Hint: Use the methods
Ax2 + Bx + Cy2 + Dy + E
=
of
Prob-
0
is either a parabola, an ellipse, or an hyperbola (orpossibly 0). Hint: The
0 is essentially Problem 15, and the case A
0 is just a minor
variant. Now consider separately the cases where A and B are both positive
or negative, and where one is positive while the other is negative.
case C
17.
=
=
The symbol [x]denotes the largest integer which is s x. Thus, [2.1] [2]
2 and [-0.9]
Draw the graph of the following functions
[-1]
(theyare all quite interesting, and several will reappear frequently in other
problems).
=
=
--1.
=
=
(i) f (x) [x].
(ii) f (x) x [x].
=
=
-
(iii) f(x) fx [x].
(iv) f (x) [x]+ )x [x].
=
-
=
,
18.
(v) f (x)
=
(vi) f (x)
=
-
1
-
.
x
1
-
.
Graph the following functions.
(i) f (x)
=
{x),where {x}is defined to be the distance from x
integer.
(ii) f (x) {2x}.
=
(iii) f (x) {x}+ {2x).
(iv) f (x) {4x).
+ i {4x}.
(v) f(x) {x}+ ({2x)
=
=
=
to the nearest
4. Graphs 73
Many 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 23, your intuitive notion of infmite decimals should suffice to carry
you through the following problem, and others which occur before Chapter 23.
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 O's (e.g.,
1.23999...
1.24000...). We will always use the one ending in 9's.
=
*19.
Describe as best you can the graphs of the following functions
picture is usually out of the question).
(a complete
the 1st number in the decimal expansion of x.
(ii) f (x) the 2nd number in the decimal expansion of x.
(iii) f (x) the number of 7's in the decimal expansion of x if this number
is finite, and 0 otherwise.
(iv) f (x) 0 if the number of 7's in the decimal expansion of x is finite,
(i) f (x)
=
=
=
=
and
1 otherwise.
by replacing all digits in the decimal
expansion
come after the first 7 (ifany) by 0.
1
if
never appears in the decimal expansion of x, and n if 1
f (x) 0
first appears in the nth place.
(v) f (x)
(vi)
*20.
=
the number
obtained
of x which
=
Let
f (x)
0,
1
=
x irrational
p rational
x
-,
q
=
-
q
in lowest terms.
(A number p/q is in lowest terms if p and q are integers with no common
factor, and q > 0). Draw the graph of f as well as you can (don'tsprinkle
2,
points randomly on the paper; consider first the rational numbers with q
then those with q
3, etc.).
=
=
21.
(x,x1
points on the graph of f(x) = x2 are the ones of the form (x.x2L
Prove that each such point is equidistant from the point (0, and the
(See Figure 30.)
graph of g(x)
line L, the graph of g(x)
y,
(b) Given a point P (a, ß) and a horizontal
show that the set of all points (x, y) equidistant from P and L is the graph
of a function of the form f (x) ax2 + bx + c.
(a) The
¼)
--i.
=
=
=
=
FIGURE
3o
*22.
(a) Show that
the square of the distance from
x2(m2 +
1) + x(-2md
-
(c, d) to (x, mx) is
2c) + d2 ‡
C
.
Using Problem 1-18 to find the minimum of these numbers,
the distance from (c, d) to the graph of f (x) mx is
show that
=
|cm-d||Jm2+1.
the distance from
this case to part (a).)
(b) Find
(c,d)
to the graph of
f (x)
=
mx
+ b. (Reduce
N
Foundations
*23.
Problem 22, show that the numbers
ure 31 are given by
',
'
x' and y'
(a) Using
1
indicated in Fig-
1
x'=-x+-y,
ngth x'
length y'
r
45
(b) Show that
FIGURE
31
the set of all
as the set of all
(x, y)
(x, y) with xy
with
=
1.
(x'/h)2
-
O'/h)2
=
1 is the same
4, Appendix1. Vectors75
APPENDIX 1.
VECTORS
Suppose that v is a point in the plane; in other words, o is a pair of numbers
v
1,
(vl
=
,
U2).
For convenience, we will use this convention that subscripts indicate the first and
second pairs of a point that has been described by a single letter. Thus, if we
that w is the pair (wi, w2)
mention the points w and z, it will be understood
while
z is the pair (zi, z2).
Instead of the actual pair of numbers (vi, v2), we often picture v as an arrow
from the origin O to this point (Figure 1), and we refer to these arrows as vectors
in the plane. Of course, we've haven't really said anything new yet, we've simply
introduced an alternate term for a point of the plane, and another mental picture.
The real point of the new terminology is to emphasize that we are going to do
some new things with points in the plane.
For example, suppose that we have two vectors (i.e.,points) in the plane,
U2)
O
v
=
(vi, v2).
Then we can define a new vector
FI GU RE 1
(1)
E
=
(El,
©2)·
point of the plane) v + w by the equation
(anew
v+w=(v1+wi,v2+w2).
Notice that all the letters on the right side of this equation
On the other
+ sign is just our usual addition of numbers.
side
the left
is new: previously, the sum of two points in the
and we've simply used equation
(1)as a defmition.
mathematician
might
want to use some new
A very fussy
defined operation, like
v + w,
or
perhaps
ve
are numbers, and the
hand, the + sign on
plane wasn't defined,
symbol
for this newly
w,
but there's really no need to insist on this; since v + w hasn't been defined before,
there's no possibility of confusion, so we might as well keep the notation simple.
Of course, any one can make new notation; for example, since it's our defmition,
1012),
we could just as well have defined o + w as (vi + wi w2, v2 +
or by some
construction
weird
real
equally
question
formula. The
other
is, does our new
have
any particular significance?
Figure 2 shows two vectors o and w, as well as the point
-
L
(vi +
wi, v2 + w2)
(vi +
(v2+ w2)
w
(vi, v2)
¯
v
"'
-
v2
wi, v2 +
for the moment, we have simply indicated in the usual way, without drawing
an arrow. Note that it is easy to compute the slope of the line L between v and
our new point: as indicated in Figure 2, this slope is just
which,
(v2
2)
-
(vi+wi)-vi
FIGURE
2
©2),
U2
2
wl'
and this, of course, is the slope of our vector w, from the origin O to
Other
words, the line L is parallel to w.
(wi, ©2). In
76 Foundations
Similarly, the slope of the line M between (wi, w2) and our new point is
w2)
(v2+
'
v2
(v1+wi)-v2
i
'
v+w
w2
-
vi
is the slope of the vector v; so M is parallel to v. In short, the new point
o + w lies on the parallelogram having v and w as sides. When we draw v + w as
an arrow (Figure 3), it points along the diagonal of this parallelogram. In physics,
vectors are used to symbolize forces, and the sum of two vectors represents the
resultant force when two different forces are applied simultaneously
to the same
which
w
i
'
o
object.
FIGURE
"w"
Figure 4 shows another way ofvisualizing the sum v+w. If we use
to denote
of at
the
starting
and
having
length,
instead
at v
but
same
an arrow parallel to w,
origin,
the
then o + w is the vector from O to the final endpoint; thus we get to
by
first
following v, and then following w.
+
v w
of
Many the properties of + for ordinary numbers also hold for this new + for
vectors. For example, the
law"
3
"commutative
v+w=w+v,
is obvious from the geometric picture, since the parallelogram spanned by v and
w is the same as the parallelogram spanned by w and v. It is also easily checked
analytically, since it states that
v + w
"w"
(vi +
and thus simply
Ÿ
U2
wl,
"
vi + wi
+
72
Similarly, unraveling
4
(El
=
Ÿ
Ul, U2
W2
wi +
+
=
defmitions, we find the
72"associative
Figure 5 indicates a method of finding v + w + z.
identity,"
The origin O (0,0) is an
"additive
=
O+v=v+O=v,
"z"
and
if we define
-v2),
-v
=
(-vi,
then we also have
"w"
(-v)
o+
w
-v
+ v
=
0.
=
Naturally we can also define
z
w
-
o
w +
=
(-v),
I
exactly
FIGURE 5
92),
i,
W2
=
[v+w]+z=v+[w+z].
o + w + z
+
depends on the commutative law for numbers:
i
FIGURE
©2)
as with numbers;
equivalently,
W
¯
U
=
(WI
-
71. W2
¯
U2).
law"
4, Appendix1. Vectors77
with
Justas
numbers,
our
definition of w
-
v simply
means
that it satisfies
"w
v" that is parallel to w o but that starts
Figure 6(a) shows v and an arrow
at the endpoint of v. As we established with Figure 4, the vector from the origin
to the endpoint of this arrow is just v + (w v)
w (Figure 6(b)).In other words,
that
we can picture w v geometrically as the arrow that goes from v to w (except
it must then be moved back to the origin).
There is also a way of multiplying a number by a vector: For a number a and
a vector o = (vi, v2), we define
v
-
-
=
-
-
(a)
"w
(avt, av2)
=
a v
-v"
w
(We sometimes simply write av instead of a v; of course, it is then especially
important to remember that v denotes a vector, rather than a number.)
The
vector a v points in the same direction as v when a > 0 and in the opposite
direction when a < 0 (Figure 7).
You can easily check the following formulas:
-
-
FIGURE
6
a
-
(b
v)
·
(ab)
=
·
v,
1-v=v,
0-v=0,
-1-o=-v.
2v
Notice that we have only defined a product of a number and a vector, we have
two vectors to get another vector.* However,
not defined a way of
there are various ways of
vectors to get numbers, which are explored
in the following problems.
'multiplying'
'multiplying'
PROBLEMS
1
2
-v
-
FIGURE
1.
7
Given a point v of the plane, let Re(v) be the result of rotating v around the
origin through an angle of 8 (Figure 8). The aim of this problem is to obtain
a formula for Ro, with minimal calculation.
(a) Show that
Ro (1,0)
Re (0, 1)
(b) Explain why we
(cos0, sin 0),
(- sin 9, cos 0).
=
=
have
Re(v+w)
Re (a w)
-
Re(v)
(c) Now show
8
Re(v)+ Re(w),
a
that for any point
Re(x, y)
FIGURE
=
=
=
(xcos0
-
.
Re(w).
(x, y)
we
have
y sin0, x sin0 + ycos0).
*If you jump to Chapter 25, you'll find that there is an important way of defining a product, but
this is something very special for the plane-it
doesn't work for vectors in 3-space, for example, even
though the other constructions
do.
78 Foundations
(d) Use this
2.
result to
give another
solution
to Problem 4-23.
Given v and w, we define the number
this is often called the
being
a rather
'dot
product' or
old-fashioned
(a) Given v, find
such vectors
vimi +
=
v•w
U2
2,
'scalar
product' of v and w
as opposed
word for a number,
w such that v . w
a vector
=
('scalar'
to a vector).
0. Now describe the set of all
w.
(b) Show that
v.w=w·v
v.(w+z)=v•w+v·z
and that
a-(v•w)=(a-v).w=v.(a-w).
Notice that the last of these equations involves three products: the dot
product of two vectors; the product of a number and a vector; and
the ordinary product of two numbers.
(c) Show that v•v > 0, and that v•v 0 only when v 0. Hence we can
define the norm ||v|| as
-
-
-
=
=
livll E,
=
will
which
be 0 only for v
=
0. What is the geometric interpretation of
the norm?
that
(d) Prove
w Il5
Ilv+
Ilvll+ llell,
and that equality holds if and only if v
some number a > 0.
(e) Show that
=
v•w
3.
(a) Let
||v +
w ||2
-
||v
-
=
0 or w
=
0 or w
=
a
-
v
for
w ||2
4
Ro be rotation by an angle of 9 (Problem 1). Show that
Re(v)
•
Re(w)
=
v
•w.
length 1 pointing along the first axis, and
let w (cos0, sin 0); this is a vector of length 1 that makes an angle of 0
with the first axis
(compareProblem 1). Calculate that
(b) Let e
=
(1, 0) be the
of
vector
=
e.w
=
cos0.
Conclude that in general
v.w=
where 0
||v||· ||w||-cos0,
is the angle between v and w.
4, Appendix1. Vectors79
4.
L
v+ w
w
Given two vectors v and w, we'd expect to have a simple formula, involving
the coordinates
vi, v2, wi, ©2, for the area of the parallelogram they span.
Figure 9 indicates a strategy for finding such a formula: since the triangle
with vertices w, A, v + w is congruent to the triangle OBv, we can reduce
the problem to an easier one where one side of the parallelogram lies along
the horizontal axis:
line L passes through v and is parallel to w, so has slope w2/wlConclude that the point B has coordinate
(a) The
A
UlW2
and that the parallelogram
O
FIGURE
g
Wl72
-
therefore has area
det(v, w)
=
v1m2
Elv2-
-
This formula, which defines the determinantdet, certainly seems to be simple
enough, but it can't really be true that det(v, w) always gives the area. After
all, we clearly have
9
det(w, v)
so sometimes
=
det(v, w),
-
"deriva-
det will be negative!
Indeed, it is easy to see that our
tion" made all sorts of assumptions (thatw2 was positive, that B had a positive
coordinate, etc.) Nevertheless, it seems likely that det(v, w) is ± the area; the
next problem gives an independent proof.
5.
v points along the positive horizontal axis, show that det(v, w) is the
area of the parallelogram spanned by v and w for w above the horizontal
axis
of the area for w below this axis.
> 0), and the negative
If Ro is rotation by an angle of 0 (Problem 1), show that
(a) If
(w2
(b)
det(Rev, Row)
det(v, w).
=
Conclude that det(v, w) is the area of the parallelogram spanned by
and the
v and w when the rotation from v to w is counterclockwise,
negative of the area when it is clockwise.
6.
Show that
det(v, w + z)
det(v + w, z)
=
det(v, w) + det(v, z)
=
det(v, z) + det(w, z)
and that
a
w
llw|| sin0
-
9
FIGURE
7.
=
det(a v, w)
-
det(v, a w).
=
-
Using the method of Problem 3, show that
det(v, w)
v
10
det(v, w)
Which
=
||v|| ||wll sin0,
-
-
is also obvious from the geometric interpretation (Figure 10).
80 Foundations
APPENDIX 2.
THE CONIC SECTIONS
Although we will be concerned almost exclusively with figures in the plane,
defined formally as the set of all pairs of real numbers, in this Appendix we want
to consider three-dimensional space, which we can describe in terms of triples of
real numbers, using a
(0,0, 1)
"three-dimensional
coordinate
system,"
consisting
of three
lines intersecting at right angles (Figure 1). Our hongontaland vedical axes
now mutate to two axes in a horizontal plane, with the third axis perpendicular to
both.
One of the simplest subsets of this three-dimensional space is the (infinite)
cone
illustrated in Figure 2; this cone may be produced by rotating a
line,"
of slope C say, around the third axis.
straight
0, 1, 0)
(1, 0, 0)
"generating
FIGURE
l
slope C
(x, y, z)
.(x,y,0)
FIGURE
2
For any given first two coordinates
plane has distance
x and y, the point
x2 + y2 from the origin, and thus
(x,y, z) is in the
(1)
cone if and only if z
=
(x, y, 0) in the horizontal
+ y2.
±CÁx2
We can descend from these three-dimensional vistas to the more familiar twodimensional one by asking what happens when we intersect this cone with some
plane P (Figure 3).
C
P
FIGURE
3
4, Appendix2. The ConicSections 81
C
C
L
If the plane is parallel to the horizontal plane, there's certainly no mystery-the
intersection is just a circle. Otherwise, the plane P intersects the horizontal plane
in a straight line. We can make things a lot simpler for ourselves if we rotate
everything so that this intersection line points straight out from the plane of the
paper, while the first axis is in the usual position that we are familiar with. The
on," so that all we see (Figure 4) is its intersection L
plane P is thus viewed
with the plane of the first and third axes; from this view-point the cone itself simply
appears as two straight lines.
In the plane of the first and third axes, the line L can be described as the
COÏÎCCtiOn
of all points of the form
"straight
FIGURE
4
(x,Mx+B),
where M
is the slope of L. For an arbitrary point (x, y, z) it follows that
P if and only if :
(x, y, z) is in the plane
(2)
Combining
(1)and (2),we
see that
=
Mx + B.
(x, y, z) is in the intersection
of the cone and
the plane if and only if
Mx + B
(*)
L
>
,
=
±C
Now we have to choose coordinate axes in the plane P. We can choose L as the
first axis, measuring distances from the intersection Q with the horizontal plane
(Figure 5); for the second axis we just choose the line through Q parallel to our
original second axis. If the first coordinate of a point in P with respect to these
axes is x, then the first coordinate of this point with respect to the original axes
can be written in the form
ax +
ax +
FIGURE
5
ß
Jx2+ y2.
ß
for some a and ß. On the other hand, if the second coordinate of the point with
respect to these axes is y, then y is also the second coordinate with respect to the
Original
RXCS.
Consequently, (*)says that the point lies on the intersection of the plane and the
cone if and only if
M(ax +
B
ß) +
Although this looks fairly complicated,
Œ2
2
=
±C
J(ax + ß)
after squaring
+ y2.
we can write this as
2+a2(M2-A2)x2+Ex+F=0
«2
simplifies this
for some E and F that we won't bother writing out. Dividing by
to
C2y2 +
K2
-
M2)x2 + Gx + H
=
0.
Now Problem 4-16 indicates that this is either a parabola, an ellipse, or an
hyperbola. In fact, looking a little more closely at the solution (andinterchanging
82 Foundations
the roles of x and y), we see that the values of G and H are irrelevant:
(1)If
M
(2)If C2
=
>
(3)If C2 <
±C we obtain a parabola;
M2 we obtain an ellipse;
M2 we obtain an hyperbola.
These analytic conditions are easy to interpret geometrically (Figure 6):
plane is parallel to one
a parabola;
(2)If our plane slopes less than
intersection omits one half of
(3)If our plane slopes more than
hyperbola.
(1)If our
FIGURE
of the generating
lines of the cone we obtain
the generating line of the cone (sothat our
the cone) we obtain an ellipse;
the generating line of the cone we obtain an
6
"conic
sections" are related to this description.
In fact, the very names of these
the same root
The word parabolacomes from a Greek root meaning
that appears in parable, not to mention paradigm, paradox, paragon, paragraph,
paralegal, parallax, parallel, even parachute. Elhpse comes from a Greek root
meaning
or omission, as in ellipsis (anomission,
or the dots that inmeaning
beyond,' or
from
Greek
dicate it). And hyperbolacomes
root
a
and
of
words
like hyperactive, hypersensitive,
hypervenexcess. With the currency
tilate, not to mention hype, one can probably say, without risk of hyperbole, that
this root is familiar to almost everyone.*
'alongside,'
'defect,'
.
.
.
'throwing
¯
¯
PROBLEMS
1.
FIGURE
7
Consider a cylinder with a generator perpendicular to the horizontal plane
(Figure 7); the only requirement for a point (x, y, z) to lie on this cylinder is
*Although the correspondence between these roots and the geornetric picture correspond so beautifully, for the sake of dull accuracy it has to be reported that the Greeks originally applied the words
to describe features of certain equations involving the conic sections.
4, Appendix2. The ConicSections 83
that
(x, y) lies on
a circle:
x2
4
2
C2
=
Show that the intersection of a plane with this cylinder can be described by
an equation of the form
(ax + ß)2
P
_
2
What possibilities are there?
2.
F2
In Figure 8, the sphere Si has the same diameter as the cylinder, so that its
equator C1 lies along the cylinder; it is also tangent to the plane P at F1.
Similarly,the equator C2 of S2 lies along the cylinder, and S2 is tangent to P
z be any point on the intersection of P and the cylinder. Explain
why the length of the line from to Flis equal to the length of the vertical
z
line L from z to Ct.
(b) By proving a similar fact for the length of the line from z to F2, show that
the distance from z to F1 plus the distance from z to F2 is a constant, so
that the intersection is an ellipse, with foci Fi and F2-
(a) Let
C2
FIGURE
2
8
3.
Similarly, use Figure 9 to prove geometrically that the intersection of a plane
and a cone is an ellipse.
FIGURE
9
84 Foundations
APPENDIX 3.
length r
POLAR COORDINATES
In this chapter we've been acting all along as if there's only one way to label
points in the plane with pairs of numbers. Actually, there are many different
system." The usual coordinates
ways, each giving rise to a difFerent
of a point are called its cartesian coordinates, after the French mathematician
who first introduced the idea of
and philosopher René Descartes
(1596-1650),
coordinate
systems. In many situations it is more convenient to introduce polar
which are illustrated in Figure 1. To the point P we assign the polar
coordinates,
coordinates
(r, 0), where r is the distance from the origin O to P, and 0 is the
angle between the horizontal axis and the line from O to P. This angle can be
measured either in degrees or in radians (Chapter 15), but in either case 0 is not
determined unambiguously.
For example, with degree measurement, points on
the right side of the horizontal axis could have either 8
0 or 9 360; moreover,
8 is completely ambiguous at the origin O. So it is necessary to exclude some ray
through the origin if we want to assign a unique pair (r, Ð) to each point under
"coordinate
e
FIGURE
i
=
=
consideration.
On the other hand, there is no problem associating a unique point to any pair
(r, 0). In fact, we can even associate a point to (r, 0) when r < 0, according to
the scheme indicated in Figure 2. Thus, it always makes sense to talk about
point with polar coordinates (r, 9)," even though there is some ambiguity when
polar coordinates" of a given point.
we talk about
"the
"the
P is the point with polar coordinates (r, 01)
and also the point with polar coordinates
FIGURE
2
It is clear from Figure 1 (andFigure 2) that the point with polar coordinates
(r, 0) has cartesian coordinates (x, y) given by
x
=
r cos0,
y
=
r sin0.
4, Appendix3. Polar Coordinates85
coordinates
Conversely, if a point has cartesian
(r, 0) satisfy
(x, y),
ordinates
r
tan0
±Áx2
=
2
=
x
a
then
(anyof)
its polar co-
2
if x ¢ 0.
f is a function. Then by the graph of f in polar cothe
collection of all points P with polar coordinates (r, 0)
we mean
=
r
f (6). In other words, the graph of f in polar coordinates is the
collection of all points with polar coordinates
(f (0), 8). No special significance
should be attached to the fact that we are considering pairs ( (0),0), with f (0)
f
ÍÌTSt,
as opposed to pairs (x,f(x)) in the usual graph of f; it is purely a matter of
convention that r is considered the first polar coordinate and 0 is considered the
Now suppose that
ordinates
satisfying
FIGURE
3
second.
The graph of f in polar coordinates is often described as
"the
graph of the
equation r = f (0)." For example, suppose that f is a constant function, f (0) = a
for all 9. The graph of the equation r = a is simply a circle with center O and
radius a (Figure 3). This example illustrates, in a rather blatant way, that polar
coordinates are likely to make things simpler in situations that involve symmetry
with respect to the origin O.
The graph of the equation r = Ð is shown in Figure 4. The solid line corresponds
to all values of 0 > 0, while the dashed line corresponds to values of 0 5 0.
F I GURE
Spiral of Archimedes
4
As another example involving both positive and negative r, consider the graph of
the equation r
cos0. Figure 5(a)shows the part that corresponds to 0 5 0 5 90
[with0 in degrees]. Figure 5(b) shows the part corresponding to 90 5 0 5 180;
here r < 0. You can check that no new points are added for 9 > 180 or 0 < 0.
It is easy to describe this same graph in terms of the cartesian coordinates of its
points. Since the polar coordinates of any point on the graph satisfy
=
FIGURE
s
r
=
cos0,
86 Foundations
and
hence
r2
=
r
cosÛ,
its cartesian coordinates satisfy the equation
2
x2
which
describes a circle (Problem 3-16). [Conversely, it is clear that if the cartesian
coordinates of a point satisfy x2 + y2 x, then it lies on the graph of the equation
cos0.]
r
Although we've now gotten a circle in two different ways, we might well be
hesitant about trying to find the equation of an ellipse in polar coordinates. But
it turns out that we can get a very nice equation if we choose one of the focias
the origin. Figure 6 shows an ellipse with one focus at 0, with the sum of the
distances of all points from O and the other focus f being 2a. We've chosen f to
the left of O, with coordinates written as
=
=
(-2ea, 0).
(We have 0
_<
e
<
1, since we must have 2a
>
distance from f to 0).
(x, y)
2a
f
FIGURE
=
r
-
r
O
(-2ea, 0)
6
The distance r from (x, y) to O is given by
r2
(1)
By assumption,
the distance from
(2a
-
r)2
=
x2 + y2
to f is 2a
(x, y)
-
r,
(x [-2ea])2+
=
-
hence
y2,
or
(2)
Subtracting
4a2
-
4ar + r2
(1)from (2),and
=
x2
+ 4eax + 4e2a2 + y2
dividing by 4a, we get
a-r=ex+e2a,
4, Appendir3. Polar Coordinates87
or
r=a-ex-eia
(1
=
e2)a
-
ex,
-
which we can write as
(3)
r
=
A
-
for A
ex,
(1
=
e2¾
-
Substituting r cos 0 for x, we have
A
=
r
-
r(1+ecos9)
er cos0,
A,
=
and thus
A
(4)
r
=
.
1+ ecos0
In Chapter 4 we found that
2
2
(5)
+
=
1
is the equation in cartesian coordinates for an ellipse with 2a as the sum of the
distances to the foci, but with the foci at (-c, 0) and (c,0), where
Ja2-c2,
b=
Since the distance between the foci is 2c, this corresponds to the ellipse (4)when
(With equation
we take c
ea or 8 = CD
(3)determining A). Conversely, given
the ellipse described by (4),for the corresponding equation (5)the value of a is
determined by (3),
=
A
a=
and again using c
b=
=
1-e2'
sa, we get
a2-C2=
a2_g2a2=a
Î-€2-
Thus, we can obtain a and b, the lengths of the major and minor axes, immediately
from e and A.
The number
e
c
Ja2
a
a
-
-
b2
1-
-
b
2
,
a
"shape"
of the ellipse
the eccentncip of the ellipse, determines the
(theratio of the
major and minor axes), while the number A determines its
as shown by (4).
"size,"
88 Foundations
PROBLEMS
1.
If two points have polar coordinates (ri, Bi) and (72,4), show that the distance d between them is given by
d2
rt2
=
+ r2
2rir2 Cos(Û] Û2)-
-
What does this say geometrically?
2.
Describe the general features of the graph of f in polar coordinates
if
is even.
f is odd.
(i) f
(ii)
(iii) f (9) f (0 +
=
3.
180)
[when0 is measured
in degrees].
Sketch the graphs of the following equations.
r
=
a sin0.
r
=
a sec0.
(iii) r
=
(i)
(ii)
(iv)
(v)
(vi)
r
=
r
=
r
=
Hint: It is a very simple graph!
cos 20. Good luck on this onel
cos 30.
| cos 20|.
| cos 30
.
4.
Find equations for the cartesian
and (iii)
in Problem 3.
coordinates
of points on the
5.
Consider a hyperbola, where the difference of the distance between the two
foci is the constant 2a, and choose one focus at O and the other at (-2ea, 0).
(In this case, we must have e > 1). Show that we obtain the exact same
equation in polar coordinates
graphs
(i),(ii)
A
r
as we obtained
6.
=
l+scos0
for an ellipse.
Consider the set of points (x, y) such that the distance (x, y) to O is equal to
the distance from (x, y) to the line y
a (Figure 7). Show that the distance
conclude
and
that the equation can be written
to the line is a
r cos 9,
=
-
(x,y)
a=r(l+cos0).
y=a
Notice that this equation for a parabola is again of the same from as
0
7.
Now, for any A and e, consider the graph in polar coordinates of the equation (4),which implies (3).Show that the points satisfying this equation satisfy
(1 e2)x2 +
-
FIGURE7
(4).
y2
=
A2
-
2Aex.
Using Problem 4-16, show that this is an ellipse for e
1, and a hyperbola for e > 1.
e
=
<
1, a parabola for
4, Appendix3. Polar Coordinates89
8.
graph of the cardioid r
1 sin 0.
sin0.
also
of
the
graph
that
Show
it
is
r
(b)
that
the
equation
it can be described by
(c) Show
(a) Sketch the
=
-
-1
=
12
and conclude that
2
2
2_L
it can be described by the equation
2
(x2 + y2 + y)2
9.
2
Sketch the graphs of the following equations.
(i)
P
-
r
1
=
-
( sin 0.
(ii) r 1-2sin0.
(iii) r 2 + cos 0.
(a) Sketch the graph
=
=
10.
of the lemniscate
r2
(-a,
(a, 0)
0)
(b) Find an equation
it is
(c) Show that
a2.
did2
FIGURE
8
=
2a2 cos 20.
for its cartesian
coordinates.
the collection of all points P in Figure 8 satisfying
=
the shape of the curves formed by the set of all P
b, when b > a2 and when b < a2.
Make a guess about
(d) SatiSfying
did2
=
CHAPTER
LIMITS
The concept of a limit is surely the most important, and probably the most difficult
one in all of calculus. The goal of this chapter is the definition of limits, but we
are, once more, going to begin with a provisional definition; what we shall define
is not the word
but the notion of a function approaching a limit.
"limit"
PROVISIONAL DEFINITION
The function f approaches the limit l near a, if we can make
like to l by requiring that x be sufficiently close to, but unequal
f(x) as close as we
to, a.
Of the six functions graphed in Figure 1, only the first three approach I at a.
Notice that although g(a) is not defined, and h(a) is defined
wrong way," it
is still true that g and h approach I near a. This is because we explicitly ruled
the value of the function
out, in our definition, the necessity of ever considering
at a-it is only necessary that f(x) should be close to l for x close to a, but unequal
to a. We are simply not interested in the value of f(a), or even in the question of
whether
f (a) is defined.
"the
l-
t-
i-
/
FIGURE
i
1
One convenient way of picturing the assertion that f approaches l 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.
90
5. Limits 91
Now consider a function f whose drawing looks like Figure 3. Suppose we ask
that f(x) be close to I, say within the open interval B which has been drawn
in Figure 3. This can be guaranteed if we consider only the numbers x in the
interval A of Figure 3. (In this diagram we have chosen the largest interval which
(a) f(x)
-2
=
will work; any smaller
c
-1
1
-2
-1
2
0
(b) f(x)
FIGURE
2
2
-
x.
interval containing a could have been 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 f, but in
this case the interval B must be drawn on the vertical axis, and the set A on the
horizontal axis. The fact that f (x) is in B when x is in A means that the 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 large.
In order to apply our definition to a particular function, let us consider f(x)
x sin 1/x (Figure 6). Despite the erratic behavior of this function near 0 it is clear,
at least intuitively, that f approaches 0 near 0, and it is certainly to be hoped
that our definition will allow us to reach the same conclusion. In the case we are
considering,
both a and I of the defmition are 0, so we must ask if we can get
sin
f (x) x 1/x as close to O as desired if we require that x be sufficiently close
of0. This
to 0, but ¢ 0. To be specific, suppose we wish to get x sin 1/x within
=
=
FIGURE
3
means
we want
-
or, more succinctly,
|x sin 1/x|
I
sin
FIGURE
--
1
10
<
1
-
x
.
<
1
.
<
x sm
-
x
<
--,
1
10
Now this is easy. Since
1,
for all x ¢ 0,
4
(a)
FIGURE
5
(b)
92 Fdundations
we
,
have
x sin
/
r
.
f(x)=xsm-
,'
',
1
2
1
x
for all x ¢ 0.
|x|,
5
-
(
and x ¢ 0, then |x sin 1/x| <
This means that if |x| <
in other words,
of 0 provided that x is within
of 0, but ¢ 0. There is
x sin 1/x is within
nothing special about the number
; it isjust as easy to guarantee that |f (x)-0| <
require that |x| <
, but x ¢ 0. In fact, if we take any positive
number e we can make |f (x)
0| < e simply by requiring that |x! < e, and
x ¢ 0.
x2 sin 1/x (Figure 7) it
For the function f(x)
seems even clearer that f approaches 0 near 0. If, for example, we want
i
þ;
-simply
'
s
'
'
-
FIGURE 6
=
x2
1
.
sm
then we certainly need only require
and consequently
that |x2| <
<
-
-,
x
that
|x|
1
10
and x
<
¢ 0,
since this
implies
g
x2sin- 1
1
100
<\x2|<-<-.
¯¯
x
1
10
/
f(x)
/
FIGURE
x" sin
=
-
7
'
,
f( )
=
v'
=
(We could do even better, and allow |x| < 1/K and x ¢ 0, but there is no
particular virtue in being as economical as possible.) In general, if e > 0, to
ensure that
1
x2
sin
FIGURE
8
We
need only require
-
<
e,
that
|x| <
e
and
x
¢ 0,
5. Limits 93
f(x)
=
sin
-
A
FIGURE
+
9
provided that e 5 1. If we are given an s which is greater than 1 (itmight be, even
thought it is "small"
s's which are of interest), then it does not sufRce to require
that |x| < e, but it certainly sufEces to require that lx| < 1 and x
¢ 0.
sin 1/x (Figure 8). In
As a third example, consider the function f (x)
1
=
f(x)
=
sin
order
to make
|Ssin
1/x|
<
s we can require
e2
|x| <
(thealgebra
and
that
x
M
¢0
is left to you).
Finally, let us consider the function f (x) sin 1/x (Figure 9). For this function
it is falsethat f approaches O near 0. This amounts to saying that it is not true
for every number e > 0 that we can get |f (x) 0 < e by choosing x sufficiently
small, and ¢ 0. To show this we simply have to find
onee > 0 for which the
condition
<
0|
cannot
be guaranteed, no matter how small we require
e
|f (x)
will do: it is impossible to
fx| to be. In fact, e
ensure that |f(x)| <
no
matter
how small we require [x| to be; for if A is any interval containing 0, there
is some number x = 1/(90+360n) 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 f does not approach
any number near 0. To show this we must again find, for any particular number l,
some number e > 0 so that |f (x) li < e is not true, no matter how small x is
required to be. The choice e =
works for any number l; that is, no niatter how
small we require
x| to be, we cannot ensure that | (x)
l| <
The reason
f
is, that for any interval A containing 0 there is some xi
1/(90 + 360n) in this
interval, so that
f (xi) 1,
=
-
-
FIGURE
10
=
{
¼
--
(
-
=
=
and also some x2
=
1/(270 + 360m) in this interval, so that
-1.
f (x2)
=
J.
94 Foundations
-1
(
)
But the interval from l
cannot contain both
to I +
length is only 1; so we cannot have
-
|1
•
*
.
I
.
x, x rational
_
¯
o, x irrational
.
.
-
l|
<
(
and also
l|
-
<
1, since its total
¼,
what l is.
no matter
The phenomenon exhibited by
If we consider the function
f(x)
•
.
f (x)
,
,
sin
=
1/x near 0 can occur in many ways.
0, x irrational
1, x rational,
=
then, no matter what a is, f does not approach any number i near a. In fact, we
cannot make f (x) l| <
no matter how close we bring x to a, because in any
interval around a there are numbers x with f(x)
0, and also numbers x with
need
would
and
also
that
|0 l| <
|1 l| <
we
f (x) 1, so
An amusing variation on this behavior is presented by the function shown in
i
-
FIGURE
|-1
and
Il
=
=
¼
-
¼.
-
Figure 11:
f (x)
=
x,
x rational
0,
x
irrational.
sin 1/x; it apThe behavior of this function is "opposite"
to that of g(x)
proaches 0 at 0, but does not approach any number at a, if a ¢ 0. By now you
should have no difficulty convincing yourself that this is true.
As a contrast to the functions considered so far, which have been quite pathological, we will now examine some of the simplest functions.
If f (x) c, then f approaches c near a, for every number a. In fact, to ensure
that |f (x) c| < e one does not need to restrict x to be near a at all; the condition
is automatically satisfied (Figure 12).
As a slight variation, let f be the function shown in Figure 13:
=
----------'-i-!
----f(x)
-
c-----¯^¯¯¯¯
-¯¯¯¯¯¯
-¯
~^¯
-T-;
=
FIGURE
12
-
-1,
f (x)
=
1,
x
x
<
>
0
0.
If a > 0, then f approaches
1 near a: indeed, to ensure that
certainly suffices to require that |x al < a, since this implies
|f (x)
-
1|
<
e it
-
1
i
f(x)
1, x
=
>
o
-a<x-a
0<x
or
b
a
-1,
/W
-
«<
oi
-i
-1
1. Similarly, if b < 0, then f approaches
near b: to ensure
Finally, as you may
<
that |f(x) (-1)| < e it suffices to require that x
easily check, f does not approach any number near 0.
The function f (x) = x is easily dealt with. Clearly f approaches a near a: to
so that
f (x)
=
-b|
-
-a|
FIGURE
13
-b.
-a|
<
e.
s we just have to require that |x
requires
x2
work.
show
that
The function f (x)
little
To
more
a
near a, we must decide how to ensure that
ensure that
|f(x)
<
=
a2
X2
__
2
f
approaches
5. Limits 95
Factoring looks like the most promising pmcedure:
[x
al
-
-
|x+al
<
we want
e.
Obviously the factor |x + al is the one that will cause trouble. On the other
hand, there is no need to make |x + a l 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 |x a| < e/1,000,000.
Therefore, to begin with, let us require that [x
< 1
(anypositive number other
than 1 would do just as well); presumably this will ensure that x is not too large,
and consequently that
lx + a l is not too large. As a matter of fact, Problem 1-12
-
-al
shows that
jx -Jals|x-al<1,
SO
lx[ < l
+ a|,
and consequently
+1.
|x+al5lx[+|a|<2|a
-al
Now we need only the additional
requirement
that
|x
<
s/(2|a|+
1). In other
words,
if (x a|
--
<
min
1,
2|a + 1
then
,
lx2
-
a2|
<
s.
+ 1)) will just be e/(2|al + 1) for small e.
Precisely the same sort of trick will show that if f (x) x3, then
a3 near a. In fact,
Naturally, min(1, e/(2|a
=
if |x
-
a|
<
min
1,
'
(1 + |a )2 + ja (1 + |a|) + |a|2
,
then
f
|x3
-
approaches
a3|
<
e.
The proof of this assertion will show where the weird denominator comes from:
< 1, then |x| < |a|+ 1, and consequently
If x
--al
X2+ax‡a2
2‡
<
aj·|x|+|a2
(1 + |a|)2 + ja (1 + |a|) +
a|2.
Therefore
|x3-a3|= |x--al |x2+ax+a2
-
<
-
(1 + \aj)2+ \a\(1+ laj)+ \a\2
[(1+ |a|)2 + |a|(1 + la|) + [a|2]
The time has now come to point out that of the many demonstrations about
limits 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 limit
This definition is simply not sufficiently precise to be
is even more vulnerable.
used in proofs. It is hardly clear how one
f (x) close to I (whatever
means) by
close "suffix to be sufficiently close to a (however
"makes"
"close"
"requiring"
96 Foundations
ciently"
is supposed to be). Despite the criticisms of our definition you may
feel
hope you do) that our arguments were nevertheless quite convincorder
ing. In
to present any sort of argument at all, we have been practically forced
real
definition. It is possible to arrive at this definition in several steps,
to invent the
each one clarifying some obscure phrase which still remains. Let us begin, once
again, with the provisional definition:
close
(I certainly
The function f approaches the limit l near a, if we can make f(x) as close
as we like to I by requiring that x be sufficiently close to, but unequal to, a.
The very first change which we made in this defmition was to note that making
f (x) close to l meant making |f (x) l| small, and similarly for x and a:
-
The function f approaches the limit I near a, if we can make |f (x) l| as
small as we like by requiring that |x a| be sufficiently small, and x ¢ a.
-
-
"as
small as
The second, more crucial, change was to note that making [f (x) l|
we like" means making |f (x) l| < e for any e > 0 that happens to be given us:
-
-
The function
can make
¢
x
approaches
f
If (x)
-
l|
the limit l near a, if for every number e > 0 we
a| be sufficiently small, and
e by requiring that x
<
-
a.
There is a common pattern to all the demonstrations about limits which we have
given. For each number e > 0 we found some other positive number, 8 say, with
the property that if x ¢ a and |x a| < ô, then f(x) l| < e. For the function
0), the number 8 was just the number e;
0, I
f (x) x sin 1/x (witha
x2 it
l/x, it was e2; for f(x)
for f(x)
was the minimum of 1 and
e/(2|a l + 1). In general, it may not be at all clear how to find the number ô,
< 8 which expresses how small
given E, but it is the condition |x
small must be:
-
=
-
=
=
-
Ñsin
=
-a|
"sufficiently"
The function f approaches the limit l near a, if for every e
& > 0 such that, for all x, if |x a| < 8 and x ¢ a, then |f
-
0 there is some
(x) l| < e.
>
-
This is practically the defmition we will adopt. We will make only one trivial
change, noting that
a| < ô and x ¢ a" can just as well be expressed "O <
"|x
-
|x-a|<ô."
DEFINITION
The function f approaches
the lirnit I near a means: for every e > 0 there
such
that,
all
>
8
0
for
is some
x, if 0 < |x a| < 8, then |f(x) l| < e.
-
-
5. Limits 97
This definition is so important (everything
we do from now on depends on it) that
proceeding any further without knowing it is hopeless. If necessary memorize it,
like a poem! That, at least, is better than stating it incorrectly; if you do this you
are doomed to give incorrect proofs. A good exercise in giving correct proofs is to
review every fact already demonstrated about functions approaching
limits, giving
real proofs of each. This requires writing down the correct definition of what you
the algebraic work has been done already.
are proving, but not much morwall
When proving that f does not approach l at a, be sure to negate the defmition
correctly:
If it is not true that
for every e
then
|f (x)
0 there is some 8
>
l
-
<
0 such that, for all x, if 0
>
<
[x al
-
<
8,
e,
then
0 such that for every & > 0 there is some x which satisfies
8 but not |f (x) l| < e.
there is some e
0
|x
<
-
a|
<
>
--
Thus, to show that the function f (x) sin 1/x does not approach 0 near 0, we
and note that for every ô > 0 there is some x with 0 < |x 0 < 8
consider e =
but not Isin 1/x 0 | < j-namely,
an x of the form 1/(90 + 360n), where n is
that
<
8.
1/(90
360n)
+
so large
As an illustration of the use of the definition of a function approaching a limit,
we have reserved the function shown in Figure 14, a standard example, but one
=
(
--
-
of the most complicated:
f(x)
0,
=
x
1/q,
x
irrational, O < x < 1
p/q in lowest terms, 0
=
<
x
1.
<
(Recall that p/q is in lowest terms if p and q are integers with no common
and q > 0.)
1
e
-
x irrational
f(x)
I-
e
1-
-
e
e
10,
-,
1
x
=
p
--
.
m lowest terms
e
I-eeee
Ip-eees
e
ee
II
FIGURE
14
I
|
I
i
I
il
e
i
factor
98 Foundations
0 at a. To prove
a, with 0 < a < 1, the function f approaches
natural
number
number
this, consider any
so large that 1/n
e > 0. Let n be a
e.
Notice that the only numbers x for which |f (x) 0| < s could be false are:
For any number
_<
-
-;
1 12
13
-;
-,
1234
-;
-,
-,
-,
-;
1
...;
-,
n-1
-,...,
n
(If a is rational, then a might be one of these numbers.) However many of these
numbers there may be, there are, at any rate, only finitely many. Therefore, of all
these numbers, one is closest to a; that is, |p/q a| is smallest for one p/q among
these numbers. (If a happens to be one of these numbers, then consider only the
values |p/q
al for p/q ¢ a.) This closest distance may be chosen as the 8. For
if 0 < |x a| < 8, then x is not one of
-
-
-
n-1
1
-,....
n
This completes the proof. Note that our
description of the 8 which works for a given e is completely adequate-there
is no
why
of
give
8
in
formula
for
terms
we must
reason
a
s.
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.
and therefore
THEOREM
l
e is true.
<
-
A function cannot approach two different limits near a. In other words, if
approaches
PROOF
|f (x) 0|
i near a, and
approaches
f
m near a, then I
=
f
m.
Since this is our first theorem about limits it will certainly be necessary to translate
the hypotheses according to the definition.
Since f approaches l near a, we know that for any e > 0 there is some number
1 > 0 such that, for all x,
if 0
<
|x
-
a
We also know, since f approaches
for all x,
l<
81, then
|f (x)
-
l|
m near a, that there
<
e.
is some â2 > 0 such that,
if0<|x-al<ô2,then|f(x)--m|<e.
We have had to use two numbers, 81 and 82, since there is no guarantee that the 8
which works in one definition will work in the other. But, in fact, it is now easy to
conclude that for any e > 0 there is some ô > 0 such that, for all x,
if0<|x-a|<8,then|f(x)-l|<eand
we simply
choose 8
=
min(8i,
82).
|f(x)-m|<e;
5. Limits 99
To complete the proof we just have to pick a particular e
>
0 for which the two
conditions
and
|f(x)-lj<s
if(x)-ml<s
cannot both hold, if i ¢ m. The proper choice is suggested by Figure 15. If
l ¢ m, so that |l m) > 0, we can choose |l ml/2 as our e. It follows that there
is a 8 > 0 such that, for all x,
--
--
if 0
length
FIGURE
15
t-m
lx
<
al
-
8, then
<
and
-
length
This implies that for 0
|l
-
ml
<
|x
a|
-
<
If (x) li
<
m|
<
-
|f (x)
-
.
2
8 we have
_<
|l f (x) + f (x)
=
-
-
m|
ll f (x)| + |f (x)
-
II
<
mi
-
2
+
il
-
m!
mi
--
2
=|l--m,
a contradiction.
|
The number I which
approaches near a is denoted by lim f (x) (read:
the limit
of f(x) as x approaches a). This definition is possible only because of Theorem 1,
which ensures that lim f(x) never has to stand for two different numbers.
The
f
equation
lim f (x)
has exactly the same meaning
as the
f
=
l
phrase
approaches l near a.
The possibility still remains that f does not approach l near a, for any l, so that
lim f (x) l is false for every number l. This is usually expressed by saying that
=
"lim
X->G
f (x) does not
exist."
Notice that our new notation introduces an extra, utterly irrelevant letter x,
which could be replaced by t, y, or any other letter which does not alæady
appear-the
symbols
lim
x-va
f(x),
lim
t
a
f(t),
lim
y-wa
f(y),
all
denote precisely the same number, which depends on f and a, and has nothing
to do with x, t, or y (theseletters, in fact, do not denote anything at all). A more
logical symbol would be something like lim f, but this notation, despite its brevity,
is so infuriatingly rigid that almost no one has seriously tried to use it. The notation
Iim f (x) is much more useful because a function f often has no simple name, even
100
Foundations
though it might be possible to express
the short symbol
simple
f (x) by a
formula involving x. Thus,
lim (x2 -I- sin x)
could
be paraphrased
only
by the awkward expression
lim f, where f(x)
Another advantage
of the standard
=x2+sinx.
symbolism
is illustrated by the expressions
limx+t3,
lim x + t3
The first means the number
which
f (x)
the second means the number
=
f
t3,
x+
which
f (t)
=
approaches
f
near a when
for all x;
approaches near a when
x + t*,
for all t.
You should have little difficulty (especially
if you consult Theorem 2) proving that
limx+t3=a+t3,
x->a
limx+t3=x+a3
These examples illustrate the main advantage of our notation, which is its flexibility. In fact, the notation lim f (x) is so flexible that there is some danger of
forgetting 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 prove the
equality
of the expressions
lim
x-va
f (x)
and
lim f (a + h).
h->O
An important part of this chapter is the proof of a theorem which will make it
easy to find many limits. The proof depends upon certain properties ofinequalities
and absolute values, hardly surprising when one considers the definition of limit.
Although these facts have already been stated in Problems 1-20, 1-21, and 1-22,
because of their importance they will be presented once again, in the form of a
lemma (alemma is an auxiliary theorem, a result that justifiesits existence only
by virtue of its prominent role in the proof of another theorem). The lemma says,
roughly, that if x is close to xo, and y is close to yo, then x + y will be close to
xo + yo, and xy will be close to xoyo, and 1/y will be close to 1/yo. This intuitive
than the precise estimates of the lemma,
statement is much easier to remember
of Theorem 2 first, in order to see just
read
and it is not unreasonable
the
proof
to
estimates
used.
how these
are
5. Limits 101
LEMMA
(Î)If
jx
xo|
-
and
<
|y
yo|
--
<
,
then
(x +
y)
yo) |
(xo+
-
<
e.
(2)If
|x
-
xo|
<
S
(
min
1,
then
2(lyol + 1)
|xy
(3)If yo ¢
0
xoyol
-
<
yo|
-
.
<
mm
¢ 0 and
-,
(1)
|(x +
y)
10)|
(10
-
-
€
<
2(ixol + 1)'
e.
1
1
y
yo
e|yo|2
2
2
xo|
<
,
<e.
I(X -X0)
=
f
(2)Since |x
yo|
-
|yo
---
PROOF
Iy
and
ly
then y
and
+ (7
70)
-
e
e
lx--xcl+Iy-yol<-+-=e.
2 2
l we have
x|
xo| 5
-
|x
xo|
-
1,
<
so that
lxl < 1+ |xo|.
Thus
xoyal
lxy
-
|x(y yo) + yo(x xo) |
5|xl
ly-yol+lyol·lx-xol
=
-
-
8
<
8
+
(1+ |xoD
2(|xo + 1) lyol 2( yo| + 1)
-
e
e
2
2
<-+-=e.
-
(3)We have
yo|
so
|y| > |yo||2. In particular,
y
|y| 5 |y
-
¢
yo|
-
<
10
--,
1
2
lyI
lyol
Thus
1
1
y
yo
|yo y
|y|· lyol
-
2
0, and
2
1
lyol lyo|
s|yo[2
2
102 Foundations
THEOREM
2
If lim f (x)
=
l and lim g (x)
m, then
=
(1) lim( f +
(2) lim( f
·
g)(x)
l + m;
=
g)(x)
l m.
=
-
Moreover, if m ¢ 0, then
lim
(3) x-va
PROOF
1
(1
(x)
-
g
The hypothesis means that for every e
=
-.
m
0 there are Si, 82
>
0 such that, for
>
all x,
if0<|x-a|<ô1,then|f(x)-l|<e,
-al
and
if 0
-m|
lx
<
ô2, then
<
|g(x)
<
after all, e/2 is also a positive number)
This means (since,
such that, for all x,
and
Now let ô
0 < |x a|
=
<
-
if 0
<
|x
if 0
<
|x
-
-
are true.
<
81, then
|f (x) l|
a|
<
82, then
|g(x)
62). If 0 < |x
82 are both true, so both
&
|f(x)-l|<But by part
(1)of the
-
and
2
a|
<
and
Again let &
=
<
|x
if 0
<
\x
min(81,
-
-
<
ô, then 0
<
a|
-
81 and
<
2
-
a\
<
ô2, then \g(x) m\
-
l|
-
|x
<
--
part
min
<
<
(2)of
1,
2(|m[ + 1)
and
2(|m|+ 1)
,
2(\l\ + 1)
€
|g(x)-m|<
·
.
if0<|x-a|<ô,then|g(x)-m|<min
--,
.
2(|l + 1)
(2).
|m| s |m|
.
2
2
<
s.
the lemma. If
a l < 8, then
S
-
|x
8
l m| < e, and this proves
So, by the lemma, |(f g)(x)
Finally, if e > 0 there is a 8 > 0 such that, for all x,
-
.
lemma this implies that |(f + g) (x) (l + m)|
ô1, then |f (x)
(
-
|g(x)-m|<-
<
1,
|f(x)-l|<min
,
m|
al
ô2). If 0
<
-
This proves (1).
To prove (2)we proceed similarly, after consulting
e > 0 there are Bi, ô2 > 0 such that, for all x,
if 0
that there are Bi, 82 > 0
a|
min(ål,
e.
5. Limits 103
But according to part
(3)of the
lemma this means, first, that g(x) ¢ 0, so (1/g)(x)
makes sense, and second that
(1
(x)--
-
g
This proves
1
<e.
m
(3).|
Using Theorem 2 we can prove, trivially, such facts as
a3+7aß
x3+7x6
.
hm x2
x-a
+ 1
without
begin
=
a2
,
+ 1
going through the laborious process of finding a ô, given an e. We must
with
lim 7
=
7,
lim 1
=
1,
lim x
=
x->a
X-wa
a,
but these are easy to prove directly. If we want to find the 8, 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 o such that, for all x,
if 0
<
|x
-
a|
8, then
<
lx2+
x
(a2 +
-
a)|
<
e.
Consulting the proof of Theorem 2(1), we see that we must first find
such that, for all x,
and
if 0
<
|x
if 0
<
|x
-
-
|x2
a
<
ôi, then
a
<
02, then x
Since we have already given proofs that lim x2
to do this:
-
--
a2|
a|
<
<
8
1
min
=
2
(
1,
2
2|a l + 1
Thus we can take
&
=
min(Bi, ô2)
=
min
min
l'
2|a + 1
and 82 > 0
2
.
a2 and lim x
·-
i
'
=
a, we know how
104 Foundations
If a ¢ 0, the same method
if 0
<
can be used to find a 8
|x
a|
-
1
8, then
<
-
0 such that, for all x,
>
1
<
-
x
e.
a
The proof of Theorem 2(3) shows that the second condition
a 8 > 0 such that, for all x,
will
follow if we find
,ela|4
if0<|x-al<8,then|x2-a
|<min
Thus we can take
.
mm
ô
min
=
1,
--,
|a|2
e|a|4
--
2
2
2|a| + 1
Naturally, these complicated expressions for 8 can be simplified considerably, after
they have been derived.
One technical detail in the proof of Theorem 2 deserves some discussion. In
order for lim f (x) to be defined it is, as we know, not necessary for f to be defined
at a, nor is it necessary for f to be defined at all points x ¢ a. However, there
must be some ô > 0 such that f (x) is defmed for x satisfying 0 < |x a l < 8;
otherwise the clause
-
(a)
"if
would
0
<
|x
-
a|
8, then
<
|f (x)
l|
-
e"
<
make
no sense at all, since the symbol f (x) would make no sense for
some x's. If f and g are two functions for which the definition makes sense,
it is easy to see that the same is true for f + g and f 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
established in the proof of Theorem 2(3).
i
-
(b)
FIGURE
l6
There are times when we would like to speak of the limit which f 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)
l
=
x->a+
or
lim f (x)
l.
=
x‡a
(The symbols on the left are read: the limit of f (x) as x approaches a from above.)
These
from above" are obviously closely related to ordinary limits, and the
definition is very similar: lim f (x) l means that for every e > 0 there is a 8 > 0
"limits
=
I-
such that,
for all x,
if 0
<
x
-
a
<
ô, then
|f (x)
-
l|
<
e.
"0
< x
(The condition
a < 8" is equivalent to "O < |x a| < ô and x
"Limits from below" (Figure 17) are defined similarly: lim f(x)
X->G
-
FIGURE
17
--
>
=
a.")
l
(or
5. Limits 105
lim f (x)
x†a
=
l) means
that for every e
>
0 there is a ô
>
0 such that, for
all x,
if 0
<
-
a
x
<
8, then
|f(x)
l
--
<
e.
It is quite possible to consider limits from above and below even if f is defined
for numbers both greater and less than a. Thus, for the function f of Figure 13,
we have
-1.
lim f (x)
1 and
=
lim f (x)
=
It is an easy exercise (Problem 29) to show that lim f (x) exists if and only 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 1/x is
close to 0. This assertion is usually written
lim sin 1/x
0.
=
x-voo
FIGURE
18
"the
limit of f (x) as x approaches oo," or "as
The symbol lim f (x) is read
x becomes infinite," and a limit of the form lim f (x) is often called a limit at
X-¥OO
infinity. Figure 18 illustrates a general situation where lim f (x) l. Formally,
=
lim
X->00
f (x)
=
l means that for every e
if x
>
>
0 there is a number
N, then
|f(x)
-
l|
<
N such that, for all x,
e.
The analogy with the definition of ordinary limits should be clear: whereas the
condition "O < |x al < 8" expresses the fact that x is close to a, the condition
> N" expresses the fact that x is large.
We have spent so little time on limits from above and below, and at infinity,
because the general philosophy behind the definitionsshould be clear if you understand the definition of ordinary limits (which
are by far the most important).
Many exercises on these definitionsare provided in the Problems, which also contain several other types of limits which are occasionally useful.
-
"x
106 Foundations
PROBLEMS
1.
Find the following limits. (These limits all follow, after some algebraic mafrom the various parts of Theorem 2; be sure you know which
used
in each case, but don't bother listing them.)
ones are
nipulations,
x2_)
lim
(i)
x->l
(ii)
lim
.
x+1
xa-8
x->2
lim
(iii) xs3
.
(iv)
lun
(v)
lim
.
x
x"
2
.
-
y"
-
.
x"
y"
-
.
y
-
x
Ja+h-
.
h
Find the following limits.
Inn
(i)
x->1
(ii)
lim
240
lim
(iii) x->0
3.
2
-
y->x
hm
(vi) h->o
2.
.
x
x3_g
.
1
x
-
l-x2
1-
.
x
1-J1-x2
.
X
In each of the following cases, find a
satisfying 0 < |x a| < ô.
such that
|f (x)
-
I
<
e for all x
-
(i) f (x)
=
(ii) f (x)
=
(iii) f (x)
=
(iv) f (x)
=
x4. ;
-;
g4
_
1
a
x
x
1, l
=
1
4
_
.
1.
=
_
y y
_
g
x
x
;a
1 + sin2
=
0, l
=
0.
(v) f(x)=S;a=0,l=0.
(vi) f (x) S; a 1, l 1.
=
4.
*5.
=
=
For each of the fimetions in Problem 4-17, decide for which numbers
limit lim f (x) exists.
a the
the same for each of the functions in Problem 4-19.
Same
problem, if we use infinite decimals ending in a string of O's
(b)
instead of those ending in a string of 9's.
(a) Do
5. Limits 107
6.
Suppose the functions f and g have the following property: for all e
0
>
and all x,
if 0
<
if 0
<
For each e
<
sin
[x 2|
<
e2, then
--
(ii)
(iii) if 0 < |x
if0<
2|
-
<
8, then
<
8, then
x-2|<ô,then
<
-
<
e,
s.
and
limf(x)
f(x)g(x)
1
---
1
4
1
2
-
-
g(x)
8|
-
<
<
s.
e.
f (x)
<s.
g (x)
function f for which
Give an example of a
If f (x) l| < e when 0
0< |x-a|<ô/2.
(a) If
--
-----
-
8.
|g(x) 4|
|f (x) 2|
<b,then|f(x)+g(x)-6|<e.
-
(iv)
+ e, then
0 fmd a 8 > 0 such that, for all x,
>
if0< x-2
if 0 < x 2|
(i)
7.
2|
-
x
<
limg(x)
x
a|
--
the following assertion is false:
8, then |f (x) ll < e/2 when
<
-
do not exist, can
lim[f(x)+g(x)]
or
lim f(x)g(x)exist?
lim f (x) exists and X-¥G
lim [f (x) + g(x)] exists, must X-Þg
lim g(x) exist?
(c) If lim f(x) exists and lim g(x) does not exist, can lim[ f(x)+g(x)]exist?
(b) If
(d) If
X-¥Ø
lim f (x) exists and lim f (x)g(x) exists, does it follow that lim g(x)
x-a
x->a
x->a
exists?
9.
Prove that lim f (x)
x->a
lim
=
h->0
f (a +
h). (This is mainly
in under-
an exercise
standing what the terms mean.)
10.
(a) Prove that
l if and only if lim [f (x) l] 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 the same way.)
(b) Prove that lim f (x) lim f (x a).
lim
f (x)
=
=
x->0
(c) Prove that
(d) Give an
11.
=
-
-
x-+a
lim f(x)
=
x->0
example
where
lim f(x31
x->0
lim
f(x2) exists,
Suppose there is a & > 0 such that f(x)
but lim
f(x) does not.
g(x) when 0
-a|
8. Prove
lim
depends
f(x)
f(x) x->a
x-a
x-wa
values of f (x) for x near a-this
fact is often expressed by saying that limits
property." (It will clearly help to use ô', or some other letter,
are a
of
8, in the definition of limits.)
instead
that lim
=
limg(x).
In other
=
<
words,
|x
<
only on the
"local
12.
5 g(x) for all x. Prove that lim
provided that these limits exist.
(a) Suppose
that
f(x)
f(x)
5 limg(x),
108
Foundations
(b) How can the hypotheses by weakened?
(c) If f(x) < g(x) for all x, does it necessarily
follow that lim f(x)
<
lim g(x)?
X-¥ß
13.
lim f(x)
lim h(x). Prove that
Suppose that f(x) 5 g(x) 5 h(x) and that X-kB
X-¥ß
and
that lim g(x)
lim g(x) exists,
lim f(x) lim h(x). (Draw a picture!)
=
=
*14.
=
-/
(a) Prove that
if lim f(x)/x = l and b 0, then lim f(bx)/x
bl. Hint:
x->0
x->0
Write f (bx)/x b [f (bx)/bx].
(b) What happens if b 0?
(c) Part (a)enables us to find lim(sin2x)/x in terms of lim(sinx)/x. Find
=
=
=
x->0
x->0
this limit in another way.
15.
Evaluate the following limits in terms of the number
lim
(i)
sin
lim
(iii) x->0
2x
X
sinax
.
.
sin2 2x
.
X
sin2
h
lim
(iv) x-vo
lim
.
x
Î
COS X
-
.
x->0
.
hm
(vi) x-so
lim
(vii)x-vo
x
tan2 x + 2x
x + x2
.
1
-
sin(x2
lim
(ix) x->i
.
lun
.
x sin x
cosx
sin(x + h)
lim
(viii)h->0
h
16.
lim(sinx)/x.
x->0
x-vo smbx
lim
(v)
=
.
x->0
(ii)
a
sin x
-
.
O
-
.
x
-
1
x2(3+sinx)
(x)
xso
(xi)
lim (x2
2-1
(x + sin x)2
-
(a) Prove that
(b) Prove that
.
3
1
1)3sin
x
1
-
l, then lim |f|(x) = |l|.
x-va
l and lim g(x)
if lim f(x)
m, then lim max( f, g)(x)
max(l, m) and similarly for min.
if lim
x->a
f(x)
=
=
=
5. Limits 109
17.
(a) Prove that
lim 1/x does not exist, i.e., show that lim 1/x
every number I.
(b) Prove that lim 1/(x
l is false for
1) does not exist.
-
x->1
18.
=
& > 0 and a number M
such that |f(x)| < M if 0 < |x al < 8. (What does this mean pictorially?)
Hint: Why does it sufBceto prove that l--1 < f (x) < l+1 for 0 < |x-a| < ô?
Prove that if lim f(x)
x-a
=
l, then there is a number
-
19.
*20.
21.
0 for irrational x and
Prove that if f(x)
then lim f(x) does not exist for any a.
=
Prove that if f(x)
x for rational
exist
if a ¢ 0.
lim f (x) does not
=
lim g(x)
(a) Prove that if xw0
(b) Generalize this fact as
=
x, and
f(x)
f(x)
x->0
1 for rational x,
for irrational x, then
-x
=
0, then lim g(x) sin 1/x
=
=
0.
0 and |h(x)| < M for all x,
then lim g(x)h(x)
0. (Naturally it is unnecessary to do part (a)if you
x->0
succeed in doing part (b);actually the statement of part (b)may make it
follows: If lim g(x)
=
=
easier than
22.
(a)-that'sone
of the values of generalization.)
Consider a function f with the following property: if g is any function for
which limg(x) does not exist, then lim[f(x) + g(x)] also does not exist.
x->0
x->0
Prove that this happens if and only if lim
f (x) does exist. Hint: This is
lim f (x) does not exist leads to an
x->0
actually
very easy: the assumption
immediate contradiction
**23.
that
if you consider the right g.
This problem is the analogue of Problem 22 when f + g is replaced by f g.
In this case the situation is considerably more complex, and the analysis
requires several steps
(thosein search of an especially challenging problem
-
can attempt an independent solution).
(a) Suppose that
lim
x->O
f(x) exists
and is ¢
f(x)g(x) also does not
same result if lim |f (x)|
x->0
exist, then lim
lim g(x) does not
0. Prove that if x->0
exist.
x->0
(b) Prove the
=
oo. (The precise definition
of
this
sort of
limit is given in Problem 37.)
that
Prove
if neither of these two conditions holds, then there is a function
(c)
such
that
lim g(x) does not exist, but lim f(x)g(x) does exist.
g
x->O
x->O
Hint: Consider separately the following two cases: (1)for some e > 0
we have |f (x)| > e for all sufficiently small x. (2)For every e > 0, there
with
are arbitrarily small x
|f (x) < e. In the second case, begin by
with
choosing points x,
|x,\ < 1/n and |f(x.)| < 1/n.
*24.
Suppose that An is, for each natural number n, some fmue
set of numbers in
members
and
and
that
A,
Am have no
in common if m ¢ n. Define
[0, 1],
110
Foundations
f
as
follows:
f(x)
Prove that lim f (x)
25.
=
x in A.
in A. for any n.
x not
0 for all a in
[0,1].
Explain why the following definitions of lim f (x) l are all correct:
x-+a
For every 8 > 0 there is an e > 0 such that, for all x,
=
(i) if 0 < |x
(ii) if 0 |x
(iii) if 0 < |x
<
(iv)
*26.
1/n,
0,
=
if 0
<
|x
-
-
-
-
a|
<
a|
<
a|
<
a|
<
e, then |f (x) l|
e, then |f (x) l|
e, then |f (x) l|
s/10, then |f(x)
<
-
<
-
8.
8.
5ô.
<
-
-
l|
<
8.
Give examples to show that the following definitions of lim f(x)
x->a
=
I are not
correct.
(a) For all & >
|f (x) l | <
(b) For all e >
0 there is an e
|x
27.
--
a
0 such that if 0
|x
<
0 there is a 8
0 such that if |f (x)
>
-
li
a|
-
<
8, then
<
e, then 0
<
8.
For each of the functions in Problem 4-17 indicate for which
one-sided
*28.
|<
>
s.
-
limits lim f (x) and lim
numbers
a the
f (x) exist.
(a) Do the
same for each of the functions in Problem 4-19.
(b) Also consider what happens if decimals ending in O's are used instead of
decimals ending in 9's.
29.
Prove that lim
30.
Prove that
f (x) exists if
lim f (x)
(i) x->0+
lim f (|x|)
(ii) x->0
lim f (x2)
(iii) x->O
lim
f (x)
=
lim
f (x).
lim f (-x).
lim f (x).
=
xa0=
x->0+
lim
=
x->0+
f (x).
(These equations, and others like them, are open to several interpretations.
They might mean only that the two limits are equal if they both exist; or that
if a certain one of the limits exists, the other also exists and is equal to it; or
that if either limit exists, then the other exists and is equal to it. Decide for
yourself which interpretations are suitable.)
*31.
Suppose that lim
(Draw a picture to illustrate this assertion.) Prove that there is some 8 > 0 such that
f(x) < f(y) whenever
x < a < y and |x al < ô and |y a| < 8. Is the converse true?
f(x)
X-¥G¯
-
<
lim
X-+G
f(x).
-
5. Limits 111
*32.
33.
-+ao)/(bmx"'+-·
-+bo)
exists if and only if m > n.
Prove that lim (anx"+
x-voo
=
What is the limit when m
n? When m > n? Hint: the one easy limit is
lim 1/x* = 0; do some algebra so that this is the only information you need.
X-+0C
-
-
Find the following limits.
(i)
(ii)
x + sin3
lim
.
5x + 6
x-oo
lim
x->oo
x sin x
.
x2 +
5
lim Vx2
+ x
(iii) xooo
x2
--
x.
1+k2x)
lim
(iv) x-+oo
(x+sinx)2
.
34.
Prove that lim f (1/x)
35.
Find the following limits in terms of the number
36.
lim
œ
=
lim(sin x)/x.
sinx
(i)
x-oo
(ii)
lim x sin
x-Foo
Define
lim f (x).
=
-.
x
"
1
--.
X
lim f (x) = l."
x->-oo
(a) Find lim (anx"+ + a°)/(bmx" +
(b) Prove that lim f (x) lim f (--x).
lim f (1/x)
lim f (x).
(c) Prove that x-vox->-oo
·
-
-
-
-
-
bd
=
X-+0C
X-¥-DO
=
37.
We define lim f (x) = oo to mean that for all N there is a 8 > 0 such that,
for all x, if 0 < x a < 8, then f (x) > N. (Draw an appropriate picture!)
-
(a) Show that
(b) Prove that
lim 1/(x
3)2
--
x-+3
if
f(x)
>
e
>
lim
X->G
38.
-·
o.
0 for all x, and lim g(x)
f(x)/|g(x)
=
=
0, then
oo.
lim f (x) oo. (Or at least
oo, and x->a
convince yourself that you could write down the definitions if you had
the energy. How many other such symbols can you define?)
lim 1/x oo.
(b) Prove that x->0+
lim f (1/x) oo.
(c) Prove that lim f (x) oo if and only if X->oo
lim f (x)
(a) Define x-a+
=
lim
oo, x->a-
f (x)
=
=
=
=
x->0+
39.
Find the following limits, when they exist.
x3+4x-7
(i)
lim
x-voo
7x2
-
x +
1
=
112 Foundations
(ii)
(iii)
lim x(1+ sin2x).
X-¥OO
lim x sin x.
(iv) x(v)
(vi)
oc
;
_
,
41.
x2 ‡
lim
1400
.
x
2X
lim x( x + 2
lim
(vii)x-oo
40.
x sin
--
X
.
).
-
--.
x
(a) Find the perimeter of a regular n-gon inscribed in a circle of radius r;
use radian measure for any trigonometric functions involved. [Answer:
2rn sin(x|n).]
(b) What value does this perimeter approach as n becomes very large?
After sending the manuscript for the first edition of this book off to the printer,
a2 and lim x3
I thought of a much simpler way to prove that lim x2
--
a
-
a
/
.
a3, without
example,
Va=
FIGURE
19
-
e a
=
going through all the factoring tricks on page 95. Suppose, for
a2, where a > 0. Given
that we want to prove that lim x2
=
«2
of Ja2 + e
g
0, we simply let 8 be the minimum
a and a
a| < 8 implies that Ja2 e < x < Ja2+ e, so
(seeFigureX2 19); then |x |X2
a2] <
< a2 + E, Or
8
<
s. It is fortunate that these pages had
a
already been set, so that I couldn't make these changes, because this
is completely fallacious. Wherein lies the fallacy?
e
>
-
-
-
-
_
-
--
"proof"
FUNCTIONS
CONTINUOUS
CHAPTER
If
f
is an arbitrary function, it is not necessarily true that
lim
this4m
f (x) f (a).
=
fail to be true. For example, f might not
In fact, there are many ways
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 3,
FIGURE
even if
1
f
is defined at a and lim f(x) exists, the limit might not equal
FIGURE
(c)
(b)
(a)
f(a).
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.
Intuitively, a function f is
The term which has been adopted is
wild
oscillations. Although
continuous
contains
the
if
graph
no breaks, jumps,or
whether
usually
enable you to decide
this description will
a function is continuous
simply by looking at its graph (a skill well worth cultivating) it is easy to be fooled,
and the precise definition is very important.
"continuous."
DEFINITION
The function f is continnons
at a if
lim f (x)
x-a
/
FIGURE
a
3
=
f (a).
We will have no difficulty finding many examples of functions which are, or are
example involving limits provides an
not, continuous at some number a-every
example about continuity, and Chapter 5 certainly provides enough of these.
The 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 other hand, if
of
willing
these
extend
second
that is, if we wish to define
functions,
the
to
we are
=
=
113
114
Foundations
a new function G by
G(x)
x sin
a,
=
1/x,
¢0
x
0,
=
x
then the choice of a
G(0) can be made in such a way that G will be continuous
this
Sto
do
0 (Figure 4). This sort of
at
we can (iffact, we must) define G(0)
extension is not possible for
f; if we define
=
=
F(x)
sin
I
=
1/x,
then F will not be continuous at 0, no matter
(a)
¢0
x
x
a,
0,
=
what
lim f
a is, because x->0
(x) does
not exist.
The function
f (x)
is not continuous
=
x,
x rational
0,
x irrational
if a ¢ 0, since lim f(x) does not exist. However, lim f(x)
x->0
x-va
0 f (0), so f is continuous at precisely one point, 0.
x2 are continuous at all numThe functions f(x) c, g(x)
x, and h(x)
bers a, since
at a,
=
=
=
=
lim
f (x)
=
lim c
=
lim g(x)
=
lim h(x)
=
lim x
X-ra
FIGURE
=
c
=
f (a),
=
a
=
g(a),
x-va
4
lim x2
-
a2
=
h(a).
Finally, consider the function
f (x)
=
we showed
10,
1/q,
that lim
x
x
irrational
p/q in lowest terms.
=
In Chapter 5
f (x)
continuous
a is irrational, this fimction is
=
0 for all a. Since O f (a) only when
at a if a is irrational, but not if a is
=
rational.
It is even easier to give examples
THEOREM
1
If
f
of continuity
if we prove two simple theorems.
and g are continuous at a, then
(1) f
(2) f
+ g is continuous at a,
g is continuous at a.
Moreover, if g(a) ¢ 0, then
(3) 1/g is continuous
at a.
6. ContinuousFunctions 115
PROOF
Since f and g are continuous at a,
lim f (x)
f (a)
=
Xad
and
lim g (x)
=
X->G
g (a).
By Theorem 2(l) of Chapter 5 this implies that
lim(f +g)(x)=
which is just the assertion that
and (3)are left to you. I
f
+g)(a),
f(a)+g(a)= (f
+ g is continuous at a. The proofs of parts
Starting with the functions f(x) c and f(x) x, which are continuous
for every a, we can use Theorem 1 to conclude that a function
=
f (x)
=
b,x" + bn-1x"¯1 +
=
-
-
-
(2)
at a,
+ bo
c.xm+c._ixm-1+·-·+co
is continuous at every point in its domain. But it is harder to get much further
than that. When we discuss the sine function in detail it will be easy to prove that
sin is continuous at a for all a; let us assume this fact meanwhile.
A function like
sin2x+x2+x'sinx
f (x)
=
sin27
Sin
x + 4x2
X
proved continuous at every point in its domain. But we are still
sin(x2); we obviously
prove the continuity of a function like f(x)
need
composition
about
the
of
continuous
theorem
functions. Before stating this
a
of
continuity is worth noting. If
theorem, the following point about the definition
we translate the equation lim f (x)
f (a) according to the definition of limits,
can now be
unable to
=
=
we obtain
for every e > 0 there is & > 0 such that, for all x,
if0< |x--a|<ô,then|f(x)f(a) <e.
But in this case, where the limit is f (a), the phrase
0<|x-a|<8
may be changed to the simpler condition
x-a
since
THEOREM
2
PROOF
if x
=
a
<ô,
it is certainly true that |f (x)
f (a)|
-
<
e.
If g is continuous at a, and f is continuous at g(a), then fog is continuous at a.
(Notice that f is required to be continuous at g(a), not at a.)
Let e
>
0. We wish to find a 8
>
0 such that for all x,
og)(a)|<e,
if |x-a|<8,then
|(f og)(x)-(f
i.e., |f (g(x)) f (g(a))| < e.
-
116
Foundations
We first use continuity of f to estimate how close g(x) must be to g(a) in order
for this inequality to hold. Since f is continuous at g(a), there is a 8' > 0 such
that for all y,
if |y
(1)
-
g(a)
|<
ô', then
|f (y) f (g(a))| < e.
g(a)
|<
ô', then
|f (g(x)) f (g(a))| <
-
In particular, this means that
if
(2)
|g(x)
-
-
e.
We now use continuity of g to estimate how close x must be to a in order for the
inequality |g(x) g(a)| < 8' to hold. The number ô' is a positive number just like
any other positive number; we can therefore take ô' as the e (!)in the definition of
continuity of g at a. We conclude that there is a ô > 0 such that, for all x,
-
(3)
Combining
if |x
(2)and (3)we
if |x
-
We can now reconsider
a|
-
see that
al
<
<
8, then
|g(x)
-
g(a)
<
8'.
for all x,
8, then |f (g(x))
f (g(a))| <
-
e.
|
the function
f (x)
x sin 1/x,
=
0,
x
x
¢0
=
0.
We have already noted that f is continuous at 0. A few applications of Theorems 1
and 2, together with the continuity of sin, show that
f is also continuous at a, for
sin2(x3)))
sin(x2
should be equally easy
sin(x
Functions
like f(x)
+
+
a ¢ 0.
for you to analyze.
The few theorems of this chapter have all been related to continuity offunctions
at a single point, but the concept of continuity doesn't begin to be really interesting
until we focus our attention on functions which are continuous at all points ofsome
interval. If f is continuous at x for all x in (a, b), then f is called continuous
on (a, b). Continuity on a closed interval must be defined a little differently; a
function f is called continuous
on [a,b] if
=
(1) f
is continuous
lim f (x)
(2) x-a+
=
f
at x for all x in (a, b),
(a) and lim f (x) = f (b).
zeb-
Functions which are continuous on an interval are usually regarded as especially
behaved; indeed continuity might be specified as the first condition which a
function ought to satisfy. A continuous function is sometimes described, intuitively, as one whose graph can be drawn without lifting
your pencil
of
the
the function
from
paper. Consideration
well
"reasonable"
f (x)
shows that this
=
x sin
0,
1/x,
x
x
¢0
=
0
description is a little too optimistic, but it is nevertheless true that
there are many important results involving functions which are continuous on an
interval. There theorems are generally much harder than the ones in this chapter,
6. ContinuousFunctions 117
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 arguments, it is included
here as a preview of things to come.
THEOREM
S
is continuous at a, and f (a) > 0. Then there is a number 8 > 0 such
that f (x) > 0 for all x satisfying |x a| < 8. Similarly, if f (a) < 0, then there is
< ô.
a number ô > 0 such that f(x) < 0 for all x satisfying |x
Suppose
f
-
-al
PROOF
Consider the case f (a) > 0. Since
for all x,
f
such that,
if |x
Since f(a)
>
a|
-
is continuous at a, if e
ô, then
<
|f (x)
f (a) | <
-
>
0 there is a 8 > 0
e.
0 we can take f(a) as the e. Thus there is ô > 0 so that for all x,
if |x
a
-
|<
8, then
|f (x) f (a)| < f (a),
-
and this last inequality implies f (x) > 0.
A similar proof can be given in the case
apply the first case to the function f.
f (a)
<
0; take e
=
--
f (a). Or
one can
-
PROBLEMS
1.
For which of the following functions f is there a continuous function F with
domain R such that F(x)
f(x) for all x in the domain of f?
=
x2
(i) f (x)
-
=
x
-
4
2
.
(ii) f (x) x
(iii) f (x) 0, x irrational.
(iv) f (x) 1/q, x p/q rational
=
-.
=
=
=
which
in lowest terms.
points are the functions of Problems 4-17 and 4-19 continuous?
2.
At
3.
is a function satisfying |f (x) 5 |x for all x. Show that
at 0. (Notice that f (0) must equal 0.)
example
of such a function f which is not continuous
Give
at any
an
(b)
a ¢ 0.
(c) Suppose that g is continuous at 0 and g(0) 0, and | f(x)| 5 |g(x)|.
Prove that f is continuous at 0.
(a) Suppose that f
f is continuous
=
4.
Give an example of a function
is continuous everywhere.
5.
For each number
other points.
a,
f
such that
f
is continuous nowhere,
fmd a function which is continuous
at a,
but |f|
but not at any
118
Foundations
6.
¼,¼,¼, but continuous
(b) Find a function f which is discontinuous at 1, ¼,),¼, and at 0, but
continuous at all other points.
a function f which
all
other points.
at
(a) Find
is discontinuous at 1,
.
.
.
...
,
7.
Suppose that f satisfies f(x + y)
f(x) + f(y), and that f is continuous
at 0. Prove that f is continuous at a for all a.
8.
Suppose that f is continuous at a and f(a)
0. Prove that if a
f + a is nonzero in some open interval containing a.
9.
is not continuous at a. Prove that for some number e > 0 there
numbers
are
x arbitrarily close to a with |f(x)
f(a)| > e. Illustrate
graphically.
(b) Conclude that for some number e > 0 either there are numbers x arbitrarily close to a with f (x) < f (a) e or there are numbers x arbitrarily
close to a with f (x) > f (a) + e.
=
=
¢
0, then
(a) Suppose f
-
-
10.
at a, then so is if|.
continuous
every
f can be written f = E + 0, where E is
even and continuous and 0 is odd and continuous.
then so are max( f, g) and
Prove that if f and g are continuous,
min(
g).
(a) Prove that
(b) Prove that
(c)
if f is continuous
f,
(d) Prove that every continuous f can be written f
are nonnegative
11.
Prove Theorem
f (x)
*12.
=
=
g
h, where g and h
-
and continuous.
1(3) by using Theorem 2 and continuity
of the
if f is continuous at l and limg(x)
l, then lim f(g(x))
f (l). (You can go right back to the definitions, but it is easier to consider
the function G with G(x)
g(x) for x ¢ a, and G(a)
l.)
continuity
assumed,
that
of
then
l
if
it is not generally
f at is not
(b) Show
true that lim f (g(x)) f (limg(x)). Hint: Try f (x) 0 for x ¢ l, and
(a) Prove that
=
=
=
=
=
f (l)
13.
function
1/x.
=
1.
=
if f is continuous on [a,b], then there is a function g which
is continuous on R, and which satisfies g(x)
f(x) for all x in [a,b].
Hint: Since you obviously have a great deal of choice, try making g
(a) Prove that
=
constant on (-oo, a] and [b,oo).
(b) Give an example to show that this assertion is false if
by (a, b).
14.
g and h are continuous
g(x) if x > a and h(x)
(a) Suppose that
f(x)
to be
if
[a,b] is replaced
at a, and that g(a)
xs a. Prove that
h(a). Define
is
f continuous
=
at a.
g is continuous on [a,b] and h is continuous on [b,c] and
h(b). Let f(x) be g(x) for x in [a,b] and h(x) for x in [b,c].
(b) Suppose
g(b)
=
6. ContinuousFunctions 119
Show that f is continuous
"pasted
15.
on
together".)
(Thus, continuous
[a,c].
functions can be
continuity":
following version of Theorem 3 for "right-hand
and
that
lim
there
>
is a number
Suppose
f (x) f (a),
f (a) 0. Then
(a) Prove the
=
x->a+
0 such that f (x) > 0 for all x satisfying 0 Ex
a < 8. Similarly,
such
number
that
>
then
there
<
8
0
if f (a) 0,
is a
f (x) < 0 for all x
satisfying 0 5 x
a < 8.
version
of
Theorem 3 when lim f (x) f (b).
Prove
a
(b)
x->b8
>
-
-
=
16.
If lim f (x) exists, but is ¢ f (a), then
xua
continuity
is said to have a rernovable
dis-
at a.
sin 1/x for x ¢ 0 and
discontinuity at 0? What if f (x)
(a) If f (x)
f
f (0)
=
x sin
=
1, does f have a
removable
1/x for x ¢ 0, and f (0) l?
f(x) for
(b) Suppose f has a removable discontinuity at a. Let g(x)
continuous
that
and
g(a)
at
lim
Prove
is
let
a. (Don't
g
x ¢ a,
f(x).
=
=
=
=
x->a
hard; this is quite easy.)
=
Let
(c) f (x) 0 if x is irrational, and let f (p/q)
terms. What is the function g defined by g(x)
work very
*(d)
=
=
1/q if p/q is in lowest
lim f(y)?
y->x
f be a function with the property that every point of discontinuity
removable
discontinuity. This means that lim f (y) exists for all x,
is a
y->x
infinitely many) numbers x.
but f may be discontinuous at some (even
lim f(y). Prove that g is continuous. (This is not quite
Define g(x)
Let
=
y->x
**(e)
(b).)
so easy as part
Is there a function f which is discontinuous at every point, and which has
only removable discontinuities? (It is worth thinking about this problem
now, but mainly as a test of intuition; even if you suspect the correct
will almost certainly be unable to prove it at the present
answer,
you
time. See Problem 22-33.)
CHAPTER
THREE HARD THEOREMS
This chapter is devoted to three theorems about continuous functions, and some
of their consequences.
The proofs of the three theorems themselves will not be
given until the next chapter, for reasons which are explained at the end of this
chapter.
THEOREM
I
If f is continuous
such that
f (x) =
on
0.
[a,b]
and
f (a)
<
0
<
f (b), then
there is some x in
[a,b]
(Geometrically, this means that the graph of a continuous function which starts
below the horizontal axis and ends above it must cross this axis at some point, as
in Figure 1.)
THEOREM
2
is continuous on [a,b], then f is bounded above on
some number N such that f (x) 5 N for all x in [a,b].
If
f
(Geometrically, this theorem means that the graph of
allel to the horizontal axis, as in Figure 2.)
THEOREM
s
f
that is, there is
lies below some line par-
is continuous on [a,b], then there is some number
f (y) f (x) for all x in [a,b] (Figure 3).
If
[a,b],
f
y in
[a,b]
such that
>
from the theorems of Chapter 6. The
These three theorems differ markedly
of
always
those theorems
hypotheses
involved continuity at a single point, while
I
I
I
I
I
a
b
a
b
I
I
I
FIGURE
l
FIGURE
120
2
FIGURE
3
I
I
a
y
I
b
7. ThreeHard Theorems 121
the hypotheses of the present theorems require continuity on a whole interval
[a,b]--if continuity fails to hold at a single point, the conclusions may fail. For
example, let f be the function shown in Figure 4,
e----e
-1,
i
05x<Ã
f(x)=
i
2
Ásxs2.
1,
Then f is continuous at every point of [0,2] except Ã, and f (0) < 0 < f (2),
but there is no point x in [0,2] such that f (x) 0; the discontinuity at the single
point à is sufficient to destroy the conclusion of Theorem 1.
Similarly, suppose that f is the function shown in Figure 5,
=
FIGURE
4
-y
f (x)
N
x
0,
x
=
f is continuous at every point of [0, 1] except 0, but f
[0, 1]. In fact, for any number N > 0 we have f (1/2N)
Then
----------
0
0.
i1/x,
=
on
is not bounded above
=
2N
>
N.
example also shows that the closed
interval [a,b] in Theorem 2 cannot be
by the open interval (a, b), for the function f is continuous on (0, 1), but
is not bounded there.
Finally, consider the fimction shown in Figure 6,
This
replaced
X2
f (x)
FIGURE
=
f
0,
x
>
l.
On the interval [0,1] the function f is bounded above, so f does satisfy the
conclusion
of Theorem 2, even though f is not continuous
on [0,1]. But f
of
satisfy
the
conclusion
Theorem 3-there is no y in [0,1] such that
does not
f(y) > f (x) for all x in [0, 1]; in fact, it is certainly not true that f (1) > f (x) for
all x in [0,1] so we cannot choose y
1, nor can we choose 0 y < l because
number
with
if
<
<
x < 1.
y
f (y) f (x) x is any
This example shows that Theorem 3 is considerably stronger than Theorem 2.
Theorem 3 is often paraphrased by saying that a continuous function on a closed
interval
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 interval; such
properties of a function are always significantly more difficult
than
properties, and are correspondingly of much greater power.
to prove
usefulness
of Theorems 1, 2, and 3, we will soon deduce some imillustrate
To
the
portant consequences, but it will help to first mention some simple generalizations
Of fl1€SC Sleorems.
5
_<
=
"takes
"global"
FIGURE
|
¡
.
y
x
1
"local"
6
THEOREM
«
If
f
is continuous on
such that
f (x)
=
c.
[a,b] and f (a) <
c
<
f (b), then
there is some x in
[a,b]
122 Foundations
PROOF
Let g
is some
THEOREM
s
f
=
c. Then g is continuous,
b] such that g(x) =
-
in
x
[a,
If f is continuous
such that
f (x)
The function
PROOF
on
=
and
=
0
<
=
f (a)
is continuous
orem 4 there is some x in
f (x)
g(b). By Theorem 1, there
0. But this means that f(x) c.
<
>
c
>
f (b), then
there is some x in
[a,b]
c.
f
-
[a,b]
and g(a)
on
[a,b]
and
[a,b] such that
-
-
f (a)
f (x)
--c
<
<
-c,
=
f (b). By Thewhich means that
-
c.
Theorems 4 and 5 together show that f takes on any value between f (a)
and f (b). We can do even better than this: if c and d are in [a,b], then f
takes on any value between f (c) and f (d). 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 in
between; this slight generalization of Theorem 1 is often called the Intermediate
Value Theorem.
THEOREM
6
If f is continuous on
some number
PROOF
The function
such that
in
-
[a,b], so
[a,b], then f is bounded below on [a,b],
f(x) > N for all x in [a,b].
that is, there is
N such that
f is continuous
-
for all x in
-M.
let N
f(x) SM
we can
[a,b], so by Theorem 2 there is a number M
[a,b]. But this means that f(x) > -M for all x
on
-
2 and 6 together show that a continuous function f on [a,b] is
bounded on [a,b], that is, there is a number N such that |f (x) 5 N for all x in
[a,b]. In fact, since Theorem 2 ensures the existence of a number Ni such that
f (x) 5 N1 for all x in [a,b], and Theorem 6 ensures the existence of a number
max(|N1|, |N2|).
N2 such that f (x) > N2 for all x in [a,b], we can take N
Theorems
=
THEOREM
7
If
f is continuous
on
[a, b],
then there is some y in
[a,b]
such that
for all x in [a,b].
(A continuous function on a closed interval takes on its minimum
interval.)
PROOF
The function
-
f is continuous
f (x) for
such that
f (y) >
all x in [a,b].
-
-
on
all x
[a,b]; by Theorem
in
f (y) 5 f(x)
value
on that
3 there is some y in
[a,b], which means
that
f(y)
5
[a,b]
f (x) for
Now that we have derived the trivial consequences of Theorems 1, 2, and 3, we
can begin proving a few interesting things.
THEOREM
a
Every positive number has a square root. In other words, if a
some number
x2
x such that
-
a
>
0, then there is
7. ThreeHard Theorems 123
PROOF
x2, which is certainly continuous.
=
Notice that the
"the
of
number a has a
the theorem can be expressed in terms of f:
statement
root" means that
of
value
takes
proof
the
The
this
fact about f
square
a.
on
f
will be an easy consequence
of Theorem 4.
illustrated in Figure 7);
There is obviously a number b > 0 such that f (b) > a
Consider the function f (x)
(as
1 we can take b
a, while if a < 1 we can take b = 1. Since
applied
<
to [0,b] implies that for some x (in[0,b]),
f (0) a < f (b), Theorem 4
x2
=
œ,
have
i.e.,
a.
we
f (x)
in fact, if a
>
=
=
f(x)
.
-
can be used to prove that a positive number has
nth
natural
number
for
root,
n. If n happens to be odd, one can do
an
any
better: every number has an nth root. To prove this we just note that if the positive
number a has the nth root x, i.e., if x" = a, then (-x)" =
n is odd), so
assertion,
odd
number
nth
The
that
root
for
has the
n any
a has an nth
root, is equivalent to the statement that the equation
Precisely the same argument
x=
_
-a
-a
(since
-x.
x"-a=0
has a root if n is odd.
FIGURE
Expressed in this way the result is susceptible
of great
gentrâÏÎZatÎOD.
7
THEOREM
9
If n is odd, then any equation
x"+an-ix"¯1+---+ao=0
has a root.
PROOF
We obviously want to consider the function
f(x)=x"+an-ix"¯'+·
·+¾
like to prove that f is sometimes positive and sometimes negative. The
x" and,
intuitive idea is that for large |x|, the function is very much like g(x)
negative
and
since n is odd, this function is positive for large positive x
for large
negative x. A little algebra is all we need to make this intuitive idea work.
The proper analysis of the function f depends on writing
we would
=
f (x)
=
x" + an-1x"¯ I +
-
·
-
+ ao
=
1 + a,
x"
x"
x
Note that
as_i
-
x
an-2
+ x2 +
---
·
·
·
+
GO
--
x"
00|
Un-1
5
|x|
+
+
Consequently, if we choose x satisfying
(*)
then
|x|> 1, 2nla,_i|,...,2n|ao
,
|xk| > |x| and
Xk|
|x|
2n|a._g|
E'
----.
|x"|
124 Foundations
so
an_2
lan-1
-+-+
ao
··+-
x2
x
1
2n
1
2n
<-+···+--=-.
¯
yn
1
2
n terms
In
other words,
---<-+---+--<-,
1
2¯
a..I
1
ao
x"¯¯2
x
which implies that
1
2
a.
-51+-+-·-+-.
Therefore, if we choose an xi
(xi)"
2
so that
f(xi)
>
_
ao
x=
i
x
>
0 which satisfies
a._i
(xi)" 1+
(*),then
ao
-+
5
+
xi
0. On the other hand, if x2
(x2)n> (x2)" 1 +
an-1
----
2
x2
+
-
<
-
-
=
(xi)"
0 satisfies
+
GO
(*),then
=
(x2)"
f(x1),
(x2)" < 0
and
f (x2),
so that
f (x2)< 0.
Now applying Theorem 1 to the interval
in [x2,xi] such that f(x) 0.
[x2,x1]
we conclude that there
is an x
=
Theorem 9 disposes of the problem of odd degree equations so happily that it
would be frustrating to leave the problem of even degree equations completely
undiscussed. At first sight, however, the problem seems insuperable. Some equations, like x2
1 = 0, have a solution, and some, like x2 + 1 0, do not-what
more is 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
=
-
x" +
I
I
a..tx"¯I
+
-
-
-
+ ao
0,
=
let us ask about the possibility of solving the equations
x"+a._ix"¯1+---+ao=c
FIGURE
8
for all possible numbers c. This amount to allowing the constant term ao to vary.
The information which can be given concerning the solution of these equations
depends on a fact which is illustrated in Figure 8.
with n even, contains,
The graph of the function f (x) x"+a._ix"¯I+words,
other
the
point.
there is a
least
have
drawn
it,
lowest
In
at
way we
a
number y such that f (y) < f (x) for all numbers x-the
function f takes on a
minimum value, not just on each closed interval, 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 [a,b], and
obtain a point yo such that f (yo)is the minimum value of f on [a,b]; but if [a,b]
happens to be the interval shown in Figure 8, for example, then the point ye will
value for the whole line. In the next
not be the place where f has its minimum
-+ao,
=
-
7. ThreeHard Theorerns 125
theorem the entire point of the proof is to choose an interval
that this cannot happen.
THEOREM
10
If n is even and
that
PROOF
f (x)
x" + a,_ixn-1
=
f (y) 5 f (x) for all
+
.
-
-
[a,b] in such
+ ao, then there is a number
a way
y such
x.
As in the proof of Theorem 9, if
M
then for all x with
max(1,
=
2n an-il,
2n aal),
...,
à M, we have
x
I
2¯
a._i
-<1+--+
a0
··+--.
x"
x
Since n is even, x" 2 0 for all x, so
--
----------
----
-
x"
-Ex"
an-1
ao
x
x"
1+-+·-·+-
2
=f(x),
provided that |x à M. Now consider the number f (0). Let b > 0 be a number
b" à 2 f (0) and also b > M. Then, if x à b, we have (Figure 9)
such that
x"
f (x) >
>
--
¯2¯2-
b"
--
f (0).
>
--b,
then
Similarly, if x 5
FIGURE
9
x"
f(X)
>
(--b)"
>
--
¯2¯
b"
=
>
-
2
2¯
f(Û).
Summarizing:
-b,
if x
then
b or x 5
>
Now apply Theorem 7 to the function
that there is a number y such that
f
f(x) 2 f(0).
on the interval
[-b, b].
-b
(1)
if
5 xs
b, then f(y) 5
f(x).
In particular, f (y) 5 f (0). Thus
-b
if xs
Combining
(2)
(1)and (2)we
or x
see that
à b, then
f (x) 2 f (0) 2 f (y).
f (y) 5 f (x) for all x.
Theorem 10 now allows us to prove the following result.
THEOREM
11
Consider the equation
(*)
x" + a,_ix"¯1
+
-
-
-
+ ao
=
c,
We conclude
126 Foundations
and suppose
c à m and has no solution for c
PROOF
m such that
n is even. Then there is a number
(*)has a
solution
for
< m.
Let f (x) x" + an-ix"¯' +
+ ao (Figure 10).
there
According to Theorem 10
is a number y such that f(y) 5 f(x) for all x.
Let m
f(y). If c < m, then the equation (*)obviously has no solution, since
the left side always has a value à m. If c
m, then (*)has y as a solution.
number
such
that b > y and f (b) > c. Then
>
Finally, suppose c m. Let b be a
f(y) m < c < f(b). Consequently, by Theorem 4, there is some number x in
[y,b] such that f (x) c, so x is a solution of (*).
=
·
·
·
=
=
=
=
b
y
FIGURE
These consequences of Theorems 1, 2, and 3 are the only ones we will derive
theorems will play a fundamental role in everything we do later, hownow (these
ever). 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
P1-P12) a proof is impossible.There are several ways of con(namely,
vincing
ourselves that this gloomy conclusion is actually the case. For example,
the proof of Theorem 8 relies only on the proof of Theorem 1; if we could prove
Theorem 1, then the proof of Theorem 8 would be complete, 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 Pl-P12. Again, suppose we consider the
funCtiOn
10
f(x)=
1
X2
-
2
If there were no number x with x2
2, then f would be continuous, since the
denominator would never
0. But f is not bounded on [0,2]. So Theorem 2
essentially
the
existence
of numbers other than rational numbers, and
depends
on
therefore on some property of the real numbers other than P1-P12.
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 difficulties suggest a
way to discover that property: let us try to construct a proof of Theorem 1, for
=
=
A
x
a
a
x'
b
example, and see what goes wrong.
One idea which seems promising is to locate the first point where f(x) 0, that
is, the smallest x in [a,b]such that f(x) = 0. To find this point, first consider
the set A which contains all numbers x in [a,b] such that f is negative on [a,x].
In Figure 11, x is such a point, while x' is not. The set A itself is indicated by a
heavy line. Since f is negative at a, and positive at b, the set A contains some
points greater than a, while all points sufficiently close to b are not in A. (We are
here using the continuity of f on [a,b], as well as Problem 6-15.)
=
FIGURE
11
7. ThreeHard Theorems 127
f(x) <
0 for all x in this interval
is the smallest number which is greater than all members of A;
clearly a < a < b. We claim that f (œ)
0, and to prove this we only have to
eliminate the possibilities f (a) < 0 and f (a) > 0.
Suppose first that f (a) < 0. Then, by Theorem 6-3, f (x) 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 every member
of A, since the larger numbers would also be in A. Consequently, f (a) < 0 is
false.
On the other hand, suppose f(a) > 0. Again applying Theorem 6-3, we see that
f(x) would be positive for all x in a small interval containing a, in particular for
some numbers smaller than « (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 tempted to say, QE.D.
We know, however, that something must be wrong, since no new properties of R
Were
ever used, and it does not require much scrutiny to find the dubious point.
It is clear that we can choose a number a which is greater than all members of A
(forexample, we can choose a = b), but it is not so clear that we can choose a
smallest one. In fact, suppose A consists of all numbers x > 0 such that x2 < 2.
If the number à did not exist, there would not be a least number greater than
all the members of A; for any y > Á we chose, we could always choose a still
Now suppose
œ
=
o
ontain
FIGU RE 12
=
A could really
only this bi
be
f(x) > o for au
in this interval
FIGURE
13
2
smaller
one.
Now that we have discovered the fallacy, it is almost obvious what additional
property of the real numbers we need. All we must do is say it properly and use it.
That 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 f might have these properties even if f is not continuous,
and even if the interval is not a closed interval.)
(i) f (x)
(ii) f(x)
(iii) f (x)
(iv) f (x)
(v) f (x)
=
=
x2
on
(-1, 1).
x3 on
(-1, 1).
=
x2
=
x2
on R.
on
x2,
[0,oo).
=
[ a+2,
x
5 a
x
>a
on
(-a
-
1, a + 1). (It will be necessary
consider several possibilities for a.)
x2,
(vi) f (x)
=
(vii)f (x)
=
f[ a+2,
0,
1/q
x
<
x>a
a
on
[-a
-
1, a + 1].
x irrational
p|q in lowest terms
x
=
on
[0,1].
to
128 Foundations
(viii)f (x)
=
(ix) f (x)
=
(x) f (x)
=
x irrational
p/q in lowest terms
x
(1,
(1,
I
1/q
-1/q
x
x
irrational
p/q in lowest terms
x,
x rational
0
x irrational
on
sin2(cosx
a2)
[0,1].
[0,1].
on
=
+ Ja +
(xi) f(x)
(xii)f (x) [x]on [0,a].
=
on
=
[0,a].
on
[0,a3p
=
2.
For each of the following polynomial functions
f (x) 0 for some x between n and n + 1.
f,
fmd an integer n such that
=
(i) f (x)
(ii) f(x)
(iii) f (x)
(iv) f (x)
3.
x3 x + 3.
x5 + 5x'+ h + 1
x + x + 1.
=
-
=
=
4x2
=
4x + 1.
--
Prove that there is some number x such that
163
Il79
gig
(i)
l+x2+sin2
(ii) sin x x 1.
__
=
4.
-
This problem is a continuation
(a) IËn
of
Problem 3-7.
k is even, and > 0, find a polynomial function of degree n with
exactly k roots.
A
m
(b) root a of the polynomial fimction f is said to have multiplicity
if f(x) (x a)mg(x), where g is a polynomial function that does not
-
=
--
have a as a root. Let f be a polynomial function of degree n. Suppose
that f has k roots, counting multiplicities, i.e., suppose that k is the sum
of the multiplicities of all the roots. Show that n k is even.
--
[a,b] and
f (x) is always
that
rational.
5.
Suppose that f is continuous on
can be said about f?
6.
Suppose that f is a continuous function on [--1,1] such that x2+(f(x))2
y
for all x. (This means that (x,f (x)) always lies on the unit circle.) Show that
x2 for all X.
either f(x) /1 x2 for all x, or else f (x)
What
_
-)1
=
7.
=
-
--
How many continuous functions f are there which satisfy (f (x))2
2
for
all x?
8.
Suppose that f and g are continuous, that f2 g2, and that
all x. Prove that either f (x)
g(x) for all x, or else f (x)
f(x) ¢ 0 for
-
-g(x)
=
9.
for all x.
=
(a) Suppose that f is continuous, that f (x) 0 only for x
f (x) > 0 for some x > a as well as for some x < a. What
=
about
f(x)
for all x
¢
a?
=
a, and that
can be said
7. ThreeHard Theorems l 29
(b) Again
assume
but suppose,
some x
*(c)
10.
<
a.
f is continuous and that f (x)
instead, that f (x) > 0 for some x >
Now what can be said about f (x) for
that
0 only for x
=
a and
¢
x
f (x)
<
=
a,
0 for
a?
Discuss the sign of x3 + x2y + xy2 + y3 when x and y are not both 0.
Suppose f and g are continuous on [a,b] and that f(a) < g(a), but f(b) >
g(x) for some x in [a,b]. (If your proof isn't very
g(b). Prove that f(x)
short, it's not the right one.)
=
11.
(1,i)
's
[0,1]
=
',
FIGURE
Suppose that f is a continuous function on [0,1] and that f (x) is in
for each x (drawa picture). Prove that f (x) x for some number x.
12.
14
11 shows that f intersects the diagonal of the square in Figline). Show that f must also intersect the other (dashed)
ure 14 (solid
diagonal.
(b) Prove the following more general fact: If g is continuous on [0,1] and
g(0)
0, g(l)
1 or g(0)
1, g(1)
0, then f(x) g(x) for some x.
(a) Problem
=
13.
=
(a) Let f (x)
*(b)
*(c)
sin
=
=
=
=
1/x for x ¢ 0 and let f (0)
=
0. Is f continuous
on
maximum
|f |
[-1, 1]? Show that f satisfies the conclusion of the Intermediate Value
Theorem on [-1, 1]; in other words, if f takes on two values somewhere
on [-1, 1], it also takes on every value in between.
Suppose that f satisfies the conclusion of the Intermediate Value Theorem, and that f takes on each value only once. Prove that f is continuous.
Generalize to the case where f takes on each value only finitely many
times.
14.
f is a continuous
on [0,1].
If
function on
[0,1], let ||f || be the
for any number c we have ||cf|| |c| ||f||.
Prove that | f + g|| 5 ||f|| + ||g||. Give an example where
(a) Prove that
*(b)
Ilfll +
=
llg||.
-
*15.
-
|f
+ g||
¢
-g||+||g
(c) Prove that |\h f||
y
value of
5
||h
Suppose that ¢ is continuous
-
and
f||.
lim ¢(x)/x"
X->OD
=
0
=
lim ¢(x)/x".
X->-DO
if n is odd, then there is a number x such that x" + ¢(x) 0.
(b) Prove that if n is even, then there is a number y such that y" + ¢(y) 5
x" + ¢(x) for all x.
(a) Prove that
's,
"
=
Hint: Of which proofs does this problem test your understanding?
FIGURE
*16.
Let f be any polynomial function. Prove that there is some number
that |f(y)| 5 |f(x)| for all x.
*17.
Suppose that
15
is a continuous
function with
y such
0 for all x, and
lim f (x) 0
lim f (x). (Draw a picture.) Prove that there is some
X->OD
X4-00
number y such that f (y) > f (x) for all x.
=
f
=
f (x)
>
l 30 Foundations
*18.
is continuous on [a,b], and let x by any number. Prove
that there is a point on the graph of f 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)) is 5 distance from (x, 0) to (z,f (z)) for all z in [a,b]. (See
Figure 15.)
that this same assertion is not necessarily true if [a,b] is replaced
Show
(b)
by (a, b) throughout.
(c) Show that the assertion as true if [a,b] is replaced by R throughout.
(d) In cases (a)and (c),let g(x) be the minimum distance from (x, 0) to a
yl, and conclude
point on the graph of f. Prove that g(y) 5 g(x) +|x
that g is continuous.
that there are numbers xo and xi in [a,b] such that the distance
Prove
(e)
from (xo,0) to (x1,f(xi)) is 5 the distance from (xo',0) to (xt', f(xi'))
for any xo', Il' in [a,b].
(a) Suppose that f
-
**19.
(a) Suppose that f
is continuous on [0, 1] and f(0)
number.
Prove that there is some number
shown
in Figure 16 for n = 4. Hint:
(x+1/n), as
natural
f
g(x)
=
f(x) f(x
(b) Suppose
-
0
<
a
<
FIGURE16
+
n. Find a function f which is continuous
satisfies f(0) = f(1), but which does not satisfy
1
f(l).
Let n be any
x such that
f (x)
=
Consider the function
+ 1/n); what would be true if g(x) ¢ 0 for all x?
1, but that a is not equal to 1/n for any natural
number
x
=
[0, 1] and
on
f(x)
=
f(x
which
+ a)
for
any x.
**20.
there does not exist a continuous function f defined on R
which takes on every value exactly twice. Hint: If f (a)
f (b) for
either
all
then
>
<
<
b,
for
in
b)
a
x
(a, or f (x) f (a) for
f (x) f (a)
values
close to f(a), but slightly
the
all x in (a, b). Why? In
first case all
larger than f (a), are taken on somewhere in (a, b); this implies that
f (x) < f (a) for x < a and x > b.
(b) Refine part (a)by proving that there is no continuous function f 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 f has either a maximum or a minimum value (which
must be taken
said
about
values
close
maximum
value?
twice).
What
the
be
to
on
can
which
continuous
takes
value
exactly
function
Find
3 times.
on every
a
f
(c)
which
value
exactly
takes on every
More generãlly, find one
n times, if
odd.
is
n
(d) Prove that if n is even, then there is no continuous f which takes on
4, for example,
every value exactly n times. Hint: To treat the case n
Í(X4).
>
let f(x1)
Then
either
f(x2) f(13)
f(x) 0 for all x in
of
the
three
x3),
x4),
intervals (xi, x2), (x2,
two
(x3, or else f (x) < 0 for
all x in two of these three intervals.
(a) Prove that
=
=
=
=
=
LEAST UPPER BOUNDS
CHAPTER
NeverThis chapter reveals the most important property of the real numbers.
merely
sequel
which
theless, it is
must be followed has
to Chapter 7; the path
a
already been indicated, and further discussion would be useless delay.
DEFDHTION
A set A of real numbers is bounded
x
Such a number
x
>
a
above if there is a number x such that
for every a in A.
is called an upper
bound for A.
Obviously A is bounded above if and only if there is a number x which is an
upper bound for A (andin this case there will be lots of upper bounds for A); we
often say, as a concession to idiomatic English, that "A has an upper bound" when
which is an upper bound for A.
we mean that there is a number
above" has now been used in two ways--first,
in
Notice that the term
Chapter 7, in reference to functions, and now in reference to sets. This dual usage
should cause no confusion, since it will always be clear whether we are talking
about a set of numbers or a function. Moreover, the two definitions are closely
connected:
if A is the set {f (x) : a Ex 5 b}, then the function f is bounded
above on [a,b] if and only if the set A is bounded above.
The entire collection R of real numbers, and the natural numbers N, are both
examples of sets which are not bounded above. An example of a set which is
bounded above is
A=(x:05x<1}.
"bounded
To show that A is bounded above we need only name some upper bound for A,
is easy enough; for example, 138 is an upper bound for A, and so are 2,
l¼, 1¼, and 1. Clearly, 1 is the least upper bound of A; although the phrase
just introduced is self explanatory, in order to avoid any possible confusion (in
means), we
particular, to ensure that we all know what the superlative of
explicitly.
define this
which
"less"
DEFINITION
A number
x
is a least upper
and
131
bound of A if
(1) x is an upper bound of A,
(2) if y is an upper bound of A, then
x 5 y.
132
Foundations
"a"
in this definition was merely a concession
The use of the indefinite article
that
ignorance.
Now
have made a precise definition, it is easily
we
to temporary
and
that
if x
seen
y. Indeed, in this
y are both least upper bounds of A, then x
=
case
xs
y,
and ys
x,
since y
since x
is an upper bound, and x is a least upper bound,
is an upper bound, and y is a least upper bound;
it follows that x
y. For this reason we speak of the least upper bound of A.
of A is synonymous and has one advantage. It abbreviates
The term supremum
nicely
quite
to
=
sup A
and saves us
"soup
(pronounced
A")
from the abbreviation
lub A
(whichis 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 below if
there is
a number
x such that
x 5 a
for every a in A.
Such a number x is called a lower bound for A. A number x is the greatest
lower bound of A if
is a lower bound of A,
if
(2)
y is a lower bound of A, then x
(1)
and
x
>
y.
The greatest lower bound of A is also called the infimum of A, abbreviated
inf A;
some authors use the abbreviation
glb A.
One detail has been omitted from our discussion so far-the question of which
sets 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).
If A 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.
then
If A 0,
A is bounded above. Indeed, any number x is an upper bound
for 0:
x > y for every y in 0
=
simply
because there is no y in Ø. Since every number is an upper bound for 0,
there is surely no least upper bound for Ø. With this trivial exception however,
8. Irast UpperBounds 133
is true-and very important, definitely important enough to warrant
of details. We are finally ready to state the last property of the real
our assertion
consideration
which we need.
numbers
(The least upper bound property) If A is a set of real numbers,
A ¢ Ø, and A is bounded above, then A has a least upper bound.
(Pl3)
b
a
a
but that is actually one of its
Property P13 may strike you as anticlimactic,
virtues.
To complete our list of basic properties for the real numbers we require 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 property is not
really so innocent as all that; after all, it does not hold for the rational numbers
Q.
For example, if A is the set of all rational numbers x satisfying x2 < 2, then there
is no rahonal number y which is an upper bound for A and which is less than or
equal to every other rational number which is an upper bond for A. It will become
clear only gradually how significant Pl3 is, but we are already in a position to
demonstrate its power, by supplying the proofs which were omitted in Chapter 7.
1
FIGURE
THEOluuW
7.1
If
in
PROOF
f is continuous
[a,b]
on
such that
[a,b]
f (x)
=
0.
and
f(a)
<
0
<
f(b),
then there is some number
x
Our proof is merely a rigorous version of the outline developed at the end of
Chapter 7-we will locate the smallest number x in [a,b] with f(x) 0.
Defme the set A, shown in Figure 1, as follows:
=
f(x) o for all x
in this interval
<
A
-
{x a 5 x
:
5 b, and
on the
f is negative
interval
·ý
a x,
I U
Clearly A 0, since a is in A; in fact, there is some 8 > 0 such that A contains
all points x satisfying a 5 x < a + 8; this follows from Problem 6-15, since f is
continuous on [a,b] and f (a) < 0. Similarly, b is an upper bound for A and, in
fact, there is a 8 > 0 such that all points x satisfying b ô < x 5 b are upper
bounds for A; this also follows from Problem 6-15, since f (b) > 0.
From these remarks it follows that A has a least upper bound a and that
0, by eliminating the possibila < œ < b. We now wish to show that f(a)
<0and
ities f(a)
f(a)>0.
Suppose first that f (a) < 0. By Theorem 6-3, there is a 8 > 0 such that
f (x) < 0 for « ô < x < a + & (Figure 2). Now there is some number xo in A
which satisfies a
otherwise a would not be the least upper
8 < xo < a (because
negative
that
This
of
A).
is
on the whole interval [a,xo]. But if
bound
means
f
and
number
between a
a + ô, then f is also negative on the whole interval
xi is a
f is negative on the interval [a,xi], so x1 is in A. But this
[xo,xt]. Therefoæ that
contradicts
that
the fact
a is an upper bound for A; our original assumption
f(a) < 0 must be false.
Suppose, on the other hand, that f (a) > 0. Then there is a number ô > 0 such
that f (x) > 0 for a 8 < x < a + 8 (Figure 3). Once again we know that there is
SRtiSfying
ô < xo < a; but this means that f is negative on [a,xo],
a
an XO in A
-
FIGURE2
=
-
-
$ta
i
a
i
o
Ex) > c for an x
in this interval
-
FIGURE
3
[a,x]}.
-
134
Foundations
which is impossible, since
leaving f
a contradiction,
f (xo) > 0. Thus the assumption f (a)
(a) 0 as the only possible alternative.
>
0 also leads to
=
The proofs of Theorems 2 and 3 of Chapter 7 require a simple preliminary
play much the same role as Theorem 6-3 played in the previous
proof
result, which will
THEOREM
1
is continuous at a, then there is a number ô
above on the interval (a 8, a + 8)
(seeFigure 4).
If
f
>
0 such that
f
is bounded
-
PROOF
Since lim f (x) =
f (a), there
if |x
-
is, for every e
a
8, then
|<
>
0, a 8
>
0 such that, for all x,
|f (x) f (a)| <
--
e.
It is only necessary to apply this statement to some particular e (anyone will do),
for example, e 1. We conclude that there is a 8 > 0 such that, for all x,
=
if|x-a|<8,then|f(x)-f(a)1<l.
It follows, in particular, that if |x a| < 8, then f (x) f (a) < l. This completes
the proof: on the interval (a
8, a + 8) the function f is bounded above by
1.
+
f (a)
--
-
-
It should hardly be necessary to add that we can now also prove that f is
bounded below on some interval (a ô, a + 8), and, finally, that f is bounded on
some open interval containing a.
A more significant point is the observation that if lim f (x) f (a), then there
-
=
x-a+
I
I
I
I
I
I
I
I
I
I
I
I
I
i
aFIGURE
4
|
ô a
a‡
8. Least UpperBounds 135
0 such that f is bounded on the set {x : a 5 x < a + ô}, and a similar
observation holds if lim f (x)
f (b). Having made these observations (and
is a ô
>
=
assuming
THEOREM
7-2
PROOF
that you will supply the proofs), we tackle our second major theorem.
If f is continuous
[a,b], then f
on
is bounded above on
[a,b].
Let
A
and
{x a Exsb
:
=
above on
f is bounded
[a,x]}.
Clearly A ¢ Ø (since
a is in A), and A is bounded above (byb), so A has a least
above" both
Notice
that we are here applying the term
bound
upper
a.
visualized
to the set A, which can be
as lying on the horizontal axis, and to f, i.e.,
which
the
:
x},
to
sets {f(y) a sy 5
can be visualized as lying on the vertical axis
(Figure 5).
b. Suppose, instead, that
Our first step is to prove that we actually have a
such
that
there
>
<
8
0
is
By
Theorem
1
is
bounded
b.
on («-8, a+ô). Since
a
f
of
satisfying
a-ô < xo < a. This
A there is some xo in A
a is the least upper bound
means that f is bounded on [a,xo]. But if xi is any number with a < xi < a + 8,
then f is also bounded on [xo,xi]. Therefore f is bounded on [a,x1], so xi is
the fact that a is an upper bound for A. This contradiction
in A, contradicting
shows that a
b. One detail should be mentioned: this demonstration implicitly
assumed that a < a [sothat f would be defined on some interval (a 8, a + 8)];
the possibility a
a can be ruled out similarly, using the existence of a ô > 0 such
that f is bounded on {x: a Ex < a + 8}.
The proof is not quite complete-we only know that f is bounded on [a,x] for
every x < b, not necessarily that f is bounded on [a,b]. However, only one small
argument needs to be added.
There is a ô > 0 such that f is bounded on {x : b 8 < x 5 b}. There is xo
in A such that b 8 < xo < b. Thus f is bounded on [a,xo] and also on [xo,b],
so f is bounded on [a,b].
"bounded
'
=
--------
U(y):a sys
x)
=
-
=
a
FIGURE
A
5
-
--
To prove the third important theorem we resort to a trick.
THEOREM
7-3
PROOF
is continuous
f (x) for all x in
If
f
on
[a,b], then
there is a number
y in
[a,b].
