A 564870 SEL ENGIN. LIB QA 152 R563 1919 1 ..... in an identitie the letters гер 1837 ARTES SCIENTIA In VERITAS LIBRARY OF THE sta UNIVERSITY OF MICHIGAN | UNUN TUEBOR а Ca ди ан only Ro SI QUÆRIS PENINSULAM-AMŒNAM. CIRCUMSPICE COLLEGE OF ENGINEERING THE GIFT OF Robert H Marquis كم Ca.. letter alas hat не verse. его a³+b³ = (a+b) a²=ab + b² c 3 a = b² = (a - b) a² +ab+b² а a²-b² - (a-b) (a + b) (a + b)³ = a³ + 3a²b+3ab+6 3 заг За (a - b) ³½ a³-3a² b + зab² b² a²b 3 FA IFB = F(AFB) а 2 тат (a² + a + 1) (α ²= a + 1) =a+a+1 a" tut RIM requis едий Ch 3 3 requis In multiplying radicals of different indices. first Change to pactional experment, Find the SC.Mof 14 We denominators Plot f (V) X-1 EWGREERIMG LDRARY QA 152 R563 1919 Presented to the Slay American Mathematical Series E. J. TOWNSEND GENERAL EDITOR COLLEGE ALGEBRA BY H. L. RIETZ, PH.D. PROFESSOR OF MATHEMATICS UNIVERSITY OF Iowa AND A. R. CRATHORNE, PH.D. ASSISTANT PROFESSOR OF MATHEMATICS UNIVERSITY OF ILLINOIS REVISED EDITION NEW YORK HENRY HOLT AND COMPANY COPYRIGHT, 1909, 1919, BY HENRY HOLT AND COMPANY. 3 for Eng like. Im Noft H Inc. git 14-1928 squis PREFACE THIS book is designed primarily for use as a text-book in the freshman year of colleges and technical schools. Special atten- tion is directed to the following features: (1) The method of reviewing the algebra of the secondary schools. (2) The selection and omission of material. (3) The explicit statement of assumptions upon which the proofs are based. (4) The application of algebraic methods to physical problems. For the majority of college freshmen, a considerable period of time elapses between the completion of the high school algebra and the beginning of college mathematics. The review of the secondary school algebra is written for these students. This part of the book is, however, more than a hasty review. While the student is reviewing a first course, he is at the same time making a distinct advance by seeing the subject-matter from new view points, which his added maturity enables him to appreciate. For example, the functional notation, graphs, and determinants are introduced and used to advantage in the review. The extensions of the number concept receive fuller treatment than is usual in a college algebra. The various classes of numbers from positive integers to complex numbers are treated in the order in which they are demanded by the equation. The application of algebra in the more advanced courses in mathematics has been an important factor in determining the subject-matter. Not only are some of the topics usually treated in the traditional course in algebra entirely omitted, but in each chapter the material is restricted to the development of those central points which experience has shown to be essential. While a complete discussion of limits and infinite series does not prop- erly belong in a course in algebra, it has been thought best to ïii iv PREFACE include an introduction to these subjects which covers in con siderable detail only the theory necessary to a discussion of the comparison and ratio tests. From the experience of the authors, a great deal is gained by thus taking a very elementary first course in limits and series. While it is out of place in a book of this character to attempt a critical study of fundamentals, great care has been taken to point out just what is proved and what is assumed in so far as a first-year student can be expected to appreciate the necessity of assumptions. Without trying to teach physics or engineering, many problems are introduced in which the principles of algebra are applied to physical problems, but no technical knowledge is assumed on the part of the student. Rules for the mechanical guidance of stu- dents in solving problems have been used sparingly. The authors take great pleasure in acknowledging their in- debtedness to their colleagues in the mathematical and engineer- ing departments of the University of Illinois. We are indebted to Professors Haskins and Young for suggestions during the preparation of the manuscript as well as for a critical reading of the manuscript; to Professors Townsend, Goodenough, Miller, Wilczynski, Dr. Lytle, and to Professor Kuhn of Ohio State Uni- versity for suggestions upon the manuscript; to Professor Watson for some of the practical problems; and again to Professor Good- enough for assistance in seeing the book through the press. H. L. RIETZ. A. R. CRATHORNE. PREFACE TO REVISED EDITION Suggested by experience in the class-room, a number of topics have been simplified in treatment. In response to requests of numerous teachers, several hundred new exercises and problems have been introduced. The book has thus been freshened with regard to both text and problems without however changing its general character. JANUARY, 1919. H. L. R. A. R. C. CONTENTS CHAPTER I INTRODUCTION ARTICLE 1. Numbers 2. Graphical Representation of Real Numbers 3. Greater and Less 4. Definitions and Assumptions 5. Derived Properties of the Numbers of Algebra 6. Positive Integral Exponents 7. Meaning of aª 1 p 8. Meaning of aª 9. Limitation on the Value of a 10. Meaning of a° D 11. Meaning of am when m is Negative 12. Radicals CHAPTER II PAGE 33 3 T 7 1 1 12 10 020 12 12 13 13 • : 14 ALGEBRAIC REDUCTIONS 13. Algebraic Expressions 14. Removal of Parentheses 15. Complex Fractions 16. Factoring 17. Radicals and Irrational Numbers 18 19 19 · 21 22 18. Reduction of Expressions containing Radicals to the Simplest Form 25 19. Addition and Subtraction of Radicals 20. Multiplication and Division of Radicals 21. Evaluation of Formulas 22. Imaginary Numbers • 26 26 27 30 • · CHAPTER III VARIABLES AND FUNCTIONS 23. Constants and Variables 24. Definition of a Function 31 • 31 V vi CONTENTS ARTICLE 25. Functional Notation 26. System of Coördinates 27. Graph of a Function. • 28. Function defined at Isolated Points 29. Zeros of a Function CHAPTER IV THE EQUATION 30. Equalities. 31. Definitions . 32. Solution of an Equation 33. Equivalent Equations 34. Operations that lead to Redundant Equations 35. An Operation that leads to Defective Equations 36. Clearing an Equation of Fractions CHAPTER V • • PAGE 32 • • 33 • 35 36 • 39 • 41 42 43 44 45 46 • 47 LINEAR EQUATIONS 37. Type Form • 49 38. Simultaneous Linear Equations • 50 39. Graphical Solution of a System of Linear Equations 40. Determinants of the Third Order 52 53 41. Solution of Three Equations with Three Unknowns 54 CHAPTER VI QUADRATIC EQUATIONS 42. Type Form 43. Solution of the Quadratic Equation 44. Solution by Factoring 45. Equations in the Quadratic Form 46. Theorems concerning the Roots of Quadratic Equations 47. Number of Roots 48. Special or Incomplete Quadratics 49. Nature of the Roots 50. Sum and Product of the Roots 51. Graph of the Quadratic Function • 59 60 62 63 • 65 €6 €6 €8 • 69 70 CONTENTS vii CHAPTER VII ARTICLE SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS 52. Type Form 53. Solution of Systems of Equations Involving Quadratics CHAPTER VIII INEQUALITIES 54. Definition . 55. Absolute and Conditional Inequalities 56. Elementary Principles 57. Conditional Inequalities CHAPTER IX MATHEMATICAL INDUCTION 58. General Statement 59. Meaning of r! 60. Binomial Theorem; Positive Integral Exponents CHAPTER X VARIATION 61. Direct Variation 62. Inverse Variation 63. Joint Variation • 64. Combined Variation CHAPTER XI PROGRESSIONS 65. Arithmetical Progressions • 66. Elements of an Arithmetical Progression 67. Relations among the Elements 68. Arithmetical Means • 69. Geometrical Progressions 70. Elements of a Geometrical Progression 71. Relations among the Elements 72. Geometrical Means 73. Number of Terms Infinite. • • PAGE 75 76 86 86 86 88 90 92 A 93 13 97 A 97 98 98 101 101 101 • 102 103 103 • 103 O ↓ 104 • 105 viii CONTENTS ARTICLE 74. Series 75. Repeating Decimals 76. Harmonical Progressions 77. Harmonical Means • • CHAPTER XII COMPLEX NUMBERS 78. Number Systems 79. Graphical Representation of Complex Numbers. 80. Equal Complex Numbers 81. Addition and Subtraction of Complex Numbers 82. Multiplication of Complex Numbers 83. Conjugate Complex Numbers 84. De Moivre's Theorem 85. Roots of Complex Numbers 86. Division of Complex Numbers • • CHAPTER XIII · THEORY OF EQUATION 87. The Polynomial of the nth Degree 88. Remainder Theorem 89. Synthetic Division 90. Rule for Synthetic Division 91. Graphs of Polynomials • 92. General Equation of Degree n 93. Factor Theorem 94. Number of Roots of an Equation • • PAGE 106 • 106 107 107 • 110 111 112 113 114 • 115 116 • 117 • 119 • • • 95. Graphs of ao(x −rı) (x − r2)··· (x —Ñ„) · 96. Complex Roots 97. Graphs of f(x) when Some Linear Factors are Imaginary. 98. Transformations of Equations. 99. Descartes's Rule of Signs. 100. Location of Roots by Graph 101. Equation in p-Form . 102. Rational Roots 103. Irrational Roots. Horner's Method 104. Negative Roots 105. Summary. 106. Algebraic Solution of Equations 107. The Cubic Equation 121 121 • · 122 • 124 • 125 • • 126 126 • 126 128 129 • 130 • 131 • 135 · 137 139 • 139 141 • 145 145 • 148 149 CONTENTS ix ARTICLE 108. The Biquadratic Equation 109. Coefficients in Terms of Roots. 110. Variable Coefficients and Roots PAGE 151 153 • 153 CHAPTER XIV LOGARITHMS 111. Generalization of Exponents 112. Definition of a Logarithm 113. Derived Properties of Logarithms 114. Common Logarithms 115. Characteristic • 156 156 157 159 * 160 116. Use of Tables 161 • 117. To find the Logarithm of a Given Number 164 118. To find the Number which Corresponds to a Given Logarithm 164 119. Computation by Means of Logarithms 165 • 120. Change of Base 169 • 121. Graph of loga X 171 • 122. Exponential and Logarithmic Equations 123. Calculation of Logarithms 172 175 • 124. Introduction 125. Case I 126. Case II 127. Case III 128. Case IV CHAPTER XV PARTIAL FRACTIONS • • 177 178 179 180 • 180 CHAPTER XVI PERMUTATIONS AND COMBINATIONS 129. Introduction 130. Meaning of a Permutation 183 183 131. Permutations of Things All Different 184 132. Permutations of n Things not All Different 184 • • 133. Combinations 134. Combinations of Things All Different 135. Binomial Coefficients 186 • 186 187 136. Total Number of Combinations 187 X CONTENTS CHAPTER XVII PROBABILITY ARTICLE PAGE 137. Meaning of Probability 138. Probability derived from Observation 139. Expectation of Money 190 191 • 191 140. Independent, Dependent, and Mutually Exclusive Events 141. Repeated Trials 192 • 194 CHAPTER XVIII DETERMINANTS 142. Extension of the Determinant Notation 143. Properties of Determinants 196 • 198 144. Development by Minors 200 145. Theorem • 203 146. Systems of Linear Equations containing the Same Number of Equations as Unknowns 203 147. Systems of Equations containing More Unknowns than Equa- tions. 207 148. Systems of Equations containing Fewer Unknowns than Equa- tions 208 149. Common Roots of Quadratic and Higher Degree Equations in One Unknown 209 150. Definition CHAPTER XIX LIMITS 151. Infinitesimals • 152. Theorems concerning Limits 153. Both Numerator and Denominator with Limit O 154. Infinity 155. Limiting Value of a Function 156. Indeterminate Forms 213 214 215 • • 216 • 217 • 218 218 CHAPTER XX INFINITE SERIES 221 221 157. Definition 158. Convergence and Divergence CONTENTS xi ARTICLE SERIES WITH POSITIVE TERMS 159. Fundamental Assumption 160. Comparison Test for Convergence 161. Comparison Test for Divergence 162. Summary of Standard Test Series 163. Ratio Test for Convergence and Divergence PAGE 223 224 227 • • 228 229 SERIES WITH BOTH NEGATIVE AND POSITIVE TERMS 164. Theorem 165. Ratio Test Extended 166. Alternating Series • 167. Approximate Value of a Series . 168. Power Series 169. Binomial Series 170. Exponential Series 171. Logarithmic Series ANSWERS INDEX • • • 232 232 234 234 236 238 240 • 241 243 265 A LIST OF SIGNS AND SYMBOLS +, read plus. 2 ×, or, read times. ÷, read divided by. =, read is equal to. read minus. read is identical with. , read is not equal to. →, read approaches. <, read is less than. " read is greater than. ≤, read is less than or equal to. ≥, read is greater than or equal to. a! or a, read factorial a. () Parentheses. [ ] Brackets. { } Braces. Vinculum. | Bar. Signs of aggregation. These signs are used to collect together symbols which are to be treated in operations as one symbol. a,, read a subscript r, or a sub r. x', x'' ……., read x prime, x second ... respectively. lim x, read limit of x. x→ ∞, read x becomes infinite, or x increases beyond bound. log.n, read logarithm of n to the base a. Ja, read absolute value of a. a”, read a to the nth power, or a exponent n. Va, read square root of a. Va, read nth root of a. f(x), p(x), etc., read "f" function of x, "p" function of x, etc. „Pr, read number of permutations of n things taken r at a time. nCr, read number of combinations of n things taken r at a time. (x, y), read point whose coordinates are x and y. xii GREEK ALPHABET Letters Names Alpha Letters a ΓΥ Beta Gamma Lambda Letters Upsilon TE TE TE A η Delta Epsilon Zeta Eta Theta Omicron COLLEGE ALGEBRA CHAPTER I INTRODUCTION 1. Numbers. In counting the objects of a group the child makes his first acquaintance with numbers. These are the num- bers called positive integers. He next employs the number we call a rational fraction, probably thinking of it first as an aliquot part, and later as the quotient of two integers. Perhaps in the fall of a ther- mometer below zero, the student had his first experience in the use of negative numbers, even if he was not taught to use the word negative. He may also come early to the convenient use of the negative number to represent debit when the corresponding positive number means credit. To express the length of the diagonal of a square of side one unit, or to find a number which multiplied by itself gives some integer, not a perfect square, say 2, he uses a number which is neither an integer nor a rational fraction, and employs a radi- cal sign to represent it by writing √2 where V2 x √2 = 2. Such numbers belong to a class of numbers known as irrational numbers. (See p. 23.) 2. Graphical representation of real numbers. The four classes of numbers mentioned in Art. 1 belong to the so-called "real numbers" used in arithmetic and algebra. They may be repre- sented by the points of a straight line as follows: Let X'X be this line. (Fig. 1.) Choose a point O on this line and call it the zero point or origin. Adopt some unit of measurement OA. Beginning at O and proceeding in both directions, apply the unit of measure to divide OX and OX' at equal intervals, thus 1 2 [CHAP. I. INTRODUCTION forming a scale of indefinite length. The positive and negative integers may then be conveniently represented by the points marking the intervals. Similarly, corresponding to any fraction α (a and b integers), there can be constructed a point on X'X b such that the fraction denotes the distance and direction of the point from 0. In fact, we assume that by means of this scale represent conveniently all real numbers, and we we are able to - 8 -7 -6 -5 -3 -2 -1 2 3 4 5 6 7 8 9 X'd P₁ A P FIG. 1. can say, to any point P on the line, there corresponds a number * which indicates the distances and direction of P from O, and con- versely, we assume that to every real number there corresponds a point of this line. In addition to the real numbers, we shall find it desirable to deal with so-called "imaginary" and complex numbers. A graphi- cal representation is given for these numbers in Art. 80. EXERCISES 1. What numbers are represented by the following points ? (a) The point midway between 4 and 5. (b) Points of trisection of the interval – 3 to 4. - (c) Points of quadrisection of the interval 2 to 3. 