CALCULUS
Michael Spivak
CALCULUS
Fourth Edition
Copyright
All
©
1967, 1980, 1994, 2008 by Michael Spwak
rights reserved
Library of Congress Catalog Card
Number 80-82517
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I
.
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.
in the
(
nited States ofAmerica
ISBN 978-0-914098-91-1
Dedicated
to the
Memory of
Y. P.
PREFACE
tan a debi
to his profession,
from the which as
so ought they
men of course
of duty
to
endeavour
way of amends,
and
themselves by
to
be a help
ornament thereunto.
FRANCIS BACON
PREFACE TO THE FIRST EDITION
Every aspect of
book was influenced by
this
merely as a prelude
to but as the
real
first
the desire to present calculus not
encounter with mathematics.
the foundations of analysis provided the arena in which
Since
modern modes of math-
ematical thinking developed, calculus ought to be the place in which to expect,
rather than avoid, the strengthening of insight with logic.
In addition to devel-
oping the students' intuition about the beautiful concepts of
them
equally important to persuade
to intuition,
nor ends
and
but the natural
in themselves,
and think about mathematical
that precision
analysis,
it
is
surely
rigor are neither deterrents
medium
in
which
to formulate
questions.
This goal implies a view of mathematics which, in a sense, the entire book
No
attempts to defend.
how
matter
well particular topics
may
be developed, the
goals of this book will be realized only if it succeeds 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
ters in the
book, two (starred)
comment, however. Of the twenty-nine chapchapters are optional, and the three chapters com-
V have been included only for the benefit of those students who might
examine on their own a construction of the real numbers. Moreover, the
appendices to Chapters 3 and 1 1 also contain optional material.
prising Part
want
to
The order of the remaining
purpose of the book
is
chapters
collection of "topics. " Since the
until Part III,
less
it
is
intentionally quite inflexible, since the
to present calculus as the evolution of
one
idea, not as a
most exciting concepts of calculus do not appear
I and II will probably require
should be pointed out that Parts
time than their length suggests
— although
the entire
book covers a one-year
any uniform rate. A rather
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
corresponds to the traditional organization of most calculus courses, but I feel
that it will only diminish the value of the book for students who have seen a
small amount of calculus previously, and for bright students with a reasonable
course, the chapters are not
meant
to
be covered
at
natural dividing point does occur between Parts II
background.
The problems have been designed
with
this particular
audience
in
mind. They
range from straightforward, but not overly simple, exercises which develop basic
techniques and
test
understanding of concepts, to problems of considerable
diffi-
culty and, I hope, of comparable interest. There are about 625 problems in all.
Those which emphasize manipulations usually contain many examples, numbered
viii
Preface
with small
Roman
in other problems.
so
guide
Some
and double
starring
and
numerals, while small
many
starring, but there are so
hints have
been provided,
not completely reliable.
is
letters are
used to label interrelated parts
indication of relative difficulty
Many
many
is
provided by a system of
criteria for
judging
difficulty,
especially for harder problems, that this
problems are so
hints are not consulted, that the best of students will
difficult, especially if
those which especially interest them; from the less difficult problems
should be
it
The
easy to select a portion which will keep a good class busy, but not frustrated.
answer section contains solutions
to
the
probably have to attempt only
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
problems often
I
am grateful
and a
refer,
a Suggested Reading
is
list,
to
which the
glossary of symbols.
mention the many people
for the opportunity to
to
whom I owe my
thanks. Jane Bjorkgren performed prodigious feats of typing that compensated for
my fitful production of the
which provides
manuscript. Richard Serkey helped collect the material
historical sidelights in the
problems, and Richard Weiss supplied
the answers appearing in the back of the book.
friends Michael
I
am
especially grateful to
my
Freeman, Jay Goldman, Anthony Phillips, and Robert Wells for
and the relentlessness with which they criticized, a
the care with which they read,
preliminary version of the book. Needles to
deficiencies
which remain, especially since
would have made
the
book appear
my admiration
express
I
say,
they are not responsible for the
sometimes rejected suggestions which
suitable for a larger
for the editors
and
staff
of
group of students.
W A. Benjamin,
Inc.,
I
must
who were
always eager to increase the appeal of the book, while recognizing the audience
for
which
The
it
was intended.
inadequacies which preliminary editions always involve were gallantly en-
dured by a rugged group of freshmen
in the
honors mathematics course
University during the academic year 1965-1966.
About
at
Brandeis
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
after
all,
the class
is
students themselves,
unlikely to rise
it
seems
to
up
in a
body and
me, deserve the
is
always safe-
protest publicly
—but the
right to assign credit for the thor-
oughness with which they absorbed an impressive amount of mathematics.
I
am
content to hope that some other students will be able to use the book to such good
purpose, and with such enthusiasm.
Waltham, Massachusetts
February
1967
michael spivak
PREFACE TO THE SECOND EDITION
I
have often been told that the
title
of this book should really be something
Introduction to Analysis," because the book
is
usually used in courses
students have already learned the mechanical aspects of calculus
standard in Europe, and they are becoming more
After thirteen years
it
seems too
late to
common
change the
title,
like
"An
where the
— such courses are
in the
United
States.
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 discussed more 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 1 60 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 Shifrin. 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. Joseph Lipman also told me
of this proof, together with the similar trick for the proof of the last theorem in
the Appendix to Chapter 1 1 which went unproved in the first edition. Roy O.
Davies told me the trick for Problem 1 1-66, which previously was proved only in
Problem 20-8 [21-8 in 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
,
new
edition their contributions weren't too badly botched.
MICHAEL SPIVAK
PREFACE TO THE THIRD EDITION
The most
significant
change
in this third edition
Chapter 17 on planetary motion,
in
is
the inclusion of a
which calculus
employed
is
new
(starred)
for a substantial
physics problem.
In preparation for
dinates are
now
the old
this,
three Appendices: the
first
Appendix
to
Chapter 4 has been replaced by
two cover vectors and conic
sections, while polar coor-
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.
Another large change
is
merely a rearrangement of old material: "The Cos-
mopolitan Integral," previously a second Appendix
Appendix
ter 18,
to the chapter
now Chapter
the material
19);
on "Integration
in
to
Chapter
13,
is
now an
Elementary Terms" (previously Chap-
moreover, those problems from that chapter which used
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.
and somewhat dismayed when I realized that after allowfirst and second editions of the book, I have
allowed another 14 years to elapse before this third edition. During this time I
seem to have accumulated a not-so-short list of corrections, but no longer have
the original communications, and therefore cannot properly thank the various individuals involved (who by now have probably lost interest anyway). I have had
time to make only a few changes to the Suggested Reading, which after all these
I
ing
was both
1
startled
3 years to elapse between the
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 SPIVAK
PREFACE TO THE FOURTH EDITION
I noted that it was 13 years
and second editions, and then another 14 years before the third,
expressing the hope that the next edition would appear sooner. Well, here it is
another 14 years later before the fourth, and presumably final, edition.
Although small changes have been made to some material, especially in Chapters 5 and 20, this edition differs mainly in the introduction of additional problems,
a complete update of the Suggested Reading, and the correction of numerous errors. These have been brought to my attention over the years by, among others,
Nils von Barth; Philip Loewen; Fernando Mejias; Lance Miller, who provided a
long list, particularly for the answer book; and Michael Maltenfort, who provided
an amazingly extensive list of misprints, errors, and criticisms.
Most of all, however, I am indebted to my friend Ted Shifrin, who has been
using the book for the text in his renowned course at the University of Georgia
for all these years, and who prodded and helped me to finally make this needed
revision. I must also thank the students in his course this last academic year, who
Promises, promises! In the preface to the third edition
between the
first
served as guinea pigs for the
new
edition, resulting, in particular, in the current
proof in Problem 8-20 for the Rising Sun
Lemma, far
simpler than Reisz's original
proof, or even the proof in [38] of the Suggested Reading,
which
itself
has
now
been updated considerably, again with great help from Ted.
MICHAEL SPIVAK
9
5
CONTENTS
PREFACE
PART I
PART II
VI
Prologue
1
Basic Properties of
2
Numbers
Numbers
of Various Sorts
21
Foundations
3
39
Functions
54
Appendix. Ordered Pairs
4
Graphs
56
75
Appendix
1.
Vectors
Appendix
2.
The Conic
Appendix
3. Polar Coordinates
5
Limits
90
6
Continuous Functions
7
Three Hard Theorems
8
Least
80
Sections
Upper Bounds
1
84
1
122
133
Appendix. Uniform Continuity
III
3
144
D< jrivatives and Integrals
9
10
11
Derivatives
149
Differentiation
168
Significance of the Derivative
Appendix. Convexity and Concavity
12
Inverse Functions
188
21
230
Appendix. Parametric Representation of Curves
13
Integrals
Appendix. Riemann Sums
14
244
253
282
The Fundamental Theorem
of Calculus
285
1
XIV
Contents
15
The Trigonometric
*16
7r
*17
Planetary Motion
18
19
is
Irrational
Functions
303
324
330
The Logarithm and Exponential Functions
Integration in Elementary Terms 363
402
Appendix. The Cosmopolitan Integral
PART IV
Infinite
20
*21
Infinite Series
Approximation by Polynomial Functions
e is Transcendental
442
Infinite
23
Infinite Series
24
Uniform Convergence and Power
Complex Numbers 526
26
27
Sequences
Series
Epilogue
28
Fields
29
Construction of the Real
30
Uniqueness of the Real Numbers
581
(to selected
problems)
Glossary of Symbols
669
Numbers
609
Suggested Reading
Index
1
471
Complex Functions 541
Complex Power Series 555
Answers
4
452
22
25
PART V
Sequences and
339
665
619
588
601
499
CALCULUS
PROLOGUE
To
you
to
be conscious that
are ignorant
is
a great step
knowledge.
BENJAMIN DISRAELI
.
CHAPTER
OF NUMBERS
BASIC PROPERTIES
I
The
of
title
this
required to read
what
is
chapter expresses in a few words the mathematical knowledge
book. In
this
meant by
short chapter
and
chapter
theless, this
we
—addition and
of which
all
—are already
and
inequalities,
Never-
familiar to us.
Despite the familiarity of the subject, the
not a review.
is
simply an explanation of
division, solutions of equations
and other algebraic manipulations
factoring
is
the "basic properties of numbers,"
multiplication, subtraction
survey
fact, this
are about to undertake will probably
seem
quite novel;
it
does not aim
an extended review of old material, but to condense this knowledge
a few simple and obvious properties of numbers. Some may even seem too
to present
into
number of diverse and important facts turn
we shall emphasize.
which we shall study in this chapter, the first nine are
obvious to mention, but a surprising
out to be consequences of the ones
Of the
twelve properties
concerned with the fundamental operations of addition and multiplication. For
the
moment we
of numbers
may
—
consider only addition:
sum a + b
this
operation
performed on a pair
is
any two given numbers a and b (which
possibly be the same number, of course). It might seem reasonable to regard
the
exists for
addition as an operation which can be performed
consider the
more
sum
a\
+
•
•
•
+ a„
of n numbers a\,
on
.
.
several
,
convenient, however, to consider addition of pairs of
define other
sums
in
terms of sums of
this type.
numbers
at once,
a„ as a basic concept.
For the
numbers
sum of
and
It is
and to
numbers
add b and c,
only,
three
may be done in two different ways. One can first
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
and
a, b,
c, this
obtaining b
property
(PI)
we
c,
shall
list:
If a, b,
and
any numbers, then
c are
a
The
statement of
three
a
numbers
+ (b + c) =
more
(a
this
+
(b
+ c) =
(a
+ b) + c.
property clearly renders a separate concept of the
superfluous;
+ b) + c.
we
simply agree that a
+
b
+
Addition of four numbers requires
involved, considerations.
The symbol
(1)
or
(2)
or
(3)
or
(4)
or
(5)
a
+b+c+d
+ c) + d,
+ + c)) + d,
a + «b + c) + d),
a + (b + (c + d)),
(a+b) + (c + d).
((a+b)
(a
(b
c denotes the
similar,
is
sum of
number
though
defined to
slightly
mean
4
Prologue
This definition
fact
is
need not be
listed.
unambiguous
we know from PI
For example,
(a+b) +
and
it
follows immediately that
a direct consequence of P 1
is
(one must
=
(3)
It is
(4)
b
let
=
+c
feasible,
+
a
(2)
sums of an
possible
may be
proof (and
The
are equal.
may
this
five
equality of
(2)
and
(3)
not be apparent at
first sight
The
equalities
c).
this
is
it
prove the equality
will also suffice to
numbers, but
it
may not be so
clear
how we
so without actually listing these 14 sums.
would be
totally
if
we
considered collections
inadequate to prove the equality
numbers
a\,
an This
worry about the
worth worrying about once) a reasonable approach is outlined in
arbitrary finite collection of
taken for granted, but for those
it is
+ c),
(b
but would soon cease to be
of six, seven, or more numbers;
all
this
simple to prove.
can reasonably arrange a proof that
fact
=
c
probably obvious that an appeal to PI
is
equal. Fortunately,
play the role of b in PI, and d the role of
(5) are also
Such a procedure
all
from the property PI already
that
although
,
follows
it
and
(1)
of the 14 possible ways of summing
of
numbers are
since these
listed separately, since
who would
.
.
.
,
.
like to
Problem 24. Henceforth, we shall usually make a tacit appeal to the results of this
problem and write sums a\ +
a n with a blithe disregard for the arrangement
\-
of parentheses.
The number
(P2)
a
If
has one property so important that
is
important role
is
For every
(P3)
list it
next:
any number, then
a
An
we
also played
number
+0 = + a — a.
by
in the third
a, there
a
+
is
(— a)
a
=
property of our
number —a such
(—a)
+a =
list:
that
0.
Property P2 ought to represent a distinguishing characteristic of the
and
if
a
it is
comforting to note that
number x
we
are already in a position to prove
any one number a, then x
numbers
a).
from both
The
—
+x =
a
(and consequently
this
proof of this assertion involves nothing
sides of the equation, in other words,
following detailed proof shows,
this
0,
Indeed,
satisfies
a
for
number
this.
all
equation also holds for
more than
adding —a
then
a
hence
hence
hence
+x =
+ (a + x) =
((—a) + a) + x =
+x =
x —
(-a)
both
sides; as the
three properties P1-P3 must be used
operation.
If
to
a,
(-a)
0;
0;
0.
+ a = 0;
all
subtracting a
to justify
Basic Properties ofNumbers
1.
As we have just hinted, it is convenient
derived from addition: we consider a — b
is
to regard subtraction as
an operation
be an abbreviation for a
to
5
+
(— b).
It
then possible to find the solution of certain simple equations by a series of steps
(each justified by PI, P2, or P3) similar to the ones just presented for the equation
a
+x =
a.
For example:
If
+
jc
then
(jc+3)
hence
x
+
(3
+
+
hence
=
3
=
=
=
(-3))
+
x
hence
5,
= 5 + (-3);
(-3)
x
5
-
3
=
2;
2;
2.
Naturally, such elaborate solutions are of interest only until
that they can always be supplied. In practice,
an equation by indicating so
solve
you become convinced
usually just a waste of time to
it is
explicitly the reliance
on properties PI, P2, and
P3 (or any of the further properties 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
If a
(P4)
and b are any numbers, then
a
The
+b—
b
+ a.
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
statement of
For example, subtraction does not have
so well behaved.
a
is
—
b
— a.
b
7^
amusing
we might ask just when
how powerless we are if we
In passing
to discover
—
=
b
—
b
a only
from properties P1-P4;
fully
it
is
and determine which
be able
—
when a
—
this
property: usually
b does equal b
rely only
—
a,
and
it
on properties P1-P4
Algebra of the most elementary variety shows that
to justify our manipulations.
a
a
Nevertheless,
b.
instructive to
step(s)
to justify all steps in detail
it
is
impossible to derive this fact
examine the elementary algebra carejustified by P1-P4. We will indeed
cannot be
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 ad-
dition that
should be
by a
b,
(P5)
little
clear.
comment
will
be needed; both the meaning and the consequences
(As in elementary algebra, the product of a
or simply ab.)
If a, b,
and
c are
any numbers, then
a
(P6)
If a
is
(b
c)
=
(a
b)
a
= a.
any number, then
a
1
=
1
•
c.
and b
will
be denoted
6
Prologue
1^0.
Moreover,
(The assertion that
because there
list it,
number, namely,
For every
(P7)
no way
is
other properties listed
may seem
^
1
—
If a
/
number a
0, there
in P7.
_1
we have
to
hold
all
if
there were only one
number
a
a~
=
a
•
a
-1
such that
1.
and b are any numbers, then
This condition
number
would
is
=
a~
is
=
b
which deserves emphasis
detail
but
0.)
a
One
list,
could possibly be proved on the basis of the
it
these properties
a
(P8)
a strange fact to
b
the appearance of the condition
is
quite necessary; since
-1
satisfying
=
•
a.
Ob —
for all
numbers
a^O
b, there
is
no
This restriction has an important consequence
1.
was defined in terms of addition, so division is
Since
is
defined in terms of multiplication: The symbol o/b means a b
meaningless, a/0 is also meaningless
division by
is always undefined.
Property P7 has two important consequences. If a b = a c, it does not
necessarily follow that b = c; for if a = 0, then both a b and 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
for division. Just as subtraction
.
—
•
•
follows:
a
If
then
a'
hence
(a~
a)
1
•
hence
It is
also a
consequence of P7 that
b
b
=
1
b
=
c.
b)
(a
hence
—
—
=
b
•
a
if
1
then
a"
hence
(a~
b)
(a
•
•
hence
b
a)
1
•
b
•
b
hence
b
(It may happen that a =
and b =
when we say "either a — or b =
and a
c
a"
•
(a
(a~
=
=
—
—
=
^
0,
c);
a)
c\
c;
= 0,
a b
if
a
then either a
and
=
or b
0. In fact,
a^0,
0;
0;
0;
0.
are both true; this possibility
0"; in
=
mathematics "or"
is
is
not excluded
always used
in the
sense of "one or the other, or both.")
This latter consequence of
P7
is
constantly used in the solution of equations.
Suppose, for example, that a number x
(x
Then
it
follows thai either
x
—
1
-
=
1)(jc
is
known
-2) =
or x
—
2
—
to satisfy
0.
0;
hence
v
=
1
or x
=
2.
Basic Properties ofNumbers
/.
On
the basis of the eight properties listed so far
Listing the next property,
little.
it
is still
7
possible to prove very
which combines the operations of addition and
multiplication, will alter this situation drastically.
(P9)
If a, b,
and c are any numbers, then
a
(Notice that the equation (b
As an example of the
b
+ c) = a
(b
•
+ c)
usefulness of
P9 we
b
a
will
+a
+c
c.
•
a
is
by
also true,
P8.)
now determine just when
—
a
b
—
— a:
If
then
(a
hence
hence
Consequently
a
a — b =
- b) + b =
a =
a + a =
b
=
—
b
+
(1
•
1)
and therefore
A
=
a
b
•
second use of P9
is
a
— a,
{b - a) + b = b +
b + b — a:
(b + b — a) + a —
•
(1
+
(b
b
-
+
a);
b.
1),
b.
=
the justification of the assertion a
which we have
already made, and even used in a proof on page 6 (can you find where?). This
was not
fact
one of the basic properties, even though no proof was offered
listed as
when it was first mentioned. With P1-P8 alone a proof was not possible,
number appears only in P2 and P3, which concern addition, while the
in question involves multiplication.
not obvious:
We
With P9 the proof
we have
sides) that
A
series
already noted,
a
=
this
this,
more
help explain the somewhat myste-
two negative numbers
is
To begin
positive.
(—a)
b
=
—(a
with,
To
b).
note that
b
+a
follows immediately (by adding
Now
may
easily acceptable assertion that
(-a)
It
immediately implies (by adding —(a -0) to both
of further consequences of P9
will establish the
prove
a-0;
0.
rious rule that the product of
we
assertion
simple, though perhaps
have
=
as
is
since the
•
b
—(a
=
=
=
[(-a)
+ a]
•
b
0b
0.
b) to both sides) that
(— a) b
•
note that
(-a)
(-b)
+
[-{a
b)\
= (-a) (-b) + (-a)
= (-a)-[(-b) + b]
= (-a)-0
= 0.
b
= — (a
•
b).
8
Prologue
Consequently, adding (a
we
b) to both sides,
=a
(-a)- (-b)
The
fact that the
obtain
b.
product of two negative numbers
of P1-P9. In other words,
if we
want PI
P9
to
positive
is
is
thus a consequence
be true, the rule for the product of two
to
negative numbers is forced upon us.
The
various consequences of P9 examined so
portant,
do not
although interesting and im-
far,
really indicate the significance of P9; after
each of these properties separately. Actually, P9
is
all,
we could have
we have shown how
algebraic manipulations. For example, although
listed
the justification for almost
all
to solve the
equation
(xwe can
0,
hardly expect to be presented with an equation in
likely to
this
form.
We
are
more
be confronted with the equation
x
The
-2) =
l)(x
"factorization"
(a-
-
1)
•
x
(x
—
-
3.x:
2)
2
-
+2=
=
x
—
(jc
(.v
•
3.x
-
+
=
2
—
l)(.v
2)
+
0.
2)
(-1)
is
really a triple use of P9:
-
(x
2)
= jc.jc+x-(-2) + (-1).jc +
= 2 +a-[(-2) + (-1)] + 2
= x 2 - 3x + 2.
(-1).(-2)
a-
A final illustration of the importance of P9
is
the fact that this property
is
actually
used every time one multiplies arabic numerals. For example, the calculation
13
x24
52
26
312
is
a concise arrangement for the following equations:
13-24=
13
=
=
13
(Note that moving 26 to the
26
•
10.)
The
left in
multiplication 13
•
4
13-4
•
(2
2-
26- 10
10
+ 4)
10+ 13-4
+ 52.
the above calculation
=
52 uses P9
also:
=
10 + 3) -4
=
10-4 + 3-4
4=
10+ 12
= 4- 10+
10 +
= (4+ 1)- 10 + 2
= 5- 10 + 2
= 52.
•
l
.
I
•
2
is
the
same
as writing
Basic Properties of Numbers
1.
The
ber,
list
properties
P1-P9 have
descriptive
names which
We
but which are often convenient for reference.
P1-P9
properties
together and indicate the
are not essential to
9
remem-
opportunity to
will take this
names by which they
are
commonly
designated.
law for addition)
(PI)
(Associative
(P2)
(Existence of an additive
+
+
a
a
(b
+ c) — (a + b) + c.
= + a — a.
identity)
(P3)
(Existence of additive inverses)
a
(P4)
(Commutative law
a
(P5)
(Associative
for addition)
law for multiplica-
+ (—a) — (—a) + a — 0.
+ b = b + a.
(b c) = (a b) c.
a
tion)
a
(Existence of a multiplicative
(P6)
•
1
=
a
= a;
—
a~
a
b
a.
1
•
1/0.
identity)
(Existence of multiplicative
(P7)
a
a~
a
b
x
=
1
,
for a
^ 0.
inverses)
(Commutative law
(P8)
for multi-
—
plication)
The
a
(Distributive law)
(P9)
•
(b
+ c) =
a
b
+
a
c.
three basic properties of numbers which remain to be listed are concerned
with inequalities. Although inequalities occur rarely in elementary mathematics,
they play a prominent role in calculus.
(a is less
than
b)
and a > b
(a is
The two
greater than
notions of inequality, a
are intimately related: a
b),
<
<
b
b
means the same as b > a (thus 1 < 3 and 3 > 1 are merely two ways of writing
the same assertion). The numbers a satisfying a >
are called positive, while
those numbers a satisfying a <
are called negative. While positivity can thus
be defined in terms of <, it is possible to reverse the procedure: a < b can be
defined to
mean
collection of
all
all
—
that b
a
is
positive.
In
fact,
numbers, denoted by P,
positive
it
is
convenient to consider the
as the basic concept,
and
properties in terms of P:
(P10)
(Trichotomy law) For every number a, one and only one of the
following holds:
(PI
1)
(PI 2)
(i)
a
=
(ii)
a
is
(iii)
—a
0,
in the collection P,
is
in the collection P.
(Closure under addition) If a and b are in P, then a
+
b
is
(Closure under multiplication) If a and b are in P, then a
in P.
in P.
b
is
state
10
Prologue
These three properties should be complemented with the following
> b
a < b
a > b
a < b
a
>
Note, in particular, that a
a
—b
if
b
>
a;
if
a
>
b or a
if
a
< b
if
if
and only
in P;
is
or a
a
if
definitions:
is
= b;
= b*
in P.
however elementary they may seem, are
consequences of P10-P12. For example, if a and b are any two numbers, then
precisely one of the following holds:
All the familiar facts about inequalities,
Using the
A
<
a
a-b = 0,
(ii)
a
(iii)
—(a
definitions just
more
slightly
b
b, so that
then a
(i)
+c <
—
b
made,
a
if
in the collection P,
Similarly,
a
in the collection P.
is
(i)
a
=
(ii)
a
(iii)
b
>
>
b,
b,
a.
from the following manipulations.
+ c) —
and
c
so
c
—a=
b
is
in P,
is
in P,
<
— b) +
(c
c,
then a
(b
<
c.
— a)
<
The
ab >
following assertion
The
0.
The symbol
Similarly
ab >
c almost built
c.
is
in P\ thus, if
is
If
<
a
b,
Then
in P.
(The two inequalities a < b and
< b <
c,
obvious: If a
<
b < c are usually written in the abbreviated form a
inequality a
+ c)
{a
suppose a < b and b <
—a
—b
< b and
one of the following holds:
follows that precisely
it
then surely (b
in P,
is
b
This shows that
is
— b) = b — a
interesting fact results
+ c.
b
—
which has the third
in.)
somewhat
is
less
and b <
0,
then
only difficulty presented by the proof is the unraveling of definitions.
<
a
—b
> a, which means — a = —a is in P.
definition,
and consequently, by P12, (— a){— b) — ab is in P. Thus
means, by
in P,
is
0.
The
ab >
b
>
in
our
has one
and also if a < 0, Z? <
>
if a / 0.
Thus squares of nonzero numbers are
special consequence: a
always positive, and in particular we have proved a result which might have seemed
fact that
if
a
>
0,
2
sufficiently
*
There
is
elementary to be included
one
slightly
1
are both true, even though
ond. This
of the symbol
and
b,
soi
is
i
<>f
ihin«
<.
;
a
statemenl
like
Theorem
while inequality holds for other values.
1
>
(since
1
=
I
).
The statements
+ 1 <3
+ 1 <2
we know that < could be replaced l>\
is used with
hound to occur when
is
revealed by
of properties:
> and
perplexing feature of the symbols
1
e<
list
I
in
the
specific
first,
and
l>\
=
in
the
numbers; the usefulness
here equalitj holds for some values of a
Basic Properties of Numbers
1.
The
fact that
—a >
extremely important role
value
a <
if
Note that
=
number
we
a,
define the
absolute
of a as follows:
\a\
—
\a\
3, |7|
the basis of a concept which will play an
is
book. For any
in this
11
\a\ is
7,
\\
a,
a
>
-a,
a
<
0.
when a =
0.
always positive, except
+ y/2-V3\ = 1+V2-V3,
and
\\
we have — 3| =
= v/T5-v/ 2-l.
For example,
+ V2 -
>/l0|
|
In general, the most straightforward approach to any problem involving absolute
values requires treating several cases separately, since absolute values are defined
by cases
important
THEOREM
1
For
all
This approach
to begin with.
fact
We
numbers a and
will
be used to prove the following very
about absolute values.
b,
we have
\a
PROOF
may
+ b\ <
\a\
+
\b\.
consider 4 cases:
(3)
a>0,
a<0,
(4)
a
(2)
In case
(1)
we
have a
also
+b >
b>0
b<0
b>0
b<0
>0,
a
(1)
<0,
and the theorem
0,
+ b\ = a + b =
\a
\a
\
+
is
obvious; in fact,
\b\,
so that in this case equality holds.
In case
(4)
we have
\a
In case
(2),
when
a
+b<
+b\
a
>
=
0,
and again equality
-(a +b)
and b <
\a
may
This case
= -a +
(-b)
holds:
=
we must prove
0,
+ b\ <
a
—
\a\
+
\b\.
that
b.
therefore be divided into two subcases. If a
+b >
0,
then
we must
prove that
a
+ b < a — b,
b
i.e.,
which
a
certainly true since b
is
+b <
0,
we must prove
<
and hence —b >
a
i.e.,
is
On
the other hand,
if
<
<
a
-
b,
a.
> and hence —a < 0.
may be disposed of with no
certainly true since a
Finally,
ing case
0.
that
b
which
< -b,
note that case
(2)
with a and
(3)
/;
interchanged.
|
additional work, by apply-
1
2
Prologue
Although
ious cases)
this
is
method of treating absolute
sometimes the only approach
may be
ods which
Theorem
1
In
used.
fact,
possible to give a
is
it
Ifll
=
Vfl 2
defined only
is
much
shorter proof of
.
(Here, and throughout the book, sfx denotes the
symbol
meth-
proof is motivated by the observation that
this
;
values (separate consideration of var-
available, there are often simpler
(\a
when x >
We may now
0.)
+ b\) 2 = (a+ b) 2 =
positive
square root of x;
this
observe that
2
+ lab + b 2
< a 2 + 2\a\- \b\ + b 2
= \a\ 2 + 2\a\ \b\ + \b\ 2
= (\a + \b\) 2
a
\
From
we can conclude
this
that
+b\ <
\a
+
\a\
\b\
One
close
observation
final
may
be
made about
a and b have the same sign
the two
+ b\ =
0,
is
(i.e.,
a and b are of opposite
We
will
inclusion
conclude
is
<
implies x
y,
to the reader
we have just
proved: a
+
\a\
1^1
are both positive or both negative), or
if
one of
this
+b\ <
+
\a\
\b\
signs.
chapter with a subtle point, neglected until now, whose
required in a conscientious survey of the properties of numbers. After
stating property P9,
by
v
is left
while
\a
if
the theorem
fact
examination of either proof offered shows that
\a
if
<
because x
provided that x and y are both nonnegative; a proof of this
(Problem 5).
we proved
that a
—b—
b
—a
implies a
=
b.
The proof began
establishing that
fl.(l
from which we concluded that a
+
=
+
+l) =
=
b.
fc.(l
+
l)
This result
>
obtained from the equation
is
by dividing both sides by
+ 1. Division by should
it must therefore be admitted that the validity of the
argument depends on knowing that 1 + / 0. Problem 25 is designed to convince
you that this fact cannot possibly be proved from properties P1-P9 alone! Once
P10, Pll, and PI 2 are available, however, the proof is very simple: We have
already seen that 1 > 0; it follows that 1 + 1 > 0, and in particular
+ 1^0.
a
-
(1
1)
b
•
(1
1)
1
be avoided scrupulously, and
1
1
This
last
demonstration has perhaps only strengthened your feeling that
absurd to bother proving such obvious
facts,
present situation will help justify serious consideration of such details.
chapter
we have assumed
merely
explicit statements
be
difficult,
however, to
that
numbers are
familiar objects,
deal of this
book
is
and
that PI
-PI 2 are
It
how
to
justify this
devoted
assumption. Although one learns
to elucidating the
is
In this
of obvious, well-known properties of numbers.
with" numbers in school, just what numbers
it
but an honest assessment of our
are,
remains rather vague.
would
"work
A
great
concept of numbers, and by the end
1
we
of the book
.
have become quite well acquainted with them. But
will
necessary to work with numbers throughout the book.
It is
way numbers
say that, in whatever
properties PI
Most of
it
will
be
therefore reasonable
admit frankly that we do not yet thoroughly understand numbers; we
to
13
Basic Properties of Numbers
may
still
are finally defined, they should certainly have
PI 2.
chapter has been an attempt to present convincing evidence that
this
we should assume
of numbers. Some of the problems
deduce
PI PI 2 are indeed basic properties which
in order to
other familiar properties
(which indicate the
numbers from PI -PI 2) are offered as further evidence. It is still a crucial question whether PI -PI 2 actually account for all properties of numbers. As a matter of fact, we shall soon see that they do not. In the
next chapter the deficiencies of properties PI -PI 2 will 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
derivation of other facts about
subtle, quite unlike
PI -P12. The discovery of this crucial property
work of Part II of this book. In the remainder of Part I we
why some additional property is required; in order to investigate
to consider a little more carefully what we mean by "numbers."
the
will require all
will
begin to see
this
we
will
have
PROBLEMS
1.
Prove the following:
If
(i)
(ii)
x
2
-
2
=
y
3
y
(v)
x"
(vi)
x3
+
x
3
do
x"
What
is
y
2
—
-
(iv)
2.
=
x
If
(iii)
ax
y"
this,
+
y"
a for
some number
=
-
(x
«^0,
+ y).
y)(x
=
,
then x
y or x
2
).
l
l
).
using
(iv),
and
whenever n
wrong with
is
it
will
show you how
the following "proof"? Let x
(x
2
2
-y
+ y)(x - y)
x
+
v
2y
2
Prove the following:
£
=
b
a
(U)
i
+
~
be
c
l=
if*,c#0.
ad
+ be
.
.
-M--* b -**°-
= xy,
= xy - y 2
= y(x — y),
= y,
= y,
,
=
1.
easy
way
to
to find a factorization for
odd.)
x
(i)
1.
= —y.
= (x — y)(x + xy + y 2
~
= (x - y)(x n ~ + x n ~ 2 y +
+ xy"~ 2 + y n ).
= (x + y)(x — xy + y 2 (There is a particularly
y
2
x
3.
=
then x
=
y.
Then
1
14
Prologue
=
1
(ab)
(iii)
l
a
[
b
if
,
/
a,b
rrs- fM "
a
(iv)
(To do
0.
this
you must remember the
l
defining property of (ab)~
.)
ac
c
ft
£-g,if». c .^o.
5/
4.
Find
a
b
b
a
all
numbers x
for
a
—
c
=—
b
d
if
(ix)
(;c-7z-)(jf+ 5) (jc
(x)
(jc
-
(xi)
2*
<
(xii)
x
(xiii)
-+
(iii)
(iv)
(v)
(vi)
(vii)
if
=
ad
&c. Also
determine when
is
a product of two numbers positive?)
jc
1
+
-3) >
- V2) >
l/2)(x
0.
0.
8.
3'
x
(xiv)
and only
which
(viii)
(ii)
1
<
—
4.
>
0.
x
i^l > o.
x
+
Prove the following:
(i)
If
(ii)
If
(iii)
If
(iv)
If
(v)
If
(vi)
If
(vii)
If
(viii)
If
(ix)
(x)
6.
then
0,
4-jc<3-2jc.
5 - x 2 < 8.
5 - x 2 < -2.
(x — l)(x — 3) > 0. (When
2
x - 2x + 2 > 0.
2
x + x + 1 > 2.
2
-x + 10> 16.
2
> 0.
x +x +
(i)
5.
^
b,d
\{
(vi)
(a)
If
<
a <
a <
fl <
a <
a >
<
<
<
a
If a,
/?
fr,
and
c
<
d, then a
£ and c
b and c
>
<
then a
b
—
<
if
x
then
/?c.
a.
a
2
<
a
2
<
< b
<
2
,
if
Prove that
if
x"
.
(Use
then a
is
y" and n
y"
and
ac < bd.
d, then
2
y, then x"
«
d.
a.
<
c
< &
Prove that
if jc"
<
ac >
>
(c)
=
=
<
c
0,
(b)
Prove that
—
/?c.
x < y and
(d)
+ d.
then ac
0,
< 1, then
<
Z? and
fl
a < b, then
b >
and a 2
a
Prove that
b
—b<—a.
then
b and c > d, then a
1,
+c <
/?
(viii).)
<
<
b.
(Use
y",
n
=
odd, then x"
<
is
odd, then x
is
even, then x
(ix),
backwards.)
1, 2,
3
y".
= y.
=y
or
.v
= — v.
.
1.
7.
Prove that
< a <
if
b,
Basic Properties of Numbers
15
then
>
a
— —+—b <
a
< V ab <
-
b.
2
Notice that the inequality -Jab
alization of this fact occurs in
*8.
<
(a
+ b)/2
holds for
P
of
positive
all
>
0.
A gener-
Problem 2-22.
Although the basic properties of inequalities were stated
lection
a, b
all
numbers, and < was defined
in
in
terms of the
col-
terms of P,
this
procedure can be reversed. Suppose that P10-P12 are replaced by
For any numbers a and b one, and only one, of the
(P'10)
following holds:
(P'l
(i)
a
=
(ii)
a
(iii)
b
<
<
b,
b,
a.
For any numbers a,
1)
<
a
a
+c<
b
c, if
a
<
b and b
b,
and
c, if
a
<
b,
b,
and
c, if
a
< b and
<
c,
then
<
c,
then
then
+ c.
For any numbers a,
(P'l 3)
and
c.
For any numbers a,
(P'12)
b,
ac < be.
Show
9.
that
P10-P12 can then be deduced
Express each of the following with
as theorems.
at least
one
less
pair of absolute value
signs.
(ii)
\V2 + y/3-V5 + y/7\.
\(\a+b\-\a\-\b\)\.
(iii)
\(\a+b\
(i)
+ \c\ - \a + b + c\)\.
\x
-2xy + y 2
|(|V2 + V3|-|a/5-a/7|)|.
2
(iv)
(v)
10.
\.
Express each of the following without absolute value
cases separately
1 1
when
necessary.
+ b\-\b\.
(i)
\a
(ii)
|(|*|-D|.
(iii)
(iv)
l*|-|*
fl -|(fl-| fl |)|.
Find
all
2
|-
numbers x
- 31 =
(i)
|x
(ii)
\x
(iii)
\x
(iv)
|jc-
(v)
|jr
-3| <
+4\ <
-
1|
1
1
+
+
for
which
8.
8.
2.
|jc
\x
-2| >
+ <
1
1
1.
2.
signs, treating
various
.
1
6
Prologue
12.
(vi)
|jc-1|
(vii)
\x
-
1
(viii)
\x
—
II
+
1
+ 1|<1.
+ 1 = 0.
+21 = 3.
|jc
•
\x
•
\x
1
Prove the following:
\xy\
(i)
=
\x
=
—
x
if
,
(The best way
7^ 0.
do
to
this
is
to
remember what
\x\
is.
in
if
y
^
\x\
+
\y\.
,
— y\ <
\x
(iv)
v)
(
\
x —
\
—
\y\
\
x
0.
(Give a very short proof.)
(A very short proof is possible,
~~ y\-
if
you write things
in
the right way.)
(vi)
|(|.v|
—
\y\)\
+
y
+
\x
(vii)
z\
<
<
—
\x
\x\
(Why does
y\.
+
+
\y\
follow immediately from
this
when
Indicate
\z\.
equality holds,
(v)?)
and prove
your statement.
13.
The maximum of two numbers x and y is denoted by max(x,y). Thus
max(— 1,3) = max(3, 3) = 3 and max(— 1,— 4) — max(— 4, — 1) = — 1.
The minimum of x and y is denoted by min(x, y). Prove that
max(x, y)
=
min(jc, y)
=
x
+
v
+
-
I
v
—
v
— x\
x
I
L
Derive a formula for max(x, y,
max(x,
14.
(a)
Prove that
First
cases.
a
^15.
<
\a\
=
\—a\.
y, z)
(The
—b <
a
that
— |a| <
(c)
Use
this fact to give
Prove that
if
a
<
Use Problem
Show
(x
,
trick
is
not to
>
example
z)).
become confused by
0.
Why
is
it
too
many
then obvious for
<
b
a
new proof that
if
and only
if \a\
<
b.
In particular,
|a|.
4
+ x3y
0,
\a
+ b\ <
|«|
+
\b\.
then
2
+ xy + y 2 > 0.
+ x 2 y 2 + .vy 3 + y 4 >
0.
1
that
+ y) = x n + y
o
+ y)~ = x + y
t
(x
max(x max(y,
x and y are not both
x
(a)
y,z), using, for
0?)
Prove that
x
46.
=
prove the statement for a
(b)
Hint:
and min(x,
z)
when
only when
only
x
.v
=
—
or y
or y
=
=
0,
or x
=
—v.
it
follows
1.
(b)
Using the
fact that
2
x
show
(c)
(d)
that
Use part
4x 2
0,
+ 6xy + 4v 2 >
when
find out
(b) to
when
(x
You should now be
the proof
is
+ y) 2 >
(x
unless x
(x
+
4
y)
0,
and y are both
4
=x +y
0.
4
+ y) =
5
a
5
+v
make
able to
5
.
.
From
Hint:
the assumption
a good guess as to
when
(x
(x+y)
+ y) n —
+ y";
x"
contained in Problem 11-63.
(b)
Find the
(c)
Find the
— 3x + 4. Hint: "Complete
- 3x + 4 = 2(x - 3/4) 2 + ?
i.e., write 2x
2
smallest possible value of x — 3x + 2y + 4y + 2.
smallest possible value of x + Axy + 5y — Ax — 6y + 7.
(a)
Suppose
that b~
(a)
Find the smallest possible value of 2x 2
the
2
square,"
18.
+ 2xy + y 2 =
=
+y you should be able to derive the equation x +2x y+2xy +y —
3
= x 2 y + xy 2 — xyix + y).
if xy / 0. This implies that (x + y)
Find out
x
17.
17
Basic Properties of Numbers
—
-b +
Ac >
y/b 2
0.
Show
-b -
- Ac
numbers
that the
yjb 2
- Ac
'
2
both
(b)
2
+ bx + c =
equation x
satisfy the
2
Suppose that b
x 2 + bx + c = 0;
—
Ac <
Show
0.
in fact,
x~
0.
no numbers x
that there are
+ bx +
c.
>
for
all
satisfying
Complete the
x. Hint:
square.
(c)
Use
x
(d)
2
another proof that
this fact to give
+ xy + y >
if
x and y are not both
0,
then
0.
For which numbers a
is it
true that x~
+ axy + y >
whenever x and
y are not both 0?
(e)
Find the smallest possible value of x"
The
fact that
a2
nevertheless the
ties
and of ax
+ foe + c,
for
>0.
a
19.
+ bx + c
>
numbers a, elementary as it may seem, is
fundamental idea upon which most important inequalifor all
The great-granddaddy of
are ultimately based.
Schwarz
all
inequalities
is
the
inequality:
x\y\+x2y2 < vx\
(A more general form occurs
in
+ xi
vyi +yi
Problem 2-2
1 .)
•
The
Schwarz inequality outlined below have only one thing
on the fact that a 2 > for all a.
three proofs of the
in
common
—
their
reliance
(a)
Prove that
if
x\
=
y2
=
0.
and X2 — Ay? for some number X >
Schwarz inequality. Prove the same thing
Xy\
equality holds in the
Now suppose
that y\
and y2 are not both
0,
and
0,
then
if
\|
=
is
no
that there
18
Prologue
number X such
<
(Xyi
=
^
-x\) 2
2
2
=
that x\
(>'i
+
Using Problem
18,
v2
+
(Xy 2
2
-
)
and x 2
Xy\
-x 2
—
Xy 2 Then
.
2
)
+ x2 y2 +
2X(x iyi
).
complete the proof of the Schwarz inequality.
< x~+y (how is
Prove the Schwarz inequality by using 2xy
(b)
+ x22
(xr
)
derived?)
this
with
x
Xi
=
v
.
,
-
x\ 2
for
first
=
i
2
and then
1
+ x 2 2 )(yi 2 +
(vi
for
=
i
2
V2
=
)
t
2
V
vi
-f-
v2
2
2.
Schwarz inequality by
Prove the
(c)
+ x2
y
=
2
proving that
first
+x 2 y 2 2 +
(x\y\
(x\y 2
)
-
x 2 y\)
2
.
Deduce, from each of these three proofs, that equality holds only when
(d)
yi
x2
= y2 =
= Xy 2
when
or
there
a
is
number
>
A.
such that x\
—
Xy\
and
.
In our later work, three facts about inequalities will be crucial. Although proofs
will
be supplied
problems
proof.
The
+ y will
to 1/yo-
is
on these
enlightening than a perusal of a completely worked-out
very simple:
if
x
close
is
enough
and y
to xo,
is
close
enough
to yo,
be close to *o + >'o, and xy will be close to xoyo, and 1/y will be close
"e" which appears in these propositions is the fifth letter of the
The symbol
Greek alphabet
Roman
more
statements of these propositions involve some weird numbers, but their
basic message
then x
at the appropriate point in the text, a personal assault
infinitely
is
and could just
("epsilon"),
made
however, tradition has
letter;
as well
be replaced by a
less
intimidating
the use of s almost sacrosanct in the
contexts to which these theorems apply.
20.
Prove that
if
£
< -
xq\
and
-
\y
£
y
<
\
j,
then
*21.
Prove that
jc
—
jcqI
\{x
+
y)
\(x
-
y)
-
e
'
£.
if
< min
I
,2(|y
then \xy
- Uo + yo)l <
- (xq -yo)\ <
x
yo\
<
|
+
—
,
l)'
problem
just
means
is
I
and
\y
—
\q\
J
<"
2(|jc
irrelevant at the
in
Problem
moment;
the
13, but the
1)'
first
formula provided by
inequality in the hypothesis
that
Ijc
|+
£.
(The notation "min" was defined
thai
1
— *n| <
7-
^
2(|y
l
+ D
and
|.v
—.vol
<
1;
Basic Properties of Numbers
1.
one point
at
in the
argument you
need the
will
first
and
inequality,
19
at an-
One more word of advice: since the
hypotheses only provide information about x — xq and y — yo, it is almost a
foregone conclusion that the proof will depend upon writing xy — xq\q in a
way that involves x — x$ and y — yo-)
other point you will need the second.
^22.
Prove that
if
yo
^
and
\y
then
-
y
mm /
<
|
I
— —^—
2
£|yol
Lvol
,
v/0 and
1
<
£.
yo
^23.
Replace the question marks
ing
s, xo,
If yo
/
and yo
statement by expressions involv-
and
yol
then
in the following
so that the conclusion will be true:
<
and
?
—
\x
<
xq\
?
_v/0 and
xo
<
£.
yo
This problem
is
trivial in
the sense that
and 22 with almost no work
point
is
be used
*24.
not to
first,
its
from Problems 21
solution follows
at all (notice that
x/y
=
x
1/y).
The
become confused; decide which of the two problems should
and don't panic
your answer looks
if
unlikely.
This problem shows that the actual placement of parentheses in a
irrelevant.
The
Chapter
Thus
a\
(a)
+
«i
(#2
+
if you are not fawant to tackle this problem, it can be saved
where proofs by induction are explained.
2,
(a 2
+
(fl3
h (fl„_2
H
+ ai + «3 denotes
+ («3 + «4)), etc.
+
a\
(«2
+
+
+
•
•
•
+ a„
(a n -\
«3),
will
denote
+ an))) ••)
and
ct\
+
a2
+
«3
Prove that
\-ak )
(a\ H
+a k+ = a\
\
H
hflt+i-
Use induction on k.
Prove that if n > k, then
Hint:
(b)
is
still
Let us agree, for definiteness, that a\
a\
sum
proofs involve "mathematical induction";
miliar with such proofs, but
until after
crucial
(a\ H
Hint: Use part
h
fljfc)
(a)
+
(a k+
to give a
\
H
1-
a„)
= a\
V an
-\
proof by induction on
k.
.
+
«4 denotes
.
20
Prologue
(c)
Let s{a\
.
,
.
.
,
a k ) be some
sum formed from
s(a\, ...,a k )=a\-\
Hint:
There must be two sums
\-
a\,
ak
.
.
,
Show
a^.
that
.
s'(a\, ...,«/)
and
s"(ai + \,
.
.
.
,
a k ) such
that
s(a\, ...,a k )
25.
Suppose
that
we
=
interpret
s'(a\
a
"number"
to
t
)
+ s"{a t+
mean
be the operations defined by the following two
+
,ak ).
\
or
either
1
,
and
tables.
1
1
1
1
1
Check
that properties
PI— P9
all
hold, even though
1
+ 1=0.
+
and
•
to
,
CHAPTER
mm
NUMBERS OF VARIOUS SORTS
In Chapter
we used
1
word "number" very
the
the basic properties of numbers.
now be
It will
our concern with
loosely, despite
necessary to distinguish carefully
various kinds of numbers.
The
numbers are
simplest
the "counting
numbers"
1,2,3
The fundamental
significance of this collection of numbers is emphasized by its
natural numbers). A brief glance at P1-P12 will show that our
basic properties of "numbers" do not apply to N
for example, P2 and P3 do not
make sense for N. From this point of view the system N has many deficiencies.
N
symbol
(for
—
N
Nevertheless,
important to deserve several comments before
sufficiently
is
we
consider larger collections of numbers.
The most
Suppose P(x) means
ciple of
N
basic property of
the principle of "mathematical induction."
is
P
that the property
mathematical induction
number
holds for the
states that
P(x)
true for
is
all
Then
x.
the prin-
numbers x
natural
provided that
(1)
P(l)
(2)
Whenever P{k)
Note that condition
that P(k)
(using
In
P(l)
=
infinite line
k
is
true.
P(k+ 1) under the assumption
Now,
= 2).
it
since P(2)
is
true
it
clear that each
It is
sort, so that
P(k)
then clearly every person
1,
person number
number
.v
largest
tell
2,
condition
is
true
will eventually
numbers
all
person number
any secret he hears
number) and a
secret
is
will eventually learn the secret. If
will learn the secret,
the secret to person
number
1
makes
first
n
to the
.
.
.
be
k.
.
person behind
number
P(x)
is
The
(2) is
true,
(to tell all
and
following example
There is a useful and
numbers in a simple way:
1
the assertion
then the instructions given
(1) true.
facetious use of mathematical induction.
which expresses the sum of the
3,
told to person
secrets learned to the next person) assures that condition
21
if
true (by using
is
follows that P(3)
number
true for
is
x,
all
follows that P(2)
of the reasoning behind mathematical induction envisions
one with the next
that person
then
true,
person has been instructed to
(the
1)
of people,
person number
If each
+
ensure the truth of P(x) for
is
of steps of this
A favorite illustration
him
1).
(2) in the special case
series
P{k
true,
is
merely asserts the truth of
fact, if
the special case k
reached by a
an
true.
true; this suffices to
is
(1) also holds.
(2) in
(2)
is
striking
is
telling
a less
formula
22
Prologue
+
— 2~n(n
+-+» =
!
To prove this formula, note first that
some natural number & we have
it
l)
clearly true for
is
n
=
Now
1.
assume that
for
+ ... +t = **+i>.
i
Then
+
1
•
•
•
+k+
(
k
+
1
)
—
= k(k+ 1)- + k+\
k(k
_
2
so the
formula
nomenon
one
also true for k
is
the formula for
all
natural
+
numbers
+
1)
+ 2k + 2
A:
+ 3* + 2
(*
+
!)(*
+ 2)
\.
By
n.
This particular example
the principle of induction this proves
illustrates
a phe-
that frequently occurs, especially in connection with formulas like the
just proved.
Although the proof by induction
is
often quite straightforward,
method by which the formula was discovered remains a mystery. Problems
and 6 indicate how some formulas of this type may be derived.
the
5
The principle of mathematical induction may be formulated in an equivalent
way without speaking of "properties" of a number, a term which is sufficiently
vague to be eschewed in a mathematical discussion. A more precise formulation
states that if A is any collection (or "set"
a synonymous mathematical term) of
—
natural
numbers and
(2)
then
A
is
the set of
all
k
in
is
1
(1)
+
1
is
A,
in
A whenever
natural numbers.
It
k
is
in
A,
should be clear that
adequately replaces the less formal one given previously
set
A
of natural numbers x which
of natural numbers n for which
in
A,
previous proof of
if
A'
holds for
There
follows that
is.
It
all
natural
is
this
it is
—we
A
is
A
is
the set
true that
__
formula showed that A contains
numbers
just consider the
P(x). For example, suppose
+ ... + „ =
!
Our
satisfy
formulation
this
the set of all natural numbers,
1 ,
and
i.e.,
that k
that the
+
1
is
formula
/;.
yet another rigorous formulation of the principle of mathematical in-
duction, which looks quite different. If
A
is
am
collection of natural
numbers,
it
Numbers of
2.
is
tempting to say that A must have a smallest member. Actually,
can
be true
to
fail
numbers
is
A
is
that has
A
The
0.
—
no smallest member
the only possible exception, however;
then
particularly important
no natural numbers
that contains
denoted by
collection" or "null set,"*
numbers
A
in a rather subtle way.
the collection
null set
in fact,
A
if
is
has no
it
a nonnull
is
23
Various Sorts
at
this
statement
set
of natural
all,
"empty
the
a collection of natural
members
at
This
all.
of natural numbers,
set
has a least member. This "intuitively obvious" statement,
known
as the
"well-ordering principle," can be proved from the principle of induction as follows.
Suppose
that the set
n such that
1,
.
A would have
k
k:
+1
+
1
are
all
.
A
B be the set
in B (because
has no least member. Let
,n are
A. Clearly
all not in
is
1
1
were
in A,
then
,
not in A. This shows that
k
if
then k
in B,
is
number n is in B, i.e., the numbers 1
number n. Thus A = 0, which completes the proof.
is
if
(otherwise k
that every
It
of natural numbers
member). Moreover, if I, ... ,k are not in A, surely
+ 1 would be the smallest member of A), so 1, ...
as smallest
1
not in
is
.
A
,
.
.
.
,
may be
is
1
n are not in
also possible to prove the principle of induction
principle (Problem 10). Either principle
+
A
in B.
for
follows
It
any natural
from the well-ordering
considered as a basic assumption
about the natural numbers.
There
is still
another form of induction which should be mentioned.
+
times happens that in order to prove P(k
also P(l) for all natural
numbers
complete induction":
A
then
A
the set of
is
If
is
(1)
1
is
(2)
k
+
:
<
I
In
k.
in A,
is
1
A
in
if
1
,
.
.
.
,
be found
in
fact
k are in A,
natural numbers.
all
Although the principle of complete induction
will
some-
numbers and
a set of natural
the ordinary principle of induction,
The proof of this
It
we must assume not only P(k), but
this case we rely on the "principle of
1)
is left
Problems
may appear much
to the reader, with a hint
7, 17,
stronger than
actually a consequence of that principle.
it is
(Problem
Applications
11).
20 and 22.
Closely related to proofs by induction are "recursive definitions." For example,
the
number
numbers
less
n\ (read
"n factorial")
than or equal to
n:
n\
=
1
•
defined as the product of
is
2
•
.
.
.
•
(n
—
1)
•
all
the natural
n.
This can be expressed more precisely as follows:
(1)
1!
=
(2)
n\
=n
1
{n
-
1)!.
This form of the definition exhibits the relationship between
*
Although
n\
and
(n
—
1
)!
in
an
may not strike you as a collection, in the ordinary sense of the word, the null set arises
many contexts. We frequently consider the set A, consisting of all x satisfying some
often we have no guarantee that P is satisfied by any number, so that A might be
in
it
quite naturally in
property P;
fact often
one proves that P
—
is
always
false
by showing that A
=
0.
24
Prologue
explicit
way
that
is
cursive definition; as this
for a rigorous
One
which
may be
problem shows, the recursive
which
will constantly
may
be
will usually
employ
the
more
definition
succinctly as a reis
really necessary
definition.
not be familiar involves some convenient notation
of writing
using. Instead
a\ H
we
expressed
proof of some of the basic properties of the
definition
which we
by induction. Problem 23 reviews a
ideally suited for proofs
definition already familiar to you,
Greek
\-a n
letter
£
,
(capital sigma, for
"sum") and write
n
In other words,
i
=
denotes the
/~Jfl/
sum of
the
numbers obtained by
letting
Thus
1,2, ... ,n.
n(n +
A. = + „2+---+n = —
Y,i
Y-.
\)
-
l
/=i
n
Notice that the
letter
i
really has
nothing to do with the
number denoted by
>_,
>
>
i=\
and can be replaced by any convenient symbol (except
A
n(n+
.
n, of course!):
1)
2
+
h=
iii
t" =
7(7
l)
'
2
+
1)
n
To
define
/#/
precisely really requires a recursive definition:
(1)
^2a =ai,
t
n
(2)
n
-i
Y< a = J2 ai+an
i
-
/=]
But only purveyors of mathematical austerity would
precision.
In practice,
no one ever considers
all
it
too strongly on such
symbolism arc used, and
add any words of explanation. The symbol
sorts of modifications
necessary to
insist
of
this
,
»
Numbers of
2.
Various Sorts
25
I]«"
7=1
i#4
example,
for
an obvious way of writing
is
a
or
more
The
of
this
+ (12 + A3 + « 5 + «6
i
+ ««
H
precisely,
deficiencies of the natural
chapter
may
3
n
/=1
;=5
numbers which we discovered
be partially remedied by extending
at the
beginning
system to the
this
set
of
integers
...,-2,-1,0, 1,2
This
set
is
only P7
A
denoted by
fails
Z
(from
German
"Zahl," number).
Of properties P1-P12,
for Z.
numbers is obtained by taking quotients m/n of integers
(with n / 0). These numbers are called rational numbers, and the set of all
rational numbers is denoted by Q_ (for "quotients"). In this system of numbers all
of P1-P12 are true. It is tempting to conclude that the "properties of numbers,"
which we studied in some detail in Chapter 1, refer to just one set of numbers,
namely, Q_. There is, however, a still larger collection of numbers to which properties P1-P12 apply
the set of all real numbers, denoted by R. The real numbers
include not only the rational numbers, but other numbers as well (the irrational
larger system of
still
—
numbers) which can be
examples of
shall
of v
devote
2,
all
represented by
on the other hand,
is
quite simple,
has a hypotenuse of length v 2,
investigated this question.)
irrational
it
and was known
is
The
fact.
are both
not easy
—we
irrationality
to the Greeks. (Since the
isosceles right triangle, with sides of length
is
1
not surprising that the Greeks should have
The proof depends on
number n can be
natural numbers. Every natural
some
is
of Chapter 16 of Part III to a proof of this
Pythagorean theorem shows that an
n and v 2
infinite decimals;
numbers. The proof that n
irrational
+
a few observations about the
written either in the
form 2k
some integer k (this "obvious"
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:
for
fact
integer k, or else in the
(2k)
(2k
In particular
if
n
form 2k
1
for
has a simple proof by induction (Problem
is
it
that
l)
2
=
=
4k
4k
2
=
2- (2k 2 ),
2
+4k +1=2-
follows that the converse
odd, then n
Suppose
+
2
is
odd.
v 2 were
must
is,
2
also hold:
The proof that v2
rational; that
(2k
is
+
2k)
if
n~
+
is
irrational
1.
even, then n
is
now
is
even;
quite simple.
suppose there were natural numbers p
26
Prologue
and q such
that
We
can assume that p and <y have no common divisor
could be divided out to begin with). Now we have
o
2
This shows that p
some
natural
even,
is
number
k.
common
(since all
divisors
2
and consequently p must be even;
that
is,
p
—
2k for
Then
p
2
=
Ak
=
2
2q
2
,
so
T>2
2A:
=
q
2
.
and consequently that q is even. Thus both p and g
are even, contradicting the fact that p and q have no common divisor. This
This shows that g
is
even,
contradiction completes the proof.
It is
important
to
strated that there
expressed
more
is
understand precisely what
no
number x such
rational
briefly
by saying
that
V2
is
this
We
proof shows.
=
that x
2.
irrational.
have demon-
This assertion
is
often
Note, however, that the
number (necessarily irrational)
whose square is 2. We have not proved that such a number exists and we can assert confidently that, at present, a proof is impossible for us. Any proof at this stage
would have to be based on PI -PI 2 (the only properties of R we have mentioned);
v2
use of the symbol
since
is
PI -PI 2 are
a rational
reverse
square
also true for
Q the exact same argument would show that there
number whose square
argument
is
implies the existence of some
will
not work
—
is
and
2,
this
we know
our proof that there
2 cannot be used to show that there
is
no
is
real
is
(Note that the
false.
no rational number whose
number whose square is 2,
because our proof used not only PI PI 2 but also a special property of Q_, the
that every number in
can be written p/q for integers p and q.)
fact
Q
This particular deficiency in our
of course, be corrected by adding a
list
of properties of the real numbers could,
new property which
asserts the existence
square roots of positive numbers. Resorting to such a measure
aesthetically pleasing
nor mathematically
satisfactory;
is,
we would
of
however, neither
know that
number has an
still
not
number has an nth root if n is odd, and that every positive
nth root if n is even. Even if we assumed this, we could not prove the existence of
=0 (even though there does happen to be one),
a number x satisfying x 5 + x +
since we do not know how to write the solution of the equation in terms of nlh
roots (in fact, it is known that the solution cannot be written in this form). And,
of course, we certainly do not wish to assume that all equations have solutions,
2
=0, for example). In fact,
since this is false (no real number x satisfies x +
every
1
1
this
direction of investigation
of elucidating
order to study
this
tin-
R
is
not a fruitful one.
The most useful hints about the
most compelling evidence for the necessity
Q,
property, do not come from the study of numbers alone. In
properties of the real numbers in a more profound way, we
property distinguishing
from
the
Numbers of
2.
must study more than the
real
numbers. At
we must begin
point
this
Various Sorts
27
with the
foundations of calculus, in particular the fundamental concept on which calculus
is
based
—
functions.
PROBLEMS
1.
Prove the following formulas by induction.
(i)
l-
(ii)
l
+ -..+/r =
+ --- + 3 = (l+--- +
3
2.
.
2
77)
/7
.
Find a formula for
J2 {2i ~
(i)
= 1+3 + 5 +
1)
+
•••
(2,7-
1).
/=]
Hint:
2
l
3.
2
<
k
3
<
+
---
+
fc!(n
7?\
/n
0/
\n'
for k
+ 32 +
(2«)
n{n
-
2
2
+
•
•
•
+
-
(2/7
-
.
do with
1)
•
-
{n
•
k
I
+
I
is
1
+
2
+
3
+
•
•
•
+
2n and
defined by
1)
ll
,
1
2
l)
?
*)!
=
<
5
"binomial coefficient"
n\
£/
2
l
these expressions have to
n, the
n\
and
2
=
2
1 )
What do
+2 +
If
(a)
-
]T(2i
(ii)
/C
7^
(J, 77
it!
(a special case
or k
>
n
we just
of the
first
formula
if
we
define 0!
=
1),
define the binomial coefficient to be 0.
Prove that
(The proof does not require an induction argument.)
This relation gives
cal's triangle"
rise to the
following configuration,
— a number not on one of the
numbers above
it;
the binomial coefficient
sides
is
I
J
in the
(77
+
l)st row.
1
1
1
1
13
14
1
5
10
2
1
3
1
4
6
10
1
5
1
is
the
the (k
known as "Passum of the two
+
l)st
number
28
Prologue
(b)
Notice that
part
(a)
entire
to
all
the
numbers
in Pascal's triangle are natural
prove by induction that
proof by induction
will, in
I
I
is
numbers. Use
always a natural number. (Your
a sense, be
summed up
in a glance
by
Pascal's triangle.)
(n\
(c)
Give another proof that
a natural
is
I
number by showing
that
J
...
(d)
,
n.
Prove the "binomial theorem": If a and b are any numbers and n
natural number, then
'"r
b+
W
-E("V
;=0
(e)
J,
Prove that
1
7=0
(.)
:3-G)-<»- a
D-iy(-)
;:
;=o
in
£C)-GH>-^
iv
/
4.
(a)
eC-C
even
+
G)
+ -- 2 «-]
Prove that
ti\k)\i-k)-\
Hint:
(b)
Apply the binomial theorem
to (1 +.v)"(l +.v)'
Prove that
n\
k=0
i
i
In
is
a
2.
5.
(a)
29
+ r + r" +
•
l_ r »+l
+ r" =
.2
•
1
^
r
if
=
The formula
sum
evaluating the
1,
by
this result
and
r,
r
(if
1
Derive
by
6.
Various Sorts
Prove by induction on n that
1
(b)
Numbers of
1
r
certainly presents
— +r+
S
setting
—
•
•
+ r"
no problem).
multiplying this equation
,
solving the two equations for S.
for
+
1
•
—
may
\-n
be derived as follows.
We
begin with the
formula
(k+
Writing
formula for k
this
2
3
33
3
(n
+
l)
n
+
1)3
(
Thus we can
-
(i)
l
(ii)
l
_
2
=
=
3
-k 3 =3k 2 + 3k+
=
33
3
•
1,
2
.
2
•
_
=
1
>, ^
find
+
3[i2
if
we
Use
.
.
.
1
+
1
2
+
1
+
,7
already
we
obtain
1
2j
+ 3 [! +
.
.
know 2, k
method
this
1.
n and adding,
,
+ 3+3
l
2
.
.
-/r3 = 3-n 2 + 3-n +
.
+n + „,
]
(which could have been
k=\
to find
+ .-. + n 3
4
+
+ n\
3
.
•
•
•
—— + —— +
1-2
2-3
1
...
111
1
1
.
•
•
•
+
n(n
2
2
22
•
+
\)'
In
5
3
12
^7.
3
l
k=\
in a similar way).
found
l)
3
2
2
n («
Use the method of Problem 6
to
+
+
1
l)
show
2
that /_^' ;
'
can always be written
/=l
the
form
P
(The
first
+
+ An p + Bn p
l
10 such expressions are
-
1
+Cn p - 2 +
---
.
in
30
Prologue
£* = v
+
\n
k=\
n
E*
2
k=]
E*
3
k=\
4
E*
w+
=i» + v +
= K + v +w
=
3
+
l»
i»
4
2
\»
-
30
n
k=\
±*
5
=y
+
5
1»
+ n»
4
-
-±n
k=\
kb
J2
k=\
X>
7
=W +V
=
+V
i„ 8
+I"
5
~b' + A"
+ h» 6 -
4
T4"
2
+
T2"
k=\
+v+i«
£*•-*»»
7
-^» 5 +i« 3 -a»
A?
^
E/.9
=
1
„ 10
To"
9
1
_,
3
M8
+
2
+
+
W
+l»
/7
4
;/
7„6,
~
To"
-
1
1„4
H
+
2
3
~
20
„2
n
k=\
]0
=Ti»
J2><
n
9
"
7
+
1
"
5
~V
+W».
*=1
Notice that the coefficients in the second column are always
the third
n 2 or n
column
is
the
reached.
powers of n with nonzero
The
coefficients in all but the first
and
that after
by 2 until
two columns seem to
some sort of pattern; finding it may
See Problem 27-17 for the complete
be rather haphazard, but there actually
be regarded as a super-perspicacity
i,
coefficients decrease
is
test.
story.)
8.
Prove that every natural number
9.
Prove that
whenever
10.
if
it
a
set
A
is
either even or odd.
of natural numbers contains hq and contains k
contains k, then
A
contains
all
natural
numbers >
+
1
hq.
Prove the principle of mathematical induction from the well-ordering principle.
11.
Prove the principle of complete induction from the ordinary principle of
induction. Hint: If
/?,
1
12.
(a)
A
contains
1
B
all
consider the set
and b
of
and A contains
k such that
If
a
if
a and b arc both irrational?
is
rational
is
irrational,
is
a
+
1
//
+
1
whenever
k are
all in
it
contains
A.
b necessarily irrational?
What
3
2.
If
a
(c)
Is
there a
(d)
Are there two
(b)
and b
rational
is
number
irrational,
is
a such that a
ab
is
Numbers of
necessarily irrational? (Careful!)
irrational,
is
31
Various Sorts
but a
rational?
is
numbers whose sum and product are both
irrational
ra-
tional?
13.
Prove that V3, v
(a)
and
5,
Why
14.
15.
16.
v2 and V3
is
treat
of the form 3n or 3«
V 3, for exam+ or 3« + 2.
1
v 4?
are irrational.
Prove that
(a)
V2 + v 6
is
irrational.
(b)
v2
+ V3
is
irrational.
(a)
Prove that
=
p
number, then x m
=
x
if
(b)
Prove also that
(a)
Prove that
(m
+ In)
if
/(m
(/?
+ ^fq where p and # are rational,
a + b^/q for some rational a and
m
-
m and
n are natural
+ n) >
2;
—
+ 2n)-~2
(m + n)"
(b)
Prove the same results with
(c)
Prove that
It
seems
m/n < v 2,
m/n < m'/n' < v2.
if
likely that \fn
is
m
is
a natural
b.
numbers and m</n~ <
then
2,
show, moreover, that
(m
with
and
—a - bjq.
y/q)
—
'17.
To
are irrational. Hint:
doesn't this proof work for
Prove that
(b)
V6
every integer
ple, use the fact that
all
rrr
<
2
2
=-.
ninequality signs reversed.
then there
irrational
is
another rational number m'/n'
whenever the natural number // is not
method of Problem 1
the square of another natural number. Although the
may
that
actually be used to treat
will
it
information.
A
natural
p
=
sible to write
and the other
The
first
any particular
1
ab
for
;
number p
for natural
convenience
few prime numbers are
prime, then n
(a)
—
is
we
also agree that
if it is
1
is
not
<
n; if either
way
a or b
is
proves that
>
1
we can
it
argument
into a rigorous
7
=
2
2
it
write n
proof by complete induction.
this
sure,
•
p,
not a
is
not a prime
Turn
4
is
a prime number.
2, 3, 5, 7, 11, 13, 17, 19. If n
ab, with a and b both
impos-
unless one of these
=
7.
(To
any reasonable mathematician would accept the informal argu-
ment, but
this
is
partly because
it
would be obvious
to her
how
to state
rigorously.)
fundamental theorem about
that this factorization
for
prime number
called a
product of primes. For example, 28
be
A
not clear in advance
it is
numbers a and b
can be factored similarly; continuing in this
as a
case,
always work, and a proof for the general case requires some extra
is
integers,
which we
will
not prove here, states
unique, except for the order of the factors.
Thus,
example, 28 can never be written as a product of primes one of which
.
32
Prologue
is
3,
nor can
appreciate
it
be written
why
is
1
(b)
Using
(c)
number m
Prove more
(d)
No discussion
prove that
this fact,
P2
P\
*18.
finitely
•••
Prove that
(a)
way
•
that involves 2 only
irrational unless n
*J~n is
generally that Zfn
infinitely
many
x
if
+
p\, p2, P3, ...
1
.
p n by considering
,
+...+a =0,
V6 — v 2 — V3
+ v2
is
x
irrational unless
,
Prove that v 2
k
to allude to Euclid's beautiful
fail
some integers fln _i, ... «o, then x is
(Why is this a generalization of Problem
(c)
=m
some natural
satisfies
for
Prove that
for
1.
.v"+a„_ix""
(b)
= m2
of them. Prove that there cannot be
many prime numbers
Pi,
once (now you should
irrational unless n
is
of prime numbers should
proof that there are
only
a
in
not allowed as a prime).
is
an
integer.
17?)
irrational.
is
irrational. Hint: Start
by working out the
first
6
powers of this number.
19.
(1
for
20.
— 1,
Prove Bernoulli's inequality: If h >
any natural number
n.
The Fibonacci sequence
01
fl 2
a„
Why
a\, «2>
=
1,
=
=
a„-\
1175-1250),
assumed
that
an
in
is
>
+nh
1
a 3?
is
•
+ a„_2
1, 1,
>
for n
2, 3, 5, 8,
.
.
.
was discovered by Fibonacci
,
initial
pair of rabbits gave birth to one
number
is
pair
is
born
for
born
in the
nth month
new
pair of rabbits
pair behaved similarly.
an -\
+ fl„_2,
The
because a pair of
each pair born the previous month, and moreover each
born two months ago now
interesting results
3.
connection with a problem about rabbits. Fibonacci
new
rabbits
0?
defined as follows:
per month, and that after two months each
a„ of pairs
>
this trivial if h
1,
This sequence, which begins
(circa
+/?)"
then
about
this
gives birth to another pair.
sequence
is
truly
The number of
— there
amazing
bonacci Association which publishes a journal, The Fibonacci
is
even a
Quarterly.
Fi-
Prove
that
/1+V5Y'
a„
/1-V5Y'
=
v/5
One way
of deriving
this
astonishing formula
is
presented in Problem 24-16.
.
2.
21.
The Schwarz
Numbers of
inequality (Problem 1-19) actually has a
Various Sorts
33
more general form:
n
E* £*
2
Give three proofs of this, analogous
22.
The
result in
Problem
2
-
to the three proofs in
an important generalization:
1-7 has
Problem
If a\,
.
.
1-19.
.
,
an >
0,
mean"
then the "arithmetic
Ya n
a\-\
=
A„
and "geometric mean"
= Va
G„
•
i
•
an
-
satisfy
G n < An
(a)
Suppose
say ai
>
.
< A n Then some a, satisfies a, > A„; for convenience,
Let a\ = A n and let «2 = «i + «2 — o-\. Show that
that a\
A„.
.
>
a\ci2
a\ai.
Why does repeating this process enough times eventually prove that G n <
An ?
(This
is
another place where
a good exercise to provide a formal
it is
proof by induction, as well as an informal reason.)
The
(b)
reasoning in
Using the
Gn < A n
(c)
this
proof is related
fact that
for
n
Gn <
let
=
2, prove,
.
Apply part
n.
.
,
an
,
An
.
,
2'"
The
another interesting proof.
to
A„ when n
2m >
a\,
23.
does equality
by induction on
k, that
= 2*.
For a general n,
to
When
< A„?
hold in the formula G„
.
-n
.
,
(b) to
the
2'"
numbers
An
times
prove that G„ < A„.
following
is
a recursive definition of a":
a
a"
1
+1
=
=
a,
a" -a.
Prove, by induction, that
a"
<
+m
„ll\l»
= a"
-a'",
„nm
(Don't try to be fancy: use either induction on n or induction on m, not both
at once.)
34
Prologue
24.
Suppose we know properties PI and P4
numbers, but that
for the natural
multiplication has never been mentioned.
Then
the following can be used
as a recursive definition of multiplication:
lb = b,
(a
Prove the following
a-(b
a
a
25.
1
•
b
(in
+
1)
b
•
In this chapter
a -b
+
b.
the order suggested!):
+ c) = a-b + a-c
= a,
—b
=
on
(use induction
a),
=
a (you just finished proving the case b
we began
A
the real numbers.
1).
with the natural numbers and gradually built up to
completely rigorous discussion of
this
process requires
book in itself (see Part V). No one has ever figured out how to get to
the real numbers without going through this process, but if we do accept the
real numbers as given, then the natural numbers can be defined as 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
little
(a)
A
set
A
of real numbers
(1)
1
is
in
(2)
k
+
1
is
called
inductive
if
A,
is
in
A whenever
k
is
in A.
Prove that
(i)
R
is
inductive.
The set of positive real numbers is inductive.
(iii)
The set of positive real numbers unequal to A is inductive.
The set of positive real numbers unequal to 5 is not inductive.
(iv)
If A and B are inductive, then the set C of real numbers which
(v)
are in both A and B is also inductive.
A real number n will be called a natural number if n is in every inductive
(ii)
(b)
set.
26.
There
(i)
Prove that
(ii)
Prove that k
is
1
is
+
a natural number.
1
is
a natural
is
may be moved from one
to spindle
the smallest ring
first
(Figure
ring.
I
(
,
I
RE
is
2 and the next-smallest ring
may be moved
moves, and that
this
1).
A
For example,
is
to spindle 3 also,
Prove that the entire stack of n rings can be
I
k
a natural number.
ring at the top oi a
spindle to another spindle, provided that
not placed on top of a smaller
moved
if
a puzzle consisting of three spindles, with n concentric rings of
decreasing diameter stacked on the
stack
number
cannot be done
if
the smallest ring
moved
to spindle 3, then
on top of the
moved onto
in fewer than 2"
—
next-smallest.
spindle 3 in 2"
1
it
is
moves.
—
1
2.
*27.
Numbers of
University B. once boasted 17 tenured professors of mathematics.
tion prescribed that at their
all
17,
any members
weekly luncheon meeting,
who had
35
Various Sorts
Tradi-
attended by
faithfully
discovered an error in their published work
make an announcement of this
and promptly resign. Such an announcement had never actually been made, because no professor was aware
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
department. 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
should
obtained
his
revenge
at the last
here very much," he said, "but
fact,
luncheon of the
I feel
that there
year. "I
is
have enjoyed
one thing
that
I
my visit
have to
tell
you. At least one of you has published an incorrect result, which has been
discovered by others in the department."
**28.
What happened
the next year?
After figuring out, or looking up, the answer to Problem 27, consider the
lowing:
so
how
Each member of the department already knew what
could
his
saying
it
change anything?
Prof.
fol-
X asserted,
PART
FOUNDATIONS
The statement
is
so frequently
made
that the differential calculus deals with
continuous magnitude, and yet
an explanation of this continuity
is
nowhere given;
even the most rigorous expositions
of the differential calculus do not base
their proofs upon continuity but,
with more or less consciousness of the fact,
they either appeal to geometric notions
or those suggested by geometry,
or
depend upon theorems which are never
established in a purely arithmetic manner.
Among
these, for
example,
belongs the above-mentioned theorem,
and a more
convinced
careful investigation
me
that this theorem, or
any one equivalent
in
some way as a
to
it,
can be regarded
sufficient basis
for infinitesimal analysis.
It then only
remained
origin in the elements
and thus
to
at the
to
discover
its
true
of arithmetic
same time
secure a real definition of
of continuity.
I succeeded Nov. 24, 1858, and
the essence
a few days afterward I communicated
the results
of my meditations to my dear friend
Durege with whom I had a long
and
lively discussion.
RICHARD DEDEKIND
CHAPTER
FUNCTIONS
Undoubtedly the most important concept in all of mathematics is that of a
function
in almost every branch of modern mathematics 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
we
relief to learn that, for the present,
will
be able to
restrict
our attention
functions of a very special kind; even this small class of functions will exhibit
engage our attention for quite some time.
cient variety to
with a proper definition. For the
discuss functions at length,
and
moment
PROVISIONAL DEFINITION
modern mathematical
A function
is
will
not even begin
a provisional definition will enable us to
will illustrate the intuitive
understood by mathematicians. Later, we
of the
We
to
suffi-
notion of functions, as
consider and discuss the advantages
will
definition. Let us therefore begin with the following:
a rule which assigns, to each of certain real numbers,
some other
real
number.
The
following examples of functions are
nition, which, admittedly, requires
Example
1
Example 2
The
The
some such
which
assigns to each
rule
which
assigns to
3
+
y
The
which assigns
rule
c
3
Example 4
number x
Example 5
irrational,
The
rule
which
+
-l
assigns to each
The
rule
which assigns
The
rule
1
if
which
a
is
5
•
c^
1,-1 the number
5
'
number x
to
satisfying
number a
each
rational.
assigns
to 2 the
number
to 17 the
5,
number
36
TV
39
this defi-
clarification.
each number
3c
2
and amplify
—17 < x <
tt/3
.
and the number
Example 6
2
to
c
the
+
+l
3y
+
to illustrate
number the square of that number.
each number y the number
rule
y
Example 3
meant
the
number
if
a
is
40
Foundations
to
—
the
number
28,
to
36
—
the
number
28,
TV
and
any
to
^
y
2, 17, jt~/\7,
or 36/tt, the
number
16
y
if
is
of the form a
+ bv2
Q.
for a, b in
The
Example 7
which assigns
rule
to
each number
t
the
number
3
/
+ a\
(This
on what the number x is, so we are really describing
infinitely many different functions, one for each number x
Example 8 The rule which assigns to each number z the number of 7's in the
decimal expansion of z, if this number is finite, and — n if there are infinitely many
rule depends, of course,
.)
7's in the
One
decimal expansion of
z.
thing should be abundantly clear from these examples
numbers
rule that assigns
to certain other
numbers, not
—a function
just a rule
is
any
which can
be expressed by an algebraic formula, or even by one uniform condition which
number; nor
applies to every
can actually apply
is
it
one knows,
in practice (no
Moreover, the rule
may
necessarily a rule
7r).
to
which numbers the function applies
apply
is
Example 6
applies to
it).
for example,
some numbers and
neglect
to
function in
which you, or anybody
(try to
The
set
it
else,
what
rule 8 associates
may
not even be clear
determine, for example, whether the
of numbers to which a function does
called the domain of the function.
Before saying anything else about functions
throughout
this
book we
shall frequently
we badly need some
notation. Since
be talking about functions (indeed
we
shall
else) we need a convenient way of naming funcand of referring to functions in general. The standard practice is to denote
a function by a letter. For obvious reasons the letter "/" is a favorite, thereby
making "g" and "h" other obvious candidates, but any letter (or any reasonable
symbol, for that matter) will do, not excluding "x" and "y", although these letters
hardly ever talk about anything
tions,
are usually reserved for indicating numbers. If
/
is
number
read "/ of
a function, then the
—
which / associates to a number x is denoted by f(x) this symbol is
x" and is often called the value of/ at*. Naturally, if we denote a function by x,
some other letter must be chosen to denote the number (a perfectly legitimate,
though perverse, choice would be "/," leading to the symbol x(f)). Note that the
symbol f(x) makes sense only for x in the domain of /; for other x the symbol
f(x)
not defined.
is
If the
functions defined in
we can
and
y,
(1)
f(x)=x 2
(2)
g(y)
then
h(c)
=
=
c
-
3
1
8 are denoted by /, g,
rewrite their definitions as follows:
for all x.
y
(3)
Examples
—
y—
+
L
+
—~
c
L
for
all y.
1
—
—
3c
+
1
5
for all c
#
1,
-1.
/?,
r,
.v,
#,
ax
,
;
41
3. Functions
(4)
r(x)
= x1
(5)
s(x)
=
for
-17 < x <
x such that
all
0,
x irrational
1,
x
tt/3.
rational.
x=2
5,
36
x=
17
77
i
=
9(x)
(6)
28,
r
•
71
=
T7
36
28,
71
36
—
71
x
16.
^
2, r,
or
,
and x
,
— a + b\/2
for a, b in Q_.
7T
(7)
ax (t)
(8)
v(a-)
These
tion
/
=
for
numbers
all
77,
exactly
—7T,
infinitely
indicating
is
t?
7's
?.
appear
many
7's
decimal expansion of x
in the
appear
in the
common procedure
decimal expansion of
adopted
x.
for defining a func-
what f(x) is for every number x in the domain of /. (Notice
same as indicating f(a) for every number a, or f(b) for ev-
exactly the
number
ery
+x
t
definitions illustrate the
—
that this
=
In practice, certain abbreviations are tolerated. Definition
b, etc.)
(1)
could be written simple
the qualifying phrase "for
all
the only possible abbreviation
It is
Of course,
x" being understood.
for definition (4)
is
r(x)=x 2
(4)
= x2
f(x)
(1)
-17<a<tt/3.
,
usually understood that a definition such as
k(x)
I
=
-v
^
0,
1
1
can be shortened
to
*(*)
in other
all
words,
unless the
numbers for which
You should have
domain
the definition
little
—
1
1
x
—
= -+x
1
is explicitly restricted further,
makes any sense at
difficulty
it is
understood
+
1)
g(x)
r(x
= /(*) + 2x +
= h(x)i£ 3 + 3* + 5 = 0;
+[) =
1;
jc
r(x)
of
checking the following assertions about the func-
tions defined above:
/(*
to consist
all.
+
2x
+
1
if
-17 <
x
<
^-
1;
42
Foundations
s{x
+ y) = six)
y
if
rational;
is
-G) -MS)-expression /(5(a)) looks unreasonable to you, then you are forgetting that
If the
s{a)
of
number
a
is
fact,
=
f(s(a))
any other number, so that f{s{a)) makes sense. As a matter
s(a) for all a. Why? Even more complicated expressions than
like
f(s(a)) are, after a
first
more
exposure, no
The
difficult to unravel.
expression
/(r(s(0(a 3 (j(i)))))),
formidable as
appears,
it
may be
evaluated quite easily with a
little
patience:
f(r(s(e(a 3 (yQ))))))
= /(r(j(0(a 3 (O))j))
=
/(r(*(0(3))))
= /(r(j(16)))
= /(r(D)
= /(D
=
The
this
first
few problems
at the
1.
end of this chapter
give further practice manipulating
symbolism.
The
function defined in
(1) is
a rather special example of an extremely impor-
tant class of functions, the polynomial functions.
function
if
=
fix)
(when f(x)
highest
numbers
there are real
+ a ,_]A'"~ +
an x
•
•
•
;
written in this
is
flo?
form
it
is
power of x with a nonzero
•
•
•
>
A
function
+ ajx" + a\x + ao,
usually tacitly
functions defined in
(2)
and
(3)
is
a
coefficient
is
for
assumed
called the
example, the polynomial function / defined by fix)
degree 6.
The
/
polynomial
a n such that
=
5x
all
that a„
x
^
0).
degree of
+
137a
belong to a somewhat larger
class
—
The
/; for
tt
has
of func-
rational functions; these are the functions of the form p/q where p
and q are polynomial functions (and q is not the function which is always 0). The
rational functions are themselves quite special examples of an even larger class of
tions, the
functions, very thoroughly studied in calculus,
which are simpler than many of the
The
following are examples of this kind
functions
mentioned
first
of function:
(9)
\x )
Jf(r\
—
—
X
(10)
fix)
(11)
fix)
=
=
-2 A
+ A"2 + x sin
.2
sin a + a sin a
1
1
.
\
in this chapter.
sin (a
2
).
sin(sin(A
)).
43
3. Functions
=
/(jc)
(12)
2
.
-22
x )))
-2
•
+ sin(jc sinx)
+ sin x
/-*
•
sin (sin (sin (x sin
sin
•
:
\
By what
criterion,
monstrosity like
you
may
jc
impelled to ask, can such functions, especially a
feel
be considered simple? The answer
(12),
that they can
is
be
built
up from a few simple functions using a few simple means of combining functions.
we need
In order to construct the functions (9)— (12)
function"
x
Z,
which I(x)
for
x,
often written simple sin x
is
which functions
If
/ and g
may
and the
The
.
be combined
and
g,
to start with the "identity
whose value sin(jc)
some of the important ways
"sine function" sin,
following are
at
in
produce new functions.
to
are any two functions,
sum of /
the
—
we can
new
define a
f +
function
g, called
by the equation
(f
+ g)(x) =
+ g(x).
f(x)
Note that according to the conventions we have adopted, the domain of / + g
consists of all x for which "fix) + gix)" makes sense, i.e., the set of all x in both
domain / and domain g. If A and B are any two sets, then A D B (read "A
intersect 5" or "the intersection of A and B") denotes the set of x in both A
and B; this notation allows us to write domain (/ + g) = domain / Pi domain g.
In a similar vein,
we
product / g and
define the
the
•
f
—
quotient
(or f/g) of
8
f and
g by
(/
•
=
g)(x)
f(x)
gix)
•
and
(£\™
^«== /w
g(x)
Moreover,
if
g
is
a function and c
(c
is
a number,
g)(x)
=
has the same value for
all
/
numbers x
define a
/ g
•
we
if
=
defined by fix)
,
is
new
function c
g by
g(x).
c
This becomes a special case of the notation
should also represent the function
we
called a
agree that the symbol c
c;
such a function, which
constant function.
domain / D domain g, and the domain of c g is simply
the domain of f/g is rather complicated
it
may be written domain / D domain g D {x gix) ^ 0}, the symbol {x gix) ^ 0}
denoting the set of numbers x such that gix) / 0. In general, {x
denotes
The domain of / g
•
the
domain of
g.
is
—
On the other hand,
:
:
:
.
}
.
.
the set of
all
jc
such that " ..."
is
true.
Thus
{x
:
x
+3
<
11} denotes the set of
and consequently [x x +3 < 11} = {x x < 2}.
Either of these symbols could just as well have been written using y everywhere
instead of x. Variations of this notation are common, but hardly require any
3
discussion. Any one can guess that {x >
< 8} denotes the set of positive
jt
numbers whose cube is less than 8; it could be expressed more formally as {jc
and x 3 < 8}. Incidentally, this set is equal to the set {x
< jc < 2}. One
x >
all
numbers x such
that x~
<
8,
:
:
:
:
:
44
Foundations
variation
{.v
=
x
:
1
or x
=
=
3 or x
The
but very standard.
slightly less transparent,
is
example, contains just the four numbers
and
2, 3,
1,
4;
{1,3,2,4}, for
set
can also be denoted by
it
= 4}.
2 or x
Certain facts about the sum, product, and quotient of functions are obvious con-
sequences of facts about sums, products, and quotients of numbers. For example,
very easy to prove that
it is
+ g) + h =f +
(f
The proof
is
and
—
same value
the
must be shown to have the same domain,
domain. For example, to prove that
the two functions
any number
at
+ g) + h = f + (g + h),
(f
proof which demonstrates that two
characteristic of almost every
functions are equal
[(/
+ h).
(g
in the
note that unraveling the definition of the two sides gives
+ g) + h](x) =
=
+ g)(x) + h{x)
[f(x) + g(x)]+h(x)
(f
and
+
[/
= fix) + (g + h)(x)
= f(x)+[g(x) + h(x)],
h)](x)
and f(x) + [g(x) + h(x)] is a fact about
proof the equality of the two domains was not explicitly men-
and the equality of [f(x)
numbers. In
+
(g
this
tioned because this
is
+ g(x)] + h{x)
obvious, as soon as
we
domain of (/ + g) + It and of / + (g + h)
domain h. We naturally write / + g + h for (/
the
we
as
domain /
clearly
is
+ g) + h
these equations;
—f+
(g
Pi
domain g
+ /?),
(1
precisely
did for numbers.
It is
by
no
down
begin to write
just as easy to prove that (/
/ g
•
-
The
h.
equations
f
+
•
=
g
h
g)
g
=/
+f
•
and
h),
(g
=
and f g
•
g
this
•
function
is
denoted
should also present
f
difficulty.
Using the operations +,
•
,
/
we can now
express the function f defined in
(9)
by
/=
/
+
/•/
+
/• sin
:
/
sin
:
+/
•
sin
•
sin
:
•
•
sin
quire yet another
however, that we cannot express function (10) this way. We reway of combining functions. This combination, the composition
of two functions,
is
It
should be
If
clear,
by
far the
/ and g are any two
/ and g, by
most important.
functions,
we
define a
new
function
/
o g, the
compo-
sition of
(fog)(x)
the
domain of fog
"fog"
is
and g"
this
often
is
{x
read "/
-
:
a
is
in
=
f(g(x)):
domain g and g(x)
circle g."
Compared
is
in
domain
has the advantage of brevity, of course, but there
of far greater import: there
is
much
less
/}.
The symbol
to the phrase "the composition of
chance of confusing
is
/'
/
another advantage
g with
go/, and
45
3. Functions
these
must
be confused, since they are not usually equal;
not
and g chosen
at
example). Lest
let
us hasten to point out that composition
(and the proof
a
is
triviality); this
write the functions (10),
(11)
(12)
/ =
(10)
sin o
function
oh)
(g
denoted by f
is
o
o h.
g
•
•
/),
sin) o sin o (sin
sin) o (/
•
[(sin
•
sin) o (/
•
/)])
•
+
/
•
sin o (/
/
become
has probably already
fact
name
seems
/
be "the function
to
abbreviating
/
for the function
Although
clear.
functions reveals their "structure" very clearly,
shortest
now
can
(12) as
(1 1),
sin o
One
We
(/•/),
sin o sin o (/
(sin
associative:
is
= /o
(fog) oh
/=
/=
almost any
/
random will illustrate this point (try f = I I and g — sin, for
you become too apprehensive about the operation of composition,
in fact,
such that f(x)
=
sin)
+ sin
method of
this
writing
The
hardly short or convenient.
it is
=
such that fix)
•
sin(jc ) for all
2
sin(x
for all
)
x unfortunately
x" The
need
for
clumsy description has been clear for two hundred years, but no
this
reasonable abbreviation has received universal acclaim. At present the strongest
contender for
honor
this
something
is
like
x -» sin(x
)
2
2
(read "x goes to sin(x )" or just "x arrow sin(x )"), but
ellipsis,
never use
criterion
is
As with any convention,
reasonable so long as the
you
x
— x+
+
/
3"
is
,
Even more popular
the motivating factor,
slight logical deficiencies
more
the
we
number and
probably end up adopting
is
is
For the sake of precision
an ambiguous phrase;
i.e.,
the function
/
such that f(x)
i.e.,
the function
/
such that
could
it
=
it
and
for
this
cause no confusion.
precise description
J
t
among
a certain amount of
speaking, confuses a
will
utility
occasion, confusion will arise unless a
example, "the function x
sin(jc )."
strictly
so convenient that
is
it
which,
hardly popular
it is
will tolerate
"the function sin(jc )."
this description,
a function, but
personal use.
On
—
and speak of "the function f(x)
quite drastic abbreviation:
will
we
In this book
writers of calculus textbooks.
is
used.
mean
either
for all
x
for all
t.
For
T
x
+r
or
t
As we
-> x
+
3
1
,
however, for
f(t)=x +
t
3
many
important concepts associated with functions,
calculus has a notation which contains the "jc —>" built in.
shall see,
By now we have made a
sufficiently extensive investigation
rant reconsidering our definition.
hardly clear what
is
this
means.
not easy to say whether
this
If
We
we
of functions to war-
have defined a function as a "rule," but
ask
question
"What happens
is
if
you break
it is
this rule?"
it
merely facetious or actually profound.
46
Foundations
A more
substantial objection to the use of the
word
"rule"
is
that
f(x)=x 2
and
f(x) =jc
2
+
3a-
+ 3-3(jc +
1)
by a rule we mean the actual
determining f(x); nevertheless, we want
are certainly
different rules, if
f(x)
=
x
/(jt)
=
jc
instructions given for
2
and
2
+ 3* + 3 - c(* +
1)
same function. For this reason, a function is sometimes defined as an
"association" between numbers; unfortunately the word "association" escapes the
to define the
objections raised against "rule" only because
even more vague.
it is
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.
"function"
is
The
which mathematicians have finally accepted for
example of the means by which intuitive ideas have been
definition
a beautiful
The correct question to ask about a
function is not "What is a rule?" or "What is an association?" but "What does
one have to know about a function in order to know all about it?" The answer to
for each number x one needs to know the number fix);
the last question is easy
we can imagine a table which would display all the information one could desire
incorporated into rigorous mathematics.
—
about the function f(x)
=x
:
x
fix)
1
1
-1
1
2
4
-2
4
s/2
2
-v/2
2
2
It
is
IX
7T
—n
71
not even necessary to arrange the
be impossible
if
we wanted
to
list all
numbers
in a table
(which would actually
of them). Instead of a two column array
can consider various pairs of numbers
(1.1). (-1.1).
(2.4).
(-2.4).
2
(it, it
),
(72,2),...
we
47
3. Functions
simply collected together into a
we simply
find f(n) we
To
set.*
find f{\)
number of the pair whose first member is 1; to
number of the pair whose first member is 7r We seem
.
might as well be defined
were given the following
/
=
as a collection of pairs of
to
take the second
take the second
be saying that a function
numbers. For example,
then /(3)
or f{\) —
=
7,
8.
/(2)
=
we
{(1,7), (3,7), (5,3), (4,8), (8,4)},
then f(\) = 7, /(3) = 7, /(5) = 3, /(4) = 8, /(8) = 4 and 1,3, 4,
only numbers in the domain of /. If we consider the collection
/=
if
collection (which contains just 5 pairs):
5, 8
are the
{(1,7), (3,7), (2,5), (1,8), (8,4)},
5,
=
/(8)
4;
but
it is
=7
impossible to decide whether /(l)
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
We
in this case.
are therefore led to the following definition.
DEFINITION
A
function
a collection of pairs of
is
numbers with
(a,b) and (a,c) are both in the collection, then b
must not contain two
collection
the following property:
=
different pairs with the
c;
if
other words, the
in
same
first
element.
and illustrates the format we shall always
These definitions are so important (at
least as important as theorems) that it is essential to know when one is actually
at hand, and to distinguish them from comments, motivating remarks, and casual
explanations. They will be preceded by the word DEFINITION, contain the term
being defined in boldface letters, and constitute a paragraph unto themselves.
This
is
our
first
full-fledged definition,
use to define significant
There
If
/
is
(a, b)
is
domain
in /. If
of a function that there
This unique b
With
(actually defining
two things
at once)
which can
rigorously:
a function, the
such that
concepts.
one more definition
is
now be made
DEFINITION
new
is
a
is,
is
of
is the set of all a for which there is some b
domain of /, it follows from the definition
a unique number b such that (a,b) is in /.
/
in the
in fact,
denoted by f{a).
we have reached our goal: the important thing about a
a number f(x) is determined for each number x in its domain.
we have also reached the point where an intuitive definition has
this definition
function
/
is
that
You may feel
that
been replaced by an abstraction with which the mind can hardly grapple.
consolations may be offered. First, although a function has been defined
*The
pairs occurring here are often called "ordered pairs," to
not the same pair as
ordered
this
pairs,
(4, 2).
It is
only
fair to
another undefined term.
(
warn
that
we
emphasize
that, for
example,
are going to define functions
in
Two
as a
(2.
4)
)rdered pairs can be defined, however, and an appendix
chapter has been provided lor skeptics.
is
terms of
t<
i
48
Foundations
collection of pairs, there
rule.
nothing to stop you from
is
Second, neither the
The
of thinking about functions.
chapter
all
by
thinking
of a function as a
nor the formal definition indicates the best way
intuitive
way
best
is
to
draw
pictures; but this requires a
itself
PROBLEMS
1.
/(*)= 1/(1+ x). What
Let
f(f(x))
(i)
is
which x does
(for
this
make
sense?).
'G
«>
(iii)
f(cx).
(iv)
f(x+y).
(v)
/(x)
(vi)
For which numbers c
(vii)
There are a lot more than you might think at first glance.
For which numbers c is it true that f(cx) = f(x) for two different
numbers x?
+ /(y).
there a
is
number
jt
such that f(cx)
=
f(x).
Hint:
2.
=x
Let g(x)
and
,
let
x rational
0,
*«-!,
3.
(i)
For which v
is
/z(v)
(ii)
For which y
is
/?(y)
(iii)
What is
(iv)
For which
(v)
For which e
is
y?
g(y)?
<
g(u;)
is
irrational.
- Mz)?
g(A(z))
w
<
<
x
if?
=
g(g(e))
g(£)?
Find the domain of the functions defined by the following formulas.
f(x)
= yi-.v 2
(ii)
/(.v)
= Vl-\/l-* 2
(iii)
/(x)=
f(x)=
iv
/(jc)
(v)
=
Let S(x)
=
7
y
!
>/l
x2
,
r
—
x
:
+
1
-.v 2
-
let
.
.v
«•
x
—
+
v/x
+
P(x)
z
v/jc
=
.
2
-
-l.
2.
2 V and
,
let s(jc)
=
sin*. Find each of the following.
In each case you answer should be a number.
(i)
(SoP)(y).
(ii)
(So
S )(y).
(iii)
(SoPo S )(t) +
(iv)
s(t
(soP)(t).
3
).
Express each of the following functions
+,
•
,
and
o (for
example, the answer
to
in
(i)
terms of
is
P
o s).
S,
P,
S,
using only
In each case your
1
49
3. Functions
answer should be
=
f{x) —
f(x) =
/(a) =
/(/) =
2s
ii)
iii)
iv)
v)
S1I
>
fix)
i)
a.
V
I
sin2 v
.
a (remember that
sin
2
2
.
a
sin
function.
.
(Note: a
h'
sin
always
a
an abbreviation
is
means a
(i)
;
this
for (sin*)
convention
is
).
adopted
because (a b ) c can be written more simply as a bc .)
+2"
2
vi)
f{u) =sin(2"
vii)
f{y)
=
sin (sin (sin (2
viii)
f(a)
=
2 sin2
).
2~
))).
+ sin (a 2 + 2 sin(a2+sina)
"
)
.
Polynomial functions, because they are simple, yet
role in
most
flexibility,
investigations of functions.
The
flexible,
occupy a favored
following two problems illustrate their
and guide you through a derivation of their most important elementary
properties.
6.
(a)
If x\, ...
degree n
all (a
,
—
— Xj)
x„ are distinct numbers, find a polynomial function
1
which
for j
/
is
at x,
1
and
at Xj if j
/, is
^
at Xj for j
/
/
.
i.
/)
of
Hint: the product of
(This product
is
usually denoted
by
n
Y\(x-Xj),
the symbol
n
(capital pi) playing the
same
role for products that
E
plays
for sums.)
(b)
Now find
a polynomial function
/ of degree
n
—
1
such that /(a,)
=
«,,
where a\, ... a n are given numbers. (You should use the functions
fi from part (a). The formula you will obtain is called the "Lagrange
,
interpolation formula.")
7.
(a)
Prove that for any polynomial function /, and any number a, there is a
polynomial function g, and a number b, such that f(x) = (a —a)g(x) + b
for all a.
(The idea
until a constant
is
simply to divide (a
remainder
A-
A
V
\
2
is left.
+A
X
-2
1
—x
A2
A2
into f(x)
by long
For example, the calculation
-3a +
),3
3
— a)
-3a
—A
-2a+1
-2a + 2
-
division,
.
50
Foundations
—
shows that x 3
+
3.v
Prove that
Prove that
n roots,
(d)
Show
roots,
8.
/
if
If n
if
odd
is
For which numbers a,
9.
(a)
If
=x
A
set
any
is
is
/.)
(x
—
a)g(A') for
some polynomial
/
then
,
=
numbers a with f(a)
find
one with only one
and d
b, c,
has at most
0.
for
all
x
+b
+d
which
(for
root.
will the function
ex
/(/(*))
formal proof
a polynomial function of degree n with
is
ax
satisfy
A
1.
even find a polynomial function of degree n with no
is
n
—
2)
a polynomial function of degree n
is
there are at most n
i.e.,
and
+x—
2
f(a) = 0, then f(x) =
(The converse is obvious.)
that for each n there
n roots.
l)(x
if
function g.
(c)
—
(x
on the degree of
possible by induction
(b)
=
1
equation makes sense)?
this
of real numbers, define a function C\ as follows:
1
C A (x)-
a in
.
A
A
x not in
(J,
Find expressions for Cahb and Caub and Cr.^, in terms of Ca and Cg.
(The symbol AC\ B was defined in this chapter, but the other two may
be new to you. They can be defined as follows:
AUB=
R—A=
(b)
Suppose /
there
10.
is
is
a set
:
{x
:
A
(a)
For which functions
if
For which functions
all
What
A
.v
is
in
R but x
is
this
/
is
(x(t))~
for
in
numbers
t
f =
and only
/
can certainly answer
*(d)
is
such that
/ = /
Show
(b)
x
or x
in B},
is
a function such that f(x)
(c)
that
{x
is
not in A}.)
=
or
1
each x. Prove that
for
Caif
/ = Ca
for
some
set
there a function g such that
question
if
"function"
is
A.
f =g
Hint:
You
replaced by "number."
there a function g such that
/ =
\/g?
+ b{t)x(t) + c(t) =
?
conditions must the functions a and b satisfy
if
there
is
to
be a
function x such that
a(t)x(t)
for all
11.
(a)
numbers
t?
H
is
Suppose
What
that
How many
a function
+ b(t) =0
such functions x
and
v
is
a
number such
is
H(H(H(-(H(y)-) ?
80 times
will there
that
be?
H(H(y))
=
v.
.
51
3. Functions
Same
Same
(b)
(c)
question
if
question
if
=
such that H(l)
=
7r/3, //(47)
replaced by 8 1
is
=
H{H{y))
H
Find a function
*(d)
80
H{y).
=
such that H{H{x))
=
H(2)
36,
H{x)
=
n/3, #(13)
47,
for
numbers
all
=
H(36)
and
x,
=
36, H(jc/3)
47. (Don't try to "solve" for H(x); there are
many
func-
H{H{x)) = H(x). The extra conditions on // are supposed
way of finding a suitable //
function H such that H{H{x)) — H{x) for all x, and such that
tions // with
to suggest a
Find a
*(e)
H(\)
12.
A
odd
(b)
if
/
13.
even
is
is
even
—x
fix)
=
fix)
or fix)
=
f(—x) and odd
x
or /(x)
fix)
if
—
Do
Prove that every even function
the
same
many
for
E
is
Prove that
/
is
/ can be
O
even and
with domain
/
is
cosx, while
For
/
is
=
written fix)
g(|x|), for in-
for
E and O you
functions, define
R
can be written /
= E+
O,
odd.
way of writing /
is
will
unique.
(If you try to
do part
(b) first,
probably find the solution to part
any function, define a new function
and g are
=
by |/|(x)
|/|
|/(x)|.
two new functions, max(/, g) and min(/,
max(/, g)(x)
min(/, g)(x)
=
=
(a).)
If
g),
/
by
max(/(x), g(x)),
min(/(x), g(x)).
Find an expression for max(/, g) and min(/, g)
(a)
= — /(— x).
=
functions g.
this
by "solving"
15.
/(x)
fog.
Prove that any function
where
If
if
or /(x)
|x|
•
(d)
(a)
=
sinx.
(c)
(b)
14.
if
18.
Determine whether / + g is even, odd, or not necessarily either, in the
four cases obtained by choosing / even or odd, and g even or odd. (Your
answers can most conveniently be displayed in a 2 x 2 table.)
Do the same for f g.
finitely
!
H(\l)=
l,
/
function
example,
(a)
=
.)
in
terms of
|
|.
Show that / = max(/, 0) + min(/, 0). This particular way
/ is fairly useful; the functions max(/, 0) and min(/, 0) are
of writing
called the
positive and negative parts of /.
(b)
A function /
is
nonnegative
called
if
fix)
>
for
all
x.
Prove that any
/ can be written / = g — h, where g and h are nonnegative,
in infinitely many ways. (The "standard way" is g = max(/, 0) and h =
— min(/, 0).) Hint: Any number can certainly be written as the difference
of two nonnegative numbers in infinitely many ways.
function
'16.
Suppose /
(a)
(b)
satisfies
fix
+
y)
=
fix)
+
f(y) for
all
x and
Prove that /(xi +••• + *„) = /(jq) + ••• + /(*„).
Prove that there is some number c such that f(x)
numbers x
(at this
irrational x).
y.
=
ex for
all rational
point we're not trying to say anything about fix) for
Hint:
First figure
out what c must be.
Now
prove that
)
52
Foundations
=
fix)
then
"17.
If
f{x)
and
when x
first
when x
=
also
ex,
is
the reciprocal of an integer and,
/
for all x, then
fix y)
a natural number, then
is
=
these two properties, but that fix)
all
(a)
(b)
(c)
Prove that /(l)
=
integer,
rational
jc
.
fix + y) = fix) + f(y) for all x and y,
all x and y. Now suppose that / satisfies
is
not always
0.
Prove that fix)
=
x for
1.
Prove that fix) — x
Prove that fix) >
problems
(e)
an
x, as follows:
if
been paying attention
(d)
is
finally, for all
satisfies
fiy) for
fix)
when x
numbers there
is
is
if
x >
(This part
0.
to the philosophical
> fiy)
=
rational.
two chapters, you
in the last
Prove that fix)
Prove that fix)
x
x for
if
x >
all
will
is
but
tricky,
if
you have
remarks accompanying the
know what
to do.)
y.
x. Hint:
Use the
fact that
between any two
a rational number.
"18.
what conditions must f,g,h, and
hix)kiy) for all x and y?
"19.
(a)
k satisfy in order that f(x)giy)
Precisely
Prove that there do
not exist functions
/ and g
=
with either of the following
properties:
(i)
f(x)
(ii)
fix)
Hint:
+ giy) = xy for all x and y.
giy) = x + y for all x and y.
Try
to get
some information about / or g by choosing
particular
values of x and y.
"20.
+
=
x and
(b)
Find functions / and g such that fix
(a)
Find a function /, other than a constant function, such that \fiy)
(b)
Suppose
y)
gixy) for
all
y.
—
fix)\<\y-x\.
that fiy)
imply that
—
fix) < iy
\fiy) — fix)\ <
—
x) 2 for
2
iy — x) ?)
all
x and
Prove that
/
is
y.
(Why does
this
a constant function.
Hint: Divide the interval from x to y into n equal pieces.
21.
Prove or give a counterexample for each of the following assertions:
(a)
(b)
(c)
fo(g + h) = fog + foh.
ig + h)of = gof + hof.
7
— = 7°^
f °g
(d)
f
-— = /°
f°g
22.
(a)
(b)
Suppose g
\8
=ho
/. Prove that
if
fix)
—
fiy). then g(x)
=
giy).
/ and g are two functions such that gix) = giy
whenever fix) = fiy). Prove that g = h o f for some function //. Hint:
Just try to define hiz) when z is of the form z = fix) (these arc the only z
that matter) and use the hypotheses to show that your definition will not
Conversely, suppose that
run into trouble.
.
.
53
3. Functions
23.
*24.
Suppose that /
x
^
o
=
g
/,
^
then g(x)
Prove that
(a)
if
(b)
every
(a)
Suppose g is a function with the property that g(x)
Prove that there is a function / such that / o g = I
(b)
y,
g(y);
number b can be
Suppose
/
that
is
written b
*25.
Find a function / such that g
function h with f o h = I
*26.
Suppose /
o
g
=
is
associative.
composition
(a)
(b)
and h
I
—
f(a) for some number
o
a.
o
f=
Prove that there
f=
I.
I for
some
Prove that g
g,
=
^
a.
g(y)
if
number b can be
a function such that every
— f(a) for some number
fog = I.
b
27.
= x.
where I(x)
is
x
^
y.
written
a function g such that
but such that there
h. Hint:
Use the
is
no
fact that
Suppose f(x = x + 1. Are there any functions g such that fog = go f?
Suppose / is a constant function. For which functions g does / o g =
)
8 of?
(c)
Suppose
that
fog = go f
28.
(a)
Let
and
(b)
*(c)
(d)
(e)
F be
•
for all functions g.
Show
that
/
is
the identity
= x.
function, f(x)
the set of all functions
whose domain
is
as defined in this chapter, all of properties
R. Prove
that,
using
+
PI— P9 except P7 hold
and 1 are interpreted as constant functions.
P7 does not hold.
that P10-P12 cannot hold. In other words, show that there is
no collection P of functions in F, such that P10-P12 hold for P. (It is
sufficient, and will simplify things, to consider only functions which are
except at two points *o and jci.)
Suppose we define / < g to mean that f(x) < g(x) for all x. Which of
P'IO-P'13 (in Problem 1-8) now hold?
for F,
provided
Show
Show
that
If
/ <
g,
is
h
o
f <
h
o
g?
Is
/
o
h
< g
o
h
?
54
Foundations
ORDERED PAIRS
APPENDIX.
Not only
it
is
in the definition
of a function, but in other parts of the book as well,
A
necessary to use the notion of an ordered pair of objects.
not yet been given, and
ordered pair
is
supposed
definition has
we have never even stated explicitly what
to have. The one property which we will
formally that the ordered pair
properties an
require states
should be determined by a and
{a, b)
and the
b,
order in which they are given:
if (a,
Ordered
as
may be
=
b)
then a
(c, d),
=
=
and b
c
d.
most conveniently by simply introducing {a b)
an undefined term and adopting the basic property as an axiom since this
pairs
treated
,
—
property
is
the only significant fact about ordered pairs, there
worrying about what an ordered pair "really"
satisfactory
The
rest
need read no
of
this short
is.
Those who
is
much
not
point
find this treatment
further.
appendix
for the benefit of those readers
is
uncomfortable unless ordered pairs are somehow defined so that
becomes a theorem. There
who
this basic
will feel
property
no point in restricting our attention to ordered pairs
of numbers; it is just as reasonable, and just as important, to have available the
notion of an ordered pair of any two mathematical objects. This means that our
definition
is
common to
ought to involve only concepts
The one common concept which pervades
all
all
branches of mathematics.
areas of mathematics
is
that of a
and ordered pairs (like everything 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
from {a, b} which of a or b is meant to be the first element. A more promising
set,
candidate
is
the rather startling
set:
{{a},{a,b}}.
This
{a},
set
has two members, both of which are themselves
containing the single
seem,
we
member a,
the other
are going to define (a, b) to be this
is
set.
sets;
really isn't anything else
(a, 6)
=
and b
=
DEFINITION
theorem
l
PROOF
II'
(a, b)
The
—
(c,
d), then a
hypothesis
means
—
c
The justification
{{<-/},
common
—
the set
it
may
for this choice
is
the definition works,
saying.
{{«}, {«,*}}.
d.
thai
{{a},
Now
worth
member is
the set {«, b}. Shocking as
given by the theorem immediately following the definition
and there
one
{aJ>}} contains
just
element of these two
[a,b}}
=
{{c],
two members,
members
of
{
{a},
{c.d}}.
{a}
and {a.b}; and a
{a, b}
}.
Similarly, c
is
is
the only
the unique
3, Appendix.
common member
of both
members of
{
{<?},
{c,
d)
}.
Ordered Pairs
=
Therefore a
c.
We
55
there-
fore have
{a,b}}
{{a},
and only the proof that b
Case
1.
=
b
which implies
b
2.
other.
It
d remains.
a. In this case, {a,b}
member, namely,
Case
=
^
that
a.
{a}.
d
—
—
=
{{a},
It is
a
—
In this case, b
^
a; so
b
=
convenient to distinguish 2 cases.
{
{a},
{a, b]
true of {{a},
{a,
}
really has only
d}}, so {a,d\
=
one
{a},
b.
is
in
one
member of {a}, {a, b}
in one member of
{a},
must therefore be true that b is
This can happen only if b is
d, but b
{a,d}},
{a}, so the set
The same must be
in the other.
or b
=
d. |
{
}
{
in {a, d},
but b
is
not in
but not in the
{a,
d)
{a};
}
but not
thus b
=
a
CHAPTER
GRAPHS
Mention the
numbers
real
probably form
in
mathematician and the image of a
to a
her mind, quite involuntarily.
banish nor too eagerly embrace
And most
likely
straight line will
she will neither
mental picture of the real numbers. "Geomet-
this
allow her to interpret statements about numbers in terms of this
and may even suggest methods of proving them. Although the properties
of the real numbers which were studied in Part I are not greatly illuminated by a
geometric picture, such an interpretation will be a great aid in Part II.
ric intuition" will
picture,
You
_,
i
|
|
\
1
,_
|
-1
FIGUR1
2
3
method of considering
are probably already familiar with the conventional
the straight line as a picture of the real numbers,
of associating to each real
i.e.,
number a point on a line. To do this (Figure 1) we pick, arbitrarily, a point which
we label 0, and a point to the right, which we label 1. The point twice as far to
|
the right
is
is
same distance from
labeled 2, the point the
— 1,
labeled
etc.
With
arrangement,
this
to a lies to the left of the point
numbers, such as
^, in the
a
<
b,
corresponding to
obvious way.
numbers also somehow
can be drawn as a point on the
irrational
if
fit
It is
to
but to the
1,
of 0,
left
then the point corresponding
We
b.
can also draw rational
usually taken for granted that the
into this scheme, so that every real
number
We will not make too much fuss about
method of "drawing" numbers is intended
certain abstract ideas, and our proofs will never
line.
justifying this assumption, since this
method of picturing
on these pictures (although we
solely as a
rely
Because
explain a proof).
inessential role, geometric
numbers
— thus
a
number
this
will frequently use
geometric picture plays such a prominent, albeit
terminology
frequently employed
is
sometimes called a
is
a picture to suggest or help
point,
and
R
when speaking of
is
often called the
real line.
The number
\a—b\ has a simple interpretation in terms of this geometric picture:
the distance
it is
between a and
end point and b
occurrence
— a\ <
\x
than
less
s
s.
a
MCI RE
—
e
a
a
+£
pictured as the collection of points whose distance from a
of points
set
as the points
—e
+
e,
corresponding to numbers x with a
—
is
the "interval" from a
to
a
is
may also
< x < a+s
which
s
2).
which correspond
called the
names
for them.
open interval from
is
interval.
Note that
if
to intervals arise so frequently that
The
a to
it is
desir-
denoted by
(a, b)
This notation naturally creates
some
set {x
b.
:
a
<x <
b)
is
also used to denote a pair of numbers, but in context
always clear (or can easily be
56
satisfy
may be
ambiguity, since (a, b)
an
of numbers x which
This
able to have special
and
This means, to choose an example whose frequent
as the other.
Sets of numbers
2
length of the line segment which has a as one
justifies special consideration, that the set
be described
(Figure
b, the
a
>
made
b,
clear)
then (a,b)
whether one
=
0, the set
is
it
is
talking about a pair or
with no elements; in prac-
tice,
however,
is
it
almost always assumed
whenever an
implicitly otherwise), that
the
open
(explicitly if
interval ia, b)
is
4.
Graphs
57
one has been
careful,
and
mentioned, the number a
the closed interval [a, b]
interval (a, b)
is
than
less
a
b.
")
The
the interval (—00, a)
set {x
from a
the interval (a, 00)
to b.
used for a
=
:
< x <
a
b)
This symbol
b, also.
The
is
is
denoted by
and
[a, b]
is
closed interval
it is sometimes
and [a, b] are shown
called the
usually reserved for the case a
<
b,
usual pictures for the intervals (a, b)
but
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 "infinite" intervals. The set {x
x > a} is denoted by (a, 00),
the interval (—00, a]
:
the interval [a, 00)
FKJURE
while the set {x
:
x > a}
is
denoted by
[a, 00); the sets
(—00, a) and (—00, a] are
this point a standard warning must be issued: the symbols 00
and —00, though usually read "infinity" and "minus infinity," art purely suggestive;
there is no number "00" which satisfies 00 > a for all numbers a. While the
symbols 00 and —00 will appear in many contexts, it is always necessary to define
these uses in ways that refer only to numbers. The set R of all real numbers is
also considered to be an "interval," and is sometimes denoted by (—00, 00).
defined similarly. At
3
(a,b)
(0, b)
1
-f
Of even
greater interest to us than the
method of drawing numbers
is
a
method
1
of drawing pairs of numbers. This procedure, probably also familiar to you, re-
1
.(l.D
(-1.1).
1
To
quires a "coordinate system," two straight lines intersecting at right angles.
|
1
,
(0, 0)
distinguish these straight lines,
(a, 0)
axis.
•(1.-1)
(-I.-!)'
(More prosaic terminology, such
horizontal axis,
as the "first"
and one the
and "second"
axes,
is
vertical
probably
blackboards, in the same way, so that "horizontal" and "vertical" are
more
we can
Each of the two axes could be labeled with real numbers, but
also label points on the horizontal axis with pairs (a 0) and points on the
vertical axis with pairs (0, /?), so that the intersection of the two axes, the "origin"
of the coordinate system, is labeled (0, 0). Any pair ia, b) can now be drawn as
in Figure 4, lying at the vertex of the rectangle whose other three vertices are labeled (0, 0), (a, 0), and (0, b). The numbers a and b are called the first and second
4
=
1
,
coordinates, respectively,
fix)
Our
tion
5
is
real concern, let us recall,
is
a
in this way.
method of drawing
functions. Since a func-
numbers, we can draw a function by drawing
The drawing obtained in this way is called the
graph of / contains all the points corresponding to pairs ix, fix)). Since most functions contain infinitely many pairs,
drawing the graph promises to be a laborious undertaking, but, in fact, many
graph of the
fix)=x
of the point determined
just a collection of pairs of
each of the pairs
fix)
one the
descriptive.)
fix)
FIGURE
call
preferable from a logical point of view, but most people hold their books, or at
least their
FIGURE
we
in the function.
function. In other words, the
functions have graphs which are quite easy to draw.
Not
surprisingly, the simplest functions of
have the simplest graphs.
is
I
I
.
I
RE
6
all,
the constant functions fix)
easy to see that the graph of the function fix)
a straight line parallel to the horizontal axis, at distance c from
The
I
It is
functions fix)
through
(0, 0),
—
ex also have particularly simple graphs
as in Figure 6.
A
proof of
this
fact
is
it
(Figure
= c,
=c
5).
straight lines
indicated in Figure
7:
58
Foundations
number
Let x be some
not equal to 0, and
L be
let
the straight line which passes
through the origin O, corresponding to (0,0), and through the point A, corresponding to (x,cx). A point A', with first coordinate y, will lie on L when the
BO
triangle A'
(x,
is
similar to the triangle
ex)
ABO,
A'B'
AB
OB'
OB
when
thus
=
c;
precisely the condition that A' corresponds to the pair (y, cy),
this
is
lies
on
The argument has implicitly assumed that c > 0,
The number c, which measures the
the graph of /.
other cases are treated easily enough.
the sides of the triangles appearing in the proof,
FIGURE
and a
line,
7
line parallel to this line
is
also said to
that A'
i.e.,
but the
ratio of
called the slope of the straight
is
have slope
c.
This demonstration has neither been labeled nor treated as a formal proof.
we
Indeed, a rigorous demonstration would necessitate a digression which
not at
prepared to
all
The
follow.
geometric and algebraic concepts would
stated assumption) that the points
to the real
d-b
geometry
Now the
ia,b)
length c
—
a
Aside from
numbers.
as precisely as
detailed
we
on a
this,
first
require a real proof (or a precisely
straight line
it
correspond
would be necessary
an exact way
in
to
develop plane
intend to develop the properties of real numbers.
development of plane geometry
no means a prerequisite
a beautiful subject, but
is
for the study of calculus.
We
shall use
it is
and
FIGURE
8
define
the plane to be the set of
all
pairs of real
to define straight lines as certain collections of pairs, including,
the collections {(x ex)
,
geometry with
definition
is
all
x
3.
real
number}. To provide
define
and (c, d) are two points in the plane,
the distance between (a, b) and (c, d) to be
If the motivation for this definition
—with
others,
constructed
the structure of geometry studied in high school, one
y/(a-c) 2
explanation
this artificially
is
this definition the
+ (b-d) 2
not
clear,
it is
numbers,
among
required. If (a, b)
numbers, we
real
:
by
geometric pictures
only as an aid to intuition; for our purposes (and for most of mathematics)
perfectly satisfactory to
are
rigorous proof of any statement connecting
i.e.,
more
pairs of
.
Figure 8 should serve as adequate
Pythagorean theorem has been
built into
our
geometry*
Reverting once more to our informal geometric picture,
(Figure 9) that the graph of the function
with slope
f{x)
=
ex
c,
+
passing through the point
f(x)
(0, d).
=
For
ex
this
it
+
d
is
is
not hard to see
a straight line
reason, the functions
Simple as they are, linear funcand yon should feel comfortable working with them. The
a typical problem whose solution should not cause any trouble. Given
points (a,b) and (c,d), find the linear function / whose graph goes
b) and (c, d). This amounts to saying that f(a) — b and /(c) = d. II
d are called linear functions.
tions occur frequently,
following
two
FIG
1
RE
9
through
*
is
distinct
The
(a,
might object to this definition on the grounds that nonncgalive numbers
have square roots. This objection is really unanswerable at the moment the
fastidious reader
are nol yet
known
del mil ion will
just
to
have
—
to
be accepted with reservations, until
this little
point
is
settled.
.
4.
/
is
= ax +
be of the form f(x)
to
aa
ac
=
a
therefore
{d
b)/(c
d
fix)
c
a formula most easily
Of course,
and
a)
ft
b
—
—x+b
+
+
=
=
=
ft
fi
-
b
b,
d\
b)/(c
[{d
a
c
possible only
is
account only for the straight
lines
—
the "point-slope
if
a
^
which are not
c;
conclusion
is
vertical line
function cannot contain (a, b) and (a, c)
•
r\
J
FIGURE
10
if
(a, b)
b
/
(see
Problem 6).
c.
and
all;
the
in fact, the
The
graph of a
same vertical line. This
two points on the same
—
(a, c)
and, by definition, a
Conversely,
if
a
set
of points in
on the same vertical line, then
it is surely the graph of a function.
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.
no two points
the plane has the property that
draw some of the
a picture
like
pairs in /,
Figure
i.e.,
is
1 1
'(2,4)
'(§-!)
(-1,1).
(1.1)
,'..
• <v*>
(0,0)
FKillRK
I
I
>
lie
If we
perhaps the function f(x) — x
2
obtain
some of the pairs of the form (x, x ), we
After the linear functions the simplest
(d)
at
on
immediate from the definition of a function
correspond to pairs of the form
form"
the graphs of linear functions
parallel to the vertical axis.
function can never contain even two distinct points
(b)
a
graph of any function
vertical straight lines are not the
a)]a, so
b
d -b
—a = —
(x -a) + b,
d
remembered by using
this solution
59
we must have
then
/3,
Graphs
.
60
Foundations
It is
not hard to convince yourself that
the one
shown
is
just a
the pairs (x, x 2 )
lie
along a curve
known as a parabola.
drawing on paper, made (in this case) with
in Figure 12; this
Since a graph
all
curve
like
is
printer's ink,
what the graph really looks like?" is hard to phrase in any
sensible manner. No drawing is ever really correct since the line has thickness.
Nevertheless, there are some questions which one can ask: for example, how can
you be sure that the graph does not look like one of the drawings in Figure 13?
It is easy to see, and even to prove, that the graph cannot look like (a); for if
the question "Is this
< x <
v,
then x 2
not the case in
plotting
first
(a)
.
< y2
It is
(c).
so the graph should be higher at y than at x,
also easy to see, simply
many pairs
or a "corner" as in
,
(x,
x
),
that the
12
is
by drawing a very accurate graph,
graph cannot have a large "jump"
In order to prove these assertions, however,
we
as in (b)
first
need
means for a function not to have a "jump"
to say, in
or "corner"; these ideas already involve some of the fundamental concepts of
calculus. Eventually we will be able to define them rigorously, but meanwhile you
may amuse yourself by attempting to define these concepts, and then examining
your definitions critically Later these definitions may be compared with the ones
mathematicians have agreed upon. If they compare favorably, you are certainly
a mathematical way, what
FIGURE
which
it
be congratulated!
to
The
=
numbers n, are sometimes called
power functions. Their graphs are most easily compared as in Figure 14, by
functions fix)
drawing several
The power
x", for various natural
at once.
functions are only special cases of polynomial functions, introduced
in the previous chapter.
Two
particular polynomial functions are
fix)
(b)
I
I
(
,
I
l<
E 14
=
x
graphed
in
Figure 15, while Figure 16
/(x)
=
x2
+X
meant
is
to give a general idea of the
in the case a n
>
a n x" +a„_\x"
is
/
will
have
at
of peaks and valleys
for
example, have
we
will
at
may
1
,
"peaks" or "valleys"
\
most one
valley).
much
is
a point
like (y,
smaller (the
(a
"peak"
f(y)).
power
The
functions,
Although these assertions are easy
will also
Many
will
be very
to
make,
easy).
7 illustrates the graphs of several rational functions.
tions exhibit
The
rational func-
even greater variety than the polynomial functions, but their behavior
be easy to analyze once
we can
interesting graphs can
be constructed by "piecing together" the graphs of
functions already studied.
lines.
a
not even contemplate giving proofs until Part III (once the powerful meth-
Figure
The
function
/
with
The graph
this
/
graph
use the derivative, the basic tool of Part
in Figure 18
is
made up
III.
entirely of straight
satisfies
n+\
n
n+\
f
f{x)=\.
15
most n —
actually be
ods of Part III are available, the proofs
FIGURE
\-
a point like (x, f(x)) in Figure 16, while a "valley"
number
3x
H
0.
In general, the graph of
= x3 -
graph of the
polynomial function
fix)
f(x)
61
Graphs
4.
>
1.
and is a linear function on each interval [\/(n + 1), 1/n] and [—1/n, —\/{n + 1)].
(The number is not in the domain of /.) 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.
FIGURE
16
62
Foundations
/(*)
=
/(*)
/(*)
=
=
l
(b)
(a)
/(*)
(d)
FIGURE
17
FIGURK
18
=
l+x 2
.
4.
It is
this
actually possible to define, in a
same property of oscillating
which we
angle of
it
The graph
Now
ure 20.
Chapter
means an angle
0,
As
by using the
usual,
"all the
we
sine function,
are using radian
way around"
a
circle,
an
etc.
of the sine function
is
shown
in
Figure 19.
19
consider the function fix)
To draw
this
f(x)
graph
=
f{x)=
it
.v
=
for x
=
f(x)
Notice that
k;i RE 20
= -1
when x
is
for
a-
=
The graph
sin l/x.
helps to
for
1
—
1
1
t:
1
t7T
1
is
shown
in Fig-
+
1
Z7T
^JT
1
,
large, so that \/x
/
1
—
—
of
observe that
first
^77"
I
near
15.
an angle halfway around (or 180° in layman's terms), an angle of 7r/2
a right angle,
FIGURE
Itc
63
simpler way, a function which exhibits
infinitely often
will discuss in detail in
measure, so an angle of
much
Graphs
3
is
+
A-JT
1
'
3.
small, fix)
is
also small;
when x
is
64
Foundations
FIGURE
21
"large negative," that
although fix)
An
this
<
is,
when
|.v|
large for negative x, again f{x)
is
interesting modification of this function
function
between
1
x and —x.
is
and
sketched in Figure 2 1
— 1,
.
the function fix)
Since sin
=
The behavior of the graph
x
sin
is
1
f(x)
2 2
close to 0,
jx
=
x
sin l/x.
The graph
oscillates infinitely often
of
near
l/x oscillates infinitely often between
for x large or large negative
/(*)
FIGURE
is
0.
=
9
xL
•
sin
1
is
harder to
Graphs
4.
analyze. Since sin
seems
be no
to
/x
1
telling
another question that
is
getting close to 0, while x
what the product
is
is
will do. It
getting larger
and
larger, there
possible to decide, but this
is
The graph of f(x) = x
best deferred to Part III.
65
sin
is
i/x
has also been illustrated (Figure 22).
For these infinitely oscillating functions,
The
be really "accurate."
part near
functions
(which
is
is
clear that the
to
graph cannot hope
show part of it, and
the interesting part). Actually,
it
is
easy to find
to
leave out the
much
simpler
whose graphs cannot be "accurately" drawn. The graphs of
m-\
<
x >
x
x
:
.2
1
and
<
X >
X
g(x)
1
1
1
can only be distinguished by some convention similar to that used for open and
(b)
FIGURE
it is
we can do
best
closed intervals (Figure 23).
2 3
Out
last
example
is
a function whose graph
x irrational
1,
x
x irrational
24
The graph
/ must contain infinitely many points on the
many points on a line parallel to the horizontal
of
also infinitely
(x,y)
rational.
x rational
=
0,
FIGURE
spectacularly nondrawable:
0,
fix)
/(*)
is
horizontal axis
axis,
but
it
and
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
ical
on
the line corresponding to irrational x
reason behind
this
convention, but
it
.
(There
is
actually a
mathemat-
depends on some sophisticated
ideas,
introduced in Problems 21-5 and 21-6.)
The
peculiarities exhibited
FIGURE
25
some of
by some functions are so engrossing that
it
is
easy
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 > contains, by definition, all the points
(x, y) whose distance from (a, b) is equal to r. The circle thus consists (Figure 25)
to forget
the simplest,
of all points (x,y) with
^
a)
2
+
(v
-b) 2 =
or
(jc
- a) 2 +
(y
-
b)
2
r
66
Foundations
The
circle
with center
called the unit
is
A
(0,0)
and radius
close relative of the circle
the sum of
whose
distances
same,
foci are the
1
often regarded as a sort of standard copy,
,
circle.
we
obtain a
circle.)
defined as the set of points,
is
is
a constant.
(When
the
(c, 0),
+y 2 +
V(x-(-c)) 2
or
^+
(x
c)
2
2
+
y
two
for convenience, the focus points are
If,
and the sum of the distances is taken to be la
some algebra), then (x, y) is on the ellipse if and only if
taken to be (— c, 0) and
factor 2 simplifies
the ellipse. This
is
from two "focus" points
-
y/(x
=2a-
2
c)
+ y2 =
-
y/(x
c)
2
+
(the
la
2
y
or
2
+
lex
+ c2 +
2
y
=
— 4a
4a 2
-
\/(x
c)
2
+
2
y
+ x2 -
lex
+
c
2
+ y2
or
4(cx
- a 2 = -4a V(x )
c)
2
+ y2
or
L L
,2„2
c x
— Icxa^ + a =
a (x
—
lex
+c +
y
)
or
/ 2
—
(c
2\
2
)jc
fl
—
2 2
a y~
=
2/2 — a 2\
)
a (c
or
11 —
a-
cr
This
2
cz
=
1.
usually written simply
is
- + ^=1
bL
aL
=
where b
a
2
—
c
2
>
0).
2
va —
A
c2
(since
we must
picture of an ellipse
sects the horizontal axis
when
y
=
=
FKil'RK 26
is
clearly choose
shown
0, so that
1.
a
>
in Figure 26.
±a,
it
follows that
The
ellipse inter-
c,
and
it
when x =
intersects the vertical axis
67
Graphs
4.
0, so that
„2
b2
The hyperbola
is
=
=
±b.
defined analogously, except that
and the constant
we
require the
Choosing the points (— c,
the two distances to be constant.
again,
y
1,
difference as 2a,
we
0)
and
difference
(c,
of
once
0)
obtain, as the condition that (x, y)
be on the hyperbola,
Ax + cy + y
\^
which
may
2
—v
(x
— cy +
a-2
—
cA
27
In this case, however,
/?
= vc — a
2
The
picture
2
,
is
then
we must
(x, y)
shown
ferent orders.
=
±a, but
It is
it
is
(x, y)
The hyperbola
two
sets
(Problem
=
1.
clearly choose c
2
2
a2
b2
from (— c,
if
>
a, so that
and only
—
a
c2
<
0.
If
if
two pieces, because the difference
0) and (c, 0) may be taken in two dif-
intersects the horizontal axis
never intersects the vertical
interesting to
2
on the hyperbola
compare
when
—
l/x.
y
=
0, so that
axis.
(Figure 28) the hyperbola with a
the graph of the function f(x)
the
±2a,
in Figure 27. It contains
between the distances of
x
=
be simplified to
a 2A
FIGURE
y
The drawings
=
b
= v2
and
look quite similar, and
are actually identical, except for a rotation through an angle of 7r/4
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
be reduced to the study of functions.
fi
guR
1 .
: s
Ellipses, for
example, are
can often
made up
of the
68
Foundations
graphs of two functions,
/w = Wi-uV),
—a <
x < a
and
-bJT~^ (x 2 /a 2 ),
g(x)
Of course,
example,
there are
we can
many
<x <
-a
other pairs of functions with
a.
this
same
property.
For
take
b
fix)
\/l
-
-W
2
(x /a
-
1
2
< x <
),
2
(x /a
2
),
a
-a<x<0
and
-bf\
=
g(x)
2
<
I
(x /a
yrr (x 2 /a
We
2\
< x <
),
2
-a < x <
,
a
0.
could also choose
f(x) =
—
b v
1
—b
v
I
(x"/a"),
—
1
{x" ja
),
x rational,
—a <
x
—
irrational,
a
x
< a
< x <
a
and
Six)
—b v
=
1
—
2
b\[\
But
(x J a
(x /a
),
2
x rational,
—a <
x irrational,
),
a
x
<
<x <
a.
these other pairs necessarily involve unreasonable functions which
all
around.
A proof,
or even a precise statement of this
fact,
is
jump
too difficult at present.
Although you have probably already begun to make a distinction between those
functions with reasonable graphs, and those with unreasonable graphs, you may
find
it
very
difficult to state
a reasonable definition of reasonable functions.
by no means easy, and a great deal of this
successive attempts to impose more and more conditions
mathematical definition of this concept
book may be viewed
as
that a "reasonable" function
we
will take
time out to ask
which deserve
to
A
must
if
is
satisfy.
we have
be called reasonable.
As we define some of these
really
The
succeeded
conditions,
in isolating the functions
answer, unfortunately, will always be
"no," or at best, a qualified "yes."
PROBLEMS
1
.
Indicate
Also
also
on a
straight line the set of
name each set, using
need the U sign).
(iii)
|.v-3|<l.
|jc-3| < 1.
— a\ < e.
Cw)
\x
(i)
(ii)
|.v
2
-\\<l
all
x satisfying the following conditions.
the notation for intervals
(in
some
cases
you
will
.
5
M
=
<
1
1
(vi)
+x z
1
2.
2
+
(vii)
x
(viii)
(*+
There
is
(where
(a)
>
1
2.
1)(jc
-
=
number
tb for
t?
way of describing the
< b).
<
some
What
prove that
t
/
x
is
<
t
>
b
[0, b], for
with
<
1
Prove that
0.
What
.
is
in [a, b],
then x
of the interval
Prove, conversely, that
(d)
The
=
(1
—
+
t)a
if
<
t
<
1,
[a, b]?
<
<
t
the set of
x
is
in [0, b],
tb for
then
(1
some
+
t
t(b
with
—
a).
What
is
the point 1/3 of the
—
+
tb
t)a
points of the open interval (a, b) are those of the
for
if
the significance of the
Hint: This expression can also be written as a
1.
(c)
Draw
points of the closed interval [a, b]
the mid-point of the interval [0, b]?
is
if
What is the midpoint
way from a to b?
3.
0.
as usual, that a
consider the interval
Now
<
-2) >
l)(x
a very useful
then x
(b)
a (give an answer in terms of a, distinguishing various cases).
we assume,
First
69
Graphs
4.
is
in [a, b].
form
(1
—
t)a
+
tb
1
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)
x
>
(ii)
x
+a >
(iii)
(iv)
(v)
(vi)
(vii)
y <
y <
4.
(")
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
x^-.
+
is
an
integer.
v
(x2
<
;t
Draw
(i)
+ b.
— y\ < 1.
\x + y\ < 1.
x + y is an integer.
x
(x)
y
x-.
\x
(viii)
(ix)
y.
l)
y
2
+
(_v-2)
< x
the set of
|1
Ivl
\
jc
.
<
1.
4
.
all
points (x, y) satisfying the following conditions:
= 1.
1*1 - \y\ = l.
\X- 1| =\y- 1|.
- x = \y- 1|.
2
x + y 2 = 0.
xy = 0.
2
- 2x + v 2 = 4.
T
2
x = y
1x1+
2
70
Foundations
5.
Draw
the set of all points (x, y) satisfying the following conditions:
x
(i)
y
=r
2
(in)
x
fiv)
x
Hint:
(a)
=
->
a1
1.
b<
=
=
\y\.
sin v.
You already know
Show that
is
far
— mix —
(b)
is
+ b.
y are interchanged.
c,
is
the graph of the
the equivalent expression f(x)
—
immediately clear from the point-slope form that the
it is
show
m
This formula, known as the "point-slope
m, and that the value of /
/
For a
a)
more convenient than
mx + (b — ma);
slope
when x and
the straight line through (a, b) with slope
function f(x)
form"
the answers
at
a
is
b.
that the straight line through (a,b)
and
(c,
d)
is
the
graph of the function
/(*)=
—
c
When
(x-a) +
b.
a
= mx +
are the graphs of f(x)
b and g(x)
m'x
+
b' parallel
straight lines?
7.
(a)
set
of
all (x,
y) satisfying
vertical one). Hint: First
(b)
(1,/n)
Show
conversely that every straight
described as the
8.
(a)
A and B not both 0, show that the
Ax + By + C = is a straight line (possibly a
decide when a vertical straight line is described.
For any numbers A, B, and C, with
set
of
g(x)
are perpendicular
if
K.
i
RE
in
Figure 29.
the lines intersect at the origin, as
29
(b)
Prove that the two straight
good
(Why
the squares of the lengths
this special case,
is
lines consisting
Ax + By + C =
0.
+C=
0,
A'x
are perpendicular
(a)
if
+
B'y
and only
if
+
A A'
of all
ix, y) satisfying the
BB'
=
0.
Prove, using Problem 1-19, that
>/(*i
+y\)
2
+
(x2
+ yi) 2 <
Vxi
2
where
as the general case?)
ditions
9.
0.
= mx + b,
= nx + c,
mn = — 1, by computing
of the sides of the triangle
I
Ax + By + C =
Prove that the graphs of the functions
f(x)
(1,»)
including vertical ones, can be
line,
y) satisfying
all (x,
+x 2 2 +
x/v
i
+
yi
con-
.
Prove that
(b)
V
2
-
(*3
ai
)
+
2
-
(>'3
< V
vi)
(x2
-
x\)
2
+
Interpret this inequality geometrically
When
ity").
does
(it is
2
-
(y 2
+
10.
