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English Pages 721 Year 1961
ELEMENTARY
MATHEMATICAL ANALYSIS
To Gloria
This book is in the
ADDISONWESLEY SERIES IN MATHEMATICS
ELEMENTARY
MATHEMATICAL ANALYSIS by
A. E. LABARRE, JR. Department of Mathematics University of Idaho
ADDISONWESLEY PUBLISHING COMPANY, INC. READING, MASSACHUSETTS, U.S.A. LONDON I
E~GLAND
Copyright© 1961 ADDISONWESLEY PUBLISHING COMPANY, INC. Printed in the United States of America ALL RIGHTS RESERVED.
THIS BOOK, OR PARTS THERE
OF, MAY NOT BE REPRODUCED IN ANY FORM WITHOUT WRITTEN PERMISSION OF THE PUBLISHERS.
Library of Congress Catalog Card No. 617751
PREFACE Introductory college mathematics textbooks may usually be characterized as either too formal or, more commonly, too trivial. This book represents an attempt to avoid these extreme criticisms, while providing a oneyear foundation course for the student who plans to major in mathematics, sciencr, engineering, or mathematical education. Besides treating traditional topics of algebra, trigonometry, analytic geometry, and elements of the calculus from a modern, integrated vantage point, I have dared to introduce some relatively recent concepts which help to sustain the jargon that fills the corridors at professional mathematics meetings. These include, for example, the notions of linear programming problem, vector, and linear transformation. This injection of "advanced mathematics from an elementary viewpoint" is applied with extreme caution so that, after all, the result is only an innocent inversion of Felix Klein's famous theme. In writing the book, I have continually kept in mind the diverse programs of highschool mathematical training found in this country. As a result, I hope that the book can be used to accommodate students reflecting wide ranges of background. Roughly speaking, my "platform" has been guided by these points: (1) Throughout the text attempts are made to reconcile, insofar as possible, the intuitive or geometric interpretations of concepts with their abstract counterparts. To this end, informal, sometimes na:ive, remarks are introduced when it is felt that these will offer some insight into the topic under discussion. (2) Wherever there is an obvious overlapping of the text with highschool material, I have tried to be very concise and have moved on rapidly to details of more advanced ideas. To illustrate, in Chapter 1 most highschool students will be familiar with the fact that "numbers" satisfy the axioms for a field (although perhaps they will not be acquainted with the word "field"); they will also know the "rules" for adding and multiplying "signed numbers." However, it is highly improbable they will realize that the rules mentioned may be deduced as consequences of axioms, using procedures analogous to those of highschool geometry. Accordingly, the postulates for a field are introduced briefly and emphasis is placed on deducing some simple but useful consequences of these. To illustrate further, in introducing the trigonometric functions, the definitions for a general angle are succinctly given, and it is emphasized that these functions constitute the simplest yet the most important instance of the general notion of a periodic function. v
VI
PREFACE
(;3) There is an abundance of exercises scattered throughout the book. Here too I have had an eye out for students with varying degrees of mathematical training. Some of the exercises are routine, others are moderately difficult, and still others (those marked *) will challenge the best students. Realizing that some instructors will want to encourage supplementary reading, especially on the cultural side of mathematics, I have included certain "exercises" (those marked **) which require outside reading. ~lost of these are references to E. T. Bell's famous book Men of J.ll athematics; some, hmvever, serve the purpose of developing "fringe" topics. (4) The "defeatist" attitude is invoked much less frequently than is ordinarily the case for a book at this level. Moreover, when the approach "proof of this theorem is beyond the scope of this text" must be resorted to, references are ahvays provided where proofs can be found. (5) I have follmved the usual practice of providing material for those rare instructors who find time for additional work, by starring some sections. Such sections may be omitted without serious loss of continuity. In this connection, if Section *412 is omitted, attention should be called to Theorem 42, for it is used in Section 117 to arrive at the addition theorems of trigonometry, and again in Section 176 where the formula for the area bounded by a simple closed plane curve is derived. I have had splendid cooperation from many individuals from the inception of this project. Thanks are due especially to Mrs. Kancy Hunker for her typing and to my colleague Professor H. Sagan for his reading of the manuscript; also to my students :l\Iessrs. D. Curtis, R. Simpson, and J. Tollefson. I must also mention Professors K. A. Bush, L. Maxwell, and Elna Grahn, each of ,vhom helped in no small way. Finally, I want to express my appreciation to the excellent staff of the AddisonWesley Publishing Company. A.E.L., JR.
CONTENTS CHAPTER
11 12 13 14 15 16 17 18 19 110 111 112 113 114
1.
ELEMENTARY ALGEBRA: BASIC CONCEPTS AND OPERATIONS
Logic Exercises. Field postulates Exercises . Expcinents Exercises . Factoring Exercises . Identities and equations Exercises . Fractions Exercises . Terms common in mathematical literature . Review exercises
CHAPTER
2.
THE REAL NUMBERS: THE POINTS ON A STRAIGHT LINE
The real numbers Exercises . Ordering of the real numbers . Solutions of simple inequalities The real numbers: the complete ordered field Exercises . Distance between two points on the real line Exercises .
21 22 23 24 *25 26 27 28 CHAPTER
3.
1 6 7 11 14 19 19 26 28 36 38 42 46 47 49 50 56 57 58 61 63 65 69
THE PLANE. THE NOTION OF A FUNCTION AND ITS GRAPH.
72
THE LINEAR FUNCTION
31 32 33 34 35 36 37 38 39 310 311 312 *313 314
1
Coordinates of a point . The notion of a function and its graph . Exercises . The Pythagorean theorem. Distance between two points . Exercises . The straight line Exercises . Systems of lines. Exercises . Perpendicular lines Distance separating a line and a point . An application to linear programming Exercises . Vll
72 73
80 83 85 92 93 101 103 109 110 112 118 120
Vlll
CONTENTS
CHAPTER 4.
41 42 43 44 45 46 47 48 49 410 411 *412
122 122 123 126 130 132 135 135 141 142 144 148 150 153
VECTORS. COMPLEX NUMBERS .
153 154 162 162 169 170 174 175 177 180 182 184
THE NOTION OF A LIMIT
The notion of a sequence . The arithmetic sequence The geometric sequence Exercises. Infinite geometric series Exercises. Limit of a sequence Exercises. Basic theorems on limits of sequences Exercises. Extension of the limit concept to an arbitrary function Exercises.
CHAPTER 7.
71 72 73
PLANE GEOMETRY. THE
The notion of a vector . The algebra of vectors . Exercises. Complex numbers . Exercises. The triangle inequality. Exercises. Planar point sets The locus of a line segment The locus of a triangle Exercises .
CHAPTER 6.
61 62 63 64 65 66 67 68 69 610 611 612
IN
The notion of a set . Equality of sets Operations on sets (Boolean algebra) Exercises. Sets of points in the plane. Exercises. The analytic method Exercises. The area of a triangle The finite induction principle Exercises. The area of a plane polygon .
CHAPTER 5.
51 52 53 54 55 56 57 *58 *59 *510 *511
SETS. THE ANALYTIC METHOD FINITE INDUCTION PRINCIPLE
A SPECIAL LIMIT: THE DERIVATIVE .
Definition of the derivative Exercises . Derivative of the power function
.
184 186 188 191 193 195 196 200 200 208 209 211 212 213 221 222
CONTENTS
74 75 76 77 78 79 710
Exercises . Application of the limit concept to continuity . Exercises . Some techniques in differentiation Exercises . The chain rule for differentiation Exercises .
CHAPTER 8.
81 82 83 84 *85 *86
101 102 103
POLYNOMIALS .
271
RATIONAL FUNCTIONS. TECHNIQUES
IN
CURVE SKETCHING
Symmetry Rational functions . Exercises .
CHAPTER 11.
111 112 113 114
248 257 258 264 265 269
Definition of a polynomial The division algorithm. The remainder and factor theorems . Synthetic division . Exercises. Basic theorems on polynomials Exercises. Further theorems on polynomials Exercises. Upper and lower bounds for real zeros of a polynomial Descartes' rule of signs Rational zeros of polynomials. Exercises . Real zeros of polynomials . Exercises . Graphs of polynomials Exercises .
CHAPTER 10.
225 226 233 234 239 240 246 248
APPLICATIONS OF DIFFERENTIATION
Extrema. Exercises. A sufficient condition for extrema Exercises . Inflection points Exercises .
CHAPTER 9.
91 92 93 94 95 96 97 *98 *99 910 *911 912 913 914 915 916 917
lX
PERIODIC FUNCTIONS. THE TRIGONOMETRIC FUNCTIONS .
Definition of periodic function; the trigonometric functions . Radian measure Exercises. Graphs of the trigonometric functions
271
272 27 5 276 278 279 284 284 292 293 297 302 305 306 309 310 315 316 316 321 329 333 333 338 341 343
x
CONTENTS
115 116 117 118 119 1110
Tables and linear interpolation Exercises . The addition theorems Exercises. Related graphs Exercises .
CHAPTER
12.
346 348 349 355 356 365
FURTHER PROPERTIES OF THE TRIGONOMETRIC
367 367 371 374 378 380 385 385 389
FUNCTIONS. INVERSE FUNCTIONS
Identities and equations Exercises. Derivatives of the trigonometric functions . Exercises. The inverse trigonometric functions Exercises . Inverse functions in general Exercises.
121 122 *123 *124 125 126 127 128 CHAPTER
131 132 *133 *134 135 136 137 138 139 1310 *1311
SOME APPLICATIONS OF THE TRIGONOMETRIC FUNCTIONS
The polar representation of a complex number Exercises. The nth root of a complex number Exercises. Polar coordinates Exercises. Transformations of coordinate systems Symmetry in polar coordinates Exercises. Solution of a triangle . Exercises.
CHAPTER
141 *142 143 144 145 146 147 148 149 1410 1411 *1412 *1413 *1414
13.
14.
EXPONENTS AND LOGARITHMS. APPLICATIONS
391 391 396 397 400 402 408 408 411 412 413 418
.
Definition of the exponential function and the logarithm Continuity of ax. Exercises. Tables of logarithms Exercises . Basic properties of logarithms Exercises. The binomial theorem . Exercises. The slope of the logarithm and exponential curves Exercises . Applications of the exponential function Exercises . Euler's formula
420 420 427 429 430 435 435 440 443 448 449 453 454 456 457
CONTENTS
15.
CHAPTER
151 152 153 154 155 156 157 158 159 1510 *15H *1512 1513 *1514 1515
THE CONIC SECTIONS
Xl
.
The circle Exercises. Implicit differentiation. Exercises . The parabola Exercises. The ellipse Exercises. The hyperbola . Exercises . New definitions of the conic sections; polar coordinates Exercises. Transformations of the plane . Exercises. Review exercises
16.
CHAPTER
161 162 163 164 165 166
The concept of an integral Exercises. The fundamental theorem of calculus Exercises. An application of integration to physics. Exercises.
17.
CHAPTER
171 172 173 174 175 176 177
CHAPTER
PARAMETRIC REPRESENTATION OF CURVES
Parametric equations . Exercises. The cycloid . The projectile Exercises. The calculus of parametrically represented curves Exercises.
CHAPTER
181 182 *183 *184 *185 *186
ANOTHER SPECIAL LIMIT: THE INTEGRAL
18.
DETERMINANTS AND MATRICES
Properties of thirdorder determinants Exercises. Higherordered determinants Exercises. Matrices Exercises.
19.
FUNCTIONS OF Two VARIABLES .
191 Rectangular coordinates 192 Vectors in threespace 193 Exercises.
.
459 460 465 467 468 468 476 476 482 483 489 490 493 494 500 502 504 504 513 514 522 524 527 529 529 532 532 536 538 540 545 546 546 555 556 558 559 569 572 572 574 578
Xll
194 195 196 197 198 199 1910 1911 1912 1913 1914 1915 *1916 *1917
CONTENTS
Distance The direction of a vector Exercises . The inner product of two vectors Exercises . Cylindrical surfaces Exercises . The equation of the plane The straight line Exercises . Quadric surfaces Exercises . Conic sections Exercises.
CHAPTER 20.
201 202 203 204 205 206 207 208 *209 *2010 *2011 *2012
COUNTING. AN INTRODUCTION TO PROBABILITY
578 580 584 585 588 590 592 593 595 599 600 605 606 608 611
The fundamental counting theorem . Exercises. Permutations Exercises. Combinations Exercises. Definition of probability; basic theorems Exercises . The binomial formula . Exercises . Mean and standard deviation. Exercises .
611 613 614 619 620 622 624 633 635 638 639 645
TABLE 1. NATURAL TRIGONOMETRIC FUNCTIONS
647
TABLE 2. COMMON LOGARITHMS OF NUMBERS
648
ANSWERS TO ODDNUMBERED EXERCISES .
651
INDEX .
701
CHAPTER 1 ELEMENTARY ALGEBRA: BASIC CONCEPTS AND OPERATIONS Introduction. The reader who has studied plane geometry in high school has some idea of the spirit in which presentday mathematics is being developed. That is, he is familiar with the pattern outlined by the Greek geometer Euclid (who lived about 300 years before Christ) and has seen Euclid's methodology applied to the development of plane geometry: There one begins with certain postulates and definitions and proceeds to deduce from these certain consequences called theorems. However, this procedure is not peculiar to geometry. In particular, algebra can be developed in precisely the same sort of way although, perhaps, the reader may not have realized this. It is the purpose of the present chapter to initiate such a development by: 1. laying down some postulates for elementary algebra, and
2. appealing only to these postulates to prove certain theorems which will be recognized as ''rules" formulated in highschool algebra. In addition, we emphasize and review certain techniques which we shall encounter time and again in later work. Especially important in this connection are these techniques: 1. Factoring algebraic expressions. 2. Adding, subtracting, multiplying, and dividing algebraic fractions. 3. Solving quadratic equations.
We begin with some statements of a general nature concerning the foundations of mathematics. This topic belongs to the subject known as logic. 11 Logic. Logic is, in itself, a wide and interesting field; it is our aim
here to present only a few elementary ideas indispensable to our work. In a development of mathematical theory, statements may be divided in to three categories: A. Definitions. B. Postulates. C. Theorems. These categories are exhaustive, which means that any given mathematical statement may be classified as a definition, a postulate, or a theorem. 1
2
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
These categories are mutually exclusive, meaning that any given mathematical statement can belong to only one of the three categories. It is frequently very enlightening to raise the question of whether a given statement arising in the course of development of a mathematical theory falls under the heading of a definition, postulate, or theorem. We now discuss these in some detail. A. DEFINITIONS
We do not want to raise the delicate question of what a definition is; instead we give some examples of definitions: (1) An isosceles triangle is a triangle having two equal sides. (2) "The integer a is a divisor of the integer b" means that there exists an integer x such that ax === b. All definitions can be ~tated in the form of "if and only if" statements. In particular, (1) above could be written: A triangle is isosceles if and only if it has two equal sides. It is impossible to define every term used in mathematics, just as it is impossible for a dictionary to carry independent definitions of all the words it contains.* Thus all mathematics begins with socalled primitive terms or undefined terms. At first glance, this may appear to be a necessary weakness of mathematics. On closer examination, however, it is seen that it is because of the appearance of undefined terms that mathematics is a powerful tool, for a mathematical theory may be applied to any set of objects satisfying the properties ascribed to the undefined elements by means of the postulates, to be discussed immediately. B. POSTULATES
Postulates are assumed statements, i.e., statements for which we require no proof. We shall use postulates and axioms interchangeably. It was Pythagoras (about 500 B.C.) who first realized the importance of assumptions to mathematics and insisted that the presentday . mathematical pattern of deducing all consequences from selected assumptions be followed. This pattern has not been adhered to through all the ages since Pythagoras; deviation from it was fortunate, moreover, in the development of the calculus, for the roots of this subject lie in deep considerations which were not understood by its founders and early workers and hence
* If you look up a word in a dictionary, you will find synonyms for it; if, in turn, you pursue the definitions of these synonyms, it is inevitable that you will arrive at the original word or one of the synonyms.
11]
LOGIC
3
an insistence on rigorous procedure would have hindered, if not stopped, the development. As examples of postulates, we have the following ones introduced by the Greeks in (synthetic) geometry: Two points determine a straight line. Three points which are not collinear determine a plane. All mathematics must begin with some postulates, for we have indicated that the presence of undefined terms is inescapable and, clearly, it is impossible to# prove statements concerning undefined things unless assumptions are made about these undefined things. Speaking of proving statements brings us to the concept of a theorem.
C. THEOREMS A theorem is a mathematical statement for which proof is required. It consists of two parts: a hypothesis and a conclusion. The hypothesis is what is given; it is usually preceded by the preposition if; the conclusion is the part which is to be deduced, it is usually preceded by the preposition then. In any event, a theorem can always be put in this form. For example, with the definitions made in A above, the following theorems can be proved: (I') If a triangle has two equal angles, then it is isosceles. (2') If a is a divisor of b and if b is a divisor of c, then a is a divisor of c. Let us prove (2'), by way of illustration. By hypothesis, a is a divisor of b, meaning that there exists an integer x such that ax == b. Also by hypothesis, b is a divisor of c, so that there exists an integer y such that by == c. It follows by substitution that axy == by, and so if we further substitute c for by, we obtain axy == c. But xy is an integer, hence this statement says that a is a divisor of c, which was to be proved. If the hypothesis and the conclusion of a theorem are interchanged, we obtain the converse of the theorem. For example, the theorem: "If two triangles are congruent, then their corresponding angles are equal" has the false converse: "If corresponding angles of two triangles are equal, then the triangles are congruent" [they may only be similar].
Clearly, then, careful distinction between a theorem and its converse is of paramount importance. When a theorem and its converse are both
4
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
true, they can be combined into one statement of the form "if and only if." Thus the theorem: "If two triangles are similar, then corresponding angles are equal" and its converse: "If corresponding angles of two triangles are equal, then the triangles are similar "
are both true and hence we may say: "Two triangles are similar if and only if their corresponding angles are equal." This same idea may be expressed in still another way. We can say "A necessary and sufficient condition for two triangles to be similar is that their corresponding angles be equal." More generally, if condition (i) implies condition (ii) [this means given (i), we can prove (ii)], we say that (i) is sufficient for (ii), or that (ii) is necessary for (i). In terms of theorem (1') above, we would say that "equality of two angles" is a sufficient condition for "an isosceles triangle" or, alternatively, "an isosceles triangle" is a necessary condition for ''equality of two angles." We shall frequently speak of two statements or conditions as being equivalent. To come to an understanding of this term, let A and B denote statements, each of the form: if __ , then __ . If statement A implies statement B and, conversely, if B implies A, then we say that A and B are equivalent. Simjlarly, if condition A implies condition B, and conversely, then we say that A and B are equivalent conditions. For example, Condition A: Equality of corresponding angles of two triangles and Condition B: Proportionality of corresponding sides of two triangles are equivalent conditions, for either one can be deduced from the other. Occasionally, when we are required to prove the statement "if A, then B," we will find it difficult (sometimes impossible) to prove. this directly, but we may be able to prove the equivalent statement "if not B, then not A" with less difficulty. This last statement is called the contrapositive of the first statement. A statement and its contrapositive are equivalent statements,* and consequently we may substitute a proof of the contrapositive
* For, suppose "A implies B" is given. Then "not B implies not A" follows, . . 1·ie d "A , " t h en b ecause ''A" imp . 1·ies "B , " we would h ave 1'f " not B" imp smce "not B implies B" which surely cannot be admitted. A similar argument shows that if "not B implies not A" is given, then "A implies B" follows.
11]
5
LOGIC
of a statement for a proof of the statement itself. This technique is called "proof by contradiction" or "indirect proof" and is a convenient device. In practice, the procedure for a proof by contradiction runs as follows: suppose we want to prove the statement "if A, then B." We start by supposing B not fulfilled ("not B"); we continue our reasoning until we have shown that A is not fulfilled ( "not A "). In other words, we prove the statement ''if not B, then not A." But this is tantamount to proving "if A, then B," and we have accomplished what we set out to do. It is sometimes convenient to speak of the inverse or opposite of a statement. Let a statement be of the form "if A, then B." By the inverse of this statement is meant the statement "if not A, then not B." Thus the inverse of the (true) statement "If two triangles are congruent, then their corresponding angles are equal"
is the (false) statement "If two triangles are not congruent, then their corresponding angles are not equal" [this is not true, for two similar triangles are not necessarily congruent, yet their corresponding angles are equal].
Note that the inverse of the inverse of a statement is the original statement (this fact accounts for the name "inverse"). If we use the conventional symbol ~ for implies (e.g., A ~ B means that B follows from A), and if A denotes "not A" or "denial of A," then the following table shows the interrelationships between
A~B, its inverse I: A ~ B, its converse CV: B ~ A, and its contrapositive CP:  B ~  A : a statement S:
S:
A
*
I: A
~
B:: 0 and read this "a is greater than zero." We continue by recalling 12. If a is an arbitrary positive number and n is a positive integer, an denotes the number obtained by using a as a factor n times. an is called the nth power of a, a is called the base of this power, and n is its exponent. DEFINITION
Remark 1. Unless n is a positive integer (positive whole number), this definition is meaningless. For instance, it is nonsense to say that a 3 means "use a as a factor minus three times. "
EXPONENTS
15]
15
Remark 2. If the base a is not a positive number, we encounter difficulties in later developments of the theory of exponents (see Chapter 14). In order to obviate such difficulties, we impose the restriction a > 0 from the start. In spite of its shortcomings, the last definition has certain consequences which serve as a guide in defining powers whose exponents are not positive integers. These consequences are called the laws of exponents. We shall now obtain them. Since an means the number obtained by using a as a factor n times, and am means the number obtained by using a as a factor m times, an · am is the number obtained by using a as a factor n m times. With this, we have proved
+
THEOREM 14. If n and m are positive integers and a is any positive number, then
Next consider division of powers of the same base. Put
am 
x a.
Then, on multiplying by am, we have an == (am) (ax). But according to the above theorem, it follows that am+x == an, and so because of the uniqueness of multiplication, m x == n, or x == n  m. This completes the proof of
+
THEOREM 15. If n and m are positive integers with n any positive number, then
an am

==
>
m * and a is
nm a .
Remark. The reason for the restriction n > m is to guarantee that n  m will be positive in order that anm will be meaningful. This restriction, however, is only a temporary one, for we shall remove it shortly. Finally, we have a theorem dealing with raising a power to a power: THEOREM 16. If n and m are positive integers and a is any positive number, then (an)m == anm.
* The n 
m
careful reader will want to know what n 0.
>
>
m means.
It means
16
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
To prove this, observe that (an)m means an taken as a factor m times, and then count the number of times that a occurs as a factor (it is nm times). So much for the rules governing manipulation of positive integral exponents. Now we embark on an instance of a frequently used technique in mathematics. We refer to the process of generalization. This process is important for the removal of inadequacies of mathematical systems. We have come in contact with it before, though perhaps it passed unnoticed. For example, the system of positive integers 1, 2, 3, ... is inadequate so far as the process of subtraction is concerned in the sense that if we limit ourselves to only these numbers, it is impossible to carry out the subtraction process in certain instances, e.g., 3 
3
or
3 
5.
Thus, in order to have more freedom of operation, we are led to introduce the number O and the negative integers and then we operate in the system of integers ... , 3, 2, 1, 0, 1, 2, 3, ... , where subtraction can be performed unrestrictedly. Note that in those instances where subtraction was possible in the original system of positive integers, the result is still the same: we leave the original system undisturbed, while simply expanding the system or, to put it in technical language, we imbed the positive integers in the larger system of integers. Let us mention another instance of this imbedding process which is close at hand. We refer to imbedding the integers in the fractions or rational numbers in order to make division always possible except, of course, division by zero. The inadequacy of the system of positive integral exponents becomes evident when we try to divide according to Theorem 15 in certain cases, e.g., a3 a3 , or a3 a5 and discover that the hypothesis of that theorem (n > m) is not fulfilled here. On the one hand, according to the definition of exponent, a3 a3
a·a·a a·a·a
1
and
a·a·a a· a· a· a· a
1 a2
On the other hand, if we merely ignore the condition of Theorem 15
15]
17
EXPONENTS
regarding the sizes of the exponents, we obtain 3 a a3
==
a33
==
ao
and
a 0 == 1
and
a 35 == a 2 .
This suggests defining
a 2 == 1
a2
and this is pr~cisely what we do. In addition, if we waive the requirement of positive integral exponents and apply Theorem 14 to the formal symbol a112 . a112
'
from which we obtain a112. a112
== a1 ==
a,
it is suggested that we define a 112 to be a square root of a. To avoid ambiguity, we define a 112 to mean the positive square root of a: a112
== Ya.*
Moreover, if we apply Theorem 16 blindly to obtain (a1;2)3
==
a312,
it is further suggested that a 312 should coincide with the cube of More generally, these considerations lead us to DEFINITION
a 112.
13.
(i) If a ~ 0, a0 == 1. (ii) If m, n are positive integers and
is a positive number, the nth positive root of a raised to the mth power: am/n == (al/n)m ==
c~r)l/n
a
==
v1am
==
(Ya
amfn
is
r,
1 · a min == amfn
(iii)
What this definition accomplishes is the following: it assigns a meaning to expressions of the form ax, where x is any rational number (i.e., ratio of two integers) and a is any positive number, in such a way that we are free to operate with these symbols in accordance with the rules spelled out by Theorems 14, 15, 16 (the hypothesis n > min Theorem 15 now being unnecessary). Again we emphasize that the original system in which
* By
v9
=
convention, the symbol
3, v9
=
3, ±v9
=
v says ±3.
"this is a positive number."
Thus
18
[CHAP. 1
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
the exponents were restricted to be positive integers is left intact by the extension process; we compute with these the same as from the start. 2
EXAMPLE 11. Express 2x + 3 /16x 5 as a power of 2. Since the denominator can be written as (2 4 )x 5 = 24 x20, 2 2 x +3
2
x
2
+3
= 2 x2 4x+23
24x20
15x5
EXAMPLE 12. Solve the exponential equation 2
3
i_27) . 3
4x1 =
The righthand member may be written as a power of 3: (27) 3
2
6
~ 3
=
35
.
This enables us to consider the following equivalent equation: 34x1
=
35.
Since the powers are equal and the bases are equal, it follows from the uniqueness of multiplication that the exponents are also equal:* 4x  I = 5, or x = J. That this number satisfies the original equation can be readily checked. EXAMPLE 13. Simplify: 3. g3/2]2 (1/2)l  (3a)O .
82,3 X =
[
Noting that 82/3 =
(81/3)2 =
1
g3/2
=
931 2 =
4,
1 27'
(1/2)1 = 1~2 = 2, (3a)
0
=
l,
we write 81 1225
* We
have to make an exception of the base 1, for if 14 x not imply that 4x  1 = 5.
l
15 , this does
17]
FACTORING
19
16 Exercises 1. Express as a power of 2: (4 2 • 8) 2 /16. 2. Find x such that x 2 = 34 (23) 2 • In Exercises 39, apply the theorems on exponents to simplify the given expressions. Write your answers using only positive exponents.
52 4. 51 6. (al/
3
+
l)(a
213

a
113
+
42 41 .
1).
*IO. Discuss the truth or falsity of all possible "converses" of Theorem 14: (i) If an · am = an+m, then n and m are positive integers and a is a positive number. (ii) If an · am = an+m and i£ n and m are positive integers, then a is a positive n um her. (iii) If an· am = an+m and if a is a positive number, then n and m are positive integers.
17 Factoring. Let us recall the arithmetic procedure involved in factoring positive whole numbers, and the main theoremthe unique factorization theoremsometimes referred to as the fundamental theorem of arithmetic. To factor a whole number means to write it as a product of pri"me numbers. t Now a prime number is a number different from 1 which cannot be written as a product except of itself and unity or, of course, its negative and  1. Thus the first few primes are 2, 3, 5, 7, 11, 13, 17, 23, 29, 31, ... t That there are infinitely many prime numbers is an important result in that branch of mathematics known as "Number Theory." We now prove this result in the same way that Euclid did over two thousand years ago.
t Sometimes the verb "to factor" is used in the broader sense meaning simply "to write as a product," not necessarily of prime numbers; for present purposes the narrower sense is more serviceable. i It is still an open question as to whether or not a formula can be found for determining the nth prime number.
20
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
The method of proof is reductio ad absurdum or "proof by contradiction." Suppose that we started listing all the primes: 2, 3, 5, 7, 11, 13, 17, 19, and suppose that we eventually arrived at a largest prime number, say Pn· Then all the primes could be listed as 2, 3, 5, ... , Pn· By constructing a number whose existence must be denied, we now show that this state of affairs leads to a contradiction. This is the number we could obtain by multiplying together all primes and adding the number 1 to get 2 · 3 · 5 · ... · Pn + 1. If such a number were divided by any one of the primes 2, 3, ... , Pn, we would get a remainder of 1; for example,
2 · 3 · ... · Pn 3
+ 1 ==
2 . 5 . ... . Pn
+ !3 .
Consequently, this number cannot be decomposed into prime factors. On the other hand, it cannot itself be prime since Pn is the largest prime number, by hypothesis, and 2 · 3 · 5 ... · Pn 1 certainly exceeds Pn· Thus we have constructed a number which is neither a prime nor decomposable into primes, clearly a contradiction. In order to deny the existence of such a number, we must deny the original hypothesis that there are only finitely many primes. This proves:
+
THEOREM 17. There are infinitely many prime numbers. We now factor some numbers: 21 == 7 · 3,
72 == 2 3 • 3 2
'
184 == 2 3 • 23
'
232 == 2 3 • 29.
Hence we say that 7 and 3 are pn"me factors of 21. As a matter of fact, we can say that 7 and 3 are the prime factors of 21, for, apart from order, a factorization into primes is unique. For example, the factorizations 232
==
2 3 • 29 == 2 · 29 · 2 2 == 29 · 2 3
==
2 2 • 29 · 2 == , etc.
are essentially the same. This result is so important that we include a formal proof. This theorem, too, was proved by Euclid, but in a different way; he used the "Euclidean algorithm" for finding the greatest common divisor of two integers (see Section 18, Exercise 72).
17]
21
FACTORING
18. The unique factorization theorem for integers, or the Fundamental theorem of arithmetic. If the order of prime factors is ignored, the factorization of any positive integer into prime numbers 1s umque. THEOREM
Let us assume the opposite, namely that there exist numbers which can be decomposed into prime factors in two different ways, and let N be the smallest such number* (1)
where we suppose that the primes have been arranged in ascending order, i.e., and PI < P2 < · · · < Pn (We allow the equals signs in order to accommodate the possibility of repetition of the primes.) Now PI ~ qI, for if these were equal, they could be canceled from the decompositions in (1) and we would then have a number smaller than N for which the unique factorization theorem is false, contrary to hypothesis. We lose no generality in assuming that PI < qI. Now consider the number defined by M
==
N 
Piq2q3 ... qm.
(2)
(Note the substitution of PI for qi) It can be written in two ways:
and M
==
PIP2 · · · Pn 
Piq2 · · · qm
==
PI(P2P3 · · · Pn 
q2q3 · · · qm).
(4)
What can we say about M? Looking at (2), we see that M is smaller than N; therefore the decomposition of M into primes is unique. (Remember: N is, by definition, the smallest number which can be decomposed in different ways into primes.) From (4) we see that PI is a factor of M and, going to (3), since PI is less than all the q;s, and therefore different from all of these, it must appear as a factor of qI  PI· In other words, there must be an integer k such that or
* We are using here the wellknown fact that if one considers any collection whatever of positive integers, it contains a smallest member. For example, the collection of even positive integers has the number 2 as its smallest member. In technical language, we describe this property of the positive integers by saying that they form a wellordered set.
22
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
which says that q1 is not a prime. This is a contradiction. It follows that there are no integers for which the unique factorization theorem is false. This proves the theorem. Of great importance in the present connection is the fact that an analogous theorem is valid in algebra. This comes about as follows: First of all, Theorem 18 is concerned with factoring integers into primes. The analogy of which we speak stems from the substitution of the concept of polynomial for integer and irreducible polynomial for prime. That is, the process in which we are interested is that of factoring polynomials into irreducible polynomials. This brings us to the question: "What is a polynomial?" It is a sum of terms of the type (number) . xpositive integer or O. More precisely, DEFINITION
14. A polynomial in x is an expression of the form
where the coefficients an, anl, ... , a1, a 0 are fixed numbers and the exponents are nonnegative integers. The largest exponent n (assuming an ~ 0) is the degree of the polynomial.* For example, 3x 4 2x
2

4x 
+ 7x 4x + 7, 4x 3

1,
3
are polynomials of degrees 4, 2, 1, respectively. Nonzero constants are polynomials of zero degree. However, since the factorization of constant polynomials is a purely arithmetic problem to which Theorem 18 applies, we shall accordingly focus our attention throughout the remainder of this section on factorization of nonconstant polynomials.
More generally, we come in contact with polynomials involving more than one letter, for instance x2
+ y2
and
Again, an essential pojnt in the concept of a polynomial in x and y, or in still other letters for that matter, is the requirement that the exponents
* Note carefully that we do not use the term "polynomial" in the loose sense of being an arbitrary expression of many terms; the terms must be of the type described.
17]
23
FACTORING
be nonnegative integers. Thus a polynomial in x and y is a finite sum of terms of the type . k (number) · x 1 y , with j, k nonnegative integers. Finally, as we have already mentioned, the "irreducible polynomials" assume the role of the "primes. " DEFINITION 15. A polynomial is irreducible if and only if it cannot be written as a product of two polynomials of lower degrees.
Remark. Strictly speaking, the concept of irreducibility depends on the kind of coefficients of the polynomials which are admissible. For example, if we limit ourselves to integers (whole numbers), it is impossible to factor the polynomial x 2  2. However, if we remove this restriction, we could write x 2  2 = (x v'2)(x  0).
+
Insofar as we are concerned at present, we shall deal only with polynomials whose coefficients are integers and we shall insist that all factors be polynomials of this same nature. The problem of deciding whether or not a given polynomial is irreducible can be quite troublesome, just as its counterpart in arithmetic.* It is an easy consequence of the last definition, however, that every firstdegree polynomial is irreducible (even though it may have a numerical factor). To prove this, consider the firstdegree polynomial ax b (a ~ O). The only polynomials of lower degree are the constants (their degree is O) and the product of two constants is another constant, therefore it cannot be equal to ax b. Thus
+
+
3x 
7,
x 
12y,
x  4,
are all irreducible. Note, however, that even though we are dealing with firstdegree expressions, we may have numerical factors: 21x 
28y
+7 =
7(3x 
4y
+ 1).
Now that the preliminary definitions have been introduced, we are in a position to state the following theorem.
* E.g., is 424,171 prime? If you cannot reach a decision, consult the tables by Derrick N. Lehmer, List of Prime Numbers from 1 to 10,006,721, StechertHafner, Inc., 1914. (Better still, do you know the rule for divisibility by 11? If you don't, see the book What is Mathematics? by R. Courant and H. Robbins, p. 35, Oxford University Press.)
24
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
19. The unique factorization theorem for polynomials. If the order of appearance of irreducible factors is ignored, the factorization of any polynomial into irreducible polynomials is unique.
THEOREM
Unfortunately, the proof of this theorem would take us too far afield and, consequently, proof is omitted.* Instead, we turn to the technical problem of finding the factors of a polynomial. Just as in arithmetic, where we learned to factor by reversing our multiplication process, so it is in algebra. In other words, we have as analogues of the familiar "multiplication tables," certain "type products." It is to be emphasized that the equality which decomposes a number into factors at the same time assembles a product. For example, reading from left to right: x2 
y2
==
(x
+ y)(x

y),
we interpret the equality above as a factorization of x 2  y 2 , whereas if we read from right to left, the same equality appears as a multiplication with the two factors (x y), (x  y). Thus every factorization is at the
+
same time a multiplication and, conversely, every multiplication is at the same time a factorization. Some of the more frequently arising factoriza
tions are enumerated below (we prove the first one; see if you can prove the rest by appealing to Postulates I, II, III only): A. Difference of two squares: a 2  b 2 == (a+ b)(a  b). The proof of this based on the postulates introduced above runs as follows: (a 
b)(a
+ b)
(a 
b)a
+ (a
a(a 
b)
+ b(a
b)b, by the distributive law III,

b), by commutativity of multipli

cation II(ii), a2
ab

+ ba
b2 , by III agam and associativity

of addition I(i), a2
b2 , by commutativity of multiplication and definition of 0. 
Applications: x2

(3u 
2) 2
(y 
8) 2

==
49
[x 
(y 
== (3u 
2)][x
+
(y 
8  7)(3u 
8
2)] == (x 
+ 7)
+ 2)(x + y

2).
== (3u 
15)(3u 
==
5)(3u 
1) 1).
y
3(u 
* For the advanced reader who can afford the time, however, we suggest that he study A Survey of Modern Algebra by Garrett Birkhoff and Saunders MacLane, Chapter IV, The Macmillan Company.
17]
25
FACTORING
+
B. Perfect square: a 2 ± 2ab b2 == (a ± b) 2 • (The signs here are not ambiguous: it can be checked that we must select either both or both.)
+
Applications: The expression x 2 can be written x2
+ 1 ==
2x

(x 
+
2x

l)(x 
1 is a perfect square, for it 1)
== (x  1) 2 .
1)
+ 3] 2
Similarly, (m
+
1)
2
+ 6(m + 1) + 9
+
== [(m
== (m
+ 4) 2 •
We shall often be confronted with the problem of determining a number not necessarily an integerwhich when added to a given expression will make it a perfect square. To make the expression x 2 + ax a perfect square, add the square of half the coefficient of x, that is, (a/2) 2 , for then 2
x+ax+
+
(a)2
+
2
==
(
a)2 2 .
x+
In particular, (x 3) 4  7(x 3) 2 becomes a perfect square on adding (  7/2) 2 • To see this, let us give (x 3) 2 a new name, say y:
+
y == (x
The express10n (x we write
+ 3)
4
7(x

y2 
Therefore, (x
+ 3) 4

7y
7(x
+ 3) 2 .
+ 3)
+ \9
2
then goes over to y 2
== (y 
+ 3) 2 + \9

7y and
t)2.
== [(x
+ 3) 2

tJ2.
This process of "completing the square" will prove to be a very useful tool in much of the later work. C. Common factor: ax ± ay == a(x ± y). [This is nothing more than the distributivity of multiplication with respect to addition (or subtraction); see Postulate III above.] Application: 3x 
21 
yx
+ 7y ==
The common factor is x
~
3(x 
7) 
y(x 
7) == (x 
7.
D. Sum of two cubes; difference of two cubes: a3
± b3 == (a ± b)(a 2
=i= ab+ b 2 ).
7)(3 
y).
26
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
Applications:
+
(x
+
2) 3
(x  2) 3
== [x == [x == [x
27
3(x
+
3x 
2)
6
+
+
9]
9]
== [x  2  3][(x  2) 2 + 3(x  2) + 9] == [x  5][x 2  4x + 4 + 3x  6 + 9] == [x  5][x 2  x + 7].
27

+ 2 + 3][(x + 2) 2 + 5][x 2 + 4x + 4 + 5][x 2 + x + 7].
E. Factoring by "trial and error":
(a+ b)x + ab
x2 ±
== (x ± a)(x ± b).
Applications:
+ llx + 18 == (x + 2)(x + 9). x 2 + x + 7 ~ (x + 7)(x + 1), hence x 2 + x + 7 is irreducible. (x  {3) 2 + 13(x  (3) + 12 == [(x  (3) + 12][(x  (3) + 1] x2
== [x 
+ 12][x 
{3
{3
+ 1].
18 Exercises Factor the following polynomials into irreducible factors: 1. 9  a 2 3. (2a  b) 2  (3 5. 8x 2 14x  15
2. 8m 3  n 3 4. lOxy  15yz  6xb 6. (x  a) 2  x 2
+ b) 2
+
7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.
a
2
+ 2ab + b
27m 3
2

8. 3x 2  l6x + 5 10. x 2  x  30
4
+ 8n 3
+
12. x 2
x 2  7x 12 2 21x  xy  2y 2 2
+ +
14. 16. 18. 20. 22.
2
9x 6x 1  y 3 x x 4 81a  b4 3x  y 9x 2  y2
+
+
(x
+ 2) +
7x 2
2
5(x
+ 17xy 
+ 2)y + 6y
+
+
+
6y
+
y
2
+6 + lOy + 16 
5x
+ 3ab + c + 3cb
a
a2 
2x 3 
6 
17a  60 24x 2 y 72xy 2
+
c 
c2
+
24. x y  x  y2 26. 6  3x 2  Ilx
2
2
2
12y ex  2cy 4c 3x 2 32x  5x  6 12x 2 25x  7 4 3a  3a 25  lOx  8x 2
+ 9zb
12
28. 30. 32. 34.
x2
s2

+ dx

cd
2ar
+ 2br
+ 6s
9a 2 x2
ex

36. 9ax 
 10 9a + 2 7x  18 9bx 
1
18] 37. 39. 41. 43. 45. 47. 49. 51.
27
EXERCISES
38. 21y 2  yz  2z 2 40. 3x 3  3x 2  x 1 2 2 42. 45z c  80t c
21m + 5 2 x  2x  2y + 4xy x2 y  x  y2 64(x  y) 2  36 (x + 3y) 3  12y  4x 8x 2 + 14x  15 25x 2(a  2b) + 2b  a x2n 8xny2n 16y4n
4m 2

+
+
+
53. 20u 
+
44. ab  3a 2b  6 2 46. (a  b)  lO(a  b) x3y5 x3y3 48. x5y3
+
50. 16x + 5 52. lOxy  15yz  6xb 54. (a  b)c  a+ b
3x 2 
+
7,au 
+
+ 25
6a 2 u
+ 9zb
Make each of the following expressions a perfect square by adding the appropriate term, and then write the resulting expression as a square. 55. x 2 57. 59. 61. 63.
+
(a+ c) 2 3(a 2 y  (13/2)y 4a 2  3a {3 2  21{3
65. {3 2 67. x 2 69. x 2
56. x 2 + 5x 58. (a  c) 2
5x
+ c)

3(a 
c)
60. u + 1.2u 62. 25x 2 + 49 64. {3 2  t{3 2
+ t{3 + {3x +h
2
+
~
;
66. x v 5x 2 68. y ± ky 70. x 2 ± 2Ax
2
*71. By appealing to Postulates I, II, III only, prove each of the factoring types B, D, E enumerated again:
+
B. a 2 ± 2ab b2 = (a ± b) 2 D. a3 ± b3 = (a ± b)(a 2 =F ab+ b2) E. x 2 ± (a+ b)x+ ab= (x ± a)(x ± b). 72. It was mentioned that Euclid proved the unique factorization theorem 18 by using the Euclidean algorithm for finding the greatest common divisor of two integers. As an example, the work for finding the g.c.d. of 2814 and 10605 is given below: 10 21 )210 210
3 )651 630
3 )2163 1953
1 )2814 2163
3 )10605 8442
0
21
210
651
2163
The process begins by dividing 10605 by 2814, getting a quotient of 3 and a remainder of 2163. This remainder is then divided into 2814, giving a remainder of 651 which is divided into 2163, etc. The process terminates when a zero remainder is obtained. The last divisor is the required g.c.d. Here it is 21. For a proof that this algorithm does indeed produce the g.c.d. as claimed, see College
28
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
Algebra by A. Adrian Albert, McGrawHill Book Co. find the g.c.d.'s of:
(i) 7136 and 4254.
1.
Use the algorithm to
(iv) 278 and 174.
73. Verify the factorization: x 5 Factor the following: 
1
(ii) 3564 and 12705.
(iii) 475 and 69.
(a) x 7
[CHAP.
(b) x 4

1.