2. State in words the position of points which represent ½, 1, 3 2, T. 3. Suppose the scale of Fig. 1 represents the scale of a Fahrenheit ther- mometer; estimate the readings when the end of the mercury column stands at P. At P. At point midway between 0 and P. At point midway between P and P1. 4. A square piece of paper of side 4 is laid on Fig. 1 so that one corner of the square is at O and the diagonal lies on the line OX. What number is represented by the point at the other corner of the square which lies on OX ? * For a more complete discussion, see Fine's Number System of Algebra, Second edition, p. 41. † P is read "P sub one." See list of signs and symbols at the end of the table of contents. ARTS. 2-4] DEFINITION AND ASSUMPTIONS 3 5. A circle of radius 1 rolls to the right along the line in Fig. 1, beginning at O. What number is represented by the point at which the circle touches the line after one complete turn? 3. Greater and less. The terms greater than and less than which are common to everyday life, when used in the technical sense of algebra, are easily misunderstood. For this reason we point out their geometrical significance. The real number A is said to be greater than the real number B (written A > B) if the point representing A falls to the right of the point representing B. The number A is said to be less than the number B (written ▲ < B) if the point representing A falls to the left of that representing B. Exercise. Arrange the following numbers in descending order of magnitude: - √3, 3√5, - 10, - 20, 1. 2, √2, - 4, >> 4. Definition and assumptions. Operations with numbers of arithmetic suggest certain definitions and rules for algebra. The student has probably performed algebraic operations according to rules thus suggested by arithmetic without being conscious of the assumptions which underlie these processes. We may now proceed to a formal statement of assumptions made at the outset in this algebra. Letters are used to represent numbers; a number, which is represented by a certain letter, is called the value of the letter. In the following, let a, b, c represent any numbers. The fundamental * operations of addition and multiplication of numbers are subject to the following laws I-IX: I. The sum † of two numbers is a uniquely determined number. That is, given a and b, there is one and only one number x such that a + b X. II. Addition is commutative. That is, a + b = b + a. * The operations are fundamental in that no attempt is made to define them. The "laws" are in the nature of assumptions since no attempt is made to prove them. † The sum is the result of adding; the product is the result of multiplying. 4 [CHAP. I. INTRODUCTION III. Addition is associative. That is, (a + b) + c = a + (b + c). IV. If equal numbers be added to equal numbers, the sums are equal numbers. That is, if and a = b, C d, მსა a + c = b + d. then V. The product of any two numbers is a uniquely determined number. That is, given a and b, there is one and only one number y such that ab = y. In this case, a and b are said to be factors of y. VI. Multiplication is commutative. That is, ab = ba. VII. Multiplication is associative. That is, (ab)c = a(bc). VIII. Multiplication is distributive with respect to addition. That is, a(b + c) = ab + ac. IX. If equal numbers be multiplied by equal numbers, the prod- ucts are equal numbers. That is, if and then a b, c = d, ac = bd. The following laws X and XI lead us to definitions of subtrac- tion and division, and enable us to give meanings to the symbols α 0, -a, 2, 1, and 1. X. Given a and b, there is one and only one number x, such that x + b = a. Subtraction is the process of finding the number x in x + b = a. In other words, to subtract b from a is to find a number x, called the remainder, such that the sum of a and b is a. ART. 5 . 4] DEFINITIONS AND ASSUMPTIONS By X, the number x in x + b a exists when a = b. In this case, the number x is called zero, and is written 0. In symbols, 0 + b = b. (1) From X and the definition of 0, there exists a number x, such that x + b = 0. In this case, a and b are said to be negatives of each other, and x may be replaced by (— b). If b is a positive number, a is a negative number. In symbols, (− b) + b = 0, (2) gives a definition of (— b). XI. Given a and b (b‡0*), there is one and only one number x, which satisfies bx a. a. In Division is the process of finding the number x in bx other words, to divide a by b is to find a number a, called the quotient of a by b, such that b multiplied by a gives a. а This quotient is often written, and, when thus written, is called a fraction. In dividing a by b, the number a is called the dividend and b the divisor, just as in arithmetic. Likewise, in the fraction, a is called the numerator and b the denominator. By XI, the number æ in bx = a (b + 0) exists when b = a, so that ax = a. called unity and is written 1, that is, In this case, the number x is α = 1. (3) a Further, by XI, and the definition of 1, there exists a number x, which satisfies bx 1 (b0). This value of x is called the reciprocal of b and is written 1 b *The sign stands for "is not equal to." 6 [CHAP. I. INTRODUCTION It should be noted that, by means of XI, a meaning is given to unity and to the reciprocal of any number other than zero in a manner analogous to that by which a meaning is given to zero and to the negative by means of X. The above propositions I-XI are stated for two and three num- bers for the sake of simplicity. It can be proved from the given assumptions that these propositions hold when three or more num- bers are concerned in the process of addition or multiplication. It is not to be inferred that these propositions I-XI are en- tirely independent of one another, but rather that they constitute convenient assumptions for the purposes of this course in algebra. Although further assumptions are made later in the course, the principles I-XI enable us to prove many important propositions. of algebra, and the wide application of algebra depends upon the fact that many changes in the physical world take place in ac- cordance with these laws. As shown in the following exercises, the operations of algebra are generalizations of the operations of elementary arithmetic. EXERCISES 1. Show that 4 × 2, 9 x 7, 1 x − 1, − 4 × — 6, 0 × - 2, 1 × − 1, are special cases of (x + 1)(x − 1) and of x(x − 2). - - 2. Show that 2 × 3 × 8, 5 × 6 × 11, 1 × 3 × 13, - 3 × - 2 × 3, — 1×· 0 × 5 are special cases of x(x + 1)(x + 6) and of (x - 1)x(x + 5). 12 4 21 3. Show that 6, 2, 3, 2 2 - 7 1 =윽​=3, 96 = 12, 8 x2 4 are special cases of = x + 2. х 2 96 - 12 ∞ In the course of operations with the numbers of algebra, the important question arises: Can any two given numbers be added, subtracted, multiplied, or divided? Our assumptions state that the number system of algebra is such that this question can be answered in the affirmative except in the case of division by zero. Division by zero is excluded from algebraic operations. EXERCISES 1. Can any two given numbers be added, subtracted, multiplied, or di- vided if the number system consists of positive integers only? Illustrate your answer. ARTS. 4, 5] DERIVED PROPERTIES OF NUMBERS 7 2. Can any two given numbers be added, subtracted, multiplied, or di- vided if the number system consists of positive integers and quotients of positive integers only? Illustrate your answer. 3. Where is the flaw in the following? Let Multiply both sides by x, Subtract a2 from both sides, Factor, Divide both sides by x But, by (1), By (5) and (6), (x − - x = a (x ‡()). x² = ax. x2 a2 = ax - a². a)(x + a) = a(x a(x − a). α, x + a = a. x = α. 2 α = a. : 2 = 1. 三​@@@@@日 ​Hence, 5. Derived properties of the numbers of algebra. From the fore- going definitions and assumptions, the following propositions can be proved. We shall present in detail the proofs of only a few of them. I. Adding a negative number (-a) is equivalent to subtracting a positive number a. That is, b + (− a)= b − a. (I, Art. 4.) To prove this, let b+(-a)=x. b + (− a) + a = x + a. (IV, Art. 4.) (1) (2) But (− a) + a = 0. (Eq. (2), page 5.) (3) From (2) and (3), b+0=x+u. (4) 0+ i + 0 = b. (Eq. (1), page 5.) (5) b = x + a. (6) But From (4) and (5), That is, b — a = x. (Def. of subtraction.) (7) b+(-a)=b-a. II. Subtracting a negative number (-a) is equivalent to adding the positive number a. That is, b(a)= b + a. III. The product of two numbers is 0 when and only when at least one of the numbers is 0. COROLLARY. The quotient ber different from 0. 0 α is equal to ◊ when a is any num- 8 [CHAP. I. INTRODUCTION IV. The product of a number a by a number (—b) is — ab. To prove this, let a (— b) = x. (a exists by V, Art. 4.) (1) Then a(b) + ab = x+ab. (IV, Art. 4.) (2) a[(b)+b]= x + ab. α. (VIII, Art. 4.) (3) a. 0 = x + ab.* (Eq. (2), page 5.) (4) 0=x+ab. (III, Art. 5.) (5) ab = 2. (Definition of negative.) (6) From (1) and (6), a(b) ab. V. The product of (-a) by (b) is ab. To prove this, let (a)(b) = x. (1) (− a)(—b)+ a(— b)=x ab. — (IV, Art. 4; IV, Art. 5.) (2) F (− b)[(− a)+ a] = x - ab. — (VIII, Art. 4.) (3) b.0x ab. = — 0x (Definition of zero.) (4) ab. (Why?) (5) ab = x. (Why?) (6) From (1) and (6), (— a)(— b) = ab. The statement that in multiplication like signs give plus and unlike signs give minus includes IV and V. VI. The quotient of two numbers is positive if the signs of the dividend and divisor are alike; negative if they are unlike. VII. A single parenthesis may be removed when it is preceded by a positive sign without changing the signs of the terms within it. VIII. A single parenthesis may be removed, when preceded by a negative sign if the signs of terms within it are changed. That is, − (a + b − c + d − e) = ~~ a b + c − d + e.. IX. The value of a fraction is not changed by multiplying or dividing both the numerator and denominator by the same number. That is, α ax (x0) b bx * The · is a sign of multiplication, thus, 5 · 6 = 30. • ART. 5] 9 DERIVED PROPERTIES OF NUMBERS X. Changing the sign of either the numerator or the denominator of a fraction is equivalent to changing the sign of the fraction. That is, α b a b b XI. Adding two fractions having a common denominator gives a fraction whose numerator is the sum of the numerators and whose denominator is the common denominator. That is, Likewise, Ꮯ C a انح b a + b + с C h a - h с C с XII. The sum and the difference of two fractions are expressed by α C ad bc bd α C ad + bc + d , and b d C a , respectively. bd We can reduce and to a common denominator, since, by IX, --- d α ad C bc and b bd d bd By XI, we can complete the process. XIII. The product of two fractions is a fraction whose numera- tor is the product of the numerators and whose denominator is the product of the denominators. That is, α с ас b d bd XIV. To divide one fraction by another, invert the latter and then multiply one by the other. That is, 210 b + 018 or 810101 ad bc d g Propositions III, XI, XIII, stated for two numbers in each case, are readily extended to three or more numbers. 10 [CHAP. I. INTRODUCTION INDEX LAWS 6. Positive integral exponents. Definition. The at is read "a exponent x" or the "xth power of a." a positive integer, a is a short way of writing a factors. • α expression When x is α to 2 ... Laws: I. am man = am+n¸ For, if m and n are positive integers, by the associative law of multiplication, aman = · (a · a • α = Cl a · a = am+n. ... m times) (a · a • a ••• n times) • m + n times Illustrations: 53. 54 57, 3.32.33.35 = 3¹¹. II. Illustration: (am)n = amn. (34)5 = 320. (abc...)m = ambmcm III. Illustration : (3.5.7.4)2 IV. (8) 32.52.72.42. m am bm Illustration: 317 3 = 33 73 V. am am an m-n, (m>n). Illustration: 56 52. 54 VI. am 1 an an-m' (m 0) has one and only one value. For example, 93, and not ±3. Likewise, 81-3, and not ±3. If both the positive and negative roots are meant, we shall write both signs before the radical. Without this limitation, it will be seen from the following illustration that the pth power of a qth root of a number is not necessarily equal to a given qth root of the pth power: (4³)4 = 16, while the square root of 4ª may be either +16 or —16. 10. Meaning of aº. In order that the first index law may hold for an exponent zero, it is necessary that or, am, ao.am = (10+m= = α aº = 1, if a ± 0. (1) That is, any number a with the exponent 0 is equal to 1, provided a ‡ 0. 11. Meaning of am when m is negative. Let m = is a positive number. By Law I, Art. 6, and Art. 10, . n, where n (1) Hence, an • a˜n — an~n = ɑo = ao − 1. = 1 a˜n an if a 0. 1 am a-m (2) That is, in a fraction, any factor of the numerator may be trans- ferred to the denominator, or vice versa, if the sign of the exponent of the factor is changed. EXAMPLE: 22.3-4 7-1.52 7.5-2 2-2.34 22 7.5-2.34 · 22.7-1.3-4.52. We have now found meanings for fractional, zero, and nega- tive exponents consistent with the first law of indices. To give logical completeness, it is necessary to show that the meanings are consistent * with all the laws of indices, but we shall assume * See Chrystal's Algebra, Fifth edition, Part I, p. 182. 14 [CHAP. I. INTRODUCTION this with the exception of one case which as an illustration we prove as follows: p p P To prove a b (ab), where p and q are positive integers, let • = P p x = a r⋅ b q Then, X9 = (a². ¿²)· p =(a)(i) = arb (Definition of fractional exponent.) =(ab)r. That is, x is the qth root of (ab)”, or (Law III, Art. 6.) (Law III, Art. 6.) x = (ab)". Hence, P p P aª · bª =(ab)º. We have now attached a meaning to the expression a pro- vided x is any integer, or rational fraction, positive or negative. Moreover, we assume if a is positive and x is a rational number that the number described by a exists. It is possible also to give a meaning to a when a is not a rational number. However, that is beyond our present purpose. (See Art. 111.) 12. Radicals. As stated in Art. 7, the nth root of a, written 1 Vā, has the same meaning as a". The number under the radical sign is often called the radicand and the number which indicates the root is called the index. Thus, in Va, a is the radicand and n is the index. The rules of operations with radicals may be obtained at once from the laws of indices. Thus, 1 1 1 I. Va. V/b = ab" = (ab)" — Wab. = Wa Ta α II. vo Page Missing in Original Volume Page Missing in Original Volume ART. 12] 17 RADICALS 70. If 31 and 21 are substituted for x in the expression X5 x5 — x¹ — 5 x³ + 5 x² + 6x, V show that the results reduce to the same number. show 71. Two spherical particles each one gram in mass whose centers are one centimeter apart attract each other with a force of 0.000,000,066,6 dyne. Express this number as an integer multiplied by 10 with an exponent. 72. By the use of negative exponents express a micron as a part of a (A micron is one millionth of a meter.) meter. 73. The numbers 6867, 5896, 4861, 3934 each multiplied by 10-10 give the wave lengths in meters of deep 'red, yellow, blue, and ultra violet light respectively. Express each wave length in microns. 