[a,b], which
{f (x) : x in [a,b] }
We already know that f is bounded on
[a,b]
means
such that
f(y) >
that the set
is bounded. This set is obviously not 0, so it has a least upper bound «. Since
f (y) for some y in [a,b].
a > f (x) for x in [a,b) it suffices to show that a
Suppose instead that a ¢ f (y) for all y in [a,b]. Then the function g defined
by
1
=
g(x)
=
,
a
-
f (x)
x in
[a,b]
136 Foundations
is continuous on [a,b], since the denominator of the right side is never 0. On the
other hand, a is the least upper bound of f(x) : x in [a,b]}; this means that
{
for every e
0 there is x in
>
[a,b]
with a
-
f (x) <
e.
This, in turn, means that
for every e
0 there is x in
>
But this means that g is not bounded on
[a,b] with g(x) > 1/e.
[a,b], contradicting the
previous theo-
rem.
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 provethat N is unbounded.
After the difficulttheorems proved in this chapter you may be startled to find
such an
theorem winding up our proceedings. If so, you are, perhaps,
allowing the geometrical picture of R to influence you too strongly. "Look," you
real numbers look like
may say,
"obvious"
"the
Illl
O
·--III
1
2
3
nx
n+1
x is itself an integer)."
so every number x is between two integers n, n + 1 (unless
Basing the argument on a geometric picture is not a proof, however, and even the
that if you place unit segments end-togeometric picture contains an assumption:
end you will eventually get a segment larger than any given segment. This axiom,
often omitted from a first introduction to geometry, is usually attributed
(notquite
corresponding
and
numbers,
the
that N is not
property for
justly) to Archimedes,
real
numbers.
of
the
bounded, is called the Archimedian
This
property is not
proper_;
reference
of
of
the
Suggested Reading), although
P1-P12 (see
a consequence
[17]
of
it does hold for Q,
course. Once we have P13 however, there are no longer
problems.
any
THEOREM
2
N is
not
bounded above.
-/
PROOF
Suppose N were bounded above.
bound a for N. Then
Since N
a à n
Ø, there would be a least upper
for all n in N.
Consequently,
aën+1
since n +
1 is in N if n is in N. But this means that
«
and
forallninN,
-
1à n
for all n in N,
this means that a 1 is also an upper bound for N, contradicting the fact that
the
least upper bound.
a is
-
8. Least UpperBounds 137
of Theorem 2
often
assumed
implicitly.
we have very
There is a consequence
THEOREM
s
PROOF
For any e
0 there is a natural number
>
an
(actually
n with
1/n
equivalent formulation) which
<
e.
Suppose not; then 1/n > e for all n in N. Thus n < 1/s for all n in N. But this
Theorem 2.
means that 1/s is an upper bound for N, contradicting
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 tolerated. As partial justificationfor this dishonesty we
can claim that this result was never used in the proof of a theoran,but if your faith
has been shaken, a review of all the proofs given so far is in order. Fortunately,
such
deception will not be necessary again. We have now stated every property of
the real numbers that we will ever need. Henceforth, no more lies.
PROBLEMS
1.
Find the least upper bound and the greatest lower bound (ifthey exist) of
the following sets. Also decide which sets have greatest and least elements
(i.e.,decide when the least upper bound and greatest lower bound happens
to belong to the set).
(i)
1
-
:
n
(ii)
n in N
.
1
-:ninZandn¢0.
.
n
(iii) (x : x
=
(iv) (x : 0 5
0 or x
xs
Ã
=
1/n for some n in N}.
and x is rational}.
(v) {x:x2+x+1>0}.
(vi) {x:x2+x--1<0}.
(vii){x:x<0andx2+x-1<0}.
1
(viii) n
---
2.
+
(-1)"
:
n in N
.
(a) Suppose
A ¢ Ø is bounded below. Let -A denote the set of all
A.
for x in
Prove that -A ¢ Ø, that -A is bounded above, and that
sup(-A) is the greatest lower bound of A.
(b) If A ¢ Ø is bounded below, let B be the set of all lower bounds of A.
Show that B ¢ 0, that B is bounded above, and that sup B is the greatest
lower bound of A.
-x
-
3.
Let f be a continuous function on
(a) The
with
[a,b] with f (a) <
0
<
f (b).
proof of Theorem 1 showed that there is a smallest x in [a,b]
f (x) 0. Is there necessarily a second smallest x in [a,b] with
=
l 38 Foundations
0? Show that there is a largest x in [a,b] with f (x) 0. (Try
to give an easy proof by considering a new function closely related to f.)
: a 5
upon consideration of A
(b) The proof of Theorem 1 depended
x]}.
Give another proof of Theorem 1,
x 5 b and f is negative on [a,
which depends upon consideration
of B
: a 5 x 5 b and
f (x) <
0}. Which point x in [a,b] with f (x) 0 will this proof locate? Give
f (x)
=
=
=
=
{x
(x
=
an example where the sets A and B are not the same.
*4.
(a) Suppose
Suppose
numbers
that
is continuous on [a,b] and that f (a)
f (b) 0.
0 for some xo in [a,b]. Prove that there are
sc < xo < d 5 b such that f (c) f (d) 0,
0 for all x in (c,d). Hint: The previous problem can be used
f
=
also that f (xo) >
c and d with a
but f (x) >
to good advantage.
(b) Suppose that f is continuous on
there are numbers
and
5.
f (d) f
=
=
=
=
that f (a) < f (b). Prove that
d 5 b such that f(c)
f(a)
(d) for all x in (c,d).
[a,b] and
c and d with a 5 c
and
(b)
f (a) < f (x) < f
<
=
that y
x > l. Prove that there is an integer k such that
x < k < y. Hint: Let I by the largest integer satisfying I 5 x, and
consider l + 1.
(b) Suppose x < y. Prove that there is a rational number r such that x <
then ny
Why have parts
> 1.
r < y. Hint: If 1/n < y
(Query:
and
until
set?)
this
postponed
problem
(a) (b)been
(c) 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.
(d) 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 from
(a) Suppose
-
-x,
-nx
(b)and (c).
*6.
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 set of rational numbers
and the set of irrational numbers are each dense.
that if f is continuous and f (x) 0 for all numbers x in a dense
0 for all x.
set A, then f (x)
and
g are continuous and f (x) g(x) for all x in a dense
(b) Prove that if f
g(x) for all x.
set A, then f(x)
that f (x) à g(x) for all x in A, show that f (x) à
instead
If
(c) we assume
all
g(x) for
x. Can à be replaced by > throughout?
(a) Prove
=
=
=
=
7.
Prove that if f is continuous
and f (x + y) = f (x) + f (y) for all x and y,
then there is
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 noncondnuousfunctions f satisfying
this now; in
(x + y) = (x) + (y) for all x and y, but we cannot
a number
f
f
f
prove
fact, this simple question involves ideas that are usually never mentioned
any undergraduate
course. The Suggested Reading contains references.
in
8. Least UpperBounds 139
*8.
Suppose that
ure 6).
(a) Prove that
f
is a function such that f (a)
lim f (x) and lim
x->a+
x->a-
f (x) both
f (b) whenever
_<
exist.
a
<
b (Fig-
Hint: Why is this prob-
lem in this chapter?
(b) Prove that f never has a removable discontinuity (thisterminology comes
from Problem 6-16).
(c) Prove that if f satisfies the conclusions of the Intermediate Value Theorem, then f is continuous.
*9.
If f is a bounded function on [0,1], let |||fl|| sup{|f(x)|
Prove analogues of the properties of || || in Problem 7-14.
10.
Suppose a > 0. Prove that every number x can be written uniquely in the
form x ku + x', where k is an integer, and 0 5 x' < a.
=
:
x
in
[0,1]}.
=
FIGURE
11.
6
one side of P
¯¯¯¯~2;T¯Á¯
FIGURE
7
that ai, a2, a3,
is a sequence of positive numbers with
assi y an/2. Prove that for any e > 0 there is some n with a, < s.
(b) Suppose P is a regular polygon inscribed inside a circle. If P' is the
inscribed regular polygon with twice as many sides, show that the difference between the area of the circle and the area of P' is less than half the
difference between the area of the circle and the area of P (useFigure 7).
(c) 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 area. By calculating
the areas of polygons, this method ("themethod of exhaustion") allows
computations
of x to any desired accuracy; Archimedes used it to show
< x <
But it has far greater theoretical importance:
that
*(d)
Using the fact that the areas of two regular polygons with the same number of sides have the same ratio as the square of their sides, prove that the
areas oftwo circles have the same ratios as the square oftheir radii. Hint:
Deduce a contradiction from the assumption that the ratio of the areas
is greater, or less, than the ratio of the square of the radii by inscribing
appropriate polygons.
(a) Suppose
...
).
/
12.
Suppose that A and B are two nonempty
for all x in A and all y in B.
sets of numbers
such that x 5 y
for all y in B.
(b) Prove that sup A 5 inf B.
(a) Prove that
13.
sup A
sy
Let A and B be two nonempty sets of numbers which are bounded above, and
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 sup(A+B) 5 sup A+sup B
is easy. Why? To prove that sup A + sup B 5 sup(A + B) it sufEces to prove
that sup A + sup Bs sup(A + B) + e for all e > 0; begin by choosing x in A
=
140
Foundations
and y
in B with sup A
FIGURE
I
as
8
-
I
as
x
<
e/2 and sup B
I
as
of closed
-
y
<
s/2.
I
I
I
b,
b:
61
intervals Ii
[ai,bi], 12 Ë2, b2],
Suppose that a, y a.+1 and bn+1 y b,, for all n (Figure 8). Prove that
there is a point x which is in every I..
(b) Show that this conclusion is false if we consider open intervals instead of
(a) Consider
14.
a sequence
=
=
.
.
.
.
closed intervals.
The simple result of Problem 14(a)is called the "Nested Interval Theorem." It
may be used to give alternative proofs of Theorems 1 and 2. The appropriate
reasoning, outlined in the next two problems, illustrates a general method, called
"bisection
a
*15.
argument."
Suppose f is continuous on [a,b] and f(a) < 0 < f(b). Then either
f ((a + b)/2) 0, or f has different signs at the end points of the interval
[a, (a + b)/2], or f has different signs at the end points of [(a + b)/2, b].
Why? If f((a + b)/2) ¢ 0, let 11 be one of the two intervals on which f
changes sign. Now bisect II. Either f is 0 at the midpoint, or f changes
sign on one of the two intervals. Let I2 be such an interval. Continue in this
way, to define I, for each n (unless
f is 0 at some midpoint). Use the Nested
Interval Theorem to find a point x where f (x) 0.
=
=
*16.
Suppose f were continuous on [a,b], but not bounded on [a,b]. Then f
would be unbounded
on either [a, (a + b)/2] or [(a+b)/2, b]. Why? Let 11
of
these
Proceed as in Problem 15
intervals
be one
on which f is unbounded.
to obtain a contradiction.
17.
(a) Let A
=
{x: x
< «}.
Prove the following (theyare all easy):
IfxisinAandy<x,thenyisinA.
(ii) A ¢ 0.
(iii) A ¢ R.
(iv) If x is in A, then there is some number
(i)
(b) Suppose,
conversely, that A satisfies
sup A}.
*18.
(i) iv).
x'
in A such that x
Prove that A
=
{x :
<
x
x'.
<
A number x is called an almost
upper bound for A if there are only
with
numbers
A
in
fmitely many
y
y > x. An almost lower bound is
defined similarly.
upper bounds and almost lower bounds of the sets in
Problem 1.
(b) 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.
(a) Find
all almost
8. Least UpperBounds 141
inf B exists; this number is called the limit
superior of A, and denoted by lim A or lim sup A. Find lim A for each
set A in Problem 1.
(c) It follows from part (b)that
(d) Define lim A, and
*19.
find it for all A in Problem 1.
If A is a bounded infinite set prove
(a) lim A 5 E
(b) E
A.
A 5 sup A.
(c) If E
A < sup A, then A contains a largest element.
of parts (b)and (c)for lim.
The
analogues
(d)
I
I
shadow points
FIGURE
*20.
9
Let f be a continuous function on R. A point x is called a shadow point
of f 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 the sun
rising in the east
(youare facing north). Suppose that all points of (a, b) are
shadow points, but that a and b are not shadow points.
(a) For x in (a, b), prove that f (x) sf (b). Hint: Let A {y
and f(x) 5 f(y)}. If supA were less than b, then supA
=
:
x 5 y 5 b
would
be a
shadow point. Use this fact to obtain a contradiction
to the fact that b
is not a shadow point.
Now
(b)
prove that f (a) 5 f (b). (This is a simple consequence of continu-
ity.)
(c) Finally, using the fact that a is not a shadow point, prove that f(a)
=
f (b).
This
is known as the Rising Sun Lemma. Aside from serving as
a 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 443.
result
l 42 Foundations
APPENDIX.
UNIFORM CONTINUITY
"foundations,"
Now that we've come to the end of the
it might be appropriate
to slip in one further fundamental concept. This notion is not used crucially in
the rest of the book, but it can help clarify many points later on.
x2 is continuous at
We know that the function f (x)
a for all a. In other
=
words,
y
a=
if a is any number, then
such that, for all x, if |x
_
for every e
-
al
<
0 there is some 8
>
8, then
]x2
>
0
2
Of course, 8 depends on e. But 8 also dependson a-the 8 that works at a might
not work at b (Figure 1). Indeed, it's clear that given e > 0 there is no one 8 > 0
that works for all a, or even for all positive a. In fact, the number a + 8/2 will
certainly satisfy |x a| < ô, but if a > 0, then
+ e
-
a for
FIGURE
a
and this won't
be
<
e once a
tational way of saying that f
>
e/8.
(This is just an admittedly confusing
compu-
is growing faster and faster!)
On the other hand, for any e > 0 there will be one 8 > 0 that works for all a
in any interval [-N, N]. In fact, the 8 which works at N or -N will also work
everywhere else in the interval.
As a final example, consider the function f(x) sin 1/x, or the function whose
graph appears in Figure 18 on page 62. It is easy to see that, so long as e < 1,
there will not be one 8 > 0 that works for these functions at all points a in the
open interval (0, 1).
These examples illustrate important distinctions between the behavior ofvarious
continuous functions on certain intervals, and there is a special term to signal this
distinction.
1
=
DEFINITION
continuous
The function f is uniforrnly
on an interval A if for every e
all
and
such
there is some ô > 0
that, for
x
y in A,
>
0
if |x-y|<8,then|f(x)-f(y)|<e.
We've seen that a function can be continuous on the whole line, or on an open
interval, without being uniformly continuous there. On the other hand, the function f (x) x2 did turn out to be uniformly continuous on any closed interval.
the same sort of thing that occurs when we
This shouldn't be too surprising-it's
ask whether a function is bounded on an interval-and we would be led to suspect
that any continuous function on a closed interval is also uniformly continuous on
that interval. In order to prove this, we'll need to deal first with one subtle point.
Suppose that we have two intervals [a,b] and [b,c] with the common endpoint b, and a function f that is continuous on [a,c]. Let e > 0 and suppose that
=
8, Appendix.UnformContinuig 143
the following two statements
hold:
if x and y are in
if x and y are in
(i)
(ii)
[a,b] and |x
[b,c] and |x
y I < 81, then
-
yl
--
<
82, then
|f (x) f (y)| <
|f (x) f (y)| <
--
-
e,
e.
We'd like to know if there is some & > 0 such that |f (x) f (y)| < e whenever
x and y are points in [a,c] with x
y < 8. Our first inclination might be to
choose ô as the minimum of ô1 and ô2. But it is easy to see what goes wrong
(Figure 2): we might have x in [a,b] and y in [b,c], and then neither (i)nor (ii)
tells us anything about |f (x) f (y)|. So we have to be a little more cagey, and
-
<
a
x
&
a
-
>
c
-
FIGURE
also use continuity
2
LEMMA
of
f
at b.
Let a < b < c and let
suppose that statements
f be
if x and y are in
PROOF
Since f is continuous
continuous
on the interval [a,c]. Let e > 0, and
(ii)hold. Then there is a 8 > 0 such that,
(i)and
[a,c]
and
at b, there
|x
y|
-
is a ôs
<
8, then jf
(x)
f (y)| <
-
e.
0 such that,
>
-b|
if |x
83, then
<
f(x)
--
f(b)
<
.
It follows that
(iii) if |x
-
b|
83 and
<
|y b|
<
-
83, then
|f (x)
-
f (y)|
<
e.
Choose 8 to be the minimum of ôi, 82, and 83. We claim that this ô works. In
fact, suppose that x and y are any two points in [a,c] with |x y| < 8. If x and y
are both in [a,b], then |f (x) f (y)| < e by (i);and if x and y are both in [b,c],
then |f (x) f (y)| < e by (ii).The only other possibility is that
-
-
-
x<b<y
In either case, since [x yl
f (x) f (y)| < e by (iii).
-
y<b<x.
or
ô, we also have
<
|x
-
b|
<
8 and
|y
-
b
<
8. So
-
THEOREM
1
PROOF
If
f
is continuous
on
[a,b], then f
is uniformly
continuous
on
[a,b].
It's the usual trick, but we've got to be a little bit careful about the mechanism of
the proof. For e > 0 let's say that f is e-good on [a,b] if there is some ô > 0 such
that, for all y and z in [a,b],
if
|y
-
z|
<
ô, then
|f (y) f (z)| < e.
f is e-good on [a,b] for all e
--
Then we're trying to prove that
Consider any particular e > 0. Let
A
=
{x: a 5
x
5 b and
f
>
0.
is e-good on
[a,x]}.
above (byb), so
A has a least
Then A ¢ 0 (sincea is in A), and A is bounded
really
should
write
since
and
might
A
œ
œ.
depend on e. But
We
upper bound
as,
b, no matter what e is.
we won't since we intend to prove that œ
=
144 Foundations
Suppose that we had a < b. Since f is continuous at a, there is some ho > 0
if |y a l < So, then |f (y) f (a)| < e/2. Consequently, if |y a| < Bo
and |z a| < So, then |f (y) f(z)| < s. So f is surely e-good on the interval
[a ôo, a + ôo]. On the other hand, since a is the least upper bound of A, it
is also clear that f is e-good on [a,a Bo]. Then the Lemma implies that f is
e-good on [a,a + bo], so a + Bo is in A, contradicting
the fact that a is an upper
bound.
b is actually in A. The
To complete the proof we just have to show that œ
this
practically
the
continuous
for
is
Since
is
at b, there is some
same:
argument
f
E/2.
then
<
<
> 0 such that, if |b
80,
y|
So f is e-good on
|f(y) f(b)|
ao
[b ôo, b]. But f is also e-good on [a,b bo], so the Lemma implies that f is
e-good on [a,b].
such that,
-
-
-
-
-
--
--
=
-
-
-
-
PROBLEMS
1.
following values of a is the function f (x) x" uniformly continuous on [0,oo): a = 1/3, 1/2, 2, 3?
(b) Find a function f that is continuous and bounded on (0, 1], but not
uniformly continuous on (0, 1].
Find
a function f that is continuous and bounded on [0,oo) but which
(c)
is not uniformly continuous on [0,oo).
2.
(a) Prove that
(a) For
which
of the
if
(b) Prove that if
f
=
and g are uniformly continuous on A, then so is f + g.
f and g are uniformly continuous and bounded on A, then
uniformly continuous on A.
is
fg
that this conclusion does not hold if one of them isn't bounded.
Show
(c)
(d) Suppose that f is uniformly continuous on A, that g is uniformly continuous on B, and that f (x) is in B for all x in A. Prove that gof is
uniformly
"bisection
continuous
argument"
on A.
(page140) to give another
3.
Use a
4.
Derive Theorem 7-2 as a consequence
of
Theorem 1.
proof of Theorem 1.
PART
DERIVATIVES
AND
INTEGRALS
In 1604, al the heightof
his scienttßc career, Galileoargued
that fora rectilinear motion
in rchich sped increasesproportionally
Ie distana caeered,
the law of motion should be
just thal (x
=
d2)
which he had discowred
in theinvestigationof faRing bodies.
Between 1695 and 1700
not a single ene of the monthly issues
of
Leiþng's Acta Eruditaram was
without artides of Leibniz,
þublished
the ßernoulli brothers
or the Marquis de (W&pitaltreating,
with natalfon ody slightly dyerent from
that which uw use leday,
the most varied poblemsof
calculus,
dŒtrential
ivitegralcalculus
and the calculus of variations.
Thus in the space of almost precisely
one century
infmitesimalcakulus er,
as we new call it in English,
The Cakulus,
the calculating toolpar exceRenœ,
had beenforged;
and nearly three cenlaries of
constant use ham not completdy dulled
this incomparable instrument.
N[CROLAS
BOEmHAKI
DERIVATIVES
CHAPTER
The derivative of a function is the first of the two major concepts of this section.
Together with the integral, it constitutes the source from which calculus derives
its particular flavor. 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 powerflu concepts that are truly characteristic of calculus.
would say
the interest of the ideas to be introduced
Perhaps (some
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 way. In fact, the
demands of physics were the original inspiration for these 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 discusstheir
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 fmd 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
will
results
and most powerful
about functions
be obtained only when we restrict
which
have even greater claim to be called
our attention even further, to functions
which are even better behaved than most continuous functions.
"certainly")
"reasonable,"
j(x)
=
Ex|,x > o
f(x)=x',x<0
f(x) lx\
=
F I GUREt
(a)
(c)
(b)
which continuous
Figure 1 illustrates certain types of misbehavior
functions can
unlike
the graph of
0),
display. The graphs of these fimetions are
at (0,
each
where
point. The quotation
line" at
Figure 2,
it is possible to draw a
marks have been used to avoid the suggestion that we have defined
or
"bent"
"tangent
"bent"
FIGURE
2
147
148 Derivativesand Integrals
"bent"
"tangent
line," although we are suggesting that the graph might be
at a
where
already
noticed
probably
line" cannot be drawn. You have
point
a
that a tangent line cannot be defined as a line which intersects the graph only
uch a definition would be both too restrictive and too permissive. With
onc
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
"tangent
"bent."
FIGURE
f(x)
3
=
x
(a)
FIGURE
FIGURE
4
(b)
(c)
5
A more promising approach to the definition of a tangent line might start with
lines," and use the notion of limits. If h 0, then the two distinct points
(a, f (a)) and (a + h, f (a + h)) determine, as in Figure 6, a straight line whose
slope is
f (a + h) f (a)
·ý
"secant
-
h
:a k
h, f(4 + h))
I
I
I
f(a +
h)
-
f(a)
I
FIGURE
6
As Figure 7 illustrates, the
some sense, of these
talked about a
"limit"
"tangent
line" at (a,f (a)) seems to be the limit, in
lines," as h approaches
0. We have never before
of lines, but we can talk about the limit of their slopes: the
"secant
9. Derivatives 149
slope of the tangent line though
(a, f (a)) should be
(a + h) f (a)
lim f
-
h
hoo
We are ready for a definition, and some comments.
DEFINITION
The function f is differentiable
f (a + h)
.
hm
if
at a
f (a)
-
h
hoo
.
exists.
In this case the limit is denoted byf'(a) and is called the derivative of f at a.
(We also say that f is differentiable if f is differentiable at a for every a in
the domain of f.)
on our definition is really an addendum; we define the
of f at (a, f(a)) to be the line through (a,f(a))
line
the
graph
tangent
to
with slope f'(a). This means that the tangent line at (a, f(a)) is defined only if
f is differentiable at a.
The second comment refers to notation. The symbol f'(a) is certainly reminiscent of functional notation. In fact, for any function f, we denote by f'the
function whose domain is the set of all numbers a such that f is differentiable
The first comment
I
FIGURE
7
(a,f(a))
at a, and whose
value at such
a number
a
h)
f (a +
is the collection of all pairs
f'
+ h)
lim f (a
(
a,
for which lim [f (a + h)
h->0
of
f (a)
-
h
A-+o
(To be very precise:
is
--
f (a)
h
hoo
f (a)]|h
--
exists.)
The function f' is called the derivative
f.
Our third 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
point O, and a
direction,in which distances from O shall be written as positive numbers, the distance from 0 of points in the other direction being written as negative numbers.
Let s (t) denote the distance of the particle from O, at time t. The suggestive notation s(t) has been chosen purposely; since a distance s(t) is determined for each
"origin"
motion
t=5
i
of the particle
t=4
8(a)
t=1
i-2
t=3
a
e0
FIGURE
t=0
a
I
I
I
line along which particle is moving
150 Derivativesand Integrals
t, the physical situation automatically supplies us with a certain function s.
The graph of s indicates the distance of the particle from O, on the vertical axis,
in terms of the time, indicated on the horizontal axis (Figure 8(b)).
number
"distance"
graph of s
The quotient
s(a+
I
I
"time"
FIGURE
8(b)
h)
s(a)
-
h
I
"average
velocity" of the particle
has a natural physical interpretation. It is the
time
during the
interval from a to a + h. For any particular a, this average speed
depends on h, of course. On the other hand, the limit
s(a+h)
-
s(a)
h
hoo
depends only on a (aswell as the particular function s) and there are important
physical reasons for considering this limit. We would like to speak of the
of the particle at time a," but the usual defmition of velocity is really a definition
of average velocity; the only reasonable
at time a" (so-called
definition of
velocity") is the limit
"velocity
"velocity
"instantaneous
.
lun
hoo
s(a + h)
h
-
s(a)
of the particle at a to be s'(a).
Thus we defmethe (instantaneous)velocity
s'(a)
could easily be negative; the absolute value |s'(a) is sometimes
Notice that
speed.
called the (instantaneous)
realize
that instantaneous velocity is a theoretical concept,
It is important to
which
does not correspond precisely to any observable quantity.
an abstraction
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
for any particular h, but merely the limit of these average velocities as h approaches 0. Thus, when velocities are measured in physics, what a physicist really
measures is an average velocity over some (verysmall) time interval; such a procedure cannot be expected to give an exact answer, but this is really no defect,
because physical measurements
can never be exact anyway.
velocity
of a particle is often called the
of change of its position." This
The
notion of the derivative, as a rate of change, applies to any other physical situation
of change of
in which some quantity varies with time. For example, the
where
of
mass" of a growing object means the derivative
m(t) is
the function m,
"rate
"rate
the mass at time t.
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 have a good
idea of what the derivative of a function looks like. The next chapter is devoted
exclusively to one aspect of this problemaalculating
the derivative of complicated functions. In this chapter we will emphasize the concepts, rather than the
9. Denvatives 151
calculations, by considering a few simple examples.
function, f (x) c. In this case
Simplest of all is a constant
=
f (a + h)
f (a)
-
=
h
hw0
lim
c
c
-
=
h
hoo
0.
0. This means that
Thus f is difTerentiable at a for every number a, and f'(a)
the tangent line to the graph of f always has slope 0, so the tangent line always
=
coincides with the graph.
Constant functions are not the only ones whose graphs coincide with their tangent lines-this happens for any linear function f (x) cx + d. Indeed
=
f'(a)
lim f(a+h)h
=
f(a)
h->0
c(a + h) +d
h
hoo
ch
lim
=
[ca+d]
-
=
-
h
h-vo
c;
the slope of the tangent line is c, the same as the slope of the graph of
A refreshing difference occurs for f (x) x2. Here
f.
=
f'(a)
=
+ h)
lim f (a
h
(a + h)2
h-0
slope 2.
f(x)
-
=
x2
(a, al )
lim
a2 +
-
a2
h
h->0
=
a2
-
h
2ah + h2
ho
=
f (a)
-
lim 2a + h
h->0
=
FIGURE
9
2a.
Some of the tangent lines to the graph of f are shown in Figure 9. In this picture
each tangent line appears to intersect the graph only once, and this fact can be
checked fairly easily: Since the tangent line through (a, a2) has slope 2a, it is the
graph of the function
g(x)=2a(x-a)+a2
2ax
=
Now, if the graphs of
f
and g
intersect at a point (x, f(x))
x2
or
so
or
a2.
-
=
x2
-
2ax a2
2ax + a2
-
=
0;
(x-a)2=0
x
=
a.
In other words, (a, a2) is the only point of intersection.
=
(x, g(x)), then
152 Denvativesand Integrals
The function f (x) x2 happens to be quite special in this regard; usually a
tangent line will intersect the graph more than once. Consider, for example, the
function f (x) x3. In this case
=
=
f'(a)
f(a+h)- f(a)
lim
=
h
h)3
(a +
h->o
lim
=
G3
-
h
h->0
a3+3a2h+3ahi+h3-a3
=lim
h
3a2h + 3ah2 + h3
h->0
h
lim 3a2 + 3ah + h2
h-vo
=hm
=
.
hw0
3a2.
=
Thus the tangent line to the graph of f at (a, a3) has slope 3a2. This means that
the tangent line is the graph of
g(x)
3a2(x a) + a3
3a2x 2a3.
=
-
=
The graphs of
f
-
and g intersect at the point
(x, f(x))
=
(x,g(x))
when
-2a3
x3
x3
or
y2x
3a2x + 2a3
_
-
0.
=
This equation is easily solved if we remember that one solution of the equation
has got to be x a, so that (x a) is a factor of the left side; the other factor can
then be found by dividing. We obtain
=
-
(x
f(x)
-
slope 3a'
-2a
(a,a')
x=
It so happens that x2 + ax
-
(x
-
a)
(x2+
ax
2a2 also has x
-
a)(x
--
=
0.
a as a factor; we obtain finally
=
0.
Thus, as illustrated in Figure 10, the tangent line through (a, a3) also intersects
the graph at the point (-2a,
These two points are always distinct,except
when a
0.
We have already found the derivative of sufficiently many functions to illustrate
the classical, and still very popular, notation for derivatives. For a given function f,
the derivative f' is often denoted by
-8a3).
df(x)
dx
10
--
a)(x +2a)
=
FIGURE
2a2)
-
For example, the symbol
dx2
9. Derivatives 153
denotes the derivative of the function f (x)
parts of the expression
df (x)
dx
=
x2. Needless to say, the separate
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 of two
and
other numbers
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 the notation
df (x)/dx
seems very complicated, in concrete cases it may be shorter; after all, the symbol
dx2/dx is actually more concise than the phrase
derivative of the function
"df(x)"
"dx."
"the
f(x)
=
x2."
The following formulas state in standard
that we have found so far:
Leibnizian
de
--
all the information
0,
=
dx
d(ax + b)
notation
=
a,
dx
dx2
-=2x,
dx
dx3
=
-
3x2
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
the third equation is to be true, then either df(x)/dx must denote f'(x), rather
than f', 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 is
intended; in practice df(x)/dx sometimes means f' and sometimes means f'(x),
while 2x may denote either a number or a function. Because of this ambiguity,
most authors are reluctant
to denote f'(a) by
df (x)
dx
(a);
instead f'(a) is usually denoted by the barbaric, but unambiguous,
df
symbol
(x)
dx
*Leibniz was led to this symbol by his intuitive notion of the derivative, which he considered to be,
of this quotient when h is an
quotients [f (x + h) f(x)]/ h, but the
small" quantity was denoted by dx and the corresponding "infinitely
This "infinitely
small" difference (x+dx)- (x) by df (x). Although this point of view is impossible to reconcile with
f
f
properties (Pl) Pl3) of the real numbers, some people find this notion of the derivative congenial.
not the limit of
small" number.
"value"
--
"infinitely
154 Derivadvesand Integrals
is associated with one more
to these difficulties, Leibnizian notation
ambiguity.
Although the notation dx2/dx is absolutely standard, the notation
df(x)/dx is often replaced
by df/dx. This, of course, is in conformity with the
of
confusing
practice
a function with its value at x. So strong is this tendency that
the fimction
functions are often indicated by a phrase like the following:
x2." We will sometimes follow classical practice to the extent of using
y
y
carefully distinguish between
as the name of a function, but we will nevertheless
the function and its values-thus
the
we will always say something like
x2."
=
y(x)
function
by)
In addition
"consider
=
"consider
(defined
Despite the many ambiguities of Leibnizian notation, it is used almost exclusively in older mathematical
writing, and is still used very frequently today. The
staunchest opponents of Leibnizian notation admit that it will be around for quite
some time, while its most ardent admirers would say that it will be around forever, and a good thing tool In any case, Leibnizian notation cannot be ignored
completely.
The policy adopted in this book is to disallow Leibnizian notation within the
text, but to include it 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 functions examined so far have all been differentiable. 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 first discussed in this chapter, and illustrated in Figure 1; if they
turn out to be differentiable at 0 something has clearly gone wmng.
Consider first f (x)
=
|x|. In this
case
f(0+h)
f(0)
--
[h|
h
h
-1
Now |h|/h
=
1 for h
>
0, and |h||h
for h
=
0. This shows that
<
f (h) f (0) does not
-
lim
hoo
exist.
h
In fact,
(h)
lim f
hoo+
-
f (0)
h
f (h) f (0)
-
and
.
lun
hoo-
-1.
=
h
(These two limits are sometimes called the right-hand
hand derivative, respectively, of f at 0.)
derivative and the left-
9. Denvatives 155
If a ¢ 0, then
f'(a) does exist.
I
In fact,
f'(x)
f'(x)
1
=
if x
if x
-1
=
0,
0.
>
<
The proof of this fact is left to you (itis easy if you remember the derivative of a
linear function). The graphs of f and of f' are shown in Figure 11.
For the function
x2,
f(x)=
f
<0
x
x>0,
x,
a similar difficulty arises in connection with
f'(0). We have
h2
FIGURE
11
h,
=
-
f (h) f (0)
-
h
h
h
h
h
<
0
-=1,
h>0.
Therefore,
hm f(h)
f(0)
-
.
(h)
lun f
f (0)
-
.
but
Thus
f'(0) does not
exists
for x † Rit
h->o+
=0,
h
h-+0-
=
h
1.
exist; f is not differentiableat 0. Once again, however, f'(x)
is easy to see that
f'(x)
2x,
1,
x
x
f
=
<
>
0
0.
The graphs of f and f' are shown in Figure 12.
For this function
Even worse things happen for f(x)
=
f.
&
1
h
d'
--=-
FIGURE
12
f(h) f(0)
-
_
¯
h
B
h
In this case the right-hand
-
-1/¾
what's
FIGURE
13
more,
(Figure 13).
1
h
-
N'
<
0.
limit
f (h) f (0)
does not exist; instead
h>0
1/J
1
becomes arbitrarily large as h approaches 0. And,
becomes arbitrarily large in absolute value, but negative
156 Derivativesand Integrals
The function f (x)
W, although
better behaved than this. The quotient
f (h) f (0)
-----.......
-
diEerentiable at 0, is at least a little
not
=
hij3
W
1
1
simply becomes arbitrarily large as h goes to 0. Sometimes one says that f has
derivative at 0. Geometrically this means that the graph of f has a
an
line" which is parallel to the vertical axis (Figure 14). Of course, f (x)
"infinite"
--.....
"tangent
has the same geometric property, but one would say that f has a derivative
infinity" at 0.
Remember that differentiability is 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 remains to be
-
of
FIGURE
=
14
N
"negative
noted:
THEOREM
1
If f is differentiable at a, then f is continuous at a.
PROOF
lim f (a + h)
hw0
-
f (a)
=
lim
f (a +
h
(a + h)
h
hoo
=
h)
lim f
hoo
=
f'(a) 0
=
0.
-
-
f (a)
-
h
f (a) lim h
-
ha0
-
As we pointed out in Chapter 5, the equation lim f (a+h)- f (a) = 0 is equivalent
h->0
to lim
f (x)
=
f (a); thus f
is continuous
at a.
It is very important to remember Theorem 1, and just as important to remember
that the converse is not true. A differentiable function is continuous, but a continuous function need not be differentiable (keep
in mind the function f (x) |x|,
which
and you will never forget
statement is true and which false).
The continuous functions examined so far have been differentiable at all points
with at most one exception, but it is easy to give examples of continuous functions
which are not differentiable at several points, even an infinite number (Figure 15).
Actually, one can do much worse than this. There is a function which is contmuous
=
FIGURE
15
9. Derivatives 157
FIGURE
(a)
(b)
(c)
(d)
16
dgiventiablenowhee! Unfortunately, the definition of this function will
be inaccessible to us until Chapter 24, and I have been unable to persuade the
artist to draw it (consider
carefully what the graph should look like and you will
with
sympathize
her point of view). It is possible to draw some rough approximations to the graph, however; several successively better approximations
are shown
in Figure 16.
must be postponed,
Although such spectacular examples of nondifferentiability
with
which
continuous
ingenuity,
find
function
is not differentiable
a
we can,
a little
all
such
ofwhich are in [0, 1]. One
function is illustrated in
at infmitely many points,
Figure 17. The reader is given the problem of defining it precisely; it is a straight
line version of the function
everywhere and
,r
'
,7'
,
,/
-1
-I
/
s
#
1
xsin-,
0,
FIGURE
17
1
x¢0
x
=
0.
This particular function f is itself quite sensitive to the question of differentiability.
Indeed, for h ¢ 0 we have
158 Derivativesand Integrals
h sin
1
-
---
0
I
h
h
h
Long ago we proved that lim sin 1/h does not exist, so f is not differentiable at 0.
h->0
Geometrically, one can see that a tangent line cannot exist, by noting that the
secant line through (0, 0) and (h, f (h)) in Figure 18 can have any slope between
and 1, no matter how small we require h to be.
f (h) f (0)
-
.
sin
-.
-1
x=0
0,
FIGURE
18
This finding represents something of a triumph; although continuous, the funcand we can now enunciate one mathtion f seems somehow quite unreasonable,
ematically undesirable feature of this function-it is not differentiable at 0. Nevabout the criterion of differenertheless, one should not become too enthusiastic
example,
tiability. For
the function
x2sin-,
8(x)
=
1
0,
is differentiable at 0; in fact g'(0)
lim
h-wo
g(h)
X
fÛ
x
=
x
=
0:
=
h
lim
h->o
=
=
-
h
h
lim h sin
ha0
The tangent line to the graph of g at
1
h2 sin
g(0)
-
0
1
-
h
0.
(0,0) is therefore
the horizontal axis (Fig-
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 formulate
such conditions if we introduce another set of defmitions, the last of this chapter.
ure
9. Derivatives 159
/
/
FIGURE
f(x)
=
x
x=0
0
19
For any function f, we obtain, by taking the derivative, a new function f' (whose
domain may be considerably smaller than that of f). The notion of differentiability can be applied to the function f', of course, yielding another function (f')',
whose domain consists of all points a such that f' is differentiable at a. The function (f')' is usually written simply f" and is called the second derivative of f.
If f"(a) exists, then f is said to be 2-times differentiable at a, and the number
f"(a) is called the second derivative of f at a.
In physics the second derivative is particularly important. If s(t) is the position at time t of a particle moving along a straight line, then s"(t) is called the
acceleration
at time t. Acceleration plays a special role in physics, because, as
stated in Newton's laws of motion, the force on a particle is the product of its
mass
and its acceleration. Consequently you can feel the second derivativewhen you
sit in an accelerating car.
There is no reason to stop at the second derivative-we can define f"' = (f"y,
f"" = (f'")', etc. This notation rapidly becomes unwieldy, so the following abbreviation is usually adopted
is really a recursive definition):
(it
f 0) f
f(***) U(k) r
'
=
,
--
Thus
JU)
fí2)
f*
f
(4)
-
-
=
_
J'
f"
f')'
f"'= (f")
ur
peu
-
_
,
i
etc.
The various functions f(k), for k ;> 2, are sometimes called higher-order
derivatives of f.
Usually, we resort to the notation f (k) only for k > 4, but it is convenient to
have f f*) defined for smaller k also. In fact, a reasonable definition can be made
for f(°),namely,
f(0) f
-
160 Denvativesand Integrals
Leibnizian notation for higher-order derivatives should
Leibnizian symbol for f"(x), namely,
also
be mentioned.
The
natural
(df(x)
d
is abbreviated
'
to
d2f(x)
-x,
f(x)
dx
dx
¯
(dx)2
d2f(x)
dx2
frequently to
or more
,
Similar notation is used for f(*) xL
The following example illustrates the notation f(k),and also shows, in one very
simple case, how various higher-order derivatives are related to the original funcx2. Then,
tion. Let f(x)
as we have already checked,
=
(a)
f'(x)
f"(x)
f'(x)
=
f
2x
X
2x,
2,
Û,
=
=
-
f*(x)
if k
0,
=
3.
>
Figure 20 shows the function f, together with its various derivatives.
A rather more illuminating example is presented by the following function,
whose graph is shown in Figure 21(a):
/
(b)
x2,
f (x)
f"(x)
-
2
x
>
x
<
if a
if a
>
-x
,
0
0.
It is easy to see that
f'(a)
2a
=
--2a
f'(a)
=
0,
0.
<
Moreover,
(h)
hm f
f (0)
-
.
f'(0)
(e)
=
h-+o
h
lim f(h)
h
=
-.
h->o
Now
f (x)
-
0, k
>
3
(h)
lim f
hoo+
h
h2
.
hm
=
---
0
=
-
h
how
-h2
and
(d)
FIGURE
20
f (h)
.
hm
.
hm
=
h
h-+0-
=
-
h
h--o-
SO
lim f (h) 0.
h
all
summarized
information
be
This
as follows:
can
J'(0)
=
=
-
h->o
f'(x)
=
2 |x|.
0,
9. Derivatives 161
f(x)
It follows that f"(0) does not exist! Existence of the second derivative is thus a
looking" function
strong criterion for a function to satisfy. Even a
when
reveals
examined
with
second
the
like f
derivative. This
some irregularity
suggests that the irregular behavior of the function
"smooth
rather
=
x2 sin 1
-,
g(x)
=
x
0,
x
x
¢0
=
0
be revealed by the second derivative. At the moment we know that
g'(0) = 0, but we do not know g'(a) for any a ¢ 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.
might
(a)
also
PROBLEMS
1.
J'(x)
=
2|x|
directly from the definition, that if f (x) = 1/x, then
Î |a2, for a ¢ 0.
j'(a)
that
the tangent line to the graph of f at (a, 1/a) does not intersect
(b) Prove
the graph of f, except at (a, 1/a).
(a) Prove, working
-
-
-2/a3
2.
1/x2, then f'(a)
for a ¢ 0.
1/a2)
that
the
Prove
at
line
intersects
tangent
to
f at one other
f (a,
(b)
vertical
axis.
which
the
opposite
side
of
the
lies
point,
on
3.
Pove that if f (x) J, then f'(a) 1/(24), for a > 0. (The expression
f (a)]/h will require some algebraic face lifting,
you obtain for [f (a + h)
but the answer should suggest the right trick.)
(b)
f"(x) 2, x > o
=
(a) Prove that if f(x)
=
=
=
=
-
4.
1,
For each natural number n, let S.(x) = x". Remembering that Si'(x)
3X2, conjecture
formula
S.'(x).
S2'(x)
for
Prove
2x, and S3'(x) =
a
your
conjecture. (The expression (x + h)" may be expanded by the binomial
=
=
--2,
I"(x)
theorem.)
x < 0
-
(c)
FIGURE
21
5.
Find
6.
Prove, starting from the definition
f'
if
f (x) [x].
=
(anddrawing a picture
to illustrate):
f (x) + c, then g'(x) f'(x);
cf (x), then g'(x)
cf'(x).
7. Suppose that f (x) x3
(a) What is f'(9), f'(25), f'(36)?
f'(629
(b) What is f'(32), f'(52),
r 2
f'(a2
What
is
(c)
(a) if g(x)
(b) if g(x)
=
=
=
=
=
If you do not find this problem silly, you are missing a very important point:
f'(x2) means the derivative of f at the number which we happen to be
calling x2; it is not the derivative at x of the function g(x)
f(x2). Justto
drive the point home:
=
(d) For f(x)
=
x3,
compare
f'(x2) and
g'(x) where g(x)
=
f(x2L
162 Derivativesand Integrals
8.
definition) that g'(x)
c).
picture
this.
this problem you must
illustrate
To
do
Draw
+
to
a
f'(x
write out the defmitions of g'(x) and
c)
correctly.
The purpose
f'(x +
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 g(x)
f(x + c). To emphasize this
(a) Suppose g(x)
from the
f(x+c). Prove (starting
==
=
=
point:
(b) Prove that if g(x) f(cx), then g'(x) c. J'(cx). Try to see pictorially
why this should be true, also.
(c) Suppose that f is differentiable and periodic, with period a (i.e.,
f (x + a) f (x) for all x). Prove that f' is also periodic.
=
=
=
9.
Find
or
and also
f'(x)
you
will surely
f'(x +
3) in the following cases. Be very methodical,
Consult the answers (afteryou do the
slip up somewhere.
problem, naturally).
(i) f (x)
=
(ii) f (x +
(iii) f (x +
(x + 3)*.
3)
3)
=
x5.
=
(x + 5)'.
10.
Find f'(x) if f(x)
be the same.
11.
Galileo was wrong: if a body falls a distance s(t) in t seconds,
and s' is proportional to s, then s cannot be a function of the form
s(t) = ct2.
(b) Prove that the following facts are true about s if s(t) (a/2)t2 (thefirst
fact will show why we switched from c to a/2):
g(t + x), and if
=
f(t)
=
g(t + x). The answers will not
(a) Provethat
=
(i)
s"(t)
=
a
(ii) [s'(t)]2
=
(theacceleration
is constant).
2as(t).
(c) 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 400-foot
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 moving with half that
speed?
12.
Imagine a road on which the speed limit is specified at every single point. In
other words, there is a certain function L such that the speed limit x miles
from the beginning of the road is L(x). Two cars, A and B, are driving along
this road; car A's position at time t is a(t), and car B's is b(t).
equation expresses the fact that car A always travels at the speed
limit? (The answer is not a'(t)
L(t).)
always
that
speed limit, and that B's position at
A
the
Suppose
at
(b)
goes
time t is A's position at time t 1. Show that B is also going at the speed
limit at all times.
(c) Suppose B always stays a constant distance behind A. Under what conditions will B still always travel at the speed limit?
(a) What
=
-
9. Derivatives 16 3
13.
g(a) and that the left-hand derivative of f at a equals
Suppose that f(a)
the right-hand
derivative of g at a. Define h(x)
f(x) for x 5 a, and
h(x)
g(x) for x > a. Prove that h is differentiable at a.
=
=
=
14.
*15.
Let f (x) x2 if x is rational, and f (x) 0 if x is irrational. Prove that
f is differentiable at 0. (Don't be scared by this function. Justwrite out the
definition of /'(0).)
=
=
be a function such at |f (x)| 5 x2 for all x. Prove that f is
differentiable at 0. (If you have done Problem 14 you should be able to
do this.)
result can be generalized if x2 is replaced by Ig(x), where g has
This
(b)
what property?
(a) Let f
1. If f satisfies f(x)| 5 x|a, prove that f is differentiable at 0.
16.
Let
17.
Let 0 < ß < 1. Prove that if
is not differentiable at 0.
*18.
œ >
f
satisfies
f (x)
>
|x
and
f (0)
0, then
=
f
Let f (x) 0 for irrational x, and 1/q for x
p/q in lowest terms. Prove
that f is not differentiable at a for any a. Hint: It obviously suffices to prove
this for irrational a. Why? If a
is the decimal expansion
m.aia2a3...
of a, consider [f (a + h) f (a)]/h for h rational, and also for
=
=
=
-
-0.00...0a
h
19.
ay2....
h(a), that f(x) 5 g(x) 5
f(a) g(a)
that g is differentiable
h'(a).
Prove
f'(a)
that
h(x) for all x,
at a, and that
with
of
the definition
g'(a).)
f'(a) g'(a) h'(a). (Begin
the
conclusion
omit
the hypothesis
that
not
does
follow
if
Show
we
(b)
f (a) g(a) h(a).
(a) Suppose
and that
=
=
=
=
=
=
20.
=
=
Let f be any polynomial function; we will see in the next chapter that f
is differentiable. The tangent line to f at (a, f(a)) is the graph of g(x)
f'(a)(x a) + f(a). Thus f(x) g(x) is the polynomial function d(x)
if f(x) x2, then
f(x) f'(a)(x a) f(a). We have already seen that
d(x)
(x a)2, and if f(x) x3, then d(x)
(x a)2(x + 2a).
=
-
-
-
=
=
-
=
-
-
=
=
-
d(x) when f (x) x4, and show that it is divisible by (x a)2.
(b) There certainly seems to be some evidence that d(x) is always divisibleby
Figure 22 provides an intuitive argument: usually, lines parallel
(x
to the tangent line will intersect the graph at two points; the tangent line
interSects
the graph only once near the point, so the intersection should
be a
intersection." To give a rigorous proof, first note that
(a) Find
=
-
-a)2.
FIGURE
22
"double
d(x)
f (x) f (a)
-
=
x-a
-
f'(a).
x-a
Now answer the following questions. Why is f (x) f (a) divisible
by (x a)? Why is there a polynomial function h such that h(x)
-
-
164 Derivativesand Integrals
a) for x ¢ a? Why is lim h(x)
d(x)/(x
does this solve the problem?
-
0? Why is h(a)
=
0? Why
=
--a).
21.
(a) Show that f'(a) lim[f(x)
(b) Show that derivatives are a
=
"local
some open interval containing
in computing f'(a), you can
course you can't ignore
*22.
that
(a) Suppose
f(x)
(Nothing deep here.)
f(a)]/(x
-
property": if f(x)
g'(a).
a, then f'(a)
g(x) for all x in
(This means that
ignore f (x) for any particular x ¢ a. Of
for all such x at once!)
=
=
f is differentiable at x. Prove that
(x + h) f (x h)
J'(x) hm f
--
--
.
=
Hint: Remember an old algebraic trick-a number is not changed if the
same quantity is added to and then subtracted from it.
Prove, more generally, that
**(b)
f'(x)
*23.
.
2h
hoo
lim f(x+h)- f(x-k)
h+ k
=
.
h,k->o+
Prove that if f is even, then f'(x) =
f'(--x). (In order to minimize cong'(x)
and then remember what other thing g
fusion, let g(x)
f(-x); find
is.) Draw a picture!
-
=
*24.
Prove that if
is odd, then
f
f'(x)
=
25.
Problems 23 and 24 say that f' is even if
What can therefore be said about f(k)p
26.
Find
=
x3.
=
x5.
(iii) f'(x)
(iv) f (x +
If S.(x)
3)
=
f
is even.
x3.
x", and
=
0
sk
5 n, prove that
Xn-k
=
(n
=
*29.
is odd, and odd if
x4
=
S,(k)(X)
*28.
f
f"(x) if
(i) f (x)
(ii) f (x)
27.
Once again, draw a picture.
f'(-x).
(a) Find f'(x) if f(x)
(b) Analyze f similarly
=
k)!
-
xn-k
kl
|x|3. Find f"(x). Does f"'(x) exist for all x?
if f (x) x* for x > 0 and f (x)
for xs 0.
-x*
=
=
for x > 0 and let f(x) 0 for xs 0. Prove that f("¯U exists
(andfind a formula for it), but that f(")(0)does not exist.
Let
f(x)
=
x"
=
9. Denvatives 165
30.
Interpret the following specimens of Leibnizian notation;
ment of some fact occurring in a previous problem.
(i)
-
nxn-1
=
dx
dz
(ii)
=
--
-
-
dy
(iv)
1
if z
y
d[f(x)+c]
dx
d[cf(x)]
(iii)
(v)
dx"
=
=
c
dx
dz
dy
dx
dx
-=-
dx
(vii)
.
y
df(x)
dx
df(x)
.
.
dx
x-a2
df(x+a)
df(x)
=
dx
df(cx)
dx
x=b+a
df(x)
=
df(cx)
c
.
dX
x=cb
df(y)
=
dx
dx
x=b
(viii)
(ix)
1
3a4
=
-
-.
if z=y+c.
dx3
(vi)
=
c
.
dy
y=cz
each is a restate-
DIFFERENTIATION
CHAPTER
The process of finding the derivative of a function is called dyerentiation.From the
previous chapter you may have the impression that this process is 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 be lost.
Nevertheless, in this chapter we will learn to differentiate a large number of functions, without the necessity of even recalling the defmition. 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 of a few simple functions, and then prove
theorems about the sum, products, quotients, and compositions of differentiable
of a computation
functions. The first theorem is merely a formal recognition
carried out in the previous chapter.
THEOREM
1
If f is a constant function, f (x)
f'(a)
PROOF
f'(a)
=
c, then
=
=
for all numbers
0
lim f(a+h)h
f(a)
a.
lim c-c
h
=
hoo
hoo
=
0.
The second theorem is also a special case of a computation in the last chapter.
THEOREM
2
If f is the identity function, f (x)
f'(a)
PROOF
j'(a)
=
-
x, then
=
1 for all numbers
f (a +
li
f (a)
a+h-a
.
h
hoo
=
-
h
hoo
=hm
h)
a.
h
lim
Aso h
-
=
1.
The derivative of the sum of two functions is just what one would hopathe
sum of the
166
derivatives.
10. Dg'erentiation 167
THEOREM
3
If f and g are differentiable at a, then f + g is also differentiable at a, and
g)'(a)
(f +
PROOF
(f
+g)(a+h)
(f
.
# g)'(a)
11m
=
-
h)-
f(a+h)+g(ak
lim
f(a+h)
lim
f(a)
-
f (a +
lim
hoo
h)
h
f (a) +
-
g(a+h)
+
h
hoo
g(a)
-
h
g(a + h)
lim
g(a)
---
h
hoo
+ g'(a).
J'(a)
=
[f(a)+g(a)]
h
hw0
=
+g)(a)
(f
h
hoO
=
g'(a).
f'(a) +
=
The formula for the derivative of a product is not as simple as one might wish,
but it is nevertheless pleasantly symmetric, and the proof requires only a simple
algebraic trick, which we have found useful before-a number is not changed if
the same quantity is added to and subtracted from it.
THEOREM
4
If
f
and g are difTerentiable at a, then
(f
-
g)'(a)
f'(a)
=
(f
·
g)'(a)
=
(f
Em
hoo
=
lim
f
f (a)
-
g'(a).
-g)(a)
-g)(a+h)
PROOF
g(a) +
-
(f
-
h
(a + h)g(a + h)
f (a)g(a)
-
h
hoo
-g(a)]
=
f(a+h)[g(a+h)
lim
hw0
=
=
lim f
hw0
h
g(a + h)
lim
(a + h) hoo
h
+
g(a)
---
-
[f(a+h)- f(a)]g(a)
h
+ lim
f (a +
h)
--
f (a)
h
hoo
·
lim g(a)
hw0
f(a).g'(a)+ f'(a)-g(a).
(Notice that we have used Theorem 9-1 to conclude
that lim
hw0
f(a +
h)
=
f (a).)
In one special case Theorem 4 simplifies considerably:
THEOREM
s
If g(x)
=
cf(x)
and
f is differentiable at
g'(a)
PROOF
If h(x)
=
c, so that g
=
h
-
c
-
f'(a).
then by Theorem 4,
f,
g'(a)
=
a, then g is differentiable at a, and
=
=
=
=
(h f)'(a)
-
f'(a) + h'(a) f (a)
f'(a) + 0 f (a)
h(a)
c
c
-
-
-
f'(a).
168 Denvatives and Integrals
Notice, in particular, that (- f)'(a)
(f
+
[-g])'(a) f'(a)
=
=
and consequently
f'(a),
--
(f
-
g)'(a)
=
g'(a).
-
To demonstrate what we have already achieved, we will compute the derivative
of some more special functions.
THEOREM
6
If
f(x)
In
=
for some natural number
f'(a)
PROOF
=
n, then
na"¯'
for all a.
The proof will be by induction on n. For n = 1 this is simply Theorem 2. Now
assume that the theorem is true for n, so that if f (x) x", then
=
f'(a)
Let g(x)
=
x"**.
If I(x)
=
=
f
·
I. It follows
g'(a)
=
na"¯1
for all a.
x"+1
x, the equation
g(x)=
thus g
=
=
x"
-
x can be written
forallx;
f(x)-I(x)
from Theorem 4 that
(f I)'(a)
-
f'(a)
=
-
I(a) +
=na"¯'-a+a"-1
f (a)
-
I'(a)
na" + a"
=
(n + 1)a",
for all a.
This is precisely the case n + 1 which we wished to prove.
Putting together the theorems proved so far we can now fmd f' for f of the
form
f(x)=anx"+an-1x"¯'+---+a2x2&GII
0-
We obtain
f'(x)=nanx"¯1+(n--1)an-ix"¯2+···+2a2X#at.
We can also find
f"(x)
=
f":
n(n
-
1)anx"¯2 + (n
1)(n
-
--
2)an-1x"¯3 +
-
-
-
+2a2.
This process can be continued easily. Each differentiation reduces the highest
power of x by 1, and eliminates one more at. It is a good idea to work out the
derivatives f"', f(4), and perhaps f , until the pattern becomes quite clear. The
last interesting derivative is
f
">(x)
n!an;
=
fork>nwehave
ff*)(x) 0.
=
Clearly, the next step in our program is to find the derivative of a quotient f/g.
It is quite a bit simpler, and, because of Theorem 4, obviously sufficient to find
the derivative of 1/g.
10. Djerentiation 169
THEOREM
7
¢ 0, then 1/g is differentiable at
If g is differentiable at a, and g(a)
'
-g'(a)
(1
(a)
-
g
PROOF
a, and
=
.
[g(a)]2
Before we even write
h)
(a +
(a)
-
h
is necessary to check that
be sure that this expression makes sense-it
sufficiently
requires
only two observations.
small
h.
This
defined
for
(1/g)(a+h) is
Since g is, by hypothesis, differentiable at a, it follows from Theorem 9-1 that g is
continuous at a. Since g(a) ¢ 0, it follows from Theorem 6-3 that there is some
8 > 0 such that g(a + h) ¢ 0 for |h| < 8. Therefore (1/g)(a + h) doesmake sense
for small enough h, and we can write
we must
(1
-
lim
hw0
g
(a + h)
1
1
(a)
-
-
g
1
-
=
h
lim
g(a + h)
.
hm
hw0
g(a)
h
a->o
=
--
g(a)
h[g(a)
g(a + h)
g(a + h)]
--
-
-[g(a+h)-g(a)]
l
.
=lun
g(a)g(a
h
h-so
+ h)
-[g(a+h)-g(a)]
1
.
=
lim
lun
-
h
h->O
g(a)
g(a + h)
1
-g'(a)
=
h->o
-
·
.
[g(a)]2
(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:
times derivative of top, minus top
times derivative of bottom, over bottom squared.")
"bottom
THEOREM
a
If
f
and g are
differentiable at a and g(a) ¢ 0, then
is differentiable at a,
f/g
and
f
(
-
g
'
(a)
g(a)
=
-
J'(a) f (a)
[g(a)]2
-
-
g'(a)
.
170 Derivativesand Integrals
PROOF
Since
f/g
=
f (1/g) we have
-
(a)
f
=
=
(a)
-
f'(a)
(1
g
f'(a)
=
(a) +
-
-
f'(a)
-
(a)
g
f (a)(-g'(a))
+
g(a)
f (a)
'
1
-
·
[g(a)]2
g(a)
f(a)·g'(a)
--
[g(a)]2
We can now differentiate a few more functions. For example,
x2
if f(x)
if f (x)
=
=
if f (x) =
1
then f'(x)
x2 + 1
(x2+ 1)(2x) (x2 1)(2x)
(x2+ 1)2
-
,
x
then
x2 4 1
f'(x)
,
1
then
-,
f'(x)
x
-
=
(x2 + 1)
=
--
-
x(2x)
1
1
(x2+ 1)2
(-1)x¯2.
-
xy
x2
;
(x2 })2
=
(x2 ()2
-
--
4x
-
Notice that the last example can be generalized: if
f (x)
=
x¯"
=
1
-,
for some natural number n,
x"
then
-nx"¯1
f'(x)
-
-
(-n)x¯"¯1.
x
thus Theorem 6 actually holds both for positive and negative integers. If we inter0, then
pret f (x) x to mean f (x) 1, and f'(x) 0 x¯I to mean f'(x)
is necessary because it is
Theorem 6 is true for n 0 also. (The word
0¯I
is meaningless.)
not clear how 0° should be defined and, in any case, O
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:
=
=
=
=
-
"interpret"
=
-
sin'(a)
=
cos a
cos'(a)=-sina
for all a,
foralla,
This information allows us to differentiatemany other functions. For example, if
f (x)
=
x sinx,
then
f'(x)
f"(x)
=
xcosx +sinx,
sin x + cos x + cos x
sin x + 2 cos x;
-x
=
-x
=
10. DgFerentiation171
if
g(x)
sin*x
=
sinx
=
sinx,
-
then
g'(x)
sin x cos x + cos x sin x
2 sin x cos x,
2[(sinx)(-sinx) + cosx cosx]
2[cos2x sin2x];
=
=
g"(x)
=
=
--
if
h(x)
cos2 x
=
=
cos x x,
-
then
h'(x)
sin x) +
(cosx)(-
=
sin x) cos x
(-
-2sinxcosx,
=
-2[cos2
h"(x)
=
x
-
sin2
Notice that
g'(x) + h'(x)
0,
=
1. As we would expect, we
hardly surprising, since (g + h)(x) = sin2 x + COS2 x
also have g"(x) + h"(x)
0.
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 f g h is an abbreviation for
=
=
-
(f
-
-
g) h
-
or
f (g
-
-
h).
Choosing the first of these, for example, we have
(f g h)'(x)
-
-
(f g)'(x) h(x) + (f 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).
=
-
=
=
-
-
The choice of f (g h) would, of course, have given the same result, with a
different intermediate step. The final answer is completely symmetric and easily
-
-
remembered:
sum of the three terms obtained
and multiplying by the other two.
(f g h)' is the
·
-
g, and h
by differentiating each of f,
For example, if
f(x)
x3 sinx
=
cosx,
then
f'(x)
3x2 sinx cosx + x3 COsX
=
COsI
-|-13(SinX)(-
Products of more than 3 functions can be handled similarly.
should have little difficulty deriving the formula
sinX).
For example, you
.g·h-k)'(x)
(f
=
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).
172 Derivativesand Integrals
You might even try to prove
(fi
.....
f.)'(x)
(byinduction) the general formula:
f;_1(x)fi'(x)fi+1(x)··--·fn(x)fi(x)
=
-...-
i=l
Differentiating the most interesting functions obviously requires a formula for
(fo g)'(x) in terms of f' and g'. To ensure that fog be differentiable at a, one
reasonable
hypothesis would seem to be that g be differentiable at a. Since the
behavior of fog near a depends on the behavior of f near g(a) (notnear a), it
also seems reasonable
to assume that f is differentiable at g(a). Indeed we shall
prove that if g is differentiable at a and f is differentiable at g(a), then fog is
differentiable at a, and
(fo
g)'(a)
f'(g(a))
=
·
g'(a).
This extremely important formula is called the Chain Ruk, presumable because
of functions. Notice that
a composition of functions might be called a
and
g)'
the
of
g',
is
not
practically
product
but
quite:
(fo
f' must be evaluated
f'
attempting
and
g' at a. Before
at g(a)
to prove this theorem we will try a few
applications. Suppose
"chain"
f (x)
=
Let us, temporarily, use S to denote the
f
sin x2.
function
("squaring")
sin
=
o
S(x)
=
x2.
Then
S.
Therefore we have
f'(x)
sin'(S(x))
=
=
Quitea
different result is obtained
cos
x2
-
S'(x)
2x.
-
if
f(x)
=
sin2x.
In this case
f
=
So sin,
so
f'(x)
=
=
S'(sin x) sin'(x)
2 sin x cos x.
·
-
sin sin
Notice that this agrees (asit should) with the result obtained by writing f
and using the product formula.
function,
Although we have invented a special symbol, S, to name the
it does not take much practice to do problems like this without bothering to write
down special symbols for functions, and without even bothering to write down the
particular composition which f is-one soon becomes accustomed to taking f
apart in one's head. The following differentiations may be used as practice for
such mental gymnastics-if you find it necessary to work a few out on paper, by
all means do so, but try to develop the knack of writing f' immediately after seeing
=
"squaring"
·
10. Dyerentiation 173
the definition of f; problems of this sort are so simple that, if you
the Chain Rule, there is no thought necessary.
=
sinx'
f(x)
=
sin3x
f (x)
=
sin
f (x)
=
f(x)
f(x)
if f(x)
then f'(x)
=
cosx3-3x2
f'(x)
=
3sin2x.cosx
just remember
-1
1
1
J'(x)
=
x
sin(sin x)
f'(x)
=
=
sin(x3 + 3x2)
J'(x)
=
(x3+ 3x2)53
r(x)
-
cos
-
·
-
x2
x
cos(sin x) cos x
.
=
cos(x2 + 3x2)
=
53(x3 + 3x2)52
-
(3x2+ 6x)
2
A function like
f(x)
sin2X2
=
[SinX2
=
2
which is the composition of three functions,
f=SosinoS,
can also be differentiated by the Chain Rule. It is only necessary to remember
that a triple composition fogoh means (fo g) oh or fo (go h). Thus if
f (x)
sin x
=
we can write
f
=
(So sin)
f
=
So
o
(sin
o
S,
S).
The derivative of either expression can be found by applying the Chain Rule
twice; the only doubtful point is whether the two expressions lead to equally simple
calculations. As a matter of fact, as any experienced differentiator knows, it is much
better to use the second:
f
=
So
(sin o S).
We can now write down f'(x) in one fell swoop. To begin with, note that the first
function to be differentiated is S, so the formula for f'(x) begins
Inside the parentheses we must put sin x2, the value at x of the second function,
sin o S. Thus we begin by writing
-
f'(x)
=
2 sin x2
necessary, after all). We must now multiply this
much of the answer by the derivative of sin oS at x; this part is easy-it involves a
composition of two functions, which we already know how to handle. We obtain,
(theparentheses
weren't
really
for the final answer,
f'(x)=2sinx2-cosxt2x.
174 Daivativesand Integrals
The following example is handled similarly. Suppose
f(x)
sin(sinx2L
=
Without even bothering to write down f as a composition gohok of three functions,
we can see that the left-most one will be sin, so our expression for f'(x) begins
f'(x)
cos(
=
)
·
.
Inside the parentheses we must put the value of ho k(x); this is simply sin x2 (what
x2) by deleting the first sin). So
our expression for f'(x) begins
you get from sin(sin
f'(x)
cos(sinx2
=
We can now forget about the first sin in sin(sinx2); we have to multiply what we
have so far by the derivative of the function whose value at x is sin x2-which is
again a problem we already know how to solve:
f'(x)
cos(sin x2)
=
-
cos
x2
-
2x.
Finally, here are the derivatives of some other functions which are the composition
You can probably just
S, as well as some other triple compositions.
of sin and
"see"
that the answers are correct-if
if
f(x)
=
not, try writing
f'(x)
then
sin((sinx) )
x)]2
f'(x)
f'(x)
f'(x)
f (x) [sin(sin
f (x) sin(sin(sin x))
f (x) sin2(x sin x)
=
=
=
=
as a composition:
f
out
cos((sinx)2).2sinx
-
2 sin(sin x) cos(sin x) cos x
=
cos(sin(sin x))
=
2 sin(x sin x) cos(x sin x)
·
-
-
cos(sin x)
=
sin(sin(x
sin x))
f'(x)
=
·
cos x
-
-
f (x)
cosx
=
cos(sin(x
sin x))
-
[sinx +
cos(x2 sin x)
[2xsin x +
-
x cos x]
x2 cosx].
The rule for treating compositions of four (oreven more) functions is easyput in parentheses starting from the right,
(mentally)
always
fo(go(hok)),
and start reducing the calculation
number of functions:
For example,
to the derivative of a composition
if
f(x)
sin2(sin2(x))
=
[f
=
So sin oSo sin
=
So
(sino (So
sin))]
then
f'(x)
=
2 sin(sin2 x) cos(sin2 x) 2 sin x cos x;
-
-
-
of a smaller
10. Dz§erentiation175
if
f (x)
2)
sin((sin x2
=
(f
sin
sin
=
=
oSo sin oS
(So
o
(sin o S))]
then
f'(x)
cos((sin x2)2)
=
-
2 sin x2 cos x2 2x;
-
-
if
f (x)
sin2(sin(sin
=
[fillin yourself,
x))
if necessary]
then
-cos(sin(sinx))
f'(x)
2sin(sin(sinx))
=
-
cos(sinx)
-
cosx.
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 at the end
of the chapter, and it is now high time that we proved the Chain Rule.
The following argument, while not a proof, indicates some of the tricks one
might try; as well as some of the difficulties encountered.
We begin, of course,
with the definition(fo g)(a + h) (fo g)(a)
(fo g)'(a) = lim
-
h
h->0
f(g(a+h))
lo
=
f(g(a))
-
h
hoo
Somewhere in here we would like the expression
put it in by fiat:
(g(a + h))
hm f
f (g(a))
-
.
(g(a + h))
lun f
-
.
=
h
h->o
for g'(a).
g(a + h)
hoo
--
f (g(a))
g(a)
One approach is to
g(a + h)
h
g(a)
-
.
This does not look bad, and it looks even better if we write
.
hm
og)(a+h)
(f
-
(f
og)(a)
h
hoo
=
f (g(a) + [g(a+
lim
h)
g(a + h)
h->0
g(a)])
-
-
f (g(a)) hm
h->0
.
g(a + h)
-
g(a)
g(a)
.
h
The second limit is the factor g'(a) which we want. If we let g(a + h)
(tobe precise we should write k(h)), then the first limit is
f(g(a)+k)h->o
-
-
--
g(a)
=
k
f(g(a))
k
It looks as if this limit
be f'(g(a)), since continuity of g at a implies that k
goes to 0 as h 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 h ¢ 0 we might have
g(a + h)
g(a)
by g(a + h) g(a)
0, making the division and multiplication
small
meaningless.
only
about
True, we
h, but g(a + h) g(a) could be 0
care
for arbitrarily small h. The easiest way this can happen is for g to be a constant
should
-
=
--
-
176 Derivativesand Integrals
g(a)
function, g(x)
0 for all h. In this case, fog is also
c. Then g(a + h)
the
Chain Rule does indeed hold:
a constant function, (fo g)(x)
f (c), so
=
=
-
=
(fo g)'(a)
0
=
f'(g(a))
=
-
g'(a).
However, there are also nonconstant functions g for which g(a + h) g(a)
for arbitrarily small h. For example, if a = 0, the function g might be
-
f
(x)
1
x2 sin
x
0,
0
¢0
x
-,
=
=
0.
=
x
In this case, g'(0)
0, as we showed in Chapter 9. If the Chain Rule is correct, we
0 for any differentiable f, and this is not exactly obvious.
must have (fo g)'(0)
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 use a trick.
=
=
THEOREM
9 (THE CHAM RULE)
If g is differentiable at a, and f is differentiable at g(a), then
at a, and
g)'(a)
(fo
PROOF
f'(g(a))
=
f og
is differentiable
g(a)
¢0
g(a)
=
g'(a).
-
Define a function ¢ as follows:
h))
f (g(a +
¢(h)
g(a + h)
=
f (g(a))
-
.
if
g(a + h)
,
g(a)
-
if g(a+
J'(g(a)),
-
h)
-
0.
It should be intuitively clear that ¢ is continuous at 0: When h is small,
g(a) is also small, so if g(a + h)
g(a) is not zero, then ¢(h) will
g(a + h)
be close to f'(g(a)); and if it is zero, then ¢(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 f is differentiable at g(a). This means that
-
-
f(g(a)+k)- f(g(a))
.
hm
Thus, if s
0 there is some number
>
.
(1)
if0
<
|k| <
f'(g(a)).
=
k
koo
8'
>
f(g(a)
ô', then
0 such that, for all k,
+k)
f(g(a))
-
k
-
f'(g(a))
Now g is differentiable at a, hence continuous at a, so there is a 8
for all h,
(2) if |hl < ô, then |g(a + h) g(a)| < ô'.
¢(h)
it followsfrom
=
f(g(a+h))-
(2)that
g(a + h)
k|
<
=
f(g(a))
-
¢ 0, then
f(g(a)+k)- f(g(a))
g(a + h)
-
g(a)
g(a)
8', and hence from
|¢(h) f'(g(a))|
-
k
(1)that
<e.
e.
0 such that,
>
-
Consider now any h with Ih| < ô. If k
<
'
10. Djerentiation 177
On the other hand, if g(a + h)
g(a)
---
0, then ¢(h)
=
f'(g(a)), so
=
it is surely
true that
|¢(h)
f'(g(a))| <
-
s.
We have therefore proved that
lim¢(h)
f'(g(a)),
=
h-0
so
¢ is continuous
at
0. The rest of the proof is easy. If h ¢ 0, then we have
f (g(a +
h)
-
f (g(a))
¢(h)
=
h
even
if g(a + h)
(fo g)'(a)
=
-
in that
(because
f(g(a+h))- f(g(a))
g(a)
lim
0
=
=
f'(g(a))
g(a)
-
both sides are 0). Therefore
case
-g(a)
.
lim ¢ (h) lun
=
g(a+h)
-
h
hw0
g(a + h)
h
-
h-o
hwO
h
g'(a).
-
Now that we can differentiate so many functions so easily we can take another
look at the function
x2 sin 1
-,
f (x)
=
¢0
x
x
0,
0.
=
x
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
l
-
x
+ x2 æs
1
1
-
·
-
·
-
x2
x
Thus
f'(x)
1
2x sin
=
1
-
-
x
cos
-,
¢0
x
x
0,
0.
=
x
As this formula reveals, the first derivative f' is indeed badly behaved at 0-it
not even continuous there. If we consider instead
f (x)
=
x3sin-, 1
x
x¢0
0,
x
-
is
0,
then
f'(x)
=
3x2 sin
1
-
x
0,
1
-
x cos
-,
x
x
x
¢0
=
0.
the expresIn this case f' is continuous at 0, but f"(0) does not exist (because
sion 3x2 sin l/x defines a function which is difTerentiable at 0 but the expression
cos 1/x does not).
-x
178 Derivativesand Integrals
As you may suspect,
improvement. If
increasing the power of x yet again produces another
1
f (x)
=
0,
0,
=
x
then
f'(x)
4x3 sin
=
-
1
x2
-
-,
cos
x
1
x
x
0,
x
It is easy to compute, right from the definition, that
easy to find for x ¢ 0:
f"(x)
=
12x2 sin
1
-
-
x
4x cos
1
-
x
=
0.
(f')'(0)
1
2x cos
-
¢0
-
-
x
sin
=
0, and f"(x) is
1
-,
x
0,
x
x
¢0
=
0.
In this case, the second derivative f" is not continuous at 0. By now you may have
guessed the pattern, which two of the problems ask you to establish: if
f (x)
sin
x
=
1
-,
x
x
0,
then
f'(0),
...
,
x
ff">(0)exist, but ff") is not
0,
=
continuous at 0; if
x2.+1 sin 1
x
x
0,
x
-,
f (x)
=
¢0
¢0
=
0,
then /'(0),..., ff">(0)exist, and ff") is continuous at 0, but ff") is not differentiable at 0. These examples may suggest that
functions can be
characterized
by the possession of higher-order derivatives-no matter how hard
we try to mask the infinite oscillation of f (x) sin 1/x, a derivative of sufliciently
high order seems able to reveal the underlying
irregularity. Unfortunately, we will
see later that much worse things can happen.
After all these involved calculations, we will bring this chapter to a close with
minor
remark.
It is often tempting, and seems more elegant, to write some of
a
the theorems in this chapter as equations about functions, rather than about their
"reasonable"
=
-
values.
Thus Theorem 3 might be written
(f+g)'=
f'+g',
Theorem 4 might be written as
(f·g)'=
and
f-g'+f'·g,
Theorem 9 often appears in the form
(fog)'=(f'og)-g'.
10. Deferentiation 179
Strictly speaking, these equations may be false, because the functions on the lefthand side might have a larger domain than those on the right. Nevertheless, this
is hardly worth worrying about. If f 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.
PROBLEMS
1.
As a warm up exercise, find f'(x) for each of the following f. (Don't worry
about the domain of f or f'; just get a formula for f'(x) that gives the right
answer when it makes sense.)
(i) f(x)=sin(x+x2L
(ii) f(x)=sinx+sinx2.
(iii) f(x) sin(cosx).
(iv) f (x) sin(sin x).
(v) f (x) sin x
=
=
(cosx
=
.
sin(cos x)
(vi) f (x)
(vii)f (x)
(viii)f (x)
2.
=
.
X
=
sin(x + sin x).
=
sin(cos(sin x)).
of the following functions f. (It took the author 20 minthe
derivatives for the answer section, and it should not take
utes to compute
much
longer. Although rapid calculation is not the goal of mathematics,
you
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.)
Find
f'(x) for each
(i) f (x)
(ii) f (x)
(iii) f (x)
=
=
=
(iv) f (x)
=
(v) f (x)
=
sin((x + 1)2(x + 2)).
sin3 2 + sinX).
sin2
((x + sin x)2L
sin
x
(
3
.
cos x3
sin(x sin x) + sin(sin x2).
(vi) f(x) (cosx)312.
sin2 x sin x2 sin2 x2.
(vii)f (x) sin3
(sin2(sin
x)).
(viii)f (x)
=
=
=
(ix) f (x) (x +
=
(x) f(x)
(xi) f(x)
(xii) f (x)
(xiii)f (x)
(xiv)f (x)
=
=
=
sin5
6
sin(sin(sin(sin(sinx)))).
sin((sin'x2 + BIL
(((x2+
x)3
4
=
sin(x2 + sin(x2
=
sin(6 cos(6 sin(6 cos
5
2
6x))).
180 Derivativesand Integrals
(xv) f (x)
=
(xvi)f (x)
=
sin x2 sin2 x
.
.
1+smx
1
2
x+smx
x3
f (x)
(xvii)
sin
=
x3
.
sin
.
--
smx
x
f (x)
(xviii)
3.
sin
=
(
x
.
x-sm
.
.
x
-
smx
Find the derivatives of the functions tan, cotan, sec, 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
(i)
(ii)
(iii)
(iv)
5.
6.
f (x)
f, find f'( f (x)) (not(fo f)'(x)).
1
=
.
1+ x
f (x)
=
f(x)
f (x)
=
sin x.
x2.
17.
=
For each of the following functions f, fmd f (f'(x)).
1
(i) f (x)
=
(ii) f(x)
=
(iii) f (x)
(iv) f(x)
=
Find
f'
-.
x
x2.
17.
17x.
=
in terms of g' if
(i) f (x) g(x + g(a)).
(ii) f (x) g(x g(a)).
(iii) f(x)=g(x+g(x)).
=
=
(iv) f (x)
(v) f (x)
(vi) f (x +
=
=
7.
-
a).
g(x)(x
g(a)(x
a).
3) g(x2).
-
-
=
(a) A circular
object
is increasing in size in some unspecified
manner,
but it
is known that when the radius is 6, the rate of change of the radius is 4.
Find the rate of change of the area when the radius is 6. (If r(t) and A(t)
represent the radius and the area at time t, then the functions r and A
str2;
satisfy A
use of the Chain Rule is called for.)
a straightforward
=
10. Deferentiation 181
(b) Suppose that
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
times the cube of the radius.)
(c) Now suppose that the rate of change of the area of the circular cross
section is 5 when the radius is 3. Find the rate of change of the 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).
we are now
gr
8.
The area between two varying concentric circles is at all times 9x in2. The
rate of change of the area of the larger circle is 10x in2/sec. How fast is the
circumference
of the smaller circle changing when it has area 16x in29
9.
Particle A moves along the positive horizontal axis, and particle B along the
graph of f(x)
-J, x < 0. At a certain time, A is at the point (5,0)
and moving with speed 3 units/sec;
and B is at a distance of 3 units from
the origin and moving with speed 4 units/sec.
At what rate is the distance
changing?
and
B
between A
=
10.
Let f(x) x2 sin 1/x for x ¢ 0, and let f (0)
are two functions such that
=
h'(x)
=
h(0)
=
sin2(sin(x
+ 1))
3
0. Suppose also that h and k
=
k'(x)
=
k(0)
=
f(x +
1)
0.
Find
(i) (fo h)'(0).
(ko f)'(0).
(ii) a'(x2),
where
(iii)
11.
Find
a(x)
=
h(x2). Exercise great care.
f'(0) if
f (x)
g(x) sin
-
1
-,
x
0,
x
x
¢0
=
0,
and
g(0)
=
g'(0)
=
0.
12.
Using the derivative of f (x)
by the Chain Rule.
13.
(a) Using Problem 9-3, find f'(x) for < x < 1, if f (x) Ál x2.
(b) Prove that the tangent line to the graph of f at (a, V1 a2 ) intersects
the graph only at that point (andthus show that the elementary geometry
=
1/x, as found in Problem 9-1, find (1/g)'(x)
-1
=
-
definition of the tangent line coincides with ours).
-
182 Derivativesand Integrals
14.
Prove similarly that the tangent lines to an ellipse or hyperbola intersect these
sets only once.
15.
If f + g is differentiable at a, are f and g necessarily differentiable at a?
If f g and f are differentiable at a, what conditions on f imply that g is
differentiable at a?
-
16.
(a) Prove
that if
provided that
f is differentiable
f (a) ¢ 0.
(b) Give a counterexample
(c) Prove that if f and
if
at a, then
is also differentiable at a,
|f|
0.
g are differentiable at a, then the functions
and min(f,g)
are differentiable at a, provided that f(a) ¢
max(f,g)
f (a)
=
g(a).
(d) Give a
17.
counterexample
if f(a)
=
times differentiable and
at x is defined to be
If
f is three
f'"(x)
Sf(x)=
(a) Show that
---
f'(x)
g(a).
f'(x) ¢ 0, the Schwarzianderivativeof f
3
2
2
f"(x)
.
f'(x)
(fog)=[Sfog]-g'2+By
ax + b
with ad
if f (x)
,
cx + d
quently, B(f a g) Og.
(b) Show that
=
--
¢ 0,
bc
then 9
f
=
0. Conse-
=
18.
Suppose that ff")(a)and gt">(a) exist. Prove Leibnit'sformula:
(f
-
g)(">(a)
n
k
=
k=0
*19.
n-k
(k)
Prove that if f("3(g(a))and gt">(a) both exist, then (fo g)("3(a) exists. A
little experimentation should convince you that it is unwise to seek a formula
for (fo g)(") (a). In order to prove that (fo g) (") (a) exists you will therefore
have to devise a reasonable assertion about (f og)(">(a) which can be proved
by induction. Try something like: "( fo g)(") (a) exists and is a sum of terms
each of which is a product of terms of the form
."
...
20.
(a) If f(x)
=
anx"+a,_ix.-1+-
Find another.
(b) If
find a fimction g such that g'
--+ao,
b2
b3
b,
f(x)= p+p+---+--,
find a function g with g'
f.
=
(c) Is there a function
f (x)
such that
f'(x)
=
=
anx" +
1/x?
-
·
-
+ ao +
bi
-
x
+
-
-
-
+
b.
--
xm
=
f.
10. Deferentiation 183
21.
Show that there is a polynomial function f of degree n such that
(a) f'(x)
(b) f'(x)
=
=
for precisely n 1 numbers x.
for no x, if n is odd.
for exactly one x, if n is even.
for exactly k numbers x, if n k is odd.
-
(c) f'(x)
(d) f'(x)
(a) The number
=
=
22.
0
0
0
0
f(x)
=
(x
-
-
a is called a double root
double root of f
(b) When does f(x)
the condition
23.
*24.
of the polynomial function f if
for some polynomial function g. Prove that a is a
if and only if a is a root of both f and f'.
ax2 + bx +
c (a ¢ 0) have a double root? What does
a)2g(x)
say
=
geometrically?
If f is differentiable at a, let d(x)
f (x) f'(a)(x a) f (a). Find d'(a).
connection
with
another solution for Problem 9-20.
this
gives
Problem
22,
In
=
This problem is a companion
be given numbers.
-
-
-
to Problem 3-6. Let ai,
a, and bi,
...,
bn
...,
x, are distinct numbers, prove that there is a polynomial function f of degree 2n
1, such that f (x¡) f'(x¡)
0 for j ¢ i, and
and
b¿.
Remember
Problem
22.
Hint:
f'(x¿)
f (x¿) at
(b) Prove that there is a polynomial function f of degree 2n 1 with f (xi)
a¿ and f'(x;) b¿ for all i.
(a) If xt,
...,
=
-
=
=
=
=
-
=
*25.
Suppose that a and b are two consecutive roots of a polynomial fimction
but that a and b are not double roots, so that we can write f(x)
(x a)(x b)g(x) where g(a) ¢ 0 and g(b) ¢ 0.
-
f,
=
-
(a) Prove that
g(a) and g(b) have the same sign. (Remember that a and b
roots.)
consecutive
are
that
there
is some number x with a < x < b and f'(x) 0. (Also
(b) Prove
draw a picture to illustrate this fact.) Hint: Compare the sign of f'(a)
and f'(b).
(c) Now prove the same fact, even if a and b are multiple roots. Hint: If
b)"g(x) where g(a) ¢ 0 and g(b) ¢ 0, consider the
f(x) = (x
b)"
polynomial function h(x)
f'(x)/(x a)m-1(x
=
-a)"(x
-
=
-
-
.
This theorem was proved by the French mathematician
Rolle, in connection
of
approximating
of
with the problem
roots
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 calculus. This was
such
not
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 Rolle; his name has become
attached to a much more general result, to appear in the next chapter, which
forms the basis for the most important theoretical results of calculus.
26.
xg(x) for some function g which is continuous
Suppose that f(x)
that
Prove
f is differentiable at 0, and find f'(0) in terms of g.
=
at
0.
184 Derivancesand Integrals
*27.
Suppose f is differentiable at 0, and that f (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)
28.
If
f(x)
=
=
f (x)/x?
x¯" for n in N, prove that
=
¡(*)(x) (-1)k (n+k-1)! X-n-k
(n 1)!
+ k l
(-1)kk! n
=
-
-
-n-k,
for x ¢ 0.
=
*29.
Prove that it is impossible to wiite x
f(x)g(x) where f and g are differentiable and f (0) g(0)
0. Hint: Differentiate.
=
=
30.
What is f(k)(X)
(a) f (x)
*(b)
*31.
f(x)
=
f
1/(x a)"?
p
1/(x2
-
=
,
x sin l /x if x ¢ 0, and let f (0) = 0. Prove that /'(0),
f">(0) exist, and that ff") is not continuous at 0. (You will
the
encounter
.
.
.
basic difficulty as that in Problem 19.)
Let f(x)
=
x2n+1sin1/x
f("3(0)exist,
that
f("3is
if x ¢ 0, and
continuous
at
let f(0)
0. Prove that f'(0),...,
0, and that f(") is not differentiable
=
0.
at
33.
if
Let f (x)
same
*32.
=
=
In Leibnizian notation
the Chain Rule ought to read:
df(g(x))
df(y)
dg(x)
dy
dx
¯
dx
finds the following statement:
Instead,
z
=
one usually
f(y). Then
dz
dz
dy
dx
dy
dx
"Let y
=
g(x) and
"
Notice that the z in dz/dx denotes the composite function fo g, while the z
in dz/dy denotes the fimction f; it is also understood that dz/dy will be
expression involving y," and that in the final answer g(x) must be substituted
for y. In each of the following cases, find dz/dx by using this formula; then
"an
with
compare
(i) z=siny,
(ii) z sin y,
(iii) z cos u,
(iv) z sino,
=
Problem 1.
y=x+x2.
y
=
u
=
v
=
=
=
cos x.
sin x.
cosu,
u
=
sinx.
SIGNIFICANCE OF THE DERIVATIVE
CHAPTER
One aim in this chapter is to justifythe time we have spent learning to find the
derivative of a function. As we shall see, knowing just a little about f' tells us a
lot about f. Extracting information about f from information about f' requires
some difficult work, however, and we shall begin with the one theorem which is
really easy.
This theorem is concerned with the maximum value of a function on an interval.
Although we have used this term informally in Chapter 7, it is worthwhile to be
precise, and also more general.
Let f be a function and A a set of numbers contained in the domain of f.
point for f on A if
A point x in A is a maximum
DEFINITION
f(x) > f(y)
The number
say that
i
x2
FIGURE
i
x.
f
"has
for every y in A.
value
the maximum
its maximum value on A at x").
of
f (x) itself is called
f
on A
(andwe
also
Notice that the maximum value of f on A could be f(x) for several different x
(Figure 1);in other words, a function f can have several differentmaximum points
value. Usually we shall be
on A, although it can have at most one maximum
interested in the case where A is a closed interval [a,b]; if f is continuous, then
Theorem 7-3 guarantees that f does indeed have a maximum value on [a,b].
The definition of a minimum of f on A will be left to you. (One possible
definition is the following: f has a minimum on A at x, if f has a maximum
I
x,
i
-
on A at x.)
We are now ready for a theorem which does not even depend upon the existence
of
THEOREM
1
least upper bounds.
Let f be any function defined on (a, b). If x is a maximum (ora minimum) point
for f on (a, b), and f is differentiable at x, then f'(x) 0.
(Notice that we do not assume differentiability, or even continuity, of f at other
=
points.)
PROOF
Consider the case where f has a maximum at x. Figure 2 illustrates the simple idea
drawn through points to the left of (x, f (x))
behind the whole argument--secants
>
slopes
through
points to the right of (x, f (x)) have
and
secants drawn
0,
have
slopes 5 0. Analytically, this argument proceeds as follows.
185
l 86 Derivativesand Integrals
If h is any number
such that x + h is in
f(x)
since
f
(a, b), then
f(x+h),
>
has a maximum on (a, b) at x. This means that
f (x +
Thus, if h
>
h)
f (x) 5 0.
-
0 we have
f(x+h)- f(x)
FIGURE
I
i
a
x
I
x+h
50.
h
I
b
and consequently
2
+ h)
lim f (x
<
f (x) < 0.
¯
h
h-0+
On the other hand, if h
-
0, we have
f(x+h)- f(x)
>0,
¯
h
so
f (x +
h)
-
f (x) > 0.
¯
h
hoo-
By hypothesis,
i
a
equal to
i
&
f'(x).
f
is differentiableat x, so these two limits must be equal, in fact
This means that
f'(x)
FIGURE
5 0
and
f'(x)
>
0,
from which it follows that f'(x) 0.
The case where f has a minimum at x is left to you
3
=
(givea one-line
proof).
Notice (Figure 3) that we cannot replace (a, b) by [a,b] in the statement of the
we add to the hypothesis the condition that x is in (a, b).)
(unless
Since f'(x) depends only on the values of f near x, it is almost obvious how to
get a stronger version of Theorem 1. We begin with a definition which is illustrated
in Figure 4.
theorem
Let f be a function, and A a set of numbers contained in the domain of f.
A point x in A is a local maximum
[minimum]point for f on A if
point for f on
there is some & > 0 such that x is a maximum
[minimum]
DEFINITION
An
THEOREM
2
(x
-
8, x +ô).
If f is defined on (a, b) and has a local maximum
differentiable at x, then f'(x) 0.
(orminimum)
=
PROOF
You s}iould see why this is an easy application
of
Theorem 1.
at x, and
f is
11. Segn$icance
of the Derivative 187
The converse of Theorem 2 is definitely not true-it is possible for f'(x) to be 0
even if x is not a local maximum or minimum point for f. The simplest example
is provided by the function f (x) x3; in this case f'(0)
0, but f has no local
=
or minimum anywhere.
Probably the most widespread
with the behavior of a function
=
maximum
misconceptions
about
calculus
are concerned
point made in
f near x when f'(x) = 0. The
the previous paragraph is so quickly forgotten by those who want the world to be
simpler than it is, that we will repeat it: the converse of Theorem 2 is not true-the
condition f'(x)
0 does not imply that x is a local maximum or minimum point
of f. Precisely for this reason, special terminology has been adopted to describe
numbers x which satisfy the condition f'(x) = 0.
=
DEFINITION
A critical
point
of a
function f is a number
f'(x)
The number
f(x) itselfis
=
x such that
0.
called a criticalvalue
of
f.
The critical values of f, together with a few other numbers, turn out to be the
ones which must be considered in order to find the maximum and minimum of a
given function f. To the uninitiated, finding the maximum and minimum 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 (untilyou have done your first
hundred or so).
Et us consider first the problem of finding the maximum or minimum of f
on a closed interval [a,b]. (Then, if f is continuous, we can at least be sure
and minimum
value exist.) In order to locate the maximum
and
that a maximum
minimum of f three kinds of points must be considered:
critical points of f in
end
The
points a and b.
(2)
(3)Points x in [a,b] such that
(1)The
I
[a,b].
f is not differentiable
at x.
1
point or a minimum point for f 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 f is differentiable at x; consequently f'(x) 0, by Theorem 1,
and this means that x is in the first group.
If there are many points in these three categories, finding the maximum and
minimum of f may still be a hopeless proposition, but when there are only a few
critical points, and only a few points where f is not differentiable, the procedure is
0, and
fairly straightforward:
one simply finds f(x) for each x satisfying f'(x)
f (x) for each x such that f is not differentiable at x and, finally, f (a) and f (b).
The biggest of these will be the maximum value of f, and the smallest will be the
minimum.
A simple example follows.
If x is a maximum
=
.
y
local
rummum
FIGURE
4
pomt
a local
maximum point
=
188 Derivativesand Integrals
Suppose we wish to find the maximum
and minimum
value of the
function
f(x)=x3-x
on the
interval
[-1, 2].
To begin
with,
f'(x)
so
f'(x)
=
0 when 3x2
1
-
have
3x2
=
1
-
0, that is, when
=
=
x
Ñ
we
Ñ
or
-
Ñ.
-Ñ
and
The numbers
both lie in [-1, 2], so the first group of candidates
for the location of the maximum and the minimum is
the end points of the interval,
The second group contains
(2)
1, 2.
-
The third group is empty, since f is differentiable everywhere. The final step is to
compute
f (-1) 0,
f (2) 6.
=
=
-jÑ,
occurring at
maximum
Clearly the minimum value is
, and the
occurring
is 6,
at 2.
This sort of procedure, if feasible, will always locate the maximum and minimum
value of a continuous function on a closed interval. If the function we are dealing
with is not continuous, however, or if we are seeking the maximum 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 information obtained by
this procedure may say nothing. Nevertheless, a little ingenuity will often reveal
the nature of things. In Chapter 7 we solved just such a problem when we showed
that if n is even, then the function
value
11
b
a
FIGURE
1
5
f(x)=x"+an_ixn-14...4
has a minimum
value on the whole line. This proves that the minimum
occur at some number
0
=
value must
x satisfying
f'(x)
=
nx"¯1
+
(n
-
1)a._ix"¯2 +··- + ai.
If we can solve this equation, and compare the values of f (x) for such x, we can
actually find the minimum of f. One more example may be helpful. Suppose we
wish to find the maximum and minimum,
if they exist, of the function
f(x)=
1
1-x2
11. Sénþicanceof theDerivative 189
on the open
interval (-1, 1). We have
f'(x)
2x
=
(1
-
x2) 2
-1
the
0 only for x 0. We can see immediately that for x close to 1 or
values of f (x) become arbitrarily large, so f certainly does not have a maximum.
This observation also makes it easy to show that f has a minimum at 0. We just
note (Figure 5) that there will be numbers a and b, with
so
f'(x)
=
=
-1
<
such that
a
<
0
and
0
a
and
byx<1.
b
<
1,
<
f (x) > f (0) for
-1<x
This means that the minimum of f on [a,b] is the minimum of f on all of
(-1, 1). Now on [a,b] the minimum occurs either at 0 (theonly place where
f' 0), or at a or b, and a and b have already been ruled out, so the minimum
value is f (0)
1.
solving
these problems we purposely did not draw the graphs of f (x) x3-x
In
and f (x) 1/(1 x2), but it is not cheating to draw the graph (Figure 6) as long
as you do not rely solely on your picture to prove anything. As a matter of fact, we
are now going to discuss a method of sketching the graph of a function that really
fact
gives enough information to be used in discussing maxima and minima-in
will
maxima
minima.
method
able
and
be
This
involves
to locate even local
we
of the sign of f'(x), and relies on some deep theorems.
consideration
The theorems about derivatives which have been proved so far, always yield
information about f' in terms of information about f. This is true even of Theorem 1, although this theorem can sometimes be used to determine certain information about f, namely, the location of maxima and minima. When the derivative
was first introduced, we emphasized that f'(x) is not [f (x + h) f (x)]/h for any
particular h, but only a limit of these numbers as h approaches 0; this fact becomes
painfully relevant when one tries to extract information about f from information
about f'. The simplest and most frustrating illustration of the difficulties encountered is afTorded by the following question: If f'(x)
0 for all x, must f be a
constant function? It is impossible to imagine how f could be anything else, and
this conviction is strengthened
by considering the physical interpretation-if the
of
velocity
a particle is always 0, surely the particle must be standing still! Nevertheless it is difficult even to begin a proof that only the constant functions satisfy
f'(x) 0 for all x. The hypothesis f'(x) = 0 only means that
=
=
=
=
f(x)
fi
-1
N¯3
=
x'
--
x
--
--
=
I
i
i
i
-1
1
=
+ h)
hm f (x
.
hoo
FIGURE
6
h
-
f (x)
=
0,
and it is not at all obvious how one can use the information about the limit to
derive information about the fimction.
190 Derwativesand Integrals
The fact that f is a constant function if f'(x) = 0 for all x, and many other facts
of the 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 f is differentiable on
b], then there is some x in (a, b) such that
/
,7
,7
[a,
,
./*,7'
y'
ta
FIGURE
=
b-a
Geometrically this means that some tangent line is parallel to the line between
(a, f (a)) and (b, f (b)). The Mean Value Theorem asserts that this is true-there
is some x in (a, b) such that f'(x), the instantaneous rate of change of f at x, is
change of f on [a,b], this average change
exactly equal to the average or
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
the
theorem is one of the most important theoretical tools of calculus-probably
might
result
about
conclude that the
derivatives. Fmm this statement you
deepest
hard theorems in this book
proof is difficult, but there you would be wrong-the
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 ashamed of. The first proof of the theorem
was an achievement, but today we can supply a proof which is quite simple. It
helps to begin with a very special case.
x
"mean"
7
-
FIGURE
f (b) f (a)
-
f'(x)
8
THOMM
3 (ROLLFS TŒOMM)
-
is continuous on [a,b] and differentiable on (a, b), and
there is a number x in (a, b) such that f'(x)
0.
If
f
f (a)
=
f (b), then
=
PROOF
If followsfrom the continuity of f on [a,b] that f has a maximum and a minimum
[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).
value of f occurs at some point x in (a, b).
Suppose next that the minimum
again,
Then,
f'(x) 0 by Theorem 1 (Figure 9).
Finally, suppose the maximum and minimum values both occur at the end
points. Since f (a)
f (b), the maximum and minimum values of f are equal,
|IS
SO
3 COnstant
function (Figure 10), and for a constant function we can choose
any x in (a, b).
value on
=
=
=
FIGURE
9
e
FIGU RE lo
i
Notice that we really needed the hypothesis that f is differentiable everywhere
the theorem is
on (a, b) in order to apply Theorem 1. Without this assumption
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 Theorem we will apply Rolle's
Theorem to the function which gives the length of the vertical segment shown in
11. Syngicanceof ike Derivative 191
Figure 12; this is the difference between f (x), and the height at x of the line L
between (a,f (a)) and (b,f (b)). Since L is the graph of
f (b) f (a)¯
-
g(x)
i
a
FIGURE
[
=
i
b
we want to
look at
b
a
-
(x
--
a) +
f (a),
a)
f (a).
11
f (x)
b
As it turns out, the constant
THEOREM
4 (THE MEAN VALUE
THEOREM)
If f is continuous on
in (a, b) such that
(x
--
--
a
-
-
f (a) is irrelevant.
[a,b] and
differentiable on (a, b), then there is a number
f (b) f (a)
x
-
f'(x)
PROOF
=
.
b--a
IÆt
f (b) f (a)
-
h(x)
=
h(a)=
(6, f(b))
=
=
L
-
b
[a,b] and
Clearly, h is continuous on
h(b)
f (x)
f(a),
f (b)
-
a
(x
a).
-
differentiable on (a, b), and
f (b) f (a)
-
[
-
f (a).
b
--
a
(x
a)
-
Consequently, we may apply Rolle's Theorem to h and conclude
some x in (a, b) such that
that there is
f (b) f (a)
-
O = h'(x)
FIGURE
So
12
f'(x)
-
-
b
,
-
a
that
f (b) f (a)
-
J'(x)
=
b
-
.
a
Notice that the Mean Value Theorem still fits into the pattern exhibited by
previous theorems-information about f yields information about f'. This information is so strong, however, that we can now go in the other direction.
COROLLARY
1
PROOF
If f is defined on an interval and f'(x)
constant on the interval.
=
0 for all x in the interval, then f is
Let a and b be any two points in the interval with a ¢ b. Then there is some x in
192 Derivativesand Integrals
(a, b)
such that
f (b) f (a)
-
f'(x)
But
f'(x)
=
=
b
.
a
-
0 for all x in the interval, so
f (b) f (a)
-
FIGURE
0
13
and consequently f(a)
interval is the same, i.e.,
-
'
b-a
Thus the value of f at any two points in the
f is constant on the interval.
=
f(b).
Naturally, Corollary 1 does not hold for functions defined on two or more intervals (Figure 13).
COROLLARY
2
and g are
g'(x) for all x in the
defined on the same interval, and f'(x)
number
such
that f
interval, then there is some
g + c.
c
If
f
=
=
PROOF
For all x in the intervalwe have (f g)'(x)
there is a number c such that f
c.
g
=
-
f'(x)
-
g'(x)
=
0 so, by Corollary1,
=
-
The statement of the next corollary requires some terminology, which is illustrated in Figure 14.
A function is increasing on an interval if f (a) < f (b) whenever a and b are
two numbers in the interval with a < b. The function f 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 f is increasing or decreasing, in which case the interval is
understood
to be the domain of f.)
DEFINITION
COROLLARY
3
PROOF
Íf f'(x) > Ûfor all x in an interval, then f is increasing on the interval; if f'(x)
for all x in the interval, then f is decreasing on the interval.
<
0
Consider the case where f'(x) > 0. Let a and b be two points in the interval with
a < b. Then there is some x in (a, b) with
f (b) f (a)
-
f'(x)
But f'(x)
>
=
b
-
a
0 for all x in (a, b), so
f (b) f (a)
-
b
-
>
0.
a
Since b a > 0 it follows that f (b) > f (a).
The proof when f'(x) < 0 for all x is left to you.
-
of theDenvative 193
11. Srgngicance
the converses of Corollary 1 and Corollary 2 are true (and
obvious), the converse of Corollary 3 is not true. If f is increasing, it is easy to
see that f'(x) > 0 for all x, but the equality sign might hold for some x (consider
(x) x3
Notice that although
f
=
Corollary 3 provides enough information to get a good idea of the graph of
a function with a minimal amount of point plotting. Consider, once more, the
function f (x) x3 x. We have
=
-
f'(x)
=
3x2
1.
-
Ñ
-Ñ,
and x
and it is
0 for x
We have already noted that f'(x)
also possible to determine the sign of f'(x) for all other x. Note that 3x2 1 > 0
=
=
=
-
when
precisely
3x2
x
I
x>Ñ
I
thus 3x2
(a)
-
2
or
>
1
,
x<-
;
1 < 0 precisely when
an increasing function
Thus
-Ñ, decreasing between -N and Ñ,
Ñ. Combining this information with the
is increasing for x <
and once again increasing for x
following facts
f
>
-1,
(2)f (x) 0 for x
(3)f (x) gets large
=
(b)
a decreasing
FIGURE
14
function
=
0, 1,
as x gets large, and large negative
as x gets large negative,
approximation
it is possible to sketch a pretty respectable
to the graph (Figure 15).
By the way, notice that the intervals on which f increases and decreases could
have been found without even bothering to examine the sign of f'. For example,
and
since f' is continuous, and vanishes only at
we know that f'
5,
-Ñ
(-Ñ, Ñ).
always
f(--Ñ)
>
Since
has the same sign on the interval
it follows that f decreases on this interval. Similarly, f' always has the
oo) and f (x) is large for large x, so f must be increasing on
same sign on
oo). Another point worth noting: If f' is continuous, then the sign of f'
on the interval between two adjacent critical points can be determined simply by
finding the sign of f'(x) for any one x in this interval.
x3
Our sketch of the graph of f(x)
x contains sufficient information
is a local maximum point, and
to allow us to say with confidence that
a local minimum point. In fact, we can give a general scheme for decid-
f(Ñ),
(Ñ,
(Ñ,
=
-
-Ñ
194 Derivativesand Integrals
f(x)
-1
I
I
0
I
/
I
sil
1
I
f
decreasing
sit
I
I
J increasing
I
i
I
I
I
I
FIGURE
xa
I
I
I
f increasing
=
15
ing whether a critical point is a local maximum
neither (Figure 16):
point, a local minimum point, or
> 0 in some interval to the left of x and f' < 0 in some interval to
the right of x, then x is a local maximum point.
(2)if f' < 0 in some interval to the left of x and f' > 0 in some interval to
the right of x, then x is a local minimum point.
if
(3) f' has the same sign in some interval to the left of x as it has in some
interval to the right, then x is neither a local maximum nor a local minimum
point.
(1)if f'
(There is no point in memorizing these rules-you
can always draw the pictures
yourself.)
The polynomial functions can all be analyzed in this way, and it is even possible
to describe the general form of the graph of such functions. To begin, we need a
I
f>of<o
(a)
FIGURE
16
I
f<of>o
(b)
I
f>o
I
f>o
(c)
f<o
f<o
(d)
11. Srp¢ance of theDerivance 195
result already
in Problem 3-7: If
mentioned
f (x)
=
anx" +a,_ixn-1
...4
4
"roots,"
has at most n
i.e., there are at most n numbers x such that
f (x) 0. Although this is really an algebraic theorem, calculus can be used
to give an easy proof. Notice that if x1 and x2 are roots of f (Figure 17), so that
0, then by Rolle's Theorem there is a number x between xi
f (xi) f (X2)
and x2 such that f'(x)
0. This means that if f has k different roots xi < x2 <
< xk, then f' has at least k
1 different roots: one between xi and x2, one
between x2 and xa, etc. It is now easy to prove by induction that a polynomial
then
f
=
=
=
=
-
...
functiOD
17
FIGURE
f(x)=anx"+a._ix"¯1+··-+ao
has at most n roots: The statement is surely true for n
it is true for n, then the polynomial
g(x)
could not
roots.
=
bn+1x"**+ b.x" +
·
=
1, and if we assume that
+ be
-
have more than n + 1 roots, since if it did, g' would have more than n
With this information it is not hard to describe the graph of
f (x)
=
anx" + a._ix"¯1
+
-
-
-
+ ao.
1, has at most
The derivative, being a polynomial function of degree n
critical
1
points.
Of
1
Therefore
has
at most n
roots.
course, a critin
f
cal point is not necessarily a local maximum
minimum
point, but at any rate,
or
if a and b are adjacent critical points of f, then f' will remain either positive or
negative on (a, b), since f' is continuous; consequently, f will be either increasing
or decreasing on (a, b). Thus f has at most n regions of decrease or increase.
As a specific example, consider the function
-
-
--
2
f(x)=x4_
Since
f'(x)
=
4x3
-
4x
4x (x
=
-
1)(x + 1),
-1,
the critical points of
f
are
0, and 1, and
-1,
*
f (-1)
°)
=
f (0) 0,
f (1)
=
-1.
/
\
=
-/1)
-1)
(-1
FIGURE
(
18
The behavior of f 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
j' OR Íl1€S€ ÍnÍ€rVRÎS simply be examining the formula for f'(x). On the
other hand, from the three critical values alone we can see (Figure 18) that f
increases on (-1, 0) and decreases on (0, 1). To determine the sign of f' on
196 Derivativesand Integrals
-1)
and
(-oo,
(1, oo)
we can compute
f'(-2)
f'(2)
=
=
4 (-2)3
-
4 23
-
-
-24,
4 (-2)
4 2 24,
-
-
=
=
-
-1)
and conclude that f is decreasing on (-oo,
conclusions also follow from the fact that f
negative
and increasing on
(x) is large for large
(1, oo). These
x and for large
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 determine that
f (x) 0 for x 0, ±Ã; second, it is clear that f is even, f (x) f (--x), so the
graph is symmetric with respect to the vertical axis. The fimction f(x) x3
already sketched in Figure 15, is odd, f (x)
f (-x), 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
=
=
=
-x,
=
=
(1)the critical
-
points of f,
(2)the value of f at the critical points,
the sign of f' in the regions between critical points
this
clear),
the numbers x such that f (x) = 0
possible),
the behavior of f (x) as x becomes large or large negative
(3)
(if
(if
(4)
(5)
is not already
(ifpossible).
Finally, bear in mind that a quick check, to see whether the function is odd or
even, may save a lot of work.
This sort of analysis, 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 information is
often very important. If f is not defmed at certain points (forexample, if f is a
rational
function whose denominator vanishes at some points), then the behavior
of f near these points should be determined.
For example, consider the function
x2-2x+2
f (x)
=
x
|(x)
(-Vž,
0)
(0,0)
-1)
-1)
(-1,
FIGURE
19
(1,
-
x'
1
-
-
2x*
,
11. Sén¢cance of theDerivative 197
which
is not defined at 1. We have
(x 1)(2x
=
2)
-
-
f'(x)
(x
x(x-2)
(x2 2x + 2)
-
-
1)
-
¯
(x 1)2
-
Thus
(1) the critical points
of
f
are
0, 2.
Moreover,
-2,
(2) f (0)
f (2) 2.
=
=
Because f is not defined on the whole interval (0,2), the sign of f' must be
determined separately on the intervals (0, 1) and (1,2), as well as on the intervals
(-oo, 0) and (2, oo). We can do this by picking particular points in each of these
intervals, or simply by staring hard at the formula for f'. Either way we find that
(3) f'(x) > 0 if x < 0,
f'(x)<0 if 0<x<1,
f'(x) <
f'(x)
0
0
>
if
if
1 < x < 2,
2 < x.
Finally, we must determine the behavior of f (x) as x becomes large or large
negative, as well as when x approaches 1 (this
information will also give us another
regions
which
the
increases
and decreases). To examine
on
way to determine
f
write
the behavior as x becomes large we
x2-2x+2
1
=x-l+
x-1
x---l
;
1 (andslightly larger) when x is large, and f (x) is close
1 (butslightly smaller) when x is large negative. The behavior of f near 1
to x
also
is
easy to determine; since
clearly
f (x) is close to x
-
-
lim (x2 2x + 2)
-
xel
-/-
=
1
0,
the fraction
x2-2x+2
x
-
1
becomes large as x approaches 1 from above and large negative as x approaches 1
from below.
All this information may seem a bit overwhelming, but there is only one way
that it can be pieced together (Figure 20); be sure that you can account for each
feature of the graph.
When this sketch has been completed, we might note that it looks like the graph
198 Derivativesand Integrals
I
I
I
2--
/
/
I
2
71
/
i
I
\
/
I
I
\
/
/
FIGURE
/
/
0
/
/
20
of an odd
function shoved over 1 unit, and the expression
x2-2x+2
¯
x-1
(x-1)2+1
x-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
of the graph.
a good idea of the appearance
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
There is a popular test for local maxima and minima which depends on the
behavior of the function only at its critical points.
work.
THEOREM
5
Suppose f'(a) 0. If f"(a) > 0, then f has a local minimum at a; if f"(a)
f has a local maximum at a.
=
then
PROOF
By clefmition,
.
f"(a)
Since
f'(a)
=
=
Inn
J'(a
h-wo
+ h)
h
-
f'(a)
0, this can be written
f"(a)
=
lim
h->0
f'(a+
h
h)
.
<
0,
of theDenvative 199
11. Srgngicance
Suppose now that f"(a)
small h. Therefore:
and
f'(a
f'(a
0. Then f'(a + h)/h must be positive for sufficiently
>
+ h) must be positive for sufficiently small h > 0
+ h) must be negative for sufficiently small h < 0.
This means (Corollary 3) that f is increasing in some interval to the right of a
and f is decreasing in some interval to the left of a. Consequently, f has a local
minimum
f(x)
-
at a.
The proof for the case f"(a)
-x
<
0 is similar.
Theorem 5 may be applied to the function
been considered. We have
(a)
f'(x)
f(x)
=
f"(x)
xa
At the critical points,
=
=
3x2
6x.
-
f (x)
=
x3
x, which
--
has already
1
--N and Ñ, we have
f"(-Ñ)
f"(Ñ) 6Ñ
-6
=
>
=
(b)
-Ñ
)
=
0,
0.
<
Ñ
is a local maximum point and
is a local minimum
Consequendy,
point.
Although Theorem 5 will be found quite useful for polynomial functions, for
many functions the second derivativeis so complicated that it is easier to consider
the sign of the first derivative. Moreover, if a is a critical point of f it may happen
that f"(a)
0. In this case, Theorem 5 provides no information: it is possible
that a is a local maximum point, a local minimum point, or neither, as shown
(Figure 21) by the functions
x.
=
(c)
-x4,
FIGURE
f(x)
21
f(x)
=
=
X4
5.
in each case f'(0)
f"(0) 0, but 0 is a local maximum point for the first, a
local minimum point for the second, and neither a local maximum nor minimum
point for the third. This point will be pursued further in Part IV
It is interesting to note that Theorem 5 automatically
proves a partial converse
of itself.
-
THEOREM
6
PROOF
=
Suppose f"(a) exists. If f has a local minimum
local maximum at a, then f"(a) 5 0.
at a, then
f"(a)
>
0; if f has a
has local minimum at a. If f"(a) < 0, then f would also have a
local maximum at a, by Theorem 5. Thus f would be constant in some interval
containing a, so that f"(a) 0, a contradiction. Thus we must have f"(a) > 0.
The case of a local maximum is handled similarly.
Suppose
f
=
200 Denvativesand Integrals
(This partial converse to Theorem 5 is the best we can hope for: the
signs cannot be replaced by > and <, as shown by the functions f (x)
>
=
and 5
x4 and
-x4.)
f (x)
=
The remainder of this chapter deals, not with graph sketching, or maxima and
but with three consequences of the Mean Value Theorem. The first is a
simple, but very beautiful, theorem which plays an important role in Chapter 15,
and which also sheds light on many examples which have occurred in previous
minima,
chapters.
THEOREM
7
Suppose that f is continuous at a, and that f'(x) exists for all x in some interval
containing a, except perhaps for x
a. Suppose, moreover, that lim f'(x) exists.
=
Then
f'(a) also
exists, and
f'(a)
PROOF
=
lim f'(x).
By definition,
f'(a)
hm f(akh)-
f(a)
.
=
.
h
hoo
For sufficiently small h > 0 the function f will be continuous on [a,a + h] and
differentiable on (a, a + h) (a similar assertion holds for sufliciently small h < 0).
By the Mean Value Theorem there is a number Œh in (a, a + h) such that
f(a)
f(a+h)-
=
h
NOW
lim
approaches
a as h approaches
exists, it follows that
f'(a)
e
=
+ h)
lim f (a
hoo
h
-
f (a)
h)·
0, because
Œh
f'(x)
f'(
lim
=
f'(Gh)
hoO
(It is a good idea to supply a rigorous
have treated somewhat informally.)
Œh
=
is in
lim
(a, a + h);
since
f'(X).
xsa
e-ô argument for this final step, which we
Even if f is an everywhere differentiable function, it is still possible for f' to be
discontinuous. This happens, for example, if
FIGURE
22
f(X)
x2 sin
=
1
-,
X
0,
-/
0
x
x
=
0.
According to Theorem 7, however, the graph of f' can never exhibit a discontinuity of the type shown in Figure 22. Problem 55 outlines the proof of another
beautiful theorem which gives further information about the function f', and Problem 56 uses this result to strengthen Theorem 7.
The next theorem, a generalization of the Mean Value Theorem, is of interest
mainly because of its applications.
of theDerivative 201
11. Signgicance
THEOREM
8 (THE CAUCHY MEAN
VALUE THEOREM)
If f and g are continuous on
number x in (a, b) such that
[a,b]
and
[f (b) f (a)]g'(x)
differentiable on (a, b), then there is a
[g(b)
=
-
-
g(a)]
f'(x).
(If g(b) ¢ g(a), and g'(x) ¢ 0, this equation can be written
f (b) f (a)
f'(x)
-
g(b)
_
¯
g(a)
-
g'(x)
1, and we obtain the Mean Value
Notice that if g(x)
x for all x, then g'(x)
applying
the Mean Value Theorem to f and g
Theorem. On the other hand,
separately, we find that there are x and y in (a, b) with
-
=
f (b) f (a)
f'(x)
g(b)
g'(y)'
-
g(a)
--
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 difEcult
to prove, but actually the simplest of tricks suffices.)
PROOF
Let
-g(a)]
h(x)
-g(x)[f(b)
f(x)[g(b)
-
Then h is continuous on
f(a)].
-
[a,b], differentiable on (a, b), and
h(a)=
f(a)g(b)-g(a)f(b)=
It followsfrom Rolle's Theorem that h'(x)
=
h(b).
0 for some x in (a, b), which means
that
0
=
f'(x) [g(b)
-
g(a)]
g'(x)
-
[f (b) f (a)].
-
The Cauchy Mean Value Theorem is the basic tool needed to prove a theorem
limits of the form
which facilitates evaluation of
lim
--,
f (x)
when
lim
f(x)
0
=
and
lim g(x)
=
0.
In this case, Theorem 5-2 is of no use. Every derivative is a limit of this form, and
computing derivatives frequently requires a great deal of work. If some derivatives
are known, however, many limits of this form can now be evaluated easily.
THEOREM
9 (I?HÔPITAI?S RULE)
Suppose that
lim f(x)
and suppose
also that
0
=
lim g(x)
and
=
0,
lim f'(x)/g'(x) exists. Then lim f(x)/g(x) exists, and
x-va
x-va
lim
x->.