1
=
(x 
(c) x 6
l)(x 4 + x 3 + x 2 + x + 1). (d) xn 
1.

1.
+
*7 4. Show that a 2 / 4 is the only number which when added to x 2 ax will complete the square, i.e., show that x 2 +ax+ k = (x + b) 2 implies k = a 2 /4. **75. Show that 424,171 is divisible by 11. More generally, show that if a number has the property that the sum of its digits taken with alternating signs is divisible by 11, then the number itself is divisible by 11, and conversely. Thus 42+41+71
=
11
is divisible by 11, therefore 424,171 is divisible by 11. (Reference: R. Courant and H. Robbins, What is Mathematics? p. 35.) *76. Explain why associativity of addition is needed in proving the factorization a 2  b2 = (a+ b)(a  b) in A above (Section 17). **77. Read the article "The Queen of Mathematics" by E. T. Bell in Volume 1 of the fourvolume set The World of Mathematics, edited by James Newman, Simon and Schuster Co., 1956, p. 498. Write a report.
19 Identities and equations. In mathematical work the equals sign ( ==) may be used in distinctly different situations. To make this clear, we refer to Section 12 and reiterate the postulates: I(ii). For all numbers a, b, the relation a+b==b+a holds. (iii). If a and bare given numbers, there exists a number x such that a+ x == b.
Let us investigate the significance of the equals signs in these two cases. As for the commutativity of addition [I(ii)], the equals sign in the statement a+b==b+a
has the force of what we call an identity in a and b. More precisely, if we select any values of a and b, substitute these in the two members of the ** Problems marked ** require outside reading.
19]
IDENTITIES AND EQUATIONS
29
equality and carry out the addition indicated, we will arrive at the same number. Now compare this with the behavior of the equals sign in the postulate I(iii), which asserts the solvability of the equation a+ x
==
b.
This is not an identity in the letters a, x, b appearing in the statement; that is, we are not free to choose all three of a, x, b arbitrarily. Indeed, the postulate permits free choice of only a and b and then the equality will hold provided x has a certain value. For instance, if a == 3 and b == 7, then 3+x==7 only on the conditi"on that x == 4. We refer to this equality as a (conditional) equation, the parentheses around "conditional" indicating that we some
times omit that word for sake of brevity. Many illustrations of identities are at hand; as a matter of fact, all of the examples of factoring which we worked out in Section 17 are, at the same time, identities. In general: 16. An identity in x is an equality involving x which holds for all numbers x for which it has meaning. A conditional equation in x, or simply an equation in x, is an equality involving x which is not an identity in x. DEFINITION
Sometimes it happens that apparently contradictory statements may be clarified by distinguishing carefully between identities and equations. To emphasize this distinction, identities are frequently designated by the use of the sign = instead of ==. Thus, in the statement p(x) = (x 
1)(2x
+ 3)
== 0
the sign "=" serves to define the symbol p(x); that is, p(x)
= (x

+ 3) 1) (2x + 3) == 1)(2x
holds for all numbers x, whereas (x 0 holds only if x == 1 or x == J. To bring out clearly the reason for the phrase "for which it has meaning" which occurs in the definition above, consider the identity x2
1 =x+l. x 1 
Of course, we cannot claim that this equality holds for all numbers x, for we have emphasized earlier that it has no meaning whatever for x == 1,
30
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
since this leads to division by 0. But the equality does hold for all values of x for which it has meaning. Consequently, in spite of the exceptional role played by the number 1 in this case, we call this an identity in x. In the case of a (conditional) equation, our problem is to determine the conditions under which the equality holds, i.e., to find those numbers which when substituted for the variable give a true statement. Such numbers are called the roots of the equation. Important tools in the process of finding the roots of an equation are the assumptions that equal numbers may be added to, subtracted from, and multiplied or divided by equal numbers to produce equal numbers. More precisely, if a == b and c == d then a+ c == b + d, a  c == b  d, ac == bd,
a c
b d
 ==  '
provided in the case of division that c == d ~ 0. We summarize by saying that equality is preserved under addition, subtraction, multiplication, and division. Since some equations have no solutions, we must recognize this from the start to avoid seeming contradictions. Some examples will serve to clarify this statement. EXAMPLE
14. Find all numbers x such that (x 
3) (x
+ 2)
=
1.
x3 If there is a number x ~ 3 which satisfies this equation, then we may cancel the (x  3)'s, since x  3 ~ 0. It follows that
x+2=1
or
x
=
1.
It is readily checked that this number satisfies the original equation. EXAMPLE
15. Solve
3 x+x5
3 + 5.
= x5
Any root of this equation must be different from 5, since 3/(x  5) is meaningless for x = 5. On the other hand, if x ~ 5, we may subtract 3/(x  5) from both sides of the equation, which then reads x = 5. In other words, we have shown that if there is a root, it must both be different from and equal to 5. This is nonsense. Consequently the equation has no root.
19] EXAMPLE
31
IDENTITIES AND EQUATIONS
16. Solve the equation
Vx
+3
Vx  1.
5
=
In an attempt to free the equation from radicals, we square both sides:
x
+3
=
lOV x 
25 
1
+x 
1.
Collecting terms, we find 21
=
10,Vx 
441
=
lOO(x 
x
=
1.
Squaring again,· we get 1),
or 5.41.
What we have shown is that if there exists a solution to the original equation, it is x = 5.41. Let us check to see if this number is actually a solution. We ask: Is ,vs.41 3 = s  V5.41  1 ?
+
But VS.41 = 2.9 and V 4.41 = 2.1, and hence our equation does indeed have x = 5.41 as a root.
Sometimes, in solving equations, operations are carried out which lead us to consider a new equation (called a derived equation) containing the roots of the original equation but, in addition, other roots (of the derived equations, that is). Such numbers are really foreign to the solution of the original equation and are called, for this reason, extraneous roots. It may also happen that we perform operations which cause an apparent loss of solutions; we say "apparent loss" because if we are careful in our analysis, there is no need to entertain the possibility of such losses. In any event, the possibility of occurrence of these situations only emphasizes the importance of testing suspected roots by substituting these in the original equation. Some examples will make these assertions more clear. EXAMPLE
17. Find all numbers x such that Vx
+ 2 + Vx

1
+5
=
0.
+
We can dispose of this problem quickly by noting that V x 2 and V x  1 are, by definition, nonnegative numbers. Therefore the left member of our equation is at least 5, regardless of any proposed value of x; clearly this will never equal 0, and so we conclude that the equation has no root. EXAMPLE
18. Solve x2
3x
=
0.
x(x  3)
=
0,

Factoring yields
32
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
that is, x O or x = 3. Both numbers are roots. We are through. If we had been careless and started by dividing by x without imposing the restriction that x is not O and examining the possibility (a real one!) of x = 0 being a root, we obtain only x  3 = 0, that is, the root x = 0 is lost.
There is a special equation whose solution is essential to our work because it is frequently encountered. This is the quadratic equation. First, however, let us look at a linear equation. By a linear equation in x is meant an equation of the form ax + b == 0, where a and b are given numbers and x is sought. It is a simple matter to determine the required number x, for if there is a number x such that ax+ b == 0,
then
x ==
b
a
(assuming a
This proves that if the equation ax x == b/a. But
+ b ==
~
0).
0 has a solution, it must be
and therefore x == b/a is the unique solution. [Compare with Section 13, Postulates I(iii) and II(iii) .] We now consider the problem of solving the quadratic equation ax 2
+ bx + c ==
0.
(1)
The problem can be expressed this way: Given three numbers a ~ 0, b, c, find all numbers x such that ax 2 + bx + c == 0. Of course, an adequate solution of this problem would tell how to form the numbers x from the three given numbers a, b, c, and we shall do just this. The idea in solving this equation is to effect a factorization of the left member and then by using the fact that a product is zero if and only if one of the factors is zero, to reduce the problem to the solution of two linear equations. Here are the steps: · Divide by a
~
0, and complete the square:
b2
4ac 4a 2
19]
33
IDENTITIES AND EQUATIONS
It follows that
2
+ !_) _ (Vb2 
x
(
2a
2
4ac )
2a
== 0
'
and on factoring the left member:
) (x + !_ _ (x+ !_2a + v'b2°=4ac 2a 2a
Vb2  4ac ) == O. 2a
When we set each factor equal to 0, we are led to two linear equations whose solutions are readily found to yield the wellknown "quadratic formula": b ± Vb2  4ac (2) x == 2a That these two numbers do satisfy (1) is not difficult to check. We leave this to the reader. Finally, we observe that the sum of the two roots r 1 , r 2 appearing in (2) is r1
+ r2 ==
b
+ yb2
 4ac
2a
+
b 
yb2 
4ac
2a
b
==  a '
and their product is r1
. _(b + r2 
b2
(b 2

(b  yb2 
yb2  4ac) 2a
2a
4ac)
4ac) _ £_.

4a 2
a
In summary, we have THEOREM 110. Let a, b, c, be numbers with a ~ 0. ax 2 +bx+ c === O if and only if
x ==
b
+ Vb 2

4ac
or
2a
x
==
b  v'b 2 2a

4ac
The sum of these roots is b/a and their product is c/a. 19. Solve for x: }x 2  2x + ! = 0. The roots of the given equation are the same as those of
EXAMPLE
3x 2

8x
+2
0.
=
To apply the quadratic formula, we set a = 3, b = 8, c = 2. There results
x
=
s±
v'64 6
4(3)(2)
8
±V40 6
4
±Vill 3
34
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
,vithout adding or multiplying, but merely by consulting the original equation, we predict (with certainty) that the sum of the roots is b/a = f and their product is c/ a = f.
+
110. Solve for x: v3  4x = 2 v5  2x. If there is a number x which satisfies the equation, then on squaring both sides, we find 4v5  2x 5  2x, 3  4x = 4 or 3 x = 2v5  2x. EXAMPLE
+
+
+
Squaring again and simplifying, we obtain the quadratic
x2
+ 14x 
11
=
0
with the solutions
x
1 ± 2v1s.
=
It remains to check these in the original equation. Substituting there, it is a question of whether the equality
holds or not. To check the equality with the minus sign in the radicands, i.e., y'31 
sv15
1= 2 + V19  4v15,
we compare the squares of these numbers: 31 
sv15 : 4
31 
sv15
+ 4Vrn

4vis
+
rn 
4v1s,
or
1=
23
+ 4y'19

4v15 
4v15.
In other words, does or does
But 2 
4y'19 
4vis = s 
4v15,
v 19 
4v 15 = 2 
v 15 ?
v15 is negative, whereas v means a positive number; therefore V19 
4v15 ~ 2 
v15
+
and hence x = 7 2v15 is not a root of the equation. On the other hand, a straightforward computation similar to the one above shows that x = 7  2v15 is a root.
The expression b2  4ac appearing in the quadratic formula plays a leading role in determining the socalled "character" of the roots. It is therefore called the discriminant. What is meant by "character of the
19]
IDENTITIES AND EQUATIONS
roots" is the following: One and only one of the following conditions can hold: (i) b2  4ac == 0, (ii) b2  4ac is positive, (iii) b2

4ac is negative.
If we go to Theorem 110 with condition (i), the roots of ax 2 reduce to b x==, 2a
+ bx+ .c == 0
that is, the (ordinarily) two numbers coincide in this case. Since we are assuming that a, b, c are real numbers (ordinary decimals of arithmetic), the same will be true of b/2a, and so we say that the character of the roots is that they are real and equal. In case (ii), since b2  4ac is positive, Vb 2  4ac is also positive and the roots b  Vb 2  4ac b + Vb 2  4ac x == x == 2a 2a are consequently real and unequal. As for (iii), when we come to extract the square root Vb 2  4ac of the negative radicand b2  4ac, we find that since for any real number r its square r 2 is not negative, the result cannot be a real number. We are therefore led to introduce a new kind of number, denoted by the symbol i, with the property that i2
== 1.
This number i is called the imaginary unit. It now follows that there are two quantities whose squares are equal to the negative number b2  4ac, namely, (i,Vb 2 4ac) 2 == i 2 (b 2 4ac) == b2  4ac and (i,Vb 2 4ac) 2 == (i) 2 (b 2 4ac) == b2  4ac.
+ +
+
+
We therefore set
Vb 2

4ac
== V(b 2
+ 4ac)
== i,Vb 2
+ 4ac.
Consequently, we obtain two unequal imaginary numbers (i.e., numbers of the form A + Bi with A, B real and B ~ 0) as the roots in this instance.*
* Imaginary numbers are discussed in greater detail in Chapter 5.
36
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
Thus, the character of the roots corresponding respectively to the cases (i), (ii), (iii) above is: (i) real and equal, (ii) real and unequal, (iii) imaginary and unequal. 111. Solve for x: 3x 2  2x + 1 Invoking the quadratic formula, we find
=
EXAMPLE
(2) ± V(2) 2 x = 2(3)
4(3)0)

=
0.
v=s 1 6 = 3±
2+
iV2
3 .
Note that the discriminant is negative: (2) 2

4(3)(1)
=
8
u, then a = v2

u2,
b
=
2uv,
c
=
u2
+ v2 •
determine a right triangle with integral sides. The converse is also true. Can you prove it? (Reference: Courant and Robbins, What is Mathematics?, p. 40, Oxford University Press.) 111 Fractions. The postulates we laid down in Section 13 apply to numbers in general and, in particular, to the fractions. Thus if we interpret each of the letters a, b, c appearing in Postulates I, II, III to be fractions, then we obtain valid statements; in other words, the rules for operating with fractions have already been set forth. Since, however, we often encounter fractions in computations, it is worthwhile to emphasize some of the consequences of Postulates I, II, III in this special case. We begin, first of all, with the criterion for equality of fractions:
17. only if ad == be. DEFINITION
Two fractions are equal, written a/b == c/d, if and
In practice, we frequently speak of "clearing of fractions" or "multiplying through by the L.C.D. (least common denominator)." Either phrase means "application of the definition of equality of fractions." For instance, when we clear the equation x 5 2  x 6 of fractions, we obtain 6x == 10  5x. A special case of this definition of equality of fractions is of particular importance: Suppose that a fraction a/b is known to be equal to 0. Then we may write a O b == 1 '
111]
39
FRACTIONS
from which it follows that a == a · 1 == b · 0 == 0; that is, the numerator of a/b must be 0. On the other hand, if the numerator of a fraction is 0, and its denominator is not 0, then the fraction is 0, for
Q == 0 (because O · b
==
b
O).
This proves the following THEOREM 111. If a fraction 1s 0, then its numerator is 0. On the other hand; if the numerator of a fraction is O and its denominator is not 0, then the fraction is 0. The definition of equality of fractions also provides a means for comparing two fractions. Thus, to compare l~~ and +g~, compute and compare the products (123) (703) and (187) (471); since these are obviously unequal (determine the last digit in each case), we conclude that the fractions are unequal. In order to add, subtract, multiply, and divide "general" fractions, familiarity with the arithmetic of fractions is necessary, for the algebraic operations are merely generalizations of these. For this reason, the following examples include both arithmetic and algebraic illustrations. A.
ADDITION AND SUBTRACTION
112. Consider the problem of replacing the sum y1g54 + single fraction equal to it. First completely factor the denominators: EXAMPLE
184
=
(4)(46)
=
1~6
by a
23 • 23,
176 = (2)(88) = (4)(44) = 24 · 11.
From this, it is clear that the L.C.D. is 24 · 11 · 23. Hence, referring to the decompositions above, 15 3 15(22) + 3(23) 399 184+176= 24·11·23 =24·11·23
3 · 7 · 19 24 · 11 · 23
399 4048
 = ·
EXAMPLE 113. The problem is to replace the expression F below by a fraction in lowest terms:
F
8x 2
5x  3 30x 
(4x 
+7
5x  3 1)(2x 
+ 17x 7)
+
1 20x 2
+ 
3
1 (1  4x)(5x 
3 3 
3)
+
5x
3 3 
5x
We observe now that 4x  1 and 1  4x are negatives of each other and therefore we need include only one of them (either one) in our L.C.D.; the
40
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
same applies to 3 
5x and 5x 
25x
2
2
1)(2x 
2
30x + 9  2x + 7  24x + 90x (4x  1)(2x  7)(5x  3)

2
1
3. Hence,
3) + (1)(2x  7) + 3(1)(4x (4x  1)(2x  7)(5x  3)
F = (5x 
[CHAP.
7)
21
+
x 58x  5 (4x  1)(2x  7)(5x 
3)
Since subtraction of fractions is subsumed under addition (because
~  ~ = i; + (  ~) ; see
Theorem 11), we pass on to multiplication
and division. B.
MULTIPLICATION AND DIVISION
First, an example from arithmetic: EXAMPLE
114.
Next, an example from algebra: EXAMPLE
115. Replace the following by a single, equal fraction: 3
8 F
=
x
+
F = (2 
x 2
x x2 
x)(4
2
+ 6x
+2
4  x
+ 2x + x
2
x
4
5a x(x
) .
+ 2x 2
+ 6)
(x2)(x+l)
x+2 3
2
2
x (x
2
+ 2)
5a
2
x (4 + 2x + x )(x + 6)(x + 2) 5a(x 2)(x I)
+
+
Since all factors are prime, this fraction is in its lowest terms.
The operation of division may again be reduced to multiplication, for to divide a/b by c/d means to find a number x such that c
a
x· = · d b Now multiply through by bd to get cxb = ad.
111]
41
FRACTIONS
But, on dividing by cb, we see that this is the same as ad a d x == be == b ·
c·
Now d/c is the original divisor inverted. Hence the THEOREM 112. To divide one fraction by another, invert the divisor and multiply the resulting fractions. EXAMPLE
3
+
116. Perform the division:
+
x 2x 
1
2x 1x5 EXAMPLE
x 3(2x  1) 2x  1 x  5  2x x5
7x  3
2
x  5
7x  38x 2x 2
1 5  x
2x 
+ 9x
+ 
15 5
117. Solve the equation:
7 x2 
16 
3  x 4 + x
2
+
+
12
8
x x 
x2
x4
We begin by factoring the denominators and noting that the L.C.D. is (x 4) (x  4) (3 x). If there is a number x which satisfies the equation (and therefore does not make the L.C.D. zero, that is, x ~ 4, x ~ 4, x ~ 3), then on multiplying through by this L.C.D., we have
+
7(3
+ x)
+

(3 
4)(3
x)(x 
+ x) + x 2(I)(x + 4)
8(x
+ 4)(x + 3).
Collecting terms yields
57 
2x = 96 
or
56x
Some arithmetic computation, after substituting this number in the original equation, will show that this is in fact a root of the equation. 118. Determine the constants A., B so that the following will be an identity in x: EXAMPLE
5 x2  4
=
_A_+ x
+2
B x 
2
If this is to be an identity in x, we must have, on adding the fractions on the right, 5
x2 
4
A(x (x
+
+
2) B(x 2) 2)(x  2)
+
Since the denominators are identical, the numerators must be the same: 5
=
A(x 
2)
+ B(x + 2).
(1)
42
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
Now this equality is to hold for all values of x. x = 2, we obtain 5 = 4B or B If we return to ( 1) and choose x
[CHAP.
1
In particular, if we choose _§_
4•
2, we find A
Thus 5 x2 
=
4
5/4+ 5/4 x 2 x  2
+
or 5
x
5
2 
4
4(x
+ 2)
+
5 4(x  2)
119. Simplify the expression E below to an equivalent one containing no negative exponents (you will encounter problems of this kind later on in the calculus). EXAMPLE
E
=
(2x 
3)
112
(i)(5x
2

1)
113
(10x) + (5x
2

1)
213
(!)(2x 
3)
112
(2),
2 213 2 (2x  3)1' 2ox (5x  1) 3(5x2  I) 113 + (2x  3) 112
E
(2x 
3)
112
112 2 (20x)(2x  3) + (5x 3(5x2  I) 113(2x 
+

1)
213
(3)(5x
2

1)
113
3) 112
2
20x(2x  3) 3(5x  I) 3(5x2  I) 113(2x  3) 112 2
55x  60x  3 3(5x2  I) 113(2x  3) 112
112 Exercises In Exercises 1 through 37, replace the expression given by a single, equal fraction in lowest terms: 1 ab 2. 1
1 31.
5 1 2+2
c+a
1 3 5. 1 4+5
2
4. 4 
y 
x
y+x
3. 2 
x  y
x+ ]
x+y
6.   1 yx
y
112]
7
43
EXERCISES
4 +6 _ 8 · x  2 x 2 x2  4
8.
+
c3  d3 c2 9 · c2  2cd + d 2 · c 
J{
d2 d
...!...
10.
+1 11.    m
+ 2 4x
+~
:
y y x2 x2 y2
+ + 2a 
12 . 2a  3b _ 6a ~ 4a 5b
mn
~1
m
2 3 x 4
b
2c
m+n 2
13
15
16
x + 3 x + 16 · x  4 x2  9
·
x ·
3 5x
x2 2
+6
+ 3x + 2
x2 
2x
14 . __2_ _ 4x2  1 x+ 3 x  9
2 6x
x2 2
x + 4x x · x2
...!...
+9
+~ x
+   5_ _ x 2  8x 12
+
4x 2  gy2 17.    yx 1x  2y
+3
+ +6
(
19.
l _ ab  a2) b2
+ (!__ + _1:_) a3
b3
~~~~~~~~~~
( 1  ~) b
+ (!__  !__) a2 b2
2 3 2x 6 x  27 x  3x 9 20 2(x  3) · x3 + 27 · x2 + 3x + 9
+
(a 2 [ 1. ba
+
+
J.
2 2 3 b) b a b  be+ c c3 · a2  2ab b2 . b  c
+
x  1 x  1 22.    x 1+1  x
1 23.  x 1
3x 2x  + + x2  1 1  x
12 c+ 1  24. c 1  ~ c 26. (5x
27
2

a
· [ a2
x 
+ 3a
25
4) + ( 1 
4
+2 2ab + b2 
. x
2
+ 4x + 4 + x2 
9
+ 5~)
J
+
b 1 a 4 1 · 3  3a 2 · ca cb ...!...
...!...
+
c
2
x  4 x 2  6x + 9
44
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
1
2
a+ b
b
32
34
n2
+n+1
1 · 2(x 
1
· k(b 
[
1
33
+ 1)
2(x
a) a 
2\/12(x2
36. 1
31.
1
1) 
x
35.
_
x
1
b 
x
2a 
3
·
3
a 
3a2 
1
n+l 1 1 1 1 1~ + 2~ + 3~ + 4~
J
+ y12
xv12
+ v12x + 1)
+
1
a3 
n+ 2
_
+
29 . 2a=±
2 a2  b2 1
1
]

28. a 
30.
[CHAP.
2v12(x
2
v12x

+
1)
l l
1+1+! 37.
b
+ yb2

4ac
b  Vb2 2a
2a
4ac
38. For what values of x is the fraction x 2 39. Solve for x:
1
3x 
5
40. Solve for R: P
=
41. Solve for P: (P
x
 
+ 2 3
. i:
s
=
2 ex ; Sdx _ lSde undefined? 3
6x
=
15
w(R  r) R 2
+ ~) (v
b)

42. Solve for x and then for u: u
43. Solve for
_
2
=
k ux

+ k2
0
=
3  i m 2
44. Find the value of
+
:b_ !
2
3
2
if a
=
!
i
and b
. d t h e m1ssmg . . d enommator: . 2x  1 45. Fm 3
x 
2x
?
46. Find in factored form the L.C.D. of 16x (x  4) 2
'
3 , 4x
5
x2

16
2
112]
45
EXERCISES
47. Solve for y:
1 y 3
3y 
2
c 3d 48. Solve forx:   x
49. Solve for x: x 2
2
=
x
c 
6cd+ 9d
8
x2
4 +2x+ x _ 2
2
x4 3x
x+6
+2

x2
+ 3x 
4
50. Classify ~he following as an identity or an equation. If an identity, prove this. If an equation, find the roots. (a)
x
2
x  2 x _ 2
1 51 . Show t h at ; 
x
x
=
+1 1
=
+1 x(x
(b) 1 
+
1
y  ~y 1y
4y 1
+
y
1 . .d . . + l) 1s an 1 entity m x.
In Exercises 52 through 54, determine the constants so that an identity will arise.
+
52
3x 8 = ~ · x 2(x  1) x
53
3x  2 · (2x  1)(3x
54
2x  5 · (3x  4) (2x 
+ B2 + x
C 1
x 
+
A
+ 4)
2x 
1
A 1)
3x 
1 + a 2m
= 
1 55. Solve for m: 
56. Solve for x: 1 
2
x
+1
4
B 3x
+
+4 B
2x 
1
1 2
3
= (x + l) 2
In Exercises 57 through 60, reduce the given expression to lowest terms. 3 3
4p+6
x 57. x2P+3
58.
3x u 59. 2u x
60.
+
14m a 49ma 7ma3
4
(ax3)2x ax
In each of Exercises 61 through 67, simplify the given expression to an equivalent one containing no negative exponents.
*61.
(3x 
5) 112 (4x) 
(2x 2  7)(!)(3x 3x  5
5) 112 (3)
46
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
*62.
1
1) 2/ 3
~~~~~~~~~~~~~
1)2/3
(x 
!
(2x
+ (5x 2 + 3)(2)(3x 
2
*63. (3x *64 .
3x ( i) (x 
1) 1/ 3(3) 
(x 
[CHAP.
1) (10x)
+ 5) 112 [(3x 
2 3x 
1
1)2 (3x 
(2x 1)2
1) 3(3)
+ 5)3]
+ 7) 113 (!)(4x 3  1) 112 (12x 2) + (4x3  1)112(!)(3x + 7)4/3 (3)
*65. (3x
*66.
*67.
( !)
32 3 (4x=!_) ' [(3x 2 3x3 5
(2x
+
+ 3) 11 \!) (x

I)
112

(2x
+ 5)4

(3x3 (x 
+ 3)213
(4x 
+ 5)2
1)3i 2ct) (2x
1)9x
2
J
+ 3) 213 (2)
SOME REVIEW PROBLEMS 68. How long will it take A and B, working together, to do a piece of work which A alone could do in 10 days, and B alone in 20 days? 69. It takes 3 hours to row 20 mi downstream in a river and 10 hr to return. Find the rate of the current. 70. A man traveled a distance of 90 mi at a constant rate of speed. If he had tripled his speed, he would have taken 1 hr and 12 min less time to travel the same distance. At what rate did he travel? 71. A can paint a house in 16 days; B can paint it in 12 days. If A and B work together for 4 days and then Band C complete the job in 3 days, how long would it take C to paint the house alone? 72. If one pipe can fill a tank in 3 hr and a second pipe in 4 hr, how long will it take them together to fill the tank?
113 Terms common in mathematical literature. As an aid to our mathematical reading, we mention in this section some of the abbreviations and terms frequently found in mathematical literature. All the terms listed are from the Latin, which reflects the fact that before the 19th century Latin was the universal language of scholars. e.g.: exemplar gratis (free example); translated "for example." q.e.d.: quod erat demonstrandum, translated "which was to be proved" and used to signal the end of a proof of a theorem. 1.e.: id est, translated "that is." a priori: beforehand.
114]
47
REVIEW EXERCISES
lemma: "little theorem." Actually, a lemma may be considered as a steppingstone in the proof of a theorem. A lemma is itself a theorem (in the sense that it requires proof), usually of sufficient importance and interest in itself to be set apart from the theorem for which it closes a gap. corollary: an immediate consequence of a theorem, "gift." Corollaries are theorems (proofs are required), but in light of results already obtained, their proofs are trivial. viz.: abbreviation of videlicet, translated "namely." cf.: compare. a fortiori: with the greater force, all the more. 114 Review exercises
Solve the following equations: 2
l.y
12
+Y+~+ y y 2
2. 2  x x3  1 3. x
213
+
=8
x x  1
+ 2x 113
=
l
2
4. Determine A, B such that 5x x2  4x
+4
+
A x 
2
B (x 
2)2
shall be an identity in x. 5. For what values of x is the fraction 5
x2  ax+ {3 undefined? equal to O? 6. Solve (1/p) (1/q) 1 for p; then for q. 7. (a) Using the quadratic formula, find a formula for the sum of the reciprobx c = 0. cals of the roots of ax 2 (b) Apply the formula developed in (a) to find the sum of the reciprocals of the roots of 5x 2  8x  12 = 0. 8. Write with positive exponents and simplify:
+
+
+
(!)(3x  1) [(4 4 3 '
3
4 
7x
7x)3  (3x (4  7x)2
1)(7)]·
48
ALGEBRA: BASIC CONCEPTS AND OPERATIONS
[CHAP.
1
9. Is
x a sufficient condition for ax 2
=
b ± v'b 2 2a
+ +

4ac
bx c = 0? State precisely the meaning of your answer. Is this condition necessary for ax 2 bx c = 0? 10. Is the converse of the unique factorization theorem true? 11. Write Theorem 17 so that its hypothesis and conclusion are in evidence. 12. What are the hypothesis and conclusion of Theorem 112?
+ +
CHAPTER 2 THE REAL NUMBERS: THE POINTS ON A STRAIGHT LINE Introduction. A fundamental notion in mathematics is that of a real number. It is our goal in this chapter to gain some idea of what a real number is. We shall approach the concept in three different ways. 1. The arithmetic approach. By this we mean that we take a look at the real numbers from the point of view of the student of elementary arithmetic, considering the real numbers to be the collection of all decimals. 2. The geometric approach. Here we take the standpoint of the highschool student of geometry, considering the real numbers as the collection of all points on an infinite straight line. 3. The abstract approach. This is the outlook of the mature mathematician. We have already defined a field to be any set containing a zero and a unit and satisfying Postulates I, II, III of Section 13. In the abstract definition of the real numbers, the notion of a field is a central one; in fact, the real numbers form what we call a complete ordered field. Intuitively, we can say that a field embodies a system in which it is possible to add, subtract, multiply and divide; to be a complete ordered field there must be defined in this field an order relation, i.e., a relation of "less than" which satisfies certain conditions which we shall introduce shortly. That these three approaches all lead to the same thingnamely the set of all real numbersis shown in more advanced work. All we can do at this time is to make the compatibility of the various interpretations seem plausible. Our approach, then, is from a naive standpoint. This does not mean, however, that we shall entertain false statements. On the contrary, the intuitive notion of a real number as a point on a straight line has led to ideas for proofs of theorems of farreaching consequences, and therefore has been essential to the historical development of mathematics. Indeed, it was not until the time (19th century) of great mathematicians like Augustin Cauchy, Karl Weierstrass, Georg Cantor, and Richard Dedekind that a precise meaning of real number was sought. Another basic concept introduced in this chapter is the notion of distance between two points on a line. This idea will play a leading role in all of our subsequent work, so be sure you master it. We begin our discussion with the arithmetic and geometric interpretations of the real numbers; later we take the abstract approach. 49
50
REAL NUMBERS: POINTS ON A STRAIGHT LINE
21 The real numbers.
[CHAP.
2
In arithmetic, you learned that the digits are 0, 1, 2, ... , 9. You also learned how to operate with decimals containing a finite number of digits, such as The arithmetic approach.
7.912,
53.7186,
8.33,
and so on. Let us consider the collection of all conceivable decimals, even those which contain an infinite number of digits after the decimal point. This collection of numbers constitutes the real numbers. We study this collection by distinguishing different kinds of decimals. The simplest numbers are those with all zeros after the decimal points. They are called the integers: ... , 3, 2, 1, 0, 1, 2, 3, ... The next simplest kind of decimal is one which arises when we pick two integers, say a and b ~ 0, and carry out the long division a/b. So important are numbers obtained in this way that they have been given a special name. Formally, 2:1. A rational number is any number which can be written as a ratio a/b of two integers a, b, where b ~ 0. DEFINITION
Now the question arises: Among all conceivable decimals, how can we spot those which arise from the long division of two integers? EXAMPLE 21. Show that the repeating decimal 52.713713 ... (where the three dots indicate that the period 713* of the number occurs over and over again, never ending) is a rational number. Form the equation x = 52.713713 ... ,
for convenience. Multiply by 1000 (so as to shift the decimal point to the right a number of places such that it will appear at the end of the first period): lOOOx = 52713.713713 ...
Now subtract the two equations to get 999x Hence x
=
52.713 ...
=
=
52661,
or
52661 999
x=·
52661/999 is a rational number.
A careful study of this example will reveal that it can be generalized; that is, we can follow the same sequence of steps in proving that every
* The period of a decimal consists of the smallest block of digits (when such exists) which is repeated indefinitely in the decimal representation of the number.
21]
THE REAL NUMBERS
51
repeating decimal is equal to some rational number. A formal proof could be
constructed using the idea illustrated by the example; such a proof would offer no new conceptual difficulties but would contain only problems of notation, and is therefore omitted. EXAMPLE 22. Prove that the terminating decimal 82.629 is a rational number. (By a terminating decimal is meant a decimal which contains all zeros from a certain place on.) The idea is to first convert the given number to a mixed number. What mixed number~ We read this number: 82 and 629 thousandths; thus we write
82.629 = 82
629 82629 = , 1000 1000
q.e.d.
A careful study of this example will reveal that it, too, can be generalized to the statement that every terminating decimal is equal to a rational number. Hence we have the general result that every repeating or terminating decimal is equal to a rational number. Indeed, much more can be proved: the converse of this last statement is also true, so that we can say that the repeating and terminating decimals constitute all the rational numbers. To prove that every rational number is equal to a terminating or repeating decimal, we start with a given rational number, i.e., we are given a ratio of two integers, a/b, where b ~ 0. Now imagine that the ordinary longdivision process of arithmetic is carried out. Evidently one of these two alternatives must be encountered: (1) either we eventually get a zero remainder, in which case the quotient (which will be equal to a/b) is a terminating decimal, and so we are through, or (2) we never get a zero remainder. Now in this last case every partial remainder obtained at some stage of the division process must be less than the integer b (the divisor). Also, the dividend a is an integer (and therefore when considered as a decimal will be terminating). Thus it is clear that at some stage after a finite number of steps of the longdivision process we must be confronted with the same partial remainder as at an earlier step (only b  1 different partial remainders are possible) and, furthermore, because of the terminating character of the number a, we can assume that the dividend contains only zeros to be "brought down." Consequently, when this situation is reached we will have a repetition in the division process which will be reflected in the period of the quotient obtained. This proves that every rational number can be written as a terminating or repeating decimal. If we now agree that terminating and repeating decimals shall be referred to as periodic decimals (terminating decimals have period 0), we have proved the following theorem: THEOREM 21. A number expressed as a decimal is equal to a rational number if and only if that decimal is periodic.
52
REAL NUMBERS: POINTS ON A STRAIGHT LINE
[CHAP.
2
Another way of expressing the same idea is to say that the rational numbers on the one hand, and the periodic decimals on the other, are equivalent notions. They are equivalent in the sense that
(1) if a number is a rational number, then it can be written a'S a periodic decimal and, conversely, (2) if a number can be written as a periodic decimal, then it is a rational number. In summary: If we are given a number in decimal representation, we know that it is a rational number if and only if the decimal is repeating or terminating. In this sense we say that we have characterized the rational numbers among the set of decimals; that is, we have given necessary and sufficient conditions for a decimal to be representative of a rational number. It is natural to ask: Is there another kind of number besides the rational number? Yes, consider the decimal
52.101001000100001 ... , where the three dots mean that the pattern: 1 followed by one zero, 1 followed by two zeros, 1 followed by three zeros, and so on, is to be continued indefinitely. Clearly, this decimal is not periodic as defined above and therefore does not represent a rational number. We call such a number an irrational number. Furthermore, in view of the theorem above, we can form irrational numbers at will by simply ensuring that the decimals we write down are not periodic. By simple modifications of the number displayed above many examples will come to mind. This paves the way to 22. An irrational number is any decimal which is not a rational number. A real number is any number which is rational or irrational. DEFINITION
The geometric approach. The idea of associating numbers with the points of a straight line is of commonplace occurrence; a carpenter's rule or a seamstress' tape and numerous other devices serve to illustrate this. The mathematical abstraction of these devices leads to the notion of a real number as we shall now consider it. We begin with a straight line (Fig. 21). After selecting two arbitrary points on this line, we associate the number O with the one on the left and the number 1 with the one on the right. This determines the "unit." With a compass whose points are set unit 3 2 1
0
FIGURE
1
21
2
3
21]
53
THE REAL NUMBERS
distance apart, we now determine some of the other points to be associated with the integers
... , 3, 2, 1, 0, 1, 2, 3, ... Let us digress for a moment. It was the philosopher Plato who, about 350 years before Christ, reached the apparently arbitrary decision that all constructions in geometry should be limited to straightedge and compass as tools, and no others. Constructions not adhering to these requirements were taboo and considered "base." Unfortunately, some of the problems considered by the ancients, such as trisecting an angle, cannot be solved under these conditions. On the other hand, the proof of the impossibility of these has led to some interesting and useful mathematics. In any event, it is instructive to ask, with Plato, "Is it possible, using ruler and compass, to locate the points to be associated with the rational numbers?" Indeed it is. For example, Fig. 22 illustrates how we can determine exactly the point which we associate with the rational number f. The procedure utilizes the elementary plane geometry construction for dividing a segment into 5 equal parts; it is as follows. We take the unit segment and draw any convenient angle {3 with vertex at O and transfer that angle as indicated so that its vertex is at 1. Then we measure 5 equal intervals on the terminal sides 0R and 1S of these angles and join corresponding points as shown. It is now clear from the diagram how the points corresponding to the fractions g, g, g, as well as!, are determined. More generally, this method can be applied to any rational number. We leave it to the reader to write down different rational numbers and determine how to locate the corresponding points, using ruler and compass. Thus we are able to locate exactly with ruler and compass any point which corresponds to a rational number. Now let x and y > x be any two rational numbers which we locate on the straight line, as shown in Fig. 23. The length of the segment joining
FIGURE
x
22
m FIGURE
y
23
54
REAL NUMBERS: POINTS ON A STRAIGHT LINE
[CHAP.
2
these two points is y  x. It follows that the midpoint of this segment corresponds to the number m, where m==x+yx==~+Y.
2
2
+
In other words, the average (x y)/2 of two rational numbers x and y is associated with the midpoint of the segment determined by x and y. Now it is easy to prove that if x and y are rational, then so is their average, (x y)/2, for let x == a/b, b ~ 0, and let y == e/d, d ~ 0. Then
+
x+y
2
(a/b)
+
(e/d)
2
ad+ be 2bd
But ad+ be and 2bd are integers and 2bd ~ O; therefore (x + y)/2 is rational. Thus, between any two rational points (i.e., points corresponding to rational numbers), we can find another rational point. This is often described by saying that the rational points are dense (in themselves). Because of this property, it might be suspected that the rational points "fill up the line." We now show, however, that this is not so. Consider again the irrational number 52.1010010001 Evidently there corresponds to this number a definite point on the line, to which we can approximate by the points corresponding to the following sequence of rational numbers: 52.1, 52.101, 52.101001, ... (Don't get the mistaken impression that these numbers become indefinitely large; all of them are less than 52.2, for example.) It turns out that the point corresponding to this number can be located only approximately with a compass and a ruler, whereas we have seen that rational points can be located exactly with these tools.* On the other hand, there are irrational numbers whose corresponding points can be located exactly on the line. The number 0 is one such number. t Let us now prove these asser
* That the number 52.101001 ... can be located only approximately with ruler and compass must be proved. The reader can learn how to distinguish the "constructible numbers" by consulting the book What is Mathematics? by R. Courant and H. Robbins, Oxford University Press, Ch. III. An interesting byproduct of the theory is the result that a general angle cannot be trisected using only ruler and compass. t The number v'2 is a root of the polynomial equation x 2  2 = 0; roots of polynomial equations of arbitrary degree with integral coefficients are called algebraic numbers. It can be proved that all constructible numbers are algebraic numbers (see the book mentioned in the previous footnote).
21]
THE REAL NUMBERS
55
tions. First of all, to locate the point corresponding to the number v2 we construct a right isosceles triangle with equal sides one unit long. Then, by the Pythagorean theorem, the hypotenuse will have length 0. Next, we have to prove THEOREM 22. y2 is not a rational number. Proof. If y2 were a rational number, then, by definition of a rational
number,
V2==~,
(1)
b
where we might just as well assume that the fraction a/b is in its lowest terms, since every fraction is equal to such a fraction. Now since a/bis in its lowest terms, a and b cannot both be even. Let the decompositions of a and b into primes be given by a == p 1 . . . Pn, b == q 1 . . . qm. Then, squaring (1), we get 2b 2 == a 2 •
that is,
'
Since the right and left members of this equality are the same integer, it follows from the unique decomposition theorem (Theorem 18) that precisely the same primes appear on both sides the same number of times. Therefore one of the p's is a 2 (and hence because of the "square" the right member contains a pair of twos). But then one of the q's must also be the prime 2, which means that a and b are both even, a contradiction. Thus the supposition that y2 is rational leads to a contradiction and, consequently, must be discarded. But the denial of v2 being rational is exactly what we set out to prove. Summarizing the geometric approach, the real numbers are the numbers which correspond to the points on a straight line. Further examples of irrational numbers are VS,  a) means that there exists a positive number h such that a + h == b. Geometrically, this means that a falls to the left of b on the straight line (Fig. 25). EXAMPLE 23. 3 because!+ / 4 = and h = 5 > 0.
< 8 because 3 + 11 = 8 and h = 11 > O;  ! < i i and h = 2\; > O; 7 > 12 because 12 + 5 = 7
An extremely simple but important result is THEOREM 23. If a is a real number, then a 2
>
0.
Proof. Let a be an arbitrary real number. There are three cases to consider: (iii) a < 0. (ii) a == 0, (i) a > 0, In case (i), we have that a is positive, and since the product of two positive numbers is itself positive, a · a == a 2 > 0, and we are through. In case (ii), a == 0 and hence a 2 == 0, so that, a fortiori, a 2 > 0. Finally, in case (iii) a < 0 implies a > 0. Therefore (a) (a), being the product of two positive numbers, is itself positive. But
(a)(a) consequently, a 2
>
Negative numbers
a2·
'
0, q.e.d.
Positive numbers ()
FIGURE
a
24
FIGURE
b
25
58
REAL NUMBERS: POINTS ON A STRAIGHT LINE
[CHAP.
2
24 Solutions of simple inequalities. We are interested in the problem of solving inequalities. To solve an inequality means to find all real numbers x for which the inequality is valid. To do this, we must first develop rules for manipulating inequalities. We know that an equality is preserved under
all four arithmetic operations (addition, multiplication, subtraction, and division). (See Section 19.) We now raise the question as to whether or not the sense of an inequality is preserved under the four arithmetic operations. We first prove that the sense is preserved under addition and subtraction, i.e., THEOREM
and a 
24. If a < band c is any real number, then a < b  c.
+c
0. Completing the square in x and adding and subtracting 1 yields 16x 2