74. Some authorities say that the mass of a hydrogen atom is 10-25 grain. How would this number be written in ordinary decimal notation? K CHAPTER II ALGEBRAIC REDUCTIONS Σ x = 13. Algebraic expressions. In algebra, an expression is a symbol or combination of symbols that represents a number. Thus, x²+ y² 25 and gt2 + vt are expressions, if x, y, g, t, v, represent numbers. If a 4 and y = 6, the first takes the value 27. If g = 32.2, t = 10, and v = 10, and v = 5, the second has the value 1660. For different values of the letters, an expression represents in general different numbers. On the other hand, the same numbers may be represented by many different expressions. For example, x²-4 and (x-2)(x+2) represent the same number. Expressions which are equal for all values of the symbols for which the ex- pressions are defined * are called identical expressions. The state- ment that two identical expressions are equal is called an identity. Thus, x² - 4 and (x − 2)(x + 2) represent the same number no matter what value is assigned to x, and the statement is an identity. x²-4(x-2)(x+2) Two expressions may be equal without being identical. Thus, for x= 2 or -5, the expressions x² + 2 x - 1 and 9 * are equal, but they are equal for no other values of x, and hence are not identical. Frequently, in problems with which we shall deal, the work is made easier by replacing expressions by identical but simpler expressions. Exercise. Which of the following are identical expressions ? x² — y², (x − y) (x + y), 4 x − y, 12 x2 - 3 xy 3 x 9 4 x(x + y), 4 x² + 4 xy. *This statement implies that we may not assign values to the letters which make the members meaningless. Thus, x = 1. 1 x = 1 + is excluded when 1 X 1 X 18 ARTS. 13-15] 19 COMPLEX FRACTIONS 14. Removal of parentheses. A single pair of parentheses can always be removed from an algebraic expression. However, if the sign before it is +, no change is necessary, but if, the signs. of the terms within it must be changed (Art. 5, VII, VIII). Expressions often occur with more than one pair of parentheses. When one pair occurs within another pair, other symbols besides are used as follows: [] called brackets, called braces, and called the vinculum. All parentheses may be removed by first removing the innermost pair according to the rule for a single pair; next, the innermost pair of all that remain, and so on. EXERCISES Remove the parentheses from the expressions of the first four exercises. 1. 8a+ (5b-a)-(2 a-7b). V 2. a (b − c + d + e). 2. α-(b 3. 2x-[3x − {x − (2 x 3 x + 4)} − (5 x − 2) ]. ✓ 4. 3x - {y - [y − (x + y) − { − y — (y — c — y)}]}, {Y 5. Find the numerical value of the expression in Exercise 4 when x = 2, y = - 1. 6. Without changing the value of the expression 4 c + 3 b + 4 a - 5 c² + 4 d³, write it with the last three terms in a parenthesis preceded by a minus sign. 7. Fill out the last pair of parentheses in a + b − (c + d) = a − d ). 8. Find the numerical value of when x = 3 x − (5 x + [− 4 x − y − x]) − (−x − 3 y), 2, y = 4. — 9. Annex a pair of parentheses preceded by a minus sign to 6 a + 5 b − (a + b) so that the resulting expression equals a + b. 10. Simplify {x(x + y) − y(x − y)}{x(x − y) + y(x + y)}. ― 15. Complex fractions. A complex fraction is one which has a fraction in the numerator or denominator or in both numerator and denominator. The rules for the simplification of arithmetical fractions apply to algebraic fractions no matter how complicated the numerator or denominator may be. 20 [CHAP. II. ALGEBRAIC REDUCTIONS The main principle is that (Art. 5, IX) the value of a fraction is not changed by multiplying numerator and denominator by the same number. As illustrated by the following exercises on complex fractions, a simplification is often brought about if we select for the number by which to multiply, the lowest common denominator of the fractions which are in the numerator and denominator of the com. plex fraction. Simplify: 1. х ૩ x xy EXERCISES 1 1 x2 2. 1 1 18 ช 1 α α 1 + 2 m a + b V 4. 1 α ሳን + 4 m a + b 1 X 8 6. ✓ 3. 5. a α 1 α b b α 1+ a 7. 7. X 1 x2 1 1 x + + 2 يم x Xx Y Y Z 8. ابع +- Y א 2 x m n N M 9. 10. 1 (m + n) 2 4 1 + mn 11. 18,3 y2 3|། ། + 814 y x + ? - Y + 1-118 x V Y 13. ४ 1 12. 1 1 1 1 + x x + y x² Y x +- Xx ? 1 14. 1 1 Y X 2018 y2 + 22122223 x2 ARTS. 15, 16] 21 FACTORING 15. x − 2 + x + 2 + 1 x + 2 8 1 2 16. 1 + 1 G 1 x 1 1 1 1 + x 1 17. ✓19. V 21. x2 x + 118 1 1 x2 1 ― X 1 x X 118 +1 1 x + h h ل a + x a x a X a + x 18. a + x α x + " Ꮯ a + x 22 N + N 1 n+ 1 20. 2 Ï 1 n π + 1 x + h c + h 1 x 22. h 1 1 1 x2 (x + h) 2 23. 24. *2 $2 E t 25. 3 r + R t 7 Ꭱ 1 + 3 p 26. R 6 P 8 42 a2x-3 Express the following as complex fractions and then simplify. 27. a-2b3 28. a-462 α x-1 29. a-2 ax-1 X-2 30. b-2 < a-1 b-1 31. a-1b2x-2 - 2 a-1b2y3 a-5x-3-2 a-5y-3 32. ab-2 +ba-2 a-2 a-1b-1b-2 33. x + (α² — x²) ≤ 1+ x(a² — x²)—§ 34. (a² - x (α² — x²) - * 16. Factoring. Certain useful methods of resolving algebraic expressions into their factors are illustrated in the problems which are worked out in the following list. 22 [CHAP. II. ALGEBRAIC REDUCTIONS EXERCISES Separate each of the following into two factors. 1. 6 a²bx³ + 2 ab²x¹ + 4 abæ³. 2. 10 m¹n² 15 m³n³. SOLUTION: 2 abx³ (3 a+bx+2 x²). ✔ 3. 4 a²bc + 8 a²b²c² 16 a³b2c. — 4. 14 xyz 7 x²yz + 28 xy²z. V 5. a² + ab + ac + bc. SOLUTION: a² + ab + ac + be = a(a + b) + c(a + b) = (a + c) (a + b). 6. x² + 5x + xy + 5 y. 8. xy + x 2 y 2. 10. x² - 2 x 8. ✓ 7. ax √9. m² 3 by + bx 3 ay. n² — (m — n)². 2 x Magal 8, we see that we can find SOLUTION: The result will be of the form (x + a) (x + b). This expanded is x² +(a + b)x + ab. Comparing with x² factors of the given expression if we can find two numbers a and b such that 2, ab = - 8, or a = - - 4, b = 2. Hence, x 4 and x + 2 are factors a+b of x² - 2 x - 8. √11. x²-3x+2. 13. y² — 7 y + 12. 12. 2 x2 6 x + 4. 14. a² + 7 a a²+7a- 30. √16. a²b² — 7 ab + 10. 15. s² 4s - 60. 17. 8x2 18x9. 18. a³ 3. SOLUTION: a³ — b³ = ( a − b) (a² + ab + b²). ✓ 19. a³ - 27. 21. 16 a¹ 64. — 23. r6 $6. ✔25. 81 s¹ — 49 t¹. 27. x²y² - 9 a²b². 29. (3 x — 4)² — (2 x — 2)². √31. x² + y² — 9 x²y² + 2 xy. 33. 1 343 x3. 35. a³ + b³, √37. 64x6 + 125 y³. 39. a5 05. /41. x + y2 + ₤43. 43. 1+ a² + a¹. a²+a+! ✔45. a4 + 4. a² √20. m - 1. 22. a¹-81. 24. 1 66. 26. a³ — (α —b)3. ↓ 28. x² + (b − a)x — ab. √30. a² + 2 ab + b² c2. 32. 64 a³ 27 y³. - 34. 27 a³ - 8 x3. 36. 278 x³. 38. x³y³ +216. √40. (a+b+c)² — (a — b — c)². √ 42. x9 8 x3 x6 + 8. 44. 16x4 + 4 a²b¹x² + a4b8. 3 √ 46. (3 m + 2)² + (2 m + 1)³. 17. Radicals. A rational number is one which can be repre- sented as the quotient of two integers. In operations with radi ARTS. 16, 17] 23 RADICALS cals we deal with irrational numbers; that is, with numbers which cannot be expressed as the quotient of two integers.* Any irra- tional number can always be inclosed between two rational num- bers which differ from one another by as small a number as we please. Thus, we may write, 1 < √2 < 2, 1.4 < √2 < 1.5, ་ 1.41 < √2 < 1.42, 1.414 < √2 < 1.415, Either of the sequences of numbers in the two outer columns determines the square root of 2 in the same way that the sequence .3, .33, .333, .3333,... determines the fraction. Any irrational real number may be determined in this way by a sequence of rational numbers. Besides those numbers in which a radical sign is used to express their exact values, there are many others which belong to the class of irrational numbers. For example, the num- ber is an irrational real number determined by the sequence 3, 3.1, 3.14, 3.141, 3.1415, .... Rules of operation with expressions involving radicals are given in connection with the treatment of fractional exponents (Art. 12). *To show that V2 cannot be expressed as the quotient of two integers, suppose it possible that M √2 M N where is a rational fraction in its lowest terms. At least one of the N numbers m or n is odd. Clearing of fractions and squaring both sides, we get 2 n² m². From this equation m² is an even number, hence m is an even number. If m is even, m² contains the factor 4. Hence n² is an even number, and n is itself even. This is contrary to the hypothesis that m S is a fraction in its 22 lowest terms. This proof is found in Euclid, and is supposed to be due to a much earliei mathematician than Euclid. 24. [CHAP. II. ALGEBRAIC REDUCTIONS EXERCISES Introduce the coefficients * of the following seven radicals under the radi- cal sign. 1. 2√3. SOLUTION: 2√3 = √4 · √3 = √12. • 2. 3√11.V 5. 2 V13. 3. 25. 16. 76. 4. 27. 7. 4 V3. 8. Perform operations in Exercises 1-7, using fractional exponents in- stead of radicals. Change the following seven expressions into equivalent expressions having no fractions under the radical sign. 9. 1 V3 SOLUTION: WINT √2 √2 √3 3 V3 √3 v3 10. V. 11. V. 13. VI. 14. 6V7. 10 √6. 12. √‡. 15. 7. 16. Change the radicals in Exercises 9-15 to fractional exponents and reduce each to a fraction whose denominator is free from fractional expo- nents. Change the following eight radicals into equal expressions with as small a positive integer as possible under the radical sign. 17. √8. SOLUTION: √8=√4.2 18. 32. 21. 48. √48. √8 = √4 • 2 =√4 • √2 = 2√2. 19. √18. 20. V125. 22. 81. 23. √800. 24. 2187. 25. Perform operations in Exercises 17-24, using fractional exponents. Change the following into equivalent expressions whose indices are the smallest possible positive integers. 26. VI. SOLUTION: √4=√ √₁ = √2. 27. 8000. 28. 81. √29. √169. 30. 343. √31. 64. 32. Perform operations in Exercises 26-31, using fractional exponents. *By the coefficient of a radical, we mean the number by which the radi- cal is multiplied. In general, the coefficient of any symbol is the expression by which the symbol is multiplied. ARTS. 17, 18] 25 RADICALS Change the following fractions into equivalent fractions, having no radicals in the denominators. 33. 1 1 + √2 SOLUTION: 1 1 - 1 −√2 _ \ = √2 − 1. 1+√2 1+√2 1−√2 1-84. 1 2 35. 1/36. 1 10 5 37. V5 3/2 1 – √3 √8+ 3 //38. 3 5 39. V/40. 4 v2 −√3 √5 + √3 41. Given √2 = 1.4142, √3 = 1.7321, √5 = 2.2861, evaluate the expres- sions in Exercises 33, 34, 36, 40 (to four significant * figures). 18. Reduction of expressions containing radicals to the simplest form. An expression containing radicals is said to be in its simplest form with regard to the radicals when, (a) there are no fractions under the radical signs; (b) the radicand contains no factor raised to a power whose exponent equals the index of the root; (c) the indices of the radicals are positive integers and as small as possible; (d) there are no radicals in the denominator. Reduce to the simplest form. EXERCISES 1. √ 4 + √ √ V3 100 √3 SOLUTION: 3 √4 + √¡ _ √ √↓ + 2v! _ √2 + § √3 √3 V3 V2 + 3 √3. √ 3 $108 + 2 3 √3 √3 3 2. VT. 3. V. 4. 5. V1215. 10 6. 32. V11 * In giving a result such as 2.2361 to four significant figures, we give 2.236. In giving the same result to three significant figures, we give 2.24 rather than 2.23, for 2.24 differs less from 2.236 than 2.23 differs from 2.236. In fact, it is usually desirable in giving any number of figures of an approx- imate result, to find whether the next figure beyond those to be retained in the result is less or greater than 5; for, if we should obtain a result 2.23 and know that the next figure > 5, the result should be given to three significant figures as 2.24. 26 [CHAP. II. ALGEBRAIC REDUCTIONS 7- √3 V 7. 8. 7+ √ 3 3V~ 2+5√] √2 + √5 −√3 9. 10. √ + √ } . √2 − √5 + √3 √ √ Express in terms of radicals in simplest form. 3 ž +22 11. 12. 2(5)2 + (3) 31 - 21 3(5)* — 2(3) 12 ~~ 13. a²-a-} a}_a} x3 1 14. x − 1 __ x ³ — 1 1 3 x³ +1 19. Addition and subtraction of radicals. Radicals having the same index and same number under the radical sign are called similar radicals. Similar radicals can be combined in addition and subtraction. EXERCISES Perform the operations indicated. 1. 2√6 +3√6+ √24-2√54. · SOLUTION: 2√6 + 3 √6 + √24 – 2√54 = 2√6+3 √6+2√6−6√6= √6. 2. √3 +4√3 – 2√3. 4. √48xy²+y√75x + √3x(x — 9 y)², 5. √3+ √27 +3√27 -√75+ V9. 7. b√u+x√a-y√α³. ↓ 9. 2√§ + ¦ √60 + √15 + √§. 3. √24 + √54 — √96. - 6. √128-250+ √72 - √18. | 8. √ ! + √ } = √§. 10. 64x²+2√2x + V8x³. 11. 5 Vä+ 3√ã + √a + 5 √a − 2 √ā — 7 vā. 12. 20(3)* — 4(3)*. 1 * 14. (405) —(180). V13. (243)¹ +(27) * +(48) ³ —3}. 15. (135)* — (40)³. 20. Multiplication and division of radicals. In Art. 12 we have seen that Va. Vb=ab and Va n/a In multiplying or W//b v n dividing radicals with different indices, the radicals should first be reduced to the same index. EXERCISES Perform the indicated operations. i. 3. V5. SOLUTION: Hence, √3 = √ √9 = √9. √5=√125 125. $3. √5 = V5. V125 = √1125. ARTS. 18-21] 27 EVALUATION OF FORMULAS 2. √3. 5. 3. √za. √3a². 4. 5√3.5√§• √2. 5. 4V12.3 V4. 6. 7. VII. 7. 3va.4Vб. 9. (2√3+3√Z) (3 √3 − √2). 3 11. (mn)³. + 8. (√7 −1)(VT + 1). ✔10. (11√2-4√15) (√ő + √5). 12. (V12 V19) (√12+✓19). — ✔ 13. (3√1+a-4√a) (√1+a+2√α). 14. (x²-x√2+1)(x²+x√2+1). 15. Vx. √y. Vz. 2n 3n 16. a. Vuz. 3/u³. 100/0 7039 V17. $72 7/18 3 19. √2. 4. V3. 18. √2 49 20. 2√3 4 21. У 语 ​5 レ ​22. Xx √3 la √28x 23. a 24. √42 x 1 cq 2 1 ✓ 25. 27. (18) 33 73 72 29. V12 a² ÷ √8 a³. 21. Evaluation of formulas. 26. 73 ✔28. √5 ab ÷ V125 b²a. 5 30. √6 ax² ÷ V/D a²x². 1. The speed v in feet per second of a projectile of weight w pounds, and diameter d inches is given by 1 1 + で ​2:0 td2 7000 w where vois muzzle speed with which the number of seconds after leaving the muzzle. d = 14, w = 1400. 2. H = 29 shell is projected, and t is the Find v when vo= Find v when v。 = 2800, t = 4, 400, w = 0.7, y = 32.2, and v = 80. Find H when T≈ 400, w = 0.7, g T 825 - wv²). α P+ :) (0-0) 3. T= 2:2 R Find T when a = 1222, b = 0.01869, R = 35.24, v = 1.550, P = 11000. * A letter with a subscript, say a,, is read, “ a sub r." 28 [CHAP. II. ALGEBRAIC REDUCTIONS 4. w = w' Find w when 8 (80 — 81). w' = 10.700, $ 0.0050, 80 = 8.5, S1 10.9. 5. The specific gravity S of a floating body is given by the expression W1 S W1 + (W2 03) where w₁ is the weight of the body in air, w₂ is the weight of a sinker in water, and wз is the weight in water of the body with sinker attached. Determine the specific gravity of a body when by physical measurements it is found that WI = 16.50, W2 182.3, W3 = 176.5. 6. One cubic centimeter of mercury at x° centigrade increases in volume when heated to y° by an amount given by the following formula : A(y — x) 100 Ax 1 + 100 where A = 0.018. Find the increase in volume when the temperature is raised from 12° to 126°. 7. The sag of an overhead trolley wire in an electric tramway is given by the formula 3 L(1 — L) d = 8 where d is the number of feet in the sag, I is the number of feet of wire between poles, L is the number of feet from pole to pole. Find the sag when poles are 100 feet apart and length of wire is 100.1 feet. 8. The time in seconds required for the discharge of water from one vessel to another through an orifice in the side is t 0.116. A. B. (F³ (A + B) · a where F and ƒ are the differences in the heights of water in the two vessels at the beginning and end respectively of the discharge, a is the area of the orifice, A is the area of a horizontal section of the discharging vessel, and B is that of the receiving vessel (measurements in inches). f Find t who F 128, ƒ 92, A = 96, B = 60, and the orifice is a circle one inch in diameter. ART. 21] 29 EVALUATION OF FORMULAS ✔ 9. The area in square feet of the top of a well-designed chimney is given by the formula A = 0.03_2 where is the quantity of coal in pounds used per hour and h is the height of the chimney. What should be the area of the top of a chimney 150 feet high which is connected with a furnace using 11,000 pounds of coal per hour? 10. To correct a barometer reading for temperature the following amount is subtracted from the reading : m(t − 32) — s(t — 62) B 1 + m(t − 32) where B is the barometer reading in inches, t the temperature in degrees Fahrenheit, m = 0.00010, s = 0.00010, s = 0.00001. What is the corrected reading of the barometer when the temperature is 75° and the barometer reads 29.95 ? 