71
Graphs
4.
v
yi)
(-V3
-
2
a- 2 )
+
-
(V3
2
yz)
-
called the "triangle inequal-
inequality hold?
strict
Sketch the graphs of the following functions, plotting enough points to get
good idea of the general appearance.
a
how many
a reasonable decision
meant
to
show
that a
little
thought
is
(Part of the
problem
is
to
make
"enough"; the queries posed below are
will often
be more valuable than hundreds
of individual points.)
f(x)=x-\
—
does the graph
does
1 1
it
(What happens
.
lie
suffice to
in relation to the
graph of the
consider only positive x at
=
x
in
fix)
—
X
iv
fix)
=
x
(ii)
(iii)
identify function?
Why
first?)
xz
2
1
—
~J1'
even.
is
odd.
is
nonnegative.
=
Where
1
1
is
f(x)
(iv)
x?
-
Describe the general features of the graph of
/
/
/
for large
1
fix)
(i)
and
for x near 0,
fix
+ a)
for
all
x
(a
/
if
function with this property
is
called peri-
odic, with period a.
12.
Graph
do
the functions fix)
this,
positive
—
Xfx for
m=
1,2,3,
(There
4.
is
an easy way
using Figure 14. Be sure to remember, however, that %fx
mth root of x when
m
is
means
to
the
even; you should also note that there will be
an important difference between the graphs when
m
is
even and when
/;/
is
odd.)
13.
(a)
(b)
14.
and f(x) = x
sin.v| and f(x) = sin x. (There is an important difference between the graphs, which we cannot yet even describe rigorously.
See if you can discover what it is; part (a) is meant to be a clue.)
Graph f(x)
Graph fix)
—
=
\x\
.
|
Describe the graph of g
in
terms of the graph of
/
if
72
Foundations
ij
g(x)
ii)
g(x)
= f(x)+c.
= f(x + c).
IV)
g(x)
=
f{cx).
v)
g(x)
=
/(l/x).
g(x)
=
f(\x\).
'
.
15.
{
'
(It is
make
a mistake here.)
=
(Distinguish the cases c
'
vn)
g(jc>=|/(jc)|.
viii)
g(x)
=
ix)
g(x)
= min(/, 0).
x)
gO) = max(/,
Draw
easy to
max(/,
0, c
>
0, c
<
0.)
0).
1).
the graph of /(x)
=
ax"
+ Z?x + c.
Use the methods of Prob-
Hint:
lem 1-18."
16.
Suppose
A and C
that
are not both 0.
Show that
Ax 2 + Bx + Cy 2 + Dy
is
either a parabola,
lines
an
ellipse,
+E=
or an hyperbola (or a "degenerate case": two
[either intersecting or parallel],
one
line,
a point, or
0).
Hint:
The
and the case A =
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. When do
case
C =
the set of all (x, y) satisfying
we have
17.
Problem
15,
a circle?
The symbol
all
denotes the largest integer which
[jc]
2 and [—0.9]
(they are
essentially
is
=
[—1]
= — 1. Draw
and
quite interesting,
is
<
x.
Thus,
[2.1]
=
[2]
=
the graph of the following functions
several will reappear frequently in other
problems).
f(x)
= Jx-[x].
= [x] + Jx~-
/(*)
=
/(*)
=
fix)
iv
M
[x].
n
1
I
18.
Graph
(i)
the following functions.
fix)
—
{x},
where
{x}
is
defined to be the distance from x to the nearest
integer.
(ii)
f{x)
=
(iii)
fix)
(iv)
fix)
=
=
(v)
/(.v)
=
{2x}.
{x)
+
\{2x).
{4x).
{.v}
+
i{2.v}
+
i{4.v}.
4.
Many
ber.
may
functions
Although we
Graphs
73
be described in terms of the decimal expansion of a num-
not be in a position to describe infinite decimals rigorously
will
Chapter 23, your intuitive notion of infinite 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
until
decimal ending in a string of
1
s
.23999
19.
.
.
=
.
1.24000
9's
We
).
.
.
equal to another ending in a string of 0's
is
will
always use the one ending in
is
fix)
(ii)
fix)
(iv)
(v)
(vi)
=
=
=
fix)
complete
number in the decimal expansion of x.
2nd number in the decimal expansion of x.
number of 7's in the decimal expansion of x
the 1st
the
the
is finite,
and
=
if
fix)
and
(a
usually out of the question).
(i)
(iii)
(e.g.,
9's.
Describe as best you can the graphs of the following functions
picture
*20.
.
number
if this
otherwise.
the
number of 7's
in the
decimal expansion of x
is
finite,
otherwise,
1
—
the number obtained by replacing all digits in the decimal
fix)
expansion of x which come after the first 7 (if any) by 0.
if 1 never appears in the decimal expansion of x, and n
fix) =
first appears in the nth place.
if
1
Let
/(*)
=
0,
x irrational
1
x
=P
q
rational in lowest terms.
q
lowest terms if p and q are integers with no common
0). Draw the graph of / as well as you can (don't sprinkle
points randomly on the paper; consider first the rational numbers with q = 2,
number p/q
factor, and q >
(A
21.
is
in
then those with q
=
(a)
The
3, etc.).
on the graph of fix)
points
Prove that each such point
(b)
from P and L
^22.
is
equidistant from the point (0, i)
^
fi,
show
=
Show
that the square of the distance
this set if
2
y
+
Using Problem 1-18
the distance
from
(c,
1)
+jc(-
part
(a).)
-2md -
to find the
from
2c)
(c,
(jc,
form fix)
-
minimum
(c,
d\/y/m 2
+
d) to ix,
+ d 2 + c2
x 2 ).
and the
=
(a, fi)
v) equidistant
= ax 2 + bx+c.
mx)
is
.
of these numbers, show that
d) to the graph of fix)
Find the distance from
and a point P
points
(jc,
/J?
\cm
this case to
y,
all
the graph of a function of the
is
2
=
that the set of
What
x im
(b)
x 2 are the ones of the form
graph of gix) = —\. (See Figure 30.)
Given a horizontal line L, the graph of gix)
not on L, so that y
l-KH'RE 30
is
—
= mx
is
1.
d) to the graph of fix)
mx + b.
(Reduce
.
74
Foundations
*23.
(a)
Using Problem 22, show that the numbers
x'
and
y' indicated in Fig-
ure 3 1 are given by
x
—=y,
= —=x +
V2
V2
l
l
s/2
s/l
y
(b)
FIGURE
31
Show
that the set of all {x, y) with fx'/v2
as the set of
all (jc,
y) with
xy
=
1
)
—
(y'/v 2
)
=
1 is
the
same
4,
APPENDIX
v
a point in the plane; in other words, v
is
=
v
For convenience,
we
(l)\,
a pair of numbers
is
V2).
convention that subscripts indicate the
will use this
second pairs of a point that has been described by a single
w and
mention the points
while z
z,
it
be understood that
will
w
Thus,
letter.
is
first
and
if
we
the pair (w\, W2),
the pair (z\, zi).
is
Instead of the actual pair of
from the origin
O
numbers
to this point (Figure
Of course,
in the plane.
{v\,v 2 )
75
Vectors
1.
VECTORS
1.
Suppose that
Appendix
we
(v\, V2),
1),
and we
often picture v as an arrow
refer to these arrows as vectors
new
we've haven't really said anything
yet,
we've simply
introduced an alternate term for a point of the plane, and another mental picture.
The
new terminology
real point of the
some new
V
Then we can
define a
new
Notice that
sign
the
left
is
+W=
on the
is
to
do
(V\
defined operation,
point of the plane) v
+
+
W\, Vl
+ if
by the equation
U>2).
right side of this equation are
numbers, and the
On
+
the other hand, the
in the
sign
on
plane wasn't defined,
(1) as a definition.
might want
+ w,
but there's really no need to
some new symbol
to use
or perhaps
on
insist
possibility of confusion, so
Of course,
W2).
(U>1,
for this
newly
like
v
we could just
points) in the plane,
(i.e.,
new: previously, the sum of two points
A very fussy mathematician
no
new
vector (a
vectors
W =
(V\, V2),
and we've simply used equation
there's
emphasize that we are going
our usual addition of numbers.
just
side
the letters
all
we have two
=
V
(1)
+
to
things with points in the plane.
For example, suppose that
FIGURE
is
this;
since v
we might
w,
v
+w
as well
hasn't
been defined before,
keep the notation simple.
any one can make new notation; for example, since
as well
have defined v
other equally weird formula.
The
+w
+
as (v\
real question
is,
w\
1U2,
does our
^2
it's
+w
new
our
2
\
),
definition,
or by
some
construction have
any particular significance?
Figure 2 shows two vectors v and w, as well as the point
(Ul
+
W],V2
+
W2)
(l>l
.
(l»2
+
U>2)
-
V2
(Vl
+
Wl)
we have simply indicated
an arrow. Note that
is
it
as indicated in Figure 2, this slope
V\
(t>2
and
2
this,
in the usual way,
of course,
is
other words, the line
—
+Wi) -
+
W2)
l>2
U'2
V\
W\
is
just
'
the slope of our vector w, from the origin
L
is
without drawing
easy to compute the slope of the line L between v and
(v\
FIGURE
W2),
which, for the moment,
our new point:
-
+ W\, V2 +
parallel to w.
O
to (w\, W2). In
.
76
Foundations
M between (w\, W2)
Similarly, the slope of the line
+ Wj) —
(V\ + U)\) -
(t>2
which
v
+w
M
the slope of the vector u; so
is
lies
on the parallelogram having
an arrow (Figure
3), it
Vl)i
t>2
V2
V[
is
'
is
In short, the
parallel to v.
w
and
v
and our new point
as sides.
new
point
When we draw v + w
as
points along the diagonal of this parallelogram. In physics,
and the sum of two vectors represents the
forces are applied simultaneously to the same
vectors are used to symbolize forces,
resultant force
when two
different
object.
FIGURE
Figure 4 shows another
3
an arrow
+w
by
Many
first
+
numbers
for ordinary
+W=W+
also hold for this
is
the
same
it
w and
spanned by
as the parallelogram
W\,
+
1)2
V\
—
Wi)
+
+
Similarly, unraveling definitions,
[v
Figure 5 indicates a
The
origin
O —
+
w]
+
(U)\
if
we
(0, 0) is
we
V\,
we
Wi
+
+
for
V
v
+
t>2-
+ z]
z.
identity,"
=v+O
=
v,
(— ui, -V2),
(
V
=
O.
also define
v
=w+
(— v),
exactly as with numbers; equivalently,
W—
V
=
V2),
V[,
[w
+w+
v
+ — v) — — V +
w—
5
also easily
numbers:
have
we can
It is
find the "associative law"
+z =
v
W\
W2+
define
also
Naturally
U>2
an "additive
v
then
=
=
W\
method of finding
O+
and
v.
states that
Vi
G UR E
new +
for
V,
and thus simply depends on the commutative law
FI
denote
obvious from the geometric picture, since the parallelogram spanned by v and
(V\ -f
4
to
example, the "commutative law"
analytically, since
FIGURE
"w"
is
V
w
If we use
and having the same length, but starting at v instead of at
the vector from O to the final endpoint; thus we get to
following v, and then following w.
+w
of the properties of
vectors. For
is
sum v + w.
the
parallel to w,
the origin, then v
v
way of visualizing
(W\
— Vl, W2 —
V2).
checked
4,
Just as with numbers, our definition of
v
w—
v
+
—
(w
"w —
Figure 6(a) shows v and an arrow
w—
v simply
Appendix
means
Vectors
1.
that
it
77
satisfies
w.
v)
v" that
is
parallel to
w—
v but that starts
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,
we can picture w — v geometrically as the arrow that goes from itoio (except that
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
at the
endpoint of
a vector v
=
v.
(y\, V2),
we
define
a
w—
v
=
v
(av\, CIV2)
(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.)
vector a
direction
v points in the
when
You can
FIGURE
a
easily
<
same
(Figure
when
direction as v
a
>
and
The
in the opposite
7).
check the following formulas:
a
6
•
(b
v)
•
=
(ab)
v,
v = v,
0v = O,
— 1 v = —v.
1
•
•
Notice that
not defined a
we have only defined a product of a number and a vector, we have
way of 'multiplying' two vectors, to get another vector.* However,
there are various ways of 'multiplying' vectors to get numbers, which are explored
in the following
problems.
PROBLEMS
Given a point v of the plane,
1.
origin through
FIGURE
an angle of
let
(Figure
7
Show
The aim
of this problem
is
to obtain
(b)
that
=
=
(cos#, sin 6),
Explain
why we have
R e (\,
R 9 (0,
0)
1)
[we should really write Re((l, 0)),
+ w) =
Re(a
f
Re(v)
(c)
At-
Now
etc.]
(-sin0,cos0).
Re (v
show
that for
Re(x, y)
=
w)
=
any point
(x
cosO
—
+
Re(v)
a
Ro(w),
Ro(w).
(x, y)
we have
y sin#,
xsmO +
y cosO).
jump to Chapter 25, you'll find that there is an important way of defining a product, but
something very special for the plane it doesn't work for vectors in 3-space, for example, even
though the other constructions do.
*If you
this
8
8).
a formula for Rg, with minimal calculation.
(a)
FIGURE
Re(v) be the result of rotating v around the
is
—
78
Foundations
(d)
Use
Given
v
another solution to Problem 4-23.
this result to give
and w, we define the number
V
this
is
w —
.
+
V\W\
V2W2',
and w
often called the 'dot product' or 'scalar product' of v
('scalar'
being a rather old-fashioned word for a number, as opposed to a vector).
(a)
Given
w
a vector
v, find
such that v
w —
•
Now describe
0.
the set of all
such vectors w.
(b)
Show
that
w=W V
+ z) = v W +
V
V
and
(w
'
•
(v
Notice that the
product
•
•
last
Show
=
w)
(a
that v
v
•
>
0,
and
will
w =
'
be
v
•
w).
(a
of a
•
three
products:
number and
the dot
a vector; and
that v
—
v
•
only
when
v
— O Hence we can
.
as
||u||
=
=
only for v
y/v
V,
•
What
O.
the geometric interpretation of
is
norm?
Prove that
\\v
+
and that equality holds
some number a > 0.
(e)
z
of two numbers.
•
||y||
which
v)
of two vectors; the product
define the norm
the
•
of these equations involves
the ordinary product
(d)
V
-
that
a
(c)
•
Show
<
w\\
if
||
+
i>||
and only
||w||,
v
if
—
or
w =
or
w =
a
v for
that
V
'
w —
+
||l'
w\\"
—
v
||
U)\\~
4
3.
(a)
Let Re be rotation by an angle of 6 (Problem
Re(v)
(b)
Let e
let
=
w =
with the
(1, 0)
axis
is
is
.
•
1
w =
•
w =
ii||
||
the angle between v
||
it
•
||
COS0,
and w.
1
that
first axis, and
makes an angle of
Calculate that
cos#.
•
that
pointing along the
a vector of length
1).
Show
w.
that in general
v
where
v
(compare Problem
e
Conclude
=
be the vector of length
(cos#, sin 0); this
first
Re(w)
•
1).
4,
4.
Given two vectors v and w, we'd expect
the coordinates v\,
i^i, u>2,
t>2,
Appendix
Vectors
1.
79
have a simple formula, involving
to
for the area of the parallelogram they span.
Figure 9 indicates a strategy for finding such a formula: since the triangle
with vertices
the
problem
u;,
+
A, v
u>
is
an easier one where one
to
we can reduce
congruent to the triangle 05u,
side of the parallelogram lies along
the horizontal axis:
The
(a)
L
line
Conclude
and
passes through v
that the point
B
parallel to w, so has slope wj/u)].
is
has coordinate
-
V\W2
W\V2
W2
and
that the parallelogram therefore has area
det(i\
FIGURE
w)
=
—
v\u>2
W[V2-
This formula, which defines the determinant det, certainly seems to be simple
9
enough, but
we
all,
can't really
it
clearly
be true that det(u, w) always gives the area. After
have
det(it), v)
so
sometimes det
tion"
made
all
coordinate,
etc.)
be negative! Indeed,
will
sorts
= — det(u,
of assumptions (that
Nevertheless,
it
seems
w),
easy to see that our "deriva-
it is
was
u>2
B had a positive
positive, that
likely that det(u,
w)
is
±
the area; the
next problem gives an independent proof.
(a)
If v points
along the positive horizontal
area of the parallelogram spanned by
axis (u>2
(b)
If
Re
is
>
0),
i;
axis,
and the negative of the area
rotation by an angle of
det(/?6iu,
for
for
w
(Problem
Row)
=
show
w
and
1),
that det(u,
w above the
below
show
w)
is
the
horizontal
this axis.
that
det(i>, w).
Conclude that det(i\ w) is the area of the parallelogram spanned by
v and w when the rotation from v to w is counterclockwise, and the
negative of the area
Show
and
when
•
sin#
1.
det(u,
w + z) =
det(u, w)
det(u
+ w, z) =
det(i>, z)
10
+ det(i>, z)
+ det(u;, z)
that
=
det(a
Using the method of Problem
3,
w)
=
det(i',
FIGURE
clockwise.
that
a det(u, w)
u;||
it is
which
is
also obvious
v,
w)
show
||
i»||
=
det(u, a
w).
that
||u)||
•
sin 0,
from the geometric interpretation (Figure
10).
80
Foundations
APPENDIX
THE CONIC SECTIONS
2.
Although we
will
be concerned almost exclusively with figures
defined formally as the
set
of all pairs of real numbers, in
to consider three-dimensional space,
real
this
which we can describe
in the plane,
Appendix we want
in
terms of
triples
of
numbers, using a "three-dimensional coordinate system," consisting of three
straight lines intersecting at right angles (Figure
now mutate
to
two axes
in a horizontal plane,
1).
Our
horizontal
and
vertical
axes
with the third axis perpendicular to
both.
(1,0,0)
One
of the simplest subsets of
cone
around the third
illustrated in Figure 2; this
FIGURE
of slope
C
say,
this
may
three-dimensional space
is
the (infinite) cone
be produced by rotating a "generating
line,"
axis.
1
slope
FIGURE
2
For any given
first
two coordinates x and
+
plane has distance v x
(x, y, z)
(1)
We
C
is
y
from the
on the cone
if
origin,
and thus
and only
if z
can descend from these three-dimensional
dimensional one by asking what happens
plane
y, the point (x, y, 0) in the horizontal
P
(Figure
= ±Cvx" +
vistas to the
when we
y
more
intersect this
.
familiar two-
cone with some
3).
cone
plane
FIGURE
If the
P
3
plane
intersection
is
is
parallel to the horizontal plane, there's certainly
just a circle.
Otherwise, the plane
P
no mystery
— the
intersects the horizontal plane
Appendix
4,
We
in a straight line.
can make things a
everything around the vertical axis so that
from the plane of the paper, while the
The Conic
2.
simpler for ourselves
lot
axis
—C
(Figure 4)
is its
view-point the cone
If this line
itself
L happens
then equation
(1)
simply appears as two straight
to
be
vertical, consisting
2
which
^2 2 = r 2 2
— C
C a
y
as the collection of
and
form
we
we
see
from
this
all
axes;
lines.
points (a,z) for
and the plane
some
a,
consists of
where
,
M
is
third axes, the line
first
points of the
all
(x,Mx
(x, y, z)
Combining (1) and
the plane if and only
is
(2),
P
in the plane
we
L can be described
+ B),
the slope of L. For an arbitrary point (x, y, z)
(2)
if
and only
see that (x, y, z)
is
if
z
follows that
it
= Mx +
B.
in the intersection of the
cone and
if
Mx + B = ±C V^ + r
(*)
Now we
first axis,
(Figure
axes
is
have to choose coordinate axes in the plane P.
measuring distances from the intersection
5);
original
for the
second
second axis
we just choose
If the first coordinate
axis.
x, then the
can be written
first
in the
coordinate of
for
some a and
6.
On
respect to these axes
Q
We
can choose L as the
with the horizontal plane
the line through
of a point in
P
Q
parallel to
our
with respect to these
point with respect to the original axes
this
form
ax
5
all
that
an hyperbola.
is
Otherwise, in the plane of the
FIGURE
of
says that the intersection of the cone
z
4
out
points (a, y, z) with
all
FIGURE
rotate
in the usual position that
is
The plane P is thus viewed "straight on," so
intersection L with the plane of the first and third
are familiar with.
slope
we
if
this intersection line points straight
first
81
Sections
is
the other hand,
y, then y
is
+
if
yS
the second coordinate of the point with
also the
second coordinate with respect to the
original axes.
Consequently,
cone
if
and only
(*)
says that the point
M(ax +
Although
this
lies
on the
intersection of the plane
and the
if
B)
+ B = ±cV(ax +
B)
looks fairly complicated, after squaring
2
+y 2
.
we can
write this as
C 2 y 2 - a 2 (M 2 - C 2 )x 2 + Ex + F =
some E and F that we won't bother writing out.
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, we see that the
for
82
Foundations
E and F
values of
(1)
If
(2)
If
M = ±C we obtain a parabola;
C > M~ we
C < M 2 we
(3) If
These
are irrelevant:
obtain an
ellipse;
obtain an hyperbola.
analytic conditions are easy to interpret geometrically (Figure
If
(1)
our plane
is
parallel to
one of the generating
lines
6):
of the cone
we
obtain
a parabola;
our plane slopes
If
(2)
intersection omits
(3) If
less
than the generating
line
of the cone
one half of the cone) we obtain an
our plane slopes more than the generating
line
(so that
our
ellipse;
of the cone
we
obtain an
hyperbola.
FIGURE
In
6
names of these "conic sections" are
comes from a Greek root meaning
the very
fact,
The word
parabola
related to this description.
'alongside,' the
same root
mention paradigm, paradox, paragon, paragraph,
and even parachute. Ellipse comes from a Greek root
that appears in parable, not to
paralegal, parallax, parallel,
meaning
dicate
'defect,'
or omission, as in
ellipsis
(an omission, ... or the dots that in-
And hyperbola comes from a Greek root meaning 'throwing beyond,' or
With the currency of words like hyperactive, hypersensitive, and hypervennot to mention hype, one can probably say, without risk of hyperbole, that
it).
excess.
tilate,
t
His.
root
is
familiar to almost everyone.*
PROBLEMS
1.
I
I
<
.
I
RE
Consider a cylinder with a generator perpendicular
(Figure
7
7);
the only requirement for a point {x y, z) to
,
* Although the correspondence between these roots
tifully. fi>r
io desi
iil><
to the horizontal
the sake of dull accuracy
features of
<
it
lias to
and
lie
on
this
plane
cylinder
is
die geometric picture correspond so beau-
be reported that the
ertain equations involving the conic
(
heeks originally applied the words
set lions.
4,
that
(jc,
y) lies
on a
2.
The Conic
83
Sections
circle:
=C
y
Show
Appendix
.
that the intersection of a plane with this cylinder
can be described by
an equation of the form
(ax
What
2.
In Figure 8, the sphere S\ has the
lies
Similarly, the
F2
at
8
along the cylinder;
equator Ci of S2
z
lies
it
is
as the cylinder, so that
why
the length of the line
line
L from
from
z to Fjis
P and
is
plane and a cone
cone.
is
is
an
an
ellipse,
with foci F\ and
F2
is
;
l
R E v
that
.
when
the plane intersects just
prove that the intersection
is
one half of the
an hyperbola when
the plane intersects both halves of the cone.
(
show
a constant, so
prove geometrically that the intersection of a
ellipse
Similarly, use (b) to
(a)
Explain
equal to the length of the vertical
a similar fact for the length of the line from z to F2,
Similarly, use Figure 9(a) to
l
P
z to C\.
that the intersection
l
its
at Fi.
tangent to
the cylinder.
the distance from z to F\ plus the distance from z to F2
3.
P
also tangent to the plane
along the cylinder, and S2
be any point on the intersection of
By proving
(b)
same diameter
.
Let
(a)
.
possibilities are there?
equator C\
FIGURE
+ /3) 2 +y 2 = C 2
(b)
,
84
.
Foundations
APPENDIX
POLAR COORDINATES
3.
In this chapter we've been acting
all
along as
points in the plane with pairs of numbers.
length r
ways, each giving
rise to
of a point are called
its
if
Actually, there are
In
many
to label
different
usual coordinates
cartesian coordinates, after the French mathematician
many
situations
coordinates (r,9), where r
is
1
introduced the idea of
first
more convenient to introduce polar
To the point P we assign the polar
is
it
coordinates, which are illustrated in Figure
FIGURE
The
a different "coordinate system."
and philosopher Rene Descartes (1596-1650), who
coordinate systems.
one way
there's only
.
the distance from the origin
O
and 9
to P,
is
the
1
measure, in radians, of the angle between the horizontal axis and the
O
This 9
to P.
is
have either 9
completely ambiguous at the origin O. So
is
through the origin
For example, points on the
not determined unambiguously.
right side of the horizontal axis could
if
we want
to assign a
it
is
from
line
=
or 9
=
2tt
;
moreover, 9
necessary to exclude
unique pair
(r,
some ray
9) to each point
under
consideration.
On
the other hand, there
In
(r,9).
to
(r,
fact,
when
9)
no problem
is
it
is
possible (though not
r
<
0,
associating a unique point to any pair
approved of by
all)
according to the scheme indicated
in
to associate a point
Figure
2.
Thus,
it
always makes sense to talk about "the point with polar coordinates (r,9)," (with
or without the possibility of r
talk
P
is
and
<
0),
even though there
is
some ambiguity when we
about "the polar coordinates" of a given point.
the point with polar coordinates (r,8\)
also the point with polar coordinates
(-r,92 ).
I
I
CURE
It is
(r,
2
clear
from Figure
1
(and Figure
2) that the
point with polar coordinates
6) has cartesian coordinates (x, v) given by
x
=
r
cos 9
v
=
r sin
9
4,
Conversely,
ordinates
Appendix
a point has cartesian coordinates (x,y), then (any of)
if
85
3. Polar Coordinates
polar co-
its
9) satisfy
(r,
=±v? + y
tan0
=-
4
if*#0.
x
Now
suppose that /
is
ordinates we mean the
satisfying r
=
collection of
f{9).
all
FIGURE
3
as
opposed
all
points
the
P
In other words, the graph of
graph of f
we
in polar co-
with polar coordinates
/
in polar coordinates
No
points with polar coordinates (f(9),9).
should be attached to the fact that
first,
Then by
a function.
collection of
(r,
is
9)
the
special significance
are considering pairs (f(0),0), with f(9)
graph of /; it is purely a matter of
polar coordinate and 9 is considered the
to pairs (x, f(x)) in the usual
convention that r
is
considered the
first
second.
The graph of /
equation
for
all 9.
r
=
in polar coordinates
often described as "the graph of the
is
/(#)." For example, suppose that
The graph of the equation
radius a (Figure
3).
This example
coordinates are likely to
make
r
=
a
is
a constant function, f(9) = a
simply a circle with center O and
/
illustrates, in
is
a rather blatant way, that polar
symmetry
things simpler in situations that involve
with respect to the origin O.
The graph of the equation r = 9 is shown in Figure 4. The solid line corresponds
to all values of 9 > 0, while the dashed line corresponds to values of 9 < 0.
FIGURE
Spiral of Archimedes
4
As another example involving both
the equation r
=
positive
and negative
Figure 5(b) shows the part corresponding to n/2
check that no
this
FIGURE
5
new
same graph
r,
consider the graph of
cos9. Figure 5(a) shows the part that corresponds to
in
points are
added
<
9
<
> n or 9 <
for 9
n\ here r
0.
It is
<
<
0.
9
<
it/2
You can
easy to describe
terms of the cartesian coordinates of its points. Since the polar
coordinates of any point on the graph satisfy
r
—
cos
i
86
Foundations
and hence
=
r
cartesian coordinates satisfy the equation
its
x
which describes a
circle
—
cos
2
+ y2 =
x
(Problem 4-16). [Conversely,
coordinates of a point satisfy x 2
r
r cos 0,
+ y2 =
x, then
two
circle in
clear that
it is
if
turns out that
we can
ellipse in
Figure 6 shows an ellipse with
distances of
all
points from
O
we might
polar coordinates. But
if
and the other focus f being 2a. We've chosen f
of O, with coordinates written as
left
(-2ea,0).
(We have
FIGURE
<
e
<
1,
since
we must have 2a >
distance from f to O).
6
distance r from (x, y) to
O
is
given by
=x 2 + y 2
2
r
(1)
By assumption,
the distance from (x, y) to f
(2a
-r) 2 =
-
(x
.
2a
is
[-2ea])
—
2
r,
hence
+ y2
,
or
Aa
(2)
Subtracting
(1)
from
2
- bar + r 2 =
(2),
well be
we choose one of the foci as
one focus at O, with the sum of the
get a very nice equation
the origin.
The
the cartesian
on the graph of the equation
different ways,
an
hesitant about trying to find the equation of
the
lies
0.]
Although we've now gotten a
it
it
x
2
+ 4eax + 4s 2 a 2 +
and dividing by 4a, we
a
—
r
=
ex
+e
get
9
a,
y
2
to
Appendix
4,
87
3. Polar Coordinates
or
r
= a — ex —
s a
—
—
=
which we can write
(1
2
e")a
EX,
as
r
(3)
= A-sx,
Substituting r cosO for x,
A =
for
(1
2
—
£ )a.
we have
= A — sr co$0,
r(l + e cosO) = A,
r
and thus
r
(4)
=
1
we found
In Chapter 4
+ £ COS
that
2
x2
y
+ T, =
ab~
(5)
is
}
1
the equation in cartesian coordinates for an ellipse with la as the
distances to the foci, but with the foci at (— c, 0)
b
=
y/a 2
Since the distance between the foci
units, so that the focus (c, 0)
=
take c
sa or e
=
now
is
by
(4),
where
this ellipse
we
determining A).
(3)
and again using
we can
e
(4)
left
by c
when we
Conversely, given
is
A
=
=-,
c
=
=
sa,
\l a
2
—
we
c2
-s 2
get
—
\l a
2
—
s 2a 2
= ay —
1
£2
=
yr
from
moved
(3),
1
Thus,
is
get the ellipse
for the corresponding equation (5) the value of a
a
b
when
2c,
(c, 0),
of the
.
at the origin,
c/a (with equation
the ellipse described
determined by
is
-c 2
and
sum
E'
obtain a and b, the lengths of the major and minor axes, immediately
and A.
The number
the
eccentricity
of the
major and minor
ellipse,
c
Va 2 -b 2
a
a
I
(y
V
determines the "shape" of the
axes), while the
number A determines
its
ellipse (the ratio
"size," as
of the
shown by
(4).
.
88
.
Foundations
PROBLEMS
1.
two points have polar coordinates
If
d between them
tance
d
What does
2.
/
/
(ii)
is
odd.
=
that the dis-
2r\r2cos(0\
—
Go).
/
in
polar coordinates
if
f(9+jz).
Sketch the graphs of the following equations.
(i)
r
(ii)
r
(iii)
r
(iv)
r
— a sin#.
— a sec9. Hint:
— cos 29. Good
= cos 39.
(v)
r
=
\
cos 29
1
(vi)
r
=
|
cos 39
\
It is
a very simple graph!
luck
on
this
one!
Find equations for the cartesian coordinates of points on the graphs
and
5.
even.
show
(rj, Oi)-,
say geometrically?
this
is
f(9)
(iii)
4.
1
+ r2 —
r\"
Describe the general features of the graph of
(i)
3.
O
=
and
(r\, 9\)
given by
is
(iii)
in
Problem
(i),
(ii)
3.
Consider a hyperbola, where the difference of the distance between the two
foci
and choose one focus
the constant 2a,
is
we must have
(In this case,
>
s
at
Show
1).
O
that
and the other
we
at
(— 2ea,
0).
obtain the exact same
equation in polar coordinates
A
=
r
+ s cos 9
1
as
we
obtained for an
Consider the
to the line
of points
set
from
the distance
is
a
ellipse.
—
r
(x, y)
(x, v) to the line
cos 9,
a
Now,
tion
for
(4),
this
any
A
and
s,
.
i
i<
i
;
7
r(\
7).
Show
that the equation
O
is
equal to
that the distance
can be written
+ cos 9).
which implies
(3).
Show
e )x
2
—
1
,
and a hyperbola
for e
>
again of the same form as
in
(4).
polar coordinates of the equa-
that the points satisfying this equation satisfy
+
Using Problem 4-16, show that
s
is
consider the graph
2
(
=
a (Figure
equation for a parabola
(1
i
=
(x, y) to
=a
Notice that
i
y
and conclude
(x,y)
y
such that the distance
v
2
this
1
= A2
is
an
2Asx.
ellipse for s
<
1,
a
parabola for
.
4,
8.
(a)
(b)
(c)
Sketch the graph of the
Show
Show
that
it is
that
it
and conclude
that
2
it
(x
10.
+
v
2
=
y/x
2
+ y2 -
y,
2
+
2
y
+ y) 2 =
x
2
+
2
y
Sketch the graphs of the following equations.
(i)
r
(ii)
r
(iii)
r
(a)
(b)
(c)
= l-isin0.
= l-2sin0.
= 2 + cos#.
Sketch the graph of the
Find an equation for
Show
d\d2
(d)
8
sin#.
can be described by the equation
r
FIGURE
89
sin 9.
= —1 —
r
3. Polar Coordinates
can be described by the equation
x
9.
=1—
cardioid r
graph of
also the
Appendix
that
=
Make
it
is
its
lemniscate
2
=
2
2a cos26.
cartesian coordinates.
the collection of
all
points
P
in Figure 8
satisfying
a~
a guess about the shape of the curves formed by the
satisfying d\di
=
b,
when b >
a
and when b < a 2
.
set
of
all
P
.
CHAPTER
LIMITS
The concept
one
in all
are,
once more, going
is
PROVISIONAL DEFINITION
of a limit
not the
The
of calculus.
word
surely the
The
Of the
six
at a
to
—
a.
our
it is
We
the definition of limits, but
what we
we
shall define
/
near a,
is
if
we can make f(x)
sufficiently close to,
1
,
only the
not defined, and h(a)
and h approach
/
near
a.
but unequal
This
is
is
first
as close as
three approach
defined "the
because
we
to, a.
we
/
at a
wrong way,"
it
explicitly ruled
definition, the necessity of ever considering the value of the function
only necessary that f(x) should be close to
/
for
x close
to a, but unequal
are simply not interested in the value of f(a), or even in the question of
whether f(a)
i
the limit
functions graphed in Figure
true that g
still
out, in
is
difficult
"limit" but the notion of a function approaching a limit.
Notice that although g(a)
is
goal of this chapter
by requiring that x be
/
most important, and probably the most
to begin with a provisional definition;
/ approaches
function
like to
is
is
defined.
i<;rki;
One
way of picturing the assertion that / approaches / near a is
provided by a method of drawing functions that was not mentioned in Chapter 4.
In this method, we draw two straight lines, each representing R, and arrows from
convenient
a point x
in
one, to fix)
different functions.
90
in
the other. Figure 2 illustrates such a picture for two
5. Limits
f(x)
(a)
FIGURE
Now
=
91
c
2
consider a function
that f(x) be close to
/,
/ whose drawing
open
say within the
looks like Figure
3.
B which
interval
Suppose we ask
has been drawn
we consider only the numbers x in the
A of Figure 3. (In this diagram we have chosen the largest interval which
will work; any smaller interval containing a could have been chosen instead.) If we
This can be guaranteed
in Figure 3.
if
interval
FIGURE
FIGURE
3
choose a smaller interval B' (Figure
but no matter
to
how
small
be some open interval
A
we choose
A which
4)
we will,
the
open
usually,
4
have to choose a smaller
interval B, there
is
A',
always supposed
works.
similar pictorial interpretation
is
possible in terms of the
graph of /, but
in
B must be drawn on the vertical axis, and the set A on the
horizontal axis. The fact that / (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.
this case the interval
)
92
Foundations
To
=
a
take a specific simple example,
5 (Figure
be able
to
f(x)
=
3x
5
To be
specific,
This means that
.
which we can
to
15
—
to
IT)
— <3x <
1
to require that
—
x
< —-,
1
s
j
30
or simply
\x
— 5| <
we
take any positive
—
that \x
5
<
1
There
3^;
easy to guarantee that
if
10'
Jo-
1
6
+
15
<3x ~ l5<
— wt: <
FIGURE
we require that x be sufficiently
make sure that 3x is within
of
like if
also write as
we just have
this
|3jc
—
nothing special about the number
is
15|
number
30
=
—
how
same
—
function f(x)
show
sort of argument
=
—
x
<
a\
works
a
is
\x
—
a\
<
=
3,000,000.v.
e/3, 000, 000 in
little
more
Presumably,
interesting.
we
should be
This means that we need to show
first
step
is
-
which gives us the useful
3 or 3,000,000.
3
< x
1
about
<
1.
+3 <
e
by requiring
e
2
\x
=
~9\
—
3|
how
Once
-3|
\x
—
|.v
to
3|
be small enough. The
+3|,
\x
•
Unlike the situation with the previous
factor.
examples, however, the extra factor here
something
<
9|
to write
\x
5
for the function f(x)
that f(x) approaches 9 near 3.
any given positive number
—
<
a\
to ensure the inequality
obvious
\x
e
e.
2
like
easy to see that
that
have to be 1,000,000 times as careful, choosing
|.r
for
<
3a\
5. It's just as
to require that
order to ensure that \f(x)
The
To ensure
the limit 3a at a for any a:
Naturally, the
able to
.
e/3.
\x
We just
just as
l
1
\3x
we just have
is
< LOO simply require that \x 5| < 300 In fact,
we can make \3x — 15 < s simply by requiring
;
e
jq. It
—
There's also nothing special about the choice a
/ approaches
3x with
—we ought
have
-Jo
To do
we
we want
suppose
we want
=
consider the function f(x)
limit 15 near 5
let's
Presumably / should approach the
f(x) as close to 15 as
to get
close to 5.
1
6).
is
which
|jc+3|,
But the only crucial thing
big
|jc
+
3|
So
is.
the
we've specified that
7 and we've guaranteed that
2
\x
- 91 =
l.v
-
31
•
is
—
3|
\x
+
3\
+
31
make
thing we'll
first
|.v
\x
to
isn't
a convenient constant
sure that
do
is
wc can
say
to require that
< 1, or 2 < x < 4, we have
< 7. So we now have
<
7|.v
-
31,
93
5. Limits
which shows that we have
also required that
\x
—
< min(e/7,
3|
The
—
|jc
3
\x
<
1
2
<
9|
convenient number. To
chose
|jc
—
it is
3
1
<
—
<
3|
was simply made
1
—
make
how
the
argument would read
we
if
10.
show
to
/ approaches a
that
/ approaches 9
near 3
again with the expectation that
Problem 1-12 shows
+ a\. We
|jc
first
ensure that
this will
work to show
worry a bit more
will basically
we need
near a for any a, except that
about getting the proper inequality for
fact,
We
for convenience.
— 3| < 10 or any other
|jc
3|
jq or \x
sure you understand the reasoning in the previous
<
a good exercise to figure out
Our argument
that
< e/7, provided that we've
more official: we require that
3|
look
it
1).
could just as well have specified that
paragraph,
—
£ for \x
make
Or, to
1.
specification \x
initial
—
|jc
to
require that
+a
|
is
|jc
— a\ <
not too large.
1,
In
that
<
\a\
<
\x
1.
so
x\
1
+
\a\
+
\a\
<
<
and consequently
\x+a\ <
so that
we
\x\
+
2\a\
1.
then have
2
\x
-a 2 =
\x
<
\x
\
—
—
+ a\
a\
|*
a
(2|fl|
I
+ l),
— a"\ < s for \x —a\< e/(2\a\ + 1), provided that we
— a\ < min(e/(2|a| + 1), 1).
also have \x — a\ < 1. Officially: we require that
In contrast to this example, we'll now consider the function fix) = \/x (for
x ^ 0), and try to show that / approaches 1/3 near 3. This means that we need
to show how to guarantee the inequality
which shows that we have
\x"
|.v
I
<
for
any given positive number s by requiring
£
— 3|
\x
to
be small enough.
We
begin
by writing
1
~3 =
x
giving us the nice factor
the
problem
factor
small, so that \/\x\
We
can
first
1/|.\
|.
1
1
=
3x
— 3\, and
\x
11
^ _ V
3-x
l
1
1
'
•
3
|jc
— 3|,
\x
even an extra i for good measure, along with
we
In this case,
first
need
to
make
sure that
|.v|
isn't
won't be too large.
require that
|jc
—
3|
<
1,
because
111
<
4
<
x
r
this gives
2
<x <
4, so that
too
,
94
Foundations
which not only
— < —
us that
tells
order to conclude that
—
< —
Ul
.
>
but also that x
,
We now
which
0,
important in
is
have
2
< -|,-3|
which shows that we have
—
\x
3|
<
wanted
to
show
< min(6£,
3|
If we instead
by
<
To show
—a
this
possible values for x
our
— (— 3)| <
The
a
<
7
a
—
\x
look
3
<
1
6s, provided that we've
official again:
we
require that
1,
works as before.
than
—
It's
we proceed in
more careful
\/a near a for any a
that, again,
we have
be a
to
little
not good enough simply to require that
or any other particular number, because
1,
that are negative (not to
if
a
is
close to
mention the embarrassing
0, so that f{x) isn't even defined!).
trick in this case
is
to first require that
—
\x
FIGURE
it
/ approaches — 1/3 near —3, we would begin
giving —4 < x < —2, once again implying that
initial stipulation.
less
x
s for
make
/ approaches
would allow values of x
possibility that
<
1
that
same way, except
should be
|
|jc
in general that
in formulating
1/3
Or, to
.
1/2, so that everything
basically the
\x
1
1).
stipulating that
|1/jc|
—
i/x
—
also required that \x
<
a
>
in other
words,
You should be
we
require that x be less than half as far from
able to check
first
that x
/
and
that
<
l/|.v|
as a (Figure 7).
2/|fl|,
and then work
out the rest of the argument.
With
quail at
really
you may have begun to
the prospect of tackling even more complicated functions. But that won't
all
the
work required
we
be necessary, since
will eventually
rely on. Instead of worrying
in functions like f(x)
=
behavior of
=
x —
=
We
want
it
(remember
a from consideration, so
at 0).
to
show
that
\/x
we'll turn
,
—
that
it
is
xsinl/x (Figure
get f{x)
=
x
sin
\x
— 0|
is
1
as desired
In other words, for any
I
.v
<
sin
e
sufficiently small (but
<
x
exempts
even defined
isn't
l/x as close to
/
I
Sill
/ approaches
specifically
doesn't matter that this function
we can
l/U)-0| =
=
Despite the erratic
8).
clear, at least intuitively, that
we require that x be sufficiently close to 0, but ^ 0.
number e > 0, we want to show that we can ensure that
|.v|
our attention to some
our provisional definition
if
by requiring that
we can
be even more frightening.
function near
this
near a
/
=
or f(x)
to
the function f(x)
first
have some basic theorems that
about the unpleasant algebra that might be involved
x
examples that might appear
Consider
for these simple examples,
for all
x
^
0,
0).
But
this
is
easy.
Since
95
5. Limits
we have
.v
so
we can make
|jc
near
< x
<
s simply
sin l/.v|
=
For the function f(x)
proaches
sin
0.
If,
for
-
a
-
for
,
all
0,
by requiring that
l/x (Figure 9)
sin
/
x
|a|
<
and
e
/
0.
seems even clearer that / ap-
it
example, we want
x
1
<
sin
To
then
that
/
' I ft
\
/ f(x)
=
x 2L
we
\x
certainly
2
<
\
-4y
need only require
<
\
^0,
and i
jq
since this implies
and consequently
x
l
sin
that \x
-
2
—
1
.
sin
<
<
<
\x'
100
x
(We could do even
better,
and allow
|jc
< 1/V
|
10 and x
particular virtue in being as economical as possible.)
^
0,
but there
In general,
if
s
>
is
no
0, to
ensure that
1
x sm
FIGURE
we need
9
provided that e <
though
it
that
<
As a
FIGUR E
= Vl-vhm
is
£,
"small"
but
make
it
If we are
e's
<
s
and
jc^O,
given an £ which
which are of
interest),
is
greater than
then
certainly suffices to require that
\y/\x\ sin l/.v|
1
<
£
|.v|
(the algebra
I
1.
|
third example, consider the function f(x)
order to
f(x)
e,
only require that
|jc
|a|
<
is left
to you)
we can
<
s
it
|jc
—
require that
and
x
^
j
1
does not
<
x
1
|
and
sin
(it
might be, even
suffice to require
i^O.
l/x (Figure
10).
In
96
Foundations
Finally, let us
it is
false that
/ approaches
^
and
0.
condition \f(x)
|jc|
is
In
to be.
matter
near
l/x (Figure
sin
For
1 1).
This amounts to saying that
0.
this
function
is
not true
it
that we can get \f(x) — 0| < s by choosing x sufficiently
e >
To show this we simply have to find one e >
for which the
— 0| < s cannot be guaranteed, no matter how small we require
number
for every
small,
=
consider the function fix)
how
=
fact, s
small
some number x
7 will do:
we
require
=
\/(\tt
|jc
it
is
impossible to ensure that
A
to be; for if
|
+ Inn)
which
is
|/(jc)|
any interval containing
is
in this interval,
and
for this
<
\
no
0, there
x we have
f{x)=\.
FIGURE
I
1
This same argument can be used (Figure 12) to show that
we must
\f(x) — l\ <
number near 0. To show
some number s >
so that
any
The
required to be.
small
we
that for
require
—
\
e
is
works for any number
/;
that
we cannot ensure that \f(x) — l\ <
A containing we can find both x\ and
any interval
number
no matter how small x
not true,
to be,
|.r|
not approach
again find, for any particular
this
choice s
/ does
is,
5.
/,
is
no matter how
The reason
is,
X2 in this interval
with
FI
G U RE
=
/(*i)
12
and
l
=
f(x 2 )
-l,
namely
X]
and
Kit
for large
both
1
,
since
its
1
no matter what
1
kJT
total length
—
x?
+ 2nn
enough n and m. But the
— and
1
1
I
=
/
1
<
I
from
interval
is
only
and
also
1
;
|
so
—
/
+ 2niJT
7 to
/
+
^
cannot contain
we cannot have
— —
1
/
1
<
^,
/is.
The phenomenon exhibited by f(x) =
f we consider the function
,,
.
fix)
=
[0.
,
I
sin l/.v
near
x irrational
-
.
x rational.
can occur
in
many
ways,
5. Limits
97
/ does not approach any number / near a. In fact, we
cannot make \f(x) — l\ < \ no matter how close we bring x to «, because in any
interval around a there are numbers x with fix) = 0, and also numbers x with
f(x) = 1, so that we would need |0 — /| < j and also |1 — /| < -j.
no matter what a
then,
An amusing
.-•'/(*)
=
x,
x rational
0,
x irrational
is,
on
variation
is
presented by the function shown in
Figure 13:
/(*)
The behavior
proaches
at 0,
should have no
FIGURE
behavior
this
We
13
of
this
=
function
x,
x rational
0,
.v
irrational.
but does not approach any
number
convincing yourself that
difficulty
-
1
1,
x >
If
a
>
0,
/ approaches
then
1
certainly suffices to require that
near
-a
or
f{x)
=
-1, X <
14
—
limits
1.
Similarly, if b
<
easily check,
The time
14):
—
1
<
e
it
a, since this implies
< x
< x
/ approaches —1 near b: to ensure
— b\ < —b. Finally, as you may
s it
(—1)1
/ does not approach any number near 0.
so that f{x)
that \f(x)
FIGURE
=
you
true.
indeed, to ensure that \f(x)
a:
— d\ <
|jc
a
ap-
x>0.
1,
=
is
it
<
x
,
at a, if
this
conclude with a very simple example (Figure
fix)
f(x)
= sin l/x;
^ 0. By now
"opposite" to that of g(x)
is
has
<
0,
then
suffices to require that ]x
now come
to point out that of the
which we have given, not one has been a
with our reasoning, but with our definition.
If
many
demonstrations about
The
real proof.
fault lies
not
our provisional definition of a
function was open to criticism, our provisional definition of approaching a limit
is
even more vulnerable.
used in proofs.
It
is
This definition
hardly clear
is
simply not sufficiently precise to be
how one "makes"
"close" means) by "requiring" x to be sufficiently
ciently" close
is
certainly
supposed
fix) close to / (whatever
close to a (however close "suffi-
to be). Despite the criticisms of
hope you do)
that our
(I
ing.
In order to present any sort of argument at
It is
may
arguments were nevertheless quite convinc-
feel
to invent the real definition.
our definition you
all,
we have been
practically forced
possible to arrive at this definition in several steps,
each one clarifying some obscure phrase which
still
remains. Let us begin, once
again, with the provisional definition:
The (unction / approaches the limit near a. if we can make fix) as close
as we like to / by requiring that x be sufficiently close to, but unequal to, a.
/
The
very
fix) close
change which we made
first
to
/
meant making \f(x) —
in this definition
/|
small,
and
was
to
note thai making
similarly for x
and
a:
98
Foundations
The
/ approaches
function
we
small as
like
the limit
by requiring that
\x
near a,
/
— a\
be
The second, more crucial, change was to note
we like"" means making \f(x) — l\ < s for any
The
/ approaches the limit
can make \f(x) — l\ < s by requiring
x
function
/
a.
is
a
There
common
pattern to
number
we can make 1/00 —l\
as
a.
making \f(x) — l\ "as small as
> that happens to be given us:
e
that
\x
number
for every
if
—
a\
be
>
s
we
sufficiently small,
the demonstrations about limits which
all
^
and x
that
near a,
/
if
sufficiently small,
and
we have
>
we found some other positive number, 8 say, with
the property that if .r^a and \x — a\ < 8, then \f(x) — 1\ < s. For the function
= 0), the number 8 was just the number e;
f(x) = .v sin l/.v (with a = 0,
for f(x) = yixlsin l/x, it was e
for f(x) — x it was the minimum of 1 and
s/(2\a\ + 1). In general, it may not be at all clear how to find the number 8,
given. For each
s
/
;
given
but
s,
it is
\x
—a\<8
which expresses how small "sufficiendy"
/ approaches
the limit
/
the condition
small must be:
The
8 >
This
is
function
such
that, for all x, if \x
practically the definition
change, noting that
—
\x
DEFINITION
a
The
is
I
<
—
|x
a\
<
8
we
8
and x
if
for every e
^
>
there
—
a, then \f(x)
l\
<
is
some
e.
We will make only one trivial
a" can just as well be expressed "0 <
will adopt.
and x
^
5."
function
some
u
near a,
— a\ <
8
f
>
approaches the limit / near a means: for every e >
— a\ < 8, then \f(x) — 1\ < s.
<
all x, if
such that, for
This definition
is
there
|.\"
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
formal proofs of each. In most
to
make
the arguments
has been done already.
conform
When
cases, this will
merely involve a
our formal definition
to
proving that
/ does
not
—
all
bit
limits,
giving
of rewording
the algebraic
approach
/
at a,
work
be sure to
negate the definition correctly:
If
it is
not true that
for every e
(hen \f(x)
then
>
—
there
l\
<
s,
is
some
8
>
such
that, for all
.v,
if
<
a\
<
<5,
4
u
5
5.
there
<
— a\ <
\x
>
some £
is
8
such that for
but not \f(x)
—
consider £
but not
|
=
sin
and note
\
l/x
—
|
so large that l/(n/2
As a
limit,
<
=
that for every 8
i
there
some x which
is
99
satisfies
s.
l/x does not approach
sin
>
there
—namely, an x of
+ 2mt) <
the
form \/(n/2
+
near
<
some x with
is
\x
—
0,
0|
we
< 8
2mt), where n
is
8.
of the use of the definition of a function approaching a
final illustration
we have
<
l\
Thus, to show that the function f(x)
>
every 8
Limits
reserved the function
shown
standard example, but
in Figure 15, a
one of the most complicated:
0,
Jt
irrational,
\/q,
x
=
/(*)
(Recall that
and q >
p/q
is
terms
in lowest
p/q
if
<x <
1
< x <
in lowest terms,
p and q are
no
integers with
1.
common
factor
0.)
x irrational
0,
/(*)
=
—
x
,
= —
in lowest
terms
—
j
5
FIGURE
4
3
4
4
5
15
number a, with < a < 1, the function / approaches at a. To prove
number £ > 0. Let n be a natural number so large that \/n < £.
Notice that the only numbers x for which \f(x) — 0| < s could be false are:
For any
this,
consider any
2
3
1
4'
(If
a
is
rational, then a
these numbers, one
is
values \p/q
—
—
a\
if
<
\x
5,
1
any
closest to a; that
is,
rate,
only
— a\
\p/q
finitely
is
many
of these
many. Therefore, of all
smallest for one
p/q among
a happens to be one of these numbers, then consider only the
a\ for
<
—
might be one of these numbers.) However
be, there are, at
(If
n
1
5'
numbers there may
these numbers.
3
p/q
then x
/
is
a) This
not
closest distance
one of
n-
1
may
be chosen as the
8.
For
—
100
Foundations
and therefore \f(x) — 0| < s is true. This completes the proof. Note that our
description of the
which works for a given e is completely adequate there is no
reason why we must give a formula for 8 in terms of e.
—
<5
Armed with
our definition,
we
have probably assumed the result
This theorem
is
now prepared to
are
which
along,
all
our
really a test case for
prove our
first
theorem; you
a very reasonable thing to do.
is
definition:
theorem could not be
the
if
proved, our definition would be useless.
THEOREM
1
PROOF
A
function cannot approach two different limits near a.
near a, and
approaches
/
Since
our
this
is
/ approaches
theorem about
first
m
limits
near «, then
be necessary
will certainly
it
In other words,
if
/
= m.
I
to translate
the hypotheses according to the definition.
Since
8\
>
/ approaches
such that, for
also
<
|jc
— a\ <
then \f(x)
8\,
/ approaches m near
know, since
any
that for
>
e
there
some number
is
x,
all
if
We
we know
near a,
/
—
<
l\
a, that there
s.
some
is
82
>
such that,
for all x,
<
if
We
|jc
— a\ <
8\
one definition will
there
for any e >
is
in
conclude that
if
<
—
\x
<
a\
8,
then \f(x)
—
m\
<
e.
and 82, since there is no guarantee that the 8
work in the other. But, in fact, it is now easy to
have had to use two numbers,
which works
82,
some
then \f(x)
we simply choose 8 = min(<$i, dh).
To complete the proof we just have
8
—
>
I\
such
<
e
that, for all x,
and \f(x) — m\ <
>
to pick a particular s
for
e;
which the two
conditions
—
|/0r)
is
7^
a
>
and
£
\f(x)
— m\ <
s
0,
The proper choice is suggested by Figure 16. If
we can choose — m|/2 as our s. It follows that there
such that, for
all
x,
\x
-a\<8,
m, so that
8
<
m\ >
cannot both hold,
/
/|
|/
—
if
if
/
/
<
w.
|/
then
|/(jc)
- /| <
\l
—
m\
2
-- in
length
FI
G
1
RE
16
——
-
length
and f(x)
—
I
—m\ <
\l-m\
z
-
This implies that for
|/
-m =
\
\l
— a\ <
<
\x
-
f(x)
+
8
we have
fix)
-m\<\l-
f(x)\
\l-m\
=
a contradiction.
|
|/-m|,
+
|/(.v)
\l-m\
-
m\
5. Limits
The number which / approaches near a
denoted by lim fix) (read: the
is
/
101
limit
of fix) 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
x—ni
equation
=
lim f(x)
I
x—>a
has exactly the same meaning as the phrase
/ approaches
The
possibility
=
lim fix)
/
a.
/ does not approach / near a, for any /, so
number /. This is usually expressed by saying
remains that
still
is
near
/
false for
"lim fix) does not
every
that
that
exist."
x-*a
Notice that our
new
notation introduces an extra, utterly irrelevant letter x,
which could be replaced by
—
appear
/,
y, or
lim fix),
x—>a
all
to
any other
letter
which does not already
the symbols
lim /(f),
lim /(y),
t—>a
y—>a
denote precisely the same number, which depends on / and a, and has nothing
do with x, t, or y (these letters, in fact, do not denote anything at all). A more
logical
symbol would be something
like
lim /, but this notation, despite
its
brevity,
a
is
so infuriatingly rigid that almost
much more
useful because a function
lim fix)
x—>a
is
though
might be possible
it
the short
no one has seriously
to express
/
tried to use
often has
it.
The
notation
no simple name, even
fix) by a simple formula involving x. Thus,
symbol
+ shut)
lim(x
x— a
could be paraphrased only by the awkward expression
lim
/,
where fix)
=
x
+
sin
.v.
a
Another advantage of the standard symbolism
lim*
+
lim*
+
x— a
The
first
means
the
means
the
= x +f 3
by the expressions
r\
3
t
.
little
for
,
all .v;
number which / approaches near a when
fit)= x +
You should have
illustrated
number which / approaches near a when
f(x)
the second
is
3
t
for all
,
f.
difficulty (especially if you consult
lim x
+t 3 —
a
+r\
+f 3 =
x
+a 3
x -> a
Y\mx
.
Theorem
2)
proving that
,
102
Foundations
These examples
In
ibility.
fact,
what
forgetting
which
tation,
main advantage of our
illustrate the
the notation lim f{x)
x—>a
it
will
really
Here
means.
be important
later:
notation, which
so flexible that there
is
is
is
is
its
flex-
some danger of
a simple exercise in the use of this no-
first
interpret precisely,
and then prove the
equality of the expressions
and
lim f(x)
An
it
make
to find many limits, as we promised long ago. The proof depends upon
properties of inequalities and absolute values, hardly surprising when one
important part of
easy
certain
lim f{a +h).
chapter
this
the proof of a theorem which will
is
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
be presented
will
form of a lemma (a lemma is an auxiliary theorem, a result that
justifies its existence only by virtue of its prominent role in the proof of another
theorem). The lemma says, roughly, that if x is close to xq, 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
once again,
in the
close to 1/yo- This intuitive statement
estimates of the
first,
LEMMA
in
lemma, and
order to see just
it is
how
much
is
easier to
remember than
the precise
not unreasonable to read the proof of Theorem 2
these estimates are used.
(l)If
£
< - and
--vol
\x
\y
£
-y
<
\
-,
then
\(x
+ y) - Uo +
vo)!
< £
(2) If
\x
—
.vol
< min
1,
(
2(|yol
and
+D
|
—
y
yo
I
<
2(1x01
+
1)'
then
\xy
(3) If
#
y
/
v
—
2
\(x
Since
'
2
and
(1)
(2)
e\yo\
Vol
< min
yo|
1
1
y
yo
<
proof
e.
I
and
|
then y
-xoyo <
\x
—
+
xq\
-
y)
<
1
(xo
+
yo )
I
£.
= IU - xo) + (y - yo)|
£
£
<\x- .vol + \y - vol <i+2 =£
we have
|.V
|
—
|.V()|
<
\x
— xq\ <
1
'
5. Limits
103
so that
<
\x\
+
1
|*
|.
Thus
\xy
-
xoyol
=
\x(y
<
\x\
<
(1
-
y
)
\y
-
vol
+
+
+
vol
I
kol)
2(|.v
£
2(|vol
+D
have
>
Ivl
Ivol
l)
2
-
\yo\
so
\
+ - = £.
2
We
-x
+
+
|
)\
\x
•
£
< (3)
-x
yo(x
I
^
vol/2. In particular, v
<
\y\
-vol <
l.v
I
vol
and
0,
—
2
1
l.v
<
l.vol
I
Thus
1
1
y
theorem
2
If lim
x—*a
f(x)
=
I
lyo
vo
—
\yo\
\y\
and lim g(x)
2
vl
= m,
l.vol
2
—=r- =
f|v
1
_
|
8
-
I
ivol
then
x—*a
\im(f +g)(x)=-l
(1)
+ m;
x—>-a
Moreover,
PROOF
The
all
if
m ^ 0,
hypothesis
(2)
\im(f
(3)
lim
g)(x)=l
-m.
then
means
(-)(*) = -•
that for every £
>
there are <5i,^2
>
such
that, for
x,
and
This means
such
<
\x
if
<
\x
(since, after all,
e/2
—
—
is
a\
<
8\,
then \f(x)
—
a\
<
82,
then \g(x)
— m\ <
also a positive
l\
number)
<
s,
£.
that there are 8\, 82
>
that, for all x,
and
Now
<
if
let
|jc
—
8
=
a\
<
if
<
\x
— a\ <
<5i,
then \f(x)
—
if
<
|.v
—
52,
then \g(x)
—
min(<5i, #2).
81 are
both
If
true, so
a|
<
<
|.v
—
a\
<
5,
<
- /| < -
and
\g(x)
£
-,
m\ < -.
then
both
£
\f(x)
l\
£
- m\ < -
<
\x
—
a\
<
8\
and
104
Foundations
are true. But by part
This proves
£
of the
(1)
lemma
this implies that
/ + g){x) —
|(
(I
+ m)| < e.
(1).
To prove
> there
we proceed similarly, after consulting
such that, for all x,
8\, 82 >
(2)
are
<
if
—
\x
<
a\
—
then \f{x)
8\,
part
< min
1\
of the lemma.
(2)
1,
(
2(|ro|+
and
<
if
— a\ <
\x
1)
£
— m\ <
then \g(x)
82,
If
2(|/|+1)
Again
let 8
=
—
\f(x)
<
min(<$i, 82). If
/
.
< nun
I\
£
1,
I
2(\m\
So,
by the lemma,
>
Finally, if e
|(/
g)(x)
•
there
— a\ <
\x
—\
>
and
I
—
\g(x)
,ov
2(|/|
and
e,
such
proves
this
that, for all
<
\x
— a\ <
then \g(x)
—m\ < min
1
I'"
/
I
i2\
£\fn\~
1
2
makes
sense,
to part (3) of the
and second
lemma
this
means,
2
that g(x)
/'
^
0, so (l/g)(x)
that
-
-
(x)
,8/
This proves
first,
\
-r-,
I
K
But according
l)
,y,
.
(5,
+
(2).
/i
if
£
m\ <
l)J
—I m\ <
a 8
is
+
then
5,
-
<
m
£.
(3). |
Using Theorem 2 we can prove,
trivially,
such
facts as
— + 7x— = — ++la—
x^a x- +1
,.
hm
x3
5
3
a
=
a
5
=
2
-,
1
without going through the laborious process of finding a
8,
given an
e.
We
must
begin with
lim 7
=
7,
=
1,
=
a,
x—*a
lim
1
x—*a
lim x
x—*a
directly. If wc 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 8 such that, for all x,
but these are easy to prove
if
<
|.v
-
a\
<
8,
then
\x
+x -
(a~
+
a)\
<
£.
.
5.
Consulting the proof of Theorem
such that, for
all
we have
we
see that
do
9
if
<
|.v
—
a\
<
<5i,
then
\x
if
<
\x
—
a\
<
8i,
then
|x
=
already given proofs that lim x
—a
and
>
8i
9^"
< —
|
— a\ < —
=
a" and lim x
x—*a
a,
we know how
this:
\
2
Thus we can
take
=
8
min(5i, 82)
=
min
min
I
I
1
2\a\
If
find 8\
s
x—>a
to
we must first
x,
and
Since
2(1),
105
Limits
a
^
0, the
same method can be used
<
if
I
—
.V
I
<
<5,
then
x
The proof of Theorem 2(3) shows
such that, for all .v,
a 8 >
1
>
to find a 8
1
a
+
such that, for
Thus we can
\x
— a\ <
8,
<
1
az
£.
that the second condition will follow
then
—a
\x
\
x,
1
9
z
i
<
if
all
< min
i2
or
efl
i
if
we
find
i4
take
/
I
/
/
.
min
i2
\
2
<5
=
min
fieri
|<z|
'
2
1,
'
2|a|+l
V
Naturally, these complicated expressions for
<5
can be simplified considerably,
after
they have been derived.
One
technical detail in the proof of
order for lim fix) to be defined
it is,
as
x^-a
at «,
nor
is it
must be some
necessary for
8
>
/
to
Theorem 2 deserves some discussion. In
we know, not necessary for / to be defined
be defined
such that f(x)
is
x
at all points
^
a.
defined for x satisfying
However, there
<
\x
—
a\
<
8;
otherwise the clause
"if
<
\x
-a\<8,
then \f(x)
-
1\
<
e"
would make no sense at all, since the symbol f(x) would make no sense for
some x's. If / 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 i/g, since \/g is undefined for x with g{x) = 0. However, this fact gets
established in the proof of
Theorem
2(3).
106
Foundations
There are times when we would like to speak of the limit which / approaches
at a, even though there is no 8 >
such that fix) is defined for x satisfying
—
< \x
a\ < 8. For example, we want to distinguish the behavior of the two
functions shown in Figure 17, even though they are not defined for numbers less
I—
than
a.
we
For the function of Figure 17(a)
f(x)=l
lim
x—*a +
write
or
=
lim f(x)
I.
x\.a
(a)
(The symbols on the
These
I—
"limits
definition
such
left
are read: the limit of f(x) as x approaches a from above.)
from above" are obviously
very similar:
is
=
lim fix)
x—>a +
/
closely related to ordinary limits,
means
that for every e
>
there
and the
a 8
is
>
that, for all x,
<
if
—
a
<
8"
is
equivalent to "0
A'
8,
—
then \f(x)
<
l\
e.
(b)
FIGURE
17
(The condition "0 < x
—
<
a
<
—
\x
a\
"Limits from below" (Figure 18) are defined similarly:
—
lim fix)
I)
means
>
that for every e
there
a 8
is
<
8
and x >
a.")
lim fix) = I (or
x—>a~
>
such that, for
x'fa
x,
all
<a —x <
if
FIGliRl.
18
It is
for
8,
then
|/(.v)
than
less
<
/|
e.
and below even
quite possible to consider limits from above
numbers both greater and
—
if
/
is
defined
Thus, for the function / of Figure 14,
a.
we have
lim
x^0+
fix)
=
and
1
an easy exercise (Problem 29)
It is
lim
'
fix)
=—
1,
jc->0-
to
show
that lim fix) exists
if
and only
if
x—*a
lim fix) and lim fix) both exist and are equal.
x—Ki+
.v—>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
will
be found useful. In Chapter 4
(lose to 0.
This assertion
is
it
was claimed
lim fix)
is
if
x
is
large, then sin l/.v
is
usually written
lim sin
The symbol
that
which
read "the
limit
1
/.v
=
0.
of fix) as x approaches oo," or "as x
x—*-oo
becomes
infinite,"
and
a limit of the
form lim fix)
is
often called a limit at
infinity.
5. Limits
107
Figure 19 illustrates a general situation where lim f(x) = I. Formally, lim f{x)
AT->00
X—»0O
there is a number N such that, for all x,
/ means that for every £ >
if
x
> N, then
|/(jc)
-
/|
<
=
s.
The analogy
with the definition of ordinary limits should be clear: whereas the
condition "0 < \x — a\ < 5" expresses the fact that x is close to a, the condition
>
"x
./V"
FIGURE
We
expresses the fact that x
is
large.
19
have spent so
little
time on limits from above and below, and at infinity
because the general philosophy behind the definitions should be clear
if
you un-
derstand the definition of ordinary limits (which are by far the most important).
Many
exercises
on these
definitions are provided in the Problems,
tain several other types of limits
which are occasionally
which
also con-
useful.
PROBLEMS
1.
Find the following
limits.
(These limits
all
follow, after
from the various parts of Theorem
nipulations,
ones are used in each case, but don't bother
—
x
1
lim
3
:
-8
lim
•2
mi
LV
X
lim
—
x->3
x
lim
—
x-»;y
jc
lim
—x
y-*x
Va + h —
vi
sfa
lim
Find the following
..
Inn
x-»-l
1
- v7^
limits.
2;
listing
some
algebraic
ma-
be sure you know which
them.)
.
108
Foundations
-
1
-Vi -v 2
lim
A
-
3.
-v7 -x 2
!
1
In each of the following cases, determine the limit
prove that
for
all
(i)
the limit by
it is
<
x satisfying
f( x )
=
fix)
=x 2 + 5x
-
[3
a-
cos(a
=
a
fix)
showing how
<
a\
2
)],
a
a
=
-2,
100
(m)
—
\x
to find a 8
/
for the given a,
such that \f(x)
—
1\
and
< e
8.
= 0.
2.
1.
A"
,4
= x,
fix)
iv)
A4
vi)
=
/(*)
f{x)
viii)
/(a)
+
-
I.
r
sin"
—
2
vii)
arbitrary a.
a
,
=0.
a
= vM? a = 0.
= y^", a = \.
For each of the functions in Problem 4-17, decide for which numbers a the
imit lim f(x) exists.
x—*a
*5.
Do
a)
the
same
for
Same problem,
b)
each of the functions
if
we
in
Problem 4-19.
use infinite decimals ending in a string of 0's
instead of those ending in a string of 9's.
Suppose the functions / and g have the following property:
and all a,
if
< a —
if
<
I
For each £ >
\x
21
— 2| <
find a 8
(i)
if
<
|a
(ii)
if
<
|a
- 2| <
- 2| <
if
<
|a
—
(iii)
{
}
<
21
<
>
+
sin'
e
then |g(A)
,
such
<
if
|a
—
21
<
8,
then
5,
then |/(x)g(*)
<5,
then
8,
|/(jc)
then
—
—
a\
<
|a
l\
<
<
s
8/2.
when
<
/
\x
<
4|
—
2|
<
e
>
e,
s
- 6\ <
- 8| < £.
£.
<
£.
2
for
—
<
s.
4
:
Give an example of a function
|/(.v)
—
+g(Jt)
Six)
If
then |/(a)
all
that, for all a,
g(x)
(iv)
V
;
e,
for
a\
which the following assertion
<
5,
then |/(a)
—
/|
is
false:
< £/2 when
109
5. Limits
8.
(a)
fix) and lim gix) do not exist, can lim [/(*)
x-^a
x—>a
x—>a
lim fix) gix) exist?
x—*a
(b)
If lim
If lim
+ g(x)]
Can
exist?
(c)
fix) exists and lim[/(A') + g(x)] exists, must lim g(x) exist?
x—*a
x—>a
x—>a
If lim f{x) exists and lim g(x) does not exist, can lim [f(x)-\-g(x)] exist?
(d)
If
lim fix) exists and lim f(x)g(x)
x^-a
exists,
does
follow that lim g(x)
it
x^-a
x^>-a
exists?
9.
=
Prove that lim f{x)
+
lim f(a
'
x-+a
(This
h).
is
mainly an exercise
in
under-
/,^0
standing what the terms mean.)
10.
(a)
=
=
I if and only if lim[/(.t) — /]
0. (First see why
x—*a
x—»a
the assertion is obvious; then provide a rigorous proof. In this chapter
Prove that lim f(x)
most problems which ask
(b)
(c)
—
Prove that lim f(x)
v^O
Prove that lim f(x)
—
lim fix
X—>0
Give an example where lim
Suppose there
is
that lim f(x)
—
a 8
>
f(x)
exists,
lim g(x).
x—>a
are a "local property."
12.
(a)
same
(It
but lim f(x) does not.
x—>0
g(x)
<
when
\x
— a\ <
—
How
If
Prove
x^>-a
this fact
is
often expressed
will clearly help to use 8'
that f{x)
< g(x)
provided that these limits
(c)
8.
In other words, lim f(x) depends only on the
by saying that
limits
some other
letter,
or
,
for
all
x.
Prove that lim f{x)
<
x—>a
(b)
way.)
the definition of limits.)
5, in
Suppose
=
such that f{x)
x—*a
values of f(x) for x near a
instead of
in the
).
x—>-0
11.
be treated
a).
x-*a
=
X—*0
(d)
for proofs should
lim f(x
lim g(x),
x—*a
exist.
can the hypotheses be weakened?
<
f{x)
gix) for
all
x, does
it
necessarily follow that lim fix)
<
x—*a
lim gix)?
x^a
13.
Suppose
hix) and that lim fix) = lim hix). Prove that
x—>a
x—*a
and that lim gix) — lim fix) = lim hix). (Draw a picture!)
x—>a
x-+a
x—>a
that fix)
lim gix) exists,
x—>a
44.
(a)
Prove that
gix)
<
lim fix)/x
x—>-0
=
/
and b
Write fibx)/x
=
b[f(bx)/bx].
(b)
What happens
if
b
(c)
Part
this
15.
if
<
—
^
0,
then lim f{bx)/x
bl.
Hint:
0?
enables us to find lim(sin2x)/x in terms of lim(sinx)/.t. Find
x—>0
x—»0
limit in another way.
(a)
Evaluate the following limits in terms of the
number a
—
lim (sin x) /x.
x—*0
(i)
=
x—>0
lim
™3».
x
sinajc
lim
0
sin
bx
—
110
Foundations
2a
sin
lim
iii)
x
x^-0
2a
sin
lim
iv)
A^O
z
a
—
1
,.
cos a
lim
-
x^-0
x2
vi)
km
o
vii)
lim
.
tan 2 a
+ 2x
r
x+x-
a
v^O
1
x
sin
—
COS*
+ /?) —
sin (a
a
sin
lim
viii)
h
sin (a
lim
ix)
x-
x-»l
a
2
(3
lim
x^-0 (x
16.
a)
1)
1
+ sin a)
-
+ sin a) 2
1
—
lim (a
v-1
xi)
—
2
Prove that
sin
1)
\x ,
=
lim f(x)
if
1
b)
Prove that
|/|.
.r—*a
=
lim f(x)
if
=
then lim |/|(a)
l,
x^>a
'
—
and lim g(x)
I
m, then lim max(/, g)(x)
x-*a
x—*-a
=
x-+a
max(/, m) and similarly for min.
17.
a)
Prove that lim \/x does not
exist, i.e.,
show
number
every
b)
18.
if
such that
Hint:
19.
*20.
l/(.v
|/(Jt)
<
|
Why does
M
Prove that
f(x)
if
(a)
Prove that
if
—
x
is
— a\ <
8.
\x
prove that /—
Generalize
a
1
for irrational
exist for
any
false for
a
if
lim g(x)
=
^
number
>
<5
and a number
M
(What does this mean pictorially?)
< f(x) < /+1 forO < \x—a\ < 8?
x and fix)
—
1
for rational x,
a.
and f(x)
for rational x,
= —x
for irrational x, then
0.
0,
then lim #(a
)
sin l/.v
=
0.
x—*Q
x—*0
(b)
is
exist.
then there
<
if
lim f(x) does not exist
21.
/,
suffice to
it
does not
1)
if
f{x) —
then lim f(x) does not
Prove that
—
=
lim f(x)
X—fCl
I
v^O
/.
Prove that lim
Prove that
=
that lim 1/a
x-^0
this fact as follows:
If
lim g(x)
=
and
\h(x)\
<
M for
all
a,
x—>0
then lim g(x)h(x)
—
0.
(Naturally
it
is
unnecessary to do part
(a) if
you
x—»o
succeed
in
easier than
doing part
(a)
that's
(b);
actually the statement of part (b)
one of the values of generalization.)
may make
it
111
5. Limits
22.
Consider a function / with the following property: if g is any function for
which limgOc) does not exist, then lim[/(x) + g(x)] also does not exist.
x—*0
.*—>-0
Prove that
happens
this
and only
if
Km
if
f(x)
This
Hint:
does exist.
is
jc-»-0
an
actually very easy: the assumption that lim f(x) does not exist leads to
x—*0
immediate contradiction if you consider the right g.
**23.
This problem
when f + g is replaced by f g.
more complex, and the analysis
the analogue of Problem 22
is
In this case the situation
considerably
is
•
requires several steps (those in search of an especially challenging problem
can attempt an independent
Suppose
(a)
solution).
and
that lim f(x) exists
is
^
0.
=
oo.
Prove that
if
x— ()
exist, then lim f(x)g(x) also does not exist.
lim g(x) does not
x—»0
x—>0
Prove the same result
(b)
sort
of limit
Prove that
(c)
lim
if
|,/"(jc)|
(The precise definition of
given in Problem 37.)
is
neither of these two conditions holds, then there
if
g such that lim g(x) does not
a function
is
but lim f(x)g(x) does exist.
x—*0
x—>0
Hint: Consider separately the following two cases: (1) for some £
we have
>
|/(jc)I
exist,
choosing points x n with
Suppose that A n
and
[0, 1],
/
>
s for all sufficiently small x. (2) For every e
< £. In the second
< \/n and |/(jc„)| < \/n.
are arbitrarily small x with |/(jc)|
*24.
this
\x n
\
case,
>
0, there
begin by
number n, some finite set of numbers in
A„ and A m have no members in common if m ^ n. Define
that
for each natural
is,
as follows:
v
et
f
\
Prove that lim f(x)
=
x
1/",
in
An
x not
(J,
in
A„
lor
any
n.
for all a in [0, 1].
x—*a
25.
Explain
why
For every
*26.
8
the following definitions of lim f(x)
>
there
(ii)
if
<
<
\x
— a\ <
— a\ <
(iii)
if
<
|jc
—
(iv)
if
<
\x
- a\
(i)
if
\x
Give examples
a\
to
<
<
an
is
e
>
such
e,
then \f{x)
s,
then
e,
then \f{x)
|/(.v)
— <
- 1\ <
— <
1\
I\
I
are
all
correct:
jc,
8.
S.
58.
e/10, then \f(x) -l\
show
=
x-*a
that, for all
<
8.
that the following definitions of lim f(x)
=
I
are not
x—*a
correct.
(a)
For
all 8
>
there
is
an
e
>
such that
if
<
\x
—
a\
<
8,
then
then
<
\f(x)-l\ <e.
(b)
For
I
jc
—
all
s
>
a\
<
8.
there
is
a 8
>
such that
if
|/(jc)
—
/]
<
s,
.
112
.
Foundations
27.
For each of the functions in Problem 4-17 indicate for which numbers a the
one-sided limits lim fix) and lim fix)
x—>a +
x—*a~
*28.
(a)
Do
(b)
Also consider what happens
same
the
each of the functions
for
decimals ending in
29.
Prove that lim fix)
in
Problem 4-19.
decimals ending in
if
are used instead of
O's
9's.
exists if
lim f{x)
x—«+
'
x^-a
30.
exist.
—
lim f(x).
x^-a~
Prove that
lim
(i)
=
f{x)
lim
f(—x).
lim/(|*|)= lim f(x).
(ii)
x->0 +
x^>0
lim fix
(iii)
=
)
lim f(x).
x—»0 +
x-+0
(These equations, and others
They might mean only
if
one of the
a certain
that
like
that the
them, are open to several interpretations.
two
limits are
either limit exists, then the other exists
if
equal
if
they both
limits exists, the other also exists
and
is
and
is
equal to
exist;
or that
equal to
it.
it;
or
Decide for
yourself which interpretations are suitable.)
31.
Suppose that lim fix) <
lim fix).
X—a+
x—>a~
Prove that there
sertion.)
<
x < a
32.
y and
—
\x
a\
Prove that lim ia„x"
<
(Draw a
picture to illustrate this as-
some 8 >
such that fix) < fiy) whenever
and |v — a\ < 8. Is the converse true?
is
8
h
H
a )/ib m x'"
h bo) (with
H
an
/
and bm
X-^-OQ
exists if
and only
if
m >
Hint: the one easy limit
n.
What
lim
X—>00
only information you need.
33.
Find the following
lim
x-*oo
1
limits.
•
,
3
Dx + 6
xsinx
1 *-v-»
(ii)
(iii)
is
1/jc
— + sin —x
x
(i)
is
c
1
lim
X—»00
•
Vx2 + x —
x.
x 2 i\
(iv)
lim
jr->oo
+ sin x)
~^—
+ Sinx) z
:
ix
34.
Prove that lim fi\/x)x^0+
35.
Find the following
sin
(i)
x^-oo
(ii)
x
lim
.
x
lim x sin
x-*oo
—
x
limits
lim fix).
the limit
=
0;
when m —
n?
do some algebra
^
When m >
so that this
is
0)
n?
the
. .
5.
36.
Define " lim
x-*— oo
(a)
Find
f(x)
lim (an x n
+
•
•
+ a )/(bm x m +
= lim f(—x).
•
(b)
Prove that lim f(x)
(c)
Prove that lim f{\/x)
=
lim
x->—oo
x-+0-
for
define lim f(x)
x—*a
all
x, if
(Of course,
=
+b
•
).
x—>— 00
X—f-OO
We
113
= L"
X^>— 00
37.
Limits
mean
oo to
f(x).
that for
all TV
there
is
a 8
>
such
that,
< \x — a\ <
then f{x) > N. (Draw an appropriate picture!)
we may still say that lim f(x) "does not exist" in the usual sense.)
<5,
x—>a
(a)
Show
(b)
Prove that
that lim l/(x
x—*-3
if
—
>
f(x)
3)
s
2
=
>
oo.
for all x,
x—*a
lim f(x)/\g(x)\
x—*a
38.
(a)
=
Define lim f(x)
x^>-a +
=
oo and lim f{x)
0,
then
oo.
=
oo.
(Or
at least
convince your-
x^f-a"
you could write down the definitions
many other such symbols can you define?)
self that
39.
=
and lim g{x)
—
(b)
Prove that lim
(c)
Prove that lim f(x)
Find the following
1
/x
limits,
if you
had the energy.
How
oo.
=
oo
when
if
and only
they
if
lim f{\/x)
=
oo.
exist.
3
...
1
+ 4.y-7
x^-oo l~i
lx £ — X TT
+
7
lun x(\ + sin"x).
a-
lim
1
(ii)
x—*oc
(iii)
(iv)
lim x sin x
x—»oo
lim x sin
—
x
.r—>oo
vi 2 +
(v)
lim
x—>oo
(vi)
lim jc(vx
x—»oo
2x
—
+2—
x.
a/^")-
\X
Vll
lim
j:-*oo
40.
(a)
x
Find the perimeter of a regular rc-gon inscribed
in a circle
of radius
[Answer: 2rn sin(7r/n).]
(b)
(c)
What
What
value does this perimeter approach as n becomes very large?
limit
can you guess from
this?
r.
a
114
a
Foundations
41.
For c
(a)
Hint:
More
(b)
> 1, show
Show that
generally,
that c
for
c
if
any
>
—
l/n
0,
£
'l/c
>
thought of a
much
then c
simpler
as n
1
we cannot have
Alter sending the manuscript for the
I
approaches
way
1//,?
approaches
first
to
1
c
as n
edition of this
without going through
,
the factoring tricks
all
example, that we want to prove that lim x
£
>
(see
a~
—
0,
we simply
Figure
s
19);
< x~ < a~ +
already been
FIGURE
20
is
set,
\x
£,
>
1
+e
large.
for large n.
becomes very
large.
book off to the printer,
= a 2 and lim jc 3 =
=
x—>a
on page 92. Suppose, for
a", where a > 0. Given
minimum of v a 2 + e — a and a — v a 2 — e
< 8 implies that \l 2 — e < x < \J 2 + s, so
— a < £. It is fortunate that these pages had
be the
let 8
then
becomes very
prove that lim x 2
x—>a
a
l/n
—
a\
or \x~
so that
I
couldn't
completely fallacious. Wherein
\
make
lies
these changes, because this "proof"
the fallacy?
CONTINUOUS FUNCTIONS
CHAPTER
/
If
is
an arbitrary function,
not necessarily true that
it is
lim f(x)
=
f(a).
X—X2
In
many ways
there are
fact,
even be defined
at a, in
this
can
fail
to
be
true.
For example,
/ might
which case the equation makes no sense (Figure
not
1).
Again, lim f(x) might not
even
FIGURE
/
if
exist (Figure 2). Finally, as illustrated in Figure 3,
x—*a
is defined at a and lim f(x) exists, the limit might not equal f(a).
1
S
(b)
(a)
FIGURE
(c)
2
We would like to regard all behavior of this type as abnormal and honor, with
some complimentary designation, functions which do not exhibit such peculiarities.
The term which
continuous
if
has been adopted
the graph contains
and the
DEFINITION
The
at
its
graph
precise definition
function
/
is
"continuous."
no breaks, jumps, or wild
is
(a skill well
will
have no
a function
oscillations.
worth
cultivating)
it is
is
/
is
Although
continuous
easy to be fooled,
important.
very
continuous at a
if
lim f(x)
We
Intuitively,
enable you to decide whether a function
this description will usually
simply by looking
is
difficulty finding
=
f{a).
many examples
of functions which are, or are
—
some number a every example involving limits provides an
example about continuity, and Chapter 5 certainly provides enough of these.
not, continuous at
I
I
(
,
I
K
X.
3
The
at 0,
we
function f(x)
and the same
is
=
sin
\/x
is
not continuous at 0, because
true of the function g(x)
=
x
sin i/x.
are willing to extend the second of these functions, that
115
it
is
On
is,
if
not even defined
the other hand,
we wish
if
to define
,
116
Foundations
a
new
function
G
by
G(x)
x
=
sin
jx
1
a,
then the choice of a
—
at
to
do
extension
=
G(0) can be made
we can
this
(if fact,
not possible for /;
is
F
not be continuous
will
^0
x
=
way
such a
in
0,
we must) define G(0)
we define
that
=0
G
will
(Figure
be continuous
This sort of
4).
if
sin 1/a,
x^O
a,
x
F(x)
then
x
=
0,
no matter what a
at 0,
is,
because lim f(x) does
x^0
not
exist.
The
function
/(*)
is
not continuous
=
/(0), so
The
/
at a, if
is
a
^
—
c,
x,
x rational
0,
x irrational
lim f(x) does not
0, since
continuous
functions f(x)
=
at precisely
g(x)
=
x,
one point,
—
and h{x)
exist.
However, lim f{x)
=
0.
x
are continuous at
all
num-
bers a, since
lim fix)
=
lim c
=
c
=
f(a),
lim g(x)
=
lim x
=
a
=
g(a),
lim
=
lim x
x—*a
x^a
x^*a
/i(a")
x-*a
Finally,
In Chapter 5
but you can
x
l
If
is
to give
continuous
a
for
Since
all a).
if
a
is
irrational,
if
(2)
if
g (a)
^
0,
+g
g
•
is
is
continuous
continuous
at a,
at a.
then
(3)
\/g
is
continuous
=
f{a) only
but not
we prove two
are continuous at a, then
/
/
<
<
1,
when a
is
a (actually, only for
all
examples of continuity
(1)
Moreover,
at
in lowest terms.
p/q
that lim f(x)
x—>a
easily see that this is true for
even easier
/ and g
=
=
we showed
irrational, this function
THEOREM
h(a).
x irrational
0,
=
4
It is
=
a"
consider the function
/(*)
FIGURE
=
at a.
if
a
is
a
rational.
simple theorems.
PROOF
/ and g
Since
are continuous at a,
—
lim f(x)
x
By Theorem
—a
'
of Chapter 5
2(1)
lim(/
and
is
(3)
left
+ g)(x) =
g{a).
+ g{a) =
f(a)
/ +g
is
+ g)(a),
(f
continuous
The
at a.
proofs of parts
(2)
to you. |
=
Starting with the functions f(x)
for every a,
—
lim g{x)
this implies that
just the assertion that
are
and
f{a)
.r—>a
x—*a
which
117
Continuous Functions
6.
we can
use
Theorem
which are continuous
x,
at a,
+ bn -ix n - +--- + b
c m x "+cn,_ x" -' + --- + c
bn x n
=
/(*)
=
conclude that a function
to
1
and f(x)
c
1
l
1
l
is
continuous at every point in
than
sin
When we
that.
is
continuous
at
domain. But
its
a for
all
assume
a; let us
/(*)
now be proved
it
meanwhile.
this fact
sin
=
2
in
its
=
sin(x");
translate the equation lim f(x)
we
obtain
>
But
in this case,
where the
=
there
—
<
a\
limit
is
THEOREM
2
If
g
if
is
=
x
Let s
it is
continuous
(Notice that
proof
a
>
0.
/
We
is
worth
noting. If
8
>
such
then \f(x)
limits,
that, for all x,
—
<
f(a)\
s.
\x
— a\ <
8
to the simpler condition
\x
since
is
/(a), the phrase
<
may be changed
obviously
f(a) according to the definition of
is
8,
we
still
functions. Before stating this
theorem, the following point about the definition of continuity
we
\x
function like
But we are
domain.
f(x)
like
need a theorem about the composition of continuous
<
A
a
unable to prove the continuity of a function
if
further
x
continuous at every point
for every e
much
be easy to prove that
will
x + x suijc
+,2,427
i~- 2
sin" x + 4*- sin x
•
can
harder to get
is
it
discuss the sine function in detail
— a\ <
certainly true that \f(x)
at a,
and /
is
continuous
S,
—
f(a)\
at g(a),
<
s.
then
/og
required to be continuous at g(a), not at
wish to find a
if
\x
-a\ <
8
8.
>
such that for
then \(f
i.e.,
o g)( x )
-
all
(/
is
a.)
x,
6 g)(a)\
\f(g(x))- f(g(a))\
<e.
<
e,
continuous
at a.
118
Foundations
We
/
use continuity of
first
that for
all
In particular, this
if
We now use
— g{a)\ <
-
<
g(a)\
(2)
and
if
now
8'
to hold.
We
you
order
such
s.
-
f(g(a))\
<
s.
we
(3)
|.v
—
if \x
<
a\
8,
see that for
-a\<
8,
8'
as the e
>
then \g(x)
—
is
such
g(a)\
in the definition
(!)
of
that, for all v,
<
8'.
x,
all
-
then \f(g(x))
<
f(g(a))\
e. |
reconsider the function
already noted that
fl/0. Functions
for
<
f(g(a))\
a 8
conclude that there
/
continuous
is
together with the continuity of
2,
>
how close x must be to a in order for the
The number 8' is a positive number just like
x sin \/x,
'0.
and
-
8\ then \f(g(x))
number; we can therefore take
positive
(3)
We have
8'
a
continuity of g to estimate
continuity of g at a.
can
to g(a) in
is
that
\g(x)
if
inequality \g{x)
Combining
\y-g(a)\ < 8\ then |/(j)
means
(2)
We
/
must be
close g(x)
continuous at g(a), there
is
y,
(1)
any other
how
to estimate
Since
for this inequality to hold.
=
f{x)
like
sin(.r~
*
sin(jc
^
= 0.
A few applications of Theorems
at 0.
show
sin,
+
x
that
+
/
is
1
also continuous at a, for
should be equally easy
sin (x )))
to analyze.
The few theorems of this
chapter have
all
been related
to continuity of functions
a single point, but the concept of continuity doesn't begin to be really interesting
at
we focus our attention on functions which are continuous at all points of some
If / is continuous at x for all x in (a,b), then / is called continuous
on (a, b); as a "special case", / is continuous on R = — oo, oo) [see page 57] if
until
interval.
(
it is
a
continuous
little
at
x for
(2)
all
jc
x
differently; a function
(1)
(We
all
in
R. Continuity on a closed interval must be defined
/
is
called
/is continuous
lim f{x)
=
continuous on
at
its
all
x
in (a,b),
f(a) and lim f(x)
also often simply say that a function
in
x for
is
[a, b] if
continuous
=
f(b).
if it is
continuous
at
x
for
domain.)
Functions which are continuous on an interval are usually regarded as especially
well behaved; indeed continuity might be specified as the
"reasonable" function ought to
scribed, intuitively, as
satisfy.
A
first
continuous function
one whose graph can be drawn without
from the paper. Consideration of the function
^{X)
Y %__ f
~]
'
*sinl/*,
x#0
0.
x
=
condition which a
is
sometimes de-
lifting
your pencil
shows that
there are
this
many
description
is
a
too optimistic, but
little
119
Continuous Functions
6.
it is
nevertheless true that
important results involving functions which are continuous on an
There theorems are generally much harder than the ones in this chapter,
but there is a simple theorem which forms a bridge between the two kinds of results.
The hypothesis of this theorem requires continuity at only a single point, but the
conclusion describes the behavior of the function on some interval containing the
point. Although this theorem is really a lemma for later arguments, it is included
here as a preview of things to come.
interval.
THEOREM
3
for all x in some
at a, and fia) > 0. Then fix) >
more precisely, there is a number > such that fix) >
for all x satisfying \x — a\ < 8. Similarly, if f{a) < 0, then there is a number 8 >
such that f{x) <
for all x satisfying \x — a\ < 8.
Suppose /
is
continuous
interval containing a;
PROOF
<5
Consider the case fia) >
8
>
such
0.
Since
/
|.v
— a\ <
In particular, this must hold for s
8
>
so that for
—e <
+ e=\fia)_
fia)
fia)-e=
\fia)
and
this
itself,
A
=
then -\fia)
implies that fix)
>
\
fia) >
is
more
leading to a proof that
similar
proof can be given
first
<
f(a)\
fix)
—
?/(«), since
8,
can apply the
FIGURE
every s
>
there
is
a
e,
fia)
j
<
s.
fia) >
(Figure
Thus
5).
x,
- a\ <
if \x
fia)
all
—
then \fix)
8,
i.e.,
is
at a, for
that, for all x,
if
there
continuous
is
0.
< fix) - fia) < \f(a),
(We could even have picked
elegant, but
in the case
case to the function
fia)
— /.
more confusing
<
0; take s
=
s to
be fia)
to picture.)
—\f(a). Or one
|
5
PROBLEMS
1.
For which of the following functions
domain
R such
(i)
fix)
=
(ii)
fix)
=
that Fix)
—
/
fix) for
is
all
there a continuous function
x
in the
F
with
domain of /?
-4
x-2'
2
x
\x\
X
2.
(iii)
fix)
(iv)
f(x)
=
=
0,
x irrational.
l/q, x
=
p/q
rational in lowest terms.
At which points are the functions of Problems 4-17 and 4-19 continuous?
.
1
20
Foundations
3.
(b)
/
/(0) must equal 0.)
Give an example of such a function / which is not continuous
^
a
(c)
/
that
a function satisfying \f{x)\
is
continuous
for
\x\
x.
all
that
at 0. (Notice that
any
at
0.
Suppose
that g
Prove that /
5.
Show
Suppose
is
4.
<
(a)
continuous
is
continuous
is
number
0,
and
<
|/(a)|
|g(Jt)|.
at 0.
Give an example of a function / such that /
is continuous everywhere.
For each
=
and g(0)
at
which
a, find a function
continuous nowhere, but
is
continuous
is
\f\
but not at any
at a,
other points.
6.
(a)
Find a function / which
discontinuous at
is
1, \, A,
1,
.
.
.
•
•
but continuous
at all other points.
(b)
Find a function / which
continuous
7.
Suppose
8.
9.
/
that
/
Suppose
that
/+a
nonzero
(a)
is
/
continuous
—
(b)
>
f(a)\
Conclude
>
e
s.
that for
trarily close to
Prove that every function
E
(c)
£
>
continuous
/
at a,
0.
/
but
continuous
is
Prove that
if
a
/
at a.
0,
then
Prove that for
numbers x arbithere are numbers x arbitrarily
or
continuous on
O
is
there are
|/|.
R can be written
f = E + O,
odd and continuous.
/ and g are continuous, then so are max(
if
at 0,
arbitrarily close to a with
either
then so
even and continuous and
Prove that
min(/,
(d)
is
=
numbers x
< f(a) — s
> f(a) + e.
(b)
where
and
,
Illustrate graphically.
Prove that
is
f(y), and that
not continuous
is
a with f{x)
/
•
interval containing a
some number
(a)
if
1, j? ^, \,
all a.
a and f(a)
there are
close to a with f(x)
10.
+
f(x)
a for
defined at a but
is
some number
|/(jc)
at
at
some open
in
=
y)
continuous
is
is
Suppose /
+
f(x
satisfies
Prove that
at 0.
discontinuous at
is
other points.
at all
is
/",
g)
and
g).
Prove that every continuous
/ can be
written
/
= g — h,
where g and h
are nonnegative and continuous.
:
Theorem
/(*)= l/x.
11.
Prove
12.
(a)
Prove that
1(3)
/'
if
is
by using Theorem 2 and continuity of the function
continuous at
/
and lim g(x)
=
x—*a
/(/).
(b)
G
Show
continuity of
that
if
with G(x)
true that lim f(g(x))
x—-a
/(0=
I-
'
=
=
/
g(x) for x
at
/
is
^ a,
then lim f(g(x))
it is
and G(a)
not assumed, then
/(lim g(x)). Hint: Try f(x)
x—>a
=
x—*a
(You can go right back to the definitions, but
the function
/,
easier to consider
=
it
=
I.)
is
not generally
for
x
^
I,
and
.
6.
13.
(a)
Prove that
is
/
if
continuous on
is
continuous on R, and which
then there
[a, b],
satisfies
gix)
121
Continuous Functions
—
a function g which
is
fix) for
x
all
in [a,&].
Since you obviously have a great deal of choice, try making g
constant on ( — oo,a] and [b, oo).
Hint:
(b)
Give an example
to
show
that this assertion
false if [a
is
b]
,
replaced
is
by (a,b).
14.
(a)
Suppose that g and h are continuous at a, and that g(a) — h(a). Define
f(x) to be gix) if jc > a and h(x) if x < a. Prove that / is continuous
at a.
(b)
Suppose g
g(b)
=
Show
continuous on [a.b] and h
is
/
that
continuous on
is
[b, c]
Let f(x) be g(x) for x in [a,b] and h(x) for x in
h(b).
is
continuous on
[a, c].
and
[fr,c].
(Thus, continuous functions can be
"pasted together".)
15.
Prove that
whenever
16.
(a)
if
/
\x
— a\ <
continuous
is
and
8
then for any £
at a,
\y
— a\ <
8
if
>
that
lim f(x)
x—>a +
<
0,
then there
for
a
is
<x —a < 8
version of Theorem
there
\f(x)
number
>
8
is
a 8
>
<
s.
f(y)\
Then
0.
there
< x —
x satisfying
all
—
so that
3 for "right-hand continuity":
f(a), and f(a) >
>
such that f(x)
f(a)
=
we have
Theorem
Prove the following version of
Suppose
8,
>
is
<
a
a
8.
Similarly,
<
such that f(x)
number
for all
x
satisfying
(b)
Prove a
when
3
=
lim f(x)
f(b).
x—*b~
17.
If lim
x—*a
f(x)
continuity
but
exists,
is
^
f(a), then
/
removable
said to have a
is
dis-
at a.
f{x)
—
\/x for x
sin
^
and /(0)
=
does
/ have a removable
(a)
If
(b)
x sin i/x for x / 0, and /(0) = 1?
fix)
Suppose / has a removable discontinuity at a. Let gix) = fix) for
x ^ a, and let gia) = lim fix). Prove that g is continuous at a. (Don't
What
discontinuity at 0?
1,
=
if
x-*a
work very hard;
(c)
Let fix)
terms.
*(d)
Let
is
=
What
/ be
if
is
quite easy.)
this
is
x
irrational,
is
and
let
fip/q)
the function g defined by gix)
=
=
\jq
p/q
if
is
in lowest
lim fiy)?
a function with the property that every point of discontinuity
a removable discontinuity. This means that lim fiy)
exists for all x,
V—KV
but
/ may be
Define gix)
=
discontinuous at
v—
so easy as part
**(e)
Is
some (even
lim fiy). Prove that g
*
is
infinitely
many) numbers
continuous. (This
x.
not quite
(b).)
there a function
/ which
is
discontinuous at every point, and which has
only removable discontinuities?
now, but mainly as a
test
(It is
worth thinking about
of intuition; even
if
Problem 22-33.)
this
problem
you suspect the correct
answer, you will almost certainly be unable to prove
time. See
is
it
at the
present
THREE HARD THEOREMS
CHAPTER
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
l
If
/
is
<
continuous on [a,b] and f(a) <
such that f(x)
=
fib), then there
is
some x
in [a,b]
0.
means that the graph of a continuous function which starts
horizontal axis and ends above it must cross this axis at some point, as
(Geometrically, this
below the
theorem
2
in
Figure
If
/
is
1.)
continuous on [a b]
,
some number
N
THEOREM
3
bounded above on
then
/
is
<
N
for
such that f(x)
(Geometrically, this theorem
allel to
,
means
is
continuous on [a,&], then there
f(y)
>
fix) for
all
x
,
that
is,
there
is
graph of /
lies
below some
line par-
2.)
/
If
,
x in [«,&].
all
that the
the horizontal axis, as in Figure
[a b]
is
some number
v in
[a,b] such that
in [a, b] (Figure 3).
These three theorems
differ
markedly from the theorems of Chapter
6.
The
hypotheses of those theorems always involved continuity at a single point, while
FIGl RE
FIGURE
122
2
FIGURE
*
.
7.
Three
Hard Theorems
on a whole
the hypotheses of the present theorems require continuity
[a, b]
—
continuity
if
example,
/ be
let
fails
hold
to
the function
shown
1,
Then /
FIGURE
no point x
v2
is
sufficient to destroy the conclusion
Similarly,
suppose that /
is
such that fix)
Then /
is
continuous
of
Theorem
x
^0
0,
x
= 0.
except 0, but
[0, 1]
<
/'(2),
1
in Figure 5,
\/x,
=
every point of
at
For
0; the discontinuity at the single
shown
the function
/(*)
N
=
is
in [0, 2]
<
except V2, and f(0)
[0, 2]
but there
point
4
continuous at every point of
is
fail.
0<x < Vi
V2<x <2
/(*)
2
interval
in Figure 4,
1.
4l
may
a single point, the conclusions
at
123
/
is
bounded above
not
we have f(\/2N) = 2N > N.
This example also shows that the closed interval [a, b] in Theorem 2 cannot be
replaced by the open interval (a,b), for the function / is continuous on (0, 1), but
on
is
2N
[0, 1].
In
any number
fact, for
N
>
not bounded there.
Finally,
consider the function
shown
in Figure 6,
X
X <
0,
X
/(*)
FIGURE
On
5
the interval [0,
conclusion of
does not
f(y) > fix) for
all x in [0, 1] so
fiy)
<
the function
if
all
x
x
/
/ is
Theorem 3
in [0, 1]; in fact,
we cannot choose v
s any number with
i
y
/
—
there
is
no y
does
x
<
satisfy the
[0, 1].
<
v
<
But
/
such that
in [0, 1]
nor can we choose
1,
<
so
not continuous on
certainly not true that f{\)
it is
=
1
1.
bounded above,
is
even though
2,
conclusion of
satisfy the
fi x )
1]
Theorem
>
> f(x)
1
for
because
1.
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 "takes
on
its
maximum
As a compensation
value" on that interval.
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 in-
terval;
V
X
to
1
To
such "global" properties of a function are always significantly
difficult
prove than "local" properties, and are correspondingly of much greater power.
illustrate the usefulness
portant consequences, but
FIGURE
more
of Theorems
it
1,
2,
will help to first
and 3, we will soon deduce some immention some simple generalizations
of these theorems.
6
theorem
4
If
/
is
continuous on
such that f(x)
=
c.
[a, b]
and f(a) <
c
<
f(b), then there
is
some x
in [a, b\
.
124
Foundations
PROOF
is
theorem
5
= f — c. Then
Let g
some x
=
such that /(jc)
PROOF
The
continuous, and g(a)
is
=
orem 4
/GO =
there
c.
is
this
>
>
c
<
<
means
By Theorem
g(b).
that fix)
f(b), then there
=
is
there
1,
c. |
some x
in [a,b]
c.
—/
function
But
0.
continuous on [a,b] and f(a)
/" is
If
g
in [a,b] such that g(x)
continuous on [a,b] and —f(a)
some x in [a,b] such that — fix)
< — c < —fib). By The-
is
= — c,
which means
that
I
Theorems 4 and 5 together show that / takes on any value between f(a)
and f(b). We can do even better than this: if c and d are in [a, ft], then /
takes on any value between /(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
is
1
often called the Intermediate
Value Theorem.
THEOREM
6
PROOF
/
If
is
continuous on [a b]
,
some number
N
The
—/
function
such that
,
/
is
>
N
for
such that fix)
is
bounded below on
then
continuous on
— fix) < M for all x in
we can let N = — M.
[a, b], so
[«, b].
in [a, b], so
all
But
x
[a b]
,
,
that
is,
there
is
in [a, b].
by Theorem 2 there
this
means
that f(x)
is
>
a
M
number
—M for
all
x
|
Theorems 2 and 6 together show that a continuous function / on [a b] is
bounded on [a, ft], that is, there is a number TV such that |/(x)| < /V for all x in
In fact, since Theorem 2 ensures the existence of a number N\ such that
[a b]
<
N\ for all x in [a, b], and Theorem 6 ensures the existence of a number
fix)
,
,
A^2
theorem
7
.
such that f(x)
/
If
for
is
all
M
>
continuous on
x in
[a
,
x
for all
[a, b],
in [a, b],
then there
is
we can
some y
=
take iV
in [a, b]
max(|A^i
|,
|A^|).
such that /(y)
< fix)
b]
(A continuous function on a closed interval takes on
its
minimum
value on that
interval.)
PROOF
The
function
—/
such that —fiy)
all
x
is
continuous on
> ~f(x)
for
all
by Theorem 3 there is some y in [a, b]
[a,b], which means that /(y) < fix) for
[a, b];
x
in
in [«,/?]. |
Now that we
have derived the
trivial
consequences of Theorems
1, 2,
and
3,
we
can begin proving a few interesting things.
theorem
8
Every positive number has a square
some number x such
that x~
=
a.
root.
In other words,
if
a >
0,
then there
is
Three
7.
PROOF
Hard Theorems
125
Consider the function fix) = x 2 which is certainly continuous. Notice that the
statement of the theorem can be expressed in terms of /: "the number a has a
,
square root" means that
will
/
on the value
takes
be an easy consequence of Theorem
The proof of this
a.
fact
about /
4.
number b > such that f(b)>a (as illustrated in Figure 7);
we can take b — a, while if a < 1 we can take b = 1. Since
/(0) < a < f(b), Theorem 4 applied to [0, b] implies that for some x (in [0, b]),
we have f(x) = a, i.e., x 2 = a. |
There
obviously a
is
a >
in fact, if
1
same argument can be used
Precisely the
an nth
number a
any natural number n. If n happens to be odd, one can do
number has an nth root. To prove this we just note that if the positive
has the nth root x,
has the nth root —x.
root,
i.e., if
The
=
x"
n
a, then (— x)
assertion, that for
= —a
(since n
is
odd), so
odd n any number a has an nth
equivalent to the statement that the equation
is
-a =
x"
has a root
FIGURE
number has
root, for
better: every
—a
to prove that a positive
n
if
is
Expressed in
odd.
way
this
the result
is
susceptible of great
generalization.
7
THEOREM
9
If n
is
odd, then any equation
x"
+a n -\x n ~
x
^
=
h«o
has a root.
PROOF
We
obviously want to consider the function
=
f(x)
we would
like to
intuitive idea
since n
negative x
.
this
A little
The proper
fix)
is
that for large
is
odd,
is
/
prove that
function
algebra
x"
_1
+ a„_ix"n-\
sometimes
\x\,
+a
H
and sometimes
positive
the function
is
very
is all
like
= x n + a^i.r"- +
1
•
•
•
+ a = xn
(1
+
—
l
X
Consequently,
if
Of
-
X^
we choose x
x
k
\
>
\x\
«o
+
< K-i +
•••
satisfying
>
1,
2n|fl„_i|,
.
.
,
.
2n|aol>
and
\a n -k\
;
—
+
X'
|jc|
(*)
\x
+
<
\a n -k\
<
|tf#i-*l
2n\a n - k
J_
\
In
The
=
and,
g(x)
x
idea work.
+
+
that
- +
negative.
n
x and negative for large
we need to make this intuitive
/ depends on writing
analysis of the function
a,
then
much
positive for large positive
is
V
Note
;
l«ol
«0\
x")
126
Foundations
so
a n -\
a n -2
X
_
z
X
«o
JL
]
l
n
2n
2n
2
X
In other words.
a "-
1 <
~
2
i
x
~
x"
2'
which implies that
2
Therefore,
if
we choose an
Ul)
2
so that f(x\)
so that f{xi)
>
<
0.
x\
"<u,
,
On
would be
^
more
ever,
is
>
y
(x 2
1
1
=
,
I
RE
8
x2 <
+
•••
then (x 2 )"
satisfies (*),
=
+
(x 2
to the interval [x 2 x\]
,
<
and
f(X2),
y
we conclude
that there
is
an x
0. |
—
x
1
=0,
have a solution, and some,
there to say? If
we
it
problem of even degree equations completely
however, the problem seems insuperable.
first sight,
said. Instead
+a n _
+
1
=
0,
more general
Some
do not
equa-
—what
question,
how-
of trying to solve the equation
x"- +...+fl
1
[
x
like
are willing to consider a
something interesting can be
=0,
us ask about the possibility of solving the equations
numbers
c.
+«„_,*"-
depends on a
fact
which
is
+a =
c
This amounts to allowing the constant term ao to vary.
The information which can be
<
if
1
frustrating to leave the
for all possible
I
(x\)" J
X]
9 disposes of the problem of odd degree equations so happily that
-V"
I
a°
1
•X?
x"
let
a "-
+
"fi
then
satisfies (*),
0.
undiscussed. At
tions, like
which
the other hand,
such that f(x)
Theorem
>
V
Now applying Theorem
in [x 2 x\]
)
.r"
jc
given concerning the solution of these equations
illustrated in Figure 8.
The graph of the function f{x) = Jc"+<r/„_i.Y" _1 + + «(), with n even, contains,
In other words, there is a
at least the way we have drawn it, a lowest point.
<
number y such that f(y)
the function / takes on a
f(x) for all numbers x
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 [«,/?], and
\a, b]
obtain a point yo such that /(>'o) is the minimum value of / on [a, b]; but
•
•
•
—
if
happens
to
be the interval shown
not be the place where
/
has
its
in
Figure
minimum
example, then the point yo will
value for the whole line. In the next
8, for
Three
7.
theorem the
Hard Theorems
proof is to choose an interval
entire point of the
[a, b] in
such a
127
way
that this cannot happen.
theorem
10
If n
is
that f(y)
PROOF
As
=
even and f(x)
in the
< f(x)
for
+ a n _\x n
x"
~l
+
proof of Theorem
all
x with
even, x"
>
for
xn
provided that
such that b"
\x\
all
.,
> M. Now
— b,
>
be a number
2n\ao\),
n -\\
+
•••
ao
a„_i
+ 5td + ... + ^) =
/.
consider the
number
> M. Then,
also b
if
/w
.
f(0). Let ^
x >
y^y^^
>
b,
we have
(Figure 9)
( °)-
then
x"
FIGURE
number y such
x, so
< x .(i
> 2/(0) and
x <
a
x
/.(*)
Similarly, if
is
> M, we have
\x\
2
is
then there
9, if
+
Since n
+ ao,
•
x.
all
M = max(l, 2n\a
then for
•
- b) "
„
/U)>y>-y(
n
9
b"
^m
= y>/(0).
Summarizing:
if
Now
>
jc
< -b, then f(x) >
Theorem 7 to the function / on
a number y such that
is
if
(1)
In particular, f(y)
<
(2)
Combining
(1)
and
Theorem 10 now
li
or x
apply
that there
theorem
Z?
f(0).
if
x
x
<
b,
the interval [—b, b].
then f(y)
<
f(x).
Thus
< -b
we
(2)
-b <
/(0).
or x
>
then f(x)
b,
see that /(y)
<
> /(0) >
/(jc) for all x. |
allows us to prove the following result.
Consider the equation
(*)
x"
+a n _
n-\
1
]
x'
1
+---+a =
c
/(v).
We
conclude
1
28
Foundations
m
and suppose n is even. Then there is a number
> m and has no solution for c < m.
such that
(*)
has a solution for
c
PROOF
Let /(*)
=
x"
According
Let
m =
the
left
f(y).
(Figure 10).
Theorem 10 there is a number v such that f(y) < fix) for
If c < m, then the equation (*) obviously has no solution,
> m.
If c
=
m, then
all
x.
since
has y as a solution.
(*)
suppose c > m. Let b be a number such that b > y and f(b) > c. Then
m < c < f(b). Consequently, by Theorem 4, there is some number x in
=
[y, b]
to
+oo
...
1
side always has a value
Finally,
/(v)
+ a,,-!*"- +
=
such that f(x)
c,
so x
a solution of (*).
is
|
These consequences of Theorems 1, 2, and 3 are the only ones we will derive
(these theorems will play a fundamental role in everything we do later, however). Only one task remains
to prove Theorems 1, 2, and 3. Unfortunately,
we cannot hope to do this on the basis of our present knowledge about the real
now
—
—
numbers (namely, PI— PI 2) a proof is impossible. There are several ways of conthis 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 P1-P12. Again, suppose we consider the
vincing ourselves that
,
figure:
10
function
=
fix)
1
-^r
were no number x with x = 2, then / would be continuous, since the
denominator would never = 0. But / is not bounded on [0, 2] So Theorem 2
depends essentially on the existence of numbers other than rational numbers, and
If there
.
therefore
on some property of the
Despite our inability to prove
which we want
to
be
true.
real
numbers other than PI -PI 2.
Theorems
If the pictures
nection with the mathematics
we
1, 2,
and
3,
they are certainly results
we have been drawing have any
are doing,
if
our notion of continuous function
corresponds to any degree with our intuitive notion, Theorems
got to be true.
property of
way
Since a proof of any of these
R which
and 3 have
theorems must require some new
has so far been overlooked, our present
to discover that property:
2,
1,
difficulties
us try to construct a proof of
let
con-
suggest a
Theorem
1
,
for
example, and see what goes wrong.
One
I
[GURE
is,
idea which seems promising
is
to locate the first point
the smallest x in [a,b] such that fix)
the set
A which contains
In Figure 11, x
heavy
line.
is
Since
all
numbers x
is
0.
in [a,b]
such a point, while x'
/
=
is
not.
To
find this point,
such that
The
where fix)
set
/
A
is
first
is
A
indicated by a
here using the continuity of
/ on
\a.
/?],
as well as
Problem
6-16.)
some
(We are
contains
points sufficiently close to b are not in A.
all
0, that
consider
negative on [o,*].
itself
negative at a, and positive at b, the set
points greater than a, while
=
f(x)
<
for all
x in
clearly a
a
< a <
is
We
b.
Suppose
for all
<
A
<
12);
but
this
0.
prove
to
this
A;
to
would be less than
some numbers bigger
6-3, fix)
in particular for
contradicts the fact that
a, is
bigger than every
member
<
is
we see
that
Consequently, /(or)
false.
these points
On the other hand,
12
suppose /(or)
>
0.
Again applying Theorem
6-3,
fix) would be positive for all x in a small interval containing or, in particular for
some numbers smaller than a (Figure 13). This means that these smaller numbers
are all not in A. Consequently, one could have chosen an even smaller a which
would be greater than all members of A. Once again we have a contradiction;
is also false. Hence f(a) —
and, we are tempted to say, Q.E.D.
f{a) >
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
(for example, 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 >
such that x 2 < 2.
could really be
number V 2 did
the members of A;
If the
all
f(x)
>0
for all
in this interval
x
not
exist,
for
any y >
there
least number greater than
we could always choose a still
would not be a
V2 we
chose,
smaller one.
Now
that
we have
discovered the
property of the real numbers
13
and
0,
members of
we only have
all
/(or)
of A, since the larger numbers would also be in A.
also contain
only this big.
FIGURE
greater than
is
> 0.
Then, by Theorem
and
x in a small interval containing a,
than a (Figure
FIGURE
=
claim that f(oc)
that f(ct)
first
number which
the smallest
eliminate the possibilities /(or)
all
129
this interval
Now suppose
A would
Hard Theorems
Three
7.
That
is
we
fallacy,
need. All
almost obvious what additional
is
it
we must do
say
is
it
properly and use
it.
the business of the next chapter.
PROBLEMS
For each of the following functions, decide which are bounded above or below
on
the indicated interval,
value. (Notice that
and even
if
/
and which take on
might have these properties
the interval
on (-1,1).
fix)
fix)
=x
on
/(*)
=
that
—a
f(x)
(")
f(x)
(iii)
(iv)
maximum
even
if
/
is
or
minimum
not continuous,
not a closed interval.)
is
=x 2
=x 3
=x 2
W
their
on (-1,1).
onR.
[0, oo).
x < a
x",
a
1
+ 2,
<
a
on {—a
x > a
+
1
;
it
will
—
1,0
+
1).
(We assume a > —
be necessary to consider several
1.
so
possibilities
for a.)
vi
fix)
=
x
a
vn
fix)
=
x < a
,
+ 2,
x > a
on
0,
x irrational
\/q,
x
=
p/q
[-
a
in lowest
1
,
a
+
1]
.
on
terms
(Again assume a
[0. 1].
> —
1 .)
.
130
Foundations
vin
fix)
=
1,
[
{
1/g,
[
,.,,,,
rx
/(a) =
,./
v
(xi)
fix)
(xii)
/(a)
a
=
•
,
on
,
terms
in lowest
p/q
x irrational
1,
—
=
=
on
on
a
terms
in lowest
[0, 11.
[0, 11.
m ^^
r
.irrational
sin" (cos
[a]
i
,
p/g
rational
Jf
f
=
a
l/<y,
W /W= {0,*,
\
irrational
f
\
[
/
a.
1
+ vo + a 2
)
„
,
on
[0, a'].
[0,a].
For each of the following polynomial functions /, find an integer n such that
for some a between /7 and n + 1.
fix) —
/(A)=A 3 -A + 3.
4
/(a) = a 5 + 5a + 2a +
/(A) = A 5 +A + 1.
/(a) = 4a 2 -4a+ 1.
(i)
(ii)
(iii)
(iv)
3.
Prove that there
*"" +
(i)
sin
4.
a
=
If n
—
k
—
a
This problem
(a)
some number a such
is
l^-rr- =
+ a- + sin" A
1
(ii)
1.
is
119.
1
a continuation of
and >
even,
is
that
Problem
0, find
3-7.
a polynomial function of degree n with
exactly k roots.
(b)
A
root a of the polynomial function
—
—
fix)
have a as a root. Let
if
that
/
(a
Suppose
is
/
is
a polynomial function of degree n. Suppose
/ be
that
/
is
the roots.
all
m
multiplicity
said to have
a polynomial function that does not
has k roots, counting multiplicities,
of the multiplicities of
5.
where g
a)'" gix),
continuous on
[a, b]
Show
and
i.e.,
suppose that k
—
that n
that fix)
is
k
is
the
is
sum
even.
What
always rational.
can be said about /?
6.
Suppose
for
7.
all
that
/
a. (This
is
a continuous function on [— 1,
means
that (a, fix)) always
either fix)
=V —
How many
continuous functions
1
a 2 for
all
lies
a, or else fix)
/
1]
such that x~
on the
+ (/(a))" =
unit circle.)
= — v — a2
1
for
all
Show
1
that
x.
=
are there which satisfy (/(a))"
x
for
all*?
8.
Suppose that / and g are continuous,
all
9.
(a)
a.
Prove that either fix)
Suppose
that
=
gix) for
that
all
2
f = g
,
and
x, or else fix)
that fix)
=
—gix)
^
for
for
all
only for x = a, and
/ is continuous, that /(a) =
some a > a as well as for some x < a. What can be
for
fix) >
about fix) for
all
x
^
a?
v.
that
said
1
Three
7.
some x <
that
/
Now what can be said about fix) for x / a?
sign of x + x y + xy + y when x and y are
Suppose / and g are continuous on
and
[a, b]
gib). Prove that fix) = gix) for some x in
short, it's not the right one.)
12.
a continuous function on
Suppose
that
for each
x (draw a picture). Prove that fix)
/
Problem
(a)
=
is
a,
for
a.
Discuss the
*(c)
11.
131
only for x
continuous and that fix) =
for some x > a and fix) <
but suppose, instead, that fix) >
Again assume
(b)
10.
Hard Theorems
is
shows that /
11
ure 14 (solid
line).
Show
not both
0.
< gia), but fib) >
your proof isn't very
that fia)
[a,/?].
and that fix) is in
some number x.
[0, 1]
=x
(If
[0, 1]
for
intersects the diagonal of the square in Fig-
that
/ must
also intersect the other (dashed)
diagonal.
Prove the following more general
(b)
FIGURE
=
g (0)
14
13.
0, g(l)
=
Let fix)
(a)
=
1
=
or giO)
x
sin l/x for
=
g(\)
1,
^
fact:
and
let
continuous on
If
g
0,
then fix)
is
/(0)
=
0.
=
Is
and
some x.
[0, 1]
gix) for
/ continuous on
[— 1, 1]? Show that / satisfies the conclusion of the Intermediate Value
Theorem on [— 1 1] in other words, if / takes on two values somewhere
;
,
*(b)
on [— 1,
1],
Suppose
that
also takes
it
on every value
in
between.
/ satisfies the conclusion of the Intermediate Value Theo/ takes on each value only once. Prove that / is continuous.
to the case where / takes on each value only finitely many
rem, and that
Generalize
*(c)
times.
14.
If
/
on
(a)
*(b)
is
a continuous function on
:
15.
maximum value
Prove that for any number c
Prove that
+
\\f
g\\
that
-
\\h
<
||/||
||c/||
=
\c\
/
\\f\\.
+
\\g\\.
Give an example where
f\\
<
\\h
-
g\\
+
-
\\g
Prove that
if
n
(b)
Prove that
if
n
x"
+
Hint:
||/
+
g\\
/
f\\.
continuous and lim 0(x)/a"
is
(a)
(a)
|
llsll.
Prove that
Suppose
+
we have
=
=
X—»oo
16.
of
[0, 1].
11/11
(c)
be the
[0, 1], let
c()ix)
for
is
is
Suppose that /
is
even, then there
a
is
number x such that x" + </>(.v) = 0.
number y such that y" + </>( v) <
a
x.
all
Of which
odd, then there
lim 0(.v)/.v".
x—>— oo
proofs does this problem test your understanding?
is
continuous on
(a, b)
and lim fix)
+
lim fix)
x—*a
minimum on all of ia,b).
corresponding result when a = — oo and/or
=
oo.
Prove that / has a
*17.
(I))
Prove the
Let
/ be any polynomial
that |/()0|
<
|/(*)| for
function. Prove that there
all
x.
is
b
=
oo
some number y such
1
32
Foundations
*18.
Suppose
that
lim fix)
X—*O0
=
/
*19.
(a)
Suppose
/
that
that there
Show
<
is
some y
is
for
x,
all
Prove that there
is
and
some
for all x.
continuous on
is
other words there
(b)
> f{x)
picture.)
[a, b],
a point on the graph of
is
to (y, /(y))
Figure
fix). (Draw a
lim
X^r—OO
that /(y)
number y such
>
a continuous function with fix)
is
=
and let x be any number. Prove
/ which is closest to (x,0); in
from
in [a b] such that the distance
,
distance from (x, 0) to
(x 0)
,
f(z)) for all z in [a, b]. (See
(z,
15.)
same
that this
assertion
is
not necessarily true
if [a
,
b]
is
replaced
by ia,b) throughout.
(c)
(d)
FIGURE
15
(e)
is true if [a, b] is replaced by R throughout.
and (c), let g(x) be the minimum distance from (jc,0) to a
point on the graph of /. Prove that giy) < g(x) + \x — y\, and conclude
Show
(a)
that g
continuous.
(a)
is
Prove that there are numbers *o and x\ in [a,b] such that the distance
from (jto, 0) to (jci, f(x\)) is < the distance from (xq, 0) to (x\\ f{x\'))
any
for
20.
that the assertion
In cases
xq' x\
in [a, &].
,
Suppose
/
that
continuous on
is
natural number. Prove that there
f(x + \/n),
/
=
g(x)
*(b)
<
number
FIGURE
16
+
any
*21.
(a)
shown
<
a
=
/(0)
but that a
/ which
<
b,
x
exist
then either fix)
Why?
in ia,b).
< fia)
fix)
Refine part
takes
>
In the
on
(c)
on each value
/
fia) for
all
case
all
first
(d)
continuous on
Hint:
x
If
in ia,b) or
x?
[0, 1]
=
and which
fix
+
a) for
/
fia)
fix)
=
fib) for
< fia)
for
in ia,b); this implies that
no continuous function / which
times or 2 times, i.e., which takes on exactly
either
it
What can be
generally, find
is
does take on. Hint:
said
or a
minimum
The
previous hint implies
value (which must be taken
about values close to the
/ which
takes
maximum
on every value
value?
exactly 3 times.
one which takes on every value exactly n times,
if
odd.
Prove that
if
//
is
even, then there
every value exactly
/;
times. Hint:
is
To
no continuous / which takes on
treat the case n
=
4, for
example,
for all
fix\) = fix 2) = /C*3) = /U4). Then either fix) >
two of the three intervals ix\,X2), (^2^3)) (^3^x4), or else fix) <
all x in two of these three intervals.
let
R
defined on
values close to fia), but slightly
on somewhere
maximum
has either a
twice).
More
is
is
twice.
< a and x > b.
by proving that there
Find a continuous function
//
for all
for x
(a)
twice each value that
that
/
g(x)
a continuous function
on every value exactly
takes
larger than fia), are taken
(b)
if
not equal to \/n for any natural
is
x.
which
all
Hint: Consider the function
/(l), but which does not satisfy fix)
Prove that there does not
a
= 4.
what would be true
l/«);
1,
Find a function
n.
is
in Figure 16 for n
— f{x +
f(x)
Suppose
satisfies
X X
as
and /(0) = /(l). Let n be any
some number x such that fix) —
[0, 1]
,v
in
for
.
.
LEAST UPPER BOUNDS
CHAPTER
Never-
This chapter reveals the most important property of the real numbers.
theless,
it
is
merely a sequel
Chapter
to
path which must be followed has
7; the
already been indicated, and further discussion would be useless delay.
DEFINITION
A set A
of real numbers
bounded above
is
x > a
Such a number x
called an
is
if
there
is
a
number x such
that
for every a in A.
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 (and in 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
we mean
that there
is
a
number which
is
an upper bound for
A
—
now been used in two ways first, in
and now in reference to sets. This dual usage
Notice that the term "bounded above" has
Chapter
reference to functions,
7, in
should cause no confusion, since it will always be clear whether we are
about a set of numbers or a function. Moreover, the two definitions are
connected:
above on
The
if
A
is
the set {f(x)
and only
[a, b\ if
entire collection
examples of
the
:
which
is
lj, 1^,
A
bounded
not
is
{x
:0
<x
<
1}.
is
is
any possible confusion
(in
the superlative of "less" means),
we
self-explanatory, in order to avoid
particular, to ensure that
we
all
know what
define this explicitly.
A number x
is
a least
and
33
is
bounded above we need only name some upper bound for A,
easy enough; for example, 138 is an upper bound for A, and so are 2,
and 1. Clearly, 1 is the least upper bound of A; although the phrase
that
just introduced
DEFINITION
/
and the natural numbers N, are both
bounded above. An example of a set which is
A =
To show
function
closely
R of real numbers,
which are
sets
bounded above
if
a < x < b}, then the
set A is bounded above.
talking
upper bound
(1)
x
is
(2)
if
y
of
A
if
an upper bound of A,
is
an upper bound of A, then x
<
y.
.
134
Foundations
The
to
use of the indefinite article "a" in this definition
Now
temporary ignorance.
seen that
if
that
x and v are both
least
< y
< x,
we have made
was merely a concession
a precise definition,
=
upper bounds of A, then x
it
easily
is
Indeed, in
v.
this
case
x
,
and y
since v
is
an upper bound, and x
is
a least upper bound,
since
is
an upper bound, and y
is
a least upper bound;
jc
— y. For this reason we speak of the
The term supremum of A synonymous and has one
it
follows that x
is
upper bound of A.
least
advantage.
It
abbreviates
quite nicely to
A
sup
and
saves us
(pronounced "soup A")
from the abbreviation
lub
(which
is
There
can
is
more
treated
number x such
a
authors).
a series of important definitions, analogous to those just given, which
is
now be
there
some
nevertheless used by
A
A
briefly.
set
A
of real numbers
for every a in
Such a number x is called a lower bound
lower bound of A if
(
The
1 )
(2)
greatest lower
if
that
x < a
and
bounded below
is
x
is
a lower
if
y
is
bound of A
A
A number
for A.
x
bound of A
bound of A, then x >
the greatest
,
a lower
is
is
also called the
infimum
y.
A
of
,
abbreviated
inf A;
some authors use
the abbreviation
gib A.
One
sets
detail has
have
bound.
at least
We
will
—
been omitted from our discussion so far the question of which
one, and hence exactly one, least upper bound or greatest lower
consider only least upper bounds, since the question for greatest
lower bounds can then be answered easily (Problem
If
A
is
not
cannot be expected to have a
have a
least
upper bound
mathematical induction,
If
A =
0,
2).
bounded above, then A has no upper bound
then A
is
if
least
it
upper bound.
It is
at
tempting
has some upper bound, but,
this assertion
bounded above.
can
fail
to
all,
be true
in
so
A
certainly
to say that
,4
does
like the principle
of
a rather special way.
Indeed, any number x
is
an upper bound
for 0:
x
simply because there
there
is
surely
no
is
least
>
y
for every y in
no y in 0. Since
upper bound for
every
0.
number
With
is
an upper bound
(his trivial
for 0,
exception however,
.
8.
our assertion
true
is
—and very important,
We
consideration of details.
definitely
135
Least Upper Bounds
important enough to warrant
are finally ready to state the last property of the real
numbers which we need.
(The
(PI 3)
upper bound property)
least
/4/0, and A
may
Property PI 3
is
you
strike
as anticlimactic,
To complete our list of basic
virtues.
A
If
a set of real numbers,
is
bounded above, then A has a
but that
properties for the real
least
upper bound.
particularly abstruse proposition, but only a property so simple that
foolish for having overlooked
it.
Of course,
really so innocent as all that; after
FIGURE
all, it
the least
does
not
one of
actually
is
its
numbers we require no
we might
upper bound property
is
hold for the rational numbers
feel
not
Q
A is the set of all rational numbers x satisfying x < 2, then there
is no rational 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 bound for A It will become
clear only gradually how significant PI 3 is, but we are already in a position to
For example,
1
if
.
demonstrate
theorem
7-1
/
If
is
such that f(x)
Our proof
Chapter 7
fix)
<
for
all
this interval
2
fix) >
for all
in this interval
FI
GUR
E
3
x
<
f(b), then there
is
some number x
\x
A, shown in Figure
:
a
0, since
<x <
a
and /
b,
in
is
1,
A; in
< x <
is
fact,
+
number x
in [a, b]
at the
with f(x)
=
end of
0.
as follows:
negative
there
is
on the
some
8
interval [a, x]
>
such that
}.
A
contains
Problem 6-16, since / is
continuous on [a, b] and /(«) < 0. Similarly, b is an upper bound for A and, in
such that all points x satisfying b — 8 < x < b are upper
fact, there is a 8 >
bounds for A; this also follows from Problem 6-16, since f{b) > 0.
From these remarks it follows that A has a least upper bound a and that
a < a < b. We now wish to show that f(a) = 0, by eliminating the possibilities f(a) <
and f(a) > 0.
Suppose first that /(a) < 0. By Theorem 6-3, there is a 8 >
such thai
—
for a
8 < x < a + 8 (Figure 2). Now there is some number xq in A
f(x) <
which satisfies a — 8 < xq < a (because otherwise a would not be the least upper
bound of A). This means that / is negative on the whole interval [fl,xo]. But if
X] is a number between a and a + 8, then /' is also negative on the whole interval
[xo,jci]. Therefore / is negative on the interval [a,;q], so x\ is in A. But this
contradicts the fact that a is an upper bound for A; our original assumption that
must be false.
f(cx) <
all
FIGURE
A ^
<
7.
0.
will locate the smallest
set
A =
Clearly
=
merely a rigorous version of the outline developed
is
—we
Define the
x
power, by supplying the proofs which were omitted in Chapter
continuous on [a,b] and f(a)
in [a, b]
PROOF
its
points x satisfying a
a
8; this
follows from
Suppose, on the other hand, that f(a) > 0. Then there is a number 8 >
such
that fix) >
for a — 8 < x < a + 8 (Figure 3). Once again we know that there is
an xq
in
A
satisfying
a —
8
<
xq
<
ce;
but
this
means
that
/
is
negative on [a, xq],
136
Foundations
which
is
impossible, since /(*o)
a contradiction, leaving f(a)
The
result,
proofs of
which
will
>
=
Thus
0-
the assumption f(a)
>
as the only possible alternative.
also leads to
|
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.
THEOREM
1
/
If
is
continuous
at a,
above on the interval
PROOF
Since lim f(x)
—
(a
then there
—
a
8,
f(a), there
+
is
a
8) (see
number
Figure
for every s
is,
>
8
>
such that
/
is
bounded
4).
>
0, a 8
such
that, for all x,
x-*-a
if \x
It is
for
only necessary to apply
example, e
=
1
.
We
if
It
— a\ <
this
on the
/(«)+!•
then \f(x)
|.v
-a\ <
interval {a
—
—
<
f(a)\
s.
statement to some particular s (any one will do),
conclude that there
follows, in particular, that if \x
the proof:
8,
8,
then \f(x)
— a\ <
8,
is
a
+
8,
a
<5
-
>
such
<
f(a)\
then f(x)
—
that, for all
,
1.
f{a)
8) the function
x
/
is
<
1.
This completes
bounded above by
I
It should hardly be necessary to add that we can now also prove that / is
bounded below on some interval (a — 8, a + 5), and, finally, that / is bounded on
some open interval containing a.
A more significant point is the observation that if lim f(x) = f(a), then there
.v—>a+
'IGURE
4
.
8.
is
a 8
>
/
such that
observation holds
bounded on
is
—
lim fix)
if
Least Upper Bounds
137
< x < a + 8}, and a similar
Having made these observations (and
the set {x
f(b).
:
a
x-*-b~
assuming that you
THEOREM
7-2
PROOF
/
If
is
continuous on [a b] then
/
,
,
is
our second major theorem.
tackle
bounded above on
[a b]
,
Let
x
A /
Clearly
:
a
(since a
<x <
in A),
is
is
bounded above on
and A
is
bounded above
to the sets {/(j)
:
<
a
<
y
x},
[a, x]}.
A
(by b), so
has a least
are here applying the term "bounded above" both
which can be visualized
to the set A,
(Figure
/
b and
upper bound a. Notice that we
{/(y) :a
we
supply the proofs),
will
as lying
on the horizontal
which can be visualized
as lying
axis,
and
on the
to /,
i.e.,
vertical axis
5).
Our first step is to prove that we actually have a = b. Suppose, instead, that
a < b. By Theorem 1 there is 8 > such that / is bounded on (a — 8, a +8). Since
a is the least upper bound of A there is some xo in A satisfying a— 8 < xq < a. This
means that / is bounded on [a, xo]- But if x\ is any number with a < x\ < a + 8,
then / is also bounded on [xo, jq]. Therefore / is bounded on [a,xi], so x\ is
in A, contradicting the fact that a is an upper bound for A. This contradiction
shows that a = b. One detail should be mentioned: this demonstration implicitly
assumed that a < a [so that / would be defined on some interval (a — 8, a + 8)];
the possibility a = a can be ruled out similarly, using the existence of a 8 >
such
that / is bounded on {x a < x < a + 8}.
The proof is not quite complete we only know that / is bounded on [a, x] for
every x < b, not necessarily that / is bounded on [a, b]. However, only one small
argument needs to be added.
There is a
> such that / is bounded on {x b — 8 < x < b}. There is xo
in A such that b — 8 < xq < b. Thus / is bounded on [a, xo] and also on [xo, b],
so / is bounded on [a b] |
<y <x
:
—
FIGURE
5
<5
:
,
To prove
theorem
7-3
/
If
is
PROOF
We
all
already
theorem we resort
the third important
continuous on
f{x) for
.
X
[a, b],
then there
number
a
y in [a, b]
such that
f(y)>
in [a, b].
know
/
that
is
bounded on
{/(*):*
is
bounded. This
Qf
> fix)
for
is
to a trick.
x
set
is
in [a, b\
Suppose instead
that
[a b]
,
a
suffices to
^
f(y) for
show
all
which means that the
v in
=
<x
~
—-,
fix)
bound a. Since
some y in [a, b\.
has a least upper
it
that
a.
=
[a,b].
f(y)
x
for
Then the function
by
g(x)
set
[a,b]}
in
obviously not 0, so
it
,
in [a, b\
g
defined
1
38
Foundations
is
continuous on
other hand, a
[a,
ft],
for every e
means
This, in turn,
upper bound of {/(*) x
the least
is
:
>
//;«
means
there
is
x
right side
in [a,
with a
in [a, b]
—
is
never
ft]); this
f(x)
<
0.
means
On
the
that
e.
that
for every e
But
denominator of the
since the
that g
is
>
there
is
x
in [a,
bounded on
not
[a,
with g(x)
ft]
>
1/e.
contradicting the previous theo-
ft],
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 prove that N is unbounded.
After the difficult theorems proved in this chapter you may be startled to find
such an "obvious" theorem winding up our proceedings. If so, you are, perhaps,
allowing the geometrical picture of R to influence you too strongly. "Look," you
may
say,
"the real
numbers look
12
n x
3
number x
so every
like
between two integers
is
+
n
n,
Basing the argument on a geometric picture
1
n
is
+
(unless x
1
is
itself
not a proof, however,
an integer)."
and even the
geometric picture contains an assumption: that if you place unit segments end-toend you will eventually get a segment larger than any given segment. This axiom,
often omitted from a
first
introduction to geometry,
is
usually attributed (not quite
N
is not
Archimedes, and the corresponding property for numbers, that
bounded, is called the Archimedean property of the real numbers. This property is not
a consequence of PI -P12 (see reference [14] of the Suggested Reading), although
justly) to
it
Q,
does hold for
of course.
Once we have PI 3 however,
there are
no longer
any problems.
theorem
2
PROOF
N
is
not
bounded above.
Suppose
N
bound a
for
were bounded above.
Since
N ^
0, there
would be a
least
upper
N. Then
a >
for all n in
//
N.
Consequently,
a >
since n
+
\
is
in
N
if
/;
is
in
means
this
a
the least
is
thai
a-
—
1 is also
upper bound.
-f
N. But
a
and
n
I
1
>
1
this
n
for all
means
for
all
in
//
N,
that
/;
an upper bound
in
N,
for
N, contradicting the
fact that
8.
There
3
PROOF
For any e
Suppose
means
A
often
>
assumed
there
that 1/e
>
is
e for
number
of
many
will
them some cheating was
tolerated.
can claim that
was never used
has been shaken, a review of
all
As
we
that
will
<
1/e for
n in N. But this
all
2. |
Theorem 3
Theorem 3 was not
that the result of
Of
course,
were so important that
order to give
in
partial justification for this dishonesty
proof of a
in the
the proofs given so far
We
such deception will not be necessary again.
numbers
n
show you
examples.
available at the time, but the examples
this result
Thus
s.
N, contradicting Theorem
for
Chapter 6
in the discussion
<
n with \/n
n in N.
all
an upper bound
brief glance through
the real
39
implicitly.
a natural
is
not; then 1/n
was used
1
a consequence of Theorem 2 (actually an equivalent formulation) which
is
we have very
theorem
Least Upper Bounds
have
now
ever need. Henceforth, no
theorem,
is
but
if
your
we
faith
Fortunately,
in order.
stated every property of
more
lies.
PROBLEMS
1
.
Find the
the following
(i.e.,
to
upper bound and the greatest lower bound
least
decide
when
belong to the
—
:
Also decide which
sets.
the least upper
sets
(if
have greatest and
bound and
greatest lower
they
least
exist)
of
elements
bound happens
set).
n in
N
n in
Z and
=
or x
n
—
:
n
^
n
(iv)
— \/n for some in
< x < V2 and x is rational}.
2
> 0}.
x +x +
2
< 0}.
:x + xx <
and x 2 + x — < 0}.
{x
:
[x
:
(v)
[x
(vi)
{x
(vii)
{x
A'
N}.
//
1
:
1
1
:
1
+
(viii)
2.
(a)
(-!)"
Suppose A
:
n
^
for x in A.
inN
is
bounded below.