8x
+1
(4x 
1 1) 2
(4x 
+5> +4> 1) 2
>
0, 0,
4.
Now, no matter what real number xis selected, (4x  1) 2 is nonnegative and therefore always exceeds 4.* Consequently, every real number furnishes a solution to this inequality.
Another useful rule of manipulation has to deal with inverting an inequality. The examples (i)
2 (ii) 3 (iii) 2
< <
1 3 > 1 2
(1/b). On the other hand, if a < band a and b have opposite signs, then (1/a) < (1/b). In all cases, of course, we are assuming that neither a nor bis 0.
*u
exceeds a means u
>
a.
25]
REAL NUMBERS: COMPLETE ORDERED FIELD
61
band a and b have the same sign, then ab > 0, so by the first part of Theorem 26 above, we have, on dividing the inequality a < b by ab "F 0, 1 1 a b or < · < ab ab b a Proof. If a
, ab ab i.e., 1 1 q.e.d. > , b a
*25 The real numbers : the complete ordered field. We mentioned in the introduction to this chapter that the real numbers may be characterized (that is, defined) as a complete ordered field. We now explain what this means. To begin with, it is necessary to come to a clear understanding of what is meant by an upper bound for a set of real numbers. Let A denote an arbitrary set of real numbers. We may think of A geometrically as being the set of points on the real number line corresponding to the numbers belonging to A (see Fig. 26). If a is one of the numbers belonging to the collection A, we indicate this by writing a E A and we read this: a is a member of A. Suppose that A is a set of numbers such that there exists a number u which is not less than any number belonging to A. Then u is said to be an upper bound for A. Formally, we have 24. By an upper bound (abbreviated u.b.) for the arbitrary set A of real numbers is meant any number u (if such a number exists) such that a E A implies u > a. DEFINITION
Remark. If a set A of real numbers has an upper bound u, then it has an infinite number of upper bounds, since any number greater than u will also serve as an upper bound. I 11
11111
I
I
a
FIGURE
* Sections marked * may be omitted
II
u
•x
26
without loss of continuity.
62
REAL NUMBERS: POINTS ON A STRAIGHT LINE
[CHAP.
2
28. Let A be the set of all natural numbers, i.e., positive integers. Is there an upper bound for A? Since, if xis any real number whatever, there exists a natural number greater than x, the set A of natural numbers has no upper bound. EXAMPLE
EXAMPLE
29. Let A be the set of all roots of the equation (x
+ 5)(x 
3) 2 (x 
8)
=
0.
Find an upper bound for the set A. A consists of the numbers 5, 3, 8. Hence the number 8 or any larger number is an upper bound. 210. Let A denote the set of all integers whose squares are less than 100. Find an upper bound for A. The members of A are readily found to be ±9, ±8, ... , ± 1, The number 9 is an upper bound for A; any number greater than 9 would also serve as an upper bound for A. EXAMPLE
+o.
Let us again focus attention on an arbitrary set A of real numbers. Again suppose A has an upper bound. Then it will have many upper bounds; let us denote the collection of all these upper bounds by U. The smallest number of the collection U is called the least upper bound of A. Formally, we lay down the DEFINITION
25. Let A be an arbitrary set of real numbers. By the u (if there is one) such that
least upper bound of A is meant the number
(i) u is an upper bound for A; that is, if x belongs to A, then u > x, and (ii) u is the least of the upper bounds for A ; that is, if u is some upper bound for A, then u < u. EXAMPLE
211.
What is the least upper bound of the set of numbers 1,
2, ... , 10?
The number 10 is the least upper bound of this set, since it is an upper bound for the collection, and it is the least of the upper bounds. 212. What is the least upper bound of the set of real numbers a1, a2, ... , an? The maximum of the numbers a1, a2, ... , an is the least uppe~ bound of the set. EXAMPLE
EXAMPLE 213. The least upper bound of the set of all real numbers x for which 1 < x < 1 is the number u = 1, for it satisfies both conditions (i) and (ii) required by Definition 25.
We are now in a position to say what a complete ordered field is. A complete ordered field R is a collection of objects which is: (A) A field (as defined in Section 13).
26]
63
EXERCISES
(B) Ordered, that is, a relation a < b is defined between pairs of elements of R such that (i) for any pair a, b of numbers of R, one and only one of a == b, a < b, b < a holds, (ii) if a < b and b < c, then a < c, (iii) if a < b and c is any number, then
a+ c (iv) if
a>
0 and b
>
0.
(C) Complete; that is, every set of elements of R with an upper bound has a least upper bound. An abstract definition of the real numbers consists of asserting that the real numbers are a complete ordered field. All further properties of the real numbers can be derived from (A), (B), and (C). Such a program is carried out in more advanced courses. The "completeness property" (C) is our precise way of saying that "there are no 'holes' in the real number line." Thus, observe that the set of all rational numbers satisfies the conditions (A) and (B), but not (C), since, for example, the sequence of rational numbers 1, 1.01, 1.01001, l.010010001, ... has a rational upper bound (the number 2 will do) but it has no rational least upper bound since the least upper bound is the number l .010010001 ... and this is not rational.
26 Exercises Note. Exercises 20 through 26, 29, 30 should be omitted if Section 25 is omitted. Solve the following inequalities for x:
+
1. 3 < 6x 7 3. 2 < 3x  8 5. rn)x  7 > 21 1
7
·
x2
+ 2x 
g · 2x2
+
3 5x 
8
< 1 1
> 9
2. 1 < 2x  5 < 7 4. 3 < 2x  7 < 12 6. 0.3x  t > 3
x2 
t 4
3
>
x
+8
64
REAL NUMBERS: POINTS ON A rSTRAIGHT LINE
[CHAP.
2
13. Which fraction is larger, !~ or ~~? Why? 14. Prove that x 2 2x 5 is nonnegative (i.e., > 0) no matter what number xis. (Complete the square, using the first two terms.) 15. Prove that x 2  2x 5 > 0 for all values of x. 2 16. Prove that x 2x  5 < 0 for all values of x. Can you strengthen the inequality, i.e., dismiss the possibility of equality holding? 17. Can the inequalities in 14 and 15 be strengthened? [See Exercise 16.] b2 > ±2ab for all numbers *18. By considering (a =F b) 2 , prove that a 2 a, b. 19. Prove that if (a/b) < (c/d) and (elf) < (g/h), where all the terms of the fractions are positive, then (a/b)(e/f) < (c/d)(g/h). *20. Prove that if a set A has a least upper bound, then it can have only one. Prove the analogous statement for greatest lower bound. 21. Give upper bounds (if there are any) and least upper bounds for the set of solutions of the inequalities appearing in Exercises 1 through 10. 22. Can a set A of real numbers have a "least lower bound"? "greatest upper bound"? *23. In Theorem 25, we used the fact that if c > 0, then (1/ c) > 0. Can you prove this by appealing to the postulates for a complete ordered field givyn in Section 25? [Hint: Prove first that 1 > 0, then consider c(l/c).] r 24. (a) What is a lower bound for the set of all solutions of the inequality
+ + +
+
+
x 1 >? 3  x 2. (b) Is there a greatest lower bound? (c) Does the set of solutions in (a) have an upper bound? least upper bound? 25. Does every set of numbers with an upper bound necessarily have a rational upper bound? (That is, a rational number which serves as an upper bound.) 26. If A is a set of rational numbers with a least upper bound u, is u necessarily a rational number? *27. Criticize the following argument: THEOREM: The first real number greater than O is!. Proof. Let x be the first real number greater than 0. If x were less than !, we would have O 28. if 7
Thus the refundable portion Risa function of the number of days of attendance. The graph of this function is shown in Fig. 33. We have followed the convention in this sketch that any point which is the center of a small circle does not belong to the graph of the function being represented. Thus, the point (7, 60%) does not belong to the graph whereas (7, 80%) does belong, which means that a student who has attended just 7 days is entitled to 80% refund, not just 60%. EXAMPLE 33. If C is the circumference of a circle of diameter d, then C = 1rd or, if we want to emphasize that the circumference C of a circle is a function of its diameter d, we write C(d) = 1rd. The graph is shown in Fig. 34, where we have made the plausible requirement that d > 0. EXAMPLE 34. Another simple function is the relationship between the area A of a circle and its radius r: A = 1rr 2 •
This may be considered as a prescription for passing from the radius r to the area A, the prescription being: square rand multiply by 1r to arrive at A. The
76
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
A
FIGURE
35
graph of this relation is shown in Fig. 35 (we impose the reasonable restriction that r shall be a nonnegative number).
Let us now try to consolidate these examples by selecting their salient features. We lay down the tentative definition: y is said to be a function of x, written y == f(x) if whenever x is given, y is determined. x is then called the independent variable and y is the dependent variable. Furthermore, f(a) denotes the number determined by replacing x by a.
In almost all instances with which we shall come in contact, x and y will be numbers. Of course, letters other than x, y, and f may be used in speaking of a function; always, however, the situation is the same: there are an independent variable (sometimes more than one) and a dependent variable. In Example 31, the independent variable x can be any nonnegative number (x kwh) and, corresponding to it, a number C(x) (this is the dependent variable) is determined, where C(x) > 0.65 (0.65 was the minimum charge). We have a technical name for the set of all numbers constituting the permissible values of the independent variable: it is called the domain of the function. On the other hand, we call the set of all possible values of the dependent variable the range of the fun'ction. Thus, for the examples, we have the data given in Table 31. 32. A function consists of the bringing together of three things: A set of numbers called the domain, a set of numbers called the range, and a law or correspondence which assigns some number in the range to each number in the domain. DEFINITION
This last statement is a more precise definition of a function than the tentative one given earlier. In practice, however, by applying the test
32]
77
THE NOTION OF A FUNCTION AND ITS GRAPH TABLE
Dependent variable designation
31
Example No.
Independent variable designation
31
x
C(x)
x
32
d
R(d)
d>O
33
d
C(d) or C
d>O
C>O
34
r
A(r) or A
r> 0
A>O
Domain
>
Range
0
C(x) R(d)
=
>
0.65
80, 60, 40, 20, 0
given by the earlier definition we can very readily recognize whether we are confronted with the notion of a function, and so, as a matter of expediency, we sometimes lay the emphasis there. It is also of great practical importance to keep in mind that if we consider f(x) to denote the number associated with x by means of a function, then f(a) denotes the number associated with a by means of the same function. EXAMPLE 35. If f(x) = x//_x f (2) is undefined. Furthermore, u ( ) f u = lu  21
an
21, then f(l)
d (1)~ _ f

1/x 1(1/x) 
1//1 
21 = 1, whereas
lxl xll  2xl .
_ 21 
+
1, then Every polynomial in xis a function of x. For example, if p(x) = 3x 2 x 2 )/x 2 , and since p(p(x)) means to replace x in the formula p(l/x) = (3 for p(x) by the formula itself, we have
+
p(p(x))
=
3(3x 2
+ 1) + 1 = 2
27x 4
+ 18x + 4. 2
Whenever the concept of a function arises, if we can settle the question as to domain, range, and correspondence involved, we have complete knowledge of the function. The graph of a function offers a visual approach to the concept of a function and reveals these three things very clearly. For this reason, it is important that we learn as much as we can about the graph of a given function; indeed, graphing a function is one of the major problems of elementary mathematics. We have the 33. By the graph of a function y == f(x) is meant the set of all points (x, y), where x and y are numbers which satisfy the relation Y == f(x). DEFINITION
Remark. This definition gives rise to a simple but nevertheless exceedingly useful principle: If a point (x, y) lies on the graph of the function y == f(x), then its coordinates x, y must satisfy the relation y == f(x) and, conversely, if a point
78
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
has coordinates x, y which satisfy the relation y === f (x), then the numbers x, y are the coordinates of a point on the graph of y === f (x). For instance, if we know that (a, b) is a point on the graph of y === 2x 2 , then it follows
that b === 2a 2 , and conversely. Once the graph of a function has been sketched, it is easy to determine the domain and the range of the function, for, first of all, to determine whether or not a given number x belongs to the domain all we need do is look at the graph to decide whether or not there is a point on the graph having this x as xcoordinate. If and only if there is such a point does this x belong to the domain of the function. Similarly, to test a given number y for membership in the range, we ask whether or not there is a point on the graph with this number y as ycoordinate. If there is, then y belongs to the range; if not, then it does not belong to the range. The graph of a function may furnish the key to the solution of a problem in a very striking way, as the next example will indicate. EXAMPLE
36. Consider the function f(x)
=
x2
3