11. The quantity of water in cubic feet per second flowing through a rectangular weir is given by the formula Q = 3.33 · [24]², L h] where h is the depth of water over the sill of the weir in feet, and I the length of the sill. Find Q, where L 26, h = 1.6. 12. Let P be the day of the month, the number of the month in the year, counting January and February as the 13th and 14th months of the preceding year, N the year, and n = If N 100 2. 400 5 1 ] + N n 4 1) P + 2q + [3 ( 9 + 1 ) ] + be divided by 7, the remainder will be the day of the week of a given date where Sunday counts as the first day. The expressions in brackets mean the largest integer contained in the inclosed number. Verify this formula for the present date. 13. Three equal uniform rods of weight w and the length l are jointed to- gether to form a triangle ABC; this triangle is hung up by the joint A, and a weight is attached to B and C by two strings of length compression in BC is given by し ​√2 The x = W √3 + 1 2√3 го + √3 Obtain a correct to nearest unit when W= 1000, w = 500. 14. The area of a triangle whose sides are a, b, c is given by the formula where 8= a + b + c 2 √s (s — a) (8 — b) (s — c)、 Calculate from this formula the area of a triangle whose sides are 5, 12, and 13 inches. 30 [CHAP. II. ALGEBRAIC REDUCTIONS 15. The formula for the horse power H.P. of an automobile engine is ' H.P.= Planc (24) (33000) where P is the pressure in pounds per square inch, l is the length of stroke of the piston in inches, a is the area of the end of the piston in square inches, n is the number of revolutions of the fly-wheel per minute, c is the number of cylinders in the engine. How many H.P. are developed by a six-cylinder engine if P is 72, l is 5, a is 15.56, and n is 1200 ? 22. Imaginary numbers. The square root of a negative num- ber is an example of an imaginary number mentioned in Art. 2. To perform operations with square roots of negative numbers, replace any such number, say V-a, by iva, and operate with i as with any other letter; but replace in any expression by 1. This method of operation is justified in Chapter XII. EXERCISES Perform indicated operations and simplify when possible by making ¿2 — — 1. 1. (1 + i) (2 — i). SOLUTION: 2. (2 — i) (2 + i). 4. (1+√3)2. V 6. ¿³ + 1. i + 1 (1 + i) (2 — i) = 2 + i — i², 2 + i + 1, since ¿² — — 3+ i. — 1, کا √3 2 2 8. (247) (24-7 i). 10. (12)(1 + ¿²). 12. (a + bi)3 + (α — bi)³. (a i √3 312 2 1 ✓ 3. (- 1 + 1 ) ( -1 - ³). i 5. (x+ui)(x — ai). 2 ї 2 7. (2√3) (2 + √− 3). ✓ 1+√-1 $9. 1 √-1 11. (4 + 3 i)² + (4 − 3 i)². 13. (34) + (3 — 4 i) 4. +1. // 14 (+1) + ( + + √³) + 1 4. 2 2 15. (x a — bi) (x — a + bi). i 2 CHAPTER III VARIABLES AND FUNCTIONS 23. Constants and variables. A constant is a symbol which represents the same number throughout a discussion. A variable is a symbol which may represent different numbers in the discus- sion or problem into which it enters. Many mathematical expres- sions contain both variables and constants. Except in certain geometrical and physical formulas it is customary to use the letters a, b, c, from the beginning of the alphabet for con- stants and the letters x, y, z, from the end of the alphabet for variables. ... ... Exercise. If A and B are two points in a plane and a point P moves in a circle about A as a center, which of the distances PA, PB is constant ? which variable? 24. Definition of a function. Many problems in mathematics, physics, engineering, and chemistry involve two variables which are so related that, a value of one being given, the other can be found. The relation between the variables may be exhibited in various ways. Sometimes the values of the variables are arranged in the form of a table. For example, a life insurance agent refers to a table to find the premium corresponding to a given age. Here the two variables are "premium" and "age." In algebra one variable may be connected with another in an equation or one variable may be an algebraic expression contain- ing the other. In the equation 3x 5y = 4, if a value be given to x the corresponding value of y can be found. Thus if a = 0, ‡; if x = 1, y , and so on. Y In evaluating the expression x² + x + 1, we find that x² + x + 1 = 1 when x = 0, x² + x + 1 = 3 when x = 1 and so on. Fixing the value of x in the first illustration fixes the value of y; in the second illustration fixing the value of x fixes the value of x² + x + 1. 31 32 [CHAP. III. VARIABLES AND FUNCTIONS Definition. If two variables are so related that when a value of one is given, a corresponding value of the other is determined, the second variable is called a function of the first. Thus in the equation 3 x 5y = 4, y is a function of x. The expression x² + x + 1, and in general any expression containing x, is a function of x. We may therefore and shall speak of a "function of " instead of "an expression involving the variable x." F" ... are They are read, “ƒ” discussion, ƒ (x) is a 1 and f(x + h) is 25. Functional notation. The symbols f(x), F(x), þ(x), used to represent functions of the variable a. function of x, “F” function of x, etc. If, in a function 22 + 3 x 1, then f(a) is a² + 3a (x + h)² + 3(x + h) — 1. Similarly, if then and : ¢(2) = a +5, $(x) (2)225.2, (1 + y) = (1 +y)² + 5(1 + y). These illustrations bring out an important point in the functional notation, viz. If the same functional symbol, say (), be used more than once in any discussion, it stands in each case for the same operation or series of operations on the number or expression contained in the parentheses of the functional symbol. This notation is very convenient because it enables us to indicate the value of the function for any values of the variable on which it depends. EXERCISES 1. The fact that the area of a circle may be calculated from a given radius may be expressed in the functional notation by A = f(r). Give the particular form of ƒ (r) in this illustration. V 2. The volume, V, of a cube is a function of the edge x. fact in functional notation. 3. The hypotenuse of a right angled triangle is of length 10. length of one leg y of the triangle as a function of the length leg x. Express this Express the of the other 4. Boyle's law says that P.V C, where P is the pressure of a gas, V its volume and Ca constant for a given temperature. Express P as a func- tion of V. Write in symbolic language the fact that C depends on the temperature T. ARTS. 24-26] 33 SYSTEM OF COORDINATES 5. If f(x) is x² + x + 1, find ƒ(a), ƒ(o), ƒ(1), ƒ(− 1), ƒ (— 10), ƒ(a+1). 6. Given (x) Find (2), (0), ( − 1), ø(1 − √2), ø(a+b). x + 1 = • X 1 17. y² + 1 Find F(0), F y2 1 F(1), F(VT). 7. Given F(y)= 8. If f(x) ƒ(3) =ƒ(− }). 2 x4 5 x3 5 x² + 5 x + 3, show that ƒ (1) = ƒ(− 1) = - G 9. Given ø(x) = x² + x + 1. √10. Given y = f(x) = 11. Given s = $(t) 12. Given s = $( x + 1 2 x 1 t + 2 = 3 t + 4 * t + 2 3 t + 4 • Find ☀(x²), ☀(x + 1), $($(x)). Show that f(y) reduces to x. Find o (글​). Find (8) in terms of t. 8 I P Y 1 - 2 -1 1 26. System of coordinates. Let X'X and Y'Y be two straight lines meeting at right angles. Let them be considered as two number scales with the point of intersection as the zero point of each. Let P be any point in the plane. From it drop per- pendiculars to the two lines. Let a represent the perpendicu- lar to Y'Y, and y the perpen- dicular to X'X. If P lies to the left of Y'Y, x is to be con- sidered negative. If P lies above X'X, then y is positive. It is clear that no matter where P is in the plane, there corre- sponds to it one and only one pair of perpendiculars, x and y. -3 -1 -2 -3 -4 FIG. 2. 19 84 The lines of reference X'X and Y Y are called the coordinate axes, and their intersection is called the origin. The first line is called the X-axis, and the second the Y-axis. The perpendicular to the X-axis from a given point in the plane is called the ordi- nate or y value of the point. The perpendicular to the Y-axis is called the abscissa or x value of the point. If we have two numbers given we can find one and only one point P which has the first number for its abscissa and the second 34 [CHAP. III. VARIABLES AND FUNCTIONS for its ordinate. If, for example, the numbers are 2 and -5, we measure from the origin, in the positive direction, a distance 2 on the X-axis and at this point we erect a perpendicular and meas- ure downwards a distance 5. We have then located a point whose a is 2 and whose y is -5. This point may be represented by the symbol (2,5). The symbol (a, b) denotes a point whose abscissa is a and whose ordinate is b. The symbol P(a, b) is sometimes used and is read, "the point P whose coordinates are a and b." When a point is located in the manner described above, it is said to be plotted. In plotting points and obtaining the geomet- rical pictures we are about to make, it will be convenient to use X FIG. 3. -X coordinate paper, which is made by ruling off the plane into equal squares with the sides parallel to the axes (Fig. 3). Then the side of a square may be taken as the unit of length to represent a number. To plot a point, count off from the origin along the X-axis the number of divisions required to represent the abscissa. and from the point thus determined count off the number of divi- sions parallel to the Y-axis required to represent the ordinate. ARTS. 26, 27] 35 GRAPH OF A FUNCTION EXERCISES AND PROBLEMS 1 1. Plot the points (3, 4), (3, 4), (−3, — 4), (3, −4), (6, 0), (− 5, 0). 2. Draw the triangle whose vertices are (3, 1), (0, 5), (— 4, — 2). 3. Draw the quadrilateral whose vertices are (2, −2), (−3, 4), (−6, −3), (3, 4). 4. If a point moves parallel to the X-axis, which of its coordinates re- mains constant ? 5. A line joining two points is bisected at the origin. If the coordinates of one end are (12, 3), what are the coordinates of the other end? 6. A square of side 3 has one corner at the point (2, 1). If the sides of the square are parallel to the coordinate axes, what are the coordinates of the points that may be at the other corners of the square? 7. Given a N. and S. line and an E. and W. line for reference lines (X and Y axes respectively), the following coordinates of points of a river indicate its general course. − (0, 1), (1,2), (1, -21), (2, 1), (3, 1), (4, 5), (5, 10), (1, 0), (-2, 1), (3, 2), (3, 1), (— 4, − 1), (5, Map the river from x = 5 to x + 5. 3). 8. A square of side 4 has its center at the point (1, 3). What are the coordinates of the corners (a) when the sides of the square are parallel to the coordinate axes; (b) when the diagonals are parallel to the axes? (Give answers correct to two decimal places.) 27. Graph of a function. By a method analogous to that em- ployed in Prob. 7, Art. 26, a function may be represented with reference to coordinate axes. This representation of a function is called the graph of the function. The graph of f(x) contains all points whose coordinates are (x, f(x)) and no other points. f ž EXAMPLE: Obtain the graph of 3 x + 4 for values of x between 5 and + 5. Let ƒ (x) = x + 4. The object is to present a picture which will exhibit the values of f(x) which correspond to assigned values of x. Any assigned value of x with the corresponding value of ƒ (x) determines a point whose abscissa is x and whose ordinate is ƒ (x). Assuming values for x and computing the corresponding values for ƒ (x), we obtain the following table. X 1 2|2|3 4 5 3 ƒ (x) | 4 | 22 | 731 17 10 22 2 3 + } 1 2 13 12 25 These corresponding values are plotted as coordinates of points in Fig. 4. 36 [CHAP. III. VARIABLES AND FUNCTIONS It should be noted that there is no limit to the number of corresponding values which we may compute and imagine plotted in a given interval along the X-axis, and further that to small. f(x) H 2 L 1 2 3 4 5 changes in values of x there cor- respond small changes in the values of f(x). These facts suggest the idea of a continuous curve to repre- sent f(x) much as as a continuous curve is used in mapping a river. (Prob. 7, Art. 26.) It must not, however, be assumed that all functions give continuous x graphs; Article 28, below, considers a graph made up of isolated points. The important fact for this course in algebra is that we may assume a continuous curve for all functions which are polynomials in x* and for most other functions which occur in this course, although the proof of continuity is beyond the scope of this book. That is to say, it is proved in higher analysis that a function of the type α。x² + α₁x²-1 + +a, (n a positive integer) FIG. 4. has a continuous graph. ... Hence, in finding the graph of a polynomial, when a sufficient number of points are located to suggest the general shape of a curve through them, draw a smooth curve through the points. In particular, it is proved in analytic geometry that when n=1, the graph of a function of this type is a straight line. In the problem in hand, the graph is the straight line shown in Fig. 4. Much use is made of 28. Function defined at isolated points. systems of coordinates in presenting statistical results when one set is a function of the other. The following infant mortality table is made up from the * By a polynomial in x, we mean an expression of type αox² + α₁xn−1 + ... + Ang where n is a positive integer, and ao, a1, ... an do not contain x. ARTS. 27, 28] 37 CONTINUOUS CURVES United States Life Tables of 1910. Out of 100,000 living new- born babies in each class, it shows the number of deaths during each month of the first year. MONTH OF FIRST YEAR WHITE MALES WHITE FEMALES NEGRO NEGRO CITY RURAL MALES FEMALES MALES MALES 9 1 2 3 4 5 ✪ 1-∞☺ 4844 3787 7370 6380 4969 4570 1242 991 1977 1746 1370 997 1012 850 1831 1555 1091 822 863 740 1695 1394 941 699 750 648 1561 1252 835 595 6 673 578 1425 1134 755 515 7 610 526 1290 1036 694 459 8 553 486 1153 948 640 408 503 450 1037 874 586 363 10 457 421 937 800 537 325 11 420 390 857 725 496 296 12 12 399 359 802 663 466 277 Using the numbers of the months as abscissas and the cor- responding numbers in the column headed "City Males" as ordinates, we locate the upper 5000 set of points in Fig. 5. The ver- 4500 tical unit is 500. 4000 The lower set of 3500 points is given by the column 3000 headed Males." "Rural 2500 In Fig. 2000 5 we thus present to the eye the 1500 relative infant mortality of city and country chil- dren. The graph in this case is made 1000 500 FIG. 5. M 1 City Males Rural Males 2 8 4 Б 6 7 8 9 10 11 12 Graphical representation of infant mortality of city and rural males. 38 [CHAP. III. VARIABLES AND FUNCTIONS up of 12 points. If we look upon the number of deaths as a function of the number of the month, this function is defined at only 12 points. The lines connecting the points in the figure are not necessary, but aid the eye to take in the whole situation. Where two sets of data are exhibited in the same diagram as in Fig. 5 the connecting lines prevent confusion of the two sets of points. EXERCISES Plot the graphs of the following functions. 1. 3x-7. ✓ 2. 4 x + 3. 5. +√25 — x². SOLUTION: We find the following table. 3. 3x. √4.20. x = 0 1 2 3 4 4.5 Greater than 5 √25 — x² 5 4.9 4.6 4 3 2.18 0 Imaginary x 1 2 P CO 3 M 4 - 4.5 5 Less than 5 √25 Xx2 x² = 4.9 4.6 4 3 Co 2.18 0 Imaginary √25 - x2 FIG. 6. Plotting these points and drawing a smooth curve through them, we have Fig. 6. 6. –√25 — x². ✔7. √36 → x². 8. + √9 - x². ±√9 9. x² - x - 12. √10. x² - 4. 