--
f (x)
.
=
g(x)
(Notice that Theorem 7 is a special case.)
hm
x-ma
f'(x)
gr(x)
.
202 Derivativesand Integrals
PROOF
The hypothesis that lim f'(x)/g'(x) exists contains two implicit assumptions:
8, a + ô) such that f'(x) and g'(x) exist for all x in
perhaps,
8)
ô,
for x a,
except,
(a
a+
with,
once again, the possible exception
(2)in this interval g'(x) ¢ 0
(1)there
is an interval (a
-
=
-
ofx=a.
On the other hand, f and g are not even assumed to be defined at a. If we define
the previous values of f(a) and g(a), if necessary),
0 (changing
f(a) g(a)
and
then f
g are continuous at a. If a < x < a + 3, then the Mean Value
Theorem and the Cauchy Mean Value Theorem apply to f and g on the interval
[a,x] (anda similar statement holds for a ô < x < a). First applying the Mean
Value Theorem to g, we see that g(x) ¢ 0, for if g(x)
0 there would be some xi
0, contradicting (2).Now applying the Cauchy Mean Value
in (a, x) with g'(x1)
Theorem to f and g, we see that there is a number ex in (a, x) such that
=
=
-
=
=
[f (x)
-
0]g'(ax)
[g(x)
=
-
0] f'(ax)
or
f (x)
f'(ax)
_
¯
g(x)
g'(ax)
Now œ, approaches a as x approaches
lim f'(y)/g'(y) exists, it follows that
because as is in (a, x); since
a,
yea
(x)
hm f
,
-
g(x)
x->a
(Once again, the reader
lim
=
xsa
J'(ax)
lim
=
g'(ax)
y
a
J'(y)
.
g'(y)
is invited to supply the details of this part of the argu-
ment.)
PROBLEMS
1.
For each of the following functions, find the maximum and minimum values
on the indicated intervals, 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.
(i) f(x)=x3_-x2-8x+1
x6 + x + 1
on
on
[-2,2].
[---1,1].
(ii) f (x)
(iii) f(x)=3x4--8x3+6x2
on[-Lg
(iv) f (x)
=
x5+x+1
on
[--¼,1].
(v) f(x)
=
x + 1
x2 4 1
on
[-1,
on
[0,5].
=
(vi) f (x)
1
x
=
x2
-
1
.
ofthe Daivative 203
11. Srgmyicance
2.
Now sketch the graph of each of the functions in Problem 1, and find all
local maximum and minimum points.
3.
Sketch the graphs of the following functions.
1
(i) f(x)=x+-. x
(ii) f (x)
(iii) f (x)
=
(iv) f (x)
=
3
x+
=
--.
x
x
2
.
X2
1
-
1
1 + x2
-ag)2.
4.
(a) If at
*(b)
<
···
<
a., find the minimum
value of
f(x)
(x
=
i=1
Now find the minimum
value of
f (x)
|x
=
-
a¿|. This is a problem
i=l
where
*(c)
calculus
help at all: on the intervals between the a¿'s the
won't
function f is linear, so that the minimum clearly occurs at one of the a¿,
and these are precisely the points where f is not differentiable. However,
the answer is easy to find if you consider how f (x) changes as you pass
from one such interval to another.
Let a > 0. Show that the maximum value of
f (x)
1
1+|x|
=
+
1
1+|x-a|
is (2 + a)/(1 + a). (The derivative can be found on each of the intervals
(-oo, 0), (0,a), and (a, oo) separately.)
5.
For each of the following functions, fmd all local maximum
points.
and minimum
y=3, 5, 7, 9
x,
x
5,
x=3
9,
7,
x=7
-3,
(i) f(x)
(ii) f (x)
=
=
x
=
5
9.
x irrational
p/q in lowest terms.
x
x
0,
1/q,
=
=
(iii) f (x)
x,
x rational
0,
x
(iv) f (x)
=
1,
0,
(v) f (x)
=
1,
0,
I
x
irrational.
=
1/n for some n in N
otherwise.
if the decimal expansion of x contains
otherwise.
a
5
204 Denvativesand Integrals
6.
of the plane, and let L be the graph of the fimction
f (x) = mx + b. Find the point ž such that the distance from (xo,yo) to
(ž, f (ž)) is smallest. [Notice that minimizing this distance is the same as
minimizing
its square. This may simplify the computations somewhat.]
(a) Let (xo,yo) be a point
e, y
a
p
,"-'
(b) Also find ž
(x, o)
,,"
dicular to L.
the distance from (xo,yo) to L, i.e., the distance from (xo,yo) to
Find
(c)
(ž, f (ž)). [ftwill make the computations easier if you first assume that
b
0; then apply the result to the graph of f(x)
mx and the point
(xo,yo b).] Compare with Problem 4-22.
Consider
a straight line described by the equation Ax + By + C
(d)
0 (Problem 4-7). Show that the distance from (xo,yo) to this line is
-a)w'
(o,
FIGURE
by noting that the line from (xo,yo) to (ž, f (ž)) is perpen-
23
=
=
-
=
r
Surface area is the
sum of these areas
(ÂX0
7.
ETO
/
2
g2
The previous Problem suggests the following question: What is the relationbetween the critical points of f and those of f22
ship
FIGURE
8.
A straight line is drawn from the point (0,a) to the horizontal axis, and
then back to (1,b), as in Figure 23. Pove that the total length is shortest
When
the angles a and ß are equal. (Naturally you must bring a function
into the picture: express the length in terms of x, where (x, 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 actually
finding the point (x, 0).)
9.
Prove that of all rectangles with given perimeter, the square has the greatest
24
b
,-
area.
/
Find, among all right circular cylinders of fixed volume V, the one with
smallest surface area (counting
the areas of the faces at top and bottom, as
in Figure 24).
11.
A right triangle with hypotenuse of length a is rotated about one of its legs
10 generate a right circular cone. Find the greatest possible volume of such
'
C
FIGURE
10.
25
a cone.
12.
Two hallways, of widths a and b, meet at right angles (Figure 25). What
is the greatest possible length of a ladder which can be carried horizontally
around the corner?
FIGURE
26
13.
A garden is to be designed in the shape of a circular sector (Figure 26), with
radius R and central angle 0. The garden is to have a fixed area A. For
what value of R and 0
(inradians) will the length of the fencing around the
minimized?
perimeter be
14.
Show that the sum of a positive number and its reciprocal is at least 2.
15.
Find the trapezoid of largest area that can be inscribed in a semicircle
base lying along the diameter.
radius a, with one
of
11. Sénjicanceof theDerivative 205
a
16.
A right angle is moved along the diameter of a circle of radius a, as shown
in Figure 27. What is the greatest possible length (A + B) intercepted on it
by the circle?
17.
Ecological Ed must cross a circular lake of radius 1 mile. He can row across
at 2 mph or walk around at 4 mph, or he can row part way and walk the
rest (Figure 28). What route should he take so as to
(i)
(ii)
27
FIGURE
FIGURE
see as much scenery as possible?
CEOss as guickly as possible?
18.
The lower right-hand
corner of a page is folded over so that it just touches
the left edge of the paper, as in Figure 29. If the width of the paper is a and
the page is very long, show that the minimum length of the crease is 3da/4.
19.
Figure 30 shows the graph of the dmvativeof f. Find all local maximum
minimum points of
f.
28
3
1
FIGURE
*20.
*21.
7"
is a polynomial function, f (x) x" + an-1xn-1
with critical points
critical values 6, 1, 2,
1, 2, 3, 4, and corresponding
the
of
4, 3. Sketch the graph
f, distinguishing cases n even and n odd.
Suppose that
f
/
29
the critical points of the polynomial function
(a) Suppose that
-
-1,
-+
-
and
f (x)
x" +
=
=
-
/
FIGURE
=
-1,
1, 2, 3,
f"(-1) 0, f"(1) > 0, f"(2) <
ao are
0, f"(3) 0. Sketch the graph of f as accurately as possible on the
basis of this information.
(b) Does there exist a polynomial function with the above properties, except
that 3 is not a critical point?
/
'v/
4
30
an-1xn-1+
\
and
22.
Describe the graph of a rational function (invery general terms, similar to
the text's description of the graph of a polynomial function).
23.
that two polynomial functions of degree m and n, respectively,
intersect in at most max(m, n) points.
For
(b) each m and n exhibit two polynomial functions of degree m and n
which intersect max(m, n) times.
*24.
(a) Prove
the polynomial function
has exactly k critical points and f"(x)
that n k is odd.
(a) Suppose that
-
+ ao
f (x) x" + a..ixn-1 +
all
critical
points
Show
0
for
x.
¢
=
-
-
-
206 Derivativesand Integrals
n, show that there is a polynomial function f of degree n with
k critical points if n k is odd.
a,_ix"¯I
+
+ ao
(c) Suppose that the polynomial function f(x) = x" +
has ki local maximum points and k2 local minimum points. Show that
k2 ki + 1 if n is even, and k2 ki if n is odd.
(d) Let n, ki, k2 be three integers with k2 ki + 1 if n is even, and k2 ki if
n is odd, and ki + k2 < n. Show that there is a polynomial function f of
degree n, with k1 local maximum points and k2 local minimum points.
(b) For each
-
-··
=
=
=
=
kl+k2
-(1+x2
I
-a.)
Hint: Pick ai
<
<
a2
<
·
Øki+k2
and try
f'(x)
(x
=
i=1
for an appropriate
25.
number l.
if f'(x) > M for all x in [a,b], then f (b) > f (a) + M(b a).
(b) Prove that if f'(x) EM for all x in [a,b], then f (b) f (a) + M(b a).
(c) Formulate a similar theorem when |f'(x)| EM for all x in [a,b].
Suppose that f'(x) > M > 0 for all x in [0,1]. Show that there is an interval
of length on which |f| > M/4.
(a) Prove that
--
_<
---
*26.
¾
27.
(a) Suppose that
f'(x)
>
g'(x) for all x, and that
f(x)>g(x)forx>aand
f(x)<g(x)forx
(b) Show by an example that these conclusions
hypothesis f(a) g(a).
f(a)
=
g(a).
Show that
<a.
do not follow without
the
=
28.
Find all functions f such that
(a) f'(x)
(b) f"(x)
(c) f"'(x)
29.
sin x.
=
=
=
x3.
x +x2
16t2
Although it is true that a weight dropped from rest will fall s(t)
mention
the behavior of
feet after t seconds, this experimental fact does not
weights which are thrown upwards or downwards. On the other hand, the
law s"(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 negative for falling
bodies, and all bodies fall according to the law s"(t)
=
=
--32.
=
-16t2
+ at + ß.
(a) Show that s is of the form s(t)
and
then in the formula for s',
setting
in
the
formula
for
By
0
t
s,
(b)
show that s(t)
where
+ vet + so,
so is the height from which the
body is released at time 0, and vo is the velocity with which it is released.
(c) A weight is thrown upwards with velocity v feet per second, at ground
is the maximum
level. How high will it go? ("How high" means
height for all times".) What is its velocity at the moment it achieves its
greatest height? What is its acceleration at that moment? When will it
hit the ground again? What will its velocity be when it hits the ground
=
=
--16t2
=
"what
again?
11. Sénýicanceof theDerwative 207
30.
A cannon ball is shot from the ground with velocity v at an angle a (Figof velocity v sing and a horiure 31) so that it has a vertical component
zontal component v cos a. Its distance s
(t) above the ground obeys the law
sina)t,
while
velocity remains constantly
s(t)
its
horizontal
+ (v
--16t2
=
U COS Œ.
(a) Show that
FIGURE
the path of the cannon ball is a parabola (find
the position at
each time t, and show that these points lie on a parabola).
(b) Find the angle a which will maximize the horizontal distance traveled
by the cannon ball before striking the ground.
31
31.
(a) Give
lim
x-
oo
an
example
f'(x) does not
(b) Prove that
(c) Prove that
of a
function f for which
lim f (x) exists, but
exist.
if 1400
lim f (x) and X->OO
lim f'(x) both exist, then X->OO
lim f'(x)
if lim f (x) exists and lim f"(x) exists, then lim f"(x)
(See also Problem 20-15.)
32.
0.
=
=
0.
and
g are two difTerentiable functions which satisfy
that
if a and b are adjacent zeros of f, and g(a)
fg' f'g 0. Prove
and g(b) are not both 0, then g(x)
0 for some x between a and b. (Naturally the same result holds with f and g interchanged; thus, the zeros of f
and g separate each other.) Hint: Derive a contradiction
from the assumption that g(x) ¢ 0 for all x between a and b: if a number is not 0, there is a
natural thing to do with it.
Suppose that
-
f
=
=
33.
Suppose that |f (x) f (y) [ < |x y|n for n > 1. Prove that f is constant by
considering
f'. Compare with Problem 3-20.
34.
A function
-
-
f is Lepschiteoforder
(*)
a at x
if there is a constant
f(x)- f(y)|
C
such
that
5 C|x-y|"
for all y in an interval around x. The function f is Lapschitz of onier a on an
intervalif (*)holds for all x and y in the interval.
is Lipschitz of order a > 0 at x, then f is continuous at x.
(b) If f is Lipschitz of order a > 0 on an interval, then f is uniformly
continuous on this interval (seeChapter 8, Appendix).
(c) If f is differentiable at x, then f is Lipschitz of order 1 at x. Is the
converse true?
(d) If f is differentiable on [a,b], is f Lipschitz of order 1 on [a,b]?
(e) If f is Lipschitz of order a > 1 on [a,b], then f is constant on [a,b].
(a) If f
208 Duivancesand Integrals
35.
Prove that if
ao
ai
an
1
2
n+1
-+-+--·+
=0,
then
ao+aix+··-+anx"=0
for some x in
[0,1].
36.
x3
3x +m never has two roots
Pove that the polynomial function f.(x)
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 fo and f2,
and consider where the graph of f. lies in relation to them.)
37.
Suppose that f is continuous and differentiable on [0,1], that f(x) is in
[0,1] for each x, and that f'(x) ¢ 1 for all x in [0, 1]. Show that there is
exactly one number x in [0, 1] such that f (x) x. (Half of this problem
has been done already, in Problem 7-11.)
=
-
=
38.
the function f (x) x2 cos x satisfies f (x) 0 for precisely
two numbers x.
(b) Prove the same for the function f (x) 2x2 x sin x cos2 x. (Some
preliminary estimates will be useful to restrict the possible location of the
(a) Prove that
=
=
-
=
zeros of
*39.
-
-
f.)
0 and
a twice differentiable function with f(0)
f'(0) J'(1) 0, then |f"(x)| > 4 for some x in [0,1].
In more picturesque terms: A particle which travels a unit distance in
a unit time, and starts and ends with velocity 0, has at some time an
acceleration > 4. Hint: Prove that either f"(x) > 4 for some x in
for some x in [j, 1].
[0, or else f"(x) <
that
in
fact
have
must
Show
we
|f"(x)| > 4 for some x in [0,1].
(b)
(a) Pove
f(1)
=
that if
1 and
f is
=
=
=
-4
j],
40.
Suppose that f is a function such that f'(x)
1/x for all x > 0 and f (1)
=
0. Prove that f(xy) f(x) + f(y) for all x, y > 0. Hint: Find g'(x) when
=
g(x)
*41.
=
=
f (xy).
Suppose that f satisfies
f"(x) + f'(x)g(x) f (x) 0
g. Prove that if f is 0 at two points, then f
=
-
for some function
interval between them. Hint: Use Theorem 6.
42.
Suppose that f is n-times differentiable and that f (x)
ent x. Prove that ff">(x)= 0 for some x.
43.
Let ai,
...,
an+1
be arbitrary
points in
[a,b],
n+1
Q(x)
(x
=
i=1
-
x¿).
and
let
=
is 0 on the
0 for n + 1 differ-
11. Signigcance
of theDerivative 209
Suppose that f is (n + 1)-times differentiable and that P is a polynomial
function of degree sn such that P(x¿)
f(x¿)for i 1,...,n + 1 (see
49).
that
each
for
Show
x in [a,b] there is a number c in (a, b) such
page
=
=
that
f(x)
P(x)
-
Q(x)
=
f (n+13(c)
-
.
(n + 1)!
Hint: Consider the function
F(t)=Q(x)[f(t)-P(t)]-Q(t)[f(x)-P(x)].
Show that F is zero at n + 2 different points in
44.
use
Problem 42.
Prove that
computing 5
(without
45.
[a,b], and
2 decimal places!).
to
Prove the following slight generalization of the Mean Value Theorem: If f
is continuous and differentiable on (a, b) and lim f (y) and lim f (y) exist,
y->a+
then there is some x in
yab-
(a, b) such that
lim f (y)
f'(x)
lim f (y)
-
y-+b-
y->a+
=
b
-
a
(Yourproof should begin: "This is a trivial consequence
Theorem because
of the
Mean Value
".)
...
46.
Prove that the conclusion
in the form
of the Cauchy Mean Value Theorem can be written
f (b) f (a)
f'(x)
-
g(b)
-
_
¯
g'(x)'
g(a)
that g(b)
are never simultaneously O on (a, b).
under
*47.
the additional
assumptions
¢
g(a) and that
and g'(x)
Prove that if f and g are continuous on [a,b] and differentiableon (a, b),
¢ 0 for x in (a, b), then there is some x in (a, b) with
and g'(x)
f'(x)
f (x) f (a)
g'(x)
g(b)
-
g(x)
-
Hint: Multiply out first, to see what this really says.
48.
f'(x)
What is wrong with the following use of THôpital's Rule:
.
hm
x->lX2
x3+x-2
=
-3x
+2
-4.)
(The limit is actually
lim
x-1
3x2+1
-3
2x
=
lim
x-i
-
6x
2
=
3.
210 Derivativesand Integrals
49.
Find the following limits:
x
tan
lim
(i)
x->0
.
I
cos2
50.
(ii)
lim
Find
f'(0) if
x
1
-
.
g(x)
¯¯¯
f (x)
0,
and g(0)
51.
=
g'(0)
0 and g"(0)
=
¢0
x
,
=
0,
=
x
17.
=
(nonerequiring
Prove the following forms of l'Hôpital's Rule
new reasoning).
(a) If
lim
x->a+
f (x)
lim g(x)
=
lim f(x)/g(x)
lim f(x)
lim f(x)/g(x)
->
l, then
=
for limits from below).
0, and
=
=
->
-oo,
=
a* or x
by x
(c) If lim f(x)
a¯).
-+
0, and
lim f'(x)/g'(x)
Hint:
l (and similarly for
lim g(x)
=
=
x-voo
x->oo
lim
lim f'(x)/g'(x)
x-va+
lim f'(x)/g'(x)
oo, then
oo (andsimilarly for
or if x
a is replaced
lim g(x)
=
f(x)/g(x)
X->OO
(andsimilarly
l
=
x-+a+
(b) If
0, and
=
x-va+
any essentially
=
l, then
=
x->oo
-oo).
Consider
lim f(1/x)/g(1/x).
lim g(x)
lim f(x)
X400
lim f(x)/g(x) = oo.
x-voo
(d) If
52.
0, and
=
=
X->OO
There is another
manipulations:
then lim
x-voo
(a) For
=
oo, then
form ofTHôpital's Rule which requires more than algebraic
lim f'(x)/g'(x)
lim f (x)
lim g(x)
l,
If X->QC
oo, and X-Þ00
1400
=
f (x)/g(x)
every e
lim f'(x)/g'(x)
X-+00
>
=
=
=
l. Prove this as follows.
0 there is a number
f'(x)
a such
that
-l
forx>a.
<s
g'(x)
Apply the Cauchy Mean Value Theorem to
f
and g on
>
a.
that
f (x) f (a)
-
g(x)
-
g(a)
(Why can we assume g(x)
-
-
l
g(a)
<
e
for x
¢ 0?)
[a,x]
to show
11. Syn$icanceof theDerivative 211
(b) Now write
f (x) f (a)
f (x)
-
g(x)
(whycan
-
g(a)
that
we assume
g(x)
g(a)
g(x)
f (x)
f (x) f (a)
-
-
f (x)
f (a) ¢ 0 for large
-
x?) and conclude
that
f (x)
g(x)
53.
l
<
2e
for sufficiently large x.
To complete the orgy of variations on l'Hôpital's Rule, use Problem 52 to
are so many
prove a few more cases of the following general statement (there
possibilities that you should select just a few, if any, that interest you):
f(x)/g(x)
=
can be a or a+ or a¯ or oo or
and
( ) can be l or oo or
( ). Here []
can be 0 or oo or
*54.
-
-oo,
and
-oo,
{ }
-oo.
is differentiable on [a,b]. Prove that if the minimum
of
f on [a,b] is at a, then f'(a) > 0, and if it is at b, then f'(b) < 0.
(One half of the proof of Theorem 1 will go through.)
(b) Suppose that f'(a) < 0 and f'(b) > 0. Show that f'(x) 0 for some x
in (a, b). Hint: Consider the minimum of f on [a,b]; why must it be
somewhere in (a, b)?
(c) Prove that if f'(a) < c < f'(b), then f'(x) c for some x in (a, b). (This
(a) Suppose
that
f
=
=
is known as Darboux's Theorem.) Hint: Cook up an appropriate
function to which part (b)may be applied.
result
55.
Suppose that f is differentiable in
discontinuous at a.
one-sided
interval containing
a,
but that
f'
is
and
lim f'(x) cannot both exist. (This
is just a minor variation on Theorem 7.)
(b) Neither of these one-sided limits can exist even in the sense of being +oo
Hint: Use Darboux's Theorem (Problem 54).
or
(a) The
limits lim
some
f'(x)
-oo.
*56.
It is easy to find a function f such that |f | is differentiable but f is not.
for
1 for x rational and f (x)
For example, we can choose f (x)
continuous,
nor is this a mere
x irrational. In this example f is not even
coincidence: Prove that if |f| is differentiable at a, and f is continuous at a,
then f 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?
-1
=
=
=
*57.
y ¢ 0 and let n be even. Prove that x" + y" = (x + y)" only
when x = 0. Hint: If xo" + y" = (xo + y)", apply Rolle's Theorem to
f (x) = x" + y" (x + y)" on [0,xo]-
(a) Let
-
212 Derivativesand Integrals
(b) Prove that
if y ¢ 0 and n is odd, then x" + y"
=
(x + y)" only if
x
=
0
orx=--y.
**58.
Use the method of Problem 57 to prove that if
every tangent line to f intersects f only once.
**59.
Prove even more generally that if
intersects f only once.
*60.
f'
n
is even and
f (x)
=
x", then
is increasing, then every tangent line
Suppose that f (0) 0 and f' is increasing. Prove that the function g(x)
f(x)/x is increasing on (0, oo). Hint: Obviously you should look at g'(x).
Prove that it is positive by applying the Mean Value Theorem to f on the
right interval (itwill help to remember that the hypothesisf(0)
0 is essenshown
tial, as
by the function f (x) 1 + x2L
=
=
=
=
*61.
Use derivatives to prove that if n
1, then
>
-1
(1 +
62.
x)"
>
for
1 + nx
that
(notice
equality holds for x
Let f (x)
x4
-
2
=
<
x
<
=
<
x
0).
1/x for x ¢ 0, and let f (0)
(a) Pove that 0 is a local minimum
(b) Prove that f'(0) f"(0) 0.
0 and 0
=
0 (Figure 32).
point for f.
=
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 interval to the left.
FIGURE
*63.
32
if f'(a) > 0 and f' is continuous
some interval containing a.
(a) Prove that
at a, then
f is increasing in
The next two parts of this problem show that continuity of f' is essential.
x2 sin 1/x, show that there
g(x)
are numbers
with g'(x)
with
g'(x)
1 and also
(b) If
=
-1.
=
=
x arbitrarily
close to
0
of theDerivative 213
11. Srgngicance
0 < a < 1. Let f (x)
ax + x2 sin 1/x for x ¢ 0, and let
f (0) 0 (seeFigure 33). Show that f is not increasing in any open
interval containing 0, by showing that in any interval there are points x
with f'(x) > 0 and also points x with
f'(x) < 0.
(c) Suppose
=
=
/
FIGURE
0,
x-0
33
The behavior of f for a à 1, which is much more difficult to analyze, is
discussed in the next problem.
**64.
ax + x2 sin
1/x for x ¢ 0, and let f (0) 0. In order to find
the sign of f'(x) when œ à 1 it is necessary to decide if 2x sin 1/x cos 1/x
is <
for any numbers x close to 0. It is a little more convenient to
consider the function g(y)
2(sin y)/y
cos y for y ¢ 0; we want to know
if g(y) <
for large y. This question is quite delicate; the most significant
but this happens only
part of g(y) is cos y, which does reach the value
when sin y = 0, and it is not at all clear whether g itself can have values
<
The obvious approach to this problem is to find the local minimum
values of g. Unfortunately, it is impossible to solve the equation g'(y)
0
explicitly, so more ingenuity is required.
Let f (x)
=
=
-
-1
=
-
-1
-1,
-
-1.
=
(a) Show that
if g'(y)
=
0, then
cos y
=
(siny)
and conclude that
g(y)
=
(siny)
(2-y2
(2
,
2y
+y
2y
2
.
214 Derivativesand Integrals
(b) Now show
that if g'(y)
=
0, then
4y2
sin2
4 + y4
and conclude
that
2+y2
|g(y)
=
.
4+ y*
fact that (2 + y2)/J4+ y4 > 1, show that if a
not increasing in any interval around 0.
(c) Using the
(d) Using the
f
**65.
4 + y4
fact that lim (2 + y2)|
y->oo
=
1, then
1, show that if a
=
>
is
f
1, then
is increasing in some interval around 0.
A function f is increasing
at a if there is some number 8
f (x) > f (a)
if a
<
x
<
>
0 such that
a + ô
and
f (x) < f (a) if
a
Notice that this does not mean that
&< x
-
<
a.
is increasing in the interval (a 8,
shown
example,
the
function
for
8);
in Figure 33 is increasing at 0, but
+
a
is not an increasing function in any open interval containing 0.
f
-
(a) Suppose that f is continuous on [0,1] and that f is increasing at a for
every a in [0,1]. Prove that f 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 f on [b,1] must be at b.
(b) Prove part (a)without the assumption that f is continuous, by consider:
ing for each b in [0,l] the set Sb =
f (y) > f (b) for all y in [b,x]}.
of
the
is
not necessary for the other parts.) Hint:
problem
(This part
Prove that Sb {x: b 5 x 5 1) by considering sup Sb·
(c) If f is increasing at a and f is differentiable at a, prove that f'(a) > 0
(thisis easy).
(d) If f'(a) > 0, prove that f is increasing at a (goright back to the definition
{x
=
of
f'(a)).
(e) Use parts (a)and (d)to show, without using the Mean Value Theorem,
that if f is continuous on [0, 1] and f'(a) > 0 for all a in [0,1], then f
is increasing on [0,1].
(f) Suppose that f is continuous on [0, 1] and f'(a) 0 for all a in (0, 1).
Apply part (e)to the function g(x)
f(x) + ex to show that f(l)
show
that f(1) f(0) < e by considering h(x)
Similarly,
f(0) >
ex
f (x). Conclude that f (0) f (1).
=
=
-
-e.
-
-
=
=
of theDerivative 215
11. Sagnÿicance
This particular proof that a function with zero derivative must be constant has
many points in common with a proof of H. A. Schwarz, which may be the
first rigorous proof ever given. Its discoverer, at least, seemed to think it was.
See his exuberant letter in reference [40]of the Suggested Reading.
**66.
is a constant function, then every point is a local maximum point
for f. It is quite possible for this to happen even if f is not a constant
1 for x >
0 for x < 0 and f(x)
function: for example, if f(x)
0. But prove, using Problem 8-4, that if f is continuous on [a,b] and
every point of [a,b] is a local maximum point, then f is a constant
function. The same result holds, of course, if every point of [a,b] is a
local minimum point.
(b) Suppose now that every point is either a local maximum or a local minimum point for f (butwe don't preclude the possibility that some points
Prove that f is conare local maxima while others are local minima).
that
<
Suppose
We
stant, as follows.
can assume that
f (ao) f (bo).
f(ao) < f(x) < f(bo) for ao < x < bo. (Why?) Using Theorem1
of the Appendix to Chapter 8, partition
[ao,bo] into intervals on which
also
choose
the lengths of these inter<
sup f
f (f (bo) f (ao))/2;
ao)/2.
vals to be less than (bo
Then there is one such interval [ai,bij
with ao < ai < b1 < bo and f(ai) < f(bi). (Why?) Continue inductively and use the Nested Interval Theorem (Problem 8-14) to find a
point x that cannot be a local maximum or minimum.
(a) If f
=
=
-inf
-
-
**67.
point for f on A if f (x) > f (y)
a strict maximum
with the definition of an ordinary
for all y in A with y ¢ x (compare
maximum
point is defined in the
point). A local strict maximum
maximum
obvious way. Find all local strict
points of the function
(a) A point x is called
0,
f (x)
1
-
--,
x irrational
p
x
m lowest terms.
q
.
=
-
q
that a function can have a local strict maximum
It seems quite
the above example might give one pause for
at every point (although
thought). Prove this as follows.
(b) Suppose that every point is a local strict maximum point for f. Let
ai < l such
xi be any number and choose a1 < xi < b1 with bi
Xl
>
all
in
Let
be any point in
that f(xt)
x2 Ý
f(x) for x
[ai, bi].
(ai, bi) and choose al a2 < x2 < b2 5 bi with b2 a2 < such that
f(x2) > f(x) for all x in [a2,b2]. Continue in this way, and use the
Nested Intenval Theorem (Problem 8-14) to obtain a contradiction.
unlikely
-
_<
-
216 Derivativesand Integrals
CONVEXITY AND CONCAVITY
APPENDIX.
Although the graph of a function can be sketched quite accurately on the basis
of the information provided by the derivative, some subtle aspects of the graph are
revealed only by examining the second derivative. These details weœ purposely
enough without woromitted previously because graph sketching is complicated
rying 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 definition is geometric in nature (seeFigure 1).
DEFINITION
1
A function f is convex
segment
joining(a, f
on an interval, if for all a and b in the interval, the line
(a)) and (b, f (b)) lies above the graph of f.
The geometric condition appearing in this definition can be expressed in an
way that is sometimes more useful in proofs. The straight line between
(a, f (a)) and (b. f (b)) is the graph of the function g defmedby
analytic
f (b) f (a)
-
g(x)
=
fbatax
This line lies above the graph of
(b,f(b))
(x
a) +
---
if g(x)
f (b) f (a)
(x
>
f (a).
f(x),
that is, if
-
b
(a,f (a))
-
a) +
f (a) > f (x)
a)
f (x) f (a)
a
-
or
f (b) f (a) (x
-
I
i
-
>
-
or
FIGURE
1
f (b) f (a)
-
b-a
2
A function
a<x<bwehave
f
is convex
-
x-a
definition of convexity.
We therefore have an equivalent
DEFINITION
f (x) f (a)
on an
interval if for a, x, and b in the interval with
f (x) f (a)
--
x--a
f (b) f (a)
-
b--a
11, Appendix.Convexipand Concaviþ 217
"over"
in Definition 1 is replaced by
inequality in Definition 2 is replaced by
If the word
(a,f(a))
f (x)
(b,f(b))
f (a)
-
"under"
or, equivalently, if the
f (b) f (a)
-
the definition of a concave function (Figure 2). It is not hard to see that
the concave functions are precisely the ones of the form f, where f is convex.
For this reason, the next three theorems about convex functions have immediate
corollaries about concave functions, so simple that we will not even bother to
State them.
Figure 3 shows some tangent lines of a convex function. Two things seem to be
we obtain
-
FIGURE
2
true:
(1)The graph of f lies above the tangent line
(a, f (a)) itself (thispoint is called the point
at (a, f (a)) except at the point
of contact of the tangent line).
of
then
the
slope
the
<
tangent line at (a, f(a)) is less than the slope
(2)If a b,
of the tangent line at (b, f (b));that is, f' is increasing.
As a matter of fact these observations
FIGURE
THEOREM
1
PROOF
are true, and the proofs are not difficult.
3
be convex. If f is differentiable at a, then the graph of
the tangent line through (a, f (a)), except at (a, f (a)) itself If a
differentiableat a and b, then f'(a) < f'(b).
Let
If 0
f
<
hi
A nonpictorial
a
a + hi
<
lies above
b and f is
hz, then as Figure 4 indicates,
<
(1)
<
f
<
f(a+hi)-
f(a)
hi
<
f(a+h2)-
f(a)
h2
proof can be derived immediately from Definition 2 applied
a + h2. Inequality (1)shows that the values of
f(a+h)- f(a)
h
to
218 Derivativesand Integrals
I
FIGURE
I
I
a+haa‡h,
a
4
decrease as h
->
0+. Consequently,
f'(a)
<
f(a+h)-
f(a)
h
for h
>
0
(infact f'(a) is the greatest lower bound of all these numbers). But this means that
for h > 0 the secant line through (a, f (a)) and (a + h, f (a + h)) has larger slope
than the tangent line, which implies that (a + h, f (a + h)) lies above the tangent
line (ananalytic translation of this argument is easily supplied).
For negative h there is a similar situation (Figure 5): if h2 < hi < 0, then
f(a+hi)-
f(a)
f(a+h2)-
hi
f(a)
h2
This shows that the slope of the tangent line is greater than
f(a+h)h
(infact f'(a) is the
I
FIGURE
5
forh<0
least upper bound of all these numbers), so that f (a + h) lies
< 0. This
proves the first part of the theorem.
above the tangent line if h
a‡h:
f(a)
1
1
a+h2
a
and Concavig 219
11, Appendix.Convexity
I
FIGURE
I
6
Now suppose that a
<
f'(a)
<
b. Then, as we have already seen (Figure 6),
f (a + (b
a))
-
b
f (a)
-
a
-
.
smce
b
--
a
>
0
b
<
0
f (b) f (a)
-
b
-
a
and
b))
a b
(b + (a
f'(b) > f
¯
f (a)
-
-
f (b)
-
-
f (b)
I
FIGURE
I
-
f (b) f (a)
-
¯
a-b
I
.
smce a
b-a
Combining these inequalities, we obtain f'(a)
<
f'(b).
Theorem 1 has two converses. Here the proofs will be a little more difficult.
We begin with a lemma that plays the same role in the next theorem that Rolle's
Theorem plays in the pmof of the Mean Value Theorem. It states that if f'
is increasing, then the graph of f lies below any secant line which happensto be
horizontal.
7
'
LEl
IA
f (x)
PROOF
is differentiable and f'is increasing. If a
f (a) f (b) for a < x < b.
Suppose
<
f
<
b and
f(a)
=
f(b), then
=
Suppose first that f(x) > f(a) = f(b) for some x in (a, b). Then the maximum
f on [a,b] occurs at some point xo in (a, b) with f (xo)> f (a) and, of course,
f'(xo) 0 (Figure 7). On the other hand, applying the Mean Value Theorem to
the interval [a,xo], we find that there is xi with a < xi < xo and
of
=
f (xo) f (a) >
-
f'(xi )
=
xo
-
a
0,
220 Derivativesand Integrals
contradicting
the fact that f' is increasing. This proves that
b, and it only remains to prove that f (x) f
_<
f (x)
f (a)
=
f (b)
for a < x <
(a) is also impossible
for x in (a, b).
Suppose f(x)
f(a) for some x in (a, b). We know that f is not constant on
x]
it
[a, (if were, f' would not be increasing on [a,x]) so there is (Figure 8) some
xi with a < xi < x and f (xi) < f (a). Applying the Mean Value Theorem to
[x1,x] we conclude that there is x2 with xi < x2 < x and
=
=
a
x,
x.
&
t
f(X)
FIGURE
j'(X2)
8
f(Xi)
-
>
=
x
--
0.
x1
On the other hand, f'(x)
0, since a local maximum
contradicts
the hypothesis that f' is increasing.
=
at x.
occurs
Again this
We now attack the general case by the same sort of algebraic machinations that
we used in the proof of the Mean Value Theorem.
THEOREM
2
If f is differentiable and f' is increasing, then f is convex.
Let a
PRooF
<
b. Define g by
f (b) f (a)
-
g(x)
,
f (x)
=
-
b-a
(x
-
It is easy to see that g' is also increasing; moreover, g(a)
the lemma to g we conclude that
g(x)
(a, f(a))
<
f (a)
if a
<
x
<
a).
=
g(b)
=
f(a). Applying
b.
'
In other words, if a
<
x
<
b, then
f (b) f (a) (x-a)<f(a)
-
f(x)-
.
b
--
a
or
f (x) f (a)
-
FIGURE
9
f (b) f (a)
-
b---a
x-a
Hence f is convex.
THEOREM
s
PROOF
is differentiable and the graph of
point of contact, then f is convex.
If
f
f lies above
each tangent
line except at the
b. It is clear from Figure 9 that if (b, f (b)) lies above the tangent line at
(a, f (a)), and (a, f (a)) lies above the tangent line at (b,f (b)),then the slope of
the tangent line at (b,f (b)) 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
Let a
<
g(x)=
f'(a)(x-a)+
f(a),
11, Appendix.Convexigand Concavig 221
and since
(b, f (b)) lies above the tangent line, we have
(1) f(b) > f'(a)(b-a)+
f(a).
Similarly, since the tangent line at (b,f (b)) is the graph of
h(x)
and
(a, f (a)) lies above
=
f'(b)(x
b) + f (b),
-
the tangent line at
(b, f (b)),we have
(2) f (a) > f'(b)(a
-
b) +
f (b).
It follows from (1)and (2)that f'(a) < f'(b).
It now follows from Theorem 2 that f is convex.
If a function f has a reasonable
second
derivative, the information given in these
in which f is convex or concave.
theorems can be used to discover the regions
Consider, for example, the function
f(x)
1
=
1+ x2
.
For this function,
-2x
f'(x)
Thus
f'(x)
=
0 only for x
=
0, and
-
(1 +
f (0)
f'(x)
>
f'(x)
<
0
0
=
x2)2
.
1, while
if x
if x
<
>
0,
0.
Moreover,
0 for all x,
f (x) 0 as xe oo or
f is even.
f (x) >
-+
f (x)
FIGURE
=
-oo,
1
1 ‡ x*
10
The graph of f therefore looks something like Figure 10. We now compute
222 Derivativesand Integrals
f"(x)
(1 +
=
x2)2(-2)
+ 2x
-
[2(1+ x2)
-
24
(1 + x2)4
2(3x2
-
1)
(1 + x2)3
It is not hard to determine the sign of f"(x). Note first that f"(x) 0 only when
Since f" is clearly continuous, it must keep the same sign
or
x
of
each
the sets
on
=
=
N
-N.
Since we easily compute,
for example, that
f"(-1)
f"(0)
f"(1)
=
-2
(
>
}
>
=
=
<
0,
0,
0,
we conclude that
Since f"
>
>
f"
<
0 means
(·--oo,-Ñ)
and
f is
FIGURE
f"
f'
--Ñ) and (Ñ,
(-N, Ñ).
0 on (-oo,
0 on
is increasing, it follows from Theorem 2 that f is convex on
oo), while on (-5,
f is concave (Figure 11).
(Ñ,
Ñ)
f is concave
convex
oo),
f is convex
11
(G,
Notice that at
j) the tangent line lies below the part of the graph to the
oo), and above the part of the graph to the left,
f is convex on
since f is concave on
thus the tangent line crosses the graph. In
general, a number a is called an inflection point of f if the tangent line to the
and
graph of f at (a, f (a)) crosses the graph; thus
are inflection
x2).
of
points
Note that the condition f"(a) 0 does not ensure
f (x) 1/(1 +
that a is an inflection point of f; for example, if f (x) x4, then f"(0) 0, but
f is convex, so the tangent line at (0, 0) certainly doesn't cross the graph of f. In
order for a to be an inflectionpoint of a function f, it is necessary that f" should
have diEerent signs to the left and right of a.
right, since
(Ñ,
(-Ñ, 5);
N
-Ñ
=
=
=
=
l l, Appendix.Convexipand Concavig 223
This example illustrates the procedure which may be used to analyze any function f. After the graph has been sketched, using the information provided by f',
the zeros of f" are computed and the sign of f" is determined on the intervals
between consecutive zeros. On intervals where f" > 0 the function is convex;
on intervals where f" < 0 the function is concave. Knowledge of the regions of
of other
convexity and concavity of f can often prevent absurd misinterpretation
which
data about f. Several functions,
can be analyzed in this way, are given in
the problems, which also contain some theoretical questions.
To round out our discussion of convexity and concavity, we will prove one further
result that you may already have begun to suspect.
We have seen that convex and
the
that
every tangent line intersects the graph
property
concave functions have
will
convince
probably
you that no other functions have
just once; a few drawings
this property. The proof of this assertion is rather tricky; it is closely related to the
proof of Theorem 2 of the next chapter, and is probably best deferred until after
that proof has been read.
THEOREM
4
PROOF
is differentiable on an interval and intersects each of its tangent lines just
once, then f is either convex or concave on that interval.
If
f
There are two parts to the proof.
line can intersect the graph of f in threedifferent
points. Suppose, on the contrary, that some straight line did intersect the graph
of f at (a, f (a)), (b, f (b)) and (c, f (c)), with a < b < c (Figure 12). Then we
(1)First
would
M
we claim that no straight
have
f (b) f (a)
·
¡
¡
a
b
-
b-a
c-a
Consider the function
c
) ý(g)
_
x
FIGURE
f (c) f (a)
-
(1)
-
for x in
a
[b,c].
g(c). So by Rolle's Theorem, there is some number
that g(b)
in (b,c) where 0 g'(x), and thus
Equation
12
(1)says
=
x
=
O
=
(x
a) f'(x)
-
or
j
f'(x)
=
f (x) f (a)
x
But this says (Figure 13) that the tangent
contradicting
the hypotheses.
I
I I
a
b
x
13
-
.
a
line at (x,f (x)) passes through (a, f (a)),
I
e
(2)Suppose that
ao
<
bo
x,
y,
FIGURE
-
-
.
,
[f (x) f (a)]
-
Zt
<
=
=
=
co and ai
<
bl
(1 t)ao+tal
(1 t)bo + tbi
(Î -1)CO #fCI
<
c1 are points in the interval. Let
-
-
0 yt 5 1.
224 Derivativesand Integrals
(Problem 4-2) the points x, all lie between ao
for y, and z,. Moreover,
Then xo = ao and xi = at and
and ai, with analogous statements
<
x,
y,
<
for
z,
Osts
1.
Now consider the function
f (yr) f (x,)
-
f (z,) f (x,)
-
for 0 < ts 1.
ze x,
By step (1),g(t) ¢ 0 for all t in [0,1]. So either g(t) > 0 for all t in [0,1] or
g(t) < 0 for all t in [0, 1]. Thus, either f is convex or f is concave (compare
pages 231-232).
g(t)
=
y,
-
-
-
x,
PROBLEMS
1.
Sketch, indicating regions of convexity and concavity and points of inflection,
the functions in Problem 11-1 (consider
(iv)as double starred).
2.
Figure 30 in Chapter 11 shows the graph of
3.
Find two convex functions f and g such that f(x)
an integer.
4.
is convex on an interval if and only if for all x and y in the
interval we have
Show that
f'. Sketch the graph
=
of
f.
g(x) if and only if x is
f
f(tx+(1-t)y)<tf(x)+(1-t)/(y),
for0<t<1.
(This is just a restatement of the definition, but a useful one.)
5.
and g are convex and f is increasing, then f og is convex.
(It will be easiest to use Problem 4.)
(b) Give an example where gof is not convex.
(c) Suppose that f and g are twice differentiable. Give another proof of the
œsult
of part (a)by considering second derivatives.
6.
and convex on an interval. Show that
either f is increasing, or else f is decreasing, or else there is a number c
such that f is decreasing to the left of c and increasing to the right of c.
(b) Use this fact to give another proof of the result in Problem 5(a) when f
(a) Prove that if f
(a) Suppose
that
f is differentiable
(You will have to be a little careful
f'(g(y)) for x < y.)
f'(g(x))
assuming f differentiable. You will
without
in
part (a)
(c)
several
of
keep
track
have to
different cases, but no particularly clever
ideas are needed. Begin by showing that if a < b and f (a) < f (b),then
f is increasing to the right of b; and if f (a) > f (b), then f is decreasing
to the left of a.
and g are
differentiable.
(one-time)
when comparing
Prove the result
*7.
Let
f(x)
f
>
and
be a twice-differentiable function with the following properties:
0. Prove that
0 for x :> 0, and f is decreasing, and f'(0)
=
11, Appendir. Convexigand Concavig 225
0 for some x > 0 (sothat in reasonable cases f will have an inflection point at x-an example is given by f (x) 1/(1+x2)). Every hypothesis
in this theorem is essential, as shown by f (x) 1 x2, which is not positive
for all x; by f (x) x2, which is not decreasing; and by f (x) = 1/(x + 1),
which does not satisfy f'(0)
0. Hint: Choose xo > 0 with f'(xo) < 0. We
for
have
f'(y) 5 f'(xo) all y > xo. Why not? So f'(xi) > f'(xo)for
cannot
>
xo. Consider f' on [0,xt].
some xi
f"(x)
=
=
=
-
=
=
*8.
that if f is convex, then f ([x + y]/2) < [f (x) + f (y)]/2,
that f satisfies this condition.
Suppose
Show that f(kx + (1 k)y) <
(b)
rational
whenever
number, between 0 and 1,
kf (x) + (1 k) f (y)
k is a
of the form m/2". Hint: Part (a)is the special case n
1. Use induction,
employing part (a)at each step.
(c) Suppose that f satisfies the condition in part (a)and f is continuous.
Show that f is convex.
(a) Prove
-
-
=
*9.
Let pl.
-
-
-
,
numbers with
Pn by positive
p¿
=
1.
i=1
(a) For any
numbers
xi,
x, show that
...,
p¿x¿ lies between the smallest
i=1
and the
largest x¿.
n-1
n-i
p;x;, where t
(b) Show the same for (lft)
=
i=1
mequalay: If f
(c) Prove Jensen's
f
p¿.
i=1
is convex, then
Hint: Use Problem 4, noting that p,
=
f
p¿x;
i=l
1 t. (Part
(b)is needed
-
n-1
that
(1/t)
i
p¿x;
p¿f
5
i=1
is in the domain of f if x1,
...,
(x;).
to show
x, are.)
i=1
*10.
function f, the right-hand derivative, lim [f(a + h) f (a)]/h,
h->0+
is denoted by f '(a), and the left-hand derivative is denoted by f.'(a).
The proof of Theorem 1 actually shows that f ' and f_' always exist if
f is convex. Check this assertion, and also show that fe' and f_' are
increasing, and that f_'(a) 5 f,'(a).
Show that if f is convex, then fo'(a) f_'(a) if and only if f+' is continuous at a. (Thus f is differentiable precisely when f,' is continuous.)
Hint: [f(b) f(a)]/(b a) is close to f_'(a) for b < a close to a, and
f,'(b) is less than this quotient.
(a) For any
**(b)
-
=
-
*11.
(a) Prove that
-
a convex function on R, or on any open interval, must be
continuous.
an example of a convex function on a closed interval that is not
continuous, and explain exactly what kinds of discontinuities are possible.
(b) Give
226 Derivativesand Integrals
12.
Call a function f weakly convexon an interval if for a
we have
f (x) f (a)
-
x-a
<
b
<
c in this interval
f (b) f (a)
-
5
b-a
a wealdy convex function is convex if and only if its graph
no straight line segments. (Sometimes a wealdy convex function
while convex functions in our sense are called
is simply called
(a) Show that
contains
"convex,"
a)
a
subsci
t onves
of
die pluir
"strictly
convex".)
(b) Reformulate
13.
b)
a nonconvex
FIGURE
[4
sul>set
of the plane
the theorems of this section for weakly convex functions.
A set A of points in the plane is called convexif A contains the line segment
joiningany two points in it (Figure 14). For a function f, let Ay be the set
of points (x, y) with y > f (x), so that A is the set of points on or above
y
the graph of f. Show that A is convex if and only if f is weakly convex,
in the terminology of the previous problem. Further information on convex
sets will be found in reference [19]of the Suggested Reading.
INVERSE FUNCTIONS
CHAPTER
We now have at our disposal quite powerful methods for investigating functions;
what we lack is an adequate supply of functions to which these methods may
be applied. We have studied various ways of forming new functions from oldusing these alone, we can
addition, multiplication,
division, and composition-but
rational
sine
the
the
produce only
functions
function, although frequently
(even
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 any other method
we discover.
If we recall that a function is a collection of pairs of numbers, we might hit upon
the bright idea of simply reversing all the pairs. Thus from the function
f
=
{(1,2), (3,4), (5,9), (13,8) },
g
=
{(2. 1), (4, 3), (9, 5), (8, 13) }.
we obtain
1 and g(4)
While f(1) 2 and f(3) 4, we have g(2)
Unfortunately, this bright idea does not always work. If
=
=
f
=
=
=
3.
{(1, 2), (3, 4), (5, 9), (13, 4) },
then the collection
{(2, 1), (4, 3), (9, 5), (4, 13) }
is not a function at all, since it contains both (4, 3) and (4, 13). It is clear where
the trouble lies: f (3) f (13), even though 3 ¢ 13. This is the only sort of thing
that can go wrong, and it is worthwhile giving a name to the functions for which
this does not happen.
=
DEFINITION
A function
f
is one-one
"one-to-one")
if f (a) ¢
(read
The identity function I is obviously
and so
one-one,
f (b) whenever
a
¢ b.
is the following modifica-
tion:
g(x)=
The function f (x)
=
x2
x¢3,5
3,
5,
x=5
x
is not one-one, since
g(x)=x2,
227
x,
=
3.
f (-1)
x>0
=
f (1), but if we define
228 Derivativesand Integrals
(andleave
g'(x)
=
number
2x
g undefined
>
0, for x
0), then g is one-one, because g is increasing (since
0). This observation is easily generalized: If n is a natural
for x
>
<
and
x>0,
f(x)=x",
then
f
is one-one.
If n is odd, one can do better: the function
f (x)
=
x"
for all x
is one-one (sincef'(x) = nx"¯' > 0, for all x ¢ 0).
It is particularly easy to decide from the graph of f whether f is one-one: the
condition f (a) ¢ f (b) for a ¢ b means that no horizontalline intersects the graph
of f twice (Figure 1).
a one-one function
a function which is not one-one
(b)
FIGURE
1
If we reverse all the pairs in
necessarily one-one function) f we obtain, in
It is popular to abstain from this procedure unless f 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.
any case, some collection
For any function f, the inverse of f, denoted by f¯I,
(a, b) for which the pair (b,a) is in f.
DEFINITION
THEOREM
(anot
of pairs.
1
PROOF
f-1
is a function if and only if
is the set of all pairs
is one-one.
f
Suppose first that f is one-one. Let (a, b) and (a, c) be two pairs in f-1. Then
(b,a) and (c,a) are in f, so a
f(c); since f is one-one this
f(b) and a
implies that b = c. Thus f¯l is a function.
Conversely, suppose that f
is a function. If f (b) f (c), then f contains
the pairs (b, f (b))and (c, f (c)) (c, f (b)), so (f (b),b) and (f (b),c) are in f¯I.
Since f¯1 is a function this implies that b c. Thus f is one-one.
=
=
¯1
=
=
=
12. InverseFunctions 229
The graphs of f and f¯1 are so closely related that it is possible to use the
graph of f to visualize the graph of f-1. Since the graph of f¯1 consists of all
pairs (a, b) with (b,a) in the graph of f, one obtains the graph of f¯I from the
graph of f by interchanging the horizontal and vertical axes. If f has the graph
shown in Figure 2(a),
FIGURE
I
1
I
2
I
I
2(a)
I
This procedure is awkward with books and impossible with blackboards, so it is
fortunate that there is another way of constructing the graph of f-1. The points
230 Derivativesand Integrals
and (b,a) are reflections of each other through the graph of I(x) = x,
which is called the diagonal (Figure 4). To obtain the graph of f¯I we merely
reflect the graph of f through this line (Figure 5).
(a, b)
FIGURE
FIGURE4
5
Reflecting through the diagonal twice will clearly leave us right back where we
started; this means that (f-1)-1
f, which is also clear from the definition. In
this
equation
conjunction with Theoæm 1,
has a significant consequence: if f
is a
is a one-one function, then the function f¯* is also one-one (since(f¯1
function).
There are a few other simple manipulations with inverse functions of which you
should be aware. Since (a, b) is in f precisely when (b, a) is in f¯', it follows that
=
-1
b
Thus f
x3, then
¯I
=
(b) is the
f¯*(b)is the
definition, 6.
The fact that
the same as
means
f(a)
number
(unique)
f¯I(x)
much more compact
f(f¯I(x))
a such that
y such that
is the number
form:
x,
f (a)
a'
a such that
unique number
=
a
=
=
f(y)
L
f¯
=
b; for example, if f (x)
b, and this number is, by
=
=
x can be restated in a
for all x in the domain of f¯'.
Moreover,
f¯I(f(x))
=
x,
for all x in the domain of
this follows from the previous equation
important equations can be written
upon
replacing
f;
f by
f¯I.
These two
fof¯*=1,
f¯1
(exceptthat
y
the right side will have a bigger domain if the domain of
not all of R).
f
or
f-I
is
12. InverseFunctions 231
Since many standard functions will be defined as the inverses of other functions,
it is quite important that we be able to tell which functions are one-one. We have
and
already hinted which class of functions are most easily dealt with-increasing
obviously
decreasing functions are
one-one. Moreover, if f is increasing, then f
is also increasing, and if f is decreasing, then f¯I is decreasing (theproof is left
to you). In addition, f is increasing if and only if f is decreasing, a very useful
fact to remember.
It is certainly not true that every one-one function is either increasing or decreasing. One example has already been mentioned, and is now graphed in Figure 6:
e
-
-/
FIGURE
6
g(x)=
3, 5
x,
x
3,
5,
x=5
x
3.
=
Figure 7 shows that there are even continuous one-one functions which are neither
increasing nor decreasing. But ifyou try drawing a few pictures you will soon agree
that every one-one continuous function defined on an interval is either increasing
or decreasing. It's possible to give a straightforward, but cumbersome, proof of this
much like Problem 6(c) in
fact that involves keeping track of a lot of cases (very
the previous Appendix). The following proof dispenses with all these unpleasant
details, although it is rather tricky.
THEOREM
2
If
f
is continuous and one-one
on that interval.
of an
interval, then f is either increasing or de-
creasing
PROOF
I et ao
bo be two numbers
<
in the interval. Since
f is one-one,
either
(i)
f (bo) f (ao) > 0
or
(ii)
f(bo) f(ao) <
we
know that
--
-
0.
We will assume that (i)is true, and show that the same inequality holds for any
ai < bi in the interval, so that f in increasing. (A similar argument shows that if
(ii)is true, then f is decreasing.)
Let
x,=(1-t)ao+tal
(1 t)bo + tb1
y,
=
7
for0stsl.
ao and xi = ai and the points xr all lie between ao and at
An analogous statement holds for y,. So x, and y, are all in the
Moreover, since ao < be and ai < bi, we also have
Then xo
FIGURE
-
=
x,<y,
(Problem 4-2).
domain of f.
for0stsl.
Now consider the function
g(t)=
f(yr) -f(Ir)
forO5 t 5 1.
Using Theorem 6-2, it is easy to see that g is continuous on [0, 1]. Moreover, g(t)
is never 0, since x, < y, and f is one-one. Consequently, g(t) is either positive for
all t in [0, 1] or negative for all t in
by the Intermediate Value
[0,1] (otherwise,
232 Derivativesand Integrals
Theorem it would also by 0 somewhere in [0,1]). But g(0)
> 0, which means that (i)also holds for ai, bl. I
g(1)
I
I
I
c
a
FIGURE
>
0 by
(i).So also
Henceforth we shall be concerned almost exclusively with continuous increasing
or decreasing functions which are defined on an interval. If f is such a function,
it is possible to say quite precisely what the domain of f¯1 will be like.
Suppose first that f is a continuous increasing function on the closed interval
f takes on every value between
[a,b]. Then, by the Intermediate Value Theorem,
f-I
of
and
the
closed interval [f(a), f(b)].
the
is
domain
f(a)
f(b). Therefore,
and
decreasing on [a,b], then the domain of f¯I is
Similarly,if f is continuous
b
a
[f(b),f(a)].
a
If f is a continuous increasing function on an open interval (a, b) the analysis
becomes a bit more difficult. To begin with, let us choose some point c in (a, b).
We will first decide which values > f (c) are taken on by f. One possibility is that
f takes on arbitrarily large values (Figure 8). In this case f takes on all values
>
f (c), by the Intermediate Value Theorem. If, on the other hand, f does not
take on arbitrarily large values, then A
{f (x) : c < x < b } is bounded above,
so A has a least upper bound œ (Figure 9). Now suppose y is any number with 1
a is the least
f(c) < y < œ. Then f takes on some value f(x) > y (because
actually
of
takes on
A). By the Intermediate Value Theorem, f
upper bound
the value y. Notice that f cannot take on the value a itself; for if a = f (x) for
a < x < b and we choose t with x < t < b, then f (t) > a, which is impossible.
Precisely the same arguments work for values less than f (c): either f takes on
all values less than f(c) or there is a number ß < f(c) such that f takes on all
values between ß and f (c), but not ß itself.
This entire argument can be repeated if f is decreasing, and if the domain of f
is R or (a, oo) or (-oo, a). Summarizing: if f is a continuous increasing, or
decreasing, function whose domain is an interval having one of the forms
-
=
FIGURE
9
(a, b), (-oo, b), (a, oo),
or
R,
then the domain of f ¯I is also an interval which has one of these four forms.
Now that we have completed this preliminary analysis of continuous one-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
s
PROOF
If f is continuous and one-one on an interval, then
f¯I
is also continuous.
We know by Theorem 2 that f is either increasing or decreasing. We might as
well assume that f is increasing, since we can then take care of the other case by
applying the usual trick of considering
We must show that
---
f.
-1(b)
lim
f-1
12. InverseFunctions 233
for each b in the domain of f-1. Such a number b is of the form f (a) for some
in the domain of f. For any e
if f(a)-ô<x<
>
0, we
to find a ô
want
Figure 10 suggests the way of finding 8
see the graph of f-1): since
a
0 such that, for all x,
f¯I(x)<a+e.
thena-e<
f(a)+ô,
>
that
(remember
by looking sideways you
a-s<a<a+s,
it follows that
f(a
-
f(a) 6
-
f(a)
e
-
i
I
I
-------
I
I
----
i
I
I
a-a
FIGURE
i
i
I
f(a-s)5
f(a+
e);
-
-
f(a)-ôand
f(a)+85
f(a+e).
Consequently, if
a+s
a
<
let 8 be the smaller of f (a + s) f (a) and f (a) f (a e). Figure 10 contains
the entire proof that this & works, and what follows is simply a verbal account of
the information contained in this picture.
Our choice of ô ensures that
I
i
I
f(a)
<
-
we
f(a) + a
e)
f(a)-B<x<
f(a)+8,
f(a-s)<x<
f(a+e).
then
lo
Since f is increasing,
f¯I
is also increasing, and we obtain
f¯I(f(a-e))
<
*(x)<
f
f¯I(f(a+s)),
i.e.,
f¯I(x)<a+e,
a-e<
which
(f(a), a)
-
,
/
L,
is precisely what we want.
Having successfully investigated continuity of f¯*, it is only reasonable to tackle
differentiability. Again, a picture indicates just what result ought to be true. Figure 11 shows the graph of a one-one function f with a tangent line L through
(a, f (a)). If this entire picture is reflected through the diagonal, it shows the graph
and the tangent line L' through (f(a), a). The slope of L' is the reciprocal
of the slope of L. In other words, it appears that
-1
(a,f(a))
/
L
/
FIGURE
/
/
11
(f¯')'(f(a))=
.
f'(a)
This formula can equally well be written in a way which expresses (f¯1y(b) directlÿ, (Or each b in the domain of f-I
(f
I)'(b)
continuity,
Unlike the argument for
involved when formulated analytically.
=
1
f'( f-1(b))
this pictorial
"proof"
There is another
becomes somewhat
approach
which
might
234 Derivativesand Integrals
be tried. Since we know that
f (f¯1
it is tempting to prove the desired formula by applying the Chain Rule:
-1)'(x)
f'(f¯1
1,
=
so
1
f'( f¯l(x))
f¯I
f(x)
-
is differentiable, since the Chain Rule
Unfortunately, this is not a proof that
f¯I
already
cannot be applied unless
is
known to be differentiable. But this arguis differentiable, and it can
ment does show what (f-1)'(x) will have to be if f
also be used to obtain some important preliminary information.
x
THEOREM
4
PROOF
If f is a continuous one-one function defined on an interval and
then f I is not differentiable at a.
f'(f-1(a))
=
0,
We have
f(f-1(x))=x.
1
If f¯i were differentiable at a, the Chain Rule would imply that
(a)
f'( f¯ (a)) (f¯')'(a)
-
1,
=
hence
0 (f-1)'(a)
=
-
which
is absurd.
1,
I
A simple example to which Theorem 4 applies is the function f (x) x3. Since
f'(0) 0 and 0 f¯I(0), the function f¯I is not differentiable at 0 (Figure 12).
Having decided where an inverse function cannot be differentiable, we are now
ready for the rigorous proof that in all other cases the derivative is given by the
formula which we have already "derived"
in two different ways. Notice that the
following argument uses continuig of f-1, which we have already proved.
=
=
(b)
FIGURE
12
THEOREM
5
=
Let f be a continuous one-one function defined on an interval, and suppose that
I(b), with
derivative f'(f¯1(b)) ¢ 0. Then f I is differf is differentiable at f
entiable at b, and
1
f¯I)'(b)
(
PROOF
Let b
=
f(a).
=
.
f'( f-1(b))
Then
-1(b)
lim
f¯* (b +
h)
-
h
h->o
=
lim
h->0
f
-1(b
+ h)
h
-
a
12. InverseFunctions 235
Now every number
b+h=
for a unique k
Then
really
(weshould
.
hm
I
¯
b + h in the domain of
f
can be written
in the form
f(a+k)
write k(h),
but we will stick with k for simplicity).
f~1(b+h)-a
h
hw0
f¯I(f(a+k))
-a
f(a+k)-b
A-o
k
=lim
.
f
Aso
(a + k)
f (a)
-
We are clearly on the right track! It is not hard to get an explicit expression
for k;
since
b+h=
we
f(a+k)
have
f"(b+h)=a+k
or
-1(b
k
-1(b).
+ h)
f
=
f
--
Now by Theorem 3, the function f¯1 is continuous
0 as h approaches 0. Since
at b. This means
that k
approaches
lim f(a+k)-
koo
f(a)
=
k
f'(a)
f'( f¯1(b)) ¢ 0,
=
this implies that
I
(f¯ )'(b)
1
=
f'(
.
f¯I
(b))
The work we have done on inverse functions will be amply repaid later, but here
is an immediate dividend. For n odd, let
f.(x)
for all x;
xn
=
for n even, let
f,(x)
Then f, is a continuous
one-one
x",
=
x
>
0.
function, whose inverse function is
gn(x)
=
=
x1/n
236 Derivativesand Integrals
By Theorem 5 we have, for x ¢ 0,
g,'(x)
1
=
f,'( f.-1
1
n( f.- l (x))"¯I
1
R(Xl|ngn-1
1
1
1-(I/n)
1
(1/n)-1
n
xa, and a is an integer or the reciprocal of a natural number, then
It is now easy to check that this formula is true if a is any rational
=
m/n, where m is an integer, and n is a natural number; if
a
Thus, if f (x)
f'(x)
=
number:
axa-1.
Let
=
f (x)
x"'7"
=
=
(x1/np=
then, by the Chain Rule,
j'(x)
=
I
m
l/n m-1
Î
1
(1/n)-1
n
[(m/n)-(1/n)]+[(1¡n)-l]
n
(m/n)-1
n
xa and a is rational,
x. for irrational a will have to be saved
the treatment of the function f(x)
the
for later-at
moment we do not even know the meaning of a symbol like x
Actually, inverse functions will be involved crucially in the definition of xa for
irrational a. Indeed, in the next few chapters several important functions will be
Although we now have a formula for
f'(x) when f (x)
=
=
.
defined in terms of their inverse functions.
PROBLEMS
1.
Find
f¯I
for each of the following f.
(i) f (x)
=
3
x +
(ii) f (x) (x
=
(iii) f (x)
=
-
1.
1)3.
x rational
irrational.
x
x,
.
-x,
.
-x2
(iv) f (x)
=
1--x
(v) f(x)=
(vi) f(x)
=
x
3
,
>
0
x<0.
x,
x¢a1,...,a,
ai+1
ai,
x=ag,
x=a..
x +
[x].
i=1,...,n-1
12. InverseFunctions 237
(vii)f (0.aia2a3 )
x
(viii)f (x) 1 x
.
.
.
=
Û.a2Gl
=
<
,
Describe the graph of
3.
f-I
is increasing and
is increasing and
is decreasing and
is decreasing and
(i) f
(ii) f
(iii) f
(iv) f
x
.
.
.
(Decimal
representation
is being used.)
<
1.
when
always positive.
always negative.
always positive.
always negative.
f is increasing,
Prove that if
.
-1
-
2.
ü3
then so is
f¯I,
and
similarly for decreasing
functions.
4.
If f and g are increasing, is f + g? Or
5.
(a) Proveg)¯Ithat
(fo
(b) Find
6.
7.
g? Or
g?
fo
Find
i
=
f (x)
=
¯1
f
-
if f and g are one-one, then fog is also one-one.
in terms of f¯1 and g¯1. Hint: The answer is not f¯1
in terms of f¯1 if g(x)
1 + f (x).
g¯I
Show that
f
in this case.
On which intervals
ax + b
is one-one if and only if ad
cx + d
[a,b]
will the
-
bc ¢ 0, and find
following functions be one-one?
(i) f(x)-x3-3x2.
x5 4
(ii) f(x)
=
8.
(iii) f(x)
=
(1+x2)-1.
(iv) f (x)
=
+ 1
x + 1
.
Suppose that f is differentiable with derivative f'(x)
that g
9.
x
=
f¯I
satisfies g"(x)
=
=
(1+
x3)¯I/2.
Show
(g(x)2
Suppose that f is a one-one function and that f¯1 has a derivative which is
0. Prove that f is difFerentiable. Hint: There is a one-step proof.
nowhere
10.
The Schwarzian derivative 2 f was defined in Problem 10-17.
if 9 f(x) exists for all x, then 2 f¯1(x)also exists for all x in
the domain of f
f-I(x).
Find
a formula for o
(b)
(a) Prove that
.
*11.
there is a differentiablefunction f such that [f (x)]5
4
all
that
expressed
0
for
Hint:
Show
be
inverse
x
x.
as an
f can
function. The easiest way to do this is to find f-1. And the easiest way
f¯I(y).
to do this is to set x =
(b) Find f' in terms of f, using an appropriate theorem of this chapter.
(c) Find f' in another way, by simply differentiating the equation defining f.
(a) Prove that
=
238 Derivativesand Integrals
The fimction in Problem 11 is often said to be denned irnplicitly by the
yS+y+x
0. The situation for this equation is quite special, 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 function may be defined
implicitly.
equation
12.
=
are the two diEerentiable functions f which are defined implicitly
1, i.e., which satisfy x2+[ f (x)]2 1
on (-1, 1) by the equation x2+y2
for all x in (-1, 1)? Notice that there are no solutions defined outside
(a) What
=
=
[-1, 1].
satisfy x2 +
[f(x)]2
-P
Which differentiable functions f satisfy [f (x)]3 3 f (x)
x3
will help to first draw the graph of the function g(x)
(b) Which functions f
*(c)
-
-
=
x?
=
-
Hint: It
3x.
In general, determining on what intervals a differentiable function is defined implicitly by a particular equation may be a delicate afTair, and is best discussed in the
context of advanced calculus. If we assume that f is such a differentiable solution,
however, then a formula for f'(x) can be derived, exactly as in Problem 11(c),by
differentiating both sides of the equation defining f (aprocess known as
"implicit
differentiation"):
13.
22
1. Notice that your
to the equation [f (x)]2 4
expected,
since there is more
will
this
only
involve
to be
answer
f (x); is
2
than one function defined implicitly by the equation y2
g
check that your answer works for both of the functions f found in
But
(b)
Problem 12(a).
(c) Apply this same method to [f(x)]3 3 f (x) x.
(a) Apply this
method
=
_
=
-
14.
to find f'(x) and f"(x) for the functions f
defined implicitly by the equation x3 + y3 7.
(b) One of these functions f satisfies f (-1) 2. Find f'(-1) and f"(-1)
for this f.
(a) Use implicit differentiation
=
=
15.
4
The collection of all points (x, y) such that 3x3 + 4x2y xy2 + 2y3
forms a certain curve in the plane. Find the equation of the tangent line to
this curve at the point (-1, 1).
16.
Leibnizian notation is particularly convenient for implicit differentiation. Because y is so consistently used as an abbreviation for f (x), the equation in x
and y which defmes f implicitly will automatically
stand for the equation
which
f is supposed to satisfy. How would the following computation be
-
written
in our notation?
+y3+g-1
y
4y'$+3y2Ë+y+xË=0,
dx
dx
dx
dy
--y
¯
Ï
4y3+3y2+x
=
12. InverseFunctions 239
17.
As long as Leibnizian notation has entered the picture, the Leibnizian notation for derivativesof inverse functions should be mentioned. If dy/dx
denotes the derivative of f, then the derivative of f¯I 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-1), is to be calculated when using the dx/dy
notation. What is the significance of the following computation?
x=y",
y
X1/n
=
dxil"
dy
1
1
¯
Ì
dx
nyn-1
dy
18.
Suppose that f is a differentiable one-one function with a nowhere zero
F(f¯I(x)).
xf¯1(x)
F'. Let G(x)
Prove that
derivative and that f
G'(x)
this
problem
tells
details,
(Disregarding
interesting
f¯1(x).
us a very
fact: if we know a function whose derivative is f, then we also know one
whose derivative is f-i. But how could anyone ever guess the function G?
Two different ways are outlined in Problems 14-17 and 19-15.)
=
=
-
=
19.
Suppose h is a function such that h'(x)
Find
(i)
(ii)
20.
sin2(sin(x
=
+ 1)) and h(0)
3.
=
(h¯I)'(3).
(߯I)'(3),
where
Find a formula for
ß(x)
=
h(x + 1).
(f¯I)"(x).
-l(x))
*21.
22.
Prove that if
f(k)
exists, and
then
is nonzero,
(f¯l)(k)(X)
eXists.
that an increasing and a decreasing function intersect at most once.
(b) Find two continuous increasing functions f and g such that f (x) g(x)
precisely when x is an integer.
(c) Find a continuous increasing function f and a continuous decreasing
function g, defined on R, which do not intersect at all.
(a) Prove
=
*23.
f¯I,
is a continuous function on R and f
prove that there is at
least one x such that f (x) x. (What does the condition f
f mean
geometrically?)
(b) Give several examples of continuous f such that f f¯1 and f (x) x
for exactly one x. Hint: Try decreasing f, and remember the geometric
interpretation. One possibility is f(x)
f¯I,
then
(c) Prove that if f is an increasing function such that f =
all
will
Although
the
geometric
interpretation
f (x) x for x. Hint:
2 lines) is to rule
be immediately convincing, the simplest proof (about
out the possibilities f (x) < x and f (x) > x.
(a) If f
=
¯1
=
=
=
=
-x.
=
=
*24.
Which functions have the property that the graph is still the graph of a function when
reflected
through the graph of --I
"antidiagonal")?
(the
240 Derivativesand Integrals
25. A function f is nondecreasing
if f (x)
sf (y) whenever
more precise we should stipulate that the domain of f
nonincreasing
function is defined similarly. Caution:
"increasing"
"nondecreasing,"
instead of
and
"strictly
x < y. (To be
be an interval.) A
Some writers use
increasing" for our
"increasing."
(a) Prove that
if f is nondecreasing, but not increasing, then f is constant
increasing" is not
on some interval. (Beware of unintentional
puns:
"not
the same as
"nonincreasing.")
(b) Prove that if
all x.
(c) Prove that
*26.
f
is differentiable and nondecreasing,
if f'(x)
>
then
f'(x)
0 for
>
0 for all x, then f is nondecreasing.
> 0 for all x, and that f is decreasing. Prove that
there is a continuous decreasing function g such that 0 < g(x) 5 f(x) for
(a) Suppose that f (x)
'
all x.
(b) Show that
we can even arrange
that g will satisfy lim g(x)/f(x)
X400
=
0.
of Curves 241
12, Appendix.ParametricRepresentation
APPENDIX.
PARAMETRIC REPRESENTATION OF CURVES
The material in this chapter serves to emphasize something that we noticed a
long time ago-a perfectly nice looking curve need not be the graph of a function
(Figure 1). In other words, we may not be able to describe it as the set of all points
(x, f (x)). Of course, we might be able to describe the curve as the set of all points
(f (x),x); for example, the curve in Figure 1 is the set of all points (x2,x). But
even this trick doesn't work in most cases. It won't allow us to describe the circle,
consisting of all points (x, y) with x2 + y2
1, or an ellipse, and it can't be used
to describe a curve like the one in Figure 2.
The simplest way of describing curves in the plane in general harks back to the
physical conception of a curve as the path of a particle moving in the plane. At
each time t, the particle is at a certain point, which has two coordinates;
to indicate
the dependence of these coordinates on the time t, we can call them u(t) and v(t).
Thus, we end up with two functions. Conversely, given two functions u and v, we
can consider the curve consisting of all points (u(t), v(t)). This curve is said to
by u and v, and the pair of functions u, v is called a
be represented parametrically
parametric representation of the curve. The curve represented parametrically by
=
FIGURE
1
v thus consists of all pairs
with x
and y
It is often
v(t)." Notice that the graph of a
u(t), y
described briefly as
curve x
function f can always be described parametrically, as the curve x t, y f (t).
Instead of considering a curve in the plane as defined by two functions, we
can obtain a conceptually simpler picture if we broaden our original defmition of
function somewhat. Instead of considering a rule which associates a number with
another number, we can consider a
c from real numbers to the plane,"
i.e., a rule c that associates, to each number t, a pointin theplane,which we can
denote by c(t). With this notion, a curve is just a function from some interval of
real numbers to the plane.
Of course, these two different descriptions of a curve are essentially the same:
functions u and v determines a single function c from the real
A pair of (ordinary)
numbers to the plane by the rule
u and
(x, y)
"the
=
u(t)
=
v(t).
=
=
=
=
"function
FIGURE
2
c(t)
=
(u(t), v(t)),
given a function c from the real numbers to the plane, each c(t)
the
plane,
is a point in
so it is a pair of numbers, which we can call u(t) and v(t),
unique
functions u and v satisfying this equation.
so that we have
In Appendix 1 to Chapter 4, we used the term
to describe a point in
the plane. In conformity with this usage, a curve in the plane may also be called
function." The conventions of that Appendix would lead us to
a
write c(t)
(c1(t),c2(t)), but in this Appendix we'll continue to use notation like
c(t)
(u(t), v(t)) to minimize the use of subscripts.
simple
example of a vector-valued function that is quite useful is
A
and, conversely,
"vector"
e(r)
"vector-valued
=
=
e(t)
FIGURE
3
Which
=
(cost,
sin t),
goes round and round the unit circle (Figure 3).
242 Derivativesand Integrals
functions f and g, we defined new functions f + g and f g
For two (ordinary)
by the rules
-
(1)
(2)
(f + g)(x)
=
g)(x)
=
(f
-
f (x) + g(x),
f (x) g(x).
-
Since we have defined a way of adding vectors, we can imitate the first of these definitions for vector-valued functions c and d: we define the vector-valued function
c+d by
(c+d)(t)
where the + on the right-hand
to saying that if
=
c(t) +d(t),
is now the sum ofvectors. This simply amounts
side
c(t)
=
(u(t), v(t)),
d(t)
=
(w(t), z(t)),
then
(c +d)(t)
=
v(t)) +
z(t)) (u(t)
+ w(t), v(t) + z(t)).
(w(t),
(u(t),
=
Recall that we have also defined a v for a number a and a vector v. To
extend this to vector-valued
functions, we want to consider an onlinary function
a
and a vector-valued function c, so that for each t we have a number a(t) and a
function a c by
vector c(t). Then we can define a new vector-valued
-
-
(a
-
c)(t)
where the on the right-hand
side
simply amounts to saying that
-
(«
-
c)(t)
=
a(t)
-
=
a(t)
-
c(t),
is the product of a number
(u(t), v(t))
=
(a(t) u(t),
-
a(t)
and a vector.
This
v(t)).
-
Notice that the curve a e,
-
(a
(a
,"
.
e)(t)
e)(t)=
4
a(t)sint),
-e)(t)
"graph
-
of a
in polar coordinates."
functions
generally, given any vector-valued
r and 8
function c, we can define new
by
c(t)
FIGURE
(a(t)cost,
is already quite general (Figure 4). In the notation of Appendix 3 to Chapter 4,
the point (a e)(t) has polar coordinates a(t) and t, so that (a
is the
Even more
'
-
=
r(t)
-
e(8(t)),
where r(t) is just the distance from the origin to c(t), and 0(t) is some choice of
the angle of c(t) (asusual, the function 9 isn't defined unambiguously,
so one has
careful
when
using
of
writing
arbitrary
c).
this
be
to
way
curve
an
We aren't in a position to extend (2)to vector-valued functions in general, since
However, Problems 2 and 4 of
we haven't defined the product of two vectors.
real-valued
products
Appendix 1 to Chapter 4 define two
o w and det(v, w). It
•
of Curves 243
12, Appendix.ParametricRepresentation
functions c and d, how we would define two
be clear, given vector-valued
ordinary (real-valued)
functions
should
and
c•d
det(c, d).
Beyond imitating simple arithmetic operations on functions, we can consider
more interesting problems, like limits. For c(t)
(u(t), v(t)), we can define
=
limc(t)
(*)
=
lim(u(t), v(t))
lim v(t)
(limu(t),
to be
.
Rules like
lim c + d
lim a
c
-
=
lim c + lim d,
=
lim a (t) lim c
-
follow immediately. Problem 10 shows how to give an equivalent definition that
imitates the basic definition of limits directly.
Limits lead us of course to derivatives. For
c(t)
(u(t), v(t))
=
define c' by the straightforward
we can
c'(a)
definition
v'(a)).
(u'(a),
=
We could also try to imitate the basic definition:
.
c'(a)
lun
=
c(a + h)
-
c(a)
,
h
amo
understood
where the fraction on the right-hand side is
1
-
·
[c(a+
h)
-
to mean
c(a)].
As a matter of fact, these two defmitions are equivalent, because
c(a+h)-c(a)
u(a+h)-u(a)
v(a+h)-v(a)
=
lim
hm
ho0
h-o
h
h
h
c(a+h)
.
,
c
=
lim
u(a+h)-u(a)
h
hw0
.
,
hw0
by our definition
=
c(a)
Inn
v(a+h)-v(a)
h
(*)of limits
v'(a)).
(u'(a),
Figure 5 shows c(a + h) and c(a), as well as the arrow from c(a) to c(a + h);
c(a), except
as we showed in Appendix 1 to Chapter 4, this arrow is c(a + h)
would
moved over so that it starts at c(a). As h
0, this arrow
appear to move
closer and closer to the tangent of our curve, so it seems reasonable
to defmethe
when
c'(a),
straight
along
c'(a)
of
c(a)
the
line
is moved
to be
tangent line
c at
other
words,
of
c(a).
In
we define the tangent line
c at c(a)
over so that it starts at
-
->
FIGURE
5
as the set of all points
c(a)+s·c'(a);
244 Derivativesand Integrals
1 we get c(a) + c'(a), etc. (Note,
0 we get the point c(a) itself, for s
for s
make
however, that this definition does not
0.) Problem 1
sense when c'(a)
shows that this definition agrees with the old one when our curve c is defined by
=
=
=
c(t)
so that we simply have the graph of
Once again, various old formulas
(c + d)'(a)
(a c)'(a)
-
or, as equations
(t, f (t)),
=
f.
have analogues. For example,
=
c'(a) + d'(a),
=
a'(a)
.c'(a),
.c(a)
+a(a)
involving functions,
(c+d)'=c'+d',
(a-c)'=a'-c+a-c'.
These formulas can be derived immediately from the definition in terms of the
component functions. They can also be derived from the definition as a limit,
by imitating previous proofs; for the second, we would of course use the standard
trick of writing
«(a
+ h)c(a + h)
a(a)c(a)
-
=
a(a + h)
We can also consider
-
[c(a+
h)
c(a)] +
-
[a(a +
h)
-
a(a)]
·
c(a).
the function
d(t)
c(p(t))
=
=
(co p)(t),
where p
is now an ordinary function, from numbers to numbers. The new curve d
through
the same points as c, except at different times; thus p corresponds
passes
of c. For
to a
"reparameterization"
c
(u, v),
=
d=(uop,vop),
we obtain
d'(a)
=
=
((uop)'(a),
(vo p)'(a))
p'(a)v'(p(a)))
(p'(a)u'(p(a)),
=
p'(a)•(u'(p(a)),v'(p(a)))
=
p'(a)
-
c'(p(a)),
or simply
d'=
p'-(c'o
p).
Notice that if p(a)
a, so that d and c actually pass through the same point at
time a, then d'(a) = p'(a) c'(a), so that the tangent vector d'(a) is just a multiple
of c'(a). This means that the tangent line to c at c(a) is the same as the tangent
line to the reparameterized
c(a). The one exception occurs
curve d at d(a)
when p'(a)
0, since the tangent line for d is then undefined, even though the
=
·
=
=
of Curves 245
12, Appendix.ParamekicRepresentatwn
c(t3) won't have a tangent
tangent line for c may be defined. For example, d(t)
line defined at t 0, even though it's merely a reparameterization of c.
Finally, since we can define real-valued functions
=
=
(c.d)(t)
det(c, d)(t)
=
c(t)
.d(t),
det(c(t) d (t)),
=
,
we ought to have formulas for the derivatives of these new functions. As you might
guess, the proper formulas are
(c•d)'(a)
[det(c,d)]'(a)
=
c(a)•d'(a)
=
det(c', d)(a) + det(c, d')(a).
+c'(a)•d(a),
calculations from the definitions in terms
These can be derived by straightforward
functions. But it is more elegant to imitate the proof of the ordinary product rule, using the simple formulas in Problems 2 and 4 of Appendix 1
trick" referred to above.
to Chapter 4, and, of course, the
of the component
"standard
PROBLEMS
1.
"point-slope
form" (Problem 4-6) of the tangent
function f, the
written
line at (a, f(a)) can be
as y
f'(a), so that the
f(a) (x
tangent line consists of all points of the form
(a) For a
--a)
-
=
(x,f(a)+(x-a)f'(a)).
Conclude that the tangent line consists of all points of the form
(a+ s, f (a) + s f'(a)).
(b) If
c
is the curve c(t)
(a, f (a))
at
2.
(t, f (t)), conclude
that the tangent line of c at
the
[usingour new defmition] is same as the tangent line of f
=
(a, f (a)).
Let c(t)
(f (t), t2), where f is the function shown in Figure 21 of Chapfunction
ter 9. Show that c lies along the graph of the non-differentiable
reparameterization
c'(0)
other
words,
that
h(x)
but
0.
In
a
can
|x|,
this
usually
only
For
interested
in
reason, we are
a corner.
curves c
with c' never 0.
=
=
=
"hide"
3.
v(t) is a parametric
u(t), y
Suppose that x
and that u'
0 on some interval.
=
=
representation
of a curve,
-/
that on this interval the curve lies along the graph of
(b) Show that at the point x u(t) we have
(a) Show
=
J'(x)
v'(t)
=
-.
u'(t)
f
=
vo
u--1
246 Derivativesand Integrals
In Leibnizian notation this is often written suggestively as
dy
Ã
dy
dx
(c) We also
have
f"(x)
4.
5.
u'(t)v"(t)
.
(u'(t))3
By implicit differentiation.
By considering the parametric representation
6
=
u(t), y
We've seen that the
=
cos3 t,
y
2/3
=
sin3
_
g
t
=
=
"graph
of
-e)(t)=
(f
f in polar
coordinates"
metric
is the curve
(f(t)cost, f(t)sint);
in other words, the graph of f in polar coordinates
is the curve
with
the para-
representation
x=
6.
x
x2/3
v(t) be the parametric representation of a curve, with
differentiable, and let P
(xo,70) be a point in the plane. Prove
that if the point Q = (u(i), v(i)) on the curve is closest to (xo,yo), and u'(i)
and v'(i) are not both 0, then the line from P to Q is perpendicular to the
tangent line of the curve at Q (Figure 6). The same result holds if Q is
furthest from (xo,yo).
Let x
u and v
FIGURE
v'(t)u"(t)
-
=
Consider a function f defined implicitly by the equation
Compute f'(x) in two ways:
(i)
(ii)
P
dx
f(0)cos9,
y=
f(0)sin9.
for the graph of f in polar coordinates the slope ofthe tangent
line at the point with polar coordinates (f (0),0) is
(a) Show that
--f(0)sin0
f(8)cos0 + f'(6)sin0
+ f'(0)cos0
if f (0) 0 and f is differentiable at 0, then the line through
the origin making an angle of 0 with the positive horizontal axis is a
tangent line of the graph of f in polar coordinates. Use this result to
add some details to the graph of the Archimedian spiral in Appendix 3
of Chapter 4, and to the graphs in Problems 3 and 10 of that Appendix
(b) Show that
=
as well.
that the point with polar coordinates
(f (Ð),0) is further from
the origin O than any other point on the graph of f. What can you
say about the tangent line to the graph at this point? Compare with
(c) Suppose
Problem 5.
of Curves 247
12, Appendix.ParametncRepresentation
that the tangent line to the graph of f at the point with polar
(f(0),0) makes an angle of a with the horizontal axis
0 is the angle between the tangent line and the
(Figure 7), so that a
the
point.
Show that
from
O to
ray
(d) Suppose
coordinates
/
-
_
tan(a
7.
=
f (0)
.
J'(0)
Problem 5 of Appendix 1 to Chapter 4 we found that the cardioid
2
2
1 sin 0 is also described by the equation (x2+ y2 + y)2
r
slope
of the tangent line at a point on the cardioid in two ways:
Find the
-
(i)
(ii)
By implicit differentiation.
By using the previous problem.
(b) Check that
7
0)
(a) In
=
FIGURE
-
.
at the origin the tangent lines are vertical,
as they appear
to
be in Figure 8.
FIGURE
8
The next problem uses the material from Chapter 15, in particular, radian
measure, and the inverse trigonometric functions and their properties.
8.
A gdoid is defined as the path traced out by a point on the rim of a rolling
wheel of radius a. You can see a beautiful cycloid by pasting a reflector on
the edge of a bicycle wheel and having a friend ride slowly in front of the
headlights of your car at night. Lacking a car, bicycle, or trusting friend, you
can settle instead for Figure 9.
FIGURE
9
248 Derivativesand Integrals
of the point on the rim after the
u(t) and v(t) be the coordinates
wheel has rotated through an angle of t (radians).
This means that the
wheel
rim
of
the
from P to Q in Figure 10 has length at. Since the
arc
wheel is rolling, at is also the distance from O to Q. Show that we have
of the cycloid
the parametric representation
(a) Let
=
v(t)
=
a(t
a(1
sint)
-
-
cost).
i
0
x
Q
length
FIGURE
u(t)
lo
at
Figure 11 shows the curves we obtain if the distance from the point to the
center of the wheel is (a)less than the radius or (b)greater than the radius.
In the latter case, the curve is not the graph of a function; at certain times
the point is moving backwards, even though the wheel is moving forwards!
(b)
FIGURE
11
In Figure 9 we drew the cycloid as the graph of a function, but we really
need to check that this is the case:
u'(t) and conclude that u is increasing. Problem 3 then shows
that the cycloid is the graph of f
vo u¯1, and allows us to compute
(b) Compute
=
f'(t).
It isn't possible to get an explicit formula for
f, but
we can come close.
of Curves 249
12, Appendix.ParametricRepresentation
(c) Show that
a
-
v(t)
|[2a
v(t)]v(t).
±
a
Hint: first solve for t in terms of v(t).
(d) The first half of the first arch of the cycloid is the graph of g
u(t)
,
=
a arccos
g(y)
=
aarccos
a
y
-
f(2a
---
,
where
y)y.
-
y
9.
-
Let u and v be continuous on
[a,b] and differentiable on (a, b); then u
representation
of a curve from P
give
parametric
v
a
(u(a),u(b))
to Q
(v(a),v(b)). Geometrically, it seems clear (Figure 12) that at some
point on the curve the tangent line is parallel to the line segment from P
to Q. Prove this analytically. Hint: This problem will give a geometric
interpretation for one of the theorems in Chapter 11.
and
=
=
FIGURE
12
10.
The following definition of a limit for a vector-valued
analogue of the definition for ordinary functions:
lim c(t)
l means that for every e
=
all t, if 0
|t
<
a|
-
8, then
<
||c(t)
-
>
l
function is the direct
0 there is some 8
<
0 such that, for
>
e.
Here || | is the norm, defined in Problem 2 of Appendix 1 to Chapter 4. If
I
=
then
(li, l2),
| c(t)
(a) Conclude
-
-12\
=
lu(t) lil
l ||
and
-
+
Iv(t)
.
that
|u(t) ll ] 5 | c(t)
-
and show that
also
I ||2
-
if lim c(t)
=
|v(t) l2| 5 ||c(t) l ||,
-
l according
-
to the above definition, then we
have
limu(t)
=
li
and
limv(t)
t-+a
=
l2,
ted
1 according to our definition(*)in terms of component
functions, on page 243.
(b) Conversely, show that if lim c(t) 1 according to the definition in terms
of component functions, then also lim c(t)
l according to the above
so that
lim c(t)
=
=
=
definition.
INTEGRALS
CHAPTER
"integral,"
the
The derivative does not display its full strength until allied with the
complete
second main concept of Part III. At first this topic may seem to be a
digression-in this chapter derivatives do 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 ever.
Although ultimately to be defined in a quite complicated way, the integral formalizes a simple, intuitive concept-that
of area. By now it should come as
no surprise to learn that the definition of an intuitive concept can present great
difficulties-"area" is certainly 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 squares, with
sides of length 1, which fit in the region. But this defmition is hopelessly inadequate
for any but the simplest regions. For example, a circle of radius I supposedly has
"x
squares" means.
as area the irrational number x, but it is not at all clear what
which supposedly has area 1, it is hard
Even if we consider a circle of radius 1/
,
this
circle, since it does not seem possible
unit
what
fits
in
way a
to say in
square
which
unit
to divide the
can be arranged to form a circle.
square into pieces
In this chapter we will only try to define the area of some very special regions
(Figure 1)-those which are bounded by the horizontal axis, the vertical lines
through (a, 0) and (b, 0), and the graph of a function f 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 other important
geometric figures.
The number which we will eventually assign as the area of R( f, a, b) will be
called the integralof f on [a, b]. Actually, the integral will be defined even for
functions f which do not satisfy the condition f (x) :> 0 for all x in [a,b]. If f is
the function graphed in Figure 2, the integral will represent the difference of the
area of the lightly shaded region and the area of the heavily shaded region (the
/
R(f,
b)
.,
(a,0)
(b, 0)
FIGURE
1
(4 0)
FIGURE
(b,0)
2
"algebraic
area"
of R( f, a, b)).
The idea behind the prospective definition is indicated in Figure 3. The interval
[a,b] has been divided into four subintervals
[to,tt] [tt,12] [12,t3][13,14
by means of numbers
m,
to, ti, 12, $3, $4 with
a=to<t1<t2<t3<t4-b
a
FI GURE
to
=
a
t
a
ta
4
=
b
(thenumbering of the subscripts begins with 0 so that the largest subscript
equal the number of subintervals).
250
will
13. Integrals 251
the function f has the minimum value mi and the
t¿] let the minimum value
maximum value Mi; similarly, on the ith interval [t¿_i,
of f be mi and let the maximum value be M¿. The sum
[to,ti]
On the first interval
s
represents
=
mi(ti
-
to) + m2(t2
ti) + m3(t3
-
the total area of rectangles
t2) + m4(t4
-
f3)
-
lying inside the region R( f, a, b), while the
sum
S
=
M1(t1
-
to) + M2(t2
t1) + M3(ts
-
t2) + M4(t4
-
--
f3)
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 observation
that A should satisfy
and
s<A
A<S,
represents
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.
DEFDiITION
of the interval [a,b] is a finite collection
b. A partition
[a,b], one of which is a, and one of which is b.
Let a
<
The points in a partition can be numbered
Suppose f is bounded on
The lower sum of
f
[a,b] and
mg
=
Mi
=
.
.
,
to so that
has been assigned.
that such a numbering
we shall always assume
.
<in-1<t.=b;
a=to<11<
DEFINITION
to,
of points in
P
=
{to,
...,
t.) is a partition of
[a, b].
Let
inf {f(x) : t¿_i 5 x 5 t¿},
sup{ f(x) : ti-1 5 x 5 ti}.
for P, denoted by L(f, P), is defined as
L(f, P)
mi(t;
=
-
ti_1).
i=1
The upper
sum of
f
for P, denoted by U(f, P), is defined as
U(f,P)=
M¿(ti-t¿_i).
i=1
The lower and upper sums correspond to the sums s and S in the previous
example; they are supposed to represent the total areas of rectangles lying below
and above the graph of f. Notice, however, that despite the geometric motivation,
these sums have been defined precisely without any appeal to a concept of
"area."
252 Derivativesand Integrals
The requirement that f be
Two details of the definition deserve comment.
order
all
that
the
essential
in
bounded on [a,b] is
me and M¿ be defined. Note,
numbers
also, that it was necessary to defme the
mi and Mi as inf's and sup's,
than as minima and maxima, since f was not assumed continuous.
One thing is clear about lower and upper sums: If P is any partition, then
rather
L(f, P) 5 U(f, P),
because
L( f, P)
m¿(t;
=
U(f,P)=
-
ti-1),
M¿(t;-t¿,),
i-l
and for each i we have
FIGURE
On the other hand, something less obvious oughtto be true: If Pi and P2 are
4
any two partitions of
[a,b],
then it should be the case that
L( f, Pi) 5 U(f, P2),
because L(f, Pi) should be 5 area R(f, a, b), and U(f, P2) should be > area
of R(f, a, b)" has not even
the
R( f, a, b). This remark proves nothing (since
been definedyet), but it does indicate that if there is to be any hope of defining the
area of R(f, a, b), a proof that L(f, PI) 5 U(f, P2) should corne first. The proof
which we are about to give depends upon a lemma which concerns the behaviorof
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. The picture indicates that the rectangles drawn for
the partition Q are a better approximation to the region R(f, a, b) than those for
the original partition P. To be precise:
"arca
j
--
l,,
b-1
FIGURE
&
5
LEMMA
If Q contains P
(i.e.,if all points of
P are also in
L( f, P) 5 L( f,
U(f, P) > U(f,
PROOF
Consider first the special
Q),then
Q),
Q).
case
(Figure 5) in which Q contains just one more point
P
=
Û
=
(to,
(10,
than P:
ts),
...,
--
-,
ik--1,N,Ík,•••9 Ín),
where
a=to<tt<---<tk-1<¾<Ik<'''<Ín-b.
13. Integrals 253
Let
m'
=
m"
=
inf
{f (x) : tk-1 X M}
inf {f (x) : u 5 x 5 tk̕
,
Then
m;(t;
L(f, P)=
ti-1),
-
i-1
k-I
L( f,
Q)
n
m¿(t¿
=
ti-1) +
-
m'(u
tk-1) i
-
(Ik
E
M)
-
RifÍi
i=1
¯
Íi-1)·
i-k+1
To prove that L(f, P) 5 L(f, Q) it therefore suffices to show that
mkffk
N'fM
Ik-1)
¯
Ík-1) Ÿ
-
Ik
N
N)•
-
ik} contains all the numbers in {f (x) : tk-1
X
smaller
and
u},
possibly some
ones, so the greatest lower bound of the first set
x 5
is lessthan or equal to the greatest lower bound of the second; thus
Now the set {f (x) : tk-1
mgim'.
Similarly,
m
mk
.
Therefore,
mkfÍk
¯
Ík-1)
=
mk(N
-
Ík-l)
EkfÍk
m'(N
N)
-
-
ik-1) + m"(tk
-
M)-
This proves, in this special case, that L( f, P) 5 L( f, Q). The proof that U( f, P) à
U( f, Q) is similar, and is left to you as an easy, but valuable, exercise.
The general case can now be deduced quite easily. The partition Q can be
obtained from P by adding one point at a time; in other words, there is a sequence
of partitions
P
such that Pj+1 contains
=
just one
PI,
P2,
more
P,
...,
=
Q
point than Pj. Then
L(f,P)=L(f,Pi)$L(f,P2)$--·5L(f,PŒ)=L(f,Q),
and
U( f, P)
=
U( f, P1) à U( f, P2) à
···
à U( f, Pa)
=
U( f,
Q).
The theorem we wish to prove is a simple consequence of this lemma.
THEOREM
l
Let P1 and P2 be partitions of
on [a,b]. Then
[a,b],
L(f,
PI)
and let
f be
5 U(f, P2)-
a
function which is bounded
254 Derivativesand Integrals
PROOF
There is a partition P which contains both Pi and P2
in both P1 and P2). According to the lemma,
(letP
consist of all points
L(f,Pi)5L(f,P)5U(f,P)5U(f,P2).
It follows from Theorem 1 that any upper sum U( f, P') is an upper bound for
the set of all lower sums L( f, P). Consequently, any upper sum U( f, P') is greater
than or equal to the least upper bound of all lower sums:
sup{L(
f,
P)
:
P a partition of
for every P'. This, in turn, means that
of all upper sums of f. Consequently,
[a, b]}
sup(L( f, P)}
sup{L(f, P)) 5 inf{U(f,
It is clear that both of these numbers
f for all partitions:
are
5 U( f, P'),
is a lower bound for the set
P)}.
between the lower sum and
upper
sum
of
L(f, P') 5 sup(L(f, P)} 5 U(f, P'),
L(f, P') 5 inf {U(f, P)} 5 U(f, P'),
for all partitions P'.
It may well happen that
sup{L(f, P))
inf{U(f,
=
P};
in this case, this is the only number between the lower sum and upper sum of f
for all partitions, and this number is consequently an ideal candidate for the area
of R(f, a, b). On the other hand, if
sup{L(f, P)}
f(x)
then every number
-,
x
inf{U(f,
<
P)},
between sup{L(f, P)) and inf{U(f, P)} will satisfy
L( f, P') 5 x 5 U (f, P')
a-a
FIGURE
6
a
a
t.
1
t.
6
for all partitions P'.
It is not at all clear justwhen such an embarrassment of riches will occur. The
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 all x in [a,b] (Figure 6). If P
{to, t,} is
partition
of
then
any
[a,b],
=
=
m¿=M¿=c,
so
c(t¿-t¿_i)=c(b-a),
L(f,P)=
i=1
U( f, P)
c(t;
=
i=1
--
ti-1)
=
c(b
-
a).
...,
13. Integrals 255
In this case, all lower sums and upper sums are equal, and
sup{L(f,P))=inf{U(f,P))=c(b-a).
Now consider (Figure 7) the function
0,
1,
a=to
FIGURE
•
•••
si
If P
la
in-1 f.=b
x rational.
t,} is any partition, then
{to,
=
defined by
x irrational
I
f(x)=
•
f
...,
mg
=
0, since there is an irrational number in
M¿
=
1, since there is a rational number
and
in
[ti-1,til-
'
Therefore,
7
[ti-1.til
0-(t;-t,_i)=0,
L(f,P)=
i=1
U(f,P)=
1-(t¿-ti-1)=b-a.
i=1
inf{U(f, P)}. The
Thus, in this case it is certainly not true that sup{L(f, P))
which
the
of
definition
principle upon
area was to be based pmvides insufficient
information to determine a specific area for R(f, a, b)-any number between 0
and b
a seems equally good. On the other hand, the region R(f, a, b) is so
weird that we might with justicerefuse to assign it any area at all. In fact, we can
maintain, more generally, that whenever
=
-
sup{L(f, P))
the region R(f,a,b)
peal to the word
terminology.
A function
f
sup{L(f, P)
which
:
P)},
is too unreasonable
"unreasonable"
DEFINITION
ý=inf{U(f,
to deserve having an area. As our apsuggests, we are about to cloak our ignorance in
is bounded on
P a partition of
[a,b] is integrable
[a,b]}
In this case, this common number
denoted by
=
inf{U(f, P)
:
on
[a,b] if
P a partition of
is called the integral
of
f
on
[a,b]}.
[a,b]
and is
b
(The symbol
Ï f.
is called an integralsrgn and was originally an elongated s, for
the numbers a and b are called the lower and uger limitsof integration.)
The integral
f is also called the area of R( f, a, b) when f (x) > 0 for all x
b].
in [a,
"sum;"
f
f
256 Derivativesand Integrals
If
f
is integrable, then according to this definition,
b
L(f, P) 5
I
for all partitions P of
5 U(f, P)
f
[a,b].
Moreover,
f is the unique number with this property.
This defmition merely pinpoints, and does not solve, the problem discussed
before: we do not know which fimctions are integrable (nordo we know how to
find the integral of f on [a,b] when f is integrable). At present we know only
two examples:
b
(1) if f (x)
=
c, then
f is integrable
(Notice that this integral assigns the expected
(2) if f (x)
=
0 x irrational
1, x rational,
'
and
[a,b]
on
f
=
c
·
(b
-
a).
area to a rectangle.)
then
f is not integrable
on
[a,b].
Several more examples will be given before discussing these problems further.
Even for these examples, however, it helps to have the following simple criterion
for integrability stated explicitly.
THEOREM
2
If
e
f
>
is bounded on [a,b], then f is integrable on
0 there is a partition P of [a,b] such that
U( f, P)
PROOF
Suppose first that for every e
>
L( f, P)
-
<
[a,b]
if and only if for every
e.
0 there is a partition P with
U( f, P)
L( f, P)
-
<
e.
Since
inf {U(f, P')} 5 U(f, P),
sup{L(f, P')} ;> L( f, P),
it follows that
inf (U (f, P') } sup {L( f, P') } < e.
-
Since this is true for all e
>
0, it follows that
sup{L(f, P')}
=
inf{U(
f, P')};
by definition, then, f is integrable. The proof of the converse assertion is similar:
If f is integrable, then
sup{L(f, P))
This means that for each e
>
=
inf{U(f, P)}.
0 there are partitions P', P" with
U(f, P")
-
L(f, P')
< E.
13. Integrals 257
Let P be a partition which contains both P' and P". Then, according
lemma,
to the
U(f, P) 5 U(f, P"),
L( f, P) > L( f, P');
consequently,
U( f, P)
L( f, P) 5 U( f, P")
-
L( f, P')
-
<
e.
Although the mechanics of the proof take up a little space, it should be clear
of the definition
that Theorem 2 amounts to nothing more than a restatement
of integrability. Nevertheless, it is a very convenient restatement
because there is
and
which
work
with. The next
mention
sup's
of
often
inf's,
difficult to
no
are
example illustrates this point, and also serves as a good introduction to the type
of reasoning which the complicated
definition of the integral necessitates, even in
very simple situations.
Let f be defined on
[0,2] by
x ¢ 1
1, x
1.
t,} is a partition of [0,2] with
f (x)
Suppose P
=
(to,
...,
10,
=
=
tj-1<1<tj
i,
21
t,
i
(seeFigure 8). Then
mg
FIGUREs
Mr
=
=
0
if i ¢ j,
but
and
0
=
mj
Mj
1.
=
Since
j-l
n
m¿(t¿-ti-1)+mj(tj-tj_t)+
L(f,P)=
m¡(t¿--ti-1),
i=l
i=j+1
j-1
U( f, P)
n
M¿(t¿
=
ti-1) + Mj(tj
--
tj_i) +
-
i=1
we
Mi(t;
-
ti-1)
i=j+l
have
U( f, P)
This certainly
shows that
f
L( f, P)
-
to choose
tj-1
Moreover, it
-
tj-1.
is integrable: to obtain a partition P with
U(f, P)
it is only necessary
tj
=
-
L(f, P)
<
e,
a partition with
<
l
<
ty
and
tj
-
ty_i
<
e.
is clear that
L(f, P) 5 0 5 U(f, P)
for all partitions P.
258 Derivativesand Integrals
Since
f
namely,
between all lower and upper sums,
is integrable, there is only one number
the integral of f, so
2
f
0.
=
Although the discontinuity of f was responsible for the difficultiesin this example, even worse problems arise for very simple continuous functions. For example,
let f (x)
x, and for simplicity consider an interval [0,b], where b > 0. If
t,} is a partition of [0,b], then (Figure 9)
P = {to,
=
...,
mi
=
and
ti-1
M,
=
ti
and therefore
M
L( f, P)
U( f, P)
=
ti-1(ti
=
t¿(t¿
-
ti-1)
-
ti-1)
i=l
'¯'
FIGURE
9
'
'
=ti(ti-to)+t2($2¯$1)Ÿ"
In(In"¯Ín-1)-
Neither of these formulas is particularly appealing, but both simplify considerably
for partitions P. = {to, t.} into n equal subintervals. In this case, the length
ty ti-1 of each subinterval is b/n, so
.
..
,
-
to
0,
b
=
t1=
-,
n
2b
t2 =
-,
n
etc;
in general
ib
t,=-.
n
Then
L( f, P.)
t¿
=
(t¿ t¿ i )
-
i
i=1
"
(i
-
1)b
:=1
"
--
1)
i=l
b2
=
b2
[f(i p
gj
(n-1
-
.
j=0
b
13. Integrals 259
Remembering the formula
k(k + 1)
2
1+···+k=
this can
,
be written
(n 1)(n) 62
-
L( f, Pa)
=
7
2
b2
n-1
2
n
Similarly,
U( f, P,,)
t¿(ti
=
ti-1)
-
i=1
n
ib
b
i=1
n
n
n(n +
1) b2
b2
n+1
=
¯
n
If n is very large, both L(f, P,) and U(f, Pa) are close to b2/2, and this remark
makes it easy to show that f is integrable. Notice first that
2 b2
2
n
U(f,Pn)-L(f,Pn)=-
--
This shows that there are partitions P, with U( f, P.) -L(
By Theorem 2 the fimction f is integrable. Moreover,
with only a little work. It is clear, first of all, that
f, P.)
as small as desired.
feb| may
HOW
be found
b2
L( f, Ps) 5
5 U( f, P.)
-
2
for all n.
This inequality shows only that b2/2 lies between certain special upper and lower
L(f, P.) can be made as small as
sums, but we have just seen that U(f, Pn)
desired, so there is only one number with this property. Since the integral certainly
has this pmperty we can conclude that
-
f(x)
=
b
10
2
o
b
FIGURE
b2
I
gi
Notice that this equation assigns area b2/2 to a right triangle with base and altitude b (Figure 10). Using more involved calculations, or appealing to Theorem 4,
it can be shown that
b2
a2
Ibf=---.
a
2
2
260 lhivadves and Integrals
The function f (x) = x2 presents even greater difficulties. In this case (Figure 11), if P
{to, t,} is a partition of [0,b], then
=
...,
m¿=
f(ti-1)
(ti-1)2
=
Choosing, once again, a partition P.
and
M¿=
{to,
=
·
-
-
,
f(t,)
In}
=
t
.
into n equal parts, so that
i b
-
t,
=
----
n
the lower and upper sums become
L( f, P,,)
|(x)
-
x'
(t¿_i)2 (t;
=
-
ti-1)
i=1
I
f(i
=
11
1)2
-
n-1
bs
FIGURE
--
n
-(t;-ti_i)
U(f,P.)=
t
j=1
Recalling the formula
12 +
-
-
+ k2 =
·
)k(k+ 1)(2k+ 1)
from Poblem 2-1, these sums can be written as
b3
L( f, P,,)
U(f,P.)=
=
-(n
1
1)(n)(2n
-
-
--
1),
1
b
--(n+1)(2n+1).
It is not too hard to show that
L( f, P.) 5
b3
---
3
5 U(f, P,,),
and that U( f, P.)
L( f, P.) can be made
sufficiently large. The same sort of reasoning
as small as desired, by choosing n
as before then shows that
--
la
o
f
b3
=
---.
3
This calculation already represents a nontrivial result-the
area of the region
usually
elementary
derived in
bounded by a parabola is not
geometry. Nevertheless, the result was known to Archimedes, who derived it in essentially the same
13. Integrals 261
way. The only superiority we can claim is that in the next chapter
a much simpler way to arrive at this result.
Some of our investigations can be summarized as follows:
we will
discover
b
Lf
=
b
f
=
a
(b
c
a)
-
f (x)
=
if
f (x)
=
c for all x,
a2
b2
-
if
-
-
x for all x,
3
b3
if f(x)=x2forallx.
f=---
fb
This list already reveals that the notation
f suffers from the lack of a convenient
for naming functions defmed by formulas. For this reason an alternative
analogous to the notation lim f (x), is also useful:
notation
notation,*
b
b
f (x)dx
means
precisely the same as
f.
Thus
b
Ï
Ï
Lb
cdx=c-(b-a),
b
b2
a2
2
b3
x2dx=---.
3
a3
xdx=---
a
2'
3
Notice that, as in the notation lim f(x), the symbol x can be replaced
x-a
other letter (except
f, a, or b, of course):
b
b
f (x)dx
=
a
b
f (t) dt
=
b
f (a) da
b
f (y)dy
=
by any
=
a
f (c)dc.
The symbol dx has no meaning in isolation, any more than the symbol x
has any meaning, except in the context lim f(x). In the equation
--->
*The notation fb f(x)dx is actually the older, and was for many years the only, symbol for the
integral. Leibniz used this symbol because he considered the integral to be the sum (denoted
by f)
of infinitely many rectangles with height f(x) and
small" width dx. Later writers used
xo,
x. to denote the points of a partition, and abbreviated x;
xi-1 by Axi. The integral was
"infinitely
-
...,
defined as the limit as Ax; approaches 0 of the sums
i=l
sums). The fact that the limit is obtained by changing
many people.
i
to lower and
f(xt) Ax¿ (analogous
to
f, f(x¿) to f(x), and
upper
Ax¿ to dx, delights
262 Derivativesand Integrals
the entire symbol x2 dx may be regarded
the function
f
as an abbreviation
such that
f(x)
=
for:
x2 for all x.
This notation for the integral is as flexible as the notation lim f(x). Several examples may aid in the interpretation of various types of formulas which frequently
appear; we have made use of Theorems 5 and 6.*
a(x+y)dx=
xdx+
(1)
(2)
lx
ydx=
2
x
x
ydy+
(y+t)dy=
+y(b-a).
-
2
tdy=---+t(x-a).
a
b
x(1+t)dz
dx
(3)
(1+t)(x-a)dx
=
lb
=(1+t)
(x-a)dx
a
b2
a2
2
2
----a(b--a)
=(1+t)
.
b
d(x+y)dy
x(d-c)+
dx=
(4)
=
dx
-
xdx
(b-a)+(d-c)
-
2
=
(d2
-
-
2
The computations
of
fx
dx and
f
is generally difficult or impossible. As
--
2
(b
---
a) +
(d
--
c)
a2
b2
--
---
--
.
2
2
x2 dx
may suggest that evaluating integrals
a matter of fact, the integrals of most func-
tions are impossible to determine exactly (although
theymay be computedto any degree
of accuracy desiredby calculating lower and upper sums). Nevertheless, as we shall see in
the next chapter, the integral of many functions can be computed very easily.
Even though most integrals cannot be computed exactly, it is important at least
to know when a fimetion f 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 next
Theorem gives the most useful result, but the proof given here uses material from
the Appendix to Chapter 8. If you prefer, you can wait until the end of the next
chapter, when a totally different proof will be given.
*Lest
other books, equation (1)requires an important
This equation interprets j y dx to mean the integral of the function f such that each
(x) is the number y. But classical notation often uses y for y(x), so f y dx might rnean the
chaos
overtake the reader when consulting
qualification.
value
f
integral of some arbitrary
y.
function
13. Integrals 263
THEOlŒM
3
PROOF
If f is continuous on
[a,b],
then
f is integrable
[a,b].
on
Notice, first, that f is bounded on [a,b], because it is continuous on [a,b]. To
prove that f is integrable on [a,b], we want to use Theorem 2, and show that for
every 2 > 0 there is a partition P of [a, b] such that
U( f, P)
-
L( f, P)
<
e.
Now we know, by Theorem 1 of the Appendix to Chapter 8, that f is uniformly
continuous on [a,b]. So there is some 8 > 0 such that for all x and y in [a,b],
if x
y|<8,
-
and
-
|f(x)- f(y)|< 2(b
a partition P
{to, t,} such
-
The trick is simply to choose
ô. Then for each i we have
If(x)
then
f(y)l
=
<
2(b
-
.
.
.
,
for all x, y in
a)
.
a)
that each
|t¿
-
ti-i
l<
[ti-1,ts],
it follows easily that
Mi
-
m¿
€
<
¯
2(b-a)
€
<
.
b-a
Since this is true for all i, we then have
U( f, P)
-
L( f, P)
(M¿
=
--
m¿)(t¿
-
ti-1)
i=1
t¿
<
b-a
--
ti-1
i=1
-b-a
=
b-a
which is what we wanted.
Although this theorem will provide all the information necessary for the use of
integrals in this book, it is more satisfying to have a somewhat larger supply of
integrable functions. Several problems treat this question in detail. It will help to
know the following three theorems, which show that f is integrable on [a,b], if it
is integrable on [a,c] and [c,b]; that f + g is integrable if f and g are; and that
c f is integrable if f is integrable and c is any number.
As a simple application of these theorems, recall that if f is 0 except at one
point, where its value is 1, then f is integrable. Multiplying this function by c, it
follows that the same is true if the value of f 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 integrability
in Theorem 2; as some of our previous demonstrations illustrate, the details of the
-
264 Derivativesand Integrals
often conspire to obscure the point of the proof. It is a good idea to
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.
argument
attempt
THEOREM
4
Let a < c < b. If f is integrable on [a,b], then f is integrable on [a,c] and on
[c,b]. Conversely, if f is integrable on [a,c] and on [c,b], then f is integrable
on [a,b]. Finally, if f is integrable on [a,b], then
lbf=
PROOF
Suppose f is integrable on
[a,b] such that
[a,b].
f+
If e
U( f, P)
We might as well assume
which contains to,...,t.
U( f, P)
-
Now P'
of
L( f, P)
=
{to,
...,
<
that c
and
b
c
>
f.
0, there is a partition P
L( f, P)
-
<
tj for some j. (Otherwise, let
c; then Q contains P, so U(f,
[a,c] and
[c,b] (Figure 12). Since
-
-
-
,
In} Of
s.
=
e.)
t¡} is a partition of
{to.
=
P"
=
(tj,
Q be the partition
Q)
-
L(f,
Q) 5
t,} is a partition
...,
L( f, P)
L( f, P') + L( f, P"),
U(f,P)-U(f,P')+U(f,P"),
=
we
¯¯
FIGURE
¯¯
12
have
[U( f, P')
-
L( f, P')] + [U(
f,
P")
L( f, P")]
-
=
U( f, P)
-
L( f, P)
<
e.
Since each of the terms in brackets is nonnegative, each is less than e. This shows
that f is integrable on [a,c] and [c,b]. Note also that
C
L(f, P') 5
5 U(f, P'),
f
Ïb
L(f,P")$
fšU(f,P"),
C
so that
b
c
L( f, P) 5
I
f
G
+
C
f
5 U( f, P).
Since this is true for any P, this proves that
b
b
c
f+
C
f=
U
f.
Now suppose that f is integrable on [a,c] and on [c,b]. If e
partition P' of [a,c] and a partition P" of [c,b] such that
U (f, P')
U( f, P")
-
--
L( f, P')
L(f, P")
<
s/2,
<
e/2.
>
0, there is a
13. Integrals 265
If P is the partition of
[a,b] containing
L( f, P)
U( f, P)
all the
points of P' and P", then
L( f, P') + L( f, P"),
U( f, P') + U( f, P");
=
=
consequently,
U( f, P)
J
L( f, P)
-
[U( f, P')
=
L( f, P')] + [U( f, P")
-
-
L( f, P")]
Theorem 4 is the basis for some minor notational conventions.
f was defined only for a < b. We now add the definitions
b
la
and
f=0
<
e.
The integral
a
f=-
f
ifa>b.
With these definitions, the equation f f + fbf = f f holds for all a, c, b even
if a < c < b is not true (theproof of this assertion is a rather tedious case-by-case
check).
THEOREM
s
If
f
and g are integrable on
[a,b], then f
Ib
(f+g)=
Let P
=
{to,
...,
t,} be any partition of
[a,b] and
b
f+
G
PROOF
+ g is integrable on
b
[a, b].
g.
d
Let
m;=inf{(f+g)(x):ti-15x5ti),
m¿'
m;"
and
=
=
inf {f (x) : ti-1 5 x 5 ti }
inf {g(x): ti-1 5 x 5 tiÌ.
,
define M¿, M¿', M¿" similarly. It is not necessarily true that
m¿
=
m¿' + m¿",
mg
>
m¿' + m¿".
but it is true (Problem 10) that
Similarly,
M¿ 5 M¿' + M¿".
Therefore,
L( f, P) + L(g, P) 5 L( f + g, P)
and
U(f +g, P) 5 U(f, P) +U(g, P).
Thus,
L( f, P) + L(g, P) 5 L(f + g, P) 5 U( f + g, P) 5 U( f, P) + U(g, P).