on the interval 4 < x < 3. What is an upper bound (see Definition 24) for the set A of values of f(x) assumed on this interval? We appeal to the graph of the function on the specified interval (see Fig. 36). It will be clear from this that 13 or any larger number is an upper bound for A. Note that the number 13 is not assumed by f(x) on the interval 4 < x < 3.
In all the examples of functions which we gave above, we had to deal with what is known as a singlevalued function, since in each case, if we select an element of the domain and apply the appropriate prescription to obtain the corresponding range element, we are led to exactly one such y
(4, 13)
(3, 6)
(0, 3) FIGURE
36
32]
THE NOTION OF A FUNCTION AND ITS GRAPH
79
element. Compare this with the situation where we have, for example, the function y == ±V4 x 2 • For x == 4, we obtain y == ±VW == ±2,vS. The point is this: for a given value of x, we obtain here more than one value of y. Such a function is called a multiplevalued function of x. We formalize these considerations in
+
DEFINITION 34. If to each element x of the domain of a function f(x) there corresponds exactly one value of f(x), then we say that f(x) is a singlevalued function of x. More precisely, y == f(x) is singlevalued if and only if x 1 == x 2 implies f(x 1 ) == f(x 2 ). Functions which are not singlevalued are called multiplevalued.
A quick glance at the graph of a function will usually reveal whether the function represented is singlevalued or not. For example, the graph shown in Fig. 37 is that of a multiplevalued function since, for the x indicated, there are two values of y, as is evidenced by the two points P and Q.* Thus far, we have considered only functions of one independent variable, but in many cases we have to deal with two or more independent variables. For example, the volume V of a right circular cylinder of radius r and altitude h is a function of both r and h:
The average or arithmetic mean m of a set of numbers x 1 , x 2 , x 3 , is a function of all of these n numbers:
m
==
X1
••• , Xn
+ X2 + · · · + Xn
~~~~~~~~
n
y
p
Q
FIGURE
37
* A word of caution: Our statement assumes a cartesian coordinate system. This same curve when placed in a socalled "polar coordinate" system is the graph of a singlevalued function (see Chapter 13).
80
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
The notation for functions of more than one variable is merely an extension of the first case of functions of one variable. Thus the volume V mentioned above may be denoted by V(r, h) == 7rr 2 h. Of course, the order in which the independent variables occur is usually of prime importance, as it is in this case with V(l, 2) == 27r, whereas V(2, 1) == 47r. Later we shall study geometric representations of functions of two variables (Chapter 19). An extremely important function of more than one variable is that function which assigns to a pair of points in the plane the distance between those points. To determine the functional relationship involved, we shall need the famous theorem of Pythagoras, which relates in a quantitative way the three sides of a right triangle. This theorem is proved synthetically (without the aid of numbers) in standard first courses in plane geometry. In Section 34 we offer a proof which is essentially analytic in nature.
33 Exercises 1. If f(x) f(x
+
1 x _ [l/(l
1 
+ x)],
find f(j), f((3), f(l/(3), f(x 
1),
1).
+
2
l1 _ u u, find F(1 ) , F(O ) , F (12 ), F( 1/b) . 2
2. If F(u)
=
2u 
3. If F(x)
=
1  3x , find F(i), F( 2/a), F(l) Sx _ 2
F(2), F(}).
+
1 4x . 2 _ x, find f(O), f(l/x), f(}), f(}); then 1f g(u) = u 2 3
4. If f(x) = compute f(g(x)).
5. If {3(x) 6. e(x) =
= 3
x
x2
x 
x

1, find {3(0), {3(3), {3(1), {3(x
+ ~~) x
2. Find e( }), e(l/w), e(2), e(4.1) 
7. Sketch the graph of the function ( 2,
y =
f (x)
~~, l lxl
8. If 'Y(y)
=
y
if O

!!_ 12, find ['Y(y + ~x~
*9. Sketch the graph of y
0 and d(P, Q) = 0 if and only if P = Q. (Appeal to Theorem 32 only.) (ii) d(P, Q) = d(Q, P). (Again use only Theorem 32.) (iii) d(P, Q) d(Q, R) > d(P, R). (Give a geometric argument here.) **18. Read Chapter 3 of E. T. Bell's Men of 1\1athematics, Simon and Schuster, 1937, and write an essay on the life of Rene Descartes.
+
+
t
+
37 The straight line. A considerable portion of elementary mathematics is directly or indirectly concerned with one of the following related problems: (i) Given a geometric configuration which is the graph of some function. Determine the function. Conversely: (ii) Given a function y == f(x), find the geometric configuration determined by its graph. ** Exercises marked ** require outside reading.
94
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
We shall now solve these problems for one of the simplest conceivable situations; we shall prove that if the graph of a function y is a straight line, then the function satisfies a relation of the form
ax
+ by + c ==
0,
where (x, y) are the coordinates of a point on the graph and a, b, care constants (with a == b == 0 ruled out) and, conversely, every equation of this form has a straight line as its graph. To begin with, if the graph is a vertical or horizontal line, its equation is of the form x == constant or y == constant, respectively. In any event, these equations have the form
ax
+ by + c ==
0. *
There remains the case where the line is neither vertical nor horizontal. Suppose, at first, that its graph is the straight line through the ongm shown in Fig. 320. From the similar triangles shown, we have
m y == , 1
x
or y
==
mx
as the required equation of the line. The number m is called the slope of the line. It is a measure of its "steepness." As a matter of fact, if one thinks of the xaxis as the horizontal and conceives of the line as showing the orientation of a staircase, one can, with the architect, write riser m ==   , tread y
P(x, y)
y
FIGURE
320
* For example, x = constant = k can be identified with ax by choosing a = 1, b = 0, c = k.
+ by + c
O
37]
95
THE STRAIGHT LINE
Horizontal FIGURE
321
where the meaning of "riser" and "tread" are the customary ones used in the design of stairs (see Fig. 321). The notion of slope of a line is so important to all that follows that we state the formal
35. By the slope of a line is meant the ratio of the difference of theycoordinates of two points on the line to the correspond1:ng difference in the xcoordinates: DEFINITION
where (x 1 , y 1 ) and (x 2 , y 2 ) are any two points on the line. Resuming the discussion, in the event that the line does not go through the origin but instead through (O, k), say, then the equation of the parallel line through the origin (shown in Fig. 322) is y === mx,
and since each ycoordinate on the first line is k units greater than the corresponding ycoordinate directly beneath it on the second line, it follows that y==mx+k (1) y y=mx+k I
I
lk
I
I I
Jmx I
FIGURE
322
y
= mx
96
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
is the equation sought. But if we choose a== m, b == 1, c == k, Eq. (1), namely mx  y + k == 0, becomes ax
+ by + c ==
0.
Therefore: Every straight line in the plane has an equation of the farm ax+ by+ c
== 0
where at least one of a, b is not zero. Since m is the slope of the line (1) and k is its yintercept [i.e., the line cuts the yaxis at the point (0, k)], the equation y==mx+k
is called the slopeintercept farm of the equation of a straight line. This canonical form, which gives y explicitly as a function of x, is of frequent use in the applications. EXAMPLE 314. A straight line has a slope of 3 and intercepts the yaxis at the point (O, 5). Find its equation. Obviously, k = 5, and the required equation is y = 3x  5. EXAMPLE 315. Find the equation of the line with slope the point (5, 7). The equation sought can be written
y
! and passing through
x
=
2 + k,
where k is yet to be determined. For this line to pass through the point (5, 7), the coordinates x = 5, y = 7 must satisfy the equation, i.e., 7 This implies that k
=
J.
=
!(5)
+ k.
Consequently 2y 
x 
9
0
is the required equation.
We have seen that every straight line in the plane has an equation of the form ax + by + c == 0. Let us now concentrate on the opposite problem: to show that every equation of the first degree in x and y: ax
+ by + c ==
0,
(2)
37]
THE STRAIGHT LINE
97
where not both of a, b are zero, represents a straight line. To prove this, we consider the only two possibilities:
(i) b
==
0,
(ii) b
~
0.
For (i), Eq. (2) reduces to ax+c==O
and since a ~ 0 because we are assuming that a can divide by a and write c X==, a
==
b
==
0 is false, we
which says x == constant and consequently is the equation of a vertical line. Finally, consider (2) with b ~ 0. We may then divide (2) by b to get a c y==bxb;
in other words, our equation is in the form (3)
y==mx+k
with m == (a/b), k == (c/b), and may be interpreted as the equation of the straight line with slope m and yintercept k. * Combined with the earlier result, we have succeeded in proving THEOREM 33. Every straight line in the plane has an equation of the form ax + by + c == 0 with a, b, not both zero and, conversely, every equation of the form ax by + c == 0 with a, b not both zero represents a straight line.
+
Remark. It is because of this result that we call an equation of the form ax + by + c == 0 linear in x and y. EXAMPLE
316. Sketch the graph of the equation 4x 
5y
+7
= 0.
According to the theorem above, this equation represents a straight line. We therefore need to locate only two points on this line in order to completely determine it. For this purpose, we will locate the intercepts of the line. These are the points where the line intersects the x and yaxes. To find the xintercept, we put y = 0, and get x = l Theyintercept is (0, !). The resulting graph is shown in Fig. 323. EXAMPLE
317. Sketch the graph of x 2

y 2 = 0.
* Note particularly from (3) that the slope of a straight line (whose equation is given) can be found by solving for y and picking out the coefficient of x.
98
THE PLANE. A FUNCTION AND ITS GRAPH y
[CHAP.
3
y
xy=O
x+y=O
(i, 0) FIGURE
323
FIGURE
324
We note that the left member of this equation admits of the factorization (x 
y)(x
+ y)
=
0.
Since a product is zero if and only if at least one of its factors is 0, we conclude that this equation is satisfied when either x  y = 0 or x + y = 0. Now, since the graph is, by definition, the set of all points which satisfy the equation and since each of these equations represents a straight line, it follows that the graph of x 2  y 2 = 0 consists of the two straight lines shown in Fig. 324. We can learn more by further study of this example. In particular, we learn that any equation which can be factored into linear factors so that the other member of the equation is 0, will have for its graph the set of all lines obtained by setting the factors individually equal to 0. EXAMPLE
318. Sketch the graph of x 2
+y
2
=
0.
We first observe that it is impossible to factor the left member of this equation as we did in the above example. Suppose that there is a number x different from O which satisfies this equation. Then its square, x 2 , will be positive and when this positive number is added to the nonnegative number y 2 we will surely get a positive number, not O as called for by the equation. It therefore follows that any point in the plane satisfying this equation must have x = 0. In the same manner, we argue that y = 0. Thus the only point whose coordinates satisfy this equation is the origin (0, O). This single point constitutes the graph of x 2 + y 2 = 0. EXAMPLE
319. Sketch the graph of x 2
+ y2
=
1.
Since x 2 and y 2 are both nonnegative, it follows that their sum will also be nonnegative. But the equation demands that this sum be negative. This being impossible with real numbers, the graph contains no points or, as we also say, the graph is empty.
Now consider the following problem: Given two points P 1 (x 1 , y 1 ) and P 2 (x 2 , y 2 ); find the equation of the straight line through these points.
37]
99
THE STRAIGHT LINE
y
P(x, y)
325
FIGURE
In the trivial instance where the points P 1 and P 2 fall on a vertical line, the equation is x == constant; similarly, y == constant handles the case of a horizontal line. Finally, when the line is neither horizontal nor vertical, we compute its slope m in two different ways: Y1  Y2 m == , X1 
X2
[using P 1 and P 2 (see Fig. 325)], and m ==
Y 
Y1
X 
X1
'
(using P and P 1 ) and then equate the results: m
or (4)
the required equation. It is called the pointslope form because the coordinates of a point (x 1 , y 1 ) and the slope m occur in it explicitly. 320. Find the equation of the line through (3, 5) and (4, 7). We compute the slope 7  5 12 Y2  YI · 4  (3) X2 Xl 7
EXAMPLE
The required equation is therefore y 
(7)
\l(x  4)
[using the point (4, 7)], or
7y
+ 12x + 1
0.
100
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
"\Ve leave it to the reader to verify that if we had used the point (3, 5) together with the slope \2, we would have obtained the same equation.
Let us return to the notion of slope defined above. By definition, riser Y1 s1ope ==   == tread x1 
Y2
,
X2
and this is a real number which may be positive, negative, or zero. We want to determine the geometric meaning of the sign of the slope. First of all, a zero slope means a zero fraction, i.e., 0 == riser . tread Kow a fraction is O if and only if its numerator is O and its denominator is not 0. Thus we have a riser of zero and a nonzero tread. This evidently is the slope of a horizontal line. Next we consider a positive slope. In this case, both numerator and denominator of the ratio riser/tread must have the same sign. Whether both are positive or negative, we get a straight line which slants upward to the right, so that as x increases, y increases. (See Fig. 326.) Now a function f(x) such that x 1 > x 2 implies f(x 1 ) > f(x 2 ) is called a monotonic increasing function (the graph of such a function necessarily rises as we move to the right, generating the curve). Let us then refer to a line y == mx + k with m > 0 simply as an increasing or 'inclined line. Kow for a negative slope. Here riser and tread have opposite signs, both possibilities being considered in Fig. 327. We see that as x increases, y decreases. A function f(x) such that for x 1 > x 2 we have f(x 1 ) < f(x 2 ) y
!}
t
Tread> 0
Tread< 0 Riser
f(x2) and therefore f(x) = mx k is a monotonic decreasing function of x. 22. Find the equation of the line parallel to 3x  5y 8 = 0 and passing through the point (2, 5). 23. The point (4, 2) bisects that portion of a line which is included between the coordinate axes. Find the equation of the line. 24. Find the equation of the line parallel to 4x  7 y 11 = 0 and through (1, 3).
+
+
+
+
+
39]
103
SYSTEMS OF LINES
*25. Have you ever thought of giving a precise definition of a straight line? In particular, how would you react if it were suggested that: A straight line is the set of all points (x, y) in a plane such that ax by + c = 0, where a, b, c, are fixed numbers and a 2 b2 > 0. 26. Sketch the locus of all points (x, y) for which
+
+
(x 
3y
+ 2)(x + y 
1)(5x
+ y)
=
0.
*27. Suppose that y  Yl = m(x  x1) is the equation of the line through P1(x1,,Y1) and P2(x2, Y2). Show that the same line has the equation Y 
Y2 = m(x 
x2).
28. Classify the following lines as increasing or decreasing: (a) 3x  7y
+8
=
0
(b) 3x
+ 7y + 8
=
0
(c) 3x 
7y
+8
=
0
29. Which of the graphs of Exercises 25 through 30, Section 33, are those of monotonic increasing functions? monotonic decreasing functions?
39 Systems of lines. lines:
A pair of lines.
Let us consider two straight
where the a's, b's, and e's are given numbers. slopes of these lines are different, i.e.,
If we assume that the
then the lines will intersect. We want to find their point of intersection, i.e., we seek a pair of numbers (x, y) satisfying both equations at the same time. For a point with coordinates satisfying both equations will belong to the graph of both and therefore will be a point of intersection of the graphs. We call such a point (if there is one) a simultaneous solution of the system of equations. We begin' by eliminating x; we multiply (1) by a 2 and (2) by a 1, and then subtract:
(3)
Now, in order that we may solve (3) for y, we must insist that the coefficient of y is not 0, thereby enabling division by this coefficient. But
104
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
this is precisely what the condition of inequality of the slopes guarantees, for if
then the difference of these fractions will not be 0, i.e.,
and if we multiply through by a 2 ai, we get a2b1
+ a1b2
~
0,
which is exactly the required condition for solving (3) for y. We have therefore succeeded in proving analytically that if two lines have unequal slopes, then they have a common point. This common point has coordinates which may be determined as follows: Solve (3) for y and then substitute in either (1) or (2). The point is
We leave it to the reader to show that these coordinates satisfy both of the original equations. Now suppose that the slopes of the given lines are the same, 1.e.,
The equations (1), (2) may then be written y == mx
C1 + b1'
(1)
(2)
There are only two alternatives; either or If case (i) holds, the lines (1) and (2) then coincide and hence all their points are common points. On the other hand, if (ii) prevails and we attempt to solve (1) and (2) simultaneously by subtracting, we get
SYSTEMS OF LINES
39]
105
which contradicts (ii). Therefore in case (ii) there is no point of intersection.* This completes the proof of THEOREM 34. Two straight lines
+ b1Y + c1 ( a2x + b2Y + c2 a1x
== 0, == 0
will intersect at the point
if they are not parallel, i.e., if
If, however,
then the lines are coincident in case
and noncoincident and nonintersecting if
We may write the theorem above in a more compact form and make it more easily recalled by introducing the notion of a determinant. This is a symbol representing a number defined by a
b
==
ad 
be.
c d Thus 5
2
1
6
0
8
5
7
5(6) 

40,
(1)(2) 
30+2 == 28,
and
Determinants were introduced by Leibniz and Cramer for the purpose
* Note the geometric interpretations: in case (i) the lines are coincident; in case (ii) they are parallel and distinct.
106
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
of solving systems of linear equations, although the simple notation above was not in use at that time. In the language of determinants, the theorem above becomes THEOREM
34. Cramer's Rule. The system of linear equations a1x
+ b1Y ==
a 2x
ci,
+ b2Y ==
c2
has the solution C1
x
a1
b1
== C2 b2
y
C1
== a2 C2
a1
b1
a1
b1
a2
b2
a2
b2
if the determinant of the system is not 0, i.e., if a1
b1
a2
b2
~ O.
If the determinant of the system is O and if in addition
C2
b2
then the lines are coincident and hence the system of equations has infinitely many solutions. If the determinant of the system is O and if C1
b1
C2
b2
~ O,
then the lines are parallel but different and therefore do not intersect, so that the system of equations has no solution. Remark. Note, in solving for x, how the numerator determinant may be obtained from the denominator determinant (the socalled determinant 0f the system) by replacing the coefficients of x by the ~corresponding constants appearing on the other side of the equation. A similar statement holds for y. EXAMPLE
321. Find the point of intersection of the lines 3x {4x 
* This is condition
7y
+ 12
0,
1
0.
3y 
(i) above in determinantal form.
39]
107
SYSTEMS OF LINES
We first arrange the equations so that the constants appear on the right: 3x 
7y
12,
{ 4x 
3y
1.
The determinant of this system is 3
7
4
3
19
9+ 28
~
0.
Hence we are assured of a point of intersection. The xcoordinate of this point is a fraction whose numerator and denominator are certain determinants. The denominator is the determinant calculated above; the numerator is the determinant obtained from the denominator determinant by replacing the coefficients of x (we are solving for x) by the corresponding constants. Thus
x
12
7
1
3
=
3
7
4
3
+7
36

19
43 19
Similarly,
y =
3
12
4
1
3
+ 48 19
19
The point of intersection is therefore
51 19
(ff, H).
A family of lines. From a given pair of lines, we can construct an infinite system or family of straight lines as follows: Let the given pair of (nonparallel) lines be J a 1x + b1Y + c1 == 0, \ a2x
+ b2y + c 2
== 0.
We multiply the left member of one of the equations by an arbitrary constant k and add this to the left member of the other equation, setting the result equal to O: (4)
Now each time we select a specific value (a real number) for k and substitute this in the equation above, there results a linear equation in x and y when like terms are combined:
108
THE PLANE. A FUNCTION AND ITS GRAPH
[CHAP.
3
Consequently, every such equation represents a straight line. A quantity like k which may be taken as an arbitrary constant is called a parameter. As k runs through the real numbers, (4) runs through a family of lines; thus we speak of (4) as being a oneparameter family of lines. We will now prove that the members of (4) obtained by selecting different values of k all have in common the property of passing through the point of intersection of the two lines with which we started. We do this as follows. Let P 0 (x 0 , y 0 ) be the point of intersection of the given lines. It follows that and since P 0 by its definition lies on each of these lines. But when we substitute the coordinates of P 0 in (4), we get [for the left member of (4)] (a1xo
+ b1Yo + c1) + k(a2xo + b2Yo + c2)
==
0
+ k(O)
==
0,
which means that (4) is satisfied by the coordinates of P O regardless of the value of k. This proves our assertion. 322. Fin. 17. The space is all positive integers; the relation is congruence modulo m. Let a, b, m be positive integers. Define a is congruent to b modulo m [written a= b(m)] to mean that a and b leave the same remainder when divided by the fixed positive integer m. *18. Redefine addition and multiplication of sets A, B by A+ B
=
(A 
B) U (B 
A),
AB= An B.
(Note that A+ B consists of the shaded area shown in Fig. 46.) With these definitions, which of the postulates for a field (Chapter 1) are satisfied?
132
SETS. ANALYTIC METHOD. INDUCTION PRINCIPLE
[CHAP.
4
**19. Read Chapter 23 of E.T. Bell's Men of Mathematics and write an essay on the life of George Boole. **20. Read Chapter 29 of Men of Mathematics and write an essay on the life of Georg Cantor.
45 Sets of points in the plane. In this section, we assume equations or inequalities (or both) to be given, and we raise the question as to what geometric configurations are thereby determined. Some examples will help to clarify the problem. 411. Determine the set of all points (x, y) in the plane for which 3 and x > 7. Another way of describing this problem is to set
EXAMPLE
y
7 is included. The small circle having its center at (7, 3) is used to indicate that its center is excluded from the required set of points. In Fig. 47, we have shaded the set An B = {(x, y)!Y < 3 and x > 7}. EXAMPLE
412. Describe the points in the plane such that 2x
+ 3y 
7
>
This inequality can be written Y
>
ix+ f.
y
Ix= 7
I I
I
FIGURE
47
0.
45]
133
SETS OF POINTS IN THE PLANE
y
FIGURE
48
Now, if we set YI
gx + i,
=
then YI is, for a given x, the ycoordinate of the point on the straight line (shown broken in Fig. 48) corresponding to this x. But we want those points which have ycoordinates (for a given x), greater than YI· Clearly, these are the points lying above the straight line. This problem can also be solved as follows. Since the coefficient of y in 2x 3y  7 = 0 is 3, points in the region above the line will assign a positive value to the linear expression 2x 3y  7 (Theorem 36). These are the desired points.
+
EXAMPLE
+
413. Determine the set of points in the plane for which (x 
1) 2
+ (y + 3) < 25 2
and
3x 
y
+ 5 < 0.
Let us call the first set A and the second one B. The problem then requires identifying A n B. We know that (x  1) 2 [y  (3)]2
+
is the square of the distance between the points (x, y) and (1, 3). Therefore the relation (x  1) 2 (y 3) 2 < 25
+
+
defines the set of all points (x, y) such that the square of the distance between (x, y) and (1, 3) is at most 25; this is the same as saying that the set of all points (x, y) such that the distance between (x, y) and (1, 3) is at most 5; that is, v'cx  1) 2 (y 3) 2 < 5.
+ +
These are, therefore, the points within or on the circumference of the circle with
134
SETS. ANALYTIC METHOD. INDUCTION PRINCIPLE
[CHAP.
4
y
y
I I I I I
13
++t
FIGURE
49
FIGURE
x
410
center at (1, 3) and radius 5. On the other hand, the set B of points such that
3x 
5
y 
3x 
5,
i.e., they are the points lying above or on the straight line y = 3x  5. Consequently, the points satisfying both conditions are those shaded in Fig. 49. We say that these points are characterized by the relations (x { 3x 
1) 2
y
+
(y+ 3) 2
4. Describe A n B geometrically. Recalling that lul < 2 is equivalent to 2 < u < 2, we obtain from the first inequality 2 < 3x  7 < 2 EXAMPLE
or
f
4 or y < 4 together with f < x < 3 are those shaded in Fig. 410.
47]
135
THE ANALYTIC METHOD y
FIGURE 411 EXAMPLE 415. Define A
=
{ (x,
y) I !xi
0.
Conversely, if a point P(x, y) satisfies all three of these conditions, it will fall within the shaded region. Hence the conditions (i), (ii), and (iii) are the required ones. 417. A point moves so that it is always twice as far from the point (3, 5) as it is from the line x = 1. Assuming all distances entering the problem to be positive, find the locus of this point. We begin by selecting a general point P(x, y) on the locus (see Fig. 413). Let d1 be its distance from the fixed point (3, 5) and let d2 be its (perpendicular) distance from the line x = 1. Then, according to the constraint described, 5) 2 and d2 = Ix  (l)j. There(y di/d2 = 2. But d1 = v'(x  3) 2 fore 2 2 5) = (y (x  3) 4 ' 1) 2 (x and finally, 14x  10y  30 = 0. 3x 2  y 2 EXAMPLE
+ + + +
+
+
(In Chapter 15 we shall prove that the locus is a curve known as a hyperbola.)
47]
137
THE ANALYTIC METHOD y
c
C(c, d)
~~x B(b, O) A(O, 0) FIGURE
414
FIGURE
415
Analytic methods may also be applied to prove geometric facts. Let us elaborate on this. The plane geometry developed by the Greeks and studied in high school is known as synthetic plane geometry. Virtually the entire theory is developed without the notion of number; that is, coordinate systems are not introduced in the plane to aid in development of the theory. On the other hand, in analytic plane geometry, the coordinate system is an essential tool. Both kinds of geometry offer advantages in the study of mathematics: sometimes it is simpler to obtain results by synthetic means; at other times, analytic methods are more feasible. We next prove by analytic methods a few theorems which are proved synthetically in the high schools. But before turning to this, let us add a few helpful words of a general nature. Imagine that the theorem in which we are interested concerns some plane figure which has been drawn. We are free to introduce the coordinate axes in any way we see fit and then to determine the coordinates of points of the figure in conformity with the introduced axes. It is natural in doing this to make the task as easy as possible. For example, suppose the theorem we are interested in concerns the triangle shown in Fig. 414. Axes can be introduced so that the origin appears at a vertex, say A, and one of the axes contains the side AB of the triangle, as shown in Fig. 415. The coordinates of B will then be of the form (b, 0) where b is an arbitrary positive number, while both coordinates of C will be arbitrary numbers, say (c, d), as shown in Fig. 415. The point is this: the axes are introduced so as to take advantage of the figure involved, but at the same time care is taken not to read more into the hypotheses than is warranted. Some examples will serve to bring out the ideas. 418. Prove that the perpendicular bisectors of the sides of a triangle meet at a point. We introduce the axes as shown in Fig. 416. The equation of the perpendicular bisector of OQ is c x = · (i) 2 EXAMPLE
138
SETS. ANALYTIC METHOD. INDUCTION PRINCIPLE
[CHAP.
4
y
P(a, b)
Q(c, O)
416
FIGURE
To find the equation of the perpendicular bisector of OP, we note that the slope of OP is b/ a and hence the perpendicular must have slope ( a/b). It also must pass through the midpoint of OP, i.e., the point (a/2, b/2). Its equation is therefore
~
y 
1 (
i) ·
x 
=
(ii)
Similarly, we find the equation of the perpendicular bisector of PQ to be
a(xatl
y~=c b
(iii)
Inserting (i) in (ii), we find 2
y
a 
2
ac+ b 2b
We now check to see if this value of y and x member of (iii) is
b y2
2
a 
2
ac+ b 2b
c/2 satisfy (iii). The left
=
2
a 
b 2
ac
2b
and the right member is
(c b a) (x _ a 2+ c)
c)
_ c  a(c a+   b  22ac+ a
2
2b
This proves that (i), (ii), and (iii) do meet at a point, namely, !:_ a ( 2'
2 
ac
+ b2 )
.
2b
This point C is the center of the circle through P, 0, Q. The reader will do well to check this by showing that d(P, C) = d(C, 0) = d(C, Q).
47]
139
THE ANALYTIC METHOD y
y
P(a, b) /
\
\
/ /
/ /
>/\
/ /
\ \
/ /
x
0
C(a, 0)
A(a, 0)
+ a, b)
/
\
P(xi, Y1)
x
M(c
/
/
N(c, 0)
FIGURE 418
FIGURE 417
EXAMPLE 419. Every angle inscribed in a semicircle is a right angle. We introduce the coordinate axes so that the circle with radius a, say, has its center at the origin (see Fig. 417). Let APC be the inscribed angle, then 2
2
2
(i)
XI+ YI = a ,
+
since the equation of the circle is x 2 y 2 = a 2 and the point P(xI, YI) lies on the circle. Now we compute the slopes of AP and PC and tak, the product: 0
YI 
0
(a)
XI 
a
YI XI 
2
YI xf 
(ii) o,2
But, according to (i), we have 2
YI
2
a 
2
XI,
and therefore the product of the slopes (ii) is 2
YI x2  a2 I
2
2
a 
XI
x2 
a2
I
1,
which means that AP and PC are perpendicular, q.e.d. EXAMPLE 420. The diagonals of a rhombus are perpendicular. We introduce the axes so that the origin is a vertex and the xaxis contains one side of the rhombus, as shown in Fig. 418. Let P(a, b), N(c, 0), and M (c a, b) be the remaining three vertices. The slope of P N is
+
b ac
and the slope of O.M is  b ·*
c+ a
+
* The slopes b/(a  c) and b/(c a) exist (i.e., the denominators do not vanish) since, as we see in (ii) below, c2  a 2 = (c a) (c  a) = b2 ~ O.
+
140
SETS. ANALYTIC METHOD. INDUCTION PRINCIPLE
[CHAP.
4
Hence the product of the slopes of the diagonals is
b2 a2 
(i)
c2
But our figure is a rhombus and hence all four sides are equal in length. particular,
In
OP= PM, i.e.,
va + b 2
2
=
c2 
c,
=
or
b2
a2.
(ii)
Substituting for b2 in (i), we obtain 2
2
c 
a
a2 
c2
1,
=
showing that the diagonals are perpendicular, q.e.d. EXAMPLE 421. The parallelogram law. In any parallelogram, the sum of the squares of the lengths of the diagonals is equal to twice the sum of the squares of the lengths of the sides.
We introduce the axes as shown in Fig. 419. The coordinates of P(a, b) are arbitrary positive numbers. The xcoordinate of Q(c, 0) is arbitrary. These determine the coordinates of R, for its xcoordinate must be c a, since the length of PR must be equal to the length of OQ, and its ycoordinate must be b, since PR is parallel to OQ. For the sum of the squares of the diagonals we get
+
+ a) + b + (c 2
2
(c
a) 2
+ 2a 2 + 2b 2 ,
2c 2
but the sum of the squares of the lengths of the sides is c2
+ a2 + b2,
and so the theorem is proved. y P(a, b)
R(c ~
'
"' )
0, a
so that we could just as well write az1
+ f3z2,
a
>
0, {3
+ {3 ==
1.
The resulting vector is called a convex linear combination of z 1 and z 2 • We call attention again to the fact that as we allow a and {3 to vary, always governed by the conditions of nonnegativeness and unit sum: a
>
0, {3
>
0, a
+ {3 ==
1,
we get different convex linear combinations of z 1 and z2 • As a matter of fact, if we examine Fig. 518 and the proof of the theorem above, we see that as a varies from 1 to O (and hence {3 varies from Oto 1), the endpoints of the corresponding convex linear combinations of z 1 and z 2 trace out the segment joining these points, starting at z 1 . Using the terminology introduced here, the above theorem becomes THEOREM 59. The locus of the line segment joining the ends of the vectors z 1 and z 2 is the set of all convex linear combinations of z 1 and z 2 • 513. The line segment joining the points (1, 3) and (5, 6) in Fig. 519 is the set of all convex linear combinations of the vectors 1 3i and 5 6i, that is, the set of all vectors of the form EXAMPLE
+
+
a(l
+ 3i) + {3(5 + 6i), a >
0, {3
y
FIGURE
519
>
0, a+ {3
=
1,
180
[CHAP. 5
VECTORS. COMPLEX NUMBERS
or, alternatively, (a
+ 5/3) + 3(a + 2{3)i, a >
0, {3
>
0, a
+ {3
=
1.
The meaning of this last description is as follows: Each time we select a pair of numbers a and {3 subject to the conditions that they are nonnegative and their sum is 1, we obtain a point of the segment joining (1, 3) and (5, 6); furthermore, if we exhaust all possible pairs of values for a and {3, we obtain all the points of the segment. In particular, if we take a = !, {3 = !, we obtain the point with xcoordinate 3 and ycoordinate J, which is the midpoint of the segment. Note also that when a = 1 and {3 = 0, we get the point (1, 3), whereas a = 0 and {3 = 1 leads to the point (5, 6). EXAMPLE 514. Find the coordinates of the point which is on the segment joining (1, 3) and (5, 6) and is ! of the way from the first point to the second one. The required point is the endpoint of the vector
+ 3i) + ![(5 + 6i)  (1 + 3i)], vector goes from 1 + 3i to 5 + 6i; the (1
since the bracketed The required point is (4,
i).
result is 4
+ ii. 2
2
*510 The locus of a triangle. In generalization of the problem of the preceding section, we consider the following problem: Given three noncollinear points z 1 , z 2 , z 3 • Find the locus of the triangle having these points as vertices. Let z be a point of the triangle (Fig. 520). Then, from the figure, we see that z is a convex linear combination of z 1 and w, since it is on the segment joining z 1 and w; hence, for certain a, {3, z == az1
+ {3w,
a
>
0, {3
>
y
zl ~  
FIGURE 520
0, a
+ {3
==
1.
(1)
510]
181
THE LOCUS OF A TRIANGLE
But for a similar reason, w, in turn, is a convex linear combination of z2 and z3 : (2) w == AZ 2 + µz3, X > 0, µ > 0, A + µ == 1. Therefore, substituting (2) in (1), we have z
+ /3(A.z2 + µz3) az1 + /3A.Z2 + /3µz3.
==
az1
Clearly, the coefficients of z 1 , z 2 , z 3 appearing here are nonnegative, since a, /3, A, µ are nonnegative. Furthermore, their sum is a
+ {3A. + /3µ
==
+ {3(A + µ) a + {3 (since A + µ a
1
(since a
+ /3
== 1)
== 1).
Consequently, if we define a convex linear combination of three vectors z 1 , z2 , z3 to be any vector v of the form
v == az1
+ /3z2 + VZ3,
a
> 0,
{3
> 0,
'Y
> 0,
a
+ /3 + 'Y ==
1,
then we have proved that every point of the triangle with vertices z 1 , z 2 , z 3 is a convex linear combination of these vertices. But the converse of this is also true. The proof is as follows. Take a convex linear combination of z 1 , z 2 , z 3 , say z
==
az1
+ f3z2 + 'Yz3,
a
> 0,
{3
> 0,
'Y
> 0,
a
+ {3 + 'Y ==
1.
We are to show that the vector z corresponds to a point of the triangle determined by the vertices z 1 , z 2 , z 3 . We write z
==
z
=
+ f3z2 + 'Yz3, az1 + ({3 + 'Y) rf3z2 + 'Yz3]. L f3 + 'Y
az1
(3)
The vector in brackets is a convex linear combination of z 2 and z 3 , for each of the scalars /3/({3 'Y) and 'Y /(/3 'Y) is nonnegative and their sum 1s
+
+
but a convex linear combination of z 2 and z 3 is a vector having its endpoint on that side of the triangle which joins the vertices z 2 and z 3 . Thus (3) says that z is a convex linear combination of the vertex z 1 and a vector
182
VECTORS. COMPLEX NUMBERS
[CHAP.
5
terminating on the side determined by z 2 and z 3 . Therefore z is a point of the triangle determined by z 1 , z2 , z3 . We have consequently proved THEOREM 510. The triangle having z1, z 2 , z 3 as vertices is the locus of all convex linear combinations of z 1 , z 2 , z 3 , that is, the set of all vectors of the form az1
+ f3z2 + 'Yz3,
a
>
0, {3
>
0,
'Y
>
0, a+ {3
+ 'Y ==
1.
We have been speaking of convex linear combinations of vectors. There is a connection between this concept and that of a convex set. In elementary geometry, a convex set C is defined as a set of points with the property that whenever the points P, Q belong to C, so do all the points of the segment joining P and Q. In view of Theorem 59, we may replace this geometric definition by an analytic one if we consider a point of a set to be determined by the vector beginning at the origin and terminating at the point. In this way, we arrive at DEFINITION 510. A set C of points is called convex if with every pair of points belonging to C there also belong all the convex linear combinations of these points.
*511 Exercises
+
1. (a) What is the locus of the line segment joining the points 1 3i and 1 i? (b) What is the locus of the points interior to the triangle whose vertices are ( 3  i), 5i, (2  i)? 2. Define A to be the set of all points z such that lz 31 < 2 and let B denote the set of all points z such that lz 1  ii < 1. Shade the set of points A n B, A u B, A', B', (A u B)', A' n B', (A n B)', A' u B'. 3. Prove that and jim (z)I < lzl, IRe (z)I < lzl
+
+
+
where Re (z), Im (z) denote the real and imaginary parts, respectively, of z. 4. What points of the plane satisfy 5
< lz  31
9. 12.
1 1 1
In Exercises 15 through 21, given that f(z)
+ +
15. f(3 5i) 18. IJC3 i)  JC3 19. f(lzl)
16. f(i) i)I 20. f(z)  f(z)
lz  zol < lzl > 1
r
z/(z  i), find: 17.
lf(2)l
21. f(z)f(z)
*22. Does it make sense to speak of z < w, where z, w are complex numbers? Can you write a definition for z < w so that the properties of < introduced for the reals (Section 25) hold? *23. Prove that if for z = x yi, w = u vi, we define:
+
d(z, w)
=
+
Ix  ul + IY  vi,
then d(z, w) satisfies the conditions required of a distance or metric (Section 58). (The metric defined here is the distance between two towns at z, w when only highways running eastwest and northsouth are available.) *24. Using Definition 510, show that the set of all points z such that Izl < 1 is a con vex set.
CHAPTER 6
THE NOTION OF A LIMIT Introduction. The idea of a limit is a challenging, intriguing concept. It is the root idea of the calculus and, more recently and generally, that branch of mathematics known as topology. The notion of a limit, although "inherent" in mathematical thinking (it was used by Archimedes in his "method of exhaustion" as early as about 250 years B.C.), was not even precise at the time the calculus was developed by Newton and Leibniz (late 17th century). It was only at the hands of A. Cauchy and K. Weierstrass (mid 19th century) that the crystallized, precise concept of limit emerged as we know it today. Thus it had taken over twenty centuries to capture a definition of "limit" meeting the present standards of rigor in mathematics! In our approach, the key to understanding this concept is to be found in considering the operation of "passage to the limit" to be an operation performed on what is known as a sequence (a special kind of function). Consequently, we first define a sequence, study two special and simple instances of sequences, then take up the idea of limit of a sequence and prove the basic theorems on limits. After these things have been accomplished, it is a simple matter to extend the notion of limit to arbitrary functions with the real numbers as domain. 61 The notion of a sequence. In counting, we use the sequence of positive integers: 1, 2, 3, ... , n, .... In counting by threes, we use the sequence of positive multiples of 3: 3, 6, 9, ... , 3n, .... In arithmetic, the sequence of prime numbers is of great importance: 2, 3, 5, 7, 11, 13, 17, 19, .... In general, a sequence may be described as a rule which associates with every positive integer a number; the positive integer specifies the number of the term of the sequence (first, second, third, etc.) and the associated number is called the value of this term. For instance, the value of the 3rd term in the sequence of primes is 5. More precisely, the unifying notion of a sequence is embodied in 184
61]
THE NOTION OF A SEQUENCE
185
61. A sequence is a singlevalued function having the positive integers as its domain.
DEFINITION
This is a very concise but elusive definition. Remember that a function involves the fusing of three ideas: a domain, a range, and a law or correspondence. For the special function which we call a sequence, the domain is fixed: it is the set of positive integers. As for the range, it is perfectly arbitrary, although in most instances of interest to us here, the range will consist of numbers. In each of our examples above, members of the ranges of the various functions are displayed on a horizontal line. The laws of correspondence are, in order: Positive integers: f(n) == n, Positive multiples of 3: f(n) == 3n. In the case of a sequence, special function notation 1s introduced. We write the terms of a sequence as
where the subscripts serve as the independent variables (for short, we sometimes write simply {an}). It is customary to use the subscripts instead of writing a(l), a(2), a(3), ... , a(n), ... , which would be the notation we would use if we carried the usual function notation over to the special case of a sequence. Thus it is conventional to write the sequences above in the form Positive integers: {an} == {n}, Positive multiples of three: {an} == {3n}. In the case of the prime numbers, no formula for securing the nth prime number, for arbitrary n, has been found to date. In spite of this, however, we know that there are an infinite number of primes, and we still speak of the sequence of prime numbers.* We now turn our attention to two special kinds of sequences for which the law of formation is particularly simple, i.e., the socalled arithmeti"c and geometric sequences, or arithmetic and geometric progressions.
* Actually, the method of the "sieve of Eratosthenes'' can be applied, at least in principle, to determine the nth prime number for any specific value of n. (For an explanation, consult Introduction to Number Theory by T. N agell, p. 51, John Wiley and Sons, Inc.)
186
THE NOTION OF
62 The arithmetic sequence.
LIMIT
A
[CHAP.
6
Consider the sequence
7, 11, 15, 19, 23, ...
We can pass from any term to its successor by simply adding 4. Conversely, to pass from any term to its predecessor we subtract 4. In general, we have DEFIXITIO~ 62. An arithmetic sequence (abbreviated A.S.) is a sequence of the form a, a d, a 2d, a 3d, ... ,
+
+
+
where a and d are fixed numbers. We call a the first term and d the common dijference. Remark. The name "common difference" is easily accounted for; it is the difference which is common to pairs of adjacent terms of the sequence.
Having been given an arithmetic sequence, we raise two questions: (i) What is the nth term of the sequence? (ii) How does one calculate the sum of the first n terms of the sequence? We note that in an A.S., the
+ _Q • d, ind term is a + 1 · d, Qrd term is a + i · d,
1st term is a
so evidently the ~th term is a
+ (n

1) · d,
and this is the answer to the first question. As for the second question, an interesting historical anecdote is brought to mind. There is a story to the effect that Karl F. Gauss, "the prince of mathematicians," who lived in the 19th century, amazed his teacher at a rather tender age.* It is alleged that Gauss' instructor, determined to impress his students with the power of arithmetic sequence theory before developing the theory, assigned a problem which is quite timeconsuming for an ordinary person with no knowledge of sequences. The problem he posed was essentially of the type "Determine the sum of the first 100 positive integers." Young Gauss attacked the problem as follows. He
* See E. T. Bell's Men of Mathematics, Simon and Schuster, New York. This work contains many other interesting anecdotes in the lives of famous mathematicians.
62]
187
THE ARITHMETIC SEQUENCE
let S 100 denote the required sum and then wrote the summands in reverse order: I\ I\ I\ I\ 8100 == I 1 \ + I 2 \ + I 3 \ + · · · + 1 100 \ . (a)
\100} + \ 99 j + \ 98 J + · · · + \ 1
8100 =
j
({3)
Next, he added equations (a) and ({3) ("columnwise," as indicated by the broken line parentheses), and so obtained twice the required sum on the left. Noting that 101 appeared as a term on the right 100 times, he then wrote 28100
==
100(101),
8100
== 5050,
the required number. With such a clever device, Gauss naturally had the answer at his fingertips, so he wrote it on his slate and placed the slate on schoolmaster Buttner's table. Buttner's first impression was that his new student Gauss must be an idiot! An interesting aspect in the present connection is that Gauss' method can be applied directly to the general problem of determining the sum of the first n terms of an A.S. With Gauss, we write Sn == a
+
(a
+ d) +
(a
+ 2d) + · · · + an,
(1)
where an simply denotes the nth term, that is, an == a + (n  l)d. Next, keeping in mind that the predecessor of an is an  d, and that the predecessor of an  d is an  2d, etc., we write the terms in reverse order as follows:
== an
Sn
+
(an 
d)
+
(an 
2d)
+ · · · + a.
(2)
We then add (1) and (2) ("columnwise"): 2Sn
which yields a terms in Sn:
==. (a+ an)
+ an
+ (a+ an) + · · · +
+ an),
as addend the same number of times as there are
or Sn
==
(a+ an) · n. 2
We have, then, the following THEOREM 61.
(a
The arithmetic sequence (A.S.) a, a
+ d, a + 2d, ...
has as its nth term an the number an
== a
+
(n 
l)d,
188
THE KOTIO~ OF A LIMIT
[CHAP.
6
and the sum Sn of the first n terms is Sn
=
a
+ (a + d) + · · · + an =
C~ an) n.
Remark 1. In words, the sum Sn of the first n terms of an A.S. is the average, (a an) /2, of the first and the nth terms multiplied by the
+
number n of terms. Remark 2. This theorem asserts that the formulas for an and Sn hold for all positive integers n; these formulas should also be verified by the principle of finite induction. To economize on writing, we introduce the definition n
L
ak == a1
+ a2 + · · · + an,
k=l
thus Lk=l ak means to form all of the terms obtainable by putting k successively equal to 1, 2, continued through n, and then add all these numbers (I: is the Greek equivalent of our "S"taken from the word "sum"). EXAMPLE 61. To find the sum S of the first 100 even numbers, we note that we are concerned with the sum of the first.100 terms of the A.S. 2, 4, 6, ... , in which a = 2 and d = 2. The lOOth term is 2 (100  1)2 = 2 198 = 200. Thus S = 2 200. We then calculate the average of the 1st and lOOth terms, which is (2 200)/2 = 101, and multiply this by the number of terms, to obtain S = (101) (100) = 10, 100.
+
+ ·· · + +
EXAMPLE 6
L Pk
=
62.
If Pk
=
jk  3j, find
I:f=l
+
Pk.
11  31 + 12  31 + 13  31 + 14  31 + 1s  31 + 16  31
= 2 + 1 + 0 + 1 + 2 + 3 = 9. 63 The geometric sequence. If a sequence has the property that to pass from any term to its successor one must multiply the term by a fixed
number (instead of adding as for an A.S.), then that sequence is called a geometric sequence. We formulate this as DEFINITION 63. A geometric sequence (abbreviated G.S.) or geometric progression is a sequence of the form
a, ar, ar 2 , ar 3 , ... ,
where a and r are fixed numbers. common ratio.
We call a the first term and r the
THE GEOMETRIC SEQUENCE
63]
189
Remark. The name "common ratio" is accounted for by the fact that if we take any two adjacent terms of the G.S. and divide the later one by the earlier one, we obtain the common ratio. Thus ar 3 ar2
===
ar 8 ar7
===
r,
etc.
Again we raise the questions: (i) What i_s the nth term of the sequence? (ii) What is the sum of the first n terms of the sequence? Examining the sequence, we see that the 1st term is arq, ind term is ar!, Qrd term is ar~, and, in general, the nth term is arnI. Next consider the problem of calculating the sum of the first n terms:
a
+ ar + ar 2 + · · · + arn 1•
Sn
===
Sn
=== a(l
We write
=== a [
+ r + r 2 + r 3 + · · · + rnI)
1 1 
rn] r '
as is easily seen by factoring: 1  rn === (1  r)(l Therefore we have
+ r + r 2 + ... + rn 1 ).
THEOREM 62. The geometric sequence (G.S.)
a, ar, ar 2 , ... has as its nth term an the number
an
===
ar
n1
'
and the sum Sn of the first n terms is n1
Sn
===
I: ark === a + ar + ar 2 + · · · + arnI
k=O
a  arn 1  r
Remark 1. In words, the sum Sn of the first n terms of a G.S. is the number obtained by writing down the first term, subtracting from it the
190
THE NOTION OF A LIMIT
[CHAP.
6
term which follows the last one included in the sum, then dividing by 1 minus the common ratio. Remark 2. As in the case of the previous theorem, this result may also be verified by induction on n. EXAMPLE
63.
To calculate the sum of the first twelve terms of the G.S. 3 3 3 3, 2, 4, s, · · ·,
observe th~t a = 3, r = !, and so the 12th term is 3( !) 12 term which follows this is 3(!) 12 • Hence
1
3(!) 11 . The
11
S
=
I:3(!)k
3+!+···+3(!)
=
11
k=O
3  3(!) 1  !
12
6141 1024
64. Three numbers whose sum is 15 form three terms of a G.S. If 3 is subtracted from the first, 3 added to the second, and 36 subtracted from the third, the resulting numbers form an A.S. Find the original numbers. Let x, y, z be the required numbers. Then EXAMPLE
x
+y+z =
(I)
15
and y

z
=
x
(2)
,
y
since x, y, z form a G.S. Also,
x  3, y + 3, z  36 form an A.S., and consequently (y + 3) 
(x 
3)
(z 
=
36) 
(y + 3).
(3)
Now (1), (2), (3) constitute three equations to be solved simultaneously. They may be written: (i) x y z 15,
+ +
y2
(ii)
(iii) 2y From (i), x + z
15 
x 
z
xz, 45.
y; substituting this in (iii), we get
2y 
(15 
y) = 45
or y
=
10.
64]
191
EXERCISES
Substituting in (i) and (ii), we have
x {
+z
25,
100
xz.
Solving the linear equation for x and substituting in the second equation, we find or z = 20. z = 5 Returning to (ii), we find the solutions: x
y
z
20
10
5
5
10
20
If the equations (i), (ii), (iii) are studied carefully, it will be noticed that x and z enter them symmetrically, i.e., if x and z are interchanged throughout, the system of equations will be unaltered. Note how this symmetry in x and z is reflected in the table of solutions.
64 Exercises 1. The first term of an A.S. is 1 and its 7th term is 3. Find the sum of 9 terms, beginning with the 4th. 2. Verify the formulas for an and Sn for (a) arithmetic sequences and (b) geometric sequences, using the principle of finite induction. 3. Find the 37th term of an A.S. whose 4th term is 6 and whose common difference is 5. 4. An uncle places a sum of money in a savings account for a nephew when he is born. On each succeeding birthday, the uncle deposits $10 more than on the preceding birthday. The total for the first 21 deposits is $3675. How large was the first deposit? 5. A man saves money toward retirement. The first year he saves $500 and each subsequent year he saves $50 more than the previous year. (a) How much does he save during the 30th year? (b) How much will he have saved altogether over a period of 30 years? 6. Find x, y such that the numbers 2, x, y, 9 will have the property that the first three are terms of an A .S. and the last three belong to a G .S. 7. If am = a 1 +2 4 +S···  1024 , find m(assume a~ 1, a> 0). 8. The 4th term of a G.S. is 6 and the 6th term is 24. Find the 1st term and the ratio. 9. Approximate the sum
s
=
(1.035) 1
+
(1.035) 2
+ ... +
(1.035) 10 .
192
THE NOTION OF A LIMIT
[CHAP.
6
10. Find the 53rd term of the arithmetic sequence 7, 10, 13, ... 11. Find the first four terms of the A.S. whose 2nd term is 16 and whose common difference is 5. 12. Find S if S = 2  5  12  19  26  · · ·  103. 13. Prove Theorems 61 and 62 for the situation where a, d, r are complex numbers. In Exercises 14 through 21, define an = in, where i 2 1. Find: 14. a7 18. a1
16. a4n
15. a2n
+ a2 + · · · + a15
L lak 
+ la2I + · · · + lanl 2 2 2 a1 + a2 + · · · + a5
19. la1l
7
20.
17. a123
21.
ak+ll
k=l
In Exercises 22 through 27, let 81 be the set of all points (x, y) such that x > 3  1, 82 those for which x > 3  !, and, in general, Sn those for which x > 3  (1/n). Shade the associated sets of points: 22. A3 = 81 u 82 u 83 24. An = 81 U 82 U · · · U Sn 26. A = 81 U 82 U · · · U Sn U · · ·
23. M3 = 81 n 82 n 83 25. Mn = 81 n 82 n · · · n Sn 27. M = 81 n 82 n · · · n Sn n
···
+
28. The first term of a G.S. is 1 i and its ratio is i. Find (a) the 12th term, (b) the nth term, (c) the sum of the first 12 terms, and (d) the sum of the first n terms. 29. If maxk Pk denotes the maximum of the numbers Pl, ... , Pn, where k assumes all the values 1, 2, ... , n for which Pk has been defined, and if
Pk
=
jk 
3j, k
=
1, ... , 6,
find maxk Pk· Also define and find mink Pk· Compute the sums in Exercises 30 through 34. 5
400
3o.
L (2k 
1)
31.
k=l
L
11 
12
3kl
k=O
7
34.
k~5
1)
k=l
4
32.
L (2k 
33.
k
L1k1
k=l
k
!kl' k ~ 0
*35. Denote by {x} the distance between x and the nearest integer and define
65]
193
INFINITE GEOMETRIC SERIES
Find: (a) f(0.1), (b) f(l), (c) f(0.001), (d) f(0.1315), (e) f(0.762), and
'°" 1
k
o o {10 x} (f) the graphs of {10 x} /10 ; {lOx} /10; L...J lQk
·
k=O
36. Stars are sometimes classified according to their brightness, as follows: a star of magnitude 1 is 1o0 .4 times as bright as a star of magnitude 2; a star of magnitude 2 is 10°.4 times as bright as one of magnitude 3, etc. Show that a firstmagnitude star is 100 times as bright as a sixthmagnitude star. 37. Prove, by mathematical induction on n, that for all positive integers n, 1  rn
=
(1 
r)(l
+ r + r2 + · · · + rn 1 ).
*38. Find all sequences which are simultaneously arithmetic and geometric sequences. [Hint: If a, b, c are the first three terms of such a sequence, then b  a = c  b and b/a = c/b.] **39. Read Chapter 14 of E.T. Bell's Men of Mathematics and write a report on the life of K. Gauss.
65 Infinite geometric series. We have proved that the sum Sn of the first n terms of the geometric sequence a, ar, ... is given by the formula
Sn == a
+ ar + ar 2 + · · · + arn1
==
a  arn 1 
r
·
Another way of writing this is to split Sn into two terms: n
S _ _!!:__ _ ar n1r 1r
(1)
We do this to facilitate our study of how this sum Sn varies as we include a larger number n of terms in the sum. In (1), it is seen that since we are assuming a G.S. to be given (and therefore a and r are fixed), the first term, a/(1  r), is independent of n and therefore its effect on Sn may be ignored for the time being. The second term, however, is a function of n; as a matter of fact, it is the first term multiplied by rn. This suggests that we concentrate on the behavior of rn as n takes on larger and larger values. There are two cases, according as the numerical value of r exceeds or does not exceed 1, exemplified by the values r == 2 and r == !. In the first instance, as n grows larger rn == 2n, that is, 2, 2 2 , 2 3 , 24, ... increases without bound, whereas in the second case rn == (!r; that is, !, (!) 2 , ( !) 3, (!) 4, ... approaches O as n increases. Corresponding to these two situations, a study of (1) reveals that Sn increases numerically without bound in the first case and Sn approaches a/(1  r) in the second case.
194
THE NOTION OF
>1 jrj < l jrj > 1 ~~~
A
LIMIT
[CHAP.
6
Jrl
ti)
1
I 0
FIGURE
ti)
•
.. x
0
1
61
FIGURE
62
In general, if lrl < 1 (Fig. 61), rn approach~s O as n increases and therefore Sn tends to a/(1  r). We express this by writing [cf. (1)] a lim Sn = 1  r n~oo
where "limn~oo Sn" is read "limit of Sn as n tends to infinity." On the other hand, if lrl > 1, then as n tends to infinity rn will increase numerically without bound, and we say lim Sn does not exist. n~oo An indicated sum of all the terms of a geometric sequence, written formally as a+ ar + ar 2 + · · · is called an infinite geometric series. The sum S of this infinite geometric series is, by definition, the limit, when it exists, of the sum Sn of its first n terms as n becomes infinite. Using this terminology, we have proved THEOREM 63. The infinite geometric series 00
I: ark == a + ar + ar
2
+ ···
k=O
has no sum if lrl
>
1, but has sum
S == lim Sn== __a_ n~oo 1  r if lrl
N. We a]so say that the sequence converges to a. DEFINITION
Remark 1. Observe that we do not insist that an be different from a. Remark 2. We emphasize the equivalence of the definitions 64 and
65 just offered, and shall use repeatedly the fact that it is therefore permissible to pass from one to the other. The reader too, of course, has the privilege of passing from the less accurate but suggestive language of Definition 64 to the mathematical precision afforded by Definition 65, and vice versa, as he sees fit. EXAMPLE 68. A rather trivial, but nevertheless important situation is that in which the sequence {an} is a constant sequence; that is, an = c for all n, where c is a fixed number. Such a sequence always has a limit. For, according to the definitions, c is the limit of the sequence c, c, c, ... since no matter what interval is selected with c at its center all the terms of this sequence are to be found in such an interval. EXAMPLE
69.
Consider the sequence of positive integers an a1
= 1,
a2
n:
= 2, ...
Intuitively, we say that this sequence is "headed for no particular point." In precise terms, this sequence has no limit, for there obviously is no fixed point each of whose intervals contains all the terms of the sequence from some index on, even though the index is allowed to vary with the interval.
198
THE NOTIO~ OF A LIMIT
(
f
I 0
[CHAP.
6
.. :.r
FIGURE
63
Examples 68 and 69 present all possibilities with regard to the limit of a sequence, i.e., a given sequence either has precisely one limit or none at all. Before proving this, however, let us look at several more instructive examples. EXAMPLE 610. The sequence an = l/n has O as its limit. For, given an interval with O at its center and total length 2€ (see Fig. 63), if we go far enough in the sequence all the terms from a certain place on will fall inside the given interval. As a matter of fact, hr an arbitrarily given E > 0, there is N such that (1/N) < € and hence if n > N we will have (1/n) < E. EXAMPLE
611.
The sequence an
=
(l)n:
1, 1, 1, ...
has no limit. This is intuitively clear, for this sequence "hops" back and forth from 1 to 1 and hence is "headed nowhere." Let us be more precise. First of all, it is easy to see that any number x different from+ 1 cannot serve as a limit, since such a number x can be enclosed in an interval which contains none of the terms of the given sequence. Now, 1 is not the limit of the sequence since, no matter how far we progress in the sequence, among the remaining terms are l's which do not belong to small intervals centered at 1. A similar argument shows that 1 is not the limit of the sequence. Therefore the sequence has no limit. EXAMPLE 612. The sequence an = n/(n + 1) has 1 as limit. For, on dividing the numerator and denominator of an by n, we have
1
an
=
1
+ (1/n) '
and it is intuitively clear that as n ~ oo, an ~ 1 (read this "as n tends to infinity, an tends to 1 ") . As for a formal proof, WP must show that the distance Ian  11 between an and 1 can be made small by choosing n large enough. Since
it is obvious that one can choose n sufficiently large so as to make 1/(n + 1) as small as one pleases. More precisely, let us calculate how large n must be in order that the distance Ian  11 will be smaller than, say, E = 0.01. We want
Ian  11
=
1
n=t1
199
100,
i.e.,
n
>
99.
Hence,
n
>
99
implies
Ian 
II
0, we want to
that is, 1
n+l>€
or
n
1
>  
l.
€
This means that when € > 0 is given, we compute its reciprocal, subtract 1, and then determine any index (integer) exceeding this. For this index and all succeeding ones, we will have Ian  1 j < €. In short: lim an = lim nH~
nHlO
_+n 1 n
= 1.
In our definition of limit of a sequence, we referred to the limit of a sequence. We have, accordingly, hinted as to the uniqueness of the limit, when it exists. Nevertheless, might there not be two numbers having the properties required of a limit of a sequence? Indeed not, as we now prove. THEOREM 64. If the sequence {an} has a limit, it has only one. Proof. The proof is by contradiction. Suppose there were a sequence
with two distinct limits, call them a, a'. Then since a ~ a', the distance la  a'I is positive (see Fig. 64) and so we can center intervals about a and a' which will not overlap. Now since a is a limit of the sequence, all the terms with index following some index, say N, will be found in the interval about a. On the other hand, since a' is a limit of the sequence, all the terms with index following some index, say N', will be found in the interval about a'. But what about the terms with index following both N and N'? They would have to belong to both of the disjoint intervals, which is impossible. This completes the proof.
1.[a ~ a'[1 ~
) FIGURE
( ~/ )
64
.. x
200
THE NOTION OF A LIMIT
[CHAP.
6
68 Exercises Determine the limits (if they exist) of the sequences defined in Exercises 1 through 10.
+
+
1. The arithmetic sequence a, a d, a 2d, ... 2 2. The geometric sequence a, ar, ar , ... 3. a1 0.1, a2 = 0.101, a3 = 0.101001, a4 = 0.1010010001, ... 4. a1 = 0.52, a2 = 0.52719, a3 = 0.52719719, a4 = 0.527197197, ... 1 1 1 1 1 5. a1 = , a2 = ~ , a3 = ~ , a4 = ~ , · · · an = n , · · · 3 3 3 3 3 1 2 6. 2°, 2 , 2 , ••• , 2n, .. . 7. (!)o, (!)1, (!)2, ... , (!)n, ... 8. (!)o, (!)1, (!)2, ... ' (!r, ...
9. an=
Hf (n~
1) 10. bn = {an}, where an is defined in Exercise 4 and {an} means the distance between an and the nearest integer. 11. Give an example of a sequence of rational numbers with an irrational number as limit. 12. Give an example of a sequence of irrational numbers with a rational number as limit. *13. Let {an} be a sequence of complex numbers (points in the plane). Write the definition of limit of {an} by replacing the word interval in Definition 64 by disc, interpreting "disc" here as meaning the interior of a circle. *14. Using the definitions set up in Exercise 13, show that Theorem 64 is valid when {an} is a sequence of complex numbers. **15. Read Chapter 22 of E. T. Bell's Men of Mathematics and write a report on the life of K. Weierstrass. **16. Read the article entitled "Topology" by R. Courant and H. Robbins in Volume 1 of the fourvolume set The World of Mathematics, edited by J. Newman, Simon and Schuster Co., 1956, p. 581.
69 Basic theorems on limits of sequences. Of fundamental importance to further development of the concept of a limit is the study of the interplay between the arithmetic processes, on the one hand, and the limit process, on the other. In such an investigation, a fundamental role is played by the "triangle inequality" proved in Chapter 2, namely,
la  bl+ lb 
cl
> la  cj.
The reason for this follows from the geometric interpretation of this inequality: "The distance between two points a and c is at most the sum of the distance between a and b and the distance between b and c."
BASIC THEOREMS ON LIMITS OF SEQUENCES
69]
It follows that if the points
201
a and b are "close together" (say less than
e/2 units apart) and if b and c are "close together" (less than e/2 units apart), then a and c will be "close together" (they will be less than E units apart).
Addition. From the sequences (1)
and (2)
we can form the new sequence C1 ===
a1
+ b1, C2
===
a2
+ b2, C3
===
a3
+ b3, · · · , Cn
an
===
+ bn, ... ,
(3)
by means of the arithmetic process of addition. On the other hand, we can pass to the limits (we assume they exist) of these sequences, say lim an n~oo
===
lim bn n~oo
a,
===
b,
lim Cn n~oo
===
c.
There naturally arises the question whether c === a + b or, in other words: If an is "close to" a and bn is "close to" b, does it follow that (an + bn) is "close to" (a b)? Symbolically, does
+
lim (an n~oo
+ bn)
===
lim an n~oo
+ n~oo lim bn?
We shall prove that it surely does. THEOREM 65. The limit of the sum of two sequences, each of which has a limit, is equal to the sum of the limits.
Proof. Let an interval of arbitrary length 2E centered about the point a+ b be given (see Fig. 65). What we must show is that there is a term in the sequence (3) such that all terms beyond it appear in this interval. To find such a term, we place intervals, each of total length E, centered about a and b. Now, since lim an === a, there is an index N 1 such that all terms following aN 1 fall in the interval of length E centered about a, i.e., n
>
N1
implies
In other words, since lim an === a, an can be made arbitrarily close to a by going far enough in the sequence an. E
E
~
i_
(2,2)
(2•2)
a
b FIGURE
(Ea E) a+b
65
•X
202
THE NOTION OF A LIMIT
[CHAP.
6
In the same manner, there exists an index N 2 such that all terms following aN 2 fall in the interval of length € centered about b, 1.e.,
>
n
N2
implies
lb  bnl
€
< 2·
Therefore, if we use N to denote the maximum of N 1, N 2 , we get
>
n
. 1ies . b oth N imp
(la  anl lb 
bnl
<
N implies la  anl
+
lb  bnl
< e;
but since the sum of the absolute values is at least the absolute value of the sum (the triangle inequality), we have n
>
N implies l(a
+ b)
 (an+ bn)I
O), then the point (a, f(a)) of the graph lies above the xaxis. If, in addition, the function y == f(x) is continuous at x == a, then, intuitively speaking, the graph must contain a small piece passing through the point (a, f(a)) which lies completely above the xaxis. In other words, if a function f(x) is continuous at a point x == a and is positive there, f(a) > 0, then it must be positive in an entire interval containing the point. For suppose that each interval about a, no matter how small, contained some point where f(x) < 0. Then we could select a sequence {In} of intervals about a with lengths tending to zero, and a sequence {Xn} of corresponding points, Xn E In, such that f(xn) < 0. However, on the one hand, lim f(xn) == f (lim Xn) (by continuity of f(x)), n~oo n~oo ==
f(a),
while
f(xn)
0. l)(x 7 7x 2 3x  5) 19. Write the equation of the line tangent toy = (x 3 through the point of intersection of the curve and the line x = 1. 20. Find the slope of the line tangent to the curve defined by the equation
+
(ix 
7)2y = (x 
1)3
~i +
+
+
2,
at the point corresponding to x = 9. 21. A swimming pool has the dimensions shown in Fig. 716. If water is pumped in at the rate of 3 ft 3 /min, at what rate is the level rising when the shallow end holds water to a depth of 3 ft? When the tank is ! full? 22. A light hangs 15 ft above the ground (Fig. 717). A man 6 ft tall walks away from the light at the rate of 3 ft/sec. At what rate is the length s of his shadow increasing when he is 20 ft from the projection of the light on the ground? 23. Oil is being poured at the rate of 12 ft 3 /min into an oil drum that is in an upright position. If the drum is a right circular cylinder 18 in. in diameter, at what rate is the level of the surface rising? 24. If 2 y = 3u (u  ]), u = V2x  1, find dy / dx as a function of x only. 25. A spherical balloon is being inflated at a rate of 2 ft 3 /min. What is the rate of increase of its surface area when its diameter is 8 in.? 26. A 3% salt solution enters an upright cylindrical vat of radius 5 ft at a rate of 7 ft 3 /min. At what rate is the liquid level rising when the vat contains 14 ft 3 of liquid?
15 1
FIGURE
716
FIGURE
717
CHAPTER 8 APPLICATIONS OF DIFFERENTIATION Introduction. Now that we have mastered some simple techniques in the process of differentiation, we shall discuss some interesting applications of these to the problems of maximization and minimization of a function. Such problems occur frequently and in widely diversified situations. For example, industrial management is interested in maximizing profits and minimizing raw material consumption; aircraft designers are interested in minimizing the resistance which their ships offer to air; lightbulb manufacturers seek a product which delivers maximum light flux while spending a minimum of electrical energy; and so forth. Not only man but nature too often behaves in a manner which is explained in terms of minimizing or maximizing principles. Thus the laws governing the reflection and refraction of light are embodied in the principle that "light travels between two points in a path such that the time required is a minimum." Also, a cable suspended from two points assumes a configuration which is such that its potential energy is a minimum. It is, therefore, of great importance that we make inroads in developing a theory to deal with problems of this sort. In this chapter, we shall deal with one of the simplest possible problems in this area: How can we maximize (or minimize) a function of a single variable? 81 Extrema. First, so that there will be no misunderstanding about terminology, we refer to Fig. 81. Here we have the graph of a function y y
FIGURE
248
81
=
f(x)
81]
EXTREMA
249
y == f(x) with domain a < x < b. Because the derivative f'(x) can be interpreted geometrically as the slope of the line tangent to the curve y == f(x), if we solve the equation f'(x)
==
0
for x, we obtain those xcoordinates of points on the graph at which the tangent line is horizontal, for it is here that the slope is zero. In the case of the graph of Fig. 81, we would obtain
[We would not obtain x 4 or x 5 by solvingf'(x) == 0 because the derivative does not exist at each of these points, i.e., no tangent line can be drawn at these places.] The points corresponding to x 1 and x 4 are called relaft"ve maximum points ofy == f(x), while those corresponding to x 2 and x 5 are relative minimum points. We now focus our attention on the point (xi, f(x 1 )). This point is a relative maximum in the sense that we can find an interval I on the xaxis with x 1 at its center such that for all numbers x in this interval, the corresponding values of the function (i.e., the vertical distances to the graph) are at most equal to f(x 1). So f(x 1 ) is a maximum value of the function relative to the values of x in the interval I. Analogous statements can be made concerning the relative minimum which occurs at x 2 or x 5 • The absolute maximum of y == f(x) occurs, in our illustration, at x == b. As the name would seem to indicate, this means that the value f(b) is not exceeded by any of the numbersf(x) belonging to the range of the function. The absolute minimum has an analogous and obvious meaning. Observe that in this illustration x 2 leads to a relative minimum as well as to the absolute minimum. For short, we sometimes drop the adjective "relative" and say simply "maximum" and "minimum." When this is done, to a void confusion we shall retain the terminology "absolute maximum" and "absolute minimum." Formally, we lay down The function y == f(x) is said to have a (relative) maximum at x == ~ if and only if there exists an interval I with ~ at its center such that x E I implies f(x) < f(~). DEFINITION
81.
The function y == f(x) is said to have an absolute maximum at x == if and only if !(~) > f(x) for all x in the domain of the function.
~
250
APPLICATIONS OF DIFFERENTIATION
[CHAP.
8
Remark. It is important to observe that our analytic definition of maximum (relative or absolute) of a function does not presuppose that the function is differentiable. It applies to arbitrary functions. We leave the formulation of the definitions of relative and absolute minima to the reader. Minimum and maximum points are collectively ref erred to as extrema. It is not always true that a function has a maximum and a minimum, whether we are concerned with relative or absolute extrema. For example, the function y == 1/x on the interval O < x < 1 (we exclude x == 0 because 1/0 is undefined) has no maximum, for as x ~ 0, 1/x grows arbitrarily large. Now, the difficulty does not lie in the fact that the function is undefined at x == 0, for although you may define it there in any manner you please, the resulting function still will not have a maximum. A question which therefore arises in connection with extrema is this: Under what conditions can we guarantee that a function will have a maximum and a minimum? K. Weierstrass proved the following sufficient condition:
THEOREM 81. Weierstrass' Theorem. If a function is continuous on a closed interval, then it assumes an absolute maximum and an absolute minimum on that interval. The proof would take us too far afield and is therefore omitted (see R. Courant, Differential and Integral Calculus, Vol. 1, p. 62). As a simple corollary to Weierstrass' Theorem, if a function is diff erentiable on a closed interval (and hence continuous there), one of its values will be an absolute maximum and one will be an absolute minimum. As for relative maxima and minima, these are guaranteed too, since absolute extrema are, a fortiori, relative extrema. Furthermore, if any extremum (relative or absolute) is found at an interior point of the closed interval of definition of a function (i.e., any point other than the endpoints of the interval), we can say more, as suggested by the graph of Fig. 81 and borne out by THEOREM 82. If y == f(x) is differentiable on an interval and has an extremum at x == ~' where ~ is an interior point of the interval, then f'(~) == O.*
Proof. By definition, f'(~)
lim f(~ ~x~o
+ ~x)

fa) .
~x
* If we do not insist that ~bean interior point of the interval, the theorem is not true. In Fig. 81, for example, we have a maximum at x = b, yet the de rivative is not zero there because the tangent line is not horizontal.
81]
251
EXTREMA
For definiteness, suppose that f( ~) is a maximum. Then, by definition ~x belongs of maximum, there is an interval about ~ such that when ~ to this interval, we have f(~ ~x) < !(~), or
+
+
f(~
+ ~x)
 f(~)
0,
(1)
Now if ~xis taken to be negative, then
Ja + ~x)

f(~)
~~~~~~
~x
for division of the inequality (1) by a negative number sense and, on passing to the limit as ~x ~ 0, we get
>
!'(~)
On the other hand, if ~x
~.t
0.
(2)
>
0, returning to (1) we obtain
f(~
+ ~x)

f(~)
~x
will reverse its
>
0, i.e.,
0.
~X
Now, in order for this limit to be positive, for l~xl sufficiently small the slope g(~ + ~x)  g(~) (1)
~x
of the secant line must be positive (Fig. 89), since g' ( 0 is the limit of the secant line slopes. This means that the numerator and denominator of the ratio (1) must have the same sign, so that for l~xl sufficiently small, (i) if the denominator is negative, i.e., if ~x g(~) < 0 or, alternatively,
f(x)} of points is convex (Chapter 4).
83]
259
A SUFFICIENT CONDITION FOR EXTREMA
y
y
y = f(x)
I
I I I
I
~x ~x t
++_.__+
FIGURE
(ii) if Lix
>
88
FIGURE
0, then g( ~
+ Lix)

g( ~)
>
x
89
0.
In both case (i) and case (ii) we see that the inequalities assert that for points close to t g(x) is monotonic increasing. For instance, in case (i) Lix < 0 and so ~ Lix is to the left of f Hence g(~ Lix)  gU) < 0, which says that ordinates drawn to the left of ~ are less than the ordinate g( ~). This means that g(x) is monotonic increasing. In other words,
+
+
This proves the first part of the lemma. To prove the converse, we note that our ~veps are reversible, i.e., (i) and (ii) are now given (by definition of a monotonic increasing function), and therefore
+
g(~ Lix)  _ g(~) ___ Li_x _ _
>
O.
When we pass to the limit we obtain g'U)
> o.
[The necessity for admitting the possibility of the limit being equal to O is uroved in Theorem 6lO(v).] This completes the proof of the lemma. We now apply the lemma. Referring to Fig. 88, we see that for a concave upward arc of a curve as the point x travels from left to right, starting at a and ending at b, the slope of the ~angent line continually increases: at a, it is negative; at ~ it is O; at b, it is positive. It appears then, that the slope f' (x), which is a function of x, is in this instance a monotonic increasing function of x. Therefore, by the lemma, its derivative (f'(x))' == f"(x) is positive. Conversely, if f"(x) is posit1ive, this means that f"(x) == (f'(x))' is positive, hence f'(x) is monotonic increasing
260
APPLICATIONS OF DIFFERENTIATION
[CHAP.
8
y
y
=
x2
""~I
FIGURE
x
810
according to the lemma, and therefore the graph of y upward. It is now clear that:
===
f(x) 1s concave
THEOREM 85. If f'(~) === 0 and f"(~) > 0, then f(~) is a minimum. Conversely, if f ( ~) is a minimum, then f' ( ~) === 0 and f" ( ~) > 0 provided these derivatives exist. The proof of the analogous statement regarding a maximum is similar to the proof given above and is therefore left to the reader. Mnemonic: As a device for recalling that a vanishing first derivative and a positive second derivative at a point imply a minimum, whereas a vanishing first derivative and a negative second derivative spell out a maximum, the simple example y === x 2 at x == 0 is suggested. For this function, whose graph is shown in Fig. 810, y' == 2x
at
=== 0
x
=== 0
and y"
== 2 >
(for all x). *
0
And indeed y === x 2 has a minimum at x === 0. In a similar way it can be shown that y == x 2 has a maximum at x == 0. EXAMPLE 83. Find the maximum and minimum values (if there are any) of y = v(x  1)/(2x + 3). We begin by computing the derivative y', thinking of the chain of definitions
y = u
1/2
u
x 2x
1
+3
* The notation y" means the second derivative of y with respect to x; we also use the notation of Leibniz: d 2 y / dx 2 •
83]
261
A SUFFICIENT CONDITION FOR EXTREMA
and applying Theorems 79 and 78 in that order: dy dx
dy dx
=
!
(x + )1/ 1 3
2 2x
=
2
+
+3
[2x
(2x
+
(x 3) 2
1)2],
~2x 3 5 x I· 2(2x 3)2 ·
(2)
+
Since it is impossible for the second factor to be zero, the radical must be zero, and therefore its radicand must vanish, which means that 2x
+3
0
=
or
Since from (2) it is obvious that dy/dx is always positive whenever it is real, it cannot change sign as we pass through x = ! and therefore, by Theorem 82, y has no maximum or minimum points. As a matter of fact, if we return to the definition of y, we see that x = ! does not even belong to the domain of
/x y = \J2x
1
+ 3'
since this value of x makes the denominator 0. EXAMPLE 2 bx ex
+ +
84. Find the maximum and minimum values of the cubic y = x 3 d. Differentiating and setting the result equal to zero, we find y'
= 3x 2
+ 2bx + c
+
= 0,
from which x
=
b ± Vb 2 3
3c