11. 12 + x — x². -X 12. 2x²-11 x + 5. 13. x2 2x + 1. 14. From the table on page 37, show graphically on the same diagram the infant mortality of "white males" and of "negro males." 15. Exhibit graphically on the same diagram the infant mortality of "white males" and "white females." 16. Exhibit graphically on the same diagram the infant mortality of "negro males" and "negro females." ARTS. 28, 29] 39 ZEROS OF A FUNCTION 17. The breaking strength of ordinary manila rope is given by the formula B = 7100 D2 where B is the breaking weight in pounds and D is the diameter of the rope. Exhibit this formula graphically for the diameters 1 3 3 4, b, 1, 1, 4, 7, 1, 11, 12, 18, 1, 1, 12, 13, 2 inches. 18. The following table taken from a jewelry catalogue gives the price of diamonds of the same quality for various weights. From this table give a graphical representation of the price of diamonds. Weight in carats 0.15 0.20 0.25 | 0.30 | 0.35 | 0.40 | 0.45 | 0.50 | 0.55 | 0.60 50 65 80 95 110 130 150 170 Price in dollars 30 40 Weight in carats Price in dollars 0.65 | | | | 0.70 | 0.75 | 0.80 | 0.90 | 1.00 | 1.25 | 1.50 | 1.75 | 2.00 190 210 230 250 285 325 400 500 600 700 19. The morning and evening temperatures of a pneumonia patient were as follows: 99°, 103.2°, 105°, 103.6°, 104.2°, 105°, 104°, 105°, 103°, 104.2°, 102.3°, 97.6°, 97.4°, 98.2°, 99°, 98.2°, 98.7°, 98.4°. Give a graphical repre- sentation. HINT: To save work in handling large numbers, 90 may be subtracted from each of the above numbers and the differences plotted. Or "degrees of fever" may be plotted—that is, degrees above 98.6°. 20. A soldier under the U. S. War Risk Insurance plan contributes towards his life insurance on a yearly term plan. This means that if he enters at age 21 he will pay in successive years while in the service the following amounts in premiums for $1000 of life insurance. Year 1 2 3 4 LO 5 6 7 Yearly premium in dollars 7.80 7.80 7.80 7.92 7.92 | 8.04 8.04 Year 8 9 10 11 12 13 14 Yearly premium in dollars 8.168.28 8.28 8.40 8.52 8.64 8.76 Represent graphically. 21. The postage on first-class mail matter is two cents per ounce or fraction thereof. With weights for abscissas and number of cents for ordi- nates exhibit this postage rate graphically. 29. Zeros of a function. By a “zero of ƒ (x)” is meant a value of such that the corresponding value of f(x) is zero. Thus 3 and - 1 are zeros of the function x2 2 x 3, and ± 5 are zeros of √25-22. Stated graphically, the "real zeros of f(x)" are the 40 [CHAP. III. VARIABLES AND FUNCTIONS 3 abscissas of the points where the graph crosses the X-axis. In Figs. 6 and 7 the graphs and the zeros of √25 — æ² and x² — 2 x are shown. One of the main problems of algebra is the develop- ment of methods for finding the zeros of functions. The graphical solution of this problem, so far as real zeros are concerned, con- sists in finding where the graph crosses the X-axis. One of the advantages of the graphical method of dealing with functions is that it presents to the eye the zeros of a function. EXERCISES Plot and find the real zeros of the following functions. 1. c² 2x - 3. SOLUTION: Compute the table. Ꮖ x² - 2x-3 3 2 1 1 2 3 4 5 12 5 0 3 4 3 0 5 12 Plotting these points and drawing a smooth curve through them, we have The graph crosses the X-axis at 1, and 3, which are therefore Fig. 7. x²-2x- 8 -X 1 O 3 FIG. 7. the zeros of x² - 2 x 3. √2. 3 x − 5. 3x 3. 4x + 9. √4. x² 6 x + 5. 5. x² + 4x. 6. 3x² 11x- 1. √7. 6x 5 - x². 8. x3 + 3x2 x - 3. 9. x2 25 x. 10. Between what integers does each of the real zeros of x3+5x2- 4x-1 lie? 11. Show that x2 + x + 1 has no real zeros. CHAPTER IV THE EQUATION 30. Equalities. A statement that one expression is equal to another expression is called an equality. The two expressions. are called the members of the equality. There are two classes of equalities, identical equalities or identities, and conditional equali- ties or equations. An identity is defined in Art. 13. It is there stated that the two members of an identity are equal for all values of the symbols for which the expressions are defined. Thus, 22 a² = (x - a)(x+a), 5 a = 10a - 5 a - are identities. But in the equality x 5 = 4, the two expressions x 5 and 4 are equal only when x has the value 9. An equality of this kind, in which the members can be equal only for partieu- lar values of the letters involved, that is, are not equal for all values, is sometimes called an equality of condition. In this book we shall use the term equation to mean conditional equality. When it seems necessary to indicate that an equality is an iden- tity and not a conditional equality, we shall use the sign in- stead of the sign between the members. But the sign will be used for both identities and equations when this usage can lead to no confusion. Which of the following equalities are identities? (α) x - 100. x2 @ @ (b) x + a a2 = x - α. (c) x² + 2 x = (x + 1)² − 1. (α) x² + 4 = (x + 2)² - 4 x. (e) 5 x² + 2 x − 6 = 0. (A) 1 = = = ₂ = 1 + x + x² + x³ + 1+x+x²+x³ 1 X In equalities (b) and (ƒ) may a take all real values? X¹ 1 x 41 42 [CHAP. IV. THE EQUATION 31. Definitions. In an equation there are some symbols whose values are assumed known and others whose values are unknown. These are spoken of as the knowns and unknowns. Following the conventions of elementary algebra, the first letters of the alphabet are used to represent knowns, while the last letters represent unknowns. Any expression in the form -1 -2 + α₁-1α + ɑn where n is a positive integer, and do, a1, a2, •••, a„-1, a„ represent ɑ1, ɑ2, any given numbers, is called a rational integral expression in * X, or a polynomial in x (cf. Art. 27). In other words, a rational in- tegral expression in x is the algebraic sum of terms of the type kxª, where a is restricted to take positive integral values. For example, 2 x2 5x and 3x² + 7 x − 3/3 are rational integral expressions in x. As an extension of this definition, we define a rational integral expression in x, y, z, as the algebraic sum of terms of the type ... kxay³zy ..., where a, B, y, are positive integers and k (called a coefficient) does not involve x, y, z, For example, 5 x²y + 3 xz + 3x²-1 is a rational integral expression in x, y, z. γ ... ... By the degree of a term kayẞzy in any letters x, y, z, is meant the sum « +B+y+ of the exponents of the letters in question. The degree of a rational integral expression is defined as that of a term whose degree is equal to or greater than that of any other term in the expression. Thus, 5 x²y + 3 xz + 3 x² – 1 is of degree two in x, one in y, one in z, one in y and z, two in x and z, three in x and y, three in x, y, and z. * By substituting the words "function of " for "expression in" through- out this article, we obtain the definition of an important class of functions. ARTS. 31, 32] SOLUTION OF AN EQUATION 43 The statement that two rational integral expressions are equal is called a rational integral equation. By transposing terms, such an equation can manifestly be written in the form f(x, y, z, ...) = 0, where f(x, y, z, ...) is a rational integral expression. The degree of ƒ (x, y, z, …..) in any letters is said to be the degree of the equa- tion in those letters. In this course, the term degree is applied to equations only when they are in the rational integral form. We sometimes speak of the degree of an equation without men- tioning to what letters we refer. In this case, it is to be under- stood that we mean the degree in all the unknowns. Equations of the first, second, third, fourth, and fifth degrees are called linear, quadratic, cubic, quartic, and quintic equations respectively. EXERCISES Give the degree of each of the following equations. 1. x2y3 + 5 = 0. 2. ax² + bx + c = 0. 3. x2 y2 a2 b2 1. 4. ya = ax. y. 5. Give the degree of the expression ax¹ In x and y. 4 mx2y2-3 nxy + y² in x. In −2 6. Give the degree of the equation 10 x4—4 ax²yz -3 xyz+by²=5x¹ — 2 x²y² In z. In y and z. In x and z. In x and y. In x, y, and z. in x. J In y. 32. Solution of an equation. To solve an equation in one un- known is to find values of the unknown that make the two mem- bers equal. Any such value is said to satisfy the equation and is called a solution or root of the equation. A solution of an equation in more than one unknown is a set of values of the un- knowns which satisfy the equation. Thus, a 1, y = 2, is a x = solution of y = x + 1. To solve a system of equations in any number of unknowns is to find sets of values of the unknowns which will satisfy the equations. Any such set of values is said to be a solution of the system of equations. 44 [CHAP. IV. THE EQUATION EXERCISES 1. Is 1 a root of x² - 5 x + 4 = 0 ? 2. Is 6 a root of x² 7x+5 2x + 4 = 0 ? 0 ? 3. Is 4 a root of x² 4. Is 1 + i(i² - 1) a solution of x² − 2 x + 2 = 0 ? 5. Is x = 5, y = 3, a solution of 2x + 4xy = 28 ? 6. Is x = 1, y = 4, a solution of x² + 3 xy — y² 7. Is x = 1, y = 2, z = 1, a solution of 3 x 1, a solution of 3x - 2y + 5 z − 2 = 0 ? -3? 33. Equivalent equations. Two equations or two systems of equations are said to be equivalent when they have the same solutions; that is, when each equation or each system is satisfied by the solutions of the other. Thus, the equations a · 2 = 0 and 3x-6=0 are equivalent, the second being derived from the first by multiplying both members by 3. Again, the equations. x2 − 5x+6=0, and — 10 x² 50x60, are equivalent. The second can be obtained from the first by performing the follow- ing operations on both numbers. (1) Multiply both members by - 10. (2) Add 50 x 60 to both members. It must not, however, be inferred when the same operation is performed on the two sides of an equation, that there necessarily results an equivalent equation. The following equations will show that this is an unwarranted inference. 1. Consider the equation Square both members, 3 x = x + 4. 9 x² = x² + 8x + 16. (1) (2) The equation (2) is satisfied by 2 and 1, while (1) is satisfied by 2 and not by 1. Hence, (1) and (2) are not equivalent. 2. Consider the equation 3x + 2 5x8. Multiply both members by (x − 1), (x − 1) (3 x + 2) = (x − 1) (5 x − 8). (3) (4) Equation (4) is satisfied by 1 and 5, while (3) is satisfied only by 5. Hence, (3) and (4) are not equivalent. 1. 3. Consider the equation V1 − x X First, add x to each member, then square both members. results 1-x=1-2x + x². (5) There (6) Equation (6) is satisfied by x = 0 and x = 1; but x = 0 does not satisfy (5). Hence, (5) and (6) are not equivalent. ARTS. 32-34] 45 EQUIVALENT EQUATIONS 4. Consider the system of equations x + y = 15, X y = 5. (7) Multiply the members of the first equation of (7) by x, the second by y. There results x(x + y) = 15x, y( − ) = . . (8) This system (8) is satisfied by the four pairs of numbers (10, 5),* (0, 0)). (0, — 5), (15, 0), but (10, 5) is the only one of these pairs which will satisfy (7). These simple examples show that the same operation performed on the two members of an equation does not necessarily lead to an equation equivalent to the original one. It is manifestly important to know whether an equation is equivalent to that from which it is derived; and if non-equivalent, whether it contains at least all the solutions of the original equa- tion. If a derived equation contains all the roots of the original equation and some others, we shall call it redundant. If the derived equation lacks some roots of the original equations, we shall call it defective. The student should always be on his guard against treating two equations as equivalent simply because the one has been derived from the other. The following operations which the student has often per- formed in elementary algebra lead to equivalent equations: (a) Adding the same number to or subtracting the same num- ber from both members. (b) Multiplying or dividing both members by the same known number provided this number is not equal to zero. (c) Changing the signs of all the terms. 34. Operations that lead to redundant equations. The following operations on the two members of an equation lead, in general, to redundant equations : (a) Multiplying both members by the same integral function of the unknown. *The notation (10, 5) means x = 10, y = 5. (See Art. 26.) 46 [CHAP. IV. THE EQUATION EXAMPLE 1. Consider the equation 3x + 2 = 5 x – 8. The solution is x 5. Multiplying each member by a — 1, we have (x − 1)(3x+2)=(x-1) (5x-8). (1) (2) Equation (2) has roots 5 and 1, but 1 is not a root of equa- tion (1). EXAMPLE 2. Consider the equation c − 1 = 0. (1). The solution is x = 1. Multiplying each member by a, we have x² x = 0. (2) Equation (2) has roots 0 and 1, but 0 is not a root of (1). (b) Raising both members to the same integral power. EXAMPLE 1. Take the equation 3 x = x + 4. Squaring each member, we have Equation (2) has roots 2 and -1, but 1 is not a root of (1) 9x2 x²+8x+16. (2) equation (1). EXAMPLE 2. Take the equation - V1. There is no value of x that satisfies (1). (1) Squaring both members, we have x = 1. (2) While 1 thus satisfies equation (2), it does not satisfy (1). 35. An operation that leads to defective equations. The follow- ing operation leads, in general, to defective equations. Dividing both members of an equation by the same rational integral function of the unknown, when such function is a factor of each member. EXAMPLE 1. Take the equation (x²- 6)(x-2)=3x-6. (1) The roots are 2, 3, and -3. Divide both members by x 2 and we have x² — 6 = 3, the roots of which are 3, and 3. That is, the root 2 is lost in dividing the members by x Gaming 2. (2) ARTS. 34-36] 47 CLEARING AN EQUATION EXAMPLE 2. Take the equation a3-5a2+6x=0. The roots are 0, 2, and 3. Dividing members by 2, we have 2_52+6=0, the roots of which are 2 and 3. The root 0 is lost in dividing by x. We shall call an equa- 36. Clearing an equation of fractions. tion fractional only in case some of its terms are fractions with unknowns in the denominators. When a fractional equation is cleared of fractions, the resulting equation is generally equivalent to that from which it is derived, but it may be redundant. Consider the equation X3 4x²+5x-2 =x-2. x2 x² -3x+2 (1) Clearing of fractions by multiplying both members by x²-3x+2, we obtain x²-3x+2=(x − 2) (x² − 3 x + 2), which is satisfied by x = 1, x = 2, x = 3, while (1) is not satisfied by x=1, or x = 2, as can be shown by substitution. In concluding this review of equivalent equations, it need hardly be said that we have by no means exhausted all the types of operations which it may be necessary to perform on the mem- bers of an equation, but enough has been said about a few simple operations to warn the student against proceeding blindly in de- riving equations from a given equation. Unless the operations. on the members of an equation are known to lead to equivalent equations, the student should never regard the solution as com- plete until the test of substitution has been applied. EXERCISES == 1. Form equations by multiplying the members of 2x = 5 by each of the following expressions. (a) x. (b) x - 3. (c) x2 4. What roots have the derived equations that are not roots of 2 x = 5 ? 48 [CHAP. IV. THE EQUATION ✓ 2. Form an equation by multiplying the members of x² + 2 = 3x by x - 3. X In respect to what root is the resulting equation redundant ? — 3. What root has the equation æ² x=0 which is not a root of the equa- tion derived by dividing the members of the given equation by x? √4. Form an equation by dividing the members of by x 2. (x2 6) (x − 2) · 2) = 3 x — 6, * What root has the given equation which is not a root of the de- rived equation ? 