Since f and g are integrable, there are partitions P', P" with
U( f, P')
U(g, P")
-
-
L( f, P')
L(g, P")
<
<
e/2,
e|2.
266 Derivativesand Integrals
If P contains both P' and P", then
U( f, P) + U(g, P)
[L(f, P) + L(g, P)]
-
<
e,
and consequently
U(f+g,P)-L(f+g,P)<s.
This proves that f + g is integrable on
[a,b].
Moreover,
L( f, P) + L(g, P) 5 L( f + g, P)
(1)
5
lb
(f+g)
SU(f+g,P)5U(f,P)+U(g,P);
and also
(2)
b
Ïbf+
L(f,P)+L(g,P)$
L(f, P) and U(g, p)
Since U(f, P)
desired, it follows that
-
U( f, P) + U(g, P)
-
--
gšU(f,P)+U(g,P).
L(g, P) can both be made as small as
[L( f, P) + L(g, P)]
can also be made as small as desired; it therefore follows from
lb
(f
THEOREM
6
If f is integrable on
on [a,b] and
[a,b],
b
+ g)
=
Lb
PROOF
b
f
+
g.
then for any number
cf=c-
(1)and (2)that
c, the
function cf is integrable
b
f.
The proof (which
is much easier than that of Theorem 5) is left to you. It is a good
separately
the cases c > 0 and c < 0. Why?
idea to treat
(Theorem 6 is just a special case of the more general theorem that f g is
integrable on [a,b], if f and g are, but this result is quite hard to prove (see
Problem 38).)
-
In this chapter we have acquired only one complicated definition, a few simply
theorems with intricate proofs, and one theorem which required material from the
Appendix to Chapter 8. This is not because integrals constitute a more difficult
topic than derivatives, but because powerful tools developed in previous chapters
have been allowed to remain dormant. The most significant discovery of calculus
is the fact that the integral and the derivative are intimately related-once
we
learn the connection, the integral will become as useful as the derivative, and as
easy to use. The connection between derivatives and integrals deserves a separate
chapter, but the preparations which we will make in this chapter may serve as a
hint. We first state a simple inequality concerning integrals, which plays a role in
many important theorems.
13. Integrals 267
THEOREM
7
Suppose
f
[a,b] and
is integrable on
m 5
that
for all x in
M
f(x)
Then
b
m(b
PROOF
a) 5
-
Ï
m(b-a)<L(f,P)
I
and
for every partition P. Since ff
inequality followsimmediately.
U(f,P)<M(b-a)
-
[a,b].
X
F(x)
cc
FIGURE
a).
-
sup{L(f, P)}
=
Suppose now that f is integrable on
[a, b] by
area
F(c+h)-F(c)
5 M(b
f
It is clear that
M
//
[a,b].
inf{U(f,
=
P)), the
desired
We can define a new function f on
X
f
=
f(t)dt.
=
a
+ h
(This depends on Theorem 4.) We have seen that f may be integrable even if it
not continuous, and the Problems give examples ofintegrable functions which
are quite pathological. The behavior of F is therefore a very pleasant surprise.
iS
13
THEOREM
8
If f is integrable on
F is defined on
[a, b] and
[a,b]
by
X
F(x)
I
=
f,
a
then F is continuous
PROOF
on
[a,b].
Suppose c is in [a,b]. Since f is integrable on
on [a,b]; let M be a number such that
|f (x)| 5
If h
>
[a,b] it is, by definition, bounded
for all x in
M
[a,b].
0, then (Figure 13)
c+h
F(c + h)
-
F(c)
f
f (x) 5
M
f
-
a
Since
-M i
c+h
c
I
=
f.
=
a
e
for all x,
it followsfrom Theorem 7 that
c+h
-M
-
h5
I
f
5 Mh;
c
in other words,
(1)
Fh
<
-
M hy F(c + h)
-
F(c) 5 M h.
-
-
0, a similar inequality can be derived: Note that
c
F(c + h)
-
F(c)
=
Lc+h f.
f
=
-
c+h
268 Denvativesand Integrals
-h,
Applying Theorem 7 to the interval
h, c], of length
[c+
we obtain
C
Mh i
Ï
f
c+h
-1,
multiplying
which reverses all the inequalities, we have
by
Mh 5 F(c + h)
(2)
Inequalities
(1)and (2)can
>
F(c)
-
|5
M
-
<
s/M.
F(c)
-
| < e,
This proves that
lim F(c+h)
=
h->0
in other words F is continuous
F(c);
at c.
Figure 14 compares f and F(x)
that F is always better behaved than
this is.
=
f.
|
f for
various
functions f; it appears
In the next chapter we will see how true
I
I
FIGURE
14
h|.
0, we have
|F(c + h)
provided that |h|
F(c) 5 -Mh.
-
be combined:
|F(c + h)
Therefore, if e
5 -Mh;
I
-
I
-
F
13. Integrals 269
PROBLEMS
1.
$ x3 dx
Prove that
b4/4, by considering
=
partitions into n equal subin-
i3 which was found in Problem 2-6. This
tervals, using the formula for
i=1
problem requires only a straightforward imitation of calculations in the text,
but you should write it up as a formal proof to make certain that all the fine
points of the argument are clear.
2.
*3.
Prove, similarly, that
febx*dxb5/5.
=
k?/np+1
Problem 2-7, show that the sum
(a) Using
can be made as close
k=1
to 1/(p + 1) as desired, by choosing n large enough.
bP**/(p
+ 1)
(b) Prove that xP dx
feb
*4.
=
.
Ib
xPdx
This problem outlines a clever way to fmd
for 0
<
a
<
b. (The
0 will then follow by continuity.) The trick is to use partitions
tn)
for which all ratios r
ty/ti-1 are equal, instead of using
{to,
which
all
differences t. t;_i are equal.
partitions for
result
P
for a
=
.
=
..
=
,
-
(a) Show that
for such a partition P
t¿
(b) If f(x)
=
xP,
U( f, P)
=
-
a
c'7"
show, using the
=
ap+1
y
_
have
we
for c
=
.
a
formula in Problem 2-5, that
(p+1yn
c-1/n
i
i=1
¯1¡n
1
=
(a*** bf** )c(p+1yn
-
1
=
(bp+1
-
ap+1)c?!"
--
-
c
c(p+1)/n
-
1+cl/"+---+cF/=
fmd a similar formula for L(f, P).
(c) Conclude that
and
b
Ï
x? dx
a
5.
bp+1
-
.
p + 1
Evaluate without doing any computations:
IlxJl
(ii) ll
(i)
-1
-1
--
x2 dx.
(x5+ 3))1
-
x2 dx.
ap+1
=
270 Derivativesand Integrals
6.
Prove that
2
sin t
dt
t+ 1
for all x
7.
>
0
0.
>
Decide which of the followingfunctions are integrable on
the integral when you can.
(i)
f(x)=fx,
[x-2,
fx
lx
'
(ii) f (x)
(iii) f (x)
=
x-2,
=
(iv) f (x)
=
(vii)f
x rational
0,
x
irrational.
x of the form a +
x not of this form.
=
(vi) f (x)
05x51
1<xs2.
+
1,
(v) f (x) l 0,
calculate
05x<1
15xs2.
[x].
[x],
x +
[0,2], and
bÃ
for rational a and b
1
-
=
x=0orx>1.
0,
is the function shown in Figure 15.
1
-
---
----
----
I
#
FIGURE
8.
2
1
15
Find the areas of the regions bounded by
2
the graphs of
f(x)
(i)
(ii) the graphs of f (x)
(-1, 0) and (1,0).
(iii) the graphs of f (x)
=
(v)
the graphs of
the graphs of
f(x)
f(x)
+2.
=
-x2
=
(iv)
x2 and g(x)
=
=
=
x2 and g(x)
x2 and g(x)
x2 and g(x)
x2 and g(x)
and the vertical lines through
=
=
=
=
1
1
x2.
-
-
x2
-2x
x2 and h(x) = 2.
+4 and the vertical
axis.
13. Integrals 271
(vi)
the graph of
through
f(x)
=
,
the horizontal axis, and the vertical
/2Sdx;
(2, 0). (Don't try to find
should
you
0
line
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 21.)
9.
Find
b
d
dx
f(x)g(y)dy
in terms of
vengeance;
10.
f f and f g. (This problem is an exercise in notation, with a
it is crucial that you recognize a constant when it appears.)
Prove, using the notation
m¿'+m¿"=
11.
(a) Which
upper
of
Theorem 5, that
inf {f(xt)+g(x2)
5 t;} 5 m¿.
ti_i 5 xt,x2
:
functions have the property that every lower sum equals every
sum?
(b) Which
functions have the property that some upper
sum equals some
sum?
lower
(other)
continuous
Which
(c)
functions have the property that all lower sums are
equal?
*(d)
Which integrable functions have the property that all lower sums are
equal? (Bear in mind that one such function is f (x) 0 for x irrational,
p/q in lowest terms.) Hint: You will need the
f(x) 1/q for x
notion of a dense set, introduced in Problem 8-6, as well as the results of
Problem 30.
=
=
=
b < c < d and f is integrable on
[b,c]. (Don't work hard.)
12.
If a
13.
(a) Prove that
b
<
then
(b) Prove
in
Ï
f
if f is integrable on
and
then
and g are integrable on
f
f (x) à
lb
f
a
f is integrable
0 for all x in
a
on
[a,b],
f(x)
à g(x) for all x
(b)you are
to warn
wasting time.)
b+c
b
Ï
and
g. (By now it should be unnecessary
à
Prove that
f
a
[a, b]
b
that if you work hard on part
14.
[a,b]
that
à 0.
that if
[a,b],
[a,d], prove
(x) dx
=
+c
f (x
-
c)
dx.
(The geometric interpretation should make this very plausible.) Hint: Every
partition P
{to, In} of [a,b] gives rise to a partition P'
{to+ c,
t, + c) of [a + c, b + c], and conversely.
=
...,
-
-
-,
=
272 Derivativesand Integrals
*15.
Prove that
ab
dt
dt +
dt.
=
ab
a
Ï
written
Hint: This can be
1/t dt
1/t dt. Every partition P
=
I
b
of [l.a] gives rise to a partition P'
and conversely.
{to,...,t,}
*16.
Prove that
Lcb
f (t) dt
=
{bto,...,bt.)
of
=
[b,ab],
b
=
f (ct)dt.
c
(Notice that Problem 15 is a special case.)
17.
Given that the area enclosed by the unit circle, described by the equation
2
1, is ar, use Problem 16 to show that the area enclosed by the
ellipse described by the equation x2/a2 + y2/b
1 is rab,
x2
=
=
b
18.
This problem outlines yet another way to compute
working
Cavalieri, one of the mathematicians
x" dx;
just before
it was
used
by
the invention of
calculus.
a
(a) Izt c,
(b) Problem
x" dx. Use Problem 16 to show that
=
=
c.a"**.
14 shows that
2a
Ï
a
x"
dx
=
o
Use this formula to prove that
2n+1caa"*I
=
-a
(x +
a)" dx.
2a"**
Ck·
k even
(c) Now use
x" dx
Problem 2-3 to prove that cn
=
7
1/(n + 1).
19.
Suppose that f is bounded on [a,b] and that f is continuous at each point
in [a,b] with the exception of xo in (a, b). Prove that f is integrable on
[a,b]. Hint: Imitate one of the examples in the text.
20.
Suppose that f is nondecreasing
on [a,b]. Notice that f is automatically
bounded on [a,b], because f (a) sf (x) 5 f (b) for x in [a,b].
(a) If P {to, t.) is a partition of [a,b], what is L(f, P) and U( f, P)?
(b) Suppose that ti ti-1 8 for each i. Prove that U(f, P) L(f, P)
8[ f(b) f(a)].
that f is integrable.
Prove
(c)
(d) Give an example of a nondecreasing function on [0,1] which is discon=
...,
-
=
-
tinuous at infinitely many points.
-
=
13. Integrals 273
It might be of interest to compare this problem with the followingextract
from Newton's hincapia.*
LEMMA II
a
K --L
'
'
,
A
total area b
·
f
))
av
c
n
*21.
z
If in anyfigure
AacE, terminatedby the ryht linesAa, AE, and the
curve acE,
therebe inscribedany number of parallelograms
Ab, Ec, Cd, &c., comprehended
under equal basesAB, BC, CD, &c., and
theside.s Bb, Cc, Dd, &c.,parallel
and theparallelograms
to one side Aa of thefigure;
aKbl, bLcm, cMdo, &c., are
completed: then¶tkebreadthof thoseparallelograms
be supposed to be diminished,
and theirnumber to be augmented in infinitum, I
sa3 that the ultimate ratios
which the mscribedfigure AKhLcMdD, the circumscribed}gureAalbmendoE,
and curvilinear
AabcdE, will haveto one another; are ratios ofequaßg.
figure
For the difference of the inscribed and circumscribed figures is the
sum of the parallelograms Kl, Lm, Mn, Do, that is
(fromthe equality
of all their bases), the rectangle under
of
their bases Kb and the
one
altitudes
of
their
that
Aa,
is, the rectangle ABla. But this rectangle,
sum
because its breadth AB is supposed diminished in i¢nitum, becomesless
than any given space. And therefore
(byLem. 1) the figures inscribed
and circumscribed become ultimately equal
one to the other; and much
more will the intermediate curvilinear figure be ultimately equal to either.
QE.D.
Suppose that
f
is increasing. Figure 16 suggests that
Lb
f¯I
(a) If
P
=
f"(t.)}.
[*(a)
FIGURE
b
(c) Find
22.
f-46)
-af¯I(a)
-
f.
...,
Ï
{f¯I(to),
--
-,
formula stated above.
dx for 0 5 a
<
b.
Suppose that f is a continuous increasing function with
that for a, b > 0 we have Toung'sinequaliy,
Ï
O
f (0)
=
0. Prove
b
a
ab 5
and that equality
Figure 16!
=
1,P)+U(f,P')=bf¯1(b)-af¯!(a).
the
(b) Now prove
b
16
bf-i(b)
(to, t,} is a partition of [a,b), let P'
Prove that, as suggested in Figure 17,
L(f
a
=
f (x) dx
f¯ (x) dx,
+
0
holds if and only if b
=
f(a). Hint: Draw
a picture
like
*Newton's Principia,A Revisionol'Mott's Translation, by Florian Cajori. University
of California
Press, Berkeley, California, 1946.
274 Duivatives and Integrals
23.
(a) Prove that
[a,b]
if f is integrable on
and m
5 f (x) 5 M for all x in
[a,b], then
b
f (x) dx
=
(b a)µ
-
for some number µ with m 5 µ 5 M.
a=to
FIGURE
ti /2
(b) Prove that
is i,=b
if f is continuous
on
[a,b],
then
b
17
f (x)dx
for some ( in
[a, b];
and show
=
(b
-
a) f (g)
by an example that continuity
generally, suppose that f is continuous on
integrable and nonnegative on [a,b]. Prove that
(c) More
b
(d) Deduce
18
24.
[a,b].
=
f (g)
that g is
g(x)dx
This result is called the Mean Value Theorem for
the same result if g is integrable and nonpositive on
(e) Show that
FIGURE
and
b
f (x)g(x)dx
for some ( in
Integrals.
[a,b]
is essential.
[a,b].
one of these two hypotheses for g is essential.
the graph of a function in polar coordinates
(Chapter 4, Appendix 3). Figure 18 shows a sector of a circle, with central
angle 0. When 0 is measured in radians (Chapter 15), the area of this sector
In this problem we consider
is r2
9
Now consider the region A shown in Figure 19, where the curve is
the graph in polar coordinates of the continuous fimction f. Show that
-
-.
area A
*25.
FIGURE
19
Let f be a continuous
[a,b], define
p)
=
function on
-
i=1
f(0)2d0.
[a,b].
If P
=
{to,
...,
t.] is a partition of
13. Integrals 275
The number £( f, P) represents
the length of a polygonal curve inscribed in
of
the graph
f (seeFigure 20). We define the length of f on [a,b] to be
the least upper bound of all £( f, P) for all partitions P (provided
that the set
of all such £(f, P) is bounded above).
is a linear function on [a,b], prove that the length of f is the
distance from (a, f(a)) to (b, f(b)).
(b) If f is not linear, prove that there is a partition P = {a,t, b) of [a,b]
such that £(f, P) is greater than the distance from (a, f(a)) to (b,
f(b)).
(You will need Problem 4-9.)
(c) Conclude that of all functions f on [a,b] with f(a) c and f(b) d,
the length of the linear function is less than the length of any other. (Or,
in conventional but hopelessly muddled terminology: "A straight line is
the shortest distance between two points".)
(d) Suppose that f' is bounded on [a,b]. If P is any partition of [a,b]
(a) If f
I
a
FIGURE
=
t,
I
t,
I
e,
t
t,
t.
I
-
6
20
=
=
show that
y
L(J1+(f')2,P)5£(f,P)5U(Ál+(f')2
Hint: Use the Mean Value Theorem.
(e) Why is sup{L( 1 + (f')2, P)} 5 sup{£(
P)} 5 inf (U(
show that sup(£(f,
(f) Now
that the length of
f
[a,b]
on
f,
P)}? (This is easy.)
1 + (f')2, P)}, thereby proving
r 2
1 + (f')2
is
is inte-
Hint: It suffices to show that if P' and P" are any two
partitions, then £( f, P') 5 U( 1 + (f')2, P"). If P contains the points
of both P' and P", how does £(f, P') compare to f(f, P)?
(g) Let 2(x) be the length of the graph of f on [a,x], and let d(x) be the
length of the straight line segment from (a, f (a)) to (x, f (x)). Show that
grable on
[a,b].
lun
ma
1.
=
d(x)
Hint: It will help to use a couple of Mean Value Theorems.
26.
A function s defmed on
[a,b] is called a step function if there is a partition
of
b]
t,,}
P
{to,
[a, such that s is a constant on each (ti-1, t¿) (thevalues
of s at t, may be arbitrary).
=
...,
(a) Prove that
if
function si 5
f
is integrable on
lb
with
f
a
b
with
Lb
Lb
s2
(b) Suppose that
such that
f
-
<
-
for any e
>
0 there is a step
-
si
<
e, and also a step function s2
f
e.
for all e
s2
f
[a,b], then
b
b
>
0 there are step functions si 5 f and s2 2 f
31 <
e. Prove that
f is integrable.
276 Derivativesand Integrals
b
I
(c) Find a function f which is not a step function, but which satisfies
L(f, P) for some partition P of [a,b].
*27.
Prove that if f is integrable on
b
functions g 5 f 5 h with
Ï
h
g
-
a
(a) Show that if si and s2 are step
(b) Prove, without using Theorem
(c) Use part (b)(andProblem
29.
fx
that
f
to choose x to be in
*30.
b
Ï
+ s2 is also.
[a,b], then b si
(si +
a
s2)
=
b
+
si
52a
proof of Theorem 5.
Prove that there is a number
f. Show by example
=
0 there are continuous
A picture will help immensely.
functions on
[a,b].
=
e. Hint: First get step functions
ones.
5, that
>
26) to give an alternative
Suppose that f is integrable on
b
[a,b] such
<
a
with this property, and then continuous
28.
for any e
[a,b], then
b
f
x
in
that it is not always possible
(a, b).
The purpose of this problem is to show that if f is integrable on
f must be continuous at many points in [a,b].
[a,b],
then
L(f, P) <
t.} be a partition of [a,b] with U(f, P)
b a. Prove that for some i we have Mr mg < l.
(b) Prove that there are numbers ai and b1 with a < ai < bi < b and
sup{ f (x) : ai 5 x 5 bl}
inf {f (x) : al 5 x 5 b1} < 1. (You can choose
=
[ai,b1] [ti-1,t¿] from part (a)unless i 1 or n; and in these two cases
a very simple device solves the problem.)
(c) Prove that there are numbers a2 and b2 with ai < a2 < b2 < bi and
sup{ f(x) : d2 5 x 5 b2} inf{ f(x) : a2 5 x 5 b2) <
Continue
in this way to find a sequence of intervals I.
(d)
[a.,b,] such
<
that sup{ f (x) : x in I.}
inf (f (x) : x in I.}
1/n. Apply the Nested
Intervals Theorem (Problem 8-14) to find a point x at which f is continuous.
(e) Prove that f is continuous at infinitely many points in [a,b].
(a) Let
P
{to,
=
-
...,
-
-
-
=
(.
---
=
-
b
*31.
Recall, from Problem 13, that
(a) Give an
example
where
b
[a,b], and
Ï
f
(b) Suppose f (x) >
=
Ï
f
à 0 if
f (x) à
f (x) > 0 for all
x, and
0 for all x in
f (x) > 0 for some
Lb
x
in
0.
0 for all x in
b
[a,b]
and
f is continuous
f (xo) > 0. Prove that
f > 0. Hint: It suffices
sum L( f, P) which is positive.
(c) Suppose f is integrable on [a,b] and f (x) > 0 for all
and
[a,b].
at xo
in
[a,b]
to find one lower
x
in
[a,b].
Prove
0. Hint: You will need Problem 30; indeed that was one
reason for including Problem 30.
that
f
>
13. Integrals 277
*32.
lb
is continuous on [a,b] and
fg 0 for all continuous
functions g on [a,b]. Prove that f
0. (This is easy; there is an obvious
(a) Suppose that f
=
=
g to choose.)
(b) Suppose f
is continuous
and that
[a,b]
on
Lb
fg
-=
0 for those con-
functions g on [a,b] which satisfy the extra conditions g(a)
0. Prove that f
0. (This innocent looking fact is an important
in the calculus of variations;
see the Suggested Reading for referHint: Derive a contradiction from the assumption f (xo)> 0 or
<
f(xo) 0; the g you pick will depend on the behavior of f near xo.
tinuous
g(b) =
lemma
ences.)
33.
Let
f (x)
=
=
=
for x rational and f (x)
x
0 for x irrational.
=
(a) Compute L(f, P) for all partitions P of [0,1].
(b) Find inf {U( f, P) : P a partition of [0,1]}.
*34.
Let f (x)
that
=
0 for irrational x, and 1/q if x
f is integrable
on
[0,1] and
that
f
p/q
=
=
in lowest terms. Show
0. (Every lower sum is clearly
0; you must figure out how to make upper sums small.)
*35.
Find two functions f and g which are integrable, but whose composition
is not. Hint: Problem 34 is relevant.
gof
*36.
Let f be a bounded function on [a, b] and let P be a partition of [a,b].
Let Mi and mi have their usual meanings, and let Ma' and m¿' have the
corresponding
meanings for the function |f I.
M¿' m¿'
s M¿ m¿.
integrable
that
if
is
on [a,b], then so is |f |.
f
(b) Prove
(c) Prove that if f and g are integrable on [a,b], then so are max(
min(f, g).
(d) Prove that f is integrable on [a,b] if and only if its
part" min( f, 0) are integrable on
max( f, 0) and its
(a) Prove that
-
-
"positive
"negative
37.
Prove that if
f
is integrable on
f, g)
and
part"
[a,b].
[a,b], then
f(t)dt
Lb
b
5
|f(t)|dt.
Hint: This follows easily from a certain string of inequalities; Problem 1-14
is relevant.
*38.
Suppose f and g are integrable on [a,b) and f(x), g(x) à 0 for all x in
[a,b]. Let P be a partition of [a,b]. Let Mg' and mi' denote the appropriate
sup's and inf's for f, define M¿" and m;" similarly for g, and define M¿ and mi
similarly for fg.
(a) Prove that
M¿ 5 M¿'M¿" and m¿ à mi'm;".
278 Derivancesand Integrals
(b) Show that
U(fg, P)
-
[M¿'M¿"-mi'm;"](t¿-t¿
L(fg, P) 5
i).
i=1
fact that f and g are bounded, so that
show that
b],
x in [a,
(c) Using the
U(fg, P)
M for
L(fg, P)
-
[M¿'
sM
-
m¿'](t¿
[M¿"
t¿_t) +
-
i=1
(d) Prove that fg
(e) Now eliminate
39.
|f(x)|, |g(x)| 5
Suppose that
f
-
m,"](t;
-
t¿_i)
.
i=l
is integrable.
the restriction
and g are
that
à 0 for x in
f(x), g(x)
integrable on
states that
[a,b].
b
[a,b].
The Cauchy-Schwartinequalig
b
fg
(a) Show that
the Schwarz inequality is a special case of the Cauchy-Schwarz
inequality.
(b) Give three proofs of the Cauchy-Schwarz inequality by imitating the
proofs of the Schwarz inequality in Problem 2-21. (The last one will
take some imagination.)
(c) If equality holds, is it necessarily true that f Ag for some I? What if
=
and g are coti
uou
2
(d)
.
Is this result true if 0 and 1 are
replaced by a and b?
*40.
Suppose that
f
is continuous
lim
x->oo
Hint:
The condition
and
x
e
lim f (x)
x->oo
f (x)
=
a. Prove that
*
1
-
t à some N. This means that
comparison
lim
f (t) dt
=
=
a.
a implies that
f (t)
is close to a for
N+M
f (t) dt is close
to N, then Ma/(N + M) is close to a.
to Ma. If M is large in
13, Appenda. RiemannSums 279
APPENDIX.
RIEMANN SUMS
Suppose that P
choose
{to,
=
.
some point x, in
.
.
,
[t, i,
t,,} is a partition of [a,b], and that for each i we
ty]. Then we clearly have
L(f, P) 5
Any sum
a
=
FIGURE
t
to
i
=
f(x;)(t;
f(xi)(t¿-
ti-1) 5 U(f, P).
ti_i) is called a Riemannsum of
-
for P. Figure 1 shows
f
the geometric interpretation of a Riemann sum; it is the total area of n rectangles
that lie pardy below the graph of f and partly above it. Because of the arbitrary
have been picked, we can't say for
way in which the heights of the rectangles
whether
particular
Riemann sum is less than or greater than the integral
sure
a
b
b
But it does seem that the overlaps shouldn't matter too much; if the
bases of all the rectangles are narrow enough, then the Riemann sum ought to be
close to the integral. The following theorem states this precisely.
f(x)dx.
l
THEORE1W
1
Suppose that f is integrable on [a,b]. Then for every e > 0 there is some 8 > 0
such that, if P
< 8,
{to, In) is any partition of [a,b] with all lengths ti
then
-t;-i
=
...,
b
a
f(xt)(t¿-t¿_i)-
f(x)dx
<e,
i=1
for any Riemann sum formed by choosing x, in
PROOF
[t¿_i,t¿].
As in the proof that a
First we will prove the theorem when f is continuous.
continuous function is integrable (Theorem 13-3), we will use Theorem 1 fmm
the Appendix to Chapter 8, so you might want to skip it. But if you've already
read the proof of Theorem 13-3, this part of the proof will be a snap-in
fact, it's
practically the same.
Given e > 0, choose ô > 0 so that for all x and y in [a,b]
if |x
-
y|
<
8, then
|f (x) f (y)| < 2(b
-
.
-
a)
any partition P
{to, t,} with each ta ti_i < ô, and any x; in
[t¿_i,t¿]. Then, as we saw in the proof of Theorem 13-3, we have
Now consider
(1)
=
-
...,
U( f, P)
-
L( f, P)
<
s.
But we also have
(2)
L(f, P) 5
f(xi)(t¿-t¿_i) 5
U(f, P)
i=1
and
(3)
L(f, P) 5
Ibf(x)dx
U(f, P).
280 Derivativesand Integrals
The desired inequality, for our continuous
and
(1),(2)
function f, follows immediately from
(3).
The argument in the general case is simple (though
perhaps a bit messy), using
Problem 13-27, which says that there are continuous functions g 5 f 5 h satisfying
b
Ib
(4)
b
g 5
f
b
b
h,
5
with
h-
g<e.
We have
-ti-1)
-ti-1)
g(xi)(ti
f(xi)(ti
5
i=l
h(xi)(t¿-ti_i),
5
i=l
i=1
and since the theorem holds for continuous functions, we know that for te
< ô,
and
right-hand
and
right-hand
sides
of
close
the leftthis inequality are
to the leftsides of
middle
that
the
This
implies
two
terms,
(4).
-ti-1
b
I
must be close to
skeptical reader.
fh
-
n
and
f
fbf,
Which
i
ti-1),
f (x¿)(t;
-
is small. Detailed inequalities are left to the
The moral of this tale is that anything which looks like a good approximation
to an integral really is, provided that all the lengths t¿ t¿ ¡ of the intervals in the
partition are small enough. Some of the following problems should bring home
this message with even greater force.
-
_
PROBLEMS
1.
Suppose that f and g are continuous
functions on
In} of
P
{to,
[a,b] choose a set of points x¿ in
of points ut in
[tt-1,ti]. Consider the sum
=
-
-
-
,
[a,b]. For a partition
[t¿ i, t¿] and another set
_
f(xi)g(ug)(ti ti-1)-
I
i-I
Notice that this is not a Riemann sum of fg for P. Nevertheless, show that
b
all such sums will
be within e of
Ï fg
provided that the partition P has all
ti-1 small enough. Hint: Estimate the difference between such a
sum and a Riemann sum; you will need to use uniform continuity.
lengths t¿
-
13, Appendix.RiemannSums 281
2.
This problem is similar to, but somewhat harder than, the previous one.
Suppose that f and g are continuous nonnegative functions on [a,b]. For a
partition P, consider sums
f (x¿)+
g(u¿)
(t¿
-
ti-1).
i=1
Lb
Show that these sums will be within e of
/ f + g if all te ti_i are small
enough. Hint: Use the fact that the square-root function is uniformly continuous on a closed interval [0,M].
(u(et),,(tt))
3.
Mrò, or
-
Finally, we're ready to taclde something big! (Compare Problem 13-25.)
Consider a curve c given parametrically by two functions u and v on [a,b].
t.} of [a, b] we define
For a partition P = (to,
...,
£(c, P)
[u(t¿)
-
u(ti-1)]2
-l-
[v(t¿)
-
U(ti-1)
2
i=l
FIGURE
this represents the length of an inscribed polygonal curve (Figure 2). We
define the length of c to be the least upper bound of all £( f, P), if it exists.
Prove that if u' and v' are continuous on [a,b], then the length of c is
2
b
4.
Ï
interval [00,Gi]. Show that the graph of f in
polar coordinates on this interval has the length
Let f' be continuous
on the
101
5.
Using Theorem 1, show that the Cauchy-Schwarz inequality (Problem 13-39)
is a consequence of the Schwarz inequality.
THE FUNDAMENTAL THEOREM
OF CALCULUS
CHAPTER
From the hints given in the previous chapter you may have already guessed the
first theorem of this chapter. We know that if f is integrable, then F(x)
f f is
only
ask
what
when
original
continuous; it is
the
fitting that we
happens
function f
is continuous. It turns out that F is differentiable (andits derivative is especially
=
simple).
THEOREM
1 (THE FIRST
Let f be integrable on
[a,b],
and
[a,b]
define F on
FUNDAMENTAL THEOREM
by
x
F(x)
OF CALCULUS)
If f is continuous at c in
then F is differentiable at c, and
[a,b],
F'(c)
(If c
=
f.
=
f (c).
=
to mean the right- or left-hand derivative
a or b, then F'(c) is understood
of F.)
PROOF
We will assume that c is in (a, b); the easy modifications
supplied by the reader. By definition,
F'(c)
=
F(c + h)
h
lim
h->o
Suppose first that h
>
-
for c
=
a or b may be
F(c)
0. Then
Ic+h
F(c+h)-F(c)=
f.
c
Define mh and Mh as follows (Figure 1):
mg
=
inf {f (x) : c 5 x 5 c + h}
,
Mh=SUp(f(X):Csx5c+h}.
It follows from Theorem 13-7 that
mh
-
hs
f
lc+h
š Mh h.
-
c
Therefore
F(c + h)
F(c)
5 Ma.
h
0, only a few details of the argument have to be changed.
-
mh 5
If h
<
mh=inf{f(x):c+hšx5c},
Mh=sup{f(x):c+hšxíc}.
282
Let
14. TheFundamentalTheoremof Cakulus 283
FIGURE
1
Then
C
mh
-
Ï
(-h) 5
Since
5 Mh
f
c+h
(-h).
·
ckh
F(c + h)
F(c)
--
c
I
=
f
=
-
+h
c
this yields
ma
h 2 F(c + h)
·
Since h < 0, dividing by h
as before:
F(c) 2 Mh h.
-
-
the inequality again, yielding the same result
reverses
ma 5
f
F(c + h)
h
F(c)
-
5 Mh·
This inequality is true for any integrable function, continuous
continuous at c, however,
lim mh
h->O
=
lim Mh
=
or not.
Since f is
f(C),
h->0
and this proves that
F'(c)
=
lim
F(c + h)
F(c)
-
=
h
h-vo
f (c).
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
b
G(x)
=
I
f,
284 Derivativesand Integrals
then
G(x)
Consequently, if
f
x
lbf
=
f.
-
is continuous at c, then
G'(c)
=
f (c).
-
here is very fortunate, and allows us to extend Theorem 1 to the situation where the function
The minus
sign
appearing
X
F(x)
is defined even for x
<
a. In this case we can write
F(x)
so if c
<
Ïf
laf,
=
=
-
a we have
-(-f(c))=
F'(c)=
f(c),
exactly as before.
Notice that in either case, differentiability of F at c is ensured by continuity of f
at c alone. Nevertheless, Theorem 1 is most interesting when f is continuous at
all points in [a,b]. In this case F is differentiable at all points in [a,b] and
F'
f.
=
In general, it is extremely difRcult to decide whether a given function f is the
derivative of some other function; for this reason Theorem 11-7 and
Problems 11-54 and 11-55 are particularly interesting, since they reveal certain
properties which f must have. If f is continuous, however, there is no problem
at all-according
to Theorem 1, f is the derivative of sorne function, namely the
function
F(x)
which
Theorem 1 has a simple corollary
integrals to a triviality.
COROMARY
If f is continuous on
[a,b]
and
f
f.
=
g'
=
frequently reduces
computations
for some function g, then
b
I
PROOF
f
=
g(b)
Let
g(a).
ÏX
F(x)
Then F'
-
=
f
=
g' on
[a,b].
=
f.
Consequently, there is a number
F=g+c.
c such that
of
14. The FundamentalTheoremof Calculus 285
The number
c can
be evaluated
easily: note that
0
=
F(a)
=
g(a) + c,
=
g(x)
-g(a);
so c
thus
=
F(x)
This is true, in particular, for x
b. Thus
=
Lb
f
g(a).
-
=
F(b)
g(b)
=
-
g(a).
The proof of this corollary tends, at first sight, to make the corollary seem useless:
after all, what good is it to know that
b
Ï
f
g(b)
=
b(a)
-
if g is, for example, g(x)
f ? The point, of course, is that one might happen
to know a quite different function g with this property. For example, if
f
=
g(x)
then g'(x)
=
f (x) so
and
=
3
f (x)
=
x2,
we obtain, without ever computing
lb
X2dx=---.
,
similarly;
b3
a3
3
3
lower and upper sums:
if n is a natural number and g(x)
One can treat other powers
xn+1/(n + 1), then g'(x)
x", so
=
=
an+1
bn+1
/b
x" dx
=
-
.
n+1
,
n+1
For any natural number n, the function f (x) x¯" is not bounded on any interval
containing 0, but if a and b are both positive or both negative, then
=
b
I
X-ndx
a¯"*'
b¯"**
=
-n+1
-n+1
.
a
-1.
Wedo not knowa simple expressionfor
Naturally this formula is only true for n ¢
g
Lb
-
dx.
x
The problem of computing this integral is discussed later, but it provides a good
to warn against a serious error. The conclusion of Corollary 1 is often
confused with the definition of integrals-many students think that
f is defmed
f
whose
where
g(a),
derivative is f." This
is
as:
g is a function
is useless. One reason is that a function f may be integrable
not only wrong-it
without being the derivative of another function. For example, if f (x)
0 for
not?).
=
x ¢ l and f (1) 1, then f is integrable, but f cannot be a derivative(why
There is also another reason that is much more important: If f is continuous,
opportunity
"defmition"
"g(b)
-
=
286 Denvativesand Integrals
g' for some function g; but we know this only becauseof
then we know that f
Theorem1. The function f (x) 1/x provides an excellent illustration: if x > 0,
then f (x) g'(x), where
=
=
=
lxl
-dt,
g(x)=
t
I
know of no simpler function g with this property.
The
to 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
in practice, however, is not much more useful).
(which
As we have just mentioned, a function f might be of the form g' even if f is not
continuous.
If f is integrable, then it is still true that
and we
corollary
Lb
f
=
g(b)
g(a).
-
The proof, however, must be entirely different-we
must return to the defmition of integrals.
THEOREM 2 (THE SECOND FUNDA-
If
f
is integrable on
[a,b] and f
=
cannot
use
Theorem 1, so
we
g' for some function g, then
MENTAL THEOREM OF CALCULUS)
PROOF
Let P
{to,
is a point x; in
=
...,
t,} be any partition of
[t¿.i,ti] such that
g(t¿)
[a,b].
g(t¿_i)
--
By the Mean Value Theorem there
g'(x¿)(t;
=
f(x;)(t;
=
-
-
ti-1)
ti-1)-
If
m¿
inf {f (x) : tt-1 5 x 5 ti }
=
,
M¿=sup{f(x):ti-15x5tt},
then clearly
-ti_i)
m¿(t¿--t,_i)
5
5 Mg(t¿-t¿_i),
f(x;)(t;
that is,
m¿(t;
t¿_i) 5 g(ti)
-
Adding these equations for i
m¿(t¿
-
=
1,
g(ti-1) 5 M¿(t;
-
-
n we obtain
...,
ti-1) 5 g(b)
-
g(a) 5
Mg(t¿
i-1
i=1
so that
L( f, P) 5 g(b)
foreverypartitionP.
-
g(a) 5 U( f, P)
But this means that
g(b)
-
ti_i).
g(a)
=
Ibf.
-
ti_i)
14. TheFundamental7heoran of Calculus 287
We have already used the corollary to Theorem 1 (or,equivalently,
to find the integrals of a few elementary functions:
b
Ï
b"*I
x" dx
a
f(x)
=
x.
a"**
=
n+ 1
n +
(aand
b both positive or
both negative if n > 0).
-1.
-
l
'
Theorem 2)
nj
As we pointed out in Chapter 13, 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
b
Ï
does not represent
mstead by
xadx
the area of the region
shown
in Figure 2, which is given
-x3dx+x3dx=--+-
a4
b4
4
4
=-+-.
FIGURE
2
Similar care must be exercised in finding the areas of regions which are bounded
by the graphs ofmore than one function-a problem which may frequently involve
considerable ingenuity in any case. Suppose, to take a simple example first, that
we wish to find the area of the region, shown in Figure 3, between the graphs of
the functions
f(x)=x
and
g(x)=x3
x3 i x2,
on the interval [0, 1]. If 0 sxs 1, then 0 5
so that the graph
below that of f. The area of the region of interest to us is therefore
area
1
R(f,0,1)-
g lies
1),
which is
This area could have been expressed
FIGURE3
area R(g,0,
of
as
If g(x) 5 f(x) for all x in [a,b], then this integral always gives the area bounded
by f and g, even if f and g are sometsmes negative. The easiest way to see this is shown
in Figure 4. If c is a number such that f + c and g + c are nonnegative on [a,b],
then the region Ri, bounded by f and g, has the same area as the region R2,
288 Derivativesand Integrals
bounded by f + c and g + c. Consequently,
b
area Ri
area R2
=
Lb
[(f+c)-(g+c)]
Lb
+ c)
(f
=
=
(g +
-
c)
b
(f
=
-
g).
This observation is useful in the following problem: Find the area of the region
bounded by the graphs of
f(x)=x2-x
b
a
The first necessity
and g
intersect
is to determine this region more precisely. The graphs of
x3-x2
x(x2
or
or
i
f
when
or
a
g(x)=x2.
and
x
=
-
O'
x
1)
-
=
0,
1-B
1+B
'
'
2
2
22
and on the interval
,
,
On the interval ([1 ß]/2, 0) we have x3
x2 > x3
assertions
from the
x. These
are apparent
(0, [1 + Ë]/2) we have
graphs (Figure 5),but they can also be checked easily, as follows. Since f(x) g(x)
only if x
0, [1+ Ã]/2, or [1- ß]/2, the function f
does not change sign
on the intervals ([1- ß]/2, 0) and (0, (1+Ã]/2); it is therefore only necessary
tO Observe,
for example, that
-
_
-
=
b)
FIGURE
4
-g
=
13-1-12--1<0
to conclude that
f
f
0 on ([1 ß]/2, 0),
g 5 0 on (0, [1+ Ã]/2).
g
-
-
>
-
The area of the region in question is thus
(x3
-
x
-
x2)dx
+
[x2
-
(x3
-
x)]dx.
As this example reveals, one of the major problems involved in finding the areas
of a region may be the exact determination of the region. There are, however,
have thus far defined the areas
more substantial problems of a logical nature-we
of some very special regions only, which do not even include some of the regions
whose areas have just been computed! We have simply assumed that area made
14. TheFundamentalTheoremof Calculus 289
f(x)
FIGURE
-
x
x
5
sense for these regions, and that certain reasonable properties of
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 better
approach
a
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.
It may 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 emphasized:
if f is continuous, the integral provides a function y such that
"area"
y'(x)
=
f (x).
This equation is the simplest example of a
equation"
(anequation
for a function y which involves derivatives of y). The Fundamental
Theorem
of Calculus says that this differential equation has solution,
a
if f 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, it is
very important to
realize that these functions
may be combined, like less esoteric functions, to yield
still more functions, whose derivatives
can be found by the Chain Rule.
Suppose, for example, that
"differential
f(x)
=
a
1 + sin2 t
dt.
290 Derivativesand Integrals
Although the notation
of the functions
tends to disguise the fact somewhat,
C(x)
In fact, f (x)
Rule,
=
x
x3
=
and
F(x)
dt.
1 + sin2 t
F'(x')
=
Fo C. Therefore, by the Chain
=
F'(C(x))
=
C'(x)
·
3x2
-
1
=
3x2.
sin273
1+
If f is defined, instead, as
f(x)
dt,
sin2
1+
3
then
f'(x)
1
La
=
1
=
-
-
1+
sin2x3
3x2.
If f is defined as the reverse composition,
f (x)
=
then
f'(x)
=
1
(lx
dt
1 + sin2 t
a
C'(F(x))
·
,
F'(x)
2
1
(x
Similarly, if
aux
f (x)
=
g(x)
=
h(x)
=
£6
(l'
dt,
sin2
1
dt
1+sin2t
sin
a
then
dt,
t
1
1+
.
1+sin2X
1
sin2
1+
x
f'(x)
3
\
y
dt1+sin2t
=3
,
1
=
-
1 + sin2(sin x)
cosx,
-1
g'(x)
=
h'(x)
=
·
cosx,
1 + sin2 (sinx)
2
1
cos
(
1+
sin2
composition
1
=
F(C(x)); in other words, f
f'(x)
f is the
dt
t
1
-
1+
sin2
.
X
14. TheFundamentalTheoremof Calculus 291
The formidable appearing function
f (x)
=
is also a composition; in fact, f
f'(x)
=
1 + sin2 t
Fo F. Therefore
Ja
=
F'(F(x))
-
dt
F'(x)
1
1 + sin2
\Ja
1
1+
sin2
1+
dt
sin2
/
t
As these examples reveal, the expression occurring above (orbelow) the integral
sign indicates the function which will appear on the rykt when f is written as a
composition. As a final example, consider the triple compositions
f (x)
=
sin2
1+
a
dt,
g(x)
"
=
1+=*"''
-
dt,
1 + sin2
-
t
which can be written
and
f=FoFoC
Omitting the intermediate
(whichyou
steps
g=FoFoF.
may
supply,
if you still feel insecure),
we obtain
1
f'(x)
=
1
-
1
1 + sin
a
dt
1 + sin2 t
-
1+
sin2,3
1
g'(x)
-
1 + sm
1
1
2
a
3x2,
1 + sin2 t
dt
1+
n2
\Ja
1+
sin2
dt
t
/
1
1 + sin2 x
should beLike the simpler differentiations of Chapter 10, these manipulations
and, like
much
easier
after
the
of
the
practice
problems,
provided
by
come
some
the 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
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 definitionof
the integral. Actually, this is not quite true. In order to apply Theorem 1 to a
continuous function we need to know that if f is continuous on [a,b], then f is
integrable on [a,b]. Although we've already offered one pmof of this result, there
292 Derivativesand Integrals
"elementary"
is a more elementary argument that you might prefer. Like most
it's quite tricky, but it has the virtue that it will force a review of the
arguments,
proof of Theorem 1.
If f is any bounded function on [a,b], then
and
sup{L(f, P)}
inf{U(f, P)}
will
both exist, even if f is not integrable. These numbers
integral of f on [a,b] and the upper integral of f on
will be denoted by
L
are called the lower
[a,b],
respectively,
and
b
lb
and
f
U
f.
The lower and upper integrals both have several properties which the integral
possesses. In particular, if a < c < b, then
bcb
Lbcb
Lf=Lf+LfandUf=Uf+Uf,
G
and if m 5
f (x) 5
C
M for all x in
a) 5 L
-
C
[a,b], then
b
m(b
d
G
I
b
5 U
f
f
5 M(b
--
a).
The proofs of these facts are left as an exercise, since they are quite similar to the
corresponding proofs for integrals. The results for integrals are actually a corollary
of the results for upper and lower integrals, because f is integrable precisely when
b
b
I
Lf=Uf.
We will prove that a continuous function f is integrable by showing that this
equality always holds for continuous
functions. It is actually easier to show that
L
Ïx
x
f=U
f
for all x in [a,b]; the trick is to note that most of the proof of Theorem
even depend on the fact that f was integrable!
THEOREM
13-3
PROOF
If f is continuous
on
then
[a,b],
Define functions L and U on
f is integrable
on
[a,b].
[a,b] by
IX
L(x)
Let x be in (a, b). If h
>
=
L
X
f
and
0 and
ma=inf{f(t):xstyx+h},
Mh=sup{f(t):xstšx+h},
U(x)
=
U
f.
1 didn't
14. The FundamentalTheoremof Calculus 293
then
x+h
x+h
Ï
mh-hšL
fšU
fšMa-h,
x
SO
-
ma
h 5 L(x + h)
or
L(x + h)
h
mh<
¯
If h
<
x
L(x) 5 U(x + h)
--
L(x)
-
<
U(x) 5 Mh h
-
U(x + h)
·
U(x)
-
¯
<Mh-
¯
h
0 and
ma=inf{f(t):x+hst
áx},
Mh=sup{f(t):x+hstsx},
one obtains the same inequality, precisely as in the proof of Theorem 1.
Since f is continuous at x, we have
lim mh
lim Mh
=
h->0
f(X),
=
h->0
and this proves that
L'(x)
U'(x)
=
This means that there is a number
U(x)
f (x) for x in (a, b).
=
c such that
for all x in
L(x) + c
=
[a,b].
Since
U(a)
the number
c must equal
L(a)
=
=
0,
0, so
U(x)
for all x in
L(x)
=
[a,b].
In particular,
U
lbf
b
=
U(b)
=
L(b)
=
L
a
and this means
that
a
f is integrable
on
f,
[a,b].
PROBLEMS
1.
Find the derivatives of each of the following functions.
x3
(i)
F(x)
(ii)
F(x)
sin3
=
t dt.
Ja
=
Ï
(f
an3rdt
1+sin6t#£2
3
2
(iii) F(x)
(iv) F(x)
>
1
=
5
=
8
Lb
1+t2+sin2t
dt.
1 + t2 + sin2
dt
dt
dy.
294 Derivativesand Integrals
(v)
F(x)
=
(vi)
F(x)
=
x
L
sin3
sin
sin
Ï
=
'
-
t
1
where
F(x)
FIGURE
.
(Find (F'
-
-
-
--
of
dt.
=
-
)'(x) in terms
F¯I(x).)
1
1
1
dy
dt.
x
(viii)F-1,
t dt
xy
(vii) F¯1 where F(x)
2.
dt.
1+t2+sin2t
t2
----
------
-
6
For each of the following f, if F(x) = jg f, at which points x is F'(x) =
f (x)? (Caution: it might happen that F'(x)
f (x), even if f is not contin=
uous at x.)
(i) f(x)=0ifxs1,
(ii) f (x) 0 if x <
=
f(x)=1ifx>l.
1, f (x)
=
1 if x
>
1.
(iii) f(x)=0ifxý=1, f(x)=lifx=l.
(iv) f(x) 0 if x is irrational, f(x) 1/q if x
(v) f (x) 0 if x 5 0, f (x) x if x > 0.
(vi) f (x) 0 if x 5 0 or x > 1, f (x) 1/[1/x]
=
=
=
=
3.
p/q in lowest terms.
if 0
=
(vii)f is the function shown in Figure 6.
(viii)f(x) 1 if x 1/n for some n in N, f(x)
=
=
=
=
Let f be integrable on
[a,b],
F(x)=
=
<
x 5
1.
0 otherwise.
let c be in (a, b), and let
lx
f,
a5x5b.
For each of the following statements, give either a proof or a counterexample.
(a) If f
is differentiable at c, then F is differentiable at c.
14. TheFundamentalTheoremof Calculus 295
(b) If f is differentiable at c, then F' is continuous
(c) If f' is continuous at c, then F' is continuous
4.
at c.
at c.
Show that the values of the following expressions do not depend on x:
1/x1
x1
(i)
dt +
1+t2
1
(ii)
Isinx
Find
(f
1
cosx
dt,
t2
--
dt.
14g2
0
[0,x/2].
XE
¯1)'(0)
5.
if
X
(i) f (x)
=
(ii) f(x)
=
Ï
1 + sin(sin t) dt.
O
cos(cost)dt.
1
(Don't try to evaluate
6.
f
explicitly.)
Find a function g such that
(i)
lx
tg(t)dt=x+x2.
0
tg(t) dt
(ii)
=
x
+x2
(Notice that g is not assumed continuous at 0.)
7.
Find all continuous functions
f
satisfying
X
f=(f(x))2+C.
Ï
O
for some constant
*8.
C.
Suppose that f is a differentiable function with f (0)
Prove that for all x > 0 we have
<
f' s 1.
3
Let
f (x)
Is the function F(x)
10.
0 and 0
2
x
*9.
=
=
cos
=
fe'f
(i)
(ii)
1
76
1
<
¯
7Ã
3
8
-
Ál+x2
1/2
<
¯
o
dx 5
1 x
dx 5
1+x
-
1
x
·ý
x
0
differentiable at 0? Hint: Stare at page 177.
Use Problem 13-23 to prove that
-
-,
n4
---.
1
I
-.
296 Derivativesand Integrals
11.
Find F'(x) if F(x)
f( xf (t)dt. (The answer is not xf (x); you should
perform an obvious manipulation on the integral before trying to find F'.)
12.
Prove that if f is continuous,
=
then
xf
(u)(x
u) du
-
f (t) dt
=
du.
Hint: Differentiate both sides, making use of Problem 11.
*13.
Use Problem 12 to prove that
x
f(u)(x-u)2du=2
14.
*15.
Find a function f such that f"'(x)
supposed to be easy; don't misinterpret
(a) If f
is periodic
I
O
periodic g for
only
16.
if
f
(This problem is
x.
"find.")
a,
if f (x + a)
f (x) for all x.
=
[0,a],
show that
f
-
b
for all b.
f is not periodic, but f' is. Hint: Choose
it can be guaranteed that f (x)
fag is not
such that
which
periodic.
(c) Suppose that f'
sin2
du2-
b+a
a
a
| Ál+
the word
1
period a and integrable on
with
(b) Find a function f
=
with period
A function f is periodic,
dui
f(t)dt
=
is periodic with period a. Prove that
f
is periodic if and
f (a) f (0).
=
feb
Find
dx, by simply guessing a function f with f'(x)
6, and using
the Second Fundamental Theorem of Calculus. Then check with Prob=
lem 13-21.
Cs
*17.
c
Use the Fundamental Theorem of Calculus and Problem 13-21 to derive the
in Problem 12-18.
result stated
ca
18.
Let C1, C and C2 be curves passing through the origin, as shown in Figure 7.
Each point on C can be joinedto a point of Ci with a vertical line segment
and to a point of C2 with a horizontal line segment. We will say that C bisects
Ci and C2 if the regions A and B have equal areas for every point on C.
x2, x > 0 and C is the graph of (x) 2x2,
f
>
that
and
C
Ci
C2.
C2
bisects
find
0,
so
x
(b) More generally, find C2 if Ci is the graph of f(x) x", and C is the
graph of f (x) cx" for some c > 1.
(a) If Ci is the graph of f (x)
=
=
=
FIGURE
7
=
19.
*20.
derivatives of F(x)
fi' 1/t dt and G(x)
(b) Now give a new proof for Problem 13-15.
(a) Find the
=
=
1/t dt.
fbbx
Use the Fundamental Theorem of Calculus and Darboux's Theorem (Problem 11-54) to give another proof of the Intermediate Value Theorem.
14. TheFundamentalTheoremof Calculus 297
-21.
Prove that if h is continuous,
and g are differentiable, and
f
F(x)
h(t) dt,
=
h(f(x))
then F'(x)
h(g(x)) g'(x)
f'(x). Hint: Try to reduce this to
the two cases you can already handle, with a constant either as the lower or
the upper limit of integration.
=
22.
·
·
-
Show also that the hypothesis
*23.
[0, 1] and f (0)
Suppose that f' is integrable on
[0, 1] we have
(a) Suppose
G'
differential
(*)
=
g and F'
f (0)
g(y(x))
-
y'(x)
Hint: Problem 13-39.
for all x in some interval,
f(x)
=
0. Prove that for all x in
Prove that if the function y satisfies the
f.
=
equation
0 is needed.
=
=
then there is a number c such that
G(y(x))
(**)
for all x in this interval.
F(x) + c
that if y satisfies
(b) Show, conversely,
(c) Find what
=
condition
y must satisfy
y'(x)
(**),then
if
1 + x2
1+y(x)
=
y is a solution
of
(*).
.
the resulting
1 + t and f (t) 1 + t2.) Then
(In this case g(t)
equations to find all possible solutions y (nosolution will have R as its
domain).
(d) Find what condition y must satisfy if
=
"solve"
=
-1
y'(x)
=
.
1 + 5[y(x)]4
(An appeal to Problem 12-11 will show that there are functions satisfying
the resulting equation.)
(e) Find all functions y satisfying
y(x)y'(x)
-x.
=
-1.
Find the solution y satisfying
24.
y(0)
=
In Problem 10-17 we found that the Schwarzian derivative
f'"(x)
f'(x)
was 0
whose
¯
3
2
f"(x)
f'(x)
for f (x)
(ax + b)/(cx + d). Now suppose that f is any function
Schwarzian derivative is 0.
=
298 Derivativesand Integrals
(a) f"2g'3
is a constant fimction.
the
(b) f is form f(x) (ax + b)/(cx + d). Hint: Consider u
apply the previous problem.
=
*25.
JN
The limit lim
f,
N-woo
"improper
called an
if it exists, is denoted by
and
f'
=
f°°f (orf°°f (x)dx), and
integral."
-1.
f°°xrdx,
(a) Determine
if
r <
13-15 to show that
(b) Use Problem
fi°°1/x dx does not
exist. Hint: What
1/x dx?
can you say about
>
f exists. Prove that if
(c) Suppose that f (x) 0 for x > 0 and that
0 5 g(x) 5 f(x) for all x > 0, and g is integrable on each interval
g also exists.
[0,N], then
1/(1+ x2) dx exists. Hint: Split this integral up at 1.
(d) Explain why
f
fe"
fe"
fe
26.
Decide whether
or not the following improper integrals exist.
1
Ï°°fl+x3
(ii)
I"
(iii)
l=°
(i)
dx.
0
x
1 + x3/2
i
o
o xjl
*27.
dx.
dx.
+x
The improper integral
is defined in the obvious way, as
fi f
But another kind of improper integral
it is fg°°
f +
|
(a) Explain why
(b) Explain why
f, provided
f°° f
lim
JN
·
N->-oo
is defined in a nonobvious
way:
these improper integrals both exist.
f" 1/(1 + x2) dx exists.
f°° x dx does not exist. (But notice
that lim
/_NN
N->oo
X
dx does
exist.)
(c) Prove
that if
Show moreover,
f°° f
exists, then
lim
f_NN
Nandœ
that lim
|_NN
lim
N-woo
Naco
f
exists and equals
|_NN2
f both
f°° f.
exist and equal
Can you state a reasonable generalization of these facts? (If you
will have a miserable time trying to do these special cases!)
you
f°° f.
can't,
*28.
"improper
There is another kind of
bounded, but the function
is unbounded:
(a) If
a
>
0, find lim
e->O+
f 1/Jdx.
even though the function f (x)
matter how we defme f (0).
integral" in which
the interval is
This limit is denoted by
=
1/
||1//dx,
is not bounded on
[0,a],
no
-1
(b) Find fe"x'
< r < 0.
dx if
Problem
13-15 to show that
(c) Use
a limit.
/
x¯I
dx does not make sense, even as
of Calculus 299
14. TheFundamentalTheorem
(d) Invent a reasonable
definition of
-l<r<0.
fa°|x|'dx for a
0 and compute it for
<
(1 x2)¯1¡2 dx, as a sum of two
a reasonable definition of
limits, and show that the limits exist. Hint: Why does 1(1 + x)¯I/2 dx
(1-x2)¯I/2
<
exist? How does (1+x)¯I/2
for
compare with
x < 0?
f
(e) Invent
-
|
29.
*30.
(a) If f
is continuous
(b) If f
is integrable on
on
[0, 1], compute
[0, 1] and
It is possible, finally, to combine
the integral.
(a) If ocf(x)
1/T
=
for 0 Ex
lim f
x->0
/1
lim x
---
x->0+
x
t
--1
dt.
lim x
(x) exists, compute x->0+
x
t
dt.
the two possible extensions of the notion of
5 1 and
f (x) dx (afterdeciding what
f(x)
this should
=
1/x2 for x
>
1, find
mean).
oo
-1
xr
(b) Show that
dx never makes sense.
(Distinguish the cases
<
--1.
0 and r
at oo; for r
r
<
=
<
--1
In one case things go wrong at 0, in the other case
things go wrong at both places.)
CHAPTER
THE TRIGONOMETRIC
FUNCTIONS
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 carefiilly, 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 initial point (Figure 1).
FIGURE
1
"directed
angles," which may be regarded as
More useful for trigonometry are
pairs (li, l2) of half-lines with the same initial point, visualized as in Figure 2.
FIGURE
i.
FIGURE
3
2
If for 11we always choose the positive half of the horizontal axis, a directed angle
is described completely by the second half-line (Figure 3).
Since each half line intersects the unit circle precisely once, a directed angle is
described, even more simply, by a point on the unit circle (Figure 4), that is, by a
pOint
(X, y) With x2
1.
+ y2
=
300
15. The Trgonometric
Functions 301
The sine and cosine of a directed angle can now be defined as follows (Figure 5):
a directed angle is determined by a point (x, y) with x2 + y2 = 1; the sine of the
angle is defined as y, and the cosine as x.
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.
What we have defined is the sine and cosine of a directed angle; what we want
to define is sin x and cos x for each number x. The usual procedure for doing
this depends on associating an angle to every number. The oldest method is to
the way around" is associated to 360,
angles in degrees." An angle
associated
angle
around"
is
180,
to
an
quarter way around"
an angle
to 90, etc. (Figure 6). The angle associated, in this manner, to the number x, is
called
angle of x degrees." The angle of 0 degrees is the same as the angle
of 360 degrees, and this ambiguity is purposely extended further, so that an angle
of 90 degrees is also an angle of 360 + 90 degrees, etc. One can now define a
function, which we will denote by sin°, as follows:
"measure
"all
"half-way
"a
"the
FIGURE
4
sin°(x)
>
7¯¯
si
A
=
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 Ã
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
would be sheer luck if the function sin° had mathematically
to elegant results-it
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
are of the circle beginning at (1,0) and running counterclockwise to P (Figure 7).
angle of x radians." Since the
The directed angle determined by P is called
of
whole
the
length
circle is 2x, the angle of x radians and the angle of 2x + x
cos A
"the
FIGURE
radians
5
are
identical. A fimction sin, can now be defined as follows:
sin'(x)
=
sine of the angle of x radians.
This same method can easily be adopted
sin' 2x, we can define
sin° 360
to define sin°; since we want to have
=
180
sin° x
so
so
=
sin"
2xx
360
----
=
sin' xx
-.
180
We shall soon drop the superscript r in sinr, since sin' (andnot sin°) is the only
function which will interest us; before we do, a few words of warning are advisable.
The expressions sin° x and sinr x are sometimes written
sin x°
sin x radians,
FIGURE
6
but this notation is quite misleading; a number x is simply a number-it
does not
radians."
If the meaning
degrees" or
carry a banner indicating that it is
"in
"in
302 Derivativesand Integrals
of the notation
"sin
x"
is in doubt one usually asks:
P
"Is x in degrees or radians?"
kngth x
but what one means is:
(t, 0)
FIGURE
'sinr,
'sin°'
"Do you mean
»
or
Even for mathematicians, addicted to precision, these remarks might be dispensable, were it not for the fact that failure to take them into account will lead to
incorrect answers to certain problems (anexample is given in Problem 19).
Although the function sin' is the function which we wish to denote simply by sin
involved even in the definition
use exclusively henceforth),there is a difficulty
(and
Of SÎûr. Our proposed definition depends on the concept
of the length of a curve.
several
problems, it is also easy
Although the length of a curve has been defined in
of length is
to reformulate the definition in terms of areas. (A treatment in terms
outlined in Problem 28.)
7
Suppose that x is the length of the arc of the unit circle from (1,0) to P; this are
of the unit cirle.
thus contains x/2x of the total length 2x of the circumference
shown in Figure 8; S is bounded by the unit circle, the
Let S denote the
should be
horizontal axis, and the half-line through (0, 0) and P. The area of S
x/2x times the area inside the unit circle, which we expect to be x; thus S should
have area
X
X
P
length x
"sector"
area -
2
which
We can therefore define cos x and sin x as the coordinates of the point P
determines a sector of area x/2.
sin
With these remarks as background, the rigorous definition of the functions
and cos now begins. The first definition identifies x as the area of the unit circlesemicircle (Figure 9).
more precisely, as twice the area of a
2x
F I GURE
8
11
DEFINITION
f(x)
g
=
2
Ál
.
-
x2 dx.
(This definition is not offered simply as an embellishment; to define the trigonometric functions it will be necessary to first define sinx and cosx only for
=
-1
area
=
|
Ex 5 1, the area A(x) of
The second definition is meant to describe, for
the sector bounded by the unit circle, the horizontal axis, and the half-line through
expressed (Figure 10) as the sum of
(x, Ál x2). If 0 Exy 1, this area can be
-
the area of a triangle and the area of a region under the unit circle:
x
FIGURE9
1-x2
1
+
1-t2dt.
15. The Trigonometric
Functions 303
-1
This same formula happens to work for
9
0 also. In this case (Figure 11),
5 x 5
the term
2
is negative,
the term
FIGURE
and
the area of the triangle which must be subtracted from
represents
1
1-t2dt.
10
-1
DEFINITION
If
<
x
<
1, then
x/1
A(x)
1
x2
-
|1
+
=
2
,
12
-
dt.
-1
< x < 1, then A is differentiable at x and
Notice that if
mental Theorem of Calculus),
(usingthe
Funda-
-2x
A'(x)
=
-
1
x
2
2) l
+
-
x2
-
Ál
-
x2
/1
-
-
x2
-x2
-
FIGURE
ll
1
2
Î
(1
+
X
-
-
2Ál
l-2x2-2(Î-X2
---
x2)
-Vl-x2
1-x2
-
----
-
x2
Ál
x2
-
2/l-x2
-1
2)1-
x2
Notice also (Figure 12) that on the interval
from
i
i
FIGURE
the function A decreases
$ Ál-t2dt=I 2
1-1
A(-1)=0+
-1
[---1,1]
to A(1) = 0. This followsdirectly from the definition of A, and also from the fact
that its derivative is negative on (-1, 1).
For 0 5 x 5 x we wish to define cos x and sin x as the coordinates of a point
P
(cosx, sin x) on the unit circle which determines a sector whose area is x/2
(Figure 13). In other words:
12
=
DEFINMON
If 0 5 x 5 x, then cos x is the unique number
A(cosx)
in
x
-;
=
2
and
sin x
=
1
-
(cosx)2.
[-1, 1] such
that
and Integrals
304 Derivatives
This definition actually requires a few words of justification.In order to know
x/2, we use the fact that A is continuous,
that there is a number y satisfying A(y)
and
w/2.
This tacit appeal to the Intermediate
and that A takes on the values 0
make
Value Theorem is crucial, if we want to
our preliminary definition precise.
and
made,
definition,
Having
we can now proceed quite rapidly.
justified,our
=
THEOREM
1
If 0
<
x
<
x, then
-sinx,
PROOF
If Ñ
=
cos'(x)
=
sin'(x)
=
2A, then the definition A(cosx)
cos x.
be written
x/2 can
=
B(cos x)
=
x;
in other words, cos is just the inverse of B. We have already computed that
1
A'(x)
2dl
=
-
,
-x2
P
=
sin x)
(cosx,
from which we conclude
that
1
area
-
gl-x2
2
Consequently,
cos'(x)
FIGURE
(B¯1),
=
13
Î
B'(B-1
)
1
1
1
=
-
=
-
1
[B¯*(x)]2
-
(cosx)2
-
sin x.
Since
sin x
V1
=
-
(cosx)2,
we also obtain
-2
sin'(x)
=
cos x cos'(x)
(cosx)
1
2
-
Ál
sinx
-
cosx
sinx
=
cosx.
The information contained in Theorem 1 can be used to sketch the graphs of
15. The TryonomenicFunctions 305
sin and cos on the
14
[0,x].
interval
COS
cos'(x)
=
-
Since
sin x
<
0,
O< x
<
x,
-1
the function cos decreases from cos O 1 to cos x
(Figure 14). Consequently,
unique
x].
that
the definition of cos,
To
fmd
0
for
in
cos y
a
y
y, we note
[0,
=
-1-
=
=
FIGURE
À(COSX)
14
means
=
,
that
A(0)
SO
2,
=
2
1
y
=
-t2dt.
Ál
2
It is easy to see that
0
2
-1
FIGURE
15
l
Ål-t2dt=
Ál-t2dt
JO
so we can also write
1-t2dt=
y=
.
1
Now we have
sin'(x)
=
cos x
>0,
0<x<r/2
<0,
x/2<x<x,
increases on [0,x/2] from sin 0 = 0 to sin x/2 = 1, and then decreases on
x]
[n/2, to sin x 0 (Figure 15).
The values of sin x and cos x for x not in [0,x] are most easily defined by a
two-step piecing together process:
so sin
=
2
(1)
If x 5 x 5 2x, then
-sin(2x
sinx
(a)
cosx
--x),
=
=
cos(2x
Figure 16 shows the graphs of sin and cos on
(2)
2=
.-
2
If x
=
FIGURE
16
[0,2x].
2xk + x' for some integer k, and some x' in
sin x
cos x
(b)
x).
-
=
sin x',
=
cos x'.
[0,2x],
then
Figure 17 shows the graphs of sin and cos, now defined on all of R.
Having extended the functions sin and cos to R, we must now check that the
basic properties of these functions continue to hold. In most cases this is easy. For
example, it is clear that the equation
sin2 x +
COS2
306 Derivativesand Integrals
(a)
(b)
FIGURE
17
holds for all x. It is also not hard to prove that
if x is not a
multiple
of x.
sin'(x)
=
cos'(x)
=
cos x,
-sinx,
For example,
if x
<
-sin(2x
sinx
<
x
2x, then
-x),
=
so
sin'(x)
=
=
=
-
sin'(2x
cos(2x
--
-
x)
-
(--1)
x)
cosx.
If x is a multiple of x we resort to a trick; it is only necessary to apply Theorem 11-7 to conclude that the same formulas are true in this case also.
FIGURE
18
15. The TyonometricFunctions 307
i-
The other standard
define
trigonometric
1
-1
1
secx
=
tan x
=
csex
=
cotx
=
---
cosx
sm x
x
¢
kr +x/2,
x
¢
kr.
cos x
1
i
tan
-
smx
-1-
cosx
sinx
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
the
functions, which are listed in the next theorem (thereis no need to memorize
results
of
rederived
whenever
needed.)
the theorem, since the
statement
can be
THEOREM
FIGURE
functions present no difficulty at all. We
2
If x ¢ kr + x/2, then
BCC'(X)
19
tan'(x)
=
SCCX tanX,
=
sec2 x.
If x ¢ kx, then
csc'(x)
cot'(x)
PROOF
2·
Left to you
(astraightforward
=
-
csc2
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 x, and the intervals
usually chosen are (Figure 19)
[-x/2, x/2]
[0,x]
(a)
(-x/2,
for sin,
for cos,
for tan.
x/2)
(The inverses of the other trigonometric functions are so rarely used that they will
not even be discussed here.)
The inverse of the function
-1
1
-x/2
f (x)
(b)
20
=
sin x,
5 x 5 x/2
is denoted by arcsin (Figure 20); the domain of arcsin is [-1, 1]. The notation
has been avoided because arcsin is not the inverse of sin (which
is not oneone), but of the restricted function f; sometimes this function f is denoted by Sin,
and arcsin by Sin¯1
sin¯I
FIGURE
-cscxcotx,
=
308 Derivativesand Integrals
The inverse of the function
05x5x
g(x)=cosx,
is denoted by arccos (Figure 21); the domain of arccos is
is denoted by Cos, and arccos by Cos-1
The inverse of the function
[-1, 1].
Sometimes g
--x/2
h(x)
rccos
tanx,
<
<
x
x/2
is denoted by aretan (Figure 22); arctan is one of the simplest examples of a
differentiable function which is bounded even though it is one-one on all of R.
Sometimes the function h is denoted by Tan, and arctan by Tan¯'.
The derivatives of the inverse trigonometric functions are surprisingly simple,
and do not involvetrigonometric functions at all. Finding the derivatives is a simple
matter, but to express them in a suitable form we will have to simplify expressions
like
-1
1
-1
FIGURE
=
21
cos(arcsin
sec(arctanx).
x),
2
22
FIGURE
A little picture is the best way to remember the correct simplifications. For example, Figure 23 shows a directed angle whose sine is x-the angle shown is thus an
angle of (arcsinx) radians; consequently cos(arcsin x) is the length of the other
side, namely,
Ál x2. However, in the proof of the next theorem we will not
-
resort to such pictures.
-1
FIGURE
23
THEOREM
3
If
<
x
<
1, then
arcsin'(x)
1
=
,
1-x2
-1
arecos'(x)
=
.
-x2
1
Moreover, for all x we have
arctan'(x)
=
1
1 + x2
.
15. The Trgonometric
Functions 309
arcsin'(x)
PROOF
(f¯ )'(x)
=
1
f(x))
f'(f
1
sin'(arcsin
x)
1
cos(arcsin x)
Now
x)]2
[sin(arcsin
+
x)]2
[cos(arcsin
_.,
y
that is,
x2 +
x)]2
[cos(arcsin
O
--
therefore,
cos(arcsin x)
Ál
=
x2.
-
(The positive square root is to be taken because arcsinx is in (-n/2, x/2), so
cos(arcsinx) > 0.) This proves the first formula.
The second formula has already been established (inthe 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'(x)
=
(h¯I)'(x)
1
h'(h-1(
1
tan'(arctan x)
1
¯
sec2(arctaux)
Dividing both sides of the identity
sin2
a #
COS2
=
a
Î
by cos2 a yields
tan2 a + 1
sec2
=
a.
It follows that
+ 1
[tan(arctanx)]2
=
sec2(arctanx),
or
x2
which
+ 1 = sec2(arctanx),
proves the third formula.
The traditional proof of the formula sin'(x)
given here) is outlined
different from the one
cos x (quite
in Problem 27. This proof depends upon first establishing
=
310 Derivativesand Integrals
the limit
lim
hw0
and the
"addition
sin h
1,
=
--
h
formula"
sin(x + 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'(0) = cos0. The second depends
of the functions sin and cos. In order to derive this
on a beautiful characterization
result we need a lemma whose proof involves a clever trick; a more straightforward
proof will be supplied in Part IV
LEMMA
Suppose
f
has a second derivative everywhere and that
=0,
f"+ f
f (0) 0,
f'(0) 0.
=
=
Then f
PROOF
=
0.
Multiplying both sides of the first equation by
yields
f'
f'f"+ ff'=
0.
Thus
[(f')2+ f2]'=2(f'f"+
so (f')2
the constant
ff')=0,
2
is a constant function. From f (0)
is 0; thus
[f'(x)]
+
[f (x)]
=
0
=
0 and f'(0)
=
0 it follows that
for all x.
This implies that
f (x) 0 for all x. I
=
THEOREM
4
If f has a second derivative everywhere
and
f"+ f
f (0)
f'(0)
=0,
=
=
a,
b,
then
f
(In particular, if
cos.)
then f
=
f(0)
=
=
0 and f'(0)
b sin + a cos.
-
=
-
1, then f
=
sin;
if f (0)
=
1 and f'(0)
=
0,
15. The TgonometricFunctions 311
PROOF
Let
-bsinx
g(x)=
-acosx.
f(x)
Then
g'(x)=
g"(x)
f'(x)-bcosx+asinx,
sinx + a cosx.
f"(x) +b
=
Consequently,
g"+g=0,
g(0)
g'(0)
0,
0,
=
=
which shows that
0=g(x)=
THEOREM
s
If x and y are any two numbers,
sin(x
then
+ y)
cos(x + y)
PROOF
forallx.
f(x)-bsinx-acosx,
sin x cos
y + cos x sin y,
cos x cos y sin x sin y.
=
=
-
For any pardcular number y we can define a function
f (x)
f
by
sin(x + y).
=
Then
f'(x)
f"(x)
cos(x + y)
sin(x + y).
=
=
-
Consequently,
f" + f
f(0)
f'(0)
=
=
=
0,
sin y,
cos y.
It follows from Theorem 4 that
f
=
(cosy)
-
sin +(sin y)
·
cos;
that is,
sin(x
+ y)
=
cos y sin x + sin y cos x,
for all x.
Since any number y could have been chosen to begin with, this proves the first
formula for all x and y.
The second formula is proved similarly.
As a conclusion
an alternative
to this chapter, and as a prelude to Chapter 18, we will mention
to the definition of the function sin. Since
approach
arcsin'(x)
l
=
1
-
x2
for
-
1< x
<
1,
312 Derivativesand Integrals
it followsfrom the Second Fundamental Theorem of Calculus that
x
1
arcsinx
arcsinx
arcsin0
dt.
1 t2
=
=
-
-
This equation could have been taken as the definitionof arcsin. It would follow
immediately that
I
arcsin'(x)
=
1
-
the function sin could then be defined as (arcsin)¯I
tive of an inverse function would show that
sin'(x)
Ál
=
-
;
x2
and the
formula for the deriva-
sin x,
which could be defined as cos x. Eventually, one could show that A(cos x)
x/2,
with
which
recovering
of
started.
the
end
the
the
development
definition
at
we
very
While much of this presentation would proceed more rapidly, the definition would
of the definitions would be known to
the reasonableness
be utterly unmotivated;
the author, but not to the student, for whom it was intended! Nevertheless, as
we shall see in Chapter 18, an approach of this sort is sometimes very reasonable
=
indeed.
PROBLEMS
1.
Diffërentiate each of the following functions.
(i) f (x)
=
arctan(arctan(arctan
(ii) f (x)
(iii) f (x)
=
arcsin(arctan(arccos
=
arctan
=
arcsin
(iv) f (x)
x)).
x)).
x).
(tanx arctan
1
.
1 + x2
2. Find the following limits by l'Hôpital's Rule.
.
sinx-x+x3/6
hm
(i)
x-O
(ii)
hm
.
X,
sinx-x+x3/6
.
x->0
.
cosx---1+x2/2
lun
(iii) x->o
lim
(iv) x40
.
(v)
4
x
.
x
cos x
1 + x2/2
-
'
4
x
arctanx-x+x3/3
hm
x-vo
lim
(vi) x->o
x
(1
-
-
x
1
--
smx
.
15. The Trgonometric
Functions 313
sin x
¯¯'
3.
Let
f (x)
x
=
1,
x
¢0
=
0.
(a) Find f'(0).
(b) Find f"(0).
At this point, you will almost certainly have to use l'Hôpital's Rule, but in
Chapter 24 we will be able to find f(k)(0)for all k, with almost no work
at all.
4.
Graph the following functions.
(a) f (x)
(b) f (x)
=
sin
2x.
sin(x2).
(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 sign of the
cos(x2)-2x is no easier than determining the behavior
=
derivative
=
f'(x)
of
f directly. The formula for f'(x) does indicate
0, which must be true since
one important fact,
is even, and which
however-f'(0)
f
be clear in your graph.)
(c) f(x) sinx + sin2x. (It will probably be instructive to first draw the
sin2x carefully on the same set of
sinx and h(x)
graphs of g(x)
and
what
the sum will look like. You can
from
0
2x,
to
axes,
guess
easily find out how many critical points f has on
[0,2x] by considering
the derivative of f. You can then determine the nature of these critical
points by finding out the sign of f at each point; your sketch will probably
=
should
=
=
=
suggest the answer.)
(First determine the behavior of f in (-x/2, x/2); in
the intervals (kr n/2, kr + x/2) the graph of f will look exactly the
same, except moved up a certain amount. Why?)
(e) f (x) sin x x. (The material in the Appendix to Chapter 11 will be
particularly helpful for this function.)
(d) f (x)
=
tan x
--
x.
-
=
-
sin x
(f) f (x)
¢0
¯¯'
x
1,
x=0.
=
where the zeros
enable you to determine approximately
of f' are located. Notice that f is even and continuous at 0; also consider
the size of f for large x.)
(g) f (x) = x sin x.
(Part
*5.
6.
should
(d)
a/0 in polar coordispiral is the graph of the function f(0)
The hyperbolic
this
paying
particular attention
nates (Chapter 4, Appendix 3). Sketch
curve,
to its behavior for 0 close to 0.
=
Prove the addition
formula for cos.
314 lkrivatives and Integrals
7.
(a) From
the addition formula for sin and cos derive formulas for sin 2x,
2x, sin 3x, and cos 3x.
(b) Use these formulas to find the following values of the trigonometric fimctions (usually
deduced by geometric arguments in elementary trigonomcos
etry):
sm
tan
cos
8.
cos
4
4
-
-,
4
2
1
x
=
-,
2
6
=
-
=
-
1,
=
-
.
sm
=
-
-.
6
2
A sin(x + B) can be written as a sin x + b cos x for suitable a
and b. (One of the theorems in this chapter provides a one-line proof.
You should also be able to figure out what a and b are.)
(b) Conversely, given a and b, find numbers A and B such that a sin x +
A sin(x + B) for all x.
b cos x
(c) Use part (b)to graph f (x) E sin x + cos x.
(a) Show that
=
=
9.
(a) Prove that
tan(x + y)
tan x + tan y
1 tanx tan y
=
-
provided that x, y, and x + y are not of the form kr + x/2.
addition
(Use the
sin and cos.)
formulas for
(b) Prove that
arctan x + arctan y
x + y
(
arctan
=
1
,
-
xy
indicating any necessary restrictions on x and y. Hint: Replace x by
arctan x and y by arctan y in part (a).
10.
Prove that
arcsina
+ arcsin
ß
indicating any restrictions
=
on
arcsin(ad1
œ
and
p2
-
_
ß.
11.
Prove that if m and n are any numbers,
sinmx sin nx
sin mx cos nx
cos mx cos nx
=
=
=
then
cos(m + n)x],
([cos(mn)x sin(m
n)x],
+ n)x +
([sin(m
-
-
---
+ n)x + cos(m
([cos(m
-
n)x].
2
15. The T onometricFunctions 315
12.
Prove that if
numbers,
m and n are natural
x
Ï
I
Ï
then
0,
sinmx sinnxdx
m
¢
x,
m
=
n,
0,
m
¢
n
m
=
n,
=
l
cos mx cos nx dx
=
l x,
-x
n
x
sinmx cos nx dx
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.
13.
[-x, x],
is integrable on
(a) If f
show that the minimum
lx
dx
(f(x)-acosnx)
occurs when
1
a
and the minimum
=
value of
"
f(x)cosnxdx,
-
value of
/K
(f (x)
sinnx)2
dx
f (x)sinnx
dx.
a
-
when
a
=
1
"
-
(In each case, bring a outside the integral sign, obtaining a quadratic
expression in a.)
(b) Define
=
a,
b,=-
1
"
1
"
-
f(x)cosnxdx,
n= 0,1,2,...,
f(x)sinnxdx,
n=1,2,3,....
Show that if c, and d, are any numbers,
(
f (x)
=
2
N
[f (x)]2dx
c, cos nx + d, sinnx
+
-
then
dx
n=1
-
2x
+
+ x
a.c.
-
--
+ b.d.
+
+ x
(c.
c,2 + d
+
-
a.)2
+
(d. b.)2
-
2
316 Derivanvesand Integrals
thus showing that the first integral is smallest when at
c¿ and b¿ d¿.
combinations" of the functions s.(x)
In other words, among all
sinnx and c.(x)
N, the particular function
for
1
yns
cosnx
=
=
"linear
=
=
g(x)=
a,cosnx+bssinnx
+
n=1
has the
14.
"closest
fit" to f on
[-n, x].
(a) Find a formula for sin x + sin y.
(Notice that this also gives a formula for
Hint: First find a formula for sin(a + b) + sin(a b). What
good does that do?
Also
find a formula for cos x + cos y and cos x cos y.
(b)
sin x
sin y.)
-
-
-
15.
(a) Starting from the formula for cos 2x, derive formulas for
terms of cos
in
(b) Prove that
cos
x
1+cosx
2
2
for05xix/2.
(c) Use part (a)to find
sin2
(d) Graph f (x) =
16.
c'os2
x and
2x.
=
-
sin2
f
and
sin2
x
.
sm
dx and
f
x
-
1-cosx
2
=
2
cos2
x
dx.
Find sin(arctan x) and cos(arctan x) as expressions not involving trigonosin y/ cos y
functions. Hint: y
arctan x means that x
tan y
sin2
sin y/ 1
metric
=
=
=
=
-
17.
18.
answers should
express sinu and cosu in terms of x. (Use Problem 16; the
be very simple expressions.)
(a) Prove that
sin(x + x/2)
If x
=
tanu/2,
cosx. (All along we have been drawing the
graphs of sin and cos as if this were the case.)
(b) What
19.
is arcsin(cosx)
(a) Find
t2
1
°°
(b) Find
1
1 ‡ t2
dt.
1
Find lim x sin
21.
(a) Define functions sin°
-.
x
and cos°
Find (sin°)'and
cos(xx|180).
sin° x
(b) Find
22.
and arccos(sinx)?
dt. Hint: The answer is not 45.
20.
x-oo
=
lim
x-o
x
and lim x sin°
x
oo
by sin°(x)
sin(xx/180)
and cos°(x)
(cos°)'in terms of these same functions.
=
1
-.
x
Prove that every point on the unit circle is of the form (cos0,sin0) for at
least one (andhence for infinitely many) numbers 0.
15. The Trigonomehic
Functions 317
23.
(a) Prove
that x is the maximum possible length of an interval on
is one-one, and that such an interval must be of the form
x/2,
2kx + x/2] or [2kn+ x/2, 2(k + 1)x x/2].
[2kn
(b) Suppose we let g(x) sin x for x in (2kr x/2, 2kx + x/2). What is
which
sin
-
-
=
-
(g-1p
24.
Let
f(x)
=
for 0 Exs
secx
Find the domain of
x.
f¯I
and sketch
its
graph.
25.
Prove that | sin x
-¢
y| for all numbers x
y. Hint: The same
with
replaced
<
is
straightforward
by
statement,
of a
5
consequence
a very
well-known theorem; simple supplementary considerations
then allow $ to
be improved to <
sin y|
-
|x
<
-
,
.
*26.
It is an excellent test of intuition to predict the value of
lim
Ib
f(x)sin1xdx.
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 f.
(a) Show that
(b) Show that
lim
f
sin1xdx
0, by computing
=
if s is a step function on [a,b]
lem 13-26), then lim j s(x) sin1x dx = 0.
the integral explicitly.
from
(terminology
(c) Finally, use
Prob-
Problem 13-26 to show that lim f f(x)sin1xdx
0 for
A-koo
which
result,
is integrable on [a,b]. This
like Problem 12,
any function f
plays an important role in the theory of Fourier series; it is known as the
Riemann-Lebesgue Lemma.
=
C
27.
A
This problem outlines the classical approach to the trigonometric functions.
The shaded sector in Figure 24 has area x/2.
(a) By considering
then
O
B
-
the triangles OAB and OCB prove that if 0
sin x
(1, 0)
-
2
FIGURE
24
(b) Conclude
x
4
_
2
sin x
«
2cosx
that
smx
cosx<--<1,
x
and
prove that
lim
imx
=
-
x-vo
1.
x
(c) Use this limit to fmd
.
hm
x->0
1
-
cosx
.
x
<
x
<
x/4,
318 Denvativesand Integrals
(d) Using parts (b)and (c),and
starting
*28.
from the definition
the addition formula for sin, find sin'(x),
of the
derivative.
This problem gives a treatment of the trigonometric
length, and uses Problem 13-25. Let f(x)
fl
Define Œ(x) to be the length of f on [x, 1].
functions in terms of
X
Î.
for
-x2
-1
=
(a) Show that
i
2(x)
1
=
1
t2
-
dt.
(This is actually an improper integral, as defined in Problem 14-28.)
(b) Show that
E'(x)
-1
for
-
<
l.
<
x
For 0 Ex 5 x, define cosx by 2(cosx) = x, and
and sin'(x)
V1-cos2x.Prove that cos'(x)
as 2(-1).
(c) Define x
-sinx
define sinx
=
=
cosxfor0<x<x.
*29.
1
=
=
Yet another development of the trigonometric functions was briefly menwith inverse functions defined by integrals. It
tioned in the textstarting
is convenient to begin with arctan, since this function is defmed for all x.
To do this problem, pretend that you have never heard of the trigonometric
functions.
(a) Let a(x)
=
lim a(x)
we
t2)¯I
dt.
/‡(1+
and
a(x)
Prove that a is odd and increasing, and that
both exist, and are negatives of each other. If
lim
define x
=
2 X->00
lim a(x), then a¯1 is defmed on (-x/2, x/2).
(b) Show that (a¯1)'(x)
=
1+
[a¯1(x)2
«¯I(x').
kr + x' with x' ¢ x/2 or
Then
define tanx
1//1
tan2
define cos x
+
x, for x not of the form kx+x/2 or kx
and cos(kx ±x/2)
0. Prove first that cos'(x)
tan x cos x, and then
all
that cos"(x)
cos x for
x.
(c) For x
-x/2,
=
=
-x/2,
=
=
=
*30.
=
-
-
If we are willing to assume that certain differential equations have solutions,
another approach to the trigonometric functions is possible. Suppose, in
particular, that there is some function yo which is not always 0 and which
satisfies yo" + yo
0.
=
(a) Prove that
/)2
yo2
is Constant,
and conclude
or yo'(0) ¢ 0.
that there is a function s satisfying s" + s
Prove
(b)
s'(0)
1. Hint: Try s of the form ayo + byo'.
that either yo(0)
=
0 and s(0)
=
¢0
0 and
=
s', then almost all facts about trigonoIf we define sin
s and cos
metric functions become trivial. There is one point which requires work,
=
=
15. The TryonometncFunctions 319
however-producing the number x. This is most easily done using an
exercise from the Appendix to Chapter 11:
(c) Use Problem 7 of the Appendix to Chapter
11 to prove that cos x cannot
be positive for all x > 0. It follows that there is a smallest xo > 0 with
cos xo 0, and we can define x = 2xo.
±1;
1, we have sin x/2
1. (Since sini+ cos2
(d) Prove that sin x/2
the problem is to decide why sin x/2 is positive.)
(e) Find cos x, sin x, cos 2x, and sin 2x. (Naturally you may use any addition formulas, since these can be derived once we know that sin'
cos
=
=
=
=
=
and cos'
31.
=
sin.)
-
(f) Prove that cos and
(a) After all the work
sin are periodic with
period 22.
involved in the definition of sin, it would be disconcerting to find that sin is actually a rational function. Prove that it isn't.
(There is a simple property of sin which a rational function cannot possibly have.)
(b) Prove that sin isn't even defined implicitly by an algebraic equation; that
is, there do not exist rational fimctions fo,
fn-1 such that
.
(sinx)" + f._ t (x)(sinx)"¯1 +
-
-
-
+
.
.
,
fo(x) 0 for all
=
x.
Hint: Prove that fo 0, so that sin x can be factored out. The remaining
factor is 0 except perhaps at multiples of 2x. But this implies that it is 0
for all x. (Why?) You are now set up for a proof by induction.
=
*32.
Suppose that ¢i and ¢2 satisfy
¢i" +
¢2
and that g2
>
0,
82Ò2 0,
g1¢1
=
=
g1.
(a) Show that
¢i"¢2 ¢2"¢i (x2
-
-
(b) Show that
-
gi)¢i¢2
0.
=
if ¢1(x)> 0 and ¢2(x) > 0 for all x in (a, b), then
b
[¢1"‡2 ¢2 ÈI]
-
>
0,
and conclude that
[‡1'(b)¢2(b)¢i'(a)¢2(a)]+ [¢i(b)¢2'(b)‡1(a)¢2'(a)]>
-
-
(c) Show thatsignin this
case we cannot have
of ¢i'(a) and ‡1'(b).
sider the
(d) Show that
¢2 < 0
able to
the equations
¢¡(a)
¢i(a)
¢i
¢i(b)
=
0. Hint: Con-
‡1(b) 0 are also impossible if ¢i > 0,
¢i < 0, ¢2 < 0 on (a, b). (You should be
=
=
< 0, ¢2 > 0, or
with almost no extra work.)
this
do
or
=
0.
320 Ihrivativesand Integrals
The net result of this problem may be stated as follows: if a and b are
zeros of ¢1, then ¢2 must have a zero somewhere between
This result, in a slightly more general form, is known as the
Sturm Comparison Theorem. As a particular example, any solution of
the differential equation
consecutive
a and b.
y"+(x+1)y=0
must have zeros on the
each other.
33.
(a) Using the
formula for sin x
sin(k +
(b) Conclude
-+
1
2
positive horizontal axis which are within x of
sin y
-
derived in Problem 14, show that
()x sin(k ()x
-
=
-
2 sin
cos kx.
that
cosx + cos2x +
+ cosnx
---
¼)x
sin(n +
-
.
I
.
2sm
-
2
Like two other results in this problem set, this equation is very important
in the study of Fourier series, and we also make use of it in Problems 19-42
and 23-19.
(c) Similarly, derive the formula
.
sin
sinx +sin2x
--.+sinnx
+
-
\
(n+1
2
.
In
.
x
sm
x
X
sm-
2
(A more natural derivation of these formulas will be given in Problem 27-14.)
and
(d) Use parts (b)
(c)to
fmd
the defmition of the integral.
lb
0
sin x
b
dx and
0
cos x dx directly from
JT IS IRRATIONAL
*CHAPTER
This short chapter, diverging from the main stream of the book, is included to
demonstrate that we are already in a position to do some sophisticated mathematics. This entire chapter is devoted to an elementary proof that x is irrational. Like
proofs of deep theorems, the motivation for many steps in our
many
proof cannot be supplied; nevertheless, it is still quite possible to follow the proof
"elementary"
step-by-step.
Two observations
must
be made before the proof. The first concerns
the func-
tion
f,(x)
x"(1
-
x)"
=
n!
,
which clearly satisfies
1
0<
for0<x<1.
f,(x)<--n!
An important property of the function f, is revealed by considering the expression
obtained by actually multiplying out x"(1-x)".
The lowest power of x appearing
will be n and the highest
will be 2n. Thus f, can be written in the form
power
f,(x)
where
=
-
1
2"
E c;x',
the numbers c¿ are integers. It is clear from this expression that
f,t*)(0)=0
ifk<nork>2n.
Moreover,
f.f">(x)
-[n!c,
1
=
f,(n+1)
n!
+ 1)!cn+1 + terms involving x]
n! [(n
f,<2">(x)
=
321
+ terms involving x]
n!
[(2n)!c2.].
322 Derivativesand Integrals
This means that
fn(n)(0) c.,
f.f"*)(0) (n +1)cn+1
=
=
f.f¾(0) (2n)(2n
=
where
the numbers
1)
-
-
...
on the right are all integers.
(n + 1)c2n,
Thus
f.(k)(0)is an integerfor all
k.
The relation
f,(x)
=
f,(1
x)
-
implies that
f.f*)(x) (-1)k
(k)(Î X),
=
-
therefore,
f.f*)(1) is also
an
integerfor all k.
The proof that x is irrational requires one further observation:
number, and e > 0, then for sufficiently large n we will have
if a is any
n!
To prove this, notice that if n
>
2a, then
a"+1
(n+1)!
Now let no be any natural
may
have, the succeeding
number
values
a(no+2)
I
If k is so large that
k)!
ano
<
n+1
n!
>
1 a"
2 n!
2a. Then, whatever
satisfy
2
(no+
a"
with no
(no+ 1)!
(no + 2)!
a
(no)!
a
"
>
(no + 1)!
2k (40)!
2k, then
1 1 ano
2 2 (no)!
value
16. x is Irrational 323
a(no+k)
(no+
'
k)!
which
is the desired result. Having made these observations,
one theorem in this chapter.
mEOREM
1
PROOF
we are ready
for the
The number n is irrational; in fact, x2 is irrational. (Notice that the irrationality
of x2 implies the irrationality of x, for if x were rational, then x2 certainly would
be.)
Suppose x2 were rational,
so that
a
b
2
for some positive integers a and b. Let
G(x)
(1)
b" [x
=
f,(x)
x2=-2¿"(x)
-
y2n-4
y
(4)
+
(-1)" f.fk(x)
Notice that each of the factors
b"z2n-2k
is an integer. Since
n-k
b"W2)"¯* b"
-
-
f.(*)(0)and f,(k)(1)are
n-kb*
_
integers, this shows that
G(0) and G(1) are integers.
Differentiating G twice yields
G"(x)
(2)
The last term,
2n(4)
x2"
f
is zero. Thus
(-1)" f,(2n+2)(x),
=
b"[x2"
"(x)
-
n
adding
G"(x)+x2G(x)=b"r2n+2
(3)
(2n+2)
(1)and (2)gives
2 n
_
Now let
H(x)
G'(x) sinxx
=
xG(x) cosxx.
-
Then
H'(x)
=
=
xG'(x) cosxx + G"(x) sinxx
[G"(x) +x2G(x)] sinxx
-
xG'(x) cosxx +x2G(x)
sinxx
=x2a"f,(x)sinxx,
by(3).
By the Second Fundamental Theorem of Calculus,
x2
Il
a"f,(x)sinxxdx=H(1)-H(0)
O
G'(1)sinx
xG(1)cosx
x [G(1) + G(0)].
=
-
=
Thus
-
G'(0)sinO+ xG(0)cos0
l
x
a"
f,(x) sinxx dx is an integer.
.
324 Derivativesand Integrals
On the other hand, O <
1/n! for 0
f,(x) <
O < na"
<
x
ra"
f,(x) sinux
<
<
1, so
for 0
--
n!
<
x
<
1.
Consequently,
0<x
li
a"f,(x)sinxxdx<--.
ra"
n
o
This reasoning was completely independent of the
value of n.
Now if n is large
enough, then
0<x
a"f.(x)sinxxdx<-<1.
na"
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: x2 is irrational. I
This proof is admittedly mysterious; perhaps most mysterious of all is the why
that x enters the proof-it almost looks as if we have poved n irrational without
of the pmof will show
ever mentioning
a definition of x. A close reexamination
that precisely one property of x is essentialsin(x)
0.
-
The pmof really depends on the properties of the function sin, and proves the
irrationality of the smallest positive number x with sin x = 0. In fact, very few
properties of sin are required, namely,
·
sm
cos'
sin(0)
cos(0)
r
=
=
=
=
cos,
sin,
-
0,
1.
Even this list could be shortened; as far as the proof is concerned, cos might just
as well be defined as sin'. The properties of sin required in the proof may then be
written
sin"+sin
sin(0)
sin'(0)
=
=
=
0,
0,
1.
Of course, this is not really very surprising at all, since, as we have seen in the
previous chapter, these properties characterize the function sin completely.
PROBLEMS
1.
that the areas of triangles OAB and OAC in Figure 1 are related
by the equation
(a) Prove
16. x is Irrational 325
area OAC
=
-
1-Vl-16(area0AB)2
1
2
.
2
Hint: Solve the equations xy = 2(area OAB), x2+ y2 = 1, for y.
polygon of m sides inscribed in the unit circle. If
(b) Ist Pm be the regular
of
P. show that
Am is the area
(x,y)
0
FIGURE
i
(x,0)
-
B
=
(1,0)
A
=
A2.
2
=
2
2J1
--
(2A./m)2.
--
and more complicated) expressions
This result allows one to obtain (more
with
=
A4
2, and thus to compute x as accurately
for A2., starting
Problem
8-11). Although better methods will
desired
to
as
(according
slight
variant
of this approach
yields a very
appear in Chapter 20, a
interesting expression for x:
c
2.
(a) Using the
fact that
area(OAB)
OB,
=
area(OAC)
show that
if
œm
is the distance from O to one side of Pm, then
---
A.
«m•
=
A2m
(b) Show that
2
--
«4
=
A2e
(c) Using the
8
·
·
·
·
.
Œ2k-1
.
fact that
a.
and the formula cosx/2
«4
etc.
=
=
=
cos
-,
m
1 + cos x
(Problem 15-15),prove that
2
326
Denvadves and Integrals
Together with part
product"
-=
2
x
1
1
-+--
-·
\2
11
2V2
"infinite
that 2/x can be written
(b),this shows
1
-+-
11
-+
as an
11
-·
\2
-
\2
2
2
IVI
'
to be precise, this equation means that the product of the first n factors
can be made as close to 2/x as desired, by choosing n sufficiently large.
This product was discovered by François Viète in 1579, and is only one of
later.
many fascinating expressions for x, some of which are mentioned
*CHAPTER
PLANETARY MOTION
Nature and Nature's Laws lay hid in night
God said "Let Newton be," and all was light.
AlexanderPope
Unlike Chapter 16, a short chapter diverging from the main stream of the book,
diverges from the main stream of the book to demonstrate that
we are already in a position to do some real physics.
In 1609 Kepler published his first two laws of planetary motion. The first law
describes the shape of planetary orbits:
this long chapter
Theplanets
movein ellipses,with thesun at onefocus.
The second law involves the area swept out by the segment from the sun to the
planet (the
vector from the sun to the planet') in various time intervals
(Figure 1):
'radius
Equal areas are swept out bythe radius vector in equal tünes.(Equivalently,
thearea
swept out in time t is proportional
to t.)
FIGURE
1
Kepler's third law, published in 1619, relates the motions of different planets. If a
is the major axis of a planet's elliptical orbit and T is its period, the time it takes
the planet to return to a given position, then:
Theratio a3/T2 is thesameforall planets.
Newton's great accomplishment
was to show (usinghis general law that the
force on a body is its mass times its acceleration) that Kepler's laws followfrom the
assumption
that the planets are attracted to the sun by a force (thegravitational
force of the sun) always directed toward the sun, proportional to the mass of the
planet, and satisfying an inverse square law; that is, by a force directed toward
the sun whose magnitude varies inversely with the square of the distance from the
sun to the planet and directly with the mass of the planet. Since force is mass
this is equivalent
simply to saying that the magnitude
of the
times acceleration,
acceleration
is a constant divided by the square of the distance from the sun.
Newton's analysis actually established three results that correlate with Kepler's
individual laws. The first of Newton's results concerns Kepler's second law (which
was actually discovered first, nicely preserving the symmetry of the situation):
327
328 Duivativesand Integrals
between
i.e.,if and only theforce
Kepler'ssecond lawis trueprecisely
for'central
forces',
the sun and theplanetalways lies along the linebetweenthe sun and theplanet.
p,
P2
S
(a)
Ps
R
P2
p,
S
FIGURE
2
Although Newton is revered as the discoverer of calculus, and indeed invented
calculus precisely in order to treat such problems, his derivation hardly seems to
use calculus at all. Instead of considering a force that varies continuously as the
planet moves, Newton first considers short equal time intervals and assumes that
a momentary force is exerted at the ends of each of these intervals.
To be specific, let us imagine that during the first time interval the planet moves
along the line P1P2, with uniform velocity (Figure 2a). If, during the next equal
time interval, the planet continued to move along this line, it would end up at
P3, where the length of PI P2 equals the length of P2P3. This would imply that
they have equal
the triangle SPiP2 has the same are as the triangle SP2P3 (since
bases, and the same height)-this just says that Kepler's law holds in the special
case where the force is 0.
Now suppose (Figure 2b) that at the moment the planet arrives at P2 it experiS to P2, which by itself would cause the planet
ences a force exerted along thelinefrom
with
the motion that the planet already has,
to move to the point Q. Combined
this causes the planet to move to R, the vertex opposite P2 in the parallelogram
whose sides are P2P3 and P2Q.
Thus, the area swept out in the second
time interval is actually the triangle
SP2R. But the area of triangle SP2R is equal to the area of triangle SPs P2, since
they have the same base SP2, and the same heights (sinceRP3 is parallel to SP2).
Hence, finally, the area of triangle SP2R is the same as the area of the original
triangle SPi P2! Conversely, if the triangle SRP2 has the same area as SP1P2, and
hence the same area as SPsP2, then RPs must be parallel to SP2, and this implies
that Q must lie along SP2·
Of course, this isn't quite the sort of argument one would expect to find in a
modern book, but in its own charming way it shows physically just why the result
should
be true.
To analyze planetary motion we will be using the material in the Appendix to
Chapter 12, and the
det defined in Problem 4 of Appendix 1 to
Chapter 4. We describe the motion of the planet by the parameterized curve
"determinant"
c(t)
=
r(t)(cos
0(t), sinO(t)),
gives the length of the line from the sun to the planet, while 0
gives the angle. It will be convenient to write this also as
so that r always
(1)
c(t)
r(t)
=
-
e(0(t)),
where
e(t)
=
(cost,
sin t)
is just the parameterized curve that runs along the unit circle. Note that
e'(t)
=
(-
sin t, cos t)
17. Planetay Motion 329
is also a vector of unit length, but perpendicular to e(t), and that we also have
det(e(t), e'(t))
(2)
Differentiating
(3)
(1),using
the
c'(t)
r'(t)
and combining with
to Chapter 4, we get
det(c(t), c'(t))
=
=
c(t)
=
=
1.
formulas on page 244, we obtain
·e(8(t))
+ r(t)0'(t)
with the
(1),together
e'(0(t)),
-
formulas in Problem 6 of Appendix 1
det(e(0(t)), e(G(t))) + r(t)20'(t)
det(e(0(t)), e'(0(t))),
r(t)2Ð'(t)
r(t)r'(t)
since det(v, v) is always 0. Using
(2)we
then get
det(c, c')
(4)
det(e(0(t)), e'(0(t)))
r 9'.
=
As we will see, r20' turns out to have another
important interpretation.
c(0)
Suppose that A(t) is the area swept out from time 0 to t (Figure 3). We want
to get a formula for A'(t), and, in the spirit of Newton, we'll begin by making
A(t), together with a straight line
an educated guess. Figure 4 shows A(t + h)
segment between c(t) and c(t + h). It is easy to write down a formula for the area
of the triangle A(h) with vertices O, c(t), and c(t + h): according to Problems 4
and 5 of Appendix 1 to Chapter 4, the area is
-
FI CURE
3
area(A(h))
¼det(c(t),
=
c(t + h)
--
c(t)).
Since the triangle A(h) has practically the same area as the region A(t+h)
this shows (orpractically shows) that
c(r + h)
A'(t)
A(t + h)
hoo
h
A(h)
hm area
han
h
-
.
=
hm
-
A(t),
A(t)
.
=
c(t)
=
=
O
FIGURE
4
(
|
}det(c(t), c'(t)).
det c(t), lim
hoo
c(t + h)
h
-
c(t)
A rigorous derivation, establishing more in the process, can be made using Problem 13-24, which gives a formula for the area of a region determined by the graph
of a function in polar coordinates. According to this Problem, we can write
(*)
A(t)
=
-
p(¢)2 d¢
if our parameterized curve c(t) = r(t) e(Ð(t))is the graph of the function p in
polar coordinates (herewe've used ¢ for the angular polar coordinate, to avoid
confusion with the function 0 used to describe the curve c).
330 Derivativesand Integrals
Now the function p is just
p=roß-1
[forany
particular angle ¢,
ЯI(¢)
r(0¯I(t))
is the time at which the curve c has angular
polar coordinate ¢, so
is the radius coordinate corresponding to ¢].
Although the presence of the inverse function might look a bit forbidding, it's
actually quite innocent: Applying the First Fundamental Theorem of Calculus
and the Chain Rule to
we immediately get
(4:)
A'(t)
O'O
(p(Ð(t))2
(r(t)20'(t),since
=
-
=
p
=
ro
0¯1
Briefly,
A'
Combining with
(4),we
Jr20'.
=
thus have
(5)
A'
=
}det(c, c') (r20'.
=
Now we're ready to consider Kepler's second law. Notice that Kepler'ssecond law
is equivalent to saying that A' is constant, and thus it is equivalent to A" 0. But
=
-1
A"
=
c')]'
([det(c,
i det(c, c")
det(c', c') +
=
-)
det(c, c").
=
(seepage
245)
So
Kepler's second law is equivalent
to det(c, c")
=
0.
Putting this all together we have:
THEOREM
1
Kepler's second law is true if and only if the force is central, and in this case each
planetary path c(t)
(K2)
PROOF
=
r(t)
·
e(0(t))
r20'
=
satisfies the equation
det(c, c')
=
constant.
Saying that the force is central just means that it always points along c(t). Since
c"(t) is in the direction of the force, that is equivalent to saying that c"(t) always
points along c(t). And this is equivalent to saying that we always have
det(c, c")
=
0.
We've just seen that this is equivalent to Kepler's second law.
c')]'
0, which by
Moreover, this equation implies that
[det(c,
=
(5)gives (K2).
17. PlanetmyMotion 331
Newton next showed that if the gravitational force of the sun is a central force
and also satisfies an inverse square law, then the path of any object in it will be a
conic section having the sun at one focus. Planets, of course, correspond to the
case where the conic section is an ellipse, and this is also true for comets that visit
the sun periodically; parabolas and hyperbolas represent objects that come from
outside the solar system, and eventually continue on their merry way back outside
the system.
THEOREM
2
PROOF
c(t)
If the gravitational force of the sun is a central force that satisfies an inverse square
law, then the path of any body in it will be a conic section having the sun at one
focus.
Notice that out conclusion specifies the shape of the path, not a particular parameterization.
But this parameterization is essentially determined by Theorem 1: the
hypothesis of a central force implies that the area A(t) (Figure 5) is proportional
to t, so determining c(t) is essentially equivalent to determining A for arbitrary
points on the ellipse. Unfortunately, the areas of such segments cannot be deterexplicitly.I
mined
This means that we have to determine the shape of the path
without
=
r(t)
finding its parameterization!
Since it is the function
c
e(Ð(t))
0¯1
of
which
shape
the
actually
the
describes
path
in
polar coordinates, we
ro
Shouldn't
0¯l
surprised
entering
the
into
proof.
be
to find
By Theorem 1, the hypothesis of a central force implies that
-
FlGURE
5
r20'
(K2)
det(c, c')
=
=
M
for some constant M. The hypothesis of an inverse square law can be written
c"(t)
(*)
H
=
-
-
r(t)
for some constant
e(8(t))
H. Using (K2), this can be written
c"(t)
H
=
-
-
-
O'(t)
M
e(0(t)).
Notice that the left-hand side of this equation is
[c'o Я']'(8(t)).
So if we let
D
(thisis the
be
written
main
More precisely, we can't
arcsin, etc.
c' o Я1
c' as a
function
of
Ð"), then the equation
can
as
D'(0(t))
1
consider
trick-"we
=
H
=
write
-
-
M
e(0(t))
H
=
-
sinO(t)),
-(cosÐ(t),
M
down a solution in terms of familiar
"standard
functions," like sin,
332 Derivativesand Integrals
and we can write this simply as
D'(u)
H
M
-
H
H
-(cos
=
u, sin u)
=
-
-
cos u,
M
-
-
M
sin u
[forall
u of the form 9(t) for some t, which happens to be all u], completely
eliminating
9.
equation
that we have just obtained is simply a pair of equations, for the
The
components of D, each of which we can easily solve individually; we thus find that
D(u)
(H
-
=
sin u
-
-M
for two constants A and B. Letting u
for c':
(H
c'=
0(t) again we thus have an explicit formula
=
sin 0
-
-M
H cos u
+ B
M
+ A,
H
+A,
cos &
+B
M
-
.
[Here sin0 really stands for sinoß, etc., abbreviations that we will use throughout.]
Although we can't get an explicit formula for c itself, if we substitute this equation, together with c r(cos0, sin0), into the equation
=
det(c, c')
(K2 ),
(equation
M
=
we get
¯
-
r
H cos2
H
sin 0
0 + B cos 0 +
M
M
A sin 0
-
-
=
M,
which simplifies to
r
H
B
+
cos 0
M2
M
--
-
A
--
.
sm
--
M
0
1.
=
Problem 15-8 shows that this can be written in the form
r(t)
H
-
M
+ C cos(0(t) + D)
=
1,
for some constants C and D. We can let D
0, since this simply amounts to
rotating our polar coordinate system (choosing
which ray corresponds to 0
0),
=
=
so we can write,
finally,
M2
r[1+
ecos0]
=
=
--
H
A.
But this is the formula for a conic section derived in Appendix 3 of Chapter 4.
In terms of the constant M in the equation
r20'
and the constant
A in the equation
r[1+
-
M
of the orbit
ecos0]
=
A
17. PlanetaryMotzon 333
the last equation
in our proof shows that we can rewrite
c"(t)
(**)
Recall
(page87) that
=
M2
-A
-
1
(*)as
e(0(t)).
the major axis a of the ellipse is given by
A
(a)
a=
while the minor axis
1-e2'
b is given by
b
(b)
A
=
.
1
e2
-
Consequently,
b2
(c)
=
-
a.
A
Remember that equation
(5)gives
A'(t)
=
A(t)
=
Šr20'(M,
-
and thus
¼Mt.
We can therefore interpret M in terms of the period T of the orbit. This period T
is, by definition,the value of t for which we have 0(t) 2x, so that we obtain the
complete ellipse. Hence
=
area of the ellipse
or
M
2(area of the ellipse)
T
Hence the constant M2/A in
=
A(T)
}MT,
=
2xab
by Problem 13-17.
T
(**)is
4x2a2b2
M2
¯
T2A
A
4x2a3
=
.
usmg
,
(c).
This completes the final step of Newton's analysis:
THEOREM
s
Kepler's third law is true if and only if the acceleration
on an ellipse, satisfies
1
c"(t)
=
for a constant
-G
·
r2
e(0(t))
G that does not depend on the planet.
c"(t) of any planet, moving
334 Derivativesand Integrals
It should be mentioned that the converse of Theorem 2 is also true. To prove
this, we first want to establish one further consequence
of Kepler's second law.
Recall that for
(cost, sin
e(t)
=
t)
e'(t)
=
(-
sin t, cos t).
e"(t)
=
(-
cos t,
we have
Consequently,
Now differentiating
c"(t)
=
r"(t)
·
-e(t).
sin t)
-
=
(3)gives
e(0(t))
+ r'(t)0'(t)
+ r'(t)0'(t)
-
e'(0(t))
-
e'(0(t))
+ r(t)0"(t)
-
e'(9(t))
+
r(t)Ð'(t)0'(t) e"(0(t)).
-
-e(t)
Using e"(t)
c"(t)
we get
=
=
[r"(t) r(t)O'(t)2]
-
-
e(0(t))
+
+ r(t)0"(t)]
[2r'(t)0'(t)
-
e'(0(t)).
Since Kepler's second law implies central forces, hence that c"(t) is always a multiple of c(t), and thus always a multiple of e(0(t)), the coefficient of e'(0(t))
must be 0 [asa matter of fact, we can see this directly by taking the derivative of
formula (K2)]. Thus Kepler's second law implies that
(6)
THEOREM
4
PROOF
c"(t)
=
[r"(t) r(t)0'(t)2
-
If the path of a planet moving under a central gravitational force is an ellipse with
the sun as focus, then the force must satisfies an inverse square law.
As in Theorem 2, notice that the hypothesis on the shape of the path, together
with the hypothesis of a central force, which is equivalent to Kepler's second law,
essentially determines the parameterization. But we can't write down an explicit
solution, so we have to obtain information about the acceleration without actually
knowing what it is.
Once again, the hypothesis of a central force implies that
Y2pr
(K2)
for some constant M, and the hypothesis that the path is an ellipse with the sun
as focus implies that it satisfies the equation
r[1+ecos0]
(A)
=
A,
for some e and A. For our (notespecially illuminating) proof, we will keep differentiating and substituting from these two equations.
First we differentiate (A) to obtain
r'[1 + e cos Ð]
-
er0' sin Ð
Multiplying by r this becomes
rr'[1+scos0]-er20'sin0=0.
=
0.
17. PlanetaryMotion 335
Using both (A) and (K2), this becomes
Ar'
eM sin 8
-
0.
=
Differentiating again, we get
Ar"
eMG'cos0
---
Using (K2) we get
0.
=
sM2
Ar"-
cos0
--
0,
=
r
and then using
(A) we get
Ar"
M2
-
-
A
-
r2
1
-
r
=
0.
Substituting from (K2) yet again, we get
MS
A[r"-r(0')2]+
=0,
or
r"
Comparing with
(6),we
-
M2
=
-
--.
Ar2
obtain
c"(t)
which
r(0')2
wanted
M2
=
-
e(0(t)),
to show: the force is inversely proportional to
the square of the distance from the sun to the planet.
is precisely what we
THE LOGARITHM AND
EXPONENTIAL FUNCTIONS
CHAPTER
In Chapter 15 the integral provided a rigorous formulation for a preliminary definition of the functions sin and cos. In this chapter the integral plays a more
essential mle. For certain functions even a preliminary defmition presents difficulties. For example, consider the function
f (x)
102.
=
This function is assumed to be defined for all x and to have an inverse function,
defined for positive x, which is the
to the base 10,"
"logarithm
f¯1(x) 108102=
In algebra, 102 is usually defined only for rational x, while the defmition for irx is quiedy ignored. A brief review of the definition for rational x will
not only explain this omission, but also recall an important principle behind the
definition of 102.
The symbol 10" is first defined for natural numbers n. This notation turns out
to be extremely convenient, especially for multiplying very large numbers, because
rational
10" 10"'
-
10"*'".
=
The extension of the definition of 10= to rational x is motivated by the desire
actually forces upon us the customary
to preserve this equation; this requirement
definition. Since we want the equation
10
to be true, we must define 10°
10"
-
=
*"
10"
=
1; since we want the equation
=
10¯" 10"
-
to be true, we must define 10¯"
10
=
10°
=
1
1/10"; since we want the equation
=
-....10'/"
101¡"
101/4+-.+1/" = 101 = 10
=
n times
n times
to be true, we must define 101/"
101/"
-
.
...
=
6;
101/"
=
and since we want the equation
10 17"*"
**/"
=
10'"/"
m times
m times
(6)
to be true, we must define 10'"/"
Unfortunately, at this point the program comes to a dead halt. We have been
guided by the principle that 10" should be defmed so as to ensure that 10'+>
102107; but this principle does not suggest any simple algebraic way of defining
-
.
=
336
18. The Logarithmand Exponential Functions 337
10= for irrational x. For this reason we will try some more sophisticated
finding a function f such that
(*)
f(x
+ y)
for all x and y.
f(x) f(y)
=
ways of
-
Of course, we are interested in a function which is not always zero, so we might
add the condition
f (1) 0. If we add the more specific condition f (1) 10,
then (*)will imply that f (x) 10=for rational x, and 102 could be d¢ned as f (x)
for other x; in general f (x) will equal [f (1)]*for rational x.
One way to find such a function is suggested if we try to solve an apparently
function f such that
more difficult problem: find a dà¶erentiable
-¢
=
=
f(x). f(y)
f(x+y)=
f (1)
=
foralIxand
10.
y,
Assuming that such a function exists, we can try to find f'-knowing the derivative
of f might pmvide a clue to the definition of f itself. Now
f(x+h)-
.
f'(x)
=
hm
f(x)
h
h->0
f (x) f (h)
-
.
=lun
f (x)
-
h
h->0
hm f(h)-1
f (x) h-vo
h
.
=
-
.
The answer thus depends on
f'(0)
lim
=
f(h)
-
1
;
h
h-wo
for the moment assume this limit exists, and denote it by a. Then
f'(x)
=
-
a
for all x.
f (x)
Even if a could be computed, this approach seems self-defeating. The derivative
of f has been expressed in terms of f again.
If we examine the inverse function f¯I
log1o, the whole situation appears in
a new light:
=
logio
1
'fx)
=
frg-1
1
1
¯
¯
œ-
f(f-1(x))
ax°
The derivative of f-1 is about as simple as one could ask! And, what is even
more interesting, of all the integrals
b
Ï
x-I
Lb
x"
dx examined previously, the integral
dx is the only one which we cannot evaluate. Since logio 1
have
1
--
*
1
-dt=logiox-logto1=logtox.
=
0 we should
338 Derivativesand Integrals
t-i dt.