(3)
Differentiating again yields y"
= 6x
+ 2b
Substituting the values found in (3) in y" = 2(b
= 2(3x + b).
y",
we get
± Vb2  3c + b)
=
±2Vb2  3c.
Therefore the function y has no maximum or minimum values in case b2 3c < 0 [for then the numbers in (3) are imaginary]. But if b2  3c > 0, then 2  3c]/3 by consulting y" above, we see that y has a maximum at x = [b 2 and a minimum at x = [b 3c]/3. What about the case where b2  3c = O? Here (3) degenerates to x = b/3, and since the vanishing of the discriminant of the quadratic
vb
+ vb
y' = 3x 2
+ 2bx + c
262
APPLICATIONS OF DIFFERENTIATION
[CHAP.
8
means that y' is a perfect square* and hence y' cannot change sign, x = b/3 gives neither a maximum nor a minimum. 85. (a) Is the curve y = (2x  3)/(3x  1) 2 rising or falling at x = O? (b) Is this curve concave upward or downward at x = O? Answering (a) calls for evaluation of dy/dx at x = 0. Now EXAMPLE
(3x  1) 2 (2)  (2x  3)2(3x  1)(3) (3x  1) 4
dy dx and
dyl < dx x=O
O •
Therefore the curve is f all1:ng at the point (0, 3); that is, y is monotonic decreasing here. To answer (b) we must compute the second derivative and evaluate it at x=O.Wehave
dy dx
2(3x 
1 
1)[3x 
(3x 
3(2x 
3)]
2(8 
3x)
(3x  1)3 '
1) 4
so that
(3x  1) 3 (6)  2(8  3x)3(3x  1) 2 (3) (3x  1)6 and therefore
It follows that the curve is concave downward at x = 0. 86. Some tollroad administrators estimate that for each 20¢ rise in toll charges, the number of users falfo off by 4000 vehicles per month. With the present toll at $1.50, 50,000 cars per month pass over the road. What toll charges will maximize the monthly income? The function to be maximized is the monthly income. Let x denote the number of multiples of 20¢ by which the present toll is raised. Then the monthly income I as a function of x is EXAMPLE
I(x)
(numbers of vehicles/month) (charge per vehicle) (50000 
4000x)(l.50
2
2
* Proof:
y'
3[x
2
+ 23 bx + b9  b3 + c ]
++H 3 (x
+ 0.20x).
+
~y,
2
b  3c 3
since b
2 
3c
0.
83]
263
A SUFFICIENT CONDITION FOR EXTREMA
Since we are required to maximize I(x), we differentiate I with respect to x and set the result to equal O:
I'(x)
=
(50000 
4000x)(0.20)
+ (1.50 + 0.20x)(4000)
0.
Solving for x, we find x = 2.5.
Since
I" (x) = 1600
0, there exists a o > 0 such that the conditions x
lvl.
We leave it to the reader to form precise variants of this definition, e.g., lim f(x) ==  oo.
X>a +
* Note that close to a."
x < a says
"to the left of
a"
and
Ix  al < o says
"sufficiently
102]
323
RATIONAL FUNCTIONS
How can we detect vertical asymptotes in the graph of r(x) == p(x) / q(x)? This can be accomplished by observing that r(x) == p(x)/q(x) will become infinite if one of the factors of the denominator q(x) approaches O while the numerator p(x) approaches a nonzero constant. Thus vertical asymptotes of x 1 y == 3x  5
+
may be found by setting the denominator equal to 0. Here we have 5
3x 
==
0
or
x
==
!,
which is indeed the equation of the vertical asymptote, for
. 1Im
X')5/3+
. x +1 1im 3 5 X 
x 3 (X
+ 15) == + 00 ' 
3
oo.
(1)
(2)
X'>5/3
It is now necessary that we endow the graph with the geometric properties corresponding to these calculations. This is done in Fig. 1010 as follows: the geometry corresponding to (1) and (2) is indicated by small portions of the curve with the same numbers (1) and (2). (d) Find the horizontal asymptotes. The horizontal asymptotes are horizontal lines showing the behavior of the function as "x tends to infinity." Refer to Fig. 1012 for clarification. Pictured there is the graph of a function with the line y == b as a horizontal asymptote. As x increases without bound, i.e., moves farther to the right, f(x) approaches the value b. We convey this situation by writing
lim f(x)
b.
X'>00
y
y = b
FIGURE
1012
324
RATIONAL FUNCTIONS. CURVE SKETCHING
[CHAP.
10
The precise meaning is the content of 103. We say the limit of f(x) as x tends to infinity is b, (written limxwJ f(x) == b) if and only if f(x) can be made arbitrarily close to b by merely choosing x sufficiently large. In other words, for every E > 0, there exists a number M > 0 such that DEFINITION
x
>
M
lf(x)  bl
implies
< E. *
How does one find horizontal asymptotes? It is a matter of calculating the limit of the function as x tends to infinity. In the present example,
. IIm y X?00
==
. x +1 11m 3X _ 5
===
X?00
. 1 + (1/x) 1Im _ (5/ ) 3 X
1
== 3 ·
X?00
(Cf. Sections 69 and 610, especially Exercise 11.) Furthermore, it is easy to see that if we let x ~  oo, we again get y ~ !. Thus lim y
==
!.
X?±00
Intuitively, this means that as x recedes from the origin, in either direction, y approaches the value ! and therefore y == ! is the equation of the horizontal asymptote. To pass to the geometric interpretation, we draw the line y == ! (see Fig. 1010) and bear in mind that "toward either end of the xaxis, the graph straightens out, approaching this line arbitrarily closely." (e) Locate all maximum, minimum, and 'inflection points. We begin by differentiating the quotient to get , _
y 
( x
+ 15)'
(3x 
3x 
5) (3x 
(x 5)2
+ 1)3
8 (3x 
5)2'
which is obviously never O and therefore the tangent line is never horizontal. We conclude from this that there are no maximum or minimum points, since the derivative exists everywhere except at x == ! and there couldn't possibly be a horizontal tangent here since there isn't even a point on the curve at x == l As for inflection points, we differentiate again, beginning with 5) 2 ,
y'
==
8(3x 
y"
==
16(3x  5) 3 (3).
and obtain
* Again there is a variant of the situation: limx4oo f(x) precise formulation to the reader.
=
b. We leave the
102]
RATIONAL FUNCTIONS
325
This expression for y" never vanishes; therefore the curve has no inflection points. (f) Find the domain and the range. The domain of the function
y
==
x+l 3x  5
means the set of all numbers x for which the function is defined (see Section 32). This function is meaningful except for x == l Therefore the domain consists of all real numbers except x == !. As for the range, it is the set of all values of y obtained by substituting the permissible values of x as just determined. Look at it this way: Pick a number y. Is there an x which will lead through the operations indicated by x 1 3x  5
+
to this y? To answer this question, imagine Yo to be given. Then xis sought such that x 1 Yo == 3x  5 ·
+
This last equality holds if and only if or x(3yo 
== 1 + 5Yo·
1)
y
I
I I I I I I I
~==J ___ _
x
= QI 31
FIGURE
1013
326
RATIONAL FUNCTIONS. CURVE SKETCHING
[CHAP.
10
Now this equation can be solved for x if and only if _3y_0_ _1_~_0, that is, Yo ~ l Therefore, the range of x
+1
y == 3x 
5
l
consists of all real numbers y except y ==
(g) Sketch the graph. Using the skeleton outline developed in Fig. 1010, we complete the sketch as shown in Fig. 1013. * We emphasize once and for all that it is sometimes advisable to omit one or more of the "steps" outlined above. This is certainly the case when the arithmetic calculations called for exceed the value of the geometric information to be gained. For example, ordinarily it would be considered feasible, so far as sketches are concerned, to locate intercepts between successive integers if the intercepts sought are irrational numbers which have to be approximated (see Section 914). EXAMPLE
105. Sketch the graph of
y =
2x(x  5) x+3
The reader should consult Fig. 1014 to verify the geometric interpretations of the following computations as they are made. (a) The xintercepts are obtained by putting y = 0: x = 0, 5. (b) Theyintercept is obtained by putting x = 0: y = 0. (c) The vertical asymptote arises by setting the denominator equal to 0: x 3 = 0. (d) To obtain the horizontal asymptote, we observe that the degree of the numerator is greater than the degree of the denominator, and therefore as x tends to oo , y grows large in absolute value. Moreover, let us write
+
y
=
2x(x  5) x+3
2x
2
lOx x+3 
and argue as follows. For large numerical values of x, the numerator 2x 2  lOx behaves as though it were 2x 2 , since this term dominates the expression 2x 2  lOx in the sense that for x sufficiently large numerically, the numerical value of lOx becomes negligible compared with the numerical value of the term 2x 2 • Similarly, the denominator x 3 behaves, for large numerical values of x, as though it were merely x. Therefore we write
+
y
* Later,
=
2x
2

lOx
x+ 3
2x x
~ 
2
=
2x
we shall prove that the graph is a hyperbola (Chapter 15).
102]
327
RATIONAL FUNCTIONS
y I I I
I I I
y
2x
=
I
I I I
FIGURE
1014
as an approximation for large lxl. In other words, y behaves like 2x as we recede from the origin (toward the right or left) and consequently we may draw the line y = 2x as an asymptote. Thus, although there is no horizontal asymptote, we have succeeded in obtaining an inclined asymptote (see Fig. 1014). Here's another approach: Dividing numerator and denominator by x, we have
2x y = 1
+
10 (3/x)'
from which it is clear that for !xi large, y is approximately equal to 2x. (e) To locate extrema and inflection points, we differentiate the quotient, obtaining y'
(x
+ 3)(4x 
10)  (2x (x 3) 2
+
For this to be zero,
x2
+ 6x 
2

lOx)
15
=
0,
which yields
x
=
3 ± 2v6.
2x
2
+ 12x (x + 3) 2
30
328
RATIONAL FUNCTIONS. CURVE SKETCHING
[CHAP.
10
To identify these points, we differentiate again:
y''
=
2 2 + 3) (4x + 12)  (2x + 12x  30)2(x + 3) (x + 3) 4 2 2 4[(x + 3)  (x + 6x  15)] (x + 3)
(x
3
96
(x
~
if x
+ 3) 3
3.
Since y"lx=3+2y'B > 0, a minimum occurs at (3 t_ 2v6, 8v6  22), whereas a negative y" when evaluated at x = 3  2v6 indicates a maximum at ( 3  2v6, 22  8v6). Finally, since y" = 0 is impossible, there are no inflection points. (f) Since the fraction 2x(x  5) y = x+3 leads to a definite number for every value of x except x this function consists of all real numbers except x = 3: 
00
0; hence P(r, 0) is in the third quadrant, as specified.)
+
Formulas (1) and (2) above may be applied to effect a transformation of the equation of a curve from its rectangular coordinate version to its polar coordinate version, and vice versa.
+
EXAMPLE 1310. Transform r = 2 sin O 5 cos O into rectangular coordinates. Sketch the graph from the simpler point of view.
We first replace sin O, cos O by their definitions: r = 2(y/r) Multiplying by r, we get r 2
=
x2
+ 5(x/r).
+ 5x or + y = 2y + 5x.
2y
2
(i)
410
TRIGONOMETRIC FUNCTIONS: APPLICATIONS
[CHAP.
13
y
x2
+ y2
=
2y
+ 5x
• (~, 1)
FIGURE
1317
This is obviously a circle. Completing the squares, yields (x 
!)2
+ (y 
2i.
1)2 =
Hence the circle has center at (i, 1), radius v29/2. In addition, it is clear from (i) that (0, 0) lies on the curve. Its graph is shown in Fig. 1317. EXAMPLE
r = 1
1311. Find the rectangular coordinate equation of the lima~on
+ 2 cos (} and sketch the graph, using the simpler version.
Substituting x/r for cos (}, we have
r = 1
+ 2(x/r),
r2 = r
+ 2x,
or
x
2
+y
2 
2x =
Vx2 + y 2•
Squaring both sides, we find
+
2 cos(} furnishes the simpler from which it is clear that the equation r = 1 approach to constructing a graph. We may obtain the graph most readily by considering the graph of the auxiliary equation y = 1 2 cos x, * which is obtained as follows (Fig. 1318). Sketch the function 2 cos x (amplitude 2, period 2), then lower the xaxis one unit, since y is to be increased by one unit. We can compute the xintercepts from O = 1 2 cos x or cos x = !, which implies x = 21r /3, 41r /3; the interval with these endpoints accounts for the small loop in the polar coordinate graph shown in Fig. 1319.
+
+
+
* This equation is not the rectangular coordinate version of r = 1 2 cos 0. J3y this we mean that the points with rectangular coordinates (x, y) satisfying y = 1 2 cos x do not have polar coordinates necessarily satisfying r = 1 2 cos(}, and conversely. In particular, the point with polar coordinates ( 1r, 1) satisfies the last equation but not the first, for its rectangular coordinates are (1, 0). The auxiliary equation y = 1 2 cos x serves only as an aid to sketching r = 1 2 cos (}.
+
+
+
+
138]
SYMMETRY IN POLAR COORDINATES
411
y
y
3
/
=
1
+2
cos x
I
I
21r
FIGURE 1318
FIGURE 1319
138 Symmetry in polar coordinates. Because polar coordinates of a point are not unique, extreme caution must be exercised in testing for symmetry of polar coordinate curves. A simple example will indicate the complications which arise. EXAMPLE 1312. We know that the locus r = 1 is the unit circle (see Fig. 1320). Let us apply a test for symmetry with respect to the pole. Note that we were careful to call for a test; we did not say the test for symmetry, as we could have if we were dealing with rectangular coordinates. As a matter of fact, if we think of (r, (}) as being an arbitrary point on the circle, then one set of coordinates of the symmetric point with respect to the pole would be (r, (}); another would be (r, 180° 8). Now if we apply the first of these tests to the equation r = 1 we must replace r by r, thus getting the different equation r = 1. Now here's the point: the equation has changed, but in spite of this we say merely that this test gives no conclusive answer to the question of symmetry with respect to the pole. Indeed, if we now apply the second test mentioned, we replace (} by 180 and do not change r; but (} does not occur in the equation r = 1 and hence this test leaves the equation unchanged, from which we conclude that the curve does indeed possess symmetry with respect to the pole.
+
+ (},
EXAMPLE 1313. Test r = cos 28 for symmetry with respect to the polar axis, the 90°line, and the pole.
(r, e)
FIGURE 1320
412
TRIGONOMETRIC FUNCTIONS: APPLICATIONS
[CHAP.
13
One test for symmetry with respect to the polar axis is replacement of (r, 8) by (r, 8). (Draw a figure to convince yourself that this is so.) Applying the test, we obtain r = cos 2(8) = cos 28. Therefore, there is symmetry with respect to the polar axis. A test for symmetry with respect to the 90°line is the replacement (r, 8) + (r, 180°  8), * which gives r = cos 2(180° 
8) = cos (360° 
28)
cos 28.
Hence there is symmetry with respect to the 90°line. One test for symmetry with respect to the pole is the replacement: (r, 8) + ( r, 8); this test yields r = cos 28. This equation is different from the original, and hence we say that this test is inconclusive. We try another one, say (r, 8) + (r, 180°
+ 8).
This test gives
r = cos 2(180°
+ 8)
= cos 28;
therefore there is symmetry with respect to the polar axis.
139 Exercises Transform to rectangular coordinates: 2
l. r = 1
+ 3 cos 8
4. r = 3 cos 8  2 sin 8 7. r
2
3. r
3 cos 28
2. r
1
5. r
2 
6. r
3 sin 8
4 8. r =   1 sin 8
= sin 28
10. 3r csc 8 = 4 1 13 · r = 3 2 cos 8
11. r
2
+
9. r
= cos 28
12. r
+ 1 + cos 8
15. r
l 
14. r = 5
+
1 8 sin 8 2 1  sin 8 9 3 cos 8 2
5 
sin 8
Transform to polar coordinates: 16. 18. 20. 22. 24.
2y 2  3x = 0 (x  3) 2 (y 
+
y
1) 2 = 4
= lxl
2xy = 1 2x 2 y2
+

2x
+y
+ +
17. 3x 5y  7 19. x 2 y2 = 9 21. x 2  y 2 = 1 2 23. x 2 xy y y112 = 25. x112
+
= 0
+
+
0
=
1
a112
26 through 40. Test the graphs of Exercises 1 through 15 for symmetry with respect to the pole, the polar axis, and the 90°line.
* By the replacement (r, 8) + (r, 180°  8) we mean that r is left unchanged in the relation, while 8 is replaced by 180°  8.
1310]
413
SOLUTION OF A TRIA~GLE
Transform the following equations to rectangular coordinates or polar coordinates, whichever is appropriate, and sketch the graph, using the more convenient equation: 41. 4r (sin () 44. r = 1
cos 8)
=
42. ()
7
+ sin()
7r
= ~
43. y 2
3
45. 9y 2