Prove that
—A ^
—A denote the set of all — x
—A bounded above, and that
Let
0, that
is
— sup (—A)
(b)
3.
Let
(a)
is the greatest lower bound of A.
A 7^
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.
If
/
be a continuous function on
The proof
of
Theorem
with f{x)
=
0.
is
7-1
If there
is
[a, b\
with f(a)
<
<
f{b).
showed that there is a smallest A" in \a.b\
more than one a in \a,b\ with /(a) = 0,
there necessarily a second smallest?
Show
that there
is
a largest x in
.
140
,
.
Foundations
=
\a,b] with f(x)
(Try to give an easy proof by considering a
0.
new
function closely related to /.)
(b)
The proof of Theorem 7-1 depended upon considering A = ix a <
x < b and / is negative on [a, x] }. Give another proof of Theorem 7-1
which depends upon consideration of B = {x a < x < b and f(x)<
will this proof locate? Give
0}. Which point x in [a,b] with f(x) =
an example where the sets A and B are not the same.
:
:
*4.
(a)
that
Suppose
also that f(xo)
numbers
c
/
is
>
some
for
in [a, b].
xq,
a<c<xo<d<b such
and d with
f(x) >
= f(b) = 0.
Prove that there are
continuous on [a,b] and that f(a)
Suppose
that /(c)
=
f(d)
=
0,
(b)
for
x in (c, d). Hint: The previous problem can be used
good advantage.
Suppose that / is continuous on [a, b] and that f(a) < f(b). Prove that
there are numbers c and d with a < c < d < b such that f(c) = f(a)
and f(d) = f(b) and f(a) < f(x) < f(d) for all x in (c, J).
(a)
Suppose
but
all
to
5.
(b)
x < k <
v.
consider
+
<
/
be the largest integer
Prove that there
v.
that r
<
s
*6.
A
and the
set
until this
If
/
if
is
A, then f(x)
if
is
Hint:
As a
said to be
start,
dense
A, then f(x)
all
continuous and f(x)
=
for
/ and g
we assume
g(x) for
x.
=
all
x,
and
such that
jc
<
Why have parts
set?)
you know
is
an
irrational
that there
is
an
if
every open interval contains a
Can > be
all
of rational numbers
/
is
is
a
number
for
>
g(x) for
replaced by
+
c such that f(x)
all
+ y) =
+
numbers x
in
a dense
There
=
g(x) for
all
x
in a
dense
x
in A,
show
that f(x)
>
x and
v,
> throughout?
y)
=
=
f(x)
ex for
can be demonstrated simply by combining the
lems.) Point of information:
all
x.
continuous and f(x
if
=
are continuous and f(x)
g(x) for
Prove that
fact, this
set
x.
instead that f(x)
then there
f(x
<
I
of irrational numbers are each dense.
Prove that
set
7.
problem
r
(Query:
1.
For example, Problem 5 shows that the
.
Prove that
set
(c)
number
a rational
are rational numbers. Prove that there
of real numbers
point of
(b)
satisfying
and 1
Suppose that x < y. Prove that there is an irrational number between x
and v. Hint: It is unnecessary to do any more work; this follows from
(b) and (c).
A set A
(a)
is
< y —x, then ny — nx >
number between r and s.
irrational number between
(d)
an integer k such that
is
1
been postponed
(b)
Suppose
Prove that there
1.
Let
y. Hint: If \/n
and
(a)
(c)
/
x >
Hint:
Suppose x <
r
—
that y
+
f(y) for
all
results
x.
all
(This conclusion
of two previous prob-
do exist noncontinuous functions
we cannot prove
/
satisfying
now;
in
simple question involves ideas that are usually never mentioned
in
f{x)
f(y) for
undergraduate courses
all
x and y, but
(see reference [7] in the
this
Suggested Reading).
8.
*8.
Suppose
/
that
a function such that f(a)
is
ure
6).
(a)
Prove that lim f(x) and lim f{x) both
x—>a~
lem
/ never has
from Problem
Prove that
(c)
if
If
/
a
is
6
Why
is
Hint:
exist.
«+
a removable discontinuity
satisfies
< b
(this
terminology comes
is
bounded function on
[0, 1], let
sup{
Problem
in
||
||
=
|||/|||
|/(jf)|
:
x
in [0, lj}.
7-14.
11.
(a)
Suppose
&n+\
S
a\,a2,a$,
that
.
..
is
a sequence of positive
>
Prove that for any e
«/?/2.
Suppose P
there
many
inscribed regular polygon with twice as
ence between the area of the
circle
between the area of the
difference
Prove that there
some
is
need part
(a).
sides,
<
n with a n
show
e.
If P'
is
the
that the differ-
and the area of P' is less than half the
and the area of P (use Figure 7).
circle
a regular polygon
P
inscribed in a circle with area
as close as desired to the area of the circle.
will
is
in the
numbers with
a regular polygon inscribed inside a circle.
is
The-
continuous.
Suppose a > 0. Prove that every number x can be written uniquely
< x' < a.
form x — kct + x', where k is an integer, and
(c)
prob-
this
10.
(b)
(Fig-
the conclusions of the Intermediate Value
Prove analogues of the properties of
FIGURF,
a
6-17).
/
orem, then /
*9.
< f(b) whenever
in this chapter?
Prove that
(b)
jc—
141
Least Upper Bounds
In order to do part
This was clear to the Greeks,
who
used part
(c)
you
as the
(a)
and area. By calculating
method of exhaustion") allows
basis for their entire treatment of proportion
method
the areas of polygons, this
computations of
that =jy
*(d)
<
Using the
7i
<
tz
=f.
to
("the
any desired accuracy; Archimedes used
But
it
fact that the areas
areas of two circles have the
7
Deduce a
is
to show-
has far greater theoretical importance:
of two regular polygons with the same
ber of sides have the same ratio as the square of their
FIGURE
it
same
sides,
num-
prove that the
ratios as the square of their radii. Hint:
contradiction from the assumption that the ratio of the areas
greater, or less, than the ratio of the square of the radii
by inscribing
appropriate polygons.
12.
Suppose
for all
13.
that
A and B
x in A and
all
(a)
Prove that sup
(b)
Prove that sup
are two
nonempty
sets
of numbers such that
jc
< y
v in B.
A <
A <
y for
all
y in B.
inf B.
A and B be two nonempty sets of numbers which are bounded above, and
let A + B denote the set of all numbers x+y with x in A and y in B. Prove that
sup(A + Z?) = sup A+sup B. Hint: The inequality sup(A+5) < sup A+supB
is easy. Why? To prove that sup A + sup B < sup(A + B) it suffices to prove
that sup A + sup B < sup(/4 + B) + s for all s > 0; begin by choosing x in A
Let
5
..
.
142
Foundations
and
B
v in
FIGURE
with sup
8
A —
1
|
1
1
fl3
Suppose
there
<
that a„
a n+
and b n+
\
a point x which
is
Show
(b)
1
bj,
Consider a sequence of closed intervals
(a)
e/2.
1
ai
ci\
14.
< e/2 and sup B — y <
x
is
that this conclusion
<
\
—
I\
b\
£>2
b„ for
=
\a\, b\], li
all
n (Figure
[«2> b{\,
.
.
.
Prove that
8).
in every /„.
false if
is
we
consider open intervals instead of
closed intervals.
The
simple result of Problem 14(a)
may be
is
called the "Nested Interval
used to give alternative proofs of Theorems
1
and
2.
Theorem."
The
It
appropriate
reasoning, outlined in the next two problems, illustrates a general method, called
a "bisection argument."
*15.
continuous on [a,b] and f(a) <
< f(b). Then either
b)/2) = 0, or / has different signs at the end points of the interval
Suppose /
+
(a +
f((a
[a,
Why?
is
b)/2], or
If
Now bisect
two
*16.
/
has different signs at the end points of [(a
+ b)/2) ^
f((a
Either
I\.
Let
intervals.
h
/
0, let I\
at the
is
be that
/
to find a point
x where f(x)
at
midpoint, or
interval.
=
/ changes
Continue
some midpoint). Use
each n (unless
is
be the interval on which
/
sign
+
b)/2, b].
changes
sign.
on one of the
in this way, to define /„ for
the Nested Interval
Theorem
0.
Suppose / were continuous on [<z,&], but not bounded on [«,/?]. Then /
would be unbounded on either [a, (a + b)/2] or [(a +b)/2, b]. Why? Let /]
be one of these intervals on which / is unbounded. Proceed as in Problem
1
to obtain a contradiction.
17.
(a)
Let
A =
If
(i)
x
<
a}. Prove the following (they are
is
in
A and
:
x
(iii)
A ^
A ^
(iv)
If
(ii)
(b)
{x
x
v
<
x, then y
is
all easy):
in A.
0.
R.
is
in
A, then there
Suppose, conversely, that
A
is
some number
satisfies (i)— (iv).
x' in
A
<
such that x
Prove that
A =
fv
:
x'
x <
sup A).
*18.
A number
finitely
x
is
called an
many numbers
almost upper bound for A if there are only
A with y > x. An almost lower bound is
y in
defined similarly.
(a)
Find
all
Problem
(b)
almost upper bounds and almosl lower bounds of the sds
1
A is a bounded infinite set. Prove thai the set B of
almost upper bounds of A is nonempty, and bounded below.
Suppose
in
that
all
..
8.
:
19.
follows from part
that inf
(b)
B
exists; this
number
called the limit
(c)
It
(d)
superior of A, and denoted by lim A or lim sup A. Find lim A
set A in Problem 1.
Define lim A, and find it for all A in Problem 1.
If
A
is
a
bounded
(a)
lim
A <
(b)
lim
A < sup A
(c)
If lim
(d)
The analogues
lim
infinite set
143
Least Upper Bounds
is
for
each
prove
A
A < sup A then A
,
of parts
contains a largest element.
(b)
and
(c)
for lim
.
shadow points
FIGURE
20.
Let
of
9
/ be a continuous function on R.
/ if there is a number y > x with
terminology
is
A point
/(>')
>
x
is
called a
fix).
The
shadow point
rationale for this
indicated in Figure 9; the parallel lines are the rays of the sun
Suppose that all points of (a, b) are
and b are not shadow points. Clearly, f(a) > f(b).
rising in the east (you are facing north).
shadow
(a)
points, but that a
Suppose
that f{a)
maximum
(b)
value on
Then show
f(a)
=
This
little
>
f(b).
[a, b]
Show
must be
that the point
where /
takes
on
its
a.
that this leads to a contradiction, so that in fact
we must have
fib).
result,
known
proving several beautiful
page 450.
Sun Lemma, is instrumental in
theorems that do not appear in this book; see
as the Rising
144
Foundations
UNIFORM CONTINUITY
APPENDIX.
Now
come
that we've
to the
end of the "foundations,"
it
might be appropriate
to slip in one further fundamental concept. This notion is not used crucially
the rest of the book, but it can help clarify many points later on.
b
2
We know
+e
-
that the function f{x)
x
2
is
continuous
at
a for
all a.
is
some
8
in
In other
words,
a
if
any number, then for every
is
such that, for
Of course,
depends on
8
not work at b (Figure
that
works
-
\x
But
even
<5,
for
but
8,
then
also depends
<5
Indeed,
<
a\
— a\ <
if \x
e.
1).
for all a, or
certainly satisfy
x,
all
it's
0,
—
In
8
a8
a
>
s.
the 8 that works at a might
>
there
fact,
the
>
a8,
is
no one
number
a
>
8
+8/2
will
then
8
Cl+
- w\ <
\x~
clear that given e
>
a
there
on a
positive a.
all
if
>
e
2
+
2
b^8 for b
8 for a
1
!(. i:RE
and
won't be
this
<
once a > s/8. (This
s
way of saying
tational
/
that
is
growing
is
an admittedly confusing compu-
just
faster
and
faster!)
On the other hand, for any e > there will be one 8 > that works for all a
in any interval [-#,#]. In fact, the 8 which works at N or -N will also work
1
everywhere
else in the interval.
As a final example, consider the function f{x) = sin l/x, or the function whose
graph appears in Figure 18 on page 62. It is easy to see that, so long as £ < 1,
that works for these functions at all points a in the
there will not be one 8 >
open
interval (0, 1).
These examples
illustrate
important distinctions between the behavior of various
continuous functions on certain intervals, and there
is
a special term to signal
this
distinction.
DEFINITION
The
function
/
is
some
8
>
there
is
uniformly continuous on an interval A
such
if
|.r
that, for all
-y\<8,
x and y
then |/'U)
if
for every £
>
in A,
-
f(y)\
<
s.
We've seen that a function can be continuous on the whole
line,
or on an open
without being uniformly continuous there. On the other hand, the func2
tion f(x) = x did turn out to be uniformly continuous on any closed interval.
This shouldn't be too surprising it's the same soil of thing that occurs when we
interval,
that
—
bounded on an interval and we would be led to suspect
any continuous function on a closed interval is also uniformly continuous on
ask whether a function
is
that interval. In order to prove this, we'll
Suppose
point b,
thai
and
we have two
a function
/'
that
need
intervals [a,b]
is
continuous on
to deal
and
first
[b, c]
[a, c].
with one subtle point.
with the
Let £
>
common
and suppose
endthat
145
Appendix. Uniform Continuity
8,
the following two statements hold:
(i)
if
(ii)
if
x and y are in
x and y are in
and
and
[a, b]
[b, c]
— y\ <
— y\ <
|jc
|x
8\,
then \f(x)
82,
then |/(x)
— f(y)\ <
— f(y)\ <
£,
£.
We'd like to know if there is some 8 >
such that \f(x) — f(y)\ < £ whenever
—
x and y are points in [a,c] with |jc
v < 8. Our first inclination might be to
choose 8 as the minimum of 8\ and 8j. 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 |/(jc) — f(y)\. So we have to be a little more cagey, and
|
<«5
FIGURE
also use continuity of
2
LEMMA
< b <
Let a
c
and
/
at b.
/ be continuous on the interval [a,c]. Let £ >
and (ii) hold. Then there is a 8 > such that,
let
suppose that statements
PROOF
Since
/
if
x and y are
is
continuous
(i)
and
in [a, c]
at b, there
-b\<
if \x
It
a
is
—
\x
<
y\
>
($3
if \x
Choose
8 to
—
<
b\
and
82
minimum
be the
—
\y
of
<
b\
and
8\, 82,
suppose that x and y are any two points
are both in [a, b], then \f(x) — f(y)\ < £ by
then \f(x)
—
f(y)\
<
£
by
(ii).
The
In either case, since
\x
—
/
It's
is
continuous on
<
y\
|/U)-/(v)|<eby(iii).
PROOF
<
f(b)\
e.
-.
8,
We
and
(i);
v
have
also
claim that
if
< b <
\x
—
f(y)\
|jc
<
e.
works. In
this 8
— y\ <
8.
If
x and y
x and y are both in
is
[b, c],
that
x.
b\
<
8
and
|v
b\
<
8.
So
I
/
then
[«,/?],
is
uniformly continuous on
the usual trick, but we've got to be a
>
^3.
in [a, c] with
or
we
—
then \f(x)
only other possibility
x < b < y
If
-
83,
fact,
1
<
f(y)\
such that,
then \f(x)
S3,
—
then \f(x)
8,
follows that
111
THEOREM
and
0,
little
bit careful
[a, b],
about the mechanism of
>
such
Then A ^
(since a is in A), and A is bounded above (by b), so A has
upper bound a. We really should write a B since A and a might depend on
we won't since we intend to prove that a = b, no matter what e is.
a least
the proof. For £
that, for all v
and
z in [a
if
Then
,
b]
is
£-good
on
[a, b] if
\y-z\ <8, then \f(y)-f(z)\ <
Consider any particular
{x
:
£
there
is
some
>
0.
8
,
we're trying to prove that
A =
/
say that
let's
>
/
0.
a < x
is
£-good on
e.
[a, b] for all £
Let
<
b and
/
,
is
£-good on
[a, x]}.
£.
But
.
146
Foundations
Suppose
we had a <
that
/
Since
b.
v — a\ <
continuous
is
at a, there
is
some
—
<5o
>
— a\ <
/(a?) < e/2. Consequently, if \y
So
<5o, then \f(y)
—
then
<
So
is
surely
on
the
interval
e.
e-good
\z
So,
/
f(z)\
\f(y)
[a — So, a +80]. On the other hand, since a is the least upper bound of A, it
such
that, if
—
and
is
|
/
also clear that
e-good on
I
<
ot\
+
a
[a,
e-good on
is
80], so
+
a
[a,
8q
is
a —
in
80].
Then
the
Lemma
implies that
A, contradicting the fact that a
is
/
is
an upper
bound.
To complete
argument
<5o
>
[b
—
the proof
for this
such that,
But
80, b].
e-good on
[a,
b~\.
we
show
just have to
is
practically the same: Since
if
b
/
is
—
8q
also
<
v
<
b,
e-good on
then |/(v)
[a,
b
—
that
/
—
is
a
=
b
is
actually in A.
continuous
f(b)\
80], so the
<
e/2.
at b, there
So /
Lemma
is
is
The
some
e-good on
implies that
/
is
|
PROBLEMS
1.
(a)
For which of the following values of a
formly continuous on
(b)
(c)
Find a function
Prove that
if
(b)
Prove that
if
fg
(d)
/
that
is
—
is
Show
x a uni-
(0, 1],
but not
1/3, 1/2, 2, 3?
continuous and bounded on
not uniformly continuous on
(a)
(c)
a
=
the function f{x)
Find a function / that is continuous and bounded on
uniformly continuous on (0, 1].
is
2.
[0, 00):
is
[0,
00) but which
[0, 00).
/ and g are uniformly continuous on A, then so is f + g.
/ and g are uniformly continuous and bounded on A, then
uniformly continuous on A.
that this conclusion does not hold
if
one of them
isn't
bounded.
Suppose that / is uniformly continuous on A, that g is uniformly continuous on 5, and that f(x) is in B for all x in A. Prove that g o / is
uniformly continuous on A.
3.
Use a
4.
Derive
"bisection
Theorem
argument" (page 142)
to give
another proof of Theorem
7-2 as a consequence of Theorem
1
1.
PART
DERIVATIVES
AND
INTEGRALS
In 1604, at the height of
his scientific career, Galileo argued
that for a rectilinear motion
in
which speed increases proportionally
to
distance covered,
law of motion should be
the
just that (x
=
ct 2 )
which he had discovered
in the investigation
offalling bodies.
Between 1695 and 1700
not a single one of the monthly issues
of Leipzig's Acta Eruditorum was published
without articles of Leibniz,
the Bernoulli brothers
or the
Marquis
de VHopital treating,
with notation only slightly different from
that which
the
we
use today,
most varied problems of
differential calculus, integral calculus
and
the calculus
Thus
of variations.
of almost precisely
in the space
one century
infinitesimal calculus or,
we now call
The Calculus,
as
it
in English,
the calculating tool
par
excellence,
had been forged;
and nearly
three centuries
of
constant use have not completely dulled
this
incomparable instrument.
NICHOLAS BOURBAKI
CHAPTER
DERIVATIVES
The
derivative of a function
Together with the
its
integral,
particular flavor. While
is
it
it is
the
first
of the two major concepts of
constitutes the source
true that the concept of a function
that you cannot do anything without
bounds are essential, everything we have done
limits
adequate,
this section will
this section.
from which calculus derives
is
fundamental,
or continuity, and that least upper
until
now has been
be easier than the preceding ones
—
preparation
—
if
for the really exciting
ideas to come, the powerful concepts that are truly characteristic of calculus.
Perhaps (some would say "certainly") the interest of the ideas to be introduced
in this section
cepts
and
stems from the intimate connection between the mathematical con-
certain physical ideas.
be described
in
Many
definitions,
terms of physical problems, often
and even some theorems, may
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 discuss their
significance in terms of mathematical problems.
The
collection of
functions exhibits such diversity that there
all
hope of discovering any
interesting general properties pertaining to
continuous functions form such a restricted
class,
we might
is
almost no
all.
Because
expect to find some
theorems pertaining to them, and the sudden abundance of theorems
Chapter 6 shows that this expectation is justified. But the most interesting
nontrivial
after
and most powerful
our attention even
results
about functions
further, to functions
will
be obtained only
when we
which have even greater claim
to
restrict
be called
"reasonable," which are even better behaved than most continuous functions.
= \x\,x>0
f(x)=x 2 ,x<0
f(x)
i
k,
i
k
i;
i
Figure
display.
1
illustrates certain types
The graphs of
Figure 2, where
FI
G UR E
2
it is
149
these functions are "bent" at (0,0), unlike the graph of
possible to
marks have been used
of misbehavior which continuous functions can
to
draw a "tangent
each point. The quotation
we have defined "bent" or
line" at
avoid the suggestion that
150
Derivatives
and
Integrals
"tangent line," although
we
are suggesting that the graph might be "bent" at a
point where a "tangent line" cannot be drawn.
You have probably already
noticed
that a tangent line cannot be defined as a line which intersects the graph only
FIGURE
3
— such a
would be both too restrictive and too permissive. With
such a definition, the straight line shown in Figure 3 would not be a tangent line
to the graph in that picture, while the parabola would have two tangent lines at
each point (Figure 4), and the three functions in Figure 5 would have more than
one tangent line at the points where they are "bent."
once
FIGURE
FIGURE
4
definition
5
A more promising
approach
to the definition of a tangent line
"secant lines," and use the notion of limits. If h
(a,
f(a)) and (a
slope
+
h,
f(a
+
/?))
^
0,
might
start
with
then the two distinct points
determine, as in Figure
6,
a straight line whose
is
f(a
+ h)-f(a)
h
\
FIGURE
f(a
+ h)-f(a)
6
As Figure 7 illustrates, the "tangent
some sense, of these "secant lines,"
talked about a "limit" of lines, but
line" at (a,f(a))
as h
we
approaches
seems
0.
We
to
be the
limit, in
have never before
can talk about the limit of their slopes: the
«
•
*
151
9. Derivatives
slope of the tangent line through (a, f(a)) should be
f(a+h)-f(a)
lim
h
We
DEFINITION
and some comments.
are ready for a definition,
The
/
function
differentiable at a
is
f(a+h)-f(a)
lim
exists.
h
h->0
In this case the limit
denoted by/' (a) and
is
(We also say that /
domain of /.)
a.
if
differentiable
is
is
called the derivative
/" is
if
of/
at
differentiable at a for every a
in the
The
comment on our
first
tangent line
to the
definition
graph of /
really
is
at (a,
f(a)) to
an addendum; we define the
be the
with slope f'(a). This means that the tangent line at
/
is
refers to notation.
iniscent of functional notation.
function whose
7
through
(a,
f(a))
(a,
f(a))
is
defined only
The symbol
f'(a)
is
certainly
if
differentiable at a.
The second comment
FIGURE
line
at a,
domain
and whose value
is
at
In
such a
any function /, we denote by /'the
fact, for
the set of
numbers a such
all
number
a
rem-
that
/
is
differentiable
is
f(a+h)-f(a)
lim
h
ft—
(To be very precise: /'
is
the collection of
f(a
pairs
all
+ h)-f(a)
a, lim
h
/i—
for
which lim [f(a + h)
— f(a)]/h
exists.)
The
function /'
is
called the
derivative
ft—>0
of/.
Our
comment, somewhat longer than
third
ical interpretation
the previous two, refers to the phys-
of the derivative. Consider a particle which
on which we have chosen an
straight line (Figure 8(a))
direction in
which distances from
tance from
O
O
shall
been chosen purposely;
be written as positive numbers, the
since a distance s(t)
t=0
motion of the
t
=
•
particle
5
t
<
>
—
=4
•
l(,
I
RE
8
a
t.
is
The
dis-
>_
)
t
=
along which particle
is
moving
=
3
—
suggestive nota-
determined
t=\
t
•
<
line
I
moving along a
and a
of points in the other direction being written as negative numbers.
Let s(t) denote the distance of the particle from O, at time
tion s(t) has
is
"origin" point O,
for
each
152
Derivatives
and
Integrals
number t, the physical situation automatically supplies us with a certain function s.
The graph of s indicates the distance of the particle from (9, on the vertical axis,
"distance"
graph of
s
terms of the time, indicated on the horizontal axis (Figure
in
The
8(b)).
quotient
s(a
+ h) —
s(a)
h
has a natural physical interpretation.
during the time interval from a to a
FIGURE
8(b)
depends on
of course.
h,
On
It is
+ h.
For any particular a,
s{a
+ h) —
h^O
h
We
physical reasons for considering this limit.
of the particle
at
average speed
s{a)
well as the particular function
(as
this
the other hand, the limit
hm
depends only on a
the "average velocity" of the particle
would
s)
and there are important
speak of the "velocity
like to
time a," but the usual definition of velocity
is
really a definition
of average velocity; the only reasonable definition of "velocity at time a" (so-called
"instantaneous velocity")
is
the limit
s{a
+ h) —
s(a)
lim
h
Thus we
define
the (instantaneous) velocity of the particle at a to be s'{a).
Notice that s'{a) could easily be negative; the absolute value
It is
important to realize that instantaneous velocity
an abstraction which does not correspond precisely
While
it
\s'{a)\ is
sometimes
(instantaneous) speed.
called the
would not be
with average velocity,
fair to
to
a theoretical concept,
is
any observable
quantity.
say that instantaneous velocity has nothing to
remember
that s'{t)
s(t
do
not
is
+ h)-s(t)
h
for
any particular
proaches
measures
0.
is
/z,
but merely the limit of these average velocities as h ap-
Thus, when velocities are measured
in physics,
what a
an average velocity over some (very small) time
cedure cannot be expected
to give
an exact answer, but
physicist really
interval; such a pro-
this
is
really
no
defect,
because physical measurements can never be exact anyway.
The
velocity of a particle
is
often called the "rate of change of its position." This
notion of the derivative, as a rate of change, applies to any other physical situation
in
which some quantity varies with time.
For example, the "rate of change of
mass" of a growing object means the derivative of the function m, where m{t)
the mass at time
is
/.
become familiar with the basic definitions of this chapter, we will
some time examining the derivatives of particular functions. Before
important theoretical results of Chapter 1, we want to have a good
In order to
spend quite
proving the
1
idea of what the derivative of a function looks
exclusively
i<>
one aspect of
cated functions.
this
In this chapter
—
like.
The next chapter
is
devoted
problem calculating the derivative of compliwill emphasize the concepts, rather than the
we
9. Derivatives
calculations,
by considering a few simple examples. Simplest of
function, f(x)
=
c.
f(a+h)-f(a) =
hm
hm
h^O
is
a constant
In this case
,.
Thus /
all is
153
0.
h^O
h
number
differentiable at a for every
the tangent line to the graph of
c-
/ always
a,
h
and f'(a)
=
This means that
0.
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
f'(a)
=
any linear function f(x)
+ d.
ex
Indeed
f(a+h)-f(a)
—
lim
h
c(a
—
+ h) + d —
[ca
+ d]
lim
h
——
ch
=
lim
c\
h^Q h
the slope of the tangent line
A refreshing difference
is c,
the
same
occurs for f(x)
w,
f
,
(a)
slope 2a
..
= hm
f(a
as the slope of the
= x.
graph of /.
Here
+ h)-f{a)
h
=
(a
+ h) -a 2
2
lim
h
=
—
a~
+ 2ah + h - a"
lim
lim 2a
+h
h-*0
=
Some
of the tangent lines to the graph of
each tangent
checked
FIGURE
9
2a.
line
/
are
shown
in Figure 9. In this picture
appears to intersect the graph only once, and
fairly easily:
Since the tangent line through
(a,
a
)
this fact
has slope 2a,
can be
it is
graph of the function
g(x)
Now,
if
the graphs of
/ and
=
=
or
)
is
.
g intersect
x
In other words, (a, a
— a) + a
2ax — a
2a(x
x~
so
(x
or
x
at a point (x,
= 2ax — a~
— 2ax + «"
-a) 2 =0
=
f(x))
0;
a.
the only point of intersection.
=
(x, g(x)),
then
the
.
1
54
Derivatives
and
Integrals
The
function f(x)
—
x 2 happens to be quite special in
more than
tangent line will intersect the graph
function f(x)
=
x3
regard; usually a
this
once. Consider, for example, the
In this case
.
=
lim
=
lim
f'(a)
f(a+h)-f(a)
+ h) 3 -a 3
(a
h-+0
a
3
3a 2 h+3ah 2
+
+ h 3 -a 3
lim
3a 2 h
=
+
3ah 2
h
h-»0
Thus
=
lim 3a
=
3a
+ 3ah + h
.
the tangent line to the graph of
the tangent line
of
/ and g
=
=
x
or
easily solved if
is
has got to be x
=
a, so that (x
then be found by dividing.
so
happens
that x
2
+ ax —
(x
Thus, as
We
a
—
3a"x
—
2a~
-
2
-
2a
3
2
+
la
3
=
3a x
3
-
3a x
we remember
(x, g(x))
when
one solution of the equation
left side;
the other factor can
obtain
a)(x
2
+ ax -
2a 2 also has x
—
that
a factor of the
is
=
0.
a){x
2a
—
2
)
—8a
3
).
=
0.
a as a factor;
— a)(x + 2a) =
we
obtain finally
(a,
a
0.
through
These two points are always
have already found the derivative of sufficiendy
and
the derivative /'
still
)
also intersects
distinct,
except
is
many
functions to illustrate
very popular, notation for derivatives. For a given function /,
often denoted by
dfix)
dx
ki; 10
a~
0.
the classical,
figi
.
illustrated in Figure 10, the tangent line
the graph at the point (—2a,
when
a 3 ) has slope 3a 2 This means that
—a) +
3a"(x
3
— a)
We
(x
It
at (a,
intersect at the point (x, fix))
.v
This equation
/
the graph of
is
g(x)
The graphs
+ h3
lim
For example, the symbol
dx 2
~dx
155
9. Derivatives
=
denotes the derivative of the function f(x)
x2
.
Needless to
say,
the separate
parts of the expression
df(x)
dx
are not supposed to have any sort of independent existence
bers, they cannot
be canceled, and the entire expression
other numbers "df(x)" and "dx."
This notation
is
—
the J's are not
not the
is
due
num-
quotient of two
to Leibniz (generalh
considered an independent co-discoverer of calculus, along with Newton), and
is
Although the notation df{x)/dx
may be shorter; after all, the symbol
affectionately referred to as Leibnizian notation.*
seems very complicated,
in concrete cases
dx 2 /dx
concise than the phrase "the derivative of the function
actually
is
more
it
f{x)=x 2 r
The following formulas state
we have found so far:
in
standard Leibnizian notation
all
the information
that
dc
=
0,
—
o.
dx
d( ax
+ b)
-
,
dx
dx 2
dx
dx 3
=
1„
zx
—
—
1y
DX
dx
Although the meaning of these formulas
interpretation are hindered
by the reasonable
not contain a function on one side and a
the third equation
than
is
2x
/',
.
be
true,
clear enough, attempts at literal
stricture that
number on
an equation should
the other.
For example,
if
then either df(x)/dx must denote f'(x), rather
or else 2x must denote, not a number, but the function whose value
It is
at v
one or the other of these alternatives is
practice df(x)/dx sometimes means /' and sometimes means f'(.\
really impossible to assert that
intended; in
while 2x
to
is
is
may
),
denote either a number or a function.
most authors are reluctant
to
Because of
this ambiguity,
denote f'(a) by
—
df(x)
—,
(a);
dx
instead f'(a)
is
usually denoted by the barbaric, but unambiguous, symbol
df(x)
dx
*
Leibniz was led to
this
not the limit of quotients
symbol by
\
f(x
his intuitive
+ h)- f(x)]/h,
notion of the derivative, which he considered to be,
small" number. This "infinitely small" quantity was denoted by dx
small" difference
f(x+dx)—f(x) by
when h is an
and the corresponding
but the "value" of this quotient
df(x). Although this poinl of view
properties (PI)— (PI 3) ol the real numbers,
some people
is
"inliniteK
"infinitely
impossible to reconcile with
find this notion of the derivative congenial.
1
56
Derivatives
and
Integrals
In addition to these
df(x)/dx
Leibnizian notation
difficulties,
Although the notation dx 2 /dx
ambiguity.
associated with one
is
often replaced by df/dx. This, of course,
is
practice of confusing a function with
its
more
absolutely standard, the notation
is
value at x
is
conformity with the
in
So strong
.
is
this
tendency that
functions are often indicated by a phrase like the following: "consider the function
y
=
x 2 ."
as the
We
name
the function
sometimes follow
will
classical practice to the extent
and
its
values
function (defined by) y(x)
Despite the
many
—
=
jc
2 ."
ambiguities of Leibnizian notation,
mathematical writing, and
sively in older
is
ever,
it
is
used almost exclu-
used very frequently today.
still
staunchest opponents of Leibnizian notation admit that
some
of using y
we will nevertheless carefully distinguish between
thus we will always say something like "consider the
of a function, but
it
will
The
be around for quite
time, while its most ardent admirers would say that it will be around forand a good thing too! In any case, Leibnizian notation cannot be ignored
completely.
The
text,
policy adopted in this
but to include
in the
it
book
is
to disallow Leibnizian notation within 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
preciate the significance of the derivative
examples of functions which are
three functions
first
first
is
it
discussed in this chapter,
f(x)
—
\x\.
and
=
1
for h
>
0,
+ h)-f(0)
and
\h\/h
illustrated in Figure
1
;
if
they
\h\
h
= —1
for h
<
does not
lim
0.
This shows that
exist.
h
/i—
In
candidates are the
In this case
h
\h\/h
The obvious
To fully apknow some
something has clearly gone wrong.
f(0
Now
differentiable.
equally important to
not differentiable.
turn out to be differentiable at
Consider
been
all
fact,
f(h)
..
and
-
lim
//
/(0)
=—
,
I.
h
•()
(These two limits are sometimes called the right-hand derivative and the left-
hand
derivative, respectively, of
/
at 0.)
,
9. Derivatives
If a
/
0,
then f'(a) does
In
exist.
fact,
.f(-v)=l
=
/'(*)
-1
if
x >
0,
if
x <
0.
The proof of this fact is left to you (it is easy if you remember the
linear function). The graphs of / and of /' are shown in Figure
/'
2
X
>
X
I),
We
a similar difficulty arises in connection with f'(0).
h
1
1.
H
'
[
FIGURE
derivative of a
1
For the function
/M={*X,
./"
157
f(h)
I
~
/(0)
have
2
=
h,
h
<
/?
>0.
h
I
=
1,
/;
Therefore,
/i->0-
but
Thus
/'(0) does not exist;
exists for
x
—
^
/
/?
lim
is
not differentiable at
/'(*)
of
Even worse
/ and
things
1.
0.
Once
again, however, f'(x)
easy to see that
it is
f
The graphs
=
/' are
=
shown
happen
for
2x,
x <
1,
^>0.
in Figure 12.
f(x)
=
y/\x\.
For
this
function
f'j
—=—
sfh
FIGURE
12
1
h
>
h
<0.
f(h)-f(0)
I
In this case the right-hand limit
,.
hm
does not
exist;
what's more,
(Figure 13).
FIGURE
\
}
instead 1/v/z
f(h)
-
becomes
—l/y/—h becomes
/(0)
=
1
lim —7=
arbitrarily large as h
approaches
0.
And,
arbitrarily large in absolute value, but negative
58
1
Derivatives
and
Integrals
The
fix)
=
function f(x)
behaved than
better
l/x, although not differentiable at 0,
this.
The
-
/(0)
f(h)
1
~
h
h
h
"tangent line" which
— l/x
K.I
RE
Sometimes one says that / has
means that the graph of / has a
has the same geometric property, but one would say that
Remember
14
0.
this
(Vh\
parallel to the vertical axis (Figure 14).
is
of "negative infinity"
1
Geometrically
"infinite" derivative at 0.
continuity.
little
I
/,2/3
simply becomes arbitrarily large as h goes to
an
a
at least
is
quotient
Of course,
/
f(x)
—
has a derivative
at 0.
that differentiability
This idea
is
is
supposed
to
be an improvement over mere
many examples
supported by the
of functions which are
continuous, but not differentiable; however, one important point remains to be
noted:
THEOREM
l
/
If
is
/
differentiable at a, then
PROOF
hm
/ (a
+ h) —
is
f(a)
continuous
at a.
= hm
•
f(a
v
— hm
+ h)-f(a)
•
h-+0
=
=
As we pointed out
to lim
x—>a
It is
f(x)
=
in
Chapter
f(a); thus
very important to
that the converse
is
/
is
5, the
/?
/'(fl)-0
0.
equation lim f(a
continuous
A
hm
h-+0
ll
+ h) — f(a) =
is
equivalent
at a. |
and just
as
important to remember
differentiable function
is
continuous, but a con-
remember Theorem
not true.
//
1,
tinuous function need not be differentiable (keep in mind the function f(x)
and you
will
never forget which statement
The continuous
functions
examined
with at most one exception, but
which are not
Actually,
I
I
(
.
i
RE
it is
so far
and which false).
have been differentiable
\x\,
true
at all points
easy to give examples of continuous functions
differentiable at several points,
one can do much worse than
15
is
=
this.
even an
There
is
infinite
number
a function which
(Figure 15).
is
continuous
9. Derivatives
everywhere
and
differentiable
draw
it
nowhere! Unfortunately, the definition of this function will
Chapter 24, and I have been unable to persuade the
(consider carefully what the graph should look like and you will
be inaccessible to us
artist to
159
until
sympathize with her point of view).
tions to the graph,
It is
draw some rough approximabetter approximations are shown
possible to
however; several successively
Figure 16.
in
Although such spectacular examples of nondifferentiability must be postponed,
we
can, with a
at infinitely
Figure 17.
little
ingenuity, find a continuous function
=
x
This particular function
ft/0we
/
have
is
itself
sin
—
,
x
^
x
=
x
0.
Indeed, for
not differentiable
of the function
/(*)
17
is
many points, all of which are in [0, 1]. One such function is illustrated in
The reader is given the problem of defining it precisely; it is a straight
line version
FIGURE
which
0.
quite sensitive to the question of differentiability.
1
60
Derivatives
and
Integrals
f(h)
-
h sin
/(0)
=
that lim sin
1
h
does not
/h
/
exist, so
Geometrically, one can see that a tangent line cannot
secant line through
— and
1
1
,
(0, 0)
and
(h,
is
not differentiable at
exist,
0.
by noting that the
f(h)) in Figure 18 can have any slope between
no matter how small we require h
to be.
x
sin
i^
-,
x
x
0,
FIGURE
—1
sin
h
h
Long ago we proved
i
h
=
18
This finding represents something of a triumph; although continuous, the function
/ seems somehow
ematically undesirable feature of this
ertheless,
tiability.
and we can now enunciate one mathNevfunction
it is not differentiable at 0.
quite unreasonable,
—
one should not become too enthusiastic about the
criterion of differen-
For example, the function
1
g(x)
x
=
.v^0
sin
A=0
0,
is
differentiable at 0; in fact g'(0)
=
0:
g(h)
-
g(0)
,.
lim
h
=
lim
h-y0
h-*
=
The
ure
tangent line to the graph of g
at
ll
lim h sin
/i-»0
=
sin -
—
//
0.
(0,0)
is
therefore the horizontal axis (Fig-
19).
This example suggests
function than
mere
such conditions
if
that
we should
differentiability.
we
We
introduce another
seek even
more
restrictive conditions
ean actually use the derivative
set
of definitions, the
last
to
on
a
formulate
oi this chapter.
9. Derivatives
\
161
/
\
/
\
\
/I
A
f
.
x
-
sin
jc
,
a;
fr
U/w
IP
^
,.
1<
^nA 1
J
*
^
=
1
\
1//
/
\
/
\
/
\
FIGURE
19
For any function /, we obtain, by taking the derivative, a new function /' (whose
domain may be considerably smaller than that of f). The notion of differentiability
can be applied
whose domain
tion (/')'
If
f"(a)
f"(a)
is
usually written simply /" and
is
called the
/ is said to be 2-times
second derivative of/
tion at time
t
acceleration
is
in
sit
at
time
an accelerating
There
=
f""
is
no reason
(/'")', etc.
viation
is
at a.
of a particle moving along a straight
acceleration.
its
differentiable at a.
particularly important.
line,
If s(t)
then s"(t)
the posi-
is
is
called the
Acceleration plays a special role in physics, because, as
t.
on a
stated in Newton's laws of motion, the force
and
is
called the
is
then
In physics the second derivative
(/')',
The funcsecond derivative of /.
differentiable at a, and the number
consists of all points a such that /'
exists,
another function
to the function /', of course, yielding
Consequently you can
particle
feel the
the product of its
is
mass
second derivative when you
car.
to stop at the
second derivative
—we can
define /"'
=
(/")',
This notation rapidly becomes unwieldy, so the following abbre-
usually adopted
(it is
really a recursive definition):
f(i)
f
(k+1)
_
/',
=
(f
ik)
y.
Thus
HV>
/(2)
/( 3)
W4)
=f
= f" = ify,
= f" = (fy,
= (/'")'
rllll
etc.
The
various functions
(k
f \
for k
>
2,
are sometimes called
higher-order
>
convenient to
derivatives of /.
Usually,
have
for
/
f
{k)
,
we
resort to the notation
defined for smaller k
also.
f
{k)
In
only for k
fact,
namely,
f(0)
_.
r
4,
but
it
is
a reasonable definition can be
made
162
Derivatives
and
Integrals
Leibnizian notation for higher-order derivatives should also be mentioned.
The
natural Leibnizian symbol for f"(x), namely,
'df(x)
dx
dx
(fix)
=
x<
is
abbreviated to
d 2 f(x)
or
(dx) 2
Similar notation
The
how
Let fix)
tion.
used for
f
(k)
frequently to
dx 2
(x).
following example illustrates the notation
simple case,
(a)
is
d 2 f(x)
more
=
f
{k
\
and
also shows, in
one very
various higher-order derivatives are related to the original func-
x 2 Then, as
we have
.
already checked,
= 2x,
/"(*) = 2,
f(x) = 0,
f(x)
fix)
=
2x
f
k)
(x)
=
if*>3.
0,
Figure 20 shows the function /, together with
A
more illuminating example
whose graph is shown in Figure 21(a):
rather
fix)
fix) =
2
It
is
its
various derivatives.
presented by the following function,
is
,2
x
>0
—X'
x
<
a
ifa
>0,
<0.
-
/(0)
0.
easy to see that
f'ia)
= 2a
= -la
/'(0)
=
f'(a)
if
Moreover,
(c)
lim
fih)
h->0
,.
lim
//
fih)
/i-»-0
//
Now
f
(k)
(x)=0,k>3
=
lim
/j->o+
and
(d)
I
l(,l
Rl. 20
v
lim
h->0-
f(h)
v
= hm
h
h->0-
so
/( 0) =
This information can
all
lim
——
^-*o+ h
/?
lim
h->0
m
" /;2
h
= o.
h
be summarized as follows:
./"(-v)
=
2|.v|.
=
n
U,
163
9. Derivatives
follows that /"(0) does not exist!
It
Existence of the second derivative
rather strong criterion for a function to
/(*)
=
x
2
,
x\
Even a "smooth looking" function
satisfy.
when examined with
x>0
like
jc<0
suggests that the irregular behavior of the function
/
some
reveals
irregularity
1
x
g(x)
sin
0,
the second derivative. This
x
/0
x
=
moment we know
might also be revealed by the second derivative. At the
g'iP)
=
but
0,
we do
not
know
We
will
return to
computing g"(0).
after
we have
any a
g'(a) for
thus a
is
^
0, so
it
that
hopeless to begin
is
question at the end of the next chapter,
this
perfected the technique of finding derivatives.
PROBLEMS
f\x)
=
1.
2\x\
Prove, working directly from the definition, that
(a)
= -1/a
f'(a)
(b)
2
fora
,
^
Prove that the tangent
if
=
f(x)
l/x, then
0.
line to the
graph of /
at (a,
I
/a) does not intersect
the graph of /, except at (a, \/a).
2.
Prove that
(a)
(b)
/"(*)
2,
l/x 2 then f'(a)
f(x)=
,
Prove that the tangent
which
point,
=
if
lies
=
on
/
line to
= -2 /a 3
at (a,
I
/a
)
for a
^
0.
/
intersects
one other
at
the opposite side of the vertical axis.
=
3.
\/(2s/a), for a > 0. (The expression
you obtain for [f(a + h) — f(a)]/h will require some algebraic face lifting,
but the answer should suggest the right trick.)
4.
For each natural
x >
Prove that
S2(x)
=
f(x)
if
>/x,
then f'(a)
number
=
2x, and S^'ix)
n, let S„(x)
3x
(The expression
conjecture.
=
x".
Remembering
that S\'(x)
—
1,
conjecture a formula for Sn '(x). Prove your
,
+
(x
//)"
may be expanded by
the binomial
theorem.)
f"(x)
=
-2, x <
=
5.
Find /'
6.
Prove, starting from the definition (and drawing a picture to
f(x)
if
[x].
(c)
FIGURE
2
if
(a)
(b)
7.
if
g(x)
g(x)
=
=
+
f(x)
(a)
(b)
(c)
If
What
What
What
=
f'(x
)
calling
cf'(x).
.
2
2
f(3 ), /'(5 ), f(6
2
f(a 2 ), f(x )?
is
is
problem
2
)?
you are missing a very important
means the derivative of / at the number which we happen
x
find this
2
;
it
is
drive the point
(d)
x
/'(*);
/'(9), /'(25), /'(36)?
is
you do not
2
=
cf(x), then g'(x)
Suppose that f(x)
=
then g'(x)
c,
illustrate):
For f(x)
silly,
not the derivative at
x of the function g(x)
—
f(x
2
home:
= x3
,
compare f'(x
)
and
g'{x)
where g{x)
—
f(x
2
).
).
point:
to
be
Just to
164
Derivatives
and
Integrals
(a)
= f(.x+c). Prove (starting from the definition) that g'{x) —
Draw
a picture to illustrate this. To do this problem you must
c).
+
f'(x
write out the definitions of g'(x) and fix + c) correctly. The purpose
Suppose g{x)
of Problem 7 was to convince you that although
not an utter
and there
triviality,
something
is
put prime marks into the equation g(x)
problem
this
to prove:
=
f(x
c
f'(cx).
+
is
easy,
is
you cannot simply
To emphasize
c).
it
this
point:
(b)
Prove that
why
(c)
f(cx), then g'(x)
should be true,
this
Suppose
/
that
+ a) =
fix
=
g(x)
if
fix)
or you will surely
and
+
to see pictorially
with period a
periodic,
Prove that /'
all x).
Find f'{x) and also f'(x
Try
also.
differentiable
is
for
=
(i.e.,
also periodic.
is
Be very methodical,
up somewhere. Consult the answers (after you do the
slip
3) in the following cases.
problem, naturally).
= (x+3) 5
f(x+3)=x 5
f(x+3) = (x + 5) 7
f(x)
(")
(in
10.
.
.
Find f'(x)
if
f(x)
=
g(t
.
+ x),
and
fit)
if
=
g(t
+ x). The
answers
will not
be the same.
11.
(a)
(b)
Prove that Galileo was wrong:
and
s'
s(t)
=
is
Prove that the following
—
[s'it)]
(ii)
is
then
s,
s
falls
a distance s(t) in
t
seconds,
cannot be a function of the form
facts are true
show why we switched from
s"it )
(i)
If s
a body
cf
fact will
(c)
proportional to
if
2
a (the acceleration
is
about
=
s if s(t)
(a/2)t
(the
first
c to a/2):
constant).
= lasit).
measured
in feet, the value
of a
is
32.
How many
seconds do you
way of a chandelier which falls from a 400-foot
make it, how fast will the chandelier be going when
you? Where was the chandelier when it was moving with half that
have to get out of the
ceiling? If you don't
it
hits
speed?
12.
Imagine a road on which the speed
other words, there
is
from the beginning of the road
road; car A's position at time
(a)
What
equation expresses the
limit?
(The answer
Suppose
time
/
is
that
A\
is
specified at every single point. In
is
t
a(t),
is
=
—
1
B, are driving along
bit).
fi's is
A always
travels at the
speed
Lit).)
at the
/
and car
fact that car
not a' it)
A always goes
position at time
Two cars, A and
L(jc).
is
this
(b)
limit
a certain function L such that the speed limit x miles
.
speed
Show
limit,
that
B
and
is
that
ZJ's
position at
also going at the speed
limit at all times.
(c)
B always stays a constant distance behind A. Under what conditions will B still always travel at the speed limit?
Suppose, instead,
that
165
9. Derivatives
13.
Suppose
= g(a)
that fia)
and
the right-hand derivative of g at a.
h(x)
14.
=
g(x) for
—
Let f{x)
/
is
x >
x2
if
a.
x
Prove that h
differentiable at 0. (Don't
=
Define h(x)
is
at
f(x) for x
a equals
<
and
a,
differentiable at a.
=
and f(x)
rational,
is
/
that the left-hand derivative of
be scared by
x
if
Prove that
irrational.
is
Just write out the
this function.
definition of /'(0).)
15.
differentiable at 0. (If
do
(b)
< x2
a function such at |/(x)|
/ be
Let
(a)
/
is
you have done Problem 14 you should be able
to
Prove that
for all x.
this.)
This result can be generalized
x
if
is
replaced by |g0c)|, where g has
what property?
16.
Let a >
17.
Let
is
18.
<
/
If
1.
<
Prove that
1.
not differentiable at
—
Let fix)
that
/
is
Suppose
—
Show
f(a)
20.
,
g'(a)
=
>
satisfies |/(jc)|
and \/q
for
=
x
g(a)
-
\x\?
p/q
It
=
—0.00
.
.
.
0a n+ \a n+ 2
differentiable at 0.
and /(0)
=
0,
then
/
in lowest terms.
Prove
obviously suffices to prove
expansion
....
g(a)
h'(a).
that the conclusion does not follow
=
is
= h(a), that f(x) < g(x) < h(x) for all x,
Prove that g is differentiable at a, and that
h'{a). (Begin with the definition of g'(a).)
=
=
/
prove that
Why? If a — m.axaia}, ... is the decimal
+ h) — f(a)]/h for h rational, and also for
that f(a)
that f'(a)
f'(a)
(b)
a
0.
h
and
/
for irrational x,
of a, consider [fia
(a)
if
\x\
not differentiable at a for any a. Hint:
this for irrational a.
19.
<
satisfies |/(.v)|
if
we omit
the hypothesis
h(a).
/ be any polynomial function; we will see in the next chapter that /
The tangent line to / at (a, /(«)) is the graph of g(x) =
f'(a)(x — a) + /(a). Thus f(x) — g(x) is the polynomial function d{x) =
2
fix) — f\a)ix — a) — /(a). We have already seen that if fix) = x then
2
3
then dix) — ix — a) (x + 2a).
dix) = (jc — a) and if fix) — x
Let
is
differentiable.
,
,
,
(a)
Find dix)
(b)
There
ix
when fix) — x 4 and show
,
certainly
—a) 2
.
seems
to
22
it is
divisible
be some evidence that dix)
is
intersects the
(x
— a) 2
.
always divisible by
graph
at
two
points; the tangent line
graph only once near the point, so the intersection should
be a "double intersection." To give a rigorous proof,
dix)
x
—
a
note that
fix) — fia) divisible
there a polynomial function h such that hix) =
answer the following questions.
by
—a)? Why
is
first
fix)- fia)
- fia).
x —a
Now
ix
by
Figure 22 provides an intuitive argument: usually, lines parallel
to the tangent line will intersect the
FIGURE
that
Why
is
1
66
Derivatives
and
Integrals
-
d(x)/(x
does
21.
(a)
(b)
22.
(a)
#
a) for x
that f'(a)
lim /?U)
is
=
Why
0?
is
=
ft(a)
0?
Why
\im[f(x)
—
— a).
f(a)]/(x
(Nothing deep here.)
x—*-a
that derivatives are a "local property":
if f(x) = g(x) for all x in
=
open
interval
containing
a,
then
some
g'(a). (This means that
f'(a)
in computing f'{a), you can ignore f(x) for any particular x ^ a. Of
course you can't ignore f{x) for all such x at once!)
Show
Suppose
/
that
is
differentiable at x. Prove that
=
f'(x)
f(x+h)-f(x-h)
lim
h->0
Remember an
Hint:
**(b)
=
problem?
this solve the
Show
Why
a?
2/7
old algebraic trick
— a number
same quantity is added to and then subtracted from
Prove, more generally, that
if
the
it.
+ h)-f(x-k)
h +k
f(x
lim
fix)
not changed
is
Although we haven't encountered something
like
lim
before,
its
mean-
make an appropriate
we actually have lim
ing should be clear, and you should be able to
The important
definition.
that
23.
we
Prove that
/
is
=
Draw
if
,
so
k.
—f'(—x). (In order to minimize conand then remember what other thing g
find g'(x)
f(—x);
s-8
=
even, then f'(x)
a picture!
/
24.
Prove that
25.
Problems 23 and 24 say that /'
What can
26.
that
is
are only considering positive h and
fusion, let g(x)
is.)
thing here
if
is
—
odd, then f'{x)
is
therefore be said about
Find f"(x)
f'{—x).
even
f
(k)
if
/
Once
is
again,
draw a
odd, and odd
if
picture.
/
is
even.
?
if
= x\
5
fix) = X
f(x)=x 4
fix)
ii)
iii)
iv)
27.
.
.
fix+3) = x 5
If S„(x)
—
x",
and
.
<
k
<
S„
n,
ik)
prove that
.n-k
=
ix)
(n-k)l'
=«,
28.
(a)
(b)
Find /'(*)
if
fix)
=
Analyze / similarly
if
\x\
3
.
./i-£
.
Find f"(x). Does f'"(x)
f(x)
=
4
.v
for
x >
exist for all
and f(x)
=—
4
.v
for
x?
x <
0.
9. Derivatives
29.
- xn
Let f(x)
x >
for
and
(and find a formula for
30.
it),
let
=
f(x)
but that
/
(
"}
for
(0)
x
<
0.
does not
Prove that
dx"
„
,
nx
dx
I
dy
v
d[f(x)
+ c]
dfjx)
nil
dx
dx
d[cf(x)]
=
IV
'fix)
c-
dx
dx
dz
dy
dx
dx
if
dx 3
—
vi;
dx
z
3a
=
v
+ c.
4
.
x=a i
-
df(x+a)
df(x)
Vll
dx
df(cx)
=
Vlll
dx
df(cx)
IX)
=
df(x)
dx
df(y)
C
dy
**-*(
"'\k
x=b+a
C
x=b
dx
dx k
dx
x=b
n-k
x=cb
exists
exist.
Interpret the following specimens of Leibnizian notation; each
ment of some fact occurring in a previous problem.
(n_1)
/
167
is
a restate-
CHAPTER
DIFFERENTIATION
The
process of finding the derivative of a function
previous chapter you
may have
approach
—
if
differentiation.
is
It is
true that such a procedure
is
definition of the derivative
you are
limit.
we
will learn to differentiate
From
the
usually laborious,
and depends upon
you forget the
Nevertheless, in this chapter
tions,
called
the impression that this process
requires recourse to the definition of the derivative,
recognizing some
is
successfully
often the only possible
a large
without the necessity of even recalling the definition.
A
likely to
be
lost.
number of func-
few theorems
will
provide a mechanical process for differentiating a large class of functions, which
formed from a few simple functions by the process of addition, multiplication,
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
functions. The first theorem is merely a formal recognition of a computation
are
division,
carried out in the previous chapter.
THEOREM
l
If
/
is
a constant function, fix)
f'{a)
PROOF
f
(a)
=
=
=
for
THEOREM
2
If
/
PROOF
is
numbers
all
f(a+h)-f(a)
= hm
lim
h
h->0
The second theorem
then
c,
is
f'(a)
—
f'(a)
=
= x,
1
c-c
0. I
computation
in the last chapter.
then
for
lim
=
h
/i^o
also a special case of a
the identity function, f{x)
a.
numbers
all
a.
f(fl+h)-f(a)
h
—
a
+h—
h
h^-0
=
h
lim
- =
1
.
h^o h
The derivative of the sum
sum of the derivatives.
168
a
lim
of two functions
|
is
just
what one would hope
the
10. Differentiation
THEOREM
3
/ and g
If
are differentiable at a, then
=
(f+g)'{a)
PROOF
(/
+ g)
+g
/
f{a)
= hm
(a)
+ g'{a).
-
/(a
= hm
+ h) + g(a +h)-
h^O
/(fl+/l)-/(fl)
but
it is
the
THEOREM
4
/ and g
If
f(a
=
lim
g(fl+/l)-g(fl)
=
/'(a)
h
+ g\a).
added
is
to
(/
.
g)'(fl )
=
and the proof requires only a simple
a number is not changed if
and subtracted from
=
/
g
is
f\a)-g{a)
—
it.
also differentiable at a,
+ f{a)
•
and
g'{a).
+ h)-(f-g)(a)
(f-g)(a
lim
h
/i^O
=
not as simple as one might wish,
is
useful before
are differentiable at a, then
g(a+h)-g(a)
hm
|
product
we have found
(/• g)\a)
PROOF
+ h)-f(a)
nevertheless pleasantly symmetric,
same quantity
+ g(a)]-
|
for the derivative of a
algebraic trick, which
[/(a)
h
lim
The formula
and
also differentiable at a,
is
169
f(a+h)g(a + h)- f(a)g(a}
lim
h
=
f(a
+ h)[g(a + h )-g(a)]
lim
[f(a+
h)
h
= hm
/ (a
=
(Notice that
/i^o
+
lim
h
•
f(a)-g'(a)
f(a)]g(a)
h
+ n) hm
/i—
-
•
hm g{a)
h^O
h
fi^O
/;
f(a)-g(a).
we have used Theorem
9-1 to conclude that lim
f(a
+ h) =
f(a).)
ft->0
Theorem 4
In one special case
THEOREM
5
If
g(x)
=
cf(x) and
/
is
simplifies considerably:
differentiable at a, then g
g'{a)
proof
If
h(x)
=
c,
so that g
=
h
=
f, then by
g'(a)
=
=
=
(h
c
is
differentiable at «,
f'(a).
Theorem
4,
f)'(a)
h(a)-f'(a)
c-f'(a)
= c-f'(a).
+ h'(a)-f(a)
+ 0-f(a)
|
and
|
1
70
Derivatives
and
Integrals
=
Notice, in particular, that (-f)'(a)
(f+[-g])'(a)
=
—f'(a), and consequently (/
—
g)'(a)
—
f'(a)-g'(a).
To demonstrate what we have already
we
achieved,
compute
will
the derivative
of some more special functions.
theorem
6
If
f(x)
= xn
some natural number
for
f'(a)=na
PROOF
The proof will be by
x n+
=
If I(x)
.
=f
I
follows
It
.
g'{a)
This
is
=
=
= na n ~
]
=
is
simply
f(x)
=
.v",
for
all
+l
— x"
I(x)
f(x)
this
1
if
x, the equation x"
g(x)
thus g
for all a.
true for n, so that
is
f'(a)
—
n- 1
induction on n. For n
assume that the theorem
Let g(x)
then
n,
for
Theorem
2.
Now
then
a.
all
x can be written
x;
from Theorem 4 that
(f
I)'(a)
+
precisely the case n
=
=
=
=
1(a)
f'(a)
na"- -a
+ a"
+ \)a'\
I'(a)
f(a)
1
rw"
(n
which we wished
1
+
+ a"
1
for
all a.
to prove. |
we can now
Putting together the theorems proved so far
find /' for
/
of the
form
f(x)
We
—
na„x"~
+
—
(n
n(n
-
Y)an x
n -2
This process can be continued
power of x by
derivatives /'",
H
+
+ a\-
2«2-
1''
+
>
n
\){n
-
Each
easily.
,
2)a„_i.v"-
3
+
•
•
•
+
2a 2
.
differentiation reduces the highest
.
{4)
,
,
is
f
for k
-
(n
and eliminates one more a, It is a good idea to work out the
and perhaps / (5) until the pattern becomes quite clear. The
f
1
last interesting derivative
(n)
(x)
= n\a n
;
we have
/
Clearly, the next step in
is
\)a„_\x"~
can also find f":
fix) =
It
+ ai% + a\x + aq.
H
obtain
f'(x)
We
= a„x" + a n _\x n ~
quite a
bit
the derivative
<>l
u,
(x)
our program
is
simpler, and, because of
I
/g.
=
0.
of a quotient f/g.
4, obviously sufficient to find
to find the derivative
Theorem
10. Differentiation
THEOREM
7
PROOF
If
g
and g(a)
differentiable at a,
is
Before
we even
/
then \/g
0,
is
differentiable at a,
171
and
write
(a+h)
(a)
h
we must be
(\/g)(a
+ h)
Since g
is,
8
>
is
at a.
differentiable at a,
Since g{a)
such that g(a
for small
—
it
is
necessary to check that
defined for sufficiently small h. This requires only two observations.
by hypothesis,
continuous
makes sense
sure that this expression
^
+ h) ^
0,
for
it
\h\
it
Theorem
follows from
<
from Theorem 9-1 that g
follows
<5.
is
some
make
sense
6-3 that there
Therefore (\/g)(a
+
h) does
is
enough h and we can write
,
(a
1
+ h)
(a)
g(a
lim
1
+ h)
h^O
=
—
g(a)
h
g(a)
-
lim
+ h)
-g(a
h[g(a)
g(a + h)]
~[g(a
+ h)
-g(a)]
g(a)g(a
h
=
-[g(a
+ h) -
g(a)]
lim
we have used
+
—
h^O
h)
1
lim
h
(Notice that
1
lim
g(a)
g(a
+ h)
1
n't n\
6
.
[g(a)] 2
'
continuity of g at a once again.)
|
The general formula for the derivative of a quotient is now easy to derive.
Though not particularly appealing, it is important, and must simply be memorized
(I
always use the incantation:
"bottom times derivative of
top,
minus top
times derivative of bottom, over bottom squared.")
theorem
8
If
/ and
g are differentiable
at
a and g(a)
^
0,
then f/g
and
f\
g(a)f(a)-f(a)g'(a)
is
differentiable at a.
,
172
Derivatives
and
Integrals
PROOF
Since f/g
=f
(1/g)
we have
^Yw
8
(a)
f'(a)
+
f(a)
(a)
8
fja)
f(a)(-g'(a))
8(a)
[8(a)] 2
'
f'(a)-g(a)-f(a)-g'(a)
[8(a)] 2
We
can
if/U)
now
=
X
-Y
if/(*)
2
2
=
-l
+l
few more functions. For example,
then f'(x)
2
+l
=
(x
2
+
\)(2x)
then fix)
1
then/
r
(x)
=
(x
=
-
-
2
(a-
2
+\)
+l)-x(2x)
(x 2 +l) 2
(x
X
=
x
if/(-v)
differentiate a
2
1
^==(-\)x
A
-
4x
\)(2x)
2
(.v
2
+
l)
2.
1-x
U2 + D2
2
-2
x
Notice that the
last
example can be generalized:
f(x)=x~" =
if
1
for
some natural number n
,
then
fix)
thus
Theorem 6
pret f{x)
=
-nx
=
-2n
=
(-n)x-"-
actually holds both for positive
x° to
and negative
Theorem 6 is true for n =
not clear how 0° should be
integers. If we inter-
x~ to mean f'(x) = 0, then
1, and f(x) =
also. (The word "interpret" is necessary because it is
mean f(x) =
[
•
_1
defined and, in any case,
is
•
meaningless.)
Further progress in differentiation requires the knowledge of the derivatives of
certain special functions to be studied
the
moment we
shall divulge,
and
sin' (a)
cos' (a)
later.
One
of these
= cosrt
= — sin«
This information allows us to differentiate
for
all
a
the sine function. For
for
all
a
many
without proof:
,
other functions. For example,
f(x)
= x sin
fix)
= x cos a + sin x,
= —x sin x + cos a + cosx
= —x sin x + 2 cos x
A",
then
f"(x)
is
use, the following information,
;
if
,
,
10. Differentiation
173
if
=
g(x)
2
•
sin
x
=
a
sin
sin a,
•
then
=
=
=
=
g' (x )
g"(x)
+ cos x sin x
x cos x
sin
2 sin x cos a,
2[(sin a)( — sin*)
—
2[cos~ a
+ cos a cosa]
a];
sin
if
h {x )
=
cos x
=
cos x
•
cos x
then
h'(x)
h"{x)
= (cosa)( — sin a) + (— sinA)cosA
= —2 sin a cos a,
= —2 [cos" a — sin a].
Notice that
+ h'(x)=0,
g'(x)
hardly surprising, since (g
also have g"(x)
=
+ h"(x)
The examples above
+ h)(x) =
Theorem 4
that
/ g
•
h
(/
The
•
=
=
=
h)'(x)
g
/
choice of
or
(g
•
(f
g)'(x)
[f'(x)g(x)
h(x)
+
+
(/
•
g)(x)h'(x)
f(x)g'(x)]h(x)
answer
final
can be handled
for
+
f(x)g(x)h'(x)
+
f(x)g(x)h'(x).
h) would, of course, have given the
The
it
f-(g-h).
f'(x)g(x)h(x)+ f(x)g'(x)h(x)
different intermediate step.
A function involv-
we have
of these, for example,
first
also; in fact
an abbreviation
is
(f-g)-h
Choosing the
As we would expect, we
.
involved only products of two functions.
Remember
two ways.
1
0.
ing triple products can be handled by
in
+ cos 2 a =
a
sin"
is
same
result,
with a
completely symmetric and easily
remembered:
•
g,
sum
of the three terms obtained by differentiating each of /,
and h and multiplying by the other two.
(f g
•
h)'
For example,
is
the
if
f(x)
=
x
sin
A COS A
then
/'(a)
=
3a
Products of
more than
should have
little
(f-g-h-
sin
a cos a
+ x~
cos a cos a
3 functions can be
=
f'(x)g(x)h(x)k(x)
+
handled
(sinA)(— sin a).
similarly.
For example, you
formula
difficulty deriving the
k)'(x)
+a
+
f(x)g'(x)h(x)k(x)
f(x)g(x)h'(x)k(x)
+
f(x)g(x)h(x)k'(x).
.
1
74
Derivatives
and
Integrals
You might even
(/l
•
.
•
.
.
prove (by induction) the general formula:
try to
= J2 h {x)
/,)'(*)
fi-\(x)fi'(x)fi+i(x)
/»(*).
l=\
Differentiating the
m
gY( x )
most interesting functions obviously requires a formula
terms of /' and
To ensure
/
(/
reasonable hypothesis would seem to be that g be differentiable
°
behavior of
also
/'
o
if
g
that
g be differentiable
o
g near a depends on the behavior of
seems reasonable
prove that
g'.
is
to
assume
is
Indeed we
differentiable at g(a), then
is
This extremely important formula
=
(f
called the Chain Rule,
is
practically the product of /'
g(a) and
at
applications.
g' at a.
and
us,
g
is
presumable because
Notice that
but not quite: /' must be evaluated
g',
Before attempting to prove
this
theorem we
=
sin.v
will try
a few
.
temporarily, use S to denote the ("squaring") function S(.x)
/ =
Therefore
shall
o
Suppose
f(x)
Let
/
it
f'(g(a)).g'(a).
a composition of functions might be called a "chain" of functions.
is
a),
and
(fo g y(a)
§)'
Since the
at a.
near g{a) (not near
differentiable at g(a).
and /
differentiable at a
differentiable at a,
/
that
/
for
one
at a,
=
x
2
.
Then
sin o S.
we have
f'(x)
= sin'(S(jc))
—
Qtiite a different result
is
obtained
•
S'(x)
2
cos.t' -2x.
if
f(x)
=
sin" X.
In this case
f = S
o sin,
so
f'{x)
Notice that
this
agrees
(as
=
=
^'(sinA')
2 sin x
•
sin'(.v)
cos x
should) with the result obtained by writing
it
/ =
sin
•
sin
and using the product formula.
Although we have invented a
much
special symbol, S, to
name
the "squaring" function,
do problems like this without bothering to write
and without even bothering to write down the
particular composition which / is—one soon becomes accustomed to taking /
it
does not take
down
special
symbols
apart in one's head.
practice to
for functions,
The
such mental gymnastics
all
means do
so, but try to
following differentiations
il
yon
find
it
may
be used as practice
for
necessary to work a few out on paper, by
develop the knack of writing
/"
immediately
after seeing
175
10. Differentiation
the definition of /; problems of this sort are so simple that,
the
Chain Rule, there
if
fix)
=
sinx
fix)
=
sin
fix)
—
sin
is
=
fix)
=
=
fix)
A function
sin (sin
3
3
+
is
3x~
•
/-l=-
1
—
=
x)
fix)
—
cos (sin x)
+ 3;r)
fix)
=
cos(jc
f(x)
=
53(.t
3x
2 53
)
—
cos
•
I
x
\ x-
3
3
+
•
cos*
2
3jc
+ 3.t 2
)
(3x
•
2
52
•
)
(3.x-
+ 6.x)
2
+ 6x).
like
9
9
=
=
sin"x
9
•
[sinx
9
]
,
the composition of three functions,
f = S
o sin o S,
can also be differentiated by the Chain Rule.
that a triple composition
/
o
g
It is
=
remember
Thus if
only necessary to
means if°g)oh or f
h
o
fix)
we can
cosjt
f (x)
/(x)
which
remember
just
/'(jc)=3sin x-cosx
.1—
(a-
—
then fix)
x
sinU
you
no thought necessary.
x
f(x)
if
o ig oh).
x
sin
write
f = iS o sin) o 5,
f = S o (sin o £).
The
by applying the Chain Rule
derivative of either expression can be found
twice; the only doubtful point
calculations.
whether the two expressions lead
is
As a matter of fact,
any experienced
as
to equally simple
differentiator knows,
it is
much
better to use the second:
/ = S
We
can
now
write
down fix)
function to be differentiated
is
/'(*)
Inside the parentheses
sin o S.
one
in
fell
S, so the
= 2(
swoop. To begin with, note that the
formula
sinjc
first
fix) begins
|.
,
the value at
.v
of the second function,
writing
fix)
=
2sinx"
(the parentheses weren't really necessary, after
much
for
)
we must put
Thus we begin by
o (sin o S).
of the answer by the derivative of
sin o
S
all).
We
composition of two functions, which we already
know how
for the final answer.
f{x)
=
2 sin*
must now multiply
at x; this part
cos x"
2x.
is
easy
to handle.
it
this
involves a
We
obtain,
.
176
Derivatives
and
Integrals
The
following example
is
handled
/(*)
Without even bothering
we can
to write
=
f'(x)
Inside the parentheses
you get from
sin (sin *
can
now
sin(sin* 2 ).
cos(
sin, so
our expression
the value of h °k(x); this
is
we
already
we have
in sin (sin jc);
know how
fix)
Finally,
—
to multiply
cos(sin*
/(jc)
=
=
=
f(x)
=
/(jc)
fix)
)
•
cos*"
•
is
2jc.
9
sin((sin*)
—
then
)
if
You can probably just
/'(jc)
=
cos((sin*)
2sin(sinjc)
[sin (sin jc)]
/'(jc)
sin (sin (sin x))
/'(jc)
=
=
sin
/'(jc)
=
(jc
sinjc)
/
not, try writing out
cos (sin (sin
2 sin*
•
)
as a composition:
/(x)
=
sin*))
sin (sin (*
/'(*)
=
jc))
2sin(jc sin*)
cos(sin(*
•
start
(go
cos*
•
cos(sin;c)
sin*))
•
cos*
•
*
[sin
cos(*
[2* sin *
+ * cos *]
sin*)
+ *~ cos *]
more) functions
rule for treating compositions of four (or even
/o
cos*
cos(* sin*)
•
always (mentally) put in parentheses starting from the
•
cos (sin*)
•
•
and
what we
—which
here are the derivatives of some other functions which are the composition
"see" that the answers are correct
The
sin*
is
to solve:
of sin and S, as well as some other triple compositions.
if
(what
for /'(jc) begins
|.
)
have so far by the derivative of the function whose value at x
again a problem
simply sin*
So our expression
first sin).
first sin
functions,
for f'{x) begins
|.
)
= cos(sinjc
forget about the
gohok of three
as a composition
be
by deleting the
/'(*)
We
will
we must put
)
=
down /
one
see that the left-most
Suppose
similarly.
is
easy—
right,
(h ok)),
reducing the calculation to the derivative of a composition of a smaller
number of functions:
f'(g(h(k(x))))
For example,
if
/(*)
=
sin
2
2
(sin
(*))
\
f = S
=
S
o sin o
S
o (sin o
o sin
(S o sin))
|
then
/'(*)
=
2 sin (sin *)
•
cos(sin".v)
•
2 sin *
•
cos x
:
—
177
10. Differentiation
if
=
f(x)
sin((sin x
)
=
=
[/
)
sin o
5
o sin o
S
sin o (S o (sin o 5))]
then
f'(x)
= cos((sin x
)-2sinx
)
cosx"
•
2x;
•
if
/(jc)
=
f'(x)
—
sin (sin(sinx))
in yourself, if necessary]
[fill
then
2
sin (sin (sin x))
With these examples
differentiator
—
of the chapter, and
The
might
as reference,
You can be
practice.
it is
now
•
cos (sin (sin x))
•
cos(sinx)
you require only one thing to become a master
safely turned loose on the exercises at the end
we proved
high time that
the
following argument, while not a proof, indicates
try,
some of
as well as
cos*.
•
Chain Rule.
some of the
We
the difficulties encountered.
one
tricks
begin, of course,
with the definition
=
(f °g)'(a)
lim
(f
o
g)(a
+ h))-f(g(a))
f(g(a
lim
.
h
/i->o
Somewhere
it
in
by
in here
we would
f(g(a+h))-f(g(a))
=
h
h-+0
One approach
expression for g'(a).
(fog)(a
it
+
f(g(a
lim
to
is
h))-f(g(a))
g(a
+
h)
-
g(a)
•
g(a
h^-o
This does not look bad, and
.
+ h)—g{a)
looks even better
if
we
h
write
+ h)-(fog)(a)
h
h-+0
=
f(g(a)
lim
+
h^O
The second
(to
like the
fiat:
lim
r
lim
g)(a)
o
(f
h
h-*0
=
put
+ h) -
limit
g(a
is
g(a+h)-g(a)
[g(a+h)-g(a)])-f(g(a))
+ h) — g{a)
k(h)),
lim
.
h^O
we
the factor g'{a) which
be precise we should write
•
then the
want. If we
limit
first
h
g(a
let
—
k
a implies that
A'
+ h) —
g{a)
is
f(g(a)+k)-f(g(a))
lim
.
h^O
It
looks as
goes to
precise.
if this limit
is
+
h)
—
g{a)
meaningless.
one can, and we soon
will,
make
already a problem, however, which you
are the kind of person
g{a
should be f'(g(a)), since continuity of g
as h does. In fact,
There
k
=
True,
for arbitrarily small
0,
who
does not divide
making
the division
we only care about
h. The easiest way
blindly.
Even
at
this sort
will
for h
have noticed
^
this
can happen
+
is
h)
—
if
you
we might have
and multiplication by g(a
small h, but g(a
of reasoning
+
h)
—
g(a)
g(a) could be
for g to
be a constant
,
1
78
Derivatives
and
Integrals
function, g{x)
—
Then g(a +
c.
a constant function, (/ o g){x)
=
g{a)
for
/(c), so the
= =
(fog)'(a)
However, there are
—
h)
=
Chain Rule does indeed
g(x)
=
if
a
=
x2
sin
1
—
x
^0
jc
=
X
we showed
0, as
in
g
is
also
hold:
+ h) —
g(a)
—
0, the function g might be
0.
In this case, g'(0)
o
f'(g(a))-g'(a).
nonconstant functions g for which g(a
also
for arbitrarily small h. For example,
/
In this case,
all h.
Chapter
0.
9. If the
Chain Rule
is
correct,
we
must have (/ o g)'(0) = for any differentiable /, and this is not exactly obvious.
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 CHAIN RULE)
If g
is
at a,
and /
differentiable at a,
is
differentiable at g(a), then
Define a function
=
as follows:
</>
=
0(A)
+h) -
g{a
+
h)
—
g(a)
is
also small, so
i£
g(a
+ h) -
if
gia
+ h) - gifl) =
g(a)
j=
g(a)
should be intuitively clear that
g(a
differentiable
f'{g{a))-g\a).
f(g(a+h))
It
is
and
(fog)'{a)
PROOF
fog
if
is
g(a
+
continuous
—
h)
g(a)
at
is
0.
When
0:
h
not zero, then
is
small,
</>(/?)
will
be close to f'(g(a)); and if it is zero, then (j)(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
differentiable at g{a). This
is
+ k)-f(g(a))
f(g(a)
=
lim
k^0
Thus,
>
if s
(1)
Now
for
g
all
is
there
ifO
<
|it
is
|
some number
<8
>
that
f'(g(a)).
such that, for
f(g(a)+k)-
f
all k,
f(g(a))
then
,
differentiable at a,
<
-/'(*(«))
hence continuous
at a, so there
is
a 8
>
Consider
if \h\
now any
h with
=
f(g(a
(p(h)
\h\
+
g(a
follows
from
(2)
that
|A:|
8'.
= g(a + h) — g(a) ^
0,
8,
then \g(a
<
8.
If
h))
-
f(g(a))
+h) -
<
+ h) - g(a)\ <
<
8',
k
=
f(g(a)
+ k) -
and hence from
\(f>ih)
-
then
f(g(a))
;
g(a)
k
(1) that
f\g{a))\ <e.
E.
such that,
h,
(2)
it
8'
means
179
10. Differentiation
On
the other hand,
+
g(a
if
—
h)
=
g(a)
=
then (p(h)
0,
f(g(a)), so
it is
surely
true that
<e.
\4>(h)- f'{g{a))\
We
have therefore proved that
=
lim 0(A)
so
continuous at
is
0.
The
/'(*(«)),
of the proof is
rest
f(g(a+h))-f(g(a))
easy. If
g(a
=
if
(f
(f
+ h) —
g(a
w
Now
that
at the
=
then
we have
+ h)-g(a)
h
(because in that case both sides are
+
f(g(a
..
= hm
.
o g) (a)
h))-f(g(a))
=
-
h^o
=
look
g{a)
0,
<f>(h)
n
even
/
//
g(a+h)-g(a)
lim
lim
</>(/?)
h^o
h
Therefore
0).
h^o
h
f(g(a))-g'(a).i
we can
many
differentiate so
functions so easily
we can
take another
function
x
sin
—
,
x
yt
U
x
=
0.
fix)
0,
In Chapter 9
we showed
only possible way). For x
/
=
that /'(0)
^
(x)
=
we can
„
2x
—
1
•
sin
working
0,
straight
from the
definition (the
use the methods of this chapter.
\-
x cos
—1
cos
—
2
We
have
at
—
1
xz
Thus
/'(*)
2x
sin
i^0
,
0.
0,
As
this
formula
reveals, the first derivative /'
indeed badly behaved
is
it is
not even continuous there. If we consider instead
x sm —
x
:
fix)
,
x/0
x
0,
=
0.
then
fix)
3x
x cos
sin
x
sion
3x
— xcos
sin
is
continuous at
0,
x
^
l)
x
=
0.
but ,/"(0) does not exist (because the expres-
1/x defines a function which
1/x does not).
,
x
0,
In this case /'
—
is
differentiable at
but the expression
180
Derivatives
and
Integrals
As you may
improvement.
power of x
suspect, increasing the
yet again produces another
If
x4
—
sin
/(*)
,
x
^
x
=
x
0,
0,
then
4x 3
(*) =
./
sin
-
1
2
cos
a:
0,
It is
X
^0
x
=
x
0.
easy to compute, right from the definition, that (/')'(O)
easy to find for x
^
=
0,
and f"(x)
is
0:
1
fix) =
2x 2
1
2x cos —
4x cos
sin
.v/0
sin
x
0,
In this case, the second derivative /"
not continuous at
is
0.
= 0.
By now you may have
guessed the pattern, which two of the problems ask you to establish:
if
1
x
=
/(*)
sin
—
x
,
x
0,
then
/'(()),
.
.
.
/
,
(
">(0) exist, but
=
fix)
f
jc
{n)
0,
not continuous at 0;
is
x 2«+l
_
sin
f
x
^0
jc
= 0,
X
0,
(w)
^
=
if
and /"" is continuous at 0, but / '" is not differentiable at 0. These examples may suggest that "reasonable" functions can be
characterized by the possession of higher-order derivatives
no matter how hard
then f'(0), ...,
/
(
(0) exist,
—
we
try to
mask
the infinite oscillation of
/ (x
)
=
sin
1
/x a derivative of sufficiently
,
high order seems able to reveal the underlying irregularity. Unfortunately,
see later that
After
all
much worse
It is
will
things can happen.
these involved calculations,
a minor remark.
we
often tempting,
we
will
bring
this
and seems more
chapter to a close with
elegant, to write
some of
the theorems in this chapter as equations about functions, rather than about their
values.
Thus Theorem
3
might be written
(/
Theorem 4 might be
+ *)' =
/'
+ *'.
written as
if-gy
and Theorem 9 often appears
= f-g'+f-g,
in the
if °g)'
form
= if
°g)
-g'-
.
10. Differentiation
hand
is
side
may
equations
Strictly speaking, these
be
false,
might have a larger domain than those on the
hardly worth worrying about. If
/ and g
domains, then these equations, and others
case any
because the functions on the
181
left-
right. Nevertheless, this
are differentiable everywhere in their
like
them,
are true,
and
this
is
the only-
one cares about.
PROBLEMS
1.
As a warm up exercise, find fix) for each of the following /. (Don't worry
about the domain of / or /'; just get a formula for fix) that gives the right
answer when it makes sense.)
(i)
f(x)
(ii)
f(x)
(iii)
f(x)
(iv)
f(x)
= sm(x+x 2
= sinx + sinjc
= sin (cos x).
= sin(sinx).
/(jc)
= an(— ).
,
,
.,
(vi)
2.
„
/(*)
(vii)
f(x)
(viii)
f(x)
=
.
/cosx\
•
n
(v)
).
—
x)
sin (cos
-
-.
X
—
=
+ sin x).
sin (a
sin(cos(sin a)).
Find f'(x) for each of the following functions /. (It took the author 20 mincompute the derivatives for the answer section, and it should not take
utes to
you much longer. Although rapid calculation is not the goal of mathematics,
if you hope to treat theoretical applications of the Chain Rule with aplomb,
these concrete applications should be child's play
—mathematicians
pretend that they can't even add, but most of them can
(i)
f(x)
(ii)
/ (a- )
(iii)
f{x)
=
=
=
(iv)
f(x)
=
(v)
f(x)
(vi)
f(x)
/
-\
ft
\
(vn)
/(.v)
(viii)
f{x)
(ix)
f(x)
(x)
fix)
(xi)
fix)
(xii)
fix)
(xiii)
fix)
(xiv)
fix)
+ \) 2 (x + 2)).
3
2
sin (a + sin x
sin ((x + sin a)
sm((x
)
).
.
/
\
r3
sin
cos*
3
= sin(x sin x) + sin(sin x~).
= (cos a-) 31
-2
2-22
= sin
x sm x sin x
— sin (sin (sin x)).
= (x + sin 5 x) 6
— sin(sin(sin(sin(sin x)))).
= sin((sin 7 7 + l) 7
= iHx 2 + x) 3 + x) 4 +x) 5
= sin(x 2 + sin(x 2 + sinjc 2 )).
= sin(6cos(6sin(6cos6A))).
'.
•
.
.
.v
).
.
when
like to
they have
to.)
182
Derivatives
and
Integrals
sin
f{x)
(xv)
2
sin
+
1
sin
2
x
+ sin x
x
„3
/
(xvii)
=
/(*)
x
1
=
f(x)
(xvi)
x
sin
sin
y
=
(xviii)/(*)
sin
sm
\
I
sm
^
x
3.
—
sin
.v
Find the derivatives of the functions tan, cotan,
sec, cosec.
(You don't have
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
to
symmetrical.)
4.
For each of the following functions /, find f'(f(x))
(not
(f
o f)'(x)).
1
/(*) =
1
f(x)
(id
5.
=
=
!,11
f(x)
(iv
/CO =
+x
simr.
x2
.
17.
For each of the following functions /, find f(f'(x)).
(i)
fix)
=
-.
f(x)
=
=
x2
(iii)
fix)
(iv)
f(x)=
Find /'
in
f(x)
.
17.
17.v.
terms of
=
=
=
(")
f(x)
(iii)
f(x)
(iv)
f(x) =
(v)
fix)
(vi)
f(x
g(x
g' if
+ g(a)).
g(x-g(a)).
g(x+g(x)).
g(x)(x-a).
= g{a)(x -a).
+ 3) = g(x 2
).
A
circular object
is
known
that
is
when
increasing in size in
the radius
is
some
6, the rate
unspecified manner, but
of change of the radius
is
it
4.
and A(t)
when
represent the radius and the area at time t, then the functions r and A
satisfy A = nr
a straightforward use of the Chain Rule is called for.)
Find the rate of change of the area
\
the radius
is
6. (If /•(/)
183
10. Differentiation
(b)
Suppose
that
watching
we
now informed that
are
when
Now
is
(You
6.
will clearly
need
to
a formula for the volume of a sphere; in case you have forgotten,
volume
the
(c)
the radius
we have been
Find the rate
really the cross section of a spherical object.
is
of change of the volume
know
the circular object
is
cube of the
^tt times the
radius.)
suppose that the rate of change of the area of the circular cross
section
when
is
5
when
the radius
is
Find the rate of change of the volume
3.
You should be
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).
8.
The
the radius
3.
is
area between two varying concentric circles
rate of
change of the area of the larger
circumference of the smaller
9.
able to
Particle
A moves
circle
circle
is
changing when
along the positive horizontal
at all times
10;r in"/sec.
is
it
axis,
9n
in
has area 16tt in
and
The
.
How fast
B
particle
2
is
the
?
along the
graph of f(x) = -V3i, 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
and moving with speed 4
between A and B changing?
the origin
10.
Let f(x)
=x
sin
^
i/x for x
and
0,
At what
units/sec.
let
f(0)
=
rate
Suppose
0.
is
the distance
also that h
and k
are two functions such that
h'{x)
=
2
+
sin (sin(jr
k\x)
1))
=
f(x
+
1)
k(0)=0.
/i(0)=3
Find
11.
(i)
(foh)'(O).
(ii)
(k o /)'(0).
(iii)
a'{x"),
Find /'(0)
where a(x)
—
/(*)
=
h(x~). Exercise great care.
if
g(.t)sin -,
x
^
x
=
x
0,
0.
and
g(0)
12.
Using the derivative of f{x)
by the Chain Rule.
13.
(a)
(b)
=
=
g'(0)
=
0.
l/x, as found in
Using Problem 9-3, find f'(x) for -1
Prove that the tangent
line to the
<
x
Problem
<
graph of /
the graph only at that point (and thus
show
1, if
9-1, find (l/g)'(x)
f(x)
=
-x 2
y/\
vl — a
2
.
)
intersects
that the elementary
geometry
at (a,
definition of the tangent line coincides with ours).
184
Derivatives
and
Integrals
14.
Prove similarly that the tangent
15.
an
lines to
ellipse
or hyperbola intersect these
only once.
sets
If
/+g
If
/ g and /
is
differentiable at a, are
/ and
g necessarily differentiable
what conditions on / imply
are differentiable at a,
•
at
that g
a?
is
differentiable at a?
16.
Prove that
(a)
/
if
differentiable at a, then |/|
is
^
provided that f{a)
(b)
Give a counterexample
(c)
Prove that
f{a)
if
/ and g
if
also differentiable at a,
is
0.
=
0.
are differentiable
max(/, g) and min(/, g) are
a,
at
then the functions
differentiable at a, provided that f(a)
^
8(a).
Give a counterexample
(d)
17.
f{a)
if
—
g(a).
Give an example of functions / and g such that g takes on all values, and fog
and g are differentiable, but / isn't differentiable. (The problem becomes
we
on all values; g could just be a constant
function, or a function that only takes on values in some interval (a,b), in
which case the behavior of / outside of (a, b) would be irrelevant.)
trivial if
18.
don't require that g takes
= f 2 find a formula for g' (involving /').
= (f) 2 find a formula for g' (involving /").
(a)
If
g
(b)
If
g
(c)
Suppose that the function / >
,
has the property that
iff = f +
in
terms of /.
needed
at
one
is
If
/
is
three times differentiable
at
x
is
defined to be
2>/(*)
(a)
Show
=
Show
that
if
f(x)
quently, 2)(/ o g)
Suppose
point.)
and f'(x)
/
0, the Schwarzian derivative
f'{x)
2\f'{x)
/
2
(f"(x)
3
of
that
=
ax
ex
20.
(In addition to simple calculations,
f'"(x)
®(fog) =
(b)
f
Find a formula for /"
a bit of care
19.
1
that
/
(n)
(fl)
=
[<2)
+b
+a
with
2
+®g
ad - be
#
0,
then
9/ =
2)g.
and g {n) (a)
(n
fog].g'
(f-g) Ha)
exist.
= J2(i)f
k=0
Prow
ik)
Leibniz's formula:
(")-8
(n
A,
<«>
0.
Conse-
-
.
185
10. Differentiation
*21.
22.
and g (n) (a) both exist, then (/ o g) (n) (a) exists. A
little experimentation should convince you that it is unwise to seek a formula
n)
(w)
for (/ o gY {a). In order to prove that (/ o g)
(a) exists you will therefore
have to devise a reasonable assertion about (fog)^(a) which can be proved
by induction. Try something like: "(/ o g) (n) (a) exists and is a sum of terms
each of which is a product of terms of the form ... ."
Prove that
f
if
(n)
igia))
f{x) = a n x"
Find another.
(a)
If
(b)
If
+a n _\x"~
Is
|-ao> find
-\
=
find a function g with g'
(c)
{
a function g such that g'
f.
=
a„jc
+a
H
H
h
1
jc
24.
that there
=
=
—
=
is
a polynomial function
f'{x)
(b)
/'(jc)
(c)
(d)
fix)
fix)
(a)
The number
jc
no
for
exacdy one x,
for
exacdy k numbers x,
—
a
x,
if
n
that
jc.
odd.
is
n
if
even.
is
if
n
—
&
is
odd.
double root of the polynomial function / if
some polynomial function g. Prove that a is a
called a
is
2
a) g(;c) for
(jc
fix)
double root of
/ of degree n such
numbers
1
for
=
When
—
for precisely n
(a)
(b)
—w
fix)=\/x?
such that
Show
/.
there a function
/(jc)
23.
=
/
if
does fix)
—
and only
ax 2 + bx
if
a
+c
is
(a
a root of both
^
0)
/ and
/'.
have a double root?
What
does
the condition say geometrically?
25.
*26.
= fix)-
f(a)(x -a)- fia). Find d\a).
In connection with Problem 24, this gives another solution for Problem 9-20.
If
/
is
differentiable at a, let d(x)
This problem
is
a companion to Problem 3-6. Let a\,
.
.
,
a n and b\,
...
,bn
be given numbers.
(a)
If x\,
tion
.
,
x n are
= a,
distinct
and fixi)
Prove that there
a,
*27.
.
/ of degree 2n —
/(jc,)
(b)
.
and
/'(jc,)
=
is
&,-
numbers, prove that there
such that fixf)
1,
=
b
t
.
Hint:
—
fixj)
is
a polynomial func-
=
for j
Remember Problem 24.
/ of degree In — 1
a polynomial function
for all
^
with
i,
and
/(jc,)
=
i.
and b are two consecutive roots of a polynomial function /,
but that a and b are not double roots, so that we can write /(jc) =
(jc — a)(jc — b)gix) where gia) ^
and gib) ^ 0.
Suppose
(a)
that a
Prove that gia) and gib) have the same
are consecutive roots.)
sign.
(Remember
that a
and b
186
Derivatives
and
Integrals
Prove that there
(b)
draw a picture
and f'(b).
Now
(c)
is
some number x with a < x < b and fix)
prove the same
even
fact,
if
Compare
Hint:
to illustrate this fact.)
=
0. (Also
the sign of f'(a)
a and b are multiple roots. Hint: If
and gib) ^
f{x) = (x —a)'"(x — b)"g(x) where g(a) ^
—
=
polynomial function h(x)
a)'"~ (x — /?)""'.
f'(x)/(x
0,
consider the
[
This theorem was proved by the French mathematician Rolle, in connection
with the problem of approximating roots of polynomials, but the result was
not originally stated in terms of derivatives.
mathematicians
who
never accepted the
In
new
fact,
Rolle was one of the
notions of calculus. This was
not such a pigheaded attitude, in view of the fact that for one hundred years
no one could define limits in terms that did not verge on the mystic, but on
the whole history has been particularly kind to Rolle; his name has become
attached to a much more general result, to appear in the next chapter, which
forms the
28.
29.
Suppose
basis for the
is
Suppose /
is
differentiable at 0,
some function g which
If
fix)
=
= x~ n
f
32.
Prove that
for n in
{k)
<*>,...
(x)
=
What
f
*(b)
Prove that fix)
0.
Hint:
What happens
— xg(x)
if
you
try
(-\,x*(»
that
+ *-i)L-«-*
ix)
n
(-!)**!
+k-
= g(0) =
1
=
-n-k
for
x
#
0.
f{x)g{x) where / and g are
differ-
0. Hint: Differentiate.
if
/(*)= \/(x-a)"?
fix)
Let fix)
/""(0)
same
*34.
N, prove
impossible to write x
and /(0)
(a)
*33.
is
it
{k)
continuous at
is
= 0.
that /(0)
(n-1)!
entiable
is
and
at 0.
f(x)/x?
=
*31.
=
/
to write g(x)
results of calculus.
xg(x) for some function g which is continuous
differentiable at 0, and find /'(0) in terms of g.
that f(x)
Prove that
for
30.
most important theoretical
Lei
=
-\)?
x 2 "sin
exist,
and
l/.v if
that
f
x
#
[u)
is
basic difficulty as that in
f(x)
/""(0)
atO.
=\ /(x 2
=
x 2n+l
exist, that
sin l/.v if
/
(/,)
is
x
0,
and
let
=
/(0)
not continuous at
Problem 21
^
0,
and
continuous at
let
0,
0.
0.
Prove that /'(0)
(You
will
encounter the
.)
/(0)
and
=
that
0.
Prow
/""
is
that
./'(())
not differentiable
,
.
.
10. Differentiation
35.
187
In Leibnizian notation the Chain Rule ought to read:
Instead,
Z
=
df(g(x))
df(y)
dx
dy
dgixf
-MM
dx
one usually finds the following statement:
f(y).
Then
dz
dz
dy
dx
dy
dx
"Let y
dz/dy denotes the function /;
expression involving y," and that
for y. In
z
(ii)
z
(iii)
z
(iv)
z
= siny,
= siny,
=
=
also
in the final
sin u
sin v
,
y
y
u
v
1
— x + x2
= cos*.
= sin x
= cost/,
.
u
=
sin*.
fog, while
understood that dz/dy
will
the z
be "an
answer g(x) must be substituted
each of the following cases, find dz/dx by using
compare with Problem
(i)
it is
g(x) and
„
Notice that the z in dz/dx denotes the composite function
in
=
this
formula; then
.
SIGNIFICANCE OF THE DERIVATIVE
CHAPTER
One aim
in this chapter
to justify the time
is
As we
we have
spent learning to find the
knowing just a little about /' tells us a
lot about /. Extracting information about / from information about /' requires
some difficult work, however, and we shall begin with the one theorem which is
derivative of a function.
shall see,
really easy.
This theorem
concerned with the
is
Although we have used
precise,
DEFINITION
Let
and
more
also
this
in
A
is
a
term informally
in
Chapter
7,
it is
A
a
set
worthwhile to be
of numbers contained in the domain of /.
maximum point for
/(*) > /(y)
/ on A
if
f° r every v in A.
The number f(x) itself is called the maximum value of / on A
say that / "has its maximum value on A at x").
maximum
(and
we
also
/ on A could be f(x) for several different x
1);
/ can have several different maximum points
on A, although it can have at most one maximum value. Usually we shall be
interested in the case where A is a closed interval [a b] if / is continuous, then
Theorem 7-3 guarantees that / does indeed have a maximum value on [a b]
The definition of a minimum of / on A will be left to you. (One possible
definition is the following: / has a minimum on A at x, if — / has a maximum
on A at x.)
We are now ready for a theorem which does not even depend upon the existence
Notice that the
(Figure
X2
interval.
general.
a function and
/ be
A point x
maximum value of a function on an
X3
in other
value of
words, a function
,
;
,
FIGURE
1
of least upper bounds.
THEOREM
l
/ be any function defined on (a,b). If x is a maximum (or a minimum)
for / on (a,b), and / is differentiate at x, then f'(x) = 0.
(Notice that we do not assume differentiability, or even continuity, of / at
Let
point
other
points.)
PROOF
Consider the case where / has a maximum
—
behind the whole argument
have slopes
slopes
188
<
>
0,
secants
at x. Figure 2 illustrates the simple idea
drawn through
points to the
left
and secants drawn through points to the right of
argument proceeds as follows.
0. Analytically, this
of (x, f(x))
(x,
f(x)) have
11. Significance of the Derivative
h
If
any number such that x
is
+h
in (a, b),
is
189
then
f(x)>f(x+h),
/
since
maximum on
has a
This means that
(a, b) at x.
f(x+h)-f(x)<0.
Thus,
h
if
we have
>
f(x+h)-fjx)
a
x
x
+
h
b
<0.
h
and consequently
FIGURE
2
+ h)-fix)
fix
,.
hm
On
the other hand,
if
<
h
<
(J.
h
h->0+
we have
0,
f(x
+ h)-
fix)
>0,
h
so
/U+/7)-/(x)
hm
By
/
hypothesis,
is
differentiable at x, so these
>
U.
two
must be equal,
limits
in fact
equal to fix). This means that
FIGURE
from which
3
The
follows that
it
Notice (Figure
Since
<
fix)
=
(unless
3) that
we add
DEFINITION
Let
/ be
is
some
x
is
a local
then fix)
proof
You should
A
a function, and
is left
to
you
a one-line proof).
(give
|
replace (a, b) by [a, b] in the statement of the
x
the values of
1.
We
/ near
x,
it is
is
in ia,b).)
almost obvious
begin with a definition which
is
how
to
illustrated
=
A
<5
is
>
a local
a set of
in the domain of /.
point for / on A if
[minimum] point for / on
numbers contained
maximum [minimum]
such that x
is
a
maximum
+ 8).
A nix -8,x
If
x
to the hypothesis the condition that
fix) depends only on
point x in
there
2
at
Figure 4.
A
THEOREM
0,
0.
we cannot
get a stronger version of Theorem
in
fix) >
and
where / has a minimum
case
theorem
f'(x)
maximum
or
minimum
for
/ on
(a, b)
and /
is
differentiable at x,
0.
see
why
this
is
an easy application of Theorem
1
.
|
1
90
Derivatives
and
Integrals
The
even
is
converse of Theorem 2
x
if
is
is
definitely not true
maximum
not a local
minimum
or
provided by the function f(x) — x
or minimum anywhere.
3
;
—
it is
possible for f'(x) to be
The
point for /.
=
in this case f'(0)
simplest example
but
0,
/
has no local
maximum
Probably the most widespread misconceptions about calculus are concerned
with the behavior of a function
the previous paragraph
simpler than
it is,
condition f'(x)
that
—
A
DEFINITION
critical
we
this
/
not
repeat
=
The
0.
the converse of Theorem 2
it:
imply that x
condition
a local
is
point of a function /
is
itself
is
—
fix)
a
maximum
made
point
or
is
not true
minimum
in
to
be
—
the
point
0.
number x such
=
f'(x)
The
f'(x)
by those who want the world
reason, special terminology has been adopted to describe
satisfy the
The number f(x)
when
near x
so quickly forgotten
will
does
of /. Precisely for
numbers x which
is
that
0.
called a critical
value of
/.
values of /, together with a few other numbers, turn out to be the
critical
maximum and minimum of a
maximum and minimum value
ones which must be considered in order to find the
given function /.
To
the uninitiated, finding the
of a function represents one of the most intriguing aspects of calculus, and there
no denying that problems of
hundred or so).
is
Let us consider
on a closed
that a
If
a local
minimum
FIG
1
RE
4
point
of
/
point
x
(Then,
if
value
three kinds of points
/
(3)
Points x in [a, b] such that
/
is
a
is
and
above: for
listed
and / is
means that .v
is
in the first
many points
/ may still be
minimum
of
is
if
x
is
In order to locate the
maximum and
[a,
£>]
,
then x must be
in
one
=
0,
by Theorem
1,
group.
in these
three categories, finding the
a hopeless proposition, but
/
f(x)
The biggest of these will be the maximum
minimum. A simple example follows.
is
/
be sure
not in the second or third group, then
and only a few points where /
straightforward: one simply finds f(x)
x such that
at least
of
not differentiable at x.
critical points,
for each
minimum
or
we can
differentiable at x; consequentiy f'(x)
If there are
fairly
continuous,
maximum point or a minimum point for / on
in (a, b)
this
maximum
first
must be considered:
(2)
x
is
exist.)
[a, b].
of the three classes
a local
maximum
interval [a,/?].
problem of finding the
The critical points of / in
The end points a and b.
(1)
X2
the
maximum and minimum
minimum
x\
first
are fun (until you have done your
this sort
is
when
maximum and
there are only a few
not differentiable, the procedure
for
not differentiable
each x satisfying f'(x)
at
x and,
finally,
=
0,
is
and
f{a) and fib).
value of /, and the smallest
will
be the
)
11. Significance of the Derivative
Suppose we wish
maximum and minimum
to find the
191
value of the function
f(x)=x 3 -x
on the
interval [—1,2].
To begin
with,
=3x 2 -
f(x)
—
so f'(x)
when 3x —
=0,
1
=
x
The numbers y
that
is,
y/T/3
1
/lA
group
is
,
-VT73:
contains the end points of the interval,
-
(2)
third
- vT/3.
or
— y 1/3 both lie in [— 2], so the first group of candidates
maximum and the minimum is
(1)
The second group
1,
when
1/3 and
for the location of the
The
we have
empty, since
/
is
1,
2.
differentiable everywhere.
The
final step
is
to
compute
f(y/l/3)
=
3
(7173
- 1/3 = ivM/3 - vl/3 = — §vT/3,
/(- v/r/3) = (- v/T73) 3
/(-D =
f(2) =
value
=
+ /T73 = § x/Ia
-JX/T73
6.
occurring at
6,
is
)
0.
minimum
Clearly the
(-7173
value
is
— ^y
1/3, occurring at
y
1/3,
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
with
is
not continuous, however, or
on an open
/(O)
that the
this
1
b
a
interval or the
whole
if
we
line,
are seeking the
then
maximum and minimum values exist, so
may say nothing. Nevertheless, a
we
the nature of things. In Chapter 7
maximum
we cannot even be
procedure
all
we
are dealing
or
minimum
sure beforehand
the information obtained by
little
ingenuity will often reveal
solved just such a problem
when we showed
1
that
FIGURE
maximum
and the
2.
if
n
is
5
even, then the function
f(x)=x"+a n ^x"- +.-.+ao
]
minimum value on the whole line.
occur at some number x satisfying
has a
/'(*)
we can
This proves that the
= nx n ~ + (n l
l)a„_iJc
B -2
+..
minimum value must
a\.
and compare the values of f(x) for such x, we can
actually find the minimum of /'. One more example may be helpful. Suppose we
wish to find the maximum and minimum, if they exist, of the function
If
solve this equation,
,
192
Derivatives
and
Integrals
on the open
interval
(— 1,
We
1).
have
fix)
=
2x
U-* 2 2
)
=
so f'(x)
only for x
=
become
values of f(x)
We
0.
arbitrarily large, so
This observation also makes
note (Figure
can see immediately that for x close to
5) that there will
/
certainly does not have a
easy to show that
/
has a
be numbers a and
b,
with
it
— <
a
1
<
1
and
< b <
1
<x <
1
minimum
or
—
1
the
maximum.
We just
at 0.
such that f(x) > /(0) for
—
< x < a
1
and
minimum of / on [a b\ is the minimum of / on all of
on [a,b] the minimum occurs either at
(the only place where
a or b, and a and b have already been ruled out, so the minimum
This means that the
(— 1,
1).
=
0),
/'
value
f(x)
=
3
jc
-
x
Now
or at
/(0)
is
b
=
,
1.
In solving these problems we purposely did not draw the graphs of fix) = x —x
and f(x) = 1/(1 — x ), 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
gives enough information to be used in discussing maxima and minima
in fact
we will be able to locate even local maxima and minima. This method involves
consideration of the sign of f'(x), and relies on some deep theorems.
—
The theorems about
which have been proved so far, always yield
terms of information about /. This is true even of Theo-
derivatives
information about /' in
rem
theorem can sometimes be used to determine certain information about /, 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
1
,
although
this
painfully relevant
about
tered
—
x'
/'.
The
when one
simplest
conviction
is
f'(x)
it is
=
impossible to imagine
It is
difficult
for all x.
is
always
hypothesis /'(jc)
h
and
it
is
not at
how /
for
all
x,
encoun-
must / be a
else, and
could be anything
all
must be standing
still!
—
if
the
Never-
even to begin a proof that only the constant functions satisK
The
f(x
Kl. 6
fix) =
0, surely the particle
Inn
FIGl
/ from information
strengthened by considering the physical interpretation
velocity of a particle
theless
extract information about
frustrating illustration of the difficulties
afforded by the following question: If
is
constant function?
this
tries to
and most
obvious
=
+h)-
only means that
f(x)
=
0.
•()
how one
can use the information about the
derive information about the (unction.
limit to
.
193
11. Significance of the Derivative
The
fact that
of the same
/
is
a constant function
can
sort,
all
Value Theorem, which
/
that if
is
if /'(jc)
=
for
all jc,
and many other
be derived from a fundamental theorem, called the
states
differentiable
on
much
[a, b],
stronger results. Figure 7
then there
/'(*)
fib)
=
b
is
some x
makes
in (a, b)
it
facts
Mean
plausible
such that
- f(a)
—a
this means that some tangent line is parallel to the line between
and
(b,
there
f(b)). The Mean Value Theorem asserts that this is true
f(a))
is some x in (a,b) such that fix), the instantaneous rate of change of / at jc, is
exactly equal to the average or "mean" change of / on [a, b], this average change
being [f(b) — f(a)]/[b — a]. (For example, if you travel 60 miles in one hour,
then at some time you must have been traveling exactly 60 miles per hour.) This
theorem is one of the most important theoretical tools of calculus probably the
Geometrically
—
(a,
FIGURE
7
—
deepest result about derivatives.
proof is
From
statement you might conclude that the
but there you would be wrong
difficult,
have occurred long ago, in Chapter
Value Theorem yourself you
FIGURE
this
will
7.
It is
—
the hard theorems in this
true that
probably fail, but
if
this
you
is
try to
prove the
book
Mean
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
8
helps to begin with a very special case.
THEOREM
3
(ROLLE'S
THEOREM)
If
/
is
there
PROOF
continuous on [a,b] and differentiable on
a
is
If follows
number x
in (a, b)
such that f'(x)
from the continuity of / on
[a
,
=
b] that
(a, b),
and f(a)
=
f(b), then
0.
/
has a
maximum and a minimum
value on [a b]
,
Suppose
—
/'(jc)
first
that the
by Theorem
1,
maximum
value occurs at a point x in (a.b).
and we are done (Figure
8).
Suppose next that the minimum value of / occurs
Then, again, /'(jc) = by Theorem 1 (Figure 9).
Finally,
points.
FIGURE
9
so
/
any
is
jc
suppose the
Since f(a)
=
Then
at
some point x
in (a,b).
maximum and minimum values both occur at the end
f(b), the maximum and minimum values of / are equal,
a constant function (Figure 10), and for a constant function
we can choose
in (a,b). |
Notice that
on (a,b)
in
we
really
needed the hypothesis
order to apply
Theorem
1.
that
Without
/
this
is
differentiable
everywhere
assumption the theorem
is
false (Figure 11).
FIGURE
io
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 Rollc'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
94
1
Derivatives
and
Integrals
Figure 12;
between
this
(a,
is
the difference
f(a)) and
(b, f{b)).
between f(x), and the height
Since L is the graph of
—
f(b)
gix)
b
we want
FIGURE
fib)-f(a)
fix) -
—
b
THEOREM
4
(THE
MEAN VALUE
THEOREM)
If
it
/
turns out, the constant f(a)
is
continuous on
in (a b)
,
(x-a)
a
and
[a, b]
is
Clearly, h
is
=
continuous on
fib))
(x-a)-
Consequently,
=
I
(
,
I
R
1.
on
then there
(a, b),
number x
a
is
- fia)
—a
fib)
some x
f(b)-f(a)
f(x)
and
h(a)
=
fia).
h(b)
=
f(b)-
differentiable
on
fjb)-f(a)
b
-a).
(x
b-a
(a, b),
and
(x-a)
—a
fia).
we may apply
Rolle's
Theorem
to h
and conclude
that there
is
in (a, b) such that
= ti(x) =
I
f(a).
irrelevant.
differentiable
[a, b]
=
f(a))
+ f(a),
Let
h(x)
(a,
L
a
b
(b,
line
such that
f'ix)
proof
x of the
to look at
11
As
f(a)
at
fib)
f(x)
b
—
fia)
a
so that
12
fix)
fib) = M ;
b
Mean
Notice that the
previous theorems
mation
COROLLARY
l
If
/
is
is
Value Theorem
— information about /
so strong, however, that
—
'fia)'
.
a
still
fits
into the pattern exhibited
yields information
we can now go
defined on an interval and f'(x)
|
=
about
/'.
by
This infor-
in the other direction.
for all
x
in the interval,
then
/
is
constant on the interval.
PR(
)<
)i
Let a and b be any
I
\v<>
points
in
the interval with a
^
b.
Then there
is
some x
in
,
195
11. Significance of the Derivative
(a,b) such that
b
- fia)
—a
—
a
fib)
/'(*)
But f'(x)
i
<ig
ire
—
for
x
all
in the interval, so
fib)
=
13
b
=
and consequently fia)
interval
is
the same,
Naturally, Corollary
fib).
/
i.e.,
1
Thus
fia)
the value of
constant on the interval.
is
/
at
any two points
in the
|
does not hold for functions defined on two or more
in-
tervals (Figure 13).
corollary
2
/ and g
If
PROOF
For
all
there
x
The
is
number
c
=
g'(x) for
)
all
x in the
c.
we have (/ — g)\x) = fix — g'ix =
/ — g = c. |
in the interval
a
is
on the same interval, and fix)
some number c such that / = g +
are defined
interval, then there
so,
)
by Corollary
1
such that
statement of the next corollary requires some terminology, which
is
illus-
trated in Figure 14.
A
DEFINITION
function
is
two numbers
an interval
if
increasing on an
in the interval with
f(a)
often say simply that
3
If
fix) >
for all
PROOF
x
for all
x
<
a
The
b.
<
fib) whenever a and b are
function
/
in
an
in the interval,
interval,
then
/
is
then
fix)
/
is
=
fib)
b
all
x in
increasing
b
-
—
> it follows that fib) >
The proof when fix) < for all x
a
- fia)
—a
(a, b), so
fib)
Since b
is
on the
interval; if
fix) <
decreasing on the interval.
Consider the case where fix) > 0. Let a and b be two points
< b. Then there is some x in (a, b) with
for
case the interval
/.)
a
But fix) >
decreasing on
with a < b. (We
is
> fib) for all a and b in the interval
/ is increasing or decreasing, in which
understood to be the domain of
corollary
interval if fia)
fia)
>0.
a
fia).
is left
to you. |
in the interval with
196
Derivatives
and
Integrals
Notice that although the converses of Corollary
obvious), the converse of Corollary 3
see that f'{x)
f(x)
>
is
1
and Corollary 2 are
not true. If
/
increasing,
is
true (and
it is
easy to
might hold for some x (consider
for all x, but the equality sign
= x\
Corollary 3 provides enough information to get a good idea of the graph of
a function with a minimal amount of point plotting.
function f(x)
=
x3
— x. We
Consider, once more, the
have
f'(x)
=
3x
2
-\.
have already noted that f'(x) =
for x = y 1/3 and x = — y^l/3, and it
2
also possible to determine the sign of f'(x) for all other x. Note that 3x — 1 >
We
precisely
when
3x
2
x
2
x > y 1/3
a
(a)
b
—
thus 3x
is
1
<
or
>
1
>\.
—y
x <
1/3;
when
precisely
-7173 <
an increasing function
x < yi73.
Thus / is increasing for x < — yl/3, decreasing between — yl/3 and vl/3,
and once again increasing for x > y/i/3. Combining this information with the
following facts
/(- V/I73) =
(1)
/(/I73)
(b)
=
/
^ v I73,
-|vT73,
(2)
/(x)=0forx = -l,0,
(3)
f(x) gets large as x gets large, and large negative as x gets large negative,
1,
a decreasing function
FIGl RE
14
it is
possible to sketch a pretty respectable approximation to the
graph (Figure
15).
By the way, notice that the intervals on which / increases and decreases could
have been found without even bothering to examine the sign of /'. For example,
since /'
is
continuous, and vanishes only at
— vl/3 and
yl/3, we know
always has the same sign on the interval (— \J\/3, vl/3).
f(y/\/3
),
it
follows that
/
Since
that /'
/(— vl/3) >
decreases on this interval. Similarly, /' always has the
on (y/l/3, oo) and f(x) is large for large x, so / must be increasing on
(vl/3, oo). Another point worth noting: If /' is continuous, then the sign of /'
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.
same
sign
Our
to
sketch of the graph of f(x)
=
x3
—
allow us to say with confidence that —y/\/3
Vl/3
is
a local
minimum
point. In fact,
x contains
is
we can
a local
give
sufficient
maximum
information
point,
a general scheme
and
that
for decid-
197
11. Significance of the Derivative
/
/
increasing
decreasing
i
/
increasing
/
FIGURE
15
ing whether a critical point
is
a local
maximum
point, a local
minimum
point, or
neither (Figure 16):
(1)
if
/'
>
some
in
the right of x, then x
(2)
if /'
<
the right of
(3)
if
some
in
jc,
/' has the
is
a local
in
some
interval to
x and /' >
in
some
interval to
maximum
interval to the left of
then x
same
<
and
interval to the left of x
is
a local
sign in
interval to the right, then
x
minimum
some
is
/'
point.
point.
interval to the
left
of x as
has in some
it
maximum nor a local minimum
neither a local
point.
(There
no point
is
in
memorizing these
rules
—you can always draw
the pictures
yourself.)
The polynomial
functions can
to describe the general
/'
>
/'
(a)
FIGURK
16
<
all
be analyzed in
form of the graph of such
/'
<
/'
(b)
>
this
way, and
functions.
To
it is
even possible
begin,
/'
<
we need
/'
(d)
<
a
1
98
Derivatives
and
Integrals
result already
mentioned
Problem
in
=
f(x)
/
f(x)
=
f(x\)
Although
0.
—
f(x2)
and X2 such
<
•
JCfc,
—
0,
really
is
if
=
that f'(x)
Theorem
This means that
0.
—
now
least k
A3, etc.
It is
+ <*o.
numbers x such
that
an algebraic theorem, calculus can be used
x\ and X2 are roots of / (Figure 17), so that
then by Rolle's
then /' has at
between X2 and
FIGURE
this
+
]
a„_ hx"-
there are at most n
i.e.,
an easy proof. Notice that
to give
+
a„x"
has at most n "roots,"
then
3-7: If
if
there
/
is
number
a
A'
between
has k different roots x\
x\
< a2 <
one between x\ and X2, one
prove by induction that a polynomial
different roots:
\
easy to
function
17
f(x)=aH x H + an -ix a
has at most n roots:
it is
The
statement
is
-
i
+---+ao
surely true for n
=
1
,
and
if
we assume
that
true for n, then the polynomial
=
g(x)
could not have more than
+
/;
b n+] x"
1
+l
+b
+ b„x" +
roots, since if
it
would have more than n
did, g'
roots.
With
information
this
fix)
The
n
—
derivative,
1
a„x"
+
fl,,-!*"-
1
+
•
•
•
+ oo.
being a polynomial function of degree n
—
1,
has at most
Therefore / has at most n — 1 critical points. Of course, a
not necessarily a local maximum or minimum point, but at any
is
a and b are adjacent
negative
on
critical
(a,b), since /'
is
criti-
rate,
points of /, then /' will remain either positive or
continuous; consequently,
Thus / has
or decreasing on (a,b).
As
=
roots.
cal point
if
not hard to describe the graph of
it is
/
will
be either increasing
most n regions of decrease or
at
increase.
a specific example, consider the function
f(x)=x 4 -2x 2
.
Since
f'(x)
the critical points of
/
are
= 4a- 3 - 4a =
—1,0, and
1
,
/(-1)
(0,0)
4a(a
-
1)(a
+
1
),
and
=
-1,
/(0)=0,
/(!)
(-1,-1)
l
[GURE
(1,-1)
18
The behavior of / on
=
-!•
the intervals between the critical points can be determined
by one of the methods mentioned before. In particular, we could determine the
sign of /' on these intervals simply be examining the formula for /'(a). On the
we can sec (Figure 18) that /
To determine the sign of /' on
other hand, from the three critical values alone
increases
on (—1,0) and decreases on
(0, 1).
199
11. Significance of the Derivative
(—00, —1) and
(1,
we can compute
oo)
= 4-(-2) 3 -4-(-2)
=4- 2 3 -4-2 = 24,
/'(-2)
/'(2)
and conclude
/
that
-24,
decreasing on (— oo, —1) and increasing on
is
conclusions also follow from the fact that f(x)
and
large for large x
is
These
(1, oo).
for large
negative x.
We
can already produce a good sketch of the graph; two other pieces of
mation
f(x)
provide the finishing touches (Figure 19). First,
=
graph
is
for
x
=
0,
±V2;
second,
symmetric with respect
already sketched in Figure 15,
is
it is
clear that
/
to the vertical axis.
odd, f(x)
=
it is
easy to determine that
even, f(x )
=
f(—x), so the
The function f(x) = x — x,
is
—f(—x), and
is
consequently sym-
may
metric with respect to the origin. Half the work of graph sketching
by noticing these things
infor-
be saved
in the beginning.
(V2,0)
(-1,-1)
FIGURE
(1,-D
19
Several problems in this and succeeding chapters ask you to sketch the graphs
of functions. In each case you should determine
(1)
the critical points of /,
(2)
the value of
(3)
the sign of /' in the regions between critical points
/
at the critical points,
(if
this
is
not already
clear),
numbers x such
that f(x)
(4)
the
(5)
the behavior of f(x) as x
Finally,
even,
bear
may
in
mind
=
becomes
(if
possible),
large or large negative
that a quick check, to see
(if
possible).
whether the function
is
odd or
save a lot of work.
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
This sort of analysis,
if
often very important. If
rational function
of
/
/
is
not defined at certain points
whose denominator vanishes
at
some
near these points should be determined.
For example, consider the function
/(*)
=
2jc
+2
(for
example,
points), then the
if
/
is
a
behavior
200
Derivatives
and
Integrals
which
is
not defined at
1
We
.
fix)
have
2
(jc- 1)(2jc-2)-(jc -2jc
=
(x
—
x(x
- D
2
of
/
+
2)
2)
*
C*-D 2
Thus
(1) the critical points
are 0,
2.
Moreover,
=
=
/(0)
(2)
f(2)
Because
/
is
(—oo, 0) and
2.
not defined on the whole interval
determined separately on the
intervals, or
-2,
(2, oo).
We
and
intervals (0, 1)
can do
(0, 2),
on the
(1,2), as well as
by picking particular points
this
simply by staring hard at the formula for
(3) f'{x)
the sign of /' must be
>
x <
if
f{x) <0
fix) <
fix) >
if
1
if
Either
in
each of these
way we
find that
0,
<x <
if
/'.
intervals
1,
< x <
2 < x.
2,
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
way to determine the regions on which / increases and decreases). To examine
the behavior as x becomes large we write
Finally,
+2
-2jc
jcclearly /(jc)
to
is
jc
—
1
is
close to
jc
—
1
(but slightly smaller)
= x- 1 +
1
]
x-V
when x is large, and f(x) is close
negative. The behavior of / near 1
(and slighdy larger)
when
is
jc
large
also easy to determine; since
lim(jc'
+ 2) =
2jc
1
^
0,
x->-l
the fraction
+2
2jc
I
becomes
large as
jc
approaches
1
from above and large negative
as
x approaches
1
from below.
All this information
that
it
may seem
a
bit
can be pieced together (Figure
overwhelming, but there
20);
is
only one
way
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
201
11. Significance of the Derivative
2
a-
2
1
FIGURE
- 2x + 2
20
of an odd function shoved over
x2
1
unit,
and the expression
(x-!) 2 +l
-2.y+2
x-
I
shows that
this
indeed the case. However,
is
which should be investigated only
after
this
1
is
one of those
special features
you have used the other information
to get
a good idea of the appearance of the graph.
Although the location of local maxima and minima of a function
vealed by a detailed sketch of
work. There
is
a popular
its
theorem
5
Suppose f'(a)
then
PROOF
By
/
=
0. If
has a local
f"{a)
maximum
>
it is usually unnecessary to do so much
maxima and minima which depends on the
its
critical points.
then
0,
/
has a local
minimum
at a.
definition,
f
(a)
= hm
f(a + h)-f'(a)
h
/i->0
Since f'(a)
—
0, this
can be written
,.
f
always re-
graph,
test for local
behavior of the function only at
is
(a)
= hm
h^O
f'(a
+ h)
h
at a; if
f"(a)
<
0,
202
Derivatives
and
Integrals
Suppose now
>
that f"(a)
Then
0.
+
f'ia
h)/h must be positive for
sufficiently
small h. Therefore:
f'ia
and f'ia
+
+ /?)
This means (Corollary
and /
is
= - v-
must be negative
/
3) that
for sufficiently small
increasing in
is
Theorem
<
the case f"(a)
some
0.
interval to the right of a
of a. Consequently,
left
/
has a local
similar. |
is
may
be applied to the function f(x)
been considered. We have
5
>
h <
positive for sufficiently small h
decreasing in some interval to the
minimum at a.
The proof for
fix)
must be
/?)
=
x3
— x,
which has already
(a)
/'(*)
/"(*)
if(x)
=
=
=
3x
2
-
1
6x.
xv4
At the
critical points,
—y
1/3 and
y
we have
1/3,
= -6/T73<o,
/
.r(- v T73)
f(y/l/3)=6y/t/3>0.
—y
Consequently,
(b)
1/3
is
maximum
a local
point and
y
1/3
is
a local
minimum
point.
Although Theorem 5
lf(x)=x 5
many
functions the second derivative
the sign of the
that f"{a)
that a
be found quite useful for polynomial functions, for
will
is
=
first
maximum
is
so complicated that
Moreover,
a
if
Theorem
In this case,
0.
a local
(Figure 21)
derivative.
is
a
critical
5 provides
easier to consider
/
it
no information:
minimum
point, a local
it is
point of
may happen
it
is
possible
point, or neither, as
shown
by the functions
(c)
FIGURE
2
fix)
= -x\
f(x)
fix)
= x5
;
1
in
each case f'iO)
local
minimum
of
theorem
6
=
but
0,
is
a local
This point
interesting to note that
maximum point
a local maximum
will be pursued further in Part
Theorem
for the
nor
first,
a
minimum
IV
5 automatically proves a partial converse
itself.
Suppose f"ia)
exists.
If
/
has a local
<
maximum
at a,
then f'ia)
Suppose / has
local
minimum
at
by Theorem
local
PROOF
/"(0)
point for the second, and neither
point for the third.
It is
=
local
maximum
c/,
containing a, so that f'ia)
The
case of a local
=
at
0, a
maximum
is
minimum
at «,
then f"(a)
>
0; if
/
has a
0.
\\
f'ia) < 0, then / would also have a
Thus / would be constant in some interval
contradiction. Thus we must have f'ia) > 0.
a.
5.
handled
similarly. |
203
11. Significance of the Derivative
Theorem 5
by > and <,
(This partial converse to
signs
cannot be replaced
f(x)
= -x 4
the best
as
shown by
the
for:
> and <
x 4 and
=
the functions fix)
.)
The remainder
maxima and
of this chapter deals, not with graph sketching, or
Mean
minima, but with three consequences of the
Value Theorem. The
theorem which plays an important
simple, but very beautiful,
and which
we can hope
is
also sheds light
first is
Chapter
role in
on many examples which have occurred
a
15,
in previous
chapters.
THEOREM
7
Suppose
/
that
is
continuous
and
at a,
containing a, except perhaps for x
=
that f'(x) exists for
all
x
in
some
interval
Suppose, moreover, that lim fix)
a.
exists.
X—fO
Then
and
f'ia) also exists,
=
/'(a)
PROOF
By
lim /'(*).
definition,
f(a)
=
fja+h)-
fja)
lim
h
>
For sufficiently small h
on
differentiable
By
the
Mean
a
(a,
+
the function
Theorem
Value
/
be continuous on
will
[a,
+
a
there
is
a
number
f(a+h)-f(a)
=
a/,
in (a,
a
+ h)
and
h]
h) (a similar assertion holds for sufficiently small h
<
0).
such that
/'(«*)•
h
Now
lim
x—>a
approaches a as h approaches
a/,
fix)
exists,
it
,,, ,
/ ia)
a/,
is
in (a, a
+
h);
since
=
fia
..
lim
+ h )-
fia)
=
lim
,
/ (a/,)
=
lim /
,
(.*).
*->a
/i—O
h
a good idea to supply a rigorous s-8 argument for this final step, which
somewhat
have treated
Even
if
/
is
an everywhere
differentiable function,
x
fix)
sin
—
,
x
^
x
= 0.
x
0.
to
Theorem
nuity of the type
beautiful
theorem which
lem 62 uses
The
7,
shown
this result to
in
it is still
possible for /' to be
if
2 2
According
we
informally.) |
discontinuous. This happens, for example,
FIGURE
because
follows that
/j^O
(It is
0,
U
however, the graph of /' can never exhibit a discontiFigure 22. Problem 61 outlines the proof of another
gives further information
strengthen
Theorem
next theorem, a generalization of the
mainly because of its applications.
about the function
/',
and Prob-
7.
Mean
Value Theorem,
is
of interest
204
Derivatives
and
THEOREM
8
Integrals
(THE
CAUCHY MEAN
VALUE THEOREM)
/ and g are continuous on
number x in (a, b) such that
(If
g(b)
^
g(a),
[f{b)
-
g'(x)
^
and
and
[a,b]
If
f(a)]g'(x)
=
-
f(a)
is
a
g(a)]f(x).
f'{x)
=
g(b)-g(a)
=
-
[gib)
(a,b), then there
equation can be written
0, this
f(b)
on
differentiable
g'(x)-
=
and we obtain the Mean Value
Theorem. On the other hand, applying the Mean Value Theorem to / and g
separately, we find that there are x and v in (a, b) with
Notice that
if
g(x)
x for
all
x, then g'{x)
f(b)-f(a)
but there
remarks
g'{yY
and y found in this way
the Cauchy Mean Value Theorem
no guarantee
may
to prove,
PROOF
is
fix)
=
"
g(b)-g(a)
1,
that the x
suggest that
but actually the simplest of
will
be equal. These
will
be quite
difficult
tricks suffices.)
Let
=
h{x)
Then
h
is
f(x)[g(b)
continuous on
follows from Rolle's
g(a)\
-
=
f(a)g{b)
Theorem
-
-
g(x)[f(b)
[a, b], differentiable
h(a)
It
-
on
and
(a,b),
g(a)f(b)
=
that h'(x)
—
-
g'{x)[f{b)
for
/(«)].
h(b).
some x
in (a, b),
which means
that
=
f'(x)[g(b)
g(a)]
The Cauchy Mean Value Theorem
which
facilitates
is
-
the basic tool
-
f(a)]. |
needed
to
prove a theorem
evaluation of limits of the form
lim
fix)
x^a g( X )
when
lim f(x)
In this case,
Theorem
5-2
is
=
and
lim g(x)
=
of no use. Every derivative
0.
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
(L'HOPITAL'S rule)
Suppose that
lim fix)
x—*a
and suppose
—
and
also that lim f'(x)/g'(x) exists.
x—*a
*
fix)
,.
lim
x^a g( X )
(Notice that
Theorem
7
is
a special
=
case.)
lim #(.y)
x^a
Then
0.
lim f(x)/g(x)
x—*a
fix)
lim
x^a g'(x)
=
.
exists,
and
—
11. Significance of the Derivative
PROOF
The
205
hypothesis that lim f'(x)/g'(x) exists contains two implicit assumptions:
x—>a
(1)
there
(a
(2) in
—
is
8,
this
of x
a
+
— 8, a + 8)
such that f'(x) and g'(x)
8) except, perhaps, for x = a,
an interval
(a
/
g'(x)
interval
once again,
with,
the
exist for all
possible
jc
in
exception
= a.
On the other hand, / and g are not even assumed to be defined at a. If we define
(changing the previous values of f(a) and g(a), if necessary),
f(a) = g(a) =
/ and g are continuous at a. \{a<x<a+8, then the Mean Value
Theorem and the Cauchy Mean Value Theorem apply to / and g on the interval
[a, x] (and a similar statement holds for a — 8 < 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 x\
in (a, x) with g'(x\ = 0, contradicting (2). Now applying the Cauchy Mean Value
Theorem to / and g, we see that there is a number a x in (a, x) such that
then
)
[f(x)-0]g'(a x
= [g(x)-0]f(a x
)
)
or
Now
a x approaches a
/(*)
/'(«x)
g(x)
g'iUx)
x approaches
as
assuming that lim f'(v)/g'(y)
exists,
it
because a x
a,
is
in (a,x); since
we
are
follows that
y-+a
f(x)
,.
lim
v^a g(x)
(Once again, the reader
is
=
f'(a x )
lim
x~>a g'(a x
=
)
lim
f'(y)
.
y^a g'(y)
invited to supply the details of this part of the argu-
ment.) |
PROBLEMS
1.
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.
For each of the following functions, find the
(i)
f(x)
=
x3
- x2 -
8a-
f(x)=x +x + \
4
3
f( x = 3x - 8.v +
+
1
5
(ii)
(iii)
iv
(vi)
)
/(*)
=
_-—
6.v
2
on [-2,
2].
on [-1,1].
on [-£,£].
1
XD
+
X
/W
=:
+1
Z2—T
2
x +l
f{x)
=
-^—
x —
x
L
1
+
on[-i,l],
l
1
on[-l,i].
on
[0,5].
,
206
Derivatives
and
Integrals
2.
3.
Now
sketch the graph of each of the functions in
local
maximum and minimum
1,
and
find
all
Sketch the graphs of the following functions.
i)
/(*)=* +
-.
—3
n)
f{x)
=
111
/(*)
=
LV
4.
Problem
points.
a)
/(*)
X
-
x2
1
=
1+JC
<
If a\
+
x
•
<
2"
a„, find the
minimum
value of f(x)
=
/_](x
—
«,)"
is
a problem
i=\
b)
Now find
the
minimum
value of /(jc)
= 2,
i
where calculus won't help
at
all:
on the
=
\
x
~ a i\-
This
\
intervals
between the
a,'s
the
minimum clearly occurs at one of the a
/
and these are precisely the points where / is not differentiable. However,
the answer is easy to find if you consider how f(x) changes as you pass
function
is
linear, so that the
from one such interval
Let a
>
0.
Show
(2
+ a)/( + a).
1
(—oo, 0),
5.
(0, a),
maximum
value of
I
=
1
is
+
+
IjcI
+
1
\x
—
a\
(The derivative can be found on each of the intervals
and
(a,
oo) separately.)
For each of the following functions, find
all
local
maximum and minimum
points.
i)
ii)
iii)
iv)
fix)
fix)
/(*)
fix)
=
=
=
=
x,
x^3,5,7,9
5,
x
-3,
x
9.
x
7.
x
fix)
=
3
5
7
9.
x irrational
\/q,
x
=
p/q
in lowest terms.
x,
x rational
0.
x
irrational.
x
—
0,
v)
=
=
=
=
0,
1
1,
0,
\
In for
some n
in
N
otherwise.
if
,
to another.
that the
/(*)
t
the decimal expansion of x contains a 5
otherwise.
6.
Prove the following (which
we
11. Significance of the Derivative
207
on
(a, b)
often use implicitly): If
/
is
increasing
and continuous at a and b, then / is increasing on [«,&]. In particular, if /
continuous on [a, b] and /' >
on (a,b), then / is increasing on [a, b].
is
7.
A
straight line
then back to
9(1,
when
(0,a)«
drawn from
is
(1, b), as in
the angles a and
the point (0, a) to the horizontal axis,
Figure 23.
Prove that the
are equal.
(Naturally you must bring a function
fi
into the picture: express the length in
on the horizontal
The dashed
axis.
and
total length
terms of x, where (x,0)
line in
shortest
is
is
the point
Figure 23 suggests an alternative
geometric proof; in either case the problem can be solved without actually
?''
(0,-fl), V
(JC,0)
finding the point (x, 0).)
8.
Let
(a)
(jto,
f(x)
FIGURE
(Jc,
23
yo) be a point of the plane,
= mx + b.
/(*))
its
let
the graph of the function
minimizing
same
square. This
may
this distance
is
the
as
simplify the computations somewhat.]
Also find i by noting that the line from
(b)
L be
(xq, yo) to
smallest. [Notice that
is
minimizing
and
Find the point x such that the distance from
(xo, yo) to (x,
f(x))
the distance
from
is
perpen-
dicular to L.
Find the distance from
(c)
(x, f(x)).
=
b
0;
(*0> yo
[It
~
b)-]
(Problem
(Ax
10.
Surface area
sum
is
make
+
i.e.,
the computations easier
if
then apply the result to the graph of f(x)
Compare
Consider a straight
(d)
9.
will
(xo, yo) to L,
Byo
4-7).
assume that
and the point
first
= mx
with Problem 4-22.
line
Show
described by the equation
that the distance
+ C)/y/A 2 + B 2
from
Ax + By + C =
(xo, yo) to this line
is
.
The
previous Problem suggests the following question:
ship
between the
critical
you
(xo, yo) to
points of
/ and
those of
/
What
is
the relation-
?
Prove that of all rectangles with given perimeter, the square has the greatest
area.
the
of these areas
11.
among
Find,
all
right circular cylinders of fixed
volume V, the one with
and bottom, as
smallest surface area (counting the areas of the faces at top
in Figure 24).
FKIl'RE 24
12.
A
right triangle with
hypotenuse of length a
to generate a right circular cone.
is
rotated about one of
its
legs
Find the greatest possible volume of such
a cone.
13.
14.
Show
that the
sum of a
positive
number and
its
reciprocal
Find the trapezoid of largest area that can be inscribed
is
in
at least 2.
a semicircle of
radius a, with one base lying along the diameter.
15.
Two
is
FI G U R E 2 5
hallways, of widths a
and
b,
meet
at right angles (Figure 25).
What
the greatest possible length of a ladder which can be carried horizontally
around the corner?
208
Derivatives
and
Integrals
16.
A
garden
be designed
to
is
R and
radius
what value of R and 9
shape of a circular sector (Figure
in the
The garden
central angle 6.
is
to
radians) will the length of the fencing
(in
26), with
have a fixed area A.
For
around the
perimeter be minimized?
17.
FIGURE
A
right angle
in Figure 27.
26
by the
18.
circle?
Ed must
Ecological
at
mph
2
moved along the diameter of a circle of radius a, as shown
What is the greatest possible length (A + B) intercepted on it
is
cross a circular lake of radius
What
rest (Figure 28).
much
see as
1
He
mile.
mph, or he can row
or walk around at 4
part
can row across
way and walk
the
route should he take so as to
scenery as possible?
cross as quickly as possible?
FIGURE
2 7
19.
(a)
Consider points
= LAOC
of a
AAOB
the areas of
in
(b)
terms of 9
'20.
maximum
The lower
the
and
is
ABOC
a
is
maximum?
Hint: Express things
>
3,
of all
n-gom
inscribed in a circle, the regular n-gon
area.
right-hand corner of a page
edge of the paper, as
left
the page
FIGURE
Bona circle with center O, subtending an angle
How must B be chosen so that the sum of
= LAOB.
Prove that for n
has
A and
(Figure 29).
is
folded over so that
in Figure 30. If the
very long, show that the
minimum
it
just touches
width of the paper
length of the crease
is
a and
is
3v 3or/4.
2 8
21.
Figure 31 shows the graph of the
minimum
FIGURE
FIGURE
2 9
22.
23.
3
Suppose
of /. Find
derivative
all
local
maximum and
points of /.
1
that
/
a polynomial function, f(x)
is
— 1,
=
x"
+ a„_]x"~ +
and corresponding
]
•••
+ao,
with
critical
4, 3.
Sketch the graph of /, distinguishing the cases n even and n odd.
(a)
Suppose
a
JI
0,
points
2, 3, 4,
critical
values 6, 1,2,
=
that the critical points of the polynomial function f(x)
B- 1
+
f'O) =
_iJc
1,
---
0.
+a
are-l, 1,2,3, and /"(-l)
Sketch the graph of
/
=
0,
/"(l)
>
0,
x"
+
/"(2) <
as accurately as possible
on the
basis of this information.
(b)
Does there
that 3
24.
I
I',
i
RE
30
is
exist a
not a
polynomial function with the above properties, except
critical
point?
Describe the graph of a rational function
(in
very general terms, similar to
the text's description of the graph of a polynomial function).
209
11. Significance of the Derivative
25.
(a)
most max(m,n)
intersect in at
(b)
m
For each
which
26.
and n
Suppose /
(a)
Suppose
that the
has exactly k
(b)
+
/"
+•+f
{n)
>
points
critical
m
and n
/ >
^
and f"(x)
(so n
must be
0.
—
polynomial function f(x)
for
+ an -\x
x"
all critical
+
+
•
ao
Show
points x.
odd.
is
For each n, show that
/
(c)
—k
n
that
+
/
/'
n, respectively,
times.
a polynomial function of degree n with
is
and
points.
two polynomial functions of degree
exhibit
max(m,«)
intersect
even). Prove that
*27.
m
Prove that two polynomial functions of degree
of degree n with k
— k is
n
if
odd, then there
is
a polynomial function
each of which /"
critical points, at
non-zero.
is
Suppose that the polynomial function f(x) — x" + a„-\x"~ +
+ «o
maximum points and ki local minimum points. Show that
•
•
•
has k\ local
= k\ + 1
&2
(d)
Let n,
n
is
k\,
=k\
if
n
ki be three integers with ki
—
k\
n
if
odd, and
28.
29.
M for
M for
>
<
a^^+ki
and
=
f'(x )
then
|/'(jc)|
for all x in [0, 1].
<
Show
Find
all
>
(a)
fix)
(b)
f"(x)
(c)
—
/
feet after
it
/
g'(x) for
all
x,
and
that there
=
that f(a)
—
g(a), that f'(x)
a.
Show
>
that f(x)
g'(x) for
)
+ M(b-a).
[a, £].
is
an interval
g(a).
Show
that
follow without the
>
all
x,
g(x) for
all
and
that /'(xq)
>
x > ao-
such that
is
.
true that a weight
dropped from
rest will fall s(t)
=
\6t
2
seconds, this experimental fact does not mention the behavior of
weights which are thrown upwards or downwards.
law s"(t)
+x
sin*.
= x\
f'"(x)=x+x 2
Although
(1
g(a).
some xq >
functions
of
> M/4.
f{x) > g(x) for x > a and f(x) < g(x) for x < a.
Show by an example that these conclusions do not
g'(xo) for
x in
all
(b)
—
/
f(b)<f(a) + M(b-a).
M for
Suppose
that f{a)
k\ if
1
in [a,b],
(a)
Suppose
(*~ a i)
I
=
x
/'(*)
that f'{x)
—
ki local minimum points.
ki+k2
all
if
|/|
ki
a polynomial function
then f{b) > f(a)
Prove that
>
and
in [a,b],
Formulate a similar theorem when
M
tr y
even,
is
x
(c)
>
is
n
if
all
(b)
that f'(x)
1
,
f'{x)
Suppose
+
/.
if
hypothesis f{a)
32.