3 x 4. 5. Form an equation by squaring the members of x = Show that 1 is a root of the derived equation but not of the given equation. 6. Given an equation whose members are rational integral functions of x. If you multiply the members by x — a where a is not a root of the given equation, what root is introduced into the derived equation? Illustrate with the given equation x b. 7. If x a is a factor of each member of a given equation, what root of the given equation is, in general, lacking in the equation obtained by divid- ing the members of the given equation by x a? Illustrate with the given equation x(x − a) = b (x − a). Reduce the following equations to rational integral equations and discuss the question of equivalence. 9. 3x - 5 3x-5 4 x 3x-7 7 4x 5 √11. √π-1-1= √æ — 4. - 13. √x-15=1+ √x. 5- 3x 7-9x 8. x + 1 1+ 3 x 3 1 2 10. x2 25 + x + 5 5 x 12. √x-1 : 9. x + 5 X 6 15. 14. 1 2x + 1 x 2 x2 x² 17. 16. x = 21 + √æ² 9. x² it 8 15-7x 1 8(1-x) 5+4 - 1. 37. Type form. unknown CHAPTER V LINEAR EQUATIONS Any linear equation (Art. 31) in one ax+b=0, a‡0, (1) b can be put into the form x = C The point which represents on the line of Fig. 1 may be α conveniently regarded as the locus of equation (1) in one-dimen- sional space. A linear equation in two unknowns ax+by+c=0, b‡0, can be put into the form y ax C b b (2) (3) Since in (3) we may assign to x any value and compute a cor- responding value for y, the equation defines y as a function of x in accordance with our definition of a mathematical function (Art. 24). The graph of a linear function has been discussed in Art. 27, and this graph may be conveniently regarded as the locus of equation (2) in two-dimensional space. The locus in two-dimensional space of an equation in two variables consists of all points whose coordinates satisfy the equation and of such points only. EXERCISES Plot the loci of the following equations. 1. x-y=1. SOLUTION: Putting this in form (3), we obtain y = x 1. 49 50 [CHAP. V. LINEAR EQUATIONS The graph of the function x 1 is shown in Fig. 8, and by definition this is the locus of the equation. Since the graph of any rational integral func- AL tion of degree one (Art. 27) is a straight line, the locus of any linear equation is a straight line. 2. 2xy = 1. 3. x 3y = 1. 4. 3x + 4y = 4. X 5. 5x + 3y = 0. 6. 7 x − 5 y = 0. 7. 7x+5y y = 4. 8. x + y x + Y = 8. 2 3 FIG. 8. 38. Simultaneous linear equations. 9. 4x-3y = 36. Let α₁x + b₁y = c1, avíc + b₂y = C₂, be two linear equations in two unknowns. Multiply the mem- bers of the first by b₂ and those of the second by b₁. Adding the members of the two resulting equations, we obtain (α₁b₂ — α½b₁)x = (b2C1 — B1C2), or bqc₁ - b₁c₂ x= , provided ab₂ — α₂b₁ 0. In a similar manner, by multiplying the first and second equa- tions by ɑ½ and α, respectively, we obtain y a₁₂-α2C1, provided ab₂-ab₁0. 1 We note that the denominators of the above fractions are alike. This denominator may be denoted by the symbol αι δι 02 02 ხი which is called a determinant. Since it has two rows and two columns, it is said to be of the second order. The letters a₁, b₁, a2, b₂, are called the elements of the determinant, and a₁, b₂, con- stitute the principal diagonal. A determinant of the second order then represents the number which is obtained by subtracting ARTS. 37, 38] SIMULTANEOUS LINEAR EQUATIONS 51 from the product of the terms in the principal diagonal, the prod- uct of the other two terms. Thus, x Y = x W — Y Z, Z w 1 2 3 4 4 – 6 — — 2. Using the determinant notation, we may now write the solu- tions of our equations in the form |c₁ b₁ a1 C1 C2 b2 A₂ Cz a₁ bi y = a1 bi A2 by A 2 b 2 We note that the numerator of the solution for x is obtained from the denominator by substituting in place of a, a, which are the coefficients of x in the equations to be solved, the known terms c₁, Cy. In a similar manner, in the numerator of the solution for y we replace b₁, b₂ by c1, c2 respectively. EXERCISES 1. Solve: x + y = 3, 2x + 3y = 1. = 3 1 1 3 9 1 X = 1 1 3 2 || 8, 23 1 3 2 1 1 6 У y = - 5. 1 1 3 – 2 2 3 Solve the following pairs of equations, using determinants. 2. 3x+2y= 23, 5x - 2y = 29. 4. 6x+5y = 16, 5x-12ỳ = - 19. 6. 11 х - 5 у 22 Бу 3 x + y 32 8x-3y = 23. 3. / +1/2 = 11, Y 2 5 + 4 215 7. 6y=9, 5. 3x 2x + 7y = 28. 7. x + 1 + 1 = * = 9, Y х 3 2 x 9 2 + 2 + 3 = 5. 52 [CHAP. V. LINEAR EQUATIONS 39. Graphical solution of a system of linear equations. As stated in Art. 37, the locus of any linear equation in x and y is a straight line. Any such equation is satisfied by an indefinitely large number of pairs of values of x and y. That is, by the co- ordinates of all points on its locus. In the graphical solution of the system of two equations, we seek the coordinates of points. common to the loci of the two equations. As the loci are two straight lines, three cases arise: (1) In general, two lines intersect in one and only one point. (2) Two lines may be parallel, and thus have no point in common. (3) Two lines may be coincident, and thus have an indefinitely large number of points in common. Corresponding to these three cases, a system of two linear equa- tions has, in general, one and only one solution, but it may have no solution or an indefinitely large number of solutions. When the loci are two parallel lines, there is no pair of numbers which satisfies both equations, and the equations are said to be incom- patible or inconsistent. Referring to the expressions for x and y (Art. 38) in determinant form, we see that there is one common solution of the equations unless the determinant in the denominator is zero. When the loci are two coincident lines, the two equations of the system are equivalent. ΑΓ X FIG. 9. X EXERCISES Find the solutions of the following equa- tions by determinants and by plotting the loci. 1. x 2. x−y + 1 = 0, 4 x + y − 16 = 0. See Fig. 9. x + y = 2, 3x-2y=1. 3. 2x + 3y = 18, 3x-2y=1. ARTS. 39, 40] 53 DETERMINANTS OF THIRD ORDER 4. x + 3y = 10, 3x+2y y = 1, ✓ 6. 2 x x + 3y = 8. 9. 11. 5. 2x 11, 3x - 12 y 15. 7. 3x - 12y = - 12, 20. 6x+5y=16, 5x-12y V9. = - 19.. 8100 x + 8 y = Y + 8, 5 6:8 У 1. 9 10 11. ✓ 10. 2x + 2y = 4, 3 x + 3y = 6. x + y = 2, 5x+5y=20. 13. 12. 9x-4y = 0, 5x + 7y = 3, 3x + 14y = 6. 14. 3x+8y = 7. 7x-Y y = 33, 12 y p x = 19. 15. In solving a system of equations Aric + b₁y = C1 u2x + b₂y = C2 by determinants, show that if the determinant in the denominator is zero, and the determinant in one numerator is likewise zero, then the two equa- tions are equivalent. 40. Determinants of the third order. The square array of nine numbers with bars on the sides ai bi ci a 2 by c₂ az bz cz is a convenient abbreviation for the expression ɑ₁b₂C3 + b₁€2ɑ3 + C1ɑ2b3 − C1b2A3 — α1С2b3 — b1α½C3, A1b2C3 - (1) and is called a determinant of the third order. As in the case of the determinant of the second order, the letters a₁, b₁, etc., are called the elements, and the letters a, b, c3 form the principal diagonal. C3 The expression (1) is called the expansion or development of the determinant. It is seen that each term of the expansion consists of the product of three elements, no two of which lie in the same row or in the same column. Any determinant of the third order 54 [CHAP. V. LINEAR EQUATIONS may be easily expanded as follows. Rewrite the first and second columns to the right of the determinant. The diagonals [a₁ b₁ c₁a₁ bil A ₂ b₂ c₂ α₂ b₂ c3 | Aз bз сз α з b3 running down from left to right give the positive terms. The diagonals running down from right to left give the negative terms. EXERCISES Obtain the expansions of the following determinants. 1 3 4 1. 2 7 3 1.7.5 +3.3.1 + 4.3.2 4.7.1 - 3.3.1-5.2.3 = 1. 1 3 5 10 2 8 2. 540 3. 1 ∞ ∞ 3 1 1 2 4.41 4. 235 7 1 4 5 623 -1 -3 213 2 a b 0 x 1 3 x 2 x -4 5 6.0 x y u O v 7. 2 x 4 6 31 8. 1 1 1 x 4 x 3 1 7 1 2 3 5. -2 41. Solution of three equations with three unknowns. Let the three equations be α₁x + b₁y + c₁% = d₁, AqX+boy+Cq% da, A3x+b3y + C3Z = d3. (1) (2) (3) Multiplying (1) and (2) by b, and b, respectively and adding, we get (a₁b₂ — α₂b₁)x + (b₂c1 — b1c2)≈ = d₁by — d½b1. Eliminating y in a similar manner from (1) and (3), we find (a3b1 − α1b3)x+(C3b1 − b3C1)≈ = d3b₁ — d₁b3. We now have two equations in two unknowns, a and z. nating z from these two, we find [(α₁b₂ — α2b1)(C3b1 — b3C1) — (α3b1 — α₁b3) (b2c1 — b₁₂)]x = (d₁b₂ — dab₁)(C3b1 — b3c1) — (d3b₁ — d₁bз)(b₂c₁ — b₁cº), (4) (5) Elimi- ARTS. 40, 41] EQUATIONS IN THREE UNKNOWNS 55 which after some simplification gives us X d3b1C2 d₁b3C2 d1b2c3 + d2b3C1 + dзb₁₂ — d¸bзc₂ — dzb₂C1 — dyb₁C¾¸ а₁b₂¤з + а½b3c1 + аžb₁с2 — α₁b3с2 — ɑzb₂с1 — аb₁cз ɑ3b1c2 ხვი ვხი ɑ2b1c3 The denominator is the development of the determinant in Art. 40, while the numerator is the same as the denominator with a replaced by d. Hence we can write the solution for x in the form 1 d₁ b₁ c1 dą bą cz dz bz C3 x = a₁ b₁ c1 an b₂ c₂ аz bz cz provided the determinant in the denominator is not zero. In a similar way, we can find the values of y and z. a₁ d₁ c₁ a₁ bi di ɑz d₂ C₂ Aq b₂ da az dz cz az bz, dz y= 2= α1 b₁ ci a₁ b₁ c₁ Az bz C 2 A z by Cz Az bз C3 Az bz Cz The denominators in the expressions for x, y, and z are the same, while the numerators are obtained from the denominators by replacing the coefficients of the unknown in question by the known terms. For example, in the numerator of y, the knowns d1, d2, da replace b1, b2, b3 respectively. Solve: 1. SOLUTION: EXERCISES x ? Z 2x + y + z = 0, 3x-5y +82 5 y + 8 z = 13. 6, 6 1 1 0 1 1 13 -5 8 -78 x = - 2, 1 1 1 39 2 1 1 3 -5 8 56 [CHAP. V. LINEAR EQUATIONS 2. 1 2 ∞O 123 1 -6 -1 2 0 1 3 13 8 39 Y - 1 -1 39 1 1 5 8 x + y + z = 6, 8888 1, z= V 3 x y+22=7, 4x + 3y 2-7. 3 5 8 · √ 3. 3x + 4y − 5 z = 32, 4x 5y+3z18, 5x-3y-4 z = 2. 2 1 2 3 − ∞ ∞ 1 1 0 5 13 117 = 3. 1 1 -1 39 1 1 4. 5x 4y+22=48, 5. // 6. 2x 1 1 + X У 3x + 3y4z = 5y+3z = 19. 24, 2x + 3y + z = 4, x + 2y + 2 z = 6, 1, 1 XC + 1 1, 5 x + y + 4 z — 21. 7. 2x+5y — 3 z 6 x 2y 5 % - 3, 3 x + 7y + 4 z = — - 18. 17, 2 1 - א 1 + 1. Y Z 8. x + 2y z = 1, 9. x + y + z = 10, = 3x + 2y − 4 z = 7, = x + 2y z = 3. (x + z) + y = 9, † (x-2)-2y+7=0. MISCELLANEOUS EXERCISES AND PROBLEMS a₁ bi ci V2 C2 C2 A2 A 2 b 2 1. Show that a2 b₂ c₂ = ɑ1 + 01 + C1 b3 C3 C3 A3 Az bz Az bz Cz A1 + A2 C1 а1 С1 а2 C1 2. Show that + b₁ + b 2 C 2 b1 C2 V2 C2 ai bi ci x 2 y 3. Develop a₁ b₁ ci 4. Develop 1 1 1 A 2 by C 2 C2 x 4 y x 4 1 x 8 5. Solve for x = 0. 6. Solve for x 3 4 x 3 2 = 0. 2 1 5 7. Solve for x and y the system of equations Ꮖ . 1 y x 1 1 1 1 1=0, y -1 0 0. 1 20 3 2 1 ART. 41] 57 EXERCISES AND PROBLEMS 8. A certain kind of wine contains 25 per cent of alcohol and another kind contains 30 per cent. How many gallons of each must be mixed to make 50 gallons of the mixture 27 per cent alcohol? 9. What amounts of silver 72 per cent pure and 84.8 per cent pure must be mixed to get 8 ounces of silver 80 per cent pure ? 10. The sum of the three angles of a triangle is 180°; find the two acute angles of a right-angled triangle if one of them is four times the other. 11. If the sides of a rectangular field were each increased 10 feet, the area would be increased 14,500 square feet. If the length were increased 10 feet and the width decreased 10 feet, the area would be diminished 1700 square feet. Find the area of the field. 12. Two numbers are written with the same two digits; the difference of the two numbers is 45, and the sum of the digits is 9. What are the numbers ? 13. A six-figure number has 1 for the last figure. If this last figure is removed and placed before the others, a new six-figure number is made whose value is one third the original number. What is the original number? 14. The planet Mercury makes a circuit about the sun in 3 months. Venus makes the circuit in 7 months. Find the number of months be- tween two successive times when Mercury is between Venus and the sun. 15. In Wilson and Gray's determination of the temperature of the sun the Fahrenheit reading of the temperature is 5552 more than the centigrade reading. What is the centigrade reading? 16. If h represent the height in meters above sea level, and b represent the reading of a barometer in millimeters, it is known that bk + hm, where k and m are constants. At a height 120 meters above sea level the barometer reads 751, at height 769 meters it reads 695. What is the formula showing the relation between b and h? 17. Two runners are practicing on a circular track 126 yards in circum- ference. When running in opposite directions they meet every 13 seconds. Running in the same direction, the faster passes the slower every 126 sec- onds. How many minutes does it take each to run a mile ? 18. The relation between the boiling point w of water in degrees Fahrenheit and h the height in feet above sea level is known to be of the form x wy=h, where x and y are numbers to be determined by experi- ment. It is observed at the height 2200 feet that the boiling point is 208° F. At sea level the boiling point is 212° F. What is the formula showing the relation between w and h? 19. It is required to find the amount of expansion of a brass rod for a rise in temperature of one degree centigrade, also the length of the rod at a tem- perature 0°. If c represent the expansion, and b。 the length required, it is known that b = ct + bo, where b is the length of the bar at the temperature t. When t = 20°, the length of the rod is 1000.22; when t = 60°, the length is 1001.65. 58 [CHAP. V. LINEAR EQUATIONS the other 4%. 20. A man has $35,000 at interest. For one part he receives 31%, for His income from this money is $1300 per year. How is the money divided? 21. To find the average grade of a freshman in mathematics, his grade in analytic geometry is multiplied by 5, his grade in algebra by 3, and his grade in trigonometry by 2, and the sum of the three products is divided by 10. This gives 89 for the average grade. If the grades in analytic geometry and algebra had been interchanged, his average grade would have been 91. If the three studies had all counted the same number of credits, his grade would have been 90. What are the grades in each of the three studies? 22. A cistern is filled with three pipes. The first and second will fill it in 72 minutes, the second and third in 120 minutes, and the first and third in 90 minutes. How long will it take each of the pipes to fill it? 23. Four numbers have the property, that when successively the average of three is added to the fourth, the numbers 29, 23, 21, and 17 result. What are the numbers ? 24. Five numbers are arranged in order of magnitude. The difference between any two consecutive numbers is the same number. The sum of the numbers is 60. What positive integers satisfy these conditions? 25. Some books are divided among 3 boys, so that the first had 12 less than half of all the books, the second had one less than half the remainder, and the third had 17. Find the number each received. 26. How many 5 per cent bonds of $100 each at 90 must I sell in order that by investing the proceeds in 6 per cent stock at 102 my income may be increased $300 ? 