36x 2

=
4(x
+ 1)
144 = 0
1310 Solution of a triangle. In the introduction to this chapter, it was pointed out that one of the applications of the trigonometric functions is to the study of triangles. To simplify the exposition, let us first standardize the notation by agreeing that the three sides of a triangle (Fig. 1321) shall be denoted by the small Latin letters a, b, c and the angles opposite these by A, B, C, respectively. Thus a triangle has six parts: three sides and three angles, and when these parts are known we say that the triangle is solved. Since the solution of right triangles involves merely trivial application of the definitions of the trigonometric functions, we shall assume that we are dealing with oblique triangles, i.e., riot with right triangles. To begin with, we must be given at least one side of the triangle whose solution is required, for even if all three angles of a triangle are given but none of its sides, an infinite number of similar triangles will contain the given angles. We may therefore proceed with the discussion according to the cases: (i) one side given, (ii) two sides given, (iii) three sides given. In studying the various ca~es, it is suggested that the reader imagine that he has a ruler and compass and is attempting to construct the triangle with the aid of the given parts. (i) One side given. It is easy to convince oursehres that at least two angles, in addition to the one side, must be specified. For if only one angle is given, there are infinitely many triangles containing the given angle and given side. However, if two angles are given, then, in effect, all three angles are known, since the sum of the angles is 180°. For the sake of argument, suppose side a and A, B, Care known. We then draw side a with the angles B and C at its extremities, as shown in Fig. 1322. Let the altitude drawn to side c be h. Then, by definition of the sine of an angle,
sin B
==
~a
and
sin A
A
' b
a
~b · A
c
B
==
c
FIGURE 1321
B
'' 'h', a
'' '
Frnum~ 1322
c
414
TRIGONOMETRIC FUNCTIONS: APPLICATIONS
[CHAP.
13
Solving these equations for h yields h == a sin B
h == b sin A.
and
Therefore a sin B == b sin A,
or, on dividing by ab, sin B sin A b == a·
(1)
This proves THEOREM 134. Law of sines. In any triangle, the sides are proportional to the sines of the opposite angles, i.e., sin A
sin B
a== b
sin C
== c·
(2)
Remark. Actually, by drawing the altitude to c we have proved only the first equality involving a and b. In a similar manner, the equality involving b and c can be proved by working with the altitude drawn to a. Returning to (1), it is clear that all parts entering that equation except b are known. We may therefore determine b. Then, using (2), we may calculate c. 1314. Calculate the distance a across a lake (Fig. 1323), using the information that the angles at A and B are 27° and 78°, respectively, and the distance c = 1200 ft. EXAMPLE
We first compute
c=
180° 
(27°
+ 78°)
75°.
c
Invoking the law of sines, we obtain sin 75° 1200
sin 27° a
or 0.9659a a
=
1200(0.4540),
~
563 ft.
A FIGURE
1323
(ii) Two sides gfren. It is obvious that at least one angle must be given, in addition to the two sides. The simplest situation occurs when two sides and the included angle are given. The "law of cosines" furnishes the solution. To obtain this law, draw the triangle so that the given angle, say B (see Fig. 1324), is in standard position. Now compute the square
1310]
413
SOLUTION OF A TRIANGLE
y
C(a cos B, a sin B) a
b
~~~~~~~~~__.,._~ x c OB (c, O) 1324
FIGURE
of the length of the side opposite B. To do this, use the definitions of cos B, sin B to find the coordinates of the vertex C: xcoordinate of C
==
a cos B,
ycoordinate of C
==
a sin B.
(These formulas hold whether B is an acute or an obtuse angle.) From the distance formula, it follows that b2
==
(a cos B 
b2
==
a 2 cos 2 B 
+ (a sin B  0) 2ac cos B + c + a
c) 2
2
2
,
2
sin 2 B.
(3)
But a 2 cos 2 B
+a
2
sin 2 B
==
a2 ;
therefore (3) simplifies to b2
==
a2
+c
2

2ac cos B.
(4)
This last relation is known as the law of cosines. Hence we have THEOREM 135. Law of cosines. In any triangle, the square of the length of a side is equal to the sum of the squares of the lengths of the remaining two sides minus twice the product of those two sides multiplied by the cosine of the angle between them: b2
==
a2
+c
2

2ac cos B.
Remark. This law is a generalization of the Pythagorean theorem, for it reduces to this theorem when we take B == go 0 : b2
or
==
a2
+c
2

2ac cos
go
0 ,
416
TRIGONOMETRIC FUXCTIOXS: APPLICATIONS
[CHAP.
13
c
,,,
.......
I
/
b
a,
I
I
I
I
.......
\
lh
',
\
',
\
I
I
/
',
\ \
I
, ',
\
',
A~~~'"'4 h, the analysis runs as follows (see Fig. 1325): If a < b, using C as center and a as radius, we describe an arc which will intersect the horizontal in two places, giving two triangles with the given parts. For this reason, we refer to the conditions b>a>bsinA
as the ambiguous case. On the other hand, if a is not only greater than h but in addition is greater than b, then (Fig. 1325) when we describe an
1310]
417
SOLUTION OF A TRIANGLE
arc of radius a, center C, it again intersects the horizontal line at two places, but only one of the triangles thus determined has the angle A as a part. Consequently, there is a unique triangle satisfying the condition
a>
b.
1315. Given that A = 30°, b = 50 ft, a = 35 ft. Solve the triangle (or triangles). EXAMPLE
From A = 30°, b
50 ft, we compute the altitude h to be h = 50 sin 30° = 25 ft.
Since a = 35 ft exceeds h but is less than b = 50 ft, there are two distinct triangles containing the given parts. Applying the law of sines to 6.ABC, we find 35 50 sin 30° = sin B' or sin B = From the tables, we find B
c~
~
5~
=
t
~ 0.714.
46°. It follows that
180° 
(46°
+ 30°)
104°.
Again by the law of sines, c sin 104°
35 sin 30°
~
70(0.970)
or c
=
70 sin 104°
=
67.9 ft.
This completes the solution of one triangle. The other is solved in the same manner.
(iii) Three sides given. Of course, the sum of any two sides must exceed the third. One allinclusive check on these conditions can be made by seeing if the minimum of the three numbers, representing the lengths of the proposed sides, exceeds the sum of the remaining two. If it does, there is a unique triangle; if it does not, there is no triangle having these lengths as sides. Supposing the test to show that a triangle is, in fact, determined, we may compute its angles by using the law of cosines. EXAMPLE
1316. The sides of a triangle are 4 ft, 5 ft, and 7 ft. What are
its angles? One angle, call it A, has a cosine given by
+
42 52  72 cos A =      2(4) (5)
1 5
0.2;
418
TRIGONOMETRIC FUNCTIONS: APPLICATIONS
[CHAP.
13
consequently
A ,.._, 180°  78.5°
101.5°.
Also, cos B
42
+ 72 
52
5 7
0.714,
+ 44.4°)
34.1°.
2( 4) (7) B ~ 44.4°.
Finally,
c~ *1311
180° 
(101.5°
Exercises
1. Compute the cosines of the interior angles of the triangle whose vertices are (1, 2), (5, 6), (3, 1). 2. The sides of a triangle are 7, 8, and 13 ft. Find the largest angle. 3. A tower at the top of an embankment casts a shadow 100 ft long straight down the side when the angle of elevation of the sun is 45°. The side of the embankment is inclined 15° from the horizontal. Compute the height of the tower. 4. A tower stands on a cliff 653 ft above a horizontal plane. From a point A in the plane the angles of elevation of the top and bottom of the tower are 47° and 45°. How high is the tower? 5. How high does an airplane rise in flying 8000 ft upward along a straight path inclined 23°42' from the horizontal? 6. How many triangles are there having a = 28 ft, c = 40 ft, A = 41 ° (show all steps in your reasoning). 7. Solve triangle ABC, whose parts are as follows: (a) a (b) a
= =
25, A = 38°, B = 62° 2.7, b = 7.3, C = 39°20'
(c) a = 5, b = 7, A. (d) a = 5, b = 7, B
= =
45° 60°
8. A proposition of plane geometry asserts that a triangle can be constructed when two sides and the included angle are given. Take the given angle (} to be in standard position and let the sides be of lengths a, b. Find the equations of all three sides of the corresponding triangle. 9. Two railroad tracks leaving a station form an angle of 30°. A passenger and a freight train simultaneously leave the station on these tracks with speeds of 50 and 20 mi/hr, respectively. (a) At what rate are they separating after 1! hr? (b) Consider the situation where the passenger train leaves ! hr after the freight. At what rate are the trains separating after the freight has been traveling for 3 hr? 10. What force must be exerted to drag a 200lb weight up a slope which inclines 17° from the horizontal? 11. In planning a straight railroad grade to rise 70 ft, it is decided to incline the tracks 4°27' from the horizontal. Find the horizontal projection of the grade, and its length.
1311]
419
EXERCISES
12. Find the horizontal and vertical components of a force of 2000 lb acting downward at an inclination of 35°42' from the horizontal. 13. How long a ramp must be used to roll a 500lb pipe from the ground level to a height of 8 ft if a force of 160 lb must suffice to move the pipe? 14. A ship is at point Pon a course directed toward E. A light Lis observed from Pat an angle a; when the ship is at Q, a miles away, Lis observed again, at an angle of (3. Show that the closest the ship will come to Lis d =
a . cot a  cot (3
15. (a) Prove that the area of a triangle is given by Area = fab sin C. (b) Using the result in (a), show that if (} is the central angle, in radians, subtended by a chord in a circle of radius r, then the area of the segment cut off by the chord is Area of segment = fr 2 ( (}  sin 8). 16. Prove that if the lengths of the sides of a triangle are natural numbers, then the cosines of its angles are rational numbers. Prove that for any triangle:
+ tan B + tan C + B) = sin C cos A+ cos B + cos C
17. tan A 18. sin (A 19.
tan A tan B tan C . A . B. C+ 1 4 sm  sm  sm 
2
2
2
CHAPTER 14 EXPONENTS AND LOGARITHMS.
APPLICATIONS
Introduction. This phase of our work is concerned with a study of the basic properties of the exponential function y == ax and its inverse, the logarithm x == loga y. After becoming acquainted with the most important properties of these functions, we turn to some elementary applications, for example the utilization of the logarithm to reduce the labor in problems requiring complex arithmetic computations. Two very remarkable results make an appearance for the first time: (1) The function ex, where e is a certain constant whose approximate value is 2.718, coincides with its derivative, i.e.,
d(ex) dx
x
==e·
'
furthermore, apart from a constant factor, ex is the only function with this property. (2) The intriguing discovery by L. Euler of a relation involving the number e, the imaginary number i, and, of all things, the basic trigonometric functions sine and cosine: ei()
. . (} . == cos (} f ism
141 Definition of the exponential function and the logarithm. In Chapter 1 we succeeded in defining ax for all rational values of x, where a is a fixed positive number. When we fix the value of a, say a == 10, and then select integral values of x, we arrive at the table in Fig. 141, where some of the points are also plotted. We note that insofar as these integral values of x are concerned, 10x is a monotonic increasing function of x; that is, as x increases, so does 10x. But lOx has been defined also for fractional values of x and we ask if the monotonic increasing character has been preserved in the process of extending the definition. That is, assuming m/n < p/q (rational numbers), does it follow that lOm/n
a
'
'
' ...
and the exponents of a are obviously never zero. In other words, we are not surprised to learn that the expansion of (a + n neither a positive integer nor 0, contains an infinite number of terms. Such an infinite sum is called an infinite series.
br,
148]
447
THE BINOMIAL THEOREM
Next, we observe that since the identity (a
+ bt =
[a ( 1
r
+ ~)
= an(l
+ x)n,
b a
X=
holds, the question regarding the validity of the expansion of (a+ b)n may be reduced to the same question for (1 Here is a partial answer:
+ xr.
xr
If !xi < 1, the expansion of (1 + by means of the formula of Theorem 147 is valid, even if n is not a positive integer. If, on the other hand, lxl > 1, this expansion is not valid.* 1414. Assuming the binomial expansion to be valid, find and simplify the first four terms in the expansion of (2x  3y) 112 • For what values of x, y is the expansion valid? We write the first four terms: EXAMPLE
(2xJ112
+ (i)(2x) 112 (3y) + ( /)?) (2x)312 (3y) 2 + 2, (1 h)n 2n > n If n > 4, then 2n < n ! 5n  1 is divisible by 4
+
1)
>
1
+ nh
1410]
LOGARITHM AND EXPONENTIAL CURVES
449
1410 The slope of the logarithm and exponential curves. We now want to compute the derivative dy/dx if y == loga x. We begin in the usual manner: y == loga x
y
+ D,.y ==
loga (x
+ D,.x),
and by subtraction, we find
(Theorem 145)
=
loga ( 1
+ D,.xx )1/ll.x
(Theorem 146).
To simplify matters, we carry out the artifice of multiplying and dividing the right member by x to get D,.y D,.x
x ( loga 1 x
+ D,.x) x
1
+ D,.xx )x/ll.x
  == 
=
(
xloga 1
1 / ll.x
(Theorem 146).
It follows that D,. lim _Jf_ ll.x~o D,.x
==
d _Jj_ dx
1
(
lim loga 1 x ll.x~o
== 
D,. )x/ll.x * +~ , x
and since the logarithm is a continuous function, which means that we may interchange the operations of "taking the log" and "passage to the limit," we have d 1 ( D,. )x/ll.x _Jj_ ==  loga lim 1 ~ (1) ll.x~o x dx x
+
·
We have therefore reduced the problem to that of computing D,.x)x/ ll.x lim ( 1+ll.x~o x
·
(2)
* We assume that the indicated limit exists. The justification can be found in R. Courant, Dijferent1:al and Integral Calculus, Vol. I.
450
EXPONENTS AND LOGARITHMS. APPLICATIONS
To do this, we put u
==
[CHAP.
14
jj.x/x, so that (2) goes over to
. (l + u )1/u • 1lm u~o I
We may go about obtaining a numerical estimate of this limit in the following way. Consider the sequence 1 ,
1
1
1
n'
2' 3' · · ·,
as representing successive values of u (note that u ~ 0), and let us apply the binomial theorem to the expression (1 + u) I/u with u replaced by 1/n: (1
+
!J
= In+
n(lt1
m
+ n(n 2~
1) (lt2
(~Y
llfn 
2) (lt3
(~r + ... + (ir
+ n(n 1
+ 1 + n(n

2!
(!)2
1)
n
+ n(n 
~!(n  2)
_ 1 + 1 + Ln/n][(n 
(ff + ... + (~r
1)/n]
2!
+ [n/n][(n 
1)/n][(n  2)/n] + ... +
3! (1
+
ir =
1
+1+ 1 
n
2~1/n)
+ [1 
(1/n)][l 
(2/n)] + ... +
3! As n
~
oo, it follows that 1/n lim
n~oo
(1+l)n n
(!)n,
~
==
(!)n. (3) n
0, and we suspect that 1
1
2!
3!
1+1+++···
~ 2.718+.
We say only that we "suspect" this approximation to be valid because we know that as n tends to infinity, the number of terms in the binomial expansion becomes infinite, and it is not obvious that we may take the limit of each term in (3) individually. What we have done, of course, is
LOGARITHM AND EXPONENTIAL CURVES
1410]
451
permissible; we merely raise these questions to indicate that there is a definite gap in our reasoning. Our considerations make plausible the DEFINITION 143. The number e is defined by
+ u) 11
e == Inn ( 1 u~o
11, ;
its approximate decimal value is e ~ 2.718.
We return to the derivative which we left at (1) above and now write dy
1
x
x
d ==  Ioga e. Even though we have not offered complete proof,* we state THEOREM 148. If y
== loga x, then
dy/dx
== (1/x)loga e.
In particular, if the base of logarithms is itself taken to be the number e and if we reserve the symbol In for logarithms to the "natural" base e, then we arrive at the result that if y == ln x, then dy/dx == 1/x. 1415. Find the equation of the line normal to the curve y = In x at its xintercept. EXAMPLE
Since dy/dx = 1/x and the xintercept is at x = 1, the slope of the tangent line there is 1 and therefore the normal line has slope 1; consequently its equation is y
= l(x  1)
or
y
+x
= 1.
In the applications, one frequently encounters the function y == loga u, where u is itself a function of x. To compute its derivative, we first invoke Theorem 148 to get dy u
d
l
== u loga e.
Then, by the chain rule, dy dx
dy
== du.
du dx
du == 1 (loga e) · u
* See
dx
Differential and Integral Calculus, Vol. I, p. 175, by R. Courant.
452
EXPONENTS AND LOGARITHMS. APPLICATIONS
[CHAP.
14
Therefore we have THEOREM 149. If y == loga u and if u is a differentiable function of x, then du dy == 1 . . loga e. dx u dx There is an immediate COROLLARY. If y
== ln u and if u is a differentiable function of x, then dy 1 du dx == u dx
du/dx u
==
Mnemonic. To differentiate ln u, differentiate u, then divide by u. Remark. Note that the awkward factor "loga e" becomes the silent factor "1" when the natural base e is used. For theoretical purposes, then, the base e is to be preferred over any other base. Now that we have determined the slope of the logarithm curve, we may fall back upon this to determine the slope of the (inverse) exponeutial curve. This is accomplished as follows: y == au implies u == loga y, by definition of the logarithm. Therefore
du dx
1 dy . loga e ydx '
  ==  . 
or, solving for dy/dx, we find
dy dx
y(du/dx) loga e
(4)
If we put
c == loga e,
(5)
then
ac
===
e.
Taking logs to the base e yields ln ac === 1, that is,
c ln a
=== 1
or
1
c ==   · Ina
That is, 1 loga e ==    , 1oge a
EXERCISES
1411]
453
in view of (5). Using this in (4), we have THEOREM 1410. If y
== au and if u is a differentiable function of x, then du dy == au · · ln a. dx
dx
Remark 1. In the special case where e is selected as the value of a and xis selected as u, Theorem 1410 asserts that y
==
e
x
.
1·
dy
imp ies
dx
ex .
In other words: The "natural exponential'' is identical with its derivative: x
e .
Moreover, in Section *1412 we shall prove that, except for a constant factor, ex is the only function with this remarkable property of coinciding with its derivative. Remark 2. Do not confuse the function au with un. The former is a constant raised to a variable power, while the latter is a variable raised to a constant power. EXAMPLE 1416. Find dy/dx if y = In sin 2 3x. The solution is 2
dy dx
d (sin 3x)/dx sin 2 3x
2 sin 3x cos 3x(3) sin 2 3x
6 cot 3x.
1411 Exercises 1. Find the equation of the line tangent to the curve y = In x at the point (e 2 , 2). 2. By computing the second derivative of y = In x, show that this curve is always concave downward. 3. By computing the second derivative of y = ex, show that this curve is always concave upward. 4. Find the equation of the line normal to the curve y = ex at its yintercept.
In Exercises 5 through 10, determine the maximum and minimum points and points of inflection of the functions. x
7r)
5. y
e
smx
6. y
e2x cos ( 2x  
7. y
5e 3 x sin 2x
8. y
e
9. y
=
In (x
+ VT+ x2)
10. y
x2
In sin x
3
454
EXPONENTS AND LOGARITHMS. APPLICATIONS
[CHAP.
14
In Exercises 11 through 19, differentiate the functions. 2
11. y = In cos x 14. y
1  x Inl+x
17. y

ex+ ex 2
12. y = In cos x
13. y
IOX 2
15. y
In (In x)
16. y
2x In x
18. y
log10 tan 3x
19. y = sin In x
2
y
(x, f(x))
FIGURE
1411
20. The subtangent (see Fig. 1411) to the graph of y = f(x) at the point (x, f(x)) is the length s of the segment on the xaxis intercepted by the tangent line and the perpendicular drawn to the xaxis. Show that the subtangent in the case of the function y = ax is a constant, i.e., does not depend on the value of x at which it is computed.
*1412 Applications of the exponential function. We have seen that if
y == ceax' then dy dx
ace ax ;
dy dx
ay,
in other words, (1)
which says that the rate of change of the function y with respect to its independent variable is proportional to the function y itself. Conversely, if we begin with the differential equation (1) [a differential equation is an equation involving the derivative(s) of the unknown function], we can prove that its solution is of the form y == ceax, where c is a constant, for consider this function: (2) u == ye ax ,
1412]
EXPONENTIAL FUNCTION: APPLICATIONS
455
where we suppose that y satisfies (1). On differentiating (2), we get du dx
aye ax+ eax dy dx eax
(
ay
+ dy) dx '
and since we are assuming that y satisfies (1), the parenthetical expression is O and therefore du/dx === O; it follows that u === constant, say u = c. Thus (2) becomes ax c = ye , so that y === ce
ax
.
This completes the proof of THEOREM 1411. A function y satisfies the equation dy/dx == ay, where a is a constant, if and only if y === ceax, where c is a constant. 1417. According to Newton's law of cooling, if a hot body (for example, a hot metal plate) is placed in a large cool bath of liquidso large that we can assume the bath temperature to remain constantthe rate at which the temperature y of the hot body at time t changes is proportional to the difference in temperature between the hot body and the bath: EXAMPLE
dy dt
= k(y  c),
(i)
where c is the constant temperature of the bath and k is a constant of proportionality. Find the function y(t). To simplify matters, we put u = y  c. (ii) Then du dt
dy dt
du dt
ku
and (i) becomes  =
'
which has the solution, according to Theorem 1411,
where A is a constant. Thus, substituting in (ii) and solving for y, we get y
= Ae
kt
+ c.
(iii)
456
EXPONENTS AND LOGARITHMS. APPLICATIONS
[CHAP.
14
We note that if t = 0, y(O) = A+ c, and consequently the constant A = y(O)  c has the interpretation of being the difference in the initial temperatures of the two bodies. Usually this can be measured, and if the constant k (which depends on the nature of the cooling object) can be evaluated, then (iii) is completely determined.
*1413 Exercises In the study of socalled "damped" o;::;cillations, the logarithmic spiral (polar coordinates) r = Aek 8 where A, k are constants, is encountered. Sketch the graphs for Exercises 1 through 4.
1. A = 1, k = 1. 3. A = !, k = 2.
2. A = 5, k = 3. 4. A = 2, k = f.
Exercises 5 through 13: In the study of investments, the following formula i)n, where Pis the quantity of money ("prinplays a leading role: A = P(l cipal") in vested at rate i per con version period for n con version periods, and A is the compound amount to which P accumulates at the end of the investment time. For example, if we invest $100 at 3% compounded semiannually* for 5 years, then A = 100(1 0.015) 10,
+
+
or log A
log 100 2
log A A
+ 10 log 1.015
+ 10(0.0065),
2.065, $116.10.
5. Lang has $4000 which he can invest at 2f%, compounded semiannually. How long will it take his investment to grow to $5500? 6. In how many years will a sum of money triple itself at 2% compounded semiannually? 7. A man buys a $75 bond that pays 3!% compounded annually. How long will it take this bond to amount to $100? 8. Solve the formula A = P(l i)n for P. In this form, P is sometimes called the present value and is convenient to use in answering questions of this type: How much must be invested now at 4% compounded quarterly in order to accumulate to $2000 in 12 years? Or, to put it another way: What is the present value of $2000 due in 12 years if money is worth 4% compounded quarterly? Find the answer. *9. A man is offered a house for $12,000 cash or for $2000 down and payments of $100 at the end of each month for 10 years. If money is worth 3% compounded monthly, which is the better deal for the purchaser and by how much?
+
* 3% compounded semiannually means we earn j¢ on each dollar invested for each conversion period of 6 months.
1414]
457
EULER'S FORMULA
*10. A man borrows some money and repays it by means of payments of $300 at the end of each 6 months for 5 years. If the interest rate is 4% compounded semiannually, how much did he borrow? 11. (a) Show that if $1 is invested for one year at 100% compounded annually, it increases to $2. Prove that (b) if the compounding is done monthly, 12 ~ $2.612, (c) if the compounding is done n times it increases to (1 1\ ) a year, it increases to
+
(
l
+ _!_)n
=
n
1
+ 1 + n(n 
1)
2n 2
+ n(n 
l)(n 3!n3
2)
+ ... + __!__, nn
(d) if the compounding is done continuously, it increases to lim n+X>
(1 + _!_)n n
= e dollars.
12. A corporation sets aside $40,000 a year to retire a bond issue due in 12 years. If money is worth 1% compounded annually, find the amount of the bond issue. *13. It is known that the rate dq/dt at which a quantity q(t) of radium disintegrates at a time t is proportional to the quantity q(t) present at time t. Determine q(t) as a function of t.
*1414 Euler's formula. There is a remarkable relation discovered by L. Euler, that ties in the trigonometric and exponential functions. Let us begin with deMoivre's formula (Theorem 132):
(} + i.sm. n (})n ( cos n
= cos
(} + .sm. (} .
(1)
i
Now, if we fix 0, then for large values of n, 0/n will be small and hence cos 0/n will be approximately cos O = 1. In addition, since sin 0/n = y/r (see Fig. 1412) whereas 0/n in radian measure is s/r, we have the ratio sin ( (} / n) = y_ • 0/n s As n increases so that the angle (} / n decreases, y ands become more nearly the same number. As a matter of fact, we have proved that lim sin (0/n) = 1 n+oo
oIn
'
r
'
n
FIGURE
1412
and this means, intuitively, that if 0/n is close to 0, sin 0/n is approximately equal to 0/n. But if in (1) we replace cos (0/n) and sin (0/n) by the
458
EXPONENTS AND LOGARITHMS. APPLICATIONS
[CHAP.
14
~
oo,
approximations 1 and fJ /n, respectively, and take the limit as n we obtain lim n~oo
(
1
+ :n_.f))n _
cos fJ
+ i sin fJ.
(2)
But by definition of e, we have e == lim (1 u~o
+ u) l/u.
Therefore, if we set u == ifJ/n, we have from (2) lim (1 u~o
+ u) iO/u ==
lim [(1 u~o
+ u) l/u]io ==
i
0
•
(3)
It is therefore plausible from (2) and (3) that the remarkable equality eiO
== cos fJ
+ i·sm · f)
(4)
is valid. A rigorous proof of this cannot be given at this time [see Courant's Differential and Integral Calculus, Vol. I, Ch. VIII]. If we specialize (4) by putting fJ == 1r, we obtain i1r == 1, a surprising equation connecting the num hers e, 1r, i, 1. The relation · fJ, eiO == cos fJ + i ·sm tying in the numbers e, i and the basic trigonometric functions sin fJ, cos fJ, was only one of many startling discoveries of L. Euler, although it is one of the most beautiful. Euler, perhaps the most prolific mathematician the world has seen, was obsessed with religion and offered several mathematical "proofs" of the existence of God. He was also a devoted family man (13 children). An interesting account of his life will be found in E.T. Bell's Men of Mathematics, Simon and Schuster Publishing Company.
CHAPTER 15
THE CONIC SECTIONS
Introduction. If we pass a plane so that it intersects a surface, the resulting curve of intersection is called a section. Thus, if the surface is a sphere, ~he curve of intersection of a plane with the sphere is a circle and hence we say that spherical sections are circles. In this chapter, we shall study the conic sections. These are the curves of intersection of a right circular cone and arbitrary planes. The different types of conic sections are illustrated in Fig. 151. Section ( 1), a circle, is obtained by passing a plane perpendicular to the axis of the cone; if the plane is tilted slightly, an ellipse [section (2)] is obtained; if the plane is tilted still further, so that it is parallel to an element of the cone, a parabola results [Section (3)]; finally, if the angle of tilt is increased further, we get a hyperbola [Section (4)]. These geometric l1.3finitions of the conic sections are the ones which we must make in order to account for the terminology conic section. However, until we have introduced the analytic machinery for studying surfaces (Chapter 19), these definitions are difficult to manage, and therefore we shall introduce other definitions in this chapter which will seem entirely unrelated to these but which indeed are, after all, equivalent to them. The equivalence of the definitions will be proved in Chapter 19. The idea of the equivalence proof runs as follows. (3)
I (4) I
Section (1): Circle
Section (3): Parabola
(1)
Section (2): Ellipse
Section (4): Hyperbola FIGURE
459
151
460
THE CONIC SECTIONS
[CHAP.
15
In this chapter we will show that every conic section has an equation of the second degree in x and y, i.e., an equation of the form
ax 2
+ bxy + cy 2 + dx + ey + f
=== 0,
(1)
and we will learn conditions on the coefficients a, b, c (the others are immaterial in this connection) that the equation shall represent a parabola, an ellipse, or a hyperbola. Conversely, we will show that every equation of the form (1) represents some conic section. Then, in Chapter 19, we will prove that every section of a right circular cone has an equation of the form (1). A word about the role of the conic sections is in order. These curves were first studied by the early Greeks. Interest in them increased when it was discovered that many problems involving the motion of celestial bodies find solutions in the form of conic sections; for example, the earth travels in an elliptical orbit with the sun located at a "focus" (to be defined later). Apart from their role in celestial mechanics, the conics find their way into many applications of a "mundane" nature; for instance, the surface of some automobile headlamps is a parabola revolved about its axis. We shall discuss these applications later; full appreciation of them can be obtained only after theoretical properties of the conics have been developed. This is therefore our more immediate goal.
151 The circle. The circle is perhaps the simplest conic section. We know that the equation of a circle with center at (h, k) and radius r is given by (x  h) 2 + (y  k) 2 === r 2 • (1) We note the following aspects of this equation: (i) it is of the second degree in x and y with the coefficients of x 2 and y 2 being equal, and (ii) it contains no xy term. Conversely, every equation satisfying conditions (i) and (ii) represents a circle, for an arbitrary equation satisfying (i) and (ii) can be written in the form ax 2 + ay 2 + bx + cy + d === 0. Since a ~ 0 (if a were equal to 0, the equation would not be of second degree), we may divide by it, thus obtaining x2
+ y 2 + Bx + Cy +
D === 0,
where B === b/a, C == c/a, D == d/a are constants.
By completing the
THE CIRCLE
151]
461
squares in x and y, we find
c)
B)2 ( ( x+2 + y+2
2 2 B c ==+D 4 4 .
2
(2)
Comparison with (1) suggests that we write the right member as the square of a number, i.e.,
B2
C2
4 + .f 
D
==
+ C 4D 42
B2

We must accordingly consider three cases:
+C B +C B 2 + C2
(i) B 2
2
2
2
(ii) (iii)


4D 4D 4D
> ==
0 150y = 0 16x 2  64x 2 y  5x 4y  10 < 0 3y 2  2y 5x 1 = 0 2y 2  4y x > 0
+
+
+ + +
+
+
*17. Prove that the equation of the tangent to the parabola y 2 = 4ax at the point Ct r,) is yr, = 2a(x 18. Discuss the effect on the parabola y 2 = 4ax of varying the number a, oo < a < oo. 19. Find the equation of the parabola with vertical axis and passing through (5, 1), (3, 0), and (7, 1). 20. Find all values of x for which the curves y = x 2 and x 2 y 2 = 1 have parallel tangent lines. Make a sketch. 21. Find the equation of the parabola with directrix x = 2 and focus (4, 2). 22. How high is a parabolic arch of span 24 ft and height 18 ft at a distance 8 ft from the center of the span? 23. Find the coordinates of the vertex and focns, the length of the focal chord, and sketch the parabola y 2 = 6(x 1). 24. A parabola has its focus at (2, 5) and the equation of its directrix is y = 1. Find its equation. 25. A focal radius of a parabola is a line connecting the focus and any point on the curve. Prove that the length of the focal radius of y 2 = 4ax to (t r,) is I~+ al. 26. A parabola has x = y as directrix and (1, 7) as focus. Find its equation. 27. Find the equation of the parabola with horizontal axis and passing through the points (0, 3), (1, 1), and (5, 1). 14j. [Hint: First sketchy = x 2 28. Sketch the graph of y = lx 2  5x 5x + 14.]
+ ~).
+
+
+
157 The ellipse. Carry out the following construction of an ellipse: Take a piece of cord and fasten it at two points F and F' (see Fig. 1511), but such that it is not taut [in other words, if 2a is the length of the cord, select F and F' such that FF' < 2a]. Now place a pencil at P so that the string FPF' is taut and let P trace out a curve, always ensuring that no slack is allowed in the string. The resulting curve is an ellipse. Formally,
157]
477
THE ELLIPSE
1511
FIGURE
y
,,,,,,,,.,,,b ' ' , a .,,, d ~ .,,, .,,,
.,,,_:_ 
1

1 P(x, y) ' ' I \ d2
p1~'4 \
=~~~~~.,_._~X
C
Fl(c, O)
FIGURE
F(c, 0)
1512
we lay down the 152. An ellipse is the locus of a point which moves in such a way that the sum of the distances between it and two fixed points is a constant. The fixed points are called the foci of the ellipse. DEFINITION
We now solve the problem of determining the equation of an ellipse, having been given its foci F and F' and the sum 2a of the distances between the foci and the moving point. We introduce axes as shown in Fig. 1512, so that the origin bisects the chord F'F. Let P(x, y) be the point which generates the ellipse. Then F'P
+ PF ==
(1)
2a,
or
V(x
+ c) 2 + y + V(x 2
 c) 2
+y
2
== 2a.
(2)
To simplify the equation, we transpose one of the square roots and square both sides: (x
+ c)2 + y2
== 4a 2

4av(x c)2
+ y2 + (x

c)2
+ y 2 .*
We now collect terms and divide by 4:
a2

ex== aV(x  c) 2
+ y2.
(3)
* The step (squaring) leading to this equation can be reversed to give (2) or (1) back again. (See Exercise *26, Section 158.)
478
THE CONIC SECTIONS
[CHAP.
15
Squaring again and collecting, we obtain
+ c2x 2 == a 2(x 2  2cx + c2 + y 2), * a2y2 + (a2  c2)x2 == a2(a2  c2).
a 4  2ca 2x
(4)
If we glance at Fig. 1512 and take into account that (i) the distance F'F (ii) the distance F' P
==
+
2c, and PF == 2a,
and the sum of two sides of a triangle always exceeds the third, we see immediately that 2a > 2c. Therefore a > c, hence a 2 > c 2, and so the difference a2  c2 is positive and consequently can be called the square of a number b:
a2  c2
==
b2
'
(5)
by definition of b. Using (5) in (4), we arrive at
so that on division by a 2 b2 , we have (6)
as the canonical form of the equation of an ellipse. To determine the xintercepts of this curve, we put y == 0, and so obtain from (6):
x
==
±a.
These points are the vertices of the ellipse, and the segment intercepted by them is called the major axis. We note that Eq. (6) has a graph which is symmetric with respect to the xaxis, yaxis, and the point (0, 0). The origin is a center of symmetry of the curve. To see the geometric significance of the number b which .was defined in (5) on purely analytic grounds, we determine the yintercepts of (6): if x == 0,
y == ±b.
These points are referred to as the covertices of the ellipse and the inter
* Note
that a 2  ex [the left member of (3)] is always positive, since the maximum value of xis a and a 2  ac = a(a  c) is positive. Therefore, again the squaring of (3) is reversible and so the locus remains intact.
THE ELLIPSE
157]
479
cepted segment is the minor axis because, as we see from (5), a > b. Emphasizing again that the steps above leading to (6) are reversible, we have THEOREM 156. The equation of the ellipse with foci at (c, 0), (c, O}, vertices at (a, 0), (a, 0), covertices at (0, b), (0, b), and center at (0, 0) is x2 a2
y2
+ b2
==
(6)
1,
+
where a 2 == b2 c 2 ; the length of the major axis is 2a, the length of the minor axis is 2b. Conversely, every equation of the form (6) represents an ellipse with center at the origin, and axes of length 2a and 2b. Remark. If we had taken the foci on the yaxis instead of the xaxis,
we would have obtained the equation
instead of (6). Let us free Eq. (6) from its "coordinate crutches." To do this, let F and F' be the foci, as before (Fig. 1512), and let C be the midpoint of the segment FF'. We retain the designation 2a for the sum FP PF' of the distances between the moving point P and the fixed foci F, F'. Let P' denote the projection of Pon the line containing the segment FF'. Then GP' corresponds toxin Eq. (6) and P'P corresponds toy in (6). Therefore
+
_2
_2
GP'
P'P
(7)
a2+b2==l.
We now return to coordinate systems. We shall use the result (7) to obtain the equation of an ellipse with center at (h, k) (see Fig. 1513), with a, b, c meaning the same as before. y When we replace
GP'
by
x  h,
P'P
by
y 
k,
(7) goes over to (x ;;2 h)2
+ (y
Consequently
~2
k)2 
1.
(8)
.____ _ _ _ _ _ _ _ x FIGURE
1513
THEOREM 157. The ellipse with center (h, k), semimajor axis a, semiminor axis b has an equation given by (8). The converse is also true.
480
THE CONIC SECTIONS
[CHAP.
15
Remark 1. If F, F' lie on a vertical line through (h, k), then the equation becomes (X h) 2 (y  k) 2 b2 a2 1.
+
Remark 2. On examining (8), we conclude that the equation of an ellipse with axes parallel to the coordinate axes can be recognized by the following criteria: (i) it is a seconddegree equation in x and y with no xy term, and
(ii) the coefficients of x 2 and y 2 have the same sign.* EXAMPLE
1510. Sketch the graph of 3x 2

6x
+ 2y + 5y 2
8 = 0.
Since the coefficients of x 2 and y 2 have the same sign and there is no xy term, this equation represents an ellipse. To put the equation in the form (8), we complete the squares: 3(x
2 
2x
+ 1) + 2(y2 + !Y + tt) 3(x  1)2 + 2(y + !)2 (x _ 1)2 + (y + !)2 113
+ 3 + 2l,
= 8 =
1A3
= 1.
113
24
16
Comparing with (8), we observe that the center is at C(l, point (Fig. 1514), and first measure
vVl
£).
We plot this
~ 2.2
units in the horizontal direction on both sides of C. "\Ye then measure  1113 V 16
,..J ,..J
26 •
y y
F (1, 4)

    , P(x, y) I
I I
F'(4, 1) FIGURE
1514
FIGURE
1515
* Indeed, the coefficient of x 2 in (8) is 1/a2 ; that of y 2 is 1/b 2 • Both are positive.
157]
481
THE ELLIPSE
units in the vertical direction both above and below C. Since 2.2 < 2.6, the ends of the first segment are the covertices and the ends of the second segment are the vertices of the ellipse.
So far we have considered ellipses with axes parallel to the coordinate axes, socalled "nonslant" ellipses. Let us now look at a "slant" ellipse. We begin with 1511. Find the equation of the ellipse whose foci are located at (1, 4) and (4, 1) and whose major axis is 10 units long. EXAMPLE
Let P(x, y) (see Fig. 1515) be a point on the ellipse. Then, by definition, FP
+ PF'
= 10,
or vex+ 1)2+ CY 
4) 2 +Vcx 
4)2+ CY+ 1) 2 = 10.
We simplify this equation as follows: Vex+ 1)2 + (y 2
x + 2x+ 1 + y
2
10  Vex  4) 2 +Cy+ 1)2,
4)2
Sy+ 16 = 100 

20Vcx 
+ x IOx 
lOy x 
2
x + y2 + 100 
2xy 
2
3x + 3y2 + 2xy 
y 
12y 
Sx + 16 + y2 + 2y + 1,

100
20V(x 
10
2Vcx 
20x + 20y = 4(x
12x 
2
2

4) 2 + (y+ 1) 2
4)2+ (y+ 1)2, 4)2+ CY+ 1)2,
Sx + 16 + y2 + 2y + 1),
32 = 0.
Note the presence of the xy term. This is because the major axis is not parallel to either coordinate axis, for when the axis was chosen parallel to a coordinate axis, we obtained an equation free from an xy term.
To study the equation of the general ellipse with foci F(a, (3) and F'(cr, r), and the sum of whose distances between these and the moving point is 2a, we write
V(x  a) 2 + (y 
{3) 2
+ V(x
+ (y
 cr)2

r)2
== 2a.
Proceeding as before, we go through the following sequence of steps:
(.c 
a)2
+ (y

== 4a 2 2(cr 
a)x
+ 2(r
{3)2
4aV(x  cr)2


{3)y
+
+ a 2 + {3 2
(y 
u2
r)2 
+ (x
r2

 cr) 2
+ (y

r) 2 ,
+ (y

r)2.
4a 2
== 4av'{x  cr)2
482
THE CONIC SECTIONS
[CHAP.
15
If we square both sides and transpose all terms to the righthand side, we get
cr) 2
16a 2 [(x 
r2(cr 
+ (y
a)X
r) 2 ]

+ 2(T 
{3)y +
a
+ {3 2
2

CT
2

T
2

4a 2 ]2
== 0. (9)
For the point we wish to make, it is not necessary to simplify the equation further, for we want only to indicate that Eq. (9) has the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F == 0.
(10)
{As a matter of fact, we have a) 2
A
== 16a 2
B
== 8(cr  a)(r 
C
==
16a 2


4(cr 
4(r 
==
4[4a 2

(er 
a) 2 ],
(r 
{3) 2 ].}
{3),
{3) 2
==
4[4a 2

158. Every ellipse has an equation of the second degree in x and y, i.e., Ax 2 Bxy Cy 2 + Dx + Ey + F = 0. THEOREM
+
+
158 Exercises In Exercises 1 through 10, sketch the graphs of the loci.
1. 3. 4. 6. 7. 9.
+ 4y + 4x + 20y + 15 = O 9x  27y + 2 = 0 5. 4x + y 8x + 12y = 0 + 4x  6y < 0 + 4y + 56y + 121 = 0 2x  4y  1 = 0 8. 4x + 5y + 16x  20y+ 31 > 0 2x + 18y  17 = 0 10. 2x + y 8x  4y + 10 < 0
9x 2 + I6y 2 3x 2 9y 2 2x 2 3y 2 2 25x  50x x 2 + 2y 2 x 2 + 9y 2 
+ +
2. x 2
= 25
2
2
2

2
2
2
2
2

Find the equations of the ellipses described in Exercises 11 through 15, and sketch the graphs.
v3
11. Center at (3, 2), semiaxes and 8. 12. Major axis 18, passing through (6, 4), center at the origin, and nonslant. 13. Foci at (1, 2) and (3, 1) and sum of the distances of the moving point from these equal to 5. 14. Semiaxes 3 and 7, center at (tr,). 15. Foci at (3, 1) and (5, 1) and semiminor axis 1 unit. 16. The center of an ellipse is at (2, 4), one focus is at (2, 1) and the corresponding vertex is at (2, 1). Find the equation of the ellipse and sketch its graph. 17. Without assuming that the locus is an ellipse, find the equation of a point which moves so that the sum of the distances between it and the points ( 6, O) and (6, O) is 18.
159]
483
THE HYPERBOLA
18. Find the locus of a point which moves so that the sum of the distances between it and (1, 1) and (3, 3) is always 12using only this information. 19. If A is the set of all points (x, y) such that x 2 _.:.__ 4x 4y 2 < 3 and B is the set of all points (x, y) with x 2 y 2 < 4, find A U B, A n B,
+
+
A' n B', A' U B'.
20. An arch is in the form of a semiellipse with major axis as the span. If the span is 80 ft and the height is 30 ft, find the height of the arch at a point 15 ft from the minor axis. 21. What is the condition that ax 2
+ by + ex + dy + e = 2
0
(a) cut the xaxis in two points, (b) be tangent to the xaxis, (c) not cut the xaxis. *22. (a) By differentiating (x 2 / a 2 ) (y 2 /b 2 ) = 1 term by term with respect to x, show that the slope dy/dx of the line tangent to this ellipse is given by
+
a2 y
dx
(b) Use the result in (a) to prove that the line tangent to an ellipse makes equal angles with the lines drawn from the foci to the point of contact of the tangent with the ellipse. (This is the property which accounts for the remarkable "whispering" phenomenon demonstrated in the Mormon tabernacle at Salt Lake City, Utah.) 23. Prove that the equation of the line tangent to the ellipse (x 2 /a 2 ) 2 (y /b 2 ) = 1 at the point (t 'YJ) is (xUa 2 ) (y'Y}/b 2 ) = 1. *24. Find the condition that x 2 2y 2  3x 4y k = 0 be (a) a real ellipse, (b) a null ellipse, (c) an imaginary ellipse. *25. Find the area of the largest rectangle (with sides parallel to the coordinate axes) which can be inscribed in the ellipse
+
+
(x 2 /16)
+ (y
2
+
+ +
/4) = 1.
*26. Show that the step in which Eq. (2), Section 157, is squared is reversible by showing that for the square root of the right member obtained we have l2a 
v(x 
c) 2
+ y2
J
= 2a  v(x 
c)2
+ y2.
159 The hyperbola. Definition and equation. The definition of a hyperbola is quite similar to that of the ellipse: DEFINITION 153. A hyperbola is the locus of a point constrained in such a way that the numerical value of the difference of its distances from two fixed points is a constant. The fixed points are called the foci of the hyperbola.
484
THE COXIC SECTIONS
[CHAP.
15
As with the ellipse, let the foci be F(c, O) and F'(c, 0), let P(x, y) be a point on the hyperbola, and let the constant numerical difference ld 1 d 2 j be 2a. Then ld1  d2I == 2a, IV(x
Since
la/
+ c) 2 + y 2
+ y2
c)2
V(x 

j
== 2a.
k implies a == ± k, one of the following equations must hold: V(x
+ c) 2 + y 2 
c) 2
V(x 
+y
2
==
±2a.
On transposing one of the radicals and squaring, we find (x
+ c) 2 + y 2 ===
4a
2
c) 2
± 4aV(x 
+ y 2 + (x

c)
2
+y
2
,
or xc 
a2
==
±aV(x 
c)2
+ y2.
Squaring again yields
or x2(c2 
a2) 
a2y2
==
a2(c2 
a2). *
(1)
A glance at Fig. 1512 reveals that
while
(since the sum of two sides of a triangle always exceeds the third); therefore addition yields · 1.e., c > a, so t h at c2 > a 2 . It follows that the difference c2 quently we may put

a 2 is a positive number, and conse(2)
Substituting this in (1), we obtain or
x2 a2
y2 b2

1.
*Note that all the steps leading to this equation are reversible.
(3)
159]
485
THE HYPERBOLA y
:....
.......
.....

b
vi .,,,, ,,,, ~
,. , ,,. d .,,.,...,. ;
''
a
.,,...
.......
/
FIGURE
x
......
''
~
1516
Thus ·we see that the equation (3) of the hyperbola in canonical form is similar to that of the ellipse, except for sign. The xintercepts are (±a, O) and are called the vertices of the hyperbola, and the segment intercepted by them is called the transverse axis. We note that Eq. (3) has a graph which is symmetric with respect to the xaxis, the yaxis, and the origin. The origin is a center of symmetry. To learn the significance of the number b introduced in (2), we will no,v prove that the graph of (3) can be obtained by drawing the rectangle with dimensions 2a by 2b (Fig. 1516), with center at the origin, and using its diagonals as the asymptotes of the hyperbola as shown. First of all, the equations of these diagonals may be obtained by replacing the one in the right member of (3) by a zero, for, on factoring, this giyes
and one does indeed obtain the diagonals by setting each of these factors equal to O:
x  y == a
b
0
'
To prove that these lines are the asymptotes of the hyperbola, it is sufficient to show that the vertical distance d (Fig. 1516) separating the line and the curve tends to zero as x tends to infinity. Since the ycoordinate of a point on the diagonal (x/a)  (y/b) == 0 is y ===
ab :r,
(4)
and since the ycoordinate of the nearest point vertically under it on the
THE CONIC SECTIONS
486
[CHAP.
15
hyperbola is obtained by solving (3) for a positive y: 
y==b
x2 ~ a2
or y
== b va
1
r_;;x2 
a2
( fi)
'
we have to show that the difference in the y's of (4) and (5) tends to Oas x ~ oo:
. [b
b
r_;;J
hm  x   v x 2 a a
x~oo

a2
== 0.
Since b/a can be factored out, we must show that lim [x 
VX2=
a2] === 0.
x~oo
This can be done by multiplying and dividing the expression whose limit is to be determined by the "conjugate surd." The limit will then be obvious: . (x hm x~oo