<
•
odd.
and
points
Prove that
(a)
(c)
31.
•
•
is
that there
maximum
an appropriate number
of length \ on which
30.
<
ai
Show
n.
k\ local
<
Hint: Pick a\
for
+ kj <
k\
degree n, with
even, and kj
is
=
32
is
On
the other hand, the
always true and has just enough ambiguity to account for
from any height, with any initial velocity.
measure heights upwards from ground level;
the behavior of a weight released
For simplicity
let
us agree to
in this case velocities are positive for rising
bodies,
and
all
bodies
fall
bodies and negative for falling
according to the law s"(t)
=
—32.
210
Derivatives
and
Integrals
Show
(a)
that s
= — \6t 2 +at +
of the form s(t)
is
(3.
By setting t =
in the formula for s, and then in the formula for s',
show that s(t) = —I6t 2 + vtf + sq, where so is the height from which the
body is released at time 0, and vq is the velocity with which it is released.
(b)
A
(c)
weight
thrown upwards with
is
How
ground
velocity v feet per second, at
("How high" means "what is the maximum
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
level.
high will
go?
it
again?
33.
A
cannon
ball
zontal
component of
has a vertical
it
component
— — 16?" +
s(t)
shot from the ground with velocity v at an angle
is
ure 32) so that
a
a and a
(Fig-
hori-
above the ground obeys the law
v cos a. Its distance s(t)
(vsina)t, while
velocity v sin
horizontal velocity remains constantly
its
v cos a.
Show
(a)
each time
l'Kil'RE
t,
and show
by the cannon
3 2
34.
will
ball before striking the
/
ground.
(b)
for
which
lim f(x)
~
= 0.
= 0.
/ and g are two differentiable
Prove that if f(a) =
and g(a) ^
that
f's
=
0.
an interval around
that
it
is
Suppose
a. Hint:
any
interval
— f(y)\ < \x — y\" for n >
f. Compare with Problem 3-20.
that \f(x)
A function
/
is
Lipschitz of order
an
v in
interval if (*)
where f/g
defined,
interval
holds for
aatx
if
there
1.
is
Prove that
/
is
a constant
around
all
x.
The
function
x and y in the
/
C
is
show
is
is
a
on an
interval.
Lipschitz of order
differentiable at x, then
constant by
Lipschitz of order
If
Lipschitz of order
/'
is
/
Lipschitz of order
1
at x.
is
uniformly
at x.
Is
the
converse true?
(d)
(e)
If
[f
/
/
x
a
(c)
/
all
such that
(b)
/
/
satisfy
for
is
a > at x, then / is continuous
on an interval, then /
If
is Lipschitz of order a >
continuous on this interval (see Chapter 8, Appendix).
If
-
then f(x)
0,
\f(x)-f(y)\<C\x-y\
(*)
all
On
which
functions
constant.
considering
(a)
but
exists,
exist, then lim f'(x)
x—*oo
;c—>oo
Jt—>00
Prove that if lim f(x) exists and lim f"(x) exists, then lim f"(x)
-V—>oc
x—>oo
x—*oo
(See also Problem 20-22.)
Suppose
for
(find the position at
lie on a parabola).
maximize the horizontal distance traveled
lim f'(x) does not exist.
x—>oo
Prove that if lim f(x) and lim f'(x) both
in
37.
a parabola
Give an example of a function
fg'
36.
is
(a)
(c)
35.
ball
that these points
Find the angle a which
(b)
cannon
that the path of the
on
is
(li(lcrential)le
is
Lipschitz of order
[<:/./?], is
a-
>
1
on
|<7./;>|,
then
/'
is
1
on
[a,fc]?
constant on \a,h\.
11
38.
Prove that
Significance
.
211
of the Derivative
if
an
2
1
n
=
+
0,
1
then
«o
for
39.
some x
+ «!•* +
•
•
•
+ a„x" -0
in (0, 1).
3
fm (a) = x — 3x + m
Prove that the polynomial function
40.
never has two roots
no matter what m may be. (This
Theorem. It is instructive, after giving an analytic proof, to graph fo and
and consider where the graph of /„, lies in relation to them.)
/2,
continuous and differentiable on
in
in [0, 1],
Suppose
an easy consequence of Rolle's
is
/
that
is
each x, and that f'(x)
[0, 1] for
exacdy one number x
^
1
for
x
all
in [0, 1].
=
such that f(x)
in [0, 1]
[0, 1], that
Show
(Half of
x.
f(x)
is
that there
this
is
problem
has been done already, in Problem 7-11.)
41.
(a)
Prove that the function f(x)
two numbers
(b)
= x — cos a
satisfies
f(x)
=
for precisely
x.
= x — x sin x — cosx.
f(x) — 2x — xsinx — cos
Prove the same for the function f(x)
Prove
*(c)
this also for the
function
preliminary estimates will be useful to
(Some
x.
restrict the possible location
of the
zeros of /.)
*42.
(a)
Prove that
/(l)
In
=
/
if
=
and f'(0)
1
=
a twice differentiable function with /(0)
is
more picturesque
f'{\)
=
0,
terms:
A
particle
> 4
then \f"(x)\
which
for
some x
and
in (0, 1).
travels a unit distance in
some time an
some x in (0, I),
a unit time, and starts and ends with velocity 0, has at
>
acceleration
< —4
or else f"(x)
(b)
43.
Show
Suppose
0.
that
/
is
=
for
=
f(x)
that
/
some function
interval
Suppose
g.
\f"(x)\
f(y) for
all
> 4
=
for
some x
l/.v for all
x, y
>
0.
x
in (0, 1).
>
and /(
Hint: Find g'(x)
=
1 )
when
+ f'(x)g(x) - f(x) =
Prove that
between them. Hint:
that
/
is
/ is at two
Use Theorem 6.
if
continuous on [«,b], that
and that f(x)
some x in (a, b).
(a, b),
for
in (j, 1).
satisfies
f\x)
45.
+
for
f(xy).
Suppose
for
some x
we must have
a function such that f'(x)
Prove that f(xy)
g(x)
44.
that in fact
>4
Prove that either f"(x)
4. Hint:
=
for
n+
1
it
on
the
//-times differentiable
on
points, then
is
/
different x in [a, b]. Prove that
is
/
(n)
(jc)
=
212
Derivatives
and
Integrals
46.
Let x\
,
.
.
.
x n+ be arbitrary points
,
=
Q(x)
/
that
(n
is
+
let
Y\(x-Xi).
i
Suppose
and
in [a, b],
\
=
]
and
l)-times differentiable
that
< n such that P(xj) = f{xi) for
Show that for each x in [a,b] there is
function of degree
Problem
3-6).
P
=
i
a
is
a polynomial
n
1
number
+
1
(see
c in (a,&)
such that
=
f(x)-p(.x)
Jf
Q(x)-
in+l)
(»
(c)
\;
+
.
1)!
Hint: Consider the function
F(0 = G(*)[/(0 - P(0] - G(0[/O0 -
Show
47.
that
F
is
+2
zero at n
different points in [«,&],
(without computing
Prove the following
is
v 66
66-8 <
<
decimal
to 2
is
some x
in (a, b)
lim /(y)
...
-
is
a
trivial
on
are never simultaneously
Prove that
g'(x)
if
^
/ and
f(x)
"
*'(*)'
g are continuous on
g'ix)
Hint: Multiply out
What
is
wrong with
3
Jt
,.
lim -^
jc-i
(The
first,
limit
is
x
2
to see
^
g(a) and that f'(x) and g'(x)
(a,b).
for x in (a, b), then there
fix)
51.
Value
Mean Value Theorem can be written
under the additional assumptions that g(b)
and
Mean
form
f(b)-fja)
g(b)-g(a)
50.
consequence of the
".)
Prove that the conclusion of the Cauchy
in the
lim f(y)
v—>a +
'
(Your proof should begin: "This
Theorem because
Mean Value Theorem: If /
and lim f(y) and lim /(v) exist,
of the
(a, b)
such that
y—>b
fix) =
o
places!).
slight generalization
continuous and differentiable on
then there
49.
and use Problem 45.
Prove that
9
48.
P(x)].
__
"
f(x)
-
and differentiable on
some x in (a, b) with
[a, b]
is
f{a)
g(b)-g(x)'
what
this really says.
the following use of l'Hopital's Rule:
+x-2
-3a-
+
actually —4.)
2
=
,.
lim
3v 2 +l
—
— =
x^l 2jc-3
,.
lim
x->1
6.v
—
- =
2
„
3.
(a,b),
.
11. Significance of the Derivative
52.
Find the following
(i)
lim
213
limits:
.
x^O tan x
2
cos *
(11)
x
.r^O
53.
—
1
=
lim
Find /'(0)
l
if
g(x)
,
x
#0
x
=
/(*)
0,
=
and g(0)
54.
g'(0)
=
=
and g"(0)
0,
17.
Prove the following forms of l'Hopital's Rule (none requiring any essentially
new
(a)
reasoning).
If
-
lim f(x)
x — «+
lim f(x)/g(x)
=
lim g(x) = 0, and lim f'(x)/g'(x)
x—>a +
x—« +
I (and similarly for limits from below).
=
/,
then
x—*a +
(b)
= lim g(x) = 0, and lim f'(x)/g'(x) = oo, then
x->a
x—>a
= oo (and similarly for — oo, or if x —> a is replaced
lim f{x)
x—>a
lim f{x)/g(x)
x—*a
If
by x —> a + or x — a~).
»•
(c)
lim /(*) =
X—>00
lim f(x)/g(x)
If
lim g(x)
X—>O0
— I (and
=
and
0,
lim f'{x)/g'(x)
X—>00
for — oo).
Hint:
similarly
=
/,
then
Consider
lim f(\/x)/g(l/x).
(d)
lim f(x) =
lim g(x)
x—>oo
x—>oo
lim f(x)/g(x) = oo.
=
If
and
0,
lim f'(x)/g'(x)
=
oo,
then
.*—>oo
x -> oo
55.
There
is
another form of l'Hopital's Rule which requires more than algebraic
manipulations:
If
lim f(x)
=
x-+oo
then lim f{x)/g{x) —I. Prove
lim g(x)
x—>oo
=
oo,
and lim f'(x)/g'(x)
=
I,
x—yoo
this as follows.
x-»-oo
(a)
For every e
>
there
is
a
number a such
/'OO
/
<
for
e
jc
that
>
a
*'(*)
Apply
the
Cauchy Mean Value Theorem
to
/ and g on
x
>
that
f(x)-f(a)
g(x)-g(a)
(Why can we assume g(x) —
I
g(a)
<
^
e
0?)
for
a.
[a,x] to
show
214
Derivatives
and
Integrals
Now write
(b)
fix)
fix)- fid)
fix)
g(x)
g(x)-g(a)
f(x)-f(a)
(why can we assume that f(x)
—
gjx)
f(a)
-
gja)
Six)
for large x?)
7^
and conclude
that
—
/
< 2e
for sufficiently
y
Six)
56.
lame
*
x.
To complete the orgy of variations on l'Hopital's Rule, use Problem 55 to
prove a few more cases of the following general statement (there are so many
you should
possibilities that
If lim
lim g(x)
f(x)
*-»•[
select just a few, if any, that interest you):
=
}
and lim f'(x)/g'(x)
),
(
then
lim
]
1
Here [ ] can be a or a + or a or 00 or —00, and
).
fix)/gix) — (
can be
or 00 or —00, and
can be / or 00 or —00.
(
57.
If
/ and g
are differentiable
{
}
)
and lim f(x)/g(x)
exists,
does
follow that
it
x—ya
lim f'(x)/g'(x) exists
(a
converse to l'Hopital's Rule)?
X—>fl
58.
Prove that
of
/
/'
if
increasing, then every tangent line of
is
only once. (In particular,
this
is
/
intersects the
true for the function f(x)
=
x"
graph
if
n
is
even.)
59.
Redo Problem 10-18
(c)
when
1
iff = f
(Why
*60.
(a)
is
Suppose
(c)
is
this
chapter?)
differentiable
on
[a b]
,
.
Prove that
if
the
minimum
/'
part
61.
/
that
is
2
on [a, b] is at a, then /'(a) > 0, and if it is at b, then fib) < 0.
(One half of the proof of Theorem 1 will go through.)
and f'{b) > 0. Show that fix) = for some x
Suppose that f'ia) <
in (a, b). Hint: Consider the minimum of / on [a, b]; why must it be
somewhere in (a, b)?
Prove that if f'{a) < c < fib), then fix) = c for some a: in (a, b). (This
result is known as Darboux's Theorem. Note that we are not assuming
that /' is continuous.) Hint: Cook up an appropriate function to which
of
(b)
problem
this
T
Suppose
(b)
may
that
/
be applied.
is
differentiable in
some
interval containing a, but that /'
is
discontinuous at a. Prove the following:
(a)
(b)
The
one-sided limits lim fix) and lim fix) cannot both
x—>« 4
x—ya~
is just a minor variation on Theorem 7.)
exist.
(This
These one-sided limits cannot both exist even if we allow limits with
value +00 or —00. Hint: Use Darboux's Theorem (Problem 60).
the
11
^62.
For example,
x
then
/
is
f(a)
=
0.
(a)
that |/|
differentiable but
is
215
of the Derivative
/
is
not.
we can choose f{x) — 1 for a rational and f(x) = —1 for
/ is not even continuous, nor is this a mere
Prove that if |/| is differentiable at a, and / is continuous at a,
In this example
irrational.
coincidence:
^63.
/ such
easy to find a function
It is
Significance
.
Hint:
also differentiable at a.
Why?
suffices to
It
consider only a with
In this case, what must \f\'(a) be?
/ and let n be even. Prove that x" + y" = (x + y)" only
when x — 0. Hint: If A'o" + y" = (a'o + y)", apply Rolle's Theorem to
f(x) = x" + y" - (x + y)" on [0, x ].
Prove that if y /
and n is odd, then x" + y" = (x + y)" only if x =
or x = —y.
Let y
()
(b)
64.
Suppose
f(x)/x
that f(0)
increasing
is
Prove that
and /'
on
(it
remember
as
Use
derivatives to prove that
+ a)" >
1
Let f{x)
= a 4 sin 2
1
+ nx
for
(a)
Prove that
(b)
Prove that f'(0)
is
^
fx for a
a local
=
>
n
if
=
(notice that equality holds for a
66.
=
the function f(x)
tial,
(1
Hint: Obviously
(0, oo).
will help to
shown by
increasing. Prove that the function g(x)
is
by applying the
positive
it is
right interval
65.
=
0,
=
you should look
Value Theorem
that the hypothesis /(0)
1
+
to
=
at g'{x).
/ on
is
the
essen-
a").
then
1,
- <
<
a
1
<a
and
0).
and
minimum
/"(0)
Mean
=
/(0)
let
=
(Figure 33).
point for /.
0.
Theorem 6 cannot
about maxima and minima that
This function thus provides another example to show that
be improved.
It
also illustrates a subtlety
often goes unnoticed: a function
right of a local
minimum
may not be
increasing in any interval to the
point, nor decreasing in
1
any interval
to the
left.
i
1
h
\
/
/
\
\
/
\
FIGURE
*67.
(a)
Prove that
some
The
(b)
if
f'(a)
>
/
3 3
and /'
is
continuous
at a,
then
/
is
increasing in
interval containing a.
next two parts of this problem show that continuity of /'
If #(a)
=
with g'U)
a2
=
sin 1/a,
1
and
show
that there are
also with g'(A)
=—
numbers a
1.
is
essential.
arbitrarily close to
216
Derivatives
and
Integrals
(c)
< a <
Suppose
/(0)
=
(see
interval containing 0,
with fix)
FIGURE
3
>
and
= ax +
Let f(x)
1.
Show
that
/
by showing that
in
Figure 34).
also points
jc
is
sin
1/x for x
^
0,
and
let
not increasing in any open
any interval there are points x
x with fix) <
0.
4
The behavior of /
for
a >
1,
which
much more
is
difficult to
analyze,
is
discussed in the next problem.
**68.
Let f(x)
= ax + x 2 s'm
the sign of /' (jc )
is
< —1
\/x for
when a >
—
g(y)
< —1
part of g(y)
when
< —
1
sin y
.
=
— cos_y,
0,
and
it
is
not at
The obvious approach
(a)
Show
more
that
if
ingenuity
g'(y)
let
=
0,
cosy
It
—
is
is
a
all
to this
^
In order to find
sin
\/x
—
cos
/x
1
more convenient to
0; we want to know
— 1,
but
this
happens only
whether g itself can have values
problem is to find the local minimum
impossible to solve the equation g'(y)
is
is
required.
then
=
2x
quite delicate; the most significant
2
-
2
y
(sin y)
that
g(y)
0.
if
clear
it
=
=
little
cos y for y
2y
and conclude
/(0)
which does reach the value
values of g. Unfortunately,
explicitly, so
to 0.
2(sin y)/y
for large y. This question
is
and
necessary to decide
any numbers x close
for
consider the function g(y)
if
x^O,
it is
1
2+
(sin v)
2y
v-
—
,
//
Now
(b)
show
=
that if g'(y)
and conclude
217
then
0,
2
4y
sm"
of the Derivative
Significance
.
y
4
+
2
+y
4
v
that
:
\g(y)\
V4 +
Using the
4
+ y 2 )/ v 4 +
fact that (2
v
>
y
not increasing in any interval around
Using the
(d)
+y
lim (2
fact that
y->oo
/
**69.
A
is
increasing in
/
function
some
if
there
a
< f(a)
if
a
that
/
and
/(•*)
Notice that
does
this
mean
not
shown
a +8); for example, the function
is
a
=
that
if
if
1
,
then
/
is
show
1,
a >
1,
then
0.
some number
x
<
—8 <
is
=
v
is
<
if
that
0.
around
f(x) > f(a)
show
1,
+
)/v 4
interval
increasing at a
is
2
4
such that
+8
a
<
x
>
8
a.
increasing in the interval (a
Figure 34
in
is
—
8,
increasing at 0, but
not an increasing function in any open interval containing 0.
Suppose
(a)
/
that
every a in
continuous on
is
Prove that
[0, 1].
yourself that there
prove that the
Prove part
(b)
(a)
minimum
If
(c)
/
[b, 1]
=
the set
1]
problem
{x
:
b
increasing at a
is
increasing at a for
(First
[0, 1].
convince
< b <
For
1
at b.
continuous, by consider-
Si, = [x
f(y) > f(b) for all y in [b, x] }.
not necessary for the other parts.) Hint:
:
< x < l}by
and /
is
is
Hint:
must be
/
/
on
be proved.)
to
/ on
of
that
increasing
without the assumption that
(This part of the
is
something
is
ing for each b in [0,
Prove that Sb
/
and
[0, 1]
is
is
considering sup
Si,.
>
differentiable at a, prove that f'(a)
(this is easy).
If f'{a)
(d)
>
0,
increasing at a (go right back to the definition
prove that
/
and
show, without using the
is
of /'(a)).
(e)
Use
parts
that
if
is
/
(a)
(d) to
continuous on
is
increasing
on
[0, 1]
and f'(a) >
/(0)
ex
-
> —e.
all
Value Theorem,
a
in [0, 1],
then
/
[0, 1].
Suppose that / is continuous on [0,
Apply part (e) to the function g(x)
(f)
Mean
for
Similarly,
show
1]
and f'{a)
=
f(x)
that /(l)
f(x). Conclude that /(0)
=
—
=
for
all
a
in (0,
1
).
ex to show that /(l)
—
f(0) < s by considering h(x)
=
+
/(l).
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
first
rigorous proof ever given.
See
his
exuberant
letter in
Its
discoverer, at least,
seemed
may
to think
reference [54] of the Suggested Reading.
be the
it
was.
218
Derivatives
and
Integrals
** 70.
/
If
is
a constant function, then every point
for /. It
for example, if f{x) =
But prove, using Problem 8-4,
function:
0.
is
maximum
a local
point
happen even if / is not a constant
for x <
and f{x) — 1 for x >
that if / is continuous on [a,b] and
quite possible for this to
is
is a local maximum point, then / is a constant
The same result holds, of course, if every point of [a,b] is a
local minimum point.
Suppose now that every point is either a local maximum or a local minimum point for the continuous function / (but we don't preclude the possibility that some points are local maxima while others are local minima).
Prove that / is constant, as follows. Suppose that /(ao) < fibo)- We
can assume that /(ao) < /(*) < f(bo) for ao < x < bo. (Why?) Using
Theorem 1 of the Appendix to Chapter 8, partition [ao, bo] into intervals
on which sup / — inf / < if {bo) — f(ao))/2; also choose the lengths of
these intervals to be less than (bo — ao)/2. Then there is one such interval
[ai,&l] with ao < a\ < b\ < bo and f(a\) < f{b\). (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.
every point of [a,b]
function.
(b)
A point x
** 71.
for
all
is
v in
maximum
A
^
with y
A
local strict
all
local strict
point).
obvious way. Find
=
x
=
q
It
maximum
maximum
point
if
defined in the
is
points of the function
x irrational
0,
f(x)
maximum point for / on A
f{x) > f(y)
x (compare with the definition of an ordinary
called a strict
P-
in lowest terms.
q
seems quite unlikely that a function can have a
at every point (although the
maximum
local strict
above example might give one pause
for
thought). Prove this as follows.
(b)
Suppose that every point
x\
is
a local
be any number and choose
that f{x\)
(a\, b\)
> fix)
and choose
for
a\
all
<
a\
strict
<
x\
x in [ai,&i].
aj
<
xj
<
&2
<
maximum
<
point for /.
b\ with b\
/
x\
b\ with bi
—
Let xj
—
<
a\
1
Let
such
be any point in
«2
<
i
such that
f(xi) > fix) f° r a H x m [ fl 2>^2]- Continue in this way, and use the
Nested Interval Theorem (Problem 8-14) to obtain a contradiction.
219
11, Appendix. Convexity and Concavity
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 subde aspects of the graph are
revealed only by examining the second derivative. These details were purposely
omitted previously because graph sketching
complicated enough without wor-
is
rying about them, and the additional information obtained
effort. Also,
in
often not worth the
is
correct proofs of the relevant facts are sufficiently difficult to be placed
an appendix. Despite these discouraging remarks, the information presented
is well worth assimilating, because the notions of convexity and concavity are
here
far
more important than
as
mere
graph sketching. Moreover, the proofs
aids to
have a pleasantly geometric flavor not often found
the basic definition
DEFINITION
1
A function /
is
The geometric
analytic way that
(a,
(b,
geometric in nature
convex on an
segment joining
f{a)) and
is
(a,
(b,
f{b))
fib))
is
line lies
and b in the interval, the
above the graph of /.
an
between
can be expressed
The
in proofs.
straight line
line
in
the graph of the function g defined by
= f(b)-f(a) (x -a) + fia).
—
/
above the graph of /
a
at
x
if
g(x)
>
f(x), that
is,
if
f(b))
fib)
b
(a,
theorems. Indeed,
1).
this definition
sometimes more useful
b
This
lies
condition appearing in
is
in calculus
Figure
interval, if for all a
f(a)) and
g(x)
(b,
(see
~
—
fia)
(x-a) + fia)> f(x)
a
fia))
or
fib)
b
—
f{a)
(x
a
-a) > fix)-
fia)
or
FIGURE
f(b)
I
b
We
DEFINITION
2
A
a
—
fia)
a
x) > f(
x —
fja)
.
a
therefore have an equivalent definition of convexity.
function / is convex on an
< x < b we have
f(x)
x
—
interval if for a, x,
fia)
a
<
fib)
-
and b
fia)
in the interval
with
220
Derivatives
and
Integrals
If the
word "over"
in Definition
inequality in Definition 2
fix)
(b,f(b))
x
we
replaced by "under"
is
1
obtain the definition of a
—
f{a)
f(b)
>
b
a
concave
—
this reason, the
FIGURE
state
2
the
f(a)
,
function (Figure
not hard to see that
2). It is
form
— /, where /
is
convex.
next three theorems about convex functions have immediate
about concave functions, so simple that we
corollaries
if
a
the concave functions are precisely the ones of the
For
equivalently,
or,
replaced by
is
will
not even bother to
them.
Figure 3 shows
some tangent
lines
of a convex function.
Two
things
seem
to
be
true:
(1)
The graph
[a,
f{a))
(2) If a
<
b,
of
itself (this
As a matter of fact
THEOREM
l
Let
lies
above the tangent
point
is
line at (a,
line at (b, f{b))\ that
is,
/'
these observations are true,
is
(a,
/ be
<
convex.
If
/
is
h\
<
ti2,
and
difficult.
b,
then f'(a)
<
graph of /
If
itself.
a
lies
above
< b and /
is
f'(b).
then as Figure 4 indicates,
...
a
line).
than the slope
and the proofs are not
differentiable at a, then the
fia+hO-
f(a)
(1)
A
less
3
differentiable at a
If
f{a))
is
increasing.
the tangent line through (a, /(«)), except at (a, f(a))
PROOF
f(a)) except at the point
point of contact of the tangent
called the
then the slope of the tangent line at
of the tangent
FIGURE
/
<
f(a
+
h 2 )-f(a)
.
nonpictorial proof can be derived immediately from Definition 2 applied to
< a
+ h\ <
a
+
hj. Inequality (1)
shows
that the values
f(a+h)-f(a)
h
of
11, Appendix. Convexity and Concavity
a
KKiURE
+
a
+ hi
Consequently,
.
f'(a)
for h
h\
4
decrease as h
(in fact
+
221
f'(a)
>
is
<
f(a+h)
f(fl)
for h
>
bound of all these numbers). But this means that
through (a, f{a)) and {a + h, f(a + h)) has larger slope
the greatest lower
the secant line
than the tangent
line,
which implies
line (an analytic translation of this
For negative h there
is
f(a
that (a
+ h,
argument
is
f(a
>
This shows that the slope of the tangent line
5):
+ h 2 )-
f(a
lies
above the tangent
easily supplied).
a similar situation (Figure
+ hi)-f(a)
+ h))
if
<
hi
h\
<
0,
then
f(a)
.
is
greater than
f(a+h)-f(a)
lor n
< U
h
(in fact
f'{a)
is
above the tangent
FIGURE
5
upper bound of all these numbers), so that f(a + h)
if h < 0. This proves the first part of the theorem.
the least
line
lies
222
Derivatives
and
Integrals
FIGURE
Now
6
suppose that a <
j (a)
<
Then,
b.
f(a
we have
as
already seen (Figure
+ (b-a))-f(a)
b
fib)
b
—
.
since b
—
a
> U
—
b
<
6),
a
- fid)
—a
and
(b)
/
f(b
>
+ (a-b))-f(b)
.
since a
—b
f(a)-f(b)
= f(b)-fja)
—
a
b
b — a
(J
a
_
Combining
a x\
IE
these inequalities,
we
obtain f'{a)
<
f'(b). |
xq
Theorem
7
We
Theorem
is
1
Here the proofs will be a little more difficult.
same role in the next theorem that Rolle's
proof of the Mean Value Theorem. It states that if /'
graph of / lies below any secant line which happens to be
has two converses.
begin with a
lemma
plays in the
increasing, then the
that plays the
horizontal.
lemma
Suppose /
is
fix) < f{a)
proof
and /' is
a < x < b.
differentiable
=
f(b) for
a
increasing. If
<
b and f(a)
=
f(b), then
Suppose that f{x) > fid) = f(b) for some x in (a,b). Then the maximum of
/ on [a,b] occurs at some point A'o in (a, b) with /(a'o) > fid) and, of course,
/'(a'o)
=
(Figure
7).
the interval [a, xq\,
we
On
the other hand, applying the
find that there
fl(
f
.
U'i
)
is
x\ with a
= /Up)
A'o
contradicting the
fact thai
./
"
is
increasing.
|
fja)
a
<
x\
>
0,
Mean
<
Value Theorem
xq and
to
223
11, Appendix. Convexity and Concavity
We now attack
we used
THEOREM
2
PROOF
If
/
Let a
proof of the
in the
and
differentiable
is
<
same sort of algebraic machinations
Mean Value Theorem.
the general case by the
/'
f(b)-f(a)
=
-
fix)
b
the
easy to see that g'
lemma
to g
is
In other words,
if
<x <
a
f(x)
x
8
THEOREM
3
If
/
is
convex.
differentiable
Let a <
b. It
ia, fia)),
is
and
clear
ia,
=
g{b)
—
f(a). Applying
< fia)
<x <
a
if
b.
fib)
b
-
—
fia)
fia)
—a
-
ix
a)
<
a
<
fib)
b
—
fia)
fia)
a
|
and the graph of /
point of contact, then
PROOF
a).
then
b,
or
is
-
f
(x
a
that
f(x)~
Hence /
—
also increasing; moreover, g(a)
we conclude
g(x)
FIGURE
convex.
is
Define g by
b.
g(x)
It is
/
increasing, then
is
/
above each tangent
line
except at the
convex.
is
from Figure 8 that
fia))
lies
lies
if ib,
above the tangent
fib))
lies
above the tangent
line at ib, fib)),
g(x)
since ib, fib)) lies
line,
Similarly, since the tangent line at ib, fib))
and
ia,
fia))
lies
It
follows
It
now
from
(1)
and
follows from
If a function
/
(2)
is
fib),
line at ib, fib)),
fia)> f'ib)ia-b) +
that
fia).
the graph of
= f(b)ix-b) +
above the tangent
(2)
we have
fib)> f'ia)ib-a) +
h( x )
line at
= f\a)(x-a) + f(a),
above the tangent
(1)
line at
then the slope of
must be larger than the slope of the tangent
ia, fia)). The following argument just says this with equations.
Since the tangent line at ia, fia)) is the graph of the function
the tangent line at ib, fib))
and
that
we have
fib).
fia) < fib).
Theorem 2
that
/
is
convex.
|
has a reasonable second derivative, the information given in these
theorems can be used
to discover the regions in
which /
is
convex or concave.
224
Derivatives
and
Integrals
Consider, for example, the function
/(*)
For
=
l+x 2
this function.
=
fix)
Thus
f'(x)
=
only for x
= 0,
-2x
a+x
and /(0)
=
'
2 2
1,
fix) >
if
x
<
if
a-
/'(*)
)
while
<
>
0,
0.
Moreover,
/U) >
f(x)
/
FIGURE
is
for all x,
—>
as x
of
/
even.
(l+.x-
2 2
)
(-2)
+ 2.v(1+JC
2(3jc
2
-
(1+JC
[2(1
1/3 or —y/\/3. Since /"
on each of the
We now
+.v 2
)
-
compute
2.vJ
2 4
)
1)
2 3
)
not hard to determine the sign of f"(x). Note
=y
— oo,
therefore looks something like Figure 9.
fix) =
x
oo or
9
The graph
It is
—>
is
first
clearly continuous,
sets
(-00,-/173),
(VlAoo).
that f"(x)
it
=
only
when
must keep the same
sign
Since
we
compute, for example, that
easily
r(-D =
/"(0)
>
= -2 <0,
4
we conclude
>0,
that
/" >
on (-oo, -yf\j?> ) and
/" <Oon(- v/T73, /I73).
>
Since /"
(—oo,
means
— vl/3 and
)
FIGURE
225
and Concavity
11, Appendix. Convexity
/'
is
increasing,
it
(y/V/3, oo),
follows from
(vl/3, oo), while on
(
Theorem 2
— vl/3, v
1/3
)
f
is
that
/
is
convex on
concave (Figure
10).
10
Notice that at (y 1/3, |) the tangent line
right, since
/
lies
below the part of the graph
convex on (y 1/3, oo), and above the part of the graph
is
to the
to the
left,
/ concave on (—-y/1/3, vl/3 ); thus the tangent line crosses the graph. In
general, a number a is called an inflection point of / if the tangent line to the
since
is
graph of /
at (a,
points of fix)
that a
/
is
is
an
=
1/(1
+x
2
).
Note
convex, so the tangent line at
/" has
that the condition f"{a)
inflection point of /; for
order to conclude that a
that
is
an
(0,
illustrates the
example,
if
f(x)
=
=
are inflection
does
not
x 4 then /"(0)
,
ensure
=
0,
but
0) certainly doesn't cross the graph of /. In
inflection point of a function /,
different signs to the left
This example
vl/3 and —vl/3
f(a)) crosses the graph; thus
and
we need
to
know
right of a.
procedure which
may
be used to analyze any func-
graph has been sketched, using the information provided by /',
/"
the zeros of
are computed and the sign of /" is determined on the intervals
between consecutive zeros. On intervals where /" >
the function is convex;
tion /. After the
on
intervals
where /" <
convexity and concavity of
the function
is
concave.
Knowledge of the
regions of
/ can often prevent absurd misinterpretation of other
data about /. Several functions, which can be analyzed in this way, are given in
the problems, which also contain some theoretical questions.
226
Derivatives
and
Integrals
To round out our discussion of convexity and concavity, we will prove one
result that
you
may already
We
have begun to suspect.
concave functions have the property that every tangent
few drawings
just once; a
this property,
THEOREM
4
If
/
is
once, then
PROOF
/(c)-
/
either
is
There are two
know
on an interval and intersects each of
convex or concave on that interval.
its
tangent lines just
parts to the proof.
(1) First
we
Suppose, on the contrary, that some straight
/
graph
rather tricky.
is
points.
of
line intersects the
probably convince you that no other functions have
will
but the only proof I
differentiable
further
have seen that convex and
claim that no straight line can intersect the graph of
at {a, /(a)), (b,
f{b)) and
<
/(c)), with a
(c,
/
in three different
line did intersect the
<
b
c (Figure 11).
graph
Then we
would have
f(b)-fia)
(])
b
—
f(c)-f(a)
a
c
—a
Consider the function
g(x)
Equation
FIGURE
11
(1)
=
where
in (b, c)
g'{x),
f(a)
for
=
says that g(b)
-
fix)
=
So by
g{c).
Rolle's
x
in [b, c].
Theorem, there
is
some number x
and thus
0=(x-a)f'(x)-[f(x)-f(a))
or
fix)
= fix)-
—
.v
But
this says (Figure 12) that the
f{a)
a
tangent line at (x, f(x)) passes through
(a, f{a)),
contradicting the hypotheses.
(2)
Suppose that ao < bo <
.v,
y,
zt
FIGURE
12
Then
and
=
ao an d x\
,
step
g(t)
t)ciQ
-t)b
-t)c
{\
(1
<
<
b\
+ ta\
+ tb\
+ tc\
c\
are points in the interval. Let
0<t
<
1.
y,
<
for
0<t <
between oq
x, all lie
1.
consider the function
=
,
git)
By
-
(1
a\
a\
<
x,
Now
=
=
=
and
and (Problem 4-2) the points
with analogous statements for y, and z.t- Moreover,
xo
a\,
=
co
<
(1),
for
g(l)
^
all
in
/
f(y
t
)-f(x
)
f(z,)-f(x
r
)
for
for
[0,
t
1
all
].
/
in
[0, 1|.
Thus, either
So
/
is
either g(t)
>
convex or /
is
()</<!.
lor all
concave.
in
/
|
[0. 1|
or
227
11, Appendix. Convexity and Concavity
PROBLEMS
1.
and concavity and points of inflection,
Problem 11-1 (consider (iv) as double starred).
Sketch, indicating regions of convexity
the functions in
2.
Figure 30 in Chapter
3.
Show
/
that
interval
4.
(a)
is
(c)
convex on an interval
is
+
-t)y) <tf(x)
(\
will
if
be
/ and g
(a)
Suppose
either
(c)
+
if
for all
x and y
for0<f <
-t)f(y),
(\
are convex
easiest to use
/
that
is
and /
Problem
(a)
is
in the
1.
fog
increasing, then
is
convex.
3.)
/
by considering second
is
/
is
decreasing to the
derivatives.
and convex on an
differentiable
is
increasing, or else
/
such that
(b)
and only
Give an example where g o f is not convex.
Suppose that / and g are twice differentiable. Give another proof of the
result of part
5.
if
Sketch the graph of /.
/'.
just a restatement of the definition, but a useful one.)
Prove that
(It
(b)
shows the graph of
1
we have
f(tx
(This
1
interval.
decreasing, or else there
is
of c and increasing to the right of
left
Use this fact to give another proof of the result in Problem
and g are (one-time) differentiable. (You will have to be a
when comparing f'(g(x)) and f'igiy)) for x < y.)
Prove the result
in part (a)
Show that
a number c
without assuming
/
when /
4(a)
little
differentiable.
c.
careful
You
will
have to keep track of several different cases, but no particularly clever
ideas are needed. Begin
/
is
by showing that
increasing to the right of b;
and
if
if
<
a
fia) >
b and f(a)
fib), then
/
<
fib), then
is
decreasing
to the left of a.
*6.
/ be
f{x) >
fix) =
Let
a twice-differentiable
x
and /
0,
some x >
for
tion point at
in this
x >
for
is
is
essential, as
=
function with the following properties:
decreasing,
(so that in
— an example
theorem
is
reasonable cases
given by fix)
shown by fix)
=
and f'iO)
=1/(1
=
1
/
will
-\-x )).
—x
,
Prove that
0.
have an
inflec-
Every hypothesis
which
is
not positive
=
2
all x; by fix)
l/(x + 1),
x which is not decreasing; and by fix)
with f'ixo) < 0. We
which does not satisfy /'(0) = 0. Hint: Choose xq >
for
,
cannot have f'iy) < f'ixo) for
some x\ > xq. Consider /' on
*7.
(a)
(b)
all
y
>
xo.
Why
not? So f'(x\
)
>
f'ixo) for
[0, x\\.
/ is convex, then fi[x + y]/2) < [fix) + fiy)]/2.
Suppose that / satisfies this condition. Show that f(kx + (1 — k)y) <
Prove that
if
and 1,
kfix) + (1 — k)fiy) whenever k is a rational number, between
=
of the form m/2" Hint: Part (a) is the special case n
1. Use induction,
.
employing part
(c)
Suppose
Show
that
that
/
(a)
/
is
at
each
satisfies
convex.
step.
the condition in part
(a)
and /
is
continuous.
.
228
Derivatives
and
.
Integrals
"8.
For n
>
1, let
p„ be
p\
numbers with Y^
positive
Pi
—
1-
n
(a)
For any numbers x\,
and the
(b)
Show
.
.
show
x„
,
V]
that
i=l
largest x,.
the
same
(I/O >^
for
/?/-*,,
where
f
= >^ p
;=1
(c)
Prove Jensen's
between the smallest
lies
/?/•*/
inequality: If
/
t
.
i=]
is
N
convex, then /(
< J^
/?,a",
J
Hint:
Use Problem
=
noting that p n
3,
1
—
t
(Part (b)
.
p, fjx,).
i=\
i=\
needed
is
show
to
n-\
that
*9.
(a)
2, Pi x
(1/0
i
is
i
n the domain of
/
if x\,
.
.
,
.
xn
are.)
— f(a)]/h,
For any function /, the right-hand derivative, lim [f(a + h)
is
denoted by f'+ {a), and the left-hand derivative is denoted by fL(a). The
proof of Theorem 1 actually shows that f'+ {a) and f'_{a) always exist if
/
is
and
(b)
convex on some open interval containing
also
show
Conversely, suppose that
=
with f(b)
on
convex on
<
g{b) and f'_ib)
Show
(b),
that
compare
/
if
is
and
[a, b]
Hint: [fib)
—
f'+ {b)
than
is
less
the slope of
OQ
/
is
f(a)]/(b
f'_{a) if
f'+ {a).
we
If
h
convex on
is
ib,
[b, c],
define h
fib)), as in
PO
with that of
=
<
convex on
is
13(a)).
+
this assertion,
that f'_{a)
and g
g' (b) (Figure
convex, then f'+ia)
tinuous at a. (Thus
(a)
is
are increasing,
/ on [a,b] and g on [b, c], show that
Given P and Q on opposite sides of O =
Hint:
Figure 13
*10.
/
f'_
[a,c] to be
[a, c].
(c)
and
that f'+
Check
a.
and only
if
f'+
is
con-
when f'+ is continuous.)
for b < a close to a, and
differentiable precisely
—
a)
is
close to f'_{a)
this quotient.
Prove that a convex function on R, or on any open
interval,
must be
continuous.
(a)
(b)
Give an example of a convex function on a closed interval that
is
not
continuous, and explain exactly what kinds of discontinuities are possible.
11.
Call a function
/
weakly convex
on an
interval
if
for
a
<
b
<
c in
this interval
we have
(a)
Show
that a weakly
-
fib)
-
fia)
convex function
is
convex
f(x)
fia)
^
if
and only
if its
graph
contains no straight line segments. (Sometimes a weakly convex function
is
(b)
FIG U RE
\^
simply called "convex," while convex functions in our sense are called
"strictly convex".)
(b)
Reformulate the theorems of
this section for
weakly convex functions.
1 1 , Appendix. Convexity
12.
13.
and Concavity
229
Find two convex functions / and g such that f(x) = g(x) if and only if x
is an integer. Hint: First find an example where g is merely weakly convex,
and then modify it, using the result of Problem 9 as a guide.
A
set
A
of points in the plane
joining any two points in
of points
(x, y)
with y
the graph of /.
in the
>
is
called convex
f(x), so that
that
if
A
contains the line segment
(Figure 14). For a function /,
Af
is
Af
convex
is
if
the set
and only
Af be the set
of points on or above
if
/
let
is
weakly convex,
terminology of the previous problem. Further information on convex
sets will
(a)
Show
it
be found
in reference [18]
a convex subset of the plane
FIGURE
14
of the Suggested Reading.
(b)
a non-convex subset of the plane
CHAPTER
INVERSE FUNCTIONS
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 old
addition, multiplication, division, and composition
but using these alone, we can
produce only the rational functions (even the sine function, although frequently
—
used for examples, has never been defined).
begin to construct
new
will
is
one
double the usefulness of any other method
will practically
discover.
If we recall that
a function
is
a collection of pairs of numbers,
the bright idea of simply reversing
we
we
functions in quite sophisticated ways, but there
important method which
we
In the next few chapters
Thus from
the pairs.
all
/
=
{(1,2), (3,4), (5,9), (13, 8)},
g
=
{(2,l), (4,3), (9,5), (8, 13)}.
we might
hit
upon
the function
obtain
=
While /(l)
=
2 and /(3)
4,
we have
=
g(2)
and g(4)
1
=
3.
Unfortunately, this bright idea does not always work. If
/
=
{(1,2), (3,4), (5,9), (13,4)},
then the collection
{(2,1),
is
not a function at
the trouble
that can
this
DEFINITION
lies:
all,
/(3)
since
=
go wrong, and
it
(4, 3), (9, 5), (4,
contains both (4, 3) and
/(13), even though
it is
13)
3^13.
worthwhile giving a
(4, 13).
This
name
is
It is
clear
where
the only sort of thing
to the functions for
which
does not happen.
A function
The
/
is
one-one
identity function /
(read "one-to-one")
is
if
f(a)
/
obviously one-one, and so
is
f(b) whenever a
^
the following modifica-
tion:
'
g (x)=
The
function f(x)
=
x2
is
x
^3,5
3,
x
=5
5,
x
=
not one-one, since
g(x)
230
x,
= x2
,
3.
/(— 1) =
x>0
b.
/(l), but
if
we
define
231
12. Inverse Functions
<
(and leave g undefined for x
= 2x >
number and
x >
0, for
g'(x)
0),
then g
is
one-one, because g
This observation
0).
is
one-one. If n
/' is
is
one-one
(since f'(x)
condition f(a)
/
for
>
nx"
^
/
f(b) for a
twice (Figure
0, for all
^
x
0).
from the graph of / whether / is one-one: the
b means that no horizontal line intersects the graph
a function that
FIGURE
not one-one
1
If we reverse all the pairs in (a
any
case,
less
/
is
is
(b)
(a)
some
not necessarily one-one function)
collection of pairs.
It is
no particular reason
we
obtain a definition
—
do so instead of a
and a theorem.
to
For any function /, the inverse of /, denoted by
(a, b) for
l
PROOF
/
'
is
which the pair
a function
if
(b,
a)
and only
if
is
implies that b
c.
Thus /
Conversely, suppose that
the pairs (b, f(b))
Since /"'
is
and
(c,
is
/
,
is
the set of
all
pairs
{a, c)
=
be two pairs
/(c); since
/
in
/
.
Then
one-one
is
this
a function.
_1
/(c))
f
definition
one-one.
Suppose first that / is one-one. Let (a, b) and
(b,a) and (c,a) are in /, so a = f(b) and a
=
obtain, in
in /.
is
/
/ we
popular to abstain from this procedure un-
is
one-one, but there
with restrictive conditions
theorem
x
all
1).
a one-one function
DEFINITION
a natural
is
particularly easy to decide
It is
of
—
n
easily generalized: If
odd, one can do better: the function
/(*)
is
increasing (since
x>0,
f{x)=x\
then
is
is
=
a function.
(c,
If
fib)
f(b)), so (/(/?), b)
a function this implies that b
—
c.
Thus /
=
/(c), then
and (f(b),
is
one-one.
/
contains
c) are in
|
f~
l
.
232
Derivatives
and
Integrals
The graphs of / and / _1 are so closely related that it is possible to use the
graph of / to visualize the graph of /
Since the graph of f~ consists of all
_1
pairs (a, b) with (b, a) in the graph of /, one obtains the graph of /
from the
graph of / by interchanging the horizontal and vertical axes. If / has the graph
l
.
shown
in Figure 2(a),
FIGURE
3"
o
3
1-1
CO°
3
3"
O
3
"2. oq
1-1
fl>
CTQ
3"
r-t
H
o
"-) M
l"B
"rj
o
f
L£
1
2(a)
CL
R
o
re
o o
1-1
3
3?
re
3" cr
O
N*
o
(T)
•*
3"
C
o" 3-
n'
3"
rn
3
>-!
en
pj_
re"
3- 5;
!-!
1
cr
3
3*
CO
o
3
CfQ
Cfi"
3
3
5"
CO
O
po
o o
CO
O
*-!
3~
o
c
3
P
xs
3"
o
B.
3
>-i
>-*•
§"
>-s
CL
O
CO
re
N crq
*Tl
PJ
3*
o
3
o
3"
CO
re
3-
3*
r&
"5'
O
3*
CD
O
3
CD
So
R
CO
3
0)
o
o
3
3
re
O
>-<
This procedure
H
3*
PJ
Qrq^ CfQ
po
is
fortunate that there
awkward with books and impossible with blackboards,
'.
The
is another way of constructing the graph of /
so
it
is
points
233
12. Inverse Functions
(a,b)
which
and (b,a) are reflections of each other through the graph of I(x) = x,
_1
we merely
is called the diagonal (Figure 4). To obtain the graph of /
reflect the
graph of / through
this line (Figure 5).
(a,b)
'diagonal
(c,d)
• (b,a)
(d, c)
FIGURE
FIGURE
4
5
Reflecting through the diagonal twice will clearly leave us right back
started; this
means
_1 _1
Theorem
conjunction with
is
that (/
)
1,
=
this
/, which
is
also clear
from the
where we
equation has a significant consequence:
a one-one function, then the function
/
is
also
one-one
In
definition.
/
if
_1 _1
(since
(/
is
)
a
function).
There are a few other simple manipulations with
should be aware. Since
b
Thus f~\b)
is
is
is
in
/
means
f{a)
the (unique)
x 3 then f~(b)
,
=
(a, b)
precisely
the
inverse functions of which
when
same
(b,
a)
number a such that f(a) —
number a such that a' =
the unique
/
in
= /~
a
as
is
you
follows that
(b).
b; for
b,
,
it
example,
and
this
if
f(x)
number
=
by
is,
definition, \fb.
The fact that f~ (x) is the number
much more compact form:
l
(x))
/(/
= x,
y such that f(y)
for all
x
in the
=
x can be restated in a
domain of f -l
Moreover,
f-
[
(f(x))
=
x,
for
all
x
in the
domain of
/;
from the previous equation upon replacing f by f
important equations can be written
this follows
fof~ =
of =
.
These two
l
r
l
(except that the right side will have a bigger
not
all
of R).
I.
i
domain
if
the
domain of / or f~
]
is
234
Derivatives
and
Integrals
Since
it
is
many
standard functions
we be
quite important that
will
be defined as the inverses of other functions,
able to
tell
which functions are one-one.
already hinted which class of functions are most easily dealt with
decreasing functions are obviously one-one. Moreover,
is
and
also increasing,
to you). In addition,
One example
ing.
increasing
if
f~
and only
if
[
—increasing and
increasing, then
is
f~
decreasing (the proof is
is
—/
one-one function
certainly not true that every
It is
6
/
is
decreasing, then
is
/
have
is
l
left
decreasing, a very useful
remember.
fact to
FIGURE
/
if
if
We
is
has already been mentioned, and
r(*)
=
x,
x
#3,5
3,
x
=5
5,
x
=
either increasing or decreasis
now graphed
in Figure 6:
3.
Figure 7 shows that there are even continuous one-one functions which are neither
increasing nor decreasing.
But
if
you
try
drawing a few pictures you
suspect that every one-one continuous function defined on an interval
will
is
soon
either
increasing or decreasing.
THEOREM
FIGURE
2
If
/
is
continuous and one-one on an interval, then
/
either increasing or
is
decreasing on that interval.
7
proof
The proof proceeds
(1) If
a
< b <
in three easy steps:
c are three points in the interval, then
either
(i)
or
(ii)
< f{b) <
f(a) > f(b) >
f(a)
fie)
f(c).
Suppose, for example, that fia) < f(c). If we had fib) < fia) (Figure 8), then
the Intermediate Value Theorem applied to the interval [b, c] would give an x with
FIGURE
b
<
x
<
c
Similarly, fib)
>
Naturally, the
(2) If
For
a
—
and fix)
< b <
c
fia), contradicting the fact that
/
is
fie) would lead to a contradiction, so fia)
same
< d
we can apply
8
sort of
argument works
for the case
one-one on
< fib) <
fia) > /(c).
are four points in the interval, then
cither
(i)
fia) < fib) < /(c) < fid)
or
(ii)
fia) > fib) > /(c) > fid).
(1)
to
a
<
b
<
c
and then
to
b < c < d.
[a.c].
fie).
12. Inverse Functions
235
and suppose that f(a) < f(b). Then / is
increasing: For if c and d are any two points, we can apply (2) to the collection
{a, b, c, d) (after arranging them in increasing order). |
Take any a < b
(3)
Henceforth we
in the interval,
shall
be concerned almost exclusively with continuous increasing
on an interval. If / is such a function,
it is possible to say quite precisely what the domain of
f~ will be like.
Suppose first that / is a continuous increasing function on the closed interval
[a, b] Then, by the Intermediate Value Theorem, / takes on every value between
f(a) and f{b). Therefore, the domain of f~ is the closed interval [f(a), f(b)].
Similarly, if / is continuous and decreasing on [a, b], then the domain of f~ is
or decreasing functions which are defined
[
.
FIGURE
l
9
[
[/(*), /(a)].
If
/
is
a continuous increasing function
on an
open interval (a, b) the analysis
To begin with, let us choose some point c in (a,b).
> f(c) are taken on by /. One possibility is that
/ takes on arbitrarily large values (Figure 9). In this case / takes on all values
> f(c), by the Intermediate Value Theorem. If, on the other hand, / does not
take on arbitrarily large values, then A = { f (x) c < x < b} is bounded above,
so A has a least upper bound a (Figure 10). Now suppose y is any number with
f(c) < y < a. Then / takes on some value f(x) > y (because a is the least
upper bound of A). By the Intermediate Value Theorem, / actually takes on
the value y. Notice that / cannot take on the value a itself; for if a = f(x) for
a < x < b and we choose / with x < t < b, then f(t) > a, which is impossible.
Precisely the same arguments work for values less than f{c): either / takes on
all values less than f(c) or there is a number fi < f(c) such that / takes on all
becomes a
We
bit
will first
more
difficult.
decide which values
:
FIGURE
10
and /(c), but not /3 itself.
This entire argument can be repeated if / is decreasing, and if the domain of /
is R or (a, oo) or (— oo, a).
Summarizing: if / is a continuous increasing, or
decreasing, function whose domain is an interval having one of the forms (a,b),
(—oo,b), (a, oo), or R, then the domain of f~ is also an interval which has one
of these four forms, and we can easily fit the remaining types of intervals, (a, b],
[a, b], (—oo, b], and [a, oo), into this discussion also.
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.
values between
ft
l
theorem
3
PROOF
If
/
is
continuous and one-one on an interval, then
We know
by Theorem 2 that /
well assume that / is increasing,
is
is
open, since
We
it is
interval
is
also continuous.
either increasing or decreasing.
since
applying the usual trick of considering
on any
/
we can then
— /. We
We
might as
take care of the other case
might
as well
assume our
by
interval
easy to see that a continuous increasing or decreasing function
can be extended
to
one on a larger open
must show that
\lmfx—*b
l
(x)
=
f-\b)
interval.
236
Derivatives
and
Integrals
for each b in the
domain of /
domain of
in the
if
—
f(a)
1 1
see the
graph of /
suggests the
):
way of finding
8
< f
s
(remember
< a + s.
(x)
by looking sideways you
that
since
a
—
£
-
s)
< f(a) < f(a +
<
<
a
+ s,
a
e);
be the smaller of f(a + e) — f(a) and / (a) — f (a — e) Figure 11 contains
the entire proof that this 8 works, and what follows is simply a verbal account of
let 8
.
the information contained in this picture.
b
Our
—
f(a)-S
choice of
8
ensures that
f(a
a
+e
Consequently,
-s)<
1
1
-
f(a)
8
+
and f(a)
< f(a +
8
s).
if
-8 <x <
f{a)
FIGURE
—
then a
follows that
we
f(a)+8
=
+ 8,
< x < f(a)
8
f(a
f(a)
Such a number b is of the form f(a) for some a
such that, for all x,
0, we want to find a 8 >
.
>
/. For any e
Figure
it
'
+ 8.
f(a)
then
f(a
Since
/
is
increasing,
/
is
—
e)
< x < f(a +
also increasing,
e).
and we obtain
r
f-\f(a-s)) < f-\x) <
l
(f(a
+ s)),
i.e.,
a
which
is
precisely
Having
—
< f~
s
what we want.
<
(x)
|
successfully investigated continuity of
differentiability.
of
f{a)). If this entire picture
/
-1
is
it
only reasonable to tackle
This formula can equally well be written
each b in the domain of
:•-
with a tangent
The
it
slope of L'
line
L through
shows the graph
is
the reciprocal
appears that
= -L-.
f
rectly, for
it is
reflected through the diagonal,
(/-!)'(/(*,))
12
,
/
and the tangent line L' through (f(a),a).
of the slope of L. In other words,
FIGl Kl.
/
Again, a picture indicates just what result ought to be true. Fig-
ure 12 shows the graph of a one-one function
(a,
+ s,
a
/
l\'i
in
a
(a)
way which
expresses (f~)'(b) di-
.
^A
f'(f-Hb))
Unlike the argument for continuity,
involved
when formulated
analytically.
this pictorial
There
is
"proof" becomes somewhat
another approach which might
237
12. Inverse Functions
be
Since
tried.
we know
that
f{f~\x))=x,
tempting to prove the desired formula by applying the Chain Rule:
it is
l
f'(f- (x))-(f- y(x)=
l
l,
so
I
i\',
(/"')'(*)
f'(f-Hx))'
Unfortunately
this
is
f~ is differentiable, since the Chain Rule
already known to be differentiable. But this argu-
not a proof that
cannot be applied unless
/
_1
is
x
ment does show what (/)'(*) will have to be if f~ is differentiable, and
also be used to obtain some important preliminary information.
l
fix)
=
x
3
THEOREM
4
If
/
is
PROOF
We
a continuous one-one function defined on an interval and
/
then
is
can
/'(/(#)) =
0,
not differentiable at a.
have
=
f(f-\x))
If
it
/
were differentiable at a, the
x.
Chain Rule would imply
1
f'(f- (a))'(f- )'(a)=
that
1
(a)
1,
hence
0.(/" 1 )
which
absurd.
is
A simple
/'(0)
=
=
to
/"'(O), the function
Having decided where an
1
f~
following argument uses
THEOREM
5
is
/ be
/
differentiable at
continuity
entiable at b,
f~
(b),
is
in all other cases the derivative
of
/
,
in
two
different ways.
on an
interval,
l
and
1
)'(b)
=
—
f'(f-Hb))'
Let b
=
f(a).
Then
f-\b + h)- f~-l (b)
l
lim
fc->0
-1
=
lim
is
.
Since
(Figure 13).
we
are
now
given by the
Notice that the
which we have already proved.
with derivative f'(f~ (b))
(f-
proof
=x
the function f(x)
not differentiable at
a continuous one-one function defined
Let
is
l
inverse function cannot be differentiable,
formula which we have already "derived"
3
1,
which Theorem 4 applies
ready for the rigorous proof that
FIGURE
(fl)=
|
example
and
,
f~
l
(b
+ h)-a
^
0.
and suppose
Then /~'
is
that
differ-
238
Derivatives
and
Integrals
Now
every
number b +
domain of /
h in the
+ h=
b
for a
can be written
'
in the
form
+ k)
f(a
unique k (we should really write k(h), but we
will stick
with k for simplicity).
Then
lim
f-
[
+ h)-a
(b
=
f~\f(a + k))-a
f(a + k) — b
lim
/1-+0
=
k
lim
We
are clearly
on the
right track!
It is
+
f(a
ft-»0
k)
-
f(a)
not hard to get an explicit expression for
k;
since
b
+h =
+ k)
f(a
we have
f-\b + h)=a + k
or
k=f-\b + h)-f-\b).
Now
by Theorem
approaches
as h
3,
the function
approaches
0.
this
is
continuous
This means that k
at b.
Since
=/(«) = /
lim
k^O
l
/
k
(/
(ft))
^
0,
implies that
1
l\'<
(f-'Yib)
f'(f-Hb))
is
The work we have done on inverse functions
an immediate dividend. For n odd, let
f„(x)
=
x"
will
for
all
be amply repaid
later,
x;
for n even, let
fn(x)=x
Then
/„
is
n
x>0.
,
a continuous one-one function,
whose inverse function
^=
gn (x)= v
-v
l/ ".
is
but here
239
12. Inverse Functions
By Theorem
5
we
^
have, for x
0,
'
8n (X)
"
fn'Vn-H*))
1
n{fn
-
{
{x)) n
-
{
I
1
(A /")"1
77
1
1
-(1/n)
X
77
1
=—
•
X
77
Thus,
f(x)
if
=
fix) = ax
number: Let a
a
.
x",
It is
=
and a
now
an integer or the reciprocal of a natural number, then
is
easy to check that this formula
m/n, where
m
an
is
integer,
fix) =,-"'/"
then, by the
=
and n
is
is
true
if
a
is
any rational
a natural number;
if
(x /")"\
]
Chain Rule,
f(x)
=
1
m(x i/ ")
1
.(!/«)-!
n
777
.[(w/n)-(l/n)] + [(l/«)-l]
_ ^ x (m/n)-l
77
Although we now have a formula
the treatment of the function f{x)
for later
—
at the
moment we do
Actually, inverse functions will
for /'(x)
=
when f{x) — x a and
x" for irrational a
not even
know
the meaning of a
be involved crucially
irrational a. Indeed, in the next
PROBLEMS
Find
/
(i)
/(jc)=jc 3
for each of the following /.
+
1.
(ii)
f(x)
= (x-\)\
m
/(*)
=
IV
fix)
=
x,
x rational
—.v,
x
-x 2
l-x 3
x,
(v)
fix)
=
a i+
a
(vi)
/(*)
i
\
,
=*+[*].
irrational.
x >
x<0.
,
a
is
rational,
have to be saved
symbol
in the definition
like a
of x
few chapters several important functions
defined in terms of their inverse functions.
1.
will
x^a\,
a„
x
=a
/=1,
x
=
t
an
,
.
,
n
:
a
will
.
for
be
'
240
Derivatives
and
Integrals
/(0.aifl2«3
(vii)
=
/(*)
(viii)
•
=
)
•
•
0.fl2 fl l«3
x
1
<x <
•
•
•
•
(Decimal representation
is
being used.
1.
1
2.
/
/
/
/
(i)
(ii)
(iii)
(iv)
3.
when
/
Describe the graph of
is
increasing
and always
positive.
is
increasing
and always
negative.
is
decreasing and always positive.
is
decreasing and always negative.
Prove that
/
if
increasing, then so
is
/
is
,
and
similarly for decreasing
functions.
If
/ and g
are increasing,
Prove that
(a)
o
(/
g)~
Find g~
(b)
Show
l
l
7.
'
8.
10.
f(x)
(ii)
f(x)
(iii)
f(x)
Suppose
*12.
13.
g?
l
in
l
.
l
in
1
—b
ax
=
-\-
-
;
+d
one-one
is
if
and only
if
ad — be
^
0,
and
find
intervals [a, b] will the following functions
= x 3 -3x 2
be one-one?
.
= x + x.
= (\+x 2 r
5
that
= f~
x
x
+
X1
+
/
is
l
.
1
1
differentiable with derivative f'(x)
2
satisfies
g"(x)
=
\g(x)
=
(1
+A" 3 )
-1 ^ 2
.
Show
.
has a derivative which is
Suppose that / is a one-one function and that /
nowhere 0. Prove that / is differentiable. Hint: There is a one-step proof.
As a follow up
that
11.
o
in this case.
that g
9.
•
/ and g are one-one, then fog is also one-one. Find
terms of /
and g~K Hint: The answer is not f~ o g~
terms of f~ if g(x) = + f(x).
that f{x)
On which
(i)
f + g? Or f g? Or /
if
ex
f
is
/
is
to
Problem 10-17, what additional condition on g
Find a formula for (/
Prove that
if
f'(f~
l
-1
)"(;c).
{x))
^
and f (k) {f~\x))
The Schwarzian derivative 2)/ was defined
(a)
Prove that
the
(b)
will insure
differentiable?
if
2)/(x) exists for
domain of /
.
Find a formula for 2)/~
(a).
all
x, then
in
exists,
then (/
_1
)
w
(jc) exists.
Problem 10-19.
2)/
_1
(x) also exists for
all
v
in
241
12. Inverse Functions
*14.
Prove that there
(a)
=
x
The
do
Show that /
way to do this is to
Hint:
easiest
=/
x
to set
this is
such that [/(*)] 5 + /CO +
can be expressed as an inverse
/
find
And
.
the easiest
way
_1
(v).
(b)
Find /' in terms of /, using
(c)
Find /'
The
/
a differentiable function
for all x.
function.
to
is
an appropriate theorem of this chapter.
another way, by simply differentiating the equation defining /.
in
function in Problem 14
defined implicitly by the
equation is quite special, however. As
often said to be
is
equation y +y+x = 0. The situation for this
the next problem shows, an equation does not usually define a function implicitly
on the whole
line,
and
some regions more than one function may be defined
in
implicitly.
15.
(a)
What
are the two differentiable functions
on (— 1,
for
1)
x
all
by the equation
(— 1,
in
jc
+y —
/ which
are defined implicitly
which
x +[f(x)]~
1, i.e.,
satisfy
=
1
Notice that there are no solutions defined outside
1)?
[-1,1].
(b)
*(c)
Which
Which
will
/
functions
satisfy
x2
+
differentiable functions
help to
first
draw
[f(x)]
/
2
=
satisfy
-1?
[/CO]
—
3/(jc)
=x
the graph of the function g(x)
In general, determining on what intervals a differentiable function
plicitly
by a particular equation
may be
context of advanced calculus. If
however, then a formula for
differentiating
we
and
a delicate affair,
assume that
/
It
defined im-
is
best discussed in the
is
such a differentiable solution,
is
can be derived, exactly as
/'(jc)
= x? Hint:
— 3x.
both sides of the equation defining /
in
process
(a
Problem
known
14(c),
by
as "implicit
differentiation"):
16.
(a)
Apply
this
answer
method
to the equation [/(.v)]
will involve f(x); this
is
2
+ x2 =
Notice that your
1.
only to be expected, since there
(b)
Problem
17.
(c)
Apply
(a)
Use
this
same method
to [f(x)]^
more
1.
/ found
in
for the functions
f
— 3f(x)=x.
implicit differentiation to find f'(x)
One
=
is
15(a).
defined implicitly by the equation x
(b)
2
than one function defined implicitly by the equation y + x
But check that your answer works for both of the functions
2
of these functions
/
satisfies
and f"(x)
+ y =7.
/(-l)
=
2.
Find /'(-l) and /"(-l)
for this /.
18.
The
collection of
all
forms a certain curve
this
19.
curve
at the
is
in the plane.
point (— 1,
Leibnizian notation
cause y
points (x, y) such that 3x 3
is
+
4x 2 y
— xy 2 + 2y 3 =
Find the equation of the tangent
4
line to
1).
particularly convenient for implicit differentiation. Be-
so consistendy used as
and y which defines /
an abbreviation
for f(x), the equation in
v
implicitly will automatically stand for the equation
,
242
Derivatives
and
Integrals
which /
is
supposed to
How
satisfy.
would the following computation be
written in our notation?
4
y
dy
4y
+
+ xy =
3
y
+ 3y^
dx
dx
dx
-y
dy
dx
20.
As long
1
4v
3
+
3y 2
+
;t
as Leibnizian notation has entered the picture, the Leibnizian no-
tation for derivatives of inverse functions should be mentioned.
If
dy /dx
l
denotes the derivative of /, then the derivative of f~ is denoted by dx/dy.
Write out Theorem 5 in this notation. The resulting equation will show you
why
another reason
which point (/
also explain at
What
notation.
Leibnizian notation has such a large following.
)' is
It will
be calculated when using the dx/dy
to
the significance of the following computation?
is
y
dx
x
—
x
'n
dx
i/»
'
,
dy
1
dx
dx
ny n-\
dy
21.
Suppose
if
whose
Two
22.
and
= f~
G'(x)
fact:
/
that
derivative
[
(x).
we know
derivative
different
Suppose h
is
is
that
a differentiable one-one function with a nowhere zero
f =
— xf
Let G(x)
F'.
(Disregarding details,
(x)
problem
this
is
/
.
But
how
in
(x)).
Prove that
us a very interesting
tells
/, then we also know one
could anyone ever guess the function G?
a function whose derivative
ways are outlined
— F(f~
is
Problems 14-14 and 19-16.)
a function such that ti{x)
=
sin (sin(x
+
1))
and h(0)
=
3.
Find
l
23.
(i)
(h- YO).
(ii)
(£-')'(3),
(a)
(b)
where 0(x)
=
h(x
+
1).
Prove that an increasing and a decreasing function intersect
at
most once.
Find two continuous increasing functions / and g such that f(x)
when x is an integer.
=
g(x)
precisely
(c)
Find a continuous increasing function
function g, defined
*24.
(a)
If
/
least
is
on R, which do not
one x such that f(x)
x.
a continuous decreasing
intersect at
R
all.
and /' = /
prove that there is at
_1
mean
(What does the condition / = /
a continuous function on
=
/ and
,
geometrically?)
(b)
Give several examples of continuous / such that f = f~ and f(x) — x
one x. Hint: Try decreasing /, and remember the geometric
for exactly
interpretation.
One
possibility
is
f(x)
=
—x.
243
12. Inverse Functions
(c)
Prove that
if
/
then
an increasing function such that / = /
x. Hint: Although the geometric interpretation will
is
,
fix) = x for all
be immediately convincing, the simplest proof (about 2
out the
*25.
26.
possibilities
f(x) < x and f(x) >
is
to rule
x.
Which functions have the property that the graph
tion when reflected through the graph of — / (the
A
lines)
is still
the graph of a func-
"antidiagonal")?
nondecreasing if fix) < fiy) whenever x < y. (To be
more precise we should stipulate that the domain of / be an interval.) A
nonincreasing function is defined similarly. Caution: Some writers use
function
/
is
""increasing" instead of "nondecreasing,"
and
our
"strictly increasing" for
"increasing."
(a)
Prove that
(b)
on some interval. (Beware of unintentional puns: "not increasing"
the same as "nonincreasing.")
Prove that if / is differentiable and nondecreasing, then f'(x) >
all
*27.
Prove that
(a)
Suppose
there
(b)
/
is
nondecreasing, but not increasing, then
/
is
constant
not
is
for
x.
(c)
all
if
is
if
f'(x)
that f(x)
>
>
for
all
x, then
for all x,
and
/
is
nondecreasing.
that
/
is
decreasing.
a continuous decreasing function g such that
<
g(x)
Prove that
< f(x)
for
x.
Show
that
we can even arrange
that g will satisfy
lim g(x)/f(x)
=
0.
244
Derivatives
and
Integrals
PARAMETRIC REPRESENTATION OF CURVES
APPENDIX.
The
long time ago
(Figure
1).
—a
perfectly nice looking curve
we may
In other words,
Of course, we
(x, fix)).
this trick doesn't
consisting of
all
work
not be able to describe
simplest
most
cases.
one
is
1
the set of
set
of all points
2
points (x ,x). But
all
1,
or an
and
ellipse,
can't
it
circle,
be used
in Figure 2.
way of describing
curves in the plane in general harks back to the
physical conception of a curve as the path of a particle
FIGURE
as the set of all points
it
won't allow us to describe the
It
+y =
points (x. v) with x
to describe a curve like the
The
in
noticed a
need not be the graph of a function
might be able to describe the curve as the
(f(x),x); for example, the curve in Figure
even
we
material in this chapter serves to emphasize something that
moving
in the plane.
At
1
each time
/,
the particle
at
is
a certain point, which has two coordinates; to indicate
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
be represented parametrically by u and v, and the pair of functions u, v is called a
the
parametric representation of the curve.
u
and
The
curve represented parametrically by
x
v thus consists of all pairs (x, y) with
—
described briefly as "the curve x
function
/ can
u{t), y
—
=
u{t)
and y
=
v(t).
often
It is
Notice that the graph of a
v(t)."
always be described parametrically, as the curve x
—
t,
y
=
fit).
Instead of considering a curve in the plane as defined by two functions,
can obtain a conceptually simpler picture
if
we broaden our
function somewhat. Instead of considering a rule which associates a
another number,
FIGURE
2
i.e.,
c(t).
numbers
Of course,
With
number
this notion,
a curve
is
t,
a.
point in the plane,
just a function
which we can
from some interval of
to the plane.
these two different descriptions of a curve are essentially the same:
A pair of (ordinary)
numbers
number with
consider a "function c from real numbers to the plane,"
a rule c that associates, to each
denote by
real
we can
we
original definition of
and
functions u
to the plane
v determines a single function c
from the
real
by the rule
c(t)
=
(u(t),v(t)),
and, conversely, given a function c from the real numbers to the plane, each c(t)
is
a point in the plane, so
so that
we have unique
In Appendix
1
to
it is
a pair of numbers, which
functions u
Chapter
4,
and
we used
we can
a "vector-valued function."
c(t)
A
=
—
(c\ (f
),
cjit)),
(u(t), v(t)) to
The
but in
I
(
,
I
K
I
.
*
the
term "vector"
to describe a point in
may
also
Appendix
be called
we'll continue to use notation like
minimize the use of subscripts.
simple example of a vector-valued function that
which goes round and round the
v(t),
conventions of that Appendix would lead us to
this
e(/)
I
and
v satisfying this equation.
the plane. In conformity with this usage, a curve in the plane
write c(t)
call u(t)
=
(cos/, sin
is
/),
unit circle (Figure 3).
quite useful
is
245
12, Appendix. Parametric Representation of Curves
/ and
For two (ordinary) functions
g,
we
new
defined
f + g and / g
functions
•
by the rules
(f
+ g)(x) =
f(x)
+ g(x),
=
f(x)
g(x).
(f
Since
we have
defined a
way
g)(x)
•
c
+d
we can imitate the first of these defwe define the vector-valued function
of adding vectors,
initions for vector-valued functions c
and
d:
by
+ d)(t)=c(t) + d(t),
(c
+
where the
on the right-hand
to saying that
side
now
is
the
sum of vectors. This simply amounts
if
c(t)
d(t)
=
=
(u(t),v(t)),
(w(t),z(t)),
then
(c
+ d){t) =
we have
Recall that
extend
this to
also defined a
Then we can
=
(w(t), z(t))
(u(t)
new
w(t), v(t)
to consider
so that for each
c,
define a
+
number a and
a
v for
we want
vector-valued functions,
and a vector-valued function
vector c(t).
+
(u(t), v(t))
we have
/
+ z(t)).
a vector
To
v.
an ordinary functio n a
a
vector-valued function
number
a c by
a(t)
and a
(a -c)(t) =a(t)-c(t),
where the on the right-hand
simply amounts to saying that
•
(a
c)(t)
=
a(t)
Notice that the curve a
(a
is
e)(t)
side
of a
in
the product of a
(u(t), v(t))
=
number and
u(t), a(t)
(a(t)
a vector. This
v(t)).
e,
e)(t)
already quite general (Figure
the point (a
is
=
4).
(a(t)cost, a(t)smt),
In the notation of
e){t) has polar coordinates a(t)
and
t
,
Appendix
so that (a
3 to
e)(t)
is
Chapter
4,
the "graph
polar coordinates."
Even more generally, given any vector-valued function
functions r and 9 by
c,
we can
define
new
c(t)=r(t)-e(9(t)).
where
r{t)
is
just the distance
from the origin
the angle of c(t) (as usual, the function 9
FIGURE
4
be careful when using
to
We
we
this
isn't
to c(t),
and
9(t)
is
some choice of
defined unambiguously, so one has
way of writing an
arbitrary curve
c).
aren't in a position to extend (2) to vector-valued functions in general, since
haven't defined the product of two vectors.
Appendix
1
to
Chapter 4 define two
real-valued
However, Problems 2 and 4 of
products v
•
w and
det(u, w).
It
246
Derivatives
and
Integrals
should be
given vector-valued functions c and d,
clear,
how we would
define two
ordinary (real-valued) functions
c
Beyond imitating simple
more interesting problems,
limc(f)
(*)
Rules
=
•
and
d
det(c, d).
we can consider
we can define
arithmetic operations on functions,
For c(t)
like limits.
lim(a(f), u(0)
=
(u(t), v(t)),
(\im
to De
\imv(t))
u(t),
.
like
lim c + lim d,
+ d = t—>a
t—>a
lim c
t-+a
lim a
=
c
follow immediately.
lim a(t)
lim c
t^a
t—*a
t—t-a
Problem 10 shows how
an equivalent definition that
to give
imitates the basic definition of limits directly.
Limits lead us of course to derivatives. For
c(t)
we can
define
c'
(u(t),v(t))
by the straightforward definition
c'(a)
We
=
=
(u(a), v'(a)).
could also try to imitate the basic definition:
c(a
=
c'(a)
+ h) —
c{a)
lim
//
where the
fraction
on the right-hand
-•
side
[c{a
understood to
is
+ h) -
mean
e{a)}.
n
As a matter of fact,
c(a
+
h)
c(a
these
+ h) —
two
c(a)
lim
=
definitions are equivalent, because
u{a
+ h) —
u(a)
v(a
u{a+h)-u{a)
,.
lim
Figure 5 shows c(a
+
h)
we showed in Appendix
moved over so that it starts
as
II
<
,
I
!<
E
5
(loser
and
h^Q
h
(*)
h
of limits
to
c(a), as well as the
Chapter
at c(a).
4, this
As h —>
0, this
closer to the tangent of our curve, so
arrow from c(a)
arrow
it
is
c(a
it
as the set of
starts at c(a).
all
In other words,
we
+s
h)
—
to c(a
arrow would appear
when
+
h);
c(a), except
to
move
to define the
c'(a)
is
moved
define the tangent line of c at c(a)
points
c(a)
+
seems reasonable
tangent line of c at c(a) to be the straight line along c'(a),
over so that
v(a)
(u'(a), v'(a)).
and
1
+ h) -
v(a
by our definition
c(a)
v(a)
lim
,
h^O
=
+ h) —
lim
c\a)\
12, Appendix. Parametric Representation of Curves
=
=
247
we get c(a) + c' (a), etc. (Note,
however, that this definition does not make sense when c'(a) = (0, 0).) Problem
shows that this definition agrees with the old one when our curve c is defined by
for s
we
get the point c(a)
itself,
for s
1
1
c(t)
we simply have
Once again, various
so that
=
(t,f(t)),
the graph of /.
old formulas have analogues. For example,
+ d)'(a) =
(a c)'(a) =
(c
•
+ d'(a),
a' (a) c(a) + a(a)
r
c (a)
c'(a),
equations involving functions,
or, as
+ d)' = c' + d',
{ot
c) = a
c + a.
(c
•
•
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(a
+ h)c(a + h) — a(a)c(a) =
ct(a
We
+ h)
+ h) -
[c{a
c(a)]
+
[a(a
+ h) - a(a)]
c(a).
can also consider the function
d(t)
where p
is
now an
passes through the
= c(p(t)) =
(c o p)(t),
ordinary function, from numbers to numbers.
same points
The new
curve d
p corresponds
as c, except at different times; thus
to a "reparameterization" of c. For
c
d
we
—
=
(u, v),
(u o p, v o p),
obtain
d'{a)
= (O o p)'(a), {v o p)'{a))
= (p'(a)u'(p(a)), p'{a)v\p{a)))
= p'{a) (u'{p(a)), v\p(a)))
= p'(a) c'(p(a)),
•
or simply
d'
Notice that
if
p (a)
time a, then d'(a)
of
c'(a).
=
=
= a, so
p'(a)
p'
that
(c o p).
d and
This means that the tangent
line to the
c actually pass through the
c'(a), so that the
reparameterized curve d
line
at
tangent vector d'(a)
to c at c(a)
d(a)
=
c(a).
is
the
same point
at
is
just a multiple
same
as the tangent
The one
exception occurs
248
Derivatives
and
Integrals
when
=
p'(a)
tangent line for d
0, since the
may be
tangent line for c
line defined at
=
t
Finally, since
0,
even though
we can
it's
then undefined, even though the
=
3
)
won't have a tangent
c.
define real-valued functions
det(c, d)(t)
= c(t)-d(t),
=
det(c(f), d(t)),
have formulas for the derivatives of these
to
c(f
merely a reparameterization of
(C'd)(t)
we ought
is
defined. For example, d(t)
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)
•
•
These can be derived by straightforward calculations from the definitions in terms
of the component 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
to
Chapter
4, and, of course, the "standard trick" referred to above.
PROBLEMS
1.
For a function /, the "point-slope form" (Problem 4-6) of the tangent
(a)
line at (a,
f{a)) can be written as y
—
f(a)
=
(x
—
a)f'(a), so that the
tangent line consists of all points of the form
(x,f(a)
Conclude
+ (x-a)f'{a)).
that the tangent line consists of
all
points of the
form
(a+s,f(a)+sf(a)).
(b)
If c
(a,
=
the curve c(t)
is
(f,
new
f{a)) [using our
f(t)),
conclude that the tangent
definition]
is
the
same
line
of c at
as the tangent line of
/
at (a, f(a)).
2.
=
Let c(t)
Show
ter 9.
=
h(x)
(f(t),
\x\,
t
),
3.
c'
never equal to
Suppose
and
(a)
(b)
that
that u
is
.v
=
this
u{t),
If w
differentiable
is
u{t)
=
(0, 0).
reason,
y
=
v(t)
one-one on some
that
—
the function
on
shown
in
Figure 21 of Chap-
graph of the non-diffcrentiable function
In other words, a reparameterization can
we
are usually only interested in curves c
(0, 0).
Show
x
is
that c lies along the
but that c'(0)
"hide" a corner. For
with
where /
a parametric representation of a curve,
interval.
this interval the
on
is
curve
this interval
lies
and
we have
v'(t)
fix)
u'{t)
along the graph of
//'(/)
^
0,
show
/ =
v
ou
.
that at the point
249
12, Appendix. Parametric Representation of Curves
In Leibnizian notation this
often written suggestively as
is
dy
dy_
dt
dx
dx
It
(c)
We
also
have
-
u'{t)v"{t)
fix) =
v'{t)u"{t)
(u'(t)Y
4.
5.
Consider a function / defined implicitly by the equation x 2/3
Compute f'{x) in two ways:
(i)
By
(ii)
By considering
Let x
=
u(t),
=
y
that if the point
FIGURE
6
v'(t)
the parametric representation x
v(t)
Q=
P =
let
(xo, yo)
from
=
cos
Q
be a point
on the curve
(u(t), v(t))
are not both 0, then the line from
1.
y
t,
=
sin
t.
(Figure
6).
closest to (xo, yo)>
is
P
in the plane.
to
Q
is
The same
Prove
an d
u'(t)
perpendicular to the
result holds if
Q
is
(xq, yo).
We've seen that the "graph of /
(/•e)(0
in other
^ =
be the parametric representation of a curve, with
tangent line of the curve at
furthest
2
y
implicit differentiation.
u and v differentiable, and
and
+
words, the graph of
/
=
in polar coordinates"
is
the curve
(/(') cos *,/(*) sin 0;
in polar coordinates
is
the curve with the para-
metric representation
x
(a)
= f(0)cosG,
y
Show that for the graph of /
line at the point
with polar coordinates (f(0),9)
-/(6>) sin
Show
that
if
the origin
f(0) sinO.
in polar coordinates the slope
/(#)cos(9
(b)
=
f(0)
=
(9
and /
of the tangent
is
+ / (6>)sin<9
+ /'(#) cos 6>'
/
is
differentiable at 9, then the line
making an angle of 9 with the
through
positive horizontal axis
is
a
/ in polar coordinates. Use this result to
add some details to the graph of the Archimedean spiral in Appendix 3
of Chapter 4, and to the graphs in Problems 3 and 10 of that Appendix
tangent line of the graph of
as well.
(c)
Suppose
that the point with polar coordinates (f(9), 9)
the origin
O
than any other point on the graph of /.
say about the tangent line to the graph at this point?
Problem
5.
is
further from
What can you
Compare with
250
Derivatives
and
Integrals
Suppose that the tangent
(d)
line to the
graph of /
at the point
with po-
(f(0),0) makes an angle of a with the horizontal axis
is the angle between the tangent line and the
that a —
lar coordinates
(Figure
7),
ray from
O
so
to the point.
Show
tan (or
-
that
0)
f(0)
=
f'(0)
7.
In Problem 8 of
(a)
r
=
1
—
sin
6
is
Appendix
3 to
also described
Find the slope of the tangent
(ii)
FIGURE
Check
be
7
+ y 2 + y) =
x2
+y
.
a point on the cardioid in two ways:
that at the origin the tangent lines are vertical, as they
appear
to
in Figure 8.
FIGURE
The
line at
(x
By implicit differentiation.
By using the previous problem.
(i)
(b)
Chapter 4 we found that the cardioid
by the equation
8
next problem uses the material from Chapter 15, in particular, radian
measure, and the inverse trigonometric functions and their properties.
8.
A
cycloid
is
defined as the path traced out by a point on the rim of a rolling
wheel of radius
You can
a.
see a beautiful cycloid by pasting a reflector
the edge of a bicycle wheel and having a friend ride slowly
in front
on
of the
headlights of your car at night. Lacking a car, bicycle, or trusting friend, you
can
settle
instead for Figure 9.
FI
GURE
9
12, Appendix. Parametric Representation of Curves
251
Let u{t) and v(t) be the coordinates of the point on the rim after the
(a)
wheel has rotated through an angle of
arc of the wheel rim from P
wheel
is
rolling, at
is
to
Q
in
/
This means that the
(radians).
Figure 10 has length at. Since the
also the distance
from
O
to
Q.
Show
that
we have
the parametric representation of the cycloid
—
—
u{t)
v(t)
Figure
1 1
shows the curves we obtain
center of the wheel
FIGURE
10
is (a)
is
FIGURE
In Figure 9
need
to
is
(b)
from the point
to the
greater than the radius.
not the graph of a function;
the wheel
is
at certain
times
moving forwards!
1
we drew
check that
that
the distance
moving backwards, even though
Compute
(b)
if
than the radius or
less
In the latter case, the curve
the point
— sin t)
a(\ — cos/).
a{t
the cycloid as the graph of a function, but
this
is
we
really
the case:
and conclude that u is increasing. Problem 3 then shows
the cycloid is the graph of / = v o u~\ and allows us to compute
u'{t)
fit).
Show
(c)
It isn't
(d)
that the tangent lines of the cycloid at the "vertices" are vertical.
possible to get an explicit formula for /, but
Show
that
u{t)
Hint:
we can come
first
= a arccos
solve for
/
in
a
-
v(t)
±y/[2a-v(t)]v(t).
terms of
v(t).
close.
252
Derivatives
and
Integrals
The
(e)
half of the
first
g(y)
9.
=
first
arch of the cycloid
—
a
v
aarccos
yj(2a
is
the graph of g
-
y)y.
v give a parametric representation of a curve
Q =
Geometrically,
it
point on the curve the tangent line
is
to
{u(b), v(b)).
Prove
to Q.
Hint:
this analytically.
assume that we don't have u'(x)
Problem 2).
10.
The
=
from P
seems clear (Figure
parallel to the line
This problem
interpretation for one of the theorems in Chapter
12
,
where
Let u and v be continuous on [a,b] and differentiable on (a,b); then u
and
FIGURE
l
v'(x)
—
1 1
You
.
(u(a),v(a))
12) that at
some
segment from P
will give
any x
for
=
a geometric
will also
in (a, b)
following definition of a limit for a vector-valued function
is
need
to
(compare
the direct
analogue of the definition for ordinary functions:
limc(0
=
means
/
that for every e
>
there
some
is
8
>
such
that, for
t->a
<
all t, if
Here
/
=
||
||
\t
is
(/i,/2),
-
a\
<
8,
then
<
-l\\
e.
Problem 2 of Appendix
the norm, defined in
1
to
Chapter
4.
If
then
||c(/)-/||
(a)
\\c(t)
Conclude
= HO-/il 2 + MO-/2l 2
-
that
\u(t)-h\ <
and show
2
that
and
\\c(t)-l\\
if
lim c{t)
=
I
\v(t)
-
l2
\
<
||c(0-/||,
according to the above definition, then
we
t—>a
also
have
lim u{t)
so that lim c(t)
functions,
(b)
=
I
=
l\
lim v(t)
according to our definition
on page 246.
show that if lim
Conversely,
and
c(t)
=
I
h,
(*) in
terms of component
according to the definition in terms
of component functions, then also lim c(t)
t—*a
definition.
—
=
I
according to the above
CHAPTER
INTEGRALS
The
derivative does not display
second main concept of Part
digression
—
strength until allied with the "integral," the
its full
At
III.
first this
completed, integrals
Although ultimately
to
Chapter
difficulties
14,
be defined
malizes a simple, intuitive concept
1
to
once
The
given.
reflection
—
that of area.
an
ever.
complicated way, the integral
By now
intuitive
should
it
1,
which
number
as area the irrational
Even
if
we
many
iz
,
but
unit square
In this chapter
(Figure
1)
we
will
— those which
x
in [a, b]. It
is
as
circle
all
is
in this circle, since
fig-
seldom
hopelessly inadequate
of radius
clear
is
number of squares, with
as the
this definition
not at
plane
what
"jt
1
supposedly has
squares" means.
it
1, it is
hard
does not seem possible
which can be arranged
to
form a
circle.
bounded by the horizontal axis, the vertical lines
and the graph of a function / such that f(x) >
are
convenient to indicate
these regions include rectangles
geometric
come
only try to define the area of some very special regions
through (a,0) and (b,0),
all
it is
fits
to divide the unit square into pieces
for
But
consider a circle of radius l/y/rr, which supposedly has area
what way a
to say in
2
sometimes defined
is
in the region.
any but the simplest regions. For example, a
for
FIGURE
fit
for-
concept can present great
shows that an acceptable definition of area
area of a region
sides of length
study of
preliminary work has been
In elementary geometry, formulas are derived for the areas of
little
The
no exception.
certainly
is
this
more powerful than
in a quite
to learn that the definition of
— "area"
ures, but a
be a complete
be an invaluable tool for creating new functions, and the
will
derivative will reappear in
FIGURE
may seem
chapter derivatives do not appear even once!
in this
integrals does require a long preparation, but
no surprise
topic
and
this
region by /?(/, a,
triangles, as well as
many
b).
Notice that
other important
figures.
The number which we will eventually assign as the area of R(f,a,b) will be
/ on [a b] Actually, the integral will be defined even for
for all x in [a, b]. If / is
functions / which do not satisfy the condition f(x) >
called the integral of
.
,
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
"algebraic area" of R(f,a,
The
[a, b]
b)).
idea behind the prospective definition
by means of numbers
3
numbering of
equal the
253
=
[fe.fc]
[/l.fe]
to, t\, ti,
a
FIGURE
indicated in Figure
3.
The
interval
has been divided into four subintervals
[fO.'l]
(the
is
t?,,
to
<
t\
t\
with
<
tj
<
?3
the subscripts begins with
number of subintervals).
[*3,*4]
<U
=b
so that the largest subscript will
254
Derivatives
and
Integrals
On
the
maximum
interval
first
/ be w, and
of
/
the function
[to, t\]
=
s
maximum
the
let
m\(t\
-
to)
+
t\)
[f,-_i, tf\ let
value m\ and the
minimum
the
value
The sum
value be M\.
+ ni2(t2 -
minimum
has the
value M\; similarly, on the zth interval
-
niT,(tT,
+ m\(t\ -
tn)
£3)
represents the total area of rectangles lying inside the region R(f, a, b), while the
sum
=
S
M
x
(t\
-
+ M 2 (t2
to)
M
+
~t[)
-
3 (t3
+
t2 )
M4
(t 4
-
t3 )
represents the total area of rectangles containing the region R{f, a, b).
A
ing principle of our attempt to define the area
that
A
should
of R(f, a,b)
is
The
guid-
the observation
satisfy
s
< A
A <
and
S\
and that this should be true, no matter how the interval [a, b] is subdivided. It is to be
hoped that these requirements will determine A. The following definitions begin
to formalize, and eliminate some of the implicit assumptions in, this discussion.
DEFINITION
Let a <
[a, b],
The
A
b.
partition of the interval
one of which
is
we
DEFINITION
and one of which
a,
points in a partition can be
a
shall
[a, b]
=
to
<
t\
<
a
finite collection
of points in
b.
is
numbered
<
is
tn
so that
=
b;
to
/„_]
<
tn
always assume that such a numbering has been assigned.
Suppose /
is
bounded on
[a, b]
m,
=
and P
=
inf{/(.v)
Mi =sup{/(x)
The lower
sum
of
/
for P,
:
:
{to
t„
/,_i
<x <
<x
r,_i
}
is
a partition of [« b}. Let
ti),
< \ti).
denoted by L(f, P),
is
defined as
n
L(f,P)
= J^m
(fi-ti- it-
t
i—\
The upper
sum
of
/
for P,
denoted by U(f,P),
is
= YlM {t -t -
l).
U(f,P)
i
i
i
defined as
1=1
The lower and upper sums correspond
to thr
example; they are supposed to represent the
sums
total areas
s
and S
in the
previous
of rectangles lying below
and above the graph of /. Notice, however, that despite the geometric motivation,
these sums have been defined precisely without any appeal to a concept of "area."
255
13. Integrals
Two
The requirement
of the definition deserve comment.
details
that
/ be
and M, be defined. Note,
also, that it was necessary to define the numbers m, and M, as inf's and sup's,
rather than as minima and maxima, since / was not assumed continuous.
bounded on
One
thing
[a,b]
order that
essential in
is
clear about lower
is
all
the
and upper sums:
ra,
P
If
any
is
partition, then
L(f,P)<U(f,P),
because
n
n
and
for
each
i
we have
-ti-\)
mtiti
FIGURE
On
4
any
MM -ti-i).
<
the other hand, something less obvious ought to be true: If P\
two
partitions of [a b]
,
,
then
it
and Pi are
should be the case that
L(f,Pi)<U(f,P2 ),
because L(f, P\) should be < area R(f,a,b), and U(f, Pi) should be > area
R(f,a,b). This remark proves nothing (since the "area of R(f,a, by has not even
been defined
yet),
but
does indicate that
it
if
there
be any hope of defining the
to
is
area of R(f,a,b), a proof that L(f, P\) < U(f, P2) should come first. The proof
which we are about to give depends upon a lemma which concerns the behavior of
lower and upper sums
the partition
P
when more
black and the points in grey.
the partition
FIGURE
Q
lemma
If
Q
contains
P
Q
The
contains both the points in
picture indicates that the rectangles
drawn
for
are a better approximation to the region R(f, a,b) than those for
the original partition P.
5
points are included in a partition. In Figure 4
contains the points in black, and
To be
(i.e., if all
precise:
points of
P
are also in Q), then
L(f,P)<L(f,Q),
U(f,P)>U(f,Q).
PROOF
Consider
first
the special case (Figure 5) in which
Q
contains just one
than P.
P
=
Q=
{to,...,tn ),
Uo
tk
-i,u,
tk
—
,tn ),
where
a
—
to
<
t\
<
•
•
•
<
tk-]
<
u
<
tk
<
•
•
-
<
tn
= b.
more point
256
Derivatives
and
Integrals
Let
=
m" =
m'
inf{/(.v)
t^] < x <
:
inf {f(x)
< x <
w
:
"}-
/^}.
Then
L(/,P)=£m
f
(f;-f;-i),
fe-I
£(/,
=
(?)
y^m,-(fj
(
To prove
/i
-
Now
is less
that L(f, P)
<
L(f, Q)
tk _\
:
tk
<
< m\u -
-\)
<
x
therefore suffices to
it
tk }
to
tk
-\)
contains
and possibly some smaller ones,
than or equal
^
+ m"(tk -u)+
_i)
m,-ft
-
f,-_i).
i=Jt+l
the set {f(x)
u},
A
=1
m k (tk x <
+ m'(« -r
r,-_i)
the greatest lower
show
+ m"(tk -
all
the
u).
numbers in {f(x)
bound of the
:
so the greatest lower
bound of the second;
mk
that
tk -\
<
first set
thus
<tn'.
Similarly,
mk <
m".
Therefore,
m k (t k
This proves,
U(f, Q)
The
is
tk
= m k (u -
_\)
tk -\)
+ m k (t k
< m'(u -
u)
tk
_\)
+ m"(tk -
u).
P) < L(f, Q). The proof that U(f, P) >
you as an easy, but valuable, exercise.
in this special case, that L(f,
and
similar,
to
is left
now be deduced quite easily. The partition Q can be
P by adding one point at a time; in other words, there is a sequence
general case can
obtained from
of partitions
P
such that Pj +
\
= P ,P2 ,...,Pa = Q
l
contains just one
L(f, P)
=
L(f, Pi
)
more
<
L(/,
point than Pj.
P2 ) <
•
•
-
<
Then
L(f,
Pa =
)
L(/, Q),
and
£/(/, P)
=
U(f, Pi)
The theorem we wish
THEOREM
l
Let P\
on
and P2 be
[<7, /?].
to
> U(f, P2 ) >
prove
is
••
•
>
£/(/,
Pa = U(f,
)
Q). |
a simple consequence of this lemma.
partitions of [a,b],
and
let
/ be
Then
U.f\P\)< U(f,P2 ).
a function which
is
bounded
.
257
13. Integrals
PROOF
There
in
is
a partition
both P\ and
Pi).
P which
According
L(f,
It
contains both P\ and Pj
PO <
(let
P
consist of all points
lemma,
to the
< U(f, P) < U(f, P2 ).
L(f, P)
|
from Theorem
follows
1 that any upper sum U(f, P') is an upper bound for
sums L(f, P). Consequently, any upper sum U(f, P') is greater
the least upper bound of all lower sums:
the set of all lower
than or equal to
sup{L(/, P)
for every P'
.
This, in turn,
P
:
a partition of [a, b]}
means
that sup{L(/, P)}
< U(f,
P').
a lower bound for the
is
set
of all upper sums of /. Consequently,
sup{L(/,P)} <inf {£/(/, P)}.
It is
clear that both of these
/
of
numbers are between
the lower
sum and upper sum
for all partitions:
L(f, P')
<
sup{L(/, P)}
< U{f,
P'),
L(f,P')<in£{U(f,P)} <U(f,P'),
for
all
It
partitions P'
may
well
happen
that
= inf {U(f, P};
sup{L(/, P)}
in this case, this
for
all
partitions,
of R(f,a.b).
and
On
number between the lower sum and upper sum of /
number is consequently an ideal candidate for the area
the only
is
this
the other hand,
if
sup{L(/,P)}
then every
f(x)
=
number* between
<
inf {£/(/, />)},
sup{L(/, P)} and inf {£/(/, P)}
will satisfy
L{f,P')<x<U(f,P')
c
for all partitions P'
show
that
when such an embarrassment of riches will occur. The
many which will soon apboth phenomena are possible.
Suppose
first
that f(x)
It is
a
=
t
t]
t2
tn-\
tn
=b
pear,
FIGURE
not at
all
clear just
following two examples, although not as interesting as
6
any partition of
[a b]
,
,
=
c for
*
all
in [a, b] (Figure 6).
then
irij
—
Mi
=
c,
so
L(f, P)
= J2 c
( fi
~
ti-l)
=
c(b
-
a),
(=1
U(f,P)
=
J2c(ti-ti-i)=c(b-a).
i=l
If
P =
{to,
.
.
.
,
/„} is
258
Derivatives
and
Integrals
In this case,
lower sums and upper sums are equal, and
all
sup{L(/, P)}
Now
=
inf {£/(/, P)}
/ defined by
0,
x irrational
1,
x rational.
|
P=
{to,
.
.
.
,tn }
any
is
-
c(ft
consider (Figure 7) the function
f
If
=
a).
partition, then
m,
=
0,
since there
is
an irrational number
Mj
=
\,
since there
is
a rational
in [/,_],
/,],
and
a
=
FIGURE
7
?o
t\
t2
tn-\
h
=
b
number
in
[/,-_ i, ?,].
Therefore,
n
L(f,P)
= J^0-(t -t . )=0,
i
i
1
/=i
/?
C/(/,P)
= £)l.(f --r _i)=fe-a.
i
I
i=\
Thus, in
principle
this case
it is
upon which
certainly not true that sup{L(/, P)}
the definition of area
was
On
more
generally, that
peal to the
is
insufficient
any area
at
all.
In
fact,
so
is
we can
whenever
sup{L(/, P)}
the region R(f,a,b)
it
The
— any number between
the other hand, the region R(f, a, b)
weird that we might with justice refuse to assign
maintain,
inf {(/(/, P)}.
be based provides
to
information to determine a specific area for R(f, a, b)
and b — a seems equally good.
=
# inf {£/(/, />)},
too unreasonable to deserve having an area.
word "unreasonable"
we
suggests,
As our ap-
are about to cloak our ignorance in
terminology.
DEFINITION
A function
/ which
sup{L( /, P)
:
P
In this case, this
is
bounded on
a partition of
[a, b]
[a, b]}
common number
is
=
is
integrable on
inf{c/(/,
P)
:
P
[a, b] if
a partition of
called the integral of
/ on
[a, b]}.
[a, b]
and
is
denoted by
/
f.
(The symbol J is called an integral sign and was originally an elongated s, for
"sum;" the numbers a and b are called the lower and upper limits (>/ integration.)
The
integral
in [a,
ft].
f f
is
also called the
area of R(f,
a, b)
when f(x) >
for
all
.v
13. Integrals
If
/
259
integrable, then according to this definition,
is
<
L(f, P)
f < U(f, P)
for
partitions
all
P
of
[a, b].
Ja
Moreover, f f is the unique number with this property.
This definition merely pinpoints, and does not solve, the problem discussed
we do not know which functions are integrable (nor do we know how to
/ on [a, b] when / is integrable). At present we know only
before:
find the integral of
two examples:
b
f(x)
if
(1)
=
/
then
c,
is
integrable
on
and
[a, b]
f
/
f
=
c
•
(b
— a).
r
,,
Ja
(Notice that this integral assigns the expected area to a rectangle.)
,~ N
-
(z)
Several
Even
r r
it
,
x
j (x)
=
f
x irrational
0.
.
II,
x
more examples
.
then /
<
.
.
,
.
not integrable on [a,b\.
is
rational,
will
be given before discussing these problems
for these examples, however,
further.
it
helps to have the following simple criterion
is
integrable
for integrability stated explicitly.
THEOREM
2
If
£
/
>
is
bounded on
there
is
[a, b],
a partition
P
then
/
of [a
,
b]
on
[a, b] if
and only
if
for every
such that
U(f,P)-L(f,P) <s.
PROOF
Suppose
first
>
that for every s
there
£/(/,
P)
-
is
a partition
L(f, P)
<
P
with
<
e.
e.
Since
mr{U(f,P')}<U(f,P),
sup{L(f,P')}>L(f,P),
it
follows that
inf {[/(/, P')}
Since
this
is
true for
all
e
>
0,
it
- sup{L(/,
follows that
sup{L(/, P')}
by
definition, then,
If
/
is
/
is
P')}
integrable.
= inf {U(f, />')};
The proof of the
converse assertion
integrable, then
sup{L(/, P)}
This means that for each e
>
= inf {U(f, P)}.
there are partitions P',
U(f,P")-L(f,P') <e.
P" with
is
similar:
.
260
Derivatives
and
Integrals
Let
P be
a partition which contains both P' and P"
.
Then, according
to the
lemma,
U(f,P) <U{f,P"),
L(f,P) >L(f,P');
consequently,
U(f, P)
-
L(f, P)
< U(f, P") -
L(f, P')
Although the mechanics of the proof take up a
that
Theorem 2 amounts
little
<
e.
|
space,
it
should be clear
nothing more than a restatement of the definition
to
of integrability. Nevertheless,
a very convenient restatement because there
is
it
is
no mention of sup's and inf 's, which are often difficult to work with. The next
example illustrates this point, and also serves as a good introduction to the type
of reasoning which the complicated definition of the integral necessitates, even in
very simple situations.
Let
/ be
defined on
by
[0, 2]
,,
fW=
,
10,
x
#
1.
x
=
\
Suppose P
=
tn }
{/(>
a partition of
is
tj-i
tj-i
1
tj
2
(see
Figure
8).
8
with
[0, 2]
1
<
l.
tj
Then
=
mi
FIGURE
<
1
M,
=
if
/'
^
j,
but
=
nij
Mj
and
=
1
Since
7-1
L(f, P)
= ^m,(f, -
tt -i)
+ mjOj -
tj-i)
+ J2
nii(t
'
~
f'-i)«
(=7+1
i=]
/-i
C/(/,P)
^
= 5^M ft--r -_i) + M/(f -r _i)+
i
I
/
/
Mi(ti-ti-i),
t=j+\
/=i
we have
U(f,P)-L(f,P)=tj-tj- h
This certainly shows that
/
is
integrable: to obtain a partition
U(f,P)-L(f,P)<s,
it
is
only necessary to choose a partition with
tj-i
Moreover,
it is
<
1
<
tj
and
tj
—
tj^\
<
e.
clear that
L(f, P)
<
< U(f, P)
for all partitions P.
P
with
13. Integrals
Since
/
is
integrable, there
is
number between
only one
261
lower and upper sums,
all
namely, the integral of /, so
= 0.
/„'
Jo
Although the discontinuity of / was responsible for the difficulties in this example, even worse problems arise for very simple continuous functions. For example,
let
fix)
P =
=
x,
and
/„} is
{to
for simplicity consider
a partition of
=
mi
[0, b],
tt -\
an interval
[0, &],
then (Figure
9)
and
Mi
=
+ *„_! (tn
where b >
0.
If
tt
and therefore
Lif,P)
= J^t
i
-. l
(ti
-t - 1 )
-
1
+ nit2
i
i=\
Uif,P)
=
t
=
Y,ti(ti~ti-l)
it\
)
-?!)
+
••
-t
+
---
•
-r„_i),
1=1
=
FIGURE
9
t\it\
Neither of these formulas
for partitions
t,
—
tj_\
Pn —
{to,
is
,
of each subinterval
+
-to)
t2
it 2
l
)
+
tl ,it„
-r„_i).
particularly appealing, but both simplify considerably
tn )
is
into n equal subintervals.
b/n, so
fo
= 0,
t\
=
b
-,
n
t2
=
2b
—
,
etc;
n
in general
ib
n
Then
Lif,Pn
)
= Y,h-\iU-t,-
l
)
i=l
A \{i-i)b\
i=\
1
n
1 u2
Etf-D
"-/=1
n-\
b2
E^'b
y=o
b
In this case, the length
262
Derivatives
and
Integrals
Remembering
the formula
!
this
k(k+\)
+ ••• + *
can be written
b2
(n-l)(n)
n-1
£
«
2
2
Similarly,
£/(/, />„)
= ]£*,&-?,-!)
^
i
=
n
n
\
+ l) b 2
~T~ "n 2
n+\ b 2
n(n
If n
is
makes
very large, both L(f, P„) and £/(/,
it
show
easy to
that
/
is
Pn
)
are close to & /2,
integrable. Notice
and
this
remark
that
first
2
U(f,Pn )-L(f,Pn )
2 b
= ---.
1
n
Pn
This shows that there are partitions
By Theorem 2
with only a
the function
little
work.
/
is
with U(f, P„) — L(f,
It is clear, first
)
as small as desired.
fQ f may now be found
Moreover,
integrable.
Pn
of all, that
h2
L(f,Pn )<
-j<U(f,Pn )
This inequality shows only that b /2
sums, but
fix)
=
desired, so there
xy
has
area
we have
only one number with
we can conclude that
—
2
Jo
Notice that
I
H.I K
I.
Ml
tude b (Figure
it
this
10).
between certain
just seen that £/(/, P„)
is
this property,
lies
for all h.
—
L(f,
Pn
this property.
)
special
can be
upper and lower
made
as small as
Since the integral certainly
"f^
equation assigns area b /2 to a right triangle with base and
Using more involved calculations, or appealing
can be shown that
h
I
b2
to
alti-
Theorem
4,
.
13. Integrals
The
ure
function f(x)
1 1), if
P =
{to
=
x
tn )
m =
{
presents even greater
a partition of
is
=
/fa_i)
2
Pn =
Choosing, once again, a partition
the lower
M,
.
.
=
tn
,
f(tt )
=
2
t,
.
into n equal parts, so that
)
-b
i
=
ti
{to,
In this case (Fig-
difficulties.
then
[0, b],
and
(/,_i)
263
and upper sums become
E,
n
b
ib
n*
n
(=i
FIGURE
11
;
=
1
A .b
;
=
/?
77
1
b
-
7?
1
3
"
./
=
!
Recalling the formula
2
l
from Problem
2- 1
,
these
+
+ ^2 =
---
sums can be written
L(f,
Pn ) =
l/C/,/3,,)
=
and
+
l)
as
—
.-(n - l)(n)(2n 6
n
1),
— .-(n+
6
l).
J
77
It is
±*(*+l)(2Jfc
l)(n)(2/7
+
•*
not too hard to show that
that U(f, P„)
sufficiently large.
—
L(f,
b
3
L(f,Pn )<
J
Pn
made
The same
)
can be
sort
<U(f,Pn ),
as small as desired,
of reasoning as before then shows that
b
i
'o
/=
b3
-*3
This calculation already represents a nontrivial result
bounded by a parabola
theless, the result
is
by choosing n
—
the area of the region
not usually derived in elementary geometry.
was known
to
Archimedes,
who
derived
it
Never-
in essentially the
same
.
.
264
Derivatives
and
Integrals
The only superiority we can claim is that in the next chapter we
much simpler way to arrive at this result.
Some of our investigations can be summarized as follows:
way.
a
if
=
/ =
i:
h
c
b
— a)
(b
2
a
f{x)
—
if
fix)
=x
if
/ (x ) = x
c for
all
x.
2
y-y
b3
if
will discover
a3
for all x,
for
all
x
i.
This
list
already reveals that the notation
notation for
naming
f f
suffers
from the lack of a convenient
functions defined by formulas. For this reason an alternative
notation,* analogous to the notation lim f(x),
x—>a
also useful:
is
rh
I
r>b
f{x)dx
means
same
precisely the
as
Ja
/.
/
Ja
Thus
/' cdx =
c
(b
•
—
a),
Ja
L
XdX=
Ja
u2
b
a
2
2
2
3
3
Notice that, as in the notation lim f(x), the symbol x can be replaced by any
x—>a
f,a,orb, of course):
other letter (except
pb
\
Ja
f(a)da=
f(t)dt=
Ja
The symbol dx
pb
pb
nb
f(x)dx=
rb
fiy)dy=
Ja
Ja
f{c)dc.
Ja
has no meaning in isolation, any
more than
the symbol x
->•
has any meaning, except in the context lim f(x). In the equation
/'
2
3
b>
fl
Ja
*The
notation
/ f(x)dx
is
actually the older,
and was
for
many
years the only, symbol for the
symbol because he considered the integral to be the sum (denoted by /)
of infinitely many rectangles with height f(x) and "infinitely small" width dx. Later writers used
xq,
x n to denote the points of a partition, and abbreviated x\ — .v,_i by A.v,. The integral was
integral. Leibniz used this
.
.
,
77
defined as the limit as Ax, approaches
of the sums
Y^
/
'(
v,
)
,
./
A,v,
(analogous to lower and upper
(v,
to fix),
7=1
sums).
The
many
people.
fact that the limit
is
obtained by changing
>_] to
/
)
and A
v,
todx, delights
265
13. Integrals
the
entire
symbol x dx may be regarded
the function
This notation for the integral
amples
may
appear;
for all x.
Several ex-
as flexible as the notation lim f{x).
use of Theorems 5 and 6.*
h
b
b
f
/
is
=x
that f(x)
for:
aid in the interpretation of various types of formulas which frequently
we have made
(1)
/ such
an abbreviation
as
(x
+
y) dx
f
=
f
+
x dx
2
2
b
a
—
- — +y
=
y dx
-
(b
a).
J
(2)
/
(y
+ t)dy=
Ja
ydy+
/
Ju
(3)
l!(J!
—
%-
+ t(x-a).
Ja
dx=
a+,)dt
)
(1
+t)(x
-a)dx
I.
= (1+/)
=
(4)
X
=
tdy
/
(x-a)dx
/
I
b
2
(1+/)
L\Ox+y)dy ) dx =l
The computations of f x dx and f x dx may suggest
is
— a)
a(b
generally difficult or impossible.
tions are impossible to
As a matter of fact,
determine exactly
{although they
that evaluating integrals
may
be computed
of accuracy desired by calculating lower and upper sums). Nevertheless, as
the next chapter, the integral of
Even though most
to
know when
precisely
/
is
Theorem
cannot be computed
integrable
be stated here, and we
on
[a b]
,
.
will
have
we
shall see in
computed very
exactly,
easily.
important
it is
Although
any degree
to
it
is
at least
possible to say
for integrability
is
to settle for partial results.
a
little
The
too
next
most useful result, but the proof given here uses material from
Chapter 8. If you prefer, you can wait until the end of the next
gives the
the Appendix
to
when a
totally different
*Lest chaos overtake the reader
when
qualification. This equation interprets
value f(x)
functions can be
which functions are integrable, the criterion
difficult to
chapter,
integrals
a function
many
most func-
the integrals of
is
the
number
v.
But
proof will be given.
consulting other books, equation
f ydx
classical
integral of some arbitrary Junction y.
to
mean
(1)
requires an important
the integral of the function
notation often uses y for
v(.v),
so
/ such
that each
f ydx might mean
the
266
Derivatives
and
Integrals
THEOREM
3
PROOF
If
/
continuous on
is
Notice,
every e
/
>
/
that
first,
prove that
is
/
then
[a, b],
on
integrable
is
[a, b].
bounded on [a,/?], because it is continuous on [a,£]. To
on [a, b], we want to use Theorem 2, and show that for
is
integrable
there
is
P of
a partition
such that
[a b]
,
U(f,P)-L(f,P) <e.
Now we
Theorem
know, by
continuous on [a b]
,
.
if \x
of the Appendix to Chapter
1
So there
— y\ <
The trick is simply to choose
8. Then for each
we have
is
some
>
8
such that for
—
then \f{x)
8,
a partition
P =
8, that
all
/
is
x and y
uniformly
in [a
,
b]
,
£
<
f(y)\
{to
tn
2{b-a)
such that each
]
\t,
— /,_i <
|
i
£
\f(x)-f(y)\<
and
[f,-_i,
f,-],
follows easily that
it
Since
for all x, y in
2{b-a)
this
is
true for
M
-
we
then have
all /,
[/(/,
-
P)
>"/
<
2(b-a)
<
b-a
= ^(M,- -
L(f, P)
<
b
—
a
=
which
is
what we wanted.
Although
this
theorem
integrals in this book,
it
—
ti-i)
£>-',-!
1
£
b
-
m,)(t,
b
—
a
a
|
will
is
provide
more
all
the information necessary for the use of
satisfying to
have a somewhat larger supply of
integrable functions. Several problems treat this question in detail.
It will
know
on
is
the following three theorems, which show that
integrable
on
[a, c]
and
[c, b];
that
/+g
is
/
is
integrable
integrable
if
/ and g
help to
[«./?], if
are;
and
it
that
f is integrable if / is integrable and c is any number.
except at one
As a simple application of these theorems, recall that if / is
point, where its value is 1, then / is integrable. Multiplying this function by c, it
follows that the same is true if the value of / at the exceptional point is c. Adding
c
such a function to an integrable function,
function
may
be changed arbitrarily
By breaking up
changed
The
in
the interval into
at finitely
many
at
many
we
see that the value of
an integrable
one point without destroying
subintervals,
we
integrability.
see that the value can be
points.
proofs of these theorems usually use the alternative criterion for integrability
Theorem
2; as
some of our previous demonstrations
illustrate,
the details of the
.
267
13. Integrals
argument often conspire to obscure the point of the proof It is a good idea to
attempt proofs of your own, consulting those given here as a last resort, or as a
check. This will probably clarify the proofs, and will certainly give good practice
used
in the techniques
THEOREM
4
<
Let a
on
<
c
b.
/
If
Conversely,
[c, b\.
[a b]
,
is
on
integrable
is
Suppose /
[a b]
,
is
integrable
/
on
Ja
and on
[a, c]
/
then
[c, b],
integrable
is
Jc
>
[a, b]. If £
on
integrable
is
integrable
Ja
PROOF
then
[a, b],
on [a, c] and on
integrable on [#,£], then
/
/ is
if
Finally, if
.
some of the problems.
in
0, there
is
P =
a partition
[to,
.
.
.
,
tn )
of
such that
U(f,P)-L(f,P) <e.
We might
which contains to,...,t„ and
U(f,P)-L(f,P) <s.)
Now
of
P'
[c, b]
=
{to,
=
assume that c
as well
.
.
.
,tj} is
c;
some
for
tj
Q
then
a partition of
j
(Otherwise,
.
let
Q
be the partition
—
contains P, so U(f, Q)
=
and P"
[a, c]
{tj,
.
.
/„} is
,
L(/, Q)
<
a partition
(Figure 12). Since
L(f,P)=L(f,P')
U(f,P)
=
U(f,P')
+ L(f,P"),
+ U(f,P"),
we have
U
=b
[U(f, P')
-
L(f, P')]
+
Since each of the terms in brackets
FIGURE
12
that
/
is
integrable
on
-
[U(f, P")
[a, c]
and
is
L(f, P")]
=
U(f, P)
nonnegative, each
[c, b].
Note
is
less
-
L(f, P)
than
e.
<
s.
This shows
also that
L(f,P')< f f<U(f,P'),
L(f,P")<
/
f<U{f,P"),
so that
L(f,
Since
this
is
P)<
f
f+
<
f f
U(f, P).
true for any P, this proves that
ff + Jcff=ff.
J (i
Ja
Now
suppose that
partition P' of [a, c]
/
is
on [a,
partition P" of
integrable
and a
c]
[c,
and on
[c, b].
b] such that
U{f,P')-L{f,P') <e/2,
U(f,P")-L(f,P")<e/2.
If e
>
0, there
is
a
268
Derivatives
and
Integrals
If
P
the partition of [a, b] containing
is
the points of P'
all
L(f,P)=L(f,P')
+ L(f,P")
U(f,P) =
+
U(f,P')
and P", then
)
U(j\P");
consequently,
U(f, P)
-
L(f, P)
Theorem 4
f f was
is
=
-
[U(f, P')
some minor
the basis for
defined only for a
L(f, P')]
<
[U(f, P")
Ja
With these
f
f
Ju
definitions, the
a
<
c
< b
is
The
e. |
integral
if
>
a
b.
Jb
=f f
f +f f
equation f'
t
if
<
pa
and
=
L(f, P")]
the definitions
ph
f
-
notational conventions.
We now add
b.
+
not true (the proof of this assertion
is
holds for
all
a, c, b
check).
THEOREM
5
If
/ and g
are integrable
on
[a, b],
then
PROOF
Let
P —
{to
tn )
(/+*)=
(f + g)
be any partition of
m,
and define M,,
=
inf {(/
= inf {fix)
m,"
=
inf{g(x)
M,', M,-" similarly.
it is
/+
:
It is
f,_i
on
[a, b]
and
/
Ja
[a, b\.
:r,_i
integrable
pt
rb
/
Ju
+ g)(x)
mi
rm
but
is
no
pb
.&
no
/
Ja
/+g
:
Let
?;_]
< x <
/,},
< x <t },
< x < t,},
t
not necessarily true that
= m/ + m/',
true (Problem 10) that
m, >
m{ +
mj".
Mj <
M/ +
Mi".
Similarly,
Therefore,
L(f,P)+L(g,P) <L(f
+ g,P)
and
U(f + g,P) < U(f,P) + U(g,P).
Thus,
L(f, P)
Since
+
/ and g
L(g, P)
< L(f
+ g,
P)
<U(f + g,
are integrable, there are partitions
even
a rather tedious case-by-case
P)
< U(f,
P\ P" with
U(f,P')-L(f,P') <e/2,
U(g, P")-L(g,P") <e/2.
P)
+
U(g, P).
269
13. Integrals
P
If
and P", then
contains both P'
+
P)
[/(/,
-
U(g, P)
+
[L(f, P)
L(g, P)]
<
e,
and consequently
U(f + g,P)-L(f + g,P) <s.
/+g
This proves that
L(f, P)
(1)
on
integrable
is
+
[a b\
,
P)<L{f + g,
L{g,
<
pb
rb
L(f.P)
/+/
+ L(g.P)<
Ja
Since U(f, P)
desired,
—
L(g, P) can both be
-
[L(f, P)
made
P)
+
U(g, P)
as small as desired;
pb
(/
/
+ *)=
on
[a
,
integrable
b]
as small as
on
[a, b],
L(g, P)]
Ja
then for any
(1)
and
(2)
that
fb
/+/Ja
/
Ja
is
+
therefore follows from
it
pb
/
made
follows that
it
can also be
If
<U(f,P) + U(g,P).
g
Ja
—
L(f, P) and U(g, P)
£/(/,
6
< u(f,P) + U( g ,py,
+ g ,P)
also
(2)
THEOREM
P)
+ *)
(/
J
< u(f
and
Moreover,
.
number
g-l
c,
the function cf
integrable
is
and
f-
Ja
PROOF
The proof (which
much
is
Ja
easier than that of Theorem 5)
idea to treat separately the cases c
(Theorem 6 is
on [a
Problem 38).)
integrable
In this chapter
theorems with
Appendix
to
>
and
c
just a special case of the
,
b]
,
if
/ and g
are,
we have acquired
intricate proofs,
Chapter
8.
This
but
<
0.
Why?
is left
to you. It
is
a
good
|
more general theorem
this result
is
/
that
•
g
is
quite hard to prove (see
only one complicated definition, a few simple
and one theorem which required material from the
is
not because integrals constitute a
more
difficult
topic than derivatives, but because powerful tools developed in previous chapters
The most
have been allowed to remain dormant.
is
the fact that the integral
and
the derivative are intimately related
learn the connection, the integral will
easy to use.
The connection between
chapter, but the preparations
hint.
We
first state
many important
significant discovery of calculus
become
derivatives
which we
will
— once we
as useful as the derivative,
and
make
and
as
integrals deserves a separate
in this chapter
may
serve as a
a simple inequality concerning integrals, which plays a role in
theorems.
)
270
Derivatives
and
Integrals
THEOREM
7
Suppose /
is
on
[a b]
and
m <
fix)
<
integrable
,
that
M
for
all
x
in [a,b].
Then
m(b-a)<
PROOF
It is
-a)<
for every partition P.
fa f =
Since
Suppose now that /
[a,b] by
+ h) -
F(c)
is
and
L(f, P)
inequality follows immediately.
F(c
a).
clear that
m(b
area
f < M(b -
/
[/(/,
M(b -
P) <
=
sup{L(/, P)}
a)
m£{U(f, P )},
the desired
|
on
integrable
F(x)=
We
[a, b].
f=
f
f
can define a new function / on
f(t)dt.
on Theorem 4.) We have seen that / may be integrable even if it
not continuous, and the Problems give examples of integrable functions which
(This depends
FIGURE
is
13
are quite pathological.
THEOREM
8
If
f
is
integrable
on
The behavior
[a, b]
F
and
is
of
F
is
therefore a very pleasant surprise.
defined on
by
[a, b]
Fix)
Ja
F
then
PROOF
continuous on
is
Suppose
c
on [«,&];
is
let
[a, b].
/ is integrable on
number such that
in [a, b]. Since
M be a
<
|/U)|
If
h >0, then (Figure
M
for
all
x
[a, b]
by
definition,
in [a, b].
13)
rc+h
Fic
it is,
+ h)-
re +h
nc
F{c)
f~
I
Ja
f =
Ja
f•/<'
Since
-M
it
follows
from Theorem 7
< fix) <
M
for
all
x,
that
/c+h
f < Mh:
in other
words,
(
If h
<
0,
-
1
M
-h
< Fic + h)-
Fie)
<
M
a similar inequality can be derived: Note that
.,
Fic
+ h)-
t/,
Fie)
Jc
J c+h
•
/?.
bounded
13. Integrals
Applying Theorem 7
to the interval [c
+ h,
Mh <
c],
of length —h,
we
271
obtain
f < -Mh;
Jc+h
—
multiplying by
1
which reverses
,
Mh
(2)
Inequalities
(1)
and
(2)
if e
>
0,
the inequalities,
< F(c
+h)-
+ h) -
\h\
words F
< s/M.
is
this
F
is
F(c)\
<
+
h)
FIGUKI.
I
4
s,
=
F(c);
f
for various functions /;
at c. |
/ and F(x) —
x
f'
always better behaved than /. In the next chapter
is.
/
/
\h\.
This proves that
continuous
Figure 14 compares
that
M
we have
lim F(c
in other
< -Mh.
F(c)
<
F(c)\
\F(c+h)provided that
we have
can be combined:
\F(c
Therefore,
all
/
we
it
will see
appears
how
true
272
Derivatives
and
Integrals
PROBLEMS
1.
4
3
fQ x dx — b /4, by
Prove that
considering partitions into n equal subin-
n
Tjr
using the formula for
tervals,
which was found
in
Problem
2-6.
This
i=\
problem requires only a straightforward imitation of calculations
but you should write it up as a formal proof to make certain that
all
the fine
Using Problem 2-7, show that the sum >^ k p /n p+ can be made
as close
argument are
points of the
2.
*3.
Prove, similarly, that
(a)
in the text,
clear.
4
5
f x dx = b /5.
k=l
+
to \/(p
(b)
1) as
by choosing n large enough.
desired,
=
Prove that j^x p dx
b p+l /\p
+
l)
.
b
*4.
way
This problem outlines a clever
P =
=
a
result for
/„} for
{to
partitions for
(a)
will
Show
which
then follow by
which
that for such a partition
(b)
If
f(x)
=
—
f,
—
P we
in
i/n
,
U(f,P)
=a p+ \\ -
C
- 1/
is
")^](c
,
'
,+
_
1
£P+l )c (P+l)/«
1
=
(b
p+]
Problem
2-5, that
1,/ ")'
-r_ c (p+l)/n
1
(c)
and find a similar formula
Conclude that
L
p
dx
=
(You might find Problem 5-41
b p+l_ a p+l
-
p
+
useful.)
Evaluate without doing any computations:
/.I
(i)
/
A-Vl
(ii)
f
(x
5
+
~X 2 dx.
3)y/\
-x 2 dx.
/"
for L(f, P).
I'
K
+c
[
1
b.
(The
are equal, instead of using
1
-a p+l )c p/n
<
to use partitions
|/,!
p+l
( fl
a
have
x p show, using the formula
c
trick
<
are equal.
=
=a
The
ti/tj_\
£;_]
for c
ti
x p dx for
/
Ja
continuity.)
all ratios r
differences
all
f
to find
H
he/'/"
,
.
13. Integrals
273
Prove that
sin
J
dt
f
Jo 7TT
for
all
x >
> o
0.
Decide which of the following functions are integrable on
the integral
(i)
/(*)
when you
0<x
x-2,
1
x
/(*)
x-2,
*
(iv)
f(x)
1
./•(.*•)
.v
=
+
[.r]
=
/
is
< jc <
1
< a <
2.
irrational.
+ bv 2
for rational a
and b
x not of this form.
<
A
<
1
=
the function
FIGURE
8.
2.
x of the form a
1,
x
0,
(vii)
1
x rational
,
1
f(x)
<
< a <
x
0,
(vi)
calculate
+ W0,
(v)
and
can.
x,
=
[0, 2],
=
shown
or X
in
>
1
Figure 15.
15
Find the areas of the regions bounded by
—+
(i)
the graphs of f(x)
=x
and g(x)
=
(ii)
the graphs of f(x)
=
and g(x)
= —a -
and g(x)
and #(a)
=
=
and g(x)
=x
x
2.
and the
vertical lines
through
(-1,0) and (1,0).
(iii)
the graphs of /(a)
(iv)
the graphs of f(x)
(v)
the graphs of f(x)
=a
—x
—x
1
1
—
jc
.
— x and h(x) = 2.
— 2x + 4 and the vertical
axis.
274
Derivatives
and
Integrals
=
the graph of f(x)
(vi)
through
(Don't try to find
(2, 0).
and
s/x, the horizontal axis,
s/x dx\
/
the vertical line
you should see a way of
Jo
know how
guessing the answer, using only integrals that you already
evaluate.
in
9.
The
Problem 2
questions that this example should suggest are considered
1 .)
Find
[(j:< (x)g(y)dy
in
fa f and fc
terms of
vengeance;
10.
g.
crucial that
it is
dx
)
(This problem is an exercise in notation, with a
you recognize a constant when it appears.)
Prove, using the notation of Theorem 5, that
+m" -
m,-'
11.
to
(a)
Which
+
inf{/(.xi)
gixj)
:
<
< x\,xj <t
/,_i
t ]
///,-.
sum
functions have the property that every lower
equals every
upper sum?
(b)
(c)
Which
functions have the property that
(other)
lower sum?
Which continuous
some upper sum equals some
all
lower sums are
all
lower sums are
functions have the property that
equal?
*(d)
Which
integrable functions have the property that
equal? (Bear in
=
f(x)
mind
Problem
12.
set,
Ua<b<c<d and /
(a)
(b)
work
/
p/q
/
/>
0.
Ja
Prove that if
is
/ and g
then
14.
on
integrable
integrable
is
Hint:
for x irrational,
You
will
need the
[a, d],
prove that
/
is
integrable
on
on
and f(x) >
[a, b]
are integrable
on
[a, b]
for all
x
and f(x) > g(x)
in [a, b],
for
all
x
rh
/ >
/
Ja
that if
in lowest terms.)
=
introduced in Problem 8-6, as well as the results of
pb
in [a, b],
f(x)
is
hard.)
if
Prove that
then
one such function
30.
[6,c]. (Don't
13.
=
\/q for x
notion of a dense
that
/
g.
now
(By
it
should be unnecessary to warn
Ja
you work hard on part
(b)
you are wasting
time.)
Prove that
no
/
b+c
fb+c
f(x)dx =
/
f(x-c)dx
Ja+c
Ju
(The geometric interpretation should make
partition
...,/„+
P —
c}
{/(),...,/„}
of \a
+
c,
b
of
+ c\,
this
very plausible.) Hint: Every
\ci,b] gives rise to a partition
and conversely.
P'
—
[tQ
+ c,
c
15.
For a, b
>
/
r
1
-dt +
t
J\
l
-dt
t
J\
r*abw
=
16.
pan
...,/„} of [1
=
\/tdt.
I
=
a] gives rise to a partition P'
,
[bt$,
.
.
.
,
btn
)
of
[b,
ab],
conversely.
Prove that
f
f(t)dt
=c
J cu
Given
ellipse
a special case.)
is
by the unit circle, described by the equation
use Problem 16 to show that the area enclosed by the
that the area enclosed
+y =
x
f f(ct)dt.
Ja
Problem 15
(Notice that
17.
P =
t
Jx
l/tdt
:
Every partition
1
-dt.
/a
and
275
prove that
\
r
{to,
13. Integrals
7i,
is
1,
+y
described by the equation x /a
=
/b
nab.
is
1
h
18.
This
problem
way
outlines yet another
to
f
compute
/
x" dx;
it
was used by
Ja
Cavalieri,
one of the mathematicians working
just before the invention of
calculus.
=
Let c n
(b)
Jo
Problem 14 shows that
/
formula
(x
19.
Suppose
Problem 2-3
/
that
to
in [a,b] with the exception of
[a b]
,
20.
.
.
+ a)"dx.
£ QW
prove that
bounded on
is
n+
to prove that
k
Now use
cn a
J-a
2" +1 c„a"+'=2a" +1
(c)
=
x" dx
/
pa
x"dx =
Jo
this
that
Jo
p la
\
Use
show
x" dx. Use Problem 16 to
(a)
[a, b]
.\"o
even
c„
and
=
\/{n
that
/
+
is
1).
continuous
Prove that
in (a,b).
at
each point
on
/
is
integrable
/
is
automatically
Hint: Imitate one of the examples in the text.
Suppose
/
that
bounded on
P =
(a)
If
(b)
Suppose
nondecreasing on
is
[a, b],
{r
,
.
.
that
.
,
f,
[a,fr].
Notice that
because f(a) < f(x) < f{b)
tn } is
a partition of [a, b],
—t{-\
=8
for
each
i.
for
what
x
is
Prove that
in [a, b\.
L(f, P) and U(f, P)?
U (f,
P)
—
L(f, P)
=
8[f(b)-f(a)].
(c)
Prove that
(d)
Give an example of a nondecreasing function on
/' is
integrable.
tinuous at infinitely
many
points.
[0,
1
]
which
is
discon-
276
and
Derivatives
Integrals
It
might be of interest
from Newton's
to
compare
this
problem with the following extract
Principia*
LEMMA II
If in any
any number of parallelograms Ab, Be,
AB, BC, CD, &c, and
Aa
one side
of the figure; and
D
E
number
their
and
to
Bb, Cc, Dd, &c,
the sides,
the inscribed figure
AKbLcMdD,
AabcdE,
/
say,
sum
be diminished,
to
the circumscribed figure
sum
area b
total
f
l
(b)
Lm, Mn, Do,
their bases), the rectangle
of their altitudes Aa, that
because
its
breadth
AB
is
is,
that
under one of
AalbmcndoE,
"21.
f-
l
Suppose
that
/
area a
FIGURE
f
(a)
f~
t,,}
,
Prove
(b)
Now prove
(c)
Find
22.
:
Suppose
/
that
that for a, b
af~\a)
is
and
a
K
I.
/.
that, as
let
P'
=
]
{f~
(t
),
suggested in Figure 17,
>
l
,
P)
+
U(f, P')
= bf~\b) -
= /o
t]
t2
ti
t„
=
b
I
7
af~\a).
dx
is
< a <
for
b.
a continuous increasing function with f(0)
Youngs inequality,
—
0.
Prove
we have
that equality holds
if
and only
if
l
I
b
f~ (x)dx,
—
f{a). Hint:
Draw
a picture
like
Figure 16!
Newton's
I
(b)
/
a partition of [a,b],
ab< f f{x)dxJo
.
and much
the formula stated above.
&
[
Ja
i
the figures inscribed
16
L(f~
I
less
Jf-Ha)
{t
(/„)}.
and the
becomes
to the other;
-
Ja
I
1)
l
P =
Kb
in infinitum,
ff-
l
(from the equality
increasing. Figure 16 suggests that
is
/' /"• =bf~\b)
If
the
curvilinear figure be ultimately equal to either.
(a)
(a)
is
the rectangle ABla. But this rectangle,
supposed diminished
And therefore (by Lem.
and circumscribed become ultimately equal one
more will the intermediate
QE.D.
is
their bases
than any given space.
f~\b)
are
will have to one another, are ratios of equality.
of the parallelograms Kl,
all
parallel
that the ultimate ratios
For the difference of the inscribed and circumscribed figures
of
acE,
aKbl, bLcm, cMdn, &c,
the parallelograms
be augmented in infinitum,
curvilinear figure
the curve
Cd, &c, comprehended
completed: then if the breadth of those parallelograms be supposed
which
B F C
Aa, AE^ and
under equal bases
and
A
terminated by the right lines
there be inscribed
to
M
AacE,
figure
Principia,
A
Revision ofMott's Translation, by Florian Cajori.
Press, Berkeley, California, 1946.
University of California
)
.
.
211
13. Integrals
23.
Prove that
(a)
[a, b],
if
/
m <
on [a,b] and
integrable
is
M
<
f(x)
for
all
x
in
then
/"
f{x)dx
=
(b
—
a)ii
Ja
for
some number
Prove that
(b)
if
/
is
with
/i
m <
< M.
i±
continuous on [a
,
b]
then
,
/' f(x)dx = (b-a)f($)
for
some
£ in [a b]
,
(c)
Show by an example
(d)
More
that continuity
essential.
is
continuous on [a b] and that 8
integrable and nonnegative on [a b] Prove that
generally, suppose that
/
is
,
.
,
/' f(x)g(x)dx = /($)
some
f g(x)dx
/
Ja
Ja
for
is
£ in [a b]
,
.
This result
is
called the
Mean
Value Theorem for
Integrals.
FIGURE
18
24.
(e)
Deduce
(f
Show
same
the
that
angle 9.
4,
g
is
integrable
one of these two hypotheses
we
In this problem
(Chapter
result if
6
is
consider the region
for
g
3).
measured
A shown
area
Let
[a
,
/ be
b]
,
19
in radians, the area of this sector
where the curve
in Figure 19,
A =
1
f
Show
r~
—
.
Now
is
the graph in
is
a partition of
that
2
f(0) d0.
[a, b]. If
P —
{to
/„
define
P)
is
Hl
« /
z J0o
a continuous function on
£(/,
b]
with central
circle,
n
'IGURK
,
essential.
Figure 18 shows a sector of a
polar coordinates of the continuous function /.
*25.
is
[a
consider the graph of a function in polar coordinates
Appendix
When
and nonpositive on
= J2 V^-f;-,) 2
+[/(*;) -/(f;_i)]
2
.
}
278
Derivatives
and
Integrals
The number
£(f, P) represents the length of a polygonal curve inscribed in
the graph of
/
the least
of all such £(f, P)
If
/
If
(b)
—
to
t\
h
ti
?4
=
b
/
such that £(/, P)
(You
FIGURE
will
P
{provided that the set
/
prove that the length of
,
prove that there
P =
a partition
is
is
the
{a,t,b} of [a, b]
greater than the distance from (a, f{a)) to
is
be
[a, b] to
to (b, f(b)).
f (a))
linear,
partitions
above).
,
not
is
all
a linear function on [a b]
is
length of / on
define the
i(f, P) for
bounded
is
distance from (a,
a
We
Figure 20).
(see
upper bound of all
need Problem
(b,
f{b)).
4-9.)
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
Conclude
20
that of
/ on
functions
all
=
with f(a)
[a, b]
c
the shortest distance between two points".)
Suppose that
(d)
show
/'
bounded on
is
[a b]
,
P
If
.
Use
Hint:
(e)
Why
is
(f)
Now
show
+(f)\
the
Mean
+
sup{l(\/l
(/')
grable on [#,&]. Hint:
2
how
<
P)}
,
<
2
(f')
P).
,
sup{£(/, P)}? (This
inf \U (V
+
1
2
(./")
,
is
easy.)
P)j, thereby proving
2
v 1
if vl
/
if')
Ja
suffices to show that if P' and
[a,b]
It
partitions, then £(/, P')
+
is
2
< U{J\ +
(./")
,
P
P"). If
,
+
2
(f)
P" are any two
contains the points
does i(f, P') compare to €(/, P)?
Let i£(x) be the length of the graph of / on
v
if
1
+
lim f'{x)
(/')
=
is
inte-
is
and
[a, x],
d(x) be the
let
Show
length of the straight line segment from (a, f{a)) to (x,f(x)).
if
,
Value Theorem.
/ on
of both P' and P",
U(Vl +
P) < i(f, P) <
that sup{£(/, P)}
that the length of
that
[a b]
that
L(\/\
(g)
any partition of
is
integrable
on
[a, b]
and
/'
is
continuous
at
a
(i.e.,
f(a)), then
=
lim
1.
x^a+ d{x)
Hint:
In Figure 21, the part of the
(h)
size
of the part between 5 and
the size of the part
I
!'.
i
RE
part
21
26.
A
{/<)
of
at
(a)
.v
s
defined on
tn
tj
between
does not hold for
(g)
function
P —
Mean Value Theorems.
graph of / between ^ and ^ is just
help to use a couple of
It will
]
of
may be
Prove that
if
\a, b]
[a, b\
1
,
the part
and
^
5, etc.
between
Show
g
and
4
is
half the
just half
that the conclusion of
this /.
is
step function if there
a constant on each (f,-_]
called a
such that
5'
is
,
is
a partition
/, )
(the values
arbitrary).
/
is
integrable on [a, b], then for any £
>
there
is
a step
.
279
13. Integrals
pb
pb
function
< f
s\
with
/ —
I
pb
Ja
f <
I
such that
—
S2
/
>
there are step functions
<
s\
I
/
Prove that
e.
< f and
s\
si
> f
f"
/ =
integrable.
is
Ja
Ja
/ which
Find a function
(c)
> f
also a step function S2
s.
Ju
that for all s
pb
pb
Suppose
(b)
and
e,
pb
—
S2
I
<
Ja
Ja
with
s\
/
not a step function, but which
is
satisfies
/
Ja
P
L(f, P) for some partition
*27.
Prove that
/
if
integrable
is
on
of
[a, b]
pb
< f <
functions g
h with
28.
Show
(a)
that
and
if s\
>
then for any e
/
—
//
there are continuous
/
£
g <
Hint: First get step functions
Ja
and then continuous
property,
this
,
pb
Ja
with
[a, b\.
A picture will
ones.
S2 are step functions
on
[a, b],
then
Prove, without using
Use part
(c)
29.
Suppose
[a b]
,
/
integrable
is
such that
/
/
=
*30.
(51+52)=
/
on
/ Show by example that
choose x to be in
Let
b
(b)
—
P =
tn
{to
)
be a partition of
Prove that for some
a.
it is
is
not
is
integrable
we have M, — m, <
/
—
<
1.
[«!,/?]]
a\
S2-
number x
5.
in
always possible
< x <
[ti_i
ti\
,
b\)
inf{/(jc)
from part
(a)
<
a\
:
unless
i
x
=
1
on
<
[a b]
,
-
,
then
L(f, P)
<
1.
a\
:
a
with (/(/, P)
[a, b]
<
=
I
Ja
proof of Theorem
Prove that there are numbers a\ and b\ with a
sup{/(A")
+
{a, b).
The purpose of this problem is to show that if /
/ must be continuous at many points in [a, b].
(a)
s\
/
also.
is
pb
Ja
Prove that there
[a,b].
+ S2
JX
«/</
to
/
5, that
Ja
26) to give an alternative
(and Problem
(b)
that
Theorem
s\
pb
pb
(b)
help immensely.
b\)
<
< b and
b\
(You can choose
or n; and in these two cases
a very simple device solves the problem.)