This suggests that we define logiox as (1/a)
The difficulty is that a is
J1
One way of evading this difficulty is to define
unknown.
log x
dt,
=
hope that this integral will be the logarithm to some base, which might be
determined later. In any case, the function defined in this way is surely more
reasonable,
from a mathematical
point of view, than logie. The usefulness of
role
of the number 10 in arabic notation (andthus
the
important
logie depends on
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 known to us.
and
If x
DEFINITION
0, then
>
log x
=
1
lx
-
dt.
The graph of log is shown in Figure 1. Notice that if x
if 0 < x < 1, then logx < 0, since, by our conventions,
>
1, then log x
>
0,
and
dt<0.
dt=-
For x 5 0, a number logx cannot
not bounded on [x, 1].
be defined in this way, because f(t)
=
1/t is
1
1
1
x
(b)
(a)
FIGURE
l
The justificationfor the notation
THEOREM
1
If x, y
>
"log"
comes
from the following theorem.
0, then
log(x y)
=
log x + log y.
18. The Logarithmand ExponentialFunctions 339
PROOF
Notice first that log'(x)
1/x, by the Fundamental Theorem
choose a number y > 0 and let
=
f (x)
log(xy).
=
Then
f'(x)
Thus
f'
log'(xy) y
=
-
1
=
1
-
f (x)
xy
-.
x
c such that
for all x
log x + c
=
=
y
-
log'. This means that there is a number
=
of Calculus. Now
0,
>
that is,
log(xy)
The number
c can
for all x
logx + c
=
>
be evaluated by noting that when x
0.
=
1 we obtain
log(1-y)=logl+c
Thus
log(xy)
Since this is true for all y
COROLLARY
1
If n is a natural
number
>
log x + log y
=
0, the theorem is proved.
and x
0, then
>
log(x")
PROOF
COROLLARY
2
Let to you
If x, y
>
for all x.
=
n
logx.
(useinduction).
0, then
log
(x
-logy.
logx
=
-
y
PROOF
This followsfrom the equations
log x
=
log
-
=
y
log
+ log y.
Theorem 1 provides some important information about the graph of log. The
fimction log is clearly increasing, but since log'(x)
1/x, the derivative becomes
and
consequently
small
large,
log
very
as x becomes
grows more and more slowly.
clear
whether
log is bounded or unbounded
It is not immediately
on R. Observe,
however, that for a natural number n,
=
log(2")
n log 2
=
(and log 2
>
0);
it follows that log is, in fact, not bounded above. Similarly,
log
=
log 1
--
log 2"
-n
=
log 2;
340 Derivativesand Integrals
therefore log is not bounded below on (0, 1). Since log is continuous, it actually takes on all values. Therefore R is the domain of the function log¯1. This
important function has a special name, whose appropriateness will soon become
clear.
The
DEFINITION
"exponential
function," exp, is defined as log¯1
The graph of exp is shown in Figure 2. Since log x is defined only for x > 0, we
always have exp(x) > 0. The derivative of the function exp is easy to determine.
THEOREM
2
For all numbers x,
exp'(x)
PROOF
eXp'(X)
)'(X)
(ÏOg¯
=
exp(x).
=
1
=
Iog'(log¯I
(x))
1
1
log-1
=
log¯'(x)
=
exp(x).
A second important property of exp is an easy consequence
THEOREM
3
If x and y are any two numbers,
exp
Let x'
=
exp(x) and y'
=
Theorem 1.
then
exp(x + y)
PROOF
of
=
exp(x)
-
exp(y).
so that
exp(y),
x
=
y
=
log x',
log y'.
Then
x + y
=
log x' + log y'
=
log(x'y').
This means that
exp(x + y)
FIGURE
=
x'y'
=
exp(x)
-
exp(y).
This theorem, and the discussion at the beginning of this chapter, suggest that
is particularly important. There is, in fact, a special symbol for this number.
exp(1)
2
DEFINITION
e
=
exp(1).
18. TheLogarithmand ExponentialFunctions 341
This definition is equivalent to the equation
1
log e
=
y
le
=
-
I
1
As illustrated in Figure 3,
1
/2
-
¯
'
I
r
dt
<
1,
since 1
-
dt.
(2 1) is an upper sum for
f (t) 1/t on [1,2],
-
=
and
4
y
-
1
FIGURE
2
3
dt
>
1,
since
4
(sum(2 for1) + )
-
-
f (t)
.
=
(4 2)
-
1/t on
=
1 is a lower
[1,4].
Thus
3
21
41
'l
-
t
dt
<
-
11
dt
<
-
11
dt,
which shows that
2<e<4.
In Chapter 20 we will find much better approximations for e, and also prove that
e is irrational (thepmof is much easier than the proof that x is irrational!).
As we remarked at the beginning of the chapter, the equation
exp(x + y)
exp(x)
=
·
exp(y)
implies that
exp(x)
=
=
[exp(1)]*
ex,
for all rational x.
Since exp is defined for all x and exp(x)
our earlier use of the exponential notation
DEFINITION
For any number
=
for rational x, it is consistent with
to defineex as exp(x) for all x.
e2
x,
e"
=
exp(x).
"exponential
function" should now be clear. We have sucThe terminology
ceeded in defining e2 for an arbitrary (even
irrational) exponent x. We have not
yet defmed ax, if a ¢ e, but there is a reasonable principle to guide us in the
attempt. If x is rational, then
a2
=
(el°5")2
=
exioga
But the last expression is defined for all x, so we can use it to define a*.
342 Derivativesand Integrals
If a
DEFINITION
0, then, for any real number
>
a'
(If a
f(x)
"
-
e21°E".
=
e this definition clearly agrees with the previous one.)
=
The requirement
unduly
x,
a >
since,
restrictive
0 is necessary, in order that log a be defined. This is not
for example, we would not even expect
to be defmed. (Of course, for certain rational
to the old definition; for example,
x, the symbol a2 will make sense,
according
Our definition of a2 was designed to ensure that
(e2)T
FIGURE
ei
=
for all x and y.
As we would hope, this equation turns out to be true when e is replaced by any
number a > 0. The proof is a moderately
involved unraveling of terminology. At
the same time we will prove the other important properties of ax.
4
THEOREM
4
If a
>
0, then
bc
(1) (abpc
_
(Notice that ab
-
(2)
al
(Notice that
rational x.)
=
Will
be positive, so (ab c will be defined);
automatically
a and a2+>
=
(2)implies that
(Î) (ab c
PROOF
for all b, c.
a'
-
a>
for all x, y.
this definition of ax agrees with the old one for all
clog(e**)
cloga
(Each of the steps in this string of equalities
log¯14
the fact that exp
c(bloga)
chloga
depends upon
=
a
.
last definition, or
our
=
(2)
al
=
a***
etioga
=
-
e(2+T)'°ga
eloga
-
exioga+,ioga
ex loga
-
.,yioga
_
a'
-
a?.
Figure 4 shows the graphs of f (x) a2 for several difTerent a. The behavior
of the function depends on whether a < 1, a
1, then
1, or a > 1. If a
=
=
=
18. TheLogmithmand Exponentù:lFunctions 343
f (x)
=
12 = 1. Suppose a
1. In this case log a
>
if
<
x
then
xloga
<
y,
yloga,
yloga
exloga
SO
a'
i.e.,
0. Thus,
>
<
a?.
Thus the function f (x) a2 is increasing. On the other hand, if 0 < a < 1,
a2
so that log a < 0, the same sort of reasoning shows that the function f (x)
=
is decreasing. In either case, if a > 0 and a ¢ 1, then f (x) a* is one-one.
Since exp takes on every positive value it is also easy to see that a* takes on every
positive value. Thus the inverse function is defined for all positive numbers, and
a2, then f¯1 is the function usually denoted by loga
takes on all values. If f(x)
(Figure 5).
Justas a* can be expressed in terms of exp, so loga can be expressed in terms
of log. Indeed,
=
=
=
f(x)
f(x)
=
/(x)
=
losa x
-=
lag x
logio x
(a <
1)
if
y
=
then
x
=
so
logx
or
y
loßa X,
ey
a'
yloga,
=
=
=
logx
log a
.
logx
log a
·
oga
In other words,
loga I
FIGURE
5
The derivatives of f(x)
=
ax and g(x)
f(x)=exioga,
g(x)
=
A more complicated
=
logx
loga
--,
=
loga x are both easy to find:
f'(x)=loga-exioga
so
=loga-ax,
1
so g'(x)
=
.
xloga
function like
f (x)
=
g(x)*(2)
is also easy to differentiate, if you remember
that, b_;definition,
it follows from the Chain Rule that
f'(x)
=
=
eh(x)log g(
g(x)h(x)
-
h'(x)1088(x)
+ h(x)
h'(x) logg(x) + h(x)
g(x)
.
g(x)
There is no point in remembering this formula--simply apply the principle behind
it in any specific case that arises; it does help, however, to remember that the first
factor in the derivative will be g(x)h(x)
344 Derivativesand Integrals
There is one special case of the above formula which is worth remembering.
The function f (x) xa was previously defined only for rational a. We can now
define and find the derivative of the function f (x) x" for any number a; the
result is just what we would expect:
=
=
f (x)
=
xa
=
$
=
ealogx
SO
f'(x)
e"1°52
-
S
-
-
xa
axa-1
=
x
x
Algebraic manipulations with the exponential functions will become second naremember
that all the rules which ought to work
ture after a little practicejust
actually do. The basic properties of exp are still those stated in Theorems 2 and 3:
exp'(x)
exp(x + y)
exp(x),
exp(x) exp(y).
=
=
-
In fact, each of these properties comes close to characterizing the function exp.
Naturally, exp is not the only function f satisfying f'
f, for if f ce", then
with
this property; however.
f'(x) = ce2 f (x); these functions are the only ones
=
=
=
THEOREM
5
If f is differentiable and
f'(x)
then there is a number
=
c such that
f (x)
PROOF
for all x,
f (x)
=
ce'
for all x.
Let
f(x)
g(x)=-
e2
(This is permissible, since e2 ¢ 0 for all x.) Then
exf'(x) f (x)e2
-
g'(x)
Therefore there is a number
-
-
0.
c such that
g(x)
f (x)
c
ex
for all x.
The second basic property of exp requires a more involved discussion. The
function exp is clearly not the only function f which satisfies
f(x+y)
=
f(x)· f(y).
a2 also satisfies this equation.
In fact, f (x) 0 or any function of the form f(x)
But the true story is much more complex than this-there are infinitely many other
functions which satisfy this property, but it is impossible, without appealing to more
advanced mathematics,
to prove that there is even one function other than those
=
=
18. TheLogarithmand ExponentialFunctions 345
It is for this reason that the definition of 10= is so difficult:
there are infinitely many functions f which satisfy
already
mentioned!
f(x+y)= f(x) f(y),
-
f (1)
but which are not the function f (x)
continuousfunction f satisfying
f (x +
10,
=
y)
102! One thing is true however-any
=
f (x) f (y)
=
-
be of the form f (x) a2 or f (x) 0. (Problem 38 indicates the way to
prove this, and also has a few words to say about discontinuousfunctions with this
property.)
In addition to the two basic properties stated in Theorems 2 and 3, the function
faster than any
exp has one further property which is very importantaxp
polynomial." In other words,
must
=
=
"grows
THEOREM
6
For any natural number
n,
€*
lim
x-
PROOF
=
-
Xn
oo.
The proof consists of several steps.
Step1. ex
x for all x, and consequently
>
0).
to be the case n
this
statement
To prove
lim e2
=
X->OC
oo
(thismay
be considered
=
(whichis clear
x
log x
>
for x
0) it suffices to show that
<
for all x
1 this is clearly true, since log x < 0. If x
1/t on [1,x], so logx < x
upper sum for f(t)
If x
0.
>
=
Skp 2.
lim
1, then (Figure 6) x
1 < x.
>
<
-
-
1 is an
ex
-
=
oo.
To prove this, note that
12 ex/2
-
1
FIGURE
6
x
x
x
-
2
By Step 1, the expression
shows
that lim e2/x
=
x->oo
Step3.
lim
x->oo
-
x
=
-
x/2
2
2
x
-
2
in parentheses is greater than 1, and lim ex/2
oo.
oo.
Note that
ex
(e2/")"
1
e*/"
-nn
-
n
n
"
=
oo; this
346 Derivativesand Integrals
The expression in parentheses becomes arbitrarily large, by Step 2, so the nth
power certainly becomes arbitrarily large.
It is now possible to examine carefully the following very interesting function:
f (x)
=
e¯I/2',
x
¢
0. We have
f'(x)
e¯1/x2
=
x
Therefore,
is decreasing for negative
large, then x2 is large, so
so
f
-1/x2
f'(x)
<
f'(x)
>
0
0
for x
for x
0,
0,
<
>
x and increasing for positive x. Moreover, if |x| is
is close to 0, so e¯1/x2 is close to 1 (Figure 7).
1
FIGURE
7
The behavior of f near 0 is more interesting. If x is small, then 1/x2 is large,
'
is large, so e¯1/x2 1 glix2) is small. This argument, suitably stated with
so ell
-
s's and
o's,shows
that
lim e-1¡x2
=
x->0
0.
Therefore, if we defme
f (x)
then the function
FIGURE
8
f
is continuous
e¯1/x2,
=
x ¢ 0
0,
0,
l
x
(Figure 8). In fact, f is actually differentiable
=
18. TheLogarithmand ExponentialFunctions 347
at
0: Indeed
€-l/h'
f'(0)
lim
=
1/h
lim
=
h
hoO
,
g(1/h)2
hw0
and
lim
h->0+
1/h
lim
=
¢(1/h)2
--,
x
while
g(x2)
x-voo
1/h
lim
=
g(1/h)2
hw0
lim
-
x-oo
-.
x
g(x2)
We already know that
e2
lim
=
-
x-voo
oo;
x
it is all the more true that
lim
and this means
=
---
oo,
that
lim
x
-
0.
=
g(x2)
x->oo
Thus
e¯1¡x2
f'(x)
.
2
0,
We can
¢0
x
x3'
=
0.
=
x
now compute that
f'(h) f'(0)
-
.
f"(0)
hm
=
h
a-vo
-l/h2
e
=lim
--
h3
h
h-o
2 e¯l/h2
lim
h4
hw0
1
'
2x4
·
=
lim
=
h->0
an argument similar to the one above shows that
e¯1¡x2
f"(x)
,
x4
=
+
e¯1/x2
lim
=
g1/h2
f"(0)
,
0. Thus
=
i
X6'
0,
x
¢0
x
=
0.
be continued. In fact, using induction it can be shown (Probf (0) 0 for every k. The function f is extremelyflat at 0, and
0 so quickly that it can mask many irregularities of other functions.
This argument
lem 40) that
approaches
'
xsoo
can
(k)
-
For example (Figure 9), suppose that
e¯1/x2 sin 1,
x
0,
x
-
f (x)
=
¢0
=
0.
348 Derivativesand Integrals
It can be shown (Problem 41) that for this function it is also true that f (k)(0) 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.
=
FIGURE
9
PROBLEMS
1.
Differentiate each of the following functions
).
notes a
(i) f (x) e
(ii) f (x) log(1 + log(1 +
(iii) f (x) (sinx)"i"(si"2).
dt)
=
that
(remember
a* always
de-
.
=
log(1 + e14e
=
(iv) f (x)
(v) f (x)
(vi) f(x)
=
=
=
e¯
e
sin x"i"2
log,psinx.
.
(vii)f(x)
(viii)f(x)
2.
=
=
(ix) f (x)
=
(x) f (x)
=
(a) The
arcsm
=
x
log(sin «Q
smx
(log(3+ e4 4x #
.
(arcsinx)log3
(logx)1°52.
xx.
derivative of logo f is f'/f.
This expression
to compute than
is called the logmithmic
doivativeof f. It is often easier
since
and
products
f',
powers in the expression for f
and
expression
products
the
in
become sums
for log o f. The derivative f' can then be recovered simply by multiplying by f; this process is
called logarithmicd
entiation.
(b) Use
logarithmic differentiation to find f'(x) for each of the following.
(i) f (x)
=
(1 +x)(1+e
).
18. TheLogarithmand ExponenúalFuncions 349
3.
(ii) /(x)
=
(iii) f(x)
=
(iv) f (x)
=
(3
x)U3x2
-
.
(1 x)(3 + x)24
+ (cosx)"i"2.
(sinx)cos2
-
e2
€2x
e¯'
-
.
73)
(1 4
Find
i
lb
/(t)
a
(forf
4.
>
[a,b]).
0 on
dt
Graph each of the following functions.
(a) f (x) e2*.
(b) f (x) e
(c) f(x) e'+
(d) f(x)=ex
=
"2.
=
(Compare the graph with the graphs of exp and
1 1/exp.)
e¯x4
=
-e-x
2
1
1 eh
e2 + e-2
+ 1
e + 1
Find the following limits by l'Hòpital's Rule.
e2
Us, y):
e
+ y
=
(e) f(x)
i
5.
-
e
-
---
---
.
ex-1-x-x24
cos x, sm x)
sea
e¯2
-
-
(i)
lim
.
2
(ii)
X2
x-»O
lim
e2
-
1
-
x2/2
-
x
x3/6
-
.
x-o
x
ex-1-x-x2/2
lim
(iii) x->o
y - i
(cmhx,
(v)
lim
,
log(1 + x)
--
.
log(1 +
x)
-
x
(vi) x->0
6.
x2/2
x +
.
xa0
,
x)
x + x2/2
X
log(1 + x)
11m
i
sinh
(iv)
lim
140
x
-
x2/2
x +
X3
The functions
sinh x
=
cosh x
=
tanh x
the hyperbolic
respectively
(b)
,
2
ex
-
e-2
1
ex+e-2
2
-
e22+1
,
and hyperbolic
There
and
tangent,
their
ordinary
trigonometric
analogies
and
between these functions
are many
One analogy is illustrated in Figure 10; a proof that the region
counterparts.
are called
F1 GUR Elo
x3/3
-
sine,
(butusually
hyperbolic
read
'sinch,'
cosine,
'cosh,'
'tanch').
350 Derivativesand Integrals
shown
in Figure 10(b)really has area x/2 is best deferred until the next chapdevelop methods of computing integrals. Other analogies
the
following three problems, but the deepest analogies must
are discussed in
wait until Chapter 27. If
you have not already done Problem 4, graph the
functions sinh, cosh, and tanh.
ter, when we will
7.
Prove that
sinh2
(a) cosh2
(b) tanh2 +1/ cosh2
(c) sinh(x + y) sinh x cosh x + cosh x sinh y.
(d) cosh(x + y) cosh x cosh y + sinh x sinh y.
sinh'
cosh.
(e) cosh'
-
=
=
=
(f)
=
(g) tanh'
8.
sinh.
1
=
.
cosh2
The functions sinh and tanh are one-one; their inverses sinh¯' and tanh¯I,
are defined on R and (-1, 1), respectively. These inverse functions are sometimes denoted by arg sinh and arg tanh (the
of the hyperbolic
sine and tangent). If cosh is restricted to
oo)
[0, it has an inverse, denoted
by arg cosh, or simply cosh-1, which is defined on [1,oo). Prove, using the
information in Problem 7, that
"argument"
sinh(cosh-1
(a)
cosh(sinh¯I
(b)
2
x)
1 + x2.
1
=
(c) (sinh¯1),
1+x2
1
(d) (cosh¯1)'(x)
=
(e)
9.
(tanh¯I)'(x)
1.
-
x2
-
for x
1
=
1
-
>
1.
1
for x|
<
1.
x
sinh¯1, cosh¯', and tanh¯I
an explicit formula for
sinh¯'
equation y =
x for x in terms of y, etc.).
(a) Find
(bysolving
the
(b) Find
dx,
1+x2
b
Ï
x2
dx
-
1
for a, b
>
1 or a, b
<
1,
b
g
dx for |a|, |b| < l.
1 x2
Compare your answer for the third integral with that obtained by writing
-
1
1-x2
=-
1
1
2 1-x
+
1¯
1+x
.
18. The Loganthmand ExponentialFunctions 351
10.
Show that
x
F(x)
is not bounded on
11.
Let
f
=
1
dt
log t
--
[2,oo).
be a nondecreasing
function on
and define
[1,oo),
f(t)
lx
-dt,
F(x)=
Prove that f is bounded on
12.
x>l.
i
i
[1,oo) if and
only if F/ log is bounded on
[1,oo).
Find
(a)
lim ax for 0
<
X-¥oo
a
<
1. (Remember the definition!)
x
lim
(b) x-voo
.
(logx)
(logx)"
lim
(c) x-oo
x
(-1)"
lim
(d) x->0+
x (logx)".
lim
(e) x->0+
x2.
13.
Graph
14.
(a) Find the
f (x)
Hint:
x (logx)"
log
=
.
1
x
xx for x
=
minimum
>
0. (Use Problem 12(e).)
value of
f(x)
=
e2/x"
for x
>
0, and conclude
that
f (x) > e"/n" for x n.
n)/xn+1,
e2(x
Using
prove that f'(x) >
(b) e"**/(n the expression f'(x)
+ 1)n+1 for x > n + 1, and thus obtain another proof that
lim f(x) oo.
X->OC
>
=
-
=
15.
Graph f (x)
16.
lim log(1 +
(a) Find y->0
=
ex /x".
y)/y.
(You can use PHôpital's Rule, but that would be
silly.)
(b) Find
lim x log(1 + 1/x).
lim (1 + 1/x)".
(c) Prove that e X->00
=
lim (1+ a/x)". (It is possible to derive this from part
fiddling.)
just a little
that
Prove
log b X-¥OO
lim x (b1/x 1).
(d) Prove that
with
*(e)
en
=
(c)
algebraic
=
(1 + 1/x)2 for x
-
0. (Use Problem 16(c).)
17.
Graph f (x)
18.
If a bank gives a percent interest per annum, then an initial investment I
after 1 year. If the bank compounds
the interest (counts
yields I(1+a/100)
the accrued interest as part of the capital for computing interest the next
=
>
352 Denvativesand Integrals
year), then the initial investment grows to I(1 + a/100)" after n years. Now
suppose that interest is given twice a year. The final amount after n years
is, alas, not I(1+a/100)2",
but merely I(l +a/200)
interest is
awarded twice as often, the interest must be halved in each calculation, since
the interest is a/2 per half year. This amount is larger than I(1+ a/100)",
but not that much larger. Suppose that the 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 initial investment of 1 dollar yield after
--although
1 year?
19.
log |x| for x ¢ 0. Prove that f'(x) 1/x for x ¢ 0.
(b) If f(x) ¢ 0 for all x, prove that (log|f )' f'/f.
(a) Let f (x)
=
=
=
20.
Suppose that on some interval the function f satisfies
number
f'
=
cf
for some
c.
is never 0, use Problem 19(b)to prove that |f(x)| lecx
number
for some
l (> 0). It follows that f (x) ke'2 for some k.
(b) Show that this result holds without the added assumption that f is
never 0. Hint: Show that f can't be 0 at the endpoint of an open
interval on which it is nowhere 0.
(c) Give a simpler proof that f (x) = kecx for some k by considering the
function g(x)
f(x)/ecx
that
Suppose
f'
fg' for some g. Show that f(x) kes(x) for some k.
(d)
(a) Assuming that f
=
=
=
=
*21.
=
A radioactive substance diminishes at a rate proportional to the amount
present (sinceall atoms have equal probability of disintegrating, the total
disintegration is proportional to the number of atoms remaining). If A(t)
cA(t) for some c (which
is the amount at time t, this means that A'(t)
represents the probability that an atom will disintegrate).
=
in terms of the amount Ao A(0) present at time 0.
element)
there is a number r (the
life" of the radioactive
A(t)/2.
with the property that A(t + r)
(a) Find A(t)
(b) Show that
=
"half
=
*22.
Newton'slaw of
conhng states that an object cools at a rate proportional to
the difTerence of its temperature and the temperature of the surrounding
medium.
Find the temperature T(t) of the object at time t, in terms of its
temperature To 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)'.
(T
=
X
*23.
24.
Prove that if
f(x)
=
f(t) dt,
Find all continuous functions
/I
(i)
0
f
=
ex
.
f
then
f
=
satisfying
0.
--
18. TheLogarithmand ExponentialFunctions 353
(ii)
25.
f
1
=
eh
-
.
all continuous
Given a differentiable function f, find
functions g satisfying
f(x)
fg
*26.
27.
Find all functions
satisfying
f
Find all continuous
f'(t)
>
-
f (t) +
=
1.
Ilf
0
(t) dt.
f(01+t2 dt.
=
and g be continuous
0. Suppose that
(a) Let f
C
(x))
functions f which satisfy the equation
(f(x))2
28.
g(f
=
nonnegative
functions on
[a,b],
and
let
IX
f(x)5C+
a5x5b.
fg
Prove Gronwall'sinequalig:
f(x)
5
CeÍ.*.
Hint: Consider the derivative of the function h (x) C + /* fg.
(b) Use a limiting argument to show that this result holds even for C 0.
(c) Suppose that f'(x) g(x) f(x) for some continuous function g, and that
f (0) 0. Then f 0. (Compare Problem 20.)
=
=
=
=
=
29.
(a) Prove
that
3
x2
n
1+x+-+-+---+-<e2
forx>0.
n!
21 3!
and
Hint: Use induction on n,
compare derivatives.
that
ex/x"
proof
lim
Give
new
co.
a
(b)
=
30.
Give yet another proof of this fact, using the appropriate form of l'Hôpital's
Rule. (See Problem 11-53.)
31.
(a) Evaluate
lim e¯**
xooo
e2
JO
dt. (You should be able to make an educated
guess before doing any calculations.)
(b) Evaluate the following limits.
(i)
lim e¯2°
xwoo
lx+(1/x)
Ï
lx+(logx/2x)
e" dt.
x+(logx/x)
(ii)
lim e¯"
xooo
lim
(iii) xwoo
e¯"
e" dt.
e"
dt.
354 Derivativesand Integrals
32.
This problem outlines the classical approach to logarithms and exponentials.
To begin with, we will simply assume that the function f (x) a=, defined in
an elementary way for rational x, can somehow be extended to a continuous
one-one function, obeying the same algebraic rules, on the whole line. (See
Problem 22-29 for a direct proof of this.) The inverse of f will then be
denoted by loga•
=
the defmition, that
(a) Show, directly from
loga
'(x)
(
lim loga 1 +
=
h->0
1
1/h
h
-
x
lim(1+k)1/k
=--loga
k->0
X
Thus, the whole problem has been reduced to the determination of
lim(1 + h)l/h. If we can show that this has a limit e, then log,'(x)
=
h->0
1
-
--
x
log, e
1
=
-,
exp(x).
(b) Let a.
and consequently
log¯* has derivative exp'(x)
=
=
for natural numbers n. Using the binomial theorem,
1+
=
exp
x
show that
a,=2+
1-
Conclude that
<
a,
1-k-1
1-
-...-
a.wi.
fact that 1/k! 5 1/2k-1 for k > 2, show that all an < 3. Thus,
the set of numbers {ai,a2, a3,
is bounded, and therefore has a least
that
bound
Show
for
a, < e for large
upper
e.
any e > 0 we have e
(c) Using the
...)
-
enough n.
If
(d) nsxsn
(
1+
+ 1, then
"
1
Conclude that lim
x->oo
and conclude
1+--
5
n+1
2
1
1
(
1+
-
1+
5
x
1
n+1
.
n+1
2
=
x
that lim(1 + h)l/h
e. Also show that
_
lim
-oo
x
1+
1
-
x
*
=
e,
g
h->0
*33.
.4
>>
a
A point P is moving along a line segment A B of length 107 while another
point Q moves along an infinite ray (Figure l1). The velocity of P is always
equal to the distancefrom P to B (inother words, if P(t) is the position of P
107 P(t)), while Q moves with constant velocity
at time t, then P'(t)
107.
The
distance
traveled by Q after time t is defined to be the
Q'(t)
Napierianlogmithmof the distance from P to B at time t. Thus
=
-
=
FIGURE
11
10't
=
Nap log[102
--
P(t)].
18. The Logarithmand ExponentialFunctions 355
in his
This was the definition of logarithms given by Napier (1550-1617)
canonis descnþtion(A Description of
publication of 1614, Mirgicilogarithmonum
the Wonderful Law of Logarithms); work which was done beforethe use of
exponents was invented! The number 107 was chosen because Napier's tafor astronomical and navigational calculations), listed the logables (intended
rithms of sines of angles, for which the best possible available tables extended
to seven decimal places, and Napier wanted to avoid fractions. Prove that
Nap log x
107
10'log
=
-.
x
Hint: Use the same trick as in Problem 22 to solve the equation
*34.
for P.
particular attention to the
graph of f(x)
(logx)/x (paying
behavior near 0 and oo).
(b) Which is larger, ex or x"?
(c) Prove that if 0 < xs 1, or x e, then the only number y satisfying
xi
y* is y x; but if x > 1, x ¢ e, then there is precisely one number
y ¢ x satisfying x> = y2; moreover, if x < e, then y > e, and if x > e,
then y < e. (Interpret these statements in terms of the graph in part (a)!)
that if x and y are natural numbers and x>
yx, then x
Prove
y or
(d)
(a) Sketch the
=
=
=
=
=
=
x=2,y=4,orx=4,y=2.
y2 consists of a curve and
the set of all pairs (x, y) with x>
which
find
the
intersect;
intersection
and draw a rough
straight
line
a
(e) Show that
=
sketch.
**(f)
g(x)".
For 1 < x < e let g(x) be the unique number > e with x8
Prove that g is differentiable. (It is a good idea to consider separate
functions,
=
logx
f1(x)=-,
f2(x)=
and write g
in terms of fi
g'(x)
0<x<e
x
logx,
x
and
e<x
f2. You should be
[g(x)]2
=
1
-
able to show that
log x
x2
1-logg(x)
if you do this part properly.)
*35.
This problem uses the material
from the Appendix to Chapter 11.
(a) Prove that
exp is convex and log is concave.
(b) Prove that
if
p¿
=
1 and all p¿
>
0, then
i=1
...·zn'"
zi'
<
Pizi +-··+
p.z..
(Use Problem 9 from the Appendix to Chapter 11.)
(c) Deduce another proof that G, < A. (Problem 2-22).
356 Derivativesand Integrals
36.
[a,b], and let P, be the partition of
equal
into n
intervals. Use Problem 2-22 to show that
be a positive function on
(a) Let f
1
L(log f, P.) 5 log
b-a
(b) Use the Appendix to Chapter
.
13 to conclude that for allintegrable
f
0
>
have
we
b
1
b-a
b
log f 5 log
a
f
b-a
.
is illustrated in the next part:
A more direct approach
(c) In
1
L(f, Ps)
b-a
[a,b]
Problem 35, Problem 2-22 was deduced as a special case of the in-
equality
g
p¿x;
for p¿
0,
>
p¿
pgg (xi)
5
i=1
i=l
1 and g convex. For g concave we have the reverse
=
i=l
inequality
Pig(xi)
5 &
Fixi
i=1
-
i-l
Apply this with g
log to prove the result of part (b)directly for any
integrable f.
(d) State a general theorem of which part (b)is just a special case.
=
37.
Suppose f satisfies f'
that
*38.
f
=
exp or
f
=
=
f
and
f (x +
y)
f (x)f (y) for all
=
0.
x and y.
Prove
Prove that if f is continuous and f(x + y)
f(x)f(y) for all x and y, then
all
either f
0 or f (x) [f (1)]2 for
x. Hint: Show that f (x)
[f (1)]2
for rational x, 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 discontinuousfunctions f satisfying f (x+y) =
=
=
=
=
f (x)f (y).
*39.
Prove that if f is a continuous function defined on the positive real numbers,
and f(xy)
f(x)+ f(y) for all positive x and y, then f 0 or f(x)
all
for
x > 0. Hint: Consider g(x)
f(e)]ogx
f(e2).
=
=
=
=
*40.
Prove that if f (x)
=
e¯1/x2
for x ¢ 0, and f (0)
=
0, then f (k)(0)
=
0 for
all k.
*41.
Prove that if
for all k.
f (x)
=
e¯1/x2 sin 1/x for
x
¢ 0, and f (0)
=
0, then f (k) (0)
=
0
18. The Logarithmand ExponentialFunctions 357
42.
(a) Prove that if a is a root of the equation
an-1x"¯'
+
+ a1x
(*) anx" +
-
then the function y(x)
(**)
-
-
+ ao
0,
=
e"2 satisfies the differential equation
=
anytn)+a.-iyf"~)+--.+aiy'+aoy=0.
xe"2 also satisfies (**).
Prove that if a is a double root of (*),then y(x)
Hint: Remember that if a is a double root of a polynomial equation
f(x) 0, then f'(a) 0.
xk ŒX iS & SOÏUÍiOR
Prove that if a is a root of (*)of order r, then y(x)
for05kyr-1.
*(b)
=
=
*(c)
=
=
If
(*)has n
n solutions
real numbers as roots
of (**)· yn
yl,
-
(d) Prove that
-
multiplicities),
(counting
part
(c)gives
·
in this case the function ci yi +
-
-
-
+ c.y. also satisfies
(**).
It is a theorem that in this case these are the only solutions of (**).Problem 20 and the next two problems prove special cases of this theorem,
and the general case is considered in Problem 20-19. In Chapter 27 we
will see what to do when (*)does not have n real numbers as roots.
*43.
Suppose that
as follows.
f
satisfies
f"- f
0 and f(0)
=
J'(0)
=
=
0. Prove that f
=
0
(a) Show that f2 J')2 0.
(b) Suppose that f (x) ¢ 0 for all x in some
=
-
interval (a, b). Show that either
for
in (a, b), for some constant c.
f (x) ce" or else f (x)
If f(xo) ¢ 0 for xo > 0, say, then there would be a number a such that
0 < a < xo and f (a) 0, while f (x) ¢ 0 for a < x < xo. Why? Use
this fact and part (b)to deduce a contradiction.
=
=
**(c)
all x
ce¯2
=
*44.
ae2 + be¯' for
that if f satisfies f"
0, then f (x)
f
should
what
and
and
out
b
be
in
(First
figure
b.
terms of f (0)
a
some a
and f'(0), and then use Problem 43.)
a and b.
(b) Show also that f a sinh +b cosh for some (other)
(a) Show
-
=
=
=
45.
Find all functions f satisfying
(a) f (") f ("¯1)
(b) ff") f(n-2)
=
=
*46.
This problem, a companion to Problem 15-30, outlines a treatment of the exponential function starting from the assumption that the differential equation
f' f has a nonzero solution.
=
0 with f'
each x by considering the function g(x)
f (xo) ¢ 0.
(a) Suppose there
is a function
f ¢
=
=
f. Prove that f (x) ¢ 0 for
f(xo +
x)
f(x
-
xo), where
358 Derivativesand Integrals
(b) Show that there is a functidn f satisfying f' f and f (0) 1.
(c) For this f show that f (x + y) f (x) f (y) by considering the function
=
=
=
g(x)=
f(x+y)/f(x).
and that (f¯I)'(x)
(d) Prove that f is one-one
47.
-
1/x.
=
Let f and g be continuous functions such that lim f(x)
than g (f » g) if
We say that f growsfaster
.
hm
and we say that
f
f (x)
=
--
f (x) exists
=
oo.
oo,
and g grow at the same rate
lim
lim g(x)
=
and
(f
g) if
~
is ¢ 0, oo.
For example, exp » P for any polynomial function P, and P » log" for
any positive integer n.
and g, with lim
f(x)
that one of the three conditions
(a) Given f
(b) If f
» g, then
f
+ g
~
(c) If
oo, is it necessarily true
f » g or g » f or f g holds?
lim g(x)
=
~
f.
log f
logg
>
c
>
1
for sufficiently large x, then f » g.
LX
(d) If
f
» g and F(x)
=
f,
=
G(x)
X
=
g, does it necessarily
follow
that F » G?
(e) Arrange each of the following sets of functions in increasing order of
growth (forconvenience, we indicate each function simply by giving its
value at x):
(i)
(ii)
(iii)
x3
x, X3
3),
log4x, (logx)2, x2, x + e¯52, x3
e**, x2, xiogx
x log2 x, e52, log(xx)
e2',
ex/2,
e2, x", x2,
22,
(logx)2x
x
48.
Suppose that gi, g2, gs,
are continuous functions. Show that there is a
continuous function f which grows faster than each g,.
49.
Prove that logio 2 is irrational.
...
INTEGRATION
CHAPTER
IN ELEMENTARY TERMS
Every computation of a derivative yields, according to the Second Fundamental
Theorem of Calculus, a formula about integrals. For example,
if F(x)
=
x(logx)
then F'(x)
x
-
=
logx;
consequently,
b
log x dx
=
F(b)
--
F(a)
=
b(log b)
b
-
-
[a(loga)
-
a],
O < a, b.
Formulas of this sort are simplified considerably if we adopt the notation
F(x)
We may then write
Ib
F(b)
=
F(a).
-
b
Ib
logxdx=x(logx)-x
.
£b
This evaluation of
log x dx depended on the lucky guess that log is the derivative of the function F(x)
In general, a function F satisfying F'
x (logx)
f
function f always has a
is called a primitive of f. Of course, a continuous
-x.
=
primitive,
=
namely,
F(x)
=
lx
f,
but in this chapter we will try to find a primitive 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
function is
obtained
which
multiplication,
and
composition
addition,
division,
be
by
can
one
from the rational functions, the trigonometric functions and their inverses, and the
functions log and exp.
It should be stated at the very outset that elementary primitives usually cannot
function F such that
be found. For example, there is no ekky
F'(x)
=
e¯2°
for all x
merely
ignorance; it is
a report on the present state of mathematical
such
what
theorem
that
exists).
function
And,
is even worse, you
no
a (difficult)
(thisis not
*The definition which we will give is precise, but not really accurate, or at least not quite standard.
functions, that is, functions g
Usually the elementary functions are defined to include
"algebraic"
satisfying
an equation
(a(x))"+Ín-1(x)(f(x))"¯1+.-+ fo(x)
where the
359
f¿ are
=0,
rational functions. But for our purposes these fimetions can be ignored.
360 Denvativesand Integrals
will
have no way of knowing whether or not an elementary primitive can be found
(youwill just have to hope that the problems for this chapter contain 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)
-
x arctan x
log(1 + x2)
2
-
satisfies
F'(x)
(justhow we
=
arctanx
would ever be led to such an observation is quite another matter), so
that
Lb
arctan x dx
=
x arctan x
"really"
-
ÏOg(Î+
2
X2)
b
,
evaluated
then we may feel that we have
arctan x dx.
This chapter consists of little more than methods for finding elementary primitives of given elementary functions (a process known simply as
and conventions designed to facilitate
together with some notation, abbreviations,
this procedure. This preoccupation with elementary functions can be justifiedby
three considerations:
f
"integration"),
(1)Integration
about
is a standard
topic in calculus, and everyone should know
it.
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 (forexample, you might take a (physics)
course in which you are
able
expected to be
to integrate).
of integration are actually very important theuseful
most
The
(3)
apply
all
not just elementary ones).
functions,
orems (that
to
(2)Every once
"methods"
Naturally, the last reason is the crucial one. Even if you intend to forget how
to integrate (andyou 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 obtained
simply by differentiating various well-known functions. The list given below makes
use of a standard symbol which requires some explanation. The symbol
f
"a
means
The
primitive of
symbol
ff
or
f (x)dx
"the
or, more precisely,
will often by used in stating
f"
useful in formulas like the following:
collection of all primitives of f."
theorems, while
(x) dx is most
ff
19. Integrationin ElementaryTerms 361
x4/4 satisfies F'(x)
x3. It
that the function F(x)
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 the
integration process as mechanical as possible, and we will resort to any possible
device. Another feature of the equation deserves mention. Most people write
This
"equation"
=
means
x3dx
=
+ C
=
to emphasize that the primitives of f (x) x3 are precisely the functions of the
x4/4 + C for some number C. Although it is possible (Problem 13)
form F(x)
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 appearing
after the
on the left side-thus
=
=
"d"
u3du
I
/
=
,
tx2
txdx
=
txdt
=
-,
2
xt2
-.
2
"indefinite
A function in ff (x)dx, i.e., a primitive of f, is often called an
integral" of f, while f f(x) dx is called, by way of 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 is emphasized once again:
the integral f f (x) dx is not defined as "F(b)
F(a), where F is an indefinite
of
repetitious,
integral
it is time to reread
f" (ifyou do not find this statement
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.
-
adx
/
=
ax
X"
x"dx=
dx
n+1
=
n-/-1
,
dx is often written
log x
abbreviations
table.)
e'dx
sin x
=
dx
e'
=
-
cos x
are used
for convenience;
similar
in the last two examples of this
362 Derivativesand Integrals
cosxdx
sinx
=
sec2xdx
=
tanx
secx tanx dx
Ï
/
dx
1+x2
dx
=
secx
arctanx
=
.
=
-x2
arcsmx
1
Two general formulas of the same nature are consequences of theorems about
difFerentiation:
[f(x)+g(x)]dx=
c-
g(x)dx,
f(x)dx+
f(x)dx=c-
f(x)dx.
These equations should be interpreted as meaning that a primitive of f + g can
be obtained by adding a primitive of f to a primitive of g, while a primitive of
a primitive of f by c.
c f can be obtained by multiplying
Notice the consequences of these formulas for definite integrals: If f and g are
-
continuous,
then
b[f(x)+g(x)]dx=
g(x)dx,
f(x)dx+
Ib
c·
b
f(x)dx
=
f(x)dx.
c·
a
a
These follow fmm 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 f and g are merely integrable, but recall how much
more difficult the proofs are in this case.)
The product formula for the derivative yields a more interesting theorem, which
will be written in several different ways.
THEOREM
l (INTEGRATION BY PARTS)
If
f'
and g' are continuous, then
fg'= fg
f(x)g'(x)dx f(x)g(x)
=
lb
f (x)g'(x)dx
f'g,
-
f'(x)g(x)dx,
-
b
=
(Notice that in the second equation
f (x)g(x)
b
-
f'(x)g(x)dx.
f g.)
f (x)g(x)denotes the functwn
-
19. Integrationin Elementar.yTerms 363
PROOF
The formula
(fg)'= f'g
fg'
+
can be written
fg'= (fg)'-- f'g.
Thus
fg'=
(fg)'-
f'g,
and fg can be chosen as one of the functions denoted by fg)'. This proves the
f(
first formula.
The second formula is merely a restatement of the first, and the third formula
follows immediately from either of the first two.
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 f, whose derivative is simpler than f, and another function which is obviously of the form g'.
xe'dx
=
xex
1·e2dx
-
fg'fgf'g
xex
-
x sin x
f
dx
=
x
g'
-
(-
f
e2
-
cos x)
1 (- cos x) dx
-
·
f
g
'
g
cos x + sin x
-x
-
There are two special tricks which often work with integration by parts. The
first is to consider the fimction g' to be the factor 1, which can always be written
in.
log x dx
1 log x dx
=
.
g'
x log x
=
-
gf
x(log x)
f
=
x
-
f'
g
x.
-
(1/x) dx
The second trick is to use integration by parts to find
in terms of
and then solve for h. A simple example is the calculation
fh
f
(1/x) log x dx
-
g'
which
=
f
log x log x
·
g
f
-
(1/x) log x dx,
-
f'
implies that
ll
-logxdx
2
x
=
(logx)2
g
fh again,
364 Derivativesand Integrals
or
A more complicated
Ï
=
.
2
x
calculation is often required:
ex sinx dx
f
(logx)2
l
-logxdx
e'
=
g'
-
(-
f
cos x)
ex
---
(-
-
f'
g
=-excosx+
cosx) dx
g
e2cosxdx
v'
u
-e2
[e2 (sinx)
cosx +
=
-
u
e>(sinx) dx];
-
u'
v
v
therefore,
e2 sin x dx
2
or
Ï
e' sinx
dx
e2 (sinx
=
=
e*(sin x
-
cos x)
cos x)
-
.
2
Since integration by parts depends upon recognizing that a function is of the
form g', the more functions 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
(logx)2dx
(logx)(logx)dx
=
g'
f
if we recall that f log x dx
gration by parts); we have
(logx)(logx) dx
f
=
g'
=
x(log x)
-
x
(thisformula
(logx)[x(logx)
-
f
x]
--
itself derived by inte-
was
(1/x)[x(logx)
-
f'
g
x]dx
g
-x]
=
(logx)[x(logx)
=
(logx) [x(logx)
-
-
x]
-
-x]
=
(logx)[x(logx)
=x(logx)2-2x(logx)+2x.
[logx
-
1]dx
log x dx +
1dx
-x]+x
-
[x(logx)
The most important method of integration is a consequence of the Chain Rule.
The use of this method requires considerably more ingenuity than integrating by
parts, and even the explanation of the method is more difficult. We will therefore
19. Integrationin Elementar_;Terms 365
in stages, stating the theorem for definite integrals first, and
of indefinite integrals for later.
develop this method
saving the treatment
THEOREM
(THE SUBSTITUTION
2
If
f
and g' are continuous, then
FORMULA)
g(b)
b
(fog)-g'
f=
g(b)
b
Ï
PROOF
If F is a primitive of
f(g(x))-g'(x)dx.
f(u)du=
f,
then the left side is F(g(b))
F(g(a)).
-
On the other
hand,
(Fog)'=(F'og)-g'=(fog)-g',
so Fog
is a primitive of (fo g) g' and the right side is
·
(Fo g)(b)
-
(Fo g)(a)
F(g(b))
=
F(g(a)).
-
The simplest uses of the substitution formula depend upon recognizing
given function is of the form (fo g) g'. For example, the integration of
that a
-
sin*xcosxdx
(sinx)"cosxdx
=
is facilitated by the appearance of the factor cos x, which will be the factor g'(x)
sinx; the remaining expression,
for g(x)
(sinx) , can be written as (g(x))5
us.
Thus
f(g(x)), for f(u)
=
=
g(X)
Lb
sin"
x cos x dx
f (u)
=
u
5
g(b)
Ïf
Ïsino
(g(x))g'(x)dx
u5du
f (u)du
=
sin6b
=
à"a
-
.
6
sina
f
=
SinX
b
=
The integration of
=
6
tan x dx can be treated similarly if we write
b
Lb
tanx dx
-sinx
=
Ï
-
a
-sinx
dx.
cosx
is g'(x), where g(x)
cosx; the remaining
f (cosx) for f (u) 1/u. Hence
In this case the factor
1/ cos x can then be written
=
=
g(x)
Ïa
tan x dx
=
f(u)
cosx
1
-
f(g(x))g'(x)dx=-
=-
f(u)du
cmbg
=
-
-
Osa
N
du
=
log(cos a)
-
log(cos b).
factor
366 Derivadvesand Integrals
Finally, to find
b
Ï
y
dx,
x logx
a
notice that 1/x
f (u)
=
g'(x) where
=
g(x)
logx, and that 1/logx
=
f(g(x)) for
=
1|u. Thus
g (X )
b
Ï
ÏOg X
=
1
dx
b
g(b)
I
Ïlogb
f(g(x))g'(x)dx=
=
y
=
-
E
loga
du
=
f(u)du
g(a)
a
log(log b)
log(log a).
-
Fortunately, these uses of the substitution formula can be shortened
The intermediate steps, which involve writing
considerably.
g(b)
lb
f(g(x))g'(x)dx=
f(u)du,
g(a)
a
can easily be eliminated by noticing
the following: To go from the left side to the
right side,
u for g (x)
du for g'(x)dx
(andchange the limits of integration);
substitute
the substitutions can be performed directly on the original
for the name of this theorem). For example,
function
sinb
-
Ïb
sin"
N fOr
.
x cos x dx
substitute
(accounting
SinX
u6 du,
=
du for cosxdx.
sina
and similarly
-
lb
a
SiBX
this method
=
-sinxdx
du.
-
du for
cosx
Usually we abbreviate
cosb
M fOT COSX
.
dx substitute
cosa
u
even more, and say simply:
"Let u
=
g(x)
du
=
g'(x) dx."
Thus
ÌCt
Lb
xlogx
dx
u
=
IOg X
Iog by
-du.
1
du=-dx
x
=
ga
In this chapter we are usually interested in primitives rather than definite integrals, but if we can find
f f (x)dx for all a and b, then we can certainly find
19. Integrationin Elemen y Terms 367
ff
(x) dx. For
example,
since
6b
Lb
Ï
sin"
it f'ollows that
x cos x dx
=
6
6
,
sin6x
sin x cosxdx
Similarly,
sin6a
-
=
.
6
-logcosx,
tanxdx
I
1
=
dx
xlogx
Iog(log x).
=
It is quite uneconomical to obtain primitives from the substitution formula by first
finding definite integrals. Instead, the two steps can be combined, to yield the
following procedure:
(1)Let
u
du
(afterthis
manipulation
=
g(x),
=
g'(x)dx;
only the letter u should appear, not the
letter x).
(2)Find a primitive (asan expression
(3)Substitute g(x) back for u.
involving u).
Thus, to find
sin x cos x dx,
(1)let
u
du
=
sinx,
=
cosxdx
so that we obtain
u3
du;
(2)evaluate
u6 du
(3)remember
to substitute
I
=
;
sin x back for u, so that
sin"
sin6
x cos x dx
=
.
6
368 Derivativesand Integrals
Similarly, if
logx,
=
u
du
1
=
-
x
dx,
then
1
dx
x logx
I
so that
I
To evaluate
1
becomes
-
du
=
u
1
dx
x log x
log u,
log(log x).
=
x
dx,
1 ‡ x2
let
u=1+x2,
du
2x dx;
=
the factor 2 which has just popped up causes no problem-the
l
2
-du
1
integral becomes
1
2
-logu,
=
-
u
so
dx
1+ x,
log(1+ x2L
=
(This result may be combined with integration by parts to yield
1 arctan x dx
-
=
=
x arctan x
x arctan x
dx
-
-
)
1 + x2
log(1 + x2
a formula that has already been mentioned.)
These applications of the substitution formula* illustrate the most straightthe suitable factor g'(x) is recognized,
forward and least interesting types-once
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 substitution formula is often written in the form
f(x)du=
f(g(x))g'(x)dx,
This formula cannot be taken literally
f f(u)du
of
(afterall,
mean a primitive
u=g(x).
should
mean a primitive of f and the
g'; these are certainly not equal).
However, it may be regarded as a symbolic summary of the procedure which we have developed. If
and a little fudging, the formula reads particularly well:
we use Leibniz's notation,
symbol
f f(g(x))g'(x)dx should
f(u)du
=
(fo
f(u)
g)
dx.
-
19. Integrañonin Elemener.yTerms 369
(thethird
problem has been disguised a little by some algebraic
sec2
chicanery).
x tan" x dx,
(cosx)e'"2 dx,
I
e2
1
dx.
en
-
If you have not succeeded in finding the right substitutions, you should be able to
guess them from the answers, which are (tan6x)/6, e"i"=, and arcsin ex. 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
the answer to the second be
detail is the proper positioning of the constant-should
e32/3 or 3e32? I always take care of these problems as follows. Clearly e32 dx
e32·
e3x, I get F'(x)
3e32, so the
Now if I differentiate F(x)
(something).
must be
to cancel the 3.)
=
J
=
"something"
=
=
¾,
Ï
dx
e3'dx
Ï
I
log(x + 3),
=
3
x +
=
,
sin4x
cos
=
,
4
-cos(2x
sin(2x +
/
4x dx
1)dx
dx
1 + 4x2
+ 1)
=
2
,
arctan2x
¯
2
More interesting uses of the substitution formula occur when the factor g'(x)
does not appear. There are two main types of substitutions where this happens.
Consider first
ll
1
+ex
-
dx.
e2
of the expression ex suggests
The prominent appearance
the simplifying
tion
u=e2,
du e'dx.
=
Although the expression e2 dx does not appear, it can always be put in:
1+e2
1
dx1 e2 ex
1 ex
We therefore obtain
1
du,
1--u
/1+e2
Ï1+u
--
-
¯
---e'dx.
substitu-
370 Derivativesand Integrals
which can
be evaluated by the algebraic trick
1
Il+u
Il+e2
¯
1-u
so that
1
-
ex
dx
du
2
1-u
=
1
+
-
u
-21og(1
du
=
--
-21og(1
u) + log u,
-21og(1
=
-
e") + 1oge2
=
e2) + x.
-
There is an alternative and preferable way of handling this problem, which does
not require multiplying and dividing by ex. If we write
u=e*,
x=logu,
dx
1
=
-
da
u
then
1+u
1-u
l1+ex
1-e=
dx
immediately becomes
1
¯
du.
Most substitution problems are much easier if one resorts to this trick of expressing x in terms of u, and dx in terms of du, instead of vice versa. It is not hard to
see why this trick always works (aslong as the function expressing u in terms of x
is one-one for all x under consideration):
If we apply the substitution
x=g¯I(u)
u=g(x),
dx
=
(g¯1)'(u)du
to the integral
f (g(x))dx,
we obtain
f(u)(g¯I)'(u)du.
(1)
On the other hand, if we apply the straightforward
u
=
g(x)
du
=
g'(x)dx
substitution
to the same integral,
-g'(x)dx,
f(g(x))dx=
f(g(x))-
we obtain
(2)
/
f (u)
1
·
g'(g-1)
du.
The integrals (1)and (2)are identical, since (g¯ )'(u)
As another concrete example, consider
Ï
e
ex +1
dx.
=
1/g'(g¯1
19. Integrationin Elemenky Terms 371
In this case we will go the whole hog and replace the entire expression
by one letter. Thus we choose the substitution
1
Vex+1,
=
u
Je*+
u2_741
u2
1 = ex,
-
x
=
dx
=
log(u2
-
2u
u2
du.
1
-
The integral then becomes
Ï
Thus
(u2
12
-
2u
u
Ï
du
u2-1
e
dx
=
2
=
u
2
2u3
1du
-
=
--
2u.
-
3
(e2+ 1)3/2 2(e2 + 1)1/2
-
3
e2+1
Another example, which illustrates the second
occur, is the integral
main
type ofsubstitution
that can
1-x2dx.
In this case, instead of replacing a complicated expression by a simpler one, we
sin2
will replace x by sin u, because
u
cos u. This really means that we
arcsin x, but it is the expression for x in terms of u
are using the substitution u
which helps us find the expression to be substituted for dx. Thus,
Ál
=
-
=
let x sin u,
dx=cosudu;
[u
=
arcsin x]
=
then the integral becomes
1
The evaluation
of this
-
sin2
u cos u du
cos2
=
u du.
integral depends on the equation
1 + cos 2u
2
(seethe discussion of trigonometric functions below) so that
COS2
Ï
and
Ï
cos2 u du
=
Ï
1+cos2u
du
2
arcsin x
1-x2¿y
=
-
u
2
sin2u
+
-
-
-x)1
=
,
sin(2 arcsin x)
2
4
arcsin x
1
sui(arcsin x) cos(arcsin x)
+
2
2
arcsinx
1
x2
+
2
2
4
.
=
4
-
372 Derivativesand Integrals
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 listed below. Using these
you should be able to integrate all the functions in Problems 1 to 9 (a few other
interesting tricks are explained in some of the remaining problems).
1. TRIGONOMETRIC FUNCTIONS
Since
sin2
cos2
x ‡
and
cos 2x
cos2
=
x
sin2
-
we obtain
cos2x
=
cos 2x
=
cos2x
(1
1- co
-
sin x)
-
2x)
sin2
-
2cos2X
=
1
=
x
-
-
2 sin2
Î,
x,
or
sin'
1
--
x
-
cos 2x
,
2
1 + cos 2x
2
These formulas may be used to integrate
cos2
=
x
.
sin" x dx,
cos" x dx,
if n is even. Substituting
(1
cos 2x)
-
or
2
(1 + cos 2x)
2
for sin2 x or cos2x yields a sum of terms involving lower powers of cos. For
example,
sin4
x dx
1
=
-
cos 2x
dx
dx
=
cos 2x dx +
-
and
If n is odd, n
/
=
cos2
2x dx
2k + 1, then
sin"
x dx
=
=
/1+cos4x
2
sin x (1
-
cos
dx.
x)k dx;
cos2 2x dx
19. Integrationin Elementap Terms 373
the latter expression, multiplied out, involves terms of the form sin x cos' x, all of
which can be integrated easily. The integral for cos" x is treated similarly. An
integral
sin" x cos" x dx
is handled the same way if n or m is odd. If n and m are both even, use the
formulas for sin2 x and cos2
A final important trigonometric integral is
!
1
cos x
dx
sec x dx
=
=
log(sec x + tan x).
"deriving"
this result, by means of the methAlthough there are several ways of
ods already at our disposal (Problem 12), it is simplest to check this formula by
differentiating the right side, and to memorize it.
2. REDUCTION
FORMULAS
Integration by parts yields (Problem 20)
/
Ï
sin" x dx
/
cos" x dx
=
-
-
1 sinn-1
x cos x +
n
=
--
1 cos"¯1
n
x sin x +
1
-
n
sinn-2
n
1
-
n
n
1
1
2n-3
x
dx=
+
2n 2 (x2 + 1)=¯I
2n 2
(x24 1).
--
-
cos"¯2
x
dx,
x dx,
1
dx
(x2y 1)n-1
and many similar formulas. The first two, used repeatedly, give a different method
for evaluating primitives of sin" or cos". 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
p(x) = anx" +an-1x"¯I
+
q(x)=bmx"+bm-ix"¯l+--·+bo.
-··+ao,
1. Moreover, we can assume that n < m,
b,
We might as well assume that a,
for otherwise we may express p/q as a polynomial function plus a rational function
which is of this form by dividing (thecalculation
=
u2
u-1
=
=u+1+
1
u-1
is a simple example). The integration of an arbitrary rational fimction depends
on two facts; the first follows from the "Fundamental Theorem of Algebra" (see
Chapter 26, Theorem 2 and Problem 26-3), but the second will not be proved in
this book.
374 Derivativesand Integrals
THEOREM
Every polynomial function
q(x)=x"'+bm-ix"¯'+-··+bo
can be written
as a product
X+¾)s1·...·(X2+þX+
q(x)=(x-a1)"-...·(x-ŒkŠr*(x24
"
51Ÿ···ŸSI)=M).
(whereri+-··+rk
(In this expression, identical factors have been collected together, so that all
x2 +
x «¿ and
ßgx + y, may be assumed distinct. Moreover, we assume that each
quadratic factor cannot be factored further. This means that
-
ß;2
since otherwise
we can
-4y¿
<
0,
factor
-ßt
+
X2
Áß¿24y,
Áß?
-ß,
-
-
-
2
4y;
2
into linear factors.)
THEOREM
If n
<
m and
p(x)
x" + an-1x"¯'
+
q(x)=xm+bm-ixm-1+·--+bo
=
-
+ ao,
··
(x2+ßix+yi)"
=(x-a1)"·---·(X-akf(x2+Ax+n)"'--can be written
then p(x)/q(x)
p(x)
al,1
=
q(x)
(x
-
at)
in the form
a1,ri
+···+
(x
+---
al)
-
Øk,l
+
+
Œk,ra
+
(x ag)
bl.1x+ci,i
·
-
[
+
(x
-
+-··+
(x2
+
any
b1,six+ct
2
bi,1x + ci,1
(x2 + ßix + yi)
+---
si
b
+··-+
,,,x
+ ci,,
.
(x2 + ßix +
yi)"I
"partial
This expression, known as the
fraction decomposition" of p(x)/q(x), is
so complicated that it is simpler to examine the following example, which illustrates
such an expression and shows how to find it. According to the theorem, it is
possible to write
2x2 + 8x6 + 13x5 + 20x4 + 15x3 + 16x2 + 7x + 10
(x2 + x + 1)2(x2 + 2x + 2)(x 1)2
-
=
a
x-1
+
b
(x--1)2
cx+d
+ x2+2x+2
ex+
+ x2+x+1
f
+
gx+h
.
(x2+x+1)2
19. Integrationin Elementary72rms 375
To find the numbers a, b, c, d, e, f, g, and h, write the right side as a polynomial
over the common denominator (x2+ x + 1)2(x2 + 2x + 3)(x 1)2; the numerator
becomes
-
2+b(x2+2X+2)(X2
a(x-1)(x2+2x+2)(x2
--1)2(x2+x+1)2+(ex+
2
-1)2(x2+2x+2)(x2+x+0
+(cx+d)(x
f)(x
+(gx+h)(x-1)2(x2+2x+2).
Actually multiplying this out (!)we obtain a polynomial of degree 8, whose coefficients are combinations of a,
h. Equating these coefficients with the coefficients of 2x2+8x6+13x5+20x4+15x3+16x2+7x+
10 (thecoefficient of x8 is 0)
obtain
unknowns
equations
the
eight
8
in
h. After heroic calculations
we
a,
solved
these can be
to give
...,
.
1,
a
e=0,
b
=
2,
=
.
1,
c
g=0,
,
d = 3,
h=1.
=
f=0,
.
Thus
=
3
2x2 + 5x6 + 13x3 + 20x4
+ 16x2 + 7x + 7
dx
(x24, y 1)2(x2 + 2x + 2)(x 1)2
1
2
1
dx +
dx +
dx +
Ï
/
-
(x-1)2
(x-1)
(x2+x+1)2
x + 3
x2+2x+2
dx.
(In simpler cases the requisite calculations may actually be feasible. I obtained this
particular example by stardng with the partial fraction decomposition and converting it into one fraction.)
We are already in a position to fmd each of the integrals appearing in the above
expression; the calculations will illustrate all the difficulties which arise in integrating rational functions.
The first two integrals are simple:
Ï
Ï
1
x
1
-
dx
=
dx
=
1),
-
-2
2
(x 1)2
--
The third integration depends on
x
"completing
x2
-
l'
the
square":
2
=
|
3
4
-
x +
+ 1
.
instead of we could not take the square root, but in this
quadratic factor could have been factored into linear factors.) We
(If we had obtained
case our original
log(x
-
376 Daivancesand Integrals
can now write
/
1
dx
(x2+x+1)2
=
16
9
-
1
dx.
-2
-
;
x+,
+1
4
-
-
The substitution
x + y
1
changes
this integral to
Ë4
16
9
du
(u2 + 1)2
which can be computed using the third reduction
Finally, to evaluate
x+ 3
Ï
we write
/
x+3
dx
x2+2x+2
=
-
1
2
(x2+2x+2)
2x+2
'
formula given above.
dx
2
dx +
x2+2x+2
(x+1)2+1
dx.
The first integral on the right side has been purposely constructed so that we can
evaluate it by using the substitution
u=x2+2x+2,
du
(2x +2)dx
=
The second integral on the right, which is just the
simply 2 arctan(x + 1). If the original integral were
Ï
x+3
(x2+2x+2)"
dx
1
=
2x+2
¯
(x2+2x+2)=
difference of the other two, is
2
dx +
[(x+1)2+1]
dx
'
the first integral on the right would still be evaluated by the same substitution.
The second integral would be evaluated by means of a reduction formula.
This example has probably convinced you that integration of rational functions
is a theoretical curiosity only, especially since it is necessary to find the factorization
of q (x) before you can even begin. This is only partly true. We have already seen
that simple rational functions sometimes arise, as in the integration
another
Ï
l +ex
dx;
1 e2
-
important example is the integral
Ï
1
x
2
-1
1
dx
=
-
x--1
12
x+1
dx
=
1
log(x
2
-
-
1)
-
1
log(x + 1).
2
-
19. Integrationin ElemeneryTerms 377
Moreover, if a problem has been reduced to the integration of a rational function,
it is then certain that an elementary primitive exists, even when the difficulty or
impossibility of finding the factors of the denominator may preclude writing this
primitive explicitly.
PROBLEMS
1.
This problem contains some integrals which require little more than algeand consequently
braic manipulation,
test your ability to discover algebraic
of the integration processes. Nevertricks, rather than your understanding
theless, 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 you can see when the end of an integration
process is in sight. The answer section, if you resort to it, will only reveal
what algebra you should have used.
Ï
(ii)
I
(iii)
Ï
(i)
Ñ+Vi
dx
.
x-1+)x+1
+e32
e=+e
dx
e4x
.
dx.
(iv)
(v)
dx.
tanixdx.
(Trigonometric integrals are always very touchy, because
identities that an easy
there are so many trigonometric
problem can easily look hard.)
dx
Ï
/
(viii)
/
(ix)
l8x2+6x+4
(x)
/
2
y x2
dx
dx
1+sinx
x +
1
-x2
dx.
1
dx.
2x
2.
The following integrations involve simple substitutions,
should be able to do in your head.
(i)
e2 sine2 dx.
most
of which
you
378 Denvativesand Integrals
xe¯"
(ii)
(iii)
dx.
Ilogx
Ï
(In the text this was done by parts.)
dx.
x
ex dx
+ 2ex + 1
e
e e'dx.
(v)
Ï
(vii)
Ï
xdx
(vi)
.
1--x4
-
e
dx.
1
x2 dx.
(viii)
x
(ix)
log(cos x) tan x dx.
-
log(logx)
J
3.
dx.
xlogx
Integration by parts.
(i)
x2ex dx.
(ii)
x3e"
(iii)
eax sinbxdx.
(iv)
x2 sin
x
(v)
(logx)3 dx.
(vi)
dx.
dx.
Ilog(logx)
(vii)
dx.
sec
dx. (This is a tricky and important integral that often comes
up. If you do not succeed in evaluating it, be sure to
consult the answers.)
(viii)
(ix)
(x)
cos(logx)dx.
log x dx.
x(log x)2 dx.
19. Integrationin Elemenky Terms 379
4.
The following integrations can all be done with substitutions of the form
sin u, x
x
cos u, etc. To do some of these you will need to remember
=
=
that
secxdx=log(secx+tanx)
followingformula, which can also be checked by differentiation:
as well as the
-log(cscx
esex dx
+ cot x).
=
In addition, at this point the derivatives of all the trigonometric
should be kept handy.
(i)
dx
I
.
1
.
1 + x2
I
(iv)
/
I
(vi)
Ï
tution
X
=
tan
=
sec2
u, you want to use the substi-
M.)
.
x2
1
dx
-
(The answer will be a certain inverse function that was
1 given short shrift in the text.)
.
x/x2
-
dx
x)1
x2
-
dx
.
x/1
+ x2
x3
I
(Since tan2 u + 1
dx
(iii)
(vii)
(You already know this integral, but use the substitution
sin u anyway, just to see how it works out.)
x
=
dx
(ii)
5.
x2
-
functions
1
72
_
dx.
x2 dx.
(viii)
1
(ix)
1+x2dx.
(x)
x2
-
You will need to remember the methods for
integrating powers of sin and cos.
1dx.
-
The following integrations involve substitutions of various types. There is
no substitute for cleverness, but there is a general rule to follow: substitute
for 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 express x directly
in terms of u, to find out the proper expression to substitute for dx.
I
(ii)
/
(i)
dx
1+/x+1
dx
1 + ex
.
.
380 Derivativesand Integrals
dx
I
(iv)
I
(v)
Ï
(vi)
/
(vii)
/42+1
(iii)
.
dx
(The substitution u
e2 leads to an integral requiring yet another substitution; this is all right, but both
.
1 + e'
=
substitutions
can be done at once.)
dx
2+tanx
dx
(Another place where one substitution
+ 1 do the work of two.)
.
dx.
2 +1
(viii)
(ix)
*(x)
6.
Ï
can be made to
dx.
e
Vl-x
1
dx. (In this case two successive substitutions work out best;
there are two obvious candidates for the first substitu-
E
-
tion, and either will work.)
x-1
1
dx.
1
x +
x
/
·
-
The previous problem provided gratis a haphazard selection of rational functions to be integrated. Here is a more systematic selection.
2x2 + 7x
-
/ x34x2--x-Î
(ii)
Ï
(iii)
Ï
(iv)
Ï
(v)
/
(vi)
Ï
(vii)
/
(viii) X4‡Î
/
(ix)
l
(x)
Ï
(i)
x3
1
2x + 1
3x2 + 3x
-
dx.
-
1
dx
.
x3+7x2-5x+5
1)2(x + 1)3
2x2 ‡ X + Î
dx.
(x + 3)(x 1)2
(x
dx.
-
-
x+4
dx.
x24)
x3+x+2
x4+2x2+1
dx.
3x2+3x+1
dx.
x3+2x2+2x+1
dx
.
2x
(x2+
x +
1)2
3x
2
(x
1)3
dx.
dx.
19. Integrationin ElementaryTerms 381
*7.
(No holds barred.) The following integrations involve all the
Potpourri.
of the previous problems
methods
I
(ii)
Ï
(i)
arctanx
1 + x2
dx.
x arctan x
(1 + x2)3
dx.
(iii)
log fl + x2 dx.
(iv)
x
(v)
Ï
log fl + x2 dx.
x2
¡
_
x2+1
f
dx.
x
Ï
(viii)
/
e'i">
(ix)
Jtan x dx.
(x)
g6
(vii)
dx.
41+x4
arcsin
(vi)
g
·
dx.
.
1 + smx
xcos3x-sinx
dx.
.
COS X
.
i
(To factor
X6
+ Î , Erst factor y3 + 1, using Problem 1-1.)
The following two problems provide still more practice at integration, ifyou need
it (andcan bear it). Problem 8 involves algebraic and trigonometric manipulations
and integration by parts, while Problem 9 involves substitutions.
(Of course, in
resulting
will
require
still
manipulations.)
the
integrals
further
many cases
8.
Find the following integrals.
2)
(i)
log(a2
I
(iii)
/
l + cos x
(ii)
dx.
sin2
x
x + 1
4
x2
-
dx.
(iv)
x arctan x
(v)
sin3 dx.
x
(vi)
Ï
sin3x
COS2
X
dx.
dx.
dx.
382 Denvatives and Integrals
x2 arctan
x
(vii)
xdx
(viii)
9.
dx.
.
x2-2x+2
(ix)
sec3xtanxdx.
(x)
x tan2 x
dx.
Find the following integrals.
I
dx
(a2 +
x2)2
(ii)
1
sin x
(iii)
arctan
(iv)
sin
-
dx.
dx.
fx +
1 dx.
-2
(v)
/
Vx3
dx.
(vi)
log(x +
Vx2 1) dx.
(vii)
log(x +
5)
(viii)
x
--
dx.
dx
-
x3/
.
(ix) (arcsinx)2 dx.
(x)
10.
x5 arctan(x2)
dx.
If you have done Problem 18-9, the integrals (ii)and (iii)in Problem 4 will look
cosh u often works for integrals
very familiar. In general, the substitution x
=
involving fx2 1, while x sinh u is the thing to try for integrals involving
x2 + 1. Try these substitutions
on the other integrals in Problem 4. (The
method is not really recommended;
it is easier to stick with trigonometric
=
-
substitutions.)
*11.
The world's sneakiest substitution
t
=
tan
x
-,
2
is undoubtedly
x
=
dx
=
2 arctan t,
2
dt.
1 ‡ t2
19. Integrationin ElementaryTerms 383
As we found in Problem 15-17, this substitution
2t
1+ t2,
.
=
smx
leads to the expressions
1 t2
1+t2
-
cosx
=
thus transforms any integral which involves only sin and
combined
and division, into the integral of
by addition, multiplication,
cos,
a rational function. Find
This substitution
I
(ii)
/
(iii)
Ï
(i)
*12.
.
dx
1 sin2
Ï
(In this case it is better to let t
.
tan x. Why?)
=
-
dx
,
a sin x +
sini
(iv)
(v)
dx
(Compare your answer with Problem 1(viii).)
1 + sin x
b cos x
(There is also another way to do this, using
Problem 15-8.)
dx. (An exercise to convince you that this substitution
should be used only as a last resort.)
dx
(A last resort.)
3+5sinx
x
.
Derive the formula for fsecxdxin the following two ways:
(a) By writing
1
cosx
cos2 x
cos x
¯
cos x
1
sin2
-
X
¯
=
-
1
2
cos x
_1+sinx
+
cos x
1-sinx
'
an expression obviously inspired by partial fraction decompositions. Be
log(1 sin x); the minus sign
sure to note that cos x/ (1 sin x) dx
remember
that
log a = log
is very important. And
From there
on, keep doing algebra, and trust to luck.
By using the substitution t = tan x/2. One again, quite a bit of manip-
f
-
-
-
-
j
.
(b) ulation
is required to put the answer in the desired form; the expression
tan x/2 can be attacked by using Problem 15-9, or both answers can
be expressed in terms of t. There is another expression for jsecxdx,
which is less cumbersome than log(sec x + tan x); using Problem 15-9,
we obtain
1+tan x
=log
secxdx=log
1
-
tan
tan
+
.
-
2
was actually the one first discovered, and was due,
mathematician's
cleverness, but to a curious historical accinot to any
This last expression
384 Derivancesand Integrals
dent: In 1599 Wright computed nautical tables that amounted to defmite
integrals of sec. When the first tables for the logarithms of tangents were
produced, the correspondence
between the two tables was immediately
noticed
remained unexplained
until the invention of calculus).
(but
13.
The derivation of f e2 sin x dx given in the text seems to prove that the only
eX
primitive of f (x) e' sin x is F(x)
(sinx cos x)/2, whereas F(x)
ex (sinx
cos x)/2 + C is also a primitive for any number C. Where does C
come from? (What is the meaning of the equation
=
=
=
-
-
e2 sin x dx
14.
Suppose that
f"
e2 sin x
=
ex cos x
-
ex sin x
-
dx?)
is continuous and that
/K
O
Given that
15.
f (x)
=
[f (x) + f"(x)] sinx
1, compute
dx
=
2.
f (0).
x dx, using the same trick that worked
Generalizethis trick: Find
(x)dx in terms of
with Problems 12-18 and 14-17.
(a) Find f arcsin
*(b)
¯l
ff
for log and arctan.
ff (x)dx. Compare
sin4
16.
x dx in two different ways: first using the reduction formula,
and then using the formula for sin2 x.
(b) Combine your answers to obtain an impressive trigonometric identity.
17.
Express
(a) Find f
f log(log x) dx
terms of elementary
x2e¯"
functions.)
dx in terms of
18.
Express
19.
Prove that the function
(Do not try to find it!)
20.
Prove the reduction
/
and work
x
21.
=
f (x)
=
f(logx)¯I
dx. (Neither is expressible
(e52+e2
ex
+ 1) has an elementary
I
tan u.)
formula for
x"e" dx
Prove that
primitive.
¯
(b) (logx)"dx.
*22.
coshx
t2
-
1dt
in
f e¯2° dx.
formulas in the text. For the third one write
x2 dx
dx
dx
(1 + x2)" J (1 + x2)"¯1 J (1 + x2)"
on the last integral. (Another possibility is to use the
Find a reduction
(a)
in terms of
=
(See Problem 18-6 for the significance
cosh x sinh x
2
-
x
-.
2
of this computation.)
substitution
19. Integrationin ElementaryTerms 385
23.
Prove that
b
lbf(x)dx=
f(a+b-x)dx.
(A geometric interpretation makes this clear, but it is also a good exercise in
the handling of limits of integration during a substitution.)
24.
Prove that the area of a circle of radius r is wr2. (Naturally you must remember that x is defined as the area of the unit circle.)
25.
Let ¢ be a nonnegative
li
and such that
¢
-1
integrable function such that ¢(x)
1. For h
=
=
0 for |x|
1
>
0, let
>
1
-¢(x|h).
¢h(x) h
=
h
I
¢h 1•
(a) Show that ¢h(x) 0 for |x| > h and that
(b) Let f be integrable on [-1, 1] and continuous at 0. Show that
=
=
-h
1
lim
h-W
J-1
h
¢af
=
lim
haC+ A-h
(c) Show that
lim
h-+0+
/1
h
h2 + x2
-1
¢ëf
dx
=
=
f (0).
x.
The final part of this problem might appear, at first sight, to be an exact
analogue of part (b),but it actually requires more careful argument.
(d) Let f
be integrable on
lim
[-1, 1] and
continuous at 0. Show that
h
L1
dx
xf (0).
2 f (x)
i h2 + x
Hint: If h is small, then h/(h2 + x2) will be small on most of
hoe
=
[-1, 1].
The next two problems use the formula
f (0)2d0,
derived in Problem 13-24, for the area of a region bounded by the graph of
polar coordinates.
26.
f
in
For each of the following functions, find the area bounded by the graphs in
polar coordinates. (Be careful about the proper range for 0, or you will get
results!)
nonsensical
(i) f (0)
a sin 0.
=
(ii) f(0)=2+cos9.
2COS2Ô.
(iii) f(0)2
(iv) f (0) a cos 20.
_
=
386 lkrivances and Integrals
27.
r
··
/(0)
Figure 1 shows the graph of f in polar coordinates; the region OAB thus
1 1
f(0)2d0. Now suppose that this graph also happens to be
has area
-
A
the ordinary graph of some function g. Then the region OAB also has area
area AOxiB
o
FIGURE
+
-
g
area AOxoA.
l
I
I
x2
x.
Prove analytically that these two numbers are indeed the same. Hint: The
function g is determined by the equations
X
1
g(X)
f(Û)COSÛ,
=
f(Û)SißÛ.
=
The next four problems use the formulas, derived in Problems 3 and 4 of the
Appendix to Chapter 13, for the length of a curve represented parametrically (and,
in particular, as the graph of a function in polar coordinates).
28.
Et c be a curve represented parametrically by u and v on [a,b], and let h
be an increasing function with h(ä)
b. Then on [ä,b] the
a and h(b)
representation
of another
voh
give
parametric
h,
6
functions ü
uo
a
the
traversed
at a different rate.
same curve c
curve ë; intuitively, ä is just
=
=
=
=
(a) Show, directlyfrom the
definition of length, that the length of c on
[a,b]
b].
length
[ä,
(b) Assuming differentiability of any functions required, show that the
lengths are equal by using the integral formula for length, and the apequals the
of i on
propriate substitution.
29.
Find the length of the following curves, all described as the graphs of funcwhich is represented
tions, except for (iii),
parametrically.
(i) f (x)
=
(ii) f (x)
=
(x2 + 2)3/2,
x
3
+
O 5 xs
1 5 x 5 2.
,
(iii) x a3 cos3 t,
(iv) f(x) log(cosx),
(v) f (x) log x,
=
y
a3
=
(vi) f(x)
30.
=
3
O 5 t 5 2x.
t,
x/6.
O 5 xs
=
=
1.
1 Ex 5 e.
arcsine2,
-log2
5 x 5 0.
For the following functions, find the length of the graph in polar coordinates.
(i) f(0)
(ii) f(0)
(iii) f(0)
(iv) f (0)
(v) f (0)
=
=
=
acos0.
a(1
cos0).
asin20/2.
-
=
0
=
3 sec 0
0 5 Ð s 2x.
0 s 0 y x/3.
19. Integrationin ElementaryTerms 387
31.
In Problem 8 of the Appendix to Chapter 12 we described the cycloid, which
has the parametric representation
y=v(t)=a(1-cost).
x=u(t)=a(t-sint),
length of one arch of the cycloid. [Answer: 8a.]
(b) Recall that the cycloid is the graph of vo u-1. Find the area under
substitution
in ff and
one arch of the cycloid by using the appropriate
3xa2.]
resultant
evaluating the
integral. [Answer:
(a) Find the
32.
Use induction and integration by parts to generalize Problem 14-13:
n
du
33.
=
If f' is continuous on
Lebesgue Lemma for
[a,b],
f:
du
f (t)dt
.
du..
...
integration by parts to prove the Riemann-
use
b
lim
Amoo
Ï
f (t) sin(At)dt
=
0.
a
This result is just a special case of Problem 15-26, but it can be used to prove
the same way that the Riemann-Lebesgue Lemma
was derived in Problem 15-26 from the special case in which f is a step
the general case
(inmuch
function).
34.
The Mean Value Theorem for Integrals was introduced in Problem 13-23.
The "Second Mean Value Theorem for Integrals" states the following. Supor
pose that f is integrable on [a,b] and that ¢ is either nondecreasing
nonincreasing
number
such
that
there
in
b]
is
b].
Then
on [a,
a
& [a,
Ïbf(x)¢(x)dx=¢(a)
a
b
&
f(x)dx+¢(b)
a
(
f(x)dx.
In this problem, we will assume that f is continuous and that ¢ is differentiable, with a continuous derivative ¢'.
(a) Prove that
if the result is true for nonincreasing
¢, then it is also true for
nondecreasing ¢.
that if the result is true for nonincreasing
Prove
(b)
then it is true for all nonincreasing
¢.
Thus, we can assume that ¢ is nonincreasing
we have to prove that
(c) Prove this
(d) Show that
is needed.
satisfying
¢(b)
=
¢(b)
=
f (x)¢(x) ¢(a)
=
f (x)dx.
by using integration by parts.
the hypothesis that ¢ is either nondecreasing
0,
0. In this case,
(
b
Ï
and
¢
or nonincreasing
388 Derivativesand Integrals
From this special case of the Second Mean Value Theorem for Integrals, the
general case could be derived by some approximation
arguments, just as in the case
of the Riemann-Lebesgue Lemma. But there is a more instructive way, outlined
in the next problem.
35.
(a) Given at,
(*) albi
as and
...,
bi,
b., let
...,
Sk
al ‡
=
·
·
‡ ak. Show that
-b3)
s1(bi
+---+a,b.=
b2)+s2(b2
-
-+
+·-
s.-1(bn-1
-
b.) + s.b.
This disarmingly simple formula is sometimes called "Abel's formula for summation
by parts." It may be regarded as an analogue for sums of the integration by parts formula
f'(x)g(x)dx
=
f(b)g(b)
f(a)g(a)
-
f(x)g'(x)dx,
-
a
especially if we
P
=
(to,
Riemann sums (Chapter 13, Appendix). In fact, for a partition
[a,b], the left side is approximately
use
ta) of
...,
J'(ty)g(ta_i)(tk
(1)
-
Ik-1).
k=1
while the right side
is approximately
f (b)g(b) f (a)g(a)
-
f(tk)&'fik)(Ik
-
Ik-1)
¯
k=1
which is approximately
n
f (b)g(b) f (a)g(a)
-
8(ik)
f (ik)
-
Ik
k=1
=
g(ik-1)
-
ik-1
-
(ik
Ik-1)
-
f(tk)[a(ik-1)-å(tk)
f(b)g(b) f(a)g(a)+
-
kl
-f(ik)
=
f(b)g(b) f(a)g(a)+
[f(tk) f(a)]· [g(tk-1)
-
-
k=1
8(Ik)·
g(tk-1)
k=1
g(b), this works out to be
+
Since the right-most sum is just g(a)
(2)
[f(b) f(a)]g(b)+
-
-
f(a)
-
[f(tk) f(a)] [g(tk-1)-8(tk)]·
-
k=1
.
If we choose
ak
=
l'fik)(fk
¯
Ik-l),
bk = 8(ik-1)
then
(1)
is
akbk,
k=1
which is the left side of
(*),while
k
sk
k
l'(fi)(fi
=
-
fi-1)
is appmximately
i=1
f (t¿) f (t; )
-
_i
i=1
so
n
(2)
is approximately
which is the right side of
(*).
snbn +
L sk(bk
k=1
-
bk-1).
=
f (tk) f (a),
-
19. Integrationin ElementaryTerms 389
This discussion is not meant to suggest that Abel's formula can actually be derived from
the formula for integration by parts, or vice versa. But, as we shall see, Abel's formula can
often be used as a substitute for integration by parts in situations where the functions in
question aren't differentiable.
(b) Suppose
{b,}is nonincreasing,
that
with
b, ;> 0 for each n, and that
msai+---+a,sM
for all n. Prove Abel's Lemma:
bim
s
aibi +-
+a,bn
s
blM.
(And, moreover,
bkW
Økbk
nb,
'
5 bgM,
formula which only looks more general, but really isn't.)
(c) Let f be integrable on [a,b] and let ¢ be nonincreasing on [a, b] with
¢(b) 0. Let P {to, t,} be a partition of [a,b]. Show that the
a
=
=
...,
sum
f(t¿_i)¢(ti-1)(ti
-
ti-1)
i=1
lies between the smallest and the largest of the sums
k
f(t¿_i)(t¿--ti-1)-
¢(a)
i=1
Conclude that
Lb
Ï
f (x)¢(x) dx
lies between the minimum
and the maximum
of
x
¢(a)
and that it therefore equals
36.
(a) Show that
f (t) dt,
¢(a)
f (t) dt for some ( in [a, b].
the following improper integrals both converge.
i
sin x +
(i)
dx.
1
(ii)
sin2 x +
(b) Decide which
of the
(i)
sin
(ii)
sin
dx.
following improper integrals converge.
dx.
dx.
390 Derivativesand Integrals
37.
the
(a) Compute
(b) Show that
(c) Use the
integral
(improper)
log x dx.
the improper integral
substitution
2u to show that
=
x
log(sin x) dx converges.
x/2
la
log(sin x) dx
=
x/2
2
log(cos x) dx + x log 2.
log(sia x) dx + 2
0
0
x/2
(d) Compute
log(cos x) dx.
relation
(e) Using the
38.
cos x
sin(x/2
=
x), compute
-
log(sin x) dx.
Prove the following version of integration by parts for improper integrals:
ao
ao
loo
u'(x)v(x)dx=u(x)v(x)
u(x)v'(x)dx.
-
The first symbol on the right side means, of course,
lim u(x)v(x)
u(a)v(a).
--
X->OC
*39.
One of the most important functions in analysis is the gamma function,
f
(x)
e¯'t*¯1dt.
=
JO
(a) Prove that the improper integral P(x) is defined if x > 0.
(b) Use integration by parts (moreprecisely, the improper integral
version
in the previous problem) to prove that
F (x + 1)
(c) Show that
numbers
F(1)
=
xf
=
1, and conclude
(x).
that F(n)
(n
=
-
1)! for all natural
n.
The gamma function thus provides a simple example of a continuous function
the values of n! for natural numbers n. Of course there
are infinitely many continuous functions f with f(n)
(n 1)!; there are
with
continuous
infinitely
functions
even
many
f(x + 1) xf(x) for all
f
the
x > 0. However,
gamma function has the important additional property
of
that log
is convex, a condition which expresses the extreme smoothness
of this function. A beautiful theorem due to Harold Bohr and
Johannes
Mollerup states that F is the only function f with log of convex, f (1) 1
and f(x + 1) xf(x). See the Suggested Reading for a reference.
which
"interpolates"
=
-
=
=
=
*40.
(a) Use the
reduction
lx/2
O
f sin" x dx to show
formula for
sin" x
dx
=
n
"/2
1
-
n
that
sin"*x
JO
dx.
19. Integrationin ElemenaryTerms 391
(b) Now show
that
2 4 6
3 5 7
x 1 3 5
xdx=--------...2246
2n
2n+1
2n 1
2n
l"/
/2/2
sin 2.+1 xdx=-·--·--...
o
sin
o
-
,
and conclude that
r/2
x
2
224466
i §
sin
2n
2n
2n 1 2n + 1
'
7
dx
x
=/2
-
sin2"**
x dx
JO
the quotient of the two integrals in this expression is between
and
1
1 + 1/2n, starting with the inequalities
(c) Show that
0
<
sin2n+1
x
sin
<
x
<
sin
x
for 0
<
<
x
n/2.
This result, which shows that the products
224466
1 3 3 5 5 7
2n
2n-1
2n
2n+1
can be made as close to x/2 as desired, is usually written
product, known as Wallis' product:
as an
infinite
224466
2¯133557
x
(d) Show also
that the products
1
2-4-6-...-2n
1-3-5.....(2n-1)
as desired. (This fact is used in the next
can be made as close to
and
27-19.)
problem
in Problem
**41.
It is an astonishing fact that improper integrals
f (x) dx can
often
be
b
computed in cases where ordinary
elementary formula for
manay
e¯22
integrals
f (x)dx
cannot. There is no
dx, but we can find the value of
e¯"
dx
precisely! There are
ways of evaluating this integral, but most require
some advanced techniques; the following method involves a fair amount of
work, but no facts that you do not already know.
392 Derivativesand Integrals
(a) Show that
ÏI
2 4
3 5
2n
2n+1
(1-x2)ndx=-·---...o
1
/
(1 + x2)"
o
x 1 3
dx=------...
2
,
2n
2n
2 4
3
2
-
-
.
(This can be done using reduction formulas, or by appropriate
tions, combined with the previous problem.)
(b) Prove, using the derivative, that
1
--
x2
2
e¯'
(c) Integrate
<
<
¯
e¯"
for 0
1
1+x2
<
1.
<
x
substitu-
for 0 5 x.
the nth powers of these inequalities from 0 to 1 and from 0 to
Then use the substitution y
x to show that
oo, respectively.
=
2 4
3 5
2n
2n+1
e¯"
<
dy
e¯T'
<
dy
0
1 3
2 4
x
·
2
(d) Now use
**42.
2n
2n
-
-
3
2
Problem 40(d) to show that
Ï
I
e
2
dy
=
--.
2
o
(a) Use integration by parts to show that
b sinx
dx
--
x
b
cosb
cosa
X2
b
a
COSX
dx,
a
exists. (Use the left side to investigate
and conclude that
side
and
right
oo.)
the limit as a
the
0+
for the limit as b
(b) Use Problem 15-33 to show that
->
fg°°(sinx)/xdx
()tdt
sin(n +
L"
.
S1R
-
t
=
->
x
2
for any natural number n.
(c) Prove that
lim
xxo
Ï"
sin(1
+
2
()t
1
-
-
t
,
sln-
t
dt
=
0.
2Hint: The term in brackets is bounded by Problem
Riemann-Lebesgue Lemma then applies.
-
15-2(vi); the
19. Integrationin ElementaryTerms 393
substitution
(d) Use the
¼)tand
(A +
=
u
°°
sin x
x
x
2
-dx=-.
43.
Given the value of
(sinx)/x
show that
(b)to
part
dx from Problem 42, compute
sin x
Ï°°
dx
x
o
by using integration by parts. (As in Problem 37, the formula for sin 2x will
play an important role.)
---
*44.
(a) Use the
substitution u
t2 to show
=
e¯"
f(x)=
(b) Find
*45.
that
du.
x Jo
F(j).
(a) Suppose
and that
that
-
f (x) is integrable
x
lim f (x)
A and
=
x->0
on every interval
lim
f (x)
X-
[a,b]
for 0
<
B. Prove that for all a,
=
a
ß
<
>
b,
0
we have
"
f(ax)
f(ßx) dx
-
(A
=
-
B) log
x
a
N f(ŒX)
Hint: To
lim
x->0
f (x)
f(
-
estimate
X)
dx use two different substitutions.
x
(b) Now suppose
=
instead that
X
Ja
dx converges
f (ax) f (ßx) dx
-
=
(ii)
>
0 and that
Alog ß
x
the
following
integrals:
(c) Compute
e¯""
l°°
o°°
for all a
A. Prove that
"
(i)
ß
-.
e¯**
-
X
cos(ax)
Jo
a
dx.
cos(ßx)
-
-.
dx.
X
In Chapter 13 we said, rather
blithely, that integrals may be computed to any
degme of accuracy desired by
lower and upper sums. But an applied
mathematician,
who really has to do the calculation,
rather than just talking about
the
overjoyed
of
computing
doing it, may not be
lower sums to evaluate
at
prospect
of
an integral to three decimal places, say (adegree
accuracy that might easily be
needed in certain circumstances). The next three problems show how more refined
calculating
methods
can make the calculations
much more efficient.
394 Derivativesand Integrals
We ought to mention at the outset that computing upper and lower sums might
not even be practical, since it might not be possible to compute the quantities m,
and Mi for each interval [ti-1,tt]. It is far more reasonable
simply to pick points x;
in
[t¿_i,t¿] and
consider
f(x¿) (t¡
-
-
t¿_i). This represents
the sum of the areas
i=1
which
partially overlap the graph of f-see Figure 1 in the
Appendix to Chapter 13. But we will get a much better result if we instead choose
the trapezoids shown in Figure 2.
of certain rectangles
FIGURE
2
Suppose, in particular, that
divide
we
the points
[a, b]
into n equal intervals, by means of
(b-a
=a+ih.
t¿=a+i
n
Then the trapezoid with base
[t; i,
t;] has area
f(ti_i)+ f(ti)
2
-
(ti
t¿_t)
-
and the sum of all these areas is simply
f (11)+ f (a)
E,=h
h
=
This method
[
2
+
f (t2)+ f (ti) +---+
f(a)+2Lf(a+ih)+
f(b)
,
h=
b
-
a
n
an integral is called the trapezoid ru/s. Notice that to
the old f(ti); their contribution
necessary to recompute
of approximating
E2, from E, it isn't
E2,
is just (E.. So in practice it is best to
to
46.
2
"¯'
obtain
approximations
f (b) + f (ts-1)
2
to
lb
f. In
compute
E2, E4, 28,
the next problem we will estimate
-
-
·
b
f
-
ÉO
$6É
E..
(a) Suppose that f" is continuous. Let Pi be the linear function which
with
f at tt-i and ti. Using Problem 11-43, show that if ni and
agrees
Ni are
19. Integrationin ElementaryTerms 395
and maximum
the minimum
I
(x
=
of
-
f"
on
ti-1)(x
[ti_i,ti]
and
ti) dx
-
-1
then
A
n¿ I
-2
2
(b) Evaluate
n¿h3
.
12
-l
that there is ome c inbl
term" (b a)S f"(c)/12n2 varies as 1/n2
obtained using ordinary sums varies as 1/n).
Notice that the
the eror
3
(f-P¿)<
12
(c) Conclude
FIGURE
Ni h3
4
¯
/
=
2
I to get
-<
=
N¿I
(f-Pi)2--.
_,
"error
-
(while
We can obtain still more accurate results if we approximate f by quadratic
functions rather than by linear functions. We first consider what happens when
the interval [a,b] is divided into two equal intervals (Figure 3).
47.
0 and b
2. Let P be the second degree polynomial function which agrees with f at 0, 1, and 2 (Problem 3-6). Show
(a) Suppose first that
that
=
a
=
2
P
(b) Conclude
[f (0) + 4 f (1) + f (2)].
=
that in the general case
Lb
lb
P=
b-a
a+b
f(a)+4f
6
2
+
f(b)
.
b
is a quadratic polynomial. But, remarkably enough, this same relation holds when f is a cubic polynomial!
Prove this, using Problem 11-43; note that f'" is a constant.
(c) Naturally
P
=
G
d
f
when
f
The previous problem shows that we do not have to do any new calculations
b
to compute
a +
2
b
:
we still
Q when Q is a
cubic polynomial which agrees with
f
at a, b, and
have
lb
Q=
.
b-a
6
f(a)+4f
But there is much more lee-way in choosing
a+b
2
Q, which
+
f(b)
we can use
.
to our advantage:
396 Derivativesand Integrals
48.
there is a cubic polynomial function
(a) Show that
Q(b)= f(b),
Q(a)= f(a),
Q'
Q
(a+b
2
Hint: Clearly Q(x) P(x) + A(x
(b) Prove that for every x we have
=
f (x)
Q(x) (x
=
-
-
(
a)
x
-
a+b
(a+b
2
2
a+b
f'
=
Q satisfying
.
2
a)(x
-
a + b
-
(
b) x
2
(x
2
-
b)
a+b
for some A.
-
2
p(4)
4!
for some g in [a,b]. Hint: Imitate the proof of Problem 11-43.
(c) Conclude that if f (4) is continuous, then
/*
a
f
a+b
b-a
6
=
f(a)+4f
+
2
f(b)
-
(b-a)5
f(4)(C)
2880
for some c in [a,b].
(d) Now divide [a,b] into 2n intervals by means of the points
h=
ti=a+ih,
b-a
.
2n
Prove Simþson'smle:
f
=
b
-
a
f (a) +
-
for some ö in
[a,b].
4
(b
f (t2¿_i) +
-
a)5
2880n* f
(4)
2
f (t2;) + f (b)
Integral 397
19, Appendix.The Cosmopolitan
THE COSMOPOLITAN INTEGRAL
APPENDIX.
We originally introduced integrals in order to find the area under the graph of
considerably
more versatile than that. For example,
a function, but the integral is
Problem 13-24 used the integral to express the area of a region of quite another
used to exsort. Moreover, Problem 13-25 showed that the integral can also be
as we've seen in Appendix to Chapter 13, a
press the lengths of curves-though,
consider
the general case! This result was problot of work may be necessary to
ably a little more surprising, since the integral seems, at first blush, to be a very
in quite a
two-dimensional creature. Actually, the integral makes its appearance
will
present in this Appendix. To derive these
few geometric formulas,which we
formulas we will assume some results fom elementary geometry (andallow a little
fudging).
objects, we'll begin by tackling some
Instead of going down to one-dimensional
three-dimensional ones. There are some very special solids whose volumes can
of revolution,"
be expressed by integrals. The simplest such solid V is a
>
of
region
under
0 on [a, b] around
the graph
obtained by revolving the
f
the horizontal axis, when we regard the plane as situated in space (Figure 1).
(to,--.,In}
is any partition of [a,b], and m¿ and My have their usual
If P
j
"volume
FIGURE
i
=
meanings, then
xm,2
x
I
,i
a
/
i
t
..
\i
It i
is the volume of a disc that lies inside the solid V (Figure 2). Similarly,
xMi2(t;
ti-1) is the volume of a disc that contains the part of V between t;_;
and ti. Consequently,
-
&
s
n
x
FIGURE
n
m¿2(ti
-
ti-1)
M¿2(t¿
5 volume V 5 x
ti
i).
i=1
i=·l
2
-
But the sums on the ends of this inequality are just the lower and upper sums for
f2
on
[a,b]:
x
Vt
.
L( f2, P) 5 volume V 5 x
-
U( f
,
P).
Consequently, the volume of V must be given by
volumn
This
FIGURE
3
method
V
=
x
f (x)2dx.
of finding volumes is affectionately
referred to as the
"disc
method."
ægion
Figure 3 shows a more complicated solid V obtained by revolving the
under the graph of f around the verticalaxis (V is the solid left over when we start
with the big cylinder of radius b and take away both the small cylinder of radius a
and the solid Vi sitting right on top ofit). In this case we assume a > 0 as well
398 Derivativesand Integrals
as
f
0. Figures 4 and 5 indicate some other possible shapes for V.
>
FIGURE
4
FIGURE
For a partition P
rectangle
with base
=
5
"shells"
obtained by rotating the
t.} we consider the
[t;_i,t¿] and height mi or My (Figure 6). Adding the volumes
{to,
..
.
,
of these shells we obtain
m;(t¿2
x
..-
ti-12)
-
volume
<
V
<
'
'
which
we can write
ti-1)
-
< volume
V
<
M¿(t¿+t¿_i)(t¿
x
i=l
FIGURE
ti-12
as
m;(t¿+ti-i)(t¿
x
-
i-l
i=1
.
M¿(tg2
x
-
ti-1).
i=I
Now these sums are not lower or upper sums of anything. But Problem 1 of the
Appendix to Chapter 13 shows that each sum
6
-ti_i)
-t¿_i)
and
mit¿(t;
mit¿_i(t;
i=1
i=l
b
can
Ï
be made as close as desired to
small enough.
xf(x)dx
by choosing the lengths t¿
-
t;_i
The same is true of the sums on the right, so we find that
b
volume
V
2x
=
xf
(x)dx;
i
"shell
finding volumes.
The surface area ofcertain curved regions can also be expressed in terms ofintegrals. Before we taclde complicated regions, a little review of elementary geometric
formulas may be appreciated here.
Figure 7 shows a right pyramid made up of triangles with bases of length l and
altitude s. The total surface area of the sides of the pyramid is thus
this is the so-called
FIGURE
method"
of
1
2
7
-p5,
where p
is the perimeter of the base. By choosing the base to be a regular polygon
19, Appendix.The Cosmopolitan
Integral 399
with a large number
must be
of sides we see that the area of a right circular
1
2
cone
(Figure 8)
-(2xr)s
=
xrs,
"slant
height." Finally, consider the frustum of a cone with slant
s is the
height s and radii ri and r2 shown in Figure 9(a). Completing this to a cone, as
in Figure 9(b), we have
where
-
,--
N
r
FIGURE
si
8
si + s
--
-
ri
r2
SO
ris
si=
s
r25
si+s=
,
r2
.
r2
ri
-
ri
--
Consequently, the surface area is
*
r.
Kr2(Si
+
S)
2
Es,
Krisi
-
2
__
=
=
r2
Es(rl
‡ r2)-
ri
--
(a)
Now consider the surface formed by revolving the graph of f around the horizontal axis. For a partition P
(to, t,,) we can inscribe a series of frusta of
cones, as in Figure 10. The total surface area of these frusta is
=
...,
£1
x
[f (ti-1)+ f (ti)] (ti
--
ti-1)2 +
[f (ti) f (ti-1)]2
-
r:
=
x
1+
[f(t¿_i)+ f(t¿)]
f(ti)
¿-1)
-
2
r,
By the Mean Value Theorem, this is
(b)
FIGURE
9
r(Xi)
-
É
Q
i=l
for some x; in (ti_i,ti). Appealing to Problem 1 of the Appendix to Chapter 13,
we conclude that the surface area is
f(x)Jl +
2x
f'(x)2 dx.
PROBLEMS
i
y
1.
the volume of the solid obtained by revolving the region bounded
the
graphs of f (x) x and f (x) x2 around the horizontal axis.
by
(b) Find the volume of the solid obtained by revolving this same region
around the vertical axis.
(a) Find
=
/
FIGURE
lo
2.
Find the volume of a sphere of
=
radius
r.
400
Derivativesand Integrals
3.
When the ellipse consisting of all points (x, y) with x2þ2
rotated around the horizontal axis we obtain an
(Figure 11). Find the volume of the enclosed solid.
"ellipsoid
FIGURE
4.
(x
-
1 is
of revolution"
11
Find the volume of the
-
+ y2 b2
a)2 + y2
-
b2
(a
>
"torus"
(Figure 12), obtained by
b) around the vertical axis.
rotating
the circle
b
/
a
FIGURE
5.
12
A cylindrical hole of radius a is bored through the center of a sphere of
radius 2a (Figure 13). Find the volume of the remaining solid.
FIGURE
13
19, Appendix.The Cosmopolitan
Integral 401
6.
(a) For the
solid shown
in Figure 14, find the volume by the shell method.
COS
FIGURE
14
Write down
can also be evaluated by the disc method.
the integral which must be evaluated in this case; notice that it is more
complicated. The next problem takes up a question which this might
suggest.
(b) This
7.
volume
Figure 15 shows a cylinder of height b and radius f (b), divided into three
solids, one of which, V1, is a cylinder of height a and radius f (a). If f
is one-one, then a comparison of the disk method and the shell method of
computing volumes leads us to believe that
b
abf(b)2-und2-x
Ï
f(x)2dx=
volume
V2
yf¯1(y)dy.
=2x
(a)
Prove this analytically,
using the
formula for
f¯1 from Problem 19-15.
f(b)
FIGURE
B
A
8.
FIGURE
15
16
16 shows a solid with a circular base of radius a. Each plane
perpendicular to the diameter AB intersects the solid in a square. Using
similar to those already used in this Appendix, express the
arguments
VOlurie
of the solid as an integral, and evaluate it.
(b) Same problem if each plane intersects the solid in an equilateral triangle.
(a) Figure
402 Derivativesand Integrals
9.
Find the volume of a pyramid (Figure 17) in terms of its height h and the
area A of its base.
F1GURE
10.
FIGURE
18
17
volume
of the solid which
is the intersection of the two cylinders
the
in Figure 18. Hint: Find
intersection of this solid with each horizontal
Find the
plane.
FIGURE
I9
11.
that the surfàce area of a sphere of radius r is 47tr2.
(b) Prove, more generally, that the area of the portion of the sphere shown
in Figure 19 is 2xrrh. (Notice that this depends only on h, not on the
position of the planes!)
12.
(a) Find
(b) Find
13.
The graph of
(Figure 20).
(a) Prove
the surface area of the ellipsoid of revolution in Problem 19-3.
the surface area of the torus in Problem 19-4.
(a) Find the
f (x)
volume
(b) Show that
=
1/x, x :> l is revolved amund the horizontal axis
of the enclosed
"infinite
trumpet."
the surface area is infinite.
that
we fill up the trumpet with the finite amount of paint found
(c) Suppose
in part (a).It would seem that we have thereby coated the infinite inside
surface area with only a finite amount of paint! How is this possible?
FIGURE
20
PART
INFINITE
SEQUENCES
AND
INFINITE
SERIES
One of the most remarkable serits of
algebraic analysis is the following:
m(m
1+-x+m
2
1 2
-
m (m
‡
m(m
+
-
-
1) (m
1·2-3
1)
·
-
-
·
) x2
2)
-
x=
‡
[m (n
-
-
1)
-
x
1-2--...·-····»
‡
-
•
When m is a positivewhole number
th¢ sum of the series,
which is thenfinite,
can he expressed,
as is known, by (1 + x)
When m is not an integer,
the series goes on lo infmity,and it will
conwrge or divergeamordin.g
er the quantities
m and x havethis or that value.
In this care, one writer the same equality
.
m
1
(1‡x)=-1+-x
m(m
‡
.
.
.
-
1)
-2
1
x2‡·-·«¾
It is assumed that
numerical
equality will always ocur
the
whenever the series is cetrgent, but
this has neveryet beenproved.
NŒLS
HENRIK
ABEL
·
APPROXIMATION BY
POLYNOMIAL FUNCTIONS
CHAPTER
There is one sense in which the
If p is a polynomial function,
"elementary
functions" are not elementary at all.
p(x)=ao+alx+··-+anxn
then p(x) can be computed easily for any number x. This is not at all true for
1/t dt approximately,
functions like sin, log, or exp. At present, to find log x
we must compute some upper or lower sums, and make certain that the error
involved in accepting such a sum for log x is not too great. Computing ex
=
f,'
-
log¯'(x) would be even more difficult: we would have to compute log a for many
x-then
values of a until we found a number
a such that log a is approximately
a
would be approximately e2.
In this chapter we will obtain important theoretical results which reduce the
computation of f (x), for many functions f, to the evaluation of polynomial functions. The method depends on fmding polynomial functions which are close approximations to f. In order to guess a polynomial which is appropriate, it is useful
to first examine polynomial functions themselves more thoroughly.
Suppose that
p(x)=ao+aix+-·-+anx".
It is interesting, and for our purposes very important, to note that the coefficients a¿
can be expressed in terms of the value of p and its various derivativesat 0. To
begin with, note that
p(0)
=
ao-
Differentiating the original expression for p(x) yields
p'(x)=ai+2a2x‡·--‡nanXn-1
Therefore,
p'(0)
=
pfil(0)
=
at.
Differentiating again we obtain
Yn(n-1)-anxn-2
p"(x)=2a2+3-2-43x+--
Therefore,
p"(0)
=
p*(0)
2a2.
=
In general, we will have
p(k) (0)
405
(k
=
k! ak
¯
Or
Gk
406
and IrginiteSenes
InfiniteSequences
If we agree to define 0! 1, and recall the notation p(0) p, then this formula
holds for k 0 also.
in (x
If we had begun with a function p that was written as a "polynomial
=
=
=
-a),"
p(x)
=
+ at (x
ao
a) +
-
+ an(x
-·-
a)n
-
then a similar argument would show that
p(k)
Øk
=
°
kl
Suppose now that f is a function
(notnecessarily a
ff)(a),..., ff">(a)
all exist.
Let
f(k)
ak=
and
such
that
Oskšn,
,
k!
polynomial)
define
Pn,a(x)
ao + at
=
(x
a) +
-
-
-
-
+ an(x
-
a)".
The polynomial Pn.. is called the Taylor polynomial of degree n for f at a.
(Strictly speaking, we should use an even more complicated expression, like Pn,a,f,
to indicate the dependence on f; at times this more precise notation will be useful.)
The Taylor polynomial has been defined so that
Pn,a(k)(a)
f(k)(a) for 0 sks
=
n;
in fact, it is clearly the only polynomial ofdegree
sn with this property.
Although the coefficients of P.,a,y seem to depend upon f in a fairly complicated way, the most important elementary functions have extremely simple Taylor
polynomials. Consider first the function sin. We have
sin(0)
sin'(0)
0,
=
=
cos
O
1,
0,
=
-sin0
sin"(0)
sin'"(0)
sin(4)
=
=
-1,
cos 0 =
sin 0 = 0.
=
(0)
-
=
From this point on, the derivatives repeat in a cycle of 4. The numbers
at
are
sin >(0)
=
k!
1
1
0,
91
7!
Therefore the Taylor polynomial P2n+1,o of degree 2n + 1 for sin at 0 is
0, 1, 0,
---,
1
1
0,
0,
3!
5!
-,
3
P2n+i,o(x)=x--+
(Of course,
P2n+I,0
=
xxx
3!
22n+2,0)·
5
5!
-
-,
....
7
7!
-,
+···+(-1)"
x
2n+1
-
(2n + 1)!
20. Approximation
byPolynomialFunctions 407
The Taylor polynomial P2.,o of degree 2n for cos at 0 is (thecomputations aœ
left to you)
4
x2
P2n.o(x)
=
1
-
+
-
2!
gh
6
-
-
+
-
4!
6!
-
-
(-1)"
+
-
.
(2n)!
The Taylor polynomial for exp is especially easy to compute. Since expf*)(0)
exp(0)
1 for all k, the Taylor polynomial of degree n at 0 is
=
=
x2
x
Pn,o(x)=1+-+-+-+---+---+-.
l!
2!
x3
4
n
3!
4!
n!
The Taylor polynomial for log must be computed at some point a
log is not even defined at 0. The standard choice is a = 1. Then
log'(x)
1
=
-,
log'(l)
=
log"(1)
=
¢ 0,
since
1;
x
log"(x)
1
=
-
-1;
-,,
x
log'"(x)
2
=
log"'(l)
--,
2;
=
x
in general
(-Î)k-lk(k 1)!,
-
log(k)
__
k-1(k
log(k)()
-
1)!.
Therefore the Taylor polynomial of degree n for log at 1 is
(x 1)2 + (x 1)3 +---+ (-1)"¯1 , yn
Pn,1(x)=(x-1)2
3
n
It is often more convenient to consider the function f (x) log(1 + x). In this
0. We have
case we can choose a
-
_
-
=
=
ff >(x)
=
log(k)
f (k)(0)
=
log(k)
SO
k-1
(k
1)!
-
.
Therefore the Taylor polynomial of degree n for f at 0 is
x2
P.,o(x)
=
x
-
-
X4
x3
+
-
-
-
n-1Xn
_¡
+···+
3
4
2
n
There is one other elementary function whose Taylor polynomial is importantarctan. The computations of the derivatives begin
arctan'(x)
=
1
1 + x2
arctan'(0)
=
1;
arctan"(0)
=
0;
-2x
arctan"(x)
arctan'"(x)
=
=
(1 +
(1 +
x2)2
x2)2
,
+ 2x 2(1 + x2) 2
(1 + x2)4
(-2)
.
.
,
arctan"'(Q)
and InfmiteSeries
408 InfiniteSequences
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 polynomial Pn,a,y was
simply defined so as to have the
same first n derivatives at a as f, the connection
between f and P.,4, y w°a actua%y turn out tohe much Aceper.
One line of evidence for a closer connection between f and the Taylor polynomials for f may be uncovered by examining the Taylor polynomial of degree 1,
which is
P1..(x)
=
f (a) + f'(a)(x
a).
-
Notice that
f(x)
-
PI,a(x)
f(x)
=
x-a
-
f(a)
f'(a).
-
x-a
Now, by the definition of f'(a) we have
.
hm
x
a
f (x)
=0.
x-a
P2.o(x)
FIGURE
P1,a(X)
-
=
1+ x +
x"
I
In other words, as x approaches a the difference f(x)
small, but actually becomes small even compared to x
graph of
f(x)
-
-
Pi,a(x) not only becomes
a. Figure 1 illustrates the
ex and of
=
PI,o(x)
=
f (0) + J'(0)x
1 +x,
=
which
is the Taylor polynomial of degree 1 for f at 0. The diagram also shows
the graph of
P2,o(x)
which
=
f (0) + f'(0)
+
f"(0) x2
21
x2
=
1+ x +
,
is the Taylor polynomial of degree 2 for f at 0. As x approaches 0, the
difference f(x) P2,0(x) seems to be getting small even faster than the difference
-
20. Approximation
byPolynomialFunctions 409
Pi,o(x). As it stands, this assertion is not very precise, but we are now
prepared to give it a definite meaning. We have just noted that in general
f (x)
-
f (x)
lim
x-a
For
f (x)
=
e2 and a
PI,a
-
-
x
(x)
0.
=
a
0 this means that
=
.
hm
eX-1-x
f(x)-Pl.o(x)
.
hm
=
x->o
0.
=
x-o
x
x
On the other hand, an easy double application ofTHôpital's Rule shows that
e2-1-x
lim
1
=
¢ 0.
-
x2
x-o
2
Thus, although f (x) Pi,o(x) becomes small compared to x, as x approaches 0, it
does not become small compared to x2. For P2.o(x) the situation is quite difTerent;
the extra term x2/2 provides just the right compensation:
-
x
e2-1-x--
lim
2
2
x2
x->0
ex
lim
=
x-so
and
f (x)
(x
x-va
in fact, the analogous
THEOREM
l
f"(a) exist,
P2,a(X)
-
lim
=
a)2
--
0;
assertion for Pn,a is also true.
Suppose that f is a function for which
f'(a),..., ff">(a)
all exist. Let
ag=
and
f(k)
,
k!
05ksn,
define
Pn,a(x)=ao+al(x-a)+---+a.(x-a)".
Then
.
Inn
x-a
f (x)
Pn,a(x)
--
(x
=
-
a)"
x
-
=
2
x-vo
f'(a)
l
2x
ex-1
lim
=
This result holds in general-if'
---
0.
0.
then
and InfiniteSeries
410 InfiniteSequences
PROOF
Writing out Pn,a(x) explicitly, we obtain
f(x)-"
f(x)
f (a)(x-a)'
Pn,a(x)
-
il
i=0
f(")(a)
¯
¯
(x
a)"
-
(x
n!
a)"
-
It will help to introduce the new functions
n-1
Q(x)
f (a) (x
2
=
a)'
-
and
g(x)
(x
=
'
i=0
-
a)";
now we must prove that
f(x)
f(">(a)
Q(x)
-
n!
g(x)
x->a
Notice that
Q"W)--ff')(a),
g(k)(X)
=
NÎ(X
k_<n-1,
a)n-k/(n
-
k)!
-
.
Thus
Q(x)]= f(a)- Q(a)=0,
Q'(x)]= f'(a) Q'(a)=0,
lim[f(x)-
X->d
lim[f'(x)
lim[ f
-
-
"¯©(x)
Qi"¯©(x)] f(n-2)
=
-
(n-2)(a)
_
=
0.
and
lim g(x)
lim g'(x)
=
x- a
=
-
-
(x)
hm f
Q(x)
-
.
(x
x-a
Since Q is a polynomial of degree n
(.-1)
fact, Q("¯1;
(a). Thus
x- a
f (x) Q(x)
and this
(x
-
a)n
-
0.
1 times to obtain
f("¯1)(x) g(n-1)
n! (x a)
1, its (n
-
.
=
.
-
-
lim
=
_
lim
=
a)"
--
lim g(n-2)(x)
x-va
We may therefore apply l'Hôpital's Rule n
x->a
=
-
x->a
lun
x- a
-
ff"¯1)
n!
1)st derivative is a constant;
(n-1)(a)
_
(x
in
-
a)
last limit is ff")(a)/n! by definition of ff">(a).
One simple consequence of Theorem 1 allows us to perfect the test for local
and minima which was developed in Chapter 11. If a is a critical point
of f, then, according
to Theorem 11-5, the function f has a local minimum
0 no
at a if f"(a) > 0, and a local maximum
at a if f"(a) < 0. If f"(a)
sign
of
might
conclusion was possible, but it is conceivable
give
that the
f"'(a)
maxima
=
by PolynomialFunctions 411
20. Approximation
0, then the sign of f(4)(a)
further information; and if f'"(a)
significant. Even more generally, we can ask what happens when
=
f'(a)
(*)
f
f"(a)
=
f">(a)
=
-
-
f
=
-
("¯1)(a)
=
0 might be
0,
=
¢ 0.
The situation in this case can be guessed by examining the functions
f(x)
g(x)
(x
=
a)",
-
--(x
=
a)",
-
which satisfy (*). Notice (Figure 2) that if n is odd, then a is neither
a local
minimum
maximum
the
other
point
On
local
for
hand,
if n is
nor a
f or g.
with
nth
minimum
at a, while
derivative, has a local
a positive
even, then f,
g, with a negative nth derivative, has a local maximum at a. Of all functions
satisfying
these are about the simplest available; nevertheless they indicate the
situation
exactly. In fact, the whole point of the next proof is that any
general
looks very much like one of these functions, in a sense that
fimction satisfying
(*),
(*)
is made precise by Theorem 1.
THEOREM
2
Suppose that
f'(a)=···= ff"¯1)(a)=0,
f(")(a) ¢ 0.
is even and f("3(a)> 0, then f has a local minimum at a.
(2)If n is even and f("3(a)< 0, then f has a local maximum at a.
(3)If n is odd, then f has neither a local maximum nor a local minimum
(1)If
PROOF
n
at a.
There is clearly no loss of generality in assuming that f (a) 0, since neither the
hypotheses nor the conclusion are affected if f is replaced by f
f (a). Then,
since the first n 1 derivatives of f at a are 0, the Taylor polynomial Pn,a of f is
=
-
-
P..a(x)
(a)
f (a) +
=
n odd
(n)
f'(a)
1! (x
(x
=
n!
-
-
a) +
-
-
-
+
f (") (a)
n!
(x
-
a)"
a)".
Thus, Theorem 1 states that
0
I
=
lim f(x)
(x
x-a
6
Pn,a(x)
-
=
-
a)=
f(x)
lim
(x
x-a
-
a)"
-
f("'(a)
.
n!