~
(x
a 2 ) (x
+ VX2 
+ yx2 
a2)
a2)
.
x 2  (x 2 X + yx2

a 2) a2
=== hm       x~oo
. 11m x~oo X

a2 + VX2 a2
=== 0.
This completes the proof regarding the asymptotes and Fig. 1516 indicatP,s why (3) is called a "rectangular hyperbola." Therefore THEOREM 159. The equation of the hyperbola with vertices at (±a, 0), covertices at (0, ±b), and foci at (±c, 0) is x2 a2 
y2 b2 === 1,
(6)
where c 2 === a 2 + b2 ; the length of the transverse axis is 2a, the length of the conjugate axis is 2b. Conversely, every equation of the form (6) represents a hyperbola with center at the origin, transverse axis of length 2a, conjugate axis of length 2b. The asymptotes are obtained from the equation x2 y2 a2  b2 === 0, which in turn is obtained from (6) by replacing the 1 there by a 0.
THE HYPERBOLA
159]
487
Referring to Fig. 1516, we note that the rectangle drawn there also determines another hyperbola: it has vertices at (0, ±b) and is shown as a broken curve. Its equation is x2 _ y2 _ b2 a2  1,
(7)
and it is called the conjugate hyperbola to (3). * For this reason, the quantity 2b is called the conjugate axis and the points (0, ± b) are called the covertices when reference to hyperbola (3) is made. As with the ellipse (Section 157), we may divorce Eq. (6) from its dependence upon the coordinate system and then proceed to obtain the equation of a hyperbola whose center is not necessarily at the origin. We omit the details but state the final result: THEOREM 1510. The hyperbola with center (h, k), horizontal transverse axis 2a, conjugate axis 2b has an equation given by (x  h) 2 a2
(y  k)2 b2
==
1
.
(8)
The converse is also true. Remark 1. If the transverse axis is vertical, the rest remammg the same, the equation of the hyperbola becomes (x  h) 2 b2
(y  k)2 a2
1.
(9)
Remark 2. Examining (8) and (9), we conclude that the equation of a hyperbola with axes parallel to the coordinate axes can be recognized by the following criteria: (i) It is a seconddegree equation in x and y with no xy term, and (ii) the coefficients of x 2 and y 2 have opposite signs. EXAMPLE
1512. Sketch the graph of x2

y2
+ 4x

3y 
1 = 0.
Since the equation satisfies the criteria under Remark 2 above, it r~prcsents a hyperbola. Completing the squares, we have (x
+ 2)2
(x
+ 2) 11
~
2
_ (y 
(y
+ J)2
+ 2)
3 2
11
4
=
1l,
= 1 .
(IO)
* In the particular case where a = band the rectangle consequently becomes a square, we speak of equilateral hyperbolas.
488
THE CONIC SECTIONS
[CHAP.
15
y
FIGURE
'The center is at C(2, and side
!)
1517
(Fig. 1517). We draw the square with this center 2v 1l =
vii
~ 3.3.
To determine whether the hyperbola is tangent to the vertical or horizontal sides of the square, we put x = 2 [or y = ! would also be suitable, just so long as we cause one of the terms in (10) to vanish]. Now x = 2 implies (y !) 2 = \l, and this equation says, in effect, that the negative number on the left equals the positive number on the right. Consequently the solutions for y are imaginary and there are no points on the graph of the hyperbola having x = 2. It is obvious, then, that the points of tangency of the hyperbola and the square occur on the vertical sides of the square, as shown in Fig. 1517. Note that theyintercepts are both negative, ! ± v5/2.
+
So far we have been concerned with nonslant hyperbolas. Let us turn our attention to the situation which develops when the axes of the hyperbola are not parallel to the coordinate axes. If we use Fig. 1515 as a guide in studying the equation of the hyperbola with foci at F(a, {j) and F'(u, r), we write
lv'(x
a)2
+ (y

v'(x
{1)2 
u)2
+ (y

r)2
I=
2a.
Proceeding as in Section 157 but with the minor changes occasioned by the absolute value signs indicated earlier in this section, we finally obtain 1511. Every hyperbola has an equation of the second degree in x and y:
THEOREM
Ax 2
+ Bxy + Cy + Dx + Ey + F 2
=
0.
1510]
489
EXERCISES
EXAMPLE 1513. Find the equation of the line tangent to the hyperbola
x2
3y 2

+ 5x 
y
0
=
at the origin. Differentiating each term with respect to x yields
6yy'
2x 
+5 
y'
=
0,
from which
+5 + 6y.
I 2X y = 1
Putting x
5. Hence the required equation is
0, we get y'
y
y
= 5x.
15 10 Exercises Sketch the graphs in Exercises 1 through 11.
+
+
+
1. 16x 2  9y 2 32x 36y 124 = 0 l6y  19 = 0 2. x 2  4y 2  2x 3. x 2  2y 2  6x  11 = 0 4. 16x 2  8y 2 64x  30y 5 > 0 5. x 2  4y 2 4x 20y 15 = 0 6. 9x 2  16y 2 = 25 2 2 7. 4x  9y 8x  18y = 0 8. 2x 6y 3y 2  x 2  4 0 2 2 9. 16x  9y 128x 90y  113 = 0 10. y 2  4x 2 8x  lOy 21 > 0 18x 96y  279 = 0 11. 9x 2  16y 2 12. Find the equation of the circle with a diameter coinciding with the transverse axis of the hyperbola x 2  4y 2  2x 16y  19 = 0. 13. Find the equation of the hyperbola with center at (O, 1), one focus at (0, 0), and conjugate axis of unit length. 14. Without assuming that the locus is a hyperbola, find the equation of a point which moves so that the numerical difference between its distances from the points (6, O) and (6, O) is 10. 15. One asymptote of the hyperbola (x 2 / a 2 )  (y 2 /9) = 1 is the line y = }x. Find a. 16. A point moves so that the difference (in numerical value) of its distances from (1, 8) and (1, 4) is 5. Find its equation from just this fact. 17. Find the equation of the hyperbola whose foci are at (4, 4) ancl ( 6, 4), and whose transverse axis is 8. 18. Find the center, semitransverse axis, semiconjugate axis, and equations of the asymptotes of the hyperbola
+
+ + + + + + + + + + + +
+ +
+
2
x 4 Sketch the graph.
1.
490
THE CONIC SECTIONS
[CHAP.
15
+
19. Find the equations of the asymptotes of the hyperbola 3x 2  y 2 6x 2y 3 = 0. Sketch the graph. 20. A hyperbola has foci at (0, 2) and (0, 2) and the vertices of the conjugate hyperbola are at (1, 0) and (1, 0). Find its equation. *21. Prove that the equation of the line tangent to the hyperbola (x 2 /a 2 ) (y 2 /b 2 ) = 1 at the point ( t 11) is (xU a 2 )  (y11/b 2 ) = 1. *22. Supply details in the proof of Theorem 1510. *23. Supply details in the proof of Theorem 1511. (y/q) = 1 be tangent to the *24. Find the condition that the line (x/p) hyperbola (x 2 / a 2 )  (y2 /b2) = 1. *25. Find the equation of the circle through the points of intersection of
+
+
x
2
a2
y
2
+ b2
2
=
and
1
x b2
y
2
+ a2
=
1.
In each of the Exercises 26 through 29, solve for y, identify the conic, and sketch the curve by compounding of ordinates.
*26. 16x 2

8xy
+ y2 
x = 0
±vx
± vx.
[Here (4x  y) 2 = x implies 4x  y = implies y = 4x sketch the straight line Yl = 4x and the parabola y2 ±VX(y~ add the ordinates geometrically.]
+ 2xy + y 22 + 3y + 2x + xy + y = 1
*27. x 2 *28. 3x 2
1
=
0 *29. x 2 
2xy
+ y2 
x
=
+3
Now x) and
= 0
In Exercises 30 through 34, find the points of intersection of the pairs of curves. y2 25, 31. 2x 2  xy = 24, 30. x 2 5x 2 y 2 = 144 y2 xy = 12
32.
+ + x+y x3 + y3
+
;3
33. 2x 2  5xy = 5, 5xy  3y 2 = 1
[Hint: Divide.]
[Hint: Eliminate the constant terms first.]
34. 2x  y = 7, y2 = 1 3x 2  7xy
+
*1511 New definitions of the conic sections; polar coordinates. Eccentricity definition of a conic. The definitions which we used to derive the equations of the conic sections ran as follows: A parabola is the locus of points equidistant from a fixed line and a fixed point. An ellipse is the locus of points the sum of whose distances from two fixed points is a constant.
1511]
NEW DEFINITIOKS OF THE CONIC SECTIONS
491
y
D
d2
    , P(x, y) I I
I
dl
F(p, O)
FIGURE
1518
FIGURE
1519
A hyperbola is the locus of points the difference (in numerical value) of whose distances from two fixed points is a constant. Although the definitions of the ellipse and the hyperbola are similar, that for the parabola is apparently quite unrelated to these. We shall now offer a definition for the conic sections from a unified point of view. 154. Let a line D (called the directrix) and a point F (called the focus) be given (see Fig. 1518). If a point P obeys the law that the ratio of its distance* d 1 from the focus F to its distance d 2 from the directrix is a fixed positive constant e (called the eccentricity), DEFINITION
e, and if
e
1, then the point P generates a hyperbola.
Of course, our main task is to justify this definition, i.e., prove that it is consistent with the ones given previously. To do this in a simple manner, we introduce a coordinate system such that the yaxis coincides with the directrix and the focus lies on the xaxis (Fig. 1519). Let the coordinates of F be (p, 0). Let P(x, y) be a point on the conic. Then the ratio of d 1 to d 2 will be some fixed positive constant, e: e
or
Vex 
p)2 x
+ y2
e.
* All distances involved in this definition are considered to be positive.
492
THE COXIC SECTIOXS
[CHAP.
15
Simplifying, we have e2 )x 2
(1 
+y
2

2px
+p
2
=== 0.
(I)
This is the equation of the locus, regardless of the value of e. ~ow if e < 1, then e2 < 1, and so 1  e2 will be positive and hence the coefficients of x 2 and y 2 will have the same sign. Therefore (1) does indeed represent an ellipse in this instance (see Theorem 157, Remark 2). Now suppose e === I. Then 1  e2 vanishes, leaving an equation of second degree in y, first degree in x, and consequently a parabola (see the corollary to Theorem 153). Finally, let e > I. Then the coefficients of x 2 and y 2 in (1) will have opposite signs, proving (1) to be a hyperbola (Theorem 1510, Remark 2). This completes the justification of Definition 154 for nonslant conics. EXAMPLE
x
+y
1514.
Find the equation of the conic section having the line
= 1 as directrix, focus at (4, 1), and eccentricity !.
Let P(x, y) be a point on the required locus (Fig. 1520). Then (2)
Now d1
v~= 4)2 + (y 
=
1)2
and
Thus, upon substituting in (2) and squaring, we find 2
+
(y  1) (x+y1) 2 4)
2l(x 
2
]
I
9
y
J'CP(x, y) d2 // \ d1 / ~F(4,1) ~~~+.../~~~~~~x
D FIGURE
1520
1512]
493
EXERCISES
After simplification, we have 17x 2
+ 17y
2xy

2
142x 

34y
+ 305
= 0.
(3)
In some applications of mathematics, notably to mechanics, we have to deal with the representation of conic sections in polar coordinate form. Let us then examine the situation where the focus of a conic is at the pole, the directrix is perpendicular to the polar axis and p units to the left of the pole, and the eccentricity is e (Fig. 1521). Let P(r, ()) be a point on the required locus. Then, by the definition of a conic section (Definition 154), we have The conics in polar coordinates.
e D
or
r   == p
+ r cos()
d2
e
·
P(r,
e)
d1
e
If we solve this equation for r, it is readily verified that
r ==
ep . 1  e cos()
p
(4)
0
FIGURE
1521
In other words, we have shown that the polar coordinate equation of a conic section with a focus at the pole and directrix p units to the left of the pole is given by (4). In case the directrix appears p units to the right of the pole, other conditions remaining the same, the corresponding equation is r
==
ep
1
+ e cos () '
(5)
Verification is left to the reader.
*1512 Exercises 1. Find the equation of the parabola whose vertex is at (9, 3) and whose directrix is x = 15. 2. Find the equation of the parabola which has (0, 0) as a focus and x y = 1 as its diredrix. 3. A parabola has its focus at (3, 2) and x y = 0 as its directrix. Find its equation. 4. Find the equation of the conic which has 4x  y = 0 as a directrix, (1, 2) as focus, and eccentricity = 2. Sketch the graph.
+
+
In Exercises 5 through 9, find the equations of the conics in polar coordinates satisfying the given conditions, and sketch the graphs.
494 5. 6. 7. 8. 9.
THE CONIC SECTIONS
[CHAP.
15
Vertical directrix through (2, 0°); focus at pole; e = 1. Horizontal directrix through (1, 45°); focus at pole; e = !. Horizontal directrix through (1, 31r/4); focus at pole; e = 2. Horizontal directrix through (4, 1r /2); focus at pole; e = 3. Vertical directrix through (3, 180°); focus at pole; e = }.
Describe the following conics: 10. r 11. r 12. r =
2 cos()
[Hint: Divide numerator and denominator by 5.]
3 sin()
(see Exercise 15 below)

5
+
1 3 
5 2 cos()
13. r =
6
+
2 12 cos()
14. Derive, in polar coordinates, the equation of the conic whose directrix is p units to the right of the pole and perpendicular to the polar axis, with focus at the pole and eccentricity e. 15. (a) Show that the equation in polar coordinates of the conic whose horizontal directrix is p units above the pole, with focus at the pole and eccentricity e is r = ep/(1 e sin()). (b) Show that in case the directrix is p units below the pole, other conditions remaining the same, the equation becomes
+
r
ep . 1  e sin()
=
*16. In Exercises 10 through 13, compute the derivative dr/d(). Interpret the result geometrically.
1513 Transformations of the plane. problem:
We now turn to the following
Given the equation Ax 2
+ Bxy + Cy + Dx + Ey + F 2
==
0.
(1)
Find the graph of this equation.
We have already seen that parabolas, ellipses, and hyperbolas have equations of the form (I). We shall see shortly that, conversely, every equation of the form (1) represents a conic. The idea involved in simplifying Eq. (I) so as to obtain the corresponding graph is illustrated in Fig. 1522, where, for definiteness, we assume that a tilted ellipse is actually under consideration. Observe that if we first rotate the xy reference system through the angle () so that the x' y' axes of the resulting coordinate system are parallel to the major and minor axes of the ellipse, and then translate the x'y'axes so that they take the position x"y", the equation
1513]
495
TRANSFORMATIONS OF THE PLANE
y
y y'
______ 1 P(x 1 , y 1 ) I
(x, y)
I
()
  +      +     . . . . _ _ __ _
~x' FIGURE
x
0
1522
FIGURE
1523
of the ellipse relative to this final system of coordinates will appear in a particularly simple form: 2 2 x" y" a2 + b2 == 1. We are therefore interested in studying the relations which must hold between the coordinates of a point in the plane when referred to a translated or rotated reference system.
Translations. Consider the circle of unit radius (Fig. 1523) whose center in the xysystem is at (h, k) and therefore whose equation in this system is (x  h) 2 (y  k) 2 == 1. (2)
+
If we translate the axes as shown in Fig. 1523, by putting
x
==
x'
+ h,
y
==
y'
+ k,
(3)
these substitutions in (2) will yield the equation
x'
2
+ y'
2
(4)
== 1,
which is decidedly simpler than (2). 155. Given two systems (x, y) and (x', y') of coordinates in a plane. We say that one system is a translation of the other if and only if there exist constants h, k such that the relations DEFINITION
x == x' hold.
+h
and
y
== y'
+k
496 EXAMPLE
THE CONIC SECTIONS
[CHAP.
15
1515. Simplify the equation of the parabola y = ax 2 +bx+ c
(i)
by translating the axes so that the linear term in x will be eliminated. To accomplish this, we first replace x and y by x' + handy'+ k, respectively, where hand k are yet to be determined. Equation (i) then goes over to y' + k
a(x' + h) 2 + b(x' + h) + c.
=
(ii)
We now insist that the coefficient of x' vanish, i.e., we set 2ah
+b =
0.
Solving for h, we have b 2a
h =
*
Substituting in (ii) and ignoring terms involving x' because we know their overall coefficient is 0, we obtain
ax' + a[2
y'
+k
=
y'
+k
= ax'
2
b2
;J

+b [ 
:a] + c,
4ac
(iii)
4a
Now k is still at our disposal. Studying (iii), we realize that the natural thing to do is choose k = (b 2  4ac)/4a, for then this number may be cancelled from both sides of (iii). In this new coordinate system (i) takes on the particularly simple form 2 y' = ax' .
t
[Note that this is the same a with which we started, i.e., the coefficient of the seconddegree term in x remains invariant under translation.]
Rotations. Besides translations of the plane, another important class of transformation consists of the rotations. To simplify the calculations, let us first consider the following problem:
* Since dy/dx = 2ax + b = 0 implies x = b/2a, merely the xcoordinate of the maximum or minimum as the case might be. Thus our translation moves the through the vertex of the parabola. t This value of k is, of course, the ycoordinate of the x = b/2a in (i) merely repeats the calculation above:
vertex, for substituting 2
2
b) y=a (  2a
we see that b/2a is point on the parabola, yaxis so that it passes
+b
(
b) + c =  b  4ac · 2a 4a
1513]
497
TRANSFORMATIONS OF THE PLANE
y
....
..........
' P(x, y) '\
p1(x1, yi)
\
e
\ \
~+~X
FIGURE
1524
Given: P(x, y) and P'(x', y') are two points on the circumference of a circle of radius r with center at the origin such that the central angle subtended by the arc PP' is () (see Fig. 1524). Find: The relations between the coordinates of the points. In order to find equations relating these numbers, we recall that if the complex number (or vector) r cis a is multiplied by the complex number r' cis a', we get (r cis a) (r' cis a') == rr' cis (a + a'), i.e., the product has length equal to the product of the lengths of the factors and has an angle equal to the sum of the angles of the factors (Theorem 131). In particular, if we multiply r cis a by cis (), we have (r cis a) (cis e)
== r cis (a +
e).
In other words, the length of the vector r cis a is unchanged (since it is multiplied by a unit vector cis e), but the angle a is changed by the amount e, i.e., the vector r cis a is rotated through the angle e. Thus multiplication of a vector by cis () is tantamount to rotation of that vector through the angle 8. We may therefore obtain equations relating P(x, y) and P'(x', y') by multiplying the complex number x' + iy' corresponding to P'(x', y') by cis 8, for the product will be x + iy. Therefore
(x' + iy') cis 8 == x + iy, or
== x + iy, + i(y' cos () + x' sin 8) == x +
(x' +·iy')(cos 8 + isin 8) (x' cos () 
y' sin 8)
iy.
By equating real and imaginary parts, we obtain
x = x' cos 8 
y' sin 8,
y
== x' sin 8 +
y' cos 8.
(5)
498
THE CONIC SECTIONS
[CHAP.
15
In our discussion above, we have taken the "active" point of view of a rotation, i.e., we have considered a point P' of the plane to be moved through an angle Oalong an arc of a circle with center at the origin. There is also the "passive" point of view. By this we mean that instead of thinking of the points as being "moved," we may consider them fixed and instead move only the axes, considered as rigidly attached at the origin. A study of Fig. 1524 shows that if we start with the point P(x, y) whose coordinates are measured relative to the xysystem shown, and rotate the axes through the angle 0, the coordinates of P will then be (x', y'). Therefore we say that the equations (5) define a rotation of the plane. More precisely: 156. Given two systems (x, y) and (x', y') of coordinate_s in a plane, we say that (x', y') is obtained from (x, y) by rotation through the angle O if the equations (5) hold. DEFINITION
Let us now consider the general seconddegree equation Ax
2
+ Bxy + Cy + Dx + Ey + F 2
== 0.
(6)
We will show that we can choose an angle O of rotation of the coordinate system to which (6) is referred in such a manner that the "mixed term" containing xy can be eliminated and therefore (6) represents a conic section. Substituting (5) into (6), we have A (x' cos O 
y' sin 0) 2
+ B(x' cos O  y' sin 0) (x' sin O + y' cos 0) + C(x' sin O + y' cos 0) + D(x' cos O  y' sin 0) + E(x' sin O + y' cos 0) + F == 0. 2
(*)
The coefficient of x'y' may be picked out; it is 2A sin Ocos O
+ B(sin 0 + cos 2
2
0)
+ 2C sin Ocos 0.
(7)
If we now insist that this quantity (7) be zero, then, since
2 sin Ocos O == sin 2 0 and cos 2 O 
sin 2 0 == cos 20,
we find by setting (7) equal to O that (A  C) sin 20
==
B cos 20.
(8)
1513]
Now, if A
499
TRANSFORMATIONS OF THE PLANE
~
C, we may divide by (A 
C), to obtain
sin 20 = A
1!_
C cos 20,
from which we find (assuming cos 2(}
~
0),
B tan 2(} ==    , AC
(9)
and there is alymys an angle (} such that (9) holds.* It remains to consider the case where A == C. Returning to (8), we find B cos 2(}
== 0.
But B ~ O,t hence cos2(} == O; that is,(}== 45°. We have therefore proved that (5) may always be referred to a coordinate system so as to cause the vanishing of the term containing xy, and consequently (6) always represents a conic section. We have yet to decide whether (6) represents a parabola, an ellipse, or a hyperbola. To do this, we shall first prove that if we carry out a rotation as above and suppose the new equation to be
A'x'
2
+ B'x'y' + C'y' + D'x' + E'y' + F' 2
then the expression
B'
2 
== 0,
4A'C'
is invariant under rotation, i.e., it has the same value that it had before the rotation. In other words,
B'
2
4A'C' == B 2


4AC.
(10)
This is only a matter of computation. We find from (*) above that
+ C) sin 2(} + B cos 2e, A cos + B cos(} sin(} + C sin A sin B cos (} sin O + C cos
B'
== (A
A'
==
C'
==
2
2
2
(}
(} 
2
(},
0.
It takes only some algebra to complete the verification of (10). But what does all of this mean? Suppose that we start with the general seonddegree equation (6), rotate so as to cause B' == 0, and then examine (10):
4A'C'
==
B2

4AC.
* The range of the tangent consists of all real numbers. t If B = 0, we have already seen that (6) represents a nonslant conic and there is nothing to prove.
THE CONIC SECTIONS
500
[CHAP.
15
Now if one of A' or C' is O (our criterion for a nonslant parabola), then we shall have B 2  4AC == 0. If A' and C' have the same signs (our criterion for a nonslant ellipse) then
4A'C' == B 2
4AC

>
0.
Finally, if A' and C' have opposite signs (the criterion for a nonslant hyperbola), then 4A'C' == B 2  4AC < 0. We have therefore proved the THEOREM 1512. The general seconddegree equation Ax 2
+ Bxy + Cy 2 + Dx + Ey + F
==
(**)
0
ha~ a conic section as its graph.* This conic section is a parabola, if B 2

4AC == 0;
an ellipse, if B 2

4AC
a hyperbola, if B 2

4AC
< >
0; 0.
Remark. For degenerate cases of these, see Section 1515, Exercise *27.
* 1514 Exercises In Exercises 1 through 4, translate the axes so as to eliminate the linear terms from the equations.
+ 2xy 2
1. x 2 3. 3x 2

4y
3y 2 2x 
4y 5y
+
+1 = +1 =
0 0
2. x~  3xy 4. x 2 Cy 2
+
+ 4y 2x + 12 = + Dx + Ey + F = 2

0 0
5. Reduce the equation xy = 1 to canonical form, i.e., rotate the axes so as to remove the xy term. 6. (a) Find the equations of translation which will cause the linear terms in Ax 2
+ Bxy + Cy + Dx + Ey + F 2
=
0
to vanish. (b) Show that after the translation in (a) is applied, the general seconddegree equation becomes Ax'
2
+ Bx'y' + Cy' + (Ah 2 + Bhk + Ck 2 + Dh + Ek+ F) 2
= 0.
* When the left member of the equation can be factored into linear factors, we have the degenerate case of two straight lines. See Exercise 10, Section 1514; also see Section 182, Exercise 25.
1514]
501
EXERCISES
y
y'
iP(x, y) 1\(x 1 ,y 1 ) I\
I \
I \ I \
/
I I
\ \
x'
I
0
1525
FIGURE
7. Using Fig. 1525, derive the equations of rotation (5) of Section 1513 by noting that x OP cos (0 )
+
OP [cos Ocos 
sin Osin ]
(OP cos) cos O 
x' cos O 
(OP sin) sin O
y' sin 0.
Now begin with y
= OP sin (0
+ )
= etc.
8. Show that the "cross term" Bxy in
+ Bxy + Cy 2
Ax 2
= 0
can be eliminated by completing the square in the first two terms and then selecting x' in the proper manner. More precisely, verify that
Ax
2
+ Bxy + Cy
2
(
= A x
2
2
B
+ A xy +
B 2) 4A 2 Y
+ Cy
2

B
2
2 4~4 Y
2
A (x
+
B ) 2Ay
+
(4AC  B 4A
)
2
y .
+
Now choose x' = x (B/2A)y. 9. Write the equations of rotation through 30°, 30°, and 180°. *10. (a) Solve the equation
Ax 2
+ Bxy + Cy 2 + Dx + Ey + F
(A ~ O)
= 0
(*)
for x, treating it as a quadratic in x, and assuming all other letters to be "known." (b) Show that the discriminant in (a) may be written as
(B 2

4AC)y 2
+ (2BD
 4AE)y
+D
2

4AF.
(c) The discriminant in (a) must be a perfect square in order that the solution for x obtained in (a) shall yield two linear expressions. Show that this condition
502
THE CONIC SECTIONS
[CHAP.
15
is expressed by setting the discriminant of the quadratic in (b) equal to O: 4ACF
+ EDE 
AE 2
CD 2

FB 2 = 0.

(**)
We therefore have the theorem:
A necessary and sufficient condition that the general seconddegree equation (*) be factorable into linear factors in x and y is that (**) hold. Expressed still another way: The equation (*) has as its graph two straight lines if and only if the condition (**) holds. (The lines can be imaginary, e.g., x 2 y 2 = 0.)
+
1515 Review exercises 1. A point has the property that the difference (in numerical value) of its distances from the fixed points (5, 2) and (3, 6) is 2. Using this information only, find its equation. 2. Find the equation of the nonslant conic through the points (1, 1), (2, 1), (1, O), and (3, 1). Identify. 3. Find the equations of the circles which pass through the point (5, O) and have their centers on the line x  2y 5 = 0.
+
Identify the following curves and sketch:
+ 7x = 0 + x + 2y 4y = 0 6x + y 2 + 12 = 0 
4. 2y 2 6. x 2 8. 3x 2
5y

2

5. x 2 7. 5x 2
+
+
x 2y 2  4y l2y  4 = 0
1
0
9. Find the equation of the locus of a point P(x, y) which is equidistant from the line x = 2 and the point (2, 0). 10. The endpoints of the diameter of a circle are (4, 5) and ( 2, 3). Find the equation of the circle. 11. Find the locus of a point whose distance from the point ( 1, 2) is onehalf its distance from the line 2x  3y = 5. Describe the following loci as completely as you can without making any computations: 2 2 12. (x  2)  (y  4) = 1 4 = 0 13. x  y
+
14. x 16. x
2 2

y2 + 4y
+ 2y2
18. (x ~ 4) 2
20. x
t
2


3x 
+ (y
6x = 4
= 0
y
+y y +x
15. x 17.
2
2
2
~ 1) = 0
19. 4x
2

2
= 2 = 0
6x
2
y
= 3x
21. x
2

4x
+ y + 4y 2
+ 4y
6
= 0
22. Prove that the radical axis of two nonintersecting circles is al ways perpendicular to the line joining the centers of the circles and bisects the segment of this line.
1515]
REVIEW EXERCISES
503
23. A chord passing through a focus of an ellipse, perpendicular to the major axis of the ellipse and having its ext re mi ties on the ellipse is called a lat us rectum or right focal chord. Compute the length of such a chord. *24. Show that a translation of axes or a rotation of axes leaves the distance between two points invariant, i.e., if P and Q are two points in the xyplane such that d(P, Q) = p, and if a translation or rotation of the coordinate system is effected, then the new coordinates of P and Q still satisfy the condition that d(P, Q) = p. *25. Show that angles are preserved under a translation or rotation of axes, i.e., if the angle between two lines is (} when calculated relative to an xycoordinate system and if the equations of thos2 lines are subjected to a translation or rotation, then the angle between the lines of the transformed equations remains e. **26. Read Chapters 10 and 11 of Bell's Men of Mathematics and write a report on the Ii ves of Lagrange and La place. *27. Give examples of seconddegree equations whose graphs are the following sets of points: (a) (b) (c) (d) (e)
empty, an arbitrary single point in the plane, a single straight line, a pair of distinct parallel lines, a pair of distinct intersecting lines.
These are the "degenerate" conics.
CHAPTER 16 ANOTHER SPECIAL LIMIT: THE INTEGRAL Introduction. In the introduction to Chapter 7 we mentioned two basic problems of elementary calculus: (I) The "slope problem," i.e., the problem of determining the slope of the line tangent to a given curve y == f(x) at a given point (x 0 , f(x 0 )) on that curve. This is the main problem of differential calculus. (2) The area problem: Given the curve y == f(x) and the lines x == a, x === b, and the xaxis, find the area of the geometrical figure having these as boundary. This is the main problem of integral calculus. ·we have already solved problem (1) in many special cases. It amounts to learning how to calculate the derivative f'(x) of the function y = f(x), for the derivative f'(x) is the analytic counterpart of the slope of the line tangent to the curve y == f(x). In this chapter we shall attack problem (2) and we shall learn of the very remarkable connection which exists between these two apparently unrelated problems. 161 The concept of an integral. Let us start with the elementary definition that the area A of the rectangle with sides h, b is given by
A
hb.
==
From this we can show that the area of a parallelogram is the product of its altitude h and its base b, A == hb, (1) for in Fig. 161 we see that a rectangle with these dimensions can be formed from the parallelogram by cutting along h and then moving the triangle so obtained to the position indicated on the right. It is important to note that we have used one of the fundamental properties of area: the socalled additivity, that is, if a geometrical figure is decomposed into nonoverlapping parts, then the area of the total figure is equal to the sum of the areas of the parts. Next we use (1) to determine the area A of a triangle. If hand bare the altitude and base, respectively (Fig. 162), then evidently A
==
!hb,
since two congruent triangles, each with altitude h and base b, form one parallelogram with these same characteristics. 504
161]
505
THE CONCEPT OF AN IXTEGRAL
   ;,, /
//
/
"/'
/
/
/
/
/
/
h
// // /
/ / / / /
...__._~~~~~'~
I~
b
I
FIGURE
/ /
b
161
FIGURE
162
FIGURE
163
Knowing the area of a triangle, we can calculate the area of a closed polygon (take the one in Fig. 163, for example). If the polygon is triangulated, i.e., decomposed into triangles with nonoverlapping areas, and if we assume that the area of the polygon is,iequal to the sum of the areas of the triangles, then the procedure is clear: we determine the area of each triangle according to the formula above, and add the numbers so obtained. The result is the required area of the polygon. This area was determined as a function of the consecutive vertices of the polygon in Section 412. To summarize: we have learned how to associate a unique number with every simple closed polygon in the plane, called the area of that polygon. Our present concern is to extend this procedure to more general geometrical figures. In this connection, we consider the following problem: Given: The geometrical figure bounded by the curve y == f(x), the lines x == a, x == b, and the xaxis (see Fig. 164). Find: The area of the given figure. Intuitively, our first reaction is that the area is "well defined" and the difficulty rests in the technicality of producing the number representing the area. But let us return for the moment to the determination of the area of the closed polygon given above. It was not clear that the area was "well defined" until a triangulation of the polygon was produced, thereby relating the area to that of figures (triangles) whose areas are known. Let y
y = f(:r)
I~
a
b FIGURE
164
x
506
ANOTHER SPECIAL LIMIT
[CHAP.
16
us try to carry this idea over to the present situation. In an attempt to do this, we encounter grave difficulties, for no matter how we try to decompose the figure (Fig. 164) we are always left with the problem of handling figures whose boundaries are "curves" [the curve y == f (x)] and since this is, in fact, the problem from the start, we have accomplished nothing. Thus we abandon, for the moment, the requirement of absolute precision and we replace it with the modest insistence on an "approximation" to the area of the figure. This idea goes back to Archimedes, who lived about 250 years before Christ and who applied the idea successfully. We begin by making the following crude approximations. First, if m is the minimum value of f(x) over the closed interval [a, b] (this minimum value always exists if f (x) is a continuous function on the closed interval),* we take the area of the rectangle with base b  a and height m as being an underestimate to what we intuitively think of as the area; likewise, if M is the maximum value of f(x) over [a, b], M(b  a) furnishes an overestimate. (See Fig. 165.) Thus, as conditions on the area A, we insist on the following: m(b  a)
0 is equivalent in (2) to x < 3, and this condition carries over to (3), so that what we really obtain is a halfline, as in the first instance. In other words, unless caution is used in eliminating the parameter of a pair of parametric equations, points which do not actually belong to the graph may be obtained.* EXAMPLE 171. Sketch the graph of
x = aa, y = a(l 
a = constant.
cos a),
We shall first eliminate the parameter a by solving for a in the first equation to get a = x/ a and then substituting this in the second equation: y
=
a 
x
a cos  · a
* Of course, the process of eliminating the parameter, besides sometimes producing an excess of points through carelessness, may also result in a deficiency of points. For example, if the parameter is eliminated from x = t, y = t by writing y/x = 1, then the point (0, 0) is lost in the process. Explain.
171]
531
PARAMETRIC EQUATIO:NS y
y
y=a
rrI
I
I
I
I
I I
x
21ra
1ra
I
I
y = a cos
FIGURE
ax
172
FIGURE
173
The amplitude of the wave a cos (xi a) is a and its period is 21ra; therefore its graph takes the form shown in Fig. 172. The required graph of x y = a  a cos a
may be obtained by observing, for each value of x, the vertical separation between the broken line y = a and the wave y = a cos (xi a), for this separation is the numerical value of the difference y = a  a cos (xi a). The final sketch appears in Fig. 173. To compute the area under one arch of this curve, we must find the value of the following integral: ( 211"a
lo
( 211"a (
a lo
y dx =
1 
x)
cos; dx
a[x  asin~J\:~a 2
21ra . EXAMPLE
172. Show that x
= 3 + 4 sin t, y =  7 + 5 cos t
is a parametric representation of an ellipse. First we write the representation in an equivalent way:
x  3 4
=
. sm t,
y+7

5
= cost.
If we now square the members of these equations and add, we have (x 
16 which is obviously an ellipse.
3)
2
+ (y + 7) 2 25
l,
532
PARAMETRIC REPRESENTATION OF CURVES
[CHAP.
17
172 Exercises In Exercises 1 through 7, eliminate the parameter t and sketch the graph.
+ t cos ~6 ,
y
+ t sin ~6
1. x
1
2. x
4 sin 2(}, y = 2 sin (}
4. x
3
2
=
2
+ 5 sin (},
y
=
3. x 2
! sin 2(},
y = sin
t3
4 cos (}
5. x 6 sin 0 8. Ix+ 31 < 1, IY  lj < 4, /z + 51 > 3 9. (a) The set of all endpoints of the vectors ca, where a = (1, 1, 1) [i.e., a is the vector from the origin to the point (1, 1, 1)] and O < c < 1. (b) Now consider c < 1, a the same as in (a). In Exercises 10 through 18 we are given the three vectors:
a = 3i + 4j 
5k,
b = 2i + 7j 
k,
c = i 
j + 3k.
Perform the following vector operations: 10. 14. 15. 16. 17. 18.
a+ b 11. a  b 12. 3(b  a) 13. 5a + 2b Find u such that a+ 2u = b. Compare aa + ab with a(a + b). Compare (a+ b) c with a+ (b + c). a  b c What is meant by (c + a)/2? Describe it geometrically.
+
(a 
c)
+
194 Distance. The formula for the distance between two points is of great importance in the development of threedimensional geometry. Suppose that two points P(x, y, z) and Q(u, v, w) are given (Fig. 196). These may be looked upon as two opposite vertices of the rectangular parallelepiped (ordinary rectangular "box") obtained by passing planes through each of P and Q parallel to all the coordinate planes. Referring to the figure, we see that 2
PQ == PS
2
+
SQ
2 ,
or (1)
194]
579
DISTANCE
z
z
Q
Q(u, v, w)
.,.:;.•~R P(x, y, z)
s 0 y
y FIGURE
196
FIGURE
197
In other words, the square of the length of a diagonal of a rectangular parallelepiped is equal to the sum of the squares of the lengths of its three mutually perpendicular edges. The problem of determining the distance PQ is therefore reduced to determining the lengths of the edges of the rectangular parallelepiped associated with the points P, Q. But this is a simple matter, for examination of the figure reveals the following coordinates for the points R, S: R(u, y, z) and S(u, v, z), and therefore the dimensions of the parallelepiped are PR == u 
x,
RS== v 
y,
SQ == w 
z.
Consequently, substituting in (1), we have THEOREM 191. The distance PQ between the points P(x, y, z) and Q(u, v, w) is given by the formula PQ ==
V (u

x) 2
+ (v 
y) 2
+ (w

z) 2.
This formula is entirely analogous to the corresponding one in the twodimensional case. Furthermore, if we wish to emphasize the "vectortheoretic" point of view, that is, if we consider, instead of the point P(x, y, z), the radius vector OP == xi + yj + zk, then the formula above gives nothing more than the length of the vector PQ* (see Fig. 197), for:
PQ == OQ (ui (u
OP
+ vj + wk)  x)i + (v 
+ yj + zk) y)j + (w  z)k. (xi
* We use boldface type PQ to denote the vector having initial point P, terminal point Q.
FUNCTIONS OF TWO VARIABLES
580
[CHAP.
19
Therefore, if we denote the length of the vector PQ by jPQI (it is also called the norm of the vector PQ), then we may express the content of Theorem 191 by IPQI == 'V(u  x)2 + (v  y)2 + (w  z)2. (2) In other words, since the components of PQ are u (2) can be interpreted as
x, v 
y, w 
z,
COROLLARY. The length of a vector is given by the square root of the sum of the squares of its components. 191. are collinear. EXAMPLE
Prove that the points P(4, 1, 0), Q(7, 0, 1), R(lO, 1, 2)
This can be accomplished by showing that PQ calculate the distances:
+ QR
=
PR. We therefore
+ (1  0)2 + (O  1) 2 = vii, QR = V(10  7)2 + (1  0)2 + (2  1) 2 = vii, PR= Vern  4) + c1  1) + c2  0) = V44 = PQ =
v1(4=
7)2
2
2
2
2vii,
and the rest is clear. EXAMPLE 192. Calculate the length of the radius vector OP if P has coordinates ( 1, 3, 4). The sum of the squares of the components of OP is 12 32 4 2 = 26. Therefore jOPI = v26.
+ +
195 The direction of a vector. It has been emphasized in earlier work that the notion of a vector embraces the ideas of both magnitude and direction. Calculation of the magnitude or length of a vector was undertaken in Section 194; at present we are concerned with specification of the direction of a vector. Let the radius vector OP == xi + yj + zk be given. The orientation or direction of this vector is completely specified by giving the components: (1) x: y: z, in the sense that if we start at an arbitrary point Q in space, go x units parallel to the xaxis, then y units parallel to the yaxis, and finally z units parallel to the zaxis, we shall arrive at a point R such that the vector from Q to R, QR, has the same direction as the given radius vector OP, i.e., the two are parallel (see Fig. 198). More generally, if we multiply the numbers (1) by a positive number a to get (2) ax : ay : az,
195]
581
THE DIRECTION OF A VECTOR
z
O x
y
y FIGURE
198
start at an arbitrary point Q and measure the distances ax, ay, az in the appropriate directions to arrive at a point R, then the vector QR will have the same direction as the given radius vector OP. The numbers ax:ay:az
are for this reason called direction numbers of OP. Formally, DEFINITION
195. Any set of numbers of the form ax : ay : az,
a
>
0
is called a set of direction numbers of the radius vector OP == xi
+ yj + zk.
193. Find a set of direction numbers for the vector emanating from P(1, 3, 5) and terminating at Q(4, 1, 5). EXAMPLE
If P is projected onto the xaxis, the point (1, 0, 0) is obtained; the projec tion of Q on the xaxis is (4, 0, O). Thus, in going from P to Q, one must travel from 1 to 4, i.e., units insofar as the direction parallel to the xaxis is concerned. In other words, an "xdirection number" is obtained by subtracting the xcoordinates:
+s
(xcoordinate of Q) 
(xcoordinate of P)
=
4 
(1)
=
5.
In a similar manner, ,ve obtain y and zdirection numbers; thus
5: 4: 0
will serve as a set of direction numbers for PQ.
Oftentimes it is found necessary to determine a unit vector in a given direction, that is to say, we are confronted with the problem: Given: the direction numbers x : y : z. Find: a unit vector in this direction.
582
FUNCTIONS OF TWO VARIABLES
[CHAP.
19
The problem is easily solved with the observation that if an arbitrary vector is divided by its length (in other words is multiplied by the reciprocal of its length), then since this process is multiplication by a scalar and therefore does not affect the direction of the vector, the resulting vector will have the same direction as the original one but its length will be unity. Here's a formal proof: We start with the vector xi yj zk. The magnitude of this vector is, according to the corollary to Theorem 191, the square root of the sum of the squares of its components:
+ +
vx2+y2+z2.
Now consider the vector
v'
1
x2
+ y2 + z2 (xi + yj + zk) v' x2 +
x y2
i
+ z2
+
y x2 +
y
+ z2
y2
j
+
v' x2 +
z y2
+ z2
k.
If we calculate its length by application of the same corollary, we get
+ +
Furthermore, since its components are those of xi yj zk multiplied by the positive constant l/v'x2 y2 z2, it follows that its direction yj zk. This process of replacing a vector is the same as the vector xi by a unit vector without altering its direction is called normalizing the vector.
+ + + + z
p \
xi+
yj
+ zk
\
\
z \ \pi
0~,+ x y
B
y FIGURE
199
THE DIRECTION OF
195]
VECTOR
A
583
A geometric interpretation of the direction numbers
x
vx2
.
y
z
.
+ y2 + z2 . vx2 + y2 + z2 . v~x2_+_y_2_+_z_2
can readily be found by reference to Fig. 199, as follows. Project the point Pon the xaxis to obtain the point P', as shown. Now OP'P is a right triangle (with right angle at P' and hypotenuse OP) and so if we set angle POP' === A, it follows immediately that OP' x cos A ===   ===      OP y x2 + y2 + z2
Similarly, if P" is the projection of P on the yaxis, and B === LPOP", OP'' cos B ==   == OP
,_Y_ __ y2 + z2
y x2 +
and if P"' is the projection of P on the zaxis and C === LPOP"', OP"' z cos c ==  ===      OP y x2 + y2 z2 The direction numbers
+
cos A : cos B : cos C are called the direction cosines and A, B, C are called the direction angles of the radius vector OP. We may now summarize our results: THEOREM 192. vector
The unit vector with same direction as the radius OP
== xi+ yj
+ zk
is the vector OP
IOPI
==
x
Vx2 + y2 + z2
i
+
y
vx2
+ y2 + z2
j
+
z
Vx2 + y2 + z2
whose components are the direction cosines of OP: cos A ===
x
Vx2 + y2 + z2
, cos B === _ _ _Y_ __ Vx2 + y2 + z2 z
cosC ===      ~ Vx2 y2 z2
+
+
The direction cosines of OP have the property that cos 2 A
+ cos 2 B + cos 2 C ==
1.
k
584
FUNCTIONS OF TWO VARIABLES
[CHAP.
19
EXAMPLE 194. Normalize the vector PQ, if P and Q are the points P(2, 3, 1) and Q(3, 1, 4).
The length of PQ is found to be
IPQI
=
~(2)]2
=
sV2.
+
(1 
3) 2
+ [4 
(1)]2
The normalized vector is therefore
PQ
=
IPQI
_5_i _ _4_j _ _3_k.
5V2
5V2
5V2
The direction cosines of this vector are consequently 1 4 3 ::,
V2
5V2
5V2
and the sum of their squares is 2
( 1 )
2
2 (
4 )
(
3 )
V2 + sV2 + sV2
thus verifying Theorem 192.
Sometimes we shall speak of the direction numbers or direction cosines of a straight line. By these we shall understand the direction numbers or direction cosines, respectively, of any vector parallel to the line. Here we are concerned only with the orientation of the line in space. For example, the line through P(2, 3, 1) and Q(3, 1, 4) has 5 : 4 : 3 as a set of direction numbers, also 5 : 4 : 3 would serve the purpose, since the vectors PQ and QP are considered to be parallel to this line. In other words, any set of direction numbers for a line can be multiplied by a nonzero constant (even a negative one) to produce another set of direction numbers for the same line. The role of the direction numbers of a line is analogous to that of the slope of a line in twospace.
196 Exercises In Exercises 1 through 5, calculate the distances between the pairs of points. 1. P(1, 3. P(2, 1, 4. P(1, 5. P(t 'YJ,
1, 3) and Q(l, 5, 4) 2. P(3, 1, 1) and Q(3, 0, 5) 1, 1) and Q(2, 2, 2) r) and Q(p, 0 to the effect that a2
+ b2 >
(4)
2ab.
First choosing a and b to be the numerical values of the xdirection cosines of OP and OQ, respectively, i.e., a
==
!xi
v x + y2 + z2
lul , b == ''Vu2
2
+ v2 + w2
we obtain, on substituting in (4), 2
x x2 + y2
+ z2
+
2
u u2 + v2
+ w2
> 
2
[xi
Vx2
+ y2 + z2
fuf V,u2_+_v_2_+_w_2 (5)
Next we take a and b to be the numerical values of the ydirection cosines of OP and OQ, respectively: a
IYI == ~'
V x2 +
y2
+ z2
b ==
lvl ,vu2·+v2_+_w_2
Then, according to (4), 2
2
v  y   + x2 + y2 + z2 u2 + v2 + w2
> 
IYI 2 ''
y x2
+ y2 + z2
!vi
vuJ: + v2
+ w2 (6)
588
FUNCTIONS OF TWO VARIABLES
19
[CHAP.
Finally, with
a ==
lzl V x2 + y2
+
w
, b ==
+ z2
lwl ,vu2+ v2
+ w2
,
(4) becomes 2
x2
+
z y2
+ z2
u2
2
> 2 [zf y x2 + y2
+ v2 + w2
+ z2
fwf VU2_+_v_2 _+_w_2 (7)
If we now add (5), (6), (7), it becomes apparent that 2
> 
2
lxl lul V x2 + y2
+ IYI lvl + lzl
lwl v2 + w2
'''''....:....__c__''''
+ z2
,vu2+
or
lxul
+
lyvl
+ lzwl < vx2+ y 2 + z2 vu2 + v2 + w2.
(8)
Now, by virtue of the triangle inequality for real numbers,
lxul
+ lyvl
lyvj
+
> lxu
+ yvl,
and therefore
lxul
+
lzwl > lxu
+ yvl +
lzwl.
But if we apply the triangle inequality for real numbers again to the right member of the last inequality, we get
lxu
+ yvj +
lzwl > lxu
+ yv + zwj.
Hence the left member of (8) may be replaced by the number
lxu
+
yv
+ zwl,
which it dominates, thereby leaving the sense of the inequality in (8) undisturbed and yielding
lxu
+ yv + zwl
0. Prove that (a  b, a  b) = (a, a)  2(a, b) (b, b)
+
+
+
2(a, b)
0, therefore
+ (b, b).
(f) Using (d) and (e)i prove the CauchySchwarz inequality for unit vectors:
l(c, d)I < /c/ Id/. (g) Using (f), prove the CauchySchwarz inequality for arbitrary vectors. 23. Give a fresh proof of the CauchySchwarz inequality by starting with the obvious statement (uv 
and expanding.
yu) 2
+ (xw

uz) 2
+ (yw

zv) 2
>
0,
590
FUNGrIO:NS OF TWO VARIABLES
[CHAP.
19
199 Cylindrical surfaces. Perhaps the simplest kind of surface is a cylindrical surface; the idea is contained in
197. Let a fixed plane curve C and a fixed line L intersecting C be given (Fig. 1910). By the cylindrical surface corresponding to the curve C and the line L is meant the collection of all points assumed by L as it moves parallel to its original position and always intersecting C. The line Lis called the generator of the cylindrical surface and the plane curve C is called the directing curve. DEFINITION