(c)
/?2 with a\ < 02 < ^2 < b\ and
—
<
inf
< x < ^2} < hci2
«2 S
^2}
[fix)
Continue in this way to find a sequence of intervals /„ = [a„, b n ] such
that sup{/(x)
x in /„} — inf{/(x) x m I,,} < l/n. Apply the Nested
Intervals Theorem (Problem 8-14) to find a point x at which / is con-
Prove that there are numbers a2 and
sup{/Cv)
(d)
x
:
:
:
:
tinuous.
(e)
Prove that
Let
/ be
/
is
continuous
at infinitely
many
points in [a b]
,
b
*31.
integrable
on
[#,&].
Recall,
from Problem
13, that
f
/
/ >
if
Ja
fix) >
(a)
for all x in [a,b].
Give an example where fix) >
[a,b],
and
j
Ja
f
= 0.
for all x,
and fix) >
for
some
.v
in
a
280
Derivatives
and
.
Integrals
Suppose f{x) >
(b)
and f(xo) >
for
x
all
f"
Prove that
0.
and /
in [a, b]
>
j
/
Hint:
0.
continuous
is
at
xo in [a,b]
to find
It suffices
one lower
Ja
sum
L(f, P) which
Suppose f
(c)
positive.
is
on
[a, b]
You
will
integrable
is
and f(x) >
x
for all
in [a, b\.
Prove
rb
that
/ >
/
Hint:
0.
need Problem 30; indeed
that
was one
Ja
reason for including Problem 30.
b
*32.
Suppose
(a)
/
that
continuous on
is
f
and
[a, b]
fg=0 for all continuous
/
Ja
on
functions g
[a, b].
f —0.
Prove that
(This
easy; there
is
is
an obvious
g to choose.)
h
Suppose /
(b)
is
f
continuous on [a,b] and that
/
=
fg
for those con-
J
tinuous functions g on [a,b] which satisfy the extra conditions g(a)
gib)
=
Prove that
0.
lemma
/ =
(This innocent looking fact
0.
is
an important
in the calculus of variations; see reference [22] of the
Suggested
Reading.) Hint: Derive a contradiction from the assumption /(xq)
<
or f(xo)
33.
*34.
—
Let f(x)
Compute L( f, P)
(b)
Find inf {£/(/, P)
that
/
is
depend on the behavior of / near
will
=
x for x rational and f(x)
(a)
=
Let f{x)
g you pick
0; the
P
a partition of
on
[0, 1]
and
that
of
>
*o-
irrational.
[0,1].
[0, 1]}.
and \/q
for irrational x,
integrable
P
for all partitions
:
x
for
=
x
if
—
f
/
=
p/q
in lowest terms.
(Every lower
0.
sum
is
Show
clearly
Jo
you must
0;
*35.
o
Let
/
is
to
make upper sums
Problem 34
not. Hint:
/ be
how
/ and g which
Find two functions
g
*36.
figure out
is
are integrable, but
corresponding meanings for the function
(a)
Prove that
M
(b)
Prove that
if
Prove that
if
(c)
min(/,
(d)
'
t
/
/ and g
is
M,
-
a partition of [a,/?].
M/
let
and m,' have the
|/|.
m,.
on
integrable
/
is
[a b]
,
are integrable
if
/
is
its
integrable
,
then so
on
[a, b],
is
|
/
1
then so are max(/, g) and
on
[a b]
,
if
and only
if its
"positive part"
"negative part" min(/, 0) are integrable on [a, b].
integrable
/
Hint:
P be
g).
Prove that
Prove that
- m,-' <
is
max(/, 0) and
37.
let
have their usual meanings, and
m,-
whose composition
relevant.
a bounded function on [a,b] and
Let Mi and
small.)
on
[a, b],
f{t)dt
<
/
then
\fit)\dt.
Phis follows easily from a certain string of inequalities;
relevant.
Problem 1-14
N
.
281
13. Integrals
"38.
for all x in
Suppose / and g are integrable on [a,b] and f(x),g(x) >
appropriate
M,-'
and
m,'
denote
the
[a, b]. Let P be a partition of [a, b]. Let
sup's
and
inf 's for /, define
M," and m,"
and define M, and m,
similarly for g,
similarly for fg.
(a)
Prove that
(b)
Show
t
< Mi'M" and
ra,
> mi'm"
that
Using the
x in [a
show
,
Y,[ M
P) <
/ and
fact that
b]
,
- L(fg,
F)
tf (/*,
(c)
M
>'
M
>'
~ mi'mflte -
ft_i).
<
g are bounded, so that \f(x)\, \g(x)\
M for
that
U(fg,P)-L(fg,P)
n
In^[M/
- m/](/, -
r,-_i)
39.
Prove that fg
(e)
Now
Suppose
m,"]{U
U-l)
is
\
integrable.
eliminate the restriction that f(x),g(x)
that
-
i=\
i=\
(d)
1
+ ^[M," -
/ and g
are integrable
on
>
The
[a, b].
for x in [a, b].
Cauchy-Schwarz
inequality
states that
([>)<{ >%!>
(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 2-2 1
take
(c)
some
If equality holds,
/ and g
(d)
(The
.
one
last
will
imagination.)
is it
/ = Xg
necessarily true that
for
some
What
A.?
if
are continuous?
Prove that
j
I
/
<
/
I
/"
I
•
Is this result
true
and
if
1
are
J
replaced by a and b?
*40.
Suppose that /
is
integrable
on
[0, x] for all
x
>
and lim f(x)
=
a.
Prove
that
lim
Hint:
The
condition
r f{t)dt = a.
i
,.
x ^°°
-
/
x Jo
lim f(x)
x—>oo
=
a implies that /(/)
is
close to a for
pN + M
t
> some N. This means
that
/
f(t)dtis close to Ma.
J
comparison
to
N, then
Ma/(N + M)
is
close to a.
If
M
is
large in
282
Derivatives
and
Integrals
RIEMANN SUMS
APPENDIX.
Suppose
P =
that
choose some point
t„]
{to
a partition of
is
Then we
x, in [f,_i, t{\.
and
[a, b],
that for each
we
i
have
clearly
n
-
L{f, P) <Y^f(xt){t
t
< U(f,
tt-i)
P).
i=\
n
Any sum /
—
^/CxtXfr
called a Riemann sum of
f,-_i) is
/
for P.
Figure
1
shows
Riemann sum; it is the total area of n rectangles
that lie partly below the graph of / and partly above it. Because of the arbitrary
way in which the heights of the rectangles have been picked, we can't say for
sure whether a particular Riemann sum is less than or greater than the integral
the geometric interpretation of a
a
=
to
FIGURE
X\
t\
x2
tn
=b
f(x)dx. But
J
J
1
does seem that the overlaps shouldn't matter too much;
it
bases of all the rectangles are narrow enough, then the
The
close to the integral.
THEOREM
1
the
if
CI
Suppose
/
that
such that,
if
{to,
.
.
,
.
be
to
following theorem states this precisely.
on
integrable
is
P=
Riemann sum ought
tn } is
[a b]
,
Then
.
>
for every s
any partition of
[a, b]
with
there
all
is
lengths
some
/,
8
>
— /,_i <
8,
then
"
no
>b
^/(jC/)(ft -/,_!)-
for
PROOF
any Riemann sum formed by choosing x
Since the
Riemann sum and
amounts
this
by choosing a
tt
showing that
to
{
in (7,_i,
the integral both
B,
—
/,].
between L(/, P) and U(f, P),
lie
we can make U(f, P) — L(f, P) < e
P) < £ for any partition with all lengths
any given
for
such that U(f, P)
8
<
f(x),\dx
/
Ja
i=\
L(f,
s
-ti-i <8.
The
M
for
/ being
definition of
some M.
on
integrable
[a, b] includes the
choose some particular partition P*
First
condition that
=
[uq,
,
.
.
|/|
<
.,Uk) for
which
U(f,P*)-L(f,P*)<B/2,
and then choose a
such that
8
8
For any partition
P
with
£/(/,
P)
all
-
—
t,
<
AMK
<
f,-_i
L(f, P)
8,
we can break
- YlS Mi ~
m,){ti
"
the
sum
r'- l)
/=i
into
two sums. The
first
pletely contained within
U(f, P*)
—
involves those
one of the
L(f, P*) < s/2.
some j = 1,
K — 1,
sum for these terms is <
.
.
.
,
For
/
for
(A'
1
)
•
interval [/,_i,/,]
intervals \uj-\,Uj\.
other
all
so there are at most
—
which the
2M
•
8
<
i
we
K —
s/2. |
will
1
This
have
sum
<
f,_]
is
Uj
is
com-
clearly
<
tj
<
for
of them. Consequently, the
.
13, Appendix.
The moral
to
of
an integral
this tale
really
that anything
is
provided that
is,
Some
partition are small enough.
this
message with even greater
all
which looks
the lengths
/,
like
283
Riemann Sums
a good approximation
— ?;_i
of the intervals in the
home
of the following problems should bring
force.
PROBLEMS
1.
Suppose
P =
/ and g
that
{to
of
tn }
of points
are continuous functions
[a, b]
choose
on
For a partition
[a,b].
a set of points x, in [?,_]
and another
/,]
,
set
Consider the sum
Uj in [f;_i, tf\.
Y^f(.Xi)g(Ui)(ti -ti-l).
(=1
Notice that
this
is
not a
Riemann sum of fg
for P. Nevertheless,
show
that
b
all
lengths
/,
sum and
8,
2.
—
a
f;
f
fg provided that the partition P has all
Ja
_i small enough. Hint: Estimate the difference between such a
such sums will be within e of
Riemann sum; you
/
will
need
to use
uniform continuity (Chapter
Appendix).
similar to, but somewhat harder than, the previous one.
and
/
g are continuous nonnegative functions on [a b] For a
P, consider sums
This problem
Suppose
is
that
partition
,
.
n
Y2 y/f(Xi)+g(Uj)
(ti
- f,-_i).
i=\
Show
that these
sums
will
fVf + g
be within s of
if all
— f,-_i
t,
are small
Ja
enough. Hint: Use the
fact that the square-root function
tinuous on a closed interval
3.
Finally,
For a partition
(u(t ),v(t ))
uniformly con-
M]
(Compare Problem 13-25.)
given parametrically by two functions u and v on [a, b].
we're ready to tackle something big!
Consider a curve
(wfa), v(ti))
[0,
is
£(c P)
c
P =
{to
/„}
of
= J2 \/W) -
[a, b]
«^-i)]
we
2
+
define
b>di)
-
wfe-i)]
2
;
i=\
this
IK.l'RE
2
represents the length of an inscribed polygonal curve (Figure
define the length of c to be the least upper
bound of all
£(/, P),
2).
We
if it exists.
284
Derivatives
and
Integrals
Prove that
if u'
and
v'
are continuous
f
4.
y/u' 2
Let /' be continuous on the interval
polar coordinates on
this interval
on
+
v'
[a, b],
then the length of c
is
2
.
[Go, 0\].
Show
that the
graph of /
in
has the length
.ft
5.
Using Theorem
is
1,
show that
the
Cauchy-Schwarz inequality (Problem 13-39)
a consequence of the Schwarz inequality.
THE FUNDAMENTAL THEOREM
OF CALCULUS
CHAPTER
From
continuous;
is
We know that
theorem of this chapter.
first
may
the hints given in the previous chapter you
it is
continuous.
only
It
fitting that
we
/
when
differentiable (and
is
=f f
is
the original function
/
integrable, then
is
ask what happens
F
turns out that
if
have already guessed the
its
F(x)
derivative
is
especially
simple).
THEOREM
1
(THE FIRST
/ be
Let
integrable
on
[a b]
,
,
F on
and define
FUNDAMENTAL THEOREM
OF CALCULUS)
=
F(x)
[a, b]
by
f.
f
Ja
If
/
is
continuous
at c in [a, b],
F
then
=
F'ic)
c
(If
—
a or
b,
then F'(c)
is
understood
differentiable at c,
is
and
f{c).
mean
to
the right- or left-hand derivative
of F.)
PROOF
We
will
assume
that c
supplied by the reader.
in (a, b); the easy modifications for c
is
By
=
a or b
definition,
—
F'(c)
+
F(c
h)
-
F(c)
lim
h
Suppose
first
that h
>
Then
0.
c+h
+ h) -
F(c
Define
nth
and M\
x
as follows (Figure
mi,
Mi,
It
follows from
Theorem
=
=
F(c)
f-
-I.
1):
inf \f{x)
:
sup{/(x)
c
< x <
c
+ h},
c
<
c
+ h}.
:
x
<
13-7 that
c+h
h
nu,
f <
<
>L
Mi,
h
Therefore
mh <
F(c
+ h)-F(c)
<M
h
.
h
If h
<
0,
only a few details of the argument have to be changed. Let
mi,
M
285
h
=
—
inf {f(x)
sup{/(jc)
+ h < x <c},
c + h < x < c}.
c
\
:
may be
286
Derivatives
and
Integrals
FIGURE
1
Then
mh
{-h) <
f
<M
-
h
i-h).
Jc+h
Since
pc+n
•C+/1
F(c
+ h)
-
F(c)
=
pc
f
=
-
f
Jc+h
Jc
this yields
mh
Since h
<
-h
> F(c + h)-
F(c)
>
M
h.
h
dividing by h reverses the inequality again, yielding the
0,
same
result
as before:
mh <
F(c
+ h)-F(c)
<
M
h
.
h
This inequality
continuous at
is
c,
true for any integrable function, continuous or not. Since
lim
this
/»/,
=
lim Mi,
=
h^O
f(c),
proves that
F
(c)
= hm
F(c
-
—
+ h)-F(c)
Although Theorem
1
=
/(c).
h
h-*0
limit
is
however,
h->0
and
/
deals only with the function obtained by varying the upper
of integration, a simple trick shows what happens
varied. If
G
is
|
defined by
G{x)
=
f
f,
when
the lower limit
is
287
The Fundamental Theorem of Calculus
14.
then
=
G(x)
f~ Jf
f
Jd
Consequently,
/
if
continuous at
is
/•
a
then
c,
=
G\c)
-/(c).
The minus sign appearing here is very fortunate, and
rem 1 to the situation where the function
defined even for
<a
jc
.
we can
In this case
if
c
<
write
-£>
F(x)
so
Theo-
= f f
Ja
(*)
is
allows us to extend
we have
a
=
F'(c)=-(-f(c))
f(c),
exactiy as before.
Notice that in either case, differentiability of
at c alone. Nevertheless,
all
points in [a b]
,
.
Theorem
F
In this case
is
1
Problems
it
is
extremely
of some
derivative
1 1
=
is
ensured by continuity of
when /
is
/
continuous at
and
f.
decide whether a given function
function;
for
Theorem
reason
this
/
1 1
-7
is
the
and
-60 and 11-61 are particularly interesting, since they reveal certain
properties which / must have.
at all
at c
,
difficult to
other
F
interesting
differentiable at all points in [a b]
is
F'
In general,
most
—according
to
Theorem
/
If
1
,
/
is
is
continuous, however, there
the derivative of
some
is
function,
no problem
namely the
function
Fix)
=
-r
/
/.
Ja
Theorem
1
has a simple corollary which frequently reduces computations of
integrals to a triviality.
corollary
If
/
is
continuous on
[a, b]
and f
I
proof
=
g' for
some function
g,
then
f = g(b)-g(a).
Let
Fix)
Then
F'
=f—
g'
on
[a,
ft].
=
f
./'.
Consequently, there
F = g + c.
is
a
number
c such that
288
Derivatives
and
Integrals
The number
c can
be evaluated
0=
so c
—
= g(a) + c,
F(a)
—g(a); thus
=
Fix)
This
note that
easily:
is
x
true, in particular, for
=
g(x)
- g(a).
Thus
b.
f=F(b) = g(b)-g(a).l
f
Ja
The proof of this
after
all,
corollary tends, at
what good
is it
know
to
for
g
to
know
is,
example, g(x)
the corollary
seem
useless:
that
f=g(b)-g(a)
I
if
make
first sight, to
= f f ? The
point, of course,
is
that
one might happen
a quite different function g with this property. For example,
if
~3
*(*)
then g'ix)
=
fix) so
we
=
and
=
f{x)
obtain, without ever
x
,
computing lower and upper sums:
/' x z dx =
Ja
One
x
n+]
can treat other powers
/in
+
similarly;
if
n
is
a natural
gix)
=
then g'ix) =x", so
1),
,n+l
,"+1
=
/' x" dx
For any natural
number n,
containing 0, but
if
n
1
the function fix)
—
+
x~"
is
1
not
bounded on any
interval
a and b are both positive or both negative, then
f
x
dx
b -n+\
—
-n
Ja
formula
this
+
n
Ja
Naturally
number and
is
only true for n
a -n+\
-n +
+1
/—
1
.
We
1
do not
know a
simple expression for
b
1
—
I,
The problem of computing
opportunity to warn against a
dx.
x
this integral
is
discussed
later,
but
it
provides a good
The conclusion of Corollary
many students think that f f is
serious error.
confused with the definition of integrals
—
1
is
often
defined
"gib) — gia), where g is a function whose derivative is /." This "definition" is
not only wrong—it is useless. One reason is that a function / may be integrable
as:
without being the derivative of another function.
x
^
1
There
and /(
is
1
)
=
1
,
then
/'
is
also another reason that
if
fix)
=
for
/ cannot be a derivative (why not?).
much more important: If / is continuous,
integrable, but
is
For example,
The Fundamental Theorem of Calculus
1 4.
we know that f — g' for some
Theorem 1. The function f(x) = \/x
then
then f{x)
=
g'(x),
function g; but
x >
of
0,
-dt,
/
t
and we know of no simpler function g with
this property.
Theorem 1 is so useful that it is
Fundamental Theorem of Calculus. In this book,
corollary to
frequently called the Second
that
somewhat stronger result (which in practice, however, is
As we have just mentioned, a function / might be of the
/
if
where
J\
continuous. If
this only because
provides an excellent illustration:
*(*)=
The
we know
289
is
integrable, then
is still
it
name is reserved for a
not much more useful).
form g' even if / is not
true that
= g(b)-g{a).
f f
The
must return
THEOREM 2 (THE SECOND FUNDAMENTAL THEOREM OF CALCULUS)
If
/
must be
proof, however,
is
entirely different
Let
is
on
integrable
P =
{to,
a point
[a, b]
.
.
.
,tn )
x, in [/,_i
and /
=
g' for
,
t{\
some function
g,
so
we
then
[a, b].
Mean
By
the
)(t i
-tt-i)
Value Theorem there
such that
i
)-g(t
i
_
l
=
=
)
g'(x
i
f(Xi)(ti -tt-i).
If
mt =inf{/(.v)
Mi =sup{f(x)
:/,_!
:f,-_i
<
x <
<x <
tt },
tt),
then clearly
ntiifi
-
<
ff_i)
f(Xi)(ti
-
r,_i)
<
Mtiti
-
<
Mtiti
-
tt -i),
is,
mtiti
Adding
,
f = g(b)-g(a).
be any partition of
g(t
that
1
to the definition of integrals.
I
PROOF
—we cannot use Theorem
-
u-i)
these equations for
/
<
=
g{tt)
1
-
gfa-i)
«
we
tt-i).
obtain
n
n
^]w,(r, -/,_,) <g(b)-g(a) <
^M,(/
i=l
i=i
so that
L(f, P)
for every partition P. But this
<g(b)-g(a)<U(f,P)
means
that
*(*)-*(
/'•I
(
-r,_,)
290
Derivatives
and
Integrals
We
have already used the corollary to
x" dx
,«+]
b"~
=
+
n
Ja
(or,
1
equivalently,
Theorem
2)
few elementary functions:
to find the integrals of a
/'
Theorem
n+\
1
n
,
#
and b both
(a
-1.
both negative
positive or
if
n
>
0).
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 <
< b, then
x 3 dx
does not represent the area of the region shown in Figure
instead
which
is
given
by
-(f*H +
FIGURE
2,
v
/o
Similar care must be exercised in finding the areas of regions which are bounded
by the graphs of more than one function a problem which may frequentiy involve
considerable ingenuity in any case. Suppose, to take a simple example first, that
2
—
we wish
to find the area of the region,
shown
in Figure 3,
between the graphs of
the functions
f(x)
= x2
and
g(x)
= x3
< x 3 < x 2 so that the graph of g
on the interval [0,1]. If < x < 1, then
below that of /. The area of the region of interest to us is therefore
,
/(*)
=
area
1)- area R(g,0,
/?(/, 0,
lies
1),
x^
which
is
l
1
r
I
o
x dx
—
Jo
r
,
x dx
I
3
Jo
4
This area could have been expressed as
/' (f-8)Ja
If
g(x)
< f(x)
by / and
in
K,
I
RJ
Figure
for
all
x
in [a, b],
then
g, even if
f
and g
4. If c
a
number such
that
bounded by
/'
is
then thr icgion R\,
this integral
are sometimes negative.
/ +c
and
always gives the area bounded
The
and g
g, has the
easiest
+c
way to
see this
is
shown
are nonnegative on [a,&],
same area
as the region #i.
The Fundamental Theorem of Calculus
14.
+c
bounded by /
and g
+ c.
area R[
=
291
Consequently,
pb
pb
area
R 2=
(f+c)-
/
=
(g
(g
+ c)]
+ c)
[(f
f
+ C)
/
*b
(f-8).
Ja
This observation
bounded by
useful in the following problem: Find the area of the region
is
the graphs of
The
first
and g
necessity
intersect
is
to
determine
and
g(x)
region
this
more
1
-
= X2
X
or
or
x
v5]/2) we have x
graphs (Figure
(b)
FIGURE
4
— v5]/2,
the interval ([1
+
only
if
on the
x
—
0,
The graphs
of
/
,
=
n
+ ^5 1-75
\
0,
2
(0, [1
precisely.
2
3
— x — X 0,
2
-* - 1) =0,
jc(jc
x
or
cA
On
= x'
when
x
/+
—X
x
fix)
5),
0)
intervals ([1
]
/2, or
we have x
> x —
3
but they can also
+ V5
[ 1
2
[ 1
2
—x >
x
and on the
interval
These assertions are apparent from the
be checked easily, as follows. Since f(x) = g(x)
x.
— V5
— V5]/2, 0) and
]
/2, the function
+ V5 ]/2);
(0, [1
f — g does
it is
not change sign
therefore only necessary
to observe, for example, that
(-^)
3
-(-^)-(-^) 2 = |>o.
1-
I
2
1
= -1 <0,
to conclude that
f-8>0
/-g<0
The
area of the region in question
on([l-V5]/2,0),
on
is
+ V5]/2).
thus
r—Z—
p0
I
(0, [1
(x
—
x
—
x~)
dx
+
[x
I
—
(x~
—
x)] dx.
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,
more substantial problems of a logical nature we have thus far defined he areas
of some very special regions only, which do not even include some of the regions
whose areas have just been computed! We have simply assumed that area made
As
this
—
t
292
Derivatives
and
Integrals
FIGURE
5
and that certain reasonable properties of "area" do hold.
These remarks are not meant to suggest that you should regard exercising ingenuity
to compute areas as beneath you, but are meant to indicate that a better approach
sense for these regions,
to the definition of area
The
is
available, although
advanced
calculus.
book and
historically, for the definition
really provide the best
proper place
its
is
somewhere
was the motivation, both
desire to define area
in
in this
of the integral, but the integral does not
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 pur-
pose for which they were invented, but we
The most important
other purposes.
if
/
is
soon see
will
how
essential they are for
use of integrals has already been emphasized:
continuous, the integral provides a function y such that
/(*)
This equation
is
= /(*)
the simplest example of a "differential equation" (an equation
The Fundamental Theorem
of Calculus says that this differential equation has a solution,
/ 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,
for a function y
which involves derivatives of
y).
if
and the
integral
is
one of the best ways of doing
this.
Since the differentiable functions provided by the Fundamental
Calculus
will
play such a prominent role in later work,
realize that these functions
still
more
functions,
whose
may be combined,
1
1
+
very important to
by the Chain Rule.
Suppose, for example, that
-[
is
of
like less esoteric functions, to yield
derivatives can be found
fix)
it
Theorem
sin
dt.
z
t
~
The Fundamental Theorem of Calculus
14.
Although the notation tends
to disguise the fact
somewhat, /
is
293
the composition
of the functions
C{x)=x 3
and
L-^-dt.
fr+ r-
=
=
F(x)
j
Ja
In
fact,
f(x)
—
F(C(x));
/ = F
other words,
in
sin
1
t
C. Therefore, by the Chain
o
Rule,
=
fix)
F'{Cix))-C'ix)
=F
f
(x
2
-3x
)
2
1
3x
+
1
If
/is defined,
x
sin
2
.
i
instead, as
r
l
fix)--
Jx 3
dt,
+ sin 2
1
t
then
1
fix)
+
1
If
/is defined
3x 2
--
x
sin
.
3
as the reverse composition,
fix) =
—
or
f/
+ sin
i
1
/
r
then
fix) = C'(F( x))
F\x)
=
2
1
'
'
<i.
2
+ sin"
1
t
*))
1
+ sin
x
Similarly, if
/>
fix)
=
sin a
i
r^T dt
/
Ja
s(*)
=
1
r
=
sin
>
t
l
—r
+ Sin
/
dt
*
t
1
./sin.*
h(x)
+ sin"
1
Ja
1
+
2
sin
7
r
then
f'(x\
\X )
J
—
2
1
\X )
Se'(r\
=
A
,
-1
—
—
1
/z'OO
V-U.3
+ sin" (sin x)
2
+
cos
'
•
sin (sin
(
/
\Ja
*)
1
1
+
21
sin
\
/
/
1
,
1
-2
x
+ sin
294
Derivatives
and
Integrals
The formidable appearing
function
yJ
-
fix)
ti
1
-Hsin—
1
J
i
Ja
is
/ = F
also a composition; in fact,
+
1
o
sin
dt
2
t
F. Therefore
= F\F{x))-F\x)
f'{x)
1
1
1
1
As these examples
+
"Of
1
+ siiT
which
appear on the
will
x
sin
t
reveal, the expression occurring
sign indicates the function
+
1
dt
sin
above
right
below) the integral
(or
when /
is
written as a
composition. As a final example, consider the triple compositions
(r-^-dt)
Ja l+sin"
/
\
-dt
L L +sirT
r
1
/(*)
= /
Ja
1
+ sin
=/
g(x)
dt,
f
1
1
t
dt.
1
+ sin
/
which can be written
f = F
o
FoC
and
g
may
Omitting the intermediate steps (which you
we
= FoF
o F.
supply,
if
you
still
feel insecure)
obtain
fix)
1
=
1
1
1
+ sin'
.
-o-t3
x
1
+
1
+ sin'
3x\
•
sin"
dt
+
1
sin
t
g'ix)
1
+ sin"
tecM
I
1
1
dt
1
+
sin
r
Ja
dt
1
+ sin"
t
t
I
l+sin x
Like the simpler differentiations of Chapter 10, these manipulations should be-
come much
easier after the practice provided
by some of the problems, and,
like
the problems of Chapter 10, these differentiations are simply a test of your under-
standing 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 definition of
—
the integral.
Actually, this
continuous function
integrable
on
[«, b\.
is
we need
not quite true.
to
know
that if
In order to apply
/
is
continuous on
Theorem
[«,/?],
Although we've already offered one proof of this
1
then
to a
/
is
result, there
is
a
more elementary argument
arguments,
it's
quite tricky, but
proof of Theorem
If
/
sup{L(/,
will
both
exist,
even
integral of / on [a
will be denoted by
has the virtue that
if
,
[a, b],
is not integrable. These numbers are
and the upper integral of / on [a b]
/
b]
,
The lower and upper
U
and
/
pb
and
if
m<
<
c
/+L
/
if
a
/
Ja
fix)
/.
<
b,
then
all
jc
f
integrals.
Ja
f < M{b -
/,
Jc
a).
exercise, since they are quite similar to the
The
results for integrals are actually a corollary
ff-vf
will
/
Ja
of the results for upper and lower integrals, because
We
U
an
as
left
pb
/+
/
pb
<U/
Ja
corresponding proofs for
U
then
in [a, b],
pb
proofs of these facts are
/ =
/
Ja
m(b-a)<L
The
pc
pb
U
and
/
Jc
M for
<
and
Ja
pb
pc
=L
lower
respectively,
both have several properties which the integral
integrals
possesses. In particular,
called the
,
pb
Ja
/
Ja
a review of the
inf {!/(/, P)}
pb
/
will force
it
then
and
/>)}
L
L
Like most "elementary"
prefer.
1.
any bounded function on
is
you might
that
it
295
The Fundamental Theorem of Calculus
1 4.
prove that a continuous function
/
is
integrable precisely
when
/•
integrable by showing that this
is
equality always holds for continuous functions.
/
It is
actually easier to
show
that
f
Ja
Ja
for all
x
in [a b]
,
;
the trick
even depend on the
theorem
13-3
PROOF
If
/
is
to note that
is
/ was
fact that
continuous on
[«,/?],
Define functions L and
U
L(x)
then
on
=L
/
[a b]
,
f f
most of the proof of Theorem
integrable!
integrable
is
on
by
and
U(x)
=U f
Ja
Ja
Let x be in (a,b). If h
>
[a, b\.
and
mh -
inf {f{t)
M = sup{/(0
h
x <
:
:
x
<
t
t
<x +
/?},
< x +h),
f.
1
didn't
.
296
Derivatives
and
.
Integrals
then
/x+h
rx+h
/<U/
f<M
h -h,
so
mh
< L(x +
h
-
h)
L(x) < U(x
+
- U(x) <
h)
h
M,,
or
mi,
L(x+h)-L(x)
<
U(x+h)-U(x)
<
If h
<
/
inf{/(0
Mh =
sup{/(f)
:
x
+ h < < x),
x
+h <
t
<
t
x],
proof of Theorem
we have
at x,
=
lim nih
this
:
inequality, precisely as in the
continuous
is
mh =
=
lim Mi,
h^O
and
f(x),
/i-vO
proves that
=
L'{x)
This means that there
is
U'(x)
number
a
U (x =
=
for
f(x)
x
in (a, b).
c such that
L(x)
)
+c
for all
x
in [a b]
,
Since
U(a)
the
number
c
must equal
=
L(a)
=
0,
0, so
U(x)=
L(x)
x
for
all
=
L(b)
in [a,b].
In particular,
U
and
this
means
/
that
is
f f
=
U(b)
on
integrable
=L
f
f,
[a ,b\. |
PROBLEMS
1
.
Find the derivatives of each of the following functions.
3
(i)
F(x)=
r*
/
sin
3
tdt.
Ja
(ii)
F{x)
=
Jf
sixth dt)
J
/
+ sin 6 t + t 2
1
(iii)
nrdt
F(x)
J 15
VA
+t 2 +sin z
1
1
(iv)
.
//
and
one obtains the same
Since
<M
h
h
F(x)
-I
dt.
1
+
t
2
+
sin
2
/
?
dy.
1
The Fundamental Theorem of Calculus
14.
(v)
F{x)
dt.
-f
Jla
a
(vi)
F(x)
(vii)
F
=
sin
297
1
+
t
sin
/
(
+ sin"
2
/
sin
/
(
3
1
dt
c
1
x
_,
where F(x)
,
f
=
1
-dt.
I
[Find (F
T
F -i where F CO =
(viii)
For each of the following /,
f{x)? (Caution:
uous at x.)
3.
(i)
fix)
(ii)
/(jc)
(iii)
/(jc)
(iv)
/(x)
(v)
/(jc)
(vi)
f(x)
(vii)
/
(viii)
fix)
=
=
=
=
=
=
1
x
<
1,
f(x)
if
.v
<
1,
/(*)
if
x
^
1,
/(x)
if
jc
is
if
jc
FIGURE
6
if
=
=
=
irrational,
if
x
>
1.
1 if
x
>
1.
if
jc
1
1
fix)
= 1.
= \/q
f, at which points x
—
fix), even
if
x
x
=
shown
in
Figure
io
sill
/
+
1
f
\/n for some n in N, fix)
!
ho
l
dt,
x
+r
dt.
I
vT
p/q
if
if
/
is
is
F'(x)
in lowest terms.
< x <
in (0,7r/2).
—
—
not contin-
1.
6.
that the values of the following expressions
2 °
t
=
<0, fix) =x if x >0.
or jc> 1, fix) = l/[l/x]
x <
Wx
(ii)
— fQ
F(x)
might happen that F'(x)
the function
is
if
if
if
=
Show
it
dt
vT^72
Jo
2.
F-'U).)
!
/
,
terms of
)'(;0 in
otherwise.
do not depend on
x:
298
Derivatives
and
Integrals
4.
Find
(i)
(./"-' )'(0) if
f(x )=
I
+ sm(sin t)dt.
1
Jo
(ii)
—
f( x )
cos(cosf) dt.
/
/
(Don't try to evaluate
5.
explicitly.)
Find a function g such that
(i)
+ x2
tg{t)dt=x
/
Jo
.
2
(ii)
= x+x 2
tg{t)dt
/
.
Jo
(Notice that g
6.
(a)
Find
not assumed continuous
is
continuous functions
all
/ =
/
Jo
(/(*))"
assuming that / has
(b)
>
at
on
is
[a b]
Use Problem 13-23
for
most one
some constant C
/
0.
on an
is
and any
,
satisfying
interval
(— oo,
b]
<
<
<
with
b,
but
b.
C =
Finally, for
that
7.
+C
Also find a solution that
non-zero for x
(c)
/
at 0.)
b, find
a solution
but non-zero elsewhere.
,
to
interval [a, b] with a
prove that
1
-dx <
Isfl
'/ 2
3
— "
<
Find F'(x)
i
1
F(.x)
if
^
+
/l-x
i;V
8
8.
7
J\
!
v
Jo
+x
1
-.
v2
v/3
J
^-Y
= /
- ^r4
xf{t)dt.
(The answer
perform an obvious manipulation on the
9.
Prove that
if
/
is
Hint: Differentiate both sides,
*10.
Use Problem 9
Find a function
supposed
to
xf(x); you should
integral before trying to find F' .)
f{t)dt\du.
I
making use of Problem
8.
to prove that
f f(u)(x-u) 2 du
11.
wo/
continuous, then
f(u)(x-u)du=
/
is
/
=
2
f
(f
such that f'"(x)
'
=
IT f(t)dt) duA du
1
/ v
1
be easy; don't misinterpret the word
+ sin
x.
"find.")
2
(This
.
problem
is
12.
A
/
function
(a)
If
/
is
periodic, with period
fix
/
Find a function
such that /
a periodic g for which
it
+ a) =
on
fix) for
[0, a],
show
that
not periodic, but /'
is.
Hint:
can be guaranteed that fix)
= fQ
is
x.
all
for all b.
f
f'-7
(b)
a, if
periodic with period a and integrable
is
299
The Fundamental Theorem of Calculus
1 4.
Choose
g
is
not
periodic.
(c)
If /'
periodic with period a and f(a)
is
=
/(0), then
/
is
also periodic
with period a.
"(d)
Conversely,
if
/'
is
periodic with period a
=
period not necessarily
13.
Find
fQ l/xdx, by simply
then f(a)
a),
—
and /
guessing a function
/
with fix)
=
y/x,
Then check
14.
Use the Fundamental Theorem of Calculus and Problem 13-21
result stated in Problem 12-21.
15.
Let C\,
C and
some
/(0).
the Second Fundamental Theorem of Calculus.
lem 13-21.
:
periodic (with
is
Ci be curves passing through the
origin, as
and using
with Prob-
to derive the
shown
in Figure 7.
Each point on C can be joined to a point of C\ with a vertical line segment
and to a point of C^ with a horizontal line segment. We will say that C bisects
C\ and Cj if the regions A and B have equal areas for every point on C.
(a)
(b)
and C is the graph of fix) = 2x 2
x > 0, find C2 so that C bisects C\ and CiMore generally, find C2 if C\ is the graph of fix) = x m and C is the
graph of fix) — ex" for some c > 1.
If C\
— x2
the graph of fix)
is
x >
,
,
,
1
FIGURE
7
16.
(a)
Find the derivatives of Fix)
(b)
Now
give a
=
ff \/t dt and Gix)
new proof for Problem
= fb *
\/t dt.
13-15.
17.
Use the Fundamental Theorem of Calculus and Darboux's Theorem (Problem 1 1-60) to give another proof of the Intermediate Value Theorem.
18.
Prove that
if
h
is
continuous,
F(x)
/ and g
are differentiable,
=
h(t)dt,
/
and
Jfix)
then F\x)
=
-
fix). Hint: Try to reduce this to
the two cases you can already handle, with a constant either as the lower or
h(g(x))
•
g'ix)
hifix))
the upper limit of integration.
19.
Let
/ be
integrable
on
[a, b], let c
Fix)
=
I
Ja
be in
a
f,
(a, b),
and
let
<x <b.
For each of the following statements, give either a proof or a counterexample.
F
(a)
If
/
is
differentiable at c, then
(b)
If
/
is
differentiable at c, then F'
(c)
If /'
is
continuous
at c,
then F'
is
is
is
differentiable at c.
continuous at
continuous
at c.
c.
300
Derivatives
and
Integrals
*20.
Let
r
1
/(*)
—
cos
=
x^O
,
x
0,
Is
21.
the function F(x)
Suppose
[0, 1]
that /'
=
on
integrable
is
*22.
and /(0)
[0, 1]
=
0.
page 179.
Prove that for
all
x
in
we have
uy-
<
l/(*)l
Show
0.
differentiable at 0? Hint: Stare at
f
Jq
=
A"
=
also that the hypothesis /(0)
Suppose
that
/
Prove that for
a differentiable function with /(0)
is
x
all
needed. Hint: Problem 13-39.
is
>
=
and
<
/'
<
1.
we have
/>(/,'')'
*23.
(a)
Suppose G'
differential
=
#(>'(-*))
(*)
then there
is
a
•
y'(x)
number
G{y{x))
(**)
=
g and F'
equation
=
—
f.
Prove that
if
fix)
for all
x
in
(c)
Find what condition y must
if
y
all
interval,
1
for all x in this interval.
then v
satisfies (**),
y\x)
equations to find
+c
F(x)
Show, conversely, that
=1+
some
c such that
(b)
(In this case g(t)
the function y satisfies the
is
a solution of
(*).
satisfy if
=
l+x 2
1+yU)
and f{t)
=
+
1
t~.)
Then
"solve" the resulting
possible solutions y (no solution will have
R
as
its
domain).
(d)
Find what condition y must
satisfy if
-1
y'ix)
1
(An appeal
to
Problem 12-14
+5[v(jc)]
show
will
4*
that there are functions satisfying
the resulting equation.)
(e)
Find
all
functions y satisfying
y(x)y\x)
=
Find the solution y satisfying y(0)
-x.
= — 1.
24.
Problem 10-19 we found
In
f"
(b)
/
2
3
/f'
(f'{x)
f"\x)
3
f'(x)
2\f'(x)
d).
Now
suppose that /
any function
is
a constant function.
is
form f(x)
the
is
Schwarzian derivative
that the
was
for f(x) = (ax + b)/(cx +
whose Schwarzian derivative is 0.
(a)
301
The Fundamental Theorem of Calculus
14.
=
(ax
+
b)/(cx
+
Hint:
d).
Consider u
—
f
and
apply the previous problem.
25.
The
lim
limit
called
f
x r dx,
Determine
(b)
Use Problem 13-15
Jj
if r
can you say about
Suppose
<
[0,
26.
(or
f(x)dx), and
f
to
< — 1.
show
that Jj
On
(d)
f
an "improper integral."
(a)
(c)
denoted by f
if it exists, is
f,
< f(x)
g(x)
N], then
Explain
What
Hint:
>
for
for all
x
x
>
>
0,
and that f f exists. Prove that if
and g is integrable on each interval
also exists.
fQ g
why
exist.
\/xdx?
Jj
that /(jc)
l/x dx does not
+x
f^° 1/(1
)dx
exists.
Hint: Split this integral
Decide whether or not the following improper integrals
up
at
1.
exist.
1
dx.
I
Jo
v/l+.v 3
X
L
Jo
f°°
m
/
—-=^ dx
1
(this is really
a type considered in Problem 28).
xJ\ +x
./0
27.
dx.
l+.t 3 / 2
The improper
integral
f
j°_
is
defined in the obvious way, as
But another kind of improper integral f_
it
is
Jq
/+
provided these
/.qo /,
f
improper
(a)
Explain
why
f_
1/(1
+ x~)dx
(b)
Explain
why
J^xdx
does not
is
lim
f° f.
defined in a nonobvious way:
integrals
both
exist.
exists.
exist.
(But notice that
lim f_ x
N
dx
does
exist.)
(c)
Prove that
Show
f™ f
exists,
then
moreover, that lim f_ N
J-oo /•
can't,
if
Can you
you
will
state a
lim J_ N
/"and lim
f
exists
f_ n2 f
and equals
both
exist
to
do
f.
and equal
reasonable generalization of these facts?
have a miserable time trying
f™
(If
you
these special cases!)
/
302
Derivatives
and
Integrals
28.
There is another kind of "improper
bounded, but the function is unbounded:
(a)
>
If a
0, find
lim f" l/^/xdx. This limit
=
even though the function f(x)
matter how we define /(0).
(b)
(c)
Find
(
JV </*
integral"
Use Problem 13-15
is
interval
denoted by f^ l/^/xdx,
not bounded on
is
is
[0, a],
no
<r <0.
-1
if
\/y/x
which the
in
to
show
_1
that /q
jc
dx does not make
sense,
even as
a limit.
(d)
Invent a reasonable definition of
r
dx
for a
—
x~)~
<
Invent a reasonable definition of j_\(\
limits,
exist?
and show
How does
+x)
(1
^~
Why
that the limits exist. Hint:
_1//2
]
compare with
(1
(a)
If
/
is
continuous on
*(b)
If
/
is
integrable
on
[0, 1],
[0, 1]
compute lim x
and continuous
lim x I
*^0+ Jx
30.
It is
—^2
dx, as a
does /_](1
2 _1//2
—jc
)
/"'
29.
and compute
it
for
<r <0.
-1
(e)
\x\
f
for
sum of two
+ x)~
1
dx
^
—1 < x < 0?
fit)
dt.
I
at 0,
compute
dt.
t
combine the two possible extensions of the notion of
possible, finally, to
the integral.
(a)
If
f(x)
=
1/Vjc for
< x <
1
and f(x)
=
l/x 2 for x
>
1,
find
poo
f(x)dx
I
Jo
(b)
this
should mean).
poo
Show
that
<
and
r
deciding what
(after
at oo; for r
r
x dx never makes
/
r
< —
——
1
1
.
sense.
(Distinguish the cases
In one case things go
wrong
wrong
places.)
things go
at
both
at 0, in the
—
1
<
other case
CHAPTER
THE TRIGONOMETRIC FUNCTIONS
The
definitions of the functions sin
one might suspect. For
intuitive definitions,
and cos are considerably more subtle than
some informal and
reason, this chapter begins with
this
which should not be scrutinized too
soon be replaced by the formal definitions which we
In elementary geometry an angle
common
FIGURE
More
pairs
point (Figure
they shall
simply the union of two half-lines with a
1).
1
useful for trigonometry are "directed angles,"
(l\,
FIGURE
2
If for
l\
is
initial
is
carefully, as
really intend to use.
h) of half-lines with the same
we always choose
initial
which may be regarded
the positive half of the horizontal axis, a directed angle
described completely by the second half-line (Figure
3).
Since each half-line intersects the unit circle precisely once, a directed angle
described, even
FIGURE
I
more
point (x, v) with x
303
simply,
+ y2 =
as
point, visualized as in Figure 2.
1.
by a point on the unit
circle (Figure 4), that
is,
is
by a
304
Derivatives
and
Integrals
The
and cosine of a directed angle can now be defined as follows (Figure 5):
is determined by a point (x, y) with x
+ y 2 = 1; the sine of the
defined as y, and the cosine as x.
sine
a directed angle
angle
is
we
Despite the aura of precision surrounding the previous paragraph,
and
yet finished with the definitions of sin
What we have
to define
this
defined
is
depends on associating an angle
"measure angles
in degrees."
an angle "half-way around"
to 90, etc. (Figure 6).
An
4
of 360 degrees, and
of 90 degrees
function,
The
is
The
which we
denote by
=
There are two
mean by an
example. Even
that this system,
to elegant results
manner,
this
degrees
+
90 degrees,
to
is
number
to the
same
the
x,
is
as the angle
One
etc.
can
now
define a
sin°, as follows:
sine of the angle of
with
is
associated to 360,
purposely extended further, so that an angle
is
this
x degrees.
approach. Although
angle of 90 or 45 degrees,
for
is,
difficulties
is
an angle "a quarter way around"
angle of
an angle of 360
will
what we want
usual procedure for doing
way around"
80,
1
angle;
are not
barely begun.
number. The oldest method
angle associated, in
ambiguity
this
also
to every
associated to
is
sin°(x)
degrees
The
angle "all the
called "the angle of x degrees."
FIGURE
and cosine of a directed
the sine
sin* and cos* for each number x.
is
we have
Indeed,
cos.
is
it
it
may
be clear what we
not quite clear what an angle of v 2
if this difficulty
could be circumvented,
it is
unlikely
as it does on the arbitrary choice of 360, will lead
would be sheer luck if the function sin° had mathematically
depending
—
it
pleasing properties.
"Radian measure" appears
number x choose
,
a point
5
remedy
for
both these
the unit circle such that x
the length of the
and running counterclockwise
The
P
directed angle determined by
radians are identical.
A
r
2tt
is
=
sin'
can
now be
360
=
sin
r
x
=
2ttx
sin
soon drop the superscript
function which will interest us; before
The
7).
+x
and the angle of 2n
x radians.
sine of the angle of
sin°; since
we want
to
have
7TX
= sm
T80
360
shall
(Figure
2n, we can define
sin
We
P
defined as follows:
This same method can easily be adopted to define
sin°
to
called "the angle of x radians." Since the
the angle of x radians
,
function
sin (x)
is
Given any
defects.
is
arc of the circle beginning at (1,0)
length of the whole circle
K,L RE
T
1
to offer a
P on
r in sin'
we
,
since sin
do, a few
r
(and not sin
)
is
words of warning are
the only
advisable.
r
expressions sin° x and sin x are sometimes written
sinjc
sin.v radians,
I
K.
i
RE
6
but
this
notation
is
quite misleading; a
carry a banner indicating that
ii
is
number
x
is
simply a
number
"in degrees" or "in radians."
it
If the
does not
meaning
15.
of the notation "sinx"
in
is
doubt one usually
x
"Is
ngth x
but what one means
in
asks:
you mean
or sin
sin
r
Even
for mathematicians, addicted to precision, these
able,
were
Although the function
sin'
.
remarks might be dispens-
not for the fact that failure to take them into account will lead to
it
sin'
is
Our proposed
example
the function
(and use exclusively henceforth), there
of
305
degrees or radians?"
incorrect answers to certain problems (an
7
Trigonometric Functions
is:
Do
FIGURE
The
definition
given in Problem
which we wish
to
19).
denote simply by
sin
a difficulty involved even in the definition
is
depends on the concept of the length of a curve.
Although the length of a curve has been defined
to reformulate the definition in
is
in several
problems,
it is
also easy
terms of areas. (A treatment in terms of length
is
outlined in Problem 28.)
length x
Suppose that x
is
the length of the arc of the unit circle from (1, 0) to P; this arc
thus contains x/2jt of the total length 2jt of the circumference of the unit circle.
Let S denote the "sector" shown in Figure
horizontal axis,
and the
half-line
through
x/2jt times the area inside the unit
area
circle,
bounded by the unit circle, the
and P. The area of S should be
which we expect to be jt; thus S should
5
8;
is
(0, 0)
have area
x
x
71
2t7
FIGURE
We
8
can therefore define cos x and
sin
=
2-
x as the coordinates of the point
P which
determines a sector of area x/2.
With these remarks
and cos now
more
as
The
begins.
background, the rigorous definition of the functions
first
definition identifies
precisely, as twice the area
jt
sin
as the area of the unit circle-
of a semicircle (Figure
9).
DEFINITION
(This definition
is
onometric functions
area
it
will
be necessary to
< x < jr.)
The second definition is meant to
the sector bounded by the unit circle,
(*,
V
1
—
x2
).
If
< * <
the area of a triangle
FIGURE
not offered simply as an embellishment; to define the
1,
this
define siiut
first
and
—1 < x < 1,
horizontal axis, and the
describe, for
the area A(x) of
the
half-line
through
area can be expressed (Figure 10) as the
and the area of a region under the unit
9
Vl -a 2
trig-
cosjc only for
vT
t
2
dt.
circle:
sum
of
306
Derivatives
and
Integrals
This same formula happens to work for —1
<x <
also.
In this case (Figure
1 1),
the term
Xyfl
negative,
is
and represents the area of the
which must be subtracted from
triangle
the term
FIGURE
iy \-t
10
DEFINITION
If
-
1
<x <
1
if
—
1
xfT-x 2
=
< x <
1
then
,
A
is
A'W =
[^
t
2
dt.
differentiable at x
-2x
\
and
(using the
Funda-
-/T1^
+ ^l-A- 2
2y/\-x 2
-x
2
+ (\-x 2
J\-x
I
+
mental Theorem of Calculus),
(x,Vl-x 2 )
FIGURE
dt.
then
,
A(x)
Notice that
2
)
Vl-x 2
2
1
-x 2
2>/l
1
-2.v 2 -2(l
2s/\-x
-x 2
)
2
-1
2>/1-jc 2
Notice also (Figure 12) that on the interval [— 1,
from
„
Al.(-1)
I
I
<
.
1
to
K E 12
A(l)
that
its
For
P =
=
0.
1]
the function
If
|
This follows directly from the definition of A, and also from the
derivative
is
negative on (—1,
< x < n we wish to
on the unit
(cosx, sin x)
< X <
decreases
i
= 0+/' 7l-f2^ =
define cos x
circle
7T,
then cos X
is
and
sin
x as the coordinates of a point
which determines a sector whose area
the unique
number
A (cos A
in [- -1, 1]
X
and
sin x
=
fact
1).
(Figure 13). In other words:
DEFINITION
A
vl —
(cosjc) 2
.
such that
is
x/2
.
77z£ Trigonometric Functions
307
This definition actually requires a few words of justification. In order to
know
75.
that there
and
that
is
A
a
number y
takes
Value Theorem
on the values
is
crucial, if
Having made, and justified,
THEOREM
l
<
If
x
<
jt,
=
A(y)
satisfying
and
B =
the fact that
A
is
continuous,
make our preliminary definition precise.
our definition, we can now proceed quite rapidly.
we want
to
then
= — sinx,
= cos*.
sin'U)
If
we use
tt/2. This tacit appeal to the Intermediate
cos'(jc)
PROOF
x /2,
=
2 A, then the definition A(cosx)
x/2 can be written
=
B(cosx)
in other words, cos
is
We
just the inverse of B.
=
A\x)
x;
have already computed that
1
'
2V\-x 2
P =
(cosjc, sinx)
from which we conclude
that
x
area
—
1
= -
B'(x)
2
'
Vl-x 2
Consequently,
FIGURE
(#)'(*)
cos' (*)=
1
3
1
B'(B-Hx))
1
v^l
[^U)] 2
-
= —V —
= — sin x
(COSX) 2
1
Since
sin
we
x
=v —
1
(cos*) 2
,
also obtain
.
1
,
sin (x)
= -
— 2 cos*
VI -
1
•
—
cos'C*)
(cosx) 2
cos x sin x
sinx
cosx.
The information contained
in
|
Theorem
1
can be used
to sketch the
graphs of
,
308
Derivatives
and
Integrals
sin
and cos on the
interval [0,
cosj
=
unique y
for a
Since
.
= — sinjc <
cos'(jt)
the function cos decreases
tt]
=
from cosO
To
in [0, tt].
1
< x <
0,
=—
to cos7r
we
find y,
A(cosx)
1
tt,
(Figure 14). Consequently,
note that the definition of cos,
=—
2
means
that
A(0)
=
|,
so
dt.
Jo
It is
easy to see that
/o
J\-t 2 dt =
FIGURE
r
\
Vi-t 2 dt
/
15
so
we can
Now we
also write
have
sin (x )
=
cos x
I
so sin increases
on
[tt/2, tt] to sin7r
The
[0, tt/2]
=0
>
0,
<
0,
< x <
tt/2
f
\
from sinO
=
<x <
tt/2
to sin7r/2
=
TT,
1,
and then decreases on
(Figure 15).
values of sin x and cos x for x not in [0,
tt]
are most easily defined by a
two-step piecing together process:
(1)
If
tt
< x <
27r,
then
sin
= — sin(2;r — x),
— cos(2tt — x).
sinx
cosjc
Figure 16 shows the graphs of sin and cos on
(a)
(2)
If
x
— 2nk + x'
for
some
integer k,
cos
sin*
cosjc
2tt
(b)
FIG! RE
16
the functions sin
and some
x' in [0. 2tt],
now
and cos
to
defined on
all
R, we must
of R.
now
check that the
basic properties of these functions continue to hold. In most eases this
example,
it
is
clear that the equation
sin""
x
+
then
= sin a',
= cosx'.
Figure 17 shows the graphs of sin and cos,
Having extended
[0, 2tt].
cos x
=
1
is
easy.
For
)
.
15.
The
Trigonometric Functions
309
(b)
FIGURE
17
holds for
all
x.
It is
also not
hard
to
prove that
sin' (a)
cos' (a)
if
x
is
not a multiple of
tt.
= cos a,
= — sin
For example,
sin
a
=
if
A',
n < x < 2n, then
—sin (2jt —a),
so
sin'(A)
If
a
rem
is
a multiple of
n we
resort to a trick;
11-7 to conclude that the
I'KiURK
18
= — sin'(27r — a)
= COS(27T — A)
= COS A
it
is
(—
•
1
only necessary to apply Theo-
same formulas are true
in this case also.
310
Derivatives
and
Integrals
The
other standard trigonometric functions present no difficulty
We
at all.
define
secx
1
=
COSJC
tan x
=
cos
-1
x
+
kix
x
^
kir.
sinx
+
7T/2,
cos*
cscx
—
1
=
x
sin
COSJC
sinx
The graphs
t
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
which are
functions,
listed in the
statement of the theorem, since the results
theorem
FIGURE
2
If
x
£
kit
+ tt/2,
=
=
sec' (jc)
tan'(jc)
x
^
kjr,
1- -
The
y
you
sec x.
= — cscjc cotx,
= — csCx.
cot'(;c)
Left to
sec x tan x,
then
csc'OO
PROOF
is
then
19
If
no need to memorize the
can be rederived whenever needed.)
next theorem (there
(a
straightforward computation).
|
The
inverses of the trigonometric functions are also easily differentiated.
trigonometric functions are not one-one, so
it
is
first
necessary to restrict
to suitable intervals; the largest possible length obtainable
is jt,
and the
them
intervals
1
usually chosen are (Figure 19)
1
-71
rt
2 .-^
y-\-
2
[— 7r/2,
7r/2]
for cos,
[0, 7r]
(— tt/2,
for sin,
tt/2)
for tan.
(The inverses of the other trigonometric functions are so rarely used that they
will
not even be discussed here.)
an
The
sin
inverse of the function
f(x)
is
—
denoted by arcsin (Figure
sin
20); the
has been avoided because arcsin
one), but of the restricted function /;
I
M.
I
RE
20
and
tt/2
sin x,
arcsin
by Sin
.
<x <
domain of
is
tt/2
arcsin
is
—
|
1,
lj.
The notation
not the inverse of sin (which
sometimes
this
function
/
is
is
not onc-
denoted by Sin,
The
15.
The
inverse of the function
=
g(x)
is
denoted by arccos (Figure
is
denoted by Cos, and arccos
The
JT
arccos
is
[— 1,
Sometimes g
1].
.
=
tan x
The
is
< x <
-jr/2
,
22); arctan
is
tt/2
one of the simplest examples of a
bounded even though it is one-one on
denoted by Tan, and arctan by Tan"
which
Sometimes the function h
2
domain of
by Cos"
21); the
denoted by arctan (Figure
differentiable function
FIGURE
<X <
COS*,
inverse of the function
h (x )
is
311
Trigonometric Functions
all
is
of R.
.
derivatives of the inverse trigonometric functions are surprisingly simple,
and do not involve trigonometric functions at all. Finding the derivatives is a simple
matter, but to express them in a suitable form we will have to simplify expressions
1
like
sec (arctan*).
cos(arcsinx),
FIGURE
A little
ple,
2 2
picture
is
the best
way
to
remember the
correct simplifications. For
Figure 23 shows a directed angle whose sine
is
x
—
angle of (arcsinx) radians; consequently cos(arcsin x)
side,
namely, v
1
—
x
2
.
However,
in the
2 3
THEOREM
3
If
—1 <
x <
1
,
then
arcsin'U)
=
1
J\-x 2
arccos' (x)
=
-1
'
Vl-x 2
Moreover, for
all
x we have
arctan' (x)
=
is
shown
is
\+x 2
exam-
thus an
the length of the other
proof of the next theorem we
resort to such pictures.
FIGURE
the angle
will
not
312
Derivatives
and
Integrals
PROOF
=
arcsin'(x)
(/
)'{x)
1
f'if-Hx))
l
sin'(arcsin x)
cos(arcsinx)
Now
+
[sin(arcsin x)\
that
[cos(arcsinx)]"
=
1,
is,
+
x
=
[cos(arcsinx)]
1;
therefore,
=
cos(arcsin x)
(The positive square root
cos(arcsinx)
>
0.)
x2
.
first
any
proof for the
The
difficulties.
arctan'(x)
is
in (—7t/2,7t/2), so
formula.
has already been established
also possible to imitate the
that proof presented
—
1
be taken because arcsinA'
to
This proves the
The second formula
It is
is
v
=
first
(in
Theorem
the proof of
formula, a valuable exercise
third formula
is
proved
1).
if
as follows.
(h~ )'(x)
1
h'(h- l (x))
1
tan' (arctan x)
1
2
sec (arctan x)
Dividing both sides of the identity
sin
by cos a
+
a
+
cos" a
=
1
yields
tan
It
a
2
=
1
sec" a.
follows that
[tan (arctan*)]"
+
1
=
sec (arctan
.v).
or
x
which proves the third formula.
The
traditional
given here)
is
+
1
=
sec (arctan x),
|
proof of the formula
sin'(x)
=
cosx
(quite different
outlined in Problem 27. This proof depends
upon
from the one
first
establishing
The
15.
313
Trigonometric Functions
the limit
h
sin
hm
=
1.
h
h-+0
and the "addition formula"
sin(x
+ y) =
sin
Both of these formulas can be derived
are known.
The
first is
+ cos x siny.
x cos y
now that
easily
just the special case sin'(O)
=
the derivative of sin
cosO.
and cos
The second depends
on a beautiful characterization of the functions sin and cos. In order
to derive this
we need a lemma whose proof involves a clever trick; a more straightforward
result
proof will be supplied
LEMMA
in Part
IV
Suppose / has a second derivative everywhere and that
+ f = 0,
= 0,
/'(0) = 0.
f"
/(0)
= 0.
Then /
PROOF
Multiplying both sides of the
first
equation by /' yields
f'f"
+ ff' = 0.
Thus
2
,
[(/
so
(f)
2
+ f2
the constant
is
is
)
+ / 2 = 2(/7" + //') = 0,
,
]
a constant function.
From /(0)
=
and /'(0)
=
it
follows that
0; thus
!-'/„vi2
[f(x)Y+[f(x)]
2
=
for all*.
This implies that
f(x)=0
THEOREM
4
If
/
for all*. |
has a second derivative everywhere and
r+/=
/(0)
./" (0)
=
=
o,
a,
6,
then
/=b
(In particular, if
then
/ =
cos.)
/(0)
=
and /'(0)
=
sin
+a
1,
then
cos.
/ =
sin; if
/(0)
=
1
and
/"(0)
=
0,
.
314
Derivatives
and
.
Integrals
proof
Let
g{x)
=
fix)
=
=
fix) — bcosx
f"(x) + bsinx
— bsinx —
a cosx.
Then
g'ix)
g'ix)
+ a sinx,
+ acosjc.
Consequently,
=
=
g(0)
g'iO)
o,
0,
which shows that
=
THEOREM
5
If
lix)
=
a cosx,
for
x. |
all
x and y are any two numbers, then
+ y) =
+ y) =
sin(x
cosix
PROOF
— bsinx —
fix)
For any particular
x cos y + cos x
cos x cos y — sin jc
sin
number y we can
fix)
sin
y
sin
y
define a function
=
,
/ by
+ y).
sin(x
Then
= cosU +y)
fix) — -sin(x + y).
f'ix)
Consequently,
+/=
/(0) =
/'(0) =
f"
It
follows from
Theorem 4
siny,
cosy.
that
/ =
that
0,
(cos y)
•
sin
+(sin y)
cos;
•
is,
sin(x
Since any
+
=
y)
cos y sin x
+ sin y cos
number y could have been chosen
formula for
all
x and
for
,
all
x
to begin with, this proves the
first
y.
The second formula
As a conclusion
jc
is
proved
to this chapter,
an alternative approach
similarly. |
and
as a prelude to
to the definition
1
for
arcsin'(x)
nTr
/
v
Chapter
of the function
—
1
sin.
< x <
1,
1
8,
we
Since
will
mention
.
.
The
15.
it
follows
Trigonometric Functions
from the Second Fundamental Theorem of Calculus
fX
=
arcsin x
—
arcsin x
=
arcsin
This equation could have been taken as the
315
that
1
-dt.
/
h
J\~-t 2
definition
of arcsin.
It
would follow
immediately that
.
,
arcsin (x)
1
=
;
Vl-x 2
the function sin could then be defined as (arcsin)
tive
-1
and the formula
for the deriva-
of an inverse function would show that
sin'(.v)
=
v
1
—
x,
sin
one could show that A(cosx) = x/2,
recovering at the very end of the development the definition with which we started.
While much of this presentation would proceed more rapidly, the definition would
which could be defined
be
utterly
as cosa". Eventually,
unmotivated; the reasonableness of the definitions would be
whom
the author, but not to the student, for
we
shall see in
Chapter
an approach of this
18,
indeed.
PROBLEMS
1.
Differentiate each of the following functions.
(i)
f(x )
(ii)
f(x)
(iii)
f(x)
=
=
=
arctan (arctan (arctan x
/
(iv)
=
f(x)
) )
arcsin(arctan(arccosx)).
arctan (tan x arctan x
)
\
1
arcsin
v/T+^2
Find the following
sin.t
limits
by
l'Hopital's Rule.
— x + * 3 /6
lim
x3
sin
n
x
lim
x4
.0
cosx
in
— x + x 3 /6
—
x
(
lv )
h
m
cosx
—
+ x/2
1
.
X*
arctan x
—
lim
lim
x
;
x-+0
fvi)
2
4
,v^0
(v)
+ x/2
1
lim
.V
[
—
\
x
sin
3
x
+ x 3 /3
known
to
was intended! Nevertheless, as
sort is sometimes very reasonable
it
—
316
Derivatives
and
Integrals
a
sin
3.
Let f{x)
=
x
,
#0
x=0.
1
(a)
Find/'(0).
(b)
Find/"(0).
At
this point, you will almost certainly have to use l'Hopital's Rule, but in
Chapter 24 we will be able to find f'iO) for all k, with almost no work
at
4.
all.
Graph
the following functions.
(a)
f(x)
(b)
f(x)
= sin2x.
= sin (a 2
(A pretty respectable sketch of
).
tained using only a picture of the graph of
your only hope
is
of
/
cos(a
=
Indeed, pure thought
sin.
problem, because determining the sign of the
-2x
)
The formula
directly.
however
(c)
=
fix)
derivative
in this
no
is
easier than determining the behavior
fix) does
for
should be clear in
fix) = sin a +
graphs of gix)
sin* and
axes,
from
easily find
0,
sin 2a.
=
(It
how many
=
/z(a)
sin
even,
critical
points
/
sum
points by finding out the sign of
/
at
draw
first
[0,
the
set
of
You can
will look like.
has on
fact,
and which
2x carefully on the same
You can then determine
the derivative of /.
is
probably be instructive to
will
2n, and guess what the
to
out
one important
indicate
which must be true since /
your graph.)
f'(0)
graph can be ob-
this
2n] by considering
the nature of these critical
each point; your sketch
will
probably
suggest the answer.)
fix)
(d)
—
— x.
tan a
fix)
=
n/2,
moved up
same, except
(e)
(First
—
the intervals ikn
— x.
sinjr
determine the behavior of / in (— n/2, n/2); in
kit + n/2) the graph of / will look exactly the
a certain amount.
(The material
in the
Why?)
Appendix
to
Chapter
1 1
will
be
particularly helpful for this function.)
x
sin
fix)
(0
X
*
1,
= 0.
(Part (d) should enable
you
to
of /' are located. Notice that
the size of
f(x)
(g)
The
=
x
/
sin
to
/
is
even and continuous
at 0; also
consider
for large x.)
.v.
hyperbolic spiral
nates (Chapter 4,
its
determine approximately where the zeros
is
the graph of the function f(0)
Appendix
behavior for
3).
Sketch
close to 0.
Prove the addition formula for cos.
this
=
a/0
in
polar coordi-
curve, paying particular attention
,
.
The
15.
(a)
From
(b)
Use
and cos derive formulas
the addition formula for sin
cos2jc, sin3x,
Trigonometric Functions
317
for sin2x,
and cos3x.
these formulas to find the following values of the trigonometric func-
tions (usually
deduced by geometric arguments
elementary trigonom-
in
etry):
—=
cos
4
4
re
tan
V2
— = ~2~'
71
71
.
sin
—=
.
1
4
7X
.
1
Sm
=
6
V3
7X
COS
8.
(a)
Show
A
that
sin(x
+
r
6=^-
B) can be written
as
a
sinjc
+ b cos x
for suitable a
and b. (One of the theorems in this chapter provides a one-line
You should also be able to figure out what a and b are.)
(b)
Conversely, given a and b, find
bcosx
9.
= A sin(.x +
(c)
Use part
(a)
Prove that
(b) to
B) for
all
+
y,
v)
+ tan y
tan x
=
+
and x
+
x.
1
provided that x,
that a sin x
= V3 sin x + cosx.
graph f(x)
tan(x
numbers A and B such
proof.
—
tan x tan y
y are not of the form kit
+
7r/2.
(Use the
addition formulas for sin and cos.)
(b)
Prove that
x
'
arctan x
+ arctan y =
arctan
1
and y by arctan y
10.
— xy
y.
Hint:
(a).
Prove that
arcsin
a
+ arcsin
indicating any restrictions
11.
in part
v
on x and
indicating any necessary restrictions
arctan x
+
Prove that
if
m
and
a?
fi
—
on
arcsin(av
or
and
1
—
fi~
+ ^v
1
—a
/?.
are any numbers, then
mx sin «jc =
sin mx cos «jc =
=
cos mx cos
sin
a?.v
j[cos(w
— ri)x —
cos(w +/?)-*],
+ n)x + sin (aw — n)x],
— «)jc
j [cos(m + n )x + cos(m
^[sin(m
]
),
Replace x by
,
318
Derivatives
and
'
Integrals
Prove that
12.
if
m and
n are natural numbers, then
sin
i:
cos
sin
These
///
7r,
m = n.
7^
n,
/;
0.
/
If
only in the Suggested Reading
this topic will receive serious attention
(see reference [26]), the
(a)
0,
n
relations are particularly important in the theory of Fourier series. Al-
though
13.
mx cos nx dx =
mx cosnx dx =
n,
m ^
m =
0,
mx sin nx dx =
is
next problem provides a hint as to their importance.
on [— n,
integrable
r
show
tt],
(f(x)
minimum
that the
value of
— a cosnx) dx
J —It
occurs
when
a
i
r
*
J-TT
— —\
f(x) cosnx dx,
and the minimum value of
(f(x)
— a s'mnxy dx
J —TT
/;
when
*
(In
r
= —
TT
/
Show
=
J
obtaining a quadratic
r
=—
m
*
that
f(x) cosnx dx,
n
—0,
f(x)s'mnxdx,
n
=
I
+
1,2,3
J-tt
and d are any numbers, then
if c,
t
-iv
fo
1,2, ...
J-TT
i
bn
=
sign,
a.)
i
(
nxdx.
Define
an
£
sin
each case, bring a outside the integral
expression in
(b)
f(x)
J-TT
2_. c " cos
[f(x)fdx-2iz(
nx
2
+ d„ sin nx
dx
+ ^2 °» c » + b » d»
I
+n
co
[
~y~
2
+ 5Z r " + ^"
n=]
j\f {x) f dx -J'^-+jr a,l+l,,r
+n
((vl " 7i)
+
^
""
(r "
"
)2
+
h
{d"
~
"j
.
.
.
The
15.
thus showing that the
among all
In other words,
sinnjc
and
=
cn (x)
integral
first
cosnx
319
Trigonometric Functions
smallest
is
.
when
a
{
=
c
t
and b
{
=
dj.
"linear combinations" of the functions s n (x)
for
1
<
<
n
TV,
=
the particular function
N
=
g(x )
— + v^
> a„ cos nx + b„
ao
sin
nx
n=\
has the "closest
14.
Find a formula for
(a)
x
sin
— sin v.)
good does
15.
sin
x
+ sin y.
tt].
(Notice that this also gives a formula for
+ b) + sin(a — b). What
Hint: First find a formula for sin(a
that do?
+ cos y
—
and cos x
Also find a formula for cos x
(a)
Starting from the formula for cos 2x derive formulas for sin x
(b)
Prove that
cos v
,
and cos
jc
terms of cos2x.
x
=
COS
2
<x <
for
/l+cosx
x
.
V-—
and
Sm
=
2
—
I—
V-
cos*
tv/2.
(c)
Use part
(d)
Graph f(x)
(a)
to find
fa
=
x.
sin
x dx and f cos 2 x dx
sin
Find sin(arctanx) and cos(arctanx) as expressions not involving trigonometric functions. Hint: y
sin
17.
/ on [— n,
(b)
in
16.
to
fit"
If
v/v
=
x
1
—
sin
=
arctanx means that x
=
tan v
=
sin
v/ cos v
=
v.
tanw/2, express sinw and cosw
in
terms of x. (Use Problem
16; the
answers should be very simple expressions.)
18.
(a)
graphs of
19.
(b)
What
is
(a)
Find
/
sin
~dt. Hint:
(All
along
were the
we have been drawing
the
case.)
The answer
is
not 45.
l
jdt.
/
l+f
Jo
.
1
—
20.
Find lim x
21.
(a)
Define functions
(b)
Find lim
sin
and cos by sin°(x) = sin(7rx/180) and cos°(x) =
cos(7rx/180). Find (sin )' and (cos )' in terms of these same functions.
x-s-0
22.
cos*.
as if this
l+t 2
r°°
Find
and cos
arcsin(cosx) and arccos(sinx)?
Jo
(b)
+ n/2) —
Prove that sin(x
sin°
and lim x
X
sin°
Prove that every point on the unit
least
one (and hence
for infinitely
x
.v->-oo
circle
is
of the form (cos
many) numbers
0.
0, sin 0) for at
.
320
Derivatives
and
Integrals
23.
(a)
Prove that
which
[2kn
(b)
sin
-
tz
25.
Prove that
+ tt/2] or
g(x) = sin a
tt/2, 2kit
=
Let f{x)
graph.
maximum
the
|
let
< x <
sec a for
—
x
sin
statement, with
length
possible
of an interval
<
—
replaced by
<
x
for
—
(Ikn
in
y\ for all
+
1
)n
tt/2,
numbers x
-
2kn
/
/
tt/2]
+
_1
y.
.
tt/2).
What
and sketch
*26.
It is
to
<
an excellent
The same
Hint:
f"
sin
fix)
/
get the right idea for a proof the
(b)
A.jc
dx
limit
can
easily
to intuition, but
be established
for
once you
any
inte-
/.
Show
that lim
Show
that
lem
(c)
to
of intuition to predict the value of
test
Continuous functions should be most accessible
(a)
its
.
lim
grate
is
a very straightforward consequence of a
is
,
2(k
Find the domain of
tt.
\x
<
sin y\
+ tt/2,
[2A-jr
well-known theorem; simple supplementary considerations then allow <
be improved
on
one-one, and that such an interval must be of the form
is
Suppose we
24.
is
if
s
f
is
sin
=
by computing the integral
0,
explicitly.
a step function on [a b] (terminology from Prob,
13-26), then lim
Finally, use
kx dx
f
s
sin
(jc )
Problem 13-26
Xx dx
show
to
=
that
0.
lim
f fix)sir\Xxdx —
for
any function / which is integrable on [a, b]. This result, like Problem 12,
plays an important role in the theory of Fourier series; it is known as the
Riemann-Lebesgue Lemma.
27.
This problem outlines the
The shaded
(a)
By
approach
classical
sector in Figure
24 has area a/2.
considering the triangles
OCB prove that
OAB
and
x
—
2
2 cos x
then
fi
=
sin
(l,0)
x
< — <
2
FIGl
RI".
24
(b)
Conclude
sin
<
sm
<
A
=
1.
A
*-^0
Use
x
that
hm
(c)
sin
a
that
cosjc
and prove
to the trigonometric functions.
this limit to find
1
lim
—
COS A
1.
if
< x <
7r/4,
.
The
15.
(d)
Using parts
starting
^28.
and
(b)
from the
and the addition formula
(c),
for sin, find sin'(x),
definition of the derivative.
This problem gives a treatment of the trigonometric functions in terms of
= yl — x 2
and uses Problem 13-25. Let f(x)
Define %(x) to be the length of / on [x, 1].
length,
Show
(This
is
an improper
integral, as defined in
then prove that i£(x)
(b)
Show
x <
Define
Problem 14-28, so you must
the length on [x, 1 — s] and
+
the limit of these lengths as e ->
is
=-
as i£(— 1).
tt
=
define sin*
v
for
,
< x <
For
—
1
< x <
cos x for
.)
cos
2
-
< x <
1
define cos* by i£(cosx)
tt,
Prove that cos'(x)
.v.
1
= — sin*
and
=
x,
—
tioned in the text
this
=
tt.
starting with inverse functions defined
convenient to begin with arctan, since
To do
and
sin'C*)
Yet another development of the trigonometric functions was briefly
is
1.
that
£'{x)
(c)
—1 <
—L=dt.
prove the corresponding assertion for
first
for
that
££(*)= f
^29.
321
Trigonometric Functions
function
this
is
by
men-
integrals.
defined for
all
It
x.
problem, pretend that you have never heard of the trigonometric
functions.
Let <x(x)
(a)
=
lim a(x) and
x-+oo
we
(b)
(c)
define
+ f") -1
Jq{\
tt
dt.
=
is odd and increasing, and that
and are negatives of each other. If
Prove that a
lim a(x) both
x—*— oo
exist,
2 lim a(x), then oc~
X—>00
l
is
defined on (—tt/2, tt/2).
Show that {a~ )'{x) = 1 + [cy-'U)] 2
For — tt/2 < x < tt/2, define tan.v = a~
l
.
tan;c/\/l
+
tan 2 x.
lim
(i)
sin
x
Show
=
[
(x),
and then define sinx
=
that
1
x—>n/2-
lim
(ii)
sin
x
=—
1
X^-7t/2+
sin*
111
sin'Cr)
tt/2
=
tan
^30.
If
we
sin"(x)
tt/2
and x
^
x=0
1,
IV
< x <
jc
= — sinx
for
—
tt/2
<
jc
<
tt2.
are willing to assume that certain differential equations have solutions,
another approach to the trigonometric functions
particular, that there
satisfies
yo"
+
vo
=
0.
is
some function yo which
is
is
possible.
not always
Suppose,
in
and which
:
322
Derivatives
and
Integrals
(a)
Prove that
(b)
+
yo°-
^
or yo'(0)
Prove that there
=
s'(0)
If
we
—
= 0,
become
There
trivial.
x >
all
0.
problem
Find cos7r,
is
=
=
cos'
to decide
why
(a)
After
>
a smallest xo
is
?
=
we have
1,
with
=
sin7r/2
±1;
positive.)
is
sin 27T. (Naturally
you may use any addi-
can be derived once we know that
the
sin are periodic
work involved
certing to find that sin
is
prove that cosx cannot
to
1 1
=
sin'
cos
—sin.)
Prove that cos and
(There
about trigono-
1 1
+cos
sin tt/2
and
and
2*o2
(Since sin
1.
sin7r, cos 2tt,
(f)
all
—
tt
all facts
=
one point which requires work,
is most easily done using an
is
Chapter
to
and s(0)
This
follows that there
It
tion formulas, since these
and
tt.
Chapter
to
and we can define
Prove that sin7r/2
the
/
that either yo(0)
=
s
then almost
s',
Use Problem 6 of the Appendix
cos*o
31.
—
and cos
s
+
+ byo'.
s satisfying s"
form ayo
producing the number
be positive for
(e)
a function
from the Appendix
exercise
(d)
is
—
define sin
however
and conclude
constant,
Hint: Try s of the
1.
metric functions
(c)
is
(joO
0.
with period
2tt.
of
sin,
in the definition
it
would be discon-
actually a rational function. Prove that
is
it
isn't.
a simple property of sin which a rational function cannot pos-
sibly have.)
(b)
Prove that
is,
sin isn't
do not
there
+
(sin x)"
even defined implicitly by an algebraic equation; that
exist rational functions /o,
/„_!
UMsinx)"-
=
Hint: Prove that /o
|:
32.
is
x.
Suppose
that
(a)
(b)
that #2
Show
Show
•
•
+f
so that sin x can
(Why?) You are
(f>\
and
now
>
set
up
.
(x)
.
,
/„_] such that
=
for all x.
be factored
tt.
for a
But
out.
this
The remaining
implies that
proof by induction.
satisfy
(j>2
0l"
02 "
and
•
except perhaps at multiples of
factor
for
all
5
+
1
.
+£101 =
0'
=
0'
+
5202
8\-
that
that
01 "02
-
if (p\(x)
>
/
02"0l
-
(£2
-
and 02 U) >
[0| "02
-
£l)0102
=
for
x
all
0.
in (a, b),
then
02"0l| > 0,
Ja
and conclude
[4>l(b)4>2(b)
that
- 0i
/
(a)02 (a)]
-
\4>\(b)<f> 2 '(b)
-
0,(«)02'(«)]
>
0.
it is
The
15.
(c)
Show
we cannot have
that in this case
and
sider the sign of 4>\'{a)
(d)
Show that
02 <
<
02 >
0,
4>\(a)
=
=
(p\(b)
0.
Con-
Hint:
<p\'{b).
the equations 0i (a)
or 0i
323
Trigonometric Functions
=
0i (b)
<
0, or 0i
=
are also impossible
02 <
0,
on
(a, b).
if
0i
>
0,
(You should be
able to do this with almost no extra work.)
problem may be stated as follows: if a and b are
consecutive zeros of 0i, then 02 must have a zero somewhere between
a and b. This result, in a slightly more general form, is known as the
Sturm Comparison Theorem. As a particular example, any solution of
The
net result of this
the differential equation
must have
33.
(a)
at least
(x
+
one zero
in
any interval
Using the formula
sin (A'
(b)
Conclude
+
for sin
^)x
—
a:
—
sin(/c
l)y
—
+
(«7r, (n
1
)tt).
show
derived in Problem 14,
sin v
j)x
=
2
sin
—
that
coskx.
that
1
—
=Q
+
y"
+
cos x
+
cos 2x
+
+ cos nx =
•
sin(n
+
l)x
.
x
2
2
sin
—
2
Like two other results in
in the study
of Fourier
this
series,
problem set, this equation is very important
and we also make use of it in Problems 19-43
and 23-22.
(c)
Similarly, derive the
formula
sin
sin
x
+ sin 2x +
+ sin nx
I
-y-A-jsin(-.rj
=
.
.
sin
x
—
2
(A more natural derivation of these formulas
will
be given
in
Prob-
lem 27-14.)
nb
(d)
Use
parts (b)
and
(c)
to find
/
Jo
the definition of the integral.
rb
sin
x dx and
/
Jo
cos x
dx direcdy from
*CHAPTER
^T
I
IRRATIONAL
71 IS
This short chapter, diverging from the main stream of the book,
demonstrate that
ics.
we
are already in a position to
This entire chapter
many "elementary"
is
devoted to an elementary proof that
it is
proofs of deep theorems, the motivation for
proof cannot be supplied; nevertheless,
is still
it
is
included to
do some sophisticated mathematirrational. Like
many
steps in
our
quite possible to follow the proof
step-by-step.
Two
observations must be
made
before the proof.
The
first
concerns the func-
tion
-A
MX)=
nl
which
clearly satisfies
<
f„(x)
<
-
forO <
<
jc
1.
n\
An
important property of the function /„
obtained by actually multiplying out x"(\
will
be n and the highest power
is
HI
where the numbers
c,
are integers.
—x) n The
be 2n
will
It is
/„(*)(())=
revealed by considering the expression
.
Thus
.
power of
1
from
<nork
this
expression that
>2n.
Moreover,
(x)
fn
—
— [n\c +
n
terms involving x]
n\
/;/"
+
l|
an
fn
324
(x)
=
-[(«
+
l)!c„ +
\x)=- [(2n)\c2n
y
].
jc
i
+
terms involving x]
appearing
in the
t—
clear
it'k
/„
lowest
can be written
form
16.
n
is
Irrational
325
This means that
(n)
(0)=Cn
fn
/n
(n+1)
fn
(0)
=
(n
(0)
=
(2n)(2n
{2n)
where the numbers on the
+ l)cn+1
-
1)
•
.
.
right are all integers.
{k)
fn
The
,
(0)
is
an
•
.
+
(n
l)c 2n
,
Thus
integer for all k.
relation
=/n (l ~X)
fn(x)
implies that
/B
(k)
=
(x)
(-i)*/«
w (i-*);
therefore,
(1)
/«
The proof
positive
that
tt
is
number, and
e
an
also
is
irrational requires
>
this,
notice that
n
if
r"
(n
Now let
hq be any natural
one further observation:
then for sufficiently large n
0,
—
To prove
integer for all k.
>
will
<£.
a
n
1)!
a
+
\
n
n\
number with no >
1
a"
2
n\
<
Then, whatever value
2a.
r
,"0
a
(no)!
may
have, the succeeding values satisfy
fl
(«o+D
2" (^)!
(«o+2)
(n
+ 2)\
(n
+ k)\
a »o
1
<
(no+D!
fl
fl
J
< —
2
<"
•
(n
2*
«"°
If
/:
is
so large
S that
< 2 » then
,
(no)! 8
have
2a, then
+1
+
we
if
(«o+D
(n )!'
1
1
r— < —
+l)!
2
•
—
2
a "o
(n
)!
a
is
any
326
Derivatives
and
Integrals
+k)
,(n
which
THEOREM
1
one theorem
in this chapter.
The number
tt
of
irrational; in fact,
is
implies the irrationality of
tt~
e,
k)\
Having made these observations, we are ready
the desired result.
is
<
+
(no
tt,
irrational.
is
jr
for if
were
tt
for the
(Notice that the irrationality
rational, then
tt
certainly
would
be.)
PROOF
Suppose
were
tt~
rational, so that
2
TT
a
= b
for
some
G(x)
(1)
and
positive integers a
=
b"[TT
2
"fn (x) ~
2
TT
b.
Let
+ TT 2 "- 4 fn {4)
"-\fn "(x)
(
X)
-
(- D" fn^ix)].
+
Notice that each of the factors
b"TT
is
an
integer.
Since
fn
2 "- 2k
(k)
(0)
=
2
k
b"(TT )"-
=
(|V~ -
b"
and fn ik \l) are
integers, this
G(0) and G(l) are
Differentiating
G"{X)
(2)
The
last
G
=
b"[TT
2
G"(x)
Now
k
shows that
integers.
twice yields
"fn "{x) -
term, (—l) n fn
(3)
k
a"- b
2 "- 2
TT
(,x), is
fn
zero.
+ tt 2 G(x) =
b"TT
{4)
X)
(
+ (-\Tfn a " + 2) (x)].
+
Thus, adding
(1)
2n+2
2
fn (x) =
TT
and
(2)
gives
a"f„(x).
let
—
—
ttG(x) cos7r.v.
= ttG'(x) cos ttx + G"{x) sinTr.r —
= [G"U) + 2 GU)]sin7TJc
= tt a n fn (x)sinTtx, by (3).
7rG'U)cos7r;t
H(x)
G'(x)
sin7r.v
Then
H'{x)
+ tt"G(x) sin ttx
77-
By
the
Second Fundamental Theorem of Calculus,
/
Jo
a"
2
tt
/;,
(.v )
sin ttx
dx
=
=
=
//
( 1 )
-
H (0)
G'(l)siii7r
7r[G(l)
+
-ttGCDcostt - G'(0) sinO
G(0)].
Thus
* f a"fn (x)
Jo
sin ttx
dx
is
an
integer.
+ ttG(0) cosO
—
16.
On
<
the other hand,
<
<
f„(x)
Tta"
fn {x)
< x <
\/n\ for
327
so
1,
jza"
<
sin7r.v
Irrational
it is
< x <
for
1
//!
Consequently,
1
f
< n
I
a
7r «"
n
f„(x)
smTtxdx <
,
n\
Jo
This reasoning was completely independent of the value of n
.
Now
if
n
is
large
enough, then
/* 1
<
7r
/
tl
fl"/„(jc)sin
<
ixxdx <
But
this
and
1
absurd, because the integral is an integer, and there is no integer between
Thus our original assumption must have been incorrect: ti is irrational. |
is
.
This proof
that
it
1.
n\
Jo
is
admittedly mysterious; perhaps most mysterious of
—
enters the proof
it
almost looks as
ever mentioning a definition of
that precisely
one property of
ti
it
A
.
is
really
we have proved
it
the
all is
way
irrational without
close reexamination of the
proof will show
essential
sin(7r)
The proof
if
=
0.
depends on the properties of the function
irrationality of the smallest positive
number x with
sin
x
—
sin,
0.
and proves the
In
fact,
very few
properties of sin are required, namely,
sin'
cos'
sin(0)
cos(0)
Even
this list
as well
= cos,
= — sin,
= 0,
= 1.
could be shortened; as far as the proof
be defined as
sin'.
The
is
concerned, cos might just
properties of sin required in the proof may then be
written
+ sin =
sin(0) =
sin'(0) =
sin"
Of course,
this
is
0,
0.
1.
not really very surprising at
all,
since, as
we have
seen in the
previous chapter, these properties characterize the function sin completely.
PROBLEMS
1.
(a)
For the areas of triangles
show
that
we have
OAB and O AC in Figure
1,
with
LAOB
<
tt/4,
328
Derivatives
and
Integrals
/l -v/l Ari
„ AC
= -W
O
16(area
1
area
(b)
Let
Am
(x, y)
Pm be
is
=
xy
Hint: Solve the equations
Pm show
the area of
.
2(area
the regular polygon of
OAB) 2
m
OAB), x 2 + y 2 =
1,
for y.
sides inscribed in the unit circle. If
that
=A
A 2m =
/
y V 2 - 2v
l-(2A,„/m) 2
.
This result allows one to obtain (more and more complicated) expressions
(1.0)
=C
A4 — 2, and thus to compute n as accurately
as desired (according to Problem 8-11). Although better methods will
appear in Chapter 20, a slight variant of this approach yields a very
for
A 2«,
starting with
interesting expression for
FIGURE
2.
(a)
Using the
tt
:
fact that
1
area(OAfl)
= OB,
area(OAC)
show
that
if a,„ is
=
A 2n
(b)
Show
to
one
side of P,„
,
then
a,,
that
CK4
L
(c)
O
the distance from
Using the
•
org
•
.
.
.
Q?2*-i
2A
fact that
am
=
7T
cos—,
m
yl
"I
-
COS X
(Problem 15-15), prove that
a4
"8
=
Of|6
=
1
V2
1
1
2V2'
11/11
+
+
\J
etc
+
2
2V 2
2V
2'
16.
Together with part
(b), this
7i is
Irrational
shows that 2/tt can be written
as
an
329
"infinite
product"
7T
to
be precise,
can be
made
this
equation means that the product of the
as close to
2/n
as desired,
by choosing n
fascinating expressions for
jt
,
some of which
n factors
sufficiently large.
This product was discovered by Francois Viete in 1579, and
many
first
is
only one of
are mentioned
later.
PLANETARY MOTION
*CHAPTER
Nature and Nature's Laws lay hid
God
Newton
said "Let
be,"
and
in night
all
was
light.
Alexander Pope
Unlike Chapter 16, a short chapter diverging from the main stream of the book,
long chapter diverges from the main stream of the book to demonstrate that
this
we
are already in a position to
do some
In 1609 Kepler published his
first
real physics.
two laws of planetary motion. The
first
law
describes the shape of planetary orbits:
The planets move
The second law
in ellipses,
with the sun at one focus.
involves the area swept out
by the segment from the sun
to the
planet (the 'radius vector from the sun to the planet') in various time intervals
(Figure
1):
Equal areas
swept out
FIGURE
are swept out by the radius vector in equal times.
in time
t
is
proportional
to t
(Equivalently, the area
.)
Kepler's third law, published in 1619, relates the motions of different planets. If a
1
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:
The
ratio
a
/T
is the
same for
all planets.
Newton's great accomplishment was
force
on a body
is its
mass times
its
to
show
(using his general law that the
acceleration) that Kepler's laws follow
assumption that the planets are attracted to the sun by a force
from the
(the gravitational
toward the sun, proportional to the mass of the
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
times acceleration, this is equivalent simply to saying that the magnitude of the
acceleration is a constant divided by the square of the distance from the sun.
force of the sun) always directed
planet,
Newton's analysis actually established three
individual laws.
was
330
The
first
actually discovered
results that correlate with Kepler's
of Newton's results concerns Kepler's second law (which
first,
nicely preserving the
symmetry of the
situation):
17. Planetary Motion
Kepler's second
the
law
is
true preciselyfor 'centralforces',
sun and the planet always
Although Newton
is
lies
i.e.,
331
if and only theforce between
along the line between the sun and the planet.
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
planet moves,
Instead of considering a force that varies continuously as the
all.
Newton
a momentary force
To be
line
imagine that during the
,
where the length of
and the same
case where the force
is
move along
to
as the triangle
this just says that
during the next equal
If,
this line,
it
would end up
at
This would imply that
P?,
SP1P3
(since they
have equal
Kepler's law holds in the special
0.
moment
at the
the planet arrives at P2
it
experi-
which by itself would cause the planet
the point Q. Combined with the motion that the planet already has,
sides are
S
the linefrom
move
causes the planet to
whose
—
suppose (Figure 2b) that
move
this
same area
height)
ences a force exerted along
to
that
time interval the planet moves
Pi Pi equals the length of Pj
the triangle SP\ Pj has the
bases,
Now
first
P\Pi, with uniform velocity (Figure 2a).
time interval, the planet continued to
f*3
and assumes
exerted at the ends of each of these intervals.
is
specific, let us
along the
considers short equal time intervals
first
P2,
to
to R, the vertex opposite Pj in the
parallelogram
P2P3 and P%Q.
Thus, the area swept out
in the
second time interval
actually the triangle
is
SPiR. But the area of triangle SP2R is equal to the area of triangle SP3P2, since
they have the same base SP2, and the same heights (since RP2 is parallel to SP2).
Hence, finally, the area of triangle SP2R is the same as the area of the original
triangle SP\ P2
!
Conversely,
if
SRP2
the triangle
has the same area as SP\ Pj, and
hence the same area as SP3P2, then RP3 must be parallel
FIGURE
2
Q must lie along SP2.
Of course, this isn't quite
to SP2,
and
this
implies
that
modern book, but
in
its
argument one would expect to find in a
it shows physically just why the result
the sort of
own charming way
should be true.
To
analyze planetary motion
Chapter
Chapter
12,
4.
we
will
Appendix
Problem 4 of Appendix 1
be using the material
and the "determinant" det defined in
describe the motion of the planet by the parameterized curve
=
r(0 (cos 0(0,
so that r always gives the length of the line
gives the angle,
then follows
which we
easily).
will
It will
assume
is
c(0
from the sun
increasing (the case
be convenient
e(0
just the
to
sin 0(0),
to the planet, while
where
is
decreasing
to write this also as
= r(0
•
e(0(O),
where
is
to
We
c(t)
(1)
in the
=
(cosf, sin
t)
parameterized curve that runs along the unit
e'(t)
=
(—
sinf,
cosO
circle.
Note
that
332
Derivatives
and
Integrals
is
also a vector of unit length, but perpendicular to e(0,
det(e(f),e'(f))
(2)
Differentiating
=
c'{t)
and combining with
to Chapter 4, we get
det(c(0,
c'(t))
since det(u, v)
r\t)
r(t)r'(t) det(e(0(f)), e(0(f)))
=
2
r(f) 0'(Odet(e(0(f)),e'(0(O)),
always
Using
0.
that A{t)
formula
is
segment between
=
1
and
to
e'(0(t)),
from time
the area swept out
c(t
+ h).
It is
of the triangle A(/z) with vertices O,
and 5 of Appendix
obtain
Chapter
area(A(/j))
c(t
shows
1
e'(0(f)))
Newton,
and
area
t
we'll
We
want
begin by making
(Figure
3).
A(t), together with a straight line
easy to write
c(t),
4, the
=
+ h) —
to
c(t
down
+ h):
a formula for the area
according to Problems 4
is
+
£det(c(f), c(t
h)
-
c{t)).
Since the triangle A(/z) has practically the same area as the region A(t
this
Appendix
2
r 0'.
for A'(t), and, in the spirit of
c(t )
we
then get
c')
an educated guess. Figure 4 shows A(t
3
have
out to have another important interpretation.
will see, r &' turns
to get a
FIGURE
we
(2)
det(c,
Suppose
also
+ r(t) 2 0'(t) det(e(0(f)),
=
is
we
together with the formulas in Problem 6 of
(1),
(4)
As we
+ r(t)6'{t)
e(0(t))
that
1.
using the formulas on page 247,
(1),
(3)
=
and
+ h) — A(t),
(or practically shows) that
+ h)
A'(t)
A(t
=
lim
=
lim
+ h) -
A(t)
area A(h)
h
=
c(t
4det
c(0, lim
+ h)-c(t)
h-+0
=
\det(c(t),c'(t)).
A rigorous derivation, establishing more in the process,
lem 13-24, which
gives a
formula
for the area of a region
of a function in polar coordinates. According to
this
can be
made
using Prob-
determined by the graph
Problem, we can write
0(0
I
A(0=
if
our parameterized curve
c(t)
—
polar coordinates (here we've used
o
r(t)
/
P(<P)~d<t>
e(0(t))
for the
is
the graph of the function
p
in
angular polar coordinate, to avoid
confusion with the function 9 used to describe the curve
c).
1 7. Planetary
Now
the function p
just
is
=
p
[for
any particular angle
#
</>,
_1
(0)
(t))
r
the time at which the curve c has angular
is
l
polar coordinate 0, so r{0~
the radius coordinate corresponding to 0].
is
Although the presence of the inverse function might look a
Applying the
actually quite innocent:
and the Chain Rule
to
333
Motion
(*)
A\t)
2
=
\p{e{t))
=
2
\r(t) 6'(t),
=
\r
it's
Fundamental Theorem of Calculus
First
we immediately
bit forbidding,
get
-0'(t)
since
=
p
r o
>-l
Briefly,
A'
Combining with
(4),
we
e'.
thus have
l„2/i/
A'=$ det(c, c') =
(5)
Now
is
2
\r
we're ready to consider Kepler's second law. Notice that Kepler's second law
equivalent
to
A"
=
saying that A'
\
is
constant,
=
[det(c, c')]'
=
and thus
\ det(c',
J
+
c)
it is
equivalent to A"
\ det(c, c")
(see
=
0.
But
page 248)
det(c, c").
So
Kepler's second law
Putting this
THEOREM
l
all
we
together
Kepler's second law
planetary path c(t)
true
is
—
r(t)
if
and only
Saying that the force
c"(t)
is
And
if
det(c,
this
the force
c)
means
of the force, that
in the direction
points along c{t).
=
central just
is
is
is
Moreover,
this
this
is
is
central,
—
0.
and
in this case
each
constant.
that
it
always points along
c(t).
Since
equivalent to saying that c"(t) always
equivalent to saying that
det(c, c")
We've just seen that
=
e(0(t)) satisfies the equation
.2/1/
PROOF
equivalent to det(c, c")
have:
r~0'
(Ki)
is
=
we always have
0.
equivalent to Kepler's second law.
equation implies that [det(c,
c')]
=
0,
which by
(5)
gives (Ki).
|
334
Derivatives
and
Integrals
Newton next showed
and
that
if
the gravitational force of the sun
an inverse square
also satisfies
conic section having the sun at one focus.
case
where the conic
section
an
is
is
a central force
then the path of any object in
law,
will
be a
Planets, of course, correspond to the
and
ellipse,
it
this
is
also true for
comets that
visit
and hyperbolas represent objects that come from
and eventually continue on their merry way back outside
the sun periodically; parabolas
outside the solar system,
the system.
THEOREM
2
If the gravitational force
then the path of any body in
law,
focus (more
PROOF
of the sun
precisely, either
an
is
it
a central force that satisfies an inverse square
be a conic section having the sun
will
one
at
parabola, or one branch of an hyperbola).
ellipse,
Notice that our conclusion specifies the shape of the path, not a particular parameterization.
But
this
parameterization
essentially
is
determined by Theorem
hypothesis of a central force implies that the area A{t) (Figure 5)
to
so determining c(t)
t,
on the
points
mined
c(t)
0~
r{t)
determining
:
is
proportional
A
for arbitrary
This means that
e(0(t)) without finding
we have
its
to
determine the shape of the path
parameterization! Since
it is
the function
which actually describes the shape of the path in polar coordinates,
shouldn't be surprised to find 6~ entering into the proof.
r o
the
Unfortunately, the areas of such segments cannot be deter-
ellipse.
explicitly*
=
is
essentially equivalent to
1
l
we
x
FIGURE
5
By Theorem
1
,
the hypothesis of a central force implies that
2
r 0'
[K,
for
= det(c,c') =
some constant M. The hypothesis of an
M
inverse square law can be written
H
e(0(t))
c"{t)
r(ty
for
some constant
H
.
Using
(K2), this
can be written
H
-— = -—e(6{t)).
c"{t)
M
0'(t)
Notice that the left-hand side of this equation
[c'o0-
So
if
we
x
]'{Q{t)).
let
D
(this is
is
the
main
trick
— "we consider
C o
c'
as a function of 0"), then the equation can
be written as
D'(0(t))
*More prn
an
sin, etc.
iscly.
we
=
can't write
H
M
down
e(0(O)
=
a solution in
H
M (cos 0(f),
sin 0(f)),
v
terms of familiar "standard functions,"
like sin,
1 7. Planetary
and we can write
D
[for all u
H
=
M
,
that
components of D, each of which we can
H
for
two constants
for
c'\
A and
c
[Here
M
—
B. Letting u
H
=
sin 6
Although we
can't get
with c
=
H
+ A,
thus find that
cos w
B
M
we
9{t) again
H
we
cos
M
thus have an explicit formula
+B
we will use throughout.]
if we substitute this equa-
an
explicit
formula for c
itself,
r(cos6, sin#), into the equation
det(c, c')
=
M
(equation (#2)),
get
h
—
M cos
which
2
H
M
~>
.
B
H
2
Problem 15-8 shows
some
constants
M
A
cos 6
= M,
M
C and
We
D.
is
form
can
let
D —
1,
0, since this
simply amounts to
which ray corresponds
to 6
=
0),
write, finally,
r\\
L
this
1
—i+CcosieiO + D) =
rotating our polar coordinate system (choosing
we can
.
sin
that this can be written in the
r(t)
But
— A sin 9
sin
simplifies to
r-
for
a
+ B cos 6
H
—
M
so
0.
simply a pair of equations, for the
is
easily solve individually;
+ A,
M
.
sin u
sin 6 really stands for sino#, etc., abbreviations that
tion, together
we
sin u
•
M
,
completely eliminating
t],
just obtained
D(u)
cos u
M
(
\
we have
H
H
(
=
.
(cos u sin u )
of the form 6{t) for some
The equation
335
simply as
this
(u)
Motion
+ecos01J =
H
=
A.
the formula for a conic section derived in
(together with Problems 5, 6,
In terms of the constant
and 7 of that Appendix).
M in the equation
r d' = M
2
and the constant
A
in the
equation of the orbit
/•
1
I
+ s cos 0] = A
Appendix
|
3 of Chapter 4
-
336
Derivatives
and
Integrals
we can
the last equation in our proof shows that
**
c"(t)
M—
A
=-
rewrite
while the minor axis b
as
I
r(t)
1
e(0(r)).
Recall (page 87) that the major axis a of the ellipse
is
given by
A
=
a
(*)
\-e 2
'
given by
is
A
=
b
(b)
y/l-e 2
Consequently,
=
a.
\
Remember
that equation
(5)
gives
l 9'
A'{t)
=
\r
A(t)
=
\Mt.
= \M,
and thus
We
is,
can therefore interpret
by
M in terms of the period T of the orbit.
definition, the value of
complete
t
for
which we have
6{t)
=
2tt, so that
This period T
we
obtain the
Hence
ellipse.
area of the ellipse
=
A(T)
= ^MT,
or
M=
Hence
2(area of the ellipse)
lizab
T
T
the constant
M"/A
in (**)
M
2
by Problem 13-17.
is
2
2
4jT a b
2
T2 A
4jT
2„3
cl
using
(c)
T<
This completes the
theorem
3
Kepler's third law
moving on
is
final step
true
if
of Newton's analysis:
and only
if the
accelerations c"{t) of the various planets,
ellipses, satisfy
c "( f )
= _G.— e(0(O)
r~
for a constant
G
that
does not depend on thr planet.
1 7. Planetary
It
this,
should be mentioned that the converse of
we
want
first
Theorem 2
is
337
Motion
also true.
To prove
one further consequence of Kepler's second
to establish
law.
Recall that for
e(t)
=
(cos?, sin
e'(/)
=
(—
=
(—cost,
t)
we have
sin
cos
t,
/).
Consequently,
e
Now differentiating
=
c"(t)
r"{t)
=
e(0(f))
c"(t)
=
we
—e(t)
[r"(t)
-
— sinf) =
—e(t).
gives
(3)
+ r\t)6'(t)
Using e"(t)
(t)
+ r'(t)6'(t) e'(6(t))
e'(0(f)) + r{t)0"{t)
+
e'(0(f))
•
r(t)9\t)G\t)
get
r(t)6'(t)
2
]
e(0(O)
+
[2/(00' (0
+
r(f)0"(*)]
Since Kepler's second law implies central forces, hence that c"(t)
tiple
of
[as
(A^)]
•
Thus
4
c"{t)
If the
=
As
in
1
path of a planet moving under a central gravitational force
Theorem
2, notice that the
we have
knowing what
to obtain information
lies
on a conic
an inverse square
law.
equivalent to Kepler's second law,
we
can't write
down an
explicit
about the acceleration without actually
it is.
l
)
for
some constant M, and
r G'
it
satisfies
r[l
(A)
some
entiating
= M,
the hypothesis that the path
the sun as focus implies that
First
satisfy
again, the hypothesis of a central force implies that
(K 2
for
is
determines the parameterization. But
solution, so
Once
must
hypothesis on the shape of the path, together
with the hypothesis of a central force, which
essentially
always a mul-
[r"{t)-r(t)e'{t) ]-e{6{t)).
section with the sun as focus, then the force
PROOF
is
Kepler's second law implies that
(6)
THEOREM
e'(0(f)).
and thus always a multiple of e(0(O), the coefficient of e'(6(t))
a matter of fact, we can see this directly by taking the derivative of
c(t),
must be
formula
e"(0(/)).
+ecos0]
=
A,
and A. For our (not especially illuminating)
and substituting from these two equations.
proof,
differentiate (A) to obtain
/•'[l
Multiplying by r
this
+ecos0] -er0'sin0 =0.
becomes
rr'
\ 1
+
s cos 0]
on a conic section with
the equation
s
we
lies
-
2
er 0' sin
=
0.
we
will
keep
differ-
338
Derivatives
and
Integrals
Using both
(A)
and
(K2), this
becomes
Ar' -eAfsinfl
Differentiating again,
we
=
0.
get
-e MO' cos 9 =
Ar"
0.
Using (K2) we get
eMcos#
Ar'
and then using
(A)
we
=0,
get
A
M-
=
Ar'
0.
r
Substituting from (K2) yet again,
we
get
A[r"-r(tf')
2
—
2
M-
+ iV=(X
r-
]
or
r
r(6')
=
M
2
"
Ar 2
Comparing with
(6),
we
obtain
M
c"(t)
2
Ar 2
e(0(O),
what we wanted to show: the force is
the square of the distance from the sun to the planet. |
which
is
precisely
inversely proportional to
CHAPTER
THE LOGARITHM AND
^^
I
EXPONENTIAL FUNCTIONS
In Chapter 15 the integral provided a rigorous formulation for a preliminary def-
and
inition of the functions sin
In this chapter the integral plays a
cos.
even a preliminary definition presents
essential role. For certain functions
more
difficul-
For example, consider the function
ties.
=
/(*)
assumed to be defined for all x and to have an inverse function,
positive x, which is the "logarithm to the base 10,"
This function
defined for
is
l
f~
In algebra, 10*
rational x
10*.
is
is
(x)
=log 10 x.
usually defined only for rational x, while the definition for
A
quietly ignored.
ir-
brief review of the definition for rational x will
not only explain this omission, but also recall an important principle behind the
definition of 10*.
The symbol
to
10"
10"
The
definition. Since
we want
true,
we must
be
true,
we must
define 10
1/w
•
0* to rational x
.
.
•
_"
10
=
10"
•
be
true,
we must
define 10
10
1/w
be
true,
we must
=
I"
•
m
to
motivated by the desire
is
=
10"
10°
10
=
the equation
1
we want
+l/ "
l/,?+
=
10
the equation
1
=
10
n times
1//,!
•
=
1/10"; since
n times
to
0+ "
we want
since
1;
=
10
=
10"
•
=
define 10°
10
1
10"+'".
the equation
10-"
to
This notation turns out
n.
requirement actually forces upon us the customary
10°
be
=
10'"
•
extension of the definition of
to preserve this equation; this
to
numbers
defined for natural
is first
be extremely convenient, especially for multiplying very large numbers, because
10
ly
v 10 and
"
=
10
1
/""
m
times
define 10'" 7 "
=
since
;
(
\/70
—
1
we want
hI /"
=
10"'
the equation
'"
/
times
)'"•
Unfortunately, at this point the program comes to a dead halt. We have been
guided by the principle that 10* should be defined so as to ensure that 10* +v =
A
1
339
v
1
;
but
this principle
does not suggest any simple algebraic
way of
defining
.
340
Derivatives
and
Integrals
we
10* for irrational x. For this reason
finding a function
/
f{x
(*)
Of course, we
+ y) =
(*) will
more
/
f(x)
=
\0
X
suggested
is
fix)
fix
lim
moment assume
we try
/ such
if
all
/
an apparently
that
— knowing
the derivative
Now
fih)
-
fix)
lim
•
h^0
h^0
=a
h
and denote
for
fix)
all
approach seems
this
/
the inverse function
log '»' (x)
1
lim
this limit exists,
a could be computed,
we examine
a new light:
x.
self-defeating.
f~
l
—
interest in g,
of
all
derivative
log 10 the whole situation appears in
,
J_
of/(/-'(.r))
" Vx
about as simple as one could
is
The
again.
1
derivative of /
by a. Then
it
= 7(7^0)
"
more
to solve
x and y,
itself.
fih)-
=
has been expressed in terms of
If
The
10,
f(x)
thus depends on
fix)
if
defined as
h
fix)
f'iO)
/
—
fih)
=
Even
f{\)
lim
fc-+0
of
and 10 could be
we might
+h)- fix)
fix)
=
for the
v
try to find /'
a clue to the definition of
fix)
The answer
specific condition
10.
we can
that such a function exists,
/ might provide
for
f(y)
f(\)=
of
ways of
y.
not always zero, so
is
function
differentiable
+ y) =
fix
Assuming
sophisticated
equal [/(l)]* for rational x.
will
problem: find a
x and
all
more
the
for rational x,
to find such a function
difficult
for
f(y)
•
we add
If
0.
imply that f(x)
for other x; in general
One way
f(x)
are interested in a function which
add the condition /(l)
then
some more
will try
such that
r x" dx
the integrals
examined
/
ask!
And, what
is
even
previously, the integral
Ja
x~ dx
is
the only one which
we cannot
evaluate. Since log 1()
Ja
have
1
/'
/
oi
J]
v
I
-dt
t
=
log
|
()
x
-
log
1
,
()
=
log to x
1
=
we should
18.
we
This suggests that
define log 10 * as
unknown. One way of evading
341
The Logarithm and Exponential Functions
(1 /or)
dt.
t
/
n
difficulty
is
that
a
is
define
this difficulty is to
log*
The
dt,
and 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 logi
The usefulness of
logjQ depends on the important role of the number 10 in arabic notation (and thus
ultimately on the fact that we have ten fingers), while the function log provides a
notation for an extremely simple integral which cannot be evaluated in terms of
any functions already known to us.
.
DEFINITION
If
x
>
0,
then
log X
=
/
-
dt.
The graph of log is shown in Figure 1 Notice that if x >
< x < 1, then log* < 0, since, by our conventions,
if
.
1
,
>
0,
\/t
is
then log x
and
-dt
/
J\
=-
- dt
Jx
t
< 0, a number log* cannot be defined
not bounded on [x, 1].
For x
area
FIGURE
l
If x.
0.
in this way,
because f(t)
=
log x
1
The justification
THEOREM
=
<
t
y >
0,
for the notation "log"
comes from
then
log(.vv)
=
log.v
+
log
>'.
the following theorem.
342
Derivatives
and
Integrals
PROOF
Notice
number
choose a
=
that log'(x)
first
>
y
by the Fundamental Theorem of Calculus.
l/x,
and
=
f(x)
log(xy).
11
Then
/
Thus
/'
=
log'.
(x)
=
log (xy)
This means that there
f{x)
that
Now
let
—
y
is
for
=
y
•
-.
x
number
a
+c
log*
—
xy
=
c such that
>
x
all
0,
is,
=
log(;ty)
The number
c can be evaluated
+c
log*
for
by noting
log(l
y)
•
=
=
that
log
1
x >
all
0.
when x =
1
we
obtain
+c
c.
Thus
=
log(xy)
Since
COROLLARY
1
If
n
this
is
true for
is
a natural
>
y
all
COROLLARY
2
Let to you (use induction).
If x, y
>
0,
all
x.
proved. |
=
n logx.
|
then
lo g(
PROOF
is
for
then
0,
log(jc")
PROOF
logy
theorem
0, the
number and x >
+
log*
~
=logx -logy.
)
This follows from the equations
log x
Theorem
provides
1
function log
is
very small as
jc
It is
log
-=
—
y
I
=
log
I
—
)
+ log y
some important information about
clearly increasing, but since log'C*)
becomes
large,
=
.
the graph of log.
follows that log
is,
in fact,
=
n
,
n log 2
(and log 2
>
0);
not bounded above. Similarly,
log(-)
The
becomes
and log consequently grows more and more slowly.
is bounded or unbounded on R. Observe,
number
log(2")
|
l/x, the derivative
not immediately clear whether log
however, that for a natural
it
(
:logl
-bg 2"
/;
log 2;
-
ally takes
on
is
all
not bounded below on
R
important function has a special
Since log
(0, 1).
is
continuous,
domain
name, whose appropriateness
Therefore
values.
343
The Logarithm and Exponential Functions
18.
therefore log
.
of the function log"
the
is
will
it
actu-
.
This
soon become
clear.
DEFINITION
The
"exponential function," exp,
The graph of exp
always have exp(x)
THEOREM
2
For
all
numbers x
is
>
shown
0.
The
in
defined as log"
is
Figure
Since log x
2.
is
defined only for x
derivative of the function
exp
is
>
0,
we
easy to determine.
,
exp'(x)
PROOF
exp'(x)
=
=
exp(x).
lv,
(log
=
)'U)
—
1
-1
log (log
00)
1
log-'(x)
=
A second
THEOREM
3
If
(x)
important property of exp
=
is
exp(x). |
an easy consequence of Theorem
1
x and v are any two numbers, then
exp(x
PROOF
-1
log
Let x'
=
exp(x) and
y'
=
+ y) = expU)
•
exp(y).
exp(y), so that
=
=
x
v
log*',
log
y'.
Then
x
+y =
log*'
+ logy' =
log(x'y').
This means that
exp(x
+ y) =
x'y'
This theorem, and the discussion
FIGURE
exp(l)
2
DEFINITION
is
particularly important.
exp(x)
at the
There
e
=
=
is,
•
exp(y). |
beginning of
this chapter,
in fact, a special
exp(l).
symbol
suggest that
for this
number.
.
344
Derivatives
and
Integrals
This definition
equivalent to the equation
is
1
As
=
f
=
loge
1
/
-
J\
t
dt.
illustrated in Figure 3,
— dt <
I
J]
1
since
,
1
—
(2
•
%
=
f(t)
an upper sum
1 ) is
for
l/ton[l,2],
and
n
FIGURE
>
dt
since \
1,
-
(2
•
sum
+\
1)
•
(4
=
for /(r)
- 2) =
1/r
on
1 is
a lower
[1,4].
Thus
3
,2
h
-4
,,
j
j
j
-dt <
-dt <
/
t
Ji
t
-dt,
t
Ji
which shows that
<
we
In Chapter 20
e
is
irrational (the
will find
proof is
As we remarked
e
<
much better approximations for e, and also prove
much easier than the proof that tt is irrational!).
that
beginning of the chapter, the equation
at the
exp(x
+
expU)
=
=
=
v)
exp(x)
•
exp(v)
implies that
Since exp
our
DEFINITION
is
defined for
all
[exp(l)]*
e
x
,
for all rational
x and exp(x)
=
e
x
for rational x,
earlier use of the exponential notation to define e
For any
number
The terminology
in defining e
yet defined a
attempt. If x
x
,
is
if
a
last
x
=
it
is
consistent with
as exp(jc) for all x.
exp(x).
now be
"exponential function" should
x
an arbitrary (even
for
^
rational,
e,
but there
expression
is
irrational)
clear.
exponent
x.
We have sucWe have not
a reasonable principle to guide us
then
a
But the
x
x,
e
ceeded
x
is
x
=
(e
defined for
l
° sa x
)
all
—
x, so
e
x[oga
.
we can
use
it
to define a
x
.
in the
DEFINITION
If
a
>
any
o, then, for
real
number x
ax =
(If
a = e
=
this definition clearly
The requirement a >
unduly
f(x)
=
is
restrictive since, for
345
The Logarithm and Exponential Functions
1 8.
x log a
agrees with the previous one.)
necessary, in order that logo be defined. This
we would
example,
is
not
not even expect
10*
<-i)
to
1/2
=
be defined. (Of course, for certain rational x, the symbol a x
make
will
sense,
according to the old definition; for example,
(_i)i/3
definition of a x
Our
was designed
=
(e*) y
As we would hope,
FIGURE
the
4
THEOREM
4
If
a
>
>
this
a
0,
1)
0.
to ensure that
xy
x and
for all
y.
when
e
replaced by any
is
a moderately involved unraveling of terminology. At
the other important properties of a x
.
then
{a
]
a
=a bc
b c
)
(Notice that a
(2)
e
_i.)
equation turns out to be true
The proof is
same time we will prove
number
= ^cr =
(Notice that
b
=
will
for all
c.
automatically be positive, so (a b ) c will be defined);
a and a x+y
(2)
ft,
=
ax
ay
for
all
x, y.
implies that this definition of a x agrees with the old one for
all
rational x.)
PROOF
(a
(1)
b
Y=
(Each of the steps in
the fact that exp
(2)
a
1
=
e
c]ogah
—
this string
log"
'
£
c
lo
gO Mog ")
=
e
_
c(bloga)
e
_ ^c
cb\oga
of equalities depends upon our
last definition,
or
.)
= e ll°z a = e °z a = a,
f,x+y
l
_
Ax+y)\oga
__
xloga+yloga __
Figure 4 shows the graphs of f(x)
=
of the function depends on whether a
a
<
x
vloga
_
_
vloga
x
for several different a.
1,
a
=
\,
or a
>
1
.
_
y
|
The behavior
If«=
1,
then
,
346
Derivatives
and
Integrals
f(x)=
\
x
=
Suppose a >
1.
x <
if
then
e
<
e
x
<
a
a
the function f(x)
<
so that log a
is
decreasing.
0, the
—
ax
same
>
a
if
Thus
takes
on
(Figure
all
it is
the inverse function
values. If f(x)
=
a
x
then
,
.
/
the other hand,
then f(x)
—
also easy to see that
x
^
and a
Since exp takes on every positive value
positive value.
,
v
is
1,
defined for
_1
is
< a <
if
shows that the function f(x)
sort of reasoning
In either case,
,
ylo % a
On
increasing.
is
Thus,
y,
xl °z a
i.e.,
0.
< y log a
x log a
so
Thus
>
In this case log a
1.
all
a
positive
a
x
is
=
1,
a
x
one-one.
on every
numbers, and
takes
the function usually denoted by log a
5).
Just as a
x
can be expressed
in
terms of exp, so log a can be expressed
in
terms
of log. Indeed,
=
log a x,
if
y
then
x-= a y
= e ylosa
..
= y log a
so
log x
or
y
=
x
=
log*
loga
In other words,
lo g„
FIGURE
The
5
derivatives of f(x)
f{x)
=
e
—
a x and g{x)
x[oga
f(x)
=
so g'(x)
=
so
,
log*
g(x)
more complicated
loga
—
log a x are both easy to find:
loga
also easy to differentiate, if
follows
-=
loga
a\
function like
you remember
f(x )
it
xlo ^ a
1
f(x)=g(x)"
is
e
xloga
loga
A
tog*
from the Chain Rule
fix)
_
e
(x)
that, by definition,
=e hM\o S g(x).
that
h(x) \ogg(x)
h'(x)logg(x)
+ h(x)
8(x)j
g(x)
h(x)
h'(x)logg(x)
+ h(x)
g'(x)
g(x)j
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)
.
There
The
one
is
=
function fix)
and
define
result
is
above formula which
special case of the
a
x
347
The Logarithm and Exponential Functions
18.
was previously defined only
worth remembering.
for rational a.
=
find the derivative of the function fix)
is
x
a
for
We
can
any number
now
a; the
what we would expect:
just
_
a
fix)
„a\ogx
SO
= -
fix)
a log*
ax
X
become second naall the rules which ought to work
those stated in Theorems 2 and 3:
Algebraic manipulations with the exponential functions will
—just remember that
ture after a
little
practice
actually do.
The
basic properties of
exp
fix)
THEOREM
5
If
/
=
is
ce
=
and
a
close to characterizing the function exp.
/
—
satisfying /'
= fa)
number
f, for
if
/ =
ce
x
,
then
for all x,
c such that
fix)
PROOF
exp(y).
.;
;
fix)
is
•
fix); these functions -are 'the only ones with this property, however.
clifTerentiable
then there
exp(x)
not the only function
is
still
=:exp(x),
(.x)
comes
In fact, each of these properties
x
;
+ y) —
exp(x
Naturally, exp
exp are
=
ce
x
for all x.
Let
*(*)
(This
is
permissible, since e
x
^
8 ix)
for
=
fix)
=
-
e*f'{x)
is
a
number
The second
is
=
fact,
fix)
=
=
ex
c
clearly not the only function
=
But the true story
for all x.
more
+ y) =
/ which
much more complex than
functions which satisfy this property, but
advanced mathematics,
to
involved discussion.
The
satisfies
fix)-f(y).
or any function of the form f(x)
is
0.
y
basic property of exp requires a
fix
In
x
c such that
gix)
function exp
f(x)e*
-^-
ie
Therefore there
Then
all x.)
it is
prove that there
this
=
a
x
also satisfies this equation.
—there are
infinitely
many other
impossible, without appealing to
is
more
even one function other than those
348
Derivatives
and
Integrals
already mentioned!
for this reason that the definition of 10 x
It is
many
there are infinitely
/ which
functions
f(x
+ y) =
/d)=
continuous function
/
f{x)
f(y),
10,
=
but which are not the function f(x)
so difficult:
is
satisfy
10 v
One
!
thing
is
—any
however
true
satisfying
f(x
+ y) =
f(x)-f(y)
must be of the form f(x) = a x or f(x) = 0. (Problem 38 indicates the way to
prove this, and also has a few words to say about discontinuous functions with this
property.)
Theorems 2 and
In addition to the two basic properties stated in
exp has one further property which
is
very important
— exp "grows
3,
the function
faster
than any
polynomial." In other words,
THEOREM
6
number
For any natural
n.
lim
x-»oo
proof
The proof consists
e
Step 1.
x
>
x
of several
for
oo.
x"
steps.
and consequently lim
x,
all
—=
x
=
0)
it
e
oo
(this
may be
considered
X—>-00
to
be the case n
—
To prove
statement (which
this
0).
x
If
jc
<
1
this
upper sum
is
> log*
clearly true, since
for f{t)
=
on
\/t
x <
clear for
is
for all
log* <
[1, jc],
x >
show
that
0.
.v >
<x —
If
0.
so log*
suffices to
then (Figure
1,
1
<
6)
x
—
1
is
an
x.
x
Step 2.
lim
x-»oo
To prove
e
—
= oo.
x
this,
note that
e
er
x/2
x
.
e
x/2
1
21
2
\
2
By Step
1
I
(
,
I
R
1 .
1
,
(e x ' r
the expression in parentheses
is
2
greater than
shows that lim
x
e /x
—
oo.
X—*oo
Note
1
,
and lim
X—>00
6
Step 3.
S/2
x
—=
lim
x—*oo x n
oo.
that
e
x
(e
/gx/n
x/ ")"
1
'
'
"
x"
G)"
n"
e
xl
=
oo; this
1 8.
The
349
The Logarithm and Exponential Functions
expression in parentheses becomes arbitrarily large, by Step 2, so the nth
power
certainly
It is
f(x)
now
=
{
e-
becomes
possible to
'x
\x
#
0.
arbitrarily large. |
examine
We
carefully the following very interesting function:
have
Xi
Therefore,
<
f'(x)>0
x < 0,
forx>0.
for
f'{x)
so /
is
decreasing for negative x and increasing for positive x. Moreover,
large, then
FIGURE
x
2
is
large, so
— \/x
so e
e's
and
is
close to 0, so e~
is
I*
large, so e
<5's,
=
is
more
\/(e
l/x ")
interesting.
is
Therefore,
is
close to
1
if
(Figure
|jc|
is
7).
x
is
small, then l/x
is
large,
argument, suitably stated with
shows that
if
we
l/x2
=0.
define
/(*)
=
r
1 /*
2
,
then the function /
is
continuous (Figure
1
x^O
x
0,
8
If
small. This
\ime-
FIGURE
l/x ~
7
The behavior of / near
x,x ~
2
8).
In
=
0.
fact,
/
is
actually differentiable
—
350
Derivatives
and
Integrals
Indeed
at 0:
l/h 2
=
/'(0)
\/h
=
lim
lim
and
—r—
h^o+
lim
-y
e
We
already
(X /
=
h)1
know
x ^°° e
all
the
more
= —
lim
oo
z
^U 2^
*
)
true that
l
)
=
x-*oo
means
t
— = oo;
lim
this
e {] /^
that
,(x
and
1//?
A -0-
(x2)
lim
it is
nm
while
lim
OO,
x
that
lim
2
X->-0O ^(JC )
=
0.
Thus
e- l 'x
=
fix)
x#0
-3,
jc-'
x
0,
We
can
now compute
0.
that
2
1/^2
h3
lim
2- e -lA
2
lim
lim
an argument similar
=
to the
/"(*)
one above shows
—
,4
t-
=
that f"(0)
lim
=
.^.4
+ ^^.4
*6
X
=
4
!
0,
lem 40)
that
approaches
f
{k)
(0)
—
9),
using induction
it
=
it
0.
can be shown (Prob-
The function / is extremely flat
can mask many irregularities of other
for every k.
so quickly that
For example (Figure
fact,
Thus
,#0
.v
This argument can be continued. In
0.
suppose that
,-l/.v
/(-*)
2
sin
—
,
.v
7^
x
=0.
x
0,
at 0,
and
functions.
18.
It
The Logarithm and Exponential Functions
can be shown (Problem 41) that for
for
all k.
this
function
it is
f
{k)
(0)
This example shows, perhaps more strikingly than any other, just
bad a function can
investigate
be,
even more
out behavior of this
and
still
be
IV we
In Part
infinitely differentiable.
restrictive conditions
on a function, which
=
how
will
will finally rule
sort.
e
FIGURE
also true that
351
1/jr
sin 1/jt,
x
^
x
=
9
PROBLEMS
1.
Differentiate each of the following functions
notes a
(remember
that a
b
"
always de-
(b'h
1)
f(x)
ii)
fix)
iii)
fix)
=
e
e°
.
= log(l + log(l + log(l + e [+e]+ '))).
= (smxY m(smx)
.
iv)
vi)
fix)
=
=
vii)
f(x)
=
v)
viii)
2.
f(x)
f(x)
=
=
,smv)Mn>
(sinA-)
i\ogx)
f(x)
fix)=x x
xi)
f(x)
Check
V sin x f .
4
4x
e ))e
+
(log(3
x)
(a)
x-ilogCsin^)
X
/
arcsin
ix)
=
.
log (£jV) sinx.
l
^
+
(arcsin
jc)
1o s
3
.
x
.
.
sm(x sm ^
mt)
).
that the derivative of
This expression
is
logo/
is
/"'//•
called the logarithmic derivative of /.
It is
often easier
products and powers in the expression for /
compute than
become sums and products in the expression for logo/. The deriva/', since
to
tive /'
(b)
can then be recovered simply by multiplying by /;
called logarithmic
differentiation.
Use logarithmic
differentiation to find
(i)
/(*)
=
(]
xl
+*)0 +e
).
fix)
this
process
for each of the following.
is
352
Derivatives
and
Integrals
f .. s
e
(1
f(x)
(iii)
=
(3-x) ]/3 x 2
— x)(3 + x) l
(sinx)
C0SJC
+
'
i
(cosx)
si
Find
*i
b
f(t)
(It
/
fit)
a
(for
4.
/ > Oon
[a,b]).
Graph each of the
following functions.
(a)
f(x)
(b)
f(x)
= e x+1
= e sinx
f(x)
—
(d)
f(x)
= e -e~
(e)
f(x)
=
(c)
e
x
.
.
+ e~ x
x
(ii)
lim
(iii)
lim
l
(Compare
x
-
.\
limits
lx
by
V
e
v
-
1
+
1
=
1
,2,
+
l'Hopital's Rule.
- x - x 2 /2
1
"'
~x^
6"
the graph with the graphs of exp
1/exp.)
e 2x
e
lim
.
e
Find the following
(i)
x
-
1
-
-
1
-x -x 2 /2
.v
-
2
x /2
-
3
a-
/6
x->0
log(l
(iv)
+Jt)-;t
+* 2 /2
lim
x->-0
2
log(l +jc)-.y+.x- /2
(v)
lim
x^O
log(l
(vi)
+ x) - x + x 2 /2 - x 3 /3
lim
73
x^O
6.
Find the following
-x) l/x
limits
(i)
lim(l
lim (tanjc)
tan 1>
(ii)
(iii)
Um(cosje)
/*
.
.
by
'
l'Hopital's Rule.
!
and
.
.
The Logarithm and Exponential Functions
18.
7.
The
functions
e
:x 2
+
y
2
=
353
sinh
—
x
e
-x
.y
2
1]
e
x
-X
cosh Jt
2
e
x, sin x)
tanh x
—
x
e~
X
=
+ e~
ex
1
e 2x
+
\
hyperbolic sine, hyperbolic cosine, and hyperbolic
(but usually read 'sinch,' 'cosh,' and 'tanch'). There
are many analogies between these functions and their ordinary trigonometric
are called the
tangent, respectively
One
counterparts.
shown
ter,
analogy
illustrated in Figure 10; a
is
in Figure 10(b) really has
when we
area x/2
is
proof that the region
best deferred until the next chap-
develop methods of computing integrals. Other analogies
will
are discussed in the following three problems, but the deepest analogies must
wait until Chapter 27. If you have not already done Problem 4, graph the
and tanh.
functions sinh, cosh,
8.
Prove that
2
(a)
cosh"
—
2
•
sinh"
=
1
2
2
+1/ cosh =
(b)
tanh
(c)
sinh(x
1.
(f)
+ y) = sinh cosh y + cosh x sinh y.
cosh(x + y) = cosh x cosh y + sinh x sinh y.
sinh' = cosh.
cosh' = sinh.
(g)
tanh'
(d)
(e)
jc
1
=
cosh
The
FIGURE
10
-
-
and tanh
and tanh are one-one; their inverses sinh
are defined on R and (— 1 1), respectively. These inverse functions are sometimes denoted by arg sinh and arg tanh (the "argument" of the hyperbolic
sine and tangent). If cosh is restricted to [0, oo) it has an inverse, denoted
functions sinh
,
,
by arg cosh, or simply cosh" which
information in Problem 8, that
,
sinh (cosh
x)
=
jc)
= Vl +x 2
-1
(b)
cosh (sinh
(c)
(sinh- )'(x)
,
1
=
\f x
2
—
is
1.
.
1
yr + .V
(d)
(cosh
_1
)'(jc)
=
1
for x
>
1
v^^T
(e)
(tanh
i
'/(Jt)
=
i-x 2
for
Ul
<
1.
defined on
[1, oo).
Prove, using the
.
354
Derivatives
and
Integrals
10.
Find an
(a)
(b)
formula for sinh
explicit
,
cosh
,
and tanh
(by solving the
-
=
equation y
x for x
sinh
terms of
in
y, etc.).
Find
f
b
1
dx,
Ja
+
s/\
rb
K
2
l
dx
for a b
,
>
or a,b
1
< —
1
1
r
b
l
d
for
fif.v
1--V 2
a
<
\a\, \b\
1.
Compare your answer for the
third integral with that obtained
1
1
1
l-x<
1 1
.
Show
1
1
+
-
x
+x
\
—
that
F(x)
by writing
f logr dt
=
J2
is
12.
not bounded on
Let
/ be
[2, oo).
a nondecreasing function on
F(x)
)
=
13.
/
bounded on
is
x
dt.
/
and define
>
1.
t
J\
Prove that
[1, oo),
[1,
oo)
if and
only
if
F / log is bounded on
[1
,
oo).
Find
(a)
lim a x for
<
a
<
1.
(Remember
the definition!)
.V—>00
(b)
lim
x^oo (logJC)"
(c)
lim
x—>oo
(log*)"
x
(-l)Mlog
(d)
lim x(logx) n
(e)
lim x
.
IV
Hint: x(logx)"
x
x
.
x->0 +
14.
Graph f(x) = x x
15.
(a)
Find the
for x
minimum
>
0.
(Use Problem
—
value of fix)
13(e).)
x
e /x" for x
>
0,
and conclude
that
n
(b)
fix) > e /n" for x > n.
Using the expression fix)
e"
+i
/in
+
lim fix)
16.
Graph /(*)
1)" +1
=
for
oo.
= <?7* w
.
x > n
=
+
x
1,
—
n+l
prove that fix) >
and thus obtain another proof that
e ix
n)/x
,
The Logarithm and Exponential Functions
18.
17.
(a)
+ y)/y.
Find lim log(l
\
—>o
355
(You can use l'Hopital's Rule, but that would be
silly.)
(b)
Find lim jclog(l +l/x).
(c)
Prove that e
(d)
Prove that e a
x—*oo
with just a
*(e)
18.
19.
—
=
.
=
lim x(b
[
^x
—
x> 0.
l/x) x for
+cr/100)
after
1
possible to derive this
(Use Problem
year. If the
then the
initial
suppose that interest
+
not 7(1
alas,
is,
awarded twice
17(c).)
computing
interest the next
investment grows to 7(1 +«/100)" after n years.
is
The
given twice a year.
2
but merely 7(1
<r//100) ",
as often, the interest
(c)
annum, then an initial investment /
bank compounds the interest (counts
the accrued interest as part of the capital for
year),
from part
1).
a bank gives a percent interest per
If
yields 7(1
is
.
+
(1
x
x
little
=
\/x)
lim (1 +a/x)
(It
X—»00
algebraic fiddling.)
Prove that log/?
Graph f(x)
+
lim (1
x—>oo
amount
final
+ a/200)
must be halved
2"
in
Now
after n years
— although
interest
is
each calculation, since
a/2 per half year. This amount is larger than 7(1 + a/ 100)",
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
the interest
is
but not that
much
1
20.
21.
year?
=
#
(a)
Let f{x)
(b)
If
f(x)
Suppose
number
(a)
(b)
(c)
log
for
K
22.
A
0.
Prove that f'(x)
=
l/x for x
x, prove that (log |/|)'
=
/'//.
interval the function
/
satisfies
/'
=
^
0.
cf for some
Assuming that / is never 0, use Problem 20(b) to prove that |/(jc)I = le cx
cx
for some k.
for some number / (> 0). It follows that f(x) = ke
Show that this result holds without the added assumption that / is
at the endpoint of an open
never 0. Hint: Show that / can't be
interval on which it is nowhere 0.
Give a simpler proof that f(x)
Suppose
—
that /'
f(x)/e
=
=
ke
cx
for
some
k by considering the
cx
.
fg' for
some
g.
Show
that f(x)
—
ke K{x) for some
radioactive substance diminishes at a rate proportional to the
disintegration
the
amount
all
is
at
atoms have equal probability of
proportional to the
time
t,
represents the probability
(a)
#
c.
present (since
is
all
on some
that
function g{x)
(d)
for x
|jc|
Find A(t)
in
remaining).
means that A'(t) = cA(t)
that an atom will disintegrate).
this
—
amount
disintegrating, the total
number of atoms
terms of the amount Aq
k.
A(0) present
for
at
some
time
0.
If A(t)
c (which
356
Derivatives
and
Integrals
(b)
Show
that there
is
number
a
with the property that A(t
23.
+
=
r)
A(t)/2.
Newton's law of cooling states that an object cools at a rate proportional to
the difference of
its
temperature and the temperature of the surrounding
medium. Find
the temperature T(t) of the object at time
temperature 7b
at
medium
24.
r (the "half-life" of the radioactive element)
time
/,
in
terms of
its
assuming that the temperature of the surrounding
0,
expressing Newton's law,
M. Hint: To solve the differential equation
remember that T — (T — M)'.
Prove that
fit) dt, then
is
kept at a constant,
if
f(x)
f =
0.
Jo
25.
Find
all
ff = e
Jo
(i)
/
continuous functions
satisfying
x
.
f f=\-e 2x\
(ii)
Jo
26.
Find
all
/
functions
—
satisfying f'{t)
+
f(t)
f(t)dt.
/
Jo
27.
Find
all
(fix))
28.
(a)
/ which
continuous functions
satisfy the
t
2
f
= Jo
fit)
+f
l
/ and g be continuous functions on
pose that for some C we have
Let
fix)
Prove Gronwall's
equation
<C+
,
inequality:
f(x)<CeJ«
g
.
Hint: Consider the derivative of the function h{x )
(b)
Let
/ and g be nonnegative
Suppose that f'{x)
(Compare Problem 2
tiable.
1
29.
fa)
=
=
(C
functions with g continuous
g(x)f(x) and /(0)
=
0.
+ f*
fg)e
and /
8
•><>
.
differen-
Prove that
/ =
0.
.)
Prove that
1+ *
+
X2
2!
+
X3
+
+
<e"
for
x >
0.
3!
Hint: Use induction on
(b)
g nonnegative. Sup-
a<x<b.
fg,
f
[a b] with
//,
and compare
x
Give a new proof that lim e /x
n
=
derivatives.
oo.
x—>oo
30.
Give yet another proof of this
Rule. (See
Problem
11-56.)
fact,
using the appropriate form of l'Hopital's
.
The Logarithm and Exponential Functions
18.
31.
(a)
Km
Evaluate
2
x
e
357
be able
(You should
(Vc
dt.
e'
I
f
to
make an educated
Jo
guess before doing any calculations.)
(b)
Evaluate the following
lim e~
(i)
limits.
r x+a/x)
x~
x^oo
r
e
/
Jx
dt.
px + (\ogx)/x
lim e~
(ii)
x
e'' dt.
/
x^oo
Jx
/>.v+(log.v)/2x
2
lim e~
'
e
/
x^co
dt.
Jx
This problem outlines the
32.
r2
approach
classical
to logarithms
and exponentials.
we will simply assume that the function f(x) = a x defined in
an elementary way for rational x can somehow be extended to a continuous
To begin
with,
,
,
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 / will then be
denoted by log
v
(a)
(
,
Show, direcdy from the
definition, that
l/h
,
log a (x)
=
limlog
h+
fl
-J
\. lo g;(linj(l +«'/*).
X
Thus, the whole problem has been reduced to the determination of
+
lim(l
h—»0
—
•
h)
{,h
If
.
\og e e.—
we can show
— and
,
log/U)
=
has derivative exp'(x)
=
that this has a limit e, then
consequently exp
=
log"
exp(x).
(b)
Let a„
show
=
(
1
I
—IV
1
for natural
n.
Using the binomial theorem,
+ Ei( -i)( -=)....(
1
A! Vx
k=2
l
n
7I
I
n
\x
< a ll+ \.
Using the fact that \/k\ < 1/2 A_1 for k > 2, show that all a„ < 3. Thus,
the set of numbers {a\, ai, «3,
is bounded, and therefore has a least
upper bound e. Show that for any e >
we have e — a n < s for large
Conclude
that a„
.
enough
(d)
numbers
that
—
(c)
H
If n
n.
< x <
n
+
1
,
then
.
.
}
358
Derivatives
and
Integrals
Conclude
that lim
and conclude
*33.
A
point
Q
point
H
1
I
x—»oo
B
time
1
lim
1
x-+—oo
=
+
e.
e.
'
is
=
then P'(t)
t,
10
The
.
P
to
B
7
—
P(t)), while
10
(in
other words,
distance traveled by
Napierian logarithm of the distance
Q
FIGURE
Also show that
e.
moving along a line segment AB of length 10 7 while another
moves along an infinite ray (Figure 1 1). The velocity of P is always
P
=
Q'(t)
=
I
I
+ //) 1// =
that lim(l
h-*Q
equal to the distance from
at
—
x
\
from P
Q
P(t)
if
Q moves
after
B
to
at
the position of
is
P
with constant velocity
time
t
time
t.
is
defined to be the
Thus
1
10
=Naplog[10 7 -P(/)].
7
/
This was the definition of logarithms given by Napier (1550-1617) in
publication of 1614, Mirifici logarithmonum canonis
the
Wonderful
Law
description
of Logarithms); work which was done
exponents was invented! The number 10
his
(A Description of
before
the use of
7
was chosen because Napier's taand navigational calculations), listed the logafor which the best possible available tables extended
and Napier wanted to avoid fractions. Prove that
bles (intended for astronomical
rithms of sines of angles,
decimal places,
to seven
Nap logx =
Use the same
Hint:
^34.
(a)
trick as in
Sketch the graph of f{x)
and
behavior near
(b)
Which
(c)
Prove that
xy
=
larger, e
is
x
is
= x;
Problem 23
=
equation for P.
to solve the
(log*)/* (paying particular attention to the
oo).
or
7r
< x <
if
y
n
10 7
10 log
but
if
e
?
1,
x
=
or x
>
1,
x
^
e,
e,
then the only
then there
is
number
precisely
y satisfying
one number
y
y
moreover, if x < e, then y > e, and if x > e,
y t^ x satisfying x — y
then y < e. (Interpret these statements in terms of the graph in part (a)!)
x
;
(d)
Prove that
x
(e)
=
y
2,
Show
=
x and y are natural numbers and x y
4, or x = 4, y — 2.
if
that the set of
a straight line
which
all
pairs (x, y) with x y
=
x
y
=
y
x
,
then x
=
y or
and
and draw a rough
consists of a curve
intersect; find the intersection
sketch.