Consequently, if x is sufficiently close to a, then
f(x)
¯
(b)
FIGURE
a even
2
U
,
has the same sign as
f(">(a)
.
Suppose now that n is even. In this case (x a)" > 0 for all x ¢ a. Since
has the same sign as ff"3(a)/n!for x sufficiently close to a, it follows
f(x)/(x
-
-a)"
and hrfiniteSeries
412 InßniteSequences
f"(a)/n! for
itself has the same sign as
this means that
>
0,
ff">(a)
that
f(x)
f(x)
fOr X
close to a. Consequently,
the case f(n)(a) < 0.
-Le
JM=
0,
f
0
>
=
If
has a local minimum at a. A similar proof works
that n is odd. The same argument
sufficiently close to a, then
(a)
close to a.
f(a)
for
Now suppose
x=0
x sufficiently
f (x)
(x a)"
as
before shows that if x is
has the same sign.
always
---
But (x a)" > 0 for x > a and (x a)" < 0 for x < a. Therefore f (x) has diferent
signs for x > a and x < a. This proves that f has neither a local maximum nor
a local minimum at a.
-
-
/W
-
0,
x=0
Although Theorem 2 will settle the question of local maxima and minima for
just about any function which arises in practice, it does have some theoretical
limitations, because ff*)(a)may be 0 for all k. This happens (Figure 3(a))for the
function
e-1/x2,
0
x
f (x)
0,
0,
x
(b)
-/
=
=
which has a minimum
at 0, and also for the negative of this
which has a maximum at 0. Moreover (Figure 3(c)), if
function (Figure 3(b)),
1¡x1¡P
f (x)
=
O
(x)
=
e
f(k)(0)
=
0 for all k, but f has neither a local minimum
The conclusion
FIGURE
cept of
a
at a
"order
nor a
local maximum
in terms of an important conand g are equal up to order n
of Theorem 1 is often expressed
of equality." Two fimetions
if
f (x)
x-a
(x
f
g(x)
--
=
a)"
-
0.
In the language of this definition, Theorem 1 says that the Taylor polynomial
Pn,a.y equals f up to order n at a. The Taylor polynomial might very well have
been designed to make this fact true, because there is at most one polynomial of
degree 5 n with this property. This assertion is a consequence of the following
elementary
THEOREM
s
theorem.
Let P and Q be two polynomials in (x a), of degree 5 n, and suppose that P
and Q are equal up to order n at a. Then P
Q.
-
=
PROOF
Let R
=
P
-
Q. Since
R is a polynomial of
degree 5 n, it is only necessary to
20. Approximation
by PolynomialFunctions 413
prove that if
R (x)
bo +
=
·
-
+ b,,(x
-
a)"
-
satisfies
R(x)
lim
=
(x
x->a
then R
x->a
=
R (x)
.
(x
-
0
=
a)
for 0
0 this condition reads simply lim R(x)
lim R(x)
lim [bo+ bi (x
=
=
<
n.
0; on the other hand,
=
a) +
-
i
<
-
-
-
+ b.(x
-
a)"]
bo.
=
Thus bo
0,
0. Now the hypotheses on R surely imply that
=
lim
For i
a)"
-
0 and
R(x)=bi(x-a)+-·-+b.(x-a)".
Therefore,
R(x)
x
-
=bi+b2(x-a)+--·+b.(x--a)"'¯I
a
and
lim
24.
Thus bi
=
R(x)
=
x
-
bi.
a
0 and
R(x)
=
b2(x
-
a) +
-·-
+ b,(x
--
a)".
Continuing in this way, we find that
bo=---=b.=0.
conollAny
be n-times differentiable at a, and suppose that P is a polynomial in (x a)
of degree 5 n, which equals f up to order n at a. Then P
P.,,,7.
Let
f
-
=
PROOF
Since P and Pn,a,I both equal f up to order n at a, it is easy to see that P equals
Pra,a,ÿ by the Theorem.
P.,a, ÿ up to order n at a. Consequently, P
=
At first sight this corollary appears to have unnecessarily complicated hypotheses;
it might seem that the existence of the polynomial P would automatically imply
that f is sufficiently differentiable for Pn,a,y to exist. But in fact this is not so. For
example (Figure 4), suppose that
xn+1,
f (x)
=
0,
x irrational
x rational.
0, then P is certainly a polynomial of degree 5 n which equals f up to
order n at 0. On the other hand, f'(a) does not exist for any a ¢ 0, so f"(0) is
If P(x)
=
undefined.
and InfiniteSeries
414 InfiniteSequences
O
,*
•,
..•
•.,
FIGURE
x",
x irrational
o,
x rational
4
When f does have n derivatives at a, however, the corollary may provide a
useful method for finding the Taylor polynomial of
f. In particular, remember
that our first attempt to find the Taylor polynomial for arctan ended in failure.
2
The equation
1
arctan x
dt
1 + t2
=
a promising method of fmding a polynomial
t2,
by 1 +
to obtain a polynomial plus a remainder:
suggests
1
1+t2
=
1
-
4
t2
6
_
n 2n
_
close to arctan-divide
(-1)"*E
t
1
*2
1+t2
both sides by 1 + t2,
This formula, which can be checked easily by multiplying
shows that
arctanx
=
t2 + t4
Lx
1
-
x
-
-
3
+
-
x
dt + (-1)"**
-
-
-
-
5
+
(-1)"
2n+1
+
th+2
1+t2
72n+1
5
x3
=
n 2n
_
dt
x 12n+2
(-1).+1
l+t2
dt.
According to our corollary, the polynomial which appears here will be the Taylor
polynomial of degree 2n + 1 for arctan at 0, provided that
x
t2n+2
1 + t2
.
lun
dt
0.
=
X2n+l
x->O
Since
x
t2n+2
1+t2
x
dt
5
2n+3
t2"*2 dt
=
2n+3
,
this is clearly true. Thus we have found that the Taylor polynomial of degree
2n + 1 for arctan at 0 is
=
x
-
--
3
x2"**
x5
x3
P2n+1,o(x)
+
-
-
5
-
-.
+
(-1)" 2n+1
.
20. Approximation
by PolynomialFunctions 415
By the way, now that we have discovered the Taylor polynomials of arctan,
possible to work backwards and find arctant*)(0) for all k: Since
P2n+1,o(x)
and since this
arctant
=
x
x
-
3
+
-
3
x
5
-
-
-
-
·
5
x
2n+1
(-1)" 2n+1
+
it is
,
polynomial is, by definition,
arctan(2)(0)
3(0) + arctant)(0)
we can find arctan(*)
+
+
21
simply
(0) by
arctan***)(0)
x2
-
+
-.
x
(2n + 1)!
of xk
equating the coefficients
in
**,
tÏ1eSe
LWO
polynomials:
arctan(k)
(0)
=
kl
arctan(2t+l)(0)
(-1)'
_
or
21 + 1
(21+ 1)!
0
if k is even,
arctan(2'*)(0)
=
(-1)' (2l)!.
-
A much more interesting fact emerges if we go back to the original equation
x3
arctan x
=
and remember
x
-
-
3
x5
+
-
x2n+1
-
-
-
-
5
+ (-1)"
2n+1
+ (-1)"**
x 72.42
1+t2
dt,
the estimate
t2n+2
x
1+t2
|X
dts
|2n+3
.
2n+3
When |x| 5 1, this expression is at most 1/(2n + 3), and we can make this as
small as we like simply by choosing n large enough.
In other words, for |x| 5 1
the
Taylorpolynomials
we can use
for arctan to compute arctanx as accurately as we like.
The most important theorems about Taylor polynomials extend 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,aforfixed
n, as x approaches a. Henceforth we will compare Taylor
polynomials P..a for fixedx, and different n. In anticipation of the coming theorem
we introduce some new notation.
If f is a function for which P.,,
R..a(x) by
f (x)
=
(x) exists,
define the remainder
term
P.,a(x) + R.,a(x)
f (">(a)
-a)+··-+
=
we
f(a) + f'(a)(x
n!
-a)"+
Rn,a(x)-
(x
We would like to have an expression for Rn,a (x) whose size is easy to estimate.
There is such an expression, involving an integral, just as in the case for arctan.
One way to guess this expression is to begin with the case n 0:
=
f(x)
=
f (a) +
Ro,a(x)-
416
and infmite Senes
InfmiteSequences
The Fundamental Theorem of Calculus enables us to write
X
f (x) f (a) +
=
Ï f'(t)
dt,
a
so that
Ro,a(x)
lx
f'(t) dt.
=
A similar expression for RI,a(x) can be derived from this formula
tion by parts in a rather tricky way: Let
u(t)
that
(notice
x represents
f'(t)
=
some
and
v(t)
=
t
using
integra-
x
-
fixed number in the expression for v(t), so v'(t)
1);
=
then
f'(t)
dt
f'(t)
=
1 dt
-
v'(t)
u(t)
x
x
=u(t)v(t)
Since v(x)
f"(t)(t-x)dt.
-
0, we obtain
=
X
f (x) f
=
(a) +
Ï
f'(t)
dt
X
Ï f"(t)(x-t)dt
Ï f"(t)(x
f(a)-u(a)v(a)+
=
x
f (a) + f'(a)(x
=
a) +
-
Thus
R1,a(x)
f"(t)(x
=
Ja
t) dt.
-
t) dt.
-
It is hard to give any motivation for choosing v(t)
t x, rather than v(t)
t.
It just happens to be the choice which works out, the sort of thing one might
discover after sufRciently many similar but futile manipulations. However, it is
now easy to guess the formula for R2,a(x). If
=
t)2
-(x
-
u(t)
then v'(t)
=
(x
-
=
f"(t)
and
=
,
2
t), so
x
x
Ï
v(t)
=
-
f"(t)(x
-
t)dt
=
=
u(t)v(t)
f"(a)(x
-
This shows that
x
R2,a(X)
=
_
f"'(t)
a)2
x
(3
(x
You should now have little difficulty giving a
-
dt
-
2
r"(t)
+
2
g)2
_(,
x
-
2
(x
-
t)'dt.
t)2dt.
rigorous
proof, by induction, that
byPolynomialFunctions 417
20. Approximation
if f (n+13 is continuous
on
then
[a,x],
Rn,a(x)
lx
=
(n+1)(t)
(x
-
t)" dt.
From this formula, which is called the integral form of the remainder, it is possible
(Problem 15) to derive two other important expressions for Rn,a(x): the Cauchy
form of the remainder,
Rn.a(x)
f(n+1)(t)
=
n!
(x
t)"(x
-
for some t in (a, x),
a)
-
Lagrange form of the remainder,
and the
Rn,a(x)
=
f (n+1)
(x
(n + 1)!
-
a)"**
for some t in (a, x).
In the proofof the next theorem (Taylor's Theorem) we will derive all three forms
of the remainder
from
in an entirely different way. One virtue of this proof (aside
of
and
remainder
cleverness)
the
that
the
the
forms
is
fact
Cauchy
Lagrange
its
will be proved without assuming the extra hypothesis that f("**)is continuous.
In
this way Taylor's Theorem appears as a direct generalization of the Mean Value
Theorem, to which it reduces for n 0, and which is the crucial tool used in the
proof.
These remarks may suggest a strategy for proving Taylor's Theorem. Since
Rn.a(a)
0, we might try to apply the Mean Value Theorem to the expression
=
=
Rn,a(x)
Rn,a(x)
x-a
-
Rn,a(a)
x-a
On second thought, however, this idea does not look very promising, since it is not
at all clear how f ("**) (t) is ever going to be involved in the answer. Indeed, if we
take the most straightforward
route, and differentiate both sides of the equation
which defines R.,a, we obtain
f'(x)
=
f'(a) + f"(a)(x
a) +·-
-
-
+
f (">(a)
(x
(n 1)
-
a)n-1 + R.,a'(x),
-
is useless. The proper application of the Mean Value Theorem has a lot in
common with the integration by parts proof outlined above. This proof involved
the derivative of a function in which x denoteda number which was fixed.This is just
how x will be treated in the following proof.
which
THEOREM
4 (TAYIDR'S THEOREM)
Suppose that f',
.
.
.
,
f(n+1)are defined
-a)+-·-+
f(x)= f(a)+ f'(a)(x
on
[a,x],
and that R.,,(x)
f (">(a)
n!
(x -a)"+R.,a(x).
is defmed by
418
and InfmiteSenes
Infinite Sequences
Then
(=+1)
Rn,a(x)
=
(2) Rn,a(x)
=
(1)
(x
n!
(x
Moreover, if f(n+1)is integrable on
Rn,a(x)
Ï
=
--
a)"+1
-
for some t in (a,x).
a)
for
t in
some
(a, x).
then
[a,x],
(n+1)(t)
x
(3)
t)"(x
-
(x
n!
a
t)" dt.
-
a, then the hypothesis should state that f is (n + 1)-times differentiable on
a];
number t in (1)and (2)will then be in (x, a), while (3)will remain true
the
[x,
stated,
provided that f(n+1)is integrable on [x,a].)
as
(If x
PROOF
<
For each number t in
[a,x]
we have
f (x) f (t) + f'(t)(x
=
Let us denote the number
we have
t) +
-
-
-
-
+
f (">(t)
(x
n!
-
t)" + R.,,(x).
R.,,(x) by S(t); the function S is defined on
[a,x],
and
f(">(t)
-t)+---+
(*) f(x)= f(t)+ f'(t)(x
n!
(x -t)"+S(t)
for all t in
[a,x].
We will now differentiate both sides of this equation, which asserts the equality of
two functions: the one whose value at t is f (x), and the one whose value at t is
f (">(t)
f(t)+.--+
n!
(x-t)"+S(t).
"as
parlance we are considering both sides of (*)
a function of t".)
make
notice
that if
sure that the letter x causes no confusion,
Justto
(In common
g(t)
f (x)
=
for all t,
then
g'(t)
and
0
=
for all t;
if
g(t)
f(k)(t)
=
k!
k
'
then
gr(7)
Í(k)
=
k(x
k!
=
-
f (k)
(k
-
-
t)k-1
(k+1)
(X
-
kl
(k+l)
1)! (x
--
t)k
(X
k!
-
f)k
1)&
20. Approximation
by PolynomialFunctions 419
Applying these formulas to each term of
0
=
f'(t)
+
-
f'(t)
f"(t)
+
obtain
(*),we
f(3)(t) "¯'
f"(t)
-
1!
1!
2
2!
-f">(t)
+
-
-
-
+
(n
1). (x
-
In this beautiful formula practically everything
S'(t)
=
(x
n!
f"*0(t)
+
(x
n!
sight
in
f (n+1)
-
i
t)""
-
-
t)"
cancels out, and we obtain
t)".
-
Now we can apply the Mean Value Theorem to the function S on
is some t in (a, x) such that
S(x)
S(a)
-
S'(t)
=
x
=
f (n+1)
-
(x
n!
a
-
+ S'(t).
t)
-
[a,x]:
there
.
Remember that
S(t)
=
R..,(x);
this means in particular that
S(x)
S(a)
R.,x(x)
Rn,a(x).
=
=
0,
=
Thus
0
f'"
Rn.a(x)
--
=-
"(t)
(x-t)
n!
x-a
.
or
R.,a(x)
ff"**)(t)
=
(x
n!
t)"(x
-
a);
-
this is the Cauchy form of the remainder.
To derive the Lagrange form we apply the Cauchy Mean Value Theorem
the functions S and g(t) = (x t).+1: there is some t in (a, x) such that
-
f (n+1)
-
S(x)
-
S(a)
n!
S'(t)
(x
Thus
_
f(n+1)(t)
¯
(n + 1)!
-
or
Rn,a(x)
which
is the Lagrange form.
=
ff"**)(t)
(x
(n + 1)!
t)"
°
-(n+1)(x-t)=
g(x)-g(a)
Rn,a(x)
(x a).41
-
-
a)"**,
to
and InßniteSeries
420 InßniteSequences
Finally, if f (n+1) is integrable on
[a,x],
then
x
S(x)
-
S(a)
Ï
=
(n+1)(t)
x
S'(t)
=
ni
a
Of
R..a(x)
f("+0(t)
L'
=
(x
-
n!
(x
t)"dt
-
t)" dt.
-
Although the Lagrange and Cauchy forms of the remainder are more than
theoretical curiosities (see,e.g., Problem 23-18), the integral form of the remainder
will usually be quite adequate. If this form is applied to the functions sin, cos, and
0, Taylor's Theorem yields the following formulas:
exp, with a
=
x
3
sinx=x--+--...+(-1)"
x
3!
5
x
5!
2n+1
(2n + 1)!
x
X2n
x2
x4
cosx=1--+-----+(-1)"
4!
2!
x2
t)2.+1dt,
+
dt,
(x-t)
(2n)]
o
xt
n
-(x-t)"dt.
n!
2!
-
x COS(2n+1)
(2n)!
e2=1+x+-+---+-+
sin(2n+2)
(x
(2n + 1)!
+
n!
To evaluate any of these integrals explicitly would be supreme foolishness-the
answer of course will be exactly the difTerenceof the left side and all the other terms
on the right side! To estimate these integrals, however, is both easy and worthwhile.
The first two integrals are especially easy. Since
"
| sin
we
(t)| 5 1 for all
t,
have
x
(2n+2)(t)
(2n +
1)! (x
dt
t)
-
2(x
1
<
-
¯
(2n + 1)!
t)
Since
_
fx
(x
-
t)
2n+1
dt
_
2n+2 t=x
=
2n + 2
t=0
x2n+2
¯2n+2
we conclude
that
x gg (2n+2)(t)
(x
(2n + 1)!
-
t) gy
dt
5
|x|2n+2
.
(2n + 2)!
**
dt
.
20. ApproximationbyPolynomialFunctions 421
Similarly, we can show that
x
2.+1
cos(2n+1)
(x
(2n)!
-
dt
t)
<
.
¯¯
(2n+ 1)!
These estimates are particularly interesting, because (asproved in Chapter 16) for
any e > 0 we can make
Xn
n!
by choosing n large enough
(howlarge n must be will depend on x). This enables
sin x to any degree of accuracy desired simply by evaluating the
pmper Taylor polynomial Pa.o(x). For example, suppose we wish to compute
sin 2 with an error of less than 10-4. Since
us to compute
sin 2
where
P2n+1,0(2) + R,
=
we can use P2n+l,o(2) as our answer,
|R| 5
22n+2
(2n + 2)!
,
provided that
22n+2
(2n + 2)!
10¯*.
<
search-it
obA number n with this property can be found by a straightforward
viously helps to have a table of values for n! and 2"
428).
this
In
case it
(seepage
works,
that
that
happens
5
n
so
=
sin
2
=
Pi l.o (2) + R
23
25
27
29
211
31
5!
7!
9!
10-4
11!
=2---+---+----+R
where
|R
<
'
It is even easier to calculate sin 1 approximately, since
sin 1
=
P2n+1,o(1) + R,
where
|R|
<
1
.
(2n + 2)!
To obtain an error less than s we need only find an n such that
1
(2n + 2)!
'
and this requires only a brief glance at a table of factorials. (Moreover, the inditerms of P2n+1,o(1) will be easier to handle.)
For very small x the estimates will be even easier. For example,
vidual
sin
To obtain
with n =
-
1
10
=
P2,+1,o
IRI < 10¯1o we
-
1
+ R,
10
where
|R|
<
1
102n+2(2n+ 2)!
.
4 (andwe could even get away
can clearly take n
used
actually
3). These
are
to compute tables of sin and cos.
A high-speed computer can compute P2, i,o(x) for many different x in almost no
time at all.
methods
=
422
and InfiniteSenes
InßniteSequences
Estimating the remainder
for e2 is only slightly harder. For simplicity assume
>
that x
0 (theestimates for x < 0 are obtained in Problem 10). On the interval
[0,x] the maximum value of e' is e2, since exp is increasing, so
xt
n!
(x
t)" dt
-
Since we already know that e
<
xXn+1
x
x
--
(x
---
¯
n!
t)" dt
-
=
.
(n + 1)!
4, we have
<
xx"*I
42x"*I
(n + 1)!
'
(n + 1)!
'
which can be made as small as desired by choosing n sufficiently large. How large
the factor 42 will make things more difficult).
n must be will depend on x
1, then
Once again, the estimates are easier for small x. If 0 Exs
(and
x2
e2=1+x+-+--.+-+R,
x"
n!
2!
4
where0<R<
.
(n + 1)!
(The inequality 0 < R follows immediately from the integral form for R.) In
particular, if n = 4, then
4
1
0<R<
<-,
5! 10
so
e=e1=1+1+-+-+-+R,
65
24
1
3!
1
2!
1
1
10
where0<R<-
4!
=-+R
17
24
2+--+R,
which shows that
2<e<3.
(This then shows that
0<R<
32x"**
(n + 1)!
,
allowing us to improve our estimate of R slightly.)
compute that the first 3 decimals for e are
e
=
2.718
By taking n
=
7 you can
...
(youshould check that n = 7 does give this degme of accuracy, but it would be
cruel to insist that you actually do the computations).
The fimction arctan is also important but, as you may recall, an expression for
arctanf*)(x) is hopelessly complicated,
so that the integral form of the remainder
useless.
the
other
hand, our derivation of the Taylor polynomial for arctan
is
On
automatically provided a formula for the remainder:
x3
arctanx=x¯¯
(¯·1)"22n+1
2
'
3
2n+1
Jo
(-1).
1,2n+2
1+t2
dt.
20. Approximation
byPolynomialFunctions 423
As we have already estimated,
n+lg2n+2
_¡
x
x
dt
1+t2
5
t
2n+3
2n+2
dt
=
we will consider only numbers x with
term can clearly be made as small as desired
For the moment
remainder
.
2n+3
|x|
5 1. In this case, the
by
choosing
n sufficiently
large. In particular,
1
5
1
3
arctanl=1--+--...+
(-1)"
2n+1
1
2n+3
where|R|<
+R,
.
With this estimate it is easy to find an n which will make the remainder less than
any preassigned number; on the other hand, n will usually have to be so large as to
make computations
hopelessly long. To obtain a remainder < 10 , for example,
take
>
(104
3)/2. This is really a shame, because arctan 1 = x/4,
must
n
we
polynomial
for arctan should allow us to compute x. Fortunately,
so the Taylor
there are some clever tricks which enable us to surmount these difficulties.Since
-
|R2n+1,c(x)\
<
,
2n + 3
n's will work for only somewhat smaller x's. The trick for computing
is
to express arctan 1 in terms of arctan x for smaller x; Problem 6 shows how
x
this can be done in a convenient way.
much smaller
1 is best
The Taylor polynomial for the function f (x)
log(x + 1) at a
handled in the same manner as the Taylor polynomial for arctan. Although the
integral form of the remainder for f is not hard to write down, it is difficultto
estimate.
On the other hand, we obtain a simple formula if we begin with the
=
=
equation
1
l+t
1
=
t+ t
-
-
-
-
-
(-1)"¯1,n-i q (-1)"t"
+
.
l+t
this implies that
x
log(1 + x)
1
-
1+ t
x2
dt
=
x
-
--
2
x3
+
-
-
-
-
3
-
+
(-1)"¯1
Xn
n
+
(-1)"
1+ t
dt,
-1.
for all x
>
If x
>
0, then
x
t+1
Xn+1
x
n
dt 5
t" dt
=
,
n+1
-1
< x < 0 (Problem 11).
is a slightly more complicated estimate when
For this function the remainder term can be made as small as desired by choosing
< x 5 1.
n sufficiently large, provided that
and there
-1
424
and InfiniteSeries
InßniteSequences
The behavior of the remainder terms for arctan and f(x)
another
when
matter
|x|
>
=
log(x + 1) is quite
1. In this case, the estimates
|R2n+1,o(x)|
|R..o(x)|
|x|2n+3
<
<
for arctan,
2n + 3
(x > 0) for f,
n+ 1
1 the
bounds x"'/m become large as m beare of no use, because when |x| >
unavoidable,
and is not just a deficiency of our
predicament
is
comes large. This
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 arctan,
that if t is in [0,x] (orin [x,0] if x < 0), then
1+t251+x252x2,
so
t2n+2
x
1+t2
Similarly, if x
>
0, then for
dt
à
g
2x2
if|x
note
1,
h+1
x
t2"*2dt
-
=
.
4n+6
o
t in [0,x] we have
1+tsl+xs2x,
1,
ifx
SO
t+1
dt à
t" dt
-
2x
=
.
2n+2
These estimates show that if |x| > 1, then the remainder terms become large as
n becomes large. In other words, for lx| > 1, the Taylor polynomials for arctan
and f are ofno use whatsoeverin computing arctanx 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 |x < 1.
This same situation occurs in a spectacular way for the function
e¯1/x2,
f (x)
=
0,
x
x
We have already seen that ff*)(0) 0 for every
that the Taylor polynomial Pn,0 for f is
=
=
=
f (0) + f'(0)x +
0.
natural
number
f (0)
f (") (0)
2!
n!
"
Pn,o(x)
¢0
=
k. This means
0.
In other words, the remainder term R..o(x) always equals f(x), and the Taylor
polynomial is useless for computing f (x), except for x = 0. Eventually we will be
able to ofFer some explanation for the behavior of this fimction, which is such a
disconcerting illustration of the limitations of Taylor's Theorem.
has been used so often in connection with our estimates
The word
for the remainder term, that the significance of Taylor's Theorem might be misconstrued.
It is true that Taylor's Theorem is an almost ideal computational aid
"compute"
20. ApproximationbyPolynomialFunctions 425
failure in the previous example), but it has equally important theoretical consequences. Most of these will be developed in succeeding
chapters, but two pmofs will illustrate some ways in which Taylor's Theorem may
be used. The first illustration will be particularly impressive to those who have
waded through the proof, in Chapter 16, that x is irrational.
(despiteits ignominious
THEOREM
5
PROOF
e is irrational.
We know that, for any n,
1
1
e=el=1+--+-+··-+--+R.,
1
n!
2!
11
Suppose that e were rational, say e
Choose n > b and also n > 3. Then
where0<R.<
.
a/b, where a and b are positive integers.
=
1
a
3
(n + 1)!
1
-=l+1+-+·--+-+R.,
b
2!
n!
so
n!a
n!
n!
--=nl+n!+-+---+-+n!R..
b
2!
n!
Every term in this equation other than n!R, is an integer (theleft side is an integer
because n > b). Consequently, nlR, must be an integer also. But
3
(n + 1)!
0<R,<
,
so
0<n!R.<
which is
3
<-<1,
n+1
3
4
impossible for an integer.
The second illustration is merely a straightforward
proved in Chapter 15: If
f"
demonstration of a fact
f 0,
f (0) 0,
f'(0) 0,
+
=
=
=
then
f
=
0. To prove this, observe first that f(k) exists for every k; in fact
f*
=
f(4)_
f
(5
etc.
-
(f")'
=
3r_
U(4)
f',
-
tr__
_
r
_
pr
n_L
426
Infinite Sequences
and InßniteSeries
This shows, not only that all ff*) exist, but also that there are at most 4 different
Since f(0) f'(0)
0, all ff*)(0)are 0. Now Taylor's
ones: f, f',
Theorem states, for any n, that
-f'.
-f,
=
f (x)
=
n!
|f("*)(t)|<M
add the
phrase
"and
t)" dt.
-
(sinceff"*2)exists),
Each function f(n+13
is continuous
there is a number M such that
(wecan
(x
=
so for any particular x
for0<t<x,andalln
all n" because there are only four different
f(k)
Thus
t)"
5 M
|f(x)|
ni
dt
=
M |x|"**
.
(n + 1)!
Since this is true for every n, and since |x|"/n! can be made as small as desired by
choosing n sufficiently large, this shows that
f (x)| Se for any e > 0; consequently,
0.
f (x)
=
The other uses to which Taylor's Theorem will be put in succeeding chapters
considerations
which have concerned us
are closely related to the computational
much
chapter.
remainder
of
this
the
for
If
term Rn.a(x) can be made as small as
sufficiently
desired by choosing n
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. If we are willing to add
up infinitely many terms (intheory at least!), then we ought to be able to ignore
the remainder completely. There should be
sums" like
"infmite
X3X5X'
.
smx=x---+----+-
-,
3!
5!
x2468
cosx=l---+-----+----·-,
2!
7!
4!
6!
x2
e2=1+x+-+-+-+--.,
2!
arctan x
=
x
-
x
3
-
3
+
x2
log(1+x)=x---+----+...
2
x
3!
5
--
--
8!
74
x3
x
4!
7
--
5
7
73
74
+
·
·
-
if |x| < 1,
¯
-l<xs1.
3
4
if
We are almost completely prepared for this step. Only one obstacle remainswe have never even defined an infinite sum. Chapters 22 and 23 contain the
necessary definitions.
20. ApproximationbyPolynomialFunctions 427
PROBLEMS
1.
Find the Taylor polynomials (ofthe indicated degree, and at the indicated
point) for the following functions.
(i) f (x)
=
e*; degree 3, at 0.
(ii) f (x)
=
esi"' degree 3, at 0.
(iii) sin;
degree 2n, at
.
(iv) cos; degree 2n, at z.
(v) exp; degree n, at 1.
(vi) log; degree n,
(vii)f(x)
(viii)f (x)
2.
=
x5 4,3
=
x5 + x3 + x; degree
(ix) f (x)
=
(x) f(x)
=
1
1+ x
+ x; degree 4, at 0.
2;
4, at 1.
degree 2n + 1, at 0.
1
degree n, at 0.
1+ x ;
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?)
-
(i)
2
x
x4
(ii)
5
(iii) x
-
4x
-
9.
12x3 + 44x2 + 2x + 1.
-
.
(iv) ax2 +
3.
at 2.
bx + c.
Write down a sum
numbers
to within
notation)
(using
the
specified
tion, consult the tables for 2"
(i) sin 1; error < 10¯17
(ii) sin 2; error < 10¯12
(iii) sin (;error < 10¯
(iv) e; error < 10-4
(v) e2; error < 10¯6.
.
which
equals each of the following
minimize
To
accuracy.
and n! on the next
page.
needless
computa-
and InfiniteSeries
428 InßniteSequences
2"
n
n!
l
2
3
2
4
8
16
4
5
32
6
7
8
9
10
11
12
64
128
256
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,888
1,048,576
13
14
15
16
17
18
19
20
*4.
1
2
6
24
120
720
5,040
40,430
362,880
3,628,800
39,916,800
479,001,600
6,227,020,800
87,178,291,200
1,307,674,368,000
20,922,789,888,000
355,687,428,096,000
6,402,373,705,728,000
121,645,100,408,832,000
2,432,902,008,176,640,000
This problem is similar to the previous one, except that the errors demanded
are so small that the tables cannot be used. You will have to do a little
thinking, and in some cases it may be necessary to consult the pmof, in
Chapter 16, that x"/n! can be made small by choosing n large-the pmof
actually provides a method for finding the appropriate
n. In the previous
rather
short
find
it
problem
sums; in fact, it was possible
was possible to
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
you can demonstrate that the sum
(provided
works).
(i)
(ii)
sin
1; error
e; error
<
<
10-flolo)
10¯
.
< 10¯2°.
10¯3
(iv) e ; error <
arctan
; error < 10¯(101°
(iii) sin 10; error
.
(v)
5.
Problem 11-38 you showed that the equation x2
cos x has precisely two solutions. Use the Taylor polynomial of cos to show that the
solutions are approximately
and find bounds on the error.
(b) Similarly, estimate the solutions of the equation 2x2 x sin x -E cos2 x.
(a) In
=
±Œ,
=
20. Approximation
ly PolynomialFunctions 429
6.
using Problem
(a) Prove,
15-9, that
1
1
+ aretan
2
3
1
1
x
4 arctan
arctan
4
5
239
(b) Show that x 3.14159.... (Every budding mathematician should verify a few decimalsof x, but the purpose of this exercise is not to set you
off on an immense calculation. If the second expression in part
(a)is
used, the first 5 decimals for x can be computed with remarkably little
-
x
=
4
arctan
-
=
-
-,
-
-
--.
=
work.)
7.
For every number
a, and every natural
number
n, we
define the
"binomial
coefficient"
)
and we define
(1 +
=
1, as usual. If a is not an integer, then
=
in sign for n
and alternates
for f(x)
a(a-1)-...-(œ-n+1)
x)" at
>
is never 0,
a. Show that the Taylor polynomial of degree n
0 is P.,o(x)
k,
=
and that the
Cauchy and
k-0
Lagrange forms of the remainder
are the following:
Cauchy form:
a(a-1)-...·(a-n)
R..o(x)
x(x
=
n!
a(a-1)-...-(«-n)
x(1+t)
=
n!
(n + 1)
=
«
n+1
t)n(1 + t)a-n-
-
,_1
I-I
"
1+t
x-t
x(1 + t)"¯1
1
1+t
,
t in
[0,x]
or
[x,0].
Lagrange form:
(n + 1)!
"
I,
xn+1(1
+ t)"¯"
t in [0,x] or [x,0].
n+ 1
Estimates for these remainder terms are rather difficult to handle, and are
postponed to Problem 23-18.
=
8.
Suppose that a¿ and b¿ are the coefficients in the Taylor polynomials at a of f
and g, respectively. In other words, a¿
f (a)/i! and bi g (a)/i!. Find
the coefficients c¿ of the Taylor polynomials at a of the following functions,
in terms of the ag's and bi's.
=
(i) | + g
(ii) fg
.
.
=
and InfiniteSeries
430 InfiniteSequences
(iii) J'.
(iv) h(x)
(v)
9.
k(x)
=
lxf
(t) dt.
f (t) dt.
=
(a) Prove that
the Taylor polynomial of
f (x)
sin(x2) of
=
degree 4n + 2 at 0
is
6
10
4n+2
--+---··+(-1)"
x
.
3!
5!
(2n + 1)!
Hint: If P is the Taylor polynomial of degree 2n + 1 for sin at 0, then
P(x) + R(x), where lim R(x)/x2.+1
0. What does this imply
sinx
=
=
xso
4n+2;
about lim R(x2
x->0
(b) Find f (k)(0) for all
(c) In general, if f(x)
at
10.
k.
g(x"), find
=
ff*)(0)in terms
of the
derivatives of g
0.
Pmve that if xs
0, then
x
e'
-
n!
(x
t)" dt
-
|x|n+1
(n+1)!
<
¯
.
-1
11.
Prove that if
<
0, then
xs
x
1+t
*12.
In+1
s (1+x)(n+1)
n
X
dt
(a) Show that
a|" for |x
if |g'(x)| 5 M|x
M |x a |n+1/(n+ 1) for |x a | < ô.
(b) Use part (a)to show that if lim g'(x)/(x
-
-
a|
.
8, then
<
|g(x)
-
g(a)| 5
--
--
lim
x->a
g(x)
=
(x
-
a)"+1
-
a)n
=
0, then
0.
if g(x)
f (x) Pn,a.f (x), then g'(x) f'(x) P.-1,a.r(x).
(d) Give an inductive proof of Theorem 1, without using PHôpital's Rule.
(c) Show that
=
-
=
-
13.
Deduce Theorem 1 as a corollary of Taylor's Theorem, with any form of
the remainder.
(The catch is that it will be necessary to assume one more
derivative than in the hypotheses for Theorem 1.)
14.
Deduce the Cauchy and Lagrange forms of the remainder from the integral
form, using Problem 13-23. There will be the same catch as in Problem 13.
20. Approximation
byPolynomialFunctions 431
15.
(a) Suppose
for all x
we have
that
>
f is twice differentiable on (0, oo) and that |f (x)| 5 Mo
0, while |f"(x)| 5 M2 for all x > 0. Prove that for all x > 0
2
|f'(x)| 5 -Mo
h
(b) Show that
for all x
h
+ -M2
for all h
2
>
0.
0 we have
>
|f'(x)|
5 2/MoM2.
is twice differentiable on (0,oo), f" is bounded, and f(x) approaches 0 as x
0 as x
oo, then also f'(x) approaches
oo.
and lim f"(x) exists, then lim
exists
lim
If
lim
f"(x)
f'(x)
f(x)
(d)
(c) If f
-+
->
=
x-o
x-o
x-oo
=
x-oo
0. (Compare Problem 11-31.)
16.
(a) Prove that
if f"(a) exists, then
f"(a)
=
+ h) + f (a
lim f (a
hw0
h2
-
h)
-
2f
(a)
.
The limit on the right is called the Schwarzseconddenvativeof f at a. Hint:
Use the Taylor polynomial P2,a (x) with x = a + h and with x
h.
a
x2
and
that
for x :> 0,
for xs 0. Show
(b) Let f(x)
=
-
-x2
=
.
hm
f(0+h)+f(0-h)-2f(0)
h2
hoo
exists, even though
f"(0) does not.
that if f has a local maximum at a, and the Schwarz second
derivative of f at a exists, then it is 5 0.
(d) Prove that if f"'(a) exists, then
(c) Prove
f'"(a)
3
17.
f(a+h)- f(a-h)-2hf'(x)
=lim
a->o
h3
.
Use the Taylor polynomial Pi,aj, together with the remainder, to prove a
weak form of Theorem 2 of the Appendix to Chapter 11: If f" > 0, then
the graph of f always lies above the tangent line of f, except at the point
of contact.
*18.
Problem 1843 presented a rather complicated proof that f
0 if f"- f = 0
and f (0) f'(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 19, 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.)
and InfiniteSeries
432 InfiniteSequences
**19.
Consider a function f which satisfies the differential equation
f(">=
Lajf II,
j=0
for certain numbers ao,
an-1. Several special cases have already received
either
detailed treatment,
in the text or in other problems; in particular, we
have found all functions satisfying f'
f, or f"+ f 0, or f"- f 0. The
trick in Problem 18-42 enables us to find many solutions for such equations,
but doesn't say whether these are the only solutions. This requires a uniqueness
result, which will be supplied by this problem. At the end you will find some
sketchy) remarks about the general solution.
(necessarily
.
.
.
,
(a) Derive the following formula for f("*0 (letus
f ("*)
(b) Deduce
f
j=0
=
+ an-1aj)
(aj.I
=
=
=
"a-i"
agree that
will
be 0):
f fa.
formula for f("*23.
a
The formula in part (b)is 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
size of f("**)(x).
(c) Let
N
max(1,
=
|a.-1|). Then |aj_i +
|aol,
...,
an-laj|
5 2N2; this
means that
n-1
bjl
f(n+1)
=
ffi3,
where
|b; | 5 2N2.
where
|bj \ 5
j=0
Show that
n-1
f("*2)
42ffil,
--
4N3
j=0
and, more
generally,
n-i
f
"**)
gk
--
(j),
|bj*|5
where
2Nk+1
j-0
(d) Conclude
number
(c)that,
from part
M such that
|f(n+k(x)|5
that
(e) Now suppose
|f(x)|<
M
¯
and conclude that
for any particular number
M -2kNk+t
x, there
for all k.
f (0) f'(0)
f ("¯U(0) 0. Show that
M |2Nx |n+k+1
·2A+1N**2Mn+k+1
<
=
=
-
·
=
=
-
·
¯
(n+k+1)!
f
=
0.
(n+k+1)!
'
is a
20. Approximation
byPolynomialFunctions 433
(f) Show that
if fi and
f2 are
both solutions of the differential equation
n-1
j=0
and
for 0 5 jsn
fi (0) f2(13(0)
-
1, then
-
fi
=
f2.
In other words, the solutions of this differential equation are determined
conditions" (thevalues fU)(0) for 0 5 jyn
by the
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
"initial
-
x"
an-1xn-1
-
has n distinct roots al,
f (x)
is a solution, and
-
-
--
-
ao
=
0
an, then any function of the form
...,
=
--
ci e"1" +
··
-
+ c.e.,x
·
f(0)
f'(0)
=c1+···+c.,
=aic1+···+a.c.,
f(n-1)(0)=ain-Ic1+···+a."¯1c,
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 properly, but we
will not try to prove this last assertion. (It is a purely algebraic fact, which
2 or 3.) These remarks are also true if some
you can easily check for n
of the roots are multiple roots, and even in the more general situation
considered in Chapter 27.
=
**20.
function on [a,b] with f (a)
f(b)
and that for all x in (a, b) the Schwarz second derivative of f at x is 0
(Problem 16). Show that f is constant on [a,b]. Hint: Suppose that
f (x) > f (a) for some x in (a, b). Consider the function
(a) Suppose
that
f is
a continuous
g(x)=
with g(a)
=
f(x)-s(x-a)(b-x)
sufficiently small e > 0 we will have
point y in (a, b). Now use
g(x) >
have
Problem 16(c)(theSchwarz second derivative of (x a)(b x) is simply
its ordinary second derivative).
(b) If f is a continuous function on [a,b] whose Schwarz second derivative
is O at all points of (a, b), then f is linear.
=
g(a),
g(b)
=
f(a). For
so g will
a maximum
-
-
and InßniteSenes
434 InßniteSequences
*21.
x4 sa
Ux2(-orx ¢
0, and f (0) 0. Show that
order 2 at 0, even though f"(0) does not exist.
(a) Let f (x)
=
=
f
0 up to
=
This example is slightly more complex, but also slightly more impressive,
than the example in the text, because both f'(a) and f"(a) exist for
a ¢ 0. Thus, for each number a there is another number m(a) such that
(*) f (x) f (a) + f'(a)(x
=
where
.
hm
x-a
namely, m(a)
a) +
-
R.(x)
(x
-
m(a)
2
-
a)2
0, and m(0)
(x
-
a)2 + Ra(x),
0;
0. Notice that the fimetion
continuous.
this
defined
in
is
not
way
m
(b) Suppose that f is a differentiable fimction such that (*)holds for all a,
with m(a)
m(a)
0 for
0. Use' Problem 20 to show that f"(a)
=
=
f"(a) for a ¢
=
=
=
all a.
suppose that (*)holds for all a, and that m is continuous.
that for all a the second derivative f"(a) exists and equals m(a).
(c) Now
Prove
*CHAPTER
e IS
TRANSCENDENTAL
The irrationality of e was so easy to pmve 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. Justhow a number might be even worse than irrational
is suggested by a slight rewording of definitions. A number x is irrational if it is
a/b for any integers a and b, with b ¢ 0. This is the
not possible to write x
saying
that x does not satisfy any equation
same as
=
bx-a=0
for integers a and b, except for a 0, b 0. Viewed in this light, the irrationality
of à does not seem to be such a terrible deficiency; rather, it appears that à just
barely manages to be irrational-although à is not the solution of an equation
=
=
a1x + ao
=
0,
it is the solution of the equation
x2-2=0,
of one higher
degree. Problem 2-18 shows how to produce many irrational num-
bers x which satisfy higher-degree equations
anx"+a,_ixn-1+---+ao=0,
the at are integers and ao ¢ 0 (thiscondition rules out the possibility that
equation of this sort is called
all at
0). A number which satisfies an
and practically every number we have ever encountered
an algebraic number,
of
solutions
of algebraic equations (x and e are the great exis defined in terms
mathematical
experience).
ceptions in our limited
All roots, such as
where
"algebraic"
=
are clearly algebraic numbers,
and even complicated
3+ñ+
are algebraic
we
(although
will not try to
1+ñ+
combinations,
like
6
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 e is transcendental is well within our grasp, and was theoretically
possible even before Chapter 20. Nevertheless, with the inclusion of this proof, we
can justifiablyclassify ourselves as something more than novices in the study of
while many irrationality proofs depend only on elementary
higher mathematics;
properties of numbers, the proof that a number is transcendental usually involves
435
and InßniteSeries
436 InßniteSequences
Even the dates connected with the transome really high-powered mathematics.
scendence of e are impressively recent-the
first proof that e is transcendental,
1873.
due to Hermite, dates from
The proof that we will give is a simplification,
due to Hilbert.
Before tackling the proof itself, it is a good idea to map out the strategy, which
depends on an idea used even in the proof that e is irrational. Two features of the
expression
1
1!
1
1
nl
2!
that
the
proof
were important for
e is irrational: On the one hand, the number
e=1+-+-+...+-+Rn
1
1+-+---+11
written
p/q with q
1
n!
as a fraction
can be
(sothat n! (p/q) is an integer); on the
other hand, O < R. < 3/(n + 1)! (son!R, 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
e> 0
there is a rational number r with |x r| < s; the catch, however, is that it may be
n!
<
-
necessary to allow a very large denominator for r, as large as 1/s perhaps. For e
we are assured that this is not the 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
will see that only this fact about e is ever used. The number e is
is irrational,
.
you
by no means unique in this respect: generally speaking, the bettera number can be
approximated
evidence for this assertion
by rational numbers, the worse it is (some
presented
that
transcendental
3).
is
in Problem
The proof
depends on a natural
e is
extension of this idea: not only e, but any finite number of powers e, e2,
, e",
rational
numbers.
simultaneously
approximated
well
In
especially
be
by
can
our
proof we will begin by assuming that e is algebraic, so that
...
(*)
for some integers ao,
certain integers M, Mi,
...
a,,e"+---+aie+ao=0,
,
...
an.
,
ao-j0
In order to reach a contradiction
Mn and certain
2
numbers et,
we will than find
...
,
en such that
Mi+ci
1
e=
e
"small"
M
,
M2+E2
=
M
,
Ma+En
e"=
M
Justhow small the c's must be will appear when these expressions are substituted
into the assumed equation (*).After multiplying through by M we obtain
[aoM+atM1+···+a.M.]+[61ai+-..+e.a.]=0.
21.
e
is Tanscendental 437
The first term in brackets is an integer, and we will choose the M's so that it will
necessarily be a nonero integer. We will also manage to find e's so small that
|riai
þ
+-·-+ena.\<
this will lead to the desired contradiction--the
sum of a nonzero integer and a
zero!
value
of absolute
less than
cannot be
I
number
As a basic strategy this is all very reasonable and quite straightforward. The
remarkable
part of the proof will be the way that the M's and c's are defined. In
order to read the proof you will need to know about the gamma function! (This
function was introduced in Problem 19-39.)
THEOREM
1
e is transcendental.
-/
PROOF
Suppose there were integers ao,
(*) a.e"
Define numbers M, M1,
...
=
I
,
a., with ao
+ an-1e"¯* +
Mn and ci,
,
xP-1
M
...
1)
[(x
-
=
Uk
¯
Øk
+ ao
-
(x
...
=
0.
n)]Pe¯2
-
dx,
1)!
-
[(x-1)·....(x--n)]Pe-2
"xP
(p
€k
-
e, as follows:
,
-
(p
o
MK
...
-
0, such that
k XE¯
(X
Î)
-
·
dx,
1)!
-
(X
...•
-
H)]
€¯
dx.
(p-1)!
The unspecified number p represents a prime number* which we will choose later.
Despite the forbidding aspect of these three expressions, with a little work they will
appear much more reasonable. We concentrate on M first. If the expression in
brackets,
1)
[(x
-
is actually multiplied
out, we obtain
-
...·
(x
-
n)],
a polynomial
x"+---±n!
number"
The term "prime
was defined in Problem 2-17. An important fact about prime numbers
be used in the proof, although it is not proved in this book: If p is a prime number which does
not divide the integer a, and which does not divide the integer b, then p also does not divide ab.
is crucial in proving that the
The Suggested Reading mentions references for this theorem (which
factorization of an integer into primes is unique). We will also use the result of Problem 2-17(d),
that there are infinitely many primes-the reader is asked to determine at pocisely which points this
information is required.
*
will
and InfiniteSeries
438 InfiniteSequences
with integer coefficients. When raised to the pth power this becomes an even
more complicated polynomial
x"P
+
·-
(n!)P.
±
-
Thus M can be written in the form
""
M
where
°°
1
=
(p
1)!
-
Ca
X
P¯1"e¯2
dx,
O
the C, are certain integers, and Co
±(n!)P.
=
But
oo
xk
Thus
-xdx=kl.
""
(p-1+a)!
(p-1)!
Cœ
M=
a=o
Now, for a
=
0 we
obtain
-
the term
±(n!)* (p
(p
1)!
1)!
-
-
±(n!)*
=
.
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
1 + a)!
=C,(p+a--1)(p+a-2)-...-p,
(p 1)!
which is divisible by p. Therefore M itself is an integer which is not divisible by p.
Now consider Mk. We have
Ca (p
-
-
"xp-1[(x-1)·...-(x-n)]Pe
Mk =
X
Uk
(p
°°
=
x?¯I
[(x
-
1)
·
(p
This can be transformed
-
...
-
dx
1)!
-
(x
-
n)]Pe-(x-k)
1)!
dx.
into an expression looking very much like M by the
substitution
u=x-k
du
dx.
=
The limits of integration are changed
to 0 and oo, and
l°°(u+k)?¯1[(u+k-1)·...·u-...·(u+k-n)]Pe¯"
Mk
=
o
(p
-
1)!
du.
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 uP. This means that the entire expression
(u+k)p-1[(u+k--1)-...·(u+k-n)]F
21.
with
is a polynomial
Thus
Mk
e
is Wanscendental439
integer coefficients, every term ofwhich has degree at least p.
o
Î
=
p-1"e¯"
du
D,
=
==1
a=i
the D, are certain integers. Notice that the summation begins with a
1;
in this case every term in the sum is divisible by p. Thus each Mk is an integer
which is divisible by p.
Now it is clear that
where
=
e
Substituting into
(*)and
&
Mk Ÿ
M
=
Ek
k=1,...,n.
,
multiplying
by M we obtain
[aoM+aiM+···+a.M.]+[aici+.--+a,«.]=0.
In addition to requiring that p > n let us also stipulate that p > |ao[.This means
that both M and ao are not divisible by p, so aoM is also not divisible by p. Since
each Mk is divisible by p, it follows that
aoM+a1M1+---+a.M.
is not divisible by p. In particular it is a nongero integer.
In order to obtain a contradiction to the assumed equation
prove that e is transcendental, it is only necessary to show that
la1Ei
+
(*),and
thereby
+ a.«.|
·
can be made as small as desired, by choosing p large enough; it is clearly sufficient
to show that each |eg|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 (thedegree of the assumed polynomial equation (*)).To
begin with, if 1 5 k i n, then
k
ek
Ïk
Ï"
o
Se"
(p
np-1|[(x
the maximum
of
|(x
1)
-
|<
-
(x
...-
(p-1)!
1)
(x
-
-
€"Rp-1
|Ek
dx
-1)!
o
Now let A be
€¯*
...•(X-N)
Xp-1((X-Î)
...
p
-
(p
oc
e¯* dx
<
¯
(p-1)!
e"n?¯I
AP
(p
1)!
¯
-
e"n?AP
¯
(p
-
e"(nA)F
1)!
(p
-
|e¯* dx.
n)| for x in
e¯* dx
1)!
p-1¿p
gn
n)]P
n
¯
-
-
-
1)!
[0,n].
Then
and InfiniteSeries
440 Ir¢nite Sequences
But n and A are fixed; thus
making
(nA)F/(p
-
1)! can be made as small as desired by
p sufficiently large.
This proof, like the proof that x is irrational, deserves some philosophic afall, we
terthoughts. At first sight, the argument seems quite
mathematiuse integrals, and integrals from 0 to oo at that. Actually, as many
cians have observed, integrals can be eliminated from the argument completely;
the only integrals essential to the proof are of the form
"advanced"-after
ao
Xk
-x
dx
for integral k, and these integrals can be replaced by k! whenever
Thus M, for example, could have been defined initially as
where
they occur.
C, are the coefficients of the polynomial
[(x
-
1)··..·(x
-
n)]P.
"completely
If this idea is developed consistently, one obtains a
that e is transcendental, depending only on the fact that
1
1
2! 3!
this
proof
Unfortunately,
is harder to understand than the original
onethe
whole structure of the proof must be hidden just to eliminate a few
integral signs! This situation is by no means peculiar to this specific theoremarguments are frequently more difficult than
ones. Our
proof that x is irrational is a case in point. You probably remember nothing
about this proof except that it involves quite a few complicated
functions. There is
actually a more advanced, but much more conceptual
proof, which shows that x
is transcendental,
a fact which 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
which can be
requires the construction of a line segment whose length is
,
accomplished if a line segment of length x is constructible. The Greeks were
totally 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 x is transcendental; since the
length of any segment that can be constructed with straightedge and compass can
and is therefore algebraic, this
÷, and f,
be written in terms of +,
proves
that a line segment of length x cannot be constructed.
The proof that x is transcendental requires a sizable amount of mathematics
which is too advanced to be reached in this book. Nevertheless, the pmof is not
much more difficult than the proof that e is transcendental.
In fact, the proof
for x is practically the same as the pmof for e. This last statement should certainly
1+
¢=
1
1!
elementary" proof
-+-+
-+---.
"elementary"
"elementary"
"advanced"
·,
-,
21. e is Transcendental441
surprise
you. The proof that e is transcendental seems to depend so thoroughly
particular
properties of e that it is almost inconceivable how any modifications
on
could ever be used for x; after all, what does e have to do with x?
Justwait and
see!
PROBLEMS
L
is algebraic.
if a > 0 is algebraic, then
algebraic
and
rational,
then a + r and er are
that
Prove
if
is
is
a
r
(b)
(a) Prove that
algebraic.
considerably: the sum, product,
is algebraic. This fact is too difficult
for us to prove here, but some special cases can be examined:
Part
(b)can
actually be strengthened
and quotient of algebraic
2.
*3.
numbers
Prove that
+ Ä and Ã(1 + Ã) are algebraic, by actually finding
algebraic equations which they satisfy. (You will need equations of degree 4.)
n
(a) Let
a
be an algebraic number which is not rational.
satisfies the
Suppose that a
polynomial equation
f(x)=anx"+a..ix"¯'+---+ao=0,
polynomial function of lower degree has this property. Show
0 for any rational number p/q.
Hint: Use Probf (p/q)
lem 3-7(b).
(b) Now show that |f(p/q)| ;> 1/q" for all rational numbers p/q with q > 0.
Hint: Write f (p/q) as a fraction over the common denominator q".
< 1}. Use the Mean Value Theorem
(c) Let M sup{|f'(x)|: |x
p/q| < 1, then
to prove that if p/q is a rational number with |a
max(1,
>
1/M) we have
1/Mq".
(It
follows
that
p/q|
for c
|a
all
rational
p/q.)
|a p/q l > c/q" for
and that no
-/
that
-a
=
-
=
-
-
*4.
Let
a
=
0.110001000000000000000001000...,
l's occur in the n! place, for each n. Use Problem 3 to prove that
œ is transcendental. (For each n, show that a is not the root of an equation
of degree n.)
where the
Although Problem 4 mentions only one specific transcendental number, it should
be clear that one can easily construct infinitely many other numbers a which do
not satisfy |a p/q| > c/q" for any c and n. Such numbers were first considered
and the inequality in Problem 3 is often called Liouville's
by Liouville (1809-1882),
inequality. None ofthe transcendental numbers constructed in this way happens to
be particularly interesting, but for a long time Liouville's transcendental numbers
were the only ones known. This situation was changed quite radically by the work
who showed, without exhibiting a single transcendental
of Cantor
(1845-1918),
number, that most numbers are transcendental. The next two problems provide an
-
442
and InfiniteSeries
Infmite Sequences
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 be arranged in a sequence
at, a2, a3, a4.
The obvious example
is N, the set of natural
(infact, more
•••
-
or less the Platonic ideal of) a countable set
the set of even natural numbers is also
clearly
numbers;
countable:
2,4,6,8,....
It is a little more surprising to learn that Z, the set of all integers
0) is also countable, but seeing is believing:
negative
(positive,
and
-1,
0, 1,
-2,
-3,
2,
3,
....
The 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
*5.
think and
(2)nevertheless,
some sets are not countable.
if A and B are countable, then so is A UB
{x : x is in A or
worked
the
trick
that
for Z.
same
x is in B }. Hint: Use
rational
numbers
of
the
countable.
that
(This is really
set
positive
is
(b) Show
startling,
quite
but the figure below indicates the path to enlightenment.)
(a) Show that
=
I
12
l
345
22
12
222
345
33
333
345
T2
1
that the set of all pairs (m,n) of integers is countable.
practically the same as part (b).)
(d) If Ai, A2, As, are each countable, prove that
(c) Show
(This is
...
A1UA2UAsU...
is also countable. (Again use the same trick as in part (b).)
(e) Prove that the set of all triples (l, m, n) of integers is countable. (A triple
(l, m, n) can be described by a pair (l, m) and a number n.)
an) is countable.
(If you have
(f) Prove that the set of all n-tuples (ai, a2,
using
this,
induction.)
done part (e),you can do
of
that
the
of
Prove
all
set
roots
polynomial functions of degree n with
(g)
integer coefficients is countable. (Part (f)shows that the set of all these
.
.
.
,
21.
polynomial functions can be arranged
most n roots.)
Now
use parts
(h)
is countable.
*6.
(d)and (g)to
e
is Wanscendental443
in a sequence,
and each
prove that the set of all algebraic
has at
numbers
Since so many sets turn out to be countable, it is important to note that the
In other words,
set of all real numbers between 0 and 1 is not countable.
there is no way of listing all these real numbers in a sequence
«I
=
«2
=
Œ3
=
0.aiia12a13al4
0.a2]a22a23a24
0.a31a32G33G34
-
-
-
·
-
-
-
·
·
is being used on the right). To prove that this is so, suppose
such a list were possible and consider the decimal
notation
(decimal
0.ã1iã22533Õ44
-
where
ä.,
cannot
=
5 if a.. ¢ 5 and ä.,
-
-
,
6 if a..
=
possibly be in the list, thus obtaining
=
5. Show that this
number
a contradiction.
Problems 5 and 6 can be summed up as follows. The set of algebraic numbers
is countable. If the set of transcendental numbers were also countable, then the
set of all real numbers would be countable, by Problem $(a), and consequently the
set of real 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 ("thereare 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.
*7.
Let f be a nondecreasing function on
lim f (x) and lim f (x) both exist.
x->a+
x->a-
[0,1].
Recall (Problem 8-8) that
0 prove that there are only finitely many numbers a in
[0,1] with lim f (x) lim f (x) > s. Hint: There are, in fact, at most
(a) For
any e
>
-
[f (1) f (0)]/e of them.
-
(b) Prove
f is discontinuous is countable.
0, then it is > 1/n for some natural
that the set of points at which
Hint: If lim
x->a+
f (x)
-
lim
x->a-
f (x)
>
number n.
This problem shows that a nondecreasing function is automatically continuous at most points. For differentiability the situation is more difficult
function can fail
to analyze and also more interesting. A nondecreasing
which
countable,
of
points
is
but it is
differentiable
be
set
not
to
at a
nondecreasing
still true that
functions are differentiable at most points
"most").
word
of
the
Reference [32]of the Sugdifferent
sense
(ina
using
proof,
gives
the Rising Sun Lemma of
gested Reading
a beautiful
and InfiniteSeries
444 InfiniteSequences
Problem 8-20. For those who have done Pmblem 10 of the Appendix
to Chapter 11, it is possible to provide at least one application to dif
ferentiability of the ideas already developed in this problem set: If f is
convex, then f is differentiable except at those points whem its righthand derivative fe' is discontinuous; but the function f,' is increasing,
differentiable except at a countable
so a convex function is automatically
of
points.
set
*8.
11-66 showed that if every point is a local maximum point for
a continuousfunction f, then f is a constant function. Suppose now that
the hypothesis of continuity is dropped. Prove that f takes on only a
countable set of values. Hint: For each x choose rattonal numbers ax
and by such that a, < x < b, and x is a maximum
point for f on
value
value
of f on some
maximum
the
bx).
is
Then
every
(ax,
f(x)
such
there?
How
interval (ax,bx).
intervals are
many
corollary.
Deduce
Problem
11-66(a)
as
a
(b)
(c) Prove the result of Problem 11-66(b) similarly.
(a) Problem
INFINITE SEQUENCES
CHAPTER
The idea of an infinite sequence is so natural a concept that it is tempting to
dispense with a definition altogether. One frequently writes simply
infinite
sequence
"an
ai, a2, a3, a4,
G5,
•••
,
"forever."
the three dots indicating that the numbers at continue to the right
A
rigorous definition of an infinite sequence is not hard to formulate, however; the
important point about an infinite sequence is that for each natural number, n,
there is a real number a.. This sort of correspondence is precisely what functions
are meant to formalize.
DEFINITION
An infinite sequence
of real numbers
is a function whose domain is N.
From the point of view of this definition, a sequence should be designated by a
letter like a, and particular values by
single
a(1), a(2), a(3),
...,
but the subscript notation
ai, a2, as,
...
is almost always used instead, and the sequence itselfis usually denoted by a symbol
like {an).Thus {n},{(-1)"},and (1/n} denote the sequences œ, ß, and y defined
by
a,=n,
ß,
=
(-1)",
1
n
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.
A sequence,
(a)
FIGURE
445
1
(b)
(c)
446 InfiniteSequences
and InfiniteSeries
I
FIGURE
2
A more convenient representation of a sequence is obtained by simply labeling
the points at, a2, as,
on a line (Figure 2). This sort of picture shows when
the sequence
out to infinity," the sequence
going." The sequence (an)
and
and
and
the sequence {y.)
back
forth between
1,"
(ß,}
to 0." Of the three phrases in quotation marks, the last is the crucial concept
associated with sequences, and will be defined precisely
(thedefinition is illustrated
in Figure 3).
...
"goes
"is
--l
"jumps
F1GURE
DEFINITION
"converges
3
A sequence (a.) converges to I (insymbols HNOO
lim a.
l) if for every e
is a natural number N such that, for all natural numbers n,
=
if n
>
N, then
|a,
-
l|
<
>
0 there
s.
In addition to the terminology introduced in this definition, we sometimes say
that the sequence (a,} approaches
I or has the limit 1. A sequence {a.}is said
to converge if it converges to I for some l, and to diverge if it does not converge.
To show that the sequence {y,,}
converges to 0, it suffices to observe the following.
If e > 0, there is a natural number N such that 1/N < e. Then, if n > N we
have
1
1
n
N
y,=-<-<E,
so|y,-0|<e.
The limit
lim
n+1-
=0
HNOO
probably seem reasonable after a little reflection (itjust says that jn + 1 is
practically the same as
for large n), but a mathematical proof might not be so
will
22. InfiniteSequences447
obvious.
S we can use an algebraic trick:
Jn+1-S= (Jn+1-O)(Jn+1+Á)
n+1+S
n+1--n
To estimate
Jn +
1
-
n+1+S
I
n+1+S
1 S by applying
It is also possible to estimate n +
to the function f (x) =
on the interval
the Mean Value Theorem
[n,n + 1]. We obtain
-
n + 1-
f'(x)
=
1
=
1
--,
for some x in (n,n + 1)
2
1
2f
Either of these estimates may be used to prove the above limit; the detailed proof
is left to you, as a simple but valuable exercise.
The limit
3n3+7n2+1
3
=lim
4n3
naoo
Sn + 63
4
-
should also seem reasonable,
because the terms involving n3 are the most
remember
the proof of Theorem 7-9
will
tant when n is large. If you
to guess the trick that translates this idea into a proof-dividing
by n3 yields
3n3
4n3
-
Sn + 63
8
4--
n
top and bottom
1
7
3+-+--
2
you
impor-
be able
3
63
+-n
Using this expression, the proof of the above limit is not difficult, especially if one
uses the following facts:
If H->OO
lim a, and N->OO
lim b, both exist, then
lim (a, + b.)
=
lim (a, b.)
=
8400
-
R->OO
moreover,
lim a. + B-¥QO
lim b.,
N->OD
Iim a.
B->OO
-
lim b.;
B-kcQ
if lim b, ¢ 0, then b, ¢ 0 for all n greater than some N, and
H-¥DO
lim an/b,
R->OD
=
lim a./ N->OO
lim b..
N->OO
and InßniteSeries
448 InfiniteSequences
to be utterly precise, the third statement would have to be even
complicated.
As it stands, we are considering the limit of the sequence
more
where
the numbers c, might not even be defined for certain n < N.
(c,} {as/b.),
could define c, any way we liked for such nThis doesn't really matter-we
of
the
limit
because
a sequence is not changed if we change the sequence at a
(If we wanted
=
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,
l is so similar to previous definitions of limits, especially
N->CO
lim f (x) l.
=
=
l and lim f (x)
I is
The similarity between the definitions of lim a.
x-oo
naoo
actually closer than mere analogy; it is possible to define the first in terms of the
second. If f is the function whose graph (Figure 4) consists of line segments joining
=
FIGURE
=
4
the points in the graph of the sequence
f (x) (a.si
=
a.)(x
-
that
{a.), so
n 5 x 5 n + 1,
n) + a,
-
then
lim a,
Conversely, if
f
if and only if
l
=
satisfies lim
lim
f (x)
=
l.
lim a.
l, and we set a,
l.
f (n), then XMOD
is frequently very useful. For example, suppose that
=
=
=
X->OO
This second observation
0 < a < 1. Then
lim a"
=
0.
n->oo
To prove this we note that
lim a'
lim e"°
=
x->oo
0, so that x log a is a
Notice that we actually have
since log a
<
negative
lim a"
R-¥CO
for if a
<
"
0
=
x-moo
=
0
and
if |a|
large in absolute value for large x.
<
1;
0 we can write
lim
R->OO
aN
=
lim (-1)" a|"
B->DO
=
0.
22. InfiniteSequences449
The behavior of the logarithm function also shows that if a > 1, then a" becomes arbitrarily large as n becomes large. This assertion is often written
Iim a"
=
N->OO
and it is sometimes even said that
oo,
a
>
1,
{a"}approaches oo. We also
write equations
like
-a"
lim
-m
-
R-FOO
-1,
then lim a"
and say that (-a"} approaches
Notice, however, that if a <
does not exist, even in this extended sense.
Despite this connection with a familiar concept, it is more important to visualize
convergence in terms of the picture of a sequence as points on a line (Figure 3).
There is another connection between limits of functions and limits of sequences
which is related to thispicture. This connection is somewhat less obvious, but considerably more interesting, than the one previously mentioned-instead
of defming
limits of sequences in terms of limits of functions, it is possible to reverse the pm-oo.
cedure.
THEOREM
I
Let f be a function defined in an open interval containing c, except perhaps at c
itself with
lim f (x) l.
X->C
=
Suppose that (a.) is a sequence such that
(1) each
is in the domain of f,
as
(2)
each a,
(3)
lim a,
¢
-
c,
c.
naoo
Then the sequence {f (as)) satisfies
lim
R->OO
f (a.)
l.
=
Conversely, if this is true for every sequence {a,}satisfying the above conditions,
then lim
x->c
PROOF
f (x)
=
l.
Suppose first that lim f (x) l. Then for every e > 0 there is a 8
for all x,
then|f(x)-ll<e.
if0<\x-c|<8,
=
>
0 such that,
If the sequence {a.}satisfies N-FOR
lim a.
c, then (Figure 3) there is a natural
ber N such that,
if n > N, then |a, c| < 8.
=
-
By our choice of 8, this means that
|f(a.)-l|<e,
num-
and InfiniteSerks
450 InfiniteSequences
showing that
lim f (a.)
=
l.
lim f (a.) l for every sequence {a.}with lim a.
Suppose, conversely, that B->OD
B-¥00
lim f (x) l were not true, there would be some e > 0 such that for every
c. If X->C
ô > 0 there is an x with
=
=
=
0<|x-c|<ô
but
|f(x)-l|>e.
In particular, for each n there would be a number
0
|x,
<
cl
-
1
<
but
--
n
x, such that
|f (x.)
l|
-
>
e.
Now the sequence (x.) clearly converges to c but, since If (x.) l| > e for all n,
the sequence {f(x.)) does not converge to l. This contradicts the hypothesis, so
lim f (x) l must be true.
-
=
Theorem 1 provides many examples of convergent sequences.
sequences {a,}and (b,} defined by
=
a,
bn
a,
FIGURE
a.
a.
a*
=
sin
13 +
cos
sin
1 + (-1)"
For example, the
-
,
clearly converge to sin(13) and cos(sin(1)), 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 is the basis for all other results. This criterion is stated in terms of concepts
defined for functions, which therefore apply also to sequences: a sequence {an}is
ÎncreBSÂng if am I > an for all n, nondecreasing
if an+1 > as for all n, and
M for all n; there are simbounded above if there is a number M such that a.
=
5
_<
which
ilar definitions for sequences
below.
THEOREM
2
PROOF
are
decreasing, nonincreasing,
and
bounded
If {a,}is nondecreasing and bounded above, then {an)converges (asimilar statement is true if {a.}is nonincreasing and bounded below).
The set A consisting
of all numbers an is, by assumption, bounded above, so A has
«. We claim that lim a, = a (Figure 5). In fact, if e > 0,
least upper bound
there is some DN SatiSfying
Then if n > N we have
a
HOOD
Œ
a,>aN,
This proves that n->OO
lim a.
=
-
GN
SO
œ.
<
e, since a is the least upper bound of A.
Œ-an<Œ-GN<E·
22. InfiniteSequences451
The hypothesis that {a.}is bounded above is clearly essential in Theorem 2: if
or not {a.}is nondecreasing)
{a,}is not bounded above, then (whether
(a,} clearly
might
that
consideration,
there
should
it
diverges. Upon first
be little
appear
trouble deciding whether or not a given nondecreasing
sequence {a,}is bounded
above, and consequently whether or not {an)converges. In the next chapter such
sequences will arise very naturally and, as 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
increasing) sequence is bounded above:
(obviously
1+(+(-,
1, 1+¾,
1+¼+¼+¾,
....
Although Theorem 2 treats only a very special class of sequences, it is more
useful than might appear at first, because it is always possible to extract from an
arbitrary
or else
sequence {a,} another sequence which is either nonincreasing
nondecreasing.
of the sequence {as]
To be precise, let us define a subsequence
of
the
be
form
to
a sequence
raph of a
a.,,an,,as
the nj are natural numbers
where
2 and 6 are peak points
2
FIGURE
3
4
5
6
7
8
9
ni
to
11
6
LFNMA
,...,
with
<
n2
< 83
·
•
Then every sequence contains a subsequence which is either nondecreasing
or
It is possible to become quite befuddled trying to prove this asthe proof is very short if you think of the right idea; it is worth
ftCOrding
as a lemma.
nonincreasing.
sertion, although
Any sequence
{a,}contains
a subsequence
which
is either nondecreasing
or non-
increasing.
PROOF
Call a natural number n a
m > n (Figure 6).
"peak
point" of the sequence {a,}if a,
<
as
for all
Case 1. The sequence has mfmitelymany peakpoints. In this case, if n1 < n2 <
are the peak points, then as, > a,2 > an, >
so {a.,) is the desired
na <
···
·
--,
subsequence.
(nonincreasing)
Case2. The sequence has only finitely
many peakpoints.In this case, let ni be greater
than all peak points. Since ni is not a peak point, there is some n2 > ni such that
an, > any. Since n2 is not a peak point (itis greater than ni, and hence greater
than all peak points) there is some n3 > n2 such that an, > a.,. Continuing in this
subsequence.
way we obtain the desired (nondecreasing)
If we assume that our original sequence {a,} is bounded, we can pick up an
extra corollary along the way.
COROLLARY (THE
BOLZANO-WEIERSTRASSTHEOREM)
Every bounded sequence has a convergent subsequence.
and InfiniteSeries
452 InfiniteSequences
Without some additional assumptions this is as far as we can go: it is easy to
construct sequences having many, evenly infinitely many, subsequences converging to different numbers (seeProblem 3). There is a reasonable assumption to
add, which yields a necessary and sufficient condition for convergence
of any secondition
this
will
work,
it does simplify
Although
not be crucial for our
quence.
many proofs. Moreover, this condition
plays a fundamental role in more advanced
investigations, and for this reason alone it is worth stating now.
If a sequence converges, so that the individual terms are eventually all close to
the same number, then the difference of any two such individual terms should be
l for some l, then for any e > 0 there is
very small. To be precise, if lim a,
l| < e/2 for n > N; now if both n > N and m > N, then
an N such that |a,
=
-
e
e
2
|a,-am|s|a.-l|+|l-aml<-+-=e- 2
This final inequality, |a, a.|
be used to formulate a condition
for convergence of a sequence.
-
<
e, which eliminates mention of the limit l, can
(theCauchy condition) which is clearly necessary
A sequence {an}is a Cauchy sequence
number N such that, for all m and n,
DEFINITION
if m, n
(This condition
>
is usually written
if for every e
N, then
|a,
lim |am
-
--
a,|
a,|
<
>
0 there is a natural
e.
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
{an}converges if and
only if it is a Cauchy sequence.
We have already shown that {a.}is a Cauchy sequence if it converges. The proof of
the converse assertion contains only one tricky feature: showing that every Cauchy
1 in the definition of a Cauchy sequence
sequence {a,}is bounded. If we take e
we find that there is some N such that
=
|a.
-
a,
|< 1
for m, n
>
N.
In particular, this means that
lam
-
Thus {am: m
whole sequence
>
aN+1|
<
1
for all m
>
N.
N} is bounded; since there are only finitely many other ag's the
is bounded.
22. InfiniteSequences453
The corollary to the Lemma thus implies that some subsequence
of
{a.}con-
verges.
Only one point remains, whose proof will be left to you: if a subsequence
Cauchy sequence converges, then the Cauchy sequence itself converges.
of a
PROBLEMS
Verifyeach of the following limits,
1.
(i)
(ii)
n
lim
(v)
(vi)
n!
=
g"
<
n/2.
lim
å
k
lim
=
1,
=
1.
(vii) lim Šn2+
lim Ja" +
(viii)n->oo
n
Ñ
-
=
0.
n(n
=
-
1)
max(a,
=
a, b
b),
-
hm
2.
$
k=1
=
.
nr+1
p + 1
Find the following limits.
.
(i)
hmoon+1
=
n+1
n
-
.
n
(ii) limn-Jn+adn+b.
-
R-¥DO
lim
(iii) n->oo
lim
(iv) naco
lim
(v)
naoo
(vi)
R->00
2" + (-1)"
2=+1+ (-1)"+1
.
Y
sin(n")
(-1)"
.
n+ 1
b"
an + b"
a"
lim nc",
>
<
n,
in particular, for
0.
0, where a(n) is the number
=
n
kP
--oo
k! for k
1.
=
Hint: The fact that each prime is
how small a(n) must be.
*(x)
...
0.
>
a
b"
a(n)
--
1
1 = 0. Hint: You should at least be able to prove
0. Hint: n!
R->OO
(ix) nlimoo
fn +
1-
Sn2+
lim
NNOO
-
0.
=
lim fn2 +
(iii) th4a
lim
(iv) n->oo
1.
=
1
n+ 3
lim
n->oo n3 + 4
n +
n->oo
-
.
|c|
<
l.
>
of
primes which divide n.
2 gives a very simple estimate of
454
and InfiniteSeries
InfiniteSequences
lim
(vii)naoo
3.
2"
-.
n!
can be said about the sequence {a,}if it converges and each as is
integer?
an
1,
1,
(b) Find all convergent subsequences of the sequence 1,
(There are infinitely many, although there are only two limits which
such subsequences can have.)
(c) Find all convergent subsequences of the sequence 1, 2, 1, 2, 3, 1, 2,
3, 4, 1, 2, 3, 4, 5,
(There are infinitely many limits which such
subsequences can have.)
the sequence
Consider
(d)
(a) What
-1,
.
.
.
-1,
-1,
.
...
1
1
2
.
1
2
3
2
1
3
4
1
For which numbers a is them a subsequence converging
to a?
4.
if a subsequence of a Cauchy sequence converges, then so
does the original Cauchy sequence.
(b) Prove that any subsequence of a convergent sequence converges.
5.
(a) Prove that
(a) Prove that
a < 2, then a
(b) Prove that the sequence
if 0
<
<
2.
<
converges.
lim a,
(c) Find the limit. Hint: Notice that if 8400
=
l, then lim
R->OD
N
=
J,
by Theorem 1.
6.
Let 0
<
ai
<
b1 and define
as
i
=
g,
be
=
i
a, + b.
the sequences {a,}and {b.}each converge.
(b) Prove that they have the same limit.
(a) Prove that
7.
In Problem 2-16 we saw that any rational approximation m/n to à can be
by a better approximation
(m + 2n)/(m + n). In particular, starting
with m
1, we obtain
n
œplaced
=
=
1,
(a) Prove that
3
-,
7
-,
...
.
this sequence is given recursively
a1=1,
a.
i=l+
1
1+a,
by
.
22. InfiniteSequences455
(b) Prove
that
lim a.
n.
=
B->OD
expansion
Ã
This gives the so-called continued.finction
1
1+
=
1
2+
2+
---
Hint: Consider separately the subsequences {a2,}and {a2, i}.
(c) Prove that for any natural numbers a and b,
b
a2+b=a+
.
b
2a +
8.
Identify the function
many
9.
f(x)
2a+..-
lim (lim (cosnbrx)2k).(It has been mentioned
=
k->oo
n->oo
times in this book.)
Many impressive looking limits can be evaluated easily (especially
by the
who makes them up), because they are really lower or upper sums in
person
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.)
Ñ+-·-+ Ñ
naoo
.
(ii)
lun
lim
(iv) n-oo
.
(v)
lun
W+
Ñ+·-·+ ѯ
.
n
(
(
1
n+ 1
1
-
n2
+
+
-
·
+
·
1
1)2
(,
n
+
+
(n 1)2
n->oo
lim
(vi) nuo
.
n
n->oo
lim
(iii) n->oo
10.
+
lim
(i)
(
n
n2+1
+
-
1
2n
+
.
-
-
-
n
(n + 2)2
n
+
.
(2n)2
-
·
·
á
+
n
+
(n n)2
.
n
+---+
n2+22
Although limits like lim
1
+
.
n2+n2
and
lim a" can be evaluated using facts 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
arguments for such limits are outlined here; the basic tools are inequalities
derived from the binomial theorem, notably
H->OD
N->OC
"elementary"
(1 +
and,
for part
h)"
for h
1 + nh,
>
>
0;
(e),
(1+h)" à 1+nh+
n(n
--
2
1) h2
n(n
2
-
2
1)
h2
,
forh>0.
456
and InfiniteSenes
InfiniteSequences
11.
(a) Prove that
n-woo
lim a"
(b) Prove that
HNOO
(c) Prove that
n->oo
(d) Prove that
H->OD
(e) Prove that
H->OD
lim a"
=
oo if a
=
0 if 0
<
1 if a
>
6
lim 6
lim &
lim
=
1, by setting a
a < l.
>
=
1 if 0
=
1.
1 + h, where h
=
1, by setting
<
a
=
>
0.
1+h and estimating h.
1.
<
(a) Prove that
a convergent sequence is always bounded.
(b) Suppose that lim a, 0, and that each a, > 0. Prove that the set of all
=
numbers
12.
a, actually
has a maximum
member.
(a) Prove that
1
(b) If
an
log(n + 1)
<
n+1
1+
=
1
2
1
3
-+
lim
=
n-koo
(
1+
<
1
--.
n
1
--logn,
-+-··+
n
show that the sequence {a.) is
follows that there is a number
y
log n
-
-
·
decreasing, and that each a,
-
+
l
-
log n
--
y
>
0. It
.
This number, known as Euler's number, has proved to be quite refractory;
it is not even known whether y is rational.
13.
(a) Suppose that f
is increasing on
Show that
[1,oo).
n
f(1)+---+ f(n-1)
(b) Now choose f
=
<
Ï
I
f(x)dx
<
f(2)+---+ f(n).
log and show that
-<n!<
(n + 1)"
n"
;
it follows that
lim
=-woo
--
&
n
=
1
-.
e
n/e, in the sense that the
This result shows that Ñ is approximately
ratio of these two quantities is close to 1 for large n. But we cannot
conclude that n! is close to (n/e)" in this sense; in fact, this is false. An
estimate for n! is very desirable, even for concrete computations,
because
n! cannot be calculated easily even with logarithm tables. The standard
(anddifficult) theorem which provides the right information will be found
in Problem 27-19.
22. InfiniteSequences457
14.
I
(a) Show that
the tangent line to the graph of f at (xo,f (xo))intersects the
horizontal axis at (xi, 0), where
f (xo)
xt=xo-
This intersection point may be regarded as a rough approximation to
the point where the graph of f intersects the horizontal axis. If we now
start at xi and repeat the process to get x2, then use x2 to get xs, etc.,
we have a sequence {x.}defined inductively by
I
|
«
FIGURE
x,
Xn+1
,
x.
xi
=
Xn
-
8k
for some (k in
i
9
x.
xx
a
t
xa
x'
"'
(C, Xk).
!
f (xk)
=
i
#'
f (xk)
f'($k)
=
Show that
Ok+1
/(Xk)
¯
•
J'(Xk)
Í'($k)
Conclude that
Ok+1
=
FIGURE
.
=
=
e
-
Figure 7 suggests that {x.}will converge to a number c with f (c) 0;
this is called Newton'smethod for finding a zero of f. In the remainder
of this problem we will establish some conditions under which Newton's
method works (Figures 8 and 9 show two cases where it doesn't). A few
facts about convexity may be found useful; see Chapter 11, Appendix.
(b) Suppose that f,' f" > 0, and that we choose xo with f(xo) > 0. Show
thatxoëx12x22---àc.
8k Xk c. Then
Let
(c)
7
,
f (x.)
I
I
.
8
for some Ok in
f (xt)
J'($k)j'(Xk)
(C, Xk),
·
f"(g)(Xk
¯
$k)
and then that
(*)
ôk+1
f"(Uk)
f'(x )
2
Ok
·
sup|f"| on
inf f' on [c,xi]and let M
m
Newton's method works if xo c < m/M.
(e) What is the formula for x.si when f(x) x2 A?
1.4 we get
If we take A = 2 and xo
1.4
xo
1.4142857
xi
1.4142136
x2
1.4142136,
x3
(d) Let
=
=
[c,x1].Show that
-
=
-
=
=
=
=
=
which
is already correct to 7 decimals! Notice that the number of correct
decimalsat least doubled each time. This is essentially guaranteed by the
FIGURE
9
inequality
(*)when
M/m
<
l.
458
and InßniteSeries
InfiniteSequences
15.
Use Newton's method
(i) f(x)
(ii) f (x)
(iii) f (x)
(iv) f (x)
*16.
=
tanx
=
cos x
-
-
x3 + x
=
x3
=
to estimate the zeros of the following functions.
cos2x
near 0.
x2
Û.
on [0,1].
on [0,1].
near
1
-
3x2 + 1
--
Prove that if R-¥OC
lim a.
=
l, then
.
hm
(a1+---+a.)
naoo
n
Hint: This problem is very
lem 13-40.
17.
Suppose that
lim f(x)/x
similar
to
(infact it is a
is continuous and lim f (x + 1)
0. Hint: See the previous problem.
f
=
X->OD
*18.
=l.
-
special
f (x)
case of)
Prob-
0. Prove that
=
l. Prove that
lim a. 1/a.
Suppose that a, > 0 for each n and that H->OO
lim g
l. Hint: This requires the same sort of argument that works in
naoo
1, for a > 0.
Problem 16, together with the fact that n->OO
lim
=
=
=
19.
(a) Suppose that {an}is a convergent
that lim an is also in
n-woo
(b) Find a convergent
sequence of points all in
[0,1].
sequence
{a.) of points
all in
(0, 1) such
[0,1]. Prove
that lim an
is not in (0, 1).
20.
Suppose that f is continuous
x,
and that the sequence
f(x), f(f(x)), f(f(f(x))),
"fixed
converges to l. Prove that I is a
special cases have occurred already.
21.
...
point" for f, i.e., f (l)
=
l. Hint: Two
(a) Suppose that f
is continuous on [0, 1] and that 0 5 f(x) 5 1 for all x in
7-11 shows that f has a fixed point (inthe terminology
1].
Problem
[0,
of Problem 20). If f is increasing,
a much stronger statement can be made:
For any x in [0,1], the sequence
x,
a fixed point, by Problem 20). Prove this
the behavior of the sequence for f (x) > x and
by
f (x) < x, or by looking at Figure 10. A diagram of this sort is used
in Littlewood's Mathematician'sMiscel&my
to preach the value of drawing
only
pictures: "For the professional the
proof needed is [thisFigure]."
Suppose that f and g are two continuous functions on [0,1], with 0 5
f(x) < 1 and 0 < g(x) 5 1 for all x in [0, 1], which satisfy fog
go f. Suppose, moreover, that f is increasing. Show that f and g have
has a limit
assertion,
FI GU RE lo
*(b)
f(x), f(f(x)),...
is necessarily
(which
examining
=
22. InfiniteSequences459
a common
f (l)
=
l
=
fixed point; in other words, there is a number l such that
g(l). Hint: Begin by choosing a fixed point for g.
amused themselves by asking whether the
For a long time mathematicians
conclusion of part (b)holds without the assumption that f is increasing, but
in the Noticesof the American Mathematitwo independent announcements
cal Society, Volume 14, Number 2 give counterexamples,
so it was probably
a
pretty silly problem all along.
The trick in Problem 20 is really much more valuable than Problem 20 might
point theorems" depend upon
suggest, and some of the most important
looking at sequences of the form x, f (x), f (f (x)),
A special, but representaof
such
treated
in Problem23 (forwhich the next problem
theorem is
tive, case one
is preparation).
"fixed
.
..
.
-¢
22.
(a) Use Problem 2-5 to show
c" Mm+1 +
(b) Suppose that |c| <
that if c
1, then
Cm
.
.
.
+ c"
Cn+1
-
=
.
1
-
c
1. Prove that
lim c'"+---+c"=0.
that {x.) is a sequence with
Prove that {x.}is a Cauchy sequence.
(c) Suppose
*23.
|x.
---
xn+1| 5 c", where c
<
l.
Suppose that f is a function on R such that
(*)
where c
<
|f (x) f (y)| 5
-
c|x
-
for all x and y,
y|,
1. (Such a function is called a contraction.)
(a) Prove that f is continuous.
(b) Prove that f has at most one
(c) By considering the sequence
x,
fixed point.
f(x), f(f(x)),
...,
for any x, prove that f does have a fixed point. (This result, in a more
lemma.")
general setting, is known as the
"contraction
24.
(a) Prove that
if f is differentiable and |f'[
<
1, then f has at most one
fixed point.
(b) Prove that if |f'(x)| 5 c < 1 for all x, then f has a fixed point.
(c) Give an example to show that the hypothesis |f'(x)| 5 1 is not suflicient
to insure that f has a fixed point.
25.
This problem is a sort of converse to the previous problem. Let be be a
lim b, exists
sequence defined by bi
a, b.4i
f (b.). Prove that if b HNOO
and f' is continuous
at b, then [J'(b)| 5 1. Hint: If |f'(b)| > 1, then
=
=
=
and InßniteSaies
460 InfiniteSequences
1 for all x in an interval around b, and b, will be in this interval
for large enough n. Now consider f on the interval [b,bn].
|f'(x)|
26.
>
This problem investigates for which a
>
0 the symbol
aa
makes
b
=
In other words, if we define bi
sense.
a, basi
=
=
ab
,
when
does
lim b, exist?
n->OO
b. (The situation is similar to that in
if b exists, then ab
Problem 5.)
According
to part (a),if b exists, then a can be written in the form
(b) y1/2
ylj> and conclude that
for some y. Describe the graph of g(y)
0<ase1/e
(c) Suppose that 1 5 a 5 e 16. Show that {b.}is increasing and also b. E e.
This proves that b exists (andalso that b 5 e).
(a) Prove that
=
=
The analysis for a
<
1 is more difBcult.
Problem 25, show that if b exists, then e¯1
elle
that e¯'
sa5
(d) Using
From now on we will suppose that e¯"
(e) Show that
<
a
<
sbs
e.
Then show
1
the function
a2
f(x)= logx
is decreasing on the interval (0, 1).
Let
b be the unique number such that ab
b. Show that a < b < l.
(f)
Using part (e),show that if 0 < x < b, then x < aa' < b. Conclude that
lim a2, i exists and that aa' y
l N->OD
(g) Using part (e)again, show that I b.
lim b. b.
lim a2.42
b, so that ¤-700
(h) Finally, show that R->OG
=
=
=
=
=
27.
Let {x.}be a sequence
y.
(a) Prove that
x, or
R->OD
=
is bounded, and let
sup{x., xn+1, Xn+2,
the sequence
-
-
-
Ì-
{y,) converges. The limit lim
lim sup x., and called the limit superior,
of the sequence
(b) Find 5
which
{x.}.
x, for each of the following:
(i)
x,
=
(ii)
x,
=
1
-.
n
1
(--1)"-.
n
HNDO
y. is denoted by
or upper
limit,
22. InßniteSequences461
(iii) x,
=
(iv) x,
=
(-1)" 1 +
-
1
.
n
J.
(orliminf
(c) Define lim x,
x.) and prove that
that lim x, exists if and only if
B-koQ
(d) Prove
lim x,
case N-FOO
=
4
E
00
x,
i
R-koQ
x.
=
g
x, and that in this
lim x..
=
the definition, in Problem 8-18, of E A for a bounded set A.
that
if the numbers x, are distinct, then lim x,
lim A, where
Prove
(e) Recall
=
n->oo
A={x.:ninN).
28.
In the Appendix to Chapter 8 we defined uniform continuity of a function
on an interval. If f (x) is defined only for rational x, this concept still makes
sense: we say that f is uniformly continuous on an interval if for every e > 0
there is some ô > 0 such that, if x and y are rational numbers in the interval
and
|x
-
y|
(a) Let x
<
8, then
be any
|f (x) f (y) | <
e.
-
or irrational) point in the
(rational
of rational
interval, and let (x.) be
lim xn x. Show
such that
points in the interval
a sequence
naoo
that the sequence {f (x.)) converges.
(b) Prove that the limit of the sequence {f (x.)] doesn't depend on the choice
of the sequence (x.}.
We
whole
will
denote this limit by
interval.
(c) Prove that
the extended
f(x),so
jis
function
j
that
=
is an extension
uniformly
of
continuous
to the
f
on the
inter-
val.
29.
0, and for rational x let f (x) a2, as defined in the usual elementary
algebraic way. This problem shows directly that f can be extended
to a
continuous function / on the whole line. Problem 28 provides the necessary
Let a
>
=
machinery.
for rational x < y.
10, show that for any e
numbers x close enough to 0.
(a) Show that a' <
(b) Using Problem
rational
(c) Using the
ai
equation a'
-
af
=
aT(a**I
interval f is uniformly continuous,
(d) Show that the extended function
isfies
*30.
f(x+y)=
/
>
0 we have |ax
-
1|
<
e for
1), prove that on any closed
in the sense of Problem 28.
of Problem 28 is increasing and sat-
f(x)f(y).
The Bolzano-Weierstrass Theorem is usually stated, and also proved, quite
differently than in the text-the classical statement uses the notion of limit
and InßniteSeries
462 InfiniteSequences
points. A point x is a limit point of the set A if for every e
point a in A with |x a| < e but x ¢ a.
>
0 theœ is a
-
(a) Find
all limit
(i)
points of the following sets.
I1
-:ninN
.
n
1
1
-+--:nandminN
(ii)
.
m
n
1+
(iii) (-1)"
n in N
:
.
(iv) Z.
(v) Q.
x is a limit point of A if and only if for every e > 0 there are
infinitely many points a of A satisfying x a| < e.
Prove that ÏÏmA is the largest limit point of A, and lim A the smallest.
(b) Prove that
-
(c)
The usual form of the Bolzano-Weierstrass Theorem states that if A is
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:
form already proved in the text. Hint: Since A is infinite, there
in A.
are distinct numbers xi, x2, x3,
(e) Using the Nested Intervals Theorem. Hint: If [a,b] is divided into two
intervals, at least one must contain infinitely many points of A.
(d) Using the
...
31.
(a) Use the Bolzano-Weierstrass Theorem to prove that if f is continuous
on [a,b], then f is bounded above on [a,b]. Hint: If f is not bounded
above, then there are points x, in [a,b] with f(x.) > n.
the Rolzano-Weierstrass Theorem to prove that if f is continuous on [a,b], then f is uniformly continuous on [a,b] (seeChapter 8,
(b) Also use
Appendix).
**32.
(a) Let {a,}be the
1
2'
sequence
1 2
3= 3'
1
4'
2
4·
3
4'
1
5'
2
5'
3
5'
4
5'
1
6'
2
6'
""
Suppose that 0 5 a < b 5 1. Let N(n; a, b) be the number of integers
2, and N(4;
j 5 n such that a; is in [a,b]. (Thus N(2;
) 3.)
Prove that
(,g)
,
N(n;a,b)
naoo
(b) A sequence
in
.
hm
n->oo
=
-b-a.
n
{a,}of numbers
[0,1] if
,
in
[0,1] is called
N(n; a, b)
n
=b--a
uniformly
distributed
22. InfiniteSequences463
for all a and b with 0 5 a < b i 1. Prove that if s is a step function
defined on [0,1], and {a,}is uniformly distributed in [0,1], then
s=
O
lim
s(ai)+---+s(a,,)
n->oo
n
(c) Prove that if {a.}is uniformly
on [0,1], then
i
.
f=hm
n--oo
**33.
(a) Let f
distributed in
f(ai)+---+
be a function defined on
[0, 1] and f
f(a.)
is integrable
.
n
[0, 1] such
that lim
y->a
f (y) exists for all
a
e > 0 prove that there are only finitely many points a
with | lim f (y) f (a)| > e. Hint: Show that the set of such
1]
[0,
y->a
points cannot have a limit point x, by showing that Iim f (y) could not
in
in
[0,1]. For any
-
y-ex
exist.
in the terminology of Problem 21-5, the set of points where
f is discontinuous is countable. This finally answers the question of
Problem 6-16: If f has only removable discontinuities, then f is continuous except at a countable set of points, and in particular, f cannot be
discontinuous everywhere.
(b) Prove that,
CHAPTER
INFINITE SERIES
Infinite sequences
were
introduced in the previous chapter with the specific inten-
tion of considering their
"sums"
ai + a2 ‡ a3 ¯I¯°·
·
in this chapter. This is not an entirely straightforward matter, for the sum of
infinitely many numbers is as yet completely undefined. What can be defined are
the
"partial
sums"
ss=a1+·--+as,
and the infinite sum must presumably be defined in terms of these partial sums.
Fortunately, the mechanism for formulating this definition has already been developed in the previous chapter. If there is to be any hope of computing the infinite
closer and closer apsum ai + a2 + a3 +
sums s, should represent
, the partial
proximations as n is chosen larger and larger. This last assertion amounts to little
sum" ai + a2 Ÿ G3 +
ought
more than a sloppy definition of limits: the
"sum"
necessarily
approach
will
of
leave the
to be lim s,. This
many sequences
·
·
·
"infinite
-
-·
B-FOO
undefined,
since the sequence
{s,}may
easily fail to have a limit. For example,
the
sequence
-1,
1,
with a,
=
(-1)n+1yields
-1,
1,
...
the new sequence
s1=at=1,
s2 = ai + a2
ss=ai+a2
=
0,
3=Î,
s4=al+a2+a3+G4=Û,
for which
lim s, does not exist. Although there happen to be some clever extensions of the definition suggested here (seeProblems 9 and 24-20) 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.
nooo
464
23. InfiniteSeries 465
The sequence
DEFINITION
{a,}is summable
if the sequence
s,
ai +
=
·
-
-
(s.} converges,
where
+ an.
lim s. is denoted by
In this case, HROO
(or,less formally, al
a.
+ a2 + a3 +
-
-
n=1
and
is called the sum of the sequence
{a,}.
The terminology introduced in this definition is usually replaced by less precise
expressions; indeed the title of this chapter is derived from such everyday language.
An infinite sum
an
is usually called an infmitesenes, the
word
"series"
emphasiz-
n=I
ing the connection with the infinite sequence
is not, summable
is conventionally
replaced
{a,}.The
statement
by the statement
does, or does not, converge. This terminology is somewhat
best the symbol
as
denotes a number
n=1
(soit can't
"converge"),
that
that the
{a,}is, or
series
a.
peculiar, because at
and it doesn't de-
note anything at all unless {a.}is summable. Nevertheless, this informal language
is convenient, standard, and unlikely to yield to attacks on logical grounds.
Certain elementary arithmetical operations on infinite series are direct consequences of the definition. It is a simple exercise to show that if {a.}and {b.}are
summable,
then
(an+
b.)
n=1
n=1
n=1
c·a.=c
n-1
b.,
an +
=
a..
n=1
As yet these equations are not very interesting, since we have no examples of
summable sequences (except
for the trivial examples in which the terms are eventually all 0). Before we actually exhibit a summable sequence, some general conditions for summability will be recorded.
There is one necessary and sufficient condition for summability which can be
stated immediately. The sequence (a,}is summable if and only if the sequence {s,}
lim sm
converges, which happens, according to Theorem 22-3, if and only if FN,H-¥DO
-
s.
=
0; this condition can be rephrased
in terms of the original sequence as follows.
and InfiniteSeries
466 InfiniteSequences
{a.) is summable if and
The sequence
THE CAUCHY CRITERION
only
lim a,4i +···+am
if
0.
=
NE.R->OC
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.
If {an} is summable,
THE VANISHING CONDITION
then
lim a,
0.
=
n->oo
This condition follows from the Cauchy criterion by taking m
l, then
lim s.
be proved directly as follows. If B->OD
=
n
+ 1; it can also
-
lim a,
lim (s.
=
-
s,
HNCO
8400
i)
_
=
lim s.
lim s.
-
R-FOO
N->00
_
i
=l-l=0.
lim 1/n
Unfortunately, this condition is far from sufficient. For example, B-¥OC
0,
but the sequence {l/n} is not summable; in fact, the following grouping of the
numbers 1/n shows that the sequence {s,}is not bounded:
=
1+(+
(+¾+¾+g+
-2
I
(+(+(+-··+¾+-··
.
-2
(2terms,
each à¼)
1
(4terms,
each
1
-2
()
(8terms,
each
g)
of proof used in this example, a clever trick which one might never
reveals
the
need for some more standard methods for attacking these problems.
see,
These methods shall be developed soon (oneof them will give an alternate pmof
The method
1/n does not converge)
that
but it will be necessary
to first procure a few
n=1
examples of convergent series.
The most important of all infinite series are the
"geometric
r"=1+r+r2+r3+
series"
-.
n=o
Only the cases (r| < 1 are interesting, since the individual terms do not approach 0
if |r| ;> 1. These series can be managed because the partial sums
s,=1+r+..-+r"
can be evaluated in simple terms. The two equations
s.=1+r+r2+-..+r"
rsn=
r+r2+---+r"+r"*l
23. InfiniteSenes 467
lead to
s.(1
r)
-
1
=
or
1
s.=
by
(division
lim r"
=
n->oc
1
-
r
is
rn+1
-
r.+1
-
1
-
r
valid since we are not considering
the case r
1). Now
=
0, since Ir| < 1. It follows that
°°
r"
=
1
lim
1
=
1
n->=
n-0
r"+1
-
---
1
r
|r
,
-
r
<
l.
In particular,
"/1"
"
-
n=l
2
=L
1"
2
1
I_1-1=1,
-1=
-
2
that is,
(+)+¼+¼+...=1,
an infinite
FIGURE
sum which can always
be remembered
from the picture in Figure 1.
1
Special as they are, geometric series are standard
tests for summability will be derived.
For a while we shall consider
only sequences
examples
{a,}with
from which important
each
0; such
nonnegative.
called
nonnegative
then
the seIf
is
sequence,
{a.} a
sequences are
remark,
with
combined
clearly
nondecreasing.
This
Theorem 22-2,
quence {s.}is
provides a simple-minded test for summability:
a,
>
A nonnegative sequence {a,} is summable if and only if the set of partial
sums s, is bounded.
THE BOUNDEDNESS CRITERION
By itself this criterion is not very helpful-deciding whether or not the set of
is bounded is just what we are unable to do. On the other hand, if some
convergent series are already available for comparison, this criterion can be used
to obtain a result whose simplicity belies its importance (itis the basis for almost
all s,
all other tests).
THEOREM
l
(THE COMPARISON TEST)
Suppose that
O < a. 5 b.
for all n.
and InßniteSeries
468 InßniteSequences
Then if
bn converges, so does
n=1
PROOF
a..
n=1
If
+---+as,
+---+b.,
s,=ai
tn=bi
then
Oss.st,
Now
{t.) is bounded,
since
foralln.
b, converges.
quently, by the boundedness criterion
Therefore {s.}is bounded; conse-
a, converges.
n=1
the comparison test can be used to analyze very complicated
series
which
looking
in
most of the complication is irrelevant. For example,
Quitefrequently
2 + sin3(n + 1)
2= + n2
n-1
converges because
05
2 + sin3(n + 1)
2"+n2
<
3
and
"3
"1
n=1
n=1
series.
is a convergent (geometric)
Similarly, we would expect the series
°°
1
2"
-
1 + sin2 n3
to converge, since the nth term of the series is practically 1/2" for large n, and we
would expect the series
°°
n+1
n=1
to diverge, since
n2+1
(n + 1)/(n2
+ 1) is practically 1/n for large n. These facts can
the
be derived immediately from
following theorem, another kind of
test."
"comparison
-/
THEOREM
2
If an, b,
converges.
>
0 and HNCO
lim a./b.
=
c
a, converges if and only if
0, then
n=1
b.
n=1
23. InfiniteSeries 469
00
PROOF
Suppose
Since R-FDD
lim a,|b,
b, converges.
n=1
-
a, 5 2cb,
But the sequence 2c
c, there is some N such that
for n
>
b, certainly converges.
N.
Then Theorem 1 shows that
n=N
as converges, and this
implies convergence of the whole series
an, which
n= N
has only finitely many additional terms.
The converse follows immediately, since we also have lim bs/a.
n->oo
n= I
=
1/c ¢ 0.
The comparison test yields other important tests when we use previously analyzed series as catalysts.
Choosing the geometric series
r", the convergent
n-0
series
THEOREM
3 (THE RATIO TEST)
Let a.
par excellence,we obtain the
>
important of all tests for summability.
most
0 for all n, and suppose that
Un+1
.
hm
=soo
Then
a, converges
if r
<
=
--
a.
r.
l. On the other hand, if r
1, then the terms a, do
>
n=1
not approach
0, so
a,
diverges. (Notice that it is therefore essential to compute
n=1
lim an+1/a. and not lim a./a,4i!)
n->oo
PROOF
Suppose first that r
n-woo
<
1. Choose any number s with r
=r<1
lim
nooo
as
implies that there is some N such that
n+1
forn>N.
<s
This can be written
for n
an+1 5 sa,
Thus
aN+1
SGNs
aN+2
SUN+1
aN+k
S GN,
SkaN
-
_>
N.
<
s
<
1. The hypothesis
470
and InfiniteSeries
InfiniteSequences
aNSk
Since
Sk
UN
=
k=0
the comparison test shows that
converges,
k=0
=
a.
aN+k
n=N
converges.
k=0
This implies the convergence
The case r
>
1 is even easier. If 1 < s
a.si
which means
of the whole series
>
r, then there
<
for n
s
>
N,
=
0, 1,
an.
is a number
N such that
that
NSk
GN+k
k
> GN
....
This shows that the individual terms of {a.} do not approach 0, so {a.) is not
summable.
of the ratio
As a simple application
test, consider
the series
1/n!
.
Letting
n=1
a,
1/n! we obtain
=
1
n+1
(n+1)!
1
an
1
n!
(n + 1)!
Thus
.
lun
an+1
naoo
which
0,
=
--
as
1/n! converges. If we consider instead the series
shows that the series
n=1
r"/n!,
where
r
then
is some fixed positive number,
n=1
rn+1
(n + 1)!
n
r"/n!
so
oc
r
r"
oo
=
n+ 1
converges. It follows that
n=1
lim
naoo
r"
=
-
n!
0,
=
0,
23. InfiniteSeries 471
a result already proved in Chapter 16 (theproof given there was based on the same
ideas as those used in the ratio test). Finally, if we consider the series
have
lim
(n+ 1)rn÷1
lim r
=
nr"
naoo
-
1
n +
naoo
=
nrn we
r,
n
oc
since lim
(n + 1)/n
=
1. This proves that if 0 5 r
<
nr" converges,
1, then
and consequently
lim nr"
0.
=
N->OO
-1
< r
(This result clearly holds for
0, also.) It is a useful exercise to provide a
without
using
the ratio test as an intermediary.
direct proof of this limit,
Although the ratio test will be of the utmost theoretical importance, as a practical
tool it will frequently be found disappointing. One drawback of the ratio test is
the fact that lim anoi/a. may be quite difficult to determine, and may not even
_<
exist. A more serious deficiency, which appears with maddening regularity,
is the
might
a.4i/a.
equal
precisely
that
the
the
fact
limit
1. The case B-900
lim
1 is
one
=
which
is inconclusive: {an}might not be summable
(forexample,
if a,
=
1/n),
(1/n)2
but then again it might be. In fact, our very next test will show that
n=1
converges,
even though
1
n +
lim
1.
=
This test provides a quite different method for determining convergence or divergence of infinite series-like the ratio test, it is an immediate consequence of the
comparison test, but the series chosen for comparison is quite novel.
THEOREM
4 (THE INTEGRAL TEST)
Suppose that
Then
f
is positive and decreasing on
a, converges
[1,oo),
and that
f (n)
=
if and only if the limit
n=1
A
oo
Ï
1
exists
f
=
lim
A->oo
1
f
lA
PROOF
The existence of lim
A-voo
i
f
is equivalent to convergence
2
3
f+
2
4
f+
f+.-..
of the series
as
for all
n.
and InfiniteSeries
472 InfmiteSequences
Now, since f is decreasing we have (Figure 2)
n+1
f (n + 1) <
f
f (n).
<
The first half of this double inequality shows that the series
a,
i
may be com-
n=1
n+1
oc
pared to the series
i
"
n=1
1
2
3
4
n
oo
i
(andhence
an+1
n=1
n + 1
ian,
pared to the series
L
f
A
00
2
n+1
00
The second half of the inequality shows that the series
FIGURE
an) converges
n=1
exists.
if hm
|
oo
f, proving that
proving that lim
Amoo
n=1
be com-
oc
f
i
may
ia,
must exist if
converges.
n=1
Only one example using the integral test will be given here, but it setdes the
question of convergence for infinitely many series at once. If p > 0, the convergence of
by the integral test, to the existence of
1/nf is equivalent,
n=1
°°
-
l
dx.
xP
Now
1
ÏA
¯
-
dx
(p
=
-
1
1) Ap-1
1
p
-
1
log A,
p
=
1.
A
This shows that lim
Amoo
Ï
converges precisely for p
oo
1/xP
dx exists if p
1, but not if p 5 1. Thus
>
1/n?
n=1
>
l. In particular,
1/n diverges.
n=1
The tests considered so far apply only to nonnegative sequences, but nonpositive
sequences may be handled in precisely the same way. In fact, since
-a,
as
=
-
,
.=i
.=i
all considerations about nonpositive sequences can be reduced to questions involving nonnegative sequences. Sequences which contain both positive and negative
terms are quite another story.
If
an is a sequence
with
both positive and negative
terms, one can con-
n=i
|a,|, all
sider instead the sequence
n=1
of
whose
terms are nonnegative.
Cheerfully
23. InfiniteSeries 473
ignoring the possibility that we may have thrown away all the interesting information about the original sequence, we proceed to culogize those sequences which
are converted by this procedure into convergent sequences.
The series
DEFINITION
as is absolutely
if the series
convergent
la,| is convergent.
n=1
n=1
(In more formal language, the sequence {a.) is absolutely
summable
if the
{|a.|) is summable.)
sequence
Although we have no right to expect this definition to be of any interest, it turns
out to be exceedingly important. The following theorem shows that the definition
is at least not entirely useless.
THEOREM
Every absolutely convergent series is convergent. Moreover, a series is absolutely
convergent if and only if the series formed from its positive terms and the series
formed from its negative terms both converge.
5
If
PROOF
|a.|
converges, then, by the Cauchy criterion,
n=I
lim |a.wil+---+|a.|=0.
.
m,n-oo
-
Since
a,4t|+---+|a.|,
|a,41+---+a.|5
it follows that
lim an+1+···+a.=0,
m,n->oo
which shows that
as converges.
n=1
To prove the second part of the theorem, let
a.*
so that
a
=
a,
=
Ï a.,
Í
0,
a.,
0,
if a. 2 0
if a. 5 0,
if as 5 0
if a. à 0,
is the series formed from the positive terms of
n=1
an, and
n=i
is the series formed from the negative terms.
a,*
If
n=1
and
both converge,
a,¯
then
n=1
a,
n=1
|
[an* (an¯)]
=
-
n=1
=
a.
n=1
an¯
--
n=1
a,
n=1
474
and Infinite Series
InfiniteSequences
also converges, so
a, converges
On the other hand, if
|a,|
absolutely.
converges,
then, as we have just shown,
also converges.
as
n=i
n=i
Therefore
n=1
n=1
n=1
°°
°°
and
=
1
both converge.
It followsfrom Theorem 5 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
convergent).
In order to prove this state(suchseries are called conditionally
ment we need a test for convergence which applies specifically to series with positive
and negative terms.
THEOREM
6 (LEIBNIZ'S THEOREM)
Suppose that
ai>a22a32---20,
and that
lim a,
=
0.
n->oo
Then the series
(-1).+1a,
=
ai
-
a2
3
-
a4 * US
·¯
••
n=1
converges.
PROOF
Figure 3 illustrates
relationships
(1)s2 5 s4 36
(2)siksstss2···,
'
(3)sk
FIGURE
se
3
·
>
if k is even and I is odd.
al
flllll
ss
·
between the partial sums which
se
Ja 5:s
IIIIII
sa
In fo
17
Is
is
5:
we will
establish:
23. InfiniteSeries 475
To prove the first two inequalities, observe that
(1)
s2, + a2.wi
=
s2, 2
s2.,
(2)
=
s2,43
-
a2,42
since a2.gi
s2n+1
s2n+1,
à a2,42
Ÿ 02n+3
U2n+2
-
since a2n+2 2 a2n+3·
To prove the third inequality, notice first that
s2,
=
s2.,
5
52n-l
-
a2,
since a2n
0.
This proves only a special case of (3),but in conjunction with
case is easy: if k is even and I is odd, choose n such that
2n 2 k and
2n
-
(1)and (2)the
general
1 à l;
then
Sk
52n
52,_)
Si,
which
proves (3).
Converges,
Now, the sequence {s2n}
above (bys¿ for any odd l). Let
a
because it is nondecreasing
sup(s2,}
=
and is bounded
lim s24.
=
n->oo
Similarly, let
ß
It follows from
(3)that
a 5
s2.wi
it is actually the case that
The standard
example
lim s2n+1·
=
B->OO
ß; since
s2,
-
«
inf(s2.wi}
=
=
=
ß.
a2n+1
and
lim a,
0
=
R->OO
This proves that a
=
ß
=
lim s..
HNOO
derived from Theorem 6 is the series
1-(+¾-¼+(----,
which
is convergent,
but not absolutely
convergent
(sincen=1
1/n does
not
If the sum of this series is denoted by x, the following manipulations
to quite a paradoxical œsult:
verge).
x=1-j+¾-¼+¾-(+---
(thepattern
here is one positive term followed by two negative ones)
=(1-))-¼+(j-4)-¼+(J-Å)-i+(j-¾)·-i+-
=l-¼+g-¼+A-h+Å-·rš+
con-
lead
and InßniteSeries
476 InfiniteSequences
-/-
x/2,
0. On the other hand, it is easy to see that x 0:
the partial sum s2 equals
and the proof of Leibniz's Theorem shows that x ;> s2.
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
so x
=
implying that x
=
),
sequence
contains all the numbers
in the sequence
11
1(}
11
11
11
1
1
l
of {an}in the following precise sense: each
In fact, {b,}is a rearrangernent
b,
the natural numbers,
agn) where f is a certain function which
that is, every natural number m is f (n) for precisely one n. In our example
"permutes"
=
f (2m+
1)
=
3m + 1
(theterms 1, y, y
places),
=
3m
(theterms
,
.
.
.
go into the 1st, 4th, 7th,
.
..
-¾,
f(4m)
...
f(4m + 2)
=
3m + 2
-g,
places),
(theterms
...
go into the 3ri, 6th, 9th,
-y,...
-(,
-y,
go into the 2nd, 5th, 8th,
-y,
...
places).
b, should equal
Nevertheless, there is no reason to assume that
n=1
sums are, by definition, lim bi +
·
·
naoo
order
·
+ b, and lim ai +
usoo
of the terms can quite conceivably matter.
a.:
these
n=1
.
.
.
+ a., so the particular
(-1)"**/n is not
The series
n=1
special in this regard;
solutely convergent-the
than a theorem) shows
THEOREM
7
indeed, its behavior is typical of series which are not abfollowing result (really
more of a grand counterexample
how bad conditionally convergent series are.
a, converges, but does not converge absolutely, then for any number a there
If
n=1
is a
PROOF
rearrangement
Let
p,
n=1
{b,}of {a,}such
that
b,
=
a.
denote the series formed from the positive terms of {an}and let
q.
n=1
denote the series of negative terms. It followsfrom Theorem 5 that at least one of
these series does not converge. As a matter of fact, both must fail to converge, for
if one had bounded partial sums, and the other had unbounded partial sums, then
23. InfiniteSeries 477
the original series
as would also have unbounded
partial sums, contradicting
the assumption
that it converges.
Now let a be any number. Assume, for simplicity, that
<
a
0 will be a simple modification).
there is a number
Since the series
«
>
0
(theproof
for
pn is not convergent,
N such that
N
n=1
We will choose Ni to be the smallest N with this property. This means that
NI-1
(1)
p. 5 a,
n=1
Ni
(2)
but
p,
>
a.
n=1
Then if
N1
Si
=
L Pn,
n=1
we
have
Si
IIII
I
p,+
o
FIGURE
·
+ps_,
I
a
a 5
-
pN1*
I
p,+
-
+p,._,+ps,
-
4
This relation, which is clear from Figure 4, followsimmediately from equation
(1):
N1-1
S1-a5S1-
Pn=PNin=1
To the sum S1 we now add on just enough negative terms to obtain a new sum T1
which is less than a. In other words, we choose the smallest integer Mi for which
M1
Ti=Si+
q.<a.
n=1
As before, we have
«
-
Ti 5
-qui
.
We now continue this procedure indefinitely, obtaining
and
smaller than a, each time choosing the smallest
sums alternately
Nk or Mk possible.
larger
The
478
and Infinite Series
Inßnite Sequences
sequence
pt,
pNI,91,•••,
...,
9Mi,
PN1+1=
···
s
PN2''''
of {a,}.The partial sums of this rearrangement increase to Si,
is a rearrangement
then decrease to Ti, then increase to S2, then decrease to T2, etc. To complete the
and |Tk-«| are less than or equal to pN, or -qMa,
proof we simply note that |Sk
respectively,
and that these terms, being members of the original sequence {a.},
-a|
decrease to 0, since
must
a, converges.
n=i
Together with Theorem 7, the next theorem establishes conclusively the distinction between conditionally convergent and absolutely convergent series.
THEOREM
8
a. converges absolutely,
If
and
{b.) is any
of
rearrangement
{a.},then
b.
n=1
n=I
also converges
and
(absolutely),
an
bn-
=
n=1
PRooF
Let
us
n=1
denote the partial sums of {an]by s., and the partial sums of {b,}by t..
Suppose that e
>
a, converges, there is some N such that
0. Since
n=1
E-
a.-sN
n=1
Moreover, since
|a,|
converges, we can also choose N so that
n=1
|a.|
-
·+
(lail +
)
aN
< E,
n=i
i.e., so that
GN+1
#
|aN+2
Now choose M so large that each of a1,
whenever m > M, the difference t.
sN
exduded. Consequently,
are definitely
-
|t.
-
SN
N+1
N+3
aN
...,
<
'''
E-
appear among b1,
is the sum of certain
N+2
N+3
...,
bM·
a¿, rehere ai,
·
-
Then
...,
aN
23. InfiniteSeries 479
Thus, if m
>
M, then
a.-t.
a.-sN"(Ím-SN)
=
n=1
n=1
5
an
-
IN
Ÿ Im
SN
¯
n=1
<s+e.
Since this is true for every e
>
0, the series
bn converges to
an.
==1
n=1
b, converges absolutely, note that {|b,| } is a rearrangement
To show that
n=1
of
{|a,\ };since
|a,|
converges
absolutely,
|b,| converges by the first part
of
n=1
n=1
the theorem.
is also important when we want to multiply two infinite
series. Unlike the situation for addition, where we have the simple formula
Absolute convergence
an
n=1
b,
+
n=i
(a, + b.)
=
,
n=1
there isn't quite so obvious a candidate for the product
a.
n=1
bN
-
')·
=(a1+a2+---)-(bi+b2
n=1
It would seem that we ought to sum all the products a,bj. The trouble is that these
form a two-dimensional array, rather than a sequence:
aib2
alb3
a2b1
a2b2
U2b3
a3bi
asb2
asb3
albi
---
·
•
-
-•
-
Nevertheless, all the elements of this array can be arranged in a sequence. The
picture below shows one way of doing this, and of course, there are (infinitely)
many other ways.
arbi
arba
albs
a2bi
a2b;
a2b3
aabi
aaba
a3bs
...
...
.
and InfiniteSenes
480 InfmiteSequences
Suppose that (c,} is some sequence of this sort, containing each product a¿bj
just once. Then we might naively expect to have
=
c,
be.
·
a,
n=1
n=I
n=1
But this im't true (seeProblem 8), nor is this really so surprising, since we've said
of the terms. The next theorem shows
nothing about the specific arrangement
that the result does hold when the arrangement of terms is irrelevant.
THEOREM
9
b, converge absolutely,
as and
If
n=1
and
{c.) is any
sequence
containing the
n=1
products arbj for each pair (i, j), then
=
c,
a.
n=1
PROOF
b..