27. Between two towns the road is level for half the distance. The speeds on a bicycle are 3, 6, and 9 miles an hour uphill, on the level, and downhill, respectively. It takes 5 hours to go and 4 hours to return. What are the lengths of the level and inclined parts of the road ? CHAPTER VI QUADRATIC EQUATIONS 42. Type form. Any equation of the second degree (Art. 31) in one unknown a can, by transforming and collecting terms, be written in the typical form ax² + bx + c = 0, where a, b, c do not involve a, and have any values with the one exception that a is not zero. Since the result of multiplying the members of an equation in this typical form by any given number is an equation in typical form, the a, b, c can be selected in an indefinitely large number of ways. In particular, since we can change the signs of all the terms in an equation, we may assume that a is positive when it is a real number. The function ax² + bx + c (a0) is called the typical quad- ratic function. EXERCISES Arrange the following equations in the typical form and select a, b, and c from the resulting equations. 1. 4x²-5 + 5+ 3x 2 x2 2 3 + m. By transposing and collecting terms, 10 3 x² + 3 x (5+m) = 0, 2 10 3 so that α C = - −(5+ m). 3 2. x² + (2x + 5)² = 3 x. 3x. x2 (x − 1)2 3. + = 1. 9 16 1 d 1 x2 4. x² - 2 dx x + d² + 0. 2 2 2 3x 5. 4 m²x² + 3 k²x² - 8 mx + 3 x — m + k = 0. 6. (y + 3)² + (y − 2)² — 0. 3 8. (z+2)³ -- (z — 3)³ — 1 = 0. 7. r² + (2 r−7) 10. = 9. u² +(mu + n)² = k. 59 60 [CHAP. VI. QUADRATIC EQUATIONS 43. Solution of the quadratic equation. A quadratic equation. may be solved by the process of "completing the square." For example, to solve 3x²+5x-2=0, write the equation in the form x² + ½ x = 2 36 Add (·)² = 3 is a perfect square. or to both members, and the left-hand member That is, 융​+ 융융 ​49 x² + { x + } } = 3 + }} = 18, 36 49 (x + 5)² = 18. 土 ​Extract the square root, x+ = ± 7, x 2, or . Both of these values of a satisfy the original equation, as we see on substituting them for x. Thus 3(-2)²+5(-2)-2-3.4-10-20. 3 ({})² + 5 (} ) − 2 = 3 ⋅ } + { − 2 = 0. Apply this method to the general quadratic equation ax² + bx + c = 0. Transpose c and divide by a, Add (2) 2 ს x² + с x α α : to both members to make the left-hand member a perfect square, or b x² + = x b 2 b 2 b2 4 ac x+ a + 2 α α 2 a 4 a² b12 b2 4 ac x + 2 a 4 a2 Extract the square root, or b x + 2 a x= ± √b² — 4 ac 2 a -b±√b² 4 ac G 2 a ART. 43] SOLUTION OF THE QUADRATIC EQUATION 61 The roots of the general quadratic equation ax² + bx + c = 0) b + √ b² - 4 ac b -√b² 4 ac are x1 , X2 2 a 2 a as may be verified by substitution. The expression − b ± √ b² − 4 ac 2 a may therefore be used as a formula for the solution of any quad- ratic equation. Thus, to solve the equation 3x2+5x-2=0, we substitute in the formula, a=3, b=5, c=-2 and find · 5+ √25 −4·3 · (− 2) - 5+ √49 21 1= 1 6 6 Similarly ໑ · 5 – √49 6 2. EXERCISES Solve the following equations by use of the formula, and verify by substitution. 1. 2x²+x 1 = 0. 3. 4x² 9x+2=0. 5. 2x² + 3 x − 9 = 0. 2. 2x² 3 c 2 = 0. 4. 3x² + x M 2 = 0. 6. x²-6x — 7 = 0. 8. x² + 3x + 5 = 0. 10. 2x²+7=4x. 7. 3x²+8 a 3 0. 9. 2.2 0. 11. 7 x² + 7 x + 4 = 0. 12. s² + 12 = 8 s. 13. 382 + 2 = 6 s. 15. 7y²+9Y 14. 3t2+2 t = 4. 10 = 0. 16. 3² + "= 200. 17. 6x2+5x — 1. 18. 9x²+3x= 2. 19. 7x²+ 2 x = 32. √ 2 +3 2 x 1 20. 2 x 7 2 3 1 1 x + 1 + x 21. 8x +11 + 78 68x x. 22. 7 x X + 1 X 13 x 4 62 [CHAP. VI. QUADRATIC EQUATIONS 23. x² + x√5 – √5 = 0. 25. 1 x 1 1 2 41. x 1 1 27. x2 = 2 ax + 3 x − 6 a — 0. 29. x² + 2 ax + bx + 2 ab = 0. 30. Show by substitution that are roots of ax² + bx + c = 0. x 18 24. 2 x 30. 103 26. x² + ax 2 a2 = 0. — 28. 2x² + 3 nx — 2 n² = 0. − b + √ b² — 4 ac 2 a and b√b² 4 ac 2 a - 44. Solution by factoring. When the left-hand member of a quadratic equation can be factored readily, this is the easiest method of solution. Take, for example, the equation x² - 4 x 21 = 0. x2 The factors of the left-hand member are easily found. They are (x+3) and (≈ - 7), and we may write our equation in the form (x+3)(x − 7) = 0. : Any value of x which makes either factor zero will satisfy the equation. If x=-3, we have (− 3 +3)(− 3 — 7) = 0 ⋅ (— 10) = 0. Again if x = 7, we have (7+3)(77) = 1000. Hence, 3 and 7 are solutions of the given quadratic equation. EXERCISES Solve the following by factoring. 1. x² 9x+14=0. 3. x2 X 200. apkā 5. 20 x2 + 11x-3=0. 20x² 7. 2$2 to. " J S 36 = 0. 7t2+10t-8=0. Solve by any method. 11. 3x²-15x = 42. 2. x² + 8x + 7 = 0. 4. 4x² + 4x + 1 = 0. 6. 3x2 13 x 10. = 8. 2 a² + a − 3 = 0. 10. 6y²+35 y — 6 = 0. 12. x² + 4x + 25 = 0. ARTS. 44-45] EQUATIONS IN THE QUADRATIC FORM 63 13. x²-6x + 10 = 0. 15. $x2 27x 70 = 0. 17. x²+(5 — x)² = (5 − 2 x )². 19. 3x2+2x+1 = 0. 14. 4x2 28 x + 49 = 0. 16. x² + x S a² a = 0. 18. x² 6x+4=0. (1 — e²)x² — 2 mx + m² = 14. √20. 21. √x + 16 = x 4.* 22. x + √x + 6 = 23. √10x 34 + 2√x + 4 = √2(3x + 35). = 0. 24. √27x= √ x + 2 + √3 x + 3. 25. V4x-5 + √2 x 9 = 4. 26. x + 11+ 11 + √( 2 x +5 +50 = 9. х 7 27. √x - 19. X · 19 √/28. √18x+5-2√3 x = √2. 29. √2√x − 1 + √34 1 + √34 − x = 9. 30. √x-√2 x + 1 = 1. Solve the following equations to two significant figures. 31. x2 1.83 x + 0.81 = 0. 33. 0.001 x² 0.01 x 0.1 = 0. 32. x² -0.91 x 6.66 0. M = 34. 2.1x² + 10.3 x — 5.8 = 0. If in an equation we 45. Equations in the quadratic form. may replace an expression containing the unknown by a new letter and have a quadratic equation in that letter, then the origi- nal equation is said to be in the quadratic form. Thus, in the equation x-3-√x-3-20, if we let z = √x-3, we have z² 3, we have z² — z — 2 — 0. In 2x-3+x+1=0, if we let z = x ¯, we have 2 x² + ≈ + 1 = 0. * HINT: Square both sides. In solving such problems the results should be tested in every case, remembering that the radical stands for the positive square root of the number under it. Why should the results be tested? 64 [CHAP. VI. QUADRATIC EQUATIONS EXERCISES Solve the following equations and check results. 1. √x + 10 + √x + 10 2. SOLUTION: Let The equation then becomes' 4 z = √x + 10. z² + z − 2 = 0, z = 1, or - 2. or Replacing z by its value in terms of x, we have √x + 10 1, or 4 √x + 10 = − - 2, * CHECK: x + 10 = 1, x=- - 9. x+10=16, x = 6. √9+ 10+√9+10=2, Hence, the result x = 1 + 1 = 2. √ő + 10 + √6 + 10 = 4 + 2, x = 6 does not satisfy it. 12. 2. x4. 13x² + 36 = 0. 4 x²+2 x2 2 + x2 10 2 x² + 2 3 4 2. 9 satisfies the equation to be solved, but the result 3. (4x² - 3)² +(8 x² — 6)²=80. 8. x+ √x + 6 14. Ꮖ 5 (x² + 5 x)² + x(x + 5): = 4. 2 6. (x + 1) + 4 + 1 = 12. 7. √x + 16 = 4. 9. √2x+1 X = 15 √x 17 - 2x 11. འ + 1 7+2x 17+ 2x 7 2x NIW 3 √2. 2 8a6 = 0. 28x - 8.= 10. √x=8 √12. 12. 2x²-4x+3√x²-2x+6=15. 14. x4 + 2 x3x² - 2 x − 3 = 0.* 2x³ 16. x³+7x² 8 0. 18. x-531x320. 20. x³-8=0. 22. 2-6 x−2 — x−1 − 0. = 13. x6 + 7 a³x³ 15. x4-8x³ + 23 x² - 28 x — 8. — 0 17. x-3 J x 2 +8=0. 19. 3x²+6x-4=0. 21. ax2n + bxn + c = 0. 3 23. x x 4 x 21 √x = 0. * HINT: Write the equation in the form (x4 + 2 x³ + x²) — 2(x² + x) — 3 = 0. ARTS. 45, 46] 65 ROOTS OF A QUADRATIC 24. x x² + 1 x² + 1 5 + X 2 25. (x + 1)(x + 2) (x + 3) (x + 4) = 24.* x: ļ the army a You 46. Theorems concerning the roots of quadratic equations. THEOREM I. If r is a root of the equation ax² + bx + c = : 0, (1) Conversely, if x then x ris a factor of the left-hand member. is a factor of the left-hand member, then r is ú root of the equation. If r is a root of the equation, ar²+br+c=0, (2) then ax² + bx + c = ax² + bx + c − (ar² + br+c) = a(x²―r²)+b(x − r) (3) (4) = (x − r)(ax+ ar+b). (5) Hence, xr is a factor of ax² + bx + c. If x r is a factor of ax² + bx + c, then the substitution of r for x makes the factor x r vanish, and r is a root of ax² + bx + c = 0. EXERCISES Form quadratic equations of which the following are roots. 1. 3, 1. SOLUTION: When the right-hand member of the equation to be formed is 0, the left-hand member has factors x 3 and x 1. (x − 3) (x − 1) = x² - 4 x + 3 = 0 = 3. 7, 0. is a quadratic equation with roots 1 and 3. ✓ 2. 3, — 2. V4. √5, – 5. ↓ 6. √5 – 1, √ỗ + 1. Hence, 5. √5,-√5. ✔ 7. i,† — i. * HINT: Multiply first and fourth, and second and third factors together and write in the form [(+6)+4][(x2 +5)+ 6]= 24. † In these exercises, ¿? - 1. (See Art. 22.) 66 QUADRATIC EQUATIONS [CHAP. VI. 8. 4 + 3 i, 4 — 3 i. 1 i√3 10. -+-- 2 2 1 i√3 2 2 / 9. 2 + 5 √3, 2 – 5 √3. 2+5√3, 2-5√3. 11. 2, 1. 12. α. α M n 13. n M 14. Verify by performing the indicated operations that a(x - 6+ 2 a b + √b² - 4 ac :) (x - b — √b² - 4 ac ax² + bx + c. 2 a 47. Number of roots. In order to avoid certain exceptions, an equation ƒ (x) = 0 is said to have as many roots as f(x) has factors of the type î r₁ where 71 is any number. A factor x — 11 may be repeated. For example, if (x — r₁)² is a factor of ƒ (x), we say that f(x) = 0 has two roots equal to r₁. We have shown that a quadratic equation has two roots. The question arises: has it only two or may it have more? This question is answered by the following THEOREM II. A quadratic equation has only two roots. Proof. Suppose there is, in addition to b + √ b² - 4 ac -b — √ b² 4 ac 12 2*1 2 a 2 a a third root 3, distinct from r₁ and r2, of the equation ax² + bx + c = 0. By Ex. 14, Art. 46, ax² + bx + c = a(x − r₁)(x − r₂). — — Hence if r3 is a root, α(7'3-71)(13-12)=0. 1′1) (1′3 — 1′2) = 0. But this is impossible since no one of these factors is zero. (III, Art. 5.) 48. Special or incomplete quadratics. If b or c is zero in the quadratic equation ax² + bx + c = 0, the equation is said to be incomplete. Arts. 46–48] SPECIAL OR INCOMPLETE QUADRATICS 67 I. When c= 0, ax² + bx = 0 is the typical form of the equation. We can always write this equation in the form x(ax+b)=0. b. Hence, the roots are 0 and - Conversely, if 0 is a root of ax² a a quadratic equation ax2 + bx + c = 0, then That is, a.0+b⋅0+ c = 0. c = 0. Therefore, a quadratic equation has a root equal to zero when and only when the equation has no known term. II. When b = 0, ax² + c = 0 is the typical form. с In this case, x = ± α Conversely, if the roots of a quadratic equation are arithmeti- cally equal, but opposite in sign, there is no term containing x in the first degree; for, if + and r are both roots of we have ax² + bx + c = 0, ar² + br+c= 0, ar² — br+ c = 0, 2 br=0. Since it follows that r0, b = 0. Hence, a quadratic equation has two roots arithmetically equal but opposite in sign when and only when the term in x vanishes. III. When b=0, c=0, the typical form is ax²=0. Both roots of a quadratic equation are equal to zero when and only when the known term and the term in x vanish. 68 [CHAP. VI QUADRATIC EQUATIONS EXERCISES Determine k so that each of the following equations shall have one root equal to zero. V 1. 5x2 2. 16x + 5 10x² + 14 x + 2 k k² = 0. − 16 = 0. 3. 10 y² 16 y + k² - 4k + 3 = 0. Determine k and m so that each of the following equations shall have two roots equal to zero. 4. 5x216 mx + kx 4 m + k + 6 : 0. J 5. 3 z² 8 mz + 4 kz + 6 z + 4 m + 2k + 1 = 0. 6. x² +(2 m + 1)x + 4 k² + 2 k = 0. ་ Determine k so that the roots may be arithmetically equal, but opposite in sign. 7. x² + 3 kx + x + 7 = 0. 8. 2x² + k²x 4x+3=0. Determine k so that the sum of the roots may be 1. // 9. kx2 2 kx + x --- 2 = 0. 4 0. 10. 9x2 4 kx + 3x Determine k so that the product of the roots may be 1. 11. 3 kx² - 25 kx + k + 8 =0. 12. 2x25 x + k² – 2 k − 1 = 0. J 49. Nature of the roots. In Art. 43, we found the two roots of the quadratic equation ax² + bx + c = 0 ⋅ b + √ b² 4 ac √ b² - 4 ac to be , 22 21 2 a 2 a In case a, b, c are real numbers, the numerical character of these roots depends upon the number b² - 4 ac under the radical sign. An examination of x and x2 leads at once to the following con- clusions: (1) If b² — 4 ac > 0, the roots are real and unequal. (2) If b² - 4 ac < 0, the roots are imaginary and unequal. (3) If b² - 4 ac = 0, the roots are real and equal. It should be observed that if the coefficients are real and one root is imaginary, then both roots are imaginary. The quantity b² - 4 ac is called the discriminant of the equation ax² + bx + c = 0. ARTS. 48-50] COEFFICIENTS IN TERMS OF ROOTS 69 50. Sum and product of the roots. If we add together the two roots of ax² + bx + c = 0, we have X1 + x2 = b + √b 2 2 a 4 ac 4 ac も ​+ 2 a α If we multiply the two roots together, we have 4 ac C b + √ b 2 4 ac X1X2 2 a а 2 a Hence: I. The sum of the roots of a quadratic equation in x is, equal to the coefficient of x with its sign changed, divided by the coefficient of 22. II. The product of the roots of a quadratic equation in x is equal to the known term divided by the coefficient of x². EXERCISES Determine the nature of the roots of the following equations. 1. x² + 11 x + 30 0. 3. x2 16x + 64 0. 2. x² - 8x+25= 0. 4. 2x² - 3x 2 = 0. 5. 4x2 28 x + 49 = 0. Determine the real values of k so that the roots of the following equations may be equal. 6. x² + 3 kx + k + 7 = 0. SOLUTION: In order that the roots of 9 k² − 4(k +7) = 0. the equation 9 k²-4k-280, or k values in the above equation, we get must have b² - 4 ac b2 4 ac = 9 k² this equation may be equal, we Hence, k must be a solution of 14 2, or 9 x² + 6x + 9 = (x + 3)² = 0. x² - 42 x − 1 + +7 9 7. x² + kx + 4 = 0. Substituting these (9x²-42x+49)=(3x-7)² = 0. 9. 4x² + 12 x + k = 0. 11. (k + 1)x² + kx + k + 1 = 0. 8. kx² + 6x + 1 = 0. 10. k²x²+10x + 1 = 0. √12. x² + 12x+8k = 0. 13. (4x+ k)²= 16x. b 14. x²(1 + m²) + 2 kmx + k² 15. a²(mx + k)² + b²x² = u²b². r² = 0. 70 [CHAP. VI. QUADRATIC EQUATIONS Determine by inspection the sum and product of the roots of the following equations. 16. 7x² + 4 x − 3 − 0. 18. mx² + 2x 6 = 0. - 17. 11 27 x 19. 8x2+7x 18x2 = 0. 62 = 0. 20. (1 — e²)x² — 2 mx + m² = 0. Determine the value of k in the following equations. 21. x² + kx 50, where one root is — 5. SOLUTION: Let x₁ be the second root. The product of the roots of this equation (II, Art. 50) is — 5. Hence, and, - 5x1 5, or x₁ = 1, 5+ x1 = - k, or k 4. /* 22. x² + x − 4 k = 0, where one root is - 4. 23. x2 x k - - 0, where the difference between the roots is 9. 24. 15x2+ kx 4 = 0, where one root is 3. 25. kx²+6x+5=0, where one root is 5 times the other. 26. 3x2 kx + 14 = 0, where the quotient of the two roots is 7. AY 51. Graph of the quadratic function. In Chapter III we have plotted certain quadratic functions. It can be shown, if a is positive and different from zero, that the graph of the function ax2 + bx + c has the same general characteristics as the curve in Fig. 10. This curve is called a parabola. The real roots of the equation ax²+bx+c=0 are given FIG. 10. by the abscis- sas of the points where the curve If crosses the X-axis. the curve has no point in common with the axis, then the roots of the equation are FIG. 11. imaginary. -X ARTS. 50, 51] 71 QUADRATIC FUNCTIONS For we have shown that every quadratic equation has two roots, real or imaginary. If the curve has no point in common with the X-axis, there is no real root of the equation. Hence both roots are imaginary. If the curve touches the X-axis, both roots of the equation are real and equal. These three cases are shown in Fig. 11, where the graphs of 2-2x-3, x²-2x+1, and x²-2x+5 are given. EXERCISES Construct the graphs of the functions in the following equations, and, by measurement, determine the roots if they are real. Calculate the value of the function for at least ten values of x between the limits given. Choose the vertical unit so that the graph will be of convenient proportions for the coordinate paper. 1. x2 5x + 4 = 0, from x = 0 to x = 5. 2. x²+x - 6 0, from x=- 3. 4x² + 12x + 5 = 0, from x 4 to x = 3. 4 to x 1. 4. x2 3x = 0, from x = x=- 1 to x = 4. 5. x² + 2x + 2 = 0, from x = − 3 to x = 2. 6. x2 6x + 11 = 0, from x 0 to x = 5. 7. 6 3 x ་ x20, from x 4 to x 2. 8. 2 to x = +2. = 6 to x = 1. 3x² + 2 x − 4 = 0, from x = : 9. 4-5x-x20, from x =— 10. x² - 1 = 0, from x = — 3 to x + 3. 