** (f)
For
1
< x <
Prove that g
e let g(x)
is
be the unique number >
differentiable.
(It
is
fl(x)
=
=
logx
< x <
X
logx
e
g{x)
=
g(x) x
.
a good idea to consider separate
functions.
f\(x)
e with x
< x
e
and write g
terms of
in
,,
*35.
you do
this
and
f\
You should be
fa.
[g(x)]
2
able to
show
-logx
1
xL
-logg(x)
part properly.)
This problem uses the material from the Appendix to Chapter
(a)
that
,
1
if
359
The Logarithm and Exponential Functions
18.
Prove that exp
is
convex and log
1 1.
concave.
is
n
(b)
Prove that
if
=
y_. Pi
1
and
all
>
/?,
then for
0,
all Zi
>
we have
i=\
Z\
Pi
...-Zn
P"
<
+
PlZl
•••
+
PnZn-
(Use Problem 8 from the Appendix to Chapter
(c)
36.
(a)
Deduce another proof that
Let
/ be
Gn < An
(Problem 2-22).
a positive function on [a, b], and
Use Problem 2-22
into n equal intervals.
11.)
let
Pn be the
show
to
partition of [a, b]
that
-—— L(f, P ))
1
b
(b)
—
L(logf,
a
Pn ) <
log (
Use the Appendix to Chapter 13
n
\b
conclude that for
to
.
a
all
integrable
/ >
we have
log f
/
-aJa
A more
(c)
direct
approach
is
<log(
\b-aJa/
/
illustrated in the next part:
In Problem 35, Problem 2-22 was deduced as a special case of the inequality
n
for pi
>
0,
y. Pi =
1
an d 8 convex. For g concave we have the reverse
inequality
^Pigixt)
Sg l^PiXi)
\i=\
i=\
Apply
this
with g
=
I
log to prove the result of part
(b) directly
for
any
integrable /.
(d)
37.
State a general
theorem of which part
Suppose / satisfies /' = / and f(x
that / = exp or f = 0.
(b) is
+ y) =
just a special case.
f(x)f(y)
for all
jc
and
y.
Prove
360
Derivatives
and
Integrals
Prove that
"38.
either
/
if
/ =
continuous and f(x
is
=
or f(x)
x
[f(l)]
+ y) =
f(x)f(y)
Show
for all x. Hint:
for
all
x and y, then
=
that /(*)
[f(l)]
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 discontinuous functions / satisfying f(x + y) =
for rational x,
f(x)f(y).
Prove that
"39.
/
if
a continuous function defined on the positive real numbers,
is
=
=
and f(xy)
f(x) + f(y) for all positive x and y, then /
f(e)\ogx for all x > 0. Hint: Consider g(x) = f(e x ).
all
if
=
Hint: Consider functions g(x)
Prove that
"41.
=
l/x2
for x # 0, and /(0) = 0, then / (*>(0) =
for
f(x) = e~
k (you will encounter the same sort of difficulties as in Problem 10-21).
Prove that
"40.
or f(x)
=
f(x)
if
e~
l/x2
e~
l
^x
'
\/x for x
sin
P(\/x) for a polynomial function P.
/
=
and /(0)
0,
then
0,
/
,A)
(0)
=
for all k.
42.
(a)
Prove that
a
if
is
a root of the equation
n
+ a n _ix"~
an x
(*)
=
then the function y(x)
a„y
(**)
*(b)
Prove that
"
/U) =
that
then f'{a)
0,
Prove that
if
<
a
if
a
- ])
0,
the differential equation
+
+a
x
y'
+a
y
—
=
0.
xe ax
also satisfies (**).
a double root of a polynomial equation
is
=0.
a root of
is
—
{n
satisfies
a double root of (*), then y(x)
is
(*)
of order
r,
then y(x)
—x
e
ax
is
a solution
for
<
If (*)
has n real numbers as roots (counting multiplicities), part
k
r
1.
n solutions y\, ...
(d)
+a„_iy
)
Remember
Hint:
*(c)
a
if
i
-\
ax
e
+a\x + a =
l
Prove that in
,
yn
(c)
gives
of (**).
this case the
function
+
c\ y\
•
•
•
+ cn yn
also satisfies (**).
a theorem that in this case these are the only solutions of (**). Problem 21 and the next two problems prove special cases of this theorem,
and the general case is considered in Problem 20-26. In Chapter 27 we
will see what to do when (*) does not have n real numbers as roots.
It is
Suppose
"43.
that
/
satisfies
/" - /
=
and /(0)
=
/'(0)
0.
Prove that
/
=
as follows.
(a)
Show
(b)
Suppose
f(x)
**
c)
If
that
=
a
<
this fact
2
if')
that f(x)
ce
f(xo)
<
2
f -
x
^
^
=
or else f(x)
for
X()
>
0.
for all x in
=
ce~
0, say,
x
some
for
all
x
interval (a,b).
Show
that either
some constant c.
number a such that
in (a,b), for
then there would be a
for a < x <
jco and f(a) = 0, while f(x) ^
and part (b) to deduce a contradiction.
xq.
Why? Use
*44.
Show
(a)
that
/
if
satisfies
- f =
f"
=
then f(x)
0,
some a and b. (First figure out what a and b should be
and /'(0), and then use Problem 43.)
Show
(b)
45.
Find
{,,)
(other) a
in
and
+
be' x
for
terms of /(0)
b.
satisfying
.
{
(b)
*46.
some
for
ae x
~
(n
X)
f = f
"- 2)
(n)
f = f
(a)
/
functions
all
/ = a sinh +&cosh
also that
361
The Logarithm and Exponential Functions
18.
.
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.
Suppose there
(a)
is
a function
/ /
f=
with
=
each x by considering the function g(x)
f. Prove that fix)
/(xo
^
*)/(*0
—
•*)>
and /(0)
=
1.
+
for
where
/(*o)#0.
Show
(b)
For
(c)
/ show
this
g(x)
47.
that there
=
is
a function
f(x
that
f(x
/
Let
/ and g be continuous
/
is
=f
f(y) by considering the function
f(x)
l
)'(x)
=
l/x.
functions such that lim f(x)
*—>-oo
grows faster than g (f ^> g) if
hm
and we say
=
one-one and that (f~
Prove that
say that
v)
satisfying f'
+ y)/f(x).
(d)
We
+
/
that
/ and g
lim
grow
:
=
=
lim g(x)
oo.
JC—>-oo
— — = oo,
-
at the
same
exists
and
rate
is
(f
/
For example, for any polynomial function
~
g)
if
0, oo.
P
with lim P(x)
—
oo
(i.e.,
P
X—>00
is
non-constant and has positive leading coefficient)
P
^> log" for any positive integer n.
(a)
Given / and
with lim f(x)
X—>0O
that one of the three conditions
g,
(b)
BF/»g,then/+*~/.
(c)
If
tog/
=
=
oo,
is it
S>
P and
necessarily true
.V—»oo
/
» g or g S> / or / ~ g holds?
c
>
.
>
lim g(x)
we have exp
,
1
logg
for sufficiently large x, then
(d)
If
/
that
;§>
F
g and F(jc)
» G?
=
/
Jo
/
^> g.
f,G(x)=
I
JO
g,
does
it
necessarily follow
.
362
Derivatives
and
Integrals
(e)
48.
49.
Arrange each of the following
growth
(for
value at
x):
convenience,
we
x 3 ex , x 3
(ii)
x log x, e 5x log(x v ), ex
(iii)
e
+ log(x 3 ),
log4.v, (logx)*,
,
Suppose that
x
,
x e x x ex
,
,
g\, gi, g3,
.
,
.
,
,
Prove that log 10 2
irrational.
,
x 3 logx.
,
.
are continuous functions.
/ which grows
+ e~ 5x
x x x logx (logx) A
,
continuous function
is
xx x
its
'.
,
2X e x/2 (\ogx) 2x
,
.
of functions in increasing order of
indicate each function simply by giving
(i)
,
sets
faster
than each
g,
Show
that there
is
a
<%^
CHAPTER
INTEGRATION
Every computation of a derivative
Theorem
ELEMENTARY TERMS
IX
according to the Second Fundamental
yields,
of Calculus, a formula about integrals. For example,
if
= x(logx) — x
F(x)
=
then F\x)
log x\
consequently,
b
f
\ogx dx
I
=
F(b)
—
F(a)
=
—
b(\ogb)
b
—
[a(loga)
—
0<a,b.
a],
Ja
Formulas of this
sort are simplified considerably if
=
F(x)
We may
Fib)
-
we adopt
the notation
F{a).
then write
b
is
fa
of the function Fix)
called
x(logx)
—
x
dx depended on the lucky guess that log is the derivax(\ogx)—x. In general, a function F satisfying F' = f
a primitive of /. Of course, a continuous function f always has a
This evaluation of
tive
=
\ogxdx
L
log x
=
primitive, namely,
Fix)
Ja
but in
this
chapter
we
will try to find
familiar functions like sin, log, etc.
is
called
an elementary function.
a primitive which can be written in terms of
A
function which can be written in this
To be
precise,*
one which can be obtained by addition, multiplication,
division,
from the rational functions, the trigonometric functions and
functions log and exp.
It
should be stated
at the
is
no
F'{x)
a
*
The
and composition
and the
their inverses,
elementary function
=
e~
x2
for all
F
such that
x
not merely a report on the present state of mathematical ignorance;
(difficult)
theorem that no such function
definition
which we
will give
is
exists).
And, what
is
1
where
363
the
/,
+
it
is
even worse, you
precise, but not really accurate, or at least not quite standard.
Usually the elementary functions are defined to include "algebraic" functions, that
satisfying an equation
(g(x))'
is
very outset that elementary primitives usually cannot
be found. For example, there
(this is
way
an elementary function
/„_,(x)U(.v))"-
1
+
•
••
+f
(x)
=
is,
functions g
0,
are rational functions. But for our purposes these functions can be ignored.
.
364
Derivatives
and
Integrals
will
have no
way of knowing whether
or not an elementary primitive can be found
(you will just have to hope that the problems for this chapter contain no misprints).
Because the search for elementary primitives
so uncertain, finding
is
one
is
often
peculiarly satisfying. If we observe that the function
log(l
= x arctan x
F(x)
+ x2
)
satisfies
F'{x)
how we would
(just
=
arctan
jc
ever be led to such an observation
is
quite another matter), so
that
/'
arctan x
—
dx
log(l
—
x arctan x
+x 2
)
Ja
then
we may
feel that
we have
This chapter consists of
itives
little
arctan x
f
dx
more than methods for finding elementary prim(a process known simply as "integration"),
of given elementary functions
together with
this
"really" evaluated
some
notation, abbreviations,
and conventions designed
to facilitate
procedure. This preoccupation with elementary functions can be justified by
three considerations:
(1)
Integration
(2)
Every once
about
and everyone should know
a standard topic in calculus,
is
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 (for example,
you might take a
(physics) course in
which you are
expected to be able to integrate).
(3)
The most
orems
useful
(that
"methods" of integration are actually very important
apply to
all
Naturally, the last reason
to integrate (and
is
you probably
Even
the crucial one.
some
will forget
function in terms of primitives of other functions.
list
use of a standard symbol which requires
primitive of
/"
or,
will often
useful in formulas
more
/
one
we
will
To begin
The
integrating
can be obtained
list
list
given below
makes
f(x)dx
in stating
A
x
x dx
of all primitives of /."
theorems, while
like the following:
I
time through), you
to express primitives of
precisely, "the collection
by used
how
to forget
some explanation. The symbol
OT
/'
The symbol f f
you intend
of primitives for some functions; such a
simply by differentiating various well-known functions.
means "a
if
details the first
must never forget the basic methods.
These basic methods are theorems which allow us
therefore need a
the-
functions, not just elementary ones).
/
f(x)dx
is
most
19. Integration in Elementary Terms
=
This "equation" means that the function F(x)
cannot be interpreted
but in
because the right side
literally
one context we
this
is
F'(x)
satisfies
and we
=
x3
It
.
a number, not a function,
allow such discrepancies; our aim
will
integration process as mechanical as possible,
device.
x 4 /4
365
is
will resort to
to
make
the
any possible
Another feature of the equation deserves mention. Most people write
/
.3
x'dx
X4
—4 +C
=
emphasize that the primitives of f(x) — x are precisely the functions of the
form F(x) — x 4 /4 + C for some number C. Although it is possible (Problem 14)
to
to obtain contradictions if this point
not
arise,
There
and concern
is
disregarded, in practice such difficulties
for this constant
is
merely an annoyance.
one important convention accompanying
is
'W
on
the
side
left
letter
appearing
— thus
u
/
/
u du
'
tx
dx
4
-
-=
'
4
2
tx
~2~'
xt
/
A
the letter ap-
this notation:
pearing on the right side of the equation should match with the
after the
do
2
tx dt -
f f(x)dx, i.e., a primitive of /, is often called an "indefinite
while / f(x) dx is called, by way of contrast, a "definite integral."
function in
integral" of /,
This suggestive notation works out quite well in practice, but
be led
/"
integral of
Chapter
risk
/ f{x)dx
the integral
We
At the
astray.
(if
of boring you, the following fact
is
not
you do not
defined as "Fib)
—
is
F(a), where
find this statement repetitious,
a dx
=
F
is
is
an
indefinite
time to reread
=
— dx =
n^-\
,
+
log x
1
I
/
— dx
table.)
e
x
dx
sin
=
x dx
side.
ax
is
often written
/
—
abbreviations are used in the
/
it
can verify the formulas in the following short table of indefinite integrals
n
/
to
13).
x"dx
I
important not
emphasized once again:
simply by differentiating the functions indicated on the right
I
it is
e
x
— — cos x
for convenience; similar
last
two examples of
this
366
Derivatives
and
Integrals
I
cos x
-
=
x dx
sec
I
dx = sin*
tanx
x tan x dx
sec
=
sec x
dx
'
I \+x
=
2
dx
I
arctan x
—
arcsin x
•A
Two
general formulas of the same nature are consequences of theorems about
differentiation:
f[f(x)
+ g(x)]dx =
j f(x)dx + j g(x)dx,
These equations should be interpreted as meaning that a primitive of / + g can
be obtained by adding a primitive of / to a primitive of g, while a primitive of
c / can be obtained by multiplying a primitive of / by c.
•
Notice the consequences of these formulas for definite integrals: If
/ and g
are
continuous, then
rb
nb
nb
f(x)dx+
[f(x)+g(x)]dx=
/
Ja
Ja
f
f(x)dx
c
=
g(x)dx,
J
c
ci
f f(x)dx.
These follow from the previous formulas, since each definite integral may be written as the difference of the values at a and b of a corresponding primitive. Continuity
is
required in order to
know
(Of course, the
that these primitives exist.
when / 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
formulas are also true
will
THEOREM
l
(integration BY PARTS)
be written
If /'
and
in several different ways.
g' are
continuous, then
/ fg' = fs-f f'8>
f f{x)g\x)dx
=
f(x)g{x)
nb
/
Jn
b
f(x)g'(x)dx
=
f(x)g(x)
j f'(x)g(x)dx,
/. />
/
a
f'(x)g(x)dx.
Ja
(Notice that in the second equation f(x)g(x) denotes the Junction
f
g.
19. Integration in Elementary Terms
PROOF
367
The formula
=
(/*)'
+
fg'
~
f'g-
f'g
can be written
=
fg'
(fg)'
Thus
j fg'=f(fgy-f fg,
and fg can be chosen
first
The second formula
As
merely a restatement of the
is
from either of the
follows immediately
two.
first
is
be considered
as a
xe x dx
I
—
a
=
xsinx dx
I
^
x
•
xe x —
(— cosx)
—
i
f
f
g
g'
I
e
i
is
useful
when
the func-
1
obviously of the form
is
e
•
x
g'.
dx
J
= xe x -
| |
formula
product of a function /, whose deriva-
44
fg
fg'
third
|
simpler than /, and another function which
J
and the
first,
the following examples illustrate, integration by parts
tion to be integrated can
tive
one of the functions denoted by f(fg)'. This proves the
as
formula.
i
i
f
g
x
/
(— cosx)dx
•
1
J
4,
i
r
g
= — x cos* + sin x
which often work with integration by parts. The
consider the function g' to be the factor 1 which can always be written
There are two
first is
to
special tricks
,
in.
log -Y
/
dx
=
/
The second
and then
trick
/
•
log x
|
4,
g'
f
to use integration
is
solve for
which implies
1
J
J
fh
A
.
dx
dx
=
x log x
=
g f
x(\ogx)
is
—
1414f
by parts
simple example
log x
—
I
(\/x) dx
x
g
—
to find
X.
fh
in
terms of
the calculation
log x
log x
—
(l/x)
log x
-X-
-X-
-X-
"X
-X
-X
g'
f
g
f
f
g
(1 /x)
-
I
that
2
f
/
J
•
— losxdx =
x
(logx)
2
dx
,
/h
again,
368
Derivatives
and
Integrals
or
1
-
/x
A
more complicated
I
e sin x
calculation
dx
=
£
r
4-
4-
/
(log*)
=
ax
log
5x
2
often required:
is
(— cos x) -
A
•
4-
4-
/
8
= —e
2
e
/
x
(— cosa)o?a
•
f
g
+ je< cos a dx
cos x
:
= - e -cosx +
I
i
u
v'
[,<.( 51 „.v,-/^( S in,»^];
V
u'
therefore,
2
e
/
/
x
sin
x dx
=
x
e (sin x
—
cos x)
or
—
x
/
e
x
sin
=
x dx
form
success.
the
the
It is
more functions you can already integrate,
main problem. For example, we can use
/
(log*) dx
we
=
/
J
/logx dx
by parts); we have
(\ogx)(logx)dx
I
is
of the
the greater your chances for
tackling
parts to integrate
(logx)(logx) dx
J
recall that
gration
that a function
do a preliminary integration before
frequently reasonable to
3
if
cosx)
upon recognizing
Since integration by parts depends
g',
e {?,mx
=
=
x(logx)
—x
I
I
f
g'
(this
(\ogx)[x(\ogx)
-
formula was
-
x]
derived by inte-
itself
(\/x)[x(logx)- x]dx
/
J
I
I
i
i
I
I
f
g'
f
g
f
g
—
(log x) [x (log x)
— x] —
=
(log x)[x (log x)
—a] —
=
=
(log A) [A (log A)
—
x (log A )
—
A]
j
/
—
1]
dx
\ogxdx
+
/
[log
[A(logA)
— 2a (log A + 2a
)
A"
—
A']
1
dx
+A
.
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
369
19. Integration in Elementary Terms
develop
this
method
theorem
in stages, stating the
saving the treatment of indefinite integrals for
THEOREM
2
If
/ and
g' are
for definite integrals
first,
and
later.
continuous, then
(THE SUBSTITUTION FORMULA)
J/g(a)
/=
rb
?(»)
f{u)du=
If
F
Ja
a primitive of /, then the
is
f(g(x))-g'(x)dx.
\
Jg(a]
a)
PROOF
(fog)-g'
/
Ja
left
side
—
F(g(b))
is
F(g(a)).
On
the other
hand,
=
(Fo g y
so
F
o
g
is
The
(fog)
a primitive of
(F
o
(F'og).g'
g)(b)
- (F
and the
g'
o
g)(a)
(fog).g>,
right side
F(g(b))
of the form (fog)-
is
I
J
x cos x dx
sin
a
facilitated
for g(x)
=
-
F(g(a)). |
K
cosxdx
(sinx)
by the appearance of the factor cosx, which
=
/'
u
.
be the factor g'(x)
=
= sin x
f(u)=w
g(x)
sin
x cos x dx
ro
b
=
'g(b)
rg\o)
f(g(x))g'(x)dx=
/
Ja
tan x
sin
dx can be treated
h
f
I
b
tanxdx
— sinx
is
a
sin
g'(x),
similarly
=
\/u.
f(g(x))g'(x)dx
=-
g(x)
=
f(u)
= -
write
dx.
=
dx
we
COSX
where g(x)
for f(u)
if
- sin ^
I
Ja
1/cosx can then be written /(cosx)
tan x
f
=—
Jo
In this case the factor
b
u du
-L
f
f(u)du
/
J'g(c
via)
sin b
/'
will
can be written as (g(x)) 5
,
Thus
Ja
integration of
that a
For example, the integration of
g'.
sinx; the remaining expression, (sin.v)
f(g(x)), for f(u)
The
is
depend upon recognizing
simplest uses of the substitution formula
given function
is
=
=
cosx; the remaining factor
Hence
cosx
1
Ja
'g(b)
-L
/•cos/j
J
i.
as, i
/
Jg{a)
f(u)du
i
- du
/
ft
"
=
log(cosa)
—
log(cos/?).
370
Derivatives
and
Integrals
Finally, to find
I
/'
dx,
*log*
JIa
=
notice that \/x
=
f(u)
=
where g(x)
g'(x)
=
logx, and that l/logjc
f(g(x)) for
Thus
1/m.
g(x)
=
logx
dx
L
x\ogX
a
/•gtfo
=
^S b
1
<iw
f
=
f(u)du
/
log(logb)
—
log(loga).
u
'log a
Fortunately, these uses of the substitution formula can be shortened considerably.
The
intermediate steps, which involve writing
rg(b)
>b
f f(g(x))g'(x)dx=
Ja
can
easily
f(u)du,
f
J g(a)
be eliminated by noticing the following: To go from the
left
side to the
right side,
u for g{x)
substitute
du
g'(x)dx
for
(and change the limits of integration);
performed
the substitutions can be
for the
name
directly
b
f
sin
I
original function (accounting
x cos x dx
/>sin
u for sinx
substitute
du
Ja
and
on the
of this theorem). For example,
=
cosx dx
for
b
/
u du,
J sin a
similarly
rb —
r
sin x
Ja
Usually
we
dx
COS X
abbreviate this
/•cos/?
u for cosx
substitute
du
for
—
sin
x dx
XJ
method even more, and
"Let
u
du
=
=
=
/
J cos a
1
— du
"
say simply:
g(x)
g'(x)dx"
Thus
u
let
=
log
L
a
In this chapter
tegrals, but if
we
we can
dx
*logX
1
du
/•log*
,v
I
,
— — dx
=
/
j
-du.
x
are usually interested in primitives rather than definite infind
fa f(x)dx
for all a
and
b,
then
we can
certainly find
—
,
19. Integration
f f(x)dx. For example,
Elementary Terms
371
since
b
•
.
sin
5
—
x cos x ax
—
sin
6
6
•
i
o
a
—
6
sin
-
6
1.
it
in
follows that
f sm 5 x cos jc a jc
6
—- *
sin
=
.
/
Similarly,
tan x
dx
= — log cos x
:
dx
=
/
/
J
It is
—
xlogx
,
log(logx).
quite uneconomical to obtain primitives from the substitution formula by
finding definite integrals.
two
Instead, the
can be combined,
steps
first
to yield the
following procedure:
(1)
Let
u
du
(after this
= g(x),
= g'(x)dx;
manipulation only the
letter u
should appear,
letter X).
Thus,
primitive (as an expression involving
(2)
Find
(3)
Substitute g(x) back for u.
.
\
to find
/-sin
cos x
jc
dx
,
1) let
u
du
so that
we
—
=
sinx,
cosx dx
obtain
/
(2)
evaluate
(3)
remember
to substitute sin
du;
x back for u so that
/sin 5 x
'
"
,
cos x
dx
=
—
sin
-
x
u).
not the
,
372
Derivatives
and
Integrals
Similarly, if
=
u
log x
1
= — dx,
du
x
then
/*
/
J
—
/"
1
dx
-
becomes
1
- du
/
xlogx
=
log u,
u
J
so that
1
To
xlogx
=
dx
log(logjc).
evaluate
j
/
l+.v 2
dx,
let
u
du
the factor 2 which has just
=
=
1
+
2
.v
,
2x dx;
popped up causes no problem
1
1
1 j/ u
o
I
/"
1
~ du =
lo S M
—
the integral
becomes
-
so
/ l+x
(This result
may
2
""
f/x
=
1
-io g (i
2
9
+ .A
be combined with integration by parts
I
1
•
arctan x
dx
to yield
x
=
x arctan x
=
x arctan x
-I
—
l+.v 2
1
j log(l
dx
+x
/
),
a formula that has already been mentioned.)
These applications of the substitution formula* illustrate the most straightforward and least interesting types once the suitable factor g'(x) is recognized,
the whole problem may even become simple enough to do mentally. The following
—
three problems require only the information provided by the short table of indefinite integrals at the
*The
beginning of the chapter and, of course, the right substitution
substitution formula
is
I
often written in the form
fiu)du=
j f(g(x))g'(x)dx,
u=g(x).
literally (after all, / f(u)du should mean a primitive of / and the
symbol f f(g(x))g'(x)dx should mean a primitive of (fog)- g'\ these are certainly not equal).
However, it may be regarded as a symbolic summary of the procedure which we have developed. If
This formula cannot be taken
we
use Leibniz's notation,
and a
little
fudging, the formula reads particularly well:
I fundu =
.
j
f(u)-£dx.
,
)
,
.
19. Integration in Elementary Terms
(the third
problem has been disguised a
by some algebraic chicanery).
little
x dx
sec x tan
/
/
(cosx)e
373
smx
dx.
dx.
/
-e 2x
y/\
you have not succeeded in finding the right substitutions, you should be able to
guess them from the answers, which are (tan 6 x)/6, e smx and arcsin e x At first you
may find these problems too hard to do in your head, but at least when g is of the
very simple form g(x) — ax + b you should not have to waste time writing out the
If
.
,
substitution.
detail
The
ix
I
(something).
-
following integrations should
the proper positioning of the constant
3x
/3 or 3e ?
e
e
is
be
all
(The only worrisome
clear.
— should
the answer to the second be
always take care of these problems as follows. Clearly
Now
"something" must be
differentiate
if I
cancel the
3, to
=
F(x)
e
3x
get F'(x)
I
,
=
fe
3e
ix
3x
,
dx
=
so the
3.)
logU+3),
/ +3
.v
,
3x
/
/
cos Ax
sin(2.t
/
dx
=
dx
=
e
e
sin 4;
4
+
1)
dx
+
1
2
/
does
'
cos(2a'
=
dx
More
3x
—
"3
arctan 2x
l+4.r 2
interesting uses of the substitution formula occur
not appear.
Consider
There are two main types of
when
substitutions
the factor g'(x)
where
this
happens.
first
+e x
1
t -e
dx.
1
The prominent appearance
of the expression e x suggests the simplifying substitu-
tion
u
du
Although the expression
e
/1
We
x
= e\
= e dx.
x
dx does not appear,
-
ex
dx
=
/
•
J
-
1
ex
therefore obtain
L
+u
1
•
I
11
— du.
u
can always be put
it
—
ex
e
x
dx
in:
374
Derivatives
and
Integrals
which can be evaluated by the algebraic
/
—
u
u
1+e
/ \-e
dx
1
2
f
— du =
1
+ - du = - 21og(l —
-
/
J
-
\
trick
u)
+
logM,
u
u
so that
There
is
= -2 log(l - e x + log e x = -2 log( -
an alternative and preferable way of handling
not require multiplying and dividing by e x
—
a
e
1
)
.
)
+ x.
problem, which does
If we write
x
=
logM,
dx
=
— du,
e
this
x
1
u
then
1+e
- dx
/ -e
immediately becomes
J
1
Most
substitution
ing x in terms of u,
see
is
why
this trick
one-one for
all
problems are much easier
and dx
in
if
one
I
—
u
u
resorts to this trick of express-
terms of du, instead of vice versa.
always works
It is
long as the function expressing u in terms of x
(as
x under consideration): If we apply the substitution
u
x=g-\u)
dx = (g- Y(u)du
=g(x),
l
to the integral
/
we
f
(1)
the other hand,
if
we apply
u
to the
same
l
<)(g- )'(u)du.
f(u]
the straightforward substitution
du
= g(x)
= g\x)dx
integral,
j f(g(x))dx
we
f(g(x))dx,
obtain
On
=
j f (g(x))
g'(x)dx,
g'(x)
obtain
(2)
The
integrals (1)
and
//<- )•
du.
g'(g-Hu))
arc identical, since (g~ )'(u)
[
(2)
not hard to
As another concrete example, consider
1 vV+T
dx.
=
\/g'(g~ '(«)).
r
,
19. Integration in Elementary Terms
the entire expression y/e x
we will go the whole hog and replace
by one letter. Thus we choose the substitution
In this case
=
=
u
u
u
2
e
-l=e\
2
x
x
=
integral then
f
/
J
+
1
1
+ 1,
= log(w 2 -
dx
The
+
y/e x
375
1),
2w
du.
—x
uz
—
1
2
—
[
becomes
(ir-D 2
—
•
2u
«2
n
—
du
-i
f
l
1
u
du
2M
—
—
3
zw.
3
y
Thus
e
t Ve
2x
Jx
x
+
Another example, which
occur,
is
2
=
-(e
A
+
"
l)
3/2
-2(e v
+
1/2
l)
.
3
1
illustrates the
second main type of substitution that can
the integral
/
Vl — x 2 dx.
In this case, instead of replacing a complicated expression by a simpler one,
will replace
—
x by sinw, because vl
are using the substitution u
=
which helps us find the expression
x
let
dx
=
=
=
«
sin
arcsinx, but
it is
cost/.
This really means that
sin u
cos
=
[u
,
Thus,
.
arcsin x]
udu;
then the integral becomes
/
The
v
1
—
udu —
sin~ u cos
cos u du.
I
evaluation of this integral depends on the equation
2
cos u
+ cos 1u
1
=
2
(see the discussion
of trigonometric functions below) so that
I
cos u
du
=
1
+ cos 2u
u
du
I
2
= - +
2
sin
2w
4
and
r
/
7
Vl-x' dx =
arcsin x
sin (2 arcsin
jc)
1
arcsin x
1
h
-
.
.
sin(arcsinx)
—2~x + r^^7
arcsin
1
/
we
the expression for x in terms of u
be substituted for dx
to
we
r
-
•
cos(arcsin x)
,
376
Derivatives
and
Integrals
Substitution
you have
and integration by
parts are the only fundamental
functions.
some of our examples
Nevertheless, as
reveal, success often
The most important
additional tricks.
you should be able
to integrate all the functions in
depends
are listed below. Using these
Problems
1
few other
to 10 (a
some of the remaining problems).
interesting tricks are explained in
1
can be found for a large number of
to learn; with their aid primitives
upon some
methods which
TRIGONOMETRIC FUNCTIONS
.
Since
sin
x
+
=
cos x
cos x
1
and
cos o
Lx
we
-2
— sin
x
2
,
obtain
—
=
cos lx
cos
2.*
—
cos x
(1
—
7
sin
— cos" x = 2 cos x —
-9
.9
— 2 sin
x) — sin"x =
x,
(1
1
)
1
or
—
1
sin"
cos 2x
x
?
+ cos lx
1
cos X
2
These formulas may be used
to integ rate
cos" x dx,
/
if
n
is
x dx,
sin"
/
even. Substituting
(1-
cos lx )
( 1
+ cos 2x)
or
2
for sin
x or cos
2
jc
yields a
2
sum of terms
involving lower powers of cos.
For
example,
2
1
- dx
4
C
/
cos 2x
2
and
/
If n
is
odd, n
= 2k +
/
I,
cos" 2x
=
dx
1
+
cos Ax
/
then
s'm"
x dx
=
/
sin.vd
—
k
cos x) dx\
dx
H
—/
/
cos 2x
dx
19. Integration in Elementary Terms
377
the latter expression, multiplied out, involves terms
of the form sinjccos'x, all of
which can be integrated easily. The integral for cos"x is
treated similarly.
An
integral
/
handled the same way if n or
formulas for sin 2 x and cos 2 *.
is
A final
x cos
sin"
m
is
"
=
dx
secxdx
I
jc
odd.
important trigonometric integral
/
w
dx
If n
and
m
are both even, use the
is
=
log(sec;c +tanjc).
Although there are several ways of "deriving" this result,
by means of the methods already at our disposal (Problem 13), it is simplest
to check this formula by
differentiating the right side, and to memorize it.
2.
REDUCTION FORMULAS
Integration by parts yields (Problem 21)
sm n xdx
/
= --sm 'f
/
cos" x
dx
and many
U
+l)«
J
n
l
2
+ ^—— f sin"- 2 xdx,
n
xcosx
= - cos" -1 x sin x + ^——
J
f
J
l
n
J
dx
*
n
/ cos"
-2
x dx
J
x
In-?, r
1
1
_
~2n-2(x 2 +\r-i + 2^2j~(xTTTy^ dX
similar formulas.
The
first
two, used repeatedly, give a different
for evaluating primitives of sin" or cos"
.
The
a large general class of functions, which will
third
method
very important for integrating
complete our discussion.
is
RATIONAL FUNCTIONS
3.
Consider a rational function p/q where
p(x)
q(x)
We
=
=
+ an _ix n ~ H
m
b m x +b m ^x m
+
an x
which
is
l
l
might as well assume that a„
for otherwise
n
y
...
a
+
,
b
.
=bm =
we may express p/q
as a
\. Moreover, we can assume that n <
m,
polynomial function plus a rational function
of this form by dividing (the calculation
u2
u
—
1
= u+l +
\
u
-
1
is a simple example).
The integration of an arbitrary rational function 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.
)
378
Derivatives
and
Integrals
theorem
Every polynomial function
q{x)=x m +b m - l x m
-l
+
+b
can be written as a product
q(x)
(where
r\
= (x-
+
a,)"
\-
•
+
rk
.
.
(x
•
.
- ak )
+
2(s\
rk
(x
h Si)
=
(In this expression, identical factors
x
— c, and
+ fax + y, may be
x
2
+
+
fox
S|
(x
]/,)
2
+ frx + yiy
m).
have been collected together, so that
assumed
all
Moreover, we assume that each
distinct.
quadratic factor cannot be factored further. This means that
2
A-
since otherwise
+ Aa +
x
Yi
we can
-4y; <0,
factor
2
-A + Va -4 k
=
-A
,
2
-
>/A-
-4y,
into linear factors.)
THEOREM
< m and
If n
p(x)
q(x)
= x" + a„_]X n ~ H
\-clq,
m
= x + b m _ x"- + ---+b
2
= {x- a,)''
(x - a k ) (x + fox + n
l
]
l
•
.
.
.
•
then /?(a)/<7(a) can be written
p(x)
q(x)
.
-a
U
+
+
•••
-
in the
+
_{x
+
.
.
•
.
(a
2
+ A* +
YiY\
form
+•
—
«1 n
-a
b u x+c ]A
k
_(a 2
(x
)
+ fox +
+
-ak Y
k
r
known
it is
_
(x 2
yi)
*/ IX
+
+ pix + n)
+
si
•••
_
C/.1
f fax
so complicated that
)
(x -cci)'i_
i
This expression,
S]
rk
1
+
as the "partial fraction
y /)
u 2 + A-v + y/)".
decomposition" of p(x)/q(x),
simpler to examine the following example, which illustrates
such an expression and shows
how
to find
it.
According
to the
theorem,
it
possible to write
2.v
7
+
8.v
ft
(a 2
4
+ 15a 3 + 16a 2 + 7a + 10
+a + 1) 2 (a 2 + 2a + 2)(a- l) 2
b
a
ex + d
ex + f
—,
+
+
+
—r +
A-l (a-1) 2 a 2 + 2a+2 a 2 +a+1
+
13a
5
+
is
20a-
+h
+a +
gx
(a 2
1)
2
is
.
379
19. Integration in Elementary Terms
To
find the
numbers
common denominator
over the
and
a, b, c, d, e, f, g,
(a
2
2
+x +
polynomial
h, write the right side as a
1 )
(x
2
+ 2.x + 3) (a
-
—
1
), the numerator
becomes
-
a(x
+
(ex
1)(a
2
+
2x
+ d)(x -
+ 2)(x 2 + x +
2
\)
{x
2
+x +
l)
2
1 )
2
+
+ b(x 2 + 2x + 2)(x 2 + x +
+
(ex
+
Actually multiplying
out
this
ficients are
combinations of a,
cients of 2.v
7
we
6
+ 8.v +
1
3.r
5
we
(!)
.
.
2
-
f)(x
( <?
1)
(a
2
+ 2x +
.v+/2)U-
l)
2)(x
2
1)
2
(a-
obtain a polynomial of degree
h.
,
4
Equating these
+ 20.x- + 15.x- +
3
16.y
2
+
+
8,
+x+
+
2.v
whose
10 (the coefficient of
unknowns a ,...,/?
obtain 8 equations in the eight
2
1)
2).
coef-
coefficients with the coeffi-
+
7.x
2
.
8
is
jc
0)
After heroic calculations
these can be solved to give
a=l,
= 0,
e
= 2,
f = 0,
b
c=l,
d
=
h
g
0,
= 3,
= \.
Thus
f
2x 7
+
J
=
+ 13.v 5 + 20a- 4 + 17.x- 3 + 16x 2 +
2
2
(x 2 + x+\) 2 (x + 2x + 2)(x-\)
6
5.x-
77
/
(X-I)
J
dx
+
/
(X
J
77T
2
-l)
dx
+
/
J
—i2
(A-
simpler cases the requisite calculations
(In
particular
ing
it
We
into
example by starting with the
one fraction.)
+.X
7x
+
7
dx
dx + /
1)+ T^
J
A +
~
A +
2
may actually be feasible.
partial fraction
2.v
I
3
+
dx.
2
obtained
this
decomposition and convert-
are already in a position to find each of the integrals appearing in the above
expression; the calculations will illustrate
all
the difficulties which arise in integrat-
ing rational functions.
The
first
two
integrals are simple:
1
/
x
-
dx
—
dx
=
2
/
The
log (a
—
1),
1
(X-l) 1
-2
A-l
depends on "completing the square":
third integration
A
2
+A+l
--=(X+W~ +
1
2
(If
we had obtained — ^
instead of |
we could
not take the square root, but
in this
case our original quadratic factor could have been factored into linear factors.)
We
380
Derivatives
and
Integrals
now
can
write
16
1
/
The
(a
2
+a +
1)
dx
2
h
~9~
j dx.
+
x
+
1
substitution
changes
u
=
du
=
+
x
dx,
this integral to
16
~9~
du,
+ D2
J (w
which can be computed using the third reduction formula given above.
Finally, to evaluate
l
we
(a
2
x
+
+
2x
3
+
dx
2)
write
x+3
/
The
first
evaluate
2
.v
+
2.t
integral
it
+
on
2
2 J x
-\l
the right side has
u
The second
integral
simply 2arctan(A
a
J
:
(x<
2
dx
+
/
J
(a-
+
1 )
2
+
Ja.
1
been purposely constructed so
that
we can
by using the substitution
du
2
+2
+ 2x + 2
2a
dx
+
+
2x
+
on the
+
2a
f
\
,
dx
2)"
which
is
just the difference of the other two,
If the original integral
1).
3
right,
= x + 2a + 2,
= (2a + 2) dx
-i2 J
(a
2
+
is
were
2
+ 2a + 2)"
dx
+
2
f
J
[(a
[U
+
2
1 )
+
dx.
1]"
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
the
is
first
integral
a theoretical curiosity only, especially since
of q{x) before you can even begin. This
that simple rational functions
sometimes
l
another important example
2
J x
-l
J x
it is
necessary to find the factorization
only partly true.
We
have already seen
arise, as in the integration
1
+e x
1
-e*
dx;
the integral
is
-
is
1
A+l
log(A
2
-
1)
- -log(A+
1)
.
381
19. Integration in Elementary Terms
Moreover,
it
is
if
a problem has been reduced to the integration of a rational function,
then certain that an elementary primitive
impossibility of finding the factors of the
even
exists,
denominator
when
may
the difficulty or
preclude writing
this
primitive explicitly.
PROBLEMS
1.
This problem contains some integrals which require
braic manipulation,
and consequently
test
your
little
more than
alge-
ability to discover algebraic
rather than your understanding of the integration processes. Never-
tricks,
theless,
any one of these
might be an important preliminary step
tricks
an honest integration problem. Moreover, you want
which
you can
integrals are easy, so that
process
The answer
in sight.
is
see
section, if
when
you
the
in
have some feel for
end of an integration
to
resort to
it,
will
only reveal
what algebra you should have used.
Vx*
+ yz
/
dx.
x
dx
/
-
+ v7 * +
+ e 2x + e 3>
yjx
1
dx.
iin
e 4x
iv)
1
— dx
fI
x
J b
I
tan" x
VI
dx (Trigonometric
.
+x 2
2
integrals are always very touchy, because
there are so
many
problem can
easily look hard.)
trigonometric identities that an easy
'
dx
Vll
2
-x 2
dx
1
.
ix)
1
+ sin x
8jc
/
I
2
+ 6.v+4
x+
j
dx.
1
dx,
J
2.
The
Jlx - x 2
following integrations involve simple substitutions, most of which you
should be able to do in your head.
.
/
e
x
sin e
x
dx
"
382
Derivatives
and
.
..
"
Integrals
x~
dx.
/
in
iv
f log*
dx.
/'
e
2x
+
eX
x
e
x
(In the text this
was done by
parts.
dx
2e*
+
1
e dx.
e
J
x dx
vi
vn
I
V
7
-a- 4
!
f e^*
—
x2
viii)
/
x\J 1
ix)
/
log(cosx) tan.vJx.
Jjc.
r log(iogx)
^
x\ogx
Integration by parts
dx.
(ii)
(iii)
(iv)
/VV
/"
/x
(v)
/
w
/"
J
vn
e
/
v
sin
sin
bx dx
x dx
(log^)
3
6?JC.
—-
log(logx)
sec x
dx
.
(This
up.
is
If
a tricky and important integral that often comes
you do not succeed
consult the answers.)
Vlll
/
cos(logx)dx.
ix)
/
s/x log x
x)
/
x(\ogx) dx.
dx
in evaluating
it,
be sure to
)
'
19. Integration in Elementary Terms
4.
The
x
=
following integrations can
=
sinw, x
cosw,
all
be done with substitutions of the form
To do some of these you
etc.
383
will
need
to
remember
that
sec x
/
=
dx
as well as the following formula,
/
+
log (sec x
which can
also
= — log(cscx
cscxdx
tan x
be checked by differentiation:
+cotx).
In addition, at this point the derivatives of
all
the trigonometric functions
should be kept handy.
dx
-
/
-x
(You already know
=
x
sin u
dx
/
(Since tan
v
+
1
x
tution x
this integral,
anyway, just to see
2
=
u
+
1
=
sec
2
u,
but use the substitution
how
it
works
you want
out.)
to use the substi-
tan u
dx
m
1
J
y.v 2
-
/dx
—-j=^
iv
x\x^—
(The answer
.
given short
1
will
be a certain inverse function that was
the text.)
shrift in
dx
r
'
xs/l-x 2
J
dx
r
VI
'
J
xVl+x 2
x- dx.
Vll
You
will
need
integrating
vm
5.
1
—
2
remember
powers of
sin
the
and
methods
/
71
(x)
f
y/x 2
cos.
dx.
- \dx.
following integrations involve substitutions of various types.
no
substitute for cleverness, but there
for
an expression which appears frequently or prominently;
troublesome expressions appear,
new
in
expression.
terms of
/
dx
1
+
v/.t
dx
(u)
/
And
u, to find
f
(i)
1
for
+x 2 dx.
LX
The
.v
to
+ ex
'
+
1
is
it
is
a general rule to follow: substitute
try to express
don't forget that
There
them both
in
if
two
different
terms of some
usually helps to express x directly
out the proper expression to substitute for dx.
384
Derivatives
and
Integrals
dx
C
mi
J
\fx
+
1/x
Li A.
IV!
I
(The substitution u
.
v
+
1
e*
=
e
x
leads to an integral requir-
ing yet another substitution; this
substitutions
can be done
is all
right,
but both
at once.)
dx
/ 2 + tan x
dx
VI;
(Another place where one
do the work of two
fJ v
S7*
y/x +
4' +
r 4
substitution can be
made
to
V
1
Vll
J
2*
+T
dx.
Vlll
dx
ix)
(In this case
.
Vx-
there are two obvious candidates for the
tion,
The
+ Ix - 1
x+A-2
is
2x z2
J *
3
h
3
.v
I
3x
2
1)
2
dx.
+ 3x -
+ 7.v 2 -
(X-
dx.
1
-hi
2jc
2x 2
(A-
+
5.v
+
+x +
1
1)
5
dx.
3
1
/ (A+3)(A-
l)
2
Ja.
+4
flA.
+
3
A +A + 2
rfjc.
/ A 4 + 2a 2 +
3a 2 + 3 a +
It, + 2a 2 + 2a +
A
1
VI
1
1
Vll
dx
Vlll
substitu-
x7
'
1
be integrated. Here
f
LV
first
either will work.)
previous problem provided gratis a haphazard selection of rational func-
tions to
in
and
best;
dx.
+
J V-v
two successive substitutions work out
I
a
4
+
1
2a
IX
+ A+
l)
2
dx.
3a
J (a^
+ a-
I)
3
dx.
dx.
1
a
more
systematic selection.
.
. ..
.
385
19. Integration in Elementary Terms
/dx
-
which looks a
,
little
from any of the previous
different
vV - x 2
problems. Hint:
Use a
It
helps to write (x"
form u
substitution of the
Extra Hint
Use a
2:
2
=
.
— x2
.
l/2
)
=
x(x n
to obtain
.
substitution of the
form
v
~2
—
1)
1/2
.
Extra Hint
1:
an answer involving arctan.
=
xa
to obtain
an answer
involving arcsin.
*8.
(No holds barred.)
Potpourri.
The
following integrations involve
all
the
methods of the previous problems
arctan x
/
dx.
2
+x T
1
arctan x
ft +
dx.
*2)2
in
/
logy/l+X 2 dx.
IV
I
xlog v
r
2
-
+ x 1 dx.
1
1
1
= dx.
+
VI
vii)
viii)
ix)
/
/
f
/
+
1
e
dx
:
sin
x
x cos3 x
sin v
vtan x dx
ft'
(To factor x b
+l
6
+
following two problems provide
(and can bear
Problem 9
it).
and integration by
many
dx
cos x
Ix
it
-s'mx
=
2
dx
The
dx
arcsin ^/x
/
I
yr
»
1, first
still
factor v
more practice
+x
1
I
integrals.
)dx.
+ cos X
.
2
sin"
/A.x +
T"
dx.
x
11
dx.
y/4-X 2
(iv)
/
x arctan x dx
using Problem
1-1.
at integration, if you
need
and trigonometric manipulations
Problem 10 involves substitutions. (Of course, in
cases the resulting integrals will require
log(a"
1,
involves algebraic
parts, while
Find the following
+
still
further manipulations.)
.
386
Derivatives
and
. .
Integrals
(v)
VI
/sin 3
/•
3
cos z x
2
Vll
/
dx.
arctan * dx.
.v
x dx
Vlll
10.
/
- 2* +
s]x 2
2
x tanx dx.
(ix)
I
sec
(x)
I
x tan x dx
Find the following
integrals.
I&T&-
(i)
(ii)
/
in
I
iv
v
1
—
x dx
sin
arctan >fx
Vx +
I
sin
I
log(.v
dx
\
dx.
(v)
There are obvious
VI
\/x 2
+
—
1 )
dx.
but integration by parts
Comparing
Vll
I
+
log (x
substitutions to
y/x)dx.
is
much
try,
easier.
the answers obtained
is,
perhaps, instructive.
/dx
Vlll
11.
x-x
'
3 /5
ix
/
(arcsin x)" dx.
(x)
/
x arctan (x )dx.
If
you have done Problem 18-10, the
look very familiar.
integrals involving \/x 2
involving
lem
4.
yi +
2
1.
integrals
(ii)
In general, the substitution x
—
1,
while x
Try these
(The method
is
—
sinhw
is
and
=
(iii)
The
world's sneakiest substitution
t
=
tan
—
,
x
=
is
will
on the other integrals in Probrecommended; it is easier to stick with
trigonometric substitutions.)
12.
Problem 4
the thing to try for integrals
substitutions
not really
in
cosh u often works for
undoubtedly
2 arctan
/,
dx
=
2
\+r
dt.
387
19. Integration in Elementary Terms
As we found
Problem 15-17,
in
=
sinx
expressions
this substitution leads to the
It
1+r
=
cos*
'
2
1
-V
'
l+t 2
This substitution thus transforms any integral which involves only
cos,
combined by
and
addition, multiplication,
and
sin
division, into the integral of
a rational function. Find
dx
/
Compare your answer
+ sin x
1
with Problem
dx
/
J
IV
/
+ b cos x
si
a sin
x
sin"
(In this case
it is
better to let
dx (An
jc
(There
-
(A
.
+ 5 sin x
Derive the formula for
a)
By
Why?)
tanjc.
to
do
this,
using
15-8.)
should be used only as a
13.
=
way
also another
is
Problem
exercise to convince
.
/i— —
3
t
x
dx
I-
mi
2
sin
l(viii).)
you
that
substitution
this
last resort.)
last resort.)
f sec x dx
in the following
two ways:
writing
1
cos*
cos*
cos 2 x
COS*
—
1
COS X
1
~
x
sin
:
2
_ 1
+
sin
x
+
an expression obviously inspired by
sure to note that
is
1
:
—
sin
x
_
partial fraction decompositions.
— sin x dx = — log (1 — sin x
)
f cos x/(l
And remember
very important.
and
on, keep doing algebra,
(b)
cos X
-
that ^ log
a
—
)
;
the
log y/a.
minus
From
Be
sign
there
trust to luck.
=
One
manip-
By using
the substitution
ulation
required to put the answer in the desired form; the expression
is
t
tanx/2.
again, quite a bit of
tanx/2 can be attacked by using Problem 15-9, or both answers can
be expressed in terms of t. There is another expression for fsecxdx,
which is less cumbersome than log(secx + tan*); using Problem 15-9,
we
obtain
/l
/
sec x
dx
=
last
\
=
log
log (tan
(
1
This
+ tan
-
tan
+ ;))
J
expression was actually the one
first
discovered,
and was due,
not to any mathematician's cleverness, but to a curious historical acci-
.
388
Derivatives
and
Integrals
computed
dent: In 1599 Wright
When
integrals of sec.
the
first
nautical tables that
amounted
to definite
tables for the logarithms of tangents
were
produced, the correspondence between the two tables was immediately
noticed (but remained unexplained until the invention of calculus).
14.
The
derivation of
=
primitive of fix)
e*(sinjt
/e
e
Suppose
that
/"
is
is
(a)
*(b)
is
is
seems
in the text
—
F{x)
to
prove that the only
— cosx)/2, whereas Fix)
any number C. Where does
e*(sinjc
also a primitive for
=
C
meaning of the equation
continuous and that
—
Given that fin)
x dx given
sinx
the
1,
+ f"(x)]smxdx = 2.
[f(x)
Jo
/
16.
sin
x
— cosx)/2 + C
come from? (What
15.
x
compute /(0).
Find f arcsinx dx, using the same trick that worked for log and arctan.
_1
Generalize this trick: Find / / (x) dx in terms of f fix) dx. Compare
with Problems 12-21 and 14-14.
17.
(a)
(b)
Find f sin x dx in two different ways:
and then using the formula for sin x
Combine your answers
to obtain
first
using the reduction formula,
an impressive trigonometric
18.
Express j \og(\ogx)dx in terms of /(log a)" dx. (Neither
terms of elementary functions.)
19.
Express
20.
Prove that the function fix)
(Do not try to find it!)
21.
Prove the reduction formulas
1
x
f x e~ dx
in
and work on the
.v
22.
=
+
=
x
e /ie
2
(x
ix
last integral.
1)
has an elementary primitive.
For the third one write
dx
=
J
l)»
+e x +
in the text.
f
~~
5x
f
+l)"+ l)"- " J
1
x~ dx
(jc
(Another possibility
2
+
is
1)"
to use the substitution
tan u .)
Find a reduction formula for
(a
(b)
<23.
j
(log
x)"dx.
Prove that
/>cosh.
sit
J]
(See Problem
1
expressible in
terms of f e~ x dx.
/dx
(x 2
is
identity.
1
-
cosh x sinh x
1
x
dt
2
8-7 for the significance of this computation.
19. Integration in Elementary Terms
24.
389
Prove that
pb
pb
f(x)dx=
/
+ b — x)dx.
f(a
/
Ja
Ja
(A geometric interpretation makes
but
this clear,
it is
also a
good
exercise in
the handling of limits of integration during a substitution.)
25.
Prove that the area of a
ber that
26.
n
and such
Show
(b)
circle
of radius
Let
that
that
/ be
0=1.
/
0/,(jc)
=
(Naturally you
.
= T (j)(x/h).
h
|jc
Show
1]
h and that
/
(j)
h
=
J-h
and continuous at
f=
lim
for
\x\
>
1
\.
0.
Show
that
f
fc
/=
/(O).
that
l
h-+Q+
final part
analogue of part
Let
—
rh
cf> h
v
lim
The
4>{x)
0, let
>
|
must remem-
circle.)
4>hix)
on [—1,
J
h
/ be
of
(b),
but
integrable
Hint: If h
is
f
J_
h
—1
/
x
+x
h
^dx =
TT.
problem might appear,
this
lim
The
nr
>
1
(d)
is
For h
for
integrable
lim
(c)
r
defined as the area of the unit
be a nonnegative integrable function such that
Let
(a)
is
it
actually requires
on [—1,
/
1]
more
2
+ x2
)
will
be an exact
careful argument.
and continuous
TT-^f(x)dx =
small, then h/(h
at first sight, to
at 0.
Show
that
jrf(0).
be small on most of [—1,
1].
next two problems use the formula
•0i
If
2 J
2
f(0) d0,
On
derived in Problem 13-24, for the area of a region bounded by the graph of
/
in
polar coordinates.
27.
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
nonsensical results!)
(i)
f(0)=
a
sin 9.
(iii)
/(0) = 2 + cos0.
2
2
f(G) = 2a cos 20.
(iv)
/(0)=acos20.
(ii)
.
390
Derivatives
and
Integrals
28.
Figure
1
has area
r
=
f(0)
shows the graph of /
-
Now
f(6) dO.
/
the ordinary graph of
in
suppose that
some function
AOx\B +
area
polar coordinates; the region
g
I
Then
g.
—
area
this
graph
also
happens
OAB
the region
OAB
thus
to
be
also has area
AOxqA.
JX]
Prove analytically that these two numbers are indeed the same. Hint:
function g
niU'RE
is
The
determined by the equations
x
= f (6) cosO,
=
g(x)
f(0)smO.
I
The
next four problems use the formulas, derived in Problems 3 and 4 of the
Appendix
to
Chapter
in particular, as the
29.
13, for the length
of a curve represented parametrically (and,
graph of a function
in polar coordinates).
Let c be a curve represented parametrically by u and v on
be an increasing function with h(a)
functions u
curve
— uoh,v — voh
c; intuitively,
c
is
just the
=
a and h(b)
=
b.
and
[a, b],
Then on
[a,
let
h
b] the
give a parametric representation of another
same curve
c traversed at a different rate.
(a)
Show, directly from the definition of length, that the length of
(b)
Assuming
c
on
[a, b]
equals the length of c on [a b ]
,
differentiability
of any functions required, show that the
lengths are equal by using the integral formula for length,
and the ap-
propriate substitution.
30.
Find the length of the following curves,
tions,
except for
which
(iii),
f(x)
=
-(x 2
fix)
=
x3
+ 2) 3
all
described as the graphs of func-
represented parametrically.
is
<x <
/2
,
1.
1
+
1
< x <
2.
12a
in
(iv)
(v)
(vi)
31.
= a cos
f{x) = log(cosA-),
f{x) = logx,
x
fix) = arcsin e
x
t.
= a 3 sin 3
•
V
<X <
1
',
<x <
<
t
/,
t
< In.
7T/6.
e.
— log 2 <
x <
0.
For the following functions, find the length of the graph in polar coordinates.
(i)
fie)
= a cose.
(ii)
/(0)
=a(l-cos0).
(iii)
(iv)
(v)
fie)=asm 2 ie/2).
0<0<2tt.
f(0) =
0<6><
/(0) = 3sec<9
7r/3.
.
391
19. Integration in Elementary Terms
32.
Problem 8 of the Appendix
In
to
Chapter 12 we described the
cycloid,
which
has the parametric representation
—
x
=
u{t)
—
a(t
y
sin t),
=
v(T)
=
—
a(\
cost).
(a)
Find the length of one arch of the cycloid. [Answer: 8a.]
(b)
Recall that the cycloid
is
the graph of v o u
Find the area under
.
one arch of the cycloid by using the appropriate substitution
in
f f and
evaluating the resultant integral. [Answer: 3jza-.]
33.
Use induction and integration by
za
f
34.
If /'
is
Problem 14-10:
parts to generalize
^*-jf(f(-(f «*)*)•••)*•
continuous on
Lemma
Lebesgue
use integration by parts to prove the
[a, b],
Riemann-
for /:
lim
f(t)sm(Xt)dt
/
=
0.
Problem 15-26, but it can be used to prove
the general case (in much the same way that the Riemann-Lebesgue Lemma
was derived in Problem 15-26 from the special case in which / is a step
This result
is
just a special case of
function).
35.
The Mean Value Theorem for Integrals was introduced in Problem 13-23.
The "Second Mean Value Theorem for Integrals" states the following. Supis either nondecreasing or
pose that / is integrable on [a b] and that
nonincreasing on [a b] Then there is a number £ in [a b] such that
,
.
,
f f(x)<j>(x)dx
In this problem,
tiable,
will
= <Ka)
f f(x)dx
assume
that
with a continuous derivative
Prove that
(a)
we
,
if
the result
is
/
+ <Kb) f
f{x)dx.
continuous and that
is
4> is
differen-
4>'
true for nonincreasing 0, then
it is
also true for
nondecreasing 0.
(b)
Prove that
then
Thus,
we have
it is
if
the result
true for
all
we can assume
is
true for nonincreasing
satisfying (p(b)
—
0,
nonincreasing 0.
that
is
nonincreasing and
(f)(b)
=
0.
In this case,
to prove that
f f(x)<P(x)dx=(f)(a) f f(x)dx.
Ja
by using integration by
(c)
Prove
this
(d)
Show
that the hypothesis that
is
needed.
Ja
is
parts.
either nondecreasing or nonincreasing
.
392
Derivatives
and
Integrals
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
From
in the next
36.
(a)
case of the
this special
problem.
Given
a\
...
,
,an and b\,
.
bn
,
.
=
let Sk
,
= s\(b\ -
\-a n bn
a\b\ H
(*)
.
b£i
+
a\
•
+ s 2 (b 2 -
+
•
Show
cik-
b?)
h s n _\{b n -\
H
that
-
bn )
+ sn bn
is sometimes called "Abel's formula for summation
regarded as an analogue for sums of the integration by parts formula
This disarmingly simple formula
by parts."
may be
It
rb
rb
f'(x)g(x)dx
I
= f(b)g(b)-f(a)g(a)-
f(x)g'(x)dx,
Ja
especially
P =
{to,
.
J(i
we
if
.
tn
,
)
Riemann sums (Chapter
use
of
[a, b], the left side
13, Appendix).
In
fact,
for a partition
approximately
is
n
£
(1)
-
f'(tk)g ftfc_i)ftfc
fjfc_l),
k=\
while the right side
is
approximately
n
f(b)g(b)
-
- J2
f(a)g(a)
f<fk)g'(?k)<!k
~
fc-l)
Jfc=l
which
is
approximately
f(b)g(b)
-
f(a)g(a)
-
J^
f(r k )
8W ~ 8{tk ~° Uk ~ fc-l)
'k
k=\
=
-
f(b)g(b)
f(a)g(a)
~tk-\
+ J2 f('k)[8('k-0-
8(t k
)]
k=\
n
=
f(b)g(b)
-
f(a)g(a)
+ £[/(**) -
f(a)}
[g(fft _i)
-
g(tk )]
k=\
n
+ f(a)J2s(t k -\)-g('k)k=\
sum
Since the right-most
is
just g(a)
—
g(b), this
works out
to
be
[8(fk-l)
~
n
-
[f(b)
(2)
f(a)]g(b)
+ £[/(fc) -
/(«)]
'
*('*)]
k=\
If we
choose
«k
=
/'('*)('*
h
-**-l).
=g(fk-i)
then
(1)
is
k=\
which
is
the
left
side
of (*), while
k
k
sk
=
^/'(/,)(/,(
=
-
f;_i)
is
approximately
]T
f(tj)
-
/(f,-_i)
/=1
1
so
(2)
is
approximately
sn bn
+ /^k&k ~
k=\
which
is
the right side of (*).
h k-\)>
=
f(t k
)
-
f(a),
.
19. Integration in Elementary Terms
This discussion
is
393
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
formula can
where the functions in
shall see, Abel's
often be used as a substitute for integration by parts in situations
question aren't differentiable.
(b)
Suppose
that {b n
}
m <
for all n.
>
nonincreasing, with b„
is
Prove Abel's
a\
-\
each n, and that
M
an <
\-
for
Lemma:
+
b\m < a\b\
•
•
+ a„b n <
•
b\M.
(And, moreover,
bk m
< akbk
a formula which only looks
(c)
/ be
Let
—
cp(b)
integrable
0.
P =
Let
on
a n bn
V
-\
more
[a b]
,
< b k M,
general, but really
and
let
isn't.)
be nonincreasing on
...,/„} be a partition of [«,&].
{to,
[a b] with
,
Show
sum
n
^/fe-iWfe-iXfi-fi-i)
/=1
lies
between the smallest and the
largest of the
sums
A:
d>(a)^2f(tt.i){ti-ti
Conclude
i).
that
f{x)(p{x)dx
I,
lies
maximum
between the minimum and the
0(a) f
of
f(t)dt,
Ja
and
that
it
therefore equals
</>(«)
f(t)dt for some §
/
in [a, b].
Ja
37.
(a)
Show
(i)
(ii)
(b)
that the following
f
Jo
sin
sii
(
x H
:
—
improper
)
integrals
dx
^(x + I)*,
Decide which of the following improper
(i)
(ii)
both converge.
fsing
/
sin"
I
-
dx.
I
d:
integrals converge.
that the
.
394
Derivatives
and
Integrals
38.
Compute
(a)
the (improper) integral
\ogxdx.
/
Jo
Show
(b)
that the
improper integral
log(sinx)Jx converges.
/
Jo
Use the
(c)
substitution x
—
2w to show that
'7t/2
/
=
log(sinx) dx
2
'71/2
Jo
+
\og(sinx)dx
I
2
Jo
log(cosjc)
I
dx
+ tt
log2.
Jo
nn/2
Compute
(d)
log(cos.v) dx.
/
Jo
Using the relation cosx
(e)
=
r log(sinjt)<fo.
— x), compute
sin(7r/2
/
Jo
39.
Prove the following version of integration by parts for improper
\oo
/"OO
u'(x)v(x)dx
/
=
Ja
The
right side
means, of course,
lim u(x)v(x)
*40.
One
u(x)v'(x)dx.
/
Ja
in
symbol on the
first
/»00
—
u(x)v(x)\
—
u(a)v(a).
of the most important functions in analysis
roo=
integrals:
e~'t
f
x
-
is
the
gamma
function,
]
dt.
Jo
Prove that the improper integral V(x) is defined if x > 0.
Use integration by parts (more precisely, the improper integral version
(a)
(b)
previous problem) to prove that
in the
ru+
Show
(c)
that T(l)
=
1,
=
\)
x r(x).
and conclude
that V(n)
=
(n
—
1)!
for
all
natural
numbers n
The gamma
function thus provides a simple example of a continuous function
which "interpolates" the values of
x >
0.
However, the
o
and f(x
(a)
Of course there
— 1)!; there are
r
is
A
states that
+
1)
gamma
for all
function has the important additional property
convex, a condition which expresses the extreme smoothness
this function.
Mollerup
HI,
n.
infinitely
that log
of
numbers
many continuous functions / with f{n) — (n
many continuous functions / with f(x + 1) = xf(x)
are infinitely
even
n\ for natural
=
Harold Bohr and Johannes
the only function / with logo/ convex, /(l) = 1
beautiful
V
is
theorem due
to
xf(x). See reference [43] of the Suggested Reading.
Use the reduction formula for
i.JT/2
/
$m n xdx
n-\
-
1
sin"
Jo
x dx
r
/
"
Jo
to
show
that
12
Sill
"
x dx.
)
'
19. Integration in Elementary Terms
(b)
Now
show
395
that
/
•
sin
I
2n+l
L
=—
,
x ax
3
Jo
I
71 '
2
-In
sin"
/
•
3
5
4
6
1
2«
-
1
•
'
2
Jo
and conclude
+
2n
7
5
* —i
= —
* dx
2n
6
^
—
•
2
2/2
'
that
I
7T
2
2
4
4
6
6
2
13
3
5
5
7
In
ln
-
In
2
In
sin "
x dx
Jo
+
2/i
1
71 '
/
/,
C* 12
1
sin
/
2 " +1
x^x
Jo
(c)
Show
1
and
that the quotient of the
+
1
<
This
2 " +1
<
x
2
sin "
<
x
2
4
4
6
6
13
3
5
5
7
can be made
sin
2 " -1
x
<x <
for
2n
n/2.
2n
-
In
as close to 7r/2 as desired,
known
In
1
+
1
an
usually written as
is
infinite
as Wallis' product:
n
4 4 6 6
3 '3 "5 '5 '7
2
2
~I"
2
Show
between
which shows that the products
2
(d)
is
,
result,
product,
integrals in this expression
/2n starting with the inequalities
1
sin
two
'
"
""
also that the products
1
v^
can be made
problem and
Wallis'
2-4-6-...1
3
•
5
•
.
.
.
•
2n
(2n
-
as close to y/n as desired.
1
(This fact
is
used in the next
Problem 27-19.)
in
procedure was quite
different!
He worked
1
with the integral
/ ()
(
1
— x 1 )" dx
(which appears in Problem 42), hoping to recover, from the values obtained for natural
numbers /;, a formula for
j=/"(l-.v 2 )'/4
A
complete account can be found
following
summary
1
f
,
(
in reference [49]
gives the basic idea. Wallis
in
/,.
Jo
2
first
of the Suggested Reading, but the
obtained the formula
In
4
(2-4---2n) 2
2-3-4---2n(2n
He
2"
+
1)
2n
+
then reasoned that n/4 should be
?1 (1-1)-
I
Jo
d-, 2
1/2
)
^=y^f =(^!)
2
.
(n!)
1
2
(2/i)!'
'
396
Derivatives
and
Integrals
we interpret ^! to mean T(l + \), this agrees with Problem 45, but Wallis did not
know of the gamma function (which was invented by Euler, guided principally by Wallis'
If
Since (2n)!/(n!)
work).
finding
for
)
(
the binomial coefficient
is
=q=
p
Wallis
),
I
hoped
to find \\
by
Now
1/2.
{p+qHp+q- \)-(p+\,
rand
this
makes sense even
p
if
not a natural number. Wallis therefore decided that
is
+q
(h
(*+«)'
1
2
With
+q
p
of
this interpretation
for
=
p
1/2,
true that
it is still
P
p+q + l
q+\
'P
(p
P
j
Denoting
+q
P
by W(q)
W(q +
1)
equation can be written
this
+
2q + 3
-r—W(q) = ^—-W{q),
+
2q + 2
+
^7
=
1
</
q
which leads
\
to the table
q
2
1
3
1
'
2
But, since
W(^) should be 4/n,
Next Wallis notes
(this
that
appearing
3),
if
4
2
rt
3
a4
a\
en
«3
logo(l/W)
'
6
2
3
of these
4 successive values W(g), W(g
tables,
2q
2^
5
3
;r
since
is
7
4
S
7T
tt
in either
5
'
2
a\, ai, 03, 04 are
a2
says that
3
I
2
wu''
5
Wallis also constructs the table
I
a
'/
W(q +
+q
+
1),
W(<7
+ 2),
then
+3
>
+2
7
2«/
z<?
-t- <+
zcy
+o
convex, compare the remarks before Problem 41), which
implies that
V
Wallis then argues that this should
combined
values in a
table
4
In
+4
+
jr
3
5
2«
4
4
6
in
"
\|
6
'
'
ir
3
'
5
2/;+
3
4
jr' 3
2
4
be true
6
5
'
when
a\,
a-i,
03, a$ are four successive
given both integer and half-integer values! Thus,
\),
W{n), W(n
2n
+2
2/!
+
'
5
'
is
W(n +
4
3
1
still
where q
taking as the four successive values
4
V a2
ai
a\
7
ln+
6
2n
+
3
2
1
3
1
2
|),
5
'
4
In
4
In
7
5
'
W(n +
'
+
+
2/;
6
2a;
+4
+3
4
7T
2-4-4-6-6---(2«)(2n)(2«
3-3-5-5---(2«
+
l)(2n
from which Willis" producl follows immediately.
3
2
+
2#i
which yields simply
2/1
1),
+ 2)
2n
I)
2n
+3
+2
1
he obtains
397
19. Integration in Elementary Terms
/>oo
42.
It is
an astonishing
fact that
improper
integrals
f(x)dx can
/
often be
Jo
b
computed
where ordinary
in cases
f
integrals
f(x)dx cannot. There
/
is
no
Ja
h
f
elementary formula for
e
/
_x 2
dx but we can
find the value of
...
,
Ja
f°°
/
e
_x 2
dx
J°
There are many ways of evaluating this integral, but most require
some advanced techniques; the following method involves a fair amount of
precisely!
work, but no
(a)
Show
you do not already know.
facts that
that
f
-x 2
(1
n
dx
)
=
2
4
3
5
Jo
1
(l+.t
I,
o
2 )"
=
dx
In
+
2«
7T
In
3
1
2
1
""
4
2
2n-2'
done using reduction formulas, or by appropriate
combined with the previous problem.)
(This can be
tions,
(b)
Prove, using the derivative, that
1
e
(c)
substitu-
x
_x
< e~ x
2
2
for
<x <
for
<
1
<
~
r
+x 2
1
1.
x.
Integrate the nth powers of these inequalities from
oo, respectively.
use the substitution y
4
,-2
sjn
Then
-
y/nx
to
and from
show that
1
to
In
In
5
3
=
to
+
1
<
dy <
e~
/
:
'
dy
Jo
n
TT
In
3
1
2^2-4"
(d)
Now use
Problem 4 1 (d)
I
**43.
(a)
Use
to
show
f
sin.v
dx
Ja
—
that
" y=
e
2
integration by parts to
I
show
cosfl
that
cos 6
and conclude
that j
the limit as a
—>
f
/
a
x
In
-3
-2
(sinx)/x dx
b
exists.
J
cosx
r— dx
—
x
l
tl
(Use the
+ and the right side for the
,
left
limit as
side to investigate
b —>
oo.)
398
Derivatives
and
Integrals
(b)
Use Problem 15-33
show
to
T
A)*
dt
Jo
(c)
+
sin(/z
/'
for
that
=
Ti
sin
any natural number
n.
Prove that
f
+
sm(X
lim
A—>-00 Jo
\)t
=
rff
sin
0.
2J
The term in brackets is bounded by Problem
Riemann-Lebesgue Lemma then applies.
Hint:
(d)
Use the
substitution u
=
+
(X
^)t
and part
C°° sin X
show
(b) to
15-2(vi);
the
that
71
x
JO
/•OO
44.
Given the value of
(sin
/
x)/xdx from Problem 43, compute
./O
sin
x
dx
by using integration by
play an important
*45.
(a)
Use the
parts. (As in
Problem 38, the formula
for sin
2x
will
role.)
substitution u
=
x
t
show
to
^
1
= -
r(x)
that
-" Ux
du.
/
x Jo
46.
(b)
Find r(i).
(a)
Suppose
and
fix)
that
is
that lim f(x)
integrable
= A
on every
and lim f(x)
=
< a <
interval [a, b] for
B. Prove that for
all
a,
yS
/?,
>
we have
dx
f
Jo
Hint:
To
—
(A
-
B) log
x
estimate
/'
'
f(ax)-f(Bx)
—
a
dx use two
different substitutions.
OO
(b)
Now suppose instead that
lim f{x) — A. Prove that
r
Jo
fiotx
)
f°° f(x)
/
Ja
~
x
dx converges
X
f(M ax =
A
log
8
—
a
for all a
>
and
that
399
19. Integration in Elementary Terms
Compute
(c)
the following integrals:
10
e
(i)
-<*x
f
—
e ~P
*
Jo
f°° cos(ax)
—
cosifix)
dx.
X
Jo
In Chapter 13
x
we
said, rather blithely, that integrals
may be computed
to
any
degree of accuracy desired by calculating lower and upper sums. But an applied
mathematician,
doing
may
it,
an integral
needed
who
really has to
to three
ought
decimal places, say
in [tj-\,
(a
to
mention
computing upper and lower sums might
might not be possible to compute the quantities m,
it
[/,-_i, ti\.
n
It is
and consider >^ /(*;)
(U
far
—
more reasonable simply
This represents the
?j_i).
i=\
of certain rectangles which partially overlap the graph of
Appendix
to
Chapter
the trapezoids
FIGURE
shown
13.
But we
will get
a
much
/
—
better result
to pick points
sum
we
t
of the areas
see Figure
if
x
1
in the
instead choose
in Figure 2.
2
Suppose, in particular, that
we
divide [a,b] into n equal intervals, by
the points
tj
Then
refined
at the outset that
for each interval
ti\
about
degree of accuracy that might easily be
The next three problems show how more
calculations much more efficient.
not even be practical, since
and Mi
calculation, rather than just talking
in certain circumstances).
methods can make the
We
do the
not be overjoyed at the prospect of computing lower sums to evaluate
=
a
+
the trapezoid with base [f,_i,
b
—a
.
a
i
tj]
+ ih.
has area
(t<
-ti-l)
means of
400
Derivatives
and
Integrals
and the sum of all these areas
S„
f{t\
=h
)
+
is
simply
f(a)
f(t2 )
2
+
f(ti)
fQ>)
+
2
./"(/„-!)
2
n-\
—
b
=
h
a
i=]
This method of approximating an integral
obtain
to
Tj2,i
Yj2„
is
from £„
just
isn't
it
So
2^«-
called the trapezoid
is
rule.
necessary to recompute the old /(?;); their contribution
in practice
it
is
compute
best to
Y,i,
£4, £85
/.In
/
the next
problem we
will estimate
Ja
47.
(a)
Suppose
•
•
to get
/
/ —
E„.
Ja
that
/"
continuous. Let
is
P,
be the linear function which agrees
/ at ti-\ and /,. Using Problem
minimum and maximum of /" on
with
the
•
pb
pb
approximations to
Notice that to
1
-f
/
(x
-
r,_i)(.v
1-46,
show that
and
if n,
and
jV,
are
[f,-_i, t{\
-
tj)dx
then
ml >
(b)
Evaluate
Nil
[''
/ to get
nth*
['
Nik
>
3
l2~
(c)
Conclude
that there
is
some
c in (a, b) with
b
(b-a) 3
„
1
Notice that the "error term"
(/?
—
the error obtained using ordinary
We
can obtain
still
more accurate
I.
3
the interval [a, b]
48.
(a)
is
Suppose
We
that a
< 2 which
=
and b
=
agrees with
2.
/
we approximate / by
first
Let
Conclude
P be
(while
quadratic
i^4-^)
when
3).
the polynomial function
and 2 (Problem
/(2)].
3
that in the general case
P =
Ja
2
consider what happens
at 0, 1,
p = -[/(0) 4-4/(1) +
/o
varies as \/n
varies as 1/n).
that
(b)
2
divided into two equal intervals (Figure
first
of degree
sums
results if
functions rather than by linear functions.
FIGUR
a) 3 f"{c)/\2n
3-6).
Show
19. Integration in Elementary Terms
pb
pb
Naturally
(c)
P =
J
The
a quadratic polynomial.
is
But, re-
J ci
same relation holds when / is a cubic polynomial!
using Problem 11-46; note that f" is a constant.
markably enough,
this,
f when /
j
Ja
Prove
401
this
we do
previous problem shows that
new
not have to do any
calculations
b
to
f
compute
a+b
we
Q when Q
/
a
is
cubic
polynomial which agrees with
49.
(a)
much more lee-way in
Show
Q(a)
that there
=
is
Q(b)
f(a),
+ A(x - a)(x - b)
defined on
is
P(x)
[a, b],
—
- Q(x) = (x-a)(x
2
a+b
(x
)
)
for
some A.
[a, b\
we have
x
then for every x in
a+b\ 2
/
f(x)
/
f
/
1
Prove that
'
satisfying
,(a-\
-r- =
I
Q
use to our advantage:
^) =
Q\
f(b),
=
(
we can
C
=
Hint: Clearly Q(x)
/
,
choosing Q, which
,(a+b\
if
a+b
a cubic polynomial function
<2
(b)
- byf
{4)
(^)
4!
some ^ in (a,b). Hint: Imitate the proof of Problem 11-46.
is continuous, then
Conclude that if /
for
(c)
(
'
b-a
f =
/a
r
-
f
f(a)+4f
n
+ bi.\
.
fib)
Ja
for
(d)
some
Now
(b-a) 5
2880
H4)
r\c)
c in (a,b).
divide [a b] into 2n intervals
,
t;
—
a
+ ih,
h
by means of the points
=
—a
b
2n
Prove Simpson's
b
6n
I.
rule:
n-\
"
b-a
-
(
/(«)
+4
£
f(to-i)
(b-a) 5
2880h
for
and
+ 4f -=-. + /(*)
,
f(a)
/
is
at a, b,
have
still
b-a
But there
/
some
c in (a, b).
4
/
(4)
+ 2Y. fted + f {b)
(c)
402
Derivatives
and
Integrals
THE COSMOPOLITAN INTEGRAL
APPENDIX.
We
under the graph of
originally introduced integrals in order to find the area
a function, but the integral
is
considerably
more
than
versatile
For example,
that.
Problem 1 3-24 used the integral to express the area of a region of quite another
sort.
Moreover, Problem 13-25 showed that the integral can also be used to ex-
press the lengths of curves
lot
of work
ably a
little
may be
more
— though,
as we've seen in
Appendix
surprising, since the integral seems, at
two-dimensional creature. Actually, the integral makes
few geometric formulas, which we
formulas
to
Chapter
13, a
necessary to consider the general case! This result was prob-
we will assume some
present in
will
this
first
its
blush, to be a very
appearance
in quite a
Appendix. To derive these
from elementary geometry (and allow a
results
little
fudging).
Instead of going
down
three-dimensional ones.
be expressed by
to
one-dimensional objects, we'll begin by tackling some
There are some very
The
integrals.
special solids
V
simplest such solid
a "solid of revolution,"
is
obtained by revolving the region under the graph of /
when we regard
the horizontal axis,
FIGURE
whose volumes can
>
on
[a b]
.
around
the plane as situated in space (Figure
1).
1
If
p =
{to,...,tn }
is
any
partition of [a,&],
and w, and M, have
their usual
meanings, then
7tmi
is
the
volume of a
irMi~{tj —/,_])
and
/,.
is
that lies
disc
the
volume
2
(ti
-r,_i)
the
inside
solid
(Figure
Similarly,
2).
V between
t(-\
Consequently,
n
n
tt
FIGURE
V
of a disc that contains the part of
y^mrOi -
*i-i)
2
S volume V < n ^PM, (f, -
;=1
2
f,_i).
1=1
But the sums on the ends of this inequality are just the lower and upper sums for
2
f on
[a, b]:
7T
L(f
2
,
Consequently, the volume of
P)
< volume V < n U(f 2
V must be
volume V
,
P).
given by
= n /'
I
f(x)dx.
Ja
This method of finding volumes
Figure 3 shows a
FIGURE
<
is
affectionately referred to as the "disc
more complicated
solid
V
method/'
obtained by revolving the region
under the graph of / around the vertical axis (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 V\ sitting right
on top of
it).
In this case
we assume
a
>
as well
.
19, Appendix.
as
/ >
Figures 4 and 5 indicate
0.
FIGURE
some other
The Cosmopolitan
possible shapes for
For a partition
P =
rectangle with base
shells
we
...,/„}
{to,
[f,-_i, t{\
we
tx
and height m, or
M-,
(Figure
2
-
2
r,_i
)
< volume V < n
^M^f, 2 -
2
f,-_i
),
/=i
as
n
/;
^mi{ti
+ti-\)(tt
-
tt-i)
< volume V < Ti^M^t,
(=1
Now
the volumes
n
y]m,-(f,-
which we can write
6
Adding
6).
obtain
i=l
FIGURE
5
consider the "shells" obtained by rotating the
/;
7i
V
4
FIGURE
of these
403
Integral
these
Appendix
+f,-_i)(r,-
-/,_i).
/=l
sums are not lower or upper sums of anything. But Problem
Chapter 13 shows that each sum
1
of the
to
and
^m,7,(f, -tt-i)
can be made
as close as desired to
J^mjfj_i(fj -/,-i)
xf(x)dx by choosing
/
the lengths
f,
—
f,_i
Ja
small enough.
The same
is
true of the
sums on the
right, so
we
find that
b
volume V
—
f
7.71
\
xf(x)dx;
Ja
this
is
The
the so-called "shell
method" of finding volumes.
surface area of certain curved regions can also be expressed in terms of inte-
we tackle 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 / and
altitude s The total surface area of the sides of the pyramid is thus
grals.
Before
.
1
FIGURE
ps,
7
where p
is
the perimeter of the base.
By choosing
the base to be a regular polygon
404
Derivatives
and
Integrals
with a large
number of sides we
see that the area of a right circular
cone (Figure
8)
must be
=
— (27tr)s
where
FIGURE
s
height s
8
is
and
and
radii r\
ri
frustum of a cone with slant
Finally, consider the
the "slant height."
in Figure 9(b),
Ttrs,
shown
Completing
in Figure 9(a).
this to
a cone, as
we have
s\
+
s\
.v
r?
r\
so
s\
r\s
=
>"2
Consequently, the surface area
7zri(s\
+ s) —
s\
_
r2
''1
Now consider
=
7ir\s\
r2
2
tts
cones, as in Figure 10.
£[/ft_l) +
JT
—
~
r\
r
P =
The
f(ti)]y/(ti
{tQ
tn
total surface
-
r\
+
r,-!) 2
* £[/fe-i)
+
\
2
=
ns{r\
+ ri).
\
formed by revolving
the surface
zontal axis. For a partition
~
is
r2
(a)
r->s
+s =
graph of / around the hori-
the
we can
inscribe a series of frustums of
area of these frustums
-
[f(t,)
f(U)\\
1
/ft_l)]
+
:
f{t'-
fit,)
(
t,
is
-t,-\
l)
)
i=l
By
the
Mean
Value Theorem,
this
is
(b)
n
FIGURE
9
JT
£[/fe-rl)
+
/U/)]>/l+/'(jC
2
;
)
a
-
fi-l)
1=1
for
some
we
conclude that the surface area
x, in (f,-_i, ?j).
Appealing
to
Problem
1
of the Appendix to Chapter 13,
is
2n I f(x)\/\
+ fix) 2
dx.
Ja
PROBLEMS
1.
(a)
Find the volume of the solid obtained by revolving the region bounded
by the graphs of f(x)
(b)
i<
i.
in
2.
and f(x)
=
x
around the horizontal
Find the volume of the solid obtained by revolving
around the
iii.i
=x
vertical axis.
Find the volume of a sphere of radius
r.
this
axis.
same region
3
19, Appendix.
3.
When
the ellipse consisting of
all
FIGURE
4.
1
we
volume of the enclosed
(x
a)
2
405
+
solid.
1
Find the volume of the "torus" (Figure
—
Integral
y /b — 1 iIN
obtain an "ellipsoid of revolution"
points (x, y) with x l /a~
rotated around the horizontal axis
(Figure 11). Find the
The Cosmopolitan
+y =
2
b
2
(a
>
b)
around the
12),
obtained by rotating the
circle
vertical axis.
length b
FIGURE
5.
A
12
cylindrical hole of radius a
radius 2a (Figure
[GURE
1
13).
is
bored through the center of a sphere of
Find the volume of the remaining
solid.
406
Derivatives
and
Integrals
6.
For the solid shown in Figure 14, find the volume by the
(a)
FIGURE
shell
14
This volume can also be evaluated by the disc method.
(b)
method.
must be evaluated
the integral which
complicated.
The
Write
in this case; notice that
next problem takes up a question which
it
is
this
down
more
might
suggest.
7.
Figure 15 shows a cylinder of height b and radius f(b), divided into three
solids,
one of which,
V\,
is
a cylinder of height a and radius f(a).
If
/
one-one, then a comparison of the disk method and the shell method of
is
computing volumes leads us
nbf{b)~
—
7zaf{a)
—n
to believe that
J
f(x) dx
=
volume
Vi
Ja
rfib)
f(b)
l
'
yf~ (y)dy.
I
Jf(a
fu>)
Prove
this analytically,
using the formula for
more simply by going through
I
I
(
,
I
RE
I
5
/
/
from Problem 19-16, or
the steps by which this formula
was derived.
19, Appendix.
Figure
The Cosmopolitan
Integral
6 shows a solid with a circular base of radius a
1
perpendicular to the diameter
AB
.
407
Each plane
intersects the solid in a square.
Using
arguments similar to those already used in this Appendix, express the
volume of the
(b)
9.
Same problem
if
an
integral,
and evaluate
A
of
it.
each plane intersects the solid in an equilateral triangle.
Find the volume of a pyramid (Figure 17)
area
FIGURE
solid as
in
terms of
its
height h
and the
base.
its
16
FKiURE
10.
17
Find the volume of the
solid
which
is
the intersection of the two cylinders
Hint: Find the intersection of this solid with each horizontal
in Figure 18.
plane.
FIGURE
11.
18
(a)
Prove that the surface area of a sphere of radius
(b)
Prove,
more
in Figure 19
is
2nrh. (Notice that
position of the planes.)
FIGURE
1
I
*
r
is
Ativ
.
shown
not on the
generally, that the area of the portion of the sphere
this
depends only on
h,
408
Derivatives
and
Integrals
A
circular
mud
boards of length
that
it
puddle can just be covered by a parallel collection of
of the
at least the radius
circle, as in
cannot be covered by the same boards
non-parallel configuration, as in
if
Figure 20(a). Prove
they are arranged in any
(b).
(b)
FIGURE
12.
13.
2
(a)
Find the surface area of the ellipsoid of revolution
(b)
Find the surface area of the torus
The graph of f(x) —
l/x, x
>
1
in
is
Problem
in
Problem
19-3.
19-4.
revolved around the horizontal axis
(Figure 21).
(a)
Find the volume of the enclosed
(b)
Show
(c)
that the surface area
surface
FIGURE
we
"infinite
trumpet."
infinite.
fill up the trumpet with the finite amount of paint found
would seem that we have thereby coated the infinite inside
area with only a finite amount of paint. How is this possible?
Suppose
in part
is
that
(a).
2
It
PART
INFINITE
SEQUENCES
AND
INFINITE
SERIES
One of the most remarkable
m x+ m(m-l)„
+
1
1
m(m -
5
-2
l)(m
-
2)
^
[m -
(n
-2-3
1
m(m -
1)
•
•
•
1-2
-
the
n
is
is
•
•
series,
then finite, can be expressed,
known, by
When m
•
a positive whole number
sum of the
which
as
is
^
1)]
+
When m
of
is the following:
algebraic analysis
1
series
is
(I
+
x)
m
.
not an integer,
and
the series goes on to infinity,
it
will
converge or diverge according
as the quantities
m and
x have this or that value.
In this case, one writes the same equality
+
(1
x) m
=
1
+
.
.
.
It is
x
H
m(m —
—
1)
^
'
1-2
„
x2
+
•
•
•
etc.
assumed that
the numerical equality will always occur
whenever the
ill
is
series is convergent, but
has never yet been proved.
NIELS HENRIK ABEL
APPROXIMATION BY
POLYNOMIAL FUNCTIONS
mm ^/
CHAPTER
There
is
If
a polynomial function,
p
is
one sense
in
which the "elementary functions" are not elementary
p(x)
at
all.
=a + aix + ---+a n x",
then p(x) can be computed easily for any
number
This
x.
functions like sin, log, or exp. At present, to find log*
is
not at
= fi 1/t dt
all
true for
approximately,
we must compute some upper or lower sums, and make certain that the
error
involved in accepting such a sum for log* is not too
great. Computing e x =
log" (x) would be even more difficult: we would have to
compute log a for many
values of a until we found a number a such that log a is
approximately x
then a
would be approximately
e
—
x
.
we will obtain important theoretical results which reduce the
computation of f(x), for many functions /, to the evaluation of
polynomial functions. The method depends on finding polynomial
functions which are close approximations to /. In order to guess a polynomial which is appropriate,
In this chapter
it is
useful
examine polynomial functions themselves more thoroughly.
Suppose that
to first
p(x)
and
=
+ a\x
a
H
a n x".
\-
our purposes very important, to note that the coefficients a,
can be expressed in terms of the value of p and its various derivatives
at 0. To
begin with, note that
It is
interesting,
for
p(0)
=a
.
Differentiating the original expression for p(x) yields
= a\ + 2a 2 x +
p'(x)
\-
na n x"~
l
.
Therefore,
=p (» 0)=a
p '(0)
Differentiating again
p"(x)
we
=
(
1
.
obtain
2a 2
+3
•
2
•
a3x
+
+ n{n -
1)
•
a n x"-
2
.
Therefore,
p"(0)
In general,
we
will
= p a \0) =
2a 2
.
have
P
WI
(0)=*! ai
.
or
ak
=
^.
k\
If
we
agree to define
holds for k
411
=
also.
0!
=
1,
and
recall the notation
p
(0)
=
p, then this formula
412
Infinite
Sequences and Infinite Series
If we
had begun with a function p
that
was written
= ao + a\(x - a) H
p(x)
as a
"polynomial
in (x
— a),"
a n (x -a)",
h
then a similar argument would show that
p
Suppose now
that
/
is
a)
(a)
= ~kT-
ak
a function (not necessarily a polynomial) such that
fQ\a),...,f M (a)
all exist.
Let
ak
:
——
=
7]
0<k<n,
,
and define
Pn ,a(x) —
The polynomial Pnfl
(Strictly speaking, we
to indicate the
«o
+a\(x —
a) H
a)".
called the
has been defined so that
PIIM ik) (a) = f {k) (a)
it is
-
fl„(.x-
Taylor polynomial of degree n for/ at a.
should use an even more complicated expression, like Pn ,a,f,
dependence on /; at times this more precise notation will be useful.)
is
The Taylor polynomial
in fact,
h
clearly the only
Although the
for
<
k
<
n\
polynomial of degree < n with
coefficients of
Pn ,a,f seem
to
this properly.
depend upon /
in a fairly
compli-
cated way, the most important elementary functions have extremely simple Taylor
polynomials. Consider
first
the function sin.
sin(O)
sin'(O)
sm'(0)
sin"'(0)
sin
From
this
(4
'(0)
We
have
= 0,
=cos0= 1,
= -sin = 0,
= -cos0 = -l,
= sin = 0.
point on, the derivatives repeat in a cycle of 4.
The numbers
(A)
av
=
sin
(0)
k\
are
0,1.0,-1.0.1.0.-1.0,1
Therefore the Taylor polynomial P2n+1,0 of degree 2n
3
P2n+ i,o(*) = x
(Of course, P2ll
,
i.o
=
5
+
1
x
x
x
-_
+ ---- + •• + (-
P2«+2,0)-
1
for sin at
v
is
2 " +1
)"
-
1
(2|I+1)!
20. Approximation by Polynomial Functions
The Taylor polynomial
left
Pi n ,Q of degree 2n for cos at
to you)
„2„4„6
P2n.0(x)=
-
1
— + -JT--ZT +
4!
2!
The Taylor polynomial
exp(0)
=
1
for
=
The Taylor polynomial
not even defined at
is
exp
is
+
l
+
(-1)"
for log
X2
compute. Since exp (k) (0)
log (x)
is
n
must be computed
=
=
X
X
- + X_ + _+...
+ -.
The standard
0.
at
4
3
"
(2n)!
6!
especially easy to
X
+
computations are
x 2n
-'-
Taylor polynomial of degree n
for all k, the
p„„(.v)
log
(the
is
413
choice
a
is
at
some point a
—
1.
log'(l)
=
1;
= -\,
z
log"(l)
=
-l;
= ^,
log" (l)
-,
/
0, since
Then
x
\og"(x)
x
/
log
"U)
#
= 2;
in general
log
w
(-l)*- 1 ^-!)!
(jc)
x
Therefore the Taylor polynomial of degree n for log
PnA (x)
It is
case
often
=
(jc
-
1)
15
(x-1) 2
-
-
more convenient
we can choose
a
=
f
{k)
=
(x)
+
,
+
•
•
•
is
1
(-\)"-
+
l
(x
- If
.
n
to consider the function
We
0.
U-l) 3
at
f(x)
=
log(l +jc). In this
have
\og
{k)
a+x),
so
/«(0)=tog
cfc)
=
(l)
1
(-l)*- (*-l)!.
Therefore the Taylor polynomial of degree n for /
PBi0 (x)
There
arctan.
is
=x-
4
3
—2 + x— - x—4 +
•••
+
is
(-l)"- 1 *"
.
n
3
one other elementary function whose Taylor polynomial
The computations
arctan'U)
=
arctan" (x)
(x)
+ xl
-2x
«-=-,
=—
2 2
(1
=
+.v
(\+x 2
is
important—
of the derivatives begin
x
1
arctan
x2
at
arctan'(O)
=
1;
arctan" (0)
=
0;
arctan
=
—2.
)
2
)
—
-(-2)
(1
+ 2x.2(\+x 2 )-2x
^—.
4
+x 2
)
,
(())
,
414
Infinite
Sequences and Infinite Series
It
is
computation
clear that this brute force
polynomials of arctan
will
be easy
to find after
of Taylor polynomials more closely
One
line
mials for
which
Pn ,a,f W U1
we have examined
the properties
— although
simply defined so as to have the same
between / and
never do. However, the Taylor
will
the Taylor polynomial PnM .f was
n derivatives at a as /, the connection
first
much
actually turn out to be
deeper.
of evidence for a closer connection between
/ may be uncovered by examining
/ and
the Taylor polyno-
the Taylor polynomial of degree
1
is
P La (x) = f(a) + f(a)(x-a).
Notice that
f(x)-P lM (x)
x
Now, by
—
x
—
f(a).
a
we have
the definition of f'{a)
f(x)-Pha (x)
lim
x^-a
FIGURE
f(x)-f(a)
a
x
—
=
(J.
a
1
In other words, as x approaches a the difference f(x)
becomes small even compared
x
e and of
small, but actually
graph of fix)
=
Pl,0(x)
which
is
=
f(0)
+ f(0)x=
the Taylor polynomial of degree
1
for
x
to
/
1
— P\ >a (x ) not
— a. Figure
1
only becomes
illustrates the
+x,
The diagram
at 0.
also
shows
As x approaches
0, the
the graph of
ft.o(*)
which
is
=
/<P)
+
/'(0)
+
^r*
the Taylor polynomial of degree 2 for
difference f(x)
—
P2,o(x)
seems
to
2
/
=
l
+x +
at 0.
y
•
be getting small even faster than the difference
20. Approximation by Polynomial Functions
f(x) — P\o(x). As it stands, this assertion is not very precise, but we are
prepared to give it a definite meaning. We have just noted that in general
/w-ft,w
Bm
x^-a
For f(x)
—
e
x
—
and a
this
e
x
-\-x =
x^o
Thus, although /(a)
become
=
Q
X
an easy double application of l'Hopitafs Rule shows that
hm
not
«--!-«
]im
jc->-0
,.
does
=ft
that
x
x-»0
the other hand,
means
—
small
x
l
n
1
-
^
0.
2
P[,o(x) becomes small compared to x, as * approaches 0, it
compared toi". For P2.0C*) the situation is quite different;
the extra term x /2 provides just the right compensation:
e'-\-x-hm
e'-l-x
= hm
X2
x->0
2x
x^-0
=
^—
e
=
lim
x^o
This result holds in general
—
f'(a)
if
,.
hm
.
THEOREM
l
(x
analogous assertion for
Suppose that /
is
—
PnM
is
1
0.
exist,
=
then
U;
also true.
a function for which
f "\a)
(
f'(a)
all exist.
a) 1
x
2
and f"(a)
f(x)-P2 a~ (x)
x-*a
in fact, the
now
a
/U)-/YoU) =
|im
On
—
x
415
Let
ai
= f
(k)
(a)
n < k <
n.
H
a n (x
.
~
,
k\
~
and define
Pn ,a(x) —
ao
+ a\(x - a)
\-
Then
hm
*-*•«
f(x)-Pn —
a (x)
,
(x
—
a)"
—=
_
U.
-
a)".
.
416
Infinite
Sequences and Infinite Series
PROOF
Writing out
PnM {x)
explicitly,
we
obtain
fM-T^—^ix-ay
f(x)-P„ M (x)
{x - a)"
It will
'
i=0
-
(x
new
help to introduce the
f
a)
(n)
n
{a)
n\
functions
" _1
Q( x )
now we must
Y
=
( ' ]
(a)
Lf——(x-a)
1
and
=
g(x)
-
(x
a)
n
;
prove that
hm
f(x)-Q(x)
x^a
/W(
=
fl )
.
g(x)
n\
Notice that
Q
{k
ik)
g
\a)
=f
{k
k<n-\,
\a),
k
=n\(x -a)"-
(x)
/{n
-
k)\
Thus
lim[/U) - Q(x)]
=
- Q(a) =
f(a)
0,
x -> a
-
lim[/'U)
lim[/
(
Q'(x)]
=
- Q
"- 2)
"- 2,
(.t)
(
f'ia)
(x)]
and
Km g(x) =
We may
lim g'(x)
-
=
Q'{a)
0,
= f u '- 2) (a) - Q
==
—
therefore apply l'Hopital's Rule n
lim g
(
"~ 2)
u,
- 2)
=
(x)
'-
u
f(x)-Q(x)
f"- (x)-Q
=
lim
lim
x^a
x^a
{x — a) n
n\{x—a)
l)
0.
l)
(x)
.
;
Q is a polynomial of degree
{a). Thus
\x) = f
Q
fact,
(n
{n
{
,.
lim
x-*a
and
this last limit
is
f
n
—
f{x)-Q(x)
=
(x — a)"
un
(
0,
,,
l)st
derivative
is
a constant;
in
and a
Theorem
local
U)-/ ",
,,
(«)
n\(x—a)
x-*-a
(a)/n\ by definition of
of /, then, according to
>
"-
./•
lim
1
/"(a)
—
1, its (n
{)
One simple consequence of Theorem
maxima and minima which was developed
at a if
0.
times to obtain
1
,.
Since
=
(a)
f
{n)
(a). |
allows us to perfect the test for local
in
Chapter
11-5, the function
maximum
at
a
if
11.
/
f"(a) <
If
a
is
a
critical
has a local
0.
If
point
minimum
f"(a)
=
no
—
.
417
20. Approximation by Polynomial Functions
conclusion was possible, but
and
further information;
significant.
Even more
nor a
ask
=/
--.
f(x)
=
g(x)
= -(x-a) n
(M
- 1)
= 0,
(a)
(x-a)'\
2) that if n
minimum
local
might be
(a)^0.
Notice (Figure
satisfy (*).
maximum
"}
(4)
/ (tf) =
what happens when
then the sign of
can be guessed by examining the functions
situation in this case
which
{
0,
/» =
=
/'(a)
f
=
we can
generally,
(*)
The
conceivable that the sign of f'"{a) might give
is
it
f"'{a)
if
/
point for
,
is
odd, then a
On
or g.
is
neither a local
the other hand,
if
n
is
minimum at a, while
maximum at a. Of all functions
even, then /, with a positive nth derivative, has a local
with a negative nth derivative, has a local
g,
satisfying
these are about the simplest available; nevertheless they indicate the
(*),
In fact, the whole point of the next proof
general situation exactly.
function satisfying
is
theorem
2
made
much
looks very
(*)
by Theorem
precise
one of these functions,
in
that
any
a sense that
1
Suppose that
/» =
f
is
If n
is
(3) If n
is
(2)
There
/
odd, then
no
clearly
is
(n)
{n)
even and f (a) >
even and f (n \a) <
n
(1) If
PROOF
like
is
•••
(n
~ l)
0,
n
first
—
0,
has a local
of generality in assuming that f(a)
/
derivatives of
\
P„ a( x
)
= 0,
minimum at a.
has a local maximum at a.
local maximum nor a local minimum
/
then /
then
hypotheses nor the conclusion are affected
since the
(a)
(a)^0.
has neither a
loss
=f
at
/
if
a are
= f(a)+ Lj^(x -«) +
0, since neither the
replaced by
is
0, the
=
/ —
f{a). Then,
Taylor polynomial
•••+
at a.
—^(x
J
P, ua
of
/
is
- af
'
(a)
nodd
=
Thus, Theorem
n
1
=
f^Ha)
if
x
,
is
(x
figure
2
Suppose now
f(x)/(x
— a)
n
—
lim
that n
-
n
(n)
{a)
n\
then
,
.
(x-a)
x^t-a
Cl)"
_ f
fix)
,.
sufficiently close to a,
(x
n even
.
f(x)-Pn a (x) =
,.
lim
fix)
(b)
-a)
states that
x-*-a
Consequently,
(x
;
.
has the same sign as
(n
f \a)
a)"
is
even.
In this case
has the same sign as
f
{n)
(.v
—
a)"
>
for
all
x
^
{a)/n\ for x sufficiendy close to a,
a.
it
Since
follows
418
Infinite
Sequences and Infinite Series
same
that f(x) itself has the
(n)
f
>
ia)
means
0, this
sign as f"(a)/n\ for
=
/(*) >
for
x close to
for the case
Now
e~
l/x
x#0
\
a.
f
{n)
/
Consequently,
<
ia)
is
signs for
a local
for
x > a and
as before
shows that
a.
always
has the same
y
sign.
5
— a) n <
Therefore f{x) has
(jc
<
for x
/
This proves that
a.
has neither a local
if
x
is
different
maximum
nor
because
f
{k)
ia)
maxima and minima
question of local
will settle the
about any function which
limitations,
A similar proof works
at a. |
Although Theorem 2
just
at a.
then
x > a and x <
minimum
minimum
The same argument
odd.
n
— a)" >
fia)
has a local
-™(x-a)
ix
If
0.
suppose that n
sufficiently close to a,
But
sufficiently close to a.
jc
that
arises in practice,
may
be
for all k. This
/(*)
=
e~
for
does have some theoretical
it
happens (Figure
3(a)) for
the
function
(b)
/x
[
x^O
\
jc=0,
0.
which has a minimum
which has a
and also for the negative of this function (Figure
0. Moreover (Figure 3(c)), if
at 0,
maximum
at
x
e- l '
/(*)
=
x
0,
-e~
X >
fix)
-e-
=
then
x<0
atO.
x
0,
l/x
\
The
(c)
FIGURE
{k
f \0) =
for all k, but
conclusion of
Theorem
at a
\
1
is
=
x <0,
has neither a local
Two
cept of "order of equality."
3
/
x >
,
l/x
3(b)),
minimum nor a
local
maximum
often expressed in terms of an important con-
functions
/ and g
are
equal up to order n
if
m
x-*a
(x
/ up
been designed
degree <
to
to
order n at
make
a.
this fact true,
n with this property.
—
=
o.
a)"
Theorem 1 says that the Taylor polynomial
The Taylor polynomial might very well have
In the language of this definition,
Pn.a.f equals
-* (t)
/(t)
ii
because there
This assertion
is
is
at
most one polynomial of
a consequence of the following
elementary theorem.
THEOREM
3
PROOl
Let
P and Q be two polynomials
and
Q
Let
R = P — Q.
are equal
up
in (x
to order n at a.
Since
R
is
a
—
a), of
degree
<
n,
and suppose
that
P
Then P = Q.
polynomial of degree <
//,
it
is
only necessary to
.
20. Approximation by Polynomial Functions
prove that
419
if
= b + ... + bn (x-a) n
R(x)
satisfies
lim-^=
-
x^a ( x
R =
then
Now the
0.
P(x)
hm
For
=
i
this
R
hypothesis on
*-+a (x
—
a) n
surely imply that
=0
<
for
condition reads simply lim R(x)
=
x^-a
lim R{x)
=
Thus
=
bo
+b
lim [b
=b
<
i
n.
a)'
(x
{
-
a)
0;
on the other hand
+ bn (x - a) n
+
'
]
.
and
R(x)
=
-
b (x
l
a)
+
-
a)
+ bn
(.x
-a)".
Therefore,
—— =bi +
R(x)
X
b 2 (x
+ b n {x - a) n -
+
x
CI
and
lim
x ^a
Thus
=
Z?i
Let
/ be
of degree
PROOF
=
bj.
a
in this way,
we
=b2 (x - a) 2 +
•
•
•
+ b n (x - af
find that
b
COROLLARY
—
and
R(x)
Continuing
x
=
= bn = 0.
|
n -times differentiable at a, and suppose that P is a
polynomial in (x
< n, which equals / up to order n at a. Then P
P
=
naf
P and />„,«,/ both equal / up to order n at a, is easy to see that P
Pn ,aj up to order n at a. Consequently, P = Pnaf by the Theorem.
Since
-a)
.
it
equals
|
it
At first sight
might seem
that
/
is
this corollary
appears to have unnecessarily complicated hypotheses;
polynomial P would automatically imply
that the existence of the
sufficiently differentiable for
Pn aJ
,
example (Figure
4),
=
x" +l
0,
IfP(x)
=
0,
then
at 0.
On
undefined.
P
But
in fact this
is
not
so.
For
suppose that
/(*)
order n
to exist.
,
x
irrational
x rational.
< n which equals / up to
the other hand, f'{a) does not exist for
any a
0, so /"(0) is
is
certainly a polynomial of degree
#
—
420
Infinite
Sequences and Infinite Series
x irrational
=
/(*)
x rational
0,
FIGURE
4
When /
useful
does have n derivatives at a, however, the corollary
method
that our
r*
arctan x
—
method of
suggests a promising
+ t~,
to obtain a
1
^ dt
\+t 2
finding a polynomial close to arctan
— divide
1
polynomial plus a remainder:
l-? 2 + / 4 -/ 6 +
\+f
in failure.
\
I
Jo
1
provide a
attempt to find the Taylor polynomial for arctan ended
first
The equation
by
may
Taylor polynomial of /. In particular, remember
for finding the
+
---
(-l)"r"
+
/i+l, 2/1 + 2
(-l) n+l
t
1+r
This formula, which can be checked easily by multiplying both sides by
1
+
t
,
shows that
px
arctan x
-f
l-t' +
+
t'
(-\)"r"dt
+
(-\)" +l
Jo
x3
=
According
to
v
-
x5
x
-y y
+
/
Jo
px
2n+\
2n+2
—
{
1
+
dt
{
~
2n+2
{
2/i
+
Jo
1
1
+
t
l
dt.
our corollary, the polynomial which appears here will be the Taylor
polynomial of degree In
+
for arctan at 0, provided that
1
r
x
lim
zn+
+2
,2/1
t
-
r+i
Jo
dt
.2«+l
x-+0
Since
\2n+3
l
this
is
2«+l
clearly true.
for arctan at
+
r
2
dt
2
<
dt
2/Z+3
^0
Thus we have found
that the Taylor
polynomial of degree
is
^2/i+i.oU)
=
-
1
—+X
X
x
5
,.2/i+l
+ (-D"
2//+
I
421
20. Approximation by Polynomial Functions
By
now
the way,
possible to
we have
that
P2n+l,0(x)
and
discovered the Taylor polynomials of arctan,
work backwards and
since this
polynomial
/m,_
arctan (0) (0)
find arctan
(/:)
(0) for all k: Since
X5
A3
=X- —
+—
3
5
by
is,
X
1
definition,
——a
arctan (2) (0)
+ arctan*mn (0) +
,_
find arctan*
(0)
2 " +l
+ (-1)"——,
2n +
,
2
+
"
arctan (2 +1) (0)
+
by simply equating the
2 „.
—x 2n+l
+
1)!
coefficients of
xk
(2n
2!
we can
it is
,
,
in these
two
polynomials:
ik)
arctan
(0)
-
—
(J
it
A:
even,
is
k\
arctan (2/+1) (0)
(-1)'
(2/+1)!
2/+1
A much
more
interesting fact
„3
arctan
a-
emerges
„5
=x- — + —
3
arctan
or
+
5
if
(2/+1)
(0)
we go back
x 2n+\
M
(-1) 2n + r
=
(-!)' -(2/)!.
to the original equation
p\
+
(-1)"
t
k
1
2n+2
*dt,
/
1
+
^
2
and remember the estimate
Un+l
l+t 2
Jo
When
x \2n+3
~ In +
3
< 1, this expression is at most 1/(2/7 + 3), and we can make this as
we like simply by choosing n large enough. In other words, for \x\ <
|a|
small as
we can
\
dt
1
arctan
use the Taylor polynomials for
to
compute arctan a as accurately as we
The most important theorems about Taylor polynomials extend
like.
this isolated result
and the Taylor polynomials will soon play quite a new role.
so far have always examined the behavior of the Taylor
polynomial Pna for fixed n, as a approaches a. Henceforth we will compare Taylor
polynomials PnM for fixed a, and different n. In anticipation of the coming theorem
we introduce some new notation.
to other functions,
The theorems proved
If / is a
Rn ,a(x) by
function for which
f(x)
=
P„, a (x)
=
f(a)
+
Pn>a {x)
+ RnAx)
f'(a)(x
we
define the
—
a)"
exists,
-«) + •••+
Jf
[n)
(a)
K
-L(x
remainder term
+ R nM (x).
nl
We
There
would
is
One way
like to
have an expression for R„, a (x) whose
such an expression, involving an integral, just as
to guess this expression
f(x)
is
size
f(a)
+ R iU (x).
easy to estimate.
in the case for arctan.
to begin with the case n
=
is
=
0:
422
Infinite
Sequences and Infinite Series
The Fundamental Theorem of Calculus
f(x)
enables us to write
= f(a)+
f'(t)dt,
f
so that
Ja
A similar expression for
tion
by parts
R\
a
(x)
in a rather tricky
can be derived from
and
u(t)-f'(t)
(notice that
formula using integra-
this
way: Let
= t-x
v(t)
x represents some fixed number
in the expression for v(t), so v'(t)
=
1
);
then
/ f\t)dt=
Ja
f{t)-
Jf
a
dt
1
I
I
u\t)
v'{t)
X
=
n\
f"(t){t-x)dt.
u(t)v(t)
Ja
a
Since v(x)
=
0,
we
f(x)
I
|
u'(t)
v(t)
obtain
= f(a)+
Jf
a
f'(t)dt
= f(a)-u(a)v(a)+
f f"(t)(x-t)dt
Ja
=
+ f'(a)(x-a)+ f
f(a)
f"(t)(x
-
t)dt.
Ja
Thus
Rl,a(x)=
It is
It
just
hard to give any motivation for choosing
happens
to
many
v'{t)
=
f
Jo
-
(x
— — x,
t
t),
=
rather than v(t)
sort of thing
similar but futile manipulations.
easy to guess the formula for
u(t)
then
v(t)
be the choice which works out, the
discover after sufficiendy
now
f"(t){x-t)dt.
J
f
f"it)
/?2, fl
U).
and
—
t.
one might
However,
it
is
If
2
= ~(x-t)
t/(0
,
so
f"(t)(x-t)dt
=
-(x-tyv2
u(t)v(t)\
\a
f
-
/'"(/)
dt
z
J<>
f"{a){x-aV
C
h
2
/
x
—=—
f'"(t)
(.v
-
2
/)
dt.
I.
This shows that
fO)(t)
-^—{x-n-dt.
Ja
You should now have
little
difficulty giving
a rigorous proof, by induction, that
20. Approximation by Polynomial Functions
/'" +1)
if
is
423
continuous on [a,x], then
fx f(n+l)(f\
Rn, a
(x)=
L---^-(x-trdt.
/
Ja
n.
This formula
is called the integral form
of the remainder, and we can easily estimate it in terms of estimates for p^/nl on [a,
*]. If m and
are the minimum
and maximum of f»^Jn\ on [a,x], then R (x)
satisfies
na
M
///
we can
so
(x
j
-
t)
n
< Rna (x)<M f
dt
t) dt,
write
-a) n+l
(x
=
Rn,a(x)
a
n
for
-
(jc
+
1
some number a between m and M. Since we've assumed
means that for some t in {a, x) we can also write
that
tinuous, this
n
n\
which
called the
is
familiar to those
+
(n
\
+
Lagrange form of the remainder
who have done Problem
a)
iy}
/
(
" +1 >
is
con-
'
(these manipulations will look
13-23).
The Lagrange form of the remainder is the one we
and we can even give a proof that doesn't require
will
need
in
almost
all
cases,
f^"+^ to be continuous
refinement admittedly of little importance in most
applications, where we often
assume that / has derivatives of all orders). This is the
form of
(a
we
LEMMA
the remainder that
will
choose in our statement of the next theorem (Taylor's
Theorem).
Suppose
R
that the function
(n
is
+
l)-times differentiable
{k
R \a) =
Then
for
any x
in (a, b]
R
(x-ay+i
For n
=
for
n by induction on n.
all
0, this
is
= 0,l,2,...,ii.
we have
R(x)
PROOF
forJfc
on [a,b], and
"+
]
\t)
for
-"^nyr
Mean
just the
{
some
'
Value Theorem, and
To do
we
this
use the
m
(«>*)•
we
will
prove the theorem
Cauchy Mean Value Theorem
to write
*(*)
(x
-
fl
)«+2
=
"
R'iz)
(n
+
2 )(z
n
and then apply the induction hypothesis
R(x)
(x
-a)"+ 2
(R')
1
'
n
R'(Z )
1
- a)»+i ~
+
{
to
"+ l
(z
(/!+
+ 2)!
I
^
S
° me
Z in (a
R' on the interval [a, z] to get
\t)
for
2
R {n+2) (t)
(n
+2
- a)»+l
1)!
some
t
in (a, z)
'
x)
'
424
Infinite
Sequences and Infinite Series
THEOREM
4
(TAYLOR'S
THEOREM)
Suppose
y<"+i>
that /'
denned on [a,x], and
are
that
Rn a (x)
is
,
=
f(x)
f(a)
+
Rn,a(x)
=
f'(a){x
- a) +
•
•
defined by
+ ^-^-(x - a)" + Rn a (x)
•
,
Then
f(n+\) (t)
(n+
n+\
~
7zr(x
:
a )'
some
f° r
t
in (a, x)
1)
(Lagrange form of the remainder).
PROOF
The
R na
function
Lemma
the conditions of the
satisfies
by the very definition of
the Taylor polynomial, so
-
(jr
for
some
/
a)" +l
R na — f
'
1)!
(n+1) _- /'
/•(»+D
=
a polynomial of degree
is
+
(n
(0
But
in (a, x).
Rn
since
(/l+l)
Rn
Rn.a(x)
Applying Taylor's Theorem
/;.
|
—
0,
we
remainder terms for
sin
and exp, with a
to the functions sin, cos,
obtain the following formulas:
x3
2 " +1
5
- + x-
smx=x-
+ (-D"
(2n
4
2
cos*
=
x
x
l-+-
+( _ ir
,
+
sin
l
<?
(of course,
and
=
1
+
we could
A"
+
—+
•
•
2'
'
^l + r
—
+
(2n
n\
actually go
(/?
+
-x
first
sin
(2n+2)
(0l
<
easy.
in the
Since
for all
1
f,
we have
•
(2/1
sin
{In
Similarly
we can show
+ 2),.
|2/i+.
>
(/)
+
x
9 „ +2
<
(2n+2)\
2)!
that
COS (2 " +1) (0
(2/j
+
l)!
|2/i+l
»+i
(0 ,^-
+
,.
l)!
!)!
one power higher
two are especially
„2«+2
+ 2)!
n+1
cos).
Estimates for the
(/)
(2n
l)!
(2n)!
(2 " +2,
<
(2«+
1)!
20. Approximation by Polynomial Functions
These estimates are particularly
any s >
we can make
because
interesting,
(as
proved
in
Chapter
425
16) for
xn
n\
by choosing n large enough (how large n
must be will depend on x). This enables
us to compute sin* to any degree of
accuracy desired simply by evaluating the
proper Taylor polynomial P (*). For example,
n
suppose we wish to compute
sin 2 with an error of less than J0~ 4
Since
,
.
sin
2
= P2n+ \,o{2) +
where
7?,
<
\R\
(2/i
we can
use P2n+l,o(2) as our answer, provided
that
2 2«+2
(2/i
A number n
with
this
<
+ 2)!
10
sin
-4
property can be found by a straightforward
search— it ob-
viously helps to have a table of values for
n\
happens that n = 5 works, so that
2=
Pu,o(2)
and
2" (see
+
5!
where
even easier to calculate
sin 1
sin
= P2n+l
page 432). In
7!
91
< 10~ 4
\R\
~
U!
+
error less than e
it
*'
.
approximately, since
1
(1)
+ r,
wriere \R
<
I
:
(2a?
To obtain an
this case
+R
3!
It is
+ 2)f
we need
+
2)!'
only find an n such that
1
+
(2/i
and
this
<
£
'
2)!
requires only a brief glance at a table of
factorials.
vidual terms of
P2n+l,o(l)
will
For very small x the estimates
=
sin
To
To obtain
jP2 " +1
°
(
lo
(Moreover the
indi-
be easier to handle.)
will
be even
+R
)
easier.
For example,
I
where
'
|/?l
<
10 2 "+2(2/
2
+ 2)!
\R\ < lO"^ we can clearly
take n = 4 (and we could even get
away
with n ==3). These methods can actually
be used to compute tables of sin and
cos; a high-speed computer can
compute P2 „ + i.oU) for many different x in almost
no time at all. Nowadays, computers, and
even cheap calculators, determine the
values of such functions "on-the-fly",
though by specialized methods that arc even
I
aster.
426
Infinite
Sequences and Infinite Series
Estimating the remainder for e x
>
that x
the
[0, x]
x <
(the estimates for
maximum value
of
only slightly harder. For simplicity assume
is
x
e
e' is
exp
since
,
x
e x"
R n.O <
Since
we
already
know
<
that e
4,
Once
increasing, so
D!
n+\
A\
<
as small as desired
+
(n
*
1)!
1)!'
by choosing n
depend on x (and the
4
factor
X
In particular,
if
n
+
v
-
= 4,
^ + '" + ^ + *'
How large
sufficiently large.
make things more
< x < 1, then
will
again, the estimates are easier for small x. If
1+
the interval
we have
(n+
n must be will
is
On
15).
+]
(«+
ex,n+.
which can be made
Problem
are obtained in
difficult).
< R <
where
(/i+ D!
then
4
0<R<
1
<
IO-
5i
SO
1
1
+
+
1
2!
=
2
+
1
1
+ +
+/?
< R <
where
-
—
4!
3!
17
24
which shows that
2
<
<
e
3.
(This then shows that
x
< R <
3 a
(n+
R
allowing us to improve our estimate of
compute
that the
first
(you should check that n
The
arctan
you
function arctan
(/r)
(x)
is
D!
By taking
slightly.)
n
—
1 you can
3 decimals for e are
e
cruel to insist that
»+i
=
1 does give
actually
is
= 2.718...
degree of accuracy, but
this
would be
do the computations).
important but, as you
also
it
may
recall,
an expression
for
hopelessly complicated, so that our expressions for the remainder
are pretty useless.
On
the other hand, our derivation of the Taylor polynomial for
arctan automatically provided a formula for the remainder:
+
arctan x
3
•+
(_\\n
v 2n+\
(-l)";r"
2n
+
l
rx /-i\n+lf2n+2
+
(it.
./o
1+/ 2
.
t
20. Approximation by Polynomial Functions
427
As we have already estimated
(_
x
\"+l +2n+2
1
L
l
moment we
For the
|2n+3
+v
r
<
dt
in+i
dt
2/2
Jo
only numbers x with
will consider
remainder term can clearly be made
as small as desired
\x
|
+
<
1
3
.
In this case, the
by choosing n
sufficiently
large. In particular,
arctan
(-1)"
1
1
1=1 —
+
+ 5"
3
+
2/2
1
+
where
R,
\R\
<
In
1
+
3
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~ 4 for example,
we must take n > (10 — 3)/2. This is really a shame, because arctan 1 = 7r/4,
,
compute n. Fortunately,
surmount these difficulties. Since
so the Taylor polynomial for arctan should allow us to
there are
some
clever tricks
which enable us
to
|2n+3
|/?2n+l,o(*)|
<
2/ ?
much
+
3
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
this can be done in a convenient way.
smaller n 's will
72"
From
on page 413, we
the calculations
logO
X
=
X2
X
+
X3
X
see that for x
4
(-!)"
-4 +
T T
+
>
we have
v
(-1)"
It+1
where
,n+\
(-1)"
n
and there
For
this
is
another matter
22+1
1
more complicated
n sufficiently large, provided that
The behavior
<
when — 1 < x < (Problem 16).
remainder term can be made as small as desired by choosing
a slighdy
function the
+
B+l
estimate
—1 < x <
1
of the remainder terms for arctan and f(x)
when
\x
I
>
1
.
=
log(x
+
1)
is
quite
In this case, the estimates
|2«+3
l^2«+1.0(*)l
<
2/2
+3
for arctan,
„/i+i
\R„.o(x)\
<
(x
22+1
>
0) for /,
bounds x'"/m become large as m becomes large. This predicament is unavoidable, and is not just a deficiency of our
estimates. It is easy to get estimates in the other direction which show that the
remainders actually do remain large. To obtain such an estimate for arctan, note
are of
no
use,
because
when
\x\
>
1
the
428
Infinite
—
—
.
'
Sequences and Infinite Series
that if
t
is
in [0, x] (or in [x, 0] if
+r
1
jc
<
0),
then
< 1+jc 2 <2x 2
>
]f\x\
,
I,
so
2n+2
rx
f
ITT1
Jo
To
Ix 2
get a similar estimate for log(l
I
1
+t
=
An
Jo
+ x), we
-t+r
1
\2n+\
r r" +2 dt
l
>
dt
+
+6
can use the formula
(-l)"-V'-
+
1
(-l)V
\+t
to get
2
1
1
log(l+*)= f
+
(-1)
n-\
Jo
+ i-iy
r
x
Jo
If
x >
0,
then for
t
in [0, x]
1
so
t
These estimates show that
+t <
r
+
1
1
+x
are
of no use whatsoever
<
dt> w~
" 2*
if \x\
>
+t
if x
2x,
>
1,
dt
t"
2«
J
+
\x\
>
1,
the Taylor polynomials for arctan
computing arctan x and log(.v
in
2
+
1
).
This
is
no
tragedy,
because the values of these functions can be found for any x once they are
for
all
x with
|jc
|
<
known
1
This same situation occurs in a spectacular
fix)
=
have already seen that /^'(O)
that the Taylor polynomial
Pn,0(x)
=
Pn
/(0)
$ for
+
way
-l/x 2
-
0,
We
dt.
then the remainder terms become large as
1,
n becomes large. In other words, for
and /
1
we have
—
/
Jo
t"
-
/
/
/'(0)JC
for the function
.v
#0
x
=
0.
for every natural
number
k.
This means
is
+
f"(0)
^^X
2
H
x
:
2!
0.
remainder term R„q(x) always equals f(x), and the Taylor
useless for computing f(x), except for x = 0. Eventually we will be
In other words, the
polynomial
is
able to offer
some explanation
for the
behavior of
this function,
disconcerting illustration of the limitations of Taylor's
The word "compute"
for the
has been used so often
in
which
is
such a
Theorem.
connection with our estimates
remainder term, that the significance of Taylor's Theorem might be mis-
construed.
It is
true that Taylor's
Theorem can be used
as a
computational aid
20. Approximation by Polynomial Functions
(despite
ignominious failure
its
previous example), but
in the
it
has even
429
more im-
portant theoretical consequences. Most of these will be developed in succeeding
two proofs
chapters, but
be used. The
first
waded through
THEOREM
5
PROOF
e
is
will illustrate
some ways
which Taylor's Theorem
in
be particularly impressive
illustration will
the proof, in Chapter 16, that
tt
is
to those
who
may
have
irrational.
irrational.
We know
=
e
any
that, for
e
l
=
n,
— + •••+ — + R
+-+
1
l
I
Suppose
Choose
that e
n
were
> b and
=
rational, say e
also n
>
n
< Rn <
where
,
ni
2!
1!
3.
(«+
1)!
a/b, where a and b are positive integers.
Then
a
-=
+ + rr +
b
2!
,
i
1
1
- + Rn
l
!
+
,
n\
so
—
n\a
=»! +
+
»!
n\
+
nl
n\R„.
n\
Every term
because n
in this
>
b).
equation other than n\R n
is
an integer
Consequently, n\R n must be an integer
<R n
<
(n
+
(the left side
also.
is
an integer
But
1)!'
so
3
< n\R„ <
+
n
which
is
impossible for an integer.
The second
proved
in
illustration
Chapter
is
merely a straightforward demonstration of a
15: If
+/=
/ (0) =
/'(0) =
/ =
0.
1,
4
1
|
f"
then
3
< - <
T
To prove
this,
/( 3)
/( 4)
/(5)
etc.
observe
that
first
0,
o,
o.
f
ik)
exists for
= {f y = _ /
= (/ (3)y = _ f y = _ f =
= (/ (4)y = r>
/
i
(
fi
every
k; in fact
fact
,
I
430
Infinite
Sequences and Infinite Series
This shows, not only that
ones: /, /',
Theorem
all
{k)
f
states, for
any
=
t
Each function /
in [0, x].
any particular
jc
there
is
(
/
0, all
(
*>(0) are 0.
Now
Taylor's
n, that
(
"
(»
some
=
.f (0)
f
for
but also that there are at most 4 different
exist,
-/, -/'. Since /(0)
number
a
* +1)
(0l
|/
(we can add the phrase "and
+
1)!
" +1)
(
continuous (since
is
f
M such that
m
<
+u (t)
<
for
<
t
and
jc,
all
(n+2)
exists),
so for
n
n" because there are only four different f {k) ).
all
Thus
M|.v|"
s
,/U)l
Since
this
is
=
The
J^Tv:
and since |jc|"/"! can be made as small as desired by
this shows that |/(x)| < £ for any e > 0; consequently,
true for every n,
choosing n sufficiently large,
/(*)
+1
0.
Theorem
other uses to which Taylor's
be put
will
in
succeeding chapters
are closely related to the computational considerations which have concerned us
for
much
of
RnM (x)
remainder term
If the
this chapter.
can be made
as small as
computed
any degree
desired by choosing n sufficiently large, then f(x) can be
to
Pna (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 (in theory at least!), then we ought to be able to ignore
of accuracy desired by using the polynomials
There should be
the remainder completely.
1
sin
x
=x—
JC
—
JC
(-
3!
5!
like
•
•
•
,
7!
4
2
JC
,
sums"
X7
"S
—
"infinite
6
JC
8
JC
.v
COSX=1 -2! +
4!-6! + 8!-'--'
2
cX
= 1 + A:+
JC
3
JC
+
2!
jc
=
JC
jc
2
log(l
We
+*) = *-
x
5
J
x
4
+ --
71
—H
it
••
"if +
are almost completely prepared for this step.
we have never even defined an
necessary definitions.
|jc
|
<
1
4
3
y+y
V
4!
3!
—— + —
3
arctan
+
infinite
sum.
if-l<v<l.
Only one
obstacle
remains—
Chapters 22 and 23 contain the
.
.
20. Approximation by Polynomial Functions
431
PROBLEMS
1
Find the Taylor polynomials (of the indicated degree, and
at the indicated
point) for the following functions.
i)
f(x)
—
e
ii)
f(x)
=
e
iii)
sin;
degree
iv)
cos;
degree 2n,
v)
exp; degree n
vi)
log;
vii)
f(x)
—
x
+ x + x;
viii)
f(x)
—
x
+jr +
ix)
f(x)
=
e
;
degree
2/7, at
,
degree n,
f(x)
at
-.
1
at 2.
;
degree 4, at
0.
x\ degree 4, at
1.
degree In
+ xl
l+x
—
at n.
r-;
—
3, at 0.
smx degree
3, at 0.
1
2.
*
degree
n,
s
1,
at 0.
at 0.
Write each of the following polynomials in x as a polynomial in (x
is
only necessary to compute the Taylor polynomial at
as the original polynomial.
3.
+
(i)
x2
- Ax -
9.
(ii)
x4
-
+44.r 2
(iii)
x
(iv)
ax
12.Y
3). (It
of the same degree
Why?)
+ 2x +
l.
.
+ bx + c.
down
Write
3
3,
—
numbers
a
sum
(using
2,
to within the specified accuracy.
tion, consult the tables for 2"
(i)
sin 1; error
< 10~ 17
(ii)
sin 2; error
< lO"
(iii)
sin ~; error
< 10~
(iv)
e;
(v)
e
error
;
notation) which equals each of the following
error
< 10~ 4
<
10
.
.
1
.
^.
.
and
n\
To minimize
on the next page.
needless computa-
432
Infinite
Sequences and Infinite Series
2"
n
n\
1
2
1
2
4
2
3
8
6
4
16
24
5
32
120
6
64
720
7
128
5,040
8
256
512
362,880
9
*4.
40,320
10
1,024
3,628,800
11
2,048
39,916,800
12
4,096
479,001,600
13
8,192
6,227,020,800
14
16,384
87,178,291,200
15
32,768
1,307,674,368,000
16
65,536
20,922,789,888,000
17
131,072
355,687,428,096,000
18
262,144
6,402,373,705,728,000
19
524,888
121,645,100,408,832,000
20
1,048,576
2,432,902,008,176,640,000
This problem
similar to the previous one, except that the errors
is
are so small that the tables cannot be used.
thinking,
Chapter
and
some
in
it
it
x n /n\ can be
16, that
actually provides a
problem
cases
method
may
made
You
be necessary
will
demanded
have to do a
little
to consult the proof, in
small by choosing n large
—
the proof
In the previous
for finding the appropriate n.
was possible to find rather short sums; in fact, it was possible
which makes the estimate of the remainder given by
to find the smallest n
Taylor's
specific
Theorem less than the
sum is a moral victory
desired error. But in this problem, finding any
(provided you can demonstrate that the
sum
works).
(i)
(ii)
(iii)
sin 1; error
<?;
<
error
sin 10;
,10.
(iv)
< 10
10" 1000
<
< 10
error
error
I
(v)
5.
(a)
ll0
arctan y^; error
'.
.
10~ 20
.
-30
<
r\— I0'°)
'.
lU u
1
(
cisely
you showed that the equation x = cos* has pretwo solutions. Use the third degree Taylor polynomial of cos to
show
that the solutions arc approximately
In Problem
the error.
1
1
-41
Then
use the
fifth
±^2/3, and
degree Taylor polynomial
find
bounds on
to get a better
approximation.
(b)
Similarly, estimate the solutions of the equation
1x
=
x sin*
+ cos" x.
,
433
20. Approximation by Polynomial Functions
6.
Prove, using
(a)
Problem
n
—
=
15-9, that
1
arctan
—
4
—
ify
off
that
=
it
a
arctan —
4
l
arctan
239'
(Every budding mathematician should ver-
3.14159
on an immense
first
—
5
a few decimals of
used, the
—
3
]
4
Show
arctan
2
71
(b)
1
+
but the purpose of this exercise
tt,
If the
calculation.
5 decimals for
is
not to
set
second expression in part
n can be computed
you
(a) is
with remarkably
little
work.)
7.
Suppose that a and
b,
t
and
g, respectively. In other
the coefficients
in
are the coefficients in the Taylor polynomials at a of
terms of the
c,
words, a
s
= f U) (a)/il
of the Taylor polynomials
a,'s
and
and
=
/?,
g
u)
(a)/i\
.
/
Find
a of the following functions,
at
Z?,'s.
+ g-
(i)
f
(ii)
fg-
(iii)
/'•
(iv)
h(x)= [
f{t)dt.
Ja
k(x)= f
(v)
f(t)dt.
Jo
8.
(a)
=
Prove that the Taylor polynomial of f(x)
sin(x
2
)
of degree An
+ 2 at
is
Hint: If
sinx
—
P
is
P(x)
v6
„10
3!
5!
v 4»+2
(2«
+
1)!'
the Taylor polynomial of degree 2n
+
R(x), where lim R(x)/x^" +
^
—
0.
+
1
for sin at 0, then
What
does
this
implv
about lim R(x 2 )/x 4n+2 ?
(b)
Find f a) (0) for
(c)
In general,
if
all k.
f(x)
=
g(x m ), find
f
ik)
(0) in
terms of the derivatives of g
atO.
The
ideas in this
in the next three
9.
(a)
problem can be extended
significantly, in
problems.
Problem 7
(i)
amounts
to the
*n,a.f+g
=
equation
*n.a,f
"t
*n,a,g-
Give a more direct proof by writing
/(*)= Pn,a,f(x) + Rn,a,f(x)
g(x)= Pn ,a,g(x) + R„ M ,g(x),
and using the obvious
fact
about
Rn a ,f +
,
Rn,a,g-
ways that are explored
.
434
Infinite
Sequences and Infinite Series
(b)
Problem 7
Similarly,
could be used to show that
(ii)
—
*n,a,fg
I'n.a.f
^n,a,g\ni
'
where [P] n denotes the truncation of P
P
terms of
<
of degree
P
n [with
sum
of
written as a polynomial in x
—
degree n, the
to
Again, give a more direct proof, using obvious
Rn
volving terms of the form
(c)
Prove that
about products
in-
.
p and q are polynomials
if
facts
all
a]
in
x—a and
lim R(x)/(x
— a)" —
0,
then
+ R(x)) = p(q(x)) + R(x)
p(q(x)
where
if
>
a polynomial in x
and q
n,
is
terms of p(q(x
(d)
If
=
a
The same
If g{a)
then
'n, a, fog
—
Y*n,b,f ° *n,a,g\n-
—
f(x+g(a)), G(x)
=
C*W
For the following applications of Problem
9,
=
i
[i
i
+
=
(a)
For fix)
(b)
For the same
(c)
Find
e
x
and g(x)
/ and
+
p„,,.,
fl..«.«
and just write Pn j instead of
•
•
•
+
=
sin*, find
g, find
Psjg
/^.tanG*)- Hint: First use
find ^.l/cosU)- (Answer: *
we assume
Problem 9
+y+
9
(f)
e
(e)
Find
P5J
for
=
sin;c/cos2x. (Answer: a
Calculations of this sort
cos*. (Answer:
+ x 2 )e*].
x 3 /[0
may be
to
y=-)
=
=
for simplicity,
and the value of Ps.cosOO
for /(jc)
for /(a-)
=
5
P4;/
P6J
.
.
Find
f{x)
a
n
Psj +g (x).
(d)
2x
ewr]
P„, a ,f-
3
11.
and any value of g(a). Hint:
g(x+a) and H(x) = G(x) — g(a).
+
Find
all
case.)
2
(Pn,,. e :'
(f)
then
n.
0,
g(a)
0,
= 0, then
**„.«.
10.
>
is
result actually holds for all a
Consider F(x)
(f)
having only terms of degree
a whose constant term
=
a special
is
—
a)) are of degree
=
and b
(Problem 8
(e)
—
is
0.
—a
a polynomial in x
Also note that
p
=
-a)"
lim R(x)/(x
1
+ 2x +
+
y.v 3
(Answer: a 3
used to evaluate
\x 2
-
+
+
a
limits that
4
±x 3
- ^a 4
)
5
f|^A
)
- \x 5 + |a 6
we might
)
other-
wise try to find through laborious use of l'Hopital's Rule. Find the following:
,
N
(a)
hm
X- 0
,.
e
x
-\-x-\x—
2
a
-
sin
Hint: First find
a
Pt,
Nix)
=
lim
x-»-0
o ,/v(a)
nator Nix) and D(a).
D(x)
and /^.o.oU)
for the
numerator and denomi-
20. Approximation by Polynomial Functions
435
-x
x2
2
{
(b)
lim
1+a
— sm x
x
term
Hint: For the
1
(c)
+x),
first
write 1/(1
— — x+x~— a 3 +
+x)
1
-
1
lim
2
x2
Hsin *
1
(d)
x
e /(I
—
COS(a
)
lim
r2« x
x^sin
'
1
(e)
1
lim
*->0 sin 2
sin(A 2 )
a
(sinx)(arctan.v)
(f)
lim
1
12.
—
—
x
cos(a 2 )
Let
sin
/(*)
a
A"
=
7^0
A=0.
1.
Starting with the Taylor polynomial of degree In
the estimate for the remainder
'
w=
1
+
-3T
3T
+
1
for sin
a together with
,
term derived on page 424, show
+ - + ( - 1)
+
"(25i)i
that
/?2*.0,/00
W here
|2n+l
#2«.0./'U)|
<
(2n
and use
this to
conclude that
.4
l
1
^0
Jo
-^
with an error of less than 10
13.
+ 2)!'
+
1703
^j^'
.946
=
3
Let
e
/(*)
(a)
(b)
=
x
-
1
,
x
#0
A
=
0.
Find the Taylor polynomial of degree n for / at
give an estimate for the remainder term R„,o,f.
Compute
/
f with an error of less than 10
0,
compute f {k) (0). and
.
JO
A
14.
Estimate
f°
/
Jo
2
exp(x )dx with an error of less than 10"
436
Infinite
Sequences and Infinite Series
15.
Prove that
if
x <
0,
then the remainder term R„q for e x
kl"
16.
Prove that
—1 < x <
if
+1
+
(n
satisfies
1)!
then the remainder term R„q for log(l
0,
-f a
)
satisfies
\
S
l*».°l
*17.
(a)
a+xKn+1 y
that if \g'(x)\ < M\x - a\" for
-a\" +] /(n
l)fbr\x-a\<8.
Show
-
\x
<
a\
then \g(x)
8,
-
g(a)\
<
+
M\x
(b)
x \n+l
Use part
to
(a)
show
that
=
0,
then
then g'{x)
=
fix) - Pn _ haJ ,(x).
lim g'(x)/(x
if
-a)"
x->a
gM-M =
x^a
(c)
(d)
18.
Show
that
if
=
g(x)
(
f(x)
X
-a) n+]
- Pn
.
a ,f(x),
Give an inductive proof of Theorem
Deduce Theorem
the remainder.
7
(The catch
is
that
it
will
Theorem
Lagrange's method for proving Taylor's
We
consider a fixed
f{x)
(*)
-
Theorem, with any form of
be necessary to assume one more
as a corollary of Taylor's
1
derivative than in the hypotheses for
19.
without using l'Hopital's Rule.
1,
number x and
+
f(t)
/'(*)(*
-
1
.)
Theorem used
the following device.
write
+
•
•
•
+
—
f
{n)
(t)
^-(x -
+
t)"
Sit)
IV.
for Sit)
= R nt ix). The
notation
right side as giving the value of
down
is
a tip-off that
some function
the fact the derivative of this function
function
of x and
whose value
t,
is
check that
always fix). To
is
make
we
are going to consider the
for a given
0, since
sure
it
/,
equals the constant
you understand the
if
git)
/ w (0,
= ^—ix-t)\
then
git)
= £^k(x-t) k -\-l) +
f
£-^(x-t)
kl
kl
{k)
(n_
U
,
,._,
_,)*-!
+
,
f k+]) ^ix-t)
^
it)
and then write
k
k
roles
20. Approximation by Polynomial Functions
Show
(a)
=
437
that
+
/'(*)
f'(0
+ QP-V
1!
no (x-t) + /w(o
(3)
+
(.y
-
ty
1!
+
-/"'<"„ _,
(n - 1)
+
r
(n+l),
+
.
/^£) u _„.
/?!
+ S'(0,
and notice
that everything collapses to
/(«+!)
5'(r)
=
-^— --(.v-/)".
(f )
Noting that
S(x)
5(a)
apply the Cauchy
(x
—
/)" +1
Mean
on [a,x]
to
=
=
*„.„ CO,
Value Theorem to the functions S and h{t)
=
obtain the Lagrange form of the remainder
(Lagrange actually handled
(b)
= 0,
/?„,*(*)
this
Similarly, apply the regular
part of the argument differently).