-
nl
n=l
Notice first that the sequence
L
L
i|.
pL=
|b;|
i=1
j=1
converges, since {a,}and {b.}are absolutely convergent, and since the limit of a
product is the product of the limits. So (PL} is a Cauchy sequence, which means
that for any e > 0, if L and L' are large enough, then
i|·
i=1
|bj|-
|bj|
|a¿|i=1
j=1
<
.
j=1
It follows that
L
(1)
|a;| |bj| 5
<
.
e.
i or j>L
Now suppose that N is any number so large that the terms c, for n 5 N include
every term arbj for i, j 5 L. Then the difference
N
L
L
(c.- La¿-Ebj
n=1
consists of terms a¿b; with i
>
i=1
L or
N
(2)
n-l
j
>
j=1
L, so
L
L
c.-Lag·ibj
i=l
i
5
|ar|-|bj|
i or j>L
j=l
<
e
by (1).
But since the limit of a product is the product of the limits, we also have
(3)
by
a¿·
i=1
j=1
-
bj
a¿·
i=1
j=l
<
e
23. InfiniteSeries 481
for large enough L. Consequently, if we choose L, and then N, large enough, we
will have
og
N
oo
oo
La¿-Ebj Ec,
i=1
L
-
i=1
j=1
L
oo
Lag-Ebj La¿-Eb;
5
-
i=1
i=1
j=1
L
La¿-ibj
ic,
j=1
+
-
n=1
i=l
2e
<
j=1
N
L
by
(2)and (3),
which proves the theorem.
Unlike our previous theorems, which were merely concerned with summability,
this result says something about the actual sums. Generally speaking, there is
in any simpler
no reason to presume that a given infinite sum can be
expressions
simple
equated
infinite
be
However,
to
many
sums by using
terms.
can
Taylor's Theorem. Chapter 20 provides many examples of functions for which
"evaluated"
"f(a)
f (x)
(x
=
a)' + Rn,a(x),
-
i=0
where
lim R..a(x)
0. This is precisely equivalent to
=
BROC
"
f (x)
f (a)
i
lim
=
.
(x
a)',
-
i=0
which means,
in turn, that
°°
f (x)
f (a)
=
(x
a)'.
-
i=0
As particular examples
we
have
x
.
3
3!
x
2
x
2!
X
=
x
-
,
7!
4
x
6
6!
4!
X2
X3
X4
2!
3!
4!
·,
1!
arctan x
7
x
5!
cosx=1---+---+-·-,
«'=1+-+--+-+-+
5
x
smx=x--+-----+
x
3
--
3
2
+
x
5
-
7
--
5
3
x
x
log(1+x)=x--+-----+-+---,
2
3
-
x
+
7
4
x
4
·
·
-
,
x
5
5
|x| 5 1,
-1<xs1.
and InfiniteSeries
482 InßniteSequences
(Notice that the series for arctan x and log(1+x) do not even converge for |x| > 1;
the series for log(1 + x) becomes
in addition, when x
-1,
=
does not converge.)
Some pretty impressive
which
results
with
are obtained
3
5
particular
values
of x:
7
0=×--+---+-·-,
3!
5! 7!
1
l
1
e= l+-+---+--+---,
1! 2! 3!
1
1 1
x
4
357
1 1 1
log2=1--+---+....
2 3 4
-=1--+---+---
'
More significant developments may be anticipated if we compare the series for
sin x 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 of the equation
.
smx=x---+--...
x3
75
3!
5!
term-by-term, ignoring the fact that we have never proved anything about the
derivatives of infmite sums. Likewise, if we differentiate both sides of the formula for
we obtain the formula cos'(x)
cos x formally (i.e.,without justification)
sin x, and if we differentiate the formula for e= we obtain exp'(x)
exp(x).
shall
such
of
the
chapter
that
term-by-term
next
In
differentiation infinite
we
see
sums is indeed valid in certain important cases.
=
=
-
PROBLEMS
1.
Decide whether each of the following infmite 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
(andthen again, they may not). The hint below indicates which tests may be
used.
°°
sin nÐ
n2
(i)
L
(ii)
1-¼+¾-(+·-·.
(iii)
1-(+(-)+(-j+(-(+--·.
(iv)
(-1),
n=1
•
log n
23. InfiniteSerms 483
©
1
(v)
(The summation begins with n 2 simply to avoid
I the meaningless term obtained for n 1).
=
.
n2
n=2
-
=
1
(vi)
.
n2‡l
(vii)
.
n=1
logn
n=1
"1
(ix)
--.
n=2
°°
(x)
1
L
(IOg 4) k
n=2
'
(log1n)"
°
(xii)n=2 (--1)"(ÎOg
1
8)n
n3n+
(xiii)
1
n=1
(xiv)
sin
.
"
1
(xv)
.
nlo
n=2
n
*
(xvi)L
n=2
1
.
n(logn)
=
(xvii)
1
n2(logn)
(xviii)
.
n=1
°°
(xix)
2"n!
-
.
.
484
and InßniteSeries
InßniteSequences
°°
(xx)
3"n!
--.
n=1
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 infmite series that require
more delicate analysis than those in Problem 1.
*2.
solved examples
(a) If you have successfully
a"nl/n"
it should be clear that
(xix)and (xx)from
for a
converges
<
Problem 1,
e and diverges for
n=1
a
>
e. For a
=
e the ratio test fails; show that
e"n!/n"
actually
n=1
diverges, by using Problem 22-13.
(b) Decide
when
*3.
nn/a"n!
when
converges, again resorting to Problem 22-13
n=1
the ratio test fails.
(logn)-k
Problem 1 presented the two series
which
lies between these two, is analyzed
e7/y?
(a) Show that fg°°
of which
n=2
the first diverges while the second converges.
(logn)1°5"
(logn)~",
and
n=2
The series
'
in parts
dy exists, by considering
(a)and (b).
the series
(e/n)".
n=1
(b) Show that
(logn)l°8"
converges, by using the integral test. Hint: Use an appopriate
tion and part (a).
substitu-
(c) Show that
(logn)°8½")
diverges, by using the integral test. Hint: Use the same substitution as
in part (b),and show directly that the resulting integral diverges.
23. InfiniteSeries 485
nl+1,,,
4.
Decide whether
5.
(a) Let {an}be a sequence of integers with 0 5 a. 5 9. Prove that
n=1
exists
(andlies between 0 and 1). (This, of course, is the number
we usually
or not
converges.
a,10¯"
which
·)
denote by 0.ara2a3a4··
(b) Suppose that 0 5 x 5 1. Pove that there is a sequence of integers
with
a.10¯"
0 5 as 5 9 and
=
Hint: For example,
x.
ai
=
n=1
(where[y]denotes the greatest integer which
(c) Show
a1, a2,
that if
(a.)
ak,
1, a2,
...,
is repeating,
-
·
a.10¯"
then
.,
i.e.,
{a,}
[10x]
is y y).
is of the form
ai,a2,...,aks
is a rational number
(andfind
n=I
it). The same result naturally holds if {a,}is eventually
the sequence {aN+k) is repeating for some N.
(d) Prove that if x
ing. (Justlook at
a.10-"
=
is rational,
then
repeating,
{a.}is eventually
i.e., if
repeat-
n=i
the process of finding the decimal expansion of p/qdividing q into p by long division.)
6.
Suppose that (an}satisfies the hypothesis of Leibniz's Theorem. Use the
proof of Leibniz's Theorem to obtain the following estimate:
(-1).+1an
7.
-
Prove that if a, :> 0 and lim
n-Foo
[a1-a2+-··±aN
&
=
N-
r, then
as converges
if r
<
1, and
n=i
diverges if r > 1. (The proof is very similar to that of the ratio test.) This
result is known as the "mot
test." It is easy to construct series for which the
ratio test fails, while the root test works. For example, the root test shows that
the series
i+i+(i)'+(02+(03+(03+
·
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 is
nevertheless quite an important theoretical tool, and if the ratio test works the
root test will also (byProblem 22-18). It is possible to eliminate limits from
the root test; a simple modification
of the proof shows that
a, converges
n=1
if there is some s
<
1 such that all but fmitely many
a, diverges if infinitely many
n-1
W
are
>
6
are
<
s, and that
1. This result is known as the
and InfiniteSeries
486 InfiniteSequences
"delicate
test"
root
the notation
(thereis a
similar delicate ratio test). It follows, using
00
Problem 22-27, that
of
i
a, converges if
Eg
is possible if lim
g
<
1 and
n=l
diverges if lim
&
n->oo
8.
1; no conclusion
>
{a,}and {b.),let c,
For two sequences
n-woo
akbn+1-k·
=
=
1.
(Then c. is the sum
k=1
diagonal in the picture on page 479.) The series
of the terms on the nth
c.
is called the Cauchyproductof
as and
b..
n=1
n=I
(-1)"|{,
show that
|c,|
If a,
b,
=
=
n=1
2 1, so that the Cauchy product does not con-
verge.
9.
{a.}is called
A sequence
.
hm
with Cesaro sum l,
summable,
Cesaro
si+---+s,
n->oo
if
=l
n
Uk).
Problem 22-16 shows that a summable sequence
is automatically Cesaro summable, with sum equal to its Cesaro sum. Find
a sequence which is not summable, but which is Cesaro summable.
Sk
(where
10.
-
61
'
Suppose that a,
sequence
s.
'
0 and {a,}is Cesaro summable.
>
{na,}is bounded. Prove that
a, and a,
=
'
=
s,,
i=1
11.
prove that s.
(a) Suppose that
a, à
n+
=
-
(b) Show that
-
(c) Show that
n=1
(d) Now replace
-
1as is bounded.
-
-
.
b, exists.
b, if
n=I
a.
Hint: If
proof of Theorem 8 which does not rely
-
an 5
n=1
a, converges.
of {a,},
0 for each n. Let {b,}be a rearrangement
+ b.. Show that for each n
+ a. and t. = bi +
let s,
at +
there is some m with s. E t,
absolutely,
-
i=1
This problem outlines an alternative
on the Cauchy criterion.
and
the series
Suppose also that the
n=I
b..
=
n=1
the condition a. à 0 by the hypothesis that
using the second
a, converges
n=1
part of Theorem 5.
23. InfiniteSenes 487
00
12.
as converges absolutely,
if
(a) Prove that
and
{b,}is any subsequence
of
n=1
{an},then
b, converges
(absolutely).
n=1
an does not converge absolutely.
this is false if
(b) Show that
00
I
.
*(c)
Prove that if
absolutely,
a, converges
then
-)+(a2+a4+as+---).
an=(ai+a3+a5+-
13.
Prove that if
a.
is absolutely
convergent,
then
a.
*14.
ff
FIGURE
----
-----
_
_
(sinx)/x dx converges.
f
0 for all x such that
Prove
that
f°°f (x) dx
*15.
Find a continuous
function f with
exists, but lim f(x) does not exist.
*16.
0. Recall the definition
1, and let f(0)
of £(f, P) from Problem 13-25. Show that the set of all £(f, P) for P.a
length"). Hint: Try
partition of [0, 1] is not bounded (thusf has
partitions of the form
-1
-
19-42 shows that
|(sinx)/x| dx diverges.
Problem
|an|.
5
n=i
n=1
n=1
_
s
Let
f(x)
x sin 1/x for 0
=
<
f (x) >
xs
=
"infinite
P-
17.
*18.
0
2
'
'
(2n+1)x'
2 2 2 2
1
7x' 5x' 3x' x'
be the function shown in Figure 5. Find
shaded region in Figure 5.
Let
f
In this problem we will establish the
a
(l+x)"=
"binomial
f
f,
and also the area of the
series"
x*,
|x|<1,
A-0
lim Ro,o(x) = 0. The proof is in several steps,
for any a, by showing that 8400
and uses the Cauchy and Lagrange forms as found in Problem 20-7.
(a) Use the
for Ir|
<
a
ratio test to show that the series
r*
does indeed converge
k=0
1 (thisis not to say that it necessarily
lim
follows in particular that HNDO
r"
=
converges to
0 for |r|
<
1
(1 +
r)").
It
and InfmiteSeries
488 InßniteSequences
(b) Suppose first that
0 Ex
lim R..o(x)
1. Show that B->OO
<
=
0, by using
Lagrange's form of the remainder, noticing that (1 + t).-n-1 5 1 for
n + 1 > œ.
< x < 0; the number
t in Cauchy's form of the
(c) Now suppose that
remainder
satisfies
< x < t 5 0. Show that
-1
-1
|x(1+
t)"¯'1|
where
5 |x|M,
M
max(1,
=
x)Œ-Î
(1 +
and
x-t
=
1+t
1-t|x
1+t
ix|
<
¯
Using Cauchy's form of the remainder,
a
+1
(n+1)
show that lim R.,o(x)
(a) Suppose that the partial
and the fact that
a-1
=a
,
n
0.
=
RNOO
19.
ixi.
sums of the sequence
{b,}is a sequence with b.
>
{a,}are bounded
bn+1 and lim b,
and that
DO
a,b,
0. Prove that
=
naoo
n=1
This is known as Diricklet's test. Hint: Use Abel's Lemma
(Problem 19-35) to check the Cauchy criterion.
(b) Derive Leibniz's Theorem from this result.
converges.
(c) Prove, using
Problem 15-33, that the series
converges if x
(cosnx)/n
n=1
is not of the form 2kr for any integer k
(d) Prove Abel'stest: If
a, converges and
(inwhich
{b,}is a
case it clearly diverges).
sequence which is either
n=1
nondecreasing
or nonincreasing
and which
is bounded, then
a,b,
n=1
converges.
Hint: Consider b,
-
b, where b
=
lim b..
naoo
00
*20.
Suppose {a,}is decreasing and ANOO
lim a,
2"a2. also converges
then
=
0. Prove that if
(the"Cauchy
as converges,
n=1
Condensation Theorem"). No-
n=1
1/n is a special case, for if
tice that the divergence of
n=1
1/n converged,
n=1
2"(1/2") would also converge; this remark may serve as a hint.
then
n=1
23. InfiniteSeries 489
*21.
(a) Pove
b, converge, then
n=1
(b) Prove that
*22.
a,2 and
that if
n=1
a.*
if
a.b. converges.
n=i
then
converges,
n=1
aN
N
COHV€Tg€S
fOr
any a
>
.
n=1
Suppose {a.}is decreasing and each a, à 0. Prove that if
then lim na, = 0. Hint: Write down the
R-¥OC
use the fact that (as) is decreasing.
Cauchy criterion
a, converges,
and
be sure to
oo
*23.
partial sums s. are bounded, and B->OO
lim a,
a, converges, then the
If
n=i
=
0. It
is tempting to conjecture
that boundedness of the partial sums, together with
the condition
0, implies convergence of
00
lim a,
=
n-+oo
a..
This is not true,
n=1
but finding a counterexample requires a little ingenuity. As a hint, notice that
some subsequenceof the partial sums will have to converge; you must somehow
allow this to happen, without letting the sequence itself converge.
24.
Pove that if a, à 0 and
an diverges, then
1 an
Compare the partial sums. Does the converse hold?
n=i
25.
Let b, ¢ 0. We say that the infinite product
p,
=
i=1
also diverges. Hint:
n=1
b, converge if the sequence
b; converges, and also Iim p, ¢ 0.
n->oo
(a) Prove that
(b) Prove that
(c) For a,
>
if
(1 + an)
converges, then a, approaches 0.
n=1
(1+
a.) converges
if and only if
n=1
log(1 + a.) converges.
n=1
0, pmve that
(1 +
an) converges
n=1
if and only if
a, conn=1
Hint: Use Problem 24 for one implication, and a simple estimate
for log(1 + a) for the reverse implication.
verges.
26.
(a) Compute
(b) Compute
1
-
.
(1 + x2") for
x|
<
Î.
and InßniteSeries
490 InfiniteSequences
27.
The divergence of
1/n is related to the following remarkable
fact: Any
n=l
positive rational number x can be written as a finitesum of distinctnumbers
of the form 1/n. The idea of the proof is shown by the following calculation
: Since
for
_23
2Î
_
31
1
62
_
¯
2
62
3
186
1
4'
7
156
'''
l
26
¯
186
we
27
1674
have.
Notice that the
numerators
23, 7, 1 of the differences are decreasing.
if 1/(n + 1) < x
1/n for some n, then the numerator in this
decrease; conclude that x can be written
as a finite sum of distinct numbers 1/k.
(a) Prove that
<
sort of calculation must always
(b) Now prove the
result
for all x by using the divergence of
1/n.
n=1
UNIFORM CONVERGENCE
POWER SERIES
CHAPTER
AND
The considerations at the end of the previous chapter suggest an entirely new way
of looking at infinite series. Our attention will shift from particular infinite sums
to equations like
ex
1+
=
--
x
1!
x
+
2
+
----
2!
-
-
-
which
concern sums of quantities that depend on x.
defined by equations of the form
interested in functions
In other
f(x)= f1(x)+f2(x)+f3(x)+·
x"¯1/(n
words,
we
are
·
1)!). In such a situation (f,} will be
each
obtain
of
functions; for
x we
a sequence of numbers {f.(x)},
some sequence
of
order
and f (x) is the sum
this sequence.
In
to analyze such functions it will
certainly be necessary to remember that each sum
(inthe
previous example
fn(x)
=
-
f1(x) + f2(x)+ f3(x)+
·
is, by definition, the limit of the sequence
ft(x), f1(x)+ f2(x),fi(x) +
If we defme a new sequence
of
f2(X)
Ÿ
f3(X).
-·••
functions {sn}by
=
s.
f1 +
+
fn.
1
then we can express this fact more succinctly by writing
A
la
1
f (x)
rather
I
lim s.(x).
For some time we shall therefore concentrate on functions defined as limits,
f (x)
FIGURE
=
such
than on functions defined as
=
lim
H-+OD
f.(x),
infinite sums. The total body
summed
easily:
nothing
of results about
would
functions can be
hope to be
up very
one
true actually is-instead we have a splendid collection of counterexamples. The
first of these shows that even if each f, is continuous, the function f may not be!
Contrary to what you may expect, the fimetions f. will be very simple. Figuæ 1
shows the graphs of the functions
f,(x)
491
x",
=
1,
05xs1
x > 1.
492
and InfiniteSenes
ÍnßniteSequences
h
-I
I
1 II
A
II I
These functions are all continuous,
continuous;
in fact,
A
but the function f (x)
Hm f,(x)=
I
naoo
lim f.(x) is not
05x<1
1.
10,
Î,
x
Another example of this same phenomenon is illustrated in Figure 2; the functions f. are defined by
A
I A
I
-1,
x 5
-1..
FIGURE
=
2
f,(x)
1-
--
n
1
=
nx,
-
n
1
1,
-
-
)
5 x 5
-
-
n
5 x.
for large enough n) equal to
In this case, if x < 0, then f.(x) is eventually (i.e.,
and if x > 0, then f, (x) is eventually 1, while f.(0)
0 for all n. Thus
-1,
=
lim fn(x)
=
naoo
A
I I
so, once again, the function
f(x)
0,
1,
x
x
=
>
0
0;
lim f.(x) is not continuous.
00
=
N
-1-
FIGURE
3
By rounding off the corners in the previous examples it is even possible to
produce a sequence of dgerentiablefunctions (f,} for which the function f(x)
lim f, (x) is not continuous. One such sequence is easy to defme explicitly:
=
n
fn(x)=
3
sin
--Ex5(mrx11
2
1
-
,
n
1,
-
n
2
n
5 x.
These functions are differentiable (Figure 3), but we still have
h
-1,
1
lim
8400
i
I
'
i
i
4
0,
1,
x
<
x
=
x
>
0
0
0.
Continuity and diKerentiability are, moreover, not the only properties for which
problems arise. Another difficulty is illustrated by the sequence {f,} shown in
Figure 4; on the interval [0,1/n] the graph of f. forms an isosceles triangle of
altitude n, while f, (x) 0 for x à 1/n. These functions may be defined explicitly
=
FIGURE
f.(x)
=
RS fOlÏOWS:
and PowerSeries 493
24. UniforrnConvergence
2n2
h
2n
fn(x)
=
-
05x5-
2n2x,
-
1
Exs
2n
1
2n
1
--
n
1
-sxs1.
0,
n
varies so erratically
Because this sequence
near 0, our primitive mathematical
instincts might suggest that lim fn(x) does not always exist. Nevertheless, this
lim f.(x) is even continuous.
limit does exist for all x, and the function f(x)
lim f,(x) 0; moreover, f.(0) 0
In fact, if x > 0, then f,(x) is eventually 0, so HNOO
0. In other words, f (x)
for all n, so that we certainly have lim f.(0)
A
=
=
=
=
=
0 for all x. On the other hand, the integral quickly reveals the
lim f.(x)
aœnge
behavior of this sequence; we have
=
f,(x)dx
FIGURE
5
(,
=
but
1
f (x)dx
0.
=
Thus,
1
lim
HNDO
1
fn(x)dx
=/=
lim f,(x)dx.
NACO
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
for each x in [0,1],
f(x) lim f,(x)
=
8400
"approach"
1- -
I A
/ A
the graph of f in the sense of
the graphs of the functions f, do not
lying close to it-if, as in Figure 5, we draw a strip around f of total width 2e (allowing a width of e above and below), then the graphs of f. 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, f,(x)) lies in this strip for n > N; this assertion
just amounts to the fact that lim f.(x) f (x). But it is necessary to choose larger
=
1
n->ao
and
larger N's as x is chosen closer and closer to 0, and no one 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
f,(x)
x"
=
1,
'
Osxs1
x>l.
A strip of total width 2e has been drawn around the graph of
If e
<
i,
f(x)
=
lim
naoo
f.(x).
this strip consists of two pieces, which contain no points with second
494 InfmiteSequences
and InßniteSenes
since each function f, takes on the value
coordinate equal to
the graph of
each f, fails to lie within this strip. Once again, for each point x there is some N
such that (x, f,(x)) lies in the strip for n > N; but it is not possible to pick one N
which works ibr 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 f, and a sequence of functions (fa),
all defined on some set A, such that
);
fa
j,
.
(,
f(x)
.
=
lim
n->oo
for all x in A.
f,(x)
This means that
FIGURE
7
for all e
>
0, and for all x in A, there is some N such that if n
If (x) f.(x)l
-
<
>
N, then
s.
But in each case different N's must be chosen for difTerent x's, and in each case it
is not true that
for all e
If(x)
-
>
0 there is some N such that for all x in A, if n
f,,(x)l <
>
N, then
e.
Although this condition differs from the first only by a minor displacement of the
phrase "for all x in A," it has a totally different significance. If a sequence (f,}
satisfies this second condition, then the graphs of f, eventually lie close to the
graph of f, as illustrated in Figure 7. This condition turns out to be justthe one
which makes the study of limit functions feasible.
DEFINITION
Let {f,} be a sequence of functions defined on A, and let f be a function which
limit of {f.) on A if for
is also defined on A. Then f is called the uniform
every e > 0 there is some N such that for all x in A,
if n
>
N, then
|f (x) f.(x) | < s.
We also say that (f,} convergesuniformly
f uniformly on A.
-
to
f
onA, or that f, appmaches
As a contrast to this definition, if we know only that
f(x)
=
lim f,(x)
NNOO
for each x in A,
then we say that {f.} converges pointwiseto f onA. Clearly, uniform convergence implies pointwise convergence (butnot 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 obvious that
if {f.) converges uniformly to f, then the integral of f, can be made as close
to the integral of f as desired. Expressed more precisely, we have the following
theorem.
and PowerSeries 495
24. Un¶ormConvergence
THEOREM
1
of
Suppose that {f.} is a sequence
that
(f.)
converges uniformly on
[a,b]. Then
Let e
>
b
b
Ï
PROOF
functions which are integrable on [a,b], and
[a,b] to a function f which is integrable on
f
lim
=
f..
0. There is some N such that for all n
|f(x)
Thus, if n
>
f,(x)|<
-
N we have
>
for allx in
e
[a,b].
N we have
b
Ilb
f (x) dx
b
f.(x) dx
-
[f (x) f.(x)] dx
=
-
a
Lb
|f(x)
5
-
f,(x)|dx
b
edx
5
e(b
=
Since this is true for any e
>
a).
-
0, it follows that
b
b
Ï
f
lim
=
f..
of continuity is only a little more difficult, involving an
a three-step estimate of |f (x) f (x + h)|. If {f,} is a sequence
of continuous functions which converges uniformly to f, then there is some n such
The treatment
"e/3-argument,"
-
that
(1)
|f (x) f.(x) | <
(2)
|f (x +
-
h)
--
fn(x +
,
h)
<
.
Moreover, since f, is continuous, for sufficiently small h we have
(3)
|f,(x)
f.(x
-
+ h) |
<
.
It will follow from (1),(2),and (3)that |f (x) f (x +h) < e. In order to obtain (3),
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 is pæcisely at this point
uniform
that
convergence enters the proof.
-
THEOREM
2
Suppose that {f.) is a sequence
that {f.} converges uniformly
PROOF
of functions which are continuous on
on [a,b] to f. Then f is also continuous
[a,b], and
on [a,b].
For each x in [a,b] we must prove that f is continuous at x. We will deal only
with x in (a, b); the cases x
b require the usual simple modifications.
a and x
=
=
and InfmiteSeries
496 InfmiteSequences
Let e
>
0. Since (f,} converges uniformly
to
f
on
[a,b], there
is some n such
that
|f(y)
f,(y)|
-
for all y in
<
In particular, for all h such that x + h is in
Now f, is continuous,
(1)
|f (x) f,(x) | <
(2)
|f (x + h)
If(x+h)-
<
so there is some ô
f.(x
-
>
,
+ h) |
<
.
0 such that for |h|
If,(x)- f,(x+h)|<
<
8 we have
.
ô, then
f(x)|
=
|f(x+h)
<
If(x+h)
-
-
S
<-+-+-
8
8
3
3
3
This proves that
f
f,
FIGURE
'
have
-
(3)
Thus, if |h|
[a,b], we
[a,b].
f,(x+h)+ f,(x+h)
fn(x+h)|+If,(x+h)-
-
f,(x)+ f,(x) f(x)|
-
fn(x)\+lln(x)- f(x)\
is continuous at x.
f(x)
=
|x|
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 f, is
differentiable, and if {f.} converges uniformly to f, it is still not necessarily true
that f is differentiable. For example, Figure 8 shows that there is a sequence of
differentiable functions {f.) which converges uniformly to the function f(x)
|x|.
=
and PowerSeries 497
24. UmformConvergence
Even if f is differentiable, it may not be true that
f'(x)
lim f,'(x);
=
HNOO
this is not at all surprising if we reflect that a smooth function can be approximated
by very rapidly oscillating functions. For example (Figure 9), if
f.(x)
then
(f.)
converges
uniformly
-
1 sin(n2x),
n
the
function
to
f.'(x)
and
=
=
n
lim n cos(n2x) does not always exist
H->OD
FIGURE
f (x) 0, but
=
cos(n2x),
(forexample,
it does not exist if x
=
0).
9
practicalÏy
the Fundamental Theorem of Calculus
guarof Theantees that some sort of theorem about derivatives will be a consequence
orem 1; the crucial hypothesis is that {f.') converge uniformly (tosome continuous
Despite such examples,
function).
THEOREM
s
Suppose that {f.) is a sequence of functions which are differentiable on [a,b],
integrable derivatives f.', and that {f.} converges (pointwise)
to f. Suppose,
with
moreover, that
Then
f
{f,'} converges
uniformly
f'(x)
PROOF
on
[a, b] to
some continuous function g.
is differentiableand
Applying Theorem 1 to the interval
lx
g
=
=
=
f,'(x).
we see that
[a,x],
x
lim
lim
H-¥OO
=
lim
n->OC
f(x)
f.'
[f.(x) f.(a)]
-
-
f(a).
for each x we have
and InßniteSeries
498 InfiniteSequences
Since g is continuous, it follows that f'(x)
interval [a, b].
g(x)
=
lim f,'(x) for all x in the
=
B->OO
-it
Now that the basic facts about uniform limits have been established,
how to treat functions defined as infinite sums,
is clear
f1(x)+f2(X)#f3(X)#·•·.
f(x)=
This equation means that
f (x)
lim f1(x)+
=
+
R-¥DO
fn(x)
our previous theorems apply when the new sequence
fi, fi
+h.
fi
+h+fs.
...
converges uniformly to f. Since this is the only case we shall ever be interested
in, we single it out with a definition.
DEFINITION
The series
f.
uniformly
converges
(moreformally:
n=1
uniformly
summable)
converges uniformly
to
to
f
the sequence
on A, if the sequence
It+h+h,
fi, fi+h,
f on A.
---
We can now apply each of Theorems 1, 2, and 3 to uniformly
the results may be stated in one common corollary.
COROLLAnr
Let
f.
uniformly
converge
{f,} is
to
f
on
convergent
series;
[a,b].
n=1
is continuous on [a,b], then f is continuous on
f, is integrable on [a,b], then
(1)If each f.
(2)If f
and each
b
I
Moreover, if
f.
converges
f.'
derivative f.' and
oc
f
b
f..
=
to f
(pointwise)
converges uniformly
on
[a,b], each f,
on
[a,b]
n=1
tion, then
(3)f'(x)
f,'(x)
=
n=1
for all x in
[a,b].
[a,b].
has an integrable
to some continuous
func-
and PowerSenes 499
24. UnformConvergence
PROOF
(1)If each f,
is continuous,
then so is each
-+
f,,, and f is the
uniform
limit
continuous
is
by
Theorem
2.
so f
fi, fi
f2 Í3,
(2)Since fi, fi + f2, fi + f2 + f3, converges uniformly to f, it followsfrom
Theorem 1 that
of the sequence
f2, ft +
+
fi +
·
·
.
-
·
,
...
b
Ï
b
f
a
lim
=
(fi
+
·
-
-
+
f.)
a
b
lim
=
nuco
oc
b
(Ï
a
fi+...+
a
f,
b
n=1
(3) Each function fi + + f,
and fi', fi'+ f2', fi'+ f2'+ fs',
·
-
-
...
is differentiable, with derivative fi' +
converges uniformly
to a continuous
·
·
+
fn',
function,
by hypothesis. It follows from Theorem 3 that
f'(x)
=
lim [fi'(x) +
5400
+
f.'(x)]
f.'(x).
=
n=i
At the moment this corollary is not very useful, since it seems quite difficult to
will converge uniformly. The
predict when the sequence fi, f1+f2, ft+f2+ f3,
most important condition which ensures such uniform convergence
is provided by
the following theorem; the proof is almost a triviality because of the cleverness
with which the very simple hypotheses have been chosen.
...
THEOREM
(THE WEIERSTIMSS
4
M-TEST)
Let {f.} be a sequence of functions defined on A, and suppose
sequence of numbers such that
forallxinA.
|f,(x)|5M.
Suppose moreover
that
M. converges.
Then for each x in A the series
n=1
converges
(infact, it converges
that {M,} is a
f,(x)
n=1
absolutely), and
f,
n=1
to the function
f (x)
=
f,,(x).
converges
uniformly
on A
500
and InfiniteSeries
InfiniteSequences
PROOF
For each x in A the series
quently
f,(x)
converges
n=1
f(x)
-
|f, (x)| converges, by
the comparison
Moreover, for
(absolutely).
[fi(x)+···+fn(x)]
all x in A we
test; conse-
have
,
fn(x)
=
n=N+1
Ifn(x)l
5
n=N+1
M..
5
n=N+1
Since
M. converges,
the number
M. can be made as small as desired,
n=1
n=N+1
by choosing
-3
N sufficiently large.
-2
-·-1
2
1
FIGURE
3
10
The following sequence {f.} illustrates a simple application of the Weierstrass
M-test. Let {x}denote the distance from x to the nearest integer (thegraph of
f (x) (x) is illustrated in Figure 10). Now define
(a)
=
1
10"
-{10"x}.
f,(x)=
1
The functions fi and f2 are shown in Figure 11 (butto make the drawingssimpler,
10" has been replaced by 2"). This sequence of functions has been defined so that
the Weierstrass M-test automatically applies: clearly
(b)
n(X)
FIGURE
11
-
1
10"
fOr
aÏÌ X,
and PowerSenes 501
24. UnformConvergence
1/10" converges. Thus
and
f.
n=1
uniformly;
converges
since each
f, is con-
n=1
tinuous, the corollary
implies that the function
f (x)
f,(x)
=
{10"x)
=
n=1
n=1
is also continuous. Figure 12 shows the graph of the first few partial sums fi +
+ f.. As n increases, the graphs become harder and harder to draw, and
-
-
·
the
infinitesum
f,
is quite undrawable,
as shown by the following theorem
n=1
mainly
(included
rough).
THEOREM
5
as an interesting sidelight, to be skipped if you find the going too
The function
f (x)
{10"x)
=
n=1
is continuous everywhere and differentiable nowhere!
PROOF
is continuous; this is the only part of the proof which
convergence. We will prove that f is not differentiable at a, for
any a, by the straightforward method of exhibiting
a particular sequence {hm}
approaching 0 for which
We have just shown that
uses uniform
f
lun f(a+hm)hm
f(a)
.
mooo
does not exist. It obviously suffices to consider only those numbers
0<as1.
Suppose that the decimal expansion of a is
a
=
0.aia2a3a4
·
a satisfying
-•-
-10¯=
Let hm 10-. if am ¢ 4 or 9, but let h.
these two exceptions will appear soon). Then
=
if a.
=
°°
f(a+ hm) f(a)
-
h.
(10"(a+ h.)}
1
-
(thereason
for
{l0"a}
±10-m
n=1
±10'"¯"[{10"(a
=
4 or 9
=
+ hm)}
{10"a}].
--
n=1
This infmite series is really a fmite sum, because if n
m, then
>
10"hmis an integer,
SO
{10"(a+
On the other hand, for n
10"a
=
10"(a + hm)
=
<
hm))
-
{10"a} 0.
=
m we can write
integer + 0.an+1a, 2an+3
integer + 0.an+1anç2an+3
.
.
.
.
.
.
am
·
-
·
(ami I)
·
·
•
502
and InfiniteSeries
Infinite Sequences
(inorder
for the second equation to be true it is essential that we choose h.
when a,
9). Now suppose that
-10-=
=
=
0.an+1anç2a.43...a.-··
...(a.±1)
0.a.sta,42a.43
(inthe
-10¯"
hm
A
,
Then we also have
A+h
f·¯
5 y.
=
A
special case m
when am
=
=
-
and exactly the same equation
for n < m we have
can
1
,
is true because we chose
±10,-.
=
-
(a)
5
-
n + 1 the second equation
4). This means that
{10"(a+ hm)} {l0"a}
i
-
be derived when 0.an+1a,42a,49
10"¯"[{10"(a + h.)}
{10"a}]
=
-
>
...
(. Thus,
±1.
In other words,
hm)
f (a +
+A+b
f (a)
-
h.
-1
1 numbers, each of which is ±1. Now adding +1 or
to a
number changes it from odd to even, and vice versa. The sum of m
1 numbers
each ±1 is therefore an even integerif m is odd, and an odd integerif m is even.
Consequently the sequence of ratios
is the sum of m
-
-
A+ h
h(x)
=
Max!
hm)
f (a +
f (a)
-
hm
(b)
cannot possibly converge, since it is a sequence of integers which are alternately
odd and even.
A+ b + h + A
1-
-
In addition to its role in the previous theorem, the Weierstrass M-test is an ideal
tool for analyzing functions which are very well behaved. We will give special
attention to functions of the form
f (x)
a.(x
=
-
a)",
A+A+b
A(x) M6x!
=
which can also
be described by the equation
f (x)
==0
(c)
FIGURE
12
f,(x),
=
fOT
n(X)
=
Un(X
-
G)n.
Such an infinite sum, of functions which depend only
centered
on powers of (x a), is called a power series
at a. For the sake of
simplicity, we will usually concentrate on power series centered at 0,
-
f (x)
anx".
=
n=0
and PowerSenes 503
24. UmformConvergence
One especially important group of power series are those of the form
°°
f(">(a)
where
f
called the
(x
n!
n=o
a)",
-
is some function which has derivatives of all orders at a; this series is
Taylor series for f at a. Of course, it is not necessarily true that
n(a)(x-a)";
f(x)=
·
n=o
this equation
holds only when the remainder
terms satisfy lim R..a(x)
=
naoo
a,x"
We already know that a power series
does not necessarily
0.
converge for
n=o
all x.
the power series
For example,
77
5
x3
x--+-----+---
5
3
converges
only
7
x| 5 1, while the power series
for
X2345
x--+---+-+··-
2
3
4
5
-1
converges only for
converges only for x
<
=
xs 1. It is even possible to produce a power series which
0. For example, the power series
n!x"
n=o
does not converge for x ¢ 0; indeed, the ratios
(n + 1)! (x"+1) (n+1)x
n!x"
=
are unbounded
anx"
for any x ¢ 0. If a power series
does converge
for
n=o
some xo
¢ 0 however, then
a great deal can be said about the series
anx"
for
n=o
lxl < lxol.
THEOREM
6
Suppose that the series
f (xo)
anxo"
=
n=o
converges, and let a be any number with 0
f (x)
<
a
<
anx"
=
n=o
|xo|. Then
on
[-a, a]
the series
and InfiniteSenes
504 InfiniteSequences
(andabsolutely).
converges uniformly
Moreover, the same is true for the series
g(x)
nauxn-1
-
n=1
Finally, f is differentiable and
f'(x)
naux"¯'
=
n=1
for all x with |x|
PROOF
<
|xo|.
anxo" converges, the terms anxo" approach
Since
0. Hence they are surely
n=o
bounded: there is some number M such that
|anxo" |a,| |xo"|5
is in [--a,a], then |x| 5 |a|, so
anx" |
|a, | |x"|
5 Iasi la"i
=
Now if x
=
for all n.
M
-
·
-
=
|a,| |xo|"
-
-
-
n
a
(thisis the
xo
clever step)
xo
But la/xo| <
1, so the
series
(geometric)
"-M
-a
M
n-0
Choosing M
converges.
a
XO
n=0
a/xo|" as the number
·
anx" converges uniformly on
follows that
X0
M. in the Weierstrass M-test, it
[--a,a].
n=0
To prove the same assertion for g(x)
na,x"-I
=
notice that
n=I
|na,x"¯1|
n |a. |
5 n|an\
=
=
<
--
I
|x"¯ |
|an-1
-
-
|a, |
|xo|" n
|a|
-
"
M
--
¯
|a|
n
-
"
-
a
xo
a
xo
Since |a/xo| < 1, the series
"M
n=1
converges
(thisfact was
|a|
a"
M"
a"
xo
|a| n=1
xo
proved in Chapter 23 as an application
of the ratio test).
and PowerSeries 505
24. UniformConvergence
naux"¯' converges
Another appeal to the Weierstrass M-test proves that
uni-
n=1
formly on [-a, a].
Finally, our corollary proves, first that g is continuous, and then that
f'(x)
=
g(x)
na.xn-1
=
for x in
[-a, a].
0
|xo|,this
n=1
Since we could have chosen any number
all x with
a with
<
a
<
result
holds for
ixi ixol.I
<
We are now in a position to manipulate power series with ease. Most algebraic
manipulations are fairly straightforward consequences of general theorems about
infinite series. For example,
that
suppose
f(x)
a,x"
=
and g(x)
b,x",
=
n=0
where
the two power series both converge
n=0
for some xo. Then for |x|
|xo| we
<
have
anx" +
n=0
So the series h(x)
b,x"
(anx"+ b,x")
=
n=0
(a, + b.)x" also
=
(an+ b.)x".
=
n=0
n=0
for |x|
converges
<
and h
|xo ,
-
f
+ g
n-O
for these x.
The treatment of products is just a little more involved. If |x|
know that the series
anx" and
n=0
b,x" converge
<
|xo|, then
we
absolutely. So it follows from
n=O
Theorem 23-9 that the product
a,x"
b.x" is given by
-
n=0
n=0
a¿xibjxi,
i=0 j=0
the elements a¿xibjxi
choose the arrangement
where
aobo
in any order.
are arranged
+ (aobi+aibo)x + (aob2+albi
In particular, we can
+a2bo)x2 +
-
-
-
which can be written as
c.x"
n=0
for c,
aab,_a.
=
k=0
This is the "Cauchy product" that was introduced in Problem 23-8. Thus, the
Cauchy product h(x)
c,x" also converges for
=
n=0
these x.
|x| < |xo|and
h
=
fg
for
506
and Infmite Series
InfmiteSequences
Finally, suppose that f (x)
anx", where ao
=
¢ 0,
so that
f (0)
=
ao
¢ 0.
n=o
1/f. This means'
b,x" which represents
Then we can try to find a power series
n=0
that we want to have
anx"
b,x"
·
n=o
1
=
l + 0 x + 0 x2 +
=
-
·
-
-
.
n=o
Since the left side of this equation will be given by the Cauchy product, we want
to have
aobo
1
=
aobi + a1bo = 0
aob2 Ÿ Ul bi + a2bo
0
=
Since ao ¢ 0, we can solve the first of these equations for bo. Then we can solve
the second for bi, etc. Of course, we still have to prove that the new series
b.x"
n=0
does converge for some x ¢ 0. This is left as an exercise (Problem 17).
For derivatives, Theorem 6 gives us all the informationwe need. In particular,
when we apply Theorem 6 to the infinite series
7
x5
x3
.
smx=x--+---+--···,
X
3!
5!
7!
9!
x2
x4
x6
X
cosx=1--+----+-----,
2!
4!
6!
8!
x2
x
ex=1+-+-+-+--+--·,
x3
x4
3!
4!
l!
2!
get precisely the results which are expected. Each of these converges for any xo,
hence the conclusions of Theorem 6 apply for any x:
we
sin'(x)
1
=
cos'(x)
=
exp'(x)
=
3x2
-
-
31
-
--
-
+
-
f (x)
+
--
-
=
-
=
x
-
cos x,
-
-
-
.
-
-·
-
75
3
+
-
-
smx,
situation
7
-
5
=
exp(x).
=
log(1+x) the
=
x3
arctan x
-
5!
2x 4x3 6x"
+
+
2!
4!
6!
2x 3x2
1+
For the functions arctan and
complicated. Since the series
5x4
+
--
-
7
+
·
-
-
is only
slightly
more
and PowerSeries 507
24. Unform Convergenœ
converges
for xo
=
arctan'(x)
1, it also converges for |x|
=
1
x2 + x4
-
1
6
_
1, and
<
.
.
for |x|
.
1 + x2
<
1.
--l
In this case, the series happens to converge for x
for the derivative is not correct for x
1 or x
=
also.
However, the formula
indeed the series
=
-1;
=
6
1-x2+x4_
-1.
Notice that this does not contradict Theorem 6,
1 and x
diverges for x
which
that
the
derivative
is
given by the expected formula only for |x| < |xo|.
proves
series
Since the
=
=
log(1 + x)
converges
for xo
1
1+x
=
=
x
x
--
2
-
2
1, it also converges
+
x
3
--
x
3
for |x|
4
---
-
4
x
5
---
-
-
-
-
5
1, and
<
,(1+x)=1-x+x2_
+
3
-1;
In this case, the original series does not converge for x
moreover, the
differentiated series does not converge for x = 1.
All the considerations which apply to a power series will automatically apply to
its derivative, at the points where the derivative is represented by a power series.
If
=
f (x)
anx"
=
n-o
converges for all x in some interval (-R, R), then Theorem 6 implies that
f'(x)
naux"¯I
=
n=1
for all x in (-R, R). Applying Theorem 6 once again we find that
f"(x)
n(n
=
-
1)anx"¯2,
n=2
and proceeding by induction we fmd that
f(k)(x)=
n(n-1).....(n-k+1)anx"¯
.
n=k
Thus, a function defined by a power series which converges in some interval
(-R, R) is automatically infinitely differentiable in that interval. Moreover, the
previous equation implies that
f(k)(0)
=
so that
ak
=
kl ag,
f (k)(0)
k!
508
and InfiniteSeries
InfiniteSequences
In other words, a convergent powersenes centered at 0 is always theTaylorseries at 0 of the
whkh it defmes.
function
On this happy note we could easily end our study of power series and Taylor '
series. A careful assessment of our situation will reveal some unexplained
facts,
however.
The Taylor series of sin, cos, and exp are as satisfactory as we could desire;
they converge for all x, and can be differentiated term-by-term for all x. The
Taylor series of the function f (x) = log(1 + x) is slightly less pleasing, because it
< x 5 1, but this deficiency is a necessary
converges only for
consequence of
the basic nature of power series. If the Taylor series for f converged for any xo
with |xo| > 1, then it would converge on the interval (-|xo|, |xo|), 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(1 + x).
The Taylor series for arctan is more difficult to comprehend-there
seems to
be no possible excuse for the refusal of this series to converge when |x| > 1. This
mysterious behavior is exemplified even more strikingly by the function f (x)
1/(1 + x2), an infinitely differentiable function which is the next best thing to a
polynomial function. The Taylor series of f is given by
-1
=
f(x)
1
=
=
1+ x2
1-x2+x4
_
6
8
If |x| > 1 the Taylor series does not converge at all. Why? What unseen obstacle
Asking this sort of
prevents the Taylor series from extending past 1 and
question is always dangerous, since we may have to setde for an unsympathetic
the way things are! In this case
answer: it happens because it happens-that's
there does happen to be an explanation, but this explanation is impossible to give
at the present time; although the question is about real numbers, it 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 27.
-1?
PROBLEMS
1.
For each of the following sequences {f.), determine the pointwise limit of
{f.} (ifit exists) on the indicated interval, and decide whether {f,} converges
uniformly to this function.
(i) f. (x)
=
(ii) f,(x)
=-
(iii) f.(x)
=
(iv) f.(x)
=
(v) f.(x)
=
J,
[0,1].
on
O
n,
x 5 n
x > n,
on
(1, oo).
'
-
x
ex
-,
e¯"",
,
on
on
[-1, 1].
R.
on
[a,b], and
on R.
and PowerSenes 509
24. UniformConvergence
2.
This problem asks for the same information as in Problem 1, but the functions
are not so easy to analyze. Some hints are given at the end.
(i) fn(x)
=
(ii) f.(x)
=
x"
---
[0,1].
on
x
nx
1+n+x
on
[0,oo).
(iii) f,,(x)
=
x2 +
on
[a,oo), a
(iv) f, (x)
=
x2 +
on
R.
(v) f,(x)
=
x +
(vi) f.(x)
=
(vii)f.(x)
=
(viii)f,,(x)
=
x +
-
1
-
on
n
1
---
[a, oo),
a
>
0.
on R.
-
xn+
n
on
[a,oo), a
on
[0,oo).
-
x +
n
0.
>
-
>
0.
find the maximum of |f f,| on [0, 1]. (ii)For each n,
consider |f(x)
f.(x)| for x large. (iii)Express f(x) f,(x) as a fraction
and estimate f(x)
f.(x)| for x > a. (iv)Give a separate estimate of
If(x) f,(x)| for small lx|. (vii)Use (v).
Hints:
(i)For each
n,
-
-
-
-
-
3.
4.
Find the Taylor series at 0 for each of the following functions.
(i) f (x)
=
(ii) f (x)
=
(iii) f(x)
=
(iv) f (x)
=
(v) f (x)
=
1
a
log(x
1
1
¢ 0.
a
,
x
-
a),
-
a
(1
=
---
¢ 0.
x)-l/2.
-
(Use Problem 20-7.)
x
1
.
1
x2
-
arcsin x.
Find each of the following infinite sums.
x2
(i)
(ii)
X2
(iii)
x3
x4
1-x+-------+-----·.
2!
31
1-x3+x6-x9+--..
--
3
-
2
----
3-2
4!
Hint: Whatis l-x+x2-x3+---?
4
+
---
4-3
x5
-
----
5-4
+
-
-
-
for lx| < l. Hint: Differentiate.
510
and InfiniteSenes
InfiniteSequences
5.
Evaluate the following infinite sums. (In most cases they are f(a) where a
is some obvious number and f (x) is given by some power series. To evalu-
ate the various
them until some well-known
power series, manipulate
power
series emerge.)
(-1)"22n &
(2n)!
(21n)!°
(ii)
n=o
1"
i
2n+1
(iii)
(iv)
-
2
.
n=o
1
3"(n + 1)
n=o
2n+1
2"n!
(vi) n=o
6.
If
f(x)
power
7.
·
=
(sinx)/x for
for f.
x
¢ 0 and f(0)
1, find
=
f(k)(0).Hint:
a
In this problem we deduce the binomial series (1+x).
n=o
without
all the work of
in part
(a)of that
lx
<
Problem 23-18, although
series
problem-the
=
c(1+ x)" for some constant
series. Hint: Consider g(x)
a
=
x"
|x| < 1
fact established
does converge for
1.
part (a)is of the form f (x)
c, and use this fact to establish the binomial
=
for |x|
<
satisfying
f (x)/(1 +
=
x)".
Prove that the series
n=1
converges
9.
f(x)
we will use a
x",
n
l.
(a) Prove that (1+ x)J'(x) af(x)
(b) Now show that any function f
8.
Find the
series
uniformly
(a) Prove that
on
n(1 +nx2)
R.
the series
2" sin
n=o
converges
uniformly
on
[a,oo)
for a
>
0. Hint: lim(sinh)/h
hwO
=
l.
and PowerSenes 511
24. UmformConvergence
(b) By considering
the sum from N to oo for x
series does not converge uniformly on (0, oo).
10.
(a) Prove that
=
2/(x3N),
show that tlie
the series
"
f (x)
nx
=
1 + n4x2
n=o
converges uniformly on [a, oo) for a
of nx/(1 + n4X2) OR [Û,00).
(b) Show that
>
0. Hint: First find the
and by using an integral to estimate the sum, show that
Conclude that the series does not converge uniformly
(c) What about the series
oo
f
maximum
>
1/4.
on R.
nx
1 +nSx2
11.
Problem 15-33 and the method of proof used for Dirichlet's test
(Problem 23-19) to obtain a uniform Cauchy condition for the series
(a) Use
sin nx
n
n=1
uniformly
uniformly
(b) For x
=
[e,2x
on
s], e
--
>
0, and conclude that the series converges
there.
x/N, with N large, show that
2N
N
sin kx
sin kx
=
k=N
>
N
-.
k=0
Conclude that
2N
1
2x
sin kx
¯¯
k=N
and that the series
12.
(a) Suppose
that
k
'
does not converge uniformly
f (x)
anx"
=
on
[0,2x].
converges for all x in some interval
n=o
0.
(-R, R) and that f (x) 0 for all x in (-R, R). Prove that each a,
this
remember
the
for
is easy.)
formula
(If you
as
only
that
0
know
Suppose
for some sequence {x.) with
(x.)
we
f
(b)
each
=
again
that
Prove
0.
lim x,
an = 0. Hint: First show that
=
=
=
8400
f (0)
=
ao
=
0; then that f'(0)
=
ai
=
0, etc.
and InfiniteSeries
512 InfiniteSequences
This result shows that if f(x) e¯1A' sin 1/x for x ¢ 0, then f cannot
possibly be written as a power series. It also shows that a function defined
by a power series cannot be 0 for x 5 0 but nonzero for x > Sthus a
power series cannot describe the motion of a particle which has remained
at rest until time 0, and then begins to move!
=
(c) Suppose
that
f(x)
anx" and g(x)
=
b.x" converge for all x
=
n=0
n=0
g(t.) for some sequence
in some interval containing 0 and that f(t.)
converging
that
each n. In particular,
b,
0.
Show
for
to
a,
{t,,}
npmsentation
only
senes centered at 0.
as a power
a function
can have
one
=
=
13.
Prove that if f (x)
a.x" is an even function, then a,
=
=
0 for n odd,
n-0
and
if
f
is an odd function, then a,
0 for n even.
=
-1
14.
Show that the power series for f (x) log(1 x) converges only for
s
x)] converges
log[(1 + x)/(1
x < 1, and that the power series for g(x)
only for x in (-1, 1).
=
-
=
*15.
-
Recall that the Fibonacci sequence {a.) is defined by a1
a, + a._t
=
a2
1, an+1
=
=
.
(a) Show that
ja, 5 2.
a
(b) Let
=1+x+2x2+3x"+···
anx"-I
f(x)=
.
n=1
Use the ratio test to prove that f(x) converges if |x| < 1/2.
(c) Prove that if |x| < 1/2, then
-1
f(x)=
;2+X-
Î
Î
Hint: This equation can be written f (x) xf (x) x2 f (X)
1/(x2 +x
and
the
decomposition
Use
the
partial
fraction
for
1),
power
(d)
series for 1/(x
a), to obtain another power series for f.
(e) It followsfrom Problem 12 that the two power series obtained for f must
be the same. Use this fact to show that
-
=
-
.
-
-
"
1-Ã
(1+Ê
¯
2
a.=
16.
Let
f (x)
anx" and g(x)
=
n=o
=
n=o
2
.
Suppose we merely knew that
n=o
c,x"
f(x)g(x)
b,x".
=
"
for some
c.,
but we didn't know how to multiply series
and PowerSeries 513
24. UmformConvergence
in general. Use Leibniz's formula (Pmblem 10-18) to show directly that this
series for fg must indeed be the Cauchy product of the series for f and g.
17.
Suppose that
f(x)
anx"
=
converges for some xo, and that ao
¢ 0;
n=O
for simplicity, we'll assume that ao
recursively by
bo
=
b,
=
1. Let {b.}be the sequence
=
defined
1
n-1
bkün-k·
-
k=0
The aim of this problem is to show that
b,x" also converges
for
some
n=0
x
1/f for small enough |x
¢ 0, so that it represents
(a) If all |a,xo"| 5
.
M, show that
|b,x"| 5
M
|bkXk
i
k=0
(b) Choose
M
>
Ä
with all
|b,xo"| 5
(c) Conclude
18.
M
b,x" converges
that
M. Show that
lanxo"|5
for |x| sufficiently small.
n=o
Show that the series
oo
X2n+1
2n+1
converges uniformly to
it converges to log 2!
*19.
.
Suppose that
j log(x
a. converges.
+ 1) on
X"
2n+2
[-a, a]
for 0
<
<
a
We know that the series
n=o
1, but that at 1
f(x)
anx"
=
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
example, if (x)
log(1 + x)). However, a beautiful theorem of Abel
f
(for
shows that the series doesconverge uniformly on [0,1]. Consequently, f is
-1
=
continuous
on
[0,1] and,
in particular,
a,
n=0
Theorem by noticing that if |a,.+·· +a.|
by Abel's Lemma (Problem 19-35).
=
anx".
lim
Prove Abel's
ad
<
e, then |amx"'+-
-+anx"|
<
e,
514
and InfiniteSeries
InfiniteSequences
20.
A sequence {a.} is called Abel summable
anx" exists;
if lim
x->l
Prob-
n-O
lem 19 shows that a summable sequence is necessarily Abel summable. Find
a sequence which is Abel summable, but which is not summable. Hint: Look
over the list of Taylor series until you find one which does not converge at 1,
is continuous at 1.
even though the function it represents
21.
(a) Using Problem
(i)
(ii)
(b) Let
-
19, find the following infinite sums.
1
1
1
+
2.1
4-3
3.2
1-¼+¼+---.
c,
1
+
5·4
-
-
·
.
be the Cauchy product of two convergent power series
b., and suppose merely that
c, converges.
n=o
Prove that, in fact,
n=o
it converges to the product
a.
n=0
(a) Suppose that {f,} is a sequence
b..
-
n=0
of
bounded
functions on [a, b] which converge uniformly
f is bounded
on
(notnecessarily continuous)
to f on [a,b]. Prove that
[a,b].
(b) Find a sequence of continuous functions on
wise to an unbounded
function on [a,b].
*23.
a.
n=0
n 0
and
22.
-
[a,b] which
converge
point-
Suppose that f is differentiable. Prove that the function f' is the pointwise
limit of a sequence of continuous functions. (Since we already know examples of discontinuous derivatives, this provides another example where the
pointwise limit of continuous functions is not continuous.)
24.
Find a sequence of integrable functions {f,} which converges to the (nonintegrable) function f that is 1 on the rationals and 0 on the irrationals. Hint:
Each f. will be 0 except at a few points.
25.
(a) Prove that
if f is the uniform limit of {f,} on [a,b] and each f, is
integrable on [a,b], then so is f. (So one of the hypotheses in Theorem 1
was unnecessary.)
(b) In Theorem
3 we assumed only that {f,} converges pointwise to f. Show
that the remaining hypotheses ensure that {f,} actually converges uniformly to f.
(c) Suppose that in Theorem 3 we do not assume {f.} converges to a function f, but instead assume only that f.(xo) converges for some xo in
g).
to some f (withf'
[a,b]. Show that f. does converge (uniformly)
(d) Prove that the series
=
°°
n=i
(-1)"
x+n
and PowerSeries 515
24. UmformConvergence
converges uniformly
26.
on
[0,oo).
Suppose that f. are continuous functions on
to f. Prove that
1-1/n
lim
R-¥QQ
Is this true if the convergence
27.
converge uniformly
(0, 1] that
1
f.
f.
=
isn't uniform?
that {f,} is a sequence of continuous functions on [a,b] which
approach 0 pointwise. Suppose moreover that we have f,(x) 2 f,wi(x)
> 0 for all
n and all x in [a,b]. Prove that {f.) actually approaches 0
uniformly on
[a,b]. Hint: Suppose not, choose an appropriate sequence
of points x. in [a,b], and apply the Bolzano-Weierstrass theorem.
(a) Suppose
(b) Prove Dini's Theorem: If (f,} is a nonincreasing sequence of continuous
functions on [a,b] which approach the continuous function f pointwise,
then {f.) also approaches f uniformly on [a,b]. (The same result holds
sequence.)
if {f,} is a nondecreasing
(c) Does Dini's Theorem hold if f isn't continuous? How about if
replaced by the open interval (a, b)?
28.
[a,b] is
that {f.} is a sequence of continuous functions on [a,b] that
converges uniformly to f. Prove that if x, approaches x, then f,(x.)
(a) Suppose
approaches
f (x).
(b) Is this statement true without assuming that the f, are continuous?
(c) Prove the converse of part (a):If f is continuous on [a,b] and {f,} is
that f,(x.) approaches
converges uniformly to f on
with the property
a sequence
approaches
x, then
f.
there is an e > 0 and a sequence x. with
the Bolzano-Weierstrass theorem.
29.
|f.(x.)
f(x)
whenever
x,
[a,b]. Hint: If not,
f (x.)| > e. Then use
-
This problem outlines a completely different approach to the integral; consequently, it is unfair to use any facts about integrals learned previously.
(a) Let
some
s
be a step function on
partition {to,
...,
t.} of
[a,b],
[a,b].
is constant
b
Define
I
on
n
-(ti-ti_i)
s as
gs¿
(ti-1,ti) for
where
value of s on (ti_i, ti). Show that this definition does
si is the (constant)
not depend on the partition (to,
, t.).
function on [a,b] if it is the uniform
A function f is called a regulated
.
(b)
so that s
..
limit of a sequence of step functions {s,}on [a,b]. Show that in this
case there is, for every e > 0, some N such that for m, n > N we have
|s.(x) sm(x) | < e for all x in [a,b].
-
(c) Show that the sequence of numbers
(d) Suppose that (t,}is another sequence
s,
of step
will
be a Cauchy sequence.
functions on
[a,b] which
to f. Show that for every e > 0 there is an N such
N we have |s.(x) t.(x)| < s for x in [a,b].
converges
uniformly
that for n
>
-
and InfiniteSeries
516 InßniteSequences
b
b
lim
(e) Conclude that n-Foo
Ïb
f
Ï
s,
lim
=
n->oo
t.. This means
that we can dfme
lim s, for any sequence of step functions
to be RNOC
{s.) converging
to f. The only remaining question is: Which functions are
Here is a partial answer.
Prove that a continuous function is regulated. Hint: To find a step funcs(x)| < e for all x in
tion s on [a,b] with |f(x)
[a,b], consider all y
for which there is such a step function on [a,y].
uniformly
regulated?
*(f)
-
*30.
Find a sequence {f,} approaching f uniformly on [0,1] for which
lim (lengthof f, on [0,1]) ¢ length of f on [0,1]. (Length is defined
naoo
in Problem 13-25, but the simplest example will involve functions the length
of whose graphs will be obvious.)
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. Nevertheless, the
real numbers do have a great deficiency-not every polynomial function has a
root. The simplest and most notable example is the fact that no number x can
satisfy x2 + 1 = 0. This deficiency is so severe that long ago mathematicians
felt the need to
0. For a
a number i with the property that i2 + 1
since there is no
long time the status of the
i was quite mysterious:
number x satisfying x2 + 1
0, it is nonsensical to say
i be the number
i2
admission
number i to
satisfying
of the
0." Nevertheless,
+ 1
the family of numbers seemed to simplify greatly many algebraic computations,
numbers"
especially when
a + bi (fora and b in R) were allowed, and
all the laws of arithmetical computation enumerated
in Chapter 1 were assumed
valid.
example,
equation
be
quadratic
For
to
every
"invent"
=
"number"
"let
=
"imaginary"
=
"complex
ax + bx + c
0
=
(a ¢ 0)
can be solved formally to give
-b
x
If b2
+
=
Åb24ac
2a
Áb2 4ac
-b
-
-
or
x
-
=
2a
-4ac
0, these formulas give correct solutions; when complex numbers are
allowed the formulas seem to make sense in all cases. For example, the equation
>
x2+x+1=0
has no real root, since
x2+x+1=(x+()2+(>0,
forallx.
But the formula for the roots of a quadratic equation
x
if we understand
would be
B
and
=
2
V3 (-1)
to mean
-
-
1 Ã
+
i
2
2
-
--
and
=
x
suggest
2
-
1
2
-
=
--
Ë
2
di,
then these numbers
i.
It is not hard to check that these, as yet purely formal, numbers
the equation
x2+x+1=0.
517
"solutions"
=
d.|¯1
-
the
do indeed satisfy
and InßniteSeries
518 InfiniteSequences
"solve"
whose coefficients are themselves
quadratic equations
It is even possible to
complex numbers.
For example, the equation
x2+x+1+i=0
ought
to have the solutions
--1
-1
1
±
x
where
-3
-
4(1 + i)
-
3
±
2
J-3
4i
--
'
2
a complex
the symbol
4i means
4i. In order to have
-
(a+ßi)2
number
2_
2+2aßi=--3-4i
2
2
a +
ßi
whose
square
is
we need
_
-4.
2aß
can easily
These two equations
possible solutions:
=
be solved for real a and ß; in fact, there are two
«=-1
a=1
and
ß=-2
ß=2.
-3
"square
-1
2i and
+ 2i. There is no
reasonable way to decide which one of these should be called V-3 4i, and which
makes
4i; the conventional usage of
sense only for real x :> 0, in
which case á denotes the (real)
nonnegative root. For this reason, the solution
Thus the two
roots"
of
4i are 1
-
-
-
---
/-3
-
-1±
-3---4i
x=
must
be
understood
2
for:
as an abbreviation
-1+
x
r
=
2
-3
where
,
With this understanding
r
is one of the square
roots
of
-
4i.
we arrive at the solutions
-l+1-2i
.
x
¯"
2
-l-l+2i
-1+i;
x
2
as you can easily check, these numbers
do provide formal solutions for the equation
x2+x+l+i=0.
For cubic equations complex numbers
ax3
roots
(a¢0)
b, c, and d, has, as we know, a real root a, and if we divide
we obtain a second-degree polynomial whose roots are
of ax3+bx2+cx+d
0; the roots of this second-degree polynomial
with real coefficients a,
ax3+bx2+cx+d
by x
the other
2+cx+d=0
are equally useful. Every cubic equation
-a
=
25. ComplexNumbers 519
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
is of no use at all if the coefficients
necessary to appeal to this theorem (which
of
complex);
in
the
cubic
equation
are
we can, with sufficient cleverness,
case
a
actually find a formula for all the roots. The following derivation is presented not
only as an interesting illustration of the ingenuity of early mathematicians,
but as
of
complex
evidence
numbers
they
the
importance
further
for
(whatever may be).
To solve the most general cubic equation, it obviously suffices to consider only
equations of the form
x3
+ bx2 + cx + d
0.
=
the term involving x2, by a fairly straight-forward
It is even possible to eliminate
manipulation.
If we let
b
x=y---,
3
then
x3=y3-by2+
x
2
2
=y
_
-
3
b2
g
+-,
3
,
27
9
so
0=x3+bx2+cx+d
=
=
(
y3
y* +
-
b2y
by2 +
-
b3
-
3
-
27
+
2b2
(b2
--
-
---
3
3
+ c y +
b
2
b3
2b2
+
-
3
b3
b3
bc
9
27
3
+
--
9
+ d
cy
--
b
3
-
+ d
.
The right-hand side now contains no term with y2. If we can solve the equadon
for y we can find x; this shows that it suffices to consider in the first place only
equations of the form
x3+px+q=0.
We shall see later on
0 we obtain the equation x3
complex
number
three, so that this
has
cube
that every
root, in fact it
does have a
equation has three solutions. The case p ¢ 0, on the other hand, requires quite
an ingenious step. Let
In the special case p
-q.
=
=
(*)
x=w--
p
3w
520
and InfiniteSeries
InfiniteSequences
Then
0=x3+px+q=
3
=w
3w2p
3mp2
+
--
3w
=
This equation
P
3_
w
+p
w--
+q
wp2
p3
+pw--+q
-
9w2
27w3
3w
3
+q.
27w3
can be written
27(w3
2
+ 27q(w3
in w3
which
is a quadratic equation
Thus
(!!).
-27q
(27)2 2 + 4 27p3
±
w3
3
_
-
2.27
=
q
--
-
2
q2
±
--
4
p3
+
--.
27
Remember that this really means:
w3
9
_
_
2
2
where r
+ r,
is a square root of
4
+
'
3
.
27
We can therefore write
-q
-±
w=
q2
p3
4
27
-+-;
s
y 2
-q/2+r,
this equation means that w is some cube root of
where r is some square
p3/27.
six
q2/4
of
allows
possibilities
This
for w, but when these are
root
+
substituted into (*),yielding
-q
-±
q2
p3
4
27
-+-
x=
a
y 2
p
--
,
-q
-±
3-
q2
p3
4
27
--+--
s
\
2
it turns out that only 3 different values for x will be obtained! An even moæ
feature of this solution arises when we consider a cubic equation all of
whose roots are real; the formula derived above may still involve complex numbers
in an essential way. For example, the roots of
surprising
x3-15x-4=0
25. ComplexNumbers 521
-2
are
4,
+
-15,
(withp
Ã,
=
q
-2
and
Ã.
-
-4)
On the other hand, the formula derived above
gives as one solution
=
-15
2+44-125
x=
-
3
2+lli
=
44
2+
.
-
125
15
+
3.
V2+lli
Now,
_g2
(2+i)3=23+3-22
¿3
=8+12i-6-i
=2+lli,
so one of the cube roots of 2+ l li is 2 + i. Thus, for one solution of the equation
we obtain
15
6 + 3i
x=2+i+
15
=2+i+
6 + 3i
90 45i
36 + 9
6
6
-
-
3i
3i
-
=2+i+
=
4 (!)
.
The other roots can also be found if the other cube roots of 2+ lli are 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 interpreted entirely
in terms of real numbers.
Suppose we agree, for the moment, to write all complex numbers as a + bi,
writing the real number a as a + Oi and the number i as O + li. The laws of
show that
ordinary arithmetic
and the relation i2 =
-1
(a+bi)+(c+di)=(a+c)+(b+d)i
(a + bi) (c + di) (ac bd) + (ad +
=
-
-
bc)i.
Thus, an equation like
(1 + 2i) (3 + li)
-
may
be regarded
simply as an abbreviation
1-3-2-1=1,
1.1+2-3=7.
=
1 + 7i
for the two equations
522
Infinite Sequences
and InfiniteSeries
ax2 +
The solution of the quadratic equation
could be paraphrased as follows:
u2
77
uv
(i.e.,if
y2
_
=
--
b2
-
bx + c
=
0 with real coefficients
4ac,
0,
(u +
vi)2
--
b2
44
-
-b+u
(-b+uÍ-3
a
2a
then
a
v
-
+b
\2a /
2
+c=0,
2a
+bœ=0,
2
-b+u+W
( (-b+u+vi
i.e., then a
2a
+ b
+ c
2a
=
0
.
It is not very hard to check this assertion about real numbers without writing
down a single "i," 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 still not convinced, try paraphrasing the solution of the cubic equation.) If we really intend to use complex
numbers consistently, however, it is going to be necessary
to present some reasonable
definition.
One possibility has been implicit in this whole discussion. All mathematical
properties of a complex number a + bi are determined completely by the real
numbers a and b; any mathematical
object with this same property may reasonably
used
number.
be
The obvious candidate is the ordered pair
to define a complex
(a, b) of real numbers; we shall accordingly d¢ne a complex number to be a pair
of real numbers, and likewise d¢ne what addition and multiplication
of complex
numbers is to mean.
DEFINITION
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 complex
number
=
(a, b) + (c, d)
=
(a + c, b + d)
(a,b)-(c,d)=(a·c--b-d,a·d+b·c).
(The + and appearing on the left side are new symbols being defined, while the
+ and appearing on the right side are the familiar addition 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
after all. This difficulty
is not true-a real number is not a pair of real numbers,
25. ComplexNumbers 523
is only a minor annoyance,
however. Notice that
(a, 0) + (b,0)
(a + b, O + 0) (a + b, 0),
(a,0)-(b,0)=(a-b-0-0,a-0+0-b)=(a-b,0);
=
=
this shows that the complex numbers of the form (a, 0) behave precisely the same
with respect to addition and multiplication
of complex numbers as real numbers
addition
and
multiplication.
with
their own
For this reason we will adopt the
do
convention that (a, 0) will be denoted simply by a. The
for complex numbers can now be recovered if one more
i
DEFINITION
Notice that i2
=
familiar a + bi notation
definition is made.
(Û,1).
-1
=
on our convention).
(0, 1) (0, 1)
-
(-1, 0)
=
(thelast
=
Moreover
(a, b)
=
=
equality
sign
depends
(a, 0) + (0,b)
(a, 0) + (b, 0) (0, 1)
-
=a+bi.
You may feel that our definition was merely an elaborate device for defining
complex numbers as
of the form a + bi." That is essentially correct;
it is a firmly established prejudice of modern mathematics that new objects must
be defined as something specific, not as
Nevertheless, it is inter"expressions
"expressions."
esting to note that mathematicians
were sincerely worried about using complex
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.
p/3w,
For example, the solution of the cubic equation required writing x
w
w2
solving
makes
that
1/w
sense. Moreover,
so we want to know
was found by
a
numbers
=
-
In
numerous other algebraic manipulations.
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 justifyevery step.
Fortunately this is not necessary. Since all algebraic manipulations
performed on
real numbers can be justifiedby the properties listed in Chapter 1, it is only necessary to check that these properties are also true for complex numbers. In most
quadratic
equation, which requires
cases this is quite easy, and these facts will not be listed as formal theorems.
example, the proof of Pl,
[(a,b) +
(c, d)] + (e, f)
requires only the application
of the
=
(a, b) +
[(c,d)
definition of addition
The left side becomes
([a+c]+e,
[b+d] + f),
+
For
(e, f)]
for complex
numbers.
524
and InfiniteSeries
InfiniteSequences
and the right side
becomes
(a+[c+e],b+[d+
f]);
these two are equal because P1 is true for real numbers.
It is a good idea to
check P2-P6 and PS 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
b) is, but the multiplicative inverse for (a, b) required in P7 is a little
out what
trickier: if (a, b) ¢ (0, 0), then a2 + b2 ¢ 0 and
-(a,
-b
(a, b)
(
-
a
(1,0).
=
,
a2+b2
a2+b2
This fact could have been guessed in two ways. To find (x, y) with
(a, b) (x, y)
=
-
it is only necessary
to
solve
(1, 0)
the equations
ax-by=1,
bx + ay
=
0.
(a2 + b2). It is also possible to reason
The solutions are x a/(a2 + b2Ly
that if 1/(a + bi) means anything, then it should be true that
-b
-
=
1
1
a-bi
a-bi
¯
a + bi
¯
a + bi
a
-
a2 + b2°
bi
guessing the inverse
Once the existence of inverses has actually been proved (after
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
15
6+3i
15
6 3i
6+3i
6-3i
90 45i
36+9
-
-
°
Unlike Pl-P9, the rules P10-P12 do not have analogues: it is easy to prove that
there is no set P of complexnumbers such that P10-P12 are satisfied for all complex
numbers.
In fact, if there were, then P would have to contain 1 (since1 = 12)and
i2), and this would contradict P10. The absence of P10-P12
also
(since
will not have disastrous consequences, but it does mean that we cannot define
-1
-1
=
z < w for complex z and w. Also, you may remember that for the real numbers,
P10-P12 were used to prove that 1 + 1 ¢ 0. Fortunately, the corresponding fact
for complex numbers can be reduced to this 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
real
numbers.
of
pairs
Long ago this collection was identified with the plane, and
plane." The horizontal axis,
for this reason the plane is often called the
which consists of all points (a, 0) for a in R, is often called the real axis, and the
"complex
25. ComplexNumbers 525
vertical
axis is called the imaginaryaxis. Two important definitions are also related
this
geometric picture.
to
If z
DEFINITION
of z
(withx and
x + iy is a complex number
=
is defined as
=
and the absolute
value
or modulus
length
|z|
(Notice that x2+ y2
the nonnegative
e'
FIGURE
"
"
¯
"
>
x
0, so that
y real), then the conjugate
ž
iy,
-
|z| of z is defined as
x2
=
Vx2+
2
y2 is defined unambiguously;
it denotes
root of x2 + y2)
real square
Geometrically, ž is simply the reflection of z in the real axis, while |zi is the
distance from z to (0,0) (Figure 1). 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.
1
THEOREM
1
Let z and w be complex numbers.
(1)Ë
=
(2)ž
Then
z.
z if and only if z is real
=
number
(i.e.,is
of the
form a + Oi, for some
real
a).
(3)z+w=¯+t¯.
(4)
(5)z w=z-w.
¯¯¯ž
--ž.
=
(ž)¯I,
if z ¢ 0.
(6)
(7) |z z ž.
(8) |z wl |z| |wl.
(9) |z+wl s Izl+lwl.
=
=
=
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
=
-ž,
Õ z + (-z)
ž+
=
=
-z,
--z
=
so
-)-1
1= 1
=
z (z-1)
-
-
_
.
z-1,
so
g-1
calculation.
Equations (7)and (8)may also be proved by a straightforward
The
only difficult part of the theorem is (9).This inequality has, in fact, already occurred (Problem 4-9), but the proof will be repeated here, using slightly different
terminology.
It is clear that equality holds in (9)if z 0 or w 0. It is also easy to see that (9)
separately the cases A > 0 and
is true if z = Aw for any real number 1 (consider
A < 0). Suppose, on the other hand, that z ¢ Aw for any real number A, and that
=
=
and Ir¢niteSeries
526 InfiniteSequences
w
¢ 0. Then, for all
real numbers
1,
Notice that wž + ze is real, since
wž+zã=eÏ+žë=ez+2w=wž+zã.
Thus the right side of (*)is a quadratic equation in I with real coefficients and no
real solutions; its discriminant must therefore be negative. Thus
2<0;
(wž+zã)2-4|w|2.
it follows, since wž + žw and |w| |z| are
-
real
numbers,
(wž+zã)<2|w
-
and
|w| |z| > 0, that
-
|z|.
From this inequality it follows that
|z+w|2=(z+w)
(ž+
2
Izl2
< |zl2+ Iwl2
)
=
(lzl + |wl)2,
=
which
implies that
Iz+wl<lzl+lwl-i
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 (aproof of this
Appendix 1 to Chapter 4]).
geometric fact is left to you [compare
The interpretation of multiplication
is more involved. If z
0,
0 or w
then z w
0 (a one-line computational proof can be given, but even this is
assertion has already been shown to follow from P1-P9), so we
unnecessary-the
attention
restrict
to nonzero complex numbers. We begin by putting every
may
our
DORZCTO complex number
into a special form (compare
Appendix 3 to Chapter 4).
For any complex number z ¢ 0 we can write
=
=
=
c
FIGURE
2
a
a
+
c
-
=
z
in this expression, |z| is a positive
ItI
=
--;
z
Izi
real number,
-
z
Izl
=
Izi
Izl
while
=1
'
=
25. ComplexNumbers 527
so that z/\z| is a complex number of absolute value 1. Now any complex number
a
x + iy with 1 |a| x2 + y2 can be written in the form
=
=
=
a
length
for some number
(cos0, sin 0)
=
0. Thus every nonzero
angle of 0 radians
x
FIGURE
=
r(cos0
=
cos 0 + i sin 0
complex
number
z can be written
+ i sin0)
for some r > 0 and some number 0. The number r is unique (itequals Iz|),but 8
is not unique; if Gois one possibility, then the others are 90 + 2kx for k in Z-any
of z. Figure 3 shows z in terms of r
one of these numbers is called an argument
and 8. (To find an argument 8 for z
iy
+
we may note that the equation
x
3
=
x+iy=
means
z
=
|z|(cos0+i
=
|z| cos0,
sin0)
that
x
y
=
Iz| sin0.
aretan y/x; if x
So, for example, if x > 0 we can take 0
x/2 when y > 0 and & 3x/2 when y < 0.)
8
Now the product of two nonzero complex numbers
=
=
0,
we can
take
=
=
z
=
w
=
+ i sin9),
s(cos¢+ i sin¢),
r(cos0
is
-
z w
=
=
rs(cos0
rs[(cos0
=rs[cos(0+¢)+isin(0+¢)].
+ i sin8)(cos¢
+ i sin¢)
sin0
sin
cos ¢
¢) + i(sin0 cos ¢ + cos0 sin
--
¢)]
Thus, the absolute value of a product is the product of the absolute values of the
factors, while the sum of any argument for each of the factors will be an argument
for the product. For a nonzero complex number
z
=
r(cos0
+ i sin0)
it is now an easy matter to prove by induction the following very important formula
known as De Moivre's Theorem):
(sometimes
z"
=
|z|"(cosn0+ i sinn0), for any
argument 0 of z.
This formula describes z" so explicitly that it is easy to decide just when z"
THEOREM
2
Every nonzero complex number has exactly n complex nth roots.
More precisely, for any complex number w ¢ 0, and any natural
there are precisely n different complex numbers z satisfying z" w.
=
PROOF
Let
w
=
s(cos¢
+ i sin¢)
=
number
w:
n,
and InfiniteSems
528 InfiniteSequences
for s
Iw| and
=
some number
7
satisfies z"
=
w
a complex number
¢. Then
V (COS
=
Û § Í Sin Û)
if and only if
r"(cosnO + i sinnO)
which
+ isin¢),
s(cos¢
-
happens if and only if
r"
=
cos n0 + i sin n9
=
s,
cos ¢ + i sin ¢.
From the first equation it follows that
denotes the positive real nth root of s. From the second equation it
that
follows
for some integer k we have
where
6
2kn
¢
0=0k=
¯¯"·
n
and 6
6
n
Ok for some k, then the number z
sin
satisfy
will
the number of nth roots of w,
0
i
To
determine
0)
z"
+
r (cos
w.
which
only
such z are distinct. Now any
it is therefore
necessary to determine
integer k can be written
Conversely, if we choose r
=
=
=
=
k=nq+k'
for some integer q, and some integer k' between 0 and n
cos04 +isin0
i
This shows that every z satisfying z"
cos0ks #isinÛy.
=
w can be written
=
Moreover, it is easy to see that these numbers
k 0,
1 differ by less than 2x.
n
¯'
1. Then
6(cosOk#iS1nÛk)k=0,...,n-1.
z=
=
-
are all
different, since any two OkfOr
-
...,
In the course of proving Theorem 2, we have actually developed a method for
fmding the nth mots of a complex number. For example, to find the cube roots
of i (Figure 4) note that li|
1 and that x/2 is an argument for i. The cube roots
=
FIGURE
4
of i are therefore
1- cos-+ism6
6
,
¯
1
cos
-
x
-
6
+
2x
3
--
+ i sin
x
-
6
+
-
2x
3
cos
-
.
-,
°
¯
1- cos
5x
5x
+ r sm
6
6
.
=
x
-+-
6
4x
3
.
.
+rsm
x
---+-
6
4x
3
=cos-+xsm--.
3x
2
.
.
3x
2
25. ComplexNumbers 529
Since
x(
cos
5x
6
3x
cos
2
--
cos
n2
=
-
=
--
--,
x
.
--,
'6
6
5x
sm
6
3x
Ä
1
2
1
2
=
--
sm
-,
.
2
--
=
--
=
-,
-1,
.
0,
=
sm
2
the cube roots of i are
-n+i
Ã+i
-i
'
2
'
2
In general, we cannot expect to obtain such simple results. For example, to find
and that
the cube roots of 2+ 1li, note that |2+ lli|
/22+112
is an argument for 2 + l li. One of the cube roots of 2 + l li is therefore
arctan
=
=
(
'
VÏžŠ cos
arctan
ll
2
arctan
+ i sin
3
-
11
2
3
Á
=
cos
g
arctan
(arctan
3
+ i sin
.
3
Previously we noted that 2 + i is also a cube root of 2 + lli. Since |2 + i| =
22 12 W, and since arctan
is an argument of 2 + i, we can write this
)
=
cube root as
2+ i
These two cube
roots
=
Ã(cosarctan j +
are actually
i sin arctan
the same number,
arctan
=
3
arctan
().
because
1
2
-
can check this by using the formula in Problem 15-9), 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
i was originally introduced in
case of a very important theorem. The number
(you
order to provide a solution for the equation x2 + 1 0. The FundamentalTheorem
ofAlgebrastates the remarkable fact that this one addition automatically provides
solutions for all other polynomial equations: every equation
=
z"+a,-iz""+--·+ao=0
ao,...,a,_iinC
has a complex root!
In the next chapter we shall give an almost complete proof of the Fundamental
Theorem of Algebra; the slight gap left in the text can be filled in as an exercise
(Problem 26-5). The proof of the theorem will rely on several new concepts which
come up quite naturally in a more thorough investigation of complex numbers.
and Ir¢nite Series
530 InfiniteSequences
PROBLEMS
1.
value
Find the absolute
and argument
of each of the
following.
(i) 3 + 4i.
(ii) (3 + 4i)¯*.
i)5
(iii) (1 +
(iv) V3+ 4i.
(v) 13+4il.
2.
Solve the following equations.
x2+ix+1=0.
(i)
(ii)
(iii)
2+1=0.
x4
x2 +
2ix
-
1
0.
=
(1 + i)y 3,
(2+i)x+iy=4°
ix
(v)
3.
=
-
x3_,2-x-2=0.
Describe the set of all complex numbers
(i)
ž
-z.
=
(iii) Iz al Iz bl.
(iv) |z-a|+[z-b|=c.
(v) |z| < 1 real part
=
-
-
of
-
4.
z such that
z.
Prove that \zi |ž|, and that the real part of z is (z+ž)/2, while the imaginary
part is (z ž)/2i.
=
-
w|2
2
5.
Prove that |z + w|2
geometrically.
6.
What is the pictorial relation between z and Á· zR
by NP
goes into the real axis under multiplication
7.
real and a + bi (fora and b real) satisfies
if ao,
, an-1 are
a.-iz"¯I
the equation z" +
0, then a bi also satisfies this
+ ao
+
equation. (Thus the nonreal roots of such an equation always occur in
pairs, and the number of such roots is even.)
z"+a,_iz"¯I
is divisible by z2
+···+ao
(b) Conclude that
(a) Prove that
-i-
|z
-
--
20z
+
|w|2),and
interpret this statement
P Hint: Which line
...
·
-
-
=
-
---2az+(a2+b2
coefficients
(whose
*8.
are real).
integer which is not the square of another integer. If a and b
of a + bá, denoted by a + b
are integers we define the conjugate
,
well
showing
bá.
that
the
that
conjugate
Show
defined
is
by
as a
a
number can be written a + bá,
for integers a and b, in only one way.
(b) Show that for all a and ß of the form a + bá, we have â a; ä a if
(a) Let c be an
-
=
=
25. ComplexNumbers531
-a;
and only
if a is an integer; a + ß
that if ao,
equation z" anz"¯1
this equation.
(c) Prove
9.
*10.
...
,
a,_i
+
·
-
-
a + ß; 3
=
are integersand z
0, then ž
+ ao
=
=
=
=
a
-
ß
ä
=
-
ß; and
satisfies the
a + b
b
a
also satisfies
-
Find all the 4th roots of i; express the one having smallest argument
form that does not involve any trigonometric functions.
in a
if o is an nth root of 1, then so is ok
n-1
nth root of 1 if {1,o, m2
o is called a primitive
is the set of all nth roots of 1. How many primitive nth roots of 1 are
there for n
3, 4, 5, 9?
(a) Prove that
(b) A number
=
n-1
(c) Let o
*11.
be
an nth root of
1, with w
¢ l. Prove that
i
w*
=
0.
k-0
Zk lie on one side of some straight line through 0,
0.
Hint: This is obvious from the geometric interthen z1 +
+ Zk ¢
of
pretation
addition, but an analytic proof is also easy: the assertion is
clear if the line is the mal axis, and a trick will reduce the general case
if zi,
(a) Prove that
·
·
...
,
·
to this one.
that z1-1,
through 0, so that zi-1
(b) Show further
*12.
...
,
·
Zk¯
k
all
lie on one side of a straight line
Û·
Pove that if \til Iz2\ Iz3|and zi + 22 + Z3 0, then zi, z2, and z3 are
the vertices of an equilateral triangle. Hint: It will help to assume that zi is
real, and this can be done with no loss of generality. Why?
=
=
=
COMPLEX
CHAPTER
FUNCTIONS
You will probably not be surprised to learn that a deeper investigation of 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. Arigorous
definition presents no problems (wewill not even accord it the full honors of a
formal definition): a function is a collection of pairs of complex numbers which
does not contain two distinct pairs with the same first element. Sincewe consider
real numbers to be certain complex numbers, the old definition is really a special
case of the new one. Nevertheless, we will sometimes resort to special terminology
in order to clarify the context in which a function is being considered. A function
f is called real-valued if f (z) is a real number for all z in the domain of f, and
complex-valued
Similarly,
to emphasize that it is not necessarily real-valued.
subset
of] R in
will
usually
explicitly
that
function
is
defined
state
we
a
on [a
f
those cases where the domain of f is [asubset of] R; in other cases we sometimes
mention that f is defined on [asubset 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
f (z)
=
anz" + an-1z.-1
+
-
-
-
+ ao,
where ao,
numbers.
These functions are called, as in the
, as are complex
"identity
real case, polynomial functions; they include the function f (z) =
...
z (the
function") and functions of the form f (z) a for some complex number a ("constant functions"). Another important generalization of a familiar function is the
value function" f (z)
Iz| for all z in C.
of
Two functions particular importance for complex numbers are Re (the
part function") and Im (the
part function"), defined by
=
"absolute
=
"real
"imaginary
Re(x+fy)=x,
Im(x + iy) = y,
The
"conjugate
for x and y real.
function" is defined by
f (z) ž
=
=
Re(z)
-
i Im(z).
Familiar real-valued functions defined on R may be combined in many ways to
produce new complex-valued functions defined on C-an example is the function
f (x +
532
iy)
=
ei sin(x
·-
y) + ix3 cos y.
26. Complex
Functions 533
The formula for this particular function illustrates a decomposition which is always
possible. Any complex-valued function f can be written in the form
f=u+iv
for some real-valued functions u and v-simply define u(z) as the real part of f(z),
and v(z) as the imaginary part. This decomposition is often very useful, but not
always; for example, it would be inconvenient to describe a polynomial function
in this way.
One other function will play an important role in this chapter. Recall that an
number 0 such that
argument of a nonzero complex number z is a (real)
z
|zl(cosÐ +
=
i sin Ð).
There are infinitely many arguments for z, but just one which satisfies O < Ð <
function (the
2x. If we call this unique argument 0(z), then Ð is a (real-valued)
"argument
function") on (z in C : z ¢ 0}.
"Graphs" of complex-valued functions defined on C, since they lie in 4-dimensional space, are presumably not very useful for visualization.
The alternative
used
instead: we draw two
picture of a function mentioned in Chapter 4 can be
copies of C, and arrows from z in one copy, to f (z) in the other (Figure 1).
F1GURE
I
of a complex-valued
function is proThe most common pictorial repæsentation
duced by labeling a point in the plane with the value f (z),instead of with z (which
can be estimated from the position of the point in the picture). Figure 2 shows this
sort of picture for several different functions. Certain features of the function are
For example, the absolute value function
illustrated very clearly by such a
around
concentric
circles
is constant on
0, the functions Re and Im are constant
vertical
and
respectively,
and the function f (z) z2 wraps
horizontal lines,
on the
the circle of radius r twice around the circle of radius r2.
fimetions in genDespite the problems involved in visualizing complex-valued
eral, it is still possible to defineanalogues of important properties previously defmed
for real-valued functions on R, and in some cases these properties may be easier
to visualize in the complex case. For example, the notion of limit can be defmed
as follows:
"graph."
=
lim f (z) = I means that for every
ß
>
0 such that, for all z, if 0
<
(real)number
|z
-
a|
<
e
8, then
>
0 there is a
| f (z) Il
-
<
number
(real)
s.
534
and Infinite Series
InßniteSequences
2
2
2
2
2
a
2
i
1
2
2
2
-4
-#
i-
2
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-1
-¾
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2
2
a
2.
2
1
1
1
i
1
2
I
-1
-J
0
-1
2
-#
1
-
-1
0
--¾
-f
2
i
:
1
i
i
T
1
$.
2
1
0-
Ia
1
1
-#
1
1
2
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2
Y
2
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-i
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(b)f((
2
2
1.
j-
1
¼
i
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1
-1---
-¾
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2
Y
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-i-
,
2
2
-j-
0
2
i
1
a
2
-i
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0
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i
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1
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2
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Re(t)
2
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(a)f(t)-|c|
-·-4¿
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111111
111111
¼
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0
000000000000
fi¼¼ì¾
1
4
1
4
-1
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-#-¾-#-#-J-J
(c)/(x)-Im(x)
i
(d)f(x)
i
i
T
(e)/( )-
FIGURE
2
0(
)
=z
26. ComplexFunctions 535
Although the defmitionreads precisely as before, the interpretation is slightly dif
ferent. Since |z w| is the distance between the complex numbers z and w, the
equation lim f (z)
l means that the values of f (z) can be made to lie inside
-
=
any given circle around I, provided that z is restricted to lie inside a sufficiently
small circle around a. This assertion is particularly easy to visualize using the
"two
copy" picture of a function (Figure 3).
FIGURE
3
Certain facts about limits can be proved exactly as in the real case. In particular,
lim e
lim z
=
c,
=
a,
z->a
lim[ f (z) + g(x)]
lim f(x)-g(z)
lim
z->a
1
--
lim f(z)-limg(x),
=
1
=
g(z)
lim f (z) + lim g(z),
=
lim g(z)
if lim g (z)¢ 0.
z-va
,
The essential property of absolute values upon which these results aœ based is the
inequality Iz+ wl 5 Izl+ |w|, and this inequality holds for complex 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.
THEOREM
I
u(z) +io(z) for real-valued functions u and v, and let l
real numbers a and ß. Then lim f (z) I if and only if
Let f(z)
=
=
a + iß for
=
limu(z)
=
a,
lim v(z)
=
ß.
z->a
PROOF
Suppose first that lim f (z) = l. If s
if0<|z-a|<8,
>
0, there is & > 0 such that, for all z,
then|f(z)-ll<e.
The second inequality can be written
[u(z)-a]+i[v(z)-ß]
<e,
and InfiniteSeries
536 InfiniteSequences
or
a]
[u(z)
-
[v(z) ß]
+
<
-
e2.
-ß
and v(z)
Since u(z)
are both real numbers,
inequality therefore implies that
their squares are positive; this
-a
«]2
[u(z)
-
which
2
and
[v(z) ß]2 <
e
and
|v(z) ß| <
e2
-
implies that
lu(z)
œ
-
Since this is true for all e
|<
-
e.
0, it follows that
>
lim u (z)
=
and
a
lim v (z)
Now suppose that these two equations hold. If a
for all z, if 0 < |z a| < 8, then
>
ß.
=
0, there is a ô > 0 such that,
-
e
|u(z)-a|<- 2
which
e
and
|v(z)-œ|<-, 2
implies that
-ll
-a]+i[v(z)
|f(z)
=
|[u(z)
5 |u(z) a | +
-
€
<--+-=e.
2
This proves that lim f(z)
=
ß]
|i| |v(z) ß|
-
-
-
8
2
l.
In order to apply Theorem 1 fruitfully, notice that since we already know the
limit lim z a, we can conclude that
=
lim Re(z)
=
Re(a),
lim Im(z)
=
Im(a).
A limit like
lim sin(Re(z))
sin(Re(a))
=
follows easily, using continuity of sin. Many applications
such limits as the following:
lim ž
=
lim Izl
=
lim
(x+iy)ea+bi
of these principles prove
ä,
lai,
ei sin x + ix' cos y
=
eb
SÍnG ‡ $8
COS
b.
Now that the notion of limit has been extended to complex functions, the notion
of continuity can also be extended: f is continuous
at a if lim f (z) = f (a), and
26. ComplexFunctions 537
f is continuous
is continuous
at a for all a in the domain of f. The previous
work on limits shows that all the following functions are continuous:
if
f
f(z)=anz"+a,-iz"¯!+---+ao,
f(z) ž,
f(z)= |z|,
=
f(x+iy)=e'sinx+ix3cosy.
Examples of discontinuous functions are easy to produce, and certain ones come
One particularly frustrating example is the
funcup very naturally.
real numbers (seethe
tion" Ð, which is discontinuous at all nonnegative
0 it is possible to change the discontinuin Figure 2). By suitably redefining
example
ities; for
(Figure 4), if O'(x) denotes the unique argument of z with
x/2 5 O'(z) < 5x/2, then 0' is discontinuousat ai for every nonnegative real
number a. But, no matter how 0 is redefined,
some discontinuities will always
"argument
"graph"
occur.
3r
FIGURE
4
f(z)
-
0'(:)
The discontinuity of 0 has an important bearing on the problem of defining a
function," that is, a function f such that (f (z))2 z for all z. For real
real numbers.
numbers the function f had as domain only the nonnegative
If
complex numbers are allowed, then every number has two square roots (except
0,
which has only one). Although this situation may seem better, it is in 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.
"square-root
=
and InßniteSeries
hýiniteSequences
538
One way to define
f
is the following. We set f(0)
f(z)
+isin
cos
=
0, and for e ¢0 we set
=
.
Clearly (f (z))2 z, but the function f is discontinuous, since 0 is discontinuous.
As a matter of fact, it is impossible to find a continuous f such that (f (z))2 z for
all
z. In fact, it is even impossible for f(z) to be defined for all z with |zl 1. To
we could always
prove this by contradiction, we can assume that f(1) 1 (since
replace f by f). Then we claim that for all 9 with 0
0 < 2x we have
=
=
=
=
_<
-
f (cos0 +
i sin O)
The argument for this is left to you
argument). But (*)impliesthat
(it is a
(*)
lim
0->2x
f(cos0 +
isin0)
cos
=
=
0
+ i sin
standard
cosx
0
-.
type of least upper bound
+ isinn
-1
=
¢ f (1),
2x. Thus, we have our contradic1 as 8
even though cos0 + isin0
tion. A similar argument shows that it is impossible to define continuous
-+
d-
->
"nth-root
¯
[a,b] × [c,d]
c--
functions" for any n > 2.
For continuous complex functions there are important analogues of certain theorems which describe the behavior of real-valued functionson closed intervals. A
natural analogue of the interval [a,b] is the set of all complex numbers
z x + ly
with a < x < b and c < y
d (Figure 5). This set is called a closed rectangle,
and is denoted by
[a,b] x [c,d].
If f is a continuous complex-valued
function whose domain is [a,b] × [c,d],
then it seems reasonable, and is indeed true, that f is bounded on [a,b] × [c,d].
That is, there is some real number M such that
=
_<
a
FI GURE
5
b
|f (z)| <
It does
not
make
M
sense to say that
for all z in
f has
[a,b]
×
[c,d].
a maximum
and
a minimum
value
on
[c,d], since there is no notion of order for complex numbers. If f is a
real-valued function, however, then this assertion does make sense, and is true. In
continuous function on [a,b] x [c,d], then
particular, if f is any complex-valued
is
also
continuous,
there
is some zo in [a,b] × [c,dj such that
so
|f|
[a,b] x
||(zo)|
<
|f(z)|
for all z in
[a,b]
×
[c,dj;
a similar statement is true with the inequality reversed. It is sometimes
d]."
f attains its maximum and minimum modulus on [a,b] ×
"
said that
[c,
The various facts listed in the previous paragraph will not be proved here, although proofs are outlined in Problem 5. Assuming these facts, however, we can
26. ComplexFunctions 539
give a proof of the Fundamental Theorem of Algebra, which is really quite
surprising, since we have not yet said much to distinguishpolynomial functions
now
from other continuous
THEOREM
2(THEFUNDAMENTAL
THEOREM OF ALGEBRA)
functions.
Then there is a complex
such that
z" +
PROOF
numbers.
be any complex
Let ao,...,an-1
z"¯ I + a,_2z"¯2 +
a,_i
-
-
+ ao
-
number
z
0.
=
Let
f (x) z" +
=
a,_i
z"¯ I +
-
-
-
+ g
Then f is continuous, and so is the function |fl defined by
|fl(z)
If(z)|= Iz"+an-iz"¯'+-··+ao|.
=
Our proof is based on the observation that a point zo with f (zo) 0 would clearly
be a minimum point for |f |. To prove the theorem we will first show that |f | does
indeed have a smallest value on the whole complexplane. The proof will be almost
identical to the proof, in Chapter 7, that a polynomial function of even degree
(withreal coefficients) has a smallest value on all of R; both proofs depend on the
fact that if |zi is large, then |f(z)| is large.
We begin by writing, for z ¢ 0,
=
(
f (z) z" 1 +
=
so that
an_i
+
----
·
-
.
+
-
a0
,
an-1
If(x)|= |z|"- 1+--+---+-
a0
.
Let
M
max(1,
=
2nla._i|,
2n|aol).
...,
Then for all z with |z à M, we have |zk|à |z| and
|a._g|
Iz|
|a._al
zk|
¯
1
2n'
|an_a|
¯
2nla._al
SO
an_I
+
z
which
Un-1
a0
+--<-+
z"
GO
+-<-,
z=
¯¯
z
¯
Î
2
implies that
an-1
ao
1+---+---+z
21-
z"
an-1
ao
z
z"
-+---+-
This means that
for |z| 2 M.
|f (z)| 2 2
--
In particular, if
[z|2
M and also
|z| 2
|f(z)|
V2|f (0)|, then
à
|f(0)|.
1
2
E-.
and InßniteSeries
540 InßniteSequences
[a,b] x [c,d] be a
Now let
closed rectangle
(z Iz!5
(Figure 6) which contains
max(M,
2|f(0)|)), and suppose that the minimum
attained at zo, so that
of
|f|
on
:
[a,b] x [c,d] is
forzin[a,b]x[c,d].
(1) |f(zo)|5|f(z)|
It follows,in particular, that |f (zo)| 5 |f (0)|. Thus
(2)
Combining
(1)and (2)we
J2|f(0)|),then|f(x)|E
see that
lf(0)|2|f(zo)|.
|f(zo)| 5 If(z)| for all z, so that |f|
value on the whole complex
minimum
FIGURE
if|z|2max(M,
attains its
plane at zo.
6
To complete the proof of the theorem we now show that f (zo)
to introduce the function g defined by
=
0. It is
convenient
g(z)
=
f(z +
zo).
Then g is a polynomial function of degree n, whose minimum absolute value
0.
occurs at 0. We want to show that g(0)
Suppose instead that g(0)
a ¢ 0. If m is the smallest positive power of z
which occurs in the expression for g, we can write
=
=
g(z)=a+ßz"+cm+1z"**
where
ß ¢ 0. Now, according
-
Ra",
25-2 there is a complex
to Theorem
such that
ß
number y
26. ComplexFunctions 541
Then, setting dk
=
CkŸk,We have
|g(yz)|= |a+ßy"zm+dm+1z"
|a
+duz"I
+
-arm+dmwizm+1
=
=
a
1-z"+d.wizm
tortuously arrived at, will enable us to reach a quick contradiction. Notice first that if |z| is chosen small enough, we will have
This expression,
so
dm+1
<1.
-z+---
a
If we choose, from among all z for
and positive,
then
Consequently, if 0
<
z
<
which
this inequality holds, some z which is real
1 we have
l
=
-
.
z
dm+1
z+
a
·
j
<l-zm
1.
=
This is the desired contradiction:
for such a number
z we have
|g(yz)| < |a|,
contradicting the fact that |a| is the minimum of |gl on the whole plane. Hence,
the original assumption must be incorrect, and g(0)
0. This implies, finally, that
-
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 fimctions. The next obvious
step is to define derivatives: a function f is differentiable
at a if
lim
f(a+z)
-
f(a)
.
exists,
542
and InfiniteSeries
InfiniteSequences
in which case the limit is denoted by f'(a). It is easy to prove that
f'(a)
=
f'(a)
(f+g)'(a)=
(f g)'(a)
=
=
-
0
1
f (z) c,
if f (z) z,
f'(a)+g'(a),
f'(a)g(a) + f (a)g'(a),
'
=
=
-g'(a)
(1
-/-
(a)
-
g
(fo
if
g)'(a)
if g(a)
=
[g(a)]
=
f'(g(a))
0,
g'(a);
-
the proofs of all these formulas are exactly the same as before. It follows, in
nz"¯1. These formulas only
particular, that if f (z) z", then f'(z)
prove the
of
candidates
rational
obvious
differentiability
functions however. Many other
are
not differentiable. Suppose, for example, that
=
=
f (x +
iy)
=
-
x
iy
(i.e.,f (z)
=
ž).
If f is to be differentiable at 0, then the limit
.
lun
f(x +
iy)
-
f (0)
.
lun
=
-
iy
(x+iy)-0 x + iy
x + iy
(x+iy)-o
x
must exist. Notice however, that
if y
=
if x
=
0, then x
-
x
0, then
-
iy
=
x + iy
1,
and
iy
-1;
=
x + iy
therefore this limit cannot possibly exist, since the quotient has both the values 1
and
for x + iy arbitrarily close to 0.
of this example, it is not at all clear where other differentiable functions
view
In
are to come from. If you recall the definitions of sin and exp, you will see 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 solved.
-1
PROBLEMS
1.
x + iy (sothat a is a complexvalued function defined on R). Show that œ is continuous. (This follows
immediately from a theorem in this chapter.) Show similarly that ß(y)
x + iy is continuous.
(b) Let f be a continuous function defined on C. For fixed y, let g(x) =
on R). Show
f (x + iy). Show that g is a continuous function (defined
similarly that h(y)
f(x + iy) is continuous. Hint: Use part (a).
(a) For any
real number
y, define a(x)
=
=
=
2.
is a continuous real-valued function defined on a closed
[a,b]×[c, d]. Prove that if f takes on the values f (z) and f(w)
(a) Suppose that f
rectangle
26'. ComplexFunctions 543
for z and w in [a,b] × [c,d], then f also takes all values between f(z)
and f(w). Hint: Consider g(t)
f(tz + (1 t)w) for t in [0,1].
complex-valued
function defined on [a,b) × [c,d],
If f is a continuous
the assertion in part (a)no longer makes any sense, since we cannot talk
of complex numbers between f (z) and f (w). We might conjectum that
f takes on all values on the line segment between f (z) and f (w), but
even this is false. Find an example which shows this.
=
*(b)
3.
that if ao,
complex numbers zi,
(a) Prove
...
a,_;
,
...
,
-
then there are
are any complex numbers,
necessarily distinct) such that
z.
z" + an-i za-1 +
(not
·
-
·
+ ao
(z z¿).
=
-
i=1
an-1z"¯I
that if ao,
+
+ ao can be
a._ i are real, then z" +
written as a product of linear factors z+a and quadratic factors z2+c+b
all of whose coefficients are real. (Use Problem 25-7.)
(b) Prove
4.
-
.
.
.
-
·
,
In this poblem we will consider only polynomials with real coefficients.
ifit can be written as hi2+.
a polynomial is called a sum of squares
for polynomials h, with real coefficients.
-+h,2
Such
-
if f is a sum of squares, then f (x) :> 0 for all x.
if f and g are sums of squares, then so is f g.
that
f (x) > 0 for all x. Show that f is a sum of squares. Hint:
(c) Suppose
write
First
f(x) xkg(x), where g(x) ¢ 0 for all x. Then k must be
and g(x) > 0 for all x. Now use Problem 3(b).
even (why?),
(a) Prove that
(b) Prove that
-
=
(a)
5.
A be a set of complex numbers.
real case, a limit point of the set A
A number z is called, as in the
if for every (real)e > 0, there is
<
with
a|
point
in
but
A
a
a
|z
e
z ¢ a. Prove the two-dimensional
version of the Bolzano-Weierstrass Theorem: If A is an infinite subset
of [a,b] x [c,d], then A has a limit point in [a,b] x [c,J]. Hint: First
divide [a,b] × [c,d] in half by a vertical line as in Figure 7(a). 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.
(a) Let
-
(b)
FIGURE
7
(The two-dimensional bisection argument outlined in this hint is so standard that the title "Bolzano-Weierstrass" often serves to describe the
method of proof, in addition to the theorem itself See, for example,
H. Petard, "A Contribution to the Mathematical Theory of Big Game
446--447.)
Hunting," Amer.Math. Monthly,45 (1938),
continuous
that
function on [a,b] × [c,d] is
a
(complex-valued)
(b) Prove
bounded on [a, b] x [c,d]. (Imitate Problem 22-31.)
(c) Prove that if f is a real-valued continuous function on [a,b] x [c,d],
then f takes on a maximum and minimum value on [a,b] x [c,d]. (You
can use the same trick that works for Theorem 7-3.)
and InfiniteSeries
544 InfiniteSequences
*6.
The proof of Theorem 2 cannot be considered to be completely elementary
depends on Theobecause the possibility of choosing y with y=
rem 25-2, 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" c
0.
-a/ß
-
=
-
to show that solutions of z2 c = 0 can
an explicit computation
be found for any complex number c.
(b) Explain why the solution of z" c = 0 can be reduced to the case where
(a) Make
(a)
a convex
subset
of the planc
-
-
is odd.
z" c has its minimum
Let
(c)
zo be the point where the function f(z)
absolute value. If zo 0, show that the integer m in the proof of Theorem 2 is equal to 1; since we can certainly find y with yl
-g the
remainder of the proof works for f. It therefore suffices to show that the
minimum absolute value of f does not occur at 0.
(d) Suppose instead that f has its minimum absolute value at 0. Since n is
odd, the points ±8, ±ai go under f into
Show that for
value
small 8 at least one of these points has smaller absolute
than
thereby obtaining a contradiction.
n
=
-
-¢
-
-c±a",
-c±ô"i.
-c,
(b)
a nonconvex
FIGURE8
subset
of the
plane
¯ZI)ml.---'(
•
-Zk)m
k
(a) Show that f'(z)
(b) Let
g(z)
=
=
(x zi)"t
ma(z
-
-
zŒ)¯I.
-
-
...
z
-
kym*
mg(Z
-
Za)¯
.
«=1
Show that if g(z)
cannot all lie on the same side of a straight
-
=
0, then zi,
...,
k
line through z. Hint: Use
Problem 25-11.
A
(c) subset K of the plane is convex if K contains the line segment joining
any two points in it (Figure 8). For any set A, there is a smallest convex
set containing it, which is called the convex hull of A (Figure 9); if a
FIGURE
9
26. Complex
Functions 545
point P is not in the convex hull of A, then all of A is contained on one
side of some straight line through P. Using this information,
prove that
the roots of f'(z)
0 lie within the convex hull of the set {Zl9
ZkÌ·
Further information on convex sets will be found in reference [19]of the
Suggested Reading.
=
8.
*9.
--·
9
Prove that if f is differentiable at z, then f is continuous at z.
Suppose that f
u + iv where
=
u and v are real-valued
functions.
u(x + iyo) and h(x)
v(x + iyo). Show that if
yo let g(x)
f'(xo + iyo) a + iß for real a and ß, then g'(xo) a and h'(xo) ß.
(b) On the other hand, suppose that k(y) u(xo+iy) and l(y) v(xo+iy).
Show that l'(yo)
a and k'(yo)
(c) Suppose that f'(z) 0 for all z. Show that f is a constant function.
(a) For fixed
=
=
=
=
=
-ß.
=
=
-
10.
(a) Using the
expression
f (x)
=
=
-
1
1+x2¯2i
1
1
x-i
find f (k)(x) for all k.
(b) Use this result to find arctan(k) (0) for all k.
1
.
x+r
,
CHAPTER
If you have not already guessed where differentiable complex functions are going
to come from, the title of this chapter should give the secret away: we intend to
define fimctions by means of infmite series. This will necessitate a discussion of
infinite sequences of complex numbers, and sums of such sequences, but (aswas
the case with limits and continuity) the basic definitions are almost exactly the
same as for real sequences and series.
An infinite sequence of complex numbers is, formally, a complex-valued function whose domain is N; the convenient subscript notation for sequences of real
numbers will also be used for sequences of complex numbers. A sequence (a.} of
complex numbers is most conveniently
pictured by labeling the points as in the
plane (Figure 1).
of complex
The sequence shown in Figure 1 converges to 0,
real
sequences being defined precisely as for
sequences: the sequence (a,}
converges
to l, in symbols
lim a,
l,
as'
a,.ag
a,•
a,
.
a•
.
a.
•
a'
•""
,a
a•,*a=
FI GURE
COMPLEX POWER SERIES
1
"convergence"
.a'
=
•
.a.
arf
a..
if fOr every e
GN
>
0 there is a natural
•
.
.
.
if n
.
number
N, then
>
N such that, for all n,
|a.
-
l|
<
e.
condition
FIGURE
This
means that any circle drawn around I will contain a. for all sufficiently large n (Figure 2); expressed more colloquially, the sequence is eventually
inside any circle drawn around l.
Convergence of complex sequences is not only defined precisely as for real
S€qu€nC€S,
but can even be reduced to this familiar case.
2
THEOREM
1
Let
a,
=
b, + ic,,
for
real
b,, and c,,,
and let
l
lim a,
Then U-¥OO
=
=
ß+
for real ß and y.
l if and only if
lim b,
R-FDD
PROOF
iy
=
ß
and
lim c,
RADO
=
y.
The proof is left as an easy exercise. If there is any doubt as to how to proceed,
consult the similar Theorem 1 of Chapter 26.
The sum of a sequence {a.}is defined, once again, as lim s., where
n->oo
s.=ai
546
+··-+a..
27. ComplexPowerSeries 547
Sequences for which this limit exists are summable;
the infinite series
alternatively, we may say that
if this limit exists, and diverges otherwise.
as converges
It
n=I
is unnecessary to develop any new tests for convergence of infinite series, because
of the following theorem.
THEOREM
2
Let
a,
for real b, and c..
b, + ic,
=
a, converges if and only if
Then
n=I
b, and
c, both converge, and in this
n=1
n=I
case
a.
b. + i
=
n=1
PROOF
c.
n=1
.
n=i
This is an immediate consequence of Theorem 1 applied to the sequence ofpartial
sums of
{a.).
There is also a notion of absolute
a, converges
convergence
if the series
absolutely
for complex series: the series
|a, \ converges (thisis a
series of real
n=I
n=t
numbers, and consequently one to which our earlier tests may be applied).
following theorem is not quite so easy as the preceding two.
THEOREM
3
The
Let
a,
for real b, and c..
b, + ic,
=
as converges absolutely if and only if
Then
n=1
b, and
n=1
c,
both converge
n=1
absolutely.
PROOF
b, and
Suppose first that
n=1
|c,| both
c,
both converge absolutely, i.e., that
n=t
converge. It follows that
|b,
|c.| converges. Now,
+
.=1
n=i
|a, |
=
|b, + ic. | 5 |b.| + |c,|.
It follows from the comparison test that
la.|
converges
n-1
|b,| + |c,\ are
|b,\ and
n=1
real
and nonnegative).
Thus
a, converges
n=1
(thenumbers |a,|
absolutely.
and
548
and InfiniteSeries
InfiniteSequences
Now suppose that
|a,|
converges. Since
n=1
|a. |
/b.2 +
=
c.2,
it is clear that
and
|b.| 5 |a,|
Once again, the comparison
|c,| 5 |a.|.
test shows that
|bs| and
n=1
of
Two consequences
|c,| converge.
n=1
Theorem 3 are particularly noteworthy.
If
an conn=i
verges
absolutely,
b, and
then
n=1
b, and
n=1
c, converge,
c, also converge absolutely; consequently
n=1
by Theorem 23-5, so
n=i
as converges
by Theorem 2.
n=1
In other words, absolute convergence implies convergence. Similar reasoning
series has the same
shows that any rearrangement
of an absolutely convergent
sum. These facts can also be proved directly, without using the corresponding theorems for real numbers, by first establishing an analogue of the Cauchy criterion
(seeProblem
13).
With these preliminaries safely disposed of,
power series, that is, functions of the form
we
can now consider
complex
a,(z-a)"=ao+a1(z-a)+a2(z-a)2+--·.
f(z)=
n=o
Here the numbers a and an are allowed to be complex, and we are naturally
interested in the behavior of f for complex z. As in the real case, we shall usually
consider power series centered at 0,
f(z)
anz";
=
n=0
in this case, if f (zo) converges, then f (z) will also converge for |z| < |zo|. The
proof of this fact will be similar to the proof of Theorem 24-6, but, for reasons
that will soon become clear, we will not use all the paraphernalia of uniform convergence and the Weierstrass M-test, even though they have complex analogues.
Our next theorem consequently generalizes only a small part of Theorem 24-6.
THEOREM
4
Suppose that
anzo"
n=o
=
ao + al
zo +
a2zo2
27. ComplexPowerSenes 549
for some zo ¢ 0. Then if |zl < |zo|, the two series
converges
anz"
ao + ai
=
a2x2 +
z+
-
-
-
n-0
nanzn-1
2a2z + 3a3z" +
ai +
=
-
·
n=l
both converge absolutely.
PROOF
As in the proof of Theorem 24-6, we will need only the fact that the set of numbers
a,zo" is bounded: there is a number M such that
|anzo"|5
for all n.
M
We then have
z
la.z"I lanzo"|=
-
zo
In
O
and, for z
¢ 0,
-1
Inanz"¯I
|
nianzo"
-
Izi
<-n
-
Izl
Iz/zo|" and
n
nanz"¯*
converge
zo
z
zo
.
this shows that both
Iz/zo|"converge,
n=1
n=0
and
-
"
M
¯
Since the series
|
absolutely
n=1
anz"
n=0
(theargument
z ¢ 0, but this series certainly converges for z
=
nanz""
for
assumed
that
n=1
0 also).
Theorem 4 evidently restricts greatly the possibilities for the set
z
:
anz" converges
.
For example, the shaded set A in Figure 3 cannot be the set of all z where
anz"
n=o
FIGURE
3
since
it contains z, but not the number w satisfying w| < |z|.
It seems quite unlikely that the set of points where a power series converges
of
could be anything except the set of points inside a circle. If we allow
and
radius
the power series converges only at 0)
of
radius 0" (when
oo"
series converges at all points), then this assertion is true (withone
the
(when power
complication which we will soon mention); the proof requires only Theorem 4 and
a knack for good organization.
converges,
"circles
"circles
550
and InfiniteSeries
InfiniteSequences
THEOREM
5
For any power series
anz"
a2:2 +
ao + aiz +
=
a3Z3
n=0
one of the
following three possibilities must be true:
(1)n=o
anz" converges only
(2)
anz" converges
for z
0.
=
absolutely for all z in C.
n-0
is a number
(3)There
>
0 such that
anz" converges absolutely if
diverges if Iz| > R. (Notice that we do not mention
and
R
|z| <
n=o
when
PROOF
R
|z|
what
happens
R.)
=
IÆt
S
=
x
in R
anw" converges for some w
:
with
|w
=
x
.
n=o
Suppose first that S is unbounded.
x in S such that
a number
|z| <
Then for any complex
number
z, there is
By definition of S, this means that
x.
an w"
n=o
converges for some w with
|wl
=
x
|z|. It followsfrom Theorem 4 that
>
anz"
n=o
absolutely. Thus, in this case possibility (2)is true.
Now suppose that S is bounded, and let R be the least upper bound of S. If
converges
R
anz" converges
0, then
=
only
for z
=
0,
so
possibility
(1)is true.
Suppose,
n=o
on the other hand, that R
is a number
x
>
0. Then if z is a complex number with |z| < R, there
in S with |z| < x. Once again, this
a.w"
that
means
converges
n=o
for some w with Iz|
<
|w|, so
anz" converges
that
Moreover, if
absolutely.
n=0
[z| >
anz" does not converge, since
R, then
Iz| is not in
S.
n=0
The number
anz".
n=o
R which occurs in case
In cases
(3)is called
(1)and (2)it is customary
is 0 and oo, respectively. When 0
the circle of convergence
<
R
<
the radius
of convergence
of
to say that the radius of convergence
oo, the circle
(z : |z|
=
R} is called
anz". If z is outside the circle, then, of course,
of
n=o
27. ComplexPowerSeries 551
does not converge, but actually a much stronger statement can be made:
anz"
n=o
bounded. To prove this, let w be any number with
R; if the terms anz" were bounded, then the proofof Theorem4 would
the terms anz" are not even
1z|> |w| >
anw" converges, which is
show that
false. Thus (Figure 4), inside the circle of
n=0
the terms anz- are not bounded
oo
.
anz" converges m the
the series
convergence
best possible way
and
(absolutely)
n=0
outside the circle the series diverges in the worst possible way
.,,=
What happens on the circle of convergence is a much more difficult question.