Some special examples are illuminating. In Fig. 1911 we have illustrated some of the more common cylindrical surfaces. When the curve C is taken to be a circle an~ the line L is perpendicular to the plane of this circle, we get a right circular cylinder, frequently abbreviated to simply cylinder. Also note that when C is a straight line and L is a different straight line, a plane is generated as the corresponding cylindrical surface. If the directing curve C is taken to be a sine wave and the generator L is taken perpendicular to the plane of this curve, then a surface resembling "corrugated iron" is obtained. Thus some familiar surfaces turn out to be cylindrical surfaces. Not all surfaces, however, fall in this category. A natural counterexample is the sphere, for if we select a point on a sphere, there is no straight line which lies completely in the sphere and contains the z
c
L y FIGURE
:
1,L
I
I
I
I
I I
I
I I
I
I
I
I
.J .J_ L
c
I
I I
I;
I I
;
I
I I I I
I
I/ I I
I
I
I/ / I
I
1910
L
L
I
I I ; I I I I I I I I I I / I ;
I
I
I
L
I
::....L~ ..../
c
c FIGURE
1911
c
199]
CYLINDRICAL SURFACES
591
z
L
Q
___ ,___..,_x c
FIGURE
1912
point, as would have to be the case if we were dealing with a cylindrical surface. In the special case where the directing curve C lies in a coordinate plane and the generator L is parallel to the remaining perpendicular coordinate axis, the mathematical description of the resulting cylindrical surface is particularly simple and is essentially a twodimensional problem, as we now see. Suppose, to be definite, that the directing curve C lies in the xyplane and the generator Lis parallel to the zaxis (Fig. 1912). The equation of C may be written as f(x, y) == 0 (1) if we first disregard the zaxis. [For example, C might be the parabola y  x 2 == 0 or the unit circle x 2 y 2  1 == 0; in the first case 2 f(x, y) == y  x , in the other f(x, y) == x 2 y 2  1.] We now select a point Pon the curve C. Then P, considered as a point in twospace, has coordinates (x, y), and if considered as a point in threespace, it has coordinates (x, y, O). In any event, the coordinates of P satisfy the relation
+
+
f(x, y) == 0, since Plies on C. Now we move from P along a line parallel to the generator L (hence parallel to the zaxis). In so doing, the x and ycoordina tes will not differ from those of P and consequently will continue to satisfy Eq. (1). Therefore, by definition of a graph, all the points Q on the line through P parallel to L belong to the cylindrical surface. Conversely, it is clear that any point belonging to the cylindrical surface determined by C and L may be obtained in the manner described. This proves the 196. An equation in two of the three variables x, y, z represents a cylindrical surface whose generator is parallel to the axis of the missing variable. THEOREM
592
FUNCTIONS OF TWO VARIABLES
[CHAP. 19
z
( 0, 0, 2)
y
FIGURE 1913 EXAMPLE 195. Sketch the first octant portion of the graph of x~
+ (z 2 /2)
=
2.
Since the variable y is missing, we know that the surface is a cylindrical surface whose generator is parallel to the yaxis. Taking y = 0 (so that we are limiting ourselves to the xzplane and therefore have a twodimensional situation), we sketch onequarter of the ellipse 2
2
;+;=1
(i)
(see Fig. 1913). If we select an arbitrary point P(x, 0, z) on this ellipse, as we move along a line parallel to the yaxis only the ycoordinate of P varies and, consequently, since the coordinates of P satisfy the equation (i) and y does not appear in this equation, all points along this line satisfy (i). But P was an arbitrary point on the ellipse. The conclusion is that (i) represents an elliptic cylinder with generator parallel to the yaxis.
1910 Exercises Sketch and identify the surfaces or solids given in Exercises 1 through 19.
+ +
+
1. x 2  x y2 2y = 0 2. 3x 6y = 12 2 3. 2y  4y x = 0 2 4y 2 56y 121 < 0 4. 25x  50x 5. 16x 2  9y 2 128x 90y  113 = 0 6. y 2  4x 2 8x  lOy 21 = 0 7. x 2  2x = y 8. 2x  5y < 8 9. (x  3) 2 = y 10. x 2 < 3  y 36 = 0 12. x 2 y 2  36 = 0 13. x 2  y 2 = 36 11. x 2  y 2 14. y  2 sin! = 0 15. y 2 z2 < 36 y 2  6x  lOy = 15 17. y  3x = 0 16. x 2 18. y = log (x  2) 19. z2 (x  1) 2 = 4 20. Find the equation relating x and y if x = 3  sin t, y = cost  5. Interpret this equation as (i) a locus in the xyplane, (ii) a locus in threespace.
+
+ + +
+
+
+ + + +
+
+
+
1911]
593
EQUATION OF THE PLANE
1911 The equation of the plane. We have seen that any two lines having direction numbers a 1 : b1 : c 1 and a 2 : b2 : c2, respectively, are perpendicular if and only if the inner product of the radius vectors determined by the points (a 1 , b1 , c 1 ) and (a 2 , b2 , c2 ) vanishes:
Now a plane may be determined by specifying the direction of a line N normal (i.e., perpendicular) to it, together with the point F of this normal line through which the plane passes (Fig. 1914) . It is convenient for our purposes to lay down the
FIGURE
1914
198. A plane is the collection of all points P such that there exists a fixed point F and a fixed vector FN for which FP is orthogonal to FN. DEFINITION
Remark. We use the words orthogonal, normal, and perpendicular to mean the same thing, namely, that a right angle is formed by the lines which are described as being orthogonal or normal or perpendicular. We now use (1) together with the definition of a plane to find the equation of a plane. Let the given point of the plane be F(h, k, l) and let the perpendicular to the plane have a : b : c as direction numbers. If P(x, y, z) js an arbitrary point of the plane, then the vector FP lies in the plane and it may be written as FP
== (x  h)i
+
(y 
+
bj
k)j
+
(z 
l)k.
(2)
Since FP is perpendicular to ai
+ ck,
(3)
it follows that the inner product vanishes:
a(x 
h)
+ b(y

k)
+ c(z
l) == 0.

This is the required equation. If we expand, we have
ax
+ by + cz + (ah

bk  cl) == 0.
This, however, may be written as
ax
+
by
+ cz + d ==
0,
(4)
where d is the constant ah  bk  cl. This proves that every plane has a linear equation in x, y, z of the form (4).
594
FUNCTIONS OF TWO VARIABLES
[CHAP.
19
Conversely, if an equation of the form (4) is given, it represents a plane. For, let (h, k, l) be any point satisfying (4). Then ah
+ bk + cl + d
0,
===
or d === ah 
bk 
cl.
Substituting this in (4), we obtain
ax
+ by + cz
bk  cl
 ah 
===
0,
or
+ b(y
a(x  h)
+ c(z
k)


l) === 0.
(5)
However, Eq. (5) asserts the orthogonality of the vectors (2) and (3) and consequently it represents a plane through (h, k, l) with direction numbers a : b : c for its normal. Hence: THEOREM 197. Every plane has an equation which is linear (first degree) in x, y, and z:
ax
+ by + cz + d ===
0.
Conversely, every linear equation in x, y, z represents a plane. Furthermore, the coefficients a, b, c of the variables are direction numbers a : b : c of the line normal to the plane. EXAMPLE 196. Find the equation of the plane perpendicular to the segment NQ and passing through N, where N(l, 1, 3), Q(3, 1, 2).
The direction numbers of the vector NQ are 2 : 2 : 5. If P(x, y, z) is any point on the plane, then NP J_ NQ. But the direction numbers of NP are x  1 :y 1 : z  3. Therefore, NP J_ NQ is equivalent to the vanishing of the inner product of these vectors, i.e.,
+
2(x 
1)
+ 2(y + 1) 
5(z  3) = 0.
This equation can also be written as 2x
+ 2y 
5z
+ 15
=
0.
An alternative solution is this: since the direction numbers of the normal NQ to the plane are 2 : 2 : 5, these are the coefficients of x, y, z in the required linear equation: 2x 2y  5z d = 0.
+
+
It remains to determine d. This is easily accomplished, since the plane must pass through N(l, 1, 3) and therefore
2(1)
+ 2(1)

5(3)
+d
0,
THE STRAIGHT LINE
1912]
595
15, giving again the result
which implies d
2x
+ 2y 
5z
+ 15
0.
=
1912 The straight line. We consider the problem of having been given two points P 1 (xi, y 1 , z 1) and P 2(x 2, Y2, z2), to find the conditions which points on the line through P 1 and P 2 must satisfy (Fig. 1915). We note
that the direction numbers of the required line are
For short, let us denote these by a : b : c. Let P(x, y, z) be an arbitrary point on the line. Then, in the language of vectors, (1)
where t is an arbitrary real number. Now let us switch to components. We have (2) x == x 1 at, y == Y1 bt, z == z1 ct,
+
+
+
since the components of OP are (x, y, z), those of OP 1 are (x 1 , y 1 , z 1), and those of P 1P 2 are (a, b, c). The equations (2) are called the parametric equations of the line through P 1 , P 2 , and tis the parameter. Conversely, if condition (1) holds for some value of t, then the point P(x, y, z) will lie on the line through P 1 , P 2 • Therefore: THEOREM 198. The (parametric) equations of the straight line through the points P 1 (x 1, y 1 , z 1) and P2(x2, Y2, z2) are x
==
x 1
+ at,
y
==
Y1
+ bt,
z
==
z1
+ ct,
where a : b : c are direction numbers of the line (e.g., a == x 2 b == y 2  y 1 , c == z 2  z 1 ) and t is an arbitrary real number. z
y FIGURE
1915
(2) 
x 1,
596
FUNCTIOKS OF TWO VARIABLES
[CHAP.
19
Sometimes it is convenient to eliminate the parameter in (2). This is easy to do; we simply solve each equation for t and equate the results:* Y 
Y1

(3)
b
The equations (3) are called the symmetric equations of the line. 197. Find the symmetric equations of the line through P(2, 1, 4) and Q(3, 7, 7). EXAMPLE
The direction numbers of the segment QP are 5 : 8 : 3; therefore, using P explicitly, the required equations are x  2 5
Y+l

8
z  4 3
,
or (using Q)
y7 8
x+3

5
z  7 3
·
To reconcile us to the fact that these equations yield the same line, let us compare the equations in x and y which we get by means of the first equality in each case: 8x 16 = 5y 5, 8x  24 = 5y  35.
+
+
+
We see that each of these equations says the same thing: 8x 5y  11 = 0. In the same manner, the reader should compare the remaining two sets of equations. To obtain points on the line z  4 x2 y+l , 5 8 3 we must choose values for the variables. Thus, if x = 6, we find that y =  3s7and z = J. If we take y 0, we get x = ¥ and z = 3g5. If we take x = 2, we obtain P; if we take z = 7, we obtain Q. EXAMPLE
198. Find a point on the line
3x  4 7
2z
+9 2
and determine direction numbers for this line. We shall accomplish both things simultaneously. The idea is to first convert the equations to symmetric form, for then the required direction numbers and a point on the line will be apparent. On dividing the numerators and denomina
* Of course, if one of the direction numbers a, b, c is zero, this cannot be done. We leave it to the reader to interpret the situation.
1912]
597
THE STRAIGHT LINE
z
FIGURE
1916
tors of the fractions by the appropriate numbers, we find that the equations above transform to the equivalent ones:
x! f
y+3
·
1
1
=  =
z+!
It is clear that a set of direction numbers would be in addition, the point (i, 3, !) lies on the line.
i : 1
: 1 or 7 : 3 : 3;
We now consider the following problem (see Fig. 1916): Given: The plane ax by cz d == 0 and the point P 0 (x 0 , y 0 , z 0 ). Find: The (perpendicular) distance between the point and the plane. The parametric equations of the line through the given point and perpendicular to the plane are
+ + +
x == x 0
+ at,
y == Yo
+ bt,
z
z0
==
+ ct,
(4)
since a : b : c are the direction numbers of any line perpendicular to the plane ax by cz d == 0. We may determine the point where the line (4) pierces the plane
+
+ +
ax
+ by + cz + d ==
0
(5)
by deteimining t such that the coordinates of a point on (4) satisfy (5), i.e.,
a(x 0
+ at) + b(y + bt) + c(z + ct) + d == 0
0
0
or
t(a 2
+ b2 + c2 ) + d + (axo + byo + czo)
that is,
t == axo  by 0 a2 b2
cz 0
+ + c2

d
== 0,
598
FUNCTIONS OF TWO VARIABLES
[CHAP.
19
Therefore, the point P(x, y, z) of intersection of (4) and (5) has as its coordinates the numbers
_
(ax 0

by 0

cz 0
+a
_ y  Yo
+b
(axo  byo  czo a2 + b2 + c2
+c
(axo  byo  czo a2 + b2 + c2
_
z  zo
a2
+ b2 + c2

Xo
x 
d)
'
d) '
(6)
d) .
We may calculate the distance P 0 P separating the point P 0 and the plane by using (6) to form the differences x  x 0 , y  Yo, z  z 0 , squaring these, adding, and extracting the positive square root:
V(axo  byo  cz 0  d) 2 PoP== ~~~~~~~~~~
Va2+b2+c2 laxo + byo + czo + di Va 2 + b2 + c2
This proves THEOREM 199. The distance D between the point P 0 (x 0 , y 0 , z 0 ) and the plane ax + by + cz + d == 0 is the numerical value of the number obtained by substituting the coordinates of P 0 in the left member of the equation of the plane and dividing by the length of the normal vector with components a : b : c:
+ byo + czo + di . Va 2 + b2 + c2
D == laxo
Remark. This expression is entirely analogous to the formula for the distance between a point and a line (Theorem 37). EXAMPLE
199.
Compute the distance between the planes
+z 2 3y + z + 8
x  3y
=
0,
x 
=
0.
We first obtain a point on one of the planes. For convenience, put x = 0 = y in the first equation; then z = 2. Thus (0, 0, 2) lies on the first plane. The distance D between it and the second plane is D
+ 2 + 81 V12 + (3)2 + l2 10 
3(0)
10
=·
V11
1913]
599
EXERCISES
1913 Exercises 1. Find the equations (symmetric and parametric) of the line (a) perpendicular to the plane 3x  5y 7 z  2 = 0 and through (3, 1, 4); (b) perpendicular to 3x 7y  z 2 = 0 and through the zintercept of that plane. 2. Find the equation of the plane (a) parallel to 3x  5y 2z  2 = 0 and through (5, 6, O); (b) parallel to 3x 2y  z 7 = 0 and through (1, 1, O).
+
+
+
+
+
+
Find the symmetric and parametric equations of the lines through the pairs of points given in Exercises 3, 4, and 5. 3. (2, 1, 4) and (5, 3, 1). 4. (1, 2, 4) and (3, 1, 5). Does the point (1, !, 4) lie on this line? 5. (1, 4, 3) and (5, 2, 7). Does the point (3, 1, 5) lie on this line? 7z  2 = 0 and 6. Calculate the distance between the planes 3x  5y 3x  5y 7z 8 = 0. 7. Show that any line with direction numbers 1 : 1 : 1 is parallel to the plane 8x 12y  4z 31 = 0. 8. Find the equation of the plane perpendicular to 3x  5y 7z  2 = 0 and through (2, 0, 3).
+
+ +
+
+
+
Find the equations of the planes through the points indicated in Exercises 9 and 10. 9. (1, 0, O), (0, 1, O), and (0, 0, 1). 10. (a, 0, O), (0, b, 0), and (0, 0, c). 11. Give the equations of (a) the xaxis, (b) the yaxis, and (c) the zaxis. 12. A cube of edge 4 in octant I has a vertex at the origin a_!ld the coordinate axes as edges. (a) Sketch the cube, indicating coordinates of vertices. (b) Find the direction cosines of the two diagonals originating at the zaxis. (c) Find the cosine of the angle between the diagonals in (b). Are they perpendicular? (d) What is the length of a diagonal? (e) Write the equation of the plane determined by the two diagonals not mentioned in (b). (f) Write the equation of the sphere (i) inscribed in this cube; (ii) circumscribed about this cube. 13. Find the value of c such that the plane 2x  3y cz = 4 shall be perpendicular to the plane 3x y  2z = 1. 14. Find the equations of the planes which are parallel to x  4y 8z = 0 and 5 units from it. 15. Find the equation of the plane parallel to 4x  5y 2z  12 = 0 and through (2, 0, 1).
+
+
+
+
Sketch the graphs of the given planes or lines in Exercises 16 through 20. 16. 18. 19. 20.
+ + 3z
2x y 2(x  3) x = 2 x  1 =
+
=
17. 2x
6
+ 4y

z
=
3
= 5(y  1) = 1  z 5t, y = 3  t, 2z = 3 + St z 
2 = 3y
Calculate the distance between the planes given in Exercises 21, 22, and 23. z = 3, 3x  4y z  5 = 0 21. 3x  4y 22. x  y 2z = a, 5x  5y lOz = b 23. x  y = 4, 2x  2y = 7
+
+
+ +
600
FUNCTIONS OF TWO VARIABLES
[CHAP.
19
Shade the locus of points satisfying the conditions of Exercises 24 through 27.
+ + + +
24. 2x 3y 5z < 60, z > 0, x > 0, y > 0 2 2 25. x y z2 < 1, x y z > 1 26. 5x  2y 3z = 1, 3x  5y 7z = 2, !xi > 1 27. jxj < 1, IYI < !, lzl < i 28. Find the equation of the plane which passes through the midpoint of the segment joining the points (1, 2, 4) and (4, 3, 4), and perpendicular to it. 29. Find the equations of the line through (6, 1, 3) perpendicular to each of the lines
+ +
+
~ = y  1 = z 2
3
+2
+
x  1 = y +_2 = ~=_!._ .
and
5
3
2
6
*30. What is the locus of the points of intersection of the planes x  2y 3z = 4, 2x 3y 4z = 8. [Hint: First eliminate x, then y, then solve the resulting equations for z.] *31. Develop the "normal equation of a plane": (a) Let ax by cz d = 0 be the equation of a plane. Show that the same plane may be represented by the "normal equation":
+ +
+ + +
a
x+
YA
b2
+
c2 .
b
V a2_ + b2 + c2
Y+
c
V a2 + b2 + c2
+
z d
'Va 2 + b2 + c2
0.
Show that the coefficients in this last equation, however, are direction cosines of the normal to the plane being represented. Show also that the length of the vector perpendicular to the plane and emanating from the origin and terminating on the plane is
'Va 2 + b2 + c2 [Hint: Compute the length of the projection of a radius vector OP (P on the plane) on the unit vector normal to the plane.] (b) The vector ON with direction numbers X : µ : v and length l is perpendicular to a plane passing through N. Find the equation of the plane.
1914 Quadric surfaces. Quadric surfaces play a role in threedimen
sional geometry that is analogous to the role of the conic sections in plane geometry. These are the surfaces, or graphs, of equations of the second degree in three variables x, y, z. The procedure for sketching such graphs is best explained by resorting to examples. EXAMPLE 1910. Sketch the surface 4x 2
+ 4y + z 2
2
1.
(i)
1914]
QUADRIC SURFACES
601
z
/ /
,
/
/
~' x
y
1917
FIGURE
It is helpful to find the traces of the surface, i.e., the curves of intersection of the surface with the coordinate planes. Thus the xytrace is obtained by putting z = 0, which yields 4x 2 4y 2 = 1.
+
This circle is the curve in which the surface (i) cuts the xyplane. It is drawn in Fig. 1917 (actually, we have drawn only onefourth of it in order not to clutter up the diagram; we frequently sketch only the firstoctant portion of the graph). To find the xztrace, we set y = 0: 4x 2
+z
2
=
1.
The resulting ellipse (that is, onefourth of it) is shown. Finally, setting x leads to the yztrace: 4y 2 z2 = 1,
O
+
another ellipse. If a complete picture of the surface is still lacking, one can "probe" with other planes, and study the curves of intersection of the surface with suitably chosen planes. For example, if we put z = ! in (i) we shall be insisting that the points under immediate consideration lie in a plane ! unit above the xyplane and we shall therefore obtain the section of the surface (i) in this plane z = ! ; it is 4x 2 4y 2 i = 1, or
+
+
a circle with center at (0, 0, !), of radius \/'3/4 (this section is shown as a broken curve in Fig. 1917). More generally, one can study sections made by planes passed parallel to the coordinate planes by setting the appropriate variables equal to constants. For
602
FUNCTIONS OF TWO VARIABLES
example, if we want the section intercepted by the plane x x = :! in (i): 4(/6 ) 4y 2 z2 = 1,
+
[CHAP.
!,
19
we put
+
or 4y2
+ z2
=
i,
immediately identifiable as an ellipse. To get a more general picture of sections formed by passing planes parallel to the yzplane, we put x = k (k constant) in (i): 4y 2 z2 = 1  4k 2.
+
From this last equation, we can see that ellipses are obtained for values of k such that 1  4k 2 > 0, that is, k2
< !,
k
< !.
or
! < If it happens that
4k 2
1 so that k
>!
0,
±!, we get "point" ellipses [the points (±!, 0, O)]. Finally, if 4k 2
1 i.e., if k
=
or k
< !,
1, "
E 2 is the event "the second roll resulted in a point
>
2,"
E 1 n E 2 is the event "the first roll resulted in a point
> >
1
then
and the second roll resulted in a point
2. "
The results of the experiment can be designated by ordered pairs of numbers, (j, k), where j and k denote the number of points obtained on the first and second rolls, respectively. Thus there are 6 · 6 == 36 possible outcomes. E 1 can occur in 30 ways, E 2 in 24 ways. Therefore P(E) 1
_

30
_
36 
5
6,
P(E) 2
_

24 _
36 
2
3'
We must now compute p(E 1 n E 2 ). Those ordered pairs (j, k) having at least a 2 in the first position and at least a 3 in the second position are 5 · 4 == 20 in number. Consequently,
But
and so (3) holds in this case. K ote that in the first case, the events E 1 and E 2 are intimately related; as a matter of fact, there E 2 implies E 1 . On the other hand, in the second case E 1 has nothing to do with E 2 •
207]
DEFINITION OF PROBABILITY
631
This suggests DEFINITION
207. Events E 1 and E 2 for which the relation
holds are said to be independent. We may clarify the situation further, besides making it easier to carry out computations, by introducing the notion of conditional probability. Suppose tha.t two cards are drawn in succession from a pack of 52. Let
A denote the event "the first card drawn is an ace," B denote the event "the second card drawn is an ace."
We raise the question: What is the probability that both cards drawn are aces? To compute this probability, we introduce the following notation: the result of drawing two cards in succession without replacement* will be denoted by an ordered pair of letters in which a denotes that an ace is drawn and n denotes that a nonace is drawn. Thus (n, a) represents an experiment in which the first card drawn is a nonace, the second is an ace. With this notation, the event A consists of all elements of the form (a,_), where the second entry may be either an a or an n. The event B, on the other hand, consists of all elements of the form (_, a), the first entry being arbitrary. Finally, the event A n B consists of all elements of the form (a, a). Now we have number of elements of form (a, a) p(A n B) total number of pairs possible The numerator may be evaluated by the fundamental counting theorem. There are 4 aces, any one of which may fill the first entry, and since the experiment is performed without replacements, only one of 3 aces may be selected on the second trial. In all, there are 4 · 3 === 12 elements of the form (a, a). Again by the fundamental counting theorem, 52 cards may be drawn on the first trial and 51 on the second, therefore there are 52 · 51 possible outcomes. Thus 4.3 p(A n B) 52 · 51 To analyze this result, we decompose it as follows: p(A
n B) ===
5\
·
s31 ,
(4)
* "Without replacement" means once a card is drawn, it is not returned to the pack.
632
COUNTIXG. AN I~TRODUCTION TO PROBABILITY
[CHAP.
20
and recognize that the first factor, 5\ , may be interpreted as the probability p(A) of drawing an ace on the first draw. Furthermore, the second factor, 5\ , is the probability of drawing an ace on the second trial under the condition that an ace was drawn on the first trial. To see this, note that, assuming an ace to have been drawn on the first trial, there remain only 3 aces and 51 cards, hence the probability is 5\ that an ace will be drawn on the second trial, given that one was obtained on the first trial. This suggests
208. By the conditional probability of B given that A has occurred, written p(BIA), is meant the number DEFINITION
P
(BIA)
==
p(A n B) . p(A)
In the notation introduced here, (4) may be written as
p(A n B)
==
p(A)p(BIA),
(5)
and it is in the form (5) that conditional probabilities find wide application. Remark. It is important to observe that if the events A and Bare independent, and consequently
p(A n B)
==
p(A) · p(B),
then (5), reads
p(A) · p(B) Therefore, assuming p(A)
~
==
p(A)p(BIA).
0, we have
p(B)
==
p(BIA),
which says that the probability of B is unaffected by the occurrence or nonoccurrence of A, which further justifies our definition of independence of events. EXAMPLE 2015. Two cards are drawn simultaneously from a pack of 52. What is the probability that both are spades?
We obviously lose no generality in assuming that the two cards are drawn in succession, without replacement. Letting E1 and E2 denote the events "the first card drawn is a spade" and "the second card drawn is a spade," respectively, we have according to (5), (i)
We also have
208]
633
EXERCISES
To calculate p(E2IE1), we assume that a spade has been drawn first and that a second card is being selected. Under this hypothesis, there are 12 spades and 51 cards left. Therefore
Substituting in (i), the required probability is found to be
EXAMPLE 2016. Two cards are drawn simultaneously. What is the probability that at'least one is a spade?
Let Eo denote the event "none of the cards is a spade"; let E1 mean "exactly one of the cards is a spade"; let E2 mean "exactly two of the cards are spades." Also, let E denote the event "at least one card is a spade." Then
and since E1
n
E2
~' according to Theorem 206 we may write (i)
To compute p(E1), note that exactly one spade is drawn if and only if the first card is a spade but the second one isn't, or the first card is not a spade but the second one is. The probability that one of these will occur is .!.~ • .Qj! 52
51
[by use of (5)], and therefore, by Theorem 206, p(E1) = 2 ·
g · ~i ·
Also [again by (5)], P( E2)  l52i · l51i·
Substituting these values in (i) yields p(E) = =
13
52~ (78 +
12)
15 34.
208 Exercises 1. If two integers are chosen from the integers 1 through 15, inclusive, what is the probability that (a) the sum is odd? (b) the sum is even? 2. In a lottery where 3 cars are prizes, a man has 5 tickets out of 2000 on the first car, 3 tickets out of 8000 on the second car, and 16 tickets out of 15,000 on the third car. Find the probability that he will draw (a) one car, (b) no car, (c) at least one car, (d) all three cars.
63!
COUNTING. AN INTRODUCTION TO PROBABILITY
[CHAP.
20
3. In a certain box of chocolate candy, 3 pieces out of 30 have cherry centers; in another box, 2 pieces out of 22 are of this kind. One piece of candy is chosen at random from each box. What is the probability that (a) both, (b) at least one, (c) exactly one will have a cherry center? 4. Twelve coins are thrown simultaneously. What is the probability that the number of heads obtained is greater than 5 but not greater than 7? 5. In the manufacture of a certain item the probability of its being defective is 0.01. In a sample of 50, what is the probability of finding at most 2 defective items? 6. Of 12 items in a certain "grab bag," an individual would be pleased to have 3 of these, is indifferent to owning 5 others, and would dispose of the rest. If he purchases 3 of the items, what is the probability that (a) he will be pleased with all? (b) he will not have to dispose of any? 7. A box contains 5 black and 3 white balls. Three balls are drawn in succession. What is the probability of drawing (a) 3 white balls? (b) 2 black balls and 1 white ball? 8. If 2 dice are tossed 4 times, what is the probability of obtaining a point of 9 exactly 3 times? 9. Two dice are tossed simultaneously. What is the probability of obtaining 7 or more? 10. A coin is tossed 3 times. Find the probability of exactly 3 tails. 11. A crate contains 1000 oranges, 100 of which are rotten. If you pick 3 oranges at random, what is the probability that 2 or more are rotten? 12. A box contains 3 blue flags, 2 red ones, and 1 yellow one. What is the probability of drawing 1 red, 1 blue, and 1 yellow flag if one selects 3 flags blindfolded? 13. In how many ways can a committee of 5 be selected from 8 men and 4 women (a) if no conditions are imposed? (b) so as to include exactly 1 woman? (c) so as to include at least 2 women? 14. An examination of 10 questions including one proof is to be chosen from 50 review questions (5 of which are proofs). Ten percent is to be scored for each correct answer; no partial credit will be given. In how many ways can the 10 questions be selected so that 80 or above will be the grade of a student who can correctly give all the proofs but, because of difficulties in algebra, can answer only 20 other questions? (Set up, but do not calculate.) 15. If an integer between 1 and 100 inclusive is chosen at random, what is the probability that it will end in a 7? 16. A quadratic equation is written, with coefficients chosen at random from the numbers 1, 2, 3, no number being used twice. What is the probability that the equation will have imaginary roots? 17. From a group consisting of 5 men and 7 women, a committee of 4 is to be chosen by lot. Find the probability that the committee will consist of at least 3 men. If he 18. The probability of A's winning a game whenever he plays is plays twice, find the probability that he will lose at most one game. 19. Four dice are all to be rolled at once. What is the probability that (a) at least 2 will be 3's? (b) exactly 3 will be 4's?
i
209]
635
THE BINOMIAL FORMULA
20. Along a certain busy stretch of highway, it is estimated that a driver will have an accident once in 121 times, and will sustain injuries once in 208 times. What is his probability of (a) an accidentfree round trip? (b) an accidentfree round trip daily for two weeks? (c) being hit and hurt? (d) being hit and unhurt? (e) being unhit and unhurt? [Use logs to obtain answers in decimal form.] 21. The letters of the word probability are written on cards, one letter to a card. The cards are shuffled and laid face up one after another. What is the probability that they will spell "probability"? *22. Study Definition 207 carefully. Suppose two events E1 and E2 are disjoint, i.e., mutually exclusive. Then are they also independent? What about the converse? That is, if two events are independent, does it follow that they are disjoint (study the example preceding Definition 207 in search of an answer). *23. Use mathematical induction to extend Theorem 20f to n events instead of two, i.e., prove that if
0 for i p(E)
~
j, then
=
p(E1)
+ p(E2) + · · · + p(En).
*209 The binomial formula.
The ideas here are best brought out
by an example. 2017. We know that in a single throw of a die, the probability of obtaining a four is !, and therefore the probability of obtaining a face not a four is f. A die is tossed 5 times. What is the probability of obtaining a 4 exactly 3 times? EXAMPLE
The outcome of rolling a die 5 times can be indicated by an ordered "p~ntuplc," each component of which is one of the integers 1, 2, ... , 6. For example, (4, 2, 3, 4, 4)
means that a 4 was rolled on the first, fourth, and fifth rolls, while a 2 and a 3 were rolled on the second and third rolls, respectively. What is the probability of occurrence of the event (4, n, n, 4, 4), where n denotes "not a 4"? Since the individual rolls are independent of each other, the probability p(4,
n, n, 4, 4)
is simply the product of (the probability·of rolling a 4 on the first trial) X (the probability of not rolling a 4 on the second), and so forth. In other ,vords,  .! 4 4) P (4 , n 'n' , 6
.
_Q_ • _Q_ ••1.
6
6
6
.! 6
636
COU~TING. AN INTRODUCTION TO PROBABILITY
[CHAP.
20
'\Ve may call the event (4, n, n, 4, 4) a "favorable event," in the sense that exactly three 4's were rolled. However, all favorable events can be obtained by merely permuting the symbols occurring here. Thus (n, 4, n, 4, 4) as well as (n, 4, 4, 4, n) is a favorable event in this sense; as a matter of fact, there are, all told, as many favorable events as there are permutations of 5 things taken 5 at a time with three alike (the 4's) and two others alike (the n's), i.e.,
5! 3!2! However, this number is the same as the number of combinations of 5 things taken 3 at a time:
5! (5)
3!2!=
3.
It is therefore the coefficient in the fourth term of the binomial expansion of (a+ b) 5 • The favorable events just counted are mutually exclusive; for example, the occurrence of (4, n, n, 4, 4) precludes the possibility of occurrence of (n, 4, n, 4, 4) or one of the other favorable events, and therefore the probability of occurrence of one or the other of them is the sum of the probabilities of occurrence of each of them (by Theorem 206). But this sum is simply (!)3(!) 2 used as a summand t•1mes, 1.e., . (5) 3
p (exactly three 4's in 5 rolls)
=
mGY (~Y.
Hence putting a = f and b = ! in (a+ b)5, we find that the required probability is the fourth term in the binomial expansion of (a+ b)5, for
(5 1)  (5)5
+6 6
_
6
5
(5) (1) +4. 5(5) (1)
+5 6
4
3
1
6
2!
6
2
6
Let us see if we can use our example to make a conjecture about a general theorem which might be utilized in solving problems of this kind. First of all, note that we are concerned with repeated~ trials; in the example, we repeatedly rolled a single die.* Also, the outcome on each trial is that either a certain event occurs or it doesn't and, moreover, the probabilities for each of these alternatives is fixed (in our example, ! is the probability of rolling a 4, f is the probability of not rolling a 4). Finally, the events are independent (in our example, the result of any one of the rolls ex,erts no influence on the others). Experiments satisfying
* Instead of rolling one die 5 times, we could have rolled 5 dice once, of course.
209]
THE BINOMIAL FORMULA
637
these conditions are called "binomial experiments" because the "binomial formula" applies to them: THEOREM 208. The binomial formula. Let p be the probability of occurrence of a certain event E and let 1  p == q be the probability of its nonoccurrence on each of a sequence of trials. Suppose that n repeated trials are performed. Then the probability that E occurs exactly k times in n trials is given by
We note that this number is the (k pansion
+ l)th term in
the binomial ex
Proof. Define a favorable ntuple to mean here that in n trials, E occurs exactly k times. Consider the favorable ntuple
C,, ... ,) in which we have precisely k f's (f for favorable, denoting occurrence of E) and the balance of the positions are filled with n's (n for nonfavorable, denoting nonoccurrence of E). To compute the probability of such a favorable ntuple, we have only to multiply the probabilities of the entries (since the individual outcomes are independent); we obtain (1)
since p is the probability of each of the k f's and q is the probability of each of the (n  k) n's. But how many favorable ntuples are there? As many as there are permutations of n things taken n at a time with k alike and (n  k) others alike, i.e.,
n!
k!(nk)!==
(n) k ·
(2)
However, the probability of occurrence of Ek times in n trials is the sum of the probabilities of occurrence of each of the mutually exclusive favor~ble ntuples, and since each favorable ntuple has a probability of occurrence given by qnkpk and there are all told (k) favorable ntuples, the required probability is k q (n)
nk k
which proves the theorem.
p'
638
COUNTING. AN INTRODUCTION TO PROBABILITY
[CHAP.
20
2018. A certain machine normally turns out 3% of defective items, 97% of acceptable items. What is the probability that in a sample of 12 items exactly 3 will be found defective? at most 3 will be found defective? EXAMPLE
Setting p
=
0.03, q
=
0.97, n
=
12, and invoking the theorem, we find
(132) (0.03) 3 (0.97) 9
as the probability of exactly 3 defective items in a sample of 12. To determine the probability that at most 3 will be found defective, we note that this is the same as the probability of finding exactly no defectives or exactly 1 defective or exactly 2 defectives or exactly 3 defectives. In other words, it is the sum of the first four terms of the binomial expansion of namely,
c:)
(0.97)
12
+
en
(0.97)
11
(0.97
+ 0.03) 12 ,
(0.03)
+
c:)
(0.97)
10
(0.03)
+
2
3 9 12) ( 3 (0.97) (0.03) .
*2010 Exercises 1. In the manufacture of a certain item, the probability of its being defective is 0.01. In a sample of 50, what is the probability of finding at most 2 defective items? 2. A single die is to be thrown 5 times. What is the probability of obtaining (a) a 1 each time? (b) a 3 each time? (c) a 4 exactly 3 times? (d) a 1 on the first throw, a 2 on the second, 3 on the third, etc.? 3. What is the probability that a 5card hand will contain exactly 3 spades? 4. If it is known that a single toss of 2 dice resulted in the sum of the points being odd, what is the probability that the point was 3? 5. What is the probability that at a deal of bridge a player will obtain exactly 1 ace? 2 aces? k aces? 6. What is the probability that in tossing a coin 6 times exactly 3 heads will be obtained? 7. In a lottery involving 300 tickets, there are 3 prizes. If an individual holds 12 tickets, what is his probability of obtaining at least 1 prize? 8. In many applications it is considered feasible to approximate a "binomial probability" (i.e., a probability obtained by Theorem 208) by the "Poisson formula." This relation is k
b(k;n,p)
~
A A e k!
'
where b(k; n, p) denotes the probability of k successes in n trials, p is the probability of success on each trial, and A = np. (a) Compute b(5; 500, 0.01). (b) Use the Poisson formula to approximate (a). (c) Compare (a) and (b).
2011]
639
MEAN AND STANDARD DEVIATION
*2011 Mean and Standard Deviation. One of the tasks in statistical
work is to organize the data which have accumulated so that efforts can be directed toward advancing simple descriptions of the data. We want to replace a (large) set of numbers with a few numbers, but at the same time we want the few numbers to reflect the properties of the original set as closely as possible. This brings up the question: What properties of a set S of numbers do we want to emphasize? In general, we would like to give some indication as to the sife or magnitude of the measurements constituting S and, in addition, we are interested in describing the variability or spread of these measurements. It must be borne in mind that we have not given precise meanings to the terms magnitude and variability of a set. Indeed, logically speaking, our problem is to make these terms precise. For the magnitude of the measurements, the mean, median, and mode are commonly used. Briefly, by the mean is meant the arithmetic average of the numbers; the median is the "middle" number of the list when the measurements are arranged according to magnitude on a line, and the mode is the number of the set that occurred the most often. More precisely, DEFINITION 209. Let x 1 , x 2 , ••• , set S. By the mean x of S is meant
x
===
Xn
be the numbers belonging to a
Li=l Xi • n
Assuming the numbers of S to be arranged according to magnitude, X1
15. x
2
9. x ~~ because 23 · 72
+5
2x

>
1) 2
= (x 
J
11. x
> >
>
0.
5. x
3
42
5 or x
4 since (x 
1) 2
17. Yes. See 15. 19. a/b < c/d and e/f < g/h, all terms positive, imply there exist positive numbers h, k such that (a/b) h = c/d and (e/f) k = g/h. Now multiply corresponding members of the equations.
+
+
21. For the odd numbers: 1. No upper bound. 3. 3 is upper bound and least upper bound. 5. No upper bound. 7. 0 is least upper bound. 9. No upper bound. 25. Yes, because of the denseness of the rationals in the reals. *27. The proof given is for the theorem: If there is a first real number greater than 0, then it is 1/2. This is valid. The deficiency here is that it is not shown that there is a "first nonnegative real number." Of course, such a number does not exist. 29. f is a lower bound for the set A of real numbers means that f < x for all x E A. I is the greatest lower bound for the set A of real numbers means that if f is a lower bound for A' then I >
r
31. a < band c < dimply a+ c other two statements.
0. (iii) a+ h = b and b + Therefore k h =
+
=
k < x
=
b.
< A
3 or x
tt
(c) (ABCD)
(X4 ( X3X3  X1) X2 X4 
X2) XI
[(x3  x1)/(x3 
x2)][(x4  x2)/(x4  x1)]
(BACD) 1 1 (ABCD)
23. All real numbers
~
0
25. 1
0. Furthermore, it is clear that a = c and b = d imply d(P, Q) = 0. Conversely, if d(P, Q) = 0, then a = c and b = d for (a  c) 2 > 0 and (b  d) 2 > 0. (ii) d(P, Q)
=
c) 2
V (a 
+ (b

d)2
= ~  a)2+ (d  b)2 = d(Q,P) (iii) The sum of the lengths of two sides of a triangle al ways exceeds the length of the third side. Equality occurs when the triangle collapses to a segment.
Section 38
7. (a) PQ = V89
3x 
3. y 
1. 3, decreasing (b)
i
14
=
(c) R(13, 12)
(d)
9. (c) That the graph is a straight line. 4 1 2 15 · 3, 2, 3
17. Through P: 7y  6x Through Q: y 6x Through R:
+ y+3
=
+ 31 7 0
=
=
0
0
+ 9x lOy + 16x 
5. 4y
0
7
0
=
61
=
0
659
ANSWERS TO ODDNUMBERED EXERCISES
19. Let the vertices be A(O, 0), B(a, O), C(b, c).
The midpoints of AC and The line through these c/2 and is obviously parallel to AB.
CB are (b/2, c/2) and [(a+ b)/2, c/2], respectively. points has equation y
0,
>
0, {3
>
0.
or ({3 
(b) The points inside the triangle are those of the form
+ µ(5i) + v(2

i), A+µ+ v
=
1, A
>
0, µ
>
0, v
>
0,
+ 2v) + i(A + 5µ

v), A+µ+ v
=
1, A
>
0, µ
>
0, v
>
0.
A(3 
i)
or (3A
3. Let z = x + yi. Then Re(z) = x and IRe(z)I = lxl. On the other hand, lzl = vx 2 + y 2 • But y 2 > 0 and vx 2 = lxl, so that IRe(z)I < lzl. The second part is similar. 5. Im(z) = Im(x+yi) = y; Re(iz) Consequently, Im(z) = Re(iz).
=
Re[i(x+yi)] = Re(+yix) = y.
667
ANSWERS TO ODDNUMBERED EXERCISES
7. The circle with center at (0, O), and radius 1. 9. Interior of the circle with center at zo, and radius r. 11. Interior of the circle with center at the origin, and unit radius. 13. All points outside or on the circle with center at 3i, and radius 8.
+ 2si
29 15 · 25
3
=
i ·
*23. d(z, w)
>
d(z, w) =
With
r
Vs
2
19 __I~
lzl 
17. 2__
•
lzl ~__i lzl 2
+
21.
1
lz:12 z i
0 is obvious.
Ix  ul + IY  vi
a+ bi, d(z, w) +
=
lu  xi+ Iv  YI
= d(w, z).
=
d(w,
r)
Ix  uJ + IY  vi + lu  al + > Ix  al+ IY  bl = d(z, r). =
Iv  bl
CHAPTER 6
Section 64 1. 33
7. m
3. 171 =
5. (a) $1950
(b) $36,750
683 10
(1.035) 1 1 9. (1. 035 ) l0(0.0 35 ) ~ (0. 035 ) ~ 29 (see Theorem 63) 11. 21, 16, 11, 6. 13. The proofs are identical with those given in the text. 15. (l)n
17.
19. n
i
21. 1
31. 57
33. 12
(b) f(l) *35. (a) f(0.1) = 0.1 (d) f(0.1315) = 0.1650
37. If n
=
1, 1 1 
r =
1 
rk =
=
0
(c) f(0.001) (e) f(0.762)
=
= 0.003
0.620
r. Given that (1

r) ( 1 + r
+ · · · + rk 1),
it follows that
(1 
(1 (1 
+ r + · · · + rk  1) + rk r)(l + r + · · · + rk 1) + rk(I r)(l + r + · · · + rk), q.c.d. r) ( 1
rk +1 
r)
668
ANSWERS TO ODDNUMBERED EXERCISES
Section 66 2
1.
3
2V2
9. 1 11.
0, (4  5r) and (3  7r) are negative while (7 4r) and 3r are positive. Hence there are four variations in sign in the product p(x) (x  r) while there are only three variations in sign in p(x) = 5x 3 4x 2 7x  3.
+
Section 915 1. 1.3
5. y
=
3. 0.68 f(x) is parallel to y = g(x) at x = ~ means f' (~) = g' (~). x
3, 2
Section 917 =
2)
or
(4
13. (x
1: 2
2k
>
or
>
1. If 2k 2k+l
>
2k
1)
k

+ [4(k + 1)
+ 4k +

3]
1
+ 3k + 1
(k
+ 1)(2k + 1)
(k
+
1)[2(k + 1) 
1],
q.e.d.
k, then =
k
+k
>
k
+
1
>
if k
1.
1: 5 1  1 is divisible by 4. If 5k  1 is divisible by 4, then there is an integer m such that 5k  1 = 4m, or 5k = 4m 1. Therefore, 5k • 5 = 20m + 5, so that 23. For n
=
+
5k+l 
·which says that 5k+I 
1
=
20ni
+4
1 is divisible by 4,
=
4(5m
+ 1),
q.e.d.
Section 1411 1. e2 y 
x 
e2 = 0
3. y" = ex
>
0
5. Maximum at x = 1r/4 and coterminal angles. :Minimum at x = 51r/4 and coterminal angles. Inflection points at 1r/2, 31r /2 and coterminal angles.
687
ANSWERS TO ODDNUMBERED EXERCISES
7. Minimum for x = ! arctan i and coterminal with an angle () such that 1r < 2() < 31r /2. Maximum for x = ! arctan i and coterminal with an angle () such that O < 2() < 1r /2. Inflection points at x = ! arctan 1/. 9. No maximum or minimum points 11. y'
2 tan x
13. y'
15. y'
1 x In x
17. y'
19. y'
Section
ex  ex 2
1 cos In x x
=

*1413
5. 12.8 years *13. q(t)
7. 8.383 years amount present at time t = 0, k = a constant
Aekt; A
=
CHAPTER 15 Section 152 1. (x 5.
4) 2
2v2 y 
9. (x 
1) 2
+ (y  3) = 25 x + 4V2  8 = o + (y + 1) = 8 2
3) 2 + (y 
3. (x 
7. (i,
2
2) 2
=
13
!) ; 0 i)2 + y2 = _g_~o
11. (x 
*13. Draw a figure and note that the triangle with vertices consisting of (t 'YJ), the center of the circle, and the point of tangency is a right triangle. Apply the Pythagorean theorem. 15. (x 
5) 2 + (y 
19. (x 
h) 2 (y 2 3) + (y 
21. (x 
+
17. (x+6) 2 + (y+2)2
5) 2 = 25
f 2 h  !f)2
26
83
1352
2) 2 = 36
23. If the circles are tangent, the radical axis becomes the common tangent line, since it must contain points of intersection of the two circles. If the circles are concentric, there is no radical axis (to see this, write the equations of a pair of concentric circles, and subtract). 25. (x 
!)2
+
(y
+ !)2
29. (a) x 2 + y 2 + 2x 
=
9l
2) 2 + (y 
27. (x 
4) 2 = 25
4y 
20 + k(x 2 + y 2
4x 
2y 
11)
O
2y 
11 + k(x 2 + y 2 + 2x 
4y 
20)
0

or
x2 + y2 (b) 9x 2
4x 

+ 9y
2

102x
+ 4y
=
0
688
ANSWERS TO ODDNUMBERED EXERCISES
+ (y
2) 2
31. (x 
3) 2

2
=
*33. Let P(a, (3) be a point of intersection of f(x, y) = 0 and g(x, y) = 0. Then f(a, (3) = 0, g(a, (3) = 0 and therefore
f(a, (3)
+ kg(a, (3)
= 0 + k · 0 = 0,
i.e., P(a, (3) lies on every curve of the family f(x, y)
+ kg(x, y)
= 0.
Section 154
+ 17 = 0 x+4 = 0
1. 4y 
3. 2y 
x
7. 2y , 13. y =
x
+3
= 0
5. y
9. y = 0
a
b
15. y'
11. y' 2x y
=
= 0 =
I
17. y' = 
~
Section 156 1. x 2

8x 
24 = 0
20y 
*17. 2yy' = 4a, y' = 2a/y; slope at (t 'YJ) is y' tangent to y 2 = 4ax at (t 'YJ) is
y But
'Y}
2 =
2a
11 = 
'Y]
~)
or
Y'YJ 
'YJ
2
=
2ax  2a~.
4a~ since (~, 'YJ) is on the curve y 2 = 4ax. Therefore Y'YJ 
19. Y =
(x 
2a/rJ. The equation of the
ihx 2

4a~ 7 2 oX
=
2ax 
2a~
+ 1/cf
or
Y'YJ = 2a(x
21. (x 
+ ~).
1) 2 = 12(y 
2)
23. V(1, O), F(t O) 25. The focus of y 2 = 4ax is located at F(a, 0) (see the diagram). Hence the distance between F(a, O) and (~, 'YJ) is d = a) 2 rJ 2 • But 'YJ 2 = 4a~, 2 +2a~+a 2 = ,~+al. sothatd = v(~a) 2 +4a~ =
v~
y
F (a, 0)
27. 8x = 9y 2
+ 16y + 33
v(~ 
+
689
ANSWERS TO ODDNUMBERED EXERCISES
Section 158
Equation 2
x
1.
25/3
5.
7.
+
(x _ 1)2
+
(y _ 1)2 _ 2  l
+
(y
(y
+ 6)2
1)2
3
+
2
21.
(O, O)
H~H
H,o)
(o, ± ~)
1)2 = 1
2
3)

(1 ±
v'io,
(:3,:3 ± 2 2
6)
~) 3
(1, 6 ± 2vw)
(1, 1)
c1 ± v2, o
(1 ± 2, 1)
c1, 1 ± v2)
(1, 1)
(1 ± 2V6, 1)
(1 ± 3V3, 1)
(1, 1 ± V3)
2 + (y +~ = 64 2
vw)
o, 6 ±
(1, 6)
1
(:32 ± y3 _§ }) 2
21
±
3
+ 8xy + 24y
13. 9x 2 x 17 · 81
=
40
27
(x 
Covertices
G·~) (32 35y 23)
= 1
(x  1)2 10
(x 
ll.
i/

25/9
4
9.
Vertices
2
+ ~:)2 + (y
(x 
Foci
y 25/16 = l
25/9
3.
Center
1
30x 
or
(x 
64
80y
+ 25
3)
2
2 + (y +~ 3
1
= 0
2
+
y 45
l
=
(i) The quadratic ax 2 c2

4ae
(ii)
c2 
4ae
(iii)
c2 
4ae
>
+ ex+ e
O must have two real roots, 1.e.,
0. O