Mean
Value Theorem to S to obtain the
strange hybrid formula
Rn.a(x)
(n+l)
(t)
= f
(x-t) n (x-a).
n\
This
20.
is
Cauchy form of the remainder.
called the
Deduce the Cauchy and Lagrange forms of the remainder from the integral
form on page 423, using Problem 13-23. There will be the same catch as in
Problem
18.
know of only one situation where
The next problem is preparation for
I
21.
For every
number
Cauchy form of
the
coefficient"
a (a —
'a\
and we define
and
for
,
=
(1
•
as usual. If
alternates in sign for n
f{x)
1)
.
.
.
•
—n+
(a
we
n,
+ x) a
at
>
is
a.
used.
define the "binomial
l)
a
1
is
that eventuality.
and every natural number
a,
the remainder
a
a
is
not an integer, then
Show that
(
)
is
never
0,
the Taylor polynomial of degree n
Pn ,o(x ) = ^2\
v
,
)
^ and
Lagrange forms of the remainder arc the following:
tnat
me Cauchy and
438
Infinite
Sequences and Infinite Series
Cauchy form:
Rn.0(x)
—
a(a
—
=
1)
•
.
.
.
— n)
(a
•
—T^
~ tTO + O a -"~
-*(*
,
l
IV.
a{a-
-n)
(a
...
1)
+t) a
,
-x{\
_\
n\
U
= (H
-
t
\
f
t
1
X
~
1-1
+
+'t) a""' ((fx
VI +t
l
\x(\
l.V"
+
1/
//
(x
1
M
,
(0,x)or
(.v,0).
Lagrange form:
+
(«
"
.V^Ci + O"""
+
//
1)!
-1
Mn(0,.x)orU,0).
,
1/
Estimates for these remainder terms are rather
postponed
22.
(a)
Suppose
that
for all x
>
/
twice differentiable
is
while \f"(x)\
0,
— ^2
on
for all
Taylor polynomial to prove that for any x
<
\f'(x)\
(b)
Show
handle,
difficult to
and are
Problem 23-21.
to
that for
all
x
2
and that |/(x)| < Mo
x > 0. Use an appropriate
(0,
oo)
>
we have
h
-M + -Mi
h
2
>
for all h
>
0.
we have
L/"(.v)|
<2 V
/
MM
2
.
Hint: Consider the smallest value of the expression appearing in
(c)
If
/
is
twice differentiable
proaches
(d)
If lim
0.
23.
(a)
as
f(x)
f
The
(b)
exists
if
limit
and lim f"(x)
f"(a)
(a)
=
on the
is
exists,
bounded, and f(x) apas
x
then lim f"(x)
—
—
oo.
lim
fix)
=
f(a
then
+
h)
+ f(a-h)-2f(a)
lim
.
is
called the Schwarz second derivative of
/
at a. Hint:
Use the Taylor polynomial P2, a (x) with x = a + h and with x
Let fix) = x 2 for x > 0, and — x 2 for x < 0. Show that
exists,
(a).
11-34.)
exists,
right
lim
(c)
(0, oo),
x —> oo, then also f'(x) approaches
(Compare Problem
Prove that
on
/"
f(0
=
a
— h.
+ h) + fi0-h)-2fi0)
=
even though /"(()) does not.
Prove that
derivative
/ has a local maximum at
is < 0.
of / at a exists, then
if
it
a,
and the Schwarz second
439
20. Approximation by Polynomial Functions
(d)
Prove that
if
f'"(a) exists, then
— — = hm
/'"(a)
3
24.
fifl
,.
r
+ h)- f(a-h)-
2hf'{a)
.
5
h
h-+Q
3
Use the Taylor polynomial P\ M ,f, together with the remainder, to prove a
weak form of Theorem 2 of the Appendix to Chapter 1 1 If f" > 0, then
:
the graph of
/ always
above the tangent
lies
line
of /, except at the point
of contact.
*25.
Problem 18-43 presented a rather complicated proof that / = if f"—f =
and /(0) = /'(0) = 0. Give another proof, using Taylor's Theorem. (This
problem is really a preliminary skirmish before doing battle with the general
case in
Problem 26, and
meant
is
to convince
you that Taylor's Theorem
a good tool for tackling such problems, even though tricks work out
is
more
neatly for special cases.)
**26.
Consider a function / which
for certain
numbers
ao,
.
.
.
,
satisfies
fl„_i.
the differential equation
Several special cases have already received
we
detailed treatment, either in the text or in other problems; in particular,
have found
all
functions satisfying /'
0, or f" — f = 0. The
solutions for such equations,
=
/, or f" + f
=
Problem 1 8-42 enables us to find many
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
trick in
(necessarily sketchy)
(a)
remarks about the general solution.
Derive the following formula for
(
/
" +1)
(let
us agree that "a_i" will be
0):
7=0
(b)
Deduce
a formula for
The formula
in part (b)
is
f
{n+2
\
not going to be used;
vince you that a general formula for
other hand, as part
size
of
(c)
Let
f
{n+k)
/V
means
=
(c)
shows,
it is
f
{
" +k)
is
it
was inserted only
out of the question.
to con-
On
the
not very hard to obtain estimates on the
(x).
max(l,
|oo|,
l«„-il)-
Then
|a -_j
/
+
an _ia/| <
that
/'" +1
= J2bj f u \
l
»
,/=0
where
l
\bj
\
< IN 2
.
2N 2
;
this
)
440
Infinite
Sequences and Infinite Series
Show
that
n-\
z
" +2)
(
= J2 b 2 f
(j)
^
where b/\ < 4A
<
J
\
7=0
and,
more
f
generally,
= J2 bj k f u \
(„+k)
where
\bf\
< 2 k N k+l
.
7=0
(d)
Conclude from part
number
(n+k
\f
(e)
Now suppose
"
and conclude
Show
that
=
/i
any particular number x, there
y)
(0)
=
=
•
is
a
/ =
and
/2
0)
for all*.
= / (n_1).(0) =
•
•
M
|
"
"
(n+k+l)\
that
if /i
N k+]
n+* +1
JC
|
"
k
/'(())
2*+ 1 /y*+ 2
\2Nx\"
Show
0.
that
+M
'
(n+k+\)\
0.
both solutions of the
/? are
/
and
that, for
<M -2
\x)\
that /(0)
M
l/U)l
(f
(c)
M such that
(M)
=
X>/
<
(0) for
j
<
differential
equation
0)
,
-
n
then
1,
/i
= /2
.
In other words, the solutions of this differential equation are determined
by the
"initial
means
that
to obtain
we can
find all solutions
any given
x"
has n distinct roots
is
a solution,
/ (7) (0) for
once we can
conditions" (the values
set
<
j
find
<
n
—
enough
This
1).
solutions
of initial conditions. If the equation
-a n _\x n
-
x
an then any
o-i
=0
a
,
function of the form
and
+ •••+<?„,
/(0)
=ci
f'(0)
=a\C]
/(«-D(0)
=a
M- 1
i
\-a„cn
H
,
ci+...+an M
-1
cn
.
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
you can
easily
this last assertion. (It
check for n
=
2 or
3.)
is
a purely algebraic
These remarks are
fact,
also true
which
if
some
20. Approximation by Polynomial Functions
441
of the roots are multiple roots, and even
in the more general situation
considered in Chapter 27.
** 27.
(a)
Suppose
and that
that
a continuous function on [a,b] with
f(a) = f(b)
Schwarz second derivative of / at x is
(Problem 23). Show that
/ is constant on [a,b]. Hint: Suppose that
fix) > f{a) for some x in {a, b). Consider the
function
/
for all
is
x
in (a, b) the
=
g(x)
=
with g(a)
g(x)
>
Problem
its
(b)
(a)
=
g
23(c) (the
For sufficiendy small e >
have a maximum point y in {a,
Schwarz second derivative of
ordinary second derivative).
is a continuous function
on
at all points
of
- a )(b-x)
f(a).
will
If/
is
*28.
g(b)
g(a), so
f( x )-s(x
(a, b),
then
[a, b]
/
is
(x
we
will
- a)(b - x)
whose Schwarz second
have
Now
b).
is
use
simply
derivative
linear.
Let /(*) = x 4 sin 1 /x 2 for x
# 0, and /(0) = 0.
order 2 at 0, even though /"(0) does
not exist.
Show
/=
that
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)
(*)
=
fi a)
+
f (a )ix -a) + ^f {x - a f + j^),
Ra (x)
—=O
- a) 2
..
,
where lim
x^a (X
namely, m(a)
m
(b)
defined in
=
f"{a) for a
this
way
is
±
0,
—
and m(0)
=
0.
Notice that the function
not continuous.
Suppose that / is a differentiable function such
with mia) = 0. Use Problem 27 to show
that
that
(*)
holds for
/» = m
(a )
=
all
a
for
all a.
c
Now
suppose that
that for
all
(*) holds for all a, and that m is continuous
a the second derivative f"(a) exists and
equals m(a).
Prove
^chapter
mm
The
irrationality of e
attempt a more
but actually
is
TRANSCENDENTAL
e IS
was
and
b,
=
a/b
a
any integers a and
for
except for a
V2 does not seem to
of
how
as saying that x does not satisfy
for integers a
be such a
barely manages to be irrational
=
=
0.
=
0,
b,
with b
/
0.
This
is
the
any equation
bx - a
b
Viewed
in this light, the irrationality
terrible deficiency; rather,
— although V2
C1]X
it is
will
and prove
worse. Just
suggested by a slight
we
that the number e is not merely irrational,
number might be even worse than irrational
rewording of definitions. A number x is irrational if it is
difficult feat,
much
not possible to write x
same
so easy to prove that in this optional chapter
+ (70 =
is
it
appears that
V2 just
not the solution of an equation
0,
the solution of the equation
x
2
-2 =
of one higher degree. Problem 2-18 shows
0,
how
to
produce many
irrational
num-
bers x which satisfy higher-degree equations
a„x"
where the
a,
are integers not
tion of this sort
is
called an
have ever encountered
is
+ a n _ix"- +-"+flo =
1
all 0.
A number which
0,
satisfies
an "algebraic" equa-
algebraic number, and practically every number we
defined in terms of solutions of algebraic equations
(tt
and e are the great exceptions in our limited mathematical experience). All roots,
such as
are clearly algebraic numbers,
7
73,
v/2,
v
?.
and even complicated combinations,
like
P + VE+ V1 + V2+4/6
are algebraic (although
we
will
not try to prove
this).
Numbers which cannot be
obtained by the process of solving algebraic equations are called transcendental;
the
main
result
of this chapter states that
The proof that
e
is
transcendental
is
e
is
a
number of this anomalous
well within our grasp,
and was
sort.
theoretically
Chapter 20. Nevertheless, with the inclusion of this proof, we
can justifiably classify ourselves as something more than novices in the study of
higher mathematics; while many irrationality proofs depend only on elementary
possible even before
properties of numbers, the proof that a
442
number
is
transcendental usually involves
.
21.
some
really
443
Transcendental
e is
high-powered mathematics. Even the dates connected with the tran-
scendence of e are impressively recent
—
the
first
Hermite, dates from 1873. The proof that
due
to
due
to Hilbert.
Before tackling the proof itself,
depends on an idea used even
is
it
in the
proof that e
we
a good idea to
proof that
e
is
will give
map
transcendental,
is
a simplification,
is
out the strategy, which
irrational.
Two features
of the
expression
e= l +
were important
for the
Ij
+_+
proof that e
1
is
..
.
On
irrational:
+-+
number
the one hand, the
-
+
---
nl
1!
can be written as a fraction p/q with q <
other hand,
< R n < 3/(« + 1)! (so n\R„
that e can be
- + *„
+
n\ (so that n\
is
(p/q)
is
an
integer);
on
approximated particularly well by rational numbers.
Of course,
—
number x can be approximated arbitrarily closely by rational numbers
there is a rational number r with \x —r\ < e; the catch, however, is that
it
s >
may be
necessary to allow a very large denominator for
r, as
a fraction p/q within 3/(«
not the case: there
is
is
every
if
we
are assured that this
the
not an integer). These two facts show
large as \/e perhaps. For e
+
1)!
whose denominator q is at most n\ If you look carefully at the proof that e
you will see that only this fact about e is ever used. The number e is
by no means unique in this respect: generally speaking, the better a number can be
approximated by rational numbers, the worse it is (some evidence for this assertion
is presented in Problem 3). The proof that e is transcendental depends on a natural
2
e",
extension of this idea: not only e, but any finite number of powers e, e
can be simultaneously approximated especially well by rational numbers. In our
of
is
e,
.
irrational,
,
proof we
will
begin by assuming that e
a„e"
(*)
for
some
integers
certain integers
ciq, ...
M, M\,
.
•
a„. In
,
.
+
.
,
M
n
•
•
7
c"
how
,
^
«o
we
certain "small"
numbers
€\,
.
.
will
.
,
than find
€ n such that
=
e"
into the
0.
order to reach a contradiction
e
Just
.
algebraic, so that
is
+ a\e + ciq =
and
.
M
M
+ €2
=
M
2
=
M„ +
€„
M
small the e's must be will appear when these expressions are substituted
assumed equation (*). After multiplying through by
we obtain
M
[aoM
+ a\M\
H
\-a„M„]
+
[€\a\ H
\-
€„a n ]
=
0.
444
Infinite
Sequences and Infinite Series
The
first
term
in brackets
an
is
and we
integer,
We
necessarily be a nonzero integer.
\-€ n a n
lead to the desired contradiction
this will
number of absolute
As a
value
less
basic strategy this
<
\
— the
\\
sum of
a nonzero integer
and a
The
very reasonable and quite straightforward.
remarkable part of the proof will be the way that the M's and
order to read the proof you
will
it
to find e's so small that
than ^ cannot be zero!
all
is
choose the M's so that
manage
will also
\€\a\ H
will
need
will
know about
to
the
e's
are defined. In
gamma
function! (This
function was introduced in Problem 19-40.)
THEOREM
l
proof
e
transcendental.
is
Suppose there were integers
a„e"
(*)
Define numbers
M, M\,
ao,
.
.
.
.
.
.
+ a n -.ie"~ +
" x p - [(x
l
M
-
€\,
.
.
.
\)
.
.
.
-L
(Pex
xP-
- Y
[(x
(p
xp
-
'[U- -
unspecified
such that
•
0.
€„ as follows:
,
(x
1)
- n)]Pe~ x
J
I
(x -n)ye~x
dx,
—
1)-.
D!
•
(.v
-
-n)Ye~ x
dx.
€k
The
•
-0-
M,
k
•
0,
+a =
l
M„ and
,
^
a„, with «o
,
(p-iy.
number p
represents a prime
number* which we
Despite the forbidding aspect of these three expressions, with a
appear
much more
reasonable.
We
concentrate on
M
first.
will
little
choose
later.
work they
will
If the expression in
brackets,
[(x-1).... •(*-»)],
is
actually multiplied out,
we
obtain a polynomial
x"
*The term "prime number" was denned
in
+
---±nl
Problem 2-17. An important
fact
about prime numbers
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.
will
be used
in the proof,
although
The Suggested Reading mentions
it is
references for this theorem (which
factorization of an integer into primes
that there are infinitely
information
is
many
required.
primes
is
unique).
—the reader
is
We
will also
is
crucial in proving that the
use the result of Problem 2- 17(d),
asked to determine
at
precisely
which points
this
21.
When
with integer coefficients.
power
raised to the pth
Transcendental
e is
445
becomes an even
this
more complicated polynomial
x
Thus
np
+
•
•
±
•
{n\)
p
.
M can be written in the form
"P
,-00
1
a=0
where the
Ca
= ± («!)''.
and Cq
are certain integers,
k
x e
r
x
=
dx
But
k\
Jo
Thus
^
Now,
for
=
a
we
obtain the term
±(n\f
We
now
will
divisible
by
consider only primes
On
p.
Ca
which
is
Now
divisible
(P-D!
the other hand,
(p —
r—- =
consider M*.
We
M =e
(p
-
^
=±(n\) p
.
1)!
p > n; then this term
if a > 0, then
Caip
Therefore
p.
~
+a)\
1
;
by
iP
M
+a-
itself is
is
an integer which
+ a - 2)
\)(p
.
an integer which
.
.
is
•
not
by
p.
p,
not divisible
have
k
k
•
is
\
3
- dx
(p-iy
(x - ri)¥e-<*-Q
,
xP- l [(x-
1)
•
.
.
.
,
•
dx.
-I.
This can be transformed into an expression looking very
much
like
M
by the
substitution
u
du
The
limits
=
=
of integration are changed to
Mt=
f°°
(u+k) p
-l
[(u
L
+ k-
x
—
k
dx.
and
oo,
1)....-M-
o^
and
...- (u
+ k-n)]''e-"
'"'•
There is one very significant difference between this expression and that for M.
The term in brackets contains the factor u in the kth place. Thus the pth power
contains the factor u p This means that the entire expression
.
(u+k)"- \(u+k- \)-...-(u+k
[
-/?)]''
a
446
Infinite
;
Sequences and Infinite Series
is
a polynomial with integer coefficients,
of which has degree at least p.
every term
Thus
"/'
M
k
"P
/-00
1
= Y"
Da
np
/
- x+a
= Y" D„
e-" du
m—i
Da
where the
term
by
divisible
sum
in the
Substituting into
(*)
=
K
\-an
H
>
In addition to requiring that p
Mk
M and
flo
divisible
not divisible
M
+
n]
let
by
p,
n.
M we obtain
[a\€\
-\
n en ]
\-
us also stipulate that p
are not divisible by p, so
a$M
is
>
=0.
|«ol-
also not divisible
+ a\M\
by p. In particular
Yan
H
is
by
p. Since
M
n
a nonzero integer.
it is
In order to obtain a contradiction to the assumed equation
prove that e
This means
follows that
it
ciqM
is
n
k=\,
,
M
and multiplying by
M +a\M\
[a
is
divisible
is
p.
e
each
summation begins with a = 1
by p. Thus each Mk is an integer
clear that
it is
that both
+ <*)!
are certain integers. Notice that the
in this case every
Now
1
(p-l)\
a=l
which
is
(p-
transcendental,
(*),
and thereby
only necessary to show that
it is
h«„e„|
|«iei H
made as small as desired, by choosing p large enough; it is clearly sufficient
show that each \€k\ can be made as small as desired. This requires nothing more
than some simple estimates; for the remainder of the argument remember that n
is a certain fixed number (the degree of the assumed polynomial equation (*)). To
can be
to
begin with,
if 1
<
k
<
n,
then
-
/* \xP~ [(x
LV
l
l€kl
-
eK
i
'
,n
np
-l
\(x
-
1
A be
the
)•...• (x -n)\ p e
-x
-dx.
(p-iy.
^o
let
dx
V-dI
(p 1)!
-"/'
Now
x
-n)y\e~
'
l)-...-(.v
"
maximum
of
1**1
|(.v
<
—
1)
n nP~ l
e
.
.
n
np
~l
Jo
A p [™
e
/
n
np
~l
xr
dx
Jo
Ap
(P- D!
n
e npAp
e"(nA) p
~~
:
n)\ for
r
(p-l)\
e
—
AP
(p-iy.
e
(x
.
(p-l)\
=
(p-l)\
x
in [0, n].
Then
21.
A
But n and
making p
—
are fixed; thus (nA) p /(p
1)!
can be
e is
made
$A1
Transcendental
as small as desired
by
some philosophic
af-
sufficiently large. |
This proof,
proof that
like the
tt
irrational, deserves
is
—
argument seems quite "advanced" after all, we
use integrals, and integrals from
to oo at that. Actually, as many mathematicians have observed, integrals can be eliminated from the argument completely;
the only integrals essential to the proof are of the form
terthoughts.
At
the
first sight,
00
.k
/
JO
and these
for integral k,
Thus M,
where
for
Ca
can be replaced by
integrals
example, could have been defined
A:!
whenever they
occur.
initially as
are the coefficients of the polynomial
[(X-1).... •(*->!)]'.
If this
idea
that e
is
is
developed
one obtains a "completely elementary" proof
consistently,
transcendental, depending only on the fact that
e=l+
]
+
,
v.
Unfortunately, this "elementary" proof
one
—
{
!
+
v
is
this
actually a
This situation
n
irrational
is
is
by no means peculiar
is
proof except that
it
a case in point.
difficult
classical
a fact which
is
to this specific
theorem—
involves quite a few complicated functions.
of great
Our
than "advanced" ones.
You probably remember nothing
more advanced, but much more conceptual
transcendental,
of the
----
harder to understand than the original
"elementary" arguments are frequendy more
proof that
is
+
the whole structure of the proof must be hidden just to eliminate a few
integral signs!
about
v
proof,
There
which shows that
historical, as well as intrinsic, interest.
problems of Greek mathematics was
to construct,
is
it
One
with compass
whose area is that of a circle of radius 1 This
segment whose length is y/n, which can be
requires the construction of
accomplished if a line segment of length n 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
and straightedge
alone, a square
.
a line
until 1882.
In that year
Lindemann proved
that
n
is
transcendental; since the
length of any segment that can be constructed with straightedge and compass can
be written
that a line
in terms of +, •, — -=-, and y/~, and is therefore algebraic,
segment of length n cannot be constructed.
,
The proof that
which
is
it
is
transcendental requires a sizable
this
proves
amount of mathematics
too advanced to be reached in this book. Nevertheless, the proof
much more
difficult
than the proof that e
is
transcendental.
is
not
In fact, the proof
.
448
Infinite
Sequences and Infinite Series
for
it is
practically the
on
same
e
7r;
after
This
e.
last
statement should certainly
transcendental seems to depend so thoroughly
is
particular properties of e that
could ever be used for
proof for
as the
The proof that
surprise you.
almost inconceivable
it is
all,
what does
e have to
how any
do with
7r?
modifications
and
Just wait
see!
PROBLEMS
1.
(a)
Prove that
(b)
Prove that
a > is algebraic, then >J~a. is algebraic.
a is algebraic and r is rational, then a
if
if
+
r
and
ccr
are
algebraic.
Part
can actually be strengthened considerably: the sum, product,
(b)
and quotient of algebraic numbers is algebraic. This fact is too
for us to prove here, but some special cases can be examined:
2.
*3.
V2 + V3
and v2(l + v3) are algebraic, by actually finding
algebraic equations which they satisfy. (You will need equations of degree 4.)
Prove that
(a)
Let
be an algebraic number which
a.
satisfies
and
not rational.
=
an x
n
+ a n _ lX "- +
Suppose
that
•
no polynomial function of lower degree has
for any rational number p/q.
f(p/q) /
Show
this property.
Hint:
Use Prob-
3-7(b).
(b)
Now show that
(c)
Let
\f{p/q)\ > \/q" for all rational numbers p/q with q >
Hint: Write f(p/q) as a fraction over the common denominator q"
to
a
+ oo = 0,
1
that
that
lem
is
the polynomial equation
f{x)
M
—
sup{
prove that
|/'(jc)|
if
\x
:
p/q
—
<
a\
a rational
is
1
}.
.
0.
Use the Mean Value Theorem
number with
a ~ P/q\ > \/Mq n (It follows that
\a — p/q\ > c/q" for all rational p/q.)
for c
\
*4.
difficult
=
—
p/q\ <
min(l, \/M)
|a
1,
then
we have
Let
a
= 0.1 10001000000000000000001000
where the 1 's occur
a is transcendental.
of degree
in the n
!
place, for each n
(For each n,
show
that
a
.
Use Problem 3
is
to
prove that
not the root of an equation
n.)
Although Problem 4 mentions only one
specific transcendental
number,
it
should
many other numbers a which do
Such numbers were first considered
by Liouville (1 809 882), and the inequality in Problem 3 is often called Liouville's
inequality. None of the 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
be clear that one can
not
satisfy \a
—
easily construct infinitely
n
p/q\ > c/q
for
any c and
n.
1
of Cantor (1845
1918),
who showed,
without exhibiting a single transcendental
21.
e is
449
Transcendental
number, that most numbers are transcendental. The next two problems provide an
introduction to the ideas that allow us to
definition with
if its
which we must work
elements can be arranged
in a
is
make
sense of such statements.
the following:
A
A
set
is
called
The
basic
countable
sequence
a\, Q2, #3, #4, ....
The obvious example
is
N, the
set
more or
(in fact,
less
the Platonic ideal of) a countable set
of natural numbers; clearly the
of even natural numbers
set
also
is
countable:
2,4,6,8
It is
a
and
0)
little
is
more
surprising to learn that Z, the set of all integers (positive, negative
also countable, but seeing
0,
The
believing:
1,-1,2,-2,3,-3
next two problems, which outline the basic features of countable
examples
might think and
(2)
(a)
Show
x
(b)
is
in
that
B
}.
Show that
Show
A and B
if
are countable, then so
Use the same
Hint:
that the set of
same
If A\, A2, A3,
.
.
is
x
:
x
is
countable. (This
(m,n) of integers
U A2 U A3 U
of all
Prove that the
(f)
Prove that the
set
done part
you can do
set
in
A
or
is
is
really
to enlightenment.)
countable.
(This
is
triples
(/,
...
same
trick as in part
m, n) of integers
(l,m,n) can be described by a pair
Prove that the
numbers
{
for Z.
are each countable, prove that
.
(e)
(g)
AUB =
(b).)
also countable. (Again use the
(e),
is
worked
below indicates the path
pairs
all
as part
A\
is
trick that
the set of positive rational
practically the
(d)
are
to
quite startling, but the figure
(c)
sets,
show that (1) a lot more sets are countable than one
nevertheless, some sets are not countable.
really a series of
*5.
is
(/,
set
integer coefficients
of
is
all
countable. (A triple
m) and a number
of all ^-tuples (a\, ai
this,
is
(b).)
an )
is
n.)
countable.
(If you
have
using induction.)
roots of polynomial functions of degree n with
countable. (Part
(f )
shows that the
set
of
all
these
.
.
450
Infinite
.
Sequences and Infinite Series
polynomial functions can be arranged
most n
(h)
Now
Since so
set
of
there
use parts
and
(d)
many
to
(g)
prove that the
is
turn out to be countable,
sets
numbers between
real
all
no way of listing
all
ct\
0t 2
«3
(decimal notation
such a
list
is
these
set
of
all
algebraic
numbers
where
a„„
important to note that the
it is
and 1 is not countable. In other words,
real numbers in a sequence
= O.auanauau
= 0.a21fl22«23«24
= 0.a3ifl32fl33fl34
being used on the
.
•
•
•
.
.
.
To prove
right).
so,
suppose
that this
number
that this
is
were possible and consider the decimal
1^22^33^44
O.flj
=
5
if
ann
cannot possibly be
/
5
in the
Problems 5 and 6 can be
is
at
countable.
is
*6.
and each has
in a sequence,
roots.)
and
list,
=
a, m
6
=
ann
if
Show
5.
thus obtaining a contradiction.
summed up
The
as follows.
set
numbers were
countable. If the set of transcendental
set
of all real numbers would be countable, by Problem
set
of real numbers between
of algebraic numbers
also countable, then the
5(a),
and consequently the
and 1 would be countable. But this is false. Thus,
numbers is countable and the set of transcendental numbers
are more transcendental numbers than algebraic numbers"). The
the set of algebraic
is
not ("there
remaining two problems
between
*7.
sets
/ be
Let
sets
x—>a +
important
which are
a nondecreasing function on
lim fix) and lim fix) both
(a)
how
illustrate further
which are countable and
it
can be
to distinguish
not.
Recall (Problem 8-8) that
[0, 1].
exist.
x-*a~
For any e
[0, 1]
>
prove that there are only
with lim fix)
—
lim fix)
>
£.
finitely
Hint:
many numbers
There
are, in fact, at
a
in
most
[/(l)-/(6)]/e ofthenT"
(b)
Prove that the
Hint: If
number
set
lim fix)
x—>« +
n
of points
—
at
lim fix)
x—>a~
which /
>
0,
is
then
discontinuous
it
is
> \/n
This problem shows that a nondecreasing function
is
for
to
and
also
be differen liable
more
at
a
interesting.
set
A
some
natural
is
more
difficult
nondecreasing function can
of points which
is
not countable, but
true that nondecreasing functions are differentiable at
word
countable.
automatically con-
tinuous at most points. For differentiability the situation
to analyze
is
fail
it is still
most points
(in
a
Reference [38] of the Suggested
Reading gives a proof using the Rising Sun Lemma of Problem 8-20.
different sense of the
"most").
21.
who have done Problem
For those
is
possible to provide at least
e is
Transcendental
451
9 of the Appendix to Chapter 11,
one application
it
to differentiability of the
problem set: If / is convex, then / is
differentiable except at those points where its right-hand derivative /+'
is discontinuous; but the function /+' is increasing, so a convex function
ideas already developed in this
is
(a)
automatically differentiable except at a countable
Problem
-70 showed that
if
a continuous function /, then
/
1 1
the hypothesis of continuity
countable
set
and b x such
(a x ,b x ).
of values.
that ax
Then
every point
is
Hint:
is
interval (ax b x ).
,
How many
(c)
Prove the result of Problem
maximum
a constant function. Suppose
< x < bx and x
Deduce Problem
a local
is
the
point for
now
that
is a maximum point for / on
maximum value of / on some
such intervals are there?
11 -70(a) as a corollary.
1 1
of points.
dropped. Prove that / takes on only a
For each x choose rational numbers a x
every value f{x)
(b)
is
set
-70(b) similarly.
CHAPTER
INFINITE SEQUENCES
The
idea of an infinite sequence
is
so natural a concept that
One
dispense with a definition altogether.
it
is
tempting
to
frequently writes simply "an infinite
sequence
"
«1, «2, «3, «4, «5,
the three dots indicating that the
numbers
rigorous definition of an infinite sequence
important point about an
there
are
DEFINITION
a real
is
meant
An
number a n This
sort of
.
is
sequence
•
3
continue to the right "forever."
A
not hard to formulate, however; the
is
that for each natural
correspondence
is
precisely
number,
n,
what functions
to formalize.
infinite
From
infinite
a,
••
sequence of real numbers
is
a function whose domain
is
N.
the point of view of this definition, a sequence should be designated by a
single letter like a,
and
particular values
by
a(l),a(2),a(3),
but the subscript notation
a\, ai, #3,
is
almost always used instead, and the sequence
like {a n
}.
Thus
{n}, {(—1)"},
itself is
usually denoted by a symbol
and {\/n} denote the sequences
or, /3,
and y defined
by
ot n
=
n,
fin
=
(-1)",
Yn
=
-•
1
A sequence, like any function,
can be graphed (Figure
1)
rather unrevealing, since most of the function cannot be
•
{<*«}
{ft;
(a)
FIGURE
452
I
but the graph
fit
(b)
{Yn\
(c)
on the page.
is
usually
22.
«2
<X\
•
=
•
$5
=
^3
=
=
fh
£l
=
At
=
/?6
•
•
453
Sequences
0f5
0(4
<*3
Infinite
•
K2
K4
1
1
K5 Yi
IGURE
y\
2
A more convenient
the points a\,
ci2,
representation of a sequence
«3,
•
•
on a
•
is
obtained by simply labeling
This sort of picture shows where
line (Figure 2).
The sequence {or,,} "goes out to infinity," the sequence
"jumps back and forth between —1 and 1," and the sequence {y„} "converges
the sequence "is going."
{/3„}
to 0."
Of the
three phrases in quotation marks, the last
associated with sequences,
and
will
be defined precisely
is
the crucial concept
(the definition
is
illustrated
in Figure 3).
• •
l-e
FIGURE
DEFINITION
a/v+5
>
" N+
+e
A3
«2
symbols lim a„
=
\
l
a natural
{«„}
converges to
number
/ (in
such that for
Af
n
natural
all
,
if
> N, then
\a„
converge
if
it
approaches
converges to
/
for
1)
some
/,
and
>
there
n,
this definition,
to
every e
£.
or has the limit
/
if for
numbers
-l\ <
In addition to the terminology introduced in
that the sequence {a n }
to
<2l
3
A sequence
is
•
°/v+3
A
/.
diverge
if
we sometimes
sequence
it
}
is
say
said
does not converge.
To show that the sequence {yn converges to 0, it suffices to observe
> 0, there is a natural number N such that \/N < s. Then,
If s
{a n }
the following.
if
n
>
N
we
have
1
Yn
.
The
will
=- <
—
1
<
e,
£,
lim yjn
+
N
-
0|
<
reflection
(it
so \yn
£.
limit
probably seem reasonable
practically the
same
after
a
1
—
little
yjn
as y/n for large n), but a
—
just says that
vn +
1
is
mathematical proof might not be so
,
454
Infinite
Sequences and Infinite Series
obvious.
To
+
estimate \Jn
yjn
+
—
1
—
1
(s/n+
=
\jn
we can
\Jn
1
use an algebraic
-
y/n)(y/n
7
+
v "
+
n
—
1
+ +
y/n
1
+
1
+
1
+
trick:
yfn)
\fn
n
7
V"
1
7
v
It is
=
>Jx
7
v"
1
—
:
1
y n + — \Jn by applying the Mean Value Theorem
on the interval [//,/? + 1]. We obtain
also possible to estimate
to the function f(x)
+ +
/?
/'(*)
==
I
for
some x
in (n,
n
+
1)
1
<
2V^'
Either of these estimates
is left
may be
used to prove the above
limit; the detailed
to you, as a simple but valuable exercise.
The
limit
3n 3
4«
n-><x>
3
+ 7« 2 +l
- 8// + 63
3
4
should also seem reasonable, because the terms involving n
tant
when
n
is
remember
large. If you
the proof of
Theorem
are the most impor-
7-9 you will be able
—dividing top and bottom
proof
to guess the trick that translates this idea into a
by « 3
proof
yields
7
1
«
«3
o
3n
4/7
3
3
2
+7n + l
-8/7 + 63
"
4 "
Using
this expression, the
proof of the above
8
63
"
—
+
3
n
n
L
2
limit
is
not
difficult, especially if
uses the following facts:
If
lim a„
and lim b n both
exist,
then
n—+oo
n—>-oo
lim (a„
n—+ 00
+
lim (a„
•
bn )
=
b„)
=
lim a„
n—+oo
moreover,
if
lim £„
n—+00
^
0,
lim a„
«—+00
+
•
/i—+oo
then
lim
n—+00
/?„
<:/„//?„
^
=
for all
lim
h—+00
a,,/
lim b„
n—+ oo
lim b n
n—>oo
//
\
greater than
lim &„.
n—>oo
some
tV,
and
one
.
22.
Infinite
455
Sequences
we wanted to be utterly precise, the third statement would have to be even
more complicated. As it stands, we are considering the limit of the sequence
{c n = [a„/b n }, where the numbers c„ might not even be defined for certain n < N
This doesn't really matter we could define c„ any way we liked for such n—
because the limit of a sequence is not changed if we change the sequence at a
finite number of points.)
Although these facts are very useful, we will not bother stating them as a
(If
}
—
theorem
—you should have no
=
the definition of lim a n
proving these results for yourself, because
difficulty
so similar to previous definitions of limits, especially
/ is
n-><x>
=
lim f{x)
The
/.
between the
similarity
mere analogy; it is
function whose graph
actually closer than
second. If
/
is
the
(3,fl 3
)
=
=
lim a„
and lim fix)
I
is
1
.v—>oo
n—»oo
possible to define the first in terms of the
definitions of
(Figure 4) consists of line segments joining
a5 )
(5,
(l,«l)
FIGURE
4
the points in the graph of the sequence
f(x)
=
\
-
a n ){x
=
I
if
(a n+
-
{a,,},
n)
so that
+ a„
n
< x <
n
+
1,
then
lim a„
Conversely,
if
f
satisfies
lim fix)
This second observation
< a <
1.
is
and only
=
/,
and we
this
we
=
/.
fin), then lim a n
=
I.
For example, suppose that
=
0.
note that
lim a
-^
< 0, so that x logfl
Notice that we actually have
since log a
is
x
=
we can
lim e
x—>00
a negative
lim a"
n—>oo
<
an
=
Then
\
for if a
set
frequently very useful.
lim a"
To prove
lim fix)
if
=0
xlo z a
and
if \a\
= 0,
large in absolute value for large x.
<
1;
write
lim a"
n—»oo
= lim(-l)>r =
h—<-oo
0.
456
Infinite
Sequences and Infinite Series
The behavior
comes
of the logarithm function also shows that
arbitrarily large as n
becomes
=
lim a"
n—>-oo
and
sometimes even said that
it is
large.
This assertion
>
a
oo,
1
,
then a" be-
often written
1,
We
approaches oo.
{«"}
is
>
a
if
also write equations
like
—a"
lim
n—»oo
and say
— oo.
that {—a"} approaches
= — oo,
Notice, however, that
if
< —1, then
a
lim a"
n—>oo
does not
exist,
Despite
this
even in
this
extended sense.
connection with a familiar concept,
it is
more important
convergence in terms of the picture of a sequence as points on a
There
which
another connection between limits of functions and limits of sequences
is
is
related to
siderably
this
picture. This connection
more interesting, than
is
somewhat
less
one previously mentioned
the
of sequences in terms of limits of functions,
limits
to visualize
line (Figure 3).
obvious, but con-
— instead of defining
possible to reverse the pro-
it is
cedure.
THEOREM
1
Let
/ be
itself,
a function defined in an open interval containing
c,
except perhaps at c
with
lim
x—*c
Suppose that
Then
{a n }
is
/(*)=/.
a sequence such that
(1)
each a n
is
(2)
each a n
(3)
lim a„
^ c,
= c.
domain of
in the
/,
the sequence {f(a n )} satisfies
lim f(a„)
=
I.
n—>O0
Conversely,
if this is
then lim f(x)
PROOF
Suppose
first
=
true for every sequence {a n } satisfying the above conditions,
I.
that lim f(x)
=
I.
Then
<
\x
—
c\
for every s
>
there
is
a 8
>
such
that,
x—yc
for all x,
if
If the
ber
N
sequence {a n }
such
satisfies
<
lim a„
8,
=
then \f(x)
c,
—
1\
then (Figure
n-+oc
that,
if
By our choice of
8, this
n
means
> N,
then
\a„
—
c\
that
\f(a„)-l\ <e,
<
8.
<
3)
e.
there
is
a natural
num-
22.
Infinite
457
Sequences
showing that
=
lim f(a„)
=
Suppose, conversely, that lim / (a„ )
I.
for every
/
sequence
n—X30
c.
8
If lim
x—>c
>
f(x)
there
is
=
were
/
would be some
not true, there
{a n
=
n—»oo
such that for every
>
s
with lim a„
}
an x with
<
— c\ <
|.v
In particular, for each n there
<
—
\x„
\f(x) —l\
but
8
>
e.
would be a number x n such
c\
< —
but
|/(jc„)
—
/|
that
>
s.
n
Now
the sequence
clearly converges to c but, since \f(x n
{.x,,}
the sequence {/(-f„)} does not converge to
—
lim f(x)
must be
I
Theorem
1
many examples
{b,,}
—»
-•
02
Cl\
FIGURE
#3
»
have some
>
e for
all n,
This contradicts the hypothesis, so
/.
of convergent sequences. For example, the
I
a„
=
b„
=cos
sin
and
(13
+
(sin
1
+
(-1)"
cos(sin(l)), respectively. It
important, however,
is
guaranteeing convergence of sequences which are not obvi-
criteria
ously of this sort. There
is
which
other
the basis for
is
l\
defined by
clearly converge to sin(13)
to
—
true. |
provides
sequences {a n } and
)
all
one important criterion which
results.
This criterion
is
is
very easy to prove, but
stated in terms of concepts
««»» »
defined for functions, which therefore apply also to sequences: a sequence {«„}
^4
increasing
5
if
a n+
bounded above
> an
\
if
there
for
is
a
nondecreasing if a„ + > a n
number M such that a„ < M for all n
all
n,
\
;
for
all
n,
is
and
there are sim-
sequences which are decreasing, nonincreasing, and bounded
ilar definitions for
below.
THEOREM
2
If {a n
ment
PROOF
The
}
is
is
set
nondecreasing and bounded above, then
true
A
if [a n
is
Then
if
bound
{a,,}
converges
(a similar state-
nonincreasing and bounded below).
numbers
consisting of all
a least upper
there
]
is
a.
We
some a^ satisfying
n > N we have
a„
is,
by assumption, bounded above, so A has
claim that lim a„
n—>oo
a — a^ <
a>i
>
tf/v,
This proves that lim a„
=
a. |
so
£, since
a —
a„
=
a
(Figure
a
is
the least upper
< a —
«/v
<
s.
5).
>
0,
bound of
A.
In fact,
if
e
458
Infinite
Sequences and Infinite Series
The
{a n
is
bounded above
Upon
diverges.
first
consideration,
Theorem
clearly essential in
is
not bounded above, then (whether or not
is
}
hypothesis that {a n }
{a,,} is
nondecreasing) {a„) clearly
might appear that there should be
it
trouble deciding whether or not a given nondecreasing sequence {«„}
above, and consequendy whether or not
sequences
they converge
hardly a
is
trivial
we
+l
\
Although Theorem 2
useful than might
you might
matter. For the present,
+
2
at
first,
j,
+i +
1
+
i
is
try to decide
bounded above:
••••
5<
very special class of sequences,
because
it
is
more
always possible to extract from an
it is
another sequence which
{a,,}
To be
nondecreasing.
+
treats only a
appear
arbitrary sequence
1
little
bounded
deciding whether or not
shall see,
whether or not the following (obviously increasing) sequence
1,
is
converges. In the next chapter such
{a,,}
very naturally and, as
will arise
2: if
is
either nonincreasing or else
subsequence of the sequence
precise, let us define a
{a,,}
be a sequence of the form
to
jraph of a
"71
where the
rij
are natural
89
"I
Then
10
"
'
'
"
'
<
< n3
«2
'
'
"
every sequence contains a subsequence which
11
nonincreasing.
It is
possible to
although the proof
sertion,
figure
"3
'
numbers with
2 and 6 are peak points
1234 567
"71T
)
|
is
become
very short
either nondecreasing or
is
quite befuddled trying to prove this asif
you think of the
right idea;
it is
worth
recording as a lemma.
6
LEMMA
Any sequence
{a„} contains
a subsequence which
is
either nondecreasing or non-
increasing.
PROOF
number
Call a natural
>
in
n (Figure
Case
<
«3
•
•
if
am
<
if
n\
a„ for
all
6).
The
1.
•
n a "peak point" of the sequence {«„}
sequence has infinitely
are the peak
many peak
points, then a,n
>
a„ 2
points.
>
a ny
In this case,
>
•
•
•
,
so {a„
k
}
<
n-±
<
the desired
is
(nonincreasing) subsequence.
Case
than
a,,-,
all
>
than
all
sequence has only finitely
.
Since
112
peak points)
many peak points. In
this case, let
\
is
not a peak point, there
COROLLARY (THE
1
be greater
.
3
that our original sequence
{a,,}
is
|
bounded, we can pick up an
extra corollary along the way.
BOLZANO-WEIERSTRASS THEOREM)
\
is
obtain the desired (nondecreasing) subsequence.
we assume
n
some nj > n\ such that
is not a peak point (it is greater than n\, and hence greater
there is some ut, > ni such that a„ > a„ 2 Continuing in this
peak points. Since n
a, n
way we
If
The
2.
a cry bounded sequence has a convergent subsequence.
.
22.
Without some additional assumptions
this
is
we can
as far as
459
Sequences
Infinite
go:
it is
easy to
construct sequences having many, evenly infinitely many, subsequences converg-
Problem 3). There is a reasonable assumption to
a necessary and sufficient condition for convergence of any se-
numbers
ing to different
add, which yields
quence. Although
many
investigations,
If a
the
and
N
condition plays a fundamental role in
this
for this reason alone
worth
it is
To be
such that
\a n
—l\<
-
final inequality, \a„
be used
A
DEFINITION
—
lim a n
if
I
e/2 for n > N;
<
ar
,
-
\a„
+
l\
— am <
sequence
N
{a,,}
is
a
now
\I
if
-a,
is
all
A sequence
{a,,}
and
both n >
N
and
<
=
E.
condition
all
N, then
2
condition) which
is
limit
/,
can
clearly necessary
converges
{a n
}
is
bounded.
find that there
if
and only
{a n }
is
some
is
means
\am
:
m > N}
whole sequence
is
there
is
a natural
that
a„\
it is
=
e.
0.)
also sufficient to ensure converis
very
little left
to
do
is
it is
a
Cauchy sequence.
Cauchy sequence
a
one
if it
=
1
converges.
The proof of
showing that every Cauchy
tricky feature:
in the definition
of a Cauchy sequence
such that
Qn
|
<
for
1
m, n >
N
that
\a m
Thus
if
If we take e
/V
\Q m
In particular, this
\
—
\a m
is
— am <
\a n
our preliminary work, there
the converse assertion contains only
sequence
>
for every e
if
this.
We have already shown that
we
m >
n,
lim
usually written
The beauty of the Cauchy
order to prove
m
m, n > N, then
gence of a sequence. After
PROOF
Cauchy
Cauchy sequence
such that, for
(This condition
3
close to
convergence of a sequence.
if
theorem
all
two such individual terms should be
for some /, then for any e >
there is
which eliminates mention of the
s,
\
to formulate a condition (the
number
in
now
stating
2
for
does simplify
the difference of any
precise,
\a„
This
it
more advanced
sequence converges, so that the individual terms are eventually
same number, then
very small.
an
condition will not be crucial for our work,
this
Moreover,
proofs.
(see
-a N +\\ <
bounded;
bounded.
1
for all
m >
N.
since there are only finitely
many
other
a,'s the
.
,
460
Infinite
Sequences and Infinite Series
The
Lemma
corollary to the
some subsequence of
thus implies that
{a n }
con-
verges.
Only one point remains, whose proof will be left to you: if a subsequence of a
Cauchy sequence converges, then the Cauchy sequence itself converges. |
PROBLEMS
Verify each of the following limits.
1.
(i)
lim-^- =
n->oo
(ii)
lim
„_>oo
111
n
+
^- =
ni
+ —
v n2
lim
n—>oo
0.
_|_4
+
Z/n 2
—=
lim
n—>-oo n"
1
-
Ifn 2
a
>
=
Hint:
You should
at least
be able to prove
0.
0.
lim
«—>oo
'{fa
=
1
VI
lim
'{fn
=
1
Vll
lim \fn 2
+n =
lim Vfl"
+ b" =
(viii)
+ 1=0.
\Jn
1
that lim
IV
l.
1
0.
1.
max(a,
a, b
b),
>
0.
n—>-oo
oiifi)
(ix)
=
lim
The
Hint:
how
where a(n)
0,
(x)
small a(n) must be.
lim
j-
nP +i
n-+oo
=
Find the following
lim
+
n
+
1
lim n
—
n
1
>/«
2"
in
1
.
n
n-+oo
+
limits.
fl
(i)
-.
p
lim
n^-oo 2" +1
+
+
+ aV" + b.
(-1)"
(-1)" +1
'
(-l) n v^sin(/i")
(iv)
lim
77^-00
(v)
lim
/?->-oo
(vi)
//
-f
1
.
a"
+
lim nc'\
/?"
\c\
<
the
each prime
fact that
n
?
is
1.
is
number of primes which
> 2
divide n.
gives a very simple estimate of
—
.
.
22.
Infinite
461
Sequences
2
(vii)
—-.
lim
n—»oo n\
3.
(a)
What can be
said
about the sequence
{a n } if
converges and each a„
it
is
an integer?
(b)
Find
convergent subsequences of the sequence
all
1
,
— 1,
1
many, although there are only two
.... (There are infinitely
,
— 1,
1
,
— 1,
which
limits
such subsequences can have.)
(c)
Find
all
3, 4,
1,
convergent subsequences of the sequence
2, 3, 4, 5, ...
(There are
.
1,
many
infinitely
2,
1,
2, 3,
1,
2,
which such
limits
subsequences can have.)
(d)
Consider the sequence
11212312341
2
3
'
3
'
'
4
For which numbers a
4.
(a)
Prove that
if
4
'
4
'
5
'
5
'
5
'
5
'
'
6
'
there a subsequence converging to
is
Cauchy sequence
a subsequence of a
a?
converges, then so
does the original Cauchy sequence.
5.
(b)
Prove that any subsequence of a convergent sequence converges.
(a)
Prove that
(b)
Prove that the sequence
< a <
if
yjlsfl,
y/2,
< sjla <
then a
2,
2.
V2V2V2,
converges.
(c)
Find the
limit.
by Theorem
6.
<
Let
a\
Hint: Notice that
—
/,
then lim
n—>oo
<b\ and
define
\
—
I
—
s/an bn
,
b„ +
,
i
—+—
an
=
-
bn
(a)
Prove that the sequences {a n } and {b n ) each converge.
(b)
Prove that they have the same
In Problem 2-16
we saw
that
limit.
any rational approximation k/ 1
replaced by a better approximation (k
with k
— =
I
1
,
we
Prove that
this
+
21) /{k
+
/).
obtain
I
1
'
2'
5
is
given recursively by
i
(a)
y 1a n = V2/,
1
an+
7.
lim a„
n—>oo
if
sequence
a\
=
1,
a n+
\
=
1
+
—
1
1
+ a„
.
to
V2
can be
In particular, starting
— —
462
Infinite
Sequences and Infinite Series
(b)
Prove that
—
lim a„
n—xx
This gives the so-called continued
V2.
fraction
expansion
/
v
2= +
1
.
2
+
2
+
...
Hint: Consider separately the subsequences {ai n }
(c)
Prove that for any natural numbers a and
+
V a2
=
b
many
9.
=
Identify the function /(.v)
times in
lim
{fl2«+l
b,
a H
.
2a
8.
and
lim
(
+
la
H
(cosrclTTjt)*-*). (It
has been mentioned
book.)
this
Many
impressive looking limits can be evaluated easily (especially by the
person
who makes them
With
disguise.
ing: the
list
remark
this
because they are really lower or upper sums in
up),
as hint, evaluate
each of the following. (Warn-
contains one red herring which can be evaluated by elementary
considerations.)
l/e
..
+
+
lfe 2
---
+
Ve"
lim
.
lim
lim
iii)
«->-oo
1
(
/
lim
iv)
«
+
y n
+
+
...
1
1
H
V n
+
(n
hm
vi)
10.
Although
\(n
In
+
+
—7T H
\)
2
n
n-V5b
1
1
—z2
lim
—
-,
(2ny
n
] )
2
n
2)
lim
dissatisfying,
2
(n
'
++
n
-j-— +
limits like
+
(n
n
'±fn
+
n) 2
n
2
and lim a" can be evaluated using
the behavior of the logarithm
vaguely
—
and exponential
functions, this
Some
out using the exponential function.
and, for part
(1
approach
1
is
of the standard "elementary" ar-
+/?)"
>
1
lor//
+////.
>
tools are inequalities
0;
(e),
+h)" >
about
because integral roots and powers can be defined with-
guments for such limits are outlined here; the basic
derived from the binomial theorem, notably
(1
facts
+/;//
+
—-
-h
2
>
—^--/*
2
,
for//
>
0.
.
.
,
.
22.
(a)
Prove that
Km
=
11
a
oo
>
a
if
by
1,
=1+
a
setting
n—>00
(b)
Prove that
Km
=
a"
< a <
if
Infinite
re,
463
Sequences
where h >
0.
1
n—*oo
(c)
Prove that lim %/a
=
(d)
Prove that lim y/a
=
1
—
1
>
a
if
1
by
1,
< a <
if
setting
=
'±J~a
1+re and estimating
re.
1
/;—>-oo
11.
(e)
Prove that lim
(a)
Prove that a convergent sequence
(b)
Suppose
all
12.
(a)
lim a„
that
numbers
"-Jn
<
+
some
a„
>
Prove that the
0.
set
of
0.
It
maximum member.
+
log(«
1)
—
log/z
<
n
If
show
= +
l
This number,
is
not even
Suppose
/(l)
Now
is
=
y
it
H
{
a
2
,
J
decreasing,
is
lim
I
+
1
•
•
/
is
•
+ ••• + /(/!- 1)<
/ =
on
increasing
that each a n
log
>
log n
H
[1, oo).
and show
<
Show
f f(x)dx < /(2)
quite refractory;
that
+
•
•
•
+ /(«).
that
<
nl
1)" +1
+
follows that
Vn!
,.
lim
n->-oo
This result shows that V/i!
/;
is
ratio of these
two quantities
conclude that
n\
estimate for n\
is
is
1
= e
approximately n/e,
is
close to
1
in the sense that the
for large n
.
But we cannot
close to (n/e)" in this sense; in fact, this
is
false.
An
very desirable, even for concrete computations, because
cannot be calculated
easily
The
standard
right information will
be found
even with logarithm
difficult) theorem which provides the
Problem 27-19.
(and
in
and
number
e n-\
re!
/;
;?
(n
it
log
1
known as Euler's number, has proved to be
known whether y is rational.
that
choose
l
l
+
r
that the sequence {«„}
follows that there
(b)
always bounded.
1
1
(a)
is
that
Prove that
fl«
13.
and
0,
n—>oo
a„ actually has a
n
(b)
=
tables.
.
.
464
Infinite
Sequences and Infinite Series
14.
(a)
Show that
the tangent line to the graph of
/
at (x\,
f(x\)) intersects the
horizontal axis at (X2, 0), where
X2= X\ —
f'(xi)
may
This intersection point
be regarded as a rough approximation
to
where the graph of / intersects the horizontal axis. If we now
start at X2 and repeat the process to get #3, then use X3 to get X4, etc.,
we have a sequence {x„ defined inductively by
the point
}
f(Xn)
x n+l
Xn
f'(Xn)
Figure 7 suggests that
this
is
we
of this problem
FIGURE
method works
7
facts
(b)
Suppose
(c)
Let
(Figures 8
8k
that /,'
>
=
>
xo
Xk
—
some conditions under which Newton's
and 9 show two cases where it doesn't). A few
will establish
about convexity
that x\
converge to a number c with f(c) — 0;
In the remainder
will
{x,,}
called Newton's method for finding a zero of /.
may
/" >
and
0,
>
>
Then
that
useful; see
we choose
Chapter
1 1
,
Appendix.
0.
Show
[c, x\].
Show
x\ with f(x\)
>
c.
xt,
c.
be found
f(Xk)
8k
for
some
£* in (c, Xk).
Show
that
<$*+i =
/'(&)
X\ X3
X5
X 2 X4
Conclude
f'(Xk)
that
f(x k )
f"{m)(xk-$k)
FIGURE
8
for
some
rjk
in (c, Xk),
and then
(*)
s k +i
that
<
8k
l
.
f'iXk)
(d)
Let
that
(e)
m =
=
on [c,A|] and
Newton's method works if x\ — c
f'(c)
inf /'
x3
x4
which
is
=
=
=
=
.
1
RE
9
x2
— A?
1.4
1.4142136
1.4142136,
already correct to 7 decimals! Notice that the
inequality
(*)
when M/in <
1
on
1.4142857
decimals at least doubled each time.
(
= sup/"
< m/M
\
x2
I
M
What is the formula for xn+ when f(x) =
If we take A = 2 and x\ = 1.4 we get
X]
I
let
This
is
essentially
number of correct
guaranteed by the
22.
15.
Use Newton's method
f{x)
=
—
(iii)
f(x)
= x +x -
(iv)
/U)=x
fix)
(i)
(ii)
16.
cos*
3
Prove that
3
near
0.
near
0.
on
on
+l
=
lim a n
if
x
1
-3.r 2
465
to estimate the zeros of the following functions.
— cos
— x2
tanx
Sequences
Infinite
[0, 1].
[0,1].
then
/,
n—»co
—
lim
=
w—>-oo
problem
Problem 13-40.
Hint: This
17.
(a)
Prove that
is
very similar to
lim a n+
n—>oo
if
\
/.
n
—
=
a„
(in fact,
/,
then
actually a special case of)
it is
lim a n /n
h—>oo
=
See the
Hint:
/.
previous problem.
(b)
Suppose that /
lim f(x)/x
that
A
+
[n,n
*18.
Suppose
lim
x
(a)
—
>
20.
that
=
{a,,} is
is
1
for a
,
— f(x) =
I.
be the inf and sup of
not in
Prove
/ on
sort of
I.
argument
Prove that
that
works
in
>
0.
all
in [0, 1].
Prove
also in [0, 1].
{a,,}
of points
all
in (0, 1)
such that lim a n
(0, 1).
/
that
is
continuous and that the sequence
x,
converges to
same
=
lim a n+ \/a n
n—>co
a convergent sequence of points
Find a convergent sequence
Suppose
b„
1)
except using multiplication instead of addition, together with
16,
Suppose
is
and
each n and that
for
Hint: This requires the
I.
that lim a n
n—>oo
(b)
Hint: Let a„
I.
+
1].
the fact that lim \fa
n—>oo
19.
=
lim f(x
x—*-cx>
CO
that a„
'{fa^
Problem
—
continuous and
is
/.
fix), fif(x)), fififix))),
Prove that
/
is
...
a "fixed point" for /,
i.e.,
/(/)
=
/.
Hint:
Two
special cases have occurred already.
21.
(a)
< fix) < 1 for all x in
Suppose that / is continuous on [0, 1] and that
[0, 1]. Problem 7-11 shows that / has a fixed point (in the terminology
of Problem 20). If / is increasing, a much stronger statement can be made:
For any x in [0,1], the sequence
x, fix),
fifix)),...
is necessarily a fixed point, by Problem 20). Prove this
by examining the behavior of the sequence for / (.v > x and
fix) < x, or by looking at Figure 10. A diagram of this sort is used
in Littlewood's Mathematician's Miscetlany to preach the value of drawing
."
pictures: "For the professional the only proof needed is [this Figure]
has a limit (which
assertion,
riiil'RE 10
)
466
Infinite
Sequences and Infinite Series
*(b)
/ and g are two continuous functions on [0, I], with <
< g(x) <
for all x in [0, 1], which satisfy f o g =
f(x) < 1 and
o
g /. Suppose, moreover, that / is increasing. Show that / and g have
a common fixed point; in other words, there is a number / such that
Suppose
that
1
= =
/(/)
g(l). Hint:
/
Begin by choosing a fixed point for
g.
For a long time mathematicians amused themselves by asking whether the
conclusion of part
holds without the assumption that
(b)
two independent announcements
cal Society,
a pretty
Volume
Number
14,
problem
silly
/
is
increasing, but
American Mathematicounterexamples, so it was probably
in the Notices of the
2 give
along.
all
Problem 20 is really much more valuable than Problem 20 might
suggest, and some of the most important "fixed point theorems" depend upon
looking at sequences of the form x, f(x), f(f(x)), .... A special, but representa-
The
tive,
is
trick in
theorem
case of one such
is
Problem 23
treated in
(for
which the next problem
preparation).
22.
(a)
Use Problem 2-5
c
m
show
to
that
if
c
^
1,
then
+c m+\ + ... +c n =
1
(b)
Suppose
that
<
\c\
1.
Prove that
lim
c
m
m,n—* oo
(c)
Suppose
that
Prove that
23.
Suppose
that
/
c
<
1
.
Prove that
(b)
Prove that
(c)
By
/
/
—
•
•
+ c n = 0.
\x„
— xn+ i\ <
where
c",
<
c
<
1.
R such
c\x
is
—
that
for all
y\,
x and
y,
called a contraction.)
continuous.
is
has at most one fixed point.
considering the sequence
for
any
f(x), /(/(*))
/ does have a fixed point. (This
known as the "contraction lemma.")
x, prove that
general setting,
(a)
<
f(y)\
x,
24.
•
a sequence with
(Such a function
(a)
+
Cauchy sequence.
a
a function on
is
\f(x)
(*)
where
{x,,} is
{x,,} is
-C
Prove that
is
/
if
is
differentiable
and
\f'\
<
1,
then
/
result, in a
more
has at most one
fixed point.
(b)
Prove that
(c)
Give an example
to
25.
if
\f'(x)\
ensure that
This problem
is
/
lo
<c<\
show
for
all .v,
then
/
has a fixed point.
not sufficient
problem.
Let b„ be a
=
lim b„ exists
1
s
has a fixed point.
a sort of converse to the previous
sequence defined by
<
i-
that the hypothesis |/'0O|
b\
=
a, b„ +
\
=
f(h„).
Prove that
if
h
22.
and
/'
continuous at
is
=
have b n
b for some
an interval around
/ on
consider
26.
Hint: If \f'{b)\
n).
and b n
b,
be
will
the interval [b,
b
=
(a)
Note
lim b„ exist?
n—<-oo
If
b
exists,
>
Suppose
that
—
(c)
e
<
we
< a <
e
more
is
if
b
b n+
for
all
n.
x
in
Now
is
<
/y
for
< a <
that
that {b„}
\
=
a
",
when does
by Problem 20.
form y
(and also that b
1
1
enough
some
e
y.
Describe
l/e
increasing
.
and
also b„
<
e.
e).
difficult.
b
then e~
exists,
< b <
l
e.
Then show
]/e
.
From now on we
Show
then a
in the
Show
.
— a,
=b
define b\
exists,
Using Problem 25, show that
e
1,
don't already
the symbol
and conclude
^e
x
exists
analysis for a
that e~
(d)
b
if
1/v
y
<a <
1
This proves that b
The
if
then a can be written
the graph of g(y)
(b)
that
>
467
b,,].
In other words,
sense.
we
then |/'(jc)I >
(provided that
1
in this interval for large
This problem investigates for which a
makes
<
then \f'(b)\
b,
Sequences
Infinite
will
<
suppose that e~ e
a
<
1.
that the function
ax
=
fix)
logjc
is
(e)
decreasing on the interval
Using part
/
(f)
(g)
27.
=
(e),
lim «2»+l exists and that a°
n—KX
Using part
Finally,
(e)
show
show
again,
that
that lim «2„ + 2
Let {x„} be a sequence which
yn
(a)
(0, 1).
number such that a b = b. Show that a < b < 1.
show that if < x < b, then jc < a a < b. Conclude that
Let b be the unique
=
is
=
I
=
= b.
b, so that
{y,,}
lim b n
bounded, and
sup{x„,x„ +I
Prove that the sequence
/.
,.v„ + 2
converges.
=
b.
let
}.
The
limit
lim yn
is
denoted by
II—>00
lim xn or lim sup x n
of the sequence
(b)
,
and
called the limit
{x,,}.
Find lim x„ for each of the following:
n—>oo
(i)
xn
=
-.
n
(ii)
x„
=
(-l)"I.
n
superior, or upper limit,
468
Infinite
Sequences and Infinite Series
*„
(iii)
1
=
(-!)"
+-
1
n
x„
(ivj
=
s/n.
Define lim x„ (or liminf xn ) and prove that
(c)
<
lim x„
lim xn
—*oo
.
II
(d)
Prove that lim x„
=
case lim x„
—
lim x n
if
lim x n
=
n—>oo
Recall the definition, in Problem 8-18, of lim
(e)
Prove that
the
if
lim x n and that in this
lim x„.
n—*oo
n-+oo
and only
exists if
numbers
x„ are distinct, then
A
for a
lim x„
bounded
=
A.
set
lim A, where
II—>oo
A =
28.
{*„
n in N}.
:
In the
Appendix
on an
interval. If
Chapter 8 we defined uniform continuity of a function
f(x) is defined only for rational x, this concept still makes
sense:
we
/
there
and
say that
some
is
\x
to
5
- y\ <
>
8,
is
uniformly continuous on an interval
such that,
then \f(x)
if
-
if
for every e
x and y are rational numbers
<
f(y)\
in the interval
e.
Let x be any (rational or irrational) point in the interval, and
(a)
a sequence of
rational
>
points in the interval such that lim xn
let {x n
=
}
be
Show
x.
that the sequence {f(x n )} converges.
(b)
Prove that the limit of the sequence {f(x n )} doesn't depend on the choice
of the sequence {x„}.
We
whole
(c)
denote
will
this limit
by f(x), so that /
is
an extension of /
to the
interval.
/
Prove that the extended function
is
uniformly continuous on the
inter-
val.
29.
Let a
>
x
f(x) = a as defined in the usual elementary
This problem shows directly that / can be extended to a
and for
0,
algebraic way.
rational
/ on
continuous function
A'
let
,
the
whole
show
that a
line.
Problem 28 provides the necessary
machinery.
(a)
(b)
For rational x
Using Problem
rational
(c)
y,
10,
numbers x
show
close
Using the equation a
interval
(d)
<
Show
/
is
x
—
x
<
any
that for
enough
ay
=
<
1
and
and a x > a y
we have
>
\a
x
—
for a
1|
<
<
1.
s for
{a
x ~ y --
1
),
prove that on any closed
in the sense
of Problem 28.
/ of Problem 28 is increasing for a >
satisfies f(x + y) = f(.\)f(y).
that the extended function
for a
e
1
to 0.
ay
uniformly continuous,
and decreasing
>
a y for a
1
.
I
22.
"30.
The
Theorem
Bolzano-Weierstrass
differently than in the text
A
points.
point x
point a in
(a)
Find
A
all
(i)
with
-
a limit point of the set
|jc
— a\ <
also proved, quite
x
s but
/
A
>
for every s
if
there
a
is
a.
of the following
sets.
N
n in
:
and
469
Sequences
the classical statement uses the notion of limit
is
limit points
I
—
usually stated,
is
Infinite
.
I"
—
1+1
(-!>"
iii)
m
and
n
:
m
;/
:
N
in
N
n in
n
(b)
(iv)
Z
(v)
Q
is a limit point of A if and only if for every e >
— a\ < e.
many points a of A satisfying
Prove that lim A is the largest limit point of A and lim A the
Prove that *
infinitely
(c)
an
usual form of the Bolzano-Weierstrass
infinite set
point of
a limit point of A. Prove
is
Using the form already proved
numbers
(a)
this in
x\, X2, *3,
•
•
•
one must contain
A
is
then some
[a, b],
A
is
infinite,
there
in A.
If [a, b]
many
infinitely
is
divided into two
A
points of
Use the Bolzano-Weierstrass Theorem to prove that if / is continuous
on [a, b], then / is bounded above on [a, b]. Hint: If / is not bounded
above, then there are points x n in [a b] with
,
(b)
states that if
two ways:
in the text. Hint: Since
Using the Nested Intervals Theorem. Hint:
intervals, at least
31.
Theorem
of numbers contained in a closed interval
[a, b]
are distinct
(e)
smallest.
,
The
(d)
there are
|jc
Theorem
Also use the Bolzano-Weierstrass
uous on
[a, b],
/
then
/ (x„ >
to
)
prove that
uniformly continuous on
is
n
if
/
[a, b] (see
is
contin-
Chapter
8,
Appendix).
**32.
(a)
Let {a n } be the sequence
112123123412
2
'
3
'
Suppose that
j
<
3
4
'
'
4
'
4
< a < b <
n such that aj
is
5
•
1.
'
5
'
5
5
'
'
6
Let N{n; a, b) be the
(Thus N(2\
in {a, b).
6
'
\, |)
number of integers
= 2, and
N(4;
A,
|)
=
3.)
Prove that
N(n;a,b)
lim
n->oo
(b)
A
sequence
{«„}
—b — a.
n
of numbers
in [0, 1]
is
called
in [0, 1] if
..
lim
N(n;a,b)
=
b
—
a
uniformly distributed
470
Infinite
Sequences and Infinite Series
for
all
<
a and b with
defined on
and
[0, 1],
a
{a,,} is
l
,.
s
/
Jo
Prove that
on
[0, 1],
if
{a n
}
is
< b <
= hm
1
.
Prove that
uniformly distributed
s(a\)-\
a step function
if 5 is
in [0, 1],
then
\-s(a„)
uniformly distributed in
[0, 1]
and /
is
integrable
then
,
/
..
— hm
/(«!)
+
+ /(0.)
/•
**33.
(a)
Let
/ be
a function defined on
[0, 1]
such that lim /(y)
'
exists for all
a
y—*u
in [0, 1]. For
any
in [0, 1] with
|
e
>
prove that there are only
lim f(y)
—
y—>a
f(a)\
>
s.
Hint:
finitely
Show
many
points a
that the set of such
points cannot have a limit point x, by showing that lim f(y) could not
exist.
(b)
Prove
/
is
that, in the
terminology of Problem 21-5, the
discontinuous
Problem 6-17:
uous except
If
at a
/
is
countable.
This
finally
set
of points where
answers the question of
has only removable discontinuities, then
countable
discontinuous everywhere.
set
of points, and in particular,
/ is contin/ cannot be
CHAPTER
mm ^/
Infinite
[NFINITE SERIES
sequences were introduced
+ ai + «3 +
a\
This
the "partial
sum
=
a\
\-a n
-\
sum must presumably be
infinite
in the
a\
—
H
fl 3
proximations as n
to
be lim
sum of
What can be
defined are
,
defined in terms of these partial sums.
this definition
has already been devel-
is to be any hope of computing the infinite
sums sn should represent closer and closer apand larger. This last assertion amounts to little
previous chapter. If there
+ 02 +
more than
•
as yet completely undefined.
is
Fortunately the mechanism for formulating
oped
•
8111118"
sn
and the
•
not an entirely straightforward matter, for the
is
many numbers
infinitely
previous chapter with the specific inten-
"sums"
tion of considering their
in this chapter.
in the
is
•
?
the partial
chosen larger
a sloppy definition of limits: the "infinite
This approach
s„.
will necessarily leave
undefined, since the sequence
{s n
}
may easily
fail
sum" a\ +0.2 + 013 H
ought
the "sum" of many sequences
to
have a
limit.
For example, the
sequence
-1,
1,
with a„
—
(—
1)" +1 yields the
si
S2
st,
54
for
which lim
sn
does not
1,
-1,
new sequence
= a\ =
=
=
=
Cl[
a\
Cl\
exist.
1,
+ G2 = 0,
+ 02 + a 3 = 1,
+CL2 + «3 + CI4 =
unavoidable that some sequences
sum of
0,
Although there happen
n—>oo
tensions of the definition suggested here (see
definition of the
...
will
to
be some clever ex-
Problems 12 and 24-20)
have no sum. For
this
it
seems
reason, an acceptable
a sequence should contain, as an essential component,
terminology which distinguishes sequences for which sums can be defined from
less
471
fortunate sequences.
.
472
Infinite.
Sequences and Infinite Series
DEFINITION
The sequence
{a n
summable
is
}
the sequence {s„ } converges,
=flH
s„
is
denoted by
an
(or, less
In this case, lim sn
n—>oo
if
\-a n
where
.
oo
y
formally,
+ ai + «3 +
ci\
n=\
and
is
sum of the
called the
The terminology
sequence
introduced in
expressions; indeed the
title
{a u
).
this definition
of this chapter
is
usually replaced by less precise
is
derived from such everyday language.
00
An
infinite
sum >^ an
is
usually called an
infinite series,
the
word
emphasiz-
"series"
n=\
ing the connection with the infinite sequence
The
{a,,}.
statement that
{a,,} is,
or
oo
is
not,
summable
is
conventionally replaced by the statement that the series
does, or does not, converge.
This terminology
somewhat
is
^
an
n=\
peculiar, because at
00
best the
symbol \_. a n denotes a number
n=
note anything at
is
(so
it
can't "converge"),
and
it
doesn't de-
\
all
unless {a n }
is
summable. Nevertheless,
this
informal language
convenient, standard, and unlikely to yield to attacks on logical grounds.
Certain elementary arithmetical operations on
quences of the
definition. It
is
infinite series are direct
a simple exercise to
show
that
if
{a n
}
and
conse-
{b„} are
summable, then
+
b„)
y^c
an
y^(a«
= 2J a„ + 22 bn,
oo
=c
}
a„
n=\
n=\
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
There
is
summability
will
be recorded.
one necessary and
stated immediately.
sufficient condition for
The sequence
{a„}
is
converges, which happens, according to
1
s„
=
0; this
'
summability which can be
summable if and only
Theorem 22-3, if and
if
the sequence
only
if
lim
m./i->00
{s,,}
sm
—
condition can be rephrased in terms of the original sequence as follows.
23.
THE CAUCHY CRITERION
The sequence
{a n }
summable
is
a n+
lim
m.n—>oo
Although the Cauchy criterion
if
\
and only
+
•
THE VANISHING CONDITION
is
it is
not very useful
However, one simple
any particular sequence.
consequence of the Cauchy criterion provides a
which
0.
of theoretical importance,
is
for deciding the summability of
473
if
+a m —
•
Infinite Series
condition for summability
necessary
too important not to be mentioned explicitly.
If [a n ]
summable, then
is
=
lim a n
0.
n—>oo
This condition follows from the Cauchy criterion by taking
be proved directly as
lim a,
=
lim
—
(s„
n—»oo
= 1-1 =
Unfortunately, this condition
is
it
can also
—I, then
follows. If lim s n
«—>00
m = n + 1;
s n -\)
=
from
sufficient.
lim
n—>oo
s„
—
lim s n -\
n—*oo
0.
far
For example, lim \jn
=
n—>oo
but the sequence {l/n}
numbers l/n shows
1
+
^
is
+ 3^4 +
1
^
5
6
^
7
the following grouping of the
fact,
not bounded:
is
+
+ 9 + •• + * + ••
1
1
8
1
1
1
^2
^2
^2
terms,
(4 terms,
each >|)
each > g)
(2
The method
not summable; in
that the sequence {s n }
of proof used in
this
0,
terms,
(8
each>^)
example, a clever
trick
which one might never
see, reveals the need for some more standard methods for attacking these problems.
These methods shall be developed soon (one of them will give an alternate proof
oo
that
y_\ l/ n does not converge) but
examples of convergent
The most important
it
be necessary to
will
first
procure a few
series.
of all
infinite series are the
l+r +
r
2
"geometric series"
+ r3 +
«=o
Only
the cases
if \r\
>
1.
\r\
These
<
1
series
can be managed because the partial
sn
can be evaluated
do not approach
sums
are interesting, since the individual terms
= 1+r+
in simple terms.
s„
rs„
=
=
1
The two
+ r + r2 +
2
r + r +
•
•
•
+ r"
equations
•
+ r"
+ r" +r" +1
.
.
474
Infinite
Sequences
and
Infinite Series
lead to
s„(\-r)= l-r" +1
or
1
-r" +1
-r
1
by
(division
lim r"
=
—
1
r
0, since
is
\r\
we
valid since
<
It
1.
Y* r" =
are not considering the case r
=
Now
1).
follows that
1
-r" +]
lim
1
«=0
1
|r|
-r
<
1.
=
1,
In particular,
E(iy--V2
«=0
EG)
„=1
that
X
7
is,
2
an
1-2
+
4^8^
16
—
^
L
>
sum which can always be remembered from
infinite
the picture in Figure
1
I
FIGURE
1
Special as they are, geometric series are standard examples from which important
summability
tests for
we
For a while
will
be derived.
shall consider
only sequences
sequences are called nonnegative. If [a n
quence
{s n
}
is
A
all
itself,
sn
is
sn
is
combined with Theorem
is
summable
if
and only
this criterion
bounded
to obtain
theorem
i
(THE COMPARISON TEST)
other
such
22-2,
if
the set of partial
bounded.
is
just
is
not very helpful
—deciding whether or not
what we are unable
to do.
On
convergent series are already available for comparison,
all
0;
summability:
nonnegative sequence {a„}
sums
By
test for
>
with each an
a nonnegative sequence, then the se-
is
clearly nondecreasing. This remark,
provides a simple-minded
THE BOUNDEDNESS CRITERION
]
{a,,}
a result whose simplicity belies
its
importance
tests).
Suppose that
<
an
< bn
for
all
//
the set of
the other hand,
this criterion
(it
is
if
some
can be used
the basis for almost
'
.
Then
if
\^ b„
converges, so does
23.
Infinite Series
{s,,}
is
475
>^ a„
n=l
PROOF
If
sn
= a\
t„
=b\ +---
an
,
+ bn
,
-\
(-
then
<
Now
{t n
}
is
bounded,
sn
<
/.^n
since
for all n.
tn
Therefore
converges.
bounded; COnse-
oo
quently,
by the boundedness
criterion yj. a
converges.
>\
|
n=\
Quite frequently the comparison
test
can be used to analyze very complicated
looking series in which most of the complication
A2 +
3
+
sin (n
2"
n=\
+
is
irrelevant. For
example,
1)
n2
converges because
2
+ sin J (n +
2" + n 2
l)
3_
*
2"
and
~
OO
00
.
n=\
is
a convergent (geometric)
Similarly,
we would
series.
expect the series
00
£
n=\
to converge, since the nth
>
-
2"
term of the
2
1
+sin n 3
series
is
practically 1/2" for large n,
and we
would expect the series
£
n=\
to diverge, since (n
+
l)/(n
+
1) is
+
n2 +
n
1
\
practically \/n for large n.
These
facts
can
be derived immediately from the following theorem, another kind of "comparison
test."
THEOREM
2
(THE LIMIT COMPARISON TEST)
If a„, b„
>
and lim a n /b n
n—>oo
converges.
—
c
/
0,
then 7 a n converges
*—^
n=\
if
and only
if
7
*—'
n=\
bn
476
Infinite
Sequences and Infinite Series
PROOF
Suppose / b„ converges. Since lim a n /b n
*—^
—
there
c,
N
some
is
n—»oc
such that
n=\
< 2cb n
an
> N.
for n
00
Then Theorem
But the sequence 2c >^ b„ certainly converges.
n=N
a„ converges,
y
and
The
finitely
shows that
implies convergence of the whole series
a
2_] n
n=l
additional terms.
this
n=N
has only
1
many
we
converse follows immediately, since
also
have lim b n /a„
=
1/c
,
which
^
/I-+00
The comparison
test yields
other important
when we
tests
0. |
use previously an-
00
Choosing the geometric
alyzed series as catalysts.
we
series par excellence,
THEOREM
3
(THE ratio TEST)
Let a„
>
for all n
,
obtain the most important of all
and suppose
a n+\
,-
converges
if r
<
1
.
tests for
,
the convergent
summability.
that
hm
Then >^ a n
/_V W
series
On
=
r.
the other hand,
if
r
>
1
,
then the terms a n
00
are
unbounded, so
>_, a n diverges. (Notice that
it is
therefore essential to
compute
n=\
PROOF
lim a u+ \/a n
n—>oo
and not lim a„/a n+ \l)
Suppose
that r
first
n—»-oo
<
Choose any number
1.
=
lim
implies that there
is
some
N
r
with r
s
<
1
such that
<
lor n
s
> N.
This can be written
a n+
\
a N+
\
< sa n
for n
> N.
Thus
tf/v+2
< sa N
< sa N+ <
c'N+k
<
,
\
s
k
aN
.
s
aN
,
<
s
<
1.
The
hypothesis
23.
=
Since 2_. a NS
&N /_,
Infinite Series
^.11
converges, the comparison test shows that
s
A=0
fc=0
00
00
~N+k
N
n=
k=0
converges. This implies the convergence of the whole series 2_. a n
77=1
The
case r
>
is
1
even
<
easier. If 1
a »+l
>
<
s
r,
then there
c
lor n
s
is
a
number
TV
such that
xr
> N,
an
which means
that
(*N+k
so that the terms are
> QNS k
unbounded.
As a simple application of
=
k
0,
1
,
.
.
.
,
|
the ratio
consider the series
test,
2,
^/n\.
Letting
77=1
an
=
we
\/n\
obtain
1
(n
«77+l
+
an
n\
1)!
(n
1
+
1
n
1)1
+
1
n\
Thus
*" +1
lim
77-*OC
which shows that the
-o,
Qr
series 2_. \/n\ converges.
If
we
consider instead the series
n=\
n
2_\f /n\
,
where
r
is
some
number, then
fixed positive
77=1
r
lim
/I—>-oo
n+l
^L +
f
n
M=
,
_J_ = o.
im
n—t-oo
yi -\-
n\
so
Y^r"/n! converges.
It
follows that
77=1
r"
lim
—
77—>O0
n\
=
0,
1
478
Infinite
Sequences and Infinite Series
a result already proved in Chapter 16 (the proof given there was based on the same
00
ideas as those used in the ratio
we
Finally, if
test).
consider the series
we
nr"
}
n=\
have
(n
+
+
lim (//
n—>oo
since
\)/n
=
\)r"
+x
1
n—»oo
nr'
This proves that
1.
+
//
lim r
lim
n—»oo
<
if
<
r
1,
Ynr
^-^
then
converges,
n=\
and consequently
lim nr"
=
0.
n—foo
(This result clearly holds for
direct
proof of this
Although the
tool
it
A more
exist.
ratio test will
<
0, also.) It
a useful exercise to provide a
is
test as
an intermediary.
be of the utmost theoretical importance, as a practical
may
be quite
One drawback
might equal
difficult to
of the ratio
determine, and
which appears with maddening
serious deficiency,
fact that the limit
r
be found disappointing.
lim a n+ \/a n
the fact that
<
1
without using the ratio
limit,
will frequently
—
1.
The
—
case lim an+ i/an
1
is
may
test
is
not even
regularity,
is
precisely the
the
one
n—>oc
which
inconclusive:
is
{a,,}
might not be summable
(for
example,
=
a„
if
\/n),
00
but then again
might be. In
it
our very next
fact,
test will
show
that
}
{\/n)~~
n=\
converges, even though
n
lim
This
THEOREM
4
(THE INTEGRAL TEST)
provides a quite different
test
gence of infinite
series
comparison
but
Suppose
test,
that
/
is
—
+
1
method
like the ratio test,
the series
positive
determining convergence or diver-
for
it is
an immediate consequence of the
chosen for comparison
and decreasing on
[1, oo),
is
quite novel.
and
that /(//)
=
00
Then
>_,
a n converges
if
and only
if
the limit
n=\
/»oo
rA
exists.
A
l'K<
)( )l
I
he existence of lim
f
/
/is equivalent to convergence
jf'*jf
'£'•••
ol the series
a n for
all
//.
'
.
23.
Now,
/
since
is
we have
(Figure 2)
/(n+
<
decreasing
I)
/<
/
Infinite Series
479
/(«)•
oo
The
half of this double inequality shows that the series
first
oo
pared
to the series 2_\
n=\
i
.
lim
if
Ja
n=
i
(and hence 2_, a n) converges
n=[
i
i
\
exists.
/
A^oo J
^^a„ +
/? proving that
I
°°
°°
pn+\
x
00
n
n
+
The second
half of the inequality shows that the series
1
pn+\
2,
i
pared to the
FIGURE
m ay be COm-
y, a »+\
series >
c/ /;
,
proving that lim
/ must
/
f maY
I
De com-
Jn
a n converges.
>
exist if
|
n=\
2
Only one example using
the integral test will be given here, but
question of convergence for infinitely
many
series at once. If
p >
it
settles
0, the
the
conver-
oo
gence of
^,
\/n p
is
by the
equivalent,
integral test, to the existence of
n=\
r°°
l
dx.
xP
J\
Now
1
1
f
XP
dx
I
ap-
Cp-1)
log A
+
1
p#1
p-i'
/?
,
=
1
°°
r-A
This shows that lim
\/x p dx exists
/
if
p >
1,
but not
if
p <
1.
A-^ooJ,
Thus ) 1/h p
^—
ll=\
oo
converges precisely for p
>
In particular,
1.
2^
I/' 2 diverges.
n=\
The
tests
sequences
considered so far apply only to nonnegative sequences, but nonpositive
may be handled
in precisely the
OO
,
OO
\=l
n=\
all
same way. In
fact, since
v
;
considerations about nonpositive sequences can be reduced to questions invok-
ing nonnegative sequences.
terms are quite another
Sequences which contain both positive and negative
story.
00
If
2_j
a"
*s
a se q uence
w tn
i
both positive and negative terms,
one can con-
n=\
oo
sider instead the sequence
2,
\
a n\,
all
of whose terms are nonnegative. Cheerfully
480
Infinite
Sequences and Infinite Series
we may have thrown away all the interesting informawe proceed to eulogize those sequences which
ignoring the possibility that
tion about the original sequence,
are converted by this procedure into convergent sequences.
The
DEFINITION
(In
series
/_, a "
n=\
*s
absolutely convergent
more formal language,
sequence
the sequence
THEOREM
5
Every absolutely convergent
if
formed from
If 2_.
\
2,
absolutely
\
a n\
convergent.
is
summable
if
the
The
be of any
interest,
it
turns
following theorem shows that the definition
not entirely useless.
convergent
PROOF
is
right to expect this definition to
out to be exceedingly important.
at least
{a,,}
the series
summable.)
{|fl„|} is
Although we have no
is
if
and only
its
if
series
convergent. Moreover, a series
is
the series
formed from
its
positive
is
absolutely
terms and the
series
negative terms both converge.
Cauchy
a n\ converges, then, by the
criterion,
n=\
lim
m,n—>oo
\a n+
h \a m
-I
|
\
=
0.
\
Since
it
<
\-a,„\
\a„+i H
\a„+\\
+
follows that
lim
m,n—*oo
which shows that
To prove
2, a »
an+ \
+ am =
H
converges.
n=\
the second part of the theorem,
a„
+
_
—
f
a„,
0,
on
,
0,
so that 2_. a n^
is
0,
the series
let
>
ifa„<0,
if
a„
S
ifa„>0,
if #71
formed from the
positive
terms of /
n=\
n=\
is
the series
formed from
oo
oo
If
the negative terms.
a + and
2_[ n
y an ~ both
«=i
n=\
l
converge, then
oo
J2
k'»l
= ]O
+
fl "
~
(
a»
)]
=
^L,
a»
+
'
^2 a »
an and }
,
an
,
23.
also converges, so
>^
Infinite Series
481
a„ converges absolutely.
n=\
OO
On
the other hand,
if
00
>_.
n
=
\
an
converges, then, as
I
we have
just
shown,
2, a n
n=\
\
Therefore
also converges.
+
o(Z>
=
Cl "
Jl
n=\
+ J2
n=\
n=\
00
OO
n=\
n=\
la "
and
oo
,
1
n=\
both converge.
It
|
follows from
be used
Theorem
5 that every convergent series with positive terms can
many
to obtain infinitely
other convergent
series,
simply by putting in
random. Not every convergent series can be obtained in this way,
however there are series which are convergent but not absolutely convergent
(such series are called conditionally convergent). In order to prove this stateminus
signs at
—
ment we need a test for convergence which
applies specifically to series with positive
and negative terms.
THEOREM
6 (LEIBNIZ'S
THEOREM)
Suppose
that
^
a\
and
>
a^
>
lim a„
—
^2
•
>
0,
that
0.
tt—>oo
Then
the series
n=\
converges.
PROOF
Figure 3
illustrates relationships
(1)
s2
(2)
Si
(3) Sk
Sl
S4
FIGURE
<
>
<
s4
s3
sb
s5
<
>
if
si
S(,
3
<
>
S8
-SlO
k
S]2
between the
,
is
even and
.V
1
1
J9
.¥7
S5
/ is
S3
odd.
Sl
partial
sums which we
will establish:
482
Infinite
Sequences and Infinite Series
To prove
the
To prove
two
first
inequalities, observe that
(1)
S2n+2
=
s 2n
>
s 2n
(2)
S2n+3
=
S2n+\
<
S2„+i,
+
S2n-1
<
S2n-\
is
easy:
if
k
is
even and
>
-
«2n+3-
that
tt2n
>
since a2„
in
0.
conjunction with
(1)
and
(2)
the general
odd, choose n such that
/ is
In
a 2n+ 2
since a 2ll+ 2
first
=
>
- ain+2 + «2«+3
This proves only a special case of (3), but
case
«2«+2
since a 2 „+\
,
the third inequality, notice
$2n
~
ain+]
>
k
Sk
<
and
In
-
<
S2n-\
<
>
1
/;
then
which proves
Now,
S2n
the sequence {s2 n } converges, because
above (by
si
Sl,
(3).
any odd
for
it is
nondecreasing and
is
bounded
Let
/).
=
a
sup{s2 n }
=
inf{5 2 7+ i}
=
lim S2„n-*oo
Similarly, let
=
ft
It
follows from
(3)
that
a <
S2n+\
it is
actually the case that
The standard example
ft;
—
a
,
since
=
S2„
=
lim s 2n +i-
n—»oo
ci2n+\
lim a n
This proves that a
ft.
derived from
Theorem 6
is
=
ft
=
=
lim
s„. |
the series
4^5
2^3
1
and
oo
which
is
convergent, but not absolutely convergent (since 2_,
verge). If the
sum of this
series
is
1
/" does not con-
denoted by x, the following manipulations lead
to quite a paradoxical result:
r-1_I
+ I_ I4^5
+ I-I +
^ ...
—
2^3
*
J
—
6
I
4
I
1
J
"
2
"'"
(the pattern
I
I
1
3
6
8
here
is
one
t
_i_
W
1
J_
J_ _u I
-L
5
10
12
14
positive
I±
16
"•"
'
term followed by two negative ones)
= (i ~ - i + (5 - i> - i + <i " to) " A + (7 " A) - n +
— 2 4^6 8 ^ 10 12 ^ 14 16 ^
— 2 U 2^3 4^5 6^7 8 ^
2-)
;
—
—
J,
x,
2
•
•
.
23.
=
so x
=
x/2, implying that x
sum
the partial
On
0.
the other hand,
equals i, and the proof of Leibniz's
si
it
is
483
Infinite Series
/
easy to see that x
Theorem shows
that x
>
0:
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
sequence
—
\Vnl
contains
In
fact,
bn
—
that
=
\,
{b,,}
is
a
1
i
2'
3
3'
8'
10'
5
12'
sequence
in the
1
'4'
number
= 3m +
6'
1
1
1
1
1
J_
JL
5
6'
7'
8'
9'
10'
11
of
12'
each
in the following precise sense:
{a,,}
a certain function which "permutes" the natural numbers,
is
every natural
1)
4'
'
rearrangement
where /
/(2m +
2
numbers
[a n )
fl/(„)
is,
the
all
i,
1
m
f{n) for precisely one
is
terms
(the
1,
n. In
go into the
5, 5
our example
1st,
4th, 7th,
.
.
places),
= 3m
/(4m)
(the
.
/(4m +
2)
= 3m +
2
is
.
(the
.
Nevertheless, there
.
.
.
— 4, — g, — -rj'
terms
•
•
g° i nto tne 3rd, 6th, 9th,
•
places),
terms -\, -\,
-^
go into the 2nd, 5th, 8th,
places).
no reason
to
assume that /_^^ n should equal
n=\
sums
are,
by
definition,
lim b\
n—>oo
+ b„
+
and lim
a\
n—xx
+
+a„,
^ an
:
these
n=\
so the particular
00
order of the terms can quite conceivably matter.
special in this regard; indeed,
—
solutely convergent
7
If
Yj «»
behavior
is
how bad
series
/_\— 1)"
typical of series
the following result (really
than a theorem) shows
THEOREM
its
The
+
/n
is
not
which are not ab-
more of a grand counterexample
conditionally convergent series are.
converges, but does not converge absolutely, then for any
number a
there
00
is
a rearrangement {b„} of
{a,,}
such that
/.&„
=
a
-
n=\
PROOF
Let 2_. Pn denote the series formed from the positive terms of
series
of negative terms.
let
/_/7„
one had bounded
partial
It
follows from
Theorem
5 that at least one of
As a matter of fact, both must fail to converge, for
sums, and the other had unbounded partial sums, then
these series does not converge.
if
and
n=\
n=\
denote the
{a,,}
484
Infinite
Sequences and Infinite Series
>^ a„ would
the original series
the assumption
Now
also have
a be any number. Assume,
let
unbounded
partial sums, contradicting
n=\
that it converges.
for simplicity, that
a >
(the
proof
for
oo
a <
be a simple modification).
will
Since the series
Y^ Pn
is
not convergent,
„=]
there
a
is
N
number
such that
N
^2 Pn >
a.
,1=1
We
will
choose N\ to be the
N
smallest
with
this property.
This means that
N\-\
J2 p " - a
(i)
-
n=\
but
^T p n >
(2)
a.
n=\
Then
if
= ^Pn,
Si
n=\
we have
S\
-a
<
pi
PNi
p\ H
FIGURE
V
pn
a
-\
1
P\-\
x
p N] -i
+
Pn,
4
This relation, which
is
clear
from Figure
4, follows
immediately from equation
(1):
/V,-l
Si
-a <
Si
-
^jT p„
= p Ni
.
n=\
To
the
which
sum
is
we now add on just enough negative terms to obtain a new sum T\
than a. In other words, we choose the smallest integer M\ for which
S\
less
T\
As
before,
=
S\
+ ^2 Q» < a
-
we have
a -T\ < -q M}
We now
continue
this
procedure
indefinitely,
obtaining sums alternately larger
and smaller than a, each time choosing the smallest Nk or M^
possible.
The
.
23.
485
Infinite Series
sequence
pi, ..., pN\, q\,
is
a rearrangement of
The
{a,,}.
.
.
,
Pni + \,
<?a*i,
••
Pm,
.
sums of this rearrangement increase
partial
to S\
,
then decrease to T\, then increase to 52, then decrease to 72, etc. To complete the
proof we simply note that S* — a and 7* — a are less than or equal to p^k or —qM
|
|
respectively,
and
1
|
that these terms, being
k
members of
,
the original sequence {a n },
00
must decrease
to 0, since y_. a n converges. |
Together with Theorem
tion
THEOREM
8
If
7, the
next theorem establishes conclusively the distinc-
between conditionally convergent and absolutely convergent
\] a n
converges absolutely, and {b n }
is
any rearrangement of
series.
{a n
then
},
\]b n
n=\
also converges (absolutely),
and
CO
00
PROOF
Let us denote the partial sums of {«„} by
sn ,
and the
sums of
partial
{b n
by
}
tn
.
00
Suppose
that e
>
0.
Since
converges, there
2, a n
some
is
TV
such that
n=\
00
'
-
an
/
<
sN
£.
Moreover, since >. a n converges, we can also choose
\
TV so that
I
n=l
7,
i.e.,
~
\a,A
(l«ll H
h I«nI)
<
e,
so that
\<*N+\\
+
+
l«/V+2l
Now choose M so large that each of a\,
whenever m > M, the difference m — s^
t
are definitely excluded.
\ON+3\ H
.
is
.
,
<
£•
a# appear among
the
sum of certain
a,-,
Consequently,
\t,n
—
Sn\
<
\Qn+\
I
+
k'/V + 2l
+
|«/V + 3
1
+
•
•
•
b\
•
,
.
.
.
,
Then
but-
«>/^/? aj
,
.
.
.
,
a^
486
Infinite
Sequences and Infinite Series
Thus,
if
m > M,
then
2_,
-
an
-
SN
< 22 a n ~
SN
y^<3»
m
t
-
(t
m
-Sn)
n=\
n=\
oo
+
\t,n
-
SN
\
n=]
<
E
+ £.
oo
Since
this
is
true for every e
>
0, the series
2. b n
converges to
2, a n
oo
To show
V, b„
that
converges absolutely, note that
{
\b n
\
}
is
a rearrangement
oo
of
{
\a„\
};
since V", a »\ converges absolutely,
\
n=\
V,
\bn\
converges by the
first
part of
n=\
the theorem. |
Absolute convergence
series.
is
also
Unlike the situation for addition,
oo
oo
y^Qii
isn't
oo
+ ^2 b „ = y^dn+bn),
n=\
there
when we want to multiply two infinite
where we have the simple formula
important
n=\
n=\
quite so obvious a candidate for the product
\Y2 a ")
'
(£M
=(fl
i
+ a2 +
•••)•
(b\+b 2 +
••)•
It would seem that we ought to sum all the products ajbj. The trouble
form a two-dimensional array, rather than a sequence:
Nevertheless,
all
a\b\
a\bj
a\b?,
aib\
^2^2
#2^3
dT,b\
a^bi
fl 3 Z?3
that these
the elements of this array can be arranged in a sequence.
picture below shows one way of doing
many
is
this,
and of course, there are
other ways.
a\b\
ciT,b\
a\b2
a\b-),
02^2
#2^3
03/^2
03 bj,
The
(infinitely)
23.
Suppose that
{c,,}
is
some sequence of
Then we might
just once.
Infinite Series
487
containing each product a/bj
this sort,
naively expect to have
OO
OO
£
E
=
c»
n=\
00
a
»-£^n=\
n=\
But this isn't true (see Problem 10), nor is this really so surprising, since we've said
nothing about the specific arrangement of the terms. The next theorem shows
that the result does hold when the arrangement of terms is irrelevant.
THEOREM
9
If
E
an<^
fl "
products
conver §e absolutely, and
E^"
a- bj for
x
each pair
OO
00
E
£
=
c«
n=\
PROOF
Notice
first
«£^-
n=\
n=l
sequence
that the
L
= ^\a,\-^\bj\
pL
and
converges, since {a„}
is
>
0, if
and since the limit of a
Cauchy sequence, which means
{b„} are absolutely convergent,
So
the product of the limits.
that for any s
any sequence containing the
00
fl
L
product
is
}
then
j),
(/,
{c n
a
{/?/.} is
V are large enough,
L and
V
L
L'
then
L
<
It
follows that
£
(1)
or
i
Now suppose that
every term djbj for
TV
i,
j
\ai\-\bj\<
e
<
S.
j>L
any number so large that the terms
is
<
L.
Then
c„ for
n
<
N
the difference
N
L
L
J2 c * - J2 a '-J2 b
J
n=\
consists of
terms
cijbj
with
/
> L
E
2)
>
or j
N
(
L, so
L
L
c"
j=\
i=\
-E"'-E^
n=\
^
E
<e
But since the
limit of a
product
is
L
OO
E «E^ - E
a
>=1
by(l).
the product of the limits,
OO
(3)
tai-w
iovj>L
7=1
i=\
./
=
!
i=\
we
L
a
'-£^
7=1
<
E
also
have
include
—
488
Infinite
Sequences and Infinite Series
for large
will
enough
L. Consequently,
if
we choose
and then N,
L,
large enough,
we
have
00
J2 a >-J2 b
J2 c »
--
J
j=\
,=1
oo
00
00
J2 a '-Jl b - Jl a '-J2 b
-
J
,=1
i=l
J
N
L
L
a
bJ
+ Jl >-Jl
which proves the theorem.
by
2s
- I2 C »
n=\
7=1
i=\
<
j=\
i=\
J=\
(2)
and
(3),
|
Unlike our previous theorems, which were merely concerned with summability,
this result says
no reason
to
something about the actual sums.
presume
terms. However,
Taylor's
sum can be
that a given infinite
many simple
Generally speaking, there
"evaluated" in any simpler
expressions can be equated to infinite sums by using
Theorem. Chapter 20 provides many examples of functions
(i)
f
where lim R n
,
a
(x)
= 0.
This
is
f(x)
= hm
for
(a)
precisely equivalent to
—
-
>
n—>oo *—-^
-
(x
a)
,
1
i
i=0
which means,
in turn, that
/(*)
=
£
f n (a) (x-a)
1
,
V.
i=0
As
particular examples
we have
4
2
cos*
=
arctan x
log(l+*)
=' + v +
=
6
X
X
x
l--+-~+
X
e
X3
x
—
= ,~
is
T"
X2
+
T+
x*2
v.
X
14 +
xJ
+
+
v.
5
X
X4
X3
.-
x*
-
v
7
=-
"^
...,
+ •••
,
\x\
<
I,
5
+
X'
T T T+
- 1
<X<
1
which
23.
in addition,
= — 1,
when x
+ x)
and log(l
(Notice that the series for arctanx
do not even converge
log(l + x) becomes
the series for
_1 1
Infinite Series
i_I_
I
2
4
3
for
\x\
489
>
1;
which does not converge.)
Some
pretty impressive results are obtained with particular values of x
'
=1 +
+
+
II
2i
111
3i
:
+ --
_ = ,__ + ___ + ...,
7T
,
„
log2
More
sinjc
significant
and cos* a
have obtained
if
= l-- +
1
,
developments
i
-
5
J
may be
1
+
....
anticipated
if
we compare
the series for
little more carefully. The series for cosjc is just the one we would
we had enthusiastically differentiated both sides of the equation
sill
X
=X—
x
x
—
1
3!
5!
we have never proved anything about the
derivatives of infinite sums. Likewise, if we differentiate both sides of the formula for cosx formally (i.e., without justification) we obtain the formula cos'(x) =
— sinx, and if we differentiate the formula for e x we obtain exp'(x) = exp(x).
In the next chapter we shall see that such term-by-term differentiation of infinite
term-by-term, ignoring the fact that
sums
is
indeed valid in certain important
cases.
PROBLEMS
1
.
Decide whether each of the following
gent.
The
tools
ison, ratio,
and
which you
integral
infinite series
used.
tests.
oo
„
.
E
SU1H0
n2
l
i-i+i00
iv
n=\
Jogn
may
convergent or diver-
need are Leibniz's Theorem and the comparA few examples have been picked with malice
will
aforethought; two series which look quite similar
(and then again, they
is
not).
The
hint
may
require different
below indicates which
tests
tests
may
be
:
490
Infinite
Sequences and Infinite Series
i
V^
v)
(
.
-
n=2 ifn*
OO
1
(The summation begins with n
the meaningless term obtained
i
Ei
VI
-)
oo
(v.)
j
E^
oo
(viiT
logn
E
n=
n
\
00
,
^—
E
IX
log/?
'
oo
i
^
A
*-*> (lo
(log/?)*
S n)
oo
1
^ ^
- (log/7)'
(logn)"
n=2
00
M E-
V-1V1)
(log
"oo^r
(
oo
i
?
EH"
oo
(xiv)
,
E
sin
oo
(xv)
V-
^"
i
oo
(xvi)7
7
^—
i
'
—
on(log n
)
oo
(xvii)
7
^—
i
~
'
-^7;
2
(log/?)
/?
n=2
(
XV111
•
*-^ n log n
n=1
)Z.^7-
n=l
=2
for n
simply to avoid
=
1).
.
23.
>
(xx)
Hint:
Use the comparison
test for
(vii), (xviii), (xix),
(v), (vi), (ix), (x), (xi), (xiii),
(i),
the integral test for
(xx);
The next two problems examine, with hints, some
more delicate analysis than those in Problem 1
(a)
If
491
.
the ratio test for
*2.
Infinite Series
you have
successfully solved
examples
(viii),
(xiv), (xvii);
(xv), (xvi).
infinite series that
(xix)
and
require
(xx)
from Problem
<
and diverges
1,
00
it
n
should be clear that
a n\/n
>_.
n
converges for a
e
for
oo
a
>
=
For a
e.
diverges,
e the ratio test
fails;
show
that
"
"
2_, e n}-/ n actually
by using Problem 22-13.
00
(b)
Decide when 2_. n
"
"
converges, again resorting to Problem 22-13
/ a n}
-
n-
when
the ratio test
fails.
oo
oo
*3.
Problem
the
first
1
presented the two series >_,(log«)
n=2
diverges while the second converges.
_A
'
and V\log«)~", of which
n=2
The
series
oo
V
^
n=2
which
lies
between these two,
is
1
-
(log«) lo §"'
analyzed in parts
(a)
and
(b).
oo
(a)
Show
that f^°
(b)
Show
that
ey
y
/y dy
exists,
by considering the
00
series
\]( e / n )"
.
V
^(logrc)
!
11=1
u
converges, by using the integral
tion
(c)
and part
Show
lo g«
test.
Hint:
Use an appropriate
substitu-
(a).
that
oo
v
^—
'
diverges,
.
!
(log«) lo s (1 °g")
by using the integral
in part (b),
and show
test.
Hint:
Use the same
substitution as
directly that the resulting integral diverges.
492
Infinite
Sequences and Infinite Series
i
4.
—r—
Decide whether or not >
—.
converges.
n=\
oo
5.
(a)
Prove that
oo
n=\
*(b)
6.
Let
Show
/ be
a,,
.
n=\
that this does not hold for conditional convergence.
a continuous function on an interval around 0, and
(for large
7,
/_]a„ converges absolutely, then so does /
if
enough
=
let a„
/(1/n)
n).
oo
(a)
Prove that
} an
if
=
converges, then /(0)
"=1
0.
oo
(b)
Prove that
if
f'(0) exists
and
(c)
Prove that
if
/"(0)
and /(0)
\] a"
converges, then f'(0)
n=]
exists
/'(0)
=
a converges.
0, then
2_] "
»=1
Suppose 2_] a " converges. Must f'(0)
"=1
(e)
0.
oo
=
oo
(d)
=
exist?
00
=
Suppose /(O)
/'(O)
=
0.
Must /. a " converge?
n=\
oo
7.
(a)
Let
<
be a sequence of integers with
{a,,}
<
an
9.
Prove that
2, a n 10~"
n=\
(and
exists
we
(b)
and
between
lies
usually denote by 0. a xaia^aA,
Suppose
< x <
that
1.
(This, of course,
1).
.
.
.
is
the
number which
.)
Prove that there
a sequence of integers
is
{a,,}
00
<
with
< 9 and /Ja„ 10~" =
a„
.v.
»=1
(where [j] denotes the greatest integer
(c)
Show
that
if
{a,,}
repeating,
is
i.e.,
Hint: For example, aj
which
is
<
=
y).
of the form a\, ai,
is
[10x]
•
•
•
,dk,
00
a
i
,
ai
,
.
.
•
,
cik:,<*
l
^2
•
,
,
then
Y^ «„
1
0~"
is
a rational
number
(and find
n=]
it).
The same
result naturally holds
the sequence {a^+k}
is
repeating for
if
{a n }
eventually repeating,
is
i.e., if
some N.
00
(d)
Prove that
if
x
—
0~"
\_] a n 1
is
rational, then {a„ }
is
eventually repeat-
n=\
ing. (Just
look at the process of finding the decimal expansion of
dividing q into
8.
Suppose that
{a n
proof of Leibniz's
}
p by long
+1
Jj-1)"
n=\
the hypothesis of Leibniz's
satisfies
Theorem
«„
p/q—
division.)
Theorem.
to obtain the following estimate:
-[
a]
-a 2 +
---±a N
]
< a N+
\,
Use
the
'
.
23.
9.
Prove that
(a)
>
an
if
~
and lim ?Ja n
=
then >
r,
^—
n—»oo
493
Infinite Series
a„ converges
<
if r
1,
,;=1
and diverges
if r
>
1
(The proof is very similar to that of the
.
ratio test.)
oo
This result
known
is
More
as the "root test."
converges
2, a n
generally,
n=]
there
if
is
some
<
s
such that
1
all
but
many
finitely
<
^/a^ are
and
s,
00
a„ diverges
y
many
if infinitely
aja^ are
>
1
.
This result
known as
is
the
71=1
"delicate root test" (there
a similar delicate ratio
is
follows, using
It
test).
oo
the notation of
and diverges
Prove that
(b)
Problem 22-27,
test
1
—
chapter.
easy to construct series for which the ratio
works. For example, the root
i
<
n—>00
;
problem from the previous
It is
lim ^/a^
if
1
lim 'ifa^ > 1 no conclusion is possible if lim t/a^
n—*oo
n—>oo
the ratio test works, the root test will also. Hint: Use a
if
if
> a converges
*—^ n
that
+
i
test
test fails,
shows that the
+ a) 2 +(i) 2 +(i) 3 +(l) 3 +
while the root
series
---
converges, even though the ratios of successive terms do not approach a
limit.
is
10.
Most examples are of this rather
artificial
nature, but the root test
nevertheless quite an important theoretical tool.
For two sequences {a n ) and {b n },
let c n
=
a kbn+\-k- (Then c„
22.
is
The
of the terms on the nth diagonal in the picture on page 486.)
c„
y
is
called the Cauchy product of
/. a »
ancl
n=l
>i=\
{—\) n /y/n, show that
\cn
\
>
1,
sum
series
oo
00
00
the
/^ bn
If
•
<*n
=
bn
=
n=\
so that the
Cauchy product does not con-
verge.
11.
(a)
A
Consider the collection
of natural numbers that do
their usual (base 10) representation.
A converges. Hint:
how many between 10 and
of the numbers in
9 are
(b)
If
B
in
is
9
the
A
sequence
in
B
then the
do not have
sum
all 1
digits
of the reciprocals of
all
ten digits
representation.)
{a n }
is
called
Cesaro summable,
si
+s„
-\
lim
7!->-0O
Sk
that
converges. (So "most" integers must have
..
(where
contain a 9 in
99?; etc.
numbers
in their usual representation,
numbers
in their
12.
A?;
the collection of natural
not
Show that the sum of the reciprocals
How many numbers between 1 and
= a\ H
\-
a*).
=
with Cesaro
sum
/,
if
/
n
Problem 22-16 shows
that a
summable sequence
l
494
Infinite
Sequences and Infinite Series
is
13.
sum
automatically Cesaro summable, with
a sequence which
Suppose
>
that a n
equal to
summable, but which
not
is
and
{a,,}
its
Cesaro sum. Find
Cesaro summable.
is
Cesaro summable. Suppose
is
also that the
oo
El
sequence {na„}
bounded. Prove that the
is
n
/
14.
n=\
n
.
n
v^
and an = — > s,,
*-^
a,
prove that
sn
n
=
/_. a » converges. Hint: If
series
1
'=
+
n
-an
bounded.
is
1
1
This problem outlines an alternative proof of Theorem 8 which does not rely
on the Cauchy
(a)
criterion.
Suppose that a n >
and let s„ — a\ +
•
there
is
m
some
Show
(d)
that
n.
and
t„
+ a„
< tm
s„
Let {b n } be a rearrangement of
=
b\
+
•
•
•
+ bn Show that
for
.
{a,,},
each n
.
00
— /.
that 2_. a "
n=\
Show
each
00
^«
/, bn
if
exists.
n=\
n=\
00
(c)
•
with
00
(b)
for
•
00
2, a n =
/_. ^"
n=\
n=\
•
Now replace the condition a n >
by the hypothesis that >^ a n converges
absolutely using the second part of Theorem 5.
15.
(a)
Prove that
if
/_,
a " converges absolutely and {b n }
is
any subsequence of
n=\
00
{</„},
then 2_j^" converges (absolutely).
OO
(b)
Show
that this
is
false if
Y_]
o-n
does not converge absolutely.
n=\
00
*(c)
Prove that
2_. a » converges absolutely then
if
n=\
00
an
}
=
(a\
+
a-j
+ as H
)
+
(«2
+ «4 + «6 +
•
)•
n=\
00
00
16.
Prove that
if
V
an
is
absolutely convergent, then
/,
Problem 19-43 shows
Prove that
*18.
/
|(sin
that the
x)/x\dx
Find a continuous function
exists,
i_,
l
a "l"
improper integral f
00,
(sinx)/xdx converges.
diverges.
/
but lim f(x) does not
Jt-»-00
—
n=\
«=]
*17.
00
a"
with f(x)
exist.
>
for
all
x such that /
f(x)dx
)
.
23.
19.
—
Let f(x)
x
< x <
l/x for
sin
not bounded (thus
is
1]
21.
€(/,
P
P) for
a
2
2
we
1
1)tt
shown
Figure
in
In this problem
2
.
the function
shaded region
all
has "infinite length"). Hint: Try
/
0,
(2/?+
/ be
that the set of
2
P=
Let
Recall the definition
0.
form
partitions of the
20.
=
/(0)
let
Show
of i(f, P) from Problem 13-25.
partition of [0,
and
1,
495
Infinite Series
Find
in Figure 5.
and
/,
fQ
also the area of the
5.
"binomial series"
will establish the
a
= £;
(!+*)«
<
\x\
1,
k=0
for
any a, by showing that lim R n o(x)
=
and uses the Cauchy and Lagrange forms
(a)
Use the
for
ratio test to
<
|r|
follows in particular that lim
FIGURE
as
found
that the series 2_^
not to say that
(this is
1
show
The proof is
0.
it
in
Problem 20-21
r
does indeed converge
+
necessarily converges to (1
\r"
I
\
I
in several steps,
=
for
<
|r|
r)
a
).
It
1.
5
(b)
Suppose
< x <
that
first
Show
1.
lim R„q(x)
that
=
H—>-00
<
Lagrange's form of the remainder, noticing that
(1
+ > a.
Now suppose
Cauchy 's form of
n
(c)
+
by using
0,
a~n ~
t)
l
for
1
\
remainder
+t) a
|.v(l
—1 < x < 0; the number /
— < x < t < 0. Show that
that
satisfies
~
<
]
where
\x\M,
\
in
the
1
M = max
+ xf
(1, (1
'),
and
x
-
t
1
TT7
-
t/x
< x
\+t
Using Cauchy 's form of the remainder, and the
+
(n
a
1
Ji
show
that lim
n-+oo
22.
(a)
Rn
Suppose that the
o(x)
=
partial
+
=
a
a-
fact that
1
1
0.
sums of the sequence
{a n
}
are
bounded and
that
oo
{/?„}
is
a sequence with b n
> b n+ and lim
\
b„
=
0.
Prove that
II—>0O
Y^ a n b„
'
<
n=\
converges.
This
is
known
as Dirichlet's
(Problem 19-36) to check the Cauchy
(b)
Derive Leibniz's
Theorem from
test.
Hint:
criterion.
this result.
Use Abel's
Lemma
496
Infinite
Sequences and Infinite Series
Prove, using Problem 15-33, that the series 2~]( cosnx )/ n converges
(c)
x
if
n=\
not of the form Ikn for any integer k
is
(in
which case
clearly diverges).
it
oo
Prove
(d)
Abel's
If >_,
test:
a converges and {b n }
a sequence which
is
>i
is
either
bounded, then j
a n bn
n=\
nondecreasing or nonincreasing and which
is
n=\
—
converges. Hint: Consider b„
where b
b,
—
lim b„.
n—>oo
oo
*23.
Suppose
{a n
)
decreasing and lim a n
is
—
Prove that
0.
if
n—KX)
/
t—^1
n=\
a„ converges,
oo
2,^ a
then
a l so converges (the "Cauchy Condensation Theorem"). No-
2n
OO
tice that the
00
divergence of >^
\
jn
is
a special case, for
if
n=\
2,
l/ n converged,
n=\
oo
2; 2" (1/2") would
then
remark may serve
also converge; this
as a hint.
n=\
|:
24.
Prove that
(a)
and 2_,^ n " converge, then 2_, a nb„ converges.
2_, a n
if
»=1
n=\
/;=1
OO
Prove that
(b)
00
/_, a "
if
converges, then 2_, a n/n
a
converges for any
a.
n=\
^25.
Suppose
{a„}
then lim na n
decreasing and each a n
is
—
0.
Hint: Write
down
>
0.
Prove that
Cauchy
the
j an
if
criterion
converges,
and be sure
to
—
0.
n^-oo
use the fact that [a n ]
is
decreasing.
00
^26.
If
a„ converges, then the partial
7
sums
s„
and lim
are bounded,
a„
n=\
It is
tempting to conjecture that boundedness of the partial sums, together
oo
with the condition
lim a„
=
0, implies convergence of the series
H->-00
>
^-^
a„.
n=\
Find a counterexample to show that
subsequence of the partial
this to
happen, without
sums
will
this
have
letting the
is
not true. Hint:
to converge;
Prove that
if a n
>
and 7
^
n=\
Compare
the partial sums.
you must somehow allow
sequence of partial sums
oo
27.
Notice that some
oo
a n diverges, then 7
<-"
-
—+a„—
1
11=1
Does the converse
hold:'
-
itself
converge.
also diverges. Hint:
.
23.
28.
we
For b n >
say that the infinite product
1 b„
1
497
Infinite Series
converges if the
sequence
n=l
P"
=
l
I
converges, and also lim p„
hi
=£ 0.
i=\
(a)
Prove that
if
I
b n converges, then b n approaches
1
n=\
OO
(b)
Prove that
00
1 b„ converges
1
and only
if
Y^ log b„
if
converges.
«=]
(c)
>
For a n
0,
prove that
+ a„)
( 1
1
J
converges
if
and only
Y~, an con-
if
n=\
Use Problem 27
verges. Hint:
+ a)
for log(l
n=\
for
one implication, and a simple estimate
for the reverse implication.
The remaining
Problem show
parts of this
that the hypothesis a n
>
is
needed.
(d)
Use the Taylor
series for log(l
+ x)
show
to
that for sufficiently small x
we have
\x
2
<x
-log(l
an
> —1 and
+ x) <
\x
2
.
oo
Conclude
that
if all
2_. a " converges, then the series
n=\
00
oo
+
2_^log(l
a„) converges
if
and only
/_]a«
if
n=l
2
converges.
)i=l
00
if all
> —1 and
an
oo
+ an
converges, then V^log(l
2_. an
)
converges
oo
and only
y, a
if
>\
Use the Cauchy
converges. Hint:
criterion.
n=\
{€)
Show
that
(-1)"
^
v^
=2
n
n=2
converges, but
nc +
n=l
diverges.
(f)
Consider the sequence
In \
I""/
Similarly,
—
—
1
1
'
1
-Li
2
3
'
pair
-L
'
4
L
'
3
'
-II
4 3
3 pairs
'
-L
'
-L
L
4; 5
'
6
L
'
.V
.1
6
1
'
5 pairs
5
1
'
6'
if
'
498
Infinite
Sequences and Infinite Series
(compare Problem
Show
26).
that
diverges, but
2_, a "
n=\
oo
Y](\+a„)=--\.
n=
B=l
29.
(a)
Compute f\(\
--A
n=2
oo
(b)
Compute
J~](l
+
.v
2 ")
for
|.v|
<
1.
00
30.
The
divergence of
\]
\/ n
related to the following remarkable fact:
s
i-
Any
n=\
positive rational
of the form \/n.
number x can be written as definite sum of distinct numbers
The idea of the proof is shown by the following calculation
for j[: Since
27 _
i
31
2
23
62
3
1
_
—
_
—
23
62
7
186
186
^
4'
7
J_
_J_
186
27
_
—
•
•
•
26
'
1674
we have
27_i_
— 2 i_i,J__
"^
"^
31
3
27
|
Notice that the numerators 23,
(a)
Prove that
sort
if
\/{n
+ 1) <
jc
1_
|
"
7,
<
1
1674
of the differences are decreasing.
1/rc for
some
n,
then the numerator in
as a finite
sum of distinct numbers
\/k.
00
(b)
this
of calculation must always decrease; conclude that x can be written
Now prove
the result for
all
x by using the divergence of
V,
1
/«.
UNIFORM CONVERGENCE AND
POWER SERIES
CHAPTER
The
considerations at the end of the previous chapter suggest an entirely
of looking
Our
at infinite series.
equations
to
attention will shift
from particular
new way
infinite
sums
like
2
X
X
=!+_
+ _ + ...
«•
depend on x.
defined by equations of the form
which concern sums of quantities
interested injunctions
/ (*)
(in
the previous
=
/l(*)
=
example f„(x)
In other words,
that
+ /2OO + /3 (*) + ••
l
x"~ /(n
—
we
are
•
In such a situation {/„} will be
1)!).
some sequence of functions; for each x we obtain a sequence of numbers {/„(.*)},
and fix) is the sum of this sequence. In order to analyze such functions it will
certainly be necessary to remember that each sum
/1W + /2W+/3W + is,
by
sequence
definition, the limit of the
/!(*),
If we define a
f
l
(x)
new sequence
+ f2 (x),
of functions
sn
then
we can
express this fact
=
more
we
shall therefore
FIGURE
on functions defined
such functions can be
true actually
first
is
—
we have
of these shows that even
Contrary
to
if
=
\-
fn,
by writing
lim s n (x).
concentrate on functions defined as
=
The
body of results about
nothing one would hope to be
very
easily:
total
a splendid collection of counterexamples.
each /„
is
continuous, the function
f„ will
shows the graphs of the functions
<
=
1,
499
limits,
lim f„ix),
n—*oo
what you may expect, the functions
fnix)
....
by
as infinite sums.
summed up
instead
}
succinctly
f(x)
rather than
{s n
+ f2 (x) + Mx),
f\ H
f(x)
For some time
fi(x)
X >
A"
1.
<
1
/ may
The
not be!
be very simple. Figure
1
,
500
Infinite
.
Sequences and Infinite Series
1-
These functions are
—
continuous, but the function f(x)
all
lim f„(x)
is
not
n—>oo
continuous; in
fact,
< x <
0,
lim
fn (x)
Another example of
tions /„ are defined
same phenomenon
this
>
x
1
1
1
illustrated in Figure 2; the func-
is
by
1
x<
-1,
n
FIGURE
2
/«(*)=
1
n
n
- < x < -
nx,
'
1
1
- <
1,
x.
n
In this case,
and
if
x
>
if
0,
x <
0,
then
then f„{x)
fn (x)
is
eventually
is
eventually
lim f„(x)
so,
once again, the function f(x)
By rounding
FIGURE
3
fn (x)
is
while /„(0)
-1,
A<0
jc=0
1,
x>0:
lim f„(x)
One
enough
for
all n.
n)
equal to
— 1,
Thus
not continuous.
is
examples
it
is
even possible to
functions {/„} for which the function f(x)
dijferentiable
not continuous.
=
0,
off the corners in the previous
produce a sequence of
lim
=
=
1,
for large
(i.e.,
such sequence
easy to define explicitly:
is
1
<
x
=
n
/H7TX\
fnix)
1
1
<
1,
These functions are
<x <
sin
differentiable (Figure 3), but
lim
fn {x)
=
x.
we
still
-1,
x<0
0,
x
=
1,
x
>
have
0.
Continuity and differentiability are, moreover, not the only properties for which
problems
arise.
Another
Figure 4; on the interval
altitude n, while f„(x)
I
I
I
,
I
l<
1
.
4
as follows:
=
by the sequence {/„} shown in
the graph of /„ forms an isosceles triangle of
difficulty
[0. 1//;]
for
is
illustrated
x > \/n. These functions
may
be defined
explicitly
24. Uniform Convergence and Power Series
2n 2 x,
=
fn(x)
2n
—
<
_
<
a
—
~ In
1
1
2n 2 x,
501
to**'
n
I
<x <
0,
Because
instincts
limit
In
this
sequence varies so erratically near
0,
1.
our primitive mathematical
might suggest that lim f„(x) does not always
does
fact, if
n—*oo
x, and the function f(x)
exist for all
x >
0,
then
for all n, so that
we
fn (x)
is
=
lim f„{x)
=
lim f„(0)
0.
n—*oo
—
lim f,,(x)
for
all
On
x.
even continuous.
is
n—»oo
eventually 0, so lim f„(x)
certainly have
Nevertheless, this
exist.
=
0;
=
moreover, /„(0)
In other words, f(x)
=
the other hand, the integral quickly reveals the
n-+oo
strange behavior of this sequence;
we have
f„(x)dx
=
j,
/' /(*)</*
=
0.
Jo
FKiURE
5
but
Jo
Thus,
lim /
-*°°Jo
f,,(x)dx
y£
n
I
lim f„(x)dx.
JO "^°°
This particular sequence of functions behaves in a
imagined when we
first
considered functions defined by
way
that
limits.
we
really never
Although
it is
true
that
f(x)
=
the graphs of the functions /„
lying close to
it
—
if,
for each x in [0,
lim f„(x)
1],
n—foo
do not "approach" the graph of / in the sense of
we draw a strip around / of total width 2s (al-
as in Figure 5,
lowing a width of e above and below), then the graphs of /„ do not lie completely
within this strip, no matter how large an n we choose. Of course, for each jc there
is
some TV such that the point (x, fn (x)) lies in this
amounts to the fact that lim f„(x) = f(x). But
just
strip for n
it is
> N;
this assertion
necessary to choose larger
n—>oo
and
all
x
larger
/V's
6
x
is
chosen closer and closer to
situation actually occurs,
examples given previously. Figure 6
though
0<
1.
A strip
<
of
total
less blatantly, for
illustrates this
MX)
If e
and no one
0,
Af will
work
for
at once.
The same
FIGURK
as
X >
each of the other
point for the sequence
x <
1
1.
width 2s has been drawn around the graph of f(x)
2? this strip consists
=
lim f„(x).
n—>-oo
of two pieces, which contain no points with second
.
502
Infinite
Sequences and Infinite Series
coordinate equal to |; since each function f„ takes on the value j, the graph of
each /„ fails to lie within this strip. Once again, for each point x there is some N
such that
It is
fn (x))
(x,
which works
for
> N; but
it is
not possible to pick one
N
at once.
easy to check that precisely the same situation occurs for each of the other
we have
examples. In each case
some
defined on
all
the strip for n
lies in
x
all
a function /, and a sequence of functions
{/„},
A, such that
set
/(*)=
lim f„(x)
for
all
x
A
in
n—>oo
This means that
FIGURE
7
for
s
all
>
0,
and
\f(x)-f
But
in
for all
x
A, there
in
N
some
is
such that
n
if
> N,
then
<£.
n (x)\
each case different
TV's
must be chosen
for different x's,
and
in
each case
n
> N,
it
not true that
is
for
all
|/(x)
Although
>
e
there
-/B (x)|
this
phrase "for
some
is
N
such that for
x
if
then
<e.
condition differs from the
all
x in A,
all
in
A,"
it
first
only by a minor displacement of the
has a totally different significance.
If a
sequence {/„}
second condition, then the graphs of /„ eventually lie close to the
graph of /, as illustrated in Figure 7. This condition turns out to be just the one
which makes the study of limit functions feasible.
satisfies this
DEFINITION
Let {/„} be a sequence of functions defined on A, and let / be a function which
Then / is called the uniform limit of /„ } on A if for
is also defined on A
.
>
every e
there
{
some
is
if
We also say that
f
{/„}
n
TV
such that for
>
yv\
then
all
x
in A,
- fn (x)\ <
|/(jc)
s.
converges uniformly to/ on^4, or
that /„
approaches
uniformly on A.
As a
contrast to this definition,
f(x)=
then
we
say that {/„}
if
we know
lim f„(x)
only that
for
each x
in
A,
n—>oo
converges pointwise to f on
A Clearly, uniform conver-
gence implies pointwise convergence (but not conversely!).
Evidence for the usefulness of uniform convergence
Integrals represent a particularly easy topic; Figure 7
if
to
is
not
at all difficult to
makes
it
amass.
almost obvious that
converges uniformly to /, then the integral of /„ can be made as close
the integral of / as desired. Expressed more precisely, we have the following
{/„}
theorem.
503
24. Uniform Convergence and Power Series
THEOREM
1
Suppose
that
/„
{
[a b]
,
that {/„}
a sequence of functions which are integrable on [tf,&], and
is
converges uniformly on [a ,b] to a function
}
/
e
on
pb
/ =
lim
n
Jit
Let
integrable
is
Then
.
pb
PROOF
/ which
>
Thus,
0.
if
n
There
N
>
is
some
N
\f(x)
— fn (x)\ <
/„.
/
^°° J a
such that for
N
>
n
all
we have
for all x in [a, b].
£
we have
pb
pb
f{ X
I
)dx-
f [f(x)~ fn (x)]dx
fn (x)dx
I
Ja
Ja
Ja
4
\f(x)-fn (x)\dx
edx
=
Since
this
is
>
true for any e
0,
pb
/=
/
lim
Ja
treatment of continuity
a).
follows that
it
pb
The
—
s(b
/„.|
/
Ja
is
only a
bit
little
"e/3-argument," a three-step estimate of \f(x)
more
involving an
difficult,
— f(x + /?)!•
If {/„}
of continuous functions which converges uniformly to /, then there
is
is
a sequence
some n such
that
Moreover, since /„
(1)
\f(x)-fn (x)\ <
(2)
\f {x+h )-
|,
fn {x +
<
h)\
|.
we have
continuous, for sufficiently small h
is
~
\f„(x)
(3)
+ h)\ <
fn(x
s
3
It will
follow from
however,
we must
(1), (2),
that
THEOREM
2
(2) true,
it
is
\f(x)- f(x+h)\ <
that
of
\h\
in
a
way
that
s.
In order to obtain
cannot be predicted
therefore quite essential that there be
no matter how small
\h\
may be
—
it is
some
(3),
until n
fixed n
precisely at this point
uniform convergence enters the proof.
Suppose
that {/„}
is
a sequence of functions which are continuous on
that {/„} converges uniformly
PROOF
(3)
restrict the size
has already been chosen;
which makes
and
For each x in [a b]
,
with x
in (a,
/?);
on
[a, b] to /.
we must prove
the cases x
—
that
a and x
—
/
is
Then /
is
[a, b],
also continuous
continuous
at
x
.
We
on
and
[a, b].
will deal
only
b require the usual simple modifications.
504
Infinite
Sequences and Infinite Series
Let £
>
/ on
Since {/„} converges uniformly to
0.
[a, b],
there
is
some n such
that
1/00
In particular, for
Now
/„
is
-/»(.v)l
h such that x
all
\f(x
<
+ h)-
8,
h
is
(2)
|/(.v+//)-./;
is
some
8
all
in [a, b],
|/(*) -/„(*)!
continuous, so there
if \h\
+
for
(1)
<
)
>
v in [a,b].
we have
|.
U+
/z)|
<
|.
such that for
|/j|
<
<5
we have
\fn(x)-Mx+h)\ <-.
(3)
Thus,
<\
then
f(x)\
+ fn (x) - f(x)\
- fn (x)\ + \f„(x) - f (x)\
=
\f(x
+h)- fn (x +h) + fn (x +h)- f
<
\f( X
+h) - /„(X+A)| +
n
SEE
\fn(-X+h)
(x)
"3+3+3
=
This proves that
£.
/
is
continuous
at x. |
f(x)
FIGURE
=
\x\
S
After the two noteworthy successes provided by
the situation for differentiability turns out to
differentiable,
that
/
is
and
if
{/„} converges
Theorem
1
uniformly to /,
it
is still
{
f„
}
2,
each /„
is
which converges uniformly
If
not necessarily true
differentiable. For example, Figure 8 shows that there
differentiable functions
and Theorem
be very disappointing.
is
a sequence of
to the function
f(x)
=
\x\.
505
24. Uniform Convergence and Power Series
Even
/
if
is
differentiable,
it
may
not be true that
fix)
this
is
not at
all
by very rapidly
surprising
if
we
=
lim
n-»oo
fn '{x);
a smooth function can be approximated
example (Figure 9), if
reflect that
oscillating functions. For
1
2
fn (x) = -sin(n x),
n
then {/„} converges uniformly to the function f(x)
f„'{x)
and lim n cos(h 2 *) does not always
—
=
0,
but
n cos(n x),
exist (for
example,
it
does not exist
if
x
= 0).
fn
FIGURE
-19
Despite such examples, the Fundamental
antees that
rem
1;
some
sort
Theorem of Calculus
of theorem about derivatives
the crucial hypothesis
will
practically guar-
be a consequence of Theo-
that {/,/} converges uniformly (to some continuous
is
function).
THEOREM
3
Suppose
that {/„}
is
a sequence of functions which are differentiable
with integrable derivatives
moreover, that
Then /
is
{/„'}
/„',
and
Applying Theorem
1
[a, b] to
some continuous function
and
f'{x)
PROOF
=
lim /„'(*)
to the interval [a, x],
/
g
=
=
lim
/
we
see that for each x
/„'
lim [/„(*)- f„(a)]
n-»-oo
=
[a,£],
that {/„} converges (pointwise) to /. Suppose,
converges uniformly on
differentiable
on
f(x)-f(a).
we have
g.
.
506
Infinite
Sequences and Infinite Series
Since g
is
continuous,
=
follows that f'(x)
it
=
g(x)
lim f„'(x) for
all
x
in the
n—>oo
interval [a, b\. |
Now
how
that the basic facts about uniform limits have
been
established,
it
is
clear
to treat functions defined as infinite sums,
/(x)
=
/l(*)
+ /2(*) + /3 (*) + ••
•
This equation means that
f(x)=
lim
+
/iOO
+ /„(*);
•••
our previous theorems apply when the new sequence
/l,
+ /2, /1+/2 + /3,
/l
converges uniformly to /. Since
in,
DEFINITION
we
The
single
series
it
this
the only case
is
we
shall ever
be interested
out with a definition.
/ fn converges uniformly (more
uniformly summable) to/ on A,
h,
We can now apply each
may be
+ /2,
/1
formally: the sequence {/„}
is
the sequence
if
+ /2 + /3,
/1
...
/ on A
converges uniformly to
the results
•••
of Theorems
stated in
one
1, 2,
common
and 3
to
uniformly convergent
series;
corollary.
00
COROLLARY
Let
2. fn
/ on
converge uniformly to
[a, b\.
n=\
(1)
If
(2)
If
each /„
is
continuous on
/ and each
/„
is
[a, b],
integrable
Ja
on
then
/
[a, b],
1
is
continuous on
[a, b].
then
Ja
00
Moreover,
\] fn
if
converges (pointwise) to
/ on
[a b]
,
,
each /„ has an integrable
n=\
derivative /„'
and 2_.
fn'
converges uniformly on
n=\
tion,
then
00
(3)
f'{x)
= ^2f ,'(x)
t
n=\
for
all
x
in [a.b\.
[a, b] to
some continuous
func-
)
507
24. Uniform Convergence and Power Series
PROOF
each /„ is continuous, then so is each
of the sequence f\ f\ + fj, f\ + fi + /3,
(1)
If
,
(2)
Since f\, f\
Theorem
1
+ fn,
+ fi + f3,
f\
•
.
.
.
,
/
so
is
and /
is
the uniform limit
continuous by
converges uniformly to /,
•
it
Theorem
2.
follows from
that
f=
f
f (/,+•• +
lim
=
lim
rt—>00
tf
E/
(3)
•
\-fn ,
f\-\
Each function
/i
+
•
•
•
4- /«
/i+
/„)
-+
r/
"
/»
differentiable, with derivative
is
and f\ ', f\' + f2, f\ + fi + yV
converges uniformly
by hypothesis. It follows from Theorem 3 that
'
,
•
/'(*)= lim
[fi'(x)
/V
+
•
•
•
+
/,/,
to a continuous function,
+ /„'(*)]
+
00
n=l
At the moment
this corollary
is
not very useful, since
it
seems quite
difficult to
when the sequence f\ f\ +/2, /1+/2+/3,
will converge uniformly The
most important condition which ensures such uniform convergence is provided by
the following theorem; the proof is almost a triviality because of the cleverness
predict
•
,
•
•
with which the very simple hypotheses have been chosen.
THEOREM
4
(THE WEIERSTRASS M-TEST)
Let {/„} be a sequence of functions defined on A, and suppose that {M„}
a
is
sequence of numbers such that
\fn(x)\
< M„
x
for all
in A.
00
00
Suppose moreover that \_. M„ converges. Then
for
each
A'
in
A
the series Y_] fn U"
00
converges
(in fact,
it
converges absolutely), and 2_. fn converges uniformly on
n=\
to the function
00
/(*)
=
2
n=\
/»(*)
A
508
Infinite
Sequences and Infinite Series
PROOF
For each x
A
in
the series
/J|/n (jc)|
converges, by the comparison
test;
COnse-
oo
quently /_/«(*) converges (absolutely). Moreover, for
all
x
in
A we
have
n=\
/U)-[/lOO +
•••
+
=
/*(*)]!
oo
n = /V+l
oo
£
<
M-
n=W+l
Since 2_.
M„
number
converges, the
=
A/V
f
x
(x)
T
r
\{2x)
FIGURE
M
n
can be
made
as small as desired,
n=N+\
n=\
by choosing
1
\_.
N
sufficiendy large. |
10
T
1
The
following sequence {/„} illustrates a simple application of the Weierstrass
M-test.
(a)
f{x)
=
Let {x} denote the distance from x to the nearest integer (the graph of
{x}
is
illustrated in Figure 10).
Now
define
f2 (x) = \{Ax)
The
functions f\
10" has
and
fj are
been replaced by
shown
2").
in Figure
1 1
(but to
the Weierstrass M-test automatically applies: clearly
(b)
l/«(*)|
FIG UK
I.
I
I
make
the drawings simpler,
This sequence of functions has been defined so that
<
for
all
x
,
.
24. Uniform Convergence and Power Series
and y^ 1/10" converges. Thus /_,f" converges uniformly;
since each /„
is
509
con-
tinuous, the corollary implies that the function
00
oo
i
/(*)
= £/»<*> =
is
•
•
Figure 12 shows the graph of the
also continuous.
•
+
/„
As n
.
E lo^lO"*}
n=\
n=\
few partial sums
first
become harder and harder
increases, the graphs
to draw,
f\
+
and
oo
the infinite
sum N
/„
is
shown by
quite undrawable, as
the following
theorem
n=\
(included mainly as an interesting sidelight, to be skipped
if you
find the going too
rough).
THEOREM
5
The
function
/^» = ET7^l
10"
|
0".v|
n=\
is
PROOF
continuous everywhere and differentiable nowhere!
We
have just shown that
/
uses uniform convergence.
is
continuous;
We
this
is
the only part of the proof which
prove that
will
/
not differentiable at a, for
is
by the straightforward method of exhibiting a particular sequence
approaching
for which
any
a,
+ h m )-f(a)
f(a
..
lim
m-KX>
does not
0<a<
exist.
hm
obviously suffices to consider only those numbers a satisfying
It
1.
Suppose that the decimal expansion of a
=
a
Let h m
{h,„}
=
10
- '"
if
/
am
4 or
9,
but
{).a\a2CiT,a$ ....
let
~
km
^
= - 10"'"
hm
these two exceptions will appear soon).
f(a+h m )-f(a)
is
am
if
-
4 or 9
)}
-
(the
reason for
Then
'
±10"'"
10"
n=l
oo
=
This
infinite series
is
^±10m - M [{10"(a +h m
really a finite
sum, because
if
n
> m, then
{10"«}],
10"/?,,, is
so
{\0"(a+h„,)}-{\0"a}
On
the other hand, for n
10"<7
\0"(a
+ hm
)
=
=
<
m we
=
can write
+ 0.a n+ ian+ 2a n+ 2
integer + 0.a n+ \a n+ 2an +3
integer
0.
.
.
.
am
.
(a,„
.
±
1)
•
•
•
an
integer,
510
Infinite
Sequences and Infinite Series
(in
order for the second equation to be true
— 10 -m when
=
a,„
Now
9).
0.a n+ \a n+ 2a n+ 3
Then we
.
.
hm
m =
the special case
=—
10~ m
essential that
is
we choose
when a m =
n
4).
{10
.a h
for n
=
si-
n
+
iflm
±
1)
•
•
•
<
the second equation
1
i
m we
we chose
true because
is
This means that
(a+h m )}-{\0 n a} =
±l(y m
i
,
and exactly the same equation can be derived when 0.a n+ [Ci tl+ 2a n+
<
hm
have
also
0.a„ + \a n+2 a n +3
(in
it
suppose that
T,
•
•
•
>
\-
Thus,
have
m
10 -"[{10"(fl
+ //„,)}
-{10"«}]
=
±1.
In other words,
f(a+h m )-f(a)
sum of m — 1 numbers, each of which is ±
Now adding +1 or — 1 to a
number changes it from odd to even, and vice versa. The sum of m — 1 numbers
each ±1 is therefore an even integer if m is odd, and an odd integer if m is even.
is
fi
+
f2
+h
the
1
.
Consequently the sequence of ratios
+ h m )-f{a)
f(a
hm
cannot possibly converge, since
odd and even.
its
role in the previous
tool for analyzing functions
fi
+h+
h
a sequence of integers which are alternately
|
In addition to
A+
is
it
attention to functions of the
theorem, the Weierstrass M-test
which are very
form
well behaved.
We
is
an
ideal
will give special
oo
= J2 a n(x-a)'\
f(x)
n=0
which can
also
be described by the equation
00
«=0
FIGURE
12
for f„(x)
=
a„(.x
on powers of
simplicity,
we
(x
—
a)".
—a),
is
will usually
Such an
called a
infinite
power
sum, of functions which depend only
series centered at
a.
For the sake of
concentrate on power series centered at 0,
00
f(x)
=
y\/„.v".
n=0
—
•
511
24. Uniform. Convergence and Power Series
One
group of power
especially important
where /
of the form
some function which has derivatives of all orders at a;
Taylor series for/ at a. Of course, it is not necessarily
this series
is
called the
equation holds only
when
is
true that
"
;/<
Y
)
=
ft ,
f(x)
this
series are those
,(fl)
r
(x
-
-a)V
;
the remainder terms satisfy lim R„ M (x)
n—>oo
= 0.