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.)
abmluœly
circle of convergence
FIGURE
4
anz" are
bounded).
not
Converges
(theterms
Algebraic manipulations
Thus, if
TCâÌCRSe.
anz" and g(z)
=
baz" both have radius of
=
n=0
R, then h(z)
>
convergence
f(z)
on complex power series can be justifiedjust as in the
n-0
(a,
=
+ b.)z" also has radius
of convergence
n=0
R and h
h(z)
=
f
caz",
=
+ g inside the circle of radius R. Similarly, the Cauchy product
for c,
akb,_g, has radius of convergence
=
inside the circle of radius R. And if
f (z)
fg
=
a,z" has radius of convergence
=
>
0
n-o
-/
and ao
à R and h
k=0
n=0
b,z"
0, then we can fmd a power series
with
radius
of convergence
n=o
>
0 which
æpresents
1/f inside its circle of convergence.
But our real goal in this chapter is to poduce differentiable functions. We
therefore want to generalize the result proved for real power series in Chapter 24,
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 24, even if we were willing to intoduce uniform convergence, because no
analogue of Theorem 24-3 seems available. Instead we will use a direct argument
(whichcould also have been used in Chapter 24). Before beginning the proof,
we notice
that at least there is no problem about
produced by term-by-term
the convergence
difTerentiation. If the series
anz"
of the series
has radius of con-
n=0
vergence
R, then Theorem 4 immediately implies that the series
converges for
Iz| < R. Moreover, if |z| > R,
nanz"¯1 also
so that the terms anz" are unbounded,
552
and InfiniteSeries
InfiniteSequences
then the terms nanz"¯1 are surely unbounded,
This shows that the radius of convergence
nanz"¯1 does not converge.
so
nanz"¯I
of
is also exactly R.
n=1
THOREM
e
If the power series
f (z)
anz"
=
n=0
has radius of convergence
and
R
>
0, then
f'(z)
f
is differentiable at z for all z with |z! < R,
nanza-1
=
n=l
PROOF
"e/3
argument." The fact that the theorem is clearly true for
We will use another
polynomial functions suggests writing
(*)
f (x +
h)
f (z)
--
h
°°
-
°°
E
nanz
n-1
=
a"
n=1
°°
y
a"
((z + h)"
h
n=0
((z + h)"
z")
-
h
n=0
N
-
a"
((z +
h)"
°°
E nanz
-
,_i
n=1
((z + h)"
z")
-
h
n=0
N
z")
-
N
z")
-
1
N
og
nanz"¯
+
n=1
-
nanza-1
n=l
We will show that for any e > 0, each absolute value on the right side can be made
< e/3 by choosing N sufficiently
large and h sufficiently small. This will clearly
prove the theorem.
Only the first term in the right side of (*)will present any difficulties. To begin
with, choose some
zo with |z! < Izo| < R; henceforth we will consider only h
with |z + h| 5 |20|. The expression
((z + h)" z")/h can be written in a more
convenient way if we remember
that
-
xn_yn
n-14
n-2
n-3
n-1
24...
-y
x
Applying this to
(z + h)"
h
--
z"
¯
(z + h)"
-
z"
'
(z+h)-z
we obtain
(z+h)"-z"
h
=
(z+h)n-1+z(z+h)"
*+
··+z"¯'.
27. ComplexPowerSeries 553
Since
|(z + h)n-1 + z(z + h)"¯
we have
a.
But the series
n|a.|
((z + h)"
+
z")
-
h
··
z"¯I|
+
·
5 nlanl
n|zo|"¯1
5
,
Izol
·
y
-
|zo|"¯1converges, so if N is sufficiently large, then
·
n=l
nia.I Izol"¯'<
·
.
n=N+1
This means that
°°
i
a"
((z + h)"
z")
-
N
i
-
h
n=0
((z + h)"
a"
h
n=0
°°
=
((z + h)"
a"
z")
-
h
n=N+1
z")
--
°°
g
5
a"
((z +
h)"
-
z")
h
n=N+1
n=N+1
In short, if N is sufficiently large, then
°°
(1)
L
a"
((z + h)"
N
z")
-
a"
-
h
n=0
for all h with
((z + h)"
h| 5
3'
|zo|.
It is easy to deal with the third term on the right side of
converges,
e
'
h
n=0
|z +
z")
--
(*):Since
nanz"¯'
n=1
it follows that if N is sufficiently large, then
nanz"
(2)
I
nanz"¯1
-
n=1
n=1
Finally, choosing an N such that
N
(1)and (2)are
((z + h)"
La"
lin
-
z")
h
n=0
true, we note that
N
=
2nanz"
,
n=1
N
since the polynomial
function
g(z)
anz" is certainly differentiable. Therefore
=
n=o
N
(3)
n=0
Un((z
+ h)"
-
z")
h
N
-
i
nanz"¯
<
.
n=1
for sufficiently small h.
As we have already indicated,
(1),(2),and (3)prove the
theorem.
and InßniteSeries
554 InfiniteSequences
Theorem 6 has an obvious corollary: a function represented by a power series is
infinitely differentiable inside the circle of convergence, and the power series is its
Taylor series at 0. It follows,in particular, that f is continuous inside the circle of
since a function differentiable at z is continuous at z (Problem 26-8).
convergence,
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 promised
answers to the questions raised at the end of Chapter 24. We have already seen
that the Taylor series for the function f (z) 1/(1 + z2), namely,
=
1-Z2
4
6
for real z only when izi < 1, and consequently has radius of convergence 1. It is no accident that the circle of convergence contains the two points
i and
If this power series converged in a circle of
at which f is undefined.
radius greater than 1, then (Figure 5) it would represent a fimction which was
continuous in that circle, in particular at i and
But this is impossible, since it
z2)
z2)
unit
equals 1/(1 +
circle, and 1/(1 +
does not approach a limit
inside the
from inside the unit circle.
i or
as z approaches
The use of complex numbers also sheds some light on the strange behavior of
the Taylor series for the function
converges
---i
-i.
-i
FIGURE
5
e¯1¡x2,
x ¢ 0
0.
0,
l
x
Although we have not yet defined et for complex z, it will presumably be true that
if y is real and unequal to 0, then
f (x)
=
=
f (iy)
e¯1/cy?
=
=
l'
e
.
The interesting fact about this expression is that it becomes large as y becomes
Thus f will not even be continuous at 0 when defined for complex numbers,
0.
so it is hardly surprising that it is equal to its Taylor series only for z
The method by which we will actually define et (aswell as sin z and cos z) for
complex z should by now be clear. For real x we know that
small.
=
x
sinx=x--+
3
x
5
¯
'
31
x
2
x
cosx=1--+------,
2!
4
41
2
x
x
1!
2!
e2=1+-+-+---
For complex z we therefore define
5
z3
.
smz=z---+------,
31
5!
2
4
cosz=1---+--+---,
2!
.
4!
2
exp(z)
=
er
=
1+
+
+
-
-
-
27. ComplexPowerSeries 555
sin z, and exp'(z)
exp(z) by Theorem 6.
Then sin'(z)
cos z, cos'(z)
Moreover, if we replace z by iz in the series for e2, and make a rearrangement
of the terms (justified
by absolute convergence), something particularly interesting
happens:
=
=
e
2!
3!
2
=1+sz------+--+-+
3
it 4
3!
4!
it
z
.
2!
=
4!
it 5
5!
4
2
It is clear from the definitions
5
z
z
z----+--+···
3! 5!
+i
,
cos x + i sin z.
=
e
+··-
5!
3
z
z
1---+----··
2! 4!
(
5
(iz)2 (iz)* (iz)4
=1+iz+-+-+-+
so
=
--
(i.e.,the
power series) that
sin(-z)
=
cos(-z)
=
sin
-
z,
cos z,
so we also have
e¯"
=
cos z
i sin z.
-
From the equations for en and e¯u we can derive the formulas
eu
.
e¯¤
-
=
smz
,
.
2:
cos z
eu +
=
e¯"
2
The development of complex power series thus places the exponential function at
the very core of the development of the elementary functions-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
As a by-product of this
been discovered without the use of complex numbers.
relationship,
obtain
we
a hitherto unsuspected connection between the numbers e
and x: if in the formula
eu
we take
z
=
=
cosx + isinz
x, we obtain the remarkable
e'"
result
--1.
=
(More generally, e2"U" is an nth root of 1.)
With these remarks we will bring to a close our investigation of complex functions. And yet there are still several basic facts about power series which have not
been mentioned. Thus far, we have seldom considered power series centered at a,
f (z)
a.(z
=
n=o
-
a)",
556
and InfiniteSeries
InfiniteSequences
0. This omission was adopted partly to simplify the exposition.
except for a
For power series centered at a there are obvious versions of all the theorems in
this chapter (theproofs require only trivial modifications): there is a number R
=
0 or
(possibly
"oo")
with
|z
all
with
R, and has unbounded
a| < R the function
z
-
al
<
Iz
-
such that the series
f(z)
terms for z with
=
a
-
a|
>
R; moreover,
for
G)n
an
has derivative
R
6
|z
n=0
b-a|
FIGURE
a)" converges absolutely for z
-
n=0
R
a
a.(z
n-1
r
It is less straightforward
to investigate the possibility of representing a function
series
centered
at b, if it is already written as a power series centered
as a power
at G. If
a.(z-a)"
f(z)=
n=o
has radius of convergence R, and b is a point with |b a| < R (Figure 6), then it
is true that f (z) can also be written as a power series centered at b,
-
b,(x-b)"
f(z)=
.=o
b, are necessarily ff">(b)/n!);moreover, this series has radius of
least R |b a| (itmay be larger).
convergence
We will not prove the facts mentioned in the previous paragraph, and there are
several other important facts we shall not
prove. For example, if
(thenumbers
at
--
f(z)
-
an(z
=
-
a)"
n=o
and g(b)
and
g(z)
b.(z
=
-
b)",
n=o
a, then we would expect that fog can be written as a power series
centered at b. All such facts could be proved now without introducing any basic
new ideas, but the proofs would not be as easy as the proofs about sums, products
and reciprocals of power series. The possibility of changing a power series centered
at a into one centered at b is quite a bit more involved, and the treatment of
fog requires still more skill. Rather than end this section with a tour deforce
analysis," one of
of computations, we will instead give a preview of
where all these facts are derived as
the most beautiful branches of mathematics,
=
"complex
straightforward
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
consequences
27. ComþlexPowerSeries 557
collection
functions. The basic theorems
of differentiable complex
of complex
show that this is not at all true:
analysis
If a cornplexfunction
is d¢ned in some region A of the planeand is dj¶erentiablein A,
foreachpointa in A the
thenit i,sautomatically infmitelydfferentiablein A. Moreover,
Taylorseriesforf at a will convergeto f in any cirde containedin A (Ftgure7).
These factsare among the first to be proved in complex analysis. It is impossible
methods used are quite different
to give any idea of the proofs themselves-the
elementary
If
these
calculus.
facts are granted, however, then
from anything in
the facts mentioned before can be proved very easily.
Suppose, for example, that f and g are functions which can be written as power
series. Then, as we have shown, f and g are differentiable-it then follows from
also differentiable.
easy general theorems that f + g, f g, 1/g and fog are
Appealing to the results from complex analysis, it follows that they can be written
as power sefies.
and 1/g from
We already know how to compute the power series for f + g, f
would
also
compute
how
and
those for f
an expression
one
easy to guess
g. It is
expansions
series
when
given
series
the
we are
in (z
for f og as a power
power
·
FIGURE
7
.g
-b)
an(z-a)"
f(z)=
n=0
g(x)
bk(z
=
b)k
-
k=0
with a
=
g(b)
be, so that
=
g(z)
-
bt(z
=
a
b)
-
.
k=1
the power series
First of all, we know how to compute
I
(g(z)
-
a)'
=
(oo
bk(z
-
b)*
,
k=1
series will
and this power
begin
with
(z
-
b)'. Consequently, the coefficient
of z"
in
a;(g(z)-a)'
f(g(z))=
t=0
can
be calculated
powers of g(z)
as a
-
fmite sum, involving only coefficients
arising from the
first n
a.
Similarly, if
f (z)
a.(z
=
-
a)"
n=0
has radius of convergence R, then f is differentiable in the region A {z: Iz-a| <
R}. Thus, if b is in A, it is possible to write f as a power series centered at b,
=
558
and Inßnite Senes
InfiniteSequences
which will converge
will be ff">(b)/nl.
a.(z
a)" may
-
al. The coefficient of z"
b
in the circle of radius R
This series may actually converge in a larger circle, because
-
-
be the series for a function differentiable in a larger region
n=o
than A. For example, suppose that f (z)
1/(1 + zi). Then f is differentiable,
where
it is not defined. Thus f (z) can be written as a power
except at i and
=
-i,
i +
4),
anzn with radius
series
of convergence
1
(asa
n=o
a2,
(-1)"
=
and as
=
0 if k is odd). It is also possible to write
f (z)
bn(z
=
-
n=0
where
FIGURE
8
b,
=
matter of fact, we know that
ff"3(i)/n!. We can
()",
easily predict the radius
of convergence of this
Î ( )2, the distance from to i or
(Figure 8).
added
complex
analysis
incentive to investigate
further, one more result
As an
will be mentioned,
which lies quite near the surface, and which will be found in
serÎCS:
it is
-l-
¼
-i
any treatment of the subject.
and 1, but for complex
For real z the values of sin z always lie between
real,
all
then
this is not at
true. In fact, if z iy, for y
edd
e¯Y
e¯¤©
ei
-1
z
=
-
siniy
-
.
2i
2i
If y is large, then sin iy is also large in absolute value. This behavior of sin is typical
of functions 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:
Liouville'sTheoren: The only boundedentirefunctions
are the constantfunctions.
As a simple application
of
Liouville's Theorem, consider a polynomial function
f(z)=z"+a._iz"¯I+---+ao,
1, so that f is not a constant. We already know that f (z) is large for
large z, so Liouville's Theorem tells us nothing interesting about f. But consider
the function
1
where
n
>
g(x)
=
-.
f (z)
If f (z) were never 0, then g would be entire; since f (z) becomes large for large z,
the function g would also be bounded, contradicting Liouville's Theorem. Thus
f (z) 0 for some z, and we have proved the Fundamental Theorem of Algebra.
-
PROBLEMS
1.
Decide whether each of the following series converges, and whether
verges absolutely.
it con-
27. ComplexPowerSenes 559
°°
(i)
(1+i)n
'
nl
=1
n
°°
(ii)
L
n=1
(iii) n=1
(iv) n=i
(v)
2.
l + 2i
2"
.
(( +
(i)".
logn
n=2
+ i"
logn
--.
n
n
Use the ratio test to show that the radius of convergence of each of the
following power series is 1. (In each case the ratios of successive terms will
approach a limit < 1 if Iz! < 1, but for |z| > 1 the ratios will tend to oo or
to a limit > 1.)
(i)
.
n=1
(ii)
.
n=l
(iii)
z".
n=1
(n + 2¯")z".
(iv)
n=1
(v)
3.
2"z"'.
n=1
Use the root test (Problem 23-7) to find the radius of convergence
the following power series.
23456
zzzzzz
3
2
-+--+--+-+-+-+---.
(i)
(ii)
22
.
n=1
z".
(iii)
n=1
(iv)
(v)
z".
n=1
2"¿"'.
n=1
32
23
33
of each of
560
and Inßnite Series
InfiniteSequences
4.
The root test can always be used, in theory at least, to find the radius of
of a
convergence
power series; in fact, a close analysis of the situation leads
to a formula for the radius of convergence, known as the "Cauchy-Hadamard
formula." Suppose first that the set of numbers
is bounded.
(a) Use Problem 23-7 to
show that
if
Ïi¯
WROO
Ñ
Iz| < 1, then
anz" conn=o
verges.
(b) Also show
that if
Ïi
|z
n->oo
fin
"oo").
anz" is
of
To complete the formula, deis unbounded.
diverges for z ¢ 0, so that the
anz"
this case,
means
oo if the set of all
=
terms.
n=o
(where
00
N
that the radius of convergence
"1/0"
Ïld
anz" has unbounded
1, then
n=o
(c) Parts (a)and (b)show
1/
>
radius
Prove that in
of convergence
n=o
is 0
5.
(whichmay
"1/00").
be considered
as
Consider the following three series from Problem 2:
n=1
n=1
n=1
Prove that the first series converges everywhere on the unit circle; that the
third series converges nowhere on the unit circle; and that the second series
converges for at least one point on the unit circle and diverges for at least
one point on the unit circle.
6.
e=+= for all complex numbers
e2 e=
z and w by showing
e=**
that the infinite series for
is the Cauchy product of the series for e2
and e".
and cos(z + w)
sinzcosw + coszsinw
(b) Show that sin(z + w)
sin
and
sin
all
complex
for
w.
w
cos z cos w
z
z
(a) Prove that
-
=
=
=
-
7.
every complex number of absolute value 1 can be written eu
for some real number y.
(b) Prove that |e**U| ex for real x and y.
(a) Prove that
=
8.
(a) Prove that
exp takes on every complex value except 0.
(b) Prove that sin takes on every complex value.
9.
For each of the following functions, compute the first three nonzero
the Taylor series centered at 0 by manipulating
power series.
(i) f (z)
=
(ii) f(z)
=
tan z.
-:)¯1/2
z(1
terms of
27. ComplexPowerSenes 561
e"
(iii) f (z)
=
(iv) f(z)
=
.
log(1
sin2
(v) f (z)
=
(vi) f (z)
=
(vii)f (z)
=
(viii)f(z)
=
1
-
z2L
-
z
z2
sin(z2
10.
.
z cos2 z
1
4
2x2 + 3
°
-
[e(E¯U
-
1].
that we write a differentiable complex function f as f
u +
Let ü and ü denote the restrictions
iv, where u and v are real-valued.
of u and v to the real numbers. In other words, ü(x)
u(x) for real
numbers x
x).
Using
other
Problem 26-9, show
(butü is not defined for
that for real x we have
(a) Suppose
=
=
f'(x)
ü'(x) +iif
=
(x),
denotes the complex derivative, while ü' and D' denote the
derivatives of these real-valued functions on R.
(b) Show, more generally, that
where
f'
ordinary
satisfies the equation
(c) Suppose that f
f f") +
(*)
a,-i
f
in-1)
+
-
-
-
+ ao f
=
0,
the a; are real numbers, and where the f (k) denote higher-order
complex derivatives. Show that ü and 6 satisfy the same equation, where
üf*)and ü(k)HOW denote higher-order derivatives of real-valued functions
where
on R.
(d) Show that
if a b+ ci is a complex root of the equation z" + an-1xn-1 4
ebx sin cx and f (x) = ebx cos cx are both
0, then f (x)
=
+ ao
solutions of
-
11.
-
-
=
=
(*).
on C.
log |w|
w if and only if z x +iy with x
(b) Given w ¢ 0,
real logarithm function), and
of to.
the
argument
log
denotes
an
y
(here
*(c)
exist
continuous
there
that
function log defined for
Show
does not
a
such that exp(log(z))
nonzero complex numbers,
z for all z ¢ 0.
(Show that log cannot even be defined continuously for Iz| 1.)
(a) Show that
exp
is not one-one
show that e2 =
=
=
=
=
Since there is no way to define a continuous logarithm fimction we canlogarithm
not speak of thelogarithm of a complex number, but only of
with
meaning
of
e'
the
numbers
for w,"
tv. And
infmitely many
one
z
"a
=
562
and InfiniteSeries
InfiniteSequences
for complex numbers a and b we define ab to be a set of complex num€blop
bers, namely the set of all numberS
or, more precisely, the set of
all numbers ebt where z is a logarithm for a.
(d) If m is an integer, then am consists of only one number, the one given by
the usual elementary definition of am
(e) If m and n are integers, then the set a"/" coincides with the set of values
given by the usual elementary definition, namely the set of all bmwhere b
is an nth root of a.
If
(f) a and b are real and b is irrational, then ab contains infinitely many
members, even for a > 0.
(g) Find all logarithms of i, and find all values of i'.
(h) By (ab)cweab.mean the set of all numbers of the form ze for some number z
in the set
Show that (16)' has infinitely many values, while 1" has
only one.
12.
all values of ab-c are
(i) Show that
(a) For real x
show that we can choose
log(x + i)
log(x + i) and log(x
log(1 + x2) + i
=
1
1+x2
arctan x
-
log(x-i)=log(1+x2)--i
(It will help to note that x/2
(b) The expression
b-c
VaÏuesOf (ßbge
âlSO
(ab c
-
-
c by
i) to be
,
-arctanx
.
arctan x
-
1
2i
arctan
=
1
1
x--i
x+i
1/x for x ¢ 0.)
yields, formally,
Use part
13.
I
dx
-[log(x
=
1
-log(x
-i)
+i)].
1+x2
(a)to
check that this answer agrees with the usual one.
(a) A sequence (a,} of
complex
is called a Cauchy sequence
numbers
b, + ic,, where
if
b, and c, are
0. Suppose that a,
lim |a. a,|
real. Prove that {a,}is a Cauchy sequence if and only if {b,}and {c.}
are Cauchy sequences.
(b) Prove that every Cauchy sequence of complex numbers converges.
using theorems about real series, that an
(c) Give direct proofs, without
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.)
=
=
-
m,n-oo
14.
(a) Prove
that
e
k=l
2
=
e'2
1
_
x
sin
ei(n+1)x/2
sm
-
2
27. ComplexPowerSeries 563
the formulas for
(b) Deduce
and
coskx
sinkx
k=1
that are given in
k=1
Problem 15-33.
15.
Let (a,} be the Fibonacci sequence,
(a) If
(b) Show that r
(1 +
=
a2
1, a.
=
2
show that rn+1 = 1+ 1/r..
1 + 1/r.
lim r, exists, and r
H->OO
anyt/a.,
=
r,
al
=
=
=
a, + a,41.
Conclude that r
=
ß)/2.
anz"
that
(c) Show
of convergence
radius
has
2/(1 +
ß).
(Using
n=1
the unproved
in this chapter
theorems
and the fact that
anz"
=
n=1
--1/(Z2
+ z 1) from Problem 24-15 we could have predicted that the
radius of convergence is the smallest absolute value of the roots of z2 +
radius of convergence
z 1 0; since the roots are (-1 ± Ã)/2, the
should be (-1 +
Notice that this number is indeed equal to
-
=
-
d)/2.
Ã).)
2/(1+
16.
Since (et 1)/z can be written as the power series 1 + z/21 + z2/3! +
which is nonzero at 0, it follows that there is a power series
-
°°
er
1
-
n=o
-
-
-
b,
n!
Using the unproved theorems in this
chapter, we can even predict the radius of convergence; it is 2x, since this is
the smallest absolute value of the numbers z
2kzi for which er 1 = 0.
appearing
called
numbers
the
b,
here are
Bernoulli numbers.*
The
radius
with nonzero
of convergence.
=
(a) Clearly bo
-
1. Now show that
=
z
=--+
ex-1
«¯2+1
z
2
z e= + 1
e=-l'
e=+1
er--l'
e¯z-1
and deduce that
bt=-
(b) By finding the
,
b,=0
ifnisoddandn>1.
coefRcient of z" in the right side of the equation
z=
X=+i+¾+
*Sometimes the numbers B, = (-1)n-1b2 are called the Bernoulli numbers, because b, = 0 if n
is odd and > 1 (seepart (a))and because the numbers b2n alternate in sign, although we will not
of this nomenclature are also in use.
prove this. Other modifications
and InfiniteSerks
564 Infinite Sequences
show that
n-i
b,=0
.
forn>1.
i=0
This formula allows us to compute any bk in terms of previous ones, and
shows that each is rational. Calculate two or three of the following:
b2
*(c)
Part
=
b4
g,
(a)shows
=
b6
-y,
=
be
g,
=
--g.
that
-z/2
°°
n=o
b2n
z/2
2;I
=
(2n)!z
·
ez
et/2
1
-
e¯=/2°
-
Replace z by 2iz and show that
z cotz
(-1)"2 z
=
.
n=o
*(d)
Show that
tan z
*(e)
=
cot z
2 cot 2z.
-
Show that
tan z
(-1)"¯12 (2
=
-
1)z
.
n=1
(This series converges for |z| < x/2.)
17.
The Bernoulli numbers play an important role in a theorem which is best
introduced by some notational nonsense. Let us use D to denote the
entiation operator," so that Df denotes f'. Then Dk f wiÏÏ mean f(k) and
"differ-
e"
f
will mean
f(n)/n!(ofcourse
this series makes no sense in general,
but it will make sense if f is a polynomial function, for example). Finally,
operator"
let A denote the
for which Af (x) f(x + 1) f (x).
Now Taylor's Theorem implies, disregarding questions of convergence, that
"difference
=
f (x + 1)
-
=
no
or
(*) f(x+1)- f(x)=
n=1
we may write this symbolically as Af = (e* 1) f, where I stands for the
"identity
operator."
Even more symbolically this can be written A = e
1,
which suggests that
-
-
D-€D_)
D
'
27. ComplexPower Series 565
Thus we obviously ought to have
Dk
D=
k=0
i.e.,
(**) J'(x)
[f(k)(X
=
# Î)
f(k)(X)
-
.
k=0
The beautiful thing about all this nonsense is that it works!
if f is a polynomial function (inwhich case
the infinite sum is really a finite sum). Hint: By applying (*)to f(k),find
(k)(X # Î) f(k)
a formula for f
(x);then use the formula in Problem 16(b)
to find the coefEcient of fU)(x) in the right side of (**).
(a) Prove that (**)is literally true
-
(b) Deduce
from
(**)that
f'(0)+-·-+ f'(n)
[f(k)
=
(k)(0)].
_
k=0
(c) Show that
for any polynomial function g we have
n+1
oc
bk
+D-g(k-1)(0)].
-[g
g(0)+---+g(n)=
g(t)dt+
k=1
(d) Apply this
to g(x)
kP
=
xF
to show that
np+11
¯***.
=
k=1
Using the fact that bi
kP
=
-
,
1
k
k=1
n
show that
np+11
np-k+1
=
k
k=2
k=1
1
The first ten instances of this formula were written out in Problem 2-7,
which offered as a challenge the discovery of the general pattern. This
may now seem to be a preposterous suggestion, but the Bernoulli numbers were actually discovered in precisely this way! After writing out
these 10 formulas, Bernoulli claims (inhis posthumously printed work
Ars Comectandi,
1713): "Whoeverwill examine the series as to their regularity may be able to continue the table." He then writes down the above
formula, offering no proof at all, merely noting that the coefficients by
(whichhe denoted simply by A, B, C, ) satisfy the equation in Problem 16(b). The relation between these numbers and the coefEcients in
the power series for z/(e2
1) was discovered by Euler.
...
-
and InfiniteSenes
566 InfiniteSequences
*18.
The formula in Problem 17(c)can be generalized to the case where g is not a
polynomial function; the infinite sum must be replaced by a finite sum plus a
remainder term. In order to find an expression for the remainder, it is useful
to introduce some new functions.
(a) The
Bernoullipolynomials
y, are defined by
p,(x)
n
=
bn-kXk
k=0
The first three are
1
91(x)=x-p2(x)
x2
=
3x
2
x
2
Show that
p.(0)
y,
=
(1)
y,'(x)
p,(x)
b.,
b.
if n
ny,_i(x),
=
=
(--1)"p,(1
=
>
1,
x)
-
for n
>
1.
Hint: Prove the last equation by induction on n, starting with n 2.
RNk
(x) be the remainder term in Taylor's Theorem for f (k), on the
(b) Let
interval [x,x + 1], so that
=
N
f(k)(X
(*)
# Î)
-
(k+n)(X
f(k)(x)
=
)+
RNk(X).
n=1
Prove that
f'(x)
=
k=0
[f f*)
+ 1
-
f (*)(x)]
RN--kk(X).
-
k=0
Hint: Imitate Problem 17(a). Notice the subscript N
(c) Use the integral form of the remainder to show that
RN-kk(X)
(d) Deduce
fn(x ‡ 1
=
-
t)
(N+1)(t)
--
k on R.
dt.
the "Euler-Maclaurin Summation Formula":
g(x)+g(x+1)+---+g(x+n)
N
be
g
lx+n+1
-[gf*¯>(x
=
g(t)dt
x
+
k=l
°
+n+1)
-
g(k-1)(x)] + S,(x, n),
27. Complex
PowerSaies 567
where
x+j+1
SN(X,
N)
=
ŸN(X
Î
-
f)
(N)(t)
-
dt.
*Ì
j=0
be the periodic function, with period 1, which satisfies ‡.(t)
p,(t) for 0
st < 1. (Part (a)implies that if n > 1, then ‡. is continuous,
since y,(1)
p,(0), and also that ‡. is even if n is even and odd if n is
odd.) Show that
(e) Let ‡,
=
=
ÈN (X
SN(X,
8)
=
f ) g(N)(t)
Lx+n+1
-
-
(-1)N+1
(N)
dt
if x is an integer
(t) dt
.
Unlike the remainder in Taylor's Theorem, the remainder S.(x, n) usually does
0, because the Bernoulli numbers and functions
not satisfy lim SN(X, n)
=
N->œ
the first few examples do not suggest this).
become large very rapidly (although
Nevertheless, important information can often be obtained from the summation
formula. The general situation is best discussed within the context of a specialized
series"), but the next problem shows one particularly important
study ("asymptotic
example.
**19.
(a) Use the
Euler-Maclaurin Formula, with N
=
2, to show that
logl+-··+log(n-1)
=
(b) Show that
log
Ï"
log t dt
1
-
-
n!
(
log
(d) Problem
"
¯
an"+1/2e¯.41¡12.
Use part
(c)to show
Ý2(Í)dt.
2t2
(n!)22
.
(2n)!S
that
-2.2*
2n2n+1
lim
=
no
and conclude
that a
=
R.
2t,
2t2
19-40(d) shows that
lim
[" ‡2(t) dt.
‡2(t) dt
JI
¯
=
1 +
‡2(t)/2t2dt exists, and
=
n!
(
--
1
i
exp(ß + 11/12), then
=
1
-
+
Ï2
nn+1/2g-n+1/l2n
-
"
11
¯
(c) Explain why the improper integral ß
that if a
1
log n +
a(2n)2n+1/2g-2n
'
show
568
and InfinitzSeries
InfiniteSequences
(e) Show that
1
Ï1/2
92(t) dt
p2(t) dt
=
=
0.
O
(You can do the computations explicitly, but the result also follows imfrom Problem 18(a).)Now what can be said about the graphs
of #(x)
‡2(t)dt and #(x) fe'#(t)dt? Use this information and
integration by parts to show that
mediately
=
£
=
©
Ï
‡2(t) dt
(f) Show that
the maximum
that
value of
©
2t2
,
(g) Finally, conclude
and
|p2(x)|for x in [0, 1] is ¼,
‡2(t)dt
Ï
0.
>
2t2
n
<
--.
conclude
1
12n
that
nn+1/2e¯"
n!
<
n.41¡2e¯"**/2".
<
The final result of Problem 19, a strong form of Stirling's Formula, shows that
nn+1/2,-n, in the
n! is approximately
sense that this expression differs from n!
by an amount which is small compared to n when n is large. For example, for
10 we obtain 3598696 instead of 3628800, with an error < 1%.
n
nature of
A more general form of Stirling's Formula illustrates the
the summation formula. The same argument which was used in Problem 19 can
now be used to show that for N > 2 we have
=
"asymptotic"
-b1)nk-1
N
log
¯
n
Since
‡N
1/2e¯¤
k(k
dt.
is bounded, we can obtain estimates of the form
‡N(t)
l°°
NtN
MN
dt
_<
gN-1
n
If N is large, the constant MN Will also be large; but for very large n the factor
will make the product very small. Thus, the expression
nl-N
-bl
E
nn+1/2e¯n
-
exp
k (k
)nk-1
large
may be a very bad approximation for n! when n is small, but for large n (how
good depends on N).
depends on N) it will be an extremely good one (how
PART
EPILOGUE
There was a most ingenious Architect
who had contrived a new Method
forbuildin.gHouses,
by beginningat the Roof, and working
downw«rdsto the Foundation.
JONATHAN
SWWF
FIELDS
CHAPTER
Throughout this book a conscientious attempt has been made to define all imfor which an intuitive definition is
portant concepts, even terms like
and R, the two main protagonists of this story,
often considered sufficient. But
Q
have only been named, never defined. What has never been defined can never be
analyzed thoroughly, and
P1-Pl3 must be considered assumptions,
numbers.
theorems,
about
Nevertheless, the term
has been purposely
not
avoided, and in this chapter the logical status of Pl-P13 will be scrutinized more
"function,"
"properties"
"axiom"
carefully.
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
is useless).
names some elements of N without identifying them (andthe
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 contains references
to this problem, as well as to the other steps that are required if one wishes to
develop calculus from its basic logical starting point. The further development
of this
program would proceed with the definition of Z, in terms of N, and the
definition of Q in terms of Z. This program results in a certain well-defined
and properties P1-P12 as
set Q, certain explicitly defined operations
+ and
of R, in terms of
theorems. The final step in this program is the construction
Q.
which
construction
It is this last
concerns us. Assuming that Q has been defined,
all
and that P1-P12 have been proved for Q, we shall ultimately d¢ne R and prove
of P1-P13 for R.
Our intention of proving Pl-Pl3 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
which includes
a meaningful mathematical
way, by using the concept of a
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 (ofobjects of any sort whatoperations"
soever), together with two
+ and defined on F (thatis, two
rules which associate to elements a and b in F, other elements a+b and a.b in F)
for which the following conditions are satisfied:
Like
Q and
"etc."
-,
"field,"
"binary
·
b) + c a + (b + c) for all a, b, and c in F.
(2)There is some element 0 in F such that
(1)(a +
=
for all a, in F,
+ 0 a
(ii)for every a in F, there is some element b in F such that a + b
(i)a
571
=
=
0.
572 Epilqgue
forallaandbinF.
(3)a+b=b+a
for alla, b, and c in F.
element I in F such that 1 ¢ 0 and
(i)a 1 a for all a in F,
(ii)For every a in F with a ¢ 0, there is some element b in F such that
a.(b•c)
(4)(a•b).c=
(5)There is some
=
-
a.b=1.
for all a and b in F.
a b + a c for all a, b, and c in F.
(6)a b b a
(7)a (b + c)
=
-
-
•
=
•
•
The familiar examples of fields are, as already indicated, Q, R, and C, with
It is probably unnecessary to
+ and being the familiar operations of + and
explain why these are fields, but the explanation is, at any rate, quite brief. When
-
-
.
the rules (1),(3),(4),(6),(7)
+ and are understood to mean the ordinary + and
are simply restatements of Pl, P4, P5, P8, P9; the elements which play the role of 0
and 1 are the numbers
O and 1 (which
accounts for the choice of the symbols 0, 1);
number
and the
b in (2)or (5)is
or a¯', respectively. (For this reason, in an
arbitrary field F we denote by
the element such that a + (-a)
0, and by
a-I the element such that
a-1
for
0.)
1,
a
a
·
-
,
-a
-a
=
¢
=
·
In addition to Q, R, and C, there are several other fields which can be described
easily. One example is the collection F1 of all numbers a + bá for a, b in Q.
The operations + and will, once again, be the usual + and for real numbers.
It is necessary to point out that these operations really do produce new elements
•
-
of Fl:
(a+bÃ) + (c +dÃ) (a+c) +(b +d)Ã, which is in F1;
(a+bÃ) (c+dd)=(ac+2bd)+(bc+ad)Ã, whichisinF1=
field are obvious for F1: since these hold for
all real numbers, they certainly hold for all real numbers of the form a + bÃ.
a+bn
in Fi and, for a
Condition (2)holds because the number 0 0+0ñis
in F1 the number ß = (-a) + (-b)Ä in F1 satisfies a + ß
0. Similarly,
On
1 1+
is in F1, so (5i)is satisfied. The verification of (Sii)is the only slightly
difficult point. If a + bd ¢ 0, then
Conditions
(1),(3),(4),(6),(7)for a
=
=
=
=
a+b÷
it is therefore necessary
1
to show that 1/(a +
a
a+bn
1
-
=1;
a+bá
bÃ
(a-bÃ)(a+bÃ)
bÃ)
is in Fi. This is true because
(-b)
a
a2-2b2
+ a2-2b2
Ë
°
(The division by a bà is valid because the relation a bà 0 could be true
only if a
b
0 (sinceà is irrational) which is ruled out by the hypothesis
a + bà ¢ 0.)
The next example of a field, F2, is considerably simpler in one respect: it contains only two elements, which we might as well denote by 0 and 1. The operations
-
=
=
-
=
28. Relds 573
+ and
-
are
described by the following tables.
000
110
101
1
0
•
001
The verification of conditions
straightforward,
case-by-case checks. For
proved by checking the 8 equations obtained by
0 or 1. Notice that in this field I + 1 0; this equation may also
example, condition
setting a,
be written
1
0
+
(lp(7)are
(1)may be
b, c
1
Our final example of a field is rather silly: F3 consists of all pairs (a, a) for a
in R, and + and are defined by
=
=
-1.
=
-
(a,a)+(b,b)=(a+b,a+b),
(a,a)-(b,b)=(a-b,a-b).
(The + and appearing on the right side are ordinary addition and multiplication
for R.) The verification that F3 is a field is left to you as a simple exercise.
A detailed investigation of the properties of fields is a study in itself but for our
purposes, fields provide an ideal framework in which to discuss the properties of
numbers in the most economical way. For example, the consequences of Pl-P9
which were derived for
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 fields.
For example, it is possible for the equation 1 + 1 0 to hold in some fields, and
consequently a b b a does not necessarily imply that a
b. For the field C
the assertion 1+1 ¢ 0 was derived from the explicit description of C; for the fields
Q and R, however, this assertion was derived from further properties which do
not have analogues in the conditions for a field. There is a related concept which
does use these properties. An ordered field is a field F (withoperations + and
together with a certain subset P of F (the
elements) with the following
properties:
-
"numbers"
=
=
-
=
-
.)
"positive"
(8)For all a
(i)a
(ii)a
(iii)
-a
(9)If a
(10)If a
in F, one and only one of the following is true:
0,
is in P,
is in P.
=
and b are in P, then a + b is in P.
and b are in P, then a b is in P.
·
We have already seen that the field C cannot be made into an ordered field.
The field F2, with only two elements, likewise cannot be made into an ordered
shows that 1 must be in P; then
field: in fact, condition (8),applied to 1
(9)
implies that 1 + 1 0 is in P, contradicting
(8).On the other hand, the field Fi,
-1,
=
=
574
Epilogue
consisting
of all numbers
an ordered
bá
with a,
the set of all a +
a +
certainly can be made into
which are positive real numbers
F3 can also be made into an ordered field; the
field: let P be
the
ordinary
sense). The field
(in
description of P is left to you.
It is natural to introduce notation
b in
bÃ
Q,
for an arbitrary
sponds to that used for
Q and R: we define
a>b
a<b
asb
a>b
if
if
if
if
ordered field which corre-
a-bisinP,
b>a,
a<bora=b,
a>bora=b.
Using these definitions we can reproduce, for an arbitrary ordered field F, the
definitions of Chapter 7:
A set A of elements of F is bounded above if there is some x in F such
that xt a for all a in A. Any such x is called an upper bound for A. An
element x of F is a least upper bound for A if x is an upper bound for A
and x 5 y for every y in F which is an upper bound for A.
Finally, it is possible to state an analogue
last abstraction of this chapter:
of
property P13 for R; this leads to the
ordered
field is an ordered
A complete
which is bounded above has a least upper
field in which every nonempty
set
bound.
of fields may seem to have taken us far from the goal of
constructing the real numbers. However, we are now provided with an intelligible
means of formulating this goal. There are two questions which will be answered
in the remaining two chapters:
The consideration
1. Is there a complete ordered field?
2. Is there only one complete ordered field?
Our starting point for these considerations will be Q, assumed to be an ordered field, containing N and Z as certain subsets. At one crucial point it will be
necessary to assume another fact about Q:
Let x be an element of
in N such that nx > y.
This assumption,
Q with
x
>
0. Then for any y in
which asserts that the rational
of the real numbers,
does not follow
numbers
Q there
is some n
have the Archimedian
other
the
properties of an
from
property
conclusively
ordered field
see reference [17]
(forthe example that demonstrates this
of the Suggested Reading). The important point for us is that when
Q is explicitly
constructed, properties P1-P12 appear as theorems, and so does this additional
28. Relds 575
assumption; if we really began from the beginning, no assumptions
be necessary.
about
Q would
PROBLEMS
Let F be the set {0,1, 2} and define operations + and on F by the following
tables. (The rule for constructing these tables is as follows: add or multiply
in the usual way, and then subtract the highest possible multiple of 3; thus
2-2=4=3+1,so2-2=l.)
1.
•
+
0
1
2
•
0
1
2
0
0
1
2
0
0
0
0
1
1
2
0
1
0
1
2
2
2
0
1
2
0
2
1
be made into an ordered
Show that F is a field, and prove that it cannot
field.
Suppose now that we try to construct a field F having elements 0, 1,
2, 3 with operations + and defined as in the previous example, by adding
or multiplying in the usual way, and then subtracting the highest possible
multiple of 4. Show that F will not be a field.
2.
·
Let F
tables.
3.
=
{0,1, a, ß} and define
operations
001aß
by the following
on F
00000
1
0
ß
1
a
auß01
p
-
·01aß
+01aß
1
+ and
0
1
p
a
a0aßl
ß0ßla
ßa10
Show that F is a field.
4.
(a) Let
F be a field in which 1 + 1
can also
(b) Suppose
be
written
that a + a
b+ b
consequently
-a).
a
=
=
=
0. Show that a + a
=
0 for all a
(this
=
0 for some a ¢ 0. Show that I + 1
0 for all b).
=
0
(and
576 Epilogue
5.
(a) Show that in any
field we have
(1+--
+1)·(1+---+1)
m times
1+--.+1
=
mn times
a times
for all natural numbers m and n.
(b) Suppose that in the field F we have
1+---+1=0
m times
for some natural number
must be a prime number
of F).
6.
Show that the smallest m with this property
(thisprime number is called the characteristic
m.
Let F be any field with only finitely many elements.
(a) Show that
there must be distinct natural
m and n with
-+1.
1+---+1=1+m times
(b) Conclude
numbers
n times
that there is some natural
number
k with
1+---+1=0.
k times
7.
Let a, b, c, and d be elements of a field F with a d
the equations
-
a.x+b•y=œ,
c•x +d•y
can
8.
=
b c
•
¢ 0. Show that
ß,
be solved for x and y.
Let a be an element of a field F. A
with b2
b•b
a.
=
"square
root" of a is an element b of F
=
(a) How many square roots does 0 have?
(b) Suppose a ¢ 0. Show that if a has a square
mots, unless 1 +
9.
-
1
0, in
=
which case a
x2 +
it has two square
b x + c = 0, where b and c are elements of
b2
that
F.
field
4•c has a square root r in F. Show that
Suppose
a
r)/2
solution
of
this equation.
is a
(-b +
(b) In the field F2 of the text, both elements clearly have a squam mot.
On the other hand, it is easy to check that neither element satisfies the
equation x2 + x + 1 0. Thus some detail in part (a)must be incorrect.
What is it?
(a) Consider
an equation
has
root, then
only one.
.
-
=
10.
Let 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
28. Fields 577
-1;
this special case should
through in a special case, namely, F
R and a
guide you through this example.)
Let F' consist of all pairs (x, y) with x and y in F. If the operations on F
and
are + and , define operations
on F' as follows:
=
=
e
·
e
(x,y)O(z,w)=(x+z,y*w),
(x, y) O (z, w)
(a) Prove that
(x•z +a·y•w,y-z+x·w).
=
F', with the operations
(b) Prove that
(x, 0) e (y, 0)
(x, 0) e (y, 0)
and
e
is a field.
y, 0),
y, 0),
=
(x +
=
(x
.
e,
so that we may agree to abbreviate (x, 0) by x.
(c) Find a squam root of a (a, 0) in F'.
=
11.
Let F be the set of all four-tuples (w, x, y, z) of real numbers.
by
Define + and
-
(s,t,u,v)+(w,x,y,z)=(s+w,t+x,u+y,v+z),
(s,t,u,v)•(w,x,y,z)=(sw-tx-uy-vz,sx+tw+uz-vy,
sy+uw+vx-tz,sz+vw+ty-ux).
F satisfies all conditions for a field, except (6).At times the
become quite ornate, but the existence of multiplicative inverses is the only point requiring any thought.
(b) It is customary to denote
(a) Show that
algebra will
(0, 1, 0, 0) by i,
(0, 0, 1, 0) by j,
(0,0, 0,
1) by k.
Find all 9 products of pairs i, j, and k. The results will show in particular
that condition (6)is definitely false. This
field" F is known as the
quaternions.
"skew
CONSTRUCTION
REAL NUMBERS
CHAPTER
OF THE
The mass 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
for an ordered
have to explicitly describe one in detail; verifying conditions (1)--(10)
field will be a straightforward ordeal, but the description of the field itself of the
elements in it, is ingenious indeed.
At our disposal is the set of rational numbers, and from this raw material it is
necessary to produce the field which will ultimately be called the real 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 have had enough
experience to realize that the situation may not be quite so hopeless as that casual
consideration
suggests. The strategy to be adopted in our construction has already
been used effectively for defining functions and complex numbers.
Instead of
nature" of these concepts, we settled for a definition
trying to determine the
that described enough about them to determine their mathematical
properties
"real
completely.
A similar proposal for defining real numbers requires a description of real numbers in terms of rational numbers. The observation, that a real number ought to
be determined completely by the set of rational numbers less than it, suggests a
strikingly simple and quite attractive possibility: a real number might (andin 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
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
real number a, then A ought to have the following
property: If x is in A and y
is a rational number satisfying y < x, then y is in A. In addition to this property
the set A should have a few others. Since there should be some rational number
x < a, the set A 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
"the
contain
a greatest member.
If we temporarily regard 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, however, that we
have not failed to notice any crucial property of the set A. There appears to be
no reason for hesitating any longer.
578
29. Constructionof theReal Numbers 579
DEFINITION
A real nurnber
is a set a, of rational numbers,
with the
following four proper-
ties:
is in a and y is a rational number with y
(2)a ¢ 0.
(1)If x
(3)a ¢ Q.
(4)There is no
<
x, then y is also in a.
greatest element in a; in other words, if x is in a, then there
> x.
is some y in a with y
The set of all real numbers
is denoted by R.
Justto remind you of the philosophy behind our definition, here is an explicit
example of a real number:
a=(xinQ:x<0orx2<2}.
It should be clear that a is the real number which will eventually be known as Å,
but it is not an entirely trivial exercise to show that a actually is a real 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 essential in order to avoid
ambiguity; without it both
(xinQ:x<l}
and
{xinQ:x<1}
"real
number 1."
be candidates for the
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 rational
numbers.
This means, in particular, that a rational number (a member of Q)
is not a real number; instead every rational number x has a natural counterpart
which is a real number, namely, {yin
Q : y < x}. After completing the construcwould
tion of the real numbers, we can mentally throw away the elements of Q and
agree that Q will henceforth denote these special sets. For the moment, however,
it will be necessary to work at the same time with rational numbers, real numbers
(setsof rational numbers) and even sets of real numbers (setsof sets of rational
numbers).
Some confusion is perhaps inevitable, but proper notation should keep
this to a minimum. Rational numbers will be denoted by lower case Roman letters
(x, y, z, a, b, c) and real numbers by lower case Greek letters (a, ß, y); capital
Roman letters (A, B, C) will be used to denote sets of real numbers.
The remainder of this chapter is devoted to the definition of +,
and P for R,
and a proof that with these structures R is indeed a complete ordered field.
We shall actually begin with the definition of P, and even here we shall work
and 0 are available, we shall
backwards. We first define a < ß; later, when +,
define P as the set of all a with 0 < a, and prove the necessary properties for P.
•,
-,
580
Epikgue
The
for beginning with the definition of
reason
<
is the simplicity of this concept
in our present setup:
Definition.If a and ß are real numbers, then « < ß means that a is contained in ß
(thatis, every element of a is also an element of ß), but a ¢ ß.
defmitions of 5, >, > would be stultifying, but it is interesting
to note that 5 can now be expressed more simply than <; if a and ß are real
numbers, then œ $ ß if and only if a is contained in ß.
If A is a bounded collection of real numbers, it is almost obvious that A 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 ß, then ß is presumably sup A. In
the proof of the following theorem we check all the little details which have not
been mentioned, not least of which is the assertion that ß is a real number. (We
will not bother numbering
theorems in this chapter, since they all add up to one
complete
ordered field.)
big Theorem: There is a
A repetition
TEOREM
PROOF
of the
If A is a set of real numbers
least upper bound.
Let ß
=
numbers;
and A
¢ 0
and A
is bounded above, then A has a
x is in some a in A}. Then ß is certainly a collection of rational
the proof that ß is a real number requires checking four facts.
(x :
is in ß and y < x. The first condition means that x is in a
for some a in A. Since a is a real number, the assumption y < x implies
that y is in œ. Therefore it is certainly true that y is in ß.
(2)Since A ¢ 0, there is some a in A. Since a is a real number, there is some
x in a. This means that x is in ß, so ß ¢ 0.
(3)Since A is bounded above, there is some real number y such that « < y for
every a in A. Since y is a real number, there is some rational number x
which is not in y. Now a < y means that a is contained in y, so it is also
true that x is not in a for any a in A. This means that x is not in ß; so
(1)Suppose that
x
that x is in ß. 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 ß; thus ß does not have a greatest member.
(4)Suppose
These four observations prove that ß is a real number. The proof that ß is the
least upper bound of A is easier. If a is in A, then clearly œ is contained in ß; this
that œ $ ß, so ß is an upper bound for A. On the other hand, if y is an
means
upper bound for A, then a 5 y for every a in A; this means that a is contained
in y, for every a in A, and this surely implies that ß is contained in y. This, in
turn, means that ß $ y; thus ß is the least upper bound of A.
The definition of + is both obvious and easy, but is must be complemented with
defmition makes any sense at all.
a proof that this
"obvious"
of theReal Numbers 581
29. Construction
real
Dejînition.If a and ß are
a +
THEOREM
PROOF
ß
{x: x
=
=
If a and ß are real numbers,
then
numbers,
y + z for some y in a and some z in
then
«
+
ß is a
ß}.
real number.
Once again four facts must be verified.
(1)Suppose
for some x in a + ß. Then x y + z for some y in a and
some z in ß, which means that w < y + z, and consequently,
w y < z.
real
number).
and
is
in
This shows that w y is in ß (since
is
Since
ß,
ß a
z
w
y + (w y), it follows that w is in « + ß.
clear
that a + ß ¢ 0, since « ¢ Ø and ß ¢ 0.
It
is
(2)
Since
a ¢ Q and ß ¢ Q, there are rational numbers a and b with a
(3)
not in a and b not in ß. Any x in a satisfies x < a (forif a < x, then
condition (1)for a real number would imply that a is in a); similarly any y
in ß satisfies y < b. Thus x + y < a + b for any x in œ and y in ß. This
shows that a + b is not in a + ß, so a + ß ¢
Q.
(4)If x is in « + ß, then x y + z for y in a and z in p. There are y' in a
and z' in ß with y < y' and
z < z'; then x < y' + z' and y' + z' is in a + ß.
member.
Thus a + p has no greatest
w
<
x
=
-
-
=
-
=
By now you can see how tiresome this whole procedure is going to be. Every time
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 concentration.
There
that it will be less boring if you check the four
is really no help for this (except
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.
THEOREM
PROOF
then
If a, ß, and y are real numbers,
(a + ß) + y
=
a +
(ß +
y).
Since (x + y) + z x + (y + z) for all rational numbers x, y, and z, every member
(œ+ ß) + y is also a member of a + (ß + y), and vice versa.
=
of
THEOREM
PROOF
If a and ß are real numbers, then a +
Left to you
ß
=
ß+
a.
(eveneasier).
To prove the other properties of + we first define 0.
Defmilion.0
=
{xin Q : x
<
0}
.
It is, thank goodness, obvious that 0 is a real number, and the following theorem
is also simple.
582 Epilogue
THEOREM
PROOF
If a is a real number,
then a + 0
=
a.
If I is in Πand y is in 0, then y < 0, so x + y < x. This implies that x + y is in a.
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
Sincex=y+(x-y),whereyisina,andx-y<0(sothatx-yis
y>x.
in 0), this shows that x is in « + 0. Thus every member of a is also a member
of a + 0.
The reasonable
candidate for
would seem to
-a
{xin Q :
be the set
is not in a}
-x
-a).
> a, so that x <
But in certain
in œ means, intuitively, that
cases this set will not even be a real number. Although a real number a does not
have a greatest member, the set
(since
--x
-x
not
Q
a
--
=
{xin Q : x is not in a)
xo; when a is a real number of this kind, the set
is not in
have a greatest element
It is therefore necessary to
{x :
which comes equipped
introduce a slight modification into the definition of
with a theorem.
may have a least element
a} will
-x
-xo.
-a,
D¢înition.If a is a real number, then
-a
THEOREM
PROOF
=
{xin Q :
-x
If a is a real number,
(1)Suppose that
is not in a, but
then
is not the least element of
and y
y is in
element
a}.
-
is not in a,
clear
that
is not in a. Moreover, it is
is not the
of
since
smaller
shows
element. This
that
is a
Q a,
-a
<
x. Then
-y
Since
-x.
>
-x
-y
smallest
Q
is a real number.
-a
x is in
also
it is
true that
x
-
-y
-x
-
-a.
is some rational
(2)Since a ¢ Q, there
number y which
number in
rational
that y is not the smallest
always be replaced by any y' > y). Then
(3)Since « ¢ 0, there is some x in a. Then
assume
-y
-x
is in
-a.
cannot
is not in a. We can
Q
-
Thus
(sincey can
a
¢ 0.
-a
possibly be in
-a,
so
then
is not in a, and there is a rational number y <
is in
which is also not in a. Let be a rational number with y <
Then
z
z<
clearly
smallest
œ.
also
and
is
So
the
element
of
is
in
not
not
a,
z
Q
z
> x, this shows that
is in
Since
does not have a greatest
(4)If x
-a,
-x
-x
-x.
-
-z
-a.
-a
--z
element.
The proof that a + (-a)
are not
=
caused, as you might
The difficulties
0 is not entirely straightforward.
the
finicky details in the definition
presume, by
29. Constructionof theReal Numbers 583
Rather, at this point we require the Archimedian property of Q stated on
page 574, which does not follow from Pl-P12. This property is needed to prove
the following lemma, which plays a crucial role in the next theorem.
of
LEMMA
-œ.
Let a be a real number, and z a positive rational number. Then there are (Figure 1)
rational numbers x in a, and y not in a, such that y
x
z. Moreover, we may
smallest
of
the
that
element
is
y not
assume
Q a.
-
=
-
PROOF
Suppose first that z is in a. If the numbers
z, 2z, 3z,
...
were all in a, then everyrational number would be in a, since every rational number w satisfies w < nz for some n, by the additional assumption on page 574. This
contradicts the fact that a is a real number, so there is some k such that x = kz is
in a and y = (k+ 1)z is not in a. Clearly y x = z.
Moreover, if y happens to be the smallest element of
a, let x' > x be an
element of a, and replace x by x', and y by y + (x' x).
--
Q
-
-
If z is not in a, there is a similar proof, based on the fact that the numbers
cannot all fail to be in a.
FIGURE
THEOREM
If
Œ
1
number,
is a real
then
a +
PROOF
(-n):
(-a)
0.
=
> x.
Then
is not in a, so
Hence
Suppose x is in a and y is in
member
of
a + (-a) is in 0.
x + y < 0, so x + y is in 0. Thus every
other
> 0.
the
direction.
If z is in 0, then
in
is
little
difficult
It a
to go
more
with
According to the lemma, there is some x in a, and some y not in a,
y not the
such
smallest element of
that y x
This equation can be written
Q a,
and
this
is
in
x + (-y)
proves that z is in «+ (-a).
z. Since x is in a,
-y
-y
-a.
-z
-z.
=
-
-
-y
-a,
=
Before proceeding with multiplication,
prove a basic property:
Definition.P
=
{a in R
:
a >
we
define the
0}
.
Notice that a + ß is clearly in P if a and ß are.
"positive
elements" and
584
Epilogue
THEOREM
If a is a
(i)a
(ii)a
(iii)
-a
PROOF
real
then one and only one of the following conditions holds:
number,
0,
is in P,
is in P.
=
any positive rational number, then a certainly contains all negative
rational numbers,
so a contains 0 and a ¢ 0, i.e., a is in P. If a contains no
positive rational numbers, then one of two possibilities must hold:
If a contains
contains all negative rational numbers; then «
0.
which
negative
rational
there
number
not in «; it can be asis
is
some
x
(2)
sumed that x is not the least element of
Q a (sincex could be replaced
x);
then
contains the positive rational number
by x/2 >
so, as we
(1)a
=
-
--x,
-a
have just proved,
is in P.
-a
0, it is clearly impossible
This shows that at least one of (i)-(iii)
must hold. If a
> 0
for condition (ii)or (iii)to hold. Moreover, it is impossible that a > 0 and
both hold, since this would imply that 0 a + (-a) > 0.
=
-a
=
Recall that « > ß was defined to mean that a contains ß, but is unequal to ß.
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 «
ß > 0 is equivalent to a > ß. This is clearly a consequence of
-
the next theorem.
THEOREM
PROOF
If a, ß, and y are real numbers
and
œ >
ß, then
+ y
«
>
ß+
y.
The hypothesis a > ß implies that ß is contained in a; it followsimmediately from
the definition of + that µ+y is contained in a+y. This shows that a+y >ß+y.
We can easily rule out the possibility of equality, for if
ß+y,
a+y=
then
a
which is
false. Thus
=
«
( +Y) + (-Y)
+ y
>
߇
=
(ß +y) +(-Y)
Definition.If a and ß are real numbers
THEOREM
•ß
=
{z: z 5 0 or z
If a and ß are real numbers
=
0,
y.
Multiplication presents difficulties of its own.
defmed as follows.
a
=
and
œ,
If a, ß
>
•
ß
can
ß > 0, then
x y for some x in a and y in
·
with a,
0, then a
ß > 0, then
a
·
ß is a
ß with
x, y
real number.
>
0}.
be
of theReal.Numbers585
29. Construction
PROOF
As usual, we must check four conditions.
z, where z is in « ß. If wy 0, then w is automatically
in a•ß. Suppose that w > 0. Then z > 0, so z x y for some positive x
in « and positive y in ß. Now
(1)Suppose
w
<
•
=
wz
w------
wxy
z
w
--x
z
·
-y.
z
1, so (w/z) x is in a. Thus w is in a ß.
Since O < w < z, we have w/z
Clearly
Ø.
a ߢ
(2)
(3)If x is not in a, and y is not in ß, then x > x' for all x' in a, and y > y'
for all y' in ß. Hence xy > x'y' for all such positive x' and y'. So xy is not
in a ß; thus a ß ¢ Q.
(4)Suppose w is in a•ß, and w 5 0. There is some x in a with x > 0 and
and
xy is in «•ß
some y in ß with y > 0. Then z
z > w. Now suppose
and
positive
>
0.
Then
for
in
some positive y in ß.
x
some
a
w
xy
w
x'
x'y,
>
then
contains
Moreover, a
some
w, and z is
x; if z
z > xy
Thus a ß does not have a greatest element.
in a
<
.
-
•
•
·
=
=
=
=
•ß.
.
Notice that a•ß
of all properties of
is clearly in P if a and ß are. This completes the verification
P. To complete the definition of we first define ja|.
·
then
Defmition.If a is a real number,
II
-a,
Dginition. If a and ß are real numbers,
then
ifa=0orß=0
ifa>0,ß>0ora<0,ß<0
ifa>0,ß<0ora<0,ß>0.
0,
a.ß=
if a 0
if a 5 0.
œ,
=
ja|•|ßl,
-(|al•|ßl),
As one might suspect, the proofs of the properties of multiplication usually involve
THEOREM
PROOF
THEOREM
reduction
to the case of positive numbers.
If a, ß, and y are real numbers,
This is clear if a, ß, y
>
then a
•
(ß
•
y)
=
(a ß)
.
.
y.
0. The proof for the general case requires
separate
cases (andis simplified
If a and
ß are
real numbers,
considering
slightly if one uses the following theorem).
then a.ß
=
ß•a.
586 Epilogue
PROOF
This is clear if a, ß
0, and the other cases are easily checked.
>
D¢inition. 1 = {xin Q : x < 1}
(It is clear that 1 is a real number.)
.
THEOREM
PROOF
If a is a real number, then a
·
1
=
a.
It is easy to see that every member of «•1 is also a member of a. On the
other hand, suppose x is in a. If xs 0, then x is automatically
in «-1. If x > 0,
then there is some rational number y in a such that x < y. Then x y (x/y),
and x/y is in 1, so x is in a 1. This
proves that « 1 a if a > 0.
If a < 0, then, applying the result just proved, we have
Let «>0.
=
-(|al•|1|)
«•1
=
-
-
-
-(la!)
=
=
=
Finally, the theorem is obvious when
«
=
a.
0.
Dginition.If a is a real number and a > 0, then
{xin Q : xs 0, or x > 0 and 1/x is not in a, but 1/x is not the smallest
a-1
=
member
if a
THEOREM
PROOF
<
of
Q
-
a};
-1
0, then a-1
_
If a is a real number
_
unequal
Clearly it suffices to consider
to 0, then aonly
œ >
is a real number.
0. Four conditions must be checked.
is in «-1. If y 5 0, then y is in «-1. If y > 0, then
x > 0, so 1/x is not in a. Since 1/y > 1/x, it follows that 1/y is not in a,
and 1/y is clearly not the smallest element of
Q a, so y is in a-1
(2)Clearly a-1
(3)Since a >0, there is some positive rational number x in a. Then 1/x is not
in a-1, so a-1
(4)Suppose x is in a-1. If xs 0, there is clearly some y in a-1 with y > x
because a-1 contains some positive rationals. If x > 0, then 1/x is not in a.
Since 1/x is not the smallest member of Q a, there is a rational number
y not in a, with y < 1/x. Choose a rational number z with y < z < 1/x.
Then 1/z is in a, and 1/z > x. Thus œ-1 does not contain a largest
(1)Suppose y
<
x, and x
-
-
member.
In order to prove that a-1 is really the multiplicative inverse of a, it helps to
have another lemma, which is the multiplicative analogue of our first lemma.
LEMMA
with a > 0, and z a rational number with z > 1. Then
there are rational numbers x in a, and y not in a, such that y/x
z. Moreover,
that
the
element
of
is
least
we can assume
y not
Q a.
Let a be a real number
=
-
of theReal Numbers 587
29. Construction
PROOF
Suppose first that z is in a. Since z
z" (1 + (z
=
1 > 0 and
-
1))"
-
>
1 + n(z
1),
-
it follows that the numbers
Z,
Z2
3
z* is in a, and y zk+1iS HOt
cannot all be in a. So there is some k such that x
in a. Clearly y/x
z. Moreover, if y happens to be the least element of Q a,
let x' > x be an element of a, and replace x by x' and y by yx'/x.
If z is not in a, there is a similar proof, based on the fact that the numbers 1/zk
cannot all fail to be in a.
=
=
=
THEOREM
PROOF
If a is a real number
-
and a
¢ 0, then
·
a
a-1
y
suffices to consider only a > 0, in which case a-1 > 0. Suppose that
x is a positive rational number in a, and y is a positive rational number in a-1
Then 1/y is not in a, so 1/y > x; consequently xy < 1, which means that xy is
in 1. Since all rational numbers xs 0 are also in 1, this shows that every member
of a
is in 1.
It obviously
•«-1
To prove the converse assertion, let z be in 1. If zy 0, then clearly z is in
a-1. Suppose O <
«
z < 1. According to the lemma, there are positive rational
numbers x in a, and y not in a, such that y/x = 1/z; and we can assume that y
·
is not the smallest element of Q a. But this means that z
is in a, and 1/y is in a-1. Consequently, z is in «•«-1
-
=
x
-
(1/y),
where x
We are almost done! Only the proof of the distributive law remains. Once again
we must consider many cases, but do not despair. The case when all numbers are
positive contains an interesting point, and the other cases can all be taken care of
very
THEOREM
PROOF
neatly.
If a, ß, and y are real numbers,
then a
•
(ß +
y)
=
a
•
ß+ a
•
y.
ASSume
first that a, ß, y > 0. Then both numbers in the equation contain all
5 0. A positive rational number in a (ß + y) is of the form
x (y + z) for positive x in a, y in ß, and z in y. Since x (y + z) = x y + x z,
where x y is a positive element of a
and x z is a positive element of a y,
this number is also in a ß + a y. Thus, every element of a (ß + y) is also in
rational
numbers
·
-
·
·
·
•ß,
-
-
•
-
•
.
a•ß+g•y.
On the other hand, a positive rational number in a ß + a y is of the form
x1 y + X2 z for positive x1. x2 in a, y in ß, and z in y. If xi 5 x2, then
(x1/x2) y 5 y, so (xi/x2) y is in ß. Thus
.
•
-
·
·
x1 y+x2
z=x2[(xt/x2)y+z]
is in «•(ß + y). Of course, the same trick works if x2 5 xiTo complete the proof it is necessary to consider the cases when a, ß, and y
are not all > 0. If any one of the three equals 0, the proof is easy and the cases
588
Epilogue
involving a < 0 can be derived immediately once all the possibilities for ß and y
have been accounted for. Thus we assume «>0 and consider three cases: ß, y <0,
and ß < 0, y > 0, and ß > 0, y < 0. The first follows immediately from the case
already proved, and the third follows from the second by interchanging ß and y.
Therefore we concentrate on the case ß<0, y>0. There are then two possibilities:
(1)ß + y 0. Then
a•y
=
a•([ß+y]
+
|ßl)
=
a•(ß
+y) +a·
|ßl,
so
«•(ß+y)=-(a•|ß|)+a.y
=
(2)ß +
y
<
a•ß+a•y.
0. Then
«·|ßl=a-(|ß+y|+y)=a-|ß+yl+a•y,
so
«•(ß+y)=-(«•|ß+yl)=-(«•|ßl)+a.y=a.ß+a.y.
This proof completes the work of the chapter. Although long and frequently
tedious, this chapter contains results sufficiently important to be read in detail at
least once (andpreferably not more than once!). For the first time we know that we
is indeed a complete ordered field, the
have not been operating in a vacuum-there
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
rich
ordered fields. If this is true, then the theorems of calculus are unexpectedly
in content, but the properties P1-P13 are disappointingly incomplete. The last
chapter disposes of this possibility; properties P1-Pl3 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 chapter
might be called
algebraist's real numbers," since they were purposely defined
the
least upper bound property, which involves the ordering <,
so as to guarantee
an algebraic notion. The real number system constructed in the next problem
might be called
analyst's real numbers," since they are devised so that Cauchy
will
always
sequences
converge.
"the
"the
1.
Since every real number
of rational
sequence
numbers,
of rational
ought to be the limit of some Cauchy sequence
we might try to definea real number to be a Cauchy
numbers.
Since two Cauchy sequences might converge
to the same real number,
however, this proposal requires some modifications.
of'the RealNumbers 589
29. Construction
(a) Define two Cauchy sequences of rational numbers {a.}and {b,}to be
equivalent (denoted
by {a,} {b,})if lim (a, b.)
0. Prove that
~
=
--
E-¥ØQ
{a.} (a,}, that {b.) {a,}if {a.}
{a.} {b.}and (b.)~ {c.}.
~
~
{b,},and
~
that
{a.}
{c.}if
~
~
a is the set of all sequences equivalent to {a.), and ß is the
equivalent
0 or
to {b.). Prove that either a n ß
then
there
is some (c,}in both a and ß. Show that
a
ß. (If anß ¢ 0,
in this case a and ß both consist precisely of those sequences equivalent
(b) Suppose that
set of all sequences
=
=
to
{c.).)
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 some
fixed sequence. We define a real number to be such a collection, and
denote the set of all real numbers by R.
(c) If a and ß are real numbers, let {a.) be a sequence in a, and {b,}a
sequence in ß. Define «+ ß to be the collection of all sequences equivalent to the sequence {a.+b.}. Show that {a.+b.) is a Cauchy sequence
and also show that this definition does not depend on the particular sequences {a,}and (b,}chosen for a and ß. Check also that the analogous
definition of multiplication is well defined.
(d) Show that R is a field with these operations; existence of a multiplicative
inverse is the only interesting point to check.
(e) Define the positive real numbers P so that R will be an ordered field.
(f) Prove that every Cauchy sequence of real numbers converges. Remember that if {an)is a sequence of real numbers, then each as is itself a
collection of Cauchy sequences of rational numbers.
2.
"the
high-school student's real numThis problem outlines a construction of
bers." We define a real number to be a pair (a, {b.)),where a is an integer and {b,}is a sequence of natural numbers from 0 to 9, with the proviso that the sequence is not eventually
9; intuitively, this pair represents
b,,10¯". With this definition, a real number
a +
is a very concrete
ob-
the difficulties involved in defining addition and multiplication
are
.=1
ject, but
formidable (howdo you add infinite decimals without worrying about carrying digits infinitely far out?). A reasonable approach is outlined below; the
trick is to use least upper bounds right from the start.
(a) Define (a,{b.))< (c,{d,})if a
d, but b¡
d¡ for 1
upper bound property.
b,
<
=
<
j
<
<
if a
c and for some n we have
n. Using this definition, prove the least
=
c, or
k
(b) Given
rational
œ
=
(a, (b.)), define
Gk
=
b.10
a +
";
intuitively, Œk is the
n=1
number
obtained by changing
all
decimal places after the kth
590 Epilogue
to 0. Conversely, given a rational
number
r of the
form a +
L b.10¯",
n=1
let r' denote the real number (a, {bs')),where b,'
and b,'
0 for n > k. Now for a
(a, {b,})and ß
=
=
=
a +
ß
=
sup{(ak Ÿ
Åk)':k a
natural
=
b. for 1 5 n 5 k
(c,(ds})define
number}
If multiplication
is defined
verification
of
all conditions for a field is a straightthen the
Once more, however, existence
forward task, not highly recommended.
of multiplicative inverses will be the hardest.
(theleast
similarly,
upper
bound exists by part
(a)).
UNIQUENESS OF THE REAL NUMBERS
CHAPTER
We shall now revert to the usual notation for real numbers, reserving boldface
symbols for other fields which may turn up. Moreover, we will regard integers and
rational numbers as special kinds of real numbers,
and forget about the specific
chapter
which
real
numbers
this
were defined. In
we are interested in only
way in
one question: are there any complete ordered fields other than R? The answer
For example, the field F3 introduced in
to this question, if taken literally, is
ordered
and
complete
field,
it is certainly not R. This field is a
Chapter 28 is a
example because the pair (a, a) can be regarded as just another name for
"yes."
"silly"
the
real
a; the operations
number
(a, a) + (b,b)
(a, a) (b, b)
-
=
=
(a + b, a
(a b,
·
a
-
+ b),
b),
This sort of example shows that any intelligent
are consistent with this renaming.
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, then
for each a in R there should correspond a
f(a) in F. The notation f(a)
suggests that renaming can be formulated in terms of functions. In order to do
this we will need a concept of function much more general than any which has
occurred until now; in fact, we will require the most general notion of
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
(ofobjects of any sort) which does not contain two distinct pairs with
the same first element. The domain of a function f is the set A of all objects a
such that (a, b) is in f for some b; this
b is denoted by f(a). If f(a) is
(unique)
in the set B for all a in A, then f is called a function from A to B. For example,
"name"
"function"
if f (x) sia x for all x in R (andf is defined only for x in R), then f is a
function from R to R; it is also a function from R to [-1, 1];
=
if
f (z)
sin
=
z for all z in C, then
f
is a function from C to C;
et for all z in C, then f is a function from C to
function from C to {zin C : z ¢ 0};
if
f (z)
=
9 is a function fmm
{zin C
:
z
¢ 0} to {xin R : 0 yx
<
C; it is also a
2x};
if f is the collection of all pairs (a, (a, a)) for a in R, then f is a function
from R to F3·
591
592
Epilo.gue
Suppose that Fi and F2 are two fields; we will denote the operations in Fi by
and the operations
in F2 by +,
etc. If F2 is going to be considered
, O, etc.,
collection
of
elements
of
F1, then there should be a function
for
new names
as a
with
the
properties:
following
from F1 to F2
•,
(1)The function f should be one-one, that is, if x ¢ y, then we should have
f (x) ¢ f (y); this means that no two elements of F1 have the same name.
that is, for every element z in F2 there
(2)The function f should be
should be some x in Fi such that
z f (x); this means that every element
"onto,"
=
is used to name some element of Ft.
(3)For all x and y in Fi we should have
of F2
f(x O y)= f(x)+ f(y),
f(xey)= f(x)· f(y);
this means that the renaming
the field.
procedure is consistent with the operations
If we are also considering F1 and F2 as ordered
quirement:
(4)If x © y,
then
of
fields, we add one more re-
f (x) < f (y).
A function with these properties is called an isomorphismfrom Fi to F2. This
definition is so important that we restate it formally.
DEFINITION
If Fi and F2 are two fields, an isomorphism
from F1 to F2 with the following properties:
from F1 to F2 is a function
f
y, then f (x) ¢ f (y).
(2)If z is in F2, then z f (x) for some x in F1(3)If x and y are in F1, then
(1)If x ¢
=
f(x e y)= f(x)+ f(y),
f(xey)= f(x)•f(y).
If F1 and F2 are ordered fields we also require:
(4)If x@y,
then
f(x)< f(y).
The fields F1 and F2 are called isomorphic
ifthere is an isomorphism between
them. Isomorphic fields may be regarded as essentially the same-any
important
of
will
other.
automatically hold for the
Therefore, we can, and
property
one
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 F is isomorphic to R. In the following theorem, F will be a field, with
operations
and
elements" P; we write a < b to mean that b a
+ and
and
is in P,
so forth.
"positive
•,
-
30. Uniquenessof theReal Numbers593
THEOREla
PROOF
If F is a complete
ordered
field, then F is isomorphic to R.
Since two fields are defined to be isomorphic if there is an isomorphism between
them, we must actually construct a function f from R to F which is an isomorphism. We begin by defining f on the integers as follows:
f (0)
0,
=
forn>0,
f(n)=1+...+1
n times
f (n)
=
(1 +
-
.
.
.
for n
+ 1)
<
0.
|n| times
It is easy to check that
f(m+n)= f(m)+ f(n),
f(m-n)= f(m)•f(n),
for all integers m and n, and it is convenient
define f on the rational numbers by
f (m/n)
=
m|n
=
to denote
f (n) by
We then
n.
m•n-1
+ 1 ¢ 0 if n > 0, since F is an ordered field).
nk, so m•l
k•n, so
k/l, then ml
This definition makes sense because if m/n
m•n-1
k•l-1. It is easy to check that
that
(notice
the n-fold sum 1 +
-
-
-
=
=
=
=
f (ri +
f(ri
r2)
=
-r2)=
f (r1) + f (r2)
,
f(ri). f(r2),
for all rational numbers ri and r2, and that f (ri) < f (r2) if ri < r2.
The definition of f (x) for arbitrary x is based on the now familiar idea that
any real number is determined by the rational numbers less than it. For any x
in R, let Ax be the subset of F consisting of all f(r), for all rational numbers
r < x. The set A, is certainly not empty, and it is also bounded above, for if ro
is a rational number with ro > x, then f (ro) > f (r) for all f (r) in A,. Since F
is a complete ordered field, the set Ax has a least upper bound; we define f(x) as
sup Ax.
We now have f (x) defined in two different ways, first for rational x, and then for
any x. Before proceeding further, it is necessary to show that these two definitions
words, if x is a rational
number, we want to show
agree for rational x. In other
that
sup A,
=
f (x),
for x m/n. This is not automatic,
required.
on the completeness of F; a slight digression is thus
Since F is complete, the elements
where
f (x) here
denotes m|n,
1+...+1
n times
=
fornaturalnumbersn
but depends
594
Epilqgue
form a set which is not bounded above; the proof is exactly the same as the proof
for R (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 there is a rational
number
r such that
f(r)<b.
a<
Having made this observation, we return to the proof that the two definitions
f (x) agree for rational x. If y is a rational number with y < x, then we have
already seen that f (y)< f (x). Thus every element of Ax is < f (x). Consequently,
of
supAx 5
On
the other
f(x).
hand, suppose that we had
sup A, <
Then there would be a rational number
sup A,
<
f (x).
r such that
f (r) < f (x).
But the condition f(r) < f(x) means that r < x, which means that f(r) is in the
set A,; this clearly contradicts the condition sup A, < f (r). This shows that the
original
assumption
is false, so
sup A,
=
f(x).
We thus have a certain well-defined function f from R to F. In order to show
of the definition. We
f is an isomorphism we must verify conditions (1)--(4)
will begin with
(4).
If x and y are real numbers with x < y, then clearly A, is contained in A,.
Thus
$supA,
f(x)=supA,
f(y).
that
=
To rule out the possibility of equality, notice that there are
rational
numbers
r
and s with
x<r<s<y.
We know that f (r)
<
f (s). It follows that
f (x) 5 f (r) < f (s) $ f (y).
This poves (4).
Condition (1)follows immediately from (4):If x ¢ y, then either x < y or
<
y x; in the first case f (x) < f (y), and in the second case f (y) < f (x); in either
case
f(x) ¢ f(y).
be an element of F, and let B be the set of all rational
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 f (s)> f (r) for r in B,
which implies that s > r. Let x be the least upper bound of B; we claim that
f (x) = a. In order to prove this it suffices to eliminate the alternatives
To prove
numbers
(2),let
a
r with
f(x) <a,
a
<
f (x).
30. Uniquenessof theReal Nurnbers595
In the first case there would be a rational number
r with
f(x)< f(r)<a.
But this means that x < r and that r is in B, which contradicts the fact that
number r with
x
sup B. In the second case there would be a rational
=
f(r)< f(x).
a<
This implies that
Hence
x. Since x
r <
sup B, this means
=
that r
<
s
for some s in B.
f(r)< f(s)<a,
again a contradiction.
Thus
a, proving
and
real
numbers
let
be
x
y
(3),
either
Then
f (x) + f (y).
f(x)
=
To check
f (x +
y)
f (x) + f (y)
<
y)
f (x +
<
and
suppose
that
f (x) + f (y) < f (x +
or
In the first case there would be a rational
(2).
number
f(x
+ y)
¢
y).
r such that
f (r) < f (x) + f (y).
But this would mean that
x+y<r.
Therefore r could be written as the sum of two rational
=
r
where x
r1 + r2,
Then, using the facts checked about
f (r)
=
f (r1+
r2)
=
<
numbers
ri and y < r2.
f for rational numbers, it would
f (r1)+ f (r2)> f (X) + f (7),
follow that
The other case is handled similarly.
Finally, if x and y are positive real numbers, the same sort of reasoning
a contradiction.
shows
that
f(x-y)= f(x). f(y);
the general case is then a simple consequence.
This theorem brings to an end our investigation of the real numbers, and resolves
any doubts about them: There is a complete ordered field and, up to isomorphism,
educaonly one complete ordered field. It is an important part of a mathematical
real
numbers
construction
of
tion to follow a
the
in detail, but it is not necessary
refer
construction.
again
this
particular
is utterly irrelevant that a real
It
to
to
ever
numbers,
number happens to be a collection of rational
and such a fact should
never enter the proof of any important theorem about the real numbers. Reasonable proofs should use only the fact that the real numbers are a complete ordered
field, because this property of the real numbers characterizes 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.
596
Epilogue
PROBLEMS
1.
Let
f
be an isomorphism from Fi to F2-
I. (Here 0 and I on the left denote
0 and f(1)
elements in Fi, while 0 and 1 on the right denote elements of F2.)
f(a) and f(a-1) f(a)-1, for a ¢ 0.
(b) Show that f(-a)
(a) Show that f(0)
=
=
2.
-
-
=
Here is an opportunity
to convince yourself that any significant property of
shared
field
is
by
a
any field isomorphic to it. The point of this problem is
write
to
out very formal proofs until you are certain that all statements of
this sort are obvious. F1 and F2 will be two fields which are isomorphic; for
simplicity we will denote the operations in both by + and
Show that:
..
(a) If the
equation x2 + 1
=
0 has a solution in Fi, then it has a solution
in F2.
0 with
+ ao
(b) If every polynomial equation x" + a._i x"¯I +
then
equation
F1,
in F1, has a root in
every polynomial
ao,...,a,_i
-
=
---
x"+b._i•x"¯I+--·+bo=0withbo,...,b,_iinF2hasarootinF2.
(c) If
+ 1 (summed
m times) 0 in Fi, then the same is true in F2.
fields (andthe isomorphism f satisfies f (x) <
(d) If
f (y) for x < y) and F1 is complete, then F2 is complete.
3.
1+
-
-
=
-
F1 and F2 are ordered
f be an isomorphism from Fi to F2 and g an isomorphism from F2
g(f(x)).
from F1 to F3 by (go f)(x)
F3.
Define the function gof
to
that
Show
gof is an isomorphism.
Let
=
4.
Suppose that F is a complete ordered field, so that there is an isomorphism f
from R to F. Show that there is actually only one isomorphism from R
R, this is Problem 3-17. Now if f and g are two
to F. Hint: In case F
isomorphisms from R to F consider g¯1
=
5.
Find an isomorphism from C to C other than the identity function.
SUGGESTED
READING
A
aught
a read
þrt as inttination leads Em;
for what he reads as a task
will do him little good.
inan
SAMITEL JOHNSON
One 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 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
independent reading, the more standard a text, the less likely it is to
encourage
here.
In some cases, this philosophy may seem to have been carried to
appear
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
which can be read now, and I can't believe that it hurts to have
many selections
of
what
lies ahead.
some idea
Many of these books have gone through numerous editions and printings, which
will be reflected in more recent publication dates. Many of the books with older
publications dates are out of print, though that generally doesn't apply to books
from the redoubtable Dover Publications, or from the Mathematical Association of
America. Those that are no longer in print can still often be found in well-stocked
academic libraries.
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
number theory. Few books have won so enthusiastic an
any book on elementary
audience as
[1]
An Introductionto the Theoryof Numbers(fifth
edition), by G. H. Hardy and
E. M. Wright; Oxford University Press, 1980.
The Pergamon Press published a series, Popular Lectures in Mathematics, with
several titles worth investigating, among them
[2]
A Selectronof Problemsin the Theoryof Numbers,by W. Sierpinski; Macmillan
(Pergamon), 1964.
Finally, I will mention
[3]
an intriguing little book, now out of print I fear,
ThreePearlsofNumberTheory,by A. Khinchin; Graylock Press, 1952.
The subject of irrational numbers straddles the fields of number
analysis. An excellent introduction will be found in
[4]
theory and
IrratzonalNumbers,
by I. M. Niven; Mathematical Association of America,
1956.
Together with many historical notes, there are references to some fairly elementary
articles in journals. There is also a proof that yr is transcendental (seealso [51])
and, finally, a proofof the "Gelfond-Schneider theorem": If a and b are algebraic,
with a ¢ 0 or 1, and b is irrational, then ab is transcendental.
All the books listed so far begin with natural numbers, but whenever necessary
take for granted the irrational numbers, not to mention the integers and rational
599
Reading
600 Suggested
numbers.
Several books present a construction of the rational numbers from the
natural numbers, but one of the most lucid treatments is still to be found in
[5]
edition),
Foundations ofAnalysis(second
by E. Landau; Chelsea, 1960.
Incidentally, the original German edition,
[6]
edition), by E. Landau; Chelsea, 1965.
GrundlagenderAnalysis(fourth
has been printed in paper back, together with a complete German-English dictionary (ofabout 300 words) for the whole book-an excellent way to begin reading
mathematical
German. The basic idea for constructing the real numbers is derived
from Dedekind, whose contributions can be found in
[7]
Essayson the Theoy ofNumbers,by R. Dedekind; Dover, 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
[8]
NaiveSet Theoy, by P. R. Halmos; Springer-Verlag, 1991.
Another very good introduction is
[9]
Theoy of Sets,by E. Kamke; Dover, 1950.
"new
math" that set theory
Perhaps it is necessary to assure some victims of the
mathematical
content (infact, some very deep theorems). Using
does have some
these deep results, Kamke proves that there is a discontinuous function f such
that f (x + y)
f (x) + f (y) for all x and y. For those who enjoy reading the
classics, the most important notions of set theory were first introduced by Cantor,
=
whose work
is reproduced
in
[10]Contributionsto theFoundingof the Theoy of TransßniteNumbers,by G. Cantor;
Dover, 1952.
Inequalities, which were treated as an elementary topic in Chapters 1 and 2,
actually form a specialized field. A good elementary
introduction is provided by
[11]AnalyticInequalities,by N. Kazarinoff;
Mathematical Association of Amer-
ica, 1961.
Twelve different proofs that the geometric mean is less than or equal to the aritheach based on a different principle, can be found in the beginning of
the more advanced book
metic mean,
[12]AnIntroductiontoInequalitùs,by E. Beckenbach and
ical Association
of
R. Bellman; Mathemat-
America, 1961.
The classic work on inequalities is
edition),
[13]Inequalities(second
by G. H. Hardy,
Cambridge University Press, 1988.
J. E. Littlewood, and
G. Polya;
Suggested
Reading 601
Each of the authors of this triple collaboration
has provided his own contribution
of
mathematical
about
the
thinking, written from a
nature
to the sparse literature
mathematician's
point of view. My favorite is
[14]A Mathematician'sApology,by
G. H. Hardy; Cambridge University Press,
1992.
Littlewood's anecdotal
selections
are entitled
[15]A Mathematician'sMiscellany,by J. E. Littlewood;
Polya's contribution
[16]
Methuen, 1953.
is pedagogy at the highest level:
Mathematicsand PlausibleReasoning(Vol.I: Inductionand AnalogyinMathematics;
Vol. II: PatternsofPlausible Inference),by G. Polya; Princeton University Press,
1990.
Geometry is the other main field which can be considered as background for
work, but should
Euclid's Elements is still a masterful mathematical
perhaps be postponed until some preparation has been made, with a modern work
geometry," like
on
calculus.
"classical
[17]ElementaryGeometry
froman AdvancedStandpoint(secondedition),
by E. Moise;
Addison-Wesley, 1974.
This beautiful book provides excellent historical perspectives and contains a thorough discussionof the role of the "Archimedean axiom" in geometry; in addition,
Chapter 28 describes an ordered field in which the Archimedean axiom does not
hold. Speaking of beautiful geometry books, all sorts of fascinating things can be
found in
edition),
[18]Introductionto Geometry(second
by H. S. Coxeter; Wiley, 1989.
Almost all treatments of geometry at least mention convexity, which forms another specialized topic. I cannot imagine a better introduction to convexity, or a
experience in general, than reading and working through
better mathematical
[19]ConvexFaures, by I. M. Yaglom and
W. G. Boltyanskii; Holt, Rinehart and
Winston, 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] CombinatorialGeometryin the Plane, by
H. Hadwiger and H. Debrunner;
Holt, Rinehart and Winston, 1964.
Along with these two out-of-the-ordinary books, I might mention
valuable little book, also of a specialized sort,
in Analysis,by B.
[21] Counterexamples
1964.
Gelbaum and
J. Olmsted;
an extremely
Holden-Day,
602 Suggested
Reading
Many of the example in this book come from more advanced topics in analysis,
but quite a few can be appreciated by someone who knows calculus.
Of calculus books I will mention only two, each something of a classic:
[22]A Courseof PureMathematics(tenthedition),
by G. H. Hardy; Cambridge
University Press, 1952.
[23]Digerentialand IntegralCalculus(twovolumes), by R. Courant; Wiley (Interscience), 1988.
Courant is especially strong on applications to physics. Speaking of such applications, an elegant exposition of the material in Chapter 17, together with much
further discussion, can be found in the article
[24] On the Geometryof the Kepler19oblem,by JohnMilnor;
matical
Monthly,Volume 90
in TheAmericanMathe-
pp. 353-365.
(1983),
(In this paper the curve c' of Chapter 17 is denoted by v, and the derivative of
the important composition vo 0¯I (page331) is introduced quite off handedly as
dv/dB.) A
derivation of Kepler's laws, together with numerous
references, can be found in another article in this same journal,
"straight-forward"
[25] The MathematicalRelationshipBetweenKepler'sLaws and Newton'sLaws, by
Andrew T. Hyman; in The AmericanMathematical Monthly, Volume 100
(1993),pp. 932-936.
The latter parts of Volume I of Courant contain material usually found in adcalculus, including differential equations and Fourier series. An introduction to Fourier series (requiring
a little advanced calculus) will also be found in
vanced
[26]An Introductionto FourierSenesand
Integrals,by R. Seeley; W. A. Renjamin,
1966.
calculus in earnest) contains additional
The second volume of Courant (advanced
equations,
well
as
as an introduction to the calculus ofvarion difFerential
ations. A widely admired, though somewhat more advanced, book on differential
equations is
material
[27]Lectureson OrdinaryI@iirendalEquadons, by W. Hurewicz;
Dover, 1990.
I will bypass the more or less standard advanced calculus books (which
can easily
be found by the reader) since nowadays there is a movement to revise the whole
presentation of advanced calculus, basing it upon linear algebra. One of the first,
and still one of the nicest, treatments of advanced calculus using linear algebra is
[28] Calculusof VectorFunctions,by R. H. Crowell and
R. E. Williamson; Prentice-
Hall, 1962.
Several recent books on advanced calculus attempt to acquaint undergraduates
with very large areas of modern mathematics.
My favorite, of course, is
[29] Calculuson Man¶olds,by M.
Spivak; W. A. Benjamin, 1965.
Suggested
Reading 603
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 underThe purposeful study of fields and related systems is the
graduate curriculum.
domain of "algebra."
One of the favorite texts is
edition),
[30]TopicsinAlgebra(second
by I. N. Herstein; Wiley, 1975.
A more advanced book is the great classic:
[31]Algebra,by B. L. van
der Waerden; Springer-Verlag, 1990.
By the way, this book contains a proof of the partial fraction decomposition of a
function.
There are now several introductions to complex analysis, as well as many elementary books on topology. Although the latter subject has not been mentioned
before, it has really been in the 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.
The next few topics, ranging from elementary to very difficult, are included in
this bibliography because they have been alluded to in the text. The proof that a
nondecreasing function is differentiable at almost all points (andan explanation of
just what this means) receives a beautiful exposition in
rational
[32]FunctionalAnalysis,by E Riesz and
B. Sz.-Nagy; Ungar, 1955.
(After this elementary beginning, the book moves on to quite advanced material.)
The gamma function has an elegant little book devoted entirely to its properties,
most of them proved by using the theorem of Bohr and Mollerup which was
mentioned in Problem 19-39:
[33]The GammaFunction, by E. Artin; Holt,
Rinehart and Winston, 1964.
The gamma function is only one of several important improper integrals in mathe¯22
ematics. In particular, the calculation of
dx (seeProblem 19-41) is imporfe
theory,
where
the
distribution
in
probability
function"
tant
"normal
(x)
1
=
2
e¯
dy
plays a fundamental role. A classic book on pobability
[34]An Introductionto 14obabilig Theory and
W. Feller; Wiley, 1968.
theory is
lšs Applications(thirdedition), by
The impossibility of integrating certain functions in elementary terms (among
them f (x) = e¯=') is one of the most esoteric subjects in mathematics. An interesting discussion of the possibilities of integrating in elementary terms, with an
604 Suggested
Reading
outline of the impossibility proofs, and references
ville, will be found in
[35] The
to the original papers of Liou-
Integration of Functions of a Single Variable (secondedition),
by
G. H. Hardy; Cambridge University Press, 1958.
A complete presentation of the impossibility proofs will be found in
[36]Integrationin Finite Terms,by J. Ritt;
Columbia University Press, 1948.
Oddly enough, a related but seemingly more difficult problem has a much neater
There are simple differentialequations (y"+ xy
0 is a specific examwhose
solutions
of
ple)
indefinite integrals of
cannot be expressed even in terms
elementary
functions. This fact is proved on page 43 of the (60-page)
book:
solution.
=
[37]An Introductionto Ihþrential Algebra,by L Kaplansky;
Hermann, 1957.
To read this book you will need to know quite a bit of algebra, however.
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 skill). You are probably already aware that the process
of integration can be expedited by tables of indefinite integrals. For those who
enjoy pursuing tables there is a really beautiful collection, that includes indefinite
integrals, definiteimproper integrals, and a great deal more besides(ifyou should
ever happen to need the value of the thirty-fourth Bernoulli number, this is the
place to look):
[38]Tablesof Integrals,Senes,and
Pwducts,by I. S. Gradschteyn et al.; Academic
Press, 1980.
For the thrifty, there is a paperback table of integrals:
[39] Tablesof IndefiniteIntegrals,by G. Petit Bois; Dover, 1961.
The remaining references are of a somewhat different sort. They fall into three
of which the first is historical. The letter of H. A. Schwarz referred
to
in Problem 11-65 will be found in
categories,
[40] Waysof Thoughtof GreatMathematicians, by H.
Meschkowski; Holden-Day,
1964.
Some historical remarks, and an attempt
will be found in
to incorporate them into the teaching of
calculus,
[41] The Calculus:A
GeneticApproach,by O. Toeplitz; University of Chicago
Press, 1981.
An admirable textbook on the history of mathematics
is
[42]An Introductionto the HistoryofMathematics(sixthedition),
ders College Publishing, 1990.
by H. Eves; Saun-
Suggested
Reading 605
Three good scholarly works are
[43]History ofAnalyticGeometry,by C. Boyer; Scholar's Bookshelf, 1988.
[44]A Historyof the Calculus,and Rs ConceptualDevelopment,by C. Boyer; Dover,
1959.
[45] TheMathematicsof GreatAmateurs,by J. Coolidge; Oxford University Press,
1990
and extracts from original sources will be found in
[46]A SourceBookin Mathematics(2vols.),
by D. Smith; Dover, 1959.
Despite the impression that might be given by the large number of books listed
here, it is often hard to find specific concrete information about the origins of calculus. For example, it is almost impossible to find out who first proved the Mean
Wissenschaften,
VolValue Theorem (according
to the EncyklopädùderMathematischen
whose
students
of
differential geomname is familiar to
ume II, it was O. Bonnet,
etry from the "Gauss-Bonnet Theorem"). Similarly, though many history books
"complicated
method of interpolatell us that Wallis proved Wallis' formula by a
mention
what
though
tion," most never bother to
it inspired Euler's
it was, even
investigations of the gamma function (a description is given in the answer book,
along with the solution to Problem 19-40).
The second category in this final group of books might be described as
ularizations."
There are a surprisingly large number of first-rate ones by real
"pop-
mathematicians:
[47] What is Mathematics? (fourthedition),
by R. Courant and H. Robbins;
Oxford University Press, 1979.
[48] Geometryand the Imagination,by D. Hilbert and S. Cohn-Vossen; Chelsea,
1952.
[49] The EnjoymentofMathematics,by H. Rademacher and O. Toeplitz; Dover,
1990.
edition), by H. Tietze; Graylock
[50]FamousProblemsof Mathematics(second
Press, 1965.
One of the most renowned
"popularizations"
is especially
concerned
with the
teaching of mathematics:
[51]ElementaryMathematicsfroman AdvancedStan point,by E Klein (vol.1: Arithmetic, Algebra,Analysis;vol.
2: Geometry);Dover, 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.
riginal
The third category is the very opposite extrem
papers. The difficulties encountered
here are formidable, and I have only had the courage to list
such
the
one
source of the quotation for Part IV. It is not even in English,
paper,
606
Suggested
Reading
although
you do have a choice of foreign languages. The article in the original
French is in
[52] OeuvresComplètesd'Abel;Christiania. JohnsonReprint
Corporation, New
York, 1965.
It first appeared in a German translation in the journalfiir die reine und angewandte
Mathematik,Volume 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 the pupils.
ANSWERS
TO SELECTED
PROBLEMS
CHAFTER
1
1.
la)x
a¯I(ax)
1 x x.
1 a-la
(a
x2
y2, then 0 x2 y2
If
(x y)(x + y), so either x
(iii)
0, that is, either x
or x
x+ y
y.
in
Replace
by
y
y
(vi)
(iv).
One step requires dividing by x y 0.
(i)
=
=
=
=
=
=
=
·
=
_
-
y
-
0 or
=
-y
=
=
=
---
2.
3.
(i) a/b
(ii) (ad +
ab¯I
(iii)
bc)/(bd)
(a
=
(a/b)/(c/d)
(v) ad(b¯I
-
ad(bc)¯I
=
a¯I
-
(a-b¯I)(c-d
=
b¯I
I)
(ab)¯'.
=
1
=
(a-b¯I)(c-1-d)=
=
(ad)/(bc).
=
-1.
(i)
(iii) x>Áorx<-Ã.
(v) All x, since x2 2X + 2 (X Î) + Î.
x2
since 3 and
are the roots of
(vii)x > 3 or x <
< x < 3.
(ix) x > sr or
<
3.
(xi) x
(xiii)x>1or0<x<1.
(i) b a and d c are in P, so (b a) + (d c) (b +
<
x
=
-
-
-2,
-2
x
-
--5
5.
(by(iii))
=
1, so
=
-
(a/b)(c/d)¯I
=
c¯1)
-
=
bc)(bd)¯I
(ad +
=
(ac)(bc)-i (by(iii)) ac/bc.
bc)(b¯Id¯')
(ad +
=
a/b + c/d.
a-1)(b
b¯1)
=
ab(a¯Ib-1)
c¯1)
(ac)(b¯I
=
ab-1 + cd-1
4.
=
-
=
-
-
-
in P. Thus, b + d
Using (ii), <
-
a + c.
then (i)implies that a +
a)
are in P, so
ac
0 and a < 1, so a2 < a.
d)
=
6
-
-
0.
=
c)
(a +
is
>
--d;
(iii)
-c
-a)
and
(v) (b
-c
-bc
--c(b
(vii)Using (iv),a >
(ix) Substitute a for c and b for d in
9.
=
-
(i) Ã+E-E+n.
(iii) |a+b|+|c|-|a+b+c|.
(v) Ä+Ä+E-W.
and b à
(i) a if a
(-c)
b+
<
(--d).
is in P, that is, ac
>
bc.
(viii).
-b
10.
-b
0;
0;
and b 5
if a 5
-a
a+2bifat-bandb_<0;
-b
-a
-
(iii) x
-
--x--x2ifxs0.
2b if a 5
x2 if
x à 0;
and b
>
0.
-5.
11.
(i) x
(iii)
(v) No
=
11,
-6
--2.
<
x
<
x
(thedistance
--1
least 2).
(vii)x 1,
from x to 1 plus the distance from x to
is at
-1.
=
12.
(i) (|xyl)2
(xy)2
x2y2
|x|2|y|2 (\x| |y|)2;since |xy| and |x|· |y|
are both > 0, this proves that |xy| |x|· |y|.
y--1| by
(ii)) |xy il (by(i)) |x/y|.
(iii) 1x|||y| |x
(v) It followsfrom (iv)that |x| |y (y x)| 5 |y| + ly x|, so lxI lyI 5
lx yl(vii) |x + y + z| 5 |x + y| + |z 5 |x| + ly| + |z|. If equality holds, then
|x + y| |xl + |y|, so x and y have the same sign. Moreover, z must
=
=
=
=
-
=
-|y|¯-1
=
=
.
=
-
=
609
-
-
=
-
-
610
Answers to SelectedProblems
have the same sign as x + y, so x, y, and z must all have the same sign
one is 0).
(unless
CHAPTER 2
1.
(i)
Since 12 1 (2) (2 1 + 1)/6, the formula is true for n
1. Suppose
that the formula is true for k. Then
k(k + 1)(2k + 1)
2
12+---+k'+(k+1)
+(k+1) 2
6
(k + 1)
+ 1) + 6(k + 1)]
6 [k(2k
(k + 1) + 2)(2k + 3)]
6 [(k
(k + 1)(k + 2) (2[k+ 1] + 1)
6
=
-
-
=
-
=
=
'
so the
2.
formula is true for k + 1.
(i)
(2i-1)=l+3+5+---+(2n-1)
i=1
=
=
=
5.
1+2+3+---+2n-2(1+---+n)
(2n)(2n + 1)
-n(n+1)
2
n2.
(a) Since
1-r2
1+r=
the formula is true for n
=
,
1-r
1. Suppose that
1
1+r+··-+r"=
-
rn+1
.
1-r
Then
1+r+---+r"+rn+1
_
1
1
1
rn+1
-
-
-
+rn+1
r
rn+1 + rn+1(1
1-r
1-r"*2
'
1-r
(b)
S=1+r+
rS=
+r"
r+---+r"+rnel
Thus
S(1-r)=S---rS=1-r"**,
-
r)
Answersto SelectedProblems 611
SO
l
rn+1
-
¯
1-r
6.
(i)
From
(k+1)'-k'=4k3+6k2+4k+1,
k=1,...,n
we obtain
k3+6
(n+1)4-1=4
k2+4
k=l
k+n,
k=1
k=1
so
n
k3_
1)(2n + Î)
64
n(n +
(n+1)
4
R4
3
2
4
2
4
-1-6
n(n ‡ 1)
-4
-n
2
k=1
=-+-+-.
(iii) From
1
1
k
1
=
--
k+1
k=1,...,n
,
k(k+1)
we obtain
18.
1
1
n+1
k(k+1)
1 is either even or odd, in fact it is odd. Suppose n is either even or odd;
then n can be written either as 2k or 2k + 1. In the first case n + 1 2k + 1
is odd; in the second case n + 1 2k + 1 + 1 2(k + 1) is even. In either
case, n + 1 is either even or odd. (Admittedly, this looks fishy,but it is really
=
=
=
9.
12.
correct.)
Let B be the set of all natural numbers I such that no 1 + l is in A. Then
1 is in B, and l + 1 is in B if l is in B, so B contains all natural numbers,
which means that A contains all natural numbers > no.
(a) Yes, for if a + b were rational, then b (a + b) a would be rational. If
a and b are irrational, then a + b could be rational, for b could be r a
for some rational number a.
rational. But if a ¢ 0, then ab could not be rational,
If
(b) a 0, then ab isa¯1
=
would be rational.
for then b (ab)
example,
M.
(c) Yes; for
example,
and
for
Yes;
E
(d)
--
=
-
--
=
·
-d.
13.
(a) Since
(3n + 1)2
(3n + 2)2
=
=
9n2 + 6n + 1 3(3n2+ 2n) + 1,
9n* + 12n + 4 3(3n2 + 4n + 1) + 1,
=
=
it follows that if k2 is divisible by 3, then k must also be divisible by 3.
p/q where p and
Now suppose that Ë were rational, and let Ë
=
612 Answersto SelectedProblems
p2
3q2, so p2 is divisible by 3, so
q have no common factor. Then
3p' for some natural number p', and consep must be. Thus, p
42. Thus,
quently (3p')2 3q2, or 3(p')2
q is also divisible by 3, a
contradiction.
The same proofs work for à and Ã, because the equations
=
=
=
_
(Sn + 1)2 = 25n2 + 10n + 1
(Sn + 2)2 25n2 + 20n + 4
(Sn + 3)2 25n2 + 30n + 9
=
=
=
=
=
5(5n2
$(5n2 + 4n) + 4,
5(5n2 + 6n + 1) + 4,
(5n+4)2=25n2+40n+16=5(5n2+¾+B+L
equations for numbers of the form 6n + m, show
and the corresponding
that if k2 is divisible by 5 or 6, then k must be. The proof fails for Ã,
because (4n + 2)2 is divisible by 4. (For precisely this reason this proof
cannot be used to show that in general á is irrational if a is not a
have no guarantee that (an + m)2 might not be a
perfect square-we
multiple of a for some m < a. Actually, this assertion is true, but the
proof requires the information in Problem 17.)
(b) Since
(2n+1)3
=
Sn3+12n2+6n+1
2(4n3+6n2+3n)+1,
=
=
p/q where p and
it follows that if k3 is even, then k is even. If
p3
p3
2q3,
is divisible by 2, so p
so
q have no common factors, then
2p' for some natural number p', and consequently
must be. Thus, p
(2p')S
2q3, or 4(p')3 q3. Thus, q is also even, a contradiction.
=
=
=
=
The proof for
å
is similar,
using the equations
(3n+ 1)3 27n3 + 27n2 + 9n + 1
(3n + 2)3
19.
If n
=
=
=
27n* + 54n2 + 36n + 8
1, then (1 + h)"
=
(1 + h)"*
(1 +
=
3(9n3+ 9n2 + 3n) + 1,
=
3(9n3 + 18n2 + 12n + 2) + 2.
1 + nh. Suppose that (1 + h)"
=
h)(1 + h)"
=1+(n+1)h+nh2>1+(n+1)h.
>
(1 + h) (1 +
since
nh),
>
1 + nh. Then
1+ h
>
0
For h > 0, the inequality follows directly fmm the binomial theorem, since
all the other terms appearing in the expansion of (1 + h)" are positive.
-1
CHAPTER 3
1.
(i)
(x + 1)/(x + 2); the expression f (f (x)) makes
-2.
and x
¢
-1/c
if c ¢ 0).
1/(1 + cx) (forx ¢
(v) (x + y + 2)/(x + 1)(y + 1) (forx, y ¢
(vii)Only c 1, since f (x) f (cx) implies that x
true for at least one x ¢ 0.
(i) y > 0 and rational, or y > 1.
(iii) 0.
0, 1.
(v)
(iii)
-1).
=
2.
sense only when xy
---1,
=
=
cx, and this must be
Answersto SelectedProblems 613
3.
(i) {x:-1Exs1}.
(iii) {x: x ¢ 1 and x ¢
(v) 0
(i) 22y
(iii) 22=r + sin(2').
(i) Po s.
(iii) so S.
(v) Po P.
(vii)sososoPoPoPos.
(a) y.
(b) H(y).
2).
.
4.
5.
11.
(c) H(y).
12.
(a)
even
odd
even
neither
neither
odd
even
odd
(b)
even
odd
even
even
odd
odd
odd
even
(c)
f
f
even
odd
g even
even
even
g odd
even
odd
(d) Let g(x) f (x) for x > 0 and define g arbitrarily for x < 0.
(i) Let g(x) h(x) 1 and let f be a function for which f(2) ¢ f(1) +
=
21.
=
=
f (1). Then fo (g + h) ¢ fog + fo h.
(ii) [(g+h)o f](x)= (g+h)(f(x))= g(f(x))+h(f(x))
(ho f)(x) [(gof) + (ho f)](x).
=
(go f)(x)+
=
1
1
1
=
=
1/(f
¢f
1
-(g(x))
(iii) fog (x) f (g(x)) f
2 and let f be a
(iv) Let g(x)
og)
o(1/g).
=
=
-
f
og
(x).
function for which
f(j) ¢
1/f(2). Then
614 Answersto SelectedProblems
CHAFTER
a
1.
(i) (2, 4).
(iii) [2,4].
(v) (-2, 2).
(vii) (-oo, 1] U [1,oo).
3.
(i) All points below the
(iii) All points below the
graph of f (x)
graph of f (x)
=
=
x.
x2.
All points between the graphs of f(x) x + 1 and f(x) x 1.
inter(vii)A collection of straight lines parallel to the graph of f (x) =
axis
secting the horizontal
at the points (n, 0) for integers n.
radius 1 and around (1, 2).
of
the
inside
circle
All
points
(ix)
with
vertices
(1,0), (0, 1), (-1, 0), and (0,
(i) A square
(iii) The union of the graph of f (x) x and of f (x) 2 x.
(v) The point (0, 0).
(vii)The circle of radius á around (1, 0), since x2 2x + y2 (x 1)2 +
(v)
=
=
-
-x,
-1).
4.
=
=
-
=
-
-
y2-1.
6.
a) + b
that the graph of f (x) m(x
mx + (b ma)
is a straight line with slope m, which goes through the point (a, b). (The
important point about this exercise is simply to remember the point slope
form.)
(b) The straight line through (a, b) and (c,d) has slope (d b)/(c a), so
the equation follows from part (a).
(c) When m m' and b ¢ b'. In that case, there is clearly no number x with
f (x) g(x), while such a number x always exists if m ¢ m', namely,
m').
x = (b' b)/(m
(a) If B = 0 and A ¢ 0, then the set is the vertical straight line formed
If B ¢ 0, the set is the graph of
-C/A.
by all points (x, y) with x
f (x) (-A/B)x + (-C/A).
points (x, y) on the vertical line with x
The
a are precisely the ones
(b)
which satisfy 1 x + 0 y + (-a)
0. The points (x, y) on the graph of
f (x) mx+b are precisely the ones which satisfy (-m)x+ 1-y+(-b)
(a) Simply observe
=
=
-
-
-
-
=
=
-
7.
-
=
=
=
=
-
=
=
11.
0.
(i) The graph of f is symmetric with respect to the vertical axis.
(ii) The graph of f is symmetric with respect to the origin. Equivalently,
the part of the graph to the left of the vertical axis is obtained by reflecting first through the vertical axis, and then through the horizontal
RXiS.
21.
(iii) The graph of f lies above or on the horizontal axis.
0 and a over
(iv) The graph of f repeats the part between
x2)
of
the
distance
from
is
The
to (0, ¼)
(x,
square
(a)
-x2+x'-
x2+
x2-
\
4)
2
x2
=x4
2
=
(x2+
16
1)2
+
1
16
and over.
Answersto SelectedProblems 615
is the square of the distance from (x, x2) to the graph of g.
(b) The point (x, y) satisfies this condition if and only if
which
(x-a)2+Q-
2-Q-02
or
x2-2ax+a2+y2-2ßy+ß2-y2-2yy+y2,
or
1
2ß-2y
(
y=
«24p2
a
y-ß
x2+
x+
2
.
2ß-2y
for ß ¢ y, which is just the condition that P is
not on L. If P is on L, then the solution is the vertical line through P.)
(This
CHAFTER 5
1.
solution
works only
(ii)
x3
8
2
-
lim
x->2
-
x
lim(x2 + 2x + 4)
=
=
xw2
12.
(iv)
yn
lim
yn
_
limXn-1
=
xey
X
-
n-2
n-2
n-1
X->T
y
yn-i
=
n-1
pn-i
=
ny"¯1
(vi)
Ja+h-
lim
lim
=
h
h->o
(Ja+h-
lim
=
ha0
)(Ja+h+Á)
h( Ja + h +
h->o
da +
h+
)
&
l
24
3.
It is possible to fmd ô by beginning with the equation
(i)
x4
If |x
-
a|
<
-
(X
-
G
1, then
|x| <
3
|x3+ax2+a2
a)(X
-
1+
3+|a|
<
#
GX2
+ a2x +
a3L
|a|, so
X2
2
|X|+|a|3
(1 + |a|)" + |a|(1 + |a|)2 + M2(1+ |a|) + |a|3·
therefore we can choose
8
=
min
(
l'
e
[a|)3+ |a|(1 + |a|)2 + |a|2(1+ |a|) + |a|3
(1 +
It is instructive, and probably easier, to use part
shows that
|x4
-
|x2
a
_
|
<
e when
2
2(|a|2 + 1)
(2)of the
lemma. This
616 Answersto SelectedProblems
which
when
is true
min
x
(ii)
<
min
1,
=
min
1,
a|
-
(3)of the
By part
lemma, |l/x
1|
-
e when
<
e
(1
-
-
=ô.
2'2
the lemma, (x*+ 1/x)
1| < e/2. According to parts
-
2|
-
6.
(v)
(i)
1|
-
e
(1
min
<
2'Ä'
e2, since Û <
Let 8
We need |f(x)
=
0
<
x
2|
-
2|
-
82
-
|1/x
1 e
2'32
=
-
Ñ
-
sin2
+
(iii) We need
.
|g(x) 4|
-
<
mm
-,
|4|
1|
-
<
e/2
(ii)of this problem, this
< e.
implies that
e/2 and |g(x) 4| < s/2, so we
min
<
min
=
'8-2-2
|x| <
<
e
1
e when
<
(i)and
happens when
|x
8.
=
(iii) By part (1)of
|x4
2(|a 2 + 1)
2(|a| + 1)
4(|a|2 + 1)(|a[ + 1)
|x-l]<min
and
1,
-
ô.
need
1
e|4|2
-
2
2
so we need
0
9.
|x 2|
<
-
<
8e)]2
[min(2,
__
lim f (x) and define g(h)
f (a + h). Then for every e > 0 there is
a 8 > 0 such that, for all x, if 0 < |x a| < 8, then |f (x) l < e|. Now, if
< e. This inequality
0 < lh|<8, then0 < (h+a)-al < ô, so|f(a+h)-l|
written
which
<
g(h)
l,
e. Thus, lim
can be
can also be written
|g(h)
Let l
=
=
-
-
-l|
=
ho0
lim f (a + h) l. The same sort of argument shows that if lim f (a+ h) m,
h->0
then lim f (x) m. So either limit exists if the other does, and in this case
=
=
h->0
=
they are equal.
10.
can get f(x) as close to I as we like if and only if we can
get f (x)
as close to 0 as we like. The formal proof is so trivial that it
takes a bit of work to make it look like a pmof at all. To be very precise,
l and let g(x)
suppose lim f (x)
f (x) l. Then for all e > 0 there
< 8, then |f(x)
< e.
is a ô > 0 such that, for all x, if 0 < |x
This last inequality can be written |g(x) 0| < e, so lim g(x)
0. The
(a) Intuitively, we
-l
=
=
-
-al
-
argument
in the other direction is similarly uninteresting.
-l|
=
Answersto SelectedProblems 617
(b) Intuitively, making
x close to a
is the same as making x a close to 0.
Formally: Suppose that lim f(x)
l, and let g(x)
f(x a). Then
for all e > 0 there is a 8 > 0 such that, for all x, if 0 < |x a < ô,
then |f (x) l| < e. Now, if 0 < |yl < 8, then 0 < |(y + a)
a < 8, so
< E. But this last inequality
<
can be written |g(y)
f(y +a)
e.
l. The argument in the reverse direction is similar.
So lim g(y)
yw0
-
=
=
-
-
--
-
-l|
-l|
=
is close to 0 if and only if x3 is. Formally: Let lim f (x)
x->O
l. For every e > 0 there is a 8 > 0 such that if 0 < lx| < 8, then
|f (x) l| < e. Then if 0 < |x| < min(1, 8), we have O < |x3| < ô, so
lim f (x) = l. On the other hand, if we assume
|f(x3) l| < e. Thus, x->O
(c) Intuitively, x
=
-
-
f(x3) exists,
that lim
x->0
such that
if 0
|x|
<
say
<
lim f(x3)
8, then
we have 0 < | W] < ô, so
lim f (x) m.
m, then
=
x->0
if (x3)
[f ([ W]3)
-
-
for all e
>
0 there is a &
m| < e. Then if 0 <
|x|
ml < e, or |f (x) m| < e.
--
<
83,
Thus
=
-1
(d) Let f (x)
but lim
x->0
17.
=
1 for x
0, and f (x) =
>
f (x) does not
function f (x)
for x
<
0. Then lim f (x2)
x->0
=
1,
exist.
1/x cannot approach a limit at 0, since it bewhat 8 > 0 may
near 0. In fact, no matter
comes
satisfying
0 < |x| < 8, but 1/x > |l| + s, namely,
be, there is some x
min(8, 1/(|l| + s)). This x does not satisfy |l/(x
l)| < e.
x
(b) No matter what 8 > 0 may be, there is some x satisfying 0 < |x 1| < 8,
but 1/(x-1) > |l|+e, namely, x min(1+8, 1+1/(|l|+e)). This x does
< e. (It is also possible to apply Problem 10(b):
1)
not satisfy |l/(x
lim 1/x
lim 1/(x
1) if the latter exists, so this limit does not exist,
290
x->l
because of part (a).)
(i) This is the usual definition, simply calling the numbers ô and e, instead
(a) The
arbitrarily
=
large
=
-
-
=
-l|
-
=
25.
(ii)
-
of e and 8.
This is a minor modification of (i):if the condition is true for all 8
then it applies to 8/2, so there is an e > 0 such that if 0 < |x a|
then |f (x) l| < ô/2 < 8.
-
>
0,
<
e,
-
is a similar modification: apply it to 8/5 to obtain (i).
(iv) This is also a modification: it says the same thing as (i),since e/10 > 0,
and it is only the existence of some e > 0 that is in question.
If lim f(x)
lim f(x) l, then for every e > 0 there are ôi, 82 > 0 such
(iii) This
29.
=
x->.+
=
x->a-
that, for all x,
if a < x < a + ô1, then |f (x) li
then|f(x)-I|<e.
ifa-82<x<a,
--
LetB=min(8i,82).If0<|x-a|<&,theneithera--82<a-B<x<a
or else a < x < a + 8 < a + Bi, so jf (x) l| < e.
-
<
e,
618 Answersto SelectedProblems
30.
-l|
<
lim f (x), then for all e > 0 there is a ô > 0 such that |f (x)
xwD+
<
< x < 0, then 0 <
< ô, so |f(-x)
e for 0 < x < 8. If
e.
l. Similarly, if lim f (x) exists, then lim f (x)
Thus lim f (-x)
x->0xsOx->D*
exists and has the same value. (Intuitively, x is close to 0 and positive if
and only if
is close to 0 and negative.)
lim f (x), then for all e > 0 there is a 8 > 0 such that |f (x)--l| <
If I
If l
(i)
=
-8
-l|
-x