11. What are the general characteristics of the graph of the function ax² + bx + c if a is negative ? PROBLEMS 1. In the course of Steinmetz's solution of the problem of finding the current strength in a divided electric circuit, it is necessary to solve the equation a²x² as² + y² = 0 for a. His solution is $2 82 ± 9² 2 x2 where q² =√s4 - 4 r²x². Verify the result. 2. If $10,000 amounts to $11,130.25 when placed at compound interest for two years, interest being compounded annually, what is the rate of interest ? V 3. A rectangular court is 25 feet longer than it is wide and it contains 3750 square feet. What are its dimensions ? 72 [CHAP. VI. QUADRATIC EQUATIONS 4. The altitude of a triangle exceeds its base by 48 feet. The area of the triangle is 1387.5 square feet. Find the base and altitude. 5. If a ball is thrown upwards with a velocity vo, the distance d from the earth to the ball after a given time t is given by the formula d vot - gt², where g 32.2. The speed at the time t is given by (1) (2) If the ball is thrown downwards with a speed to, the above formulas become vr = vo gt. d = vot + & yt², vt = vo + gt. (3) (4) If a ball is thrown upwards with a velocity of 50 feet per second, in what time will it be just 30 feet from the ground? Explain the two answers. 6. How long will it take the ball described in Problem 5 to reach the ground? [HINT: Put d = 0 in formula (1).] 7. At what time is the velocity of the ball zero? 8. How high will the ball rise ? 9. How does the time taken in rising to the highest point compare with the whole time that the ball is in the air? 10. How far does the ball rise in the second second ? 11. How long will it take a body to fall 400 feet, if it is thrown down- ward with an initial speed of 50 feet per second ? 12. What is the velocity of a falling body at the end of 23 seconds if the initial velocity be 0 ? 13. If a body falls from rest, how far will it fall in the fifth second ? 14. The edges of a cube are each increased in length one inch. It is found that the volume of the cube is thereby increased 19 cubic inches. What is the length of the edges of the cube? 15. What is the area of a square whose diagonal is one foot longer than a side? 16. What is the area of an equilateral triangle whose side is one foot longer than the altitude? 17. Show that the equation x² + bx + c = 0 has one positive and one negative solution if c is negative. 18. In joining together two steel boiler plates with a single row of rivets, the distance p between the centers of the rivets is given by the formula (712 p = 0.56 +d, t ART. 51] 73 PROBLEMS INVOLVING QUADRATICS where t is the thickness of the plate and d the diameter of the rivet holes. In a boiler the rivets are to be placed 1½ inches apart. If the thickness of 1 the plate is inch, what is the diameter of the rivet holes? 1 C 19. Graph on the same sheet the function 2 x² x+c, where c takes the values 1, 3, 0, 10. What effect does changing the constant term in a quadratic function have on the graph? 20. Graph on the same sheet the function ax2 +3, where a takes the values 5, 1, 1, 16. Decreasing the coefficient of x2 has what effect on the graph? What is the effect on the roots of the quadratic equation ac² x + 3 = 0, if a is made to approach 0 ? 21. If s is the area in square inches of the flat end of a boiler, and t the thickness of the boiler plate in sixteenths of an inch, then the pressure p per square inch which the flat end plate can safely sustain is given by the formula 200 (t + 1)². s 6 p = 2 What should be the thickness to the nearest sixteenth of an inch of the boiler plate for the end of a boiler 20 inches in diameter to sustain a pressure of 100 pounds per square inch? 22. In a group of points every point is connected with every other point by a straight line. There are 300 straight lines. How many points are there? 23. A rectangular piece of tin is twice as long as it is wide. From each corner a 2-inch square is cut out, and the ends are turned up so as to make a box whose contents are 60 cubic inches. What are the dimensions of the piece of tin? 24. Let be the height and t the thickness (in feet) of a rectangular masonry retaining wall. For very sandy soil with a grade angle of 20°, h and t are connected by the equation t + 0.19 th 0.18 h² = 0. What should be the thickness (to the nearest inch) of a retaining wall four feet high? 25. For loam, the equation in Problem 24 would be t² + 0.14th — 0.13 h² 0. What should be the thickness of a retaining wall four feet high? 26. A long horizontal pipe is connected with the bottom of a reservoir. If I be the depth of the water in the reservoir in feet, d the diameter of the pipe in inches, L the length of the pipe in feet, and v the velocity of the water in the pipe in feet per second, then according to Cox's formula Hd L 4 v2 + 5 v — 2 1200 Find the velocity of water in a 5-inch pipe, 1000 feet long, connected with a reservoir containing 49 feet of water. 74 [CHAP. VI. QUADRATIC EQUATIONS ✓ 27. The so-called effective area of a chimney is given by E = A — 0.6 √Ā, where A is the measured area. Find A when E is 30 square feet. 28. A stone is dropped into a well, and 4 seconds afterward the report of its striking the water is heard. If the velocity of sound is taken at 1190 feet per second, what is the depth of the well? (Use g = 32.2. lem 5.) See Prob- 29. The electrical resistance of a wire depends upon the temperature of the wire according to the formula R₁ = Ro(1+ at + bt²), Ꭱ where a and b are constants depending on the material, Ro is the resistance at 0°, and R, the resistance at t°. For copper wire a = 0.00387, b = 0.00000597, and R。 = 0.02057. At what temperature is the resistance double that at 0° ? 30. The radius of a cylinder is 10 and its height 4. What value can be added to either the radius or to the height, and yet give the same increase in volume ? The following equations occur in some electrical problems. 31. g = R R² + X² Solve for R. 32. T- a (n— n') Solve for (n— n'). 1 + b (n — n')² 2 33. P r (Rr Xx) AW (r² + x²). Solve for x. 34. In making war bread a mixture of rye and corn meal was used. From a hundred pounds of rye flour a certain amount was taken and re- placed by corn meal. Later, from the mixture the same amount was re- moved and again replaced by corn meal. The resulting mixture was 16 parts rye to 9 parts corn. What were the proportions in the first mixture ? CHAPTER VII SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS 52. Type form. two unknowns is The type form of a quadratic equation in Ax² + Bxy + Cy² + Dx + Ey + F = 0, where at least one of the coefficients A, B, or C is not zero. Such an equation is satisfied by an indefinite number of pairs of values of x and y. If the pairs of real values of x and Y which satisfy the equation be considered as the coordinates of points, and these points plotted on coordinate paper, they will lie on a curve called the locus of the equation. EXERCISES Arrange the following equations in the typical form, find values of A, B, C, D, E, and F. Find at least 4 pairs of values of x and y which satisfy each equation. 1. (x + y)² = 3 x² = 3 x² − y + 2. SOLUTION: Written in the typical form this becomes where A = 2, B 2 x² + 2 xy + y² + y − 2 = 0, = 2, C = 1, D = 0, E = 1, F2. Substituting x = 1 in the equation, there results, y² + 3 y − 4 = 0, ΟΥ Hence, x = 1, y = 1; x = equation. y = 1, or - 4. 1, y - 4 are two pairs of values satisfying the Putting x = 2 in the equation, we find in the same way two other pairs of values, x = 2, y = 2. x² - xy + y² = 37. 5 + √65 2 x = 2, y = 5 - √65 2 etc. 3. 3x2 2 y2 = 4. 5. x² + y² = (x + y + 1)². 6. xy = 2 x − y + 9. 7. (x + y)(3x) = (x − y) (2 + y). 4. x² + y² = 25. 75 76 [CHAP. VII. SIMULTANEOUS QUADRATICS 53. Solution of systems of equations involving quadratics. Al- though there are an indefinite number of pairs of values of x and y which satisfy one quadratic equation, there are not more than four pairs which satisfy two non-equivalent quadratic equations. If these values are real numbers, they are the coordinates of points lying on the locus of each of the given equations. Hence, the real solutions of a pair of simultaneous quadratic equations can be represented graphically by the points of intersection of the loci of the two equations. The general problem of simulta- neous quadratics is that of finding pairs of values of a and y which satisfy two equations of the form, Д₁x² + В₁xу + С₁у² + D₁x + E₂y + F₁ = 0) Д2x² + B₂xу + С₂y² + D₂+ E₂y + F₂ = 0 f where the coefficients may have any values. As illustrated in the following example, this general problem involves the solu- tion of an equation of the fourth degree. Solve x² + y² + x − 9 = 0, 3y-8 0. x² + 2 y2 Subtracting the second from the first, we have } - − y² + 3 y + x − 1 = 0, or Substituting in the second equation, we have ΟΙ x = 1 − 3 y + y². (1 − 3 y + y²) 2 + 2 y² — 3 y − 8 = 0, 7 = 0. y4 — 6 y³ + 13 y² 9 y — 7 At this stage of our progress in algebra, we cannot solve a general equation of the fourth degree, hence we cannot proceed with the solution of this problem. Although we cannot solve. the general problem of simultaneous quadratics, yet there are some types of these equations which can easily be solved. We take up a few of the most important. CASE I. When each equation is of the form Ax² + Cy² + F = 0. If, instead of x and y, we consider x² and y² as the unknowns, the method of solution is that for linear equations. ART. 53] SOLUTION OF SIMULTANEOUS QUADRATICS 77 Solve 1. EXERCISES 16 x² + 27 y² = 576, x² + y² = 25. SOLUTION: Solving for 2 and y2 we have 576 27 25 1 99 X2 9, 16 1 27 1 11 16 576! 1 25 y2 = 18. 16 27 1 1 y = ± 1. x = ± 3. Hence, we find the following four solutions, (3, 4), (— 3, 4), (3, — 4), (— 3, 4). To show these solutions graphically, we plot the loci of the two equations. Solving each for y, we have y = ± √ 士 ​576 16 x² 27 y = ±√25 x². The first equation has for its locus the oval-shaped figure A, B, C, D, called an ellipse, the second, the circle (Fig. 12). The points of in- tersection represent graphically the four pairs of solutions. If the loci of two equations do not intersect, the solutions of the equations will be found to be imaginary. Plot and solve : Alot 2. 9 x² + 25 y² + 25 y² = 225,* x² + y² 16. 3. 9 x² + 25 y² = 225, x² + y² 9. √5. 9 x² + 25 y² = 225, x² + y² # C 4. 9 x² + 25 y² = 225, x² + y² = 4. B D FIG. 12. A 6. 9 x² + 25 y² = 225, √7. 4x² - 9 y² = 36, 25 x² + 9 y² = 225. x² + y² = 16. Obtain to two significant figures the solutions of the following: = 25. 8. 4.3x² + 9.1 y² = 89, x² + y² 9.8. 9. 1.9 x² + 0.21 y² = 3.6, x2 x² — y² — — 4.1. * The graph of 9 x² + 25 y² = 225 is an ellipse of breadth 10 and height 6. Its position with respect to the axes is similar to that of the ellipse in Fig. 12. The graph of x² + y² = 16 is a circle of radius 4, and center at the in- tersection of the axes. 78 [CHAP. VII. SIMULTANEOUS QUADRATICS CASE II. When one of the equations is linear. EXERCISES Plot and solve: 1. x² + y² - 8x 8x-4 y 5=0, 3x+4y = 5. SOLUTION: From the linear equation, x 5 4 Ay. 3 Substituting in the quadratic equation and reducing, we have Solving, we obtain 5 y²+ 4 y 28 = 0. y = 2, or — 14. Substituting these values in the linear equation, we find x = − 1, or 27. The solutions are then (1, 2), (47, — 44). B X The loci of these two equations are shown in Fig. 13. The circle with its center at the point (4, 2) and radius 5 is the locus of the first equation, while the straight line AB is the locus of the second equation.* 2. x² + y² — 8 3 x + 4y + 5 = 0. x − 4 y − 5 = 0, 5 4 ปี 5 = 0, = 0. 5. xy = 36, 3. x² + y² — 8 x 3 x + 4y + 15 4. x² + y² = 58, — X 8. ੪. 4 + 3, 5 2 x + y = 10. FIG. 13. 16. xy 40, XC y = 3. y2 9. y² - 4 x = 0, x · +1=0. x + y = 13. 7. 2 – 5 XY 5x = 1, 7x-y=1. (x + y)² = 200 X. 10. (3x+y)(y + 2) = (x + 5)(3 x − y), 2 x − y = 1. 12. 0.1y+0.125 x = y — X, y — z = 0.5 0.75 xy — 3 x. 11. x + 1 = 2 x + y, 3 3 2(x + y) = x(4xy). 13. 0.3x + 0.125 y = 3 x -Y, 3 x 0.5y= 0.5 y = 2.25 xy + 3 y. * The remaining two lines of the figure belong to the next two exercises. In Ex. 2, where the line touches (is tangent to) the circle, only one pair of numbers satisfies the system, and the solutions are said to be equal. ART. 53] SOLUTION OF SIMULTANEOUS QUADRATICS 79 Obtain to two significant figures the solutions of the following : 14. xy 0.24, 3.1 x 0.63 y = 4.3. 15. y² - 4.1 x = 0.38, x — 0.37 y = 0.29. Find the values of a, b, c, or r in the following exercises so that the straight line which is the locus of the first degree equation (1) cuts the other locus in two distinct points, (2) is tangent to the curve, (3) fails to meet the curve. 16. x² + y² = 1·², 3x + 4y = 5. The locus of the first equation is a circle with center at the origin and a radius equal to r. SOLUTION: From the second equation, we have 5-41. X 3 Substituting in the first, we find 25 y² — 40 y + 25 − 9 r² = 0. Solving for y, we obtain, 40 ± 30 √r2 — 1 y = 50 If y is real, r2 1 must be equal to or greater than zero. Furthermore, if is any number greater than 1, the two loci intersect in real and distinct points. If r = 1, there is only one value for y, and the line is tangent to the circle. If r < 1, the line does not intersect the circle. 17. x² + y² = 1•², x + y = 10. 19. x² + y² = 25, ux+6y=1. 18. x² + y² = 25, 3 x + 4y = c. 20. x² + y² = 25, 7 x — by = 3. 21. For what values of b in terms of rand m does the system of equations y = mx + 1, x² + y² = r² have equal solutions? 22. Determine the relation between a, b, and k such that the system has equal solutions. x2 a2 + y = mx + k, 3/2 .ག 1 b2 80 [CHAP. VII. SIMULTANEOUS QUADRATICS CASE III. When all the terms which contain the unknowns are of the second degree. EXERCISES Solve the following pairs of equations. 1. x² + 3 xy = 28, x² + y² = 20. SOLUTION: Let y = From the first equation, we have mæ mx and substitute in both equations. x² + 3 mx² — 28, 28 whence x2 1+ 3 m From the second equation, we have x² + m²x² = 20, 20 whence Equating these values of x2, we obtain x2 1 + m² 28 1+ 3 m 20 1 + m² Clearing of fractions and reducing, we obtain or 7 m² 15 m +20, Substituting these values of m in we find, for m = = 2, m = 2, or 4. 28 x2 1+ 3 m 4, x =+ 2, x² = y = mx + 4. For m = , we find x² = 98, x mx Y MX =mx 10 FIG. 14. ±√10 ± 4.43+, √10±0.63+. The solutions are therefore (2, 4), (— 2, — 4), (†√10, √10), (√10, -10). The loci of the two equations of this exercise are shown in X Fig. 14. The geometrical inter- pretation of the substitution y = mx is also shown in the figure. √2. x² + 3 xy = 28, 4 y² + xy = 8. 3. x² + xy + y² = 6, x² + y² = 12. Art. 53] SOLUTION OF SIMULTANEOUS QUADRATICS 81 4. 2 y² - 4 xy + 3 x² 17, y 2 x2 16. 6. x² + xy + 2 y² = 74, 2x² + 2 xy + y² = 73. x + y 8. + X Y x + y x² + y² = 45. x Y 10 3 10. 4 a² -2 ab b2 — 16, 5 a2 = 7 ab - 36. $5. 5. x² + y² = 65, ry = 28. 7. x² - 4 y² = 9, xy + 2 y² = 3. 9. x² — xy + y² = 21. y2-2xy + 15 = 0. = 4, √11. x² + xy = y² + xy = 1. Find to two significant figures the solutions of the following: 12. x²+1.6 y² = 8.3, xy = 2.3. 13. x² + 0.12 xy = 104, y² + 1.4 xy = 21. CASE IV. When the equations are symmetrical.* The typical form of a symmetrical quadratic equation in two unknowns is A(x² + y²) + Bxy + D(x + y) + F = 0. EXERCISES Solve the following systems of equations: 1. x² + y² + x + y = 8, (1) xy + x + y = 5. (2) SOLUTION: Let x = u + v, y = u — V. Substituting in the two equations we obtain after reductions u² + v² + u = = 4, (3) u2 - v2 + 2 u = 5. (4) Eliminating v2 by adding we obtain an equation in u, 2 u² + 3 