00
We
all
already
x. For
know
power
that a
example, the power
series
\x\
<
1
,
while the power series
— 1 < jc <
converges only for x
—
0.
x1
x5
X2
converges only for
a n x" does not necessarily converge for
series
x3
converges only for
>_^
1
It is
.
X4
X3
X5
even possible to produce a power
series
which
For example, the power series
Ix"
E»
n=0
does not converge for x
/
0;
indeed, the ratios
(n+
l)!(.t"
+1
)
=
(n
n xY n
+
l)x
I
!
are
unbounded
for
any x
^
0.
If a
power
series
V,a„x" does converge
for
00
some xq
\x\
theorem
6
<
|x
Suppose
7^
however, then a great deal can be said about the series
2, a n*
n
for
1
that the series
oo
/(.to)
= ^2a„x
"
«=0
converges, and
let
<
a be any number with
00
n=0
a
<
\xq\.
Then on [—«,«]
the series
1
512
Infinite
Sequences and Infinite Series
converges uniformly (and absolutely). Moreover, the same
is
true for the series
oo
g(x)
Finally,
/
is
= }na n x n ~
l
.
and
differentiable
oo
n=\
for all x with
PROOF
Since
<
|.v|
\xq\.
Yja„*o" converges,
the terms a n XQ
,i=0
bounded: there
is
some number
n
|<z„jco
Now
if
x
is
|
[—a, a], then
in
\a n x
n
\
=
\a n
<
\a n
.
\
approach
M such that
\xq"\ < M
=
\a„\
\x\
<
\x
n
0.
for
Hence
they are surely
all n.
so
\a\,
n
\
•
\a"\
•
1*0
\
a
KM
<
M
the clever step)
(this is
XQ
a
XQ
But
<
|«/jcq|
1,
so the (geometric) series
EM f
«=o
converges. Choosing
M
a
ME
=
*o
n=0
number M„
\a/xo\" as the
in the Weierstrass M-test,
it
00
follows that
j>a„x" converges uniformly on [—a,
a].
n=0
To prove
the
same
assertion for g(x)
\na n x
n
|
—
>/2a„jc"
=
n\an
<
n\a„\
M
\
\x"
\a"-
n
\x
\
1
a
n
XQ
\a\
<
'
Since \q/xq\
<
1,
M n —a
—
\a\
notice that
,1
*0
the series
Am
a
M
XQ
rr /
a
y-^
n
XQ
n=\
converges
(this fact
was proved
in
Chapter 23
as
an application of the ratio
test)
513
24. Uniform Convergence and Power Series
Another appeal
"
2_, na n x
to the Weierstrass M-test proves that
converges uni-
n=\
formly on [—a,
Finally,
a].
our corollary proves,
that g
first
continuous, and then that
is
oo
=
/"(a)
g(x)
—
2_, na nX
n~
for
[—a,
in
A"
a].
n=\
Since
all
we could have chosen any number a
< |ao|. |
x with a
|
We
are
<
with
<
a
holds for
\xo\, this result
|
now
manipulate power
in a position to
manipulations are
fairly
series
with ease. Most algebraic
straightforward consequences of general theorems about
OO
—
For example, suppose that f(x)
infinite series.
where the two power
series
00
x"
2_. a »
„=0
both converge for some
= y^b„x n
and g(x)
,
n=0
Then
ao.
for |a|
<
|ao|
we
have
OO
OO
00
00
n
"
)xn
J2 "nx + J2 b » x " = J2 {a " x + b » x ") = J2 {a " + b "
n=0
n=0
n=0
-
,1=0
oo
So the
series
for these a
The
h(x)
=
+ b n )x"
/,(««
|a|
<
|ao|,
and h
= f+g
<
then
.
treatment of products
is
just a
that the series
more
little
\^a„x n and 2_,b n x n converge
|a|
|ao|,
we
So
absolutely.
it
follows
from
n=0
m=0
OO
Theorem 23-9
involved. If
OO
oo
know
also converges for
00
V^ a n x" \\ b n x "
that the product
given by
/;=0
/i=0
oo
1S
oo
1=0 7=0
where the elements djX'bjXJ are arranged
choose the arrangement
aobo
+
(aob\
which can be written
+ a\bo)x +
(ao^2
n
for cn
n=0
the
In particular,
+ a\b\ + a2bo)x L
we can
-\
n
\cn x
/
is
any order.
as
oo
This
in
= ^Y\a
k
b„- k
.
k=0
"Cauchy product"
that
was introduced
in
Problem 23-10. Thus,
the
= fg
for
00
Cauchy product h(x)
= /
j
»=o
these a.
Cn x
n
also converges for
|a|
<
|aq|
and
//
a
514
Infinite
Sequences and Infinite Series
Finally,
— y^
suppose that f{x)
n
nx
where
,
so that /(0)
¥" 0,
ciq
=
^
ao
0.
„=0
oo
Then we can
try to find a
power
>^ b n x" which
series
represents \/f. This
means
„=0
that
we want
have
to
oo
oo
J2
"
J2 b " X =
anX*
left
+
1
2
+
.V
•
•
A"
+
•
•
•
.
„=0
,i=0
Since the
=
1
be given by the Cauchy product, we want
side of this equation will
have
to
a b
1
aob\
«0^2
Since «o
7^ 0,
we can
solve the
=
+ a\bo =
+ a\b\ + aibQ =
Then we can
of these equations for bo.
first
solve
oo
the second for b\
,
etc.
Of course, we
have
still
prove that the
to
new
>^ b„x"
series
„=o
^
does converge for some x
0.
This
Theorem 6 gives us all
when we apply Theorem 6 to the infinite
For derivatives,
J
=
1
cosjr
get precisely the results
A
v
x
+
+
we
need. In particular,
series
5
X
+
1
-a
={+
e
we
x
the information
18).
--- + X--...,
3
X
:--
sin*=
an exercise (Problem
as
is left
4
6
8
.V
4!-&
?
A
9
7
X
A3
+
v.
.V
^-"-'
+
+
4
A
+
---<
v.
y.
which are expected. Each of these converges
hence the conclusions of Theorem 6 apply
3a 2
5a
—+—
sin (a)
=
,
1
cos(A)
=
2a
4a
-+
^
exp (x)
=1+—+
for
=
—+
For the functions arctan and f(x)
=
:
3a~
=
H
log(l
...
+a)
= -smA,
exp(A).
the situation
complicated. Since the series
arctan a
=a
—— + —
A
3
ao,
cos a,
6a 5
3
-^-
any
4
r
2a
any a
for
A
5
A7
—
-\
is
only slightly
more
. .
.
24. Uniform Convergence and Power Series
=
converges for xq
1
,
=
arctan'(jc)
also converges for
it
1
|jc
—x +x —x +
•
<
|
•
1
and
,
=
•
x
+ xz
1
=
In this case, the series happens to converge forx
for the derivative
is
=
not correct for x
=—
or x
1
l-x 2 +x 4 -x 6 +
515
-1
1
also.
for \x
<
1
However, the formula
indeed the
;
\
series
...
diverges for x — 1 and x = — 1 Notice that this does not contradict Theorem 6,
which proves that the derivative is given by the expected formula only for \x\ < |*o|.
.
Since the series
"•345
X
X
X
X
log(l+ , )=x __ + ___+__...
converges for xq
1+*
—
=
1
,
it
log'(
also converges for \x
1
+x =
)
1
—
x
<
\
1
,
and
+ x — x' +
•
•
•
differentiated series does not converge for x
All the considerations
its
which apply
derivative, at the points
where the
|
= — 1;
<
1
moreover, the
1
power
to a
jc
|
In this case, the original series does not converge for x
=
for
derivative
series will automatically
is
apply to
represented by a power
series.
If
00
converges for
all
x in some interval
(
/"(a)
— R,
=
^na
nu
n
for all x in
(
— R.
R).
R), then
=
x
nx
Theorem 6
'
ni-i
\
Applying Theorem 6 once again we find
f"{x)
implies that
= ^n{n-
\)a n x"-
that
2
,
n=2
and proceeding by induction we
find that
oo
Thus, a function defined by a power
(
— R,
R)
is
series
which converges
in
automatically infinitely differentiable in that interval.
previous equation implies that
f«\0)=k\ak
so that
ai
w
= / (0)
.
,
some
interval
Moreover, the
516
Infinite
Sequences and Infinite Series
In other words, a convergent power
function which
On
happy note we could
this
A
series.
series centered at
always
is
the Taylor series at
of the
it defines.
end our study of power
easily
series
and Taylor
some unexplained
careful assessment of our situation will reveal
facts,
however.
The Taylor
series
they converge for
of
x,
all
and exp are
sin, cos,
and can be
—
differentiated term-by-term for
—
Taylor series of the function f(x)
we could
as satisfactory as
+ x)
log(l
all
slightly less pleasing,
is
desire;
x.
The
because
it
< x < 1 but this deficiency is a necessary consequence of
power series. If the Taylor series for / converged for any xo
with |xo > 1? then it would converge on the interval (— |jto|, |*ol), 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 (—
), where it equals
converges only for
1
,
the basic nature of
I
1
,
1
log(l+*).
The Taylor
series for arctan
more
is
be no possible excuse for the refusal of this
mysterious behavior
+
1/(1
x
an
),
exemplified even
is
The Taylor
f(x)
=
>
If \x
=
=
1
series
+ x2
-
1
of
2
.v
strikingly
which
/
+
is
4
-
.v
answer:
always dangerous, since
is
it
there does
|jc|
>
1.
This
by the function f(x)
=
the next best thing to a
8
.r
.
Why? What unseen
and —1? Asking
1
we may have
+
to settle for
obstacle
this sort
of
an unsympathetic
—
intelligently
Taylor
when
to
happens because it happens that's the way things are! In this case
happen to be an explanation, but this explanation is impossible to give
at the present time;
to devote
6
.v
all.
prevents the Taylor series from extending past
question
is
— there seems
given by
the Taylor series does not converge at
1
I
converge
series to
more
infinitely differentiable function
polynomial function.
comprehend
difficult to
only
although the question
when placed
two chapters
series in
in
to quite
is
about real numbers,
a broader context.
new
It will
it
can be answered
therefore be necessary
material before completing our discussion of
Chapter 27.
PROBLEMS
1.
For each of the following sequences
{/„}
(if it exists)
uniformly to
(i)
(ii)
on the indicated
—
determine the pointwise
and decide whether
this function.
fn (x)=yx~,
fn (x)
{,/*„},
interval,
I
\
x
on
— n,
[0.1].
x >
on
ii.
\a,b],
and on R.
limit
of
{/„} converges
24. Uniform Convergence and Power Series
fn (x) =
—
x
(iv)
fn {x) =
e~ nx \
(v)
/„(*)
(iii)
2.
517
on (l,oo).
,
n
on [-1,1].
= ^, onR.
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.
,
2n
fn (x) = x"-x on
(i)
[0,1].
nx
(ii)
on
[0, oo).
fn (x) = Jx 2 + -ron
2
[a, oo),
/„(*)=
+n + x
1
(iii)
a
>
0.
n
= J x2 +~2 onR
(iv)
/„(*)
(v)
fn (x) = Jx
H
/„(x)
= Jx
-\
on
[a, oo),
yfx
on
[0, oo).
a
>
0.
n
V
(vii)
/„(x)
= « Jx
H
v^ pn[a,
oo), a
(viii)
fn (x) - nlJx
A
Jx
oo)
Hints:
(i)
on
maximum
- fn (x)\
for
x
large,
separate estimate of \f(x)
— fn (x)\
Find the Taylor
for
(i)
/(*)=
x
series at
—
=
log(*
(iii)
f(x)
=
-== =
(iv)
—
-
a),
x
fix)
'
Vl-x 2
v)
fix)
=
and on
(0, oo).
(iii)
- f„\ on [0, 1].
Mean Value Theorem,
of \f
for small
(ii)
|jc|.
(vii)
Use
(v).
each of the following functions.
a^0.
,
f(x)
1
0.
a
(ii)
V
[0,
>
J
For each n, find the
consider \f(x)
3.
y/x
h
V
(vi)
-
arcsinx.
(1
a
<
-
x)~
0.
x
'2
.
(Use Problem 20-21.)
For each n,
(iv)
Give a
518
Infinite
Sequences and Infinite Series
4.
Find each of the following
(i)
l
-' +
(ii)
1
-
Hint:
i- i + i--
What
X2
.....
x
+ x 6 -x 9 +
x3
X J3
y-r^
(111)
sums.
infinite
is
+
1
-
---
x
for \x
|
+ x2 - x3
X*4
XJ
<
1.
?
H
5
4T3-5^
+
"' lor|x!<1
-
Hint: Differentiate.
5.
Evaluate the following
is
infinite
sums. (In most cases they are f(a) where a
some obvious number and f(x) is given by some power series. To evalupower series, manipulate them until some well-known power
ate the various
series emerge.)
A
^
,.x
(-1)"2 2 "tt 2 "
(2n)!
oo
-,
y
y,
^2,, +
/l\ 2 " +l
1
in
lUJ
OO
iv
n=0
oo
,
^3»(»+l)
A
^
VI
If
f{x)
power
2n
=
+
1
2»n!
(smx)/x
for x
^
and /(0)
=
1,
find
/ w (0).
series for /.
In this problem
without
all
in part (a)
the
we deduce
the binomial series (1
work of Problem 23-2
of that problem
— the
1
,
series
although
f(x)
+x) a = /
we
= Vj
<
1.
(
l.v". \x\
<
l
will use a fact established
x"
I
)
^"'
«=0
\x\
Hint: Find the
does converge for
519
24. Uniform Convergence and Power Series
(a)
(b)
Prove that
(1
+x)f'(x)
=
oef(x) for
|x|
<
1.
Now show that any function / satisfying part (a) is of the form f(x) =
a
c(l + x) for some constant c, and use this fact to establish the binomial
a
series. Hint: Consider g(x) = /(jc)/(1 + x)
.
8.
Suppose that
/„ are
nonnegative bounded functions on
A and
OO
on A, does
sup/,,. If 2_jf" conver Ses uniformly
verges
9.
(a
let
M„ =
00
it
follow that
\] M„
con-
converse to the Weierstrass M-test)?
Prove that the
series
00
y
^
x
n(\
+nx 2
)
converges uniformly on R.
10.
(a)
Prove that the series
00
,
^
3nx
n=0
converges uniformly on
(b)
By considering
series
11.
(a)
the
[a,
oo) for a
N
sum from
>
Hint:
0.
x
to oo for
does not converge uniformly on
=
\im(smh)/h
=
1.
2/(n3 N ), show that the
(0, oo).
Prove that the series
oo
nx
>
4
+ rrx-
,
,
1
converges uniformly on
of nx/(l
(b)
Show
+n
4
x
2
)
on
[a, oo) for
a >
0.
L
«3«
Hint: First find the
maximum
[0, oo).
that
'
\NJ ~
2
n>VN
and by using an integral to estimate the sum, show that f(\/N~) >
Conclude that the series does not converge uniformly on R.
(c)
What about
the series
oo
E
n=U
nx
;>
1/4.
520
Infinite
Sequences and Infinite Series
12.
(a)
Use Problem 15-33 and Abel's Lemma (Problem 19-36)
form Cauchy condition", showing that for any e > 0,
to obtain a "uni-
sin/cx
E
k=m
can be
ing
m
made
on the whole interval [e, In - e] by chooslarge enough. Conclude that the series
arbitrarily small
(and n)
00
E
converges uniformly on
(b)
For
jc
=
nx
sin
[e, 27T
—
e] for s
show
7r/N, with TV large,
>
0.
that
N
IN
Vj sin kx
>
yv
7T
k=N
Conclude
fc=0
that
2N
sin A:x
E
>
27r'
k=N
and
that the series does not converge uniformly
on
[0,
2n\.
oo
13.
(a)
(b)
Suppose
that /(*)
=
^a x
n
converges for
n
all
x in some interval
(-R, R) and that f(x) = for all x in (-/?, /?). Prove that each a„ = 0.
is easy.)
(If you remember the formula for a„ this
for some sequence {xn } with
Suppose we know only that /(*„) =
show that
lim x„ = 0. Prove again that each a„ = 0. Hint: First
n—>oo
/(0)
=
=
fl
0;
then that
;
/
(0)
=
a\
This result shows that if /(*) = e~
possibly be written as a power series.
=0,
l/x2
etc.
l/x for x
sin
/
0,
then
/ cannot
shows that a function defined
thus a
but nonzero for x >
for x <
by a power series cannot be
remained
has
which
power series cannot describe the motion of a particle
at rest until time 0, and then begins to move!
It
also
oo
oo
(c)
Suppose
in
some
that /(*)
= £>„*"
m } converging to
0.
Show
and g{x)
and
interval containing
{t
—
=
that /(/,„)
that a„
=
^b x"
n
=
converge for
g(tm ) for
b n for each
all
x
some sequence
n.
00
14.
Prove that
and
if
/
is
if
f(x)
-
^a
nx
n
is
an even function, then a„
an odd function, then a„
=
for n even.
=
for n
odd,
—
24. Uniform Convergence and Power Series
15.
Show
x
<
power series for fix) — log(l — x) converges only for —1 <
and that the power series for g(x) = log[(l + x)/(\ — x)] converges
that the
1,
only for x in (— 1,
*16.
521
1).
Recall that the Fibonacci sequence {a n }
(a)
Show
(b)
Let
that a n+ \/a n
f(x)
<
= J2 a
n
is
—
defined by a\
=
ai
\,
a n+
\
=
2.
x"-
=
]
+
1
+
a-
2
2.x-
+ 3.v 3 +
n=\
Use the
(c)
ratio test to prove that
Prove that
<
if \x\
f(x) converges
partial fraction
series for \/(x
(e)
—a),
+x —
x2
1/2.
f(x)—xf(x) — xf(x)=
decomposition for i/(x
to obtain
series
another power
+x —
1.
and the power
1 ),
series for /.
/ must be
the
same
(they are
of the function), conclude that
—
an
<
1
Since the two power series obtained for
both the Taylor
|
-1
=
Hint: This equation can be written
Use the
|jc
1/2, then
/(*)
(d)
if
.
V5
17.
Let f(x)
=
2_, a nx"
= /
and g(x)
— }
]cn x
n
Suppose we merely knew
.
that
n=0
n={)
oo
f(x)g(x)
bn x
n
for
some
cn , but
we
didn't
know how
to multiply series
n=0
series for
Use Leibniz's formula (Problem 10-20) to show directly that this
fg must indeed be the Cauchy product of the series for / and g.
Suppose
that f(x)
in general.
18.
=
2_.°nx" converges
for
some
A'o,
and
that «o
7^
0;
n=0
for simplicity, we'll
recursively
assume
that
ciq
=
1.
Let {b„} be the sequence defined
by
b
=
1
n-\
b„
=
-}b a
k
k=0
n -k-
'
522
Infinite
.
Sequences and Infinite Series
The aim
of
problem
this
show
to
is
y^b n x"
that
some
also converges for
«=0
xc
7^ 0,
so that
represents 1\/f
fc
small
// for
it
< M, show
If all |a„.vo"|
a)
enough
\x
that
n-\
|/WI<M^|W|k=0
Choose
(b)
M so that
|c/„.v () "|
< M, and
also so that
M/(M 2 -
1)
<
Show
1.
that
|/WI<M
2 ".
oo
Conclude
(c)
\]b„x n
that
converges for
|x| sufficiently
small.
n=0
oo
s
19.
Suppose
that
/_,
n=0
oo
a " conver ges
-
We know
that the series f(x)
=
>_^a„jc"
n=0
must converge uniformly on [—«,«] for
< a < 1, but it may not converge
uniformly on [—1,1]; in fact, it may not even converge at the point —1
(for
example,
f(x)
if
shows that the
=
log(l
+ x)).
However, a beautiful theorem of Abel
converge uniformly on
series does
/
is
oo
m
Theorem
and, in particular, /_, an ~ ^
n=0
by noticing that if \am + -+a„\ < s, then
by Abel's
Lemma
continuous on
Consequently,
[0, 1].
oo
[0, 1]
/. a » x
n=0
\a m x
m
'"
+
Prove Abel's
+ a„x"\ <
s,
(Problem 19-36).
00
20.
A
sequence
{a,,}
is
called
Abel summable
if
lim
\] a n x "
exists;
Prob-
n=0
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,
even though the function it represents is continuous at
1
21.
(a)
1111
Using Problem
W
2
"
•
1
19, find the following infinite sums.
3
•
2
+
4
3
5
•
4
+"
'
00
OO
(b)
Let 2_. c n be the Cauchy product of two convergent power series 2_^ a
n=0
oo
oo
and
y_\ b>n
ancl suppose merely that
00
converges to the product
y. c n
„=o
n=0
it
>,
«=0
00
2, a n /^ b n
'
n=0
«=0
.
converges. Prove that, in
fact,
.
24. Uniform Convergence and Power Series
22.
(a)
Show
that the series
x
E
+
In
converges uniformly to A
1
converges to log
it
(Problem
23.
(a)
Suppose
x n+\
2n+\
71=0
at
+
log(A'
(Why
2.
+2
In
1
1)
on [—a,
/
(b)
doesn't this contradict Abel's
that {/„}
Find a sequence of continuous functions on
unbounded
/
that
function
/ on
[a, b]
[a, b].
Prove that
which converge point-
[a b]
,
Prove that the function /'
differentiable.
is
on
to
ples of discontinuous derivatives, this provides another
pointwise limit of continuous functions
26.
Theorem
is
the pointwise
know examexample where the
sequence of continuous functions. (Since we already
limit of a
25.
but that
a sequence of bounded (not necessarily continuous)
is
on [a, b] which converge uniformly
bounded on [a, b].
Suppose
1,
19)?)
wise to an
24.
< a <
a] for
functions
is
523
is
not continuous.)
Find a sequence of integrable functions {/„} which converges to the (nonin-
on the
tegrable) function
/
Each
except at a few points.
(a)
/„ will
be
Prove that
integrable
/
if
on
that
is
is
1
rationals
and
the uniform limit of {/„}
then so
[a, b],
is
on
the irrationals. Hint:
on [a,b] and each
/. (So one of the hypotheses in
/„
is
Theorem
1
was unnecessary.)
(b)
In
Theorem
we assumed only
3
that {/„} converges pointwise to /. Show-
that the remaining hypotheses ensure that {/„} actually converges uniformly to /.
(c)
Suppose that
in
Theorem
(d)
Show
we do not assume
assume only
tion /, but instead
[a, b].
3
{/„} converges to a func-
that f„(xo) converges for
that /„ does converge (uniformly) to
Prove that the
some /
some xo
in
=
g).
(with /'
series
E (-D"
+n
,
converges uniformly on
27.
Suppose
to /.
x
[0, oo).
that /„ are continuous functions
l//i
Km
[0, 1] that
true
if
/
the convergence
/•!
./"„
"^°° Jo
Is this
on
Prove that
=
/
Jo
isn't
uniform?
/.
converge uniformly
524
Infinite
Sequences and Infinite Series
28.
(a)
Suppose
that {/„}
pointwise.
>
and
for all n
uniformly on
x
all
Suppose moreover that we have fn (x) > fn+ \(x)
Prove that {/„} actually approaches
in [a,/?].
Hint: Suppose not, choose an appropriate sequence
[a, b].
of points x„ in
(b)
a sequence of continuous functions on [a, b] which
is
approaches
[a, b],
and apply the Bolzano-Weierstrass theorem.
Prove Dini's Theorem: If {/„} is a nonincreasing sequence of continuous
on [a,b] which approaches the continuous function / point-
functions
approaches / uniformly on
a nondecreasing sequence.)
wise, then {/„} also
(c)
holds
if
Does
Dini's
is
{/„}
Theorem hold
replaced by the open interval
29.
(a)
Suppose
that {/„}
How
continuous?
if /' isn't
(The same
[a, b].
about
result
b]
if [a,
is
(a, b)?
a sequence of continuous functions on [a,b] that
is
converges uniformly to /.
Prove that
if
x n approaches x, then fn (x n )
approaches f{x).
(b)
(c)
Is this
statement true without assuming that the /„ are continuous?
/ is continuous on [a,b] and {/„} is
a sequence with the property that fn (x n ) approaches f(x) whenever x n
approaches x, then /„ converges uniformly to / on [a,/?]. Hint: If not,
and a sequence x n with \fn (x n — f(x„)\ > s for infinitely
there is an s >
many distinct x n Then use the Bolzano-Weierstrass theorem.
Prove the converse of part
If
(a):
)
.
30.
This problem outlines a completely different approach to the integral; consequently,
(a)
Let
s
unfair to use any facts about integrals learned previously.
it is
be a step function on
[a,fr], so that s
is
constant on (/,_i,r,) for
"
pb
some
partition
{to,
.
.
.
,
tn }
of
[a
b]
,
Define
.
/
V_, s
s as
i=
Si
the (constant) value of s
is
not depend on the partition
(b)
A function
limit
{to,
(f,-_i,
.
.
Show
i
)
where
l
that this definition does
regulated function on [a b] if it is the uniform
{s n } on [a,b].
Show that in this
every s > 0, some /V such that for m n > jV we have
called a
is
— ?,•_
tn ).
,
.
U).
(U
,
of a sequence of step functions
case there
|^m (jc)
/
on
i
is,
for
— sm (x)\ <
(c)
Show that
(d)
Suppose
,
£ for all
.v
in [a, b].
the sequence of numbers
that {t„}
is
N we
>
Conclude
that
s„\ will be a
I
Cauchy sequence.
another sequence of step functions on [a,b] which
converges uniformly to /.
that for n
\
have
lim /
n—*oo JI a
\s n
s„
Show
that for every e
(x)
—
=
lim
t„(x)\
oo
/;—
/
<
tn
.
s for
x
>
there
N
such
we can
k'hne
is
an
in [a, b}.
This means that
J/ a
.
b
f
I
f
to
be lim
s„
for
any sequence of step functions
{s,,}
converging
J (I
uniformly to /.
regulated?
The
only remaining question
is:
Which
functions are
.
24. Uniform Convergence and Power Series
*(f)
Prove that a continuous function
is
regulated. Hint:
To
525
find a step func-
on [a, b] with \f(x) — s(x)\ < e for all x in [a, b], consider
for which there is such a step function on [a v]
Every step function s has the property that lim s(x) and lim s(x)
tion s
all
v
,
(g)
for
all
and
x—>a +
a. Conclude that every regulated function has the
find
an integrable function that
that, conversely,
every function
lim f(x) exist for
x—>a~
*31.
all
a
is
/
is
not regulated.
with the property that
exist
x—*-a~
same
property,
(It
also true
is
lim f(x)
and
x—>a +
regulated.)
Find a sequence {/„} approaching / uniformly on
lim (length of /„ on [0, 1]) / length of / on [0,
[0, 1] for
1].
which we have
(Length
is
defined in
n—»oo
Problem 13-25, but the simplest example
whose graphs will be obvious.)
will involve functions the length
of
b
CHAPTER
#VV
COMPLEX NUMBERS
With the exception of the
few paragraphs of the previous chapter,
last
has presented unremitting propaganda for the real numbers.
this
book
Nevertheless, the
—
real
root.
numbers do have a great deficiency not every polynomial function has a
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
=0. For a
number with the property that /" +
long time the status of the "number"
was quite mysterious: since there is no
=
number x satisfying x + 1
be the number
0, it is nonsensical to say "let
2
satisfying
+ = 0." Nevertheless, admission of the "imaginary" number to
the need to "invent" a
felt
/
1
i
/
i
1
the family of
especially
all
numbers seemed
to simplify greatly
when "complex numbers"
a
+ bi
valid.
many
algebraic computations,
a and b in R) were allowed, and
(for
enumerated
the laws of arithmetical computation
be
to
i
in
Chapter
1
were assumed
For example, every quadratic equation
ax
+ bx + c =
2
(a
/
0)
can be solved formally to give
x
=
-b +
\Jb
- 4ac
2
-b - v'
=
x
or
b
2
— 4ac >
0, these
4a c
.
formulas give correct solutions;
allowed the formulas seem to
-
La
La
If
2
make
x
2
sense in
all cases.
+x +
=
1
when complex numbers
are
For example, the equation
has no real root, since
x
But the formula
2
+x +
=
1
(.v
+
2
+
\)
\
>
for
0,
for the roots of a quadratic equation suggest the "solutions"
x
-l
—
+ V-3
,
and
x
= -l-V-3
2
if
all .v.
we understand y — 3
to
mean
;
2
y/3
(
— 1) = v 3-y — = v3 i,
1
then these numbers
would be
73
-- + —-/
1
2
It is
—
and
2
1
V3
2
2
i.
not hard to check that these, as yet purely formal, numbers do indeed
the equation
x
526
—
2
+x+
1
=0.
satisfy
527
25. Complex Numbers
It is
even possible to "solve" quadratic equations whose coefficients are themselves
complex numbers. For example, the equation
x
+x + + =0
2
1
/
ought to have the solutions
±71-4(1+/)
1
where the symbol v — 3 —
—3 —
4/. In
±7-3-4/
-1
means a complex number a
4/
+
fti
whose square
is
order to have
(a
+
pi)
2
=
a2 - p2
+
lafii
2
2
a - p
=
=
-4.
= -3 - 4/
we need
2oeP
These two equations can
-3,
be solved for real a and
easily
fi\
in fact, there are
two
possible solutions:
=
a
and
= -2
Thus
-
—3 —
the two "square roots" of
reasonable
way to
v— 3 — 4/;
a
1
4/ are
1
=—
= 2.
—
1
and — 1
2/
decide which one of these should be called
which case y/x denotes the
nonnegative root. For
(real)
=
x
this
There
2/.
V —3 —
makes sense only
the conventional usage of y/x
+
4/,
is
no
and which
for real x
>
0, in
reason, the solution
± 7-3 - 4/
1
2
must be understood
x
With
this
=
an abbreviation
as
-1 +r
—
—
-
where
,
understanding
we
r
is
for:
one of the square roots of —3 —
4/.
arrive at the solutions
-1
+ -2/
1
2
-1-1+2/
x=
as
you can
easily check, these
^
= - 1+
,
.
'
;
numbers do provide formal
x
2
+x + 1 + =
i
solutions for the equation
0.
For cubic equations complex numbers are equally useful. Every cubic equation
ax
with real coefficients a,
3
+
b, c,
bx
2
and
+ ex + d =
d, has, as
we know,
(a
# 0)
a real root a, and
if
we
divide
+ bx + ex + d by — a we obtain a second-degree polynomial whose roots are
the other roots of ax +bx -\-cx+d = 0; the roots of this second-degree polynomial
ax~
.v
528
Infinite
Sequences and Infinite Series
may
be complex numbers. Thus a cubic equation
or one real root and 2 complex roots.
The
will
have either three real roots
existence of the real root
is
guaranteed
by our theorem that every odd degree equation has a real root, but it is not really
necessary to appeal to this theorem (which is of no use at all if the coefficients
are complex); in the case of a cubic equation
actually find a formula for
all
The
the roots.
we
can, with sufficient cleverness,
following derivation
is
presented not
only as an interesting illustration of the ingenuity of early mathematicians, but as
numbers (whatever they may
further evidence for the importance of complex
To
solve the
most general cubic equation,
it
be).
obviously suffices to consider only
equations of the form
x
It is
3
+ bx 2 + ex + d =
0.
even possible to eliminate the term involving x
manipulation. If we
,
by a
fairly
straight-forward
let
b
then
o
,
2
b2 v
b3
SO
= x 3 + bx 2 +
T.
Jv
3
+
1
\
The
for y
(b 2
~T
+d
2b 2
-
3
i
right-hand side
we can
cx
now
find x\ this
contains no term with y
shows that
it
suffices to
.
If
we can
solve the equation
consider in the
first
place only
equations of the form
x
+
px
+q =
0.
=
we obtain the equation x 3 = —q. We shall see later on
has three, so that this
that every complex number does have a cube root, in fact
equation has three solutions. The case p ^ 0, on the other hand, requires quite
In the special case
p
it
an ingenious
step.
Let
(*)
x
=w-
—
25. Complex Numbers
529
Then
=
x
3
+
px
+
(
a
2
3wp~
5w 2 p
-2
3
=
—
= w-
q
—
+
3w
+ pw - — +
+ p (w -
)
p~3
)
q
,2
u;
4
3
P
3
This equation can be written
27(u;
which
is
3
2
)
a quadratic equation in
+ 27^(w 3 )-/? 3 =
w
0,
(!!).
Thus
3
u;
=
-21q ± y(27)V
2-27
V 4
2
Remember
that this really
+ 4-27/i 3
27
means:
2
w
3
We
9
=——
+ r,
,
where
r
•
r
is
a square root ol
P
q
—
+—
3
can therefore write
If
=
3
N
this
equation means that
root of q~/4
+
substituted into
x
jll.
p
some cube
This allows
root of
—q/2 + r, where
six possibilities for
3
N 2
V 4
+
r
is
some square
when
these are
27
3-
3
\
it
w, but
yielding
(*),
=
if is
?*y /«!4 + £
27
/
turns out that only 3 different values for x will be obtained!
surprising feature of this solution arises
when we
whose roots are real; the formula derived above may
in an essential way. For example, the roots of
15jc-4
=
An
even more
consider a cubic equation
still
involve
all
of
complex numbers
530
Infinite
Sequences and Infinite Series
are 4,
(with
—2 + V3, and —2 — V3. On the other
p — — 15, q = —4) gives as one solution
=
^/2
hand, the formula derived above
15
+
V4"
125
V2
3-y/2
V4
+ V4-
125
V/2TT17 +
3-
^2 +
11/
Now,
(2
+ /) 3 = 2 3 + 3-2 2 + 3-2-/ 2 +/ 3
= 8+ 12/ -6-i
/
-2+
one of the cube roots of
so
we
2+ 11/
is
11/,
2+
Thus, for one solution of the equation
r.
obtain
x
=2+ +
15
i
=2+ +
6
i
=
2
+ /+
6
+
3/
15
6- -3/
+ 3/
6 - -3/
-
90
45/
+
36
9
= 4(!).
The
The
other roots can also be found
fact that
even one of these real roots
depends on complex numbers
complex numbers cannot be
is
+
11/ are
known.
obtained from an expression which
is
impressive enough to suggest that the use of
As a matter of fact,
entirely nonsense.
for the solutions of the quadratic
in
the other cube roots of 2
if
the formulas
and cubic equations can be interpreted
entirely
terms of real numbers.
Suppose we agree,
writing the real
moment, to
number a as a + 0/ and
for the
ordinary arithmetic and the relation
(a
+ bi) +
+ bi)
(a
Thus, an equation
(c
+ di) =
+ di) =
=
(a
all
complex numbers
the number /
—1 show that
+
(ac
as
+
bd)
+ d)i
+ (ad + bc)i.
=
+7/
+
c)
-
(b
like
(1
may be
(c
2
i
write
+2/)
.(3
+
1/)
1
regarded simply as an abbreviation for the two equations
1-3-2-1 =
1-1+2-3 =
1.
7.
1 /
.
as a
The
+
bi,
laws of
531
25. Complex Numbers
The
+
ax
solution of the quadratic equation
bx
+
c
=
with real coefficients
could be paraphrased as follows:
— v = b — 4ac,
uv = 0,
= b — 4a c)
i.e., if (u + vi)
u
II
u
2a
then
-b +
+ c = 0,
+b
V
J
2a
2a
u
v
L2^J
2a
then a
0,
—b +- u + vi\
,
i.e.,
=
,
/
+b\
|
[
—b +
2a
down
+ vi
.
+c=
2a
not very hard to check this assertion about real
It is
u
numbers without
a single "/," but the complications of the statement
itself
writing-
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
been implicit in this whole discussion. All mathematical
complex number a + bi are determined completely by the real
numbers a and b\ any mathematical object with this same property may reasonably
be used to define a complex number. The obvious candidate is the ordered pair
(a, b) of real numbers; we shall accordingly define a complex number to be a pair
of real numbers, and likewise define what addition and multiplication of complex
numbers is to mean.
possibility has
properties of a
DEFINITION
A complex number
is
an ordered pair of real numbers;
if
z
=
is called the real part of z, and b is called
part of z. The set of all complex numbers is denoted by C. If
are two complex numbers we define
plex number, then a
(a,b)
(The
+
+ and
and
•
•
+
(c,d)
(a
(a,b)-(c,d)
(a
appearing on the
left
(a, b) is
the
a com-
imaginary
(a, b)
and
(c,
d)
+ c,b + d)
c
—
side are
b
d,a d
+ b-c).
new symbols being defined,
while the
appearing on the right side are the familiar addition and multiplication
for real numbers.)
When complex numbers were first
introduced,
bers were, in particular, complex numbers;
is
not true
a real
number
is
if
it
was understood
our definition
is
not a pair of real numbers, after
that real
num-
taken seriously
all.
This
this
difficulty
532
Infinite
Sequences and Infinite Series
is
only a minor annoyance, however. Notice that
+
(a, 0)
(a, 0)
=
+ b,0 + 0) = (a + b, 0),
= (a b - 0, a + b) =
(b, 0)
(b, 0)
•
(a
•
•
shows that the complex numbers of the form
this
{a
•
b, 0);
behave precisely the same
{a 0)
,
with respect to addition and multiplication of complex numbers as real numbers
do with
own
their
The
complex numbers can now be recovered if one more
be denoted simply by a
will
,
for
DEFINITION
=
i
Notice that
2
i
we
addition and multiplication. For this reason
convention that (a 0)
=
(0, 1)
•
definition
will
+
is
adopt the
bi notation
made.
(0,1).
(—1,0)
(0, 1)
familiar a
.
= —1
(the last equality sign
depends
on our convention). Moreover
(a,b)
= {a,0) +
= (a,0) +
— a bi.
(0,b)
(b,0)-(0,
1)
-\-
You may
it is
our definition was merely an elaborate device for defining
feel that
complex numbers
as "expressions of the
a firmly established prejudice of
be defined as something
specific,
modern
modern mathematics
not as "expressions."
esting to note that mathematicians
numbers
form a +/?/." That
is
essentially correct;
that
new
objects
Nevertheless,
it
must
inter-
is
were sincerely worried about using complex
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.
until the
definition
For example, the solution of the cubic equation required writing x
so
we want
to
know
that
l/w makes
sense.
Moreover,
w
=w—
p/3w,
was found by solving a
quadratic equation, which requires numerous other algebraic manipulations. In
short,
real
we
are likely to use, at
numbers.
We
Fortunately this
real
is
certainly
some time or
do not want
not necessary. Since
numbers can be
justified
other,
to stop
any manipulations performed on
each time and
is
quite easy,
by the properties
and these
every
step.
algebraic manipulations performed
all
listed in
essary to check that these properties are also true for
cases this
justify
facts will
Chapter
1,
it
is
on
only nec-
complex numbers. In most
not be listed as formal theorems. For
example, the proof of PI,
|
(a, b)
+
(c,
d)]
+
(e,
f)
=
(a, b)
+
[
(c,
d)
+
{e,
/)]
requires only the application of the definition of addition for complex numbers.
The
left
side
becomes
(\a
+c] +e,
[b
+ d\ +
./').
J
,
533
25. Complex Numbers
and the
right side
becomes
{a
+
two are equal because PI
these
+ e\,b+\d + f\)\
[c
true for real numbers.
is
It is
a good idea to
check P2-P6 and P8 and P9. Notice that the complex numbers playing the role
and
of
out what
P2 and P6 are
in
1
— (a,b)
trickier: if
is,
^
(a,b)
and
(0, 0)
(1, 0), respectively. It
but the multiplicative inverse for
(0, 0),
+ b ^0
2
then a
2
not hard to figure
is
a
little
and
^—+ —
,^—b- a- + b-
(a,fc).
is
required in P7
(a, b)
=(1,0).
\aThis
fact
=
(a,b)- (x,y)
it is
The
=
solutions are x
+
bi)
—
bx
+ ay
+ b~),
a/(a~
means
ax
a
Once
y
by
—
(1,0)
+
it
a
1
a
bi
+ bi
a
the existence of inverses has actually
by some method),
it
=
=
1
0.
—bj(cr
anything, then
1
+ b~).
the inverse of a
precisely this trick
—
—
a
bi
+
bi
+ b2
been proved
is
(after
6
+
90
really valid;
is
no set
numbers. In
P
of complex numbers such that PI
fact, if
—1
=
it is
easy to prove that
satisfied for all complex
)
not have disastrous consequences, but
),
and
this
it
P would have to contain 1 (since 1 = 1 and
would contradict P10. The absence of P10-P12
will
< w
PI 2 are
there were, then
—1
i
one
—
+9
also
(since
the easiest
6
3/
Unlike P1-P9, the rules P10 PI 2 do not have analogues:
is
it is
actually being sought
- 45/
36
there
guessing the inverse
6-3/
15
3/
also possible to reason
to evaluate
15
6
2
a
bi
—
complex number
which we used
It is
should be true that
follows that this manipulation
remember when
was
find (x, y) with
only necessary to solve the equations
that if \/(a
to
To
could have been guessed in two ways.
it
does
mean
that
we cannot
define
complex z and w. Also, you may remember that for the real numbers,
z
P10 PI 2 were used to prove that 1 + ^ 0. Fortunately, the corresponding fact
for complex numbers can be reduced to this one: clearly (1, 0) + (1 0) ^ (0, 0).
for
1
,
Although we
remembering
will usually write
that the set of
all
numbers. Long ago
complex numbers
complex numbers
in the
C
is
form a
+ bi,
it
is
worth
just the collection of all
was identified with the plane, and
for this reason the plane is often called the "complex plane." The horizontal axis,
which consists of all points (a, 0) for a in R, is often called the real axis, and the
pairs of real
this collection
534
Infinite
Sequences and Infinite Series
vertical axis
If z
=
x
of
is
defined as
z
Two
important definitions are also related
geometric picture.
to this
DEFINITION
called the imaginary axis.
is
+
iy
is
complex number
a
=
z
—
x
modulus
and the absolute value or
and y
(with x
\z\
iy
real),
then the conjugate
,
of
z
is
defined as
length
z\^»z =
(x,y)
= x + iy
+y >
(Notice that x
+ y is
x^ + y
\x
0, so that
the nonnegative real square root of
•z
FIGURE
= x — ly
Geometrically, z
it
denotes
.)
simply the reflection of z in the real
is
distance from z to (0,0) (Figure
1
defined unambiguously;
axis,
while
|z|
is
the
Notice that the absolute value notation for
1).
complex numbers is consistent with that for real numbers. The distance between
two complex numbers z and w can be defined quite easily as \z — w\. The following
theorem lists all the important properties of conjugates and absolute values.
THEOREM
1
Let
and
z
(1)
l
(2)
z
w
be complex numbers. Then
= z.
= z
number
(5)
(6)
Fy =
(8)
\z
(9)
\z
(4)
z
real
is
(i.e.,
is
of the form a
+
0/',
for
some
real
z
•
•
w\
+
Assertions
(=r
=
w\
(1)
1
<
\z\
and
\w\.
•
\z\
c#o.
,if
+
(2)
\w\.
are obvious. Equations
forward calculations and
=
1
Equations
only
if
a).
+ w — z + w.
=£=-*.
w = z w.
z
(3)
PRCJOF
and only
if
(7)
difficult
and
=
(8)
1
(4)
and
= +
=
may
may
(6)
(-*)
=
+
.
also
(3)
and
(5)
may be checked by straight-
then be proved by a
so
-z,
7^\
—c =
=
so
trick:
z,
(z)"
be proved by a straightforward calculation. The
part of the theorem
is
This inequality has,
(9).
in fact,
curred (Problem 4-9), but the proof will be repeated here, using
already oc-
slightly different
terminology.
It is
is
clear that equality holds in
true
X <
0).
if z
=
^-W for
any
real
(9) if z
=
number X
Suppose, on the other hand, that
or
w=
0.
It is
also easy to see that (9)
(consider separately the cases X
:.
^ Xw
for
any
real
number
k,
>
and
and
that
535
25. Complex Numbers
w ^
Then,
0.
for
numbers
real
all
k,
=
=
,2
<
kw\
— kw)
— kw)
(z
(z
_
+
- zz
=
Notice that wz
+ zw
is
wz
Thus
its
+ zw =
follows, since
is
+ zw
this inequality
it
and
\w\
+
+ zw) <
=
=
w\
(z
+
Two complex numbers
2).
segment from
geometric
The
a a
FIGURE
2
+c
z
Thus
2
<
\z\
0;
2\w\
and
fact
w =
\z\.
\z\
(\z\
+
w\
<
\w\r,
+
\w\. I
is
its
z
—
The
numbers both have
picture for addition
w =
(a,b) and
sides the line
(c,
left to
you [compare Appendix
(a one-line
—
may
our attention
more
to
is
z
(0, 0) to z,
+w
Chapter
involved.
very simple (Fig-
(a
and the
proof of
been shown
If z
to follow
nonzero complex numbers.
this
4]).
=
or
w =
computational proof can be given, but even
the assertion has already
to
is
1
is
d) determine a paral-
segment from
(0, 0) to w; the vertex opposite (0, 0)
unnecessary
restrict
0, that
+ w) (z + w)
+ \w\ 2 + (wz + zw)
+ \w\~ + 2\w\
interpretation of multiplication
•
>
|it;|
operations of addition and multiplication of complex
lelogram having for two of
then
+ zw.
2
important geometric interpretations.
line
wz
that
\z
+w
+ zw =
are real numbers,
\z\
=
ure
wz
+ zw) 2 - 4\w\ 2
<
The
\z\
follows that
\z
which implies
+ zw)
k(wz + zw).
discriminant must therefore be negative.
wz
_
_
k(wz
a quadratic equation in k with real coefficients and no
(wz
From
\w\
+ zw =
wz
(wz
it
(z
_
ww —
+
9
k
real, since
the right side of (*)
real solutions;
k
— kw)
— kw)
(z
this
in this expression,
number
|z|
is
;^0we can
write
a positive real number, while
is
from P1-P9), so we
We begin by putting every
nonzero complex number into a special form (compare Appendix 3 to Chapter
For any complex
0,
4).
.
536
Infinite
Sequences and Infinite Series
so that z/\z\
a
= x + iy
is
number of absolute
a complex
with
=
1
=x +y
\a\
a
for
angle of
9.
=
(cos 9, sin 9)
+
cos 9
Now any
1.
i
sin
=
Z
+
r(cos9
complex number
form
in the
9
Thus every nonzero complex number
radians
z
can be written
sin#)
i
>
and some number 9. The number r is unique (it equals |z|), but 9
any
9q is one possibility, then the others are 9q + 2&?r for k in Z
one of these numbers is called an argument of z. Figure 3 shows z in terms of r
and 9. (To find an argument 9 for z = x + iy we may note that the equation
for
is
FIGURE
some number
=
value
can be written
3
some
r
not unique;
—
if
x
means
+
=z=
iy
sin 9)
i
that
x
v
So, for example,
9
+
|z|(cos 9
if
= 7r/2 when y >
Now the product
x >
and #
=
=
|z|
cos#,
|z|
s\n9.
=
we can
take 9
=
when y <
37r/2
arctany/jt;
if
x
=
0,
we can
take
0.)
of two nonzero complex numbers
=
w =
Z
+
s{cos(p +
r(cos#
i
sin^),
i
sin0),
is
z
w—
=
=
rs (cos 9
+
i
—
cos
rs [cos(6>
+ 0) +
sin
/
9 sin
sin(<9
Thus, the absolute value of a product
factors, while the
sum
now an
known
=
as
De
\z\"
This formula describes
2
)F
is
(sin
cos
+ cos
sin 0)]
0)]
the product of the absolute values of the
=
r(cos9
+
sin 9)
i
Moivre's Theorem):
(cosn9
+
i
sin«#), for any argument 9 of z.
z" so explicitly that
precisely, for
it is
any complex number
there are precisely n different
)(
+
</>)
/'
easy to decide just
Every nonzero complex number has exactly n complex nth
More
I'R(
+
easy matter to prove by induction the following very important formula
z"
THEOREM
sin
nonzero complex number
Z
(sometimes
i
</>)
of any argument for each of the factors will be an argument
for the product. For a
it is
+
sin 9) (cos
rs [(cos
w /
0,
w =
0)
+
i
sin
—
w:
roots.
z satisfying z"
,9(cos0
z"
and any natural number
complex numbers
Let
when
=
w.
n,
.
—
.
537
25. Complex Numbers
=
for s
and some number
\w\
—
Z
=w
satisfies z"
if
and only
r" (cos
which happens
if
Then a complex number
0.
+
nO
and only
first
equation
it
+
/
nO
sin
some
+
s(cos0
i
sin0),
=
=
s,
+
cos
sin 0.
i
follows that
t/s denotes the positive real
follows that for
=
nO)
sin
i
—
r
where
sinO)
if
cos nO
the
i
if
r"
From
+
r(cosO
integer k
V?,
nth root of
From
s.
the second equation
it
we have
Ikn
n
Conversely,
+
r(cos8
is
it
if
we choose
r
=
i
—
and
^/s
=
sin^) will satisfy z"
n
9k for
To determine
w.
some k, then the number z =
the number of nth. roots of w,
Now
therefore only necessary to determine which such z are distinct.
any
integer k can be written
—
k
some
for
integer q
,
and some integer
+
cos Ok
This shows that every
Z
Moreover,
k
= 0,
.
.
.
it is
n
,
—
=
>/s (cos Ok
4
+
easy to see that these
1
by
differ
less
k'
=w
than
i
-\-
k'
and n —
between
=
sin Ok
z satisfying z"
In the course of proving
FIGURE
i
nq
cos
+
0k>
i
sin
1
.
Then
Ot
can be written
k
sin Ok)
numbers
=
are
0,
all
.
.
.
,
—
n
1
different, since
any two
Ok for
2tt. |
Theorem
2,
we have
actually developed a
method
complex number. For example, to find the cube
|/| = 1 and that tt/2 is an argument for i. The cube
for
finding the «th roots of a
roots
of
i
(Figure 4) note that
roots
of
i
are therefore
r
1
[cos
n
-+
n
•
i
sin
-^
j
[
(n
2tt\
It
( tc
2tt
=
COS
57T
—
—h
:
.
/
sin
6
it
An
4tt
jt
+
6
T
=
cos
3tt
—
—h
2
57T
——
6
.
/
sin
3jt
—
2
—
538
Infinite
,
Sequences and Infinite Series
Since
x/3
71
—
cos
6
—
V3
6
2
5jt
'
Sm
"6-
sin
——
— =
0,
2
the cube roots of
/
-s/3
/
-
we cannot expect
the cube roots of 2
is
= —
1
are
\/3
arctan -y
2'
2
+
——
In general,
1
=
37T
3tt
cos
1
6=2'
2
5tt
cos
IT
Sm
?
+
for
i
—i.
,
to obtain such simple results. For
Hi, note that
an argument
+
,
|2
+
1
1/|
=
y/2
2+11/. One of the cube
(arctan U- \
2
+
2
ll
example,
= ^125
+ 1/
roots of 2
is
1
to find
and
that
therefore
I arctan -y
VT25
(arctan
cos
Previously
V2 +
2
l
2
=
we noted
V5, and
that 2
+
/
is
also a
since arctan ^
is
'
^\
/ arctan
arctan
+ 11/. Since
2 + /, we can
cube root of 2
an argument of
-y-
|2
+
/|
write this
cube root as
2
+ = v 5 (cos arctan
/
These two cube roots are
actually the
arctan
A
+
/
sin arctan 5 )
•
same number, because
—
=
1
arctan
2
3
(you can check this by using the formula in Problem 15-9), but this
sort of thing
The
one might
fact that
every complex
order to provide a solution for
number has an «th root for all n is just a special
The number was originally introduced in
=0. The Fundamental Theorem
the equation x +
/
1
of Algebra states the remarkable fact that this
all
hardly the
notice!
case of a very important theorem.
solutions for
is
one addition automatically provides
other polynomial equations: every equation
z"
+an -iz n
~
]
H
hflo
=
ao
fl„_i in
C
has a complex root!
we
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
In the next chapter
come up
shall give
quite naturally in a
more thorough
investigation of
complex numbers.
.
25. Complex Numbers
539
PROBLEMS
1
Find the absolute value and argument(s) of each of the following.
(iv)
+ 4/.
(3 + 4/r
d+/) 5
^3 + 4/.
(v)
|3+4/|.
(i)
(ii)
(iii)
2.
3
1
.
.
Solve the following equations.
(i)
(ii)
(iii)
x2
x
4
+ ix +
=
I
0.
+x +\=0.
2
+ 2ix - = 0.
ix - (1 + i)y = 3,
(2 + i)x+iy=4
x 3 - x 2 - x - 2 = 0.
x
2
1
1
(v)
3.
Describe the
})
ii)
"i)
;iv)
y)
4.
=
z
= z~
of all complex numbers
<
is (z
such that
l
.
\z—a\ = \z — b\.
\z — a\ + \z — b\ =
\z\
z
-z.
Prove that
part
5.
z
set
1
|z|
—
Prove that
—
c.
real part of z
=
\z\,
and
that the real part of z
+
—
is
(z+z)/2, while the imaginary
z)/2i.
\z
w| 2
+
\z
w\ 2
=
2{\z\
2
+
\w\
2
),
and
interpret this statement
geometrically.
6.
What is the pictorial relation between z and V zv—i ? (Note that there
may be more than one answer, because v/ and V— 7 both have two different
possible values.) Hint: Which line goes into the real axis under multiplic <ui< ui
/
•
by
7.
(a)
Prove that
if ao,
the equation z"
.
.
.
,
a„_i are
+ a„_iz" _1 +
•
real
•
•
and a
+ ao =
+ bi
0,
(for
then a
a and
—
£>
real
satisfies
bi also satisfies this
equation. (Thus the nonreal roots of such an equation always occur in
and the number of such
Conclude that z" +a n _\z"~
pairs,
roots
is
even.)
l
(b)
-\
(whose
'8.
(a)
hflo
is
divisible
by
z
2
— 2az + (a 2 + b 2
)
coefficients are real).
Let c be an integer which
is
not the square of another integer. If a and b
we define the conjugate of a + b^/c, denoted by a + by/c,
as a — b<J~c. Show that the conjugate is well defined In showing thai a
number can be written a + b*Jc, for integers a and b, in onh one way.
are integers
540
Infinite
Sequences and Infinite Series
(b)
Show
that for
and only
a rr
(c)
=
*10.
is
(a)~
a and
an
if
if ao,
+
z"
integer;
.
.
.
+
a
=
/3
a„_iz"
a„_i are
,
_1
a
•
/3;
/3;
+
•
•
and
integers
+
•
gq
=
=
z =
z
then
0,
a
+
£>\/c satisfies the
a
—
b-Jc also
satisfies
equation.
Find
all
form
that does not involve
the 4th roots of
(a)
Prove that
(b)
A number a>
is
+ b-Jc, we have a = a; a = a if
and
+ —a — —a; a ft = a
of the form a
/3
a ^0.
}
Prove that
this
9.
a
if
equation
all
if co is
the set of
there for n
is
any trigonometric functions.
an nth root of
roots of
3, 4, 5,
1,
then so
is
or.
primitive nth root of
called a
all /7th
=
express the one having smallest argument in a
i;
How many
1.
1
if
[l,eo,co
,
..., co"~
primitive nth roots of
1
l
}
are
9?
n-l
(c)
Let
co
be an «th root of
1,
with
w^
Prove that /^oo
1.
=
0.
k=0
*11.
(a)
Prove that
then
if z\, ...
+
z\
+
Zk
/
he on one side of some straight line through
Zk
,
This
0. Hint:
is
obvious from the geometric
pretation of addition, but an analytic proof
clear if the line
is
the real axis,
and a
is
0,
inter-
also easy: the assertion
is
reduce the general case
trick will
to this one.
(b)
Show
through
*12.
z\~\
further that
Prove that
0, so that
if |zi
|
=
z\~
\zi\
—
...
Zk~
,
x
H
\zs\
h
and
{
all lie
Zk~
zi
l
+
#
0.
Z2
+
the vertices of an equilateral triangle. Hint:
real,
and
this
can be done with no
loss
on one
Z3
=
It will
0,
side of a straight line
then
help to
of generality.
and Z3 are
assume that z\ is
zi, Zi,
Why?
CHAPTER
mm ^^
COMPLEX 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. A rigorous
definition presents
no problems (we
formal
a function
definition):
does not contain two
real
numbers
case of the
in
order to
to
new
not even accord
will
a collection of pairs of
is
distinct pairs
with the same
first
it
the
complex numbers which
we will sometimes
which a function
clarify the context in
number
we
element. Since
be certain complex numbers, the old definition
one. Nevertheless,
honors of a
full
is
really a special
resort to special terminology
being considered.
is
consider
A function
domain of
/
f(z)
/, and
complex-valued to emphasize that it is not necessarily real-valued. Similarly,
we will usually state explicitly that a function / is defined on [a subset of] R in
those cases where the domain of / is [a subset of] R; in other cases we sometimes
mention that / is defined on [a subset of] C to emphasize that f(z) is defined for
is
called
complex
real-valued
if
is
a real
for all z in the
z as well as real z.
Among
on C, certain ones are
important. Foremost among these are the functions of the form
the multitude of functions defined
f(z)=a n z" +a„_
where
ao, ...
,
l
z"~
+---
l
+ ao.
These functions are
a n are complex numbers.
particularly
called, as in the
polynomial functions; they include the function f(z) — z (the "identity
function") and functions of the form f(z) —a for some complex number a ("conreal case,
stant functions").
Another important generalization of a familiar function
—
"absolute value function" f(z)
Two
and Im
complex numbers are Re
Re(.r
"conjugate function"
+ />) — x,
+ iy) = y,
.
lm(x
is
x and y
=
l
=
-
Re(z)
Familiar real-valued functions defined on
i
541
real.
Im(z).
R may be combined in
produce new complex-valued functions defined on
+
.
,
tor
.
defined by
f(z )
fix
(the "real
"imaginary part function"), defined by
(the
T
the
for all z in C.
functions of particular importance for
part function")
The
\z\
is
iy)
=
e y s'm(x
—
y)
+
C — an example
ix
cos
v.
many ways
is
to
the function
542
Infinite
Sequences and Infinite Series
The formula for this particular function illustrates a decomposition which
possible. Any complex-valued function / can be written in the form
some
and v(z)
and
real-valued functions u
as the
v
— simply
define u{z) as the real part of f(z),
imaginary part. This decomposition
always; for example,
it
always
= u + iv
f
for
is
would be inconvenient
is
often very useful, but not
to describe a
polynomial function
in this way.
One
other function will play an important role in
argument of a nonzero
complex number
Z
There are
In. If
we
infinitely
call this
=
|z|(cos#
many arguments
unique argument
"argument function") on
[z in
C
z
:
z
for
is
a
+
(real)
i
sin
such that
(9).
but just one which
z,
Recall that an
this chapter.
number
0(z), then
is
<
satisfies
<
a (real-valued) function (the
^ 0}.
"Graphs" of complex-valued functions defined on C, since they lie in 4-dimenpresumably not very useful for visualization. The alternative
picture of a function mentioned in Chapter 4 can be used instead: we draw two
sional space, are
and arrows from
copies of C,
FIGURE
z in
one copy,
to f(z) in the other (Figure
1).
1
The most common
pictorial representation of a
complex-valued function
is
pro-
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
Certain features of the function are
sort of picture for several different functions.
by such a "graph." For example, the absolute value function
is constant on concentric circles around 0, the functions Re and Im are constant
on the vertical and horizontal lines, respectively, and the function f(z) = z wraps
illustrated very clearly
the circle of radius r twice
around the
circle
of radius
r
.
Despite the problems involved in visualizing complex-valued functions in general,
it is still
possible to define analogues of important properties previously defined
on R, and in some cases these properties may be easier
complex
case. For example, the notion of limit can be defined
the
lor real-valued functions
to visualize in
as follows:
lim f(z)
=
I
means
that for every (real)
number s >
there
is
a
(real)
z—*a
8
>() such
that, lor all z, if
<
\z
-
u\
<
8,
then \f(z) -l\
<S.
number
*
1
26. Complex Functions
-i'
It
f
|..
-i.
1"
-i'
§..
-i.
i"
-"
-"
i-
-i-
I"
3
1"
i
'
'
li
§1
-i.
-i-
,
,
3
3
3
3
3
2
2
2
2
7
1
1
1
1
1
1
2
2
7
?
2
2
:
1
3
>
3
3
3
3
3
i
2
2
2
2
1
1
1
1
1
1
i
i
i
2
2
7.
1
1
-1 -1 -1 -1 -1 -1 3
3
3
3
3
2
2
2
2
2
(c)
3
2
f(z)
1
1
i
1
1
1
1
2
2
2
2
2
2
3
3
3
3
3
3
:
2
2
2
2
2
2
7T
7T
TZ
It
=
„o
.1(1
U
"I
<
,1
1
.in
ni
<
,1
1
-III
2
_In
i
i
<
.
I
r
i
,
:
(e)
1
2
liy
>
«
,1
_1|,
,
_i,i
1
„0
ol
<
,,o
ni
<
1
1
I
—i
—
i
—
i
<
'
*
i
-i"
"0
"i
_In
,,n
U
ni
_^„
„
I.'
-il
lo
4
2
•
,
-i.
I
< »1
1
<
2
2
,,3
2
2
,,3
2
,,3
2
ni
2
.3
1
,
2
,
1
^
2
..I
2
ni
2
2
,11
ol
il1
li
2
71
ff
il
2
2
2
Im(z)
TT
i,l
, ,1
-1 -1 -1 -1 -1 -1
3
,1
1
Hi
2
2
Ȥ
„i
()
1
1
„
a
2
(
_i„
f(z)
3
,1
1
,,o
_i„
i
1
Pi
«
_!,,
.
T1
<
2
2
,
I
1
ol
i>n
(b)
(a)
P
T°
2
— -h --
!-.
§<
"2T
—In
>
2
|.
,
f(z)=6(z)
f(z)
=
Re(z)
.
2
543
544
Infinite
Sequences and Infinite Series
Although the definition reads precisely
—
w\
equation \imf(z)
=
ferent.
Since
\z
means
/
as before, the interpretation
is
slightly dif-
z
and w, the
between the complex numbers
the distance
is
made
that the values of f(z) can be
to
lie
inside
z—>a
any given
around
circle
small circle around a
/,
provided that
This assertion
.
copy" picture of a function (Figure
is
z
restricted to
is
lie
inside a sufficiently
particularly easy to visualize using the
"two
3).
©
FIGURE
3
Certain
facts
about
can be proved exactly as
limits
lim c
=
c,
lim z
—
a,
in the real case. In particular,
z.^>-a
+
g(z)]
lim f(z)
g(z)
lim[/(z)
=
=
1
1
g(z)
lim g(z)
lim
a
The
<
limg(z),
+ £—
>a
lim f(z)
lim g(z),
if
'
essential property of absolute values
+
lim f(z)
z—>a
Z—>a
limg(z) 7^0.
upon which
these results are based
is
the
+
\w\, and this inequality holds for complex numbers as
These facts already provide quite a few limits, but many
more can be obtained from the following theorem.
inequality
\z
w\
\z\
well as for real numbers.
THEOREM
1
Let f(z)
real
=
u(z)
+ iv(z)
numbers a and
)3.
for real-valued functions u
Then
=
and
I
if
and only
lim u{z)
=
a,
=
(3.
lim f(z)
v,
and
let
/
=
a
+
if
z->«
Z—>C1
lim v(z)
Z->a
PROOF
Suppose
first
that lim /(-)
=
/.
If e
'
>
8
>
then \f(z)
—
0, there
is
such that, for
z->a
if
The second
<
\z
—
a\
<
8.
I\
inequality can be written
\[u(z)-a\ +i[v(z)~P]\ <£,
< £
all z,
ifi
for
545
26. Complex Functions
or
[u(z)
Since u(z) —a. and v(z)
— /3
- a] 2 +
-
[v(z)
2
<e 2
P]
.
are both real numbers, their squares are positive; this
inequality therefore implies that
<
[m(z) —or]
£~
and
[u(z)
—
{5\~
and
\v(z)
—
>3
<
s
,
which implies that
—
\u(z)
Since
this
is
true for
all
s
>
<
a\
0,
it
8
s.
follows that
\imu(z)=a
and
limu(z)
Z—>a
=
/J.
z-*a
Now suppose that these two equations hold.
< \z — a\ < 8, then
for all z, if
\u(z)
<
1
—
If e
£
< —
a\
and
>
—
\v{z)
0, there
a\
S
< —
is
a
8
>
such
that,
,
which implies that
\f(z)-l\
=
\[u(z)-a]
<
|«( z
)- a
£
This proves that lim f(z)
In order to apply
limit lim z
= a, we
=
I.
Theorem
+ i[v(z)-P\\
+ |i|. \V(Z)-P\
|
£
|
1
notice that since
fruitfully,
we
already
know
the
can conclude that
lim
Re (z) =
Re(a).
=
Irn(a).
=
sin(Re(a))
Z—*a
lim Im(z)
Z-*a
A
limit like
lim sin (Re (z))
Z—*a
follows easily, using continuity of sin.
Many
applications of these principles prove
such limits as the following:
lim z
=a
,
Z->tf
lim
|z|
=
\a\,
z-*a
lim
e y sin x
+
cos y
ix
=
e
sin
a
+ ia~
cos b.
(x+iy)—*a+bi
Now that
the notion of limit has
of continuity can also be extended:
been extended
/
is
to
complex
continuous at a
if
functions, the notion
lim f(z)
=
/(«), and
'
546
Infinite
Sequences and Infinite Series
f is continuous if / is continuous at a for all a in the domain of /. The previous
work on limits shows that all the following functions are continuous:
f(z)
f(z)
f(z)
fix
=
=
=
+an -iz
an z
l
^
h«0,
z,
\z\,
v
ix cosy.
+,-3
e y sinx
+ iy)
•
Examples of discontinuous functions are easy to produce, and certain ones come
up very naturally One particularly frustrating example is the "argument function" 9, which is discontinuous at all nonnegative real numbers (see the "graph"
Figure
in
for
ities;
2).
By
suitably redefining 9
example (Figure
4),
it
is
possible to change the discontinu-
9'(z) denotes the unique argument of
if
z
with
< 0'(z) < 5tt/2, then
is discontinuous at ai for every nonnegative real
number a. But, no matter how 9 is redefined, some discontinuities will always
9'
tt/2
occur.
2^
TT
'
2
2&*
n
•
.l\n
2
'
TT
2ft»
2
'
2
2
*
*
n
2
TT
&*2\n
i*
2 5*
•2\n
'
2
' 2
'
*2±n
TT
TT
2h
*2i»
'
2 4 ?l
2
3
TT
4
2
2'
TT
TT
71
71
71
71
71
71
•2i;r
•
TT
•2j»
27r 27r 27r 2;r 2;r 2;r 2;r 27r 2;t
TT
~2
in
T
(
3jT
,
3tt
T
T
(
'
3rr
,
3tt
T
T
(
,
3*
2
3tt
T
FIGURE
The
4
f(z)
<
=
0'{Z)
discontinuity of 9 has an important bearing
on the problem of defining a
"square-root function," that is, a function / such that (f(z))~ = z for all -. For real
numbers the function \/~ had as domain only the nonnegative real numbers. If
complex numbers arc 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.
—
.
547
26. Complex Functions
One way
/
to define
is
the following.
=
j(z)
Clearly (f(z))~
=
Z,
As a matter of fact,
In
all z.
prove
fact,
it is
/
but the function
it is
set
9&
—
h
is
even impossible
/ by — /). Then we
argument). But
for this
claim that for
is
left
to
+
/
you
is
+
even though cos #
+
d—
functions
[a, b]
x
[c,
/
—>
sin ^
1
i
for
we
set
—&
/
is
discontinuous.
= Z for
— To
such that (/(z))
—
all z
with
1
|z|
.
always
a standard type of least upper
bound
that f{\)
=
9
cos
—
(since
1
<
6 with
9
<
sin
—
Itc
9
+
= cos n +
= -l
sin 9)
any n
>
—
as
A similar argument shows that
'
#
we could
we have
sin 9)
(it
for z
i
implies that
(*)
lim /(cos 9
1
sin
be defined for
all
0-^2tt
tion.
/
and
0,
discontinuous, since
for f(z) to
/(cos 9
(*)
The argument
=
/(0)
impossible to find a continuous
by contradiction, we can assume
this
replace
/TiY
\/\z\[ cos
We
it is
2^
.
i
sin
n
Thus, we have our contradic-
impossible to define continuous "nth-root
2.
d]
For continuous complex functions there are important analogues of certain theorems which describe the behavior of real-valued functions on closed intervals. A
natural analogue of the interval [a, b] is the set of all complex numbers z = x + iy
with a < x < b and c < y < d (Figure 5). This set is called a closed rectangle,
and is denoted by [a, b] x [c, d].
figure
If
5
then
That
/
it
seems reasonable, and
is,
there
is
some
does not
[a,b] x
make
is
<
/
is
in [a, b]
x
indeed true, that
number
real
|/(z)|
It
domain is
bounded on
a continuous complex-valued function whose
is
M
sense to say that
[a, b]
x
[c, d],
[a, b]
x
[c, d].
M such that
for
/
all z
has a
[c, d].
maximum and
a
minimum
value on
complex numbers. If / is a
make sense, and is true. In
if / is any complex-valued continuous function on [a, b] x [c, d], then
continuous, so there is some zo in [a, b] x [c, d] such that
[c,
d], since there
is
no notion of order
for
real-valued function, however, then this assertion does
particular,
|/|
is
also
|/(zo)l
a similar statement
"/
attains
The
its
is
<l/(z)l
for
all
-in [a,b] x [c,d];
true with the inequality reversed.
maximum and minimum modulus on
It is
[a, b\
x
sometimes said
that
[c, d~\."
various facts listed in the previous paragraph will not be proved here,
though proofs are outlined
in
Problem
5.
Assuming
these facts, however,
al-
we can
'
548
Infinite
Sequences and Infinite Series
now
Fundamental Theorem of Algebra, which
give a proof of the
surprising, since
we have
much
not yet said
is
really quite
polynomial functions
to distinguish
from other continuous functions.
THEOREM 2 (THE fundamental
THEOREM OF ALGEBRA)
Let ao,
.
.
.
,
fl„_i
be any complex numbers. Then there
number
z
such that
n
Z
PROOF
a complex
is
+ an - lZ n
-
+
]
a„_2Z
n-2
+
•
+
=
fl
0.
Let
"+a n _ lZ,«-!
+---+O0.
1
f(z)=z"
Then /
is
continuous, and so
=
\f\(z)
the function \f\ defined
is
by
= \z"+a n ^z"- +---+a
l
\f{z)\
\.
Our proof is based on the observation that a point zo with f(zo) = would clearly
minimum point for / To prove the theorem we will first show that / does
indeed have a smallest value on the whole complex plane. The proof will be almost
be a
.
1
1
|
identical to the proof, in
|
Chapter
on
(with real coefficients) has a smallest value
fact that if
We
|z| is
large, then |/(z)|
begin by writing, for
of R; both proofs depend on the
all
large.
is
^0,
z
f(z)
polynomial function of even degree
7, that a
= zn
+
1
+
•••
+
a
so that
l/(z)l
=
1
|z|
^1 +
+
Let
M = max(l, 2«|a„_i
Then
for all z with
k
> M, we have
|z|
\a„-k\
7 k\
\z
\a„-k\
<
—
\
>
|,
.
|z|
.
.
,
2n\ao\).
and
\a n -k\
<
J_
2n
2n\an .
so
a n -\
«0
0-n-\
z"
Z
which implies
<
+
•••
+
z
1
2'
that
fl
i
+
»-'
+...
+
""
7
>
~n
1
-
a„-\
clq
Z
z"
This means that
1
l/(z)l
In particular,
<
-11
if |z|
>
M and also
\z\
\.f
~l"
^
>
>
(z)l
for
|z|
> M.
(/2|/(0)|, then
>l/ [0)\.
1
> —
- 2
549
26. Complex Functions
Now let
[a, b]
x
max(M, y2|/(0)|
[c,
)},
d]
be a closed rectangle (Figure
and suppose
that the
which contains
6)
minimum
of
on
|/|
[a, b]
{z
x
:
[c,
|z|
<
d]
is
attained at zo, so that
(1)
It
|/(zo)l
<
l/(z)l
for z in [a,b]
<
follows, in particular, that |/(zo)l
(2)
if \z\
x [c,d].
Thus
1/(0)1-
> max(M, ^21/(0)1),
then |/(z )|
and (2) we see that |/(zo)l S \f(z)\
minimum value on the whole complex plane at zq-
Combining
(1)
>
|/(0)|
for all
z,
>
|/Uo)l-
so that |/| attains
its
max(M, ^21/(0)1)
l/(z)l
>
1/(0)1
for z here
FIGURE
6
To complete
the proof of the
theorem we now show
that /(zo)
=
0.
It is
convenient to introduce the function g defined by
=
g(z)
Then g
is
occurs at
0.
+ zo).
f(z
a polynomial function of degree n,
We
want
to
show
that g(0)
Suppose instead that g(0) = a
which occurs in the expression for
=
whose minimum absolute value
0.
m
/
0.
g,
we can
If
is
the smallest positive
ft
^
0.
Now, according
to
Theorem 25-2
such that
y
m
__
~
z
write
m+l
+
g(z)=a + Pzm +cm+l z
where
power of
«
fi'
---
+
there
f„;
is
a
complex number y
550
Infinite
Sequences and Infinite Series
Then,
setting dk
=
Cky
k
we have
,
= + Py m z m + dm+[Z
= \a-az'"+dm+] z"1+l
'"
\g(yz)\
+l
\c*
a[ l-,'"
+
+ dn z"\
+]
d„
+ ^±l -'"+'+.
a
a
=
-
1
I
z
m
m+\
l
+ zm
+
a
dm A
a
1
+
Z
i
a
This expression, so tortuously arrived
diction. Notice
first
that
if \z\ is
l
m+
a
If we choose,
from among
all z
enable us to reach a quick contra-
at, will
chosen small enough, we
\
z
have
this inequality holds,
some
<
+
which
for
will
1.
z
which
is
real
and positive, then
dm+
a
Consequently,
1
-
<
if
z
m
+
z
<
1
\
z+
:."
-z
a
+
'II-
<
is
\g(yz)\
the original assumption
/(zo)
1
1
-
the desired contradiction: for such a
contradicting the fact that
-
_'"
i
we have
=
This
jn
<
|a| is
must be
the
<
z"'
m
z
m
z
d„ + 1
+
+
z
a
dm +\
+
z
a
+
+ zm
number
z
we have
|a|,
minimum
incorrect,
I
of
\g\
and g(0)
on the whole
=
0.
plane. Hence,
This implies,
finally,
that
0. |
Even taking into account our omission of the proofs for the basic facts about
continuous complex functions, this proof verified a deep fact with surprisingly
little work. It is only natural to hope that other interesting developments will arise
if
we pursue
step
is
further the analogues of properties of real functions.
to define derivatives: a function
..
/
differentiable at a
f(a+z)-f(a)
lim
z-»0
is
exists,
z
The
if
next obvious
;
551
26. Complex Functions
which case the
in
limit
denoted by f'{a).
is
f(a)
=
if/(z)=c,
f(a)=\
(f+g)\a)
{f
-
g)'(a)
1\'
J
the proofs of
all
=
=
i£f(z)=Z,
+ g'(a),
f\a)gia) + f{a)g'(a),
f{a)
-g'(a)
=
(«)-t4^
(fl)
if*(«)^0.
Mtf
=
x
if
differentiability
of rational functions however.
,
/
is
to
.
+ iy) =
be differentiable
at 0,
-
x
iy
f{z)
(i.e.,
f(x+iy)-f(0)
x
(x+iy)-*-0
exist.
other obvious candidates are
=
z).
then the limit
lim
must
Many
Suppose, for example, that
fix
If
same as before. It follows, in
These formulas only prove the
these formulas are exactly the
~
nz n
f(z) = z" then f'(z)
particular, that
not differentiate.
easy to prove that
It is
=
+ iy
lim
(x+iy)^0
x-iy
x + iy
Notice however, that
if
j
=
~
x
— iy
=
+ iy
x
—
x
+ iy
x
,
then
0,
1,
and
if
x
=
then
0,
iy
=—
1
therefore this limit cannot possibly exist, since the quotient has both the values
and —
1
for x
+
1
iy arbitrarily close to 0.
where other differentiable functions
are to come from. If you recall the definitions of sin and exp, you will sec 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.
In view of this example,
it is
not at
all
clear
PROBLEMS
1.
(a)
a(x)
=
valued function defined on R).
Show
that
immediately from a theorem in
this chapter.)
For any real
x
(b)
+ iy
is
Let
/ be
f(x
+
number
(a)
rectangle
+ iy
a
is
(so that
or
is
a complex-
continuous. (This follows
Show
a continuous function defined on C.
iy).
Suppose
x
v)
=
g(x)
=
similarly that
fi(
continuous.
Show
that g
similarly that hiy)
2.
v, define
that
/
[a, b]
is
x
=
is
fix
For fixed
v. let
a continuous function (defined on R).
+
iy)
is
continuous. Hint: Use part
Show
(a).
a continuous real-valued function defined on a closed
[c, d].
Prove that
if
/
takes
on
the values f(z)
and f(w)
552
Infinite
Sequences and Infinite Series
and w in [a, /?] x [c, d], then / also takes all values between f(z)
and f(w). Hint: Consider g(t) = /(rz + (1 - r)w) for t in [0, 1].
If / is a continuous complex-valued function defined on [a, b] x [c, o"],
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 conjecture that
/ 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.
for z
"(b)
Prove that
3.
if
«o,
•
complex numbers
Z n (not necessarily distinct)
+ a,^ ]Z n
z"
complex numbers, then there are
are any
1
-
]
+...+ao = Y\(z ~
such that
zi).
i=\
(b)
Prove that
if oo,
.
.
.
on _i are
,
real,
then
z+a and
written as a product of linear factors
all
of whose coefficients are
+ a n _\z n
z"
~
+
+ «o can be
quadratic factors z 2 +az+b
{
•
•
•
(Use Problem 25-7.)
real.
In this problem we will consider only polynomials with real coefficients. Such
a polynomial
called a
is
for
polynomials
(a)
Prove that
if
(b)
Prove that
if
(c)
Suppose that
sum of squares if
it
can be written as
/?i~
+
-
+^,
•
2
;
with real coefficients.
//,
/ is a sum of squares, then f(x) > for all x.
/ and g are sums of squares, then so is f g.
for all x. Show that / is a sum of squares.
f(x) >
•
Hint:
k
First write
f(x)
=
\(x
I
—
where g(x) >
a{) g(x),
for
all
x.
Then
use
i=]
Problem
5.
(a)
(a)
3(b).
A be
Let
limit point of the
real case, a
a point a in
A
with
\z
—
<
a\
[a, b]
x
[c, d],
divide [a, b] x
is
FIGURE
infinite, at least
this in half
(b)
[c,
then
A
if
e but z
^
number
z
by a
a.
title
>
0, there
If
A
is
an
infinite subset
[c, d].
Hint: First
vertical line as in Figure 7(a). Since
many
infinitely
line, as in
Figure
is
Prove the two-dimensional
7(b).
and horizontal
A
points of A. Divide
Continue
in this
way,
lines.
(The two-dimensional bisection argument outlined
method of
called, as in the
for every (real) s
Theorem:
one half contains
by a horizontal
dard that the
is
has a limit point in [a, b] x
d] in half
alternately dividing by vertical
7
A
set
version of the Bolzano-Weierstrass
of
A
a set of complex numbers.
in this hint
is
so stan-
"Bolzano-Weierstrass" often serves to describe the
proof, in addition to the
H. Petard, "A Contribution
to the
theorem
See, for example,
itself.
Mathematical Theory of Big
Game
Hunting," Amer. Math. Monthly, 45 (1938), 446-447.)
(b)
Prove that a continuous (complex-valued) function on [a,b] x \c\d]
bounded on
(c)
Prove that
if
[a,b\ x
/
is
on a
/
can use the same
then
takes
[c, d\.
(Imitate
a real-valued continuous function on \a,b\ x
maximum and minimum
trick that
works
for
is
Problem 22-31.)
value on
Theorem
7-3.)
[a, b\
x
[c, d],
[c, d\.
(You
26. Complex Functions
*6.
The proof of Theorem 2 cannot be considered
25-2,
interest to
tion z"
(a)
—
c
a convex subset of the plane
explicit
be found
for
n
1
"
0.
Make an
why
Explain
(b)
(a)
—
computation
show
to
any complex number
—
c
=
can be reduced to the case where
odd.
is
/
absolute value. If zo
rem 2
is
equal to
1;
show
0,
—
z"
—
that the integer
m
in the
we can
since
remainder of the proof works
minimum
absolute value of
certainly find
for /.
It
c has
minimum
proof of Theo1
y with y
therefore suffices to
/ does not occur
its
= — ar/)6,
show
at 0.
small 8 at least one of these points has smaller absolute value than
8
— c,
thereby obtaining a contradiction.
a noneonvex subset of the plane
FIGURE
the
that the
Suppose instead that / has its minimum absolute value at 0. Since n is
odd, the points ±<5, ±8i go under / into — c±<5", — c±5"i. Show that for
(d)
(b)
can
c.
—c =
the solution of z"
that solutions of z
Let zo be the point where the function f(z)
(c)
be completely elementary
to
= —a/fi depends on Theoand thus on the trigonometric functions. It is therefore of some
provide an elementary proof that there is a solution for the equa-
because the possibility of choosing y with y
rem
553
7.
Let/(z)
=
(z-zi)
mj
C
for
:)'"<
mk >
mi,
0.
k
Show
that /'(z)
=
(z
-
zi)""
•
.
.
•
.
(z
-
m
Zk)
J2 nia{z ~
~l
Za)
-
a=\
Let g(z)
(b)
— }^m a (z —
za
)
all lie on the same
Problem 25-11.
of the plane
any two points
set
FIGURE
9
containing
in
it,
it
is
convex
(Figure
which
is
Show
that
if
g(z)
side of a straight line
cannot
A subset K
.
8).
if
K
For any
called the
=
0,
then
through
z.
z\, ..., Zk
Hint:
Use
contains the line segment joining
set
A, there
is
a smallest convex
convex hull of A
(Figure
9); if
a
554
Infinite
Sequences and Infinite Series
point
P
side of
is
not in the convex hull of A, then
some
the roots of f'(z)
=
all
through P. Using
straight line
lie
of
A
is
contained on one
information, prove that
this
within the convex hull of the set
Further information on convex
sets will
{z\,
.
.
.
,
Zk}-
be found in reference [18] of the
Suggested Reading.
8.
*9.
Prove that
Suppose
(a)
if
that
/
is
/ =
For fixed yo
differentiable at
u
let
then
+ iv
where u and
g(x)
=
u(x
+
/
is
continuous
=
+ iyo). Show that if
g'(xo) = a and h'(xo) = fi.
= u(xo + iy) and /(y) = v(xo + iy).
and h(x)
a and /3, then
iyo)
(c)
(a)
Using the expression
ifi
1/1
1
1
(b)
find
/ w (Jc)
Use
this result to find
+
jc
z
at z.
v are real-valued functions.
for real
f'{xQ + /yo) = a +
On the other hand, suppose that k(y)
Show that /'(yo) = a and k'(yo) = —fi.
Suppose that f'(z) — for all z. Show that /
(b)
10.
z,
1
Zi
for all*.
arctan (A) (0) for
all k.
v(x
is
a constant function.
.
COMPLEX POWER
CHAPTER
If
to
come from,
the
title
of
complex functions are going
differentiable
chapter should give the secret away:
this
we
intend to
means of infinite series. This will necessitate a discussion of
infinite sequences of complex numbers, and sums of such sequences, but (as was
the case with limits and continuity) the basic definitions are almost exacdy 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 a„ in the
define functions by
«4
a\
«6
aj*
as
«3
a5
'a 7
an
FIGURE
you have not already guessed where
SERIES
*«12
}
plane (Figure
1).
1
The sequence shown
Figure
in
converges to
1
0,
"convergence" of complex
sequences being defined precisely as for real sequences:
converges
to
lim a„
d\.
,°2
if
aN
.03
for every e
>
sequence {a n }
the
symbols
in
/,
there
is
a natural
if
number
> N, then
n
=
I,
N
such
—
\a„
that, for all n
<
l\
,
e.
«6
This condition means that any
drawn around / will contain a n for all suffiexpressed more colloquially, the sequence is eventually
ciently large n (Figure 2);
circle
drawn around /.
Convergence of complex sequences
any
inside
FIGURE
circle
sequences, but can even be reduced to
2
THEOREM
1
is
not only defined precisely as for real
this familiar case.
Let
and
a„
=
b>i
I
=
fi
+
for real b„
icn
and
cn
,
let
Then
lim a„
=
/
if
+ iy
and only
if
lim b n
=
for real
and y
/3
n—>oc
and
/3
lim c„
n—>oo
n—»oo
PROOF
The proof is
left
as
consult the similar
The
an easy
Theorem
sum of a sequence
exercise.
)
is
s„
555
is
of Chapter 26.
1
\a n
If there
=
y.
any doubt
a
i
H
h
to
|
defined, once again, as lim sn
n—>oo
=
how
as to
an
.
,
where
proceed,
556
Infinite
Sequences and Infinite Series
Sequences for which
are
this limit exists
summable;
alternatively,
we may say
that
00
the infinite series 2_. an
converges
limit exists,
if this
and diverges otherwise.
It
»=i
is
new
unnecessary to develop any
convergence of infinite
tests for
series,
because
of the following theorem.
THEOREM
2
Let
an
=
+
b„
oo
00
Then 2, a n converges
if
and only
and
for real b n
ic„
2, c n
n=\
n=\
.
oo
b n and
2,
if
cn
both converge, and
in this
n=\
case
oo
OO
a
J2
' i
=
J2
+ \Jl c "
h"
i
n=\
PROOF
This
an immediate consequence of Theorem
is
sums of
{a,,}. |
There
is
also a notion of absolute
convergence for complex
an
2>
converges absolutely
if
the series
following theorem
is
2^
\an
\
converges
(this is
a series of real
which our
to
may be
earlier tests
applied).
The
not quite so easy as the preceding two.
Let
a„
=
b„
+
cn
.
oo
Then >, a n
converges absolutely
if
and only
oo
2, bn and 2, c n
if
n=\
absolutely.
Suppose
and
for real b„
ic„
00
PROOF
the series
n=[
n=\
numbers, and consequently one
3
series:
oo
oo
THEOREM
applied to the sequence of partial
1
first
that
/_]b n and
/£„
,;=1
z,=
both converge
n=\
71=1
both converge absolutely,
i.e.,
that 2_.
\b,i\
and
|<7„|
and
;;=1
l
oo
oo
2_]
\cn
\
both converge.
It
follows that
2,
+
\bn\
\cn
\
converges.
Now,
n=\
n=\
\a n
\
=
\b„
+icn <
\b n
\
\
+
\c„\.
00
It
follows from the comparison test that 2_. a n\ converges (the
\
numbers
,;=1
oo
\b„\
+
\c n
\
are real
and nonnegative). Thus
/^an
converges absolutely.
a
557
27. Complex Power Series
Now
suppose that
2,
\
a »\ converges. Since
n=\
|0b
it is
=
I
\/^ 2
Cn 2
+
,
clear that
\b n
<
\
\a n
and
\
<
\c„\
\a n
\.
oo
Once
oo
again, the comparison test shows that 2_.
n=\
\b„\
and
\]
\
cn\ converge.
|
n=\
oo
Two
consequences of Theorem 3 are particularly noteworthy.
If \_.
a " con "
n=\
OO
00
verges absolutely, then /_]£« and
/Jc„
n=\
also converge absolutely; consequently
n=\
OO
00
00
2_\b„ and /_/'„ converge, by
Theorem
23-5, so
n=\
n=\
2_,
it
a " converges by
shows that any rearrangement of an absolutely convergent
facts
can also be proved
orems for real numbers, by
(see Problem 13).
first
With these preliminaries
power series,
that
is,
2.
=1
In other words, absolute convergence implies convergence.
sum. These
Theorem
Similar reasoning
series
has the same
directly,
without using the corresponding the-
establishing
an analogue of the Cauchy criterion
we can now
safely disposed of,
functions of the
consider
complex
form
oo
f(z)
— ^2a„(z -
a)
n
—
a
+ a\(z -
a)
+ a2(z - a) 2
-\
.
n=0
Here
the
numbers a and
a„ are allowed to be complex,
interested in the behavior of
consider
power
series
/
for
complex
z.
As
and we are naturally
in the real case,
we
shall usually
centered at 0,
oo
f( Z )
= J2 a
n
>i?
'
n=0
f(zo) converges, then f(z) will also converge for \z\ < \zo\. The
fact will be similar to the proof of Theorem 24-6, but, for reasons
in this case, if
proof of
this
that will
soon become
clear,
we
will
not use
all
the paraphernalia of uniform con-
vergence and the Weierstrass M-test, even though they have complex analogues.
Our
theorem
4
next theorem consequently generalizes only a small part of Theorem 24-6.
Suppose
that
oo
/,
n ZQ
n
=
ao
+ a\zo + a 2 zo 2
H
.
558
Infinite
Sequences and Infinite Series
converges for some zo
Then
7^ 0.
if |z|
<
the two series
\zo\,
00
^
an z
n
=
+ a\ z + aiz 2
a
H
n=0
oo
— a\ + 2a2Z + 3«3Z 2 +
na,,z"
}
j
n=\
both converge absolutely.
PROOF
As
in the
a n zo
n
is
proof of Theorem 24-6, we
bounded: there
is
a
need only the
fact that the set
of numbers
M such that
< M
for
number
all n.
\a n Zo"\
We
will
then have
\a„z
and, for
I
—
\a„zo
<
M
z^O,
\na n z
I
=
—-n\a„zo
\z\
<
M
—
n
Since the series \_. \z/zo\" and /_] n l-/"ol" converge,
n=0
n=\
00
and \_. na nZ"~
n
converge absolutely (the argument for 2_\ na nZ ~
assumed
that
n=\
n=\
7^0, but this series certainly converges for
Theorem 4
\] a nz"
n=0
OO
;
shows that both
this
z
=
also). |
evidently restricts greatly the possibilities for the set
|
z
'.
2_. a "Z" converges
|
oo
For example, the shaded
set
A
in
Figure 3 cannot be the
set
of all
z
where /
a u z"
n=0
ikuri;
i
converges, since
It
it
contains
z,
but not the
seems quite unlikely that the
set
number w
satisfying \w\
<
\z\.
of points where a power series converges
could be anything except the set of points inside a circle. If we allow "circles of
radius 0" (when the power series converges only at 0) and "circles of radius oo"
(when the power
series
complication which
we
one
and
soon mention); the proof requires only Theorem 4
converges at
will
a knack for good organization.
all points),
then
this assertion
is
true (with
.
559
27. Complex Power Series
theorem
5
For any power series
00
^
= ao+a\z + aiz 2 + a^z 3 +
a„z"
n=0
one of the following three
must be
possibilities
true:
00
2_. a nZ
(1)
n
converges only for
z
—
0.
oo
2_. a nZ" converges absolutely for
«=o
(2)
There
(3)
a
is
number R >
all z
in
C.
such that 2_, a nZ
n
converges absolutely
if |z|
< R
,i=0
and diverges
when
PROOF
Id
if
=
R.)
x
in
>
\z\
R.
(Notice that
we do not mention what happens
Let
=
S
Suppose
I
first
that
R 2, a wn
n
:
S
converges for some
unbounded. Then
is
for
w
=x
with \w\
\
any complex number
z,
there
is
oo
a
number x
in
S such that
<
|z|
By
x.
definition of S, this
means
/^ a„
that
w"
n=0
00
converges for
some w with
X >
w\
converges absolutely. Thus, in
Now
R —
on
IS
then 2_\ a nZ
n
is
number x
in
this
case possibility
bounded, and
converges only for
n=0
the other hand, that
a
for
0,
suppose that S
R >
S with
some w with
<
\z\
\w\
0.
<
Then
x.
Once
>
R, then 2_, a nZ
n
R be
let
z
=
if z is
a complex
again, this
occurs in case
that
2, anZ
n
(2) is true.
the least upper
0, so possibility (1)
is
bound of
means
2_, a nW
n=0
that
<
|z|
n
If
Suppose,
true.
number with
S.
R, there
converges
Moreover,
if
=
does not converge, since
The number R which
from Theorem 4
so that 2_, a n z " converges absolutely.
7!
|z|
follows
It
(3) is
|z|
is
not in S.
called the
|
radius of convergence of
oo
V
is
a„z"
.
and
and
(2) it is
oo, respectively.
When
In cases
(1)
customary
< R <
to say that the radius
oo, the circle {z
:
\z\
of convergence
=
R}
is
called
v
560
Infinite
Sequences and Infinite Series
Eri—Kj
/]an z n
of convergence of
the circle
If Z
.
outside the circle, then, of course,
is
n=0
much
a n z' does not converge, but actually a
stronger statement can be made:
n=0
the terms a n z" are not even bounded.
>
|z|
\w\
show
>
R;
if
2_, a nW
that
n
converges, which
^—
11=0
_
convergence the
the terms a„z" are not
of
convergence
circle
bounded
To prove this, let w be any number with
were bounded, then the proof of Theorem 4 would
the terms a n z"
y. a nZ
series
n
Thus
false.
is
(Figure
4), inside
converges in the best possible
n=0
outside the circle the series diverges in the worst possible
way
way
the circle of
and
(absolutely)
terms a„z" are
(the
not bounded).
What happens
We
will
series
on the circle of
convergence
not consider that question at
all,
which converge everywhere on the
converge nowhere on the
anything
a
much more
difficult question.
except to mention that there are power
circle
of convergence, power series which
of convergence, and power series that do just about
circle
between. (See Problem
in
is
5.)
Algebraic manipulations on complex power series can be justified just as in the
oo
oo
Thus,
real case.
FIGURE
if
—
f(z)
=
"
2_, a " z an<^ S(z)
n=0
4
2-,^" z
"
k° tri nave racnus °f
n=0
oo
convergence > R, then h(z)
> R and h
= f+g
=
+
/](fl«
b n )z" also has radius
inside the circle of radius R.
oo
h(z)
=
Similarly, the
of convergence
Cauchy product
n
— /
for c„
c
2_, nz!\
«=0
> R and
akbn -k, has radius of convergence
h
=
fg
k=0
oo
inside the circle of radius R.
And if
f(z)
—
2~]a„z" has radius of convergence
>
n=0
oo
and «o
¥" 0,
then
we can
find a
power
series
/_\b„z
n
with radius of convergence
n=0
>
which represents
But our real goal
side
// inside
its
circle
is
of convergence.
to
produce differentiable functions.
power
to generalize the result
Chapter 24, even
if
we were willing
proved for
to introduce
real
(which could also have been used in Chapter 24).
notice that at least there
produced by term-by-term
is
series in
uniform convergence, because no
analogue of Theorem 24-3 seems available. Instead we
we
We
Chapter 24,
a function defined by a power series can be differentiated term-by-term inthe circle of convergence. At this point we can no longer imitate the proof of
therefore
that
want
1
in this chapter
will use
a direct argument
Before beginning the proof,
no problem about the convergence of the
differentiation. If the series /^ctnz" has radius
n=0
vergence R, then
converges for
Theorem 4 immediately
< R. Moreover,
if |z|
>
implies that the series
series
of con-
oo
/
na n z n ~
also
R, so that the terms a„z n arc unbounded,
27. Complex Power Series
then the terms na n z"
l
are surely
unbounded,
so
y. nanZ
n
561
does not converge.
oo
n
This shows that the radius of convergence of 2_, na nl ~
also exactly R.
is
n=\
theorem
6
If the
power
series
= J^a n z n
f(z)
n=0
has radius of convergence
R >
/
then
0,
differentiable at z for all z with
is
\z\
<
R,
and
f {z) =
Y
j
na n z
n-
n=\
PROOF
We
The
another "e/3 argument."
will use
theorem
fact that the
is
clearly true for
polynomial functions suggests writing
f(z
+ h)-f(z)
}na
-
nz
A
2
n~
h
dz
A
/ na
+ hr-z")
an
n=0
A
<
A
+ h)"-z")
((z
l^ a »
n
«z
1
„=0
n=0
A
A
2^na
+ hy-z")
dz
+ l^ a »
7
n=
n=0
+ / nan z n
l
j
We will show that for any
N
s
>
nz
\
- }jia n z n
n=\
< e/3 by choosing
nz
\
+ h)"-z")
l^ a "
1
=
~
n=\
each absolute value on the right side can be
0,
and h
sufficiently large
This
sufficiently small.
made
will clearly
prove the theorem.
Only
the
term
first
+
<
convenient way
with
\z
h\
|
<
|z|
I-
-n
_
]
<
((z
v"
— = x "}
x-y
Applying
\zo\
The expression
if we remember that
co
of (*)
in the right side
with, choose some zo with
+ xn
~2
+
v
h)
n
—
we
difficulties.
will
+ h) n
z")/ h can be written in a
+ x"" 3 v 2 +
•
+
•
y"" 1
.
+ h)"
(z + h)-z
(z
h
'
obtain
<-
_i_
iiV
^-^
1
_
-"
=
+ hf- +
1
(z
z(z
+
To begin
consider only h
this to
(z
we
present any
will
R; henceforth
h)"-
2
+
+
n
z
~\
more
.
562
Infinite
Sequences and Infinite Series
Since
+ h)"- +
+h)'- 2
1
\(z
z(z
+
+
l
z"-
\
<
]
n\zo\"~
,
we have
+ h) n -z n
((z
]
<
a„
But the
series Y_]n|a„|
converges, so
\zq\"
•
,/j-l
n\a„\
N
if
\zo
sufficiently large,
is
then
n=\
n a n\
2
<
\Z0\"~
\
,
n=N+1
This means that
A
{{z
A
+ hT-z")
((z
+ h)"-z n
/;
)
//
n=0
71=0
+
((z
an
J2
n
-z")
h)"
= N+\
E
*
h
«z
+ h) n
-z")
h
n = N+\
s
< —
3
In short,
if
N
A
A
1;
is
sufficiently large,
«z
a»
+ h) n -zn
then
A
L
)
1
n=0
((z + h)"-z
"-
n
£
)
h
n=0
for all h with
It is
fl
+
\z
h\
<
\zq\.
easy to deal with the third term on the right side of (*): Since
converges,
it
follows that
N
if
sufficiently large,
Y,nan z
n~ l
- J2 na " z
"~
<
n=\
n=\
choosing an
..
N
such that
A
hm )
Hz
an
(1)
and
(2)
+ hr-z")
-
are true,
=
n=0
n
((z
+ h)"
A
>
we
na n z
note that
„_!
n=\
since the polynomial function g(z)
^a
then
N
oo
Finally,
is
y. na nZ'
n=\
-z")
=
2_\ a nZ"
n=0
^
certainly differentiable. Therefore
N
-1
(3)
n=0
is
<
£
3'
n=\
for sufficientlv small h.
As we have already indicated,
(1), (2),
and
(3)
prove the theorem.
|
27. Complex Power Series
Theorem 6
has an obvious corollary: a function represented by a power series
of convergence, and the power
infinitely differentiable inside the circle
Taylor series
/
at 0. It follows, in particular, that
convergence, since a function differentiable at
The
563
continuity of a
power
series inside
z
at z
is its
circle
of
(Problem 26-8).
of convergence helps explain the
circle
its
continuous
is
series
continuous inside the
is
is
behavior of certain Taylor series obtained for real functions, and gives the promised
answers to the questions raised
end of Chapter 24.
at the
—
that the Taylor series for the function f(z)
4
2
l-z +
5
-z +
<
z
2
),
We
have already seen
namely,
---,
and consequently has radius of converof convergence contains the two points
i and
—i at
If this power series converged in a circle of
radius greater than 1, then (Figure 5) it would represent a function which was
continuous in that circle, in particular at i and —i. But this is impossible, since it
equals 1/(1 + z 2 ) inside the unit circle, and 1/(1 + z 2 ) does not approach a limit
as z s approaches / or —i from inside the unit circle.
The use of complex numbers also sheds some light on the strange behavior of
converges for real
gence
FIGURE
when
6
z
+
1/(1
1
.
z
only
\z\
no accident that the
which / is undefined.
It is
1,
circle
the Taylor series for the function
-l/.x
e
fix)
2
Although we have not yet defined
if
y
is
real
and unequal
e z for
= 0.
x
0,
complex
z, it will
2
f (iy) = e -Wy) =e
The
interesting fact
about
Thus /
so
hardly surprising that
will
expression
this
small.
it is
it is
equal to
x
x3
=x—
=
1
—
V
x2
when
x
=
1
X
+
x5
+
x
+
(as
—
4
"
4!
x
2
+
2!
V.
therefore
e
z
5!
+
define
5
3
sinz
—z-
+
3!
=
1
-
-
2
exp(z)
=
e
z
=
1
—
~5\
74
2
cose
yi
+
+
4!
+
+
large as y
^—
becomes
defined for complex numbers,
know
x we
2!
e
becomes
Taylor series only for
its
clear. For real
cosx
we
that
2
it
will actually define e
sin
z
l/.v
that
is
not even be continuous at
The method by which we
complex z should by now be
For complex
presumably be true
to 0, then
+
well as sin z
that
z
=
0.
and cos z)
for
564
Infinite
Sequences and Infinite Series
Then
=
sin'(z)
Moreover,
if
of the terms
we
= — sinz,
cosz, cos'(z)
and
replace z by iz in the series for e
(justified
z
,
=
exp(z) by Theorem 6.
and make a rearrangement
exp'(z)
by absolute convergence), something particularly
interesting
happens:
(iz)
3
(iz)
4
.
(iz)
5
,
so
=
z
e'
It is
clear
from the
definitions
cosz
the
(i.e.,
also have
u
e
From
series) that
= — sin z,
= cosz,
cos(— z)
we
sinz.
i
power
sin(— z)
so
+
the equations for
e'
z
and
—
'
e
cosz
—
we can
c
i
sin
derive the formulas
sin
2/
cosz
2
The development
of complex power series thus places the exponential function
the very core of the development of the elementary functions
—
it
at
reveals a con-
nection between the trigonometric and exponential functions which was never
imagined when these functions were
and which could never have
been discovered without the use of complex numbers. As a by-product of this
relationship, we obtain a hitherto unsuspected connection between the numbers e
and it: if in the formula
z
e'
we
take z
=
tt,
we
=
generally, e^'^"
is
With these remarks we
tions.
And
cosz
+
i
sinz
obtain the remarkable result
e
(More
defined,
first
yet there are
been mentioned. Thus
in
=-1.
an «th root of
will
power series which have not
we have seldom considered power series centered at a.
still
far,
1.)
bring to a close our investigation of complex func-
several basic facts about
f(z)
= J^a n (z-a) n
,
27. Complex Power Series
—
except for a
0.
This omission was adopted partly to simplify the exposition.
For power series centered at a there are obvious versions of
this
chapter (the proofs require only
(possibly
with
\z
—
all
the theorems in
there
trivial modifications):
oo
is
number R
a
n
or "oo") such that the series 2_, a n(z — a) converges absolutely for z
n=0
a\ < R, and has unbounded terms for z with \z — a\ > R; moreover, for
with
all z
565
— a\ < R
\z
the function
00
f(z)=Y an (z-a)
n
/
has derivative
= J^na n (z-a) n -
f\z)
It is less
FIGURE
6
.
straightforward to investigate the possibility of representing a function
power
as a
1
series
centered at
b, if
power
already written as a
it is
series
centered
at a. If
oo
f(z)
= J2a„(z-a)"
,<=o
has radius of convergence
is
and b
/?,
a point with
is
power
true that f(z) can also be written as a
\b
series
f iz) = J2b„(:
—
a\
< R
(Figure
centered at
6),
then
it
b,
bf
n=0
numbers b n are
(the
convergence
We
at least
necessarily
R—
prove the
will not
\b
— a\
facts
several other important facts
f
[it
(n)
{b)/n\)\ moreover, this series has radius of
may
be larger).
mentioned
we
shall
f{z)=Y, aniz-a)
in the previous
paragraph, and there are
not prove. For example,
n
and
if
<Xz)=YJ b n(Z-b)\
n=0
and g(b)
=
centered at
new
and
at
a, then
b.
we would
expect that
All such facts could be
ideas, but the proofs
would not be
reciprocals of power series.
a into one centered at b
fog
more
of computations, we will
requires
still
is
skill.
The
fog
proved
can be written as a power
now
as easy as the proofs
possibility
quite a bit
series
without introducing any basic
about sums, products
of changing a power series centered
more
Rather than end
involved,
and the treatment of
this section
with a
tour deforce
instead give a preview of "complex analysis," one of
most beautiful branches of mathematics, where all these facts are derived as
straightforward consequences of some fundamental results.
the
Power
tions
series
were introduced
which are
entiable,
it is
differentiable.
in this
chapter
in
order to provide complex func-
Since these functions are actually infinitely differ-
natural to suppose that
we have
therefore selected only a very special
566
Infinite
Sequences and Infinite Series
complex
collection of differentiable
show
analysis
that this
If a complex function
then
it is
not at
is
basic theorems of
complex
all true:
defined in
is
The
functions.
some
A
region
of the plane and
A
automatically infinitely differentiable in
.
is differentiable in
A,
A
the
Moreover, for each point a in
a\
Taylor series for
These
to give
a-)
facts are
f
at
a
among
will converge
the
first
f
to
in
any
be proved
to
any idea of the proofs themselves
from anything
in
elementary calculus.
—
contained in
A
(Figure 7).
complex analysis. It is impossible
methods used are quite different
in
the
If these facts are granted,
mentioned before can be proved very
the facts
circle
however, then
easily.
Suppose, for example, that / and g are functions which can be written as power
series. Then, as we have shown, / and g are differentiable
it then follows from
—
f +
easy general theorems that
FIGURE
7
Appealing
as
power
We
to the results
f
g,
from complex
1/g and
g,
•
analysis,
it
fog
are also differentiable.
follows that they can be written
series.
know how to compute the power series for f + g, f g and \/g from
and
g. It is also easy to guess how one would compute an expression
/
fog as a power series in (z — b) when we are given the power series expansions
already
those for
for
oo
f(z)
= J2 an(z-a) n
n=0
=
g(z)
Y
J
b k(z-b)
k
,
k=0
with a
=
g{b)
=
bo, so that
Y h(z-b)
g(z)-a =
k
.
J
k=\
First
of all,
we know how
to
compute the power
(g(z)-a)
=
l
r£ b
f
and
this
power
series will
begin with
(z
—
b)
k
1
.
series
(z-b) k \
.
Consequently, the coefficient of
z"
in
=
f(g(z))
J2cii(g(z)-a)
1
1=0
can be calculated as a
powers of g(z)
Similarly,
—
finite
sum, involving only
coefficients arising
from the
first
n
a.
if
oo
f(z)
has radius of convergence
R).
Thus,
if
b
is
in
A,
it
/?,
is
then
= J^an (z-a) n
/
is
differentiable in the region
possible to write
/
as a
power
A =
series
{z
:
\z—a\
centered
<
at b,
567
27. Complex Power Series
which
will
converge in the
be f (,1) {b)/n\
will
.
This
circle
series
R —
of radius
may
—
\b
The
a\.
coefficient of z
n
actually converge in a larger circle, because
00
2^ a n(z —
may be
a)"
the series for a function differentiable in a larger region
than A. For example, suppose that f(z) — 1/(1 + Z ). Then / is differentiable,
except at i and —i, where it is not defined. Thus f(z) can be written as a power
00
2_, a nZ
series
ain
—
n
with radius of convergence
(—1)" and
cik
=
if
k
is
odd).
It is
a matter of
(as
1
fact,
we know
that
also possible to write
oo
f {z) = Y^b (Z-\)\
lr
n=0
where b n
FIGURE
series:
= f
it is
8
{n)
VI +
As an added
will
(^)/n\
2
(^)
is
z the
not at
easily predict the radius
the distance from ^ to
be mentioned, which
For real
can
lies
or
—
i
of convergence of
(Figure
8).
analysis further,
quite near the surface,
this
and which
one more
will
result
be found in
subject.
values of sin z always
all true.
i
complex
incentive to investigate
any treatment of the
this
,
We
.
In
fact, if z
=
gi(iy)
between
lie
iy, for
_
real,
y
—
1
and
1
,
but for complex
z
then
g—i(iy)
sin zy
2/
If
y
is
large, then sin iy
is
2/
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's
Theorem: The only bounded
entire functions are the constant junctions.
As a simple application of Liouville's Theorem, consider a polynomial function
= z"+a n _ lZ "~ +---+a
l
f(z)
where n >
large
z,
1,
so that
so Liouville's
,
/ is not a constant. We already know that
Theorem tells us nothing interesting about
f(z)
is
large for
/. But consider
the function
g(z)
1
=
f(z)
If f(z)
were never 0, then g would be entire; since f(z) becomes large for large z,
would also be bounded, contradicting Liouville's Theorem. Thus
for some z, and we have proved the Fundamental Theorem of Algebra.
the function g
f(z)
=
PROBLEMS
1
.
Decide whether each of the following
verges absolutely.
series converges,
and whether
it
con-
*
568
Infinite
Sequences and Infinite Series
n=\
y,
+ 2/
1
E
oo
iv
)
E
(
+
5
2'')"'
—
Alogw
v
2^
+
.„log«
n
n
n=2
Use the ratio test to show that the radius of convergence of each of the
power
following
approach a
<
limit
>
to a limit
series
1
is
if
1
|z|
(In
.
<
1,
each case the ratios of successive terms
but for
|z|
>
1
will
the ratios will tend to oo or
1.)
n=\
n=\
oo
in
n=\
oo
iv
oo
(v)
3.
E
2
"—
Use the root
test
power
the following
problems
to
2
+
z
y+
oo
n
Y-z
n
.
n=\
in
Z^
nn
n=\
oo
iv
9
T-z
n
»=]
oo
E 2V
n=l
series. (In
Chapter
2
z
(i)
(Problem 23-9)
'
!
-
3
some
cases,
22.)
4
2^ +
z
to find the radius of
z
5
?
23
+
z
6
3^
+ "-
you
will
convergence of each of
need
limits derived in the
27. Complex Power Series
The
root test can always be used, in theory at
convergence of a power
to a
least, to find
569
the radius of
a close analysis of the situation leads
series; in fact,
formula for the radius of convergence, known as the "Cauchy-Hadamard
formula." Suppose
that the set of numbers \f\a n
first
bounded.
is
\
00
(a)
Use Problem 23-9
show
to
that
lim ^/\a„\
H—>0O
if
\z\
<
1,
then } a n z" con*-^
«=0
verges.
OO
(b)
Also show that
lim >/|a n
if
|
n—*oo
\z\
>
/a„z"
then
1,
has
*—~'
unbounded
terms.
n=0
00
(c)
Parts
and
(a)
show
(b)
that the radius of convergence of
>_,a„z"
is
#1=0
1/ lim vjflj (where "1/0"
means
To complete
"oo").
the formula, de-
II—>oo
lim ^/\a„
fine
\
=
oo
the set of
if
^/\a„\
all
unbounded. Prove
is
that in
n—>-oo
oo
this case,
(which
is
5.
2_, a nZ
n
diverges for z
may be
00
e
Show
E-"-
n=\
n=\
9.
converges everywhere on the unit circle; that the
nowhere on the unit circle; and that the second series
one point on the unit circle and diverges for at least
circle.
e
w
—
e
w
.
+
that sin(z
w—
w)
=
w
for
sin z sin
sin z cos
all
w +
complex
z
cos z sin
w and
some
real number y.
x+iy
x
e for real
|e
=
and
(b)
Prove that
(a)
Prove that exp takes on every complex value except
(b)
Prove that sin takes on every complex value.
|
jc
(i)
f(z)
(ii)
f(z)
=
=
tan
z.
z(\
-z)- ]/2
.
+
w)
=
1
can be written e iy
y.
For each of the following functions, compute the
the Taylor series centered at
cos(z
and w.
Prove that every complex number of absolute value
for
8.
00
„
z+w for all complex numbers
z and w by showing
z+w is the Cauchy product
that the infinite series for e
of the series for e z
cos z cos
(a)
00
„
2:
E-.
least
Prove that e z
and
7.
from Problem
£%•
one point on the unit
(b)
of convergence
first series
converges for at
(a)
series
n=l
third series converges
6.
0, so that the radius
considered as "l/oo").
Consider the following three
Prove that the
^
first
0.
three nonzero terms of
by manipulating power
series.
570
Infinite
Sequences and Infinite Series
,sin z
=
iii)
f(z)
iv)
/(z)=log(l
f(z)
=
sin
v)
f(z)
=
sin(r)
vi)
f(z)
=
vii)
).
y~
z
z cos z
4
=
/(7)
3'
-2z 2 +
:
viii)
-z 2
l[^<V^I-l)_
!j
complex function / as / = u +
/u, where u and u are real-valued. Let u and D denote the restrictions
of u and v to the real numbers. In other words, u{x) = u{x) for real
numbers x (but it is not defined for other x). Using Problem 26-9, show
Suppose
10.
that
we
write a differentiable
we have
that for real x
f'(x)
where
=
complex
/' denotes the
+
u'(x)
iv'ix),
and
derivative, while u'
v'
denote the
ordinary derivatives of these real-valued functions on R.
(b)
Show, more generally, that
(k
f \x)
(c)
Suppose that /
satisfies
/
(*)
w
u
{k)
(x)
+
iv
a)
(x).
the equation
+fl»-l/ w|
- 1,
+ -" + flO/ =
I
where the a, are real numbers, and where the f [k) denote higher-order
complex derivatives. Show that u and v satisfy the same equation, where
(k)
u
and v (k) now denote higher-order derivatives of real -valued functions
(d)
onR.
Show that
...-(-
aq
if
—
0,
solutions of
11.
(a)
Show
(b)
Given
that
complex root of the equation z" + an - z."~ +
then f(x) = e bx sin ex and f(x) = e bx coscx are both
a =b-\- ci
l
a
\
(*).
exp
w^
is
0,
is
not
show
one-one on C.
that e z
= w
if
and only
if z
(here log denotes the real logarithm function),
'(c)
Show
= x + iy
=
with x
log
|
w
\
and y an argument of w.
that there does not exist a continuous function log defined for
nonzero complex numbers, such that exp(log(z))
(Show
that log cannot even
Since there
is
no way
be defined continuously
to define a
=
Z for all j
for
|z|
=
1
/
0.
.)
continuous logarithm function
we
can-
not speak of the logarithm of a complex number, but only of "a logarithm
for w,"
meaning one of the
infinitely
many numbers
z
with e z
=
w.
And
,
.
571
27. Complex Power Series
complex numbers a and b we define a b to be a set of complex numbl ° sa
bers, namely the set of all numbers e
or, more precisely, the set of
bz
all numbers e
where z is a logarithm for a
for
(d)
If
m
is
an
then a m consists of only one number, the one given by
integer,
the usual elementary definition of a'"
(e)
(f)
(g)
(h)
.
and n are integers, then the set a m n coincides with the set of values
given by the usual elementary definition, namely the set of all b m where b
is an nth root of a.
b
If a and b are real and b is irrational, then a contains infinitely many
members, even for a > 0.
Find all logarithms of /, and find all values of /'.
By (a b ) c we mean the set of all numbers of the form z c for some number z
If
m
'
in the set a
b
Show
.
many
that (1')' has infinitely
1'
values, while
'
has
only one.
12.
(i)
Show that all values of a bc
(a)
For real x show that
(It
(b)
will
The
are also values of (a
we can choose
log(.t
+
/ )
=
log( v
logU'
—
i)
=
log(\/l
1
+ i)
log(x
+ x 2 + I— —
+ x2 —
help to note that tt/2
:
—
I
i
)
(a
and log(x —
/)
to
>
0.)
)
Is
.
arctan x
)
— — arctan x
1
i
)
a bc
=
b c
=
arctan x
b c
)
n(a c ) b ?
be
.
arctan \/x for x
expression
1/1
1
+x
1
2
\x —
2/
1
x
i
+
i
yields, formally,
/
Use part
13.
(a)
A
sequence
lim
real.
are
(b)
(c)
(a)
—
\a„,
+ x2
1
to
=
^IJogC*
check that
{a„}
a„\
-0 - log(* +i)J.
2/
this
answer agrees with the usual one.
of complex numbers
=
0.
Prove that {a„}
Suppose
a
is
that a n
called a
bn
+
if
Cauchy sequence
icn ,
where b n and
and only
if {b n
}
if
c„ are
and
{c„}
Cauchy sequences.
Prove that every Cauchy sequence of complex numbers converges.
Give direct proofs, without using theorems about real
has the same sum.
is
that
an
permitted, and in fact advisable, to use the proofs
(It is
series.)
Prove that
"
1
X^ e ikx = J*1
A=1
series,
convergent and that any rearrangement
of the corresponding theorems for real
(a)
—
Cauchy sequence
absolutely convergent series
14.
is
- e inx
- - e ix
sin
-a
V2 /
(
)
x
sin
2
J(n+l)x/2
*
572
Infinite
.
Sequences and Infinite Series
(b)
Deduce
\
the formulas for
and 2_\
cos kx
k=\
m kx
that are given
in
k=\
Problem 15-33.
15.
= a2 =
Let {a„} be the Fibonacci sequence, a\
=
If r„
(b)
Show that
(c)
Prove that the limit does
-
r„
,
=
if r
<
1
=
an+ i/an show that rn +\
(a)
and
lim
<
r„
r„ +2
+ l/r„.
r =
+
1
then
r„ exists,
1,
1
1
+ a n+
a„
/r, so that r
<
Hint: If r„
exist.
—
a„+2
=
(l
+ v5 )/2,
(1
\.
+ v5 )/2.
then
—
2
r,,
.
oo
(d)
Show
has radius of convergence 2/(1
2_, a nZ"
that
+
V5).
(Using
n=\
oo
the unproved theorems in this chapter
and the
=
"
2_, a n z
fact that
n=\
— l/(z +z —
from Problem 24-16 we could have predicted that the
1)
radius of convergence
—
Z
1
=
+
+
16.
Since (e z
which
v5)/2.
(—1
±
v5)/2, the radius of convergence
Notice that
number
this
is
indeed equal to
75).)
—
\)/z can be written as the
nonzero
is
+
the smallest absolute value of the roots of z
0; since the roots are
should be (— 1
2/(1
is
at 0,
it
power
follows that there
is
a
series
1
power
+
z/2!
+
z /3!
+
•
•
•
series
n=0
with nonzero radius of convergence. Using the unproved theorems in
we can even
chapter,
predict the radius of convergence;
the smallest absolute value of the non-zero
e
z
—
=0. The
1
numbers
numbers
z
it is
=
this
2tt, since this
2kni
b„ appearing here are called the
is
for which
Bernoulli
numbers.*
(a)
Clearly bo
—
1
.
Now show
that
z
e
z
-
2
1
+\
e -z-\
e~
and deduce
b
|
+
z
2
e
:
e
z
+
-
1
1
+
e z-\
z
ez
1
that
= —^
b„
,
=
if
n
is
odd and n >
1
_
if n
*Sometimes the numbers Bn = ( — 1)" »2« arc called the Bernoulli numbers, because />„ =
is odd and >
(see part (a)) and because the numbers bin alternate in sign, although we will not
1
prove
this.
(
)ihei
modifications of
this
nomenclature are also
in use.
27. Complex Power Series
By
(b)
573
finding the coefficient of z" in the right side of the equation
H £ £-•%
oo
show
,3
,1
+
£
+
fr
+
that
This formula allows us to compute any bk in terms of previous ones, and
shows that each
i>2
Part
*(c)
(a)
=
rational. Calculate
is
*4
\,
b 2n
Show
"2
Show
—
=
-^
z
e" 2
+ e~^ 2
2
e*/2
-
""
^/
2
oo
,
£^<-'>" 22 ";
2 "-
n=0
=
cot z
—
2 cot 2z.
that
tanz
=
,
Y
a-
'"(
Bernoulli
1)- 2 2 "(2 2 "
1
l);
2 ""
1
.
\
(This series converges for
The
^8
43.
that
oo
17.
=
by 2/z and show that
z
tan z
*(e)
e'+l
^-1
z
„ 2n
2-* (2n)!*
*(d)
^6
~^J.
shows that
yReplace
=
two or three of the following:
\z\
<
tc/2.)
numbers play an important
role in a
theorem which
is
best
D to denote the "differThen D k f will mean f and
introduced by some notational nonsense. Let us use
entiation operator," so that
Df
denotes
/'.
(k)
oo
e
D
f
but
will
it
mean /_\f
( '
n=0
will make sense
!l
/n\ (of course this series
if
/
is
makes no sense
in general,
a polynomial function, for example).
Finally,
A denote the "difference operator" for which Af(x) = f(x + 1) — f(x).
Now Taylor's Theorem implies, disregarding questions of convergence, that
let
f(x+\) = J2
f
w (x)
n=0
or
(*)
f(x+\)-f(x) = J2
n=\
f W (x)
,
574
Infinite
Sequences and Infinite Series
we may
Af =
write this symbolically as
D
Even more symbolically
"identity operator."
which suggests
(e
—
1)
where
/',
stands for the
1
can be written
this
A=
e
D
—
1
that
D
Thus we obviously ought
have
to
oo
D=
,
Y^D A.
k
fc=0
i.e.
oo
.
bi
k=0
The
(a)
beautiful thine" about
Prove that
(**)
the infinite
a formula for f
Deduce from
\x +
1
)
of
—f
(x);
f ^{x)
that
it
works!
a polynomial function
By applying
(in
(*)
which case
to
f
(k
\ find
then use the formula in Problem 16(b)
in the right side of (**).
{
(**) that
00
/'(O)
is
is
really a finite sum). Hint:
is
(k
to find the coefficient
(b)
/
literally true if
is
sum
nonsense
all this
'.->
+
•
•
u
+ f{n) = ]T -£U
(k)
(n
+
1
- / w (0)].
)
k=0
(c)
Show
g(0)
(d)
+
any polynomial function g we have
that for
...
Apply
+ g(n)=
['
g(x)
this to
=
Tk' =
£-*>
Using the
The
first
Y ^[g
[
-
])
—+ ^r^(k\k-\
p
i
+
\
'
fact that b\
(n
+
\)
-
(k
g
- ])
(0)\.
= — i,
p-k+]
v
k=\
show
that
ten instances of this formula
which offered
{k
x p to show that
P
k=\
+
g (t)dt
were written out
in
Problem
as a challenge the discovery of the general pattern.
may now seem
to
this
way!
After writing out
these 10 formulas, Bernoulli claims (in his posthumously printed
Ars Conjectandi, 1713):
may be
This
be a preposterous suggestion, but the Bernoulli num-
bers were actually discovered in precisely
larity
2-7,
"Whoever
will
examine the
able to continue the table."
He
work
series as to their regu-
then writes
down
the above
.
.
27. Complex Power Series
formula, offering no proof at
all,
merely noting that the
(which he denoted simply by A, B, C,
lem
power
the
*18.
The
16(b).
The formula
Problem
some new
(a)
The
coefficients bk
equation in Prob-
sum must be
where g
to the case
replaced by a
an expression
coefficients in
Euler.
finite
for the remainder,
is
not a
sum
plus a
it is
useful
functions.
Bernoulli polynomials
are defined by
<p n
=^\V) bn-kX
fc=0
first
)
can be generalized
to find
<Pn(x)
The
satisfy the
was discovered by
1)
infinite
remainder term. In order
to introduce
—
17(c)
polynomial function; the
.
between these numbers and the
relation
series for z/(e z
in
.
575
k
.
'
^
three are
1
cpi(x)
=x -
(p~>(x)
= x~2 — x +
-,
!
-,
o
3x
,
(p }
Show
(x)
x
=x --rT+y:
that
(p„(0)
<p n
(\)
(pn '{x)
<pn (x)
=bn
=
,
bn
if
//
>
1
=n<p„-i(x),
=
(-l)
n
<pn
-x).
Hint: Prove the last equation by induction
(b)
Ru
Let
k
be the remainder term
{x)
interval [x x
,
(*)
+
1
]
,
f«\x +
on
n.
in Taylor's
Theorem
for
(k
f \ on
so that
i)
-
—r^
f \x) = Y l
k
n=\
k
+ R»
(-
X)
-
Prove that
k=0
Hint: Imitate
(c)
Use
k=0
'
Problem
the integral
17(a).
Notice the subscript
form of the remainder
££*»VM = /'
to
show
N—k
that
+1
^±f^/<-'>«*.
on R.
the
)
576
Infinite
Sequences and Infinite Series
Deduce
(d)
+ g(x+
g(x)
=
Summation Formula":
the "Euler-Maclaurin
\)
+
f
---
+ g(x+n)
g(t)dt
Jx
+ J2^[g
lk
-
l)
(x+n +
-
1)
{k
-
]]
g
(x)]
+ SN (x,n),
'
k=\
where
+ ; +1
<p
.
Let
(e)
\jjn
=
since <p„(l)
Show
1.
iN)
g
(ndt.
X
*x+j
A J
(Part
and
^,,(0),
(a)
implies that
also that
\(/ n
is
n
if
even
which
1,
>
1,
if
n
then
satisfies \[/„(t)
%//„ is
even and odd
is
=
continuous,
if
n
is
that
r.x+n-
SN (x,n)
-{)
"
•
be the periodic function, with period
^(f)forO<? <
odd.)
+j+\
';,'
N (X
—
=-
N\
{N)
g
{t)dt
c+fl+l
+ " +l
(
=
(-l)
N+l
r
"^~-g^ N) {t)dt
if
x
is
an integer!
Unlike the remainder in Taylor's Theorem, the remainder Sn(x,h) usually
does not
become
satisfy
lim Sn(x,h)
=
0,
because the Bernoulli numbers and functions
large very rapidly (although the
first
few examples do not suggest
Nevertheless, important information can often be obtained from the
formula.
The
general situation
is
this).
summation
best discussed within the context of a specialized
study ("asymptotic series"), but the next problem shows one particularly important
example.
**19.
(a)
log
1
N=
Use the Euler-Maclaurin Formula, with
+
•
•
•
+ log(n —
2, to
show
that
1
/*"
1/1
\
2 (t)
f\
= / U «""-2 l08 " + 12U- 1 + y. "2^
j
1
(b)
Show
that
i0
(c)
\lr
Sy n n+l/2 e -n+l/a2n)J-
Explain why the improper integral
that
if
a
=
exp(fi
+
1
ft
+
=
It 2
J
]
dt
xj/2(t)/2t
I
1/12), then
n\
ttts
\2
\
-rrr^r-r
a n n+l/2 e -n+l/a2n)
J
=-
r
/
J
init)
_
2t
,
at
.
exists,
and show
.
27. Complex Power Series
(d)
Problem 19-4 1(d) shows
577
that
2
{n\)
2 2n
V^= n^-oc (ln)\y/n
i
Use part
(c)
show
to
that
a 2 n 2n+\ e -2n22n
=
s/tt
lim
a(2n) 2n+l / 2 e- 2n s/n
n ^°°
and conclude
(e)
Show
=
a
that
\J1ti
'
.
that
1/2
<f>2(t)dt
/
=
(You can do the computations
mediately from Problem
f
\J/(x)
tfr(n)
—
attention to
i/2(t)dt
of
(f)
<p2
(Figure
Noting that
for
its
0.
Conclude
that
>
forO
<0
for
Graph
Hint:
all n.
but the result also follows im-
explicitly,
18(a).)
Jo
with
=
(p2(t)dt
/
Jo
Jo
values at xq, I,
and
x\,
<x <
1/2
1/2<jc<
1,
\js on
[0, 1], paying particular
where xq an d x\ are the roots
9).
i}/{x)
=
f(x)
—\j/{\
=
— x), show
that
f(f)dt >
/
on
[0, 1],
Jo
and hence everywhere, with
(g)
Finally, use this
\[r(n)
=
f
2t 2
Jn
Using the
all
n
information and integration by parts to show that
t2it)
(h)
for
fact that the
dt
maximum
>
0.
value of |<^2U)I for x in
conclude that
o<rm«<
/
It
Jn
(i)
Finally,
2
'
\2n
conclude that
^n" +l/2 e-"
<
n\
<
>/2^/i"
+,/
V" +
1/,l2 "
)
.
[0, 1]
is
g,
578
Infinite
Sequences and Infinite Series
The
n\
is
of Problem 19, a strong form of Stirling's Formula, shows that
approximately s/lix n" + / e~", in the sense that this expression differs from n\
final result
to n when n is large. For example, for
we obtain 3598696 instead of 3628800, with an error < 1%.
A more general form of Stirling's Formula illustrates the "asymptotic" nature of
the summation formula. The same argument which was used in Problem 19 can
now be used to show that for N > 2 we have
by an amount which
n
=
is
small
compared
10
Y
b
+r
)=r^k(k-\)n
J
-'
io g
k-
\
1
n
Since
i/f/v
is
bounded, we can obtain estimates of the form
rr dt
Nt N
,
If A^
1_/v
77
is
large, the constant
will
make
Mm
will also
/2
e-
be large; but for very large n the factor
expf
a very bad approximation for
depends on N)
~ nN~v
the product very small. Thus, the expression
^n" +l
may be
Nt N
it
will
n\
V
when
^-— )
n
is
small, but for large n (how large
be an extremely good one (how good depends on N).
PART
EPILOGUE
There was a most ingenious Architect
who had
contrived a
new Method
for building Houses,
by beginning at the Roof, and working
downwards
to the
Foundation.
JONATHAN SWIFT
.
CHAPTER
AiV
Throughout
FIELDS
book a conscientious attempt has been made
this
portant concepts, even terms
like "function," for
often considered sufficient. But
which an
to define all
Q and R, the two main protagonists of
have only been named, never defined.
What
im-
intuitive definition
is
this story,
has never been defined can never be
analyzed thoroughly, and "properties" P1-P13 must be considered assumptions,
not theorems, about numbers. Nevertheless, the term "axiom" has been purposely
avoided, and in
this
chapter the logical status of P1-P13 will be scrutinized
more
carefully.
Q and
Like
talk
about
from a
all
R, the
sets
N
Z
and
have also remained undefined. True, some
four was inserted in Chapter 2, but those rough descriptions are far
definition.
To
say, for
N
example, that
consists of
1,
merely
2, 3, etc.,
names some elements of N without identifying them (and the "etc." is useless).
natural numbers can be defined, but the procedure is involved and not quite
The
pertinent to the rest of the book.
to this
The Suggested Reading
list
contains references
problem, as well as to the other steps that are required
develop calculus from
its
The
basic logical starting point.
one wishes
if
further
to
development
program would proceed with the definition of Z, in terms of N, and the
in terms of Z. This program results in a certain well-defined
set Q_, certain explicitly defined operations + and •, and properties P1-P12 as
theorems. The final step in this program is the construction of R, in terms of
It is this last construction which concerns us. Assuming that Q_ has been defined,
and that PI -PI 2 have been proved for Q, we shall ultimately define R and prove all
of
this
definition of
Q
Q
of P1-P13 for R.
Our
means that we must define not only
and multiplication of real numbers. Indeed, the
intention of proving PI -PI 3
bers, but also addition
bers are of interest only as a set together with these operations:
real
real
how
numnum-
the real
numbers behave with respect to addition and multiplication is crucial; what the
real numbers may actually be is quite irrelevant. This assertion can be expressed in
a meaningful mathematical way, by using the concept of a "field," which includes
number systems of this book. This
modern mathematics incorporates the
as special cases the three important
extraordi-
narily important abstraction of
properties
P1-P9 common
to Q_,
soever), together with
rules
which associate
in F) for
A field
is
+ b) + c — a +
(a
(2)
There
(i)
(ii)
a set
F
+
and
two "binary operations"
and b
to elements a
in
which the following conditions are
(1)
581
R, and C.
(b
+ c)
is
some element
a
+ =a
in
for
F
all
(of objects of
•
defined on
F, other elements a
any
F
sort
what-
is, two
and a • b
(that
+b
satisfied:
a, b,
and
c in F.
such that
for all a in F,
for every a in F, there
is
some element b
in
F
such that a
+ b — 0.
582
Epilogue
+b=b+a
a
(3)
(4) (a
b)
•
There
(5)
=a
c
•
a*
(i)
=a
1
a.fc=
(6)
a
*
(7)
a
•
The
+
b
and
why
in
F
for
a
•
all
a and b
b
+a
examples of
1^0
such that
/
with a
0, there
and
some element b
is
in
F
such that
in F.
for
c
•
all
and
a, b,
fields are, as
mean
are simply restatements of PI
c in F.
Q, R, and
already indicated,
+
and
•
these are fields, but the explanation
are understood to
•
F
c in F.
1.
a
—
1 in
being the familiar operations of
•
and
»
+ c)
(/?
familiar
explain
+
=
b
and
for all a in F,
For every a
(ii)
in F.
for all a, b,
c)
•
some element
is
and b
for all a
(b
•
+
the ordinary
probably unnecessary
It is
.
is,
at
and
•
any
,
C, with
rate, quite brief.
the rules
to
When
(1), (3), (4), (6), (7)
P4, P5, P8, P9; the elements which play the role of
,
and 1 are the numbers and 1 (which accounts for the choice of the symbols 0, 1);
and the number b in (2) or (5) is —a or a
respectively. (For this reason, in an
arbitrary field F we denote by —a the element such that a + (—a) — 0, and by
,
the element such that a
a
In addition to Q_, R,
a
•
-1
=
^
for a
1,
and C, there are
0.)
several other fields
which can be described
numbers a + b\2 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
One example
easily.
is
the collection F\ of
all
•
•
of Fi:
+ bV2) +
(a + bVl)-
(a
Conditions
all
real
1
=
1
(c
(1), (3), (4), (6), (7)
+ c) + (b + d)\fl, which in F\\
which
in
(ac + 2bd) + (bc + ad)V2,
(a
the
(2)
for a field are
for
all
is
fi
in F\,
difficult point. If
a
F\.
obvious for F\: since these hold for
real
numbers of
+ b\2.
= a + b\2
form a
the
number = 0+0v 2 is in F\ and, for a
= (—a) + (—b)w2 in F\ satisfies a + ji = 0. Similarly,
so (5i) is satisfied. The verification of (5ii) is the only slighdy
holds because the
number
+0v2
is
is
numbers, they certainly hold
Condition
in F\
+ dV2) =
+ dy/2) =
(c
+ b\f2 /
0,
a
then
—==
X
+ bV2
1;
a+b^/2
it is
_
1
a+bV2
(The division by a
only
a
show
therefore necessary to
if
a
=
+ by/l^
The
b
—
+ bv2
a-bs/2
— b\2
(since
is
)
_
is
in F\.
is
irrational)
which
-2b
is
is
true because
(-b)
2
valid because the relation a
v2
This
a
a2
(a-bV2)(a+bV2)
a
—
2
r-
-2b
by/2
=
2
could be true
ruled out by the hypothesis
0.)
next example of a
tains only
that \/{a
field, Fj, is
considerably simpler in one respect:
two elements, which we might
as well
denote by
and
1.
The
it
con-
operations
583
28. Fields
+
and
•
are described by the following tables.
01
+
.01
1
1
1
The
verification of conditions (1) (7) are straightforward, case-by-case checks. For
example, condition
be written
Our
— or
= — 1.
1
may
(1)
setting a, b, c
in
1
1
be proved by checking the 8 equations obtained by
Notice that in
1.
example of a field is rather
R, and + and • are defined by
final
(a, a)
+
(a, a)
+
(The
for R.)
and
The
=
(b, b) =
(b, b)
may
also
consists of all pairs (a, a) for a
Ft,
silly:
equation
0; this
(a
+ b,a + b),
(a
•
b,a
b).
•
appearing on the right side are ordinary addition and multiplication
•
verification that F3
A detailed
•
+1=
this field 1
is
a field
you
to
is left
investigation of the properties of fields
is
as a simple exercise.
a study in
but for our
itself,
purposes, fields provide an ideal framework in which to discuss the properties of
numbers
most economical way. For example, the consequences of P1-P9
in the
which were derived
for
"numbers"
in
particular, they are true for the fields
Notice that certain
For example,
C
—
the assertion
Q
fields
common properties of Q,
—
b
b
—a
+ 1/0
1
Q,
and R, however,
any
actually hold for
1
field;
in
R, and C.
possible for the equation
it is
consequently a
Chapter
R, and
+
1
1
=
C
do not hold
for
to hold in some
=
does not necessarily imply that a
all fields.
fields,
For the
b.
and
field
was derived from the explicit description of C; for the
was derived from further properties which
the conditions for a field. There is a related concept
this assertion
do not have analogues in
which does use these properties. An ordered field
and •) together with a certain subset P of F (the
is
a field
F
(with operations
+
"positive" elements) with the
following properties:
(8)
For
all
(i)
a
=
0,
(ii)
a
is
in P,
—a
(iii)
(9)
If
The
is
and b are
in P,
then a
have already seen that the
field F2,
implies that
1
consisting of
true:
+b
•
b
is
field
is
in P.
in P.
G
cannot be
made
+1=
all
(8),
is
applied to
in
numbers a
1
— — 1,
P, contradicting
+
shows that
(8).
/?v2 with a,b
in
into
made
with only two elements, likewise cannot be
condition
field: in fact,
is
in P.
a and b are in P, then a
(10) If a
We
a in F, one and only one of the following
1
an ordered
into
must be
field.
an ordered
in P;
then
(9)
On
the other hand, the field F\
Q,
certainly can be
made
,
into
.
584
.
Epilogue
an ordered
(in
field:
let
P
be the
the ordinary sense).
description of
P
The
set
of
+
a
all
field Ft,
which are
b\/7.
positive real
can also be made into an ordered
numbers
field;
the
to you.
is left
natural to introduce notation for an arbitrary ordered field which corre-
It is
sponds to that used for
Q and R:
>
<
<
>
a
a
a
a
b
we
if
define
a
—b
>
<
>
b
if
b
b
if
a
b
if
a
A
Chapter
in P,
a,
b or a
b or a
Using these definitions we can reproduce,
definitions of
is
for
=
=
b,
b.
an arbitrary ordered
field
F, the
7:
A of elements of F is bounded above if there is some x in F such
that x > 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 < y for every y in F which is an upper bound for A
set
.
Finally,
last
it is
.
possible to state an analogue of property PI 3 for R; this leads to the
abstraction of this chapter:
A complete
which
The
is
ordered field is an ordered field in which every nonempty
bounded above has a least upper bound.
consideration of fields
may seem
constructing the real numbers. However,
means of formulating
Is
1
2. Is
Our
now
provided with an
There are two questions which
will
intelligible
be answered
there a complete ordered field?
there only one complete ordered field?
starting point for these considerations will
field,
containing
N and Z as
certain subsets.
necessary to assume another fact about
Let x be an element of
in
have taken us far from the goal of
are
remaining two chapters:
in the
dered
this goal.
to
we
set
N such
that
nx >
Q with x
Q, assumed
be
At one
to
be an or-
crucial point
it
will
be
Q:
>
0.
Then
for
any y
in
Q_ there
is
some
n
y.
This assumption, which asserts that the rational numbers have the Archimedean
property of the real numbers, does not follow from the other properties of an
ordered
field (for the
example
of the Suggested Reading).
that demonstrates this conclusively see reference [14]
The important
point for us
constructed, properties PI -PI 2 appear as theorems,
is
that
and
when
Q
is
explicitly
so does this additional
28. Fields
assumption;
if
we
really
585
began from the beginning, no assumptions about Q_ would
be necessary.
PROBLEMS
Let
1.
F be the
set {0, 1, 2}
(The rule
tables.
in the usual way,
= 4 = 3+1,
2-2
and define operations
for constructing these tables
+ and
so
2
2
1
F
is
a
add or multiply
1
2
that
by the following
2-2=1.)
1
Show
F
possible multiple of 3; thus
1
1
on
as follows:
is
and then subtract the highest
•
field,
and prove
that
it
1
2
2
1
made
cannot be
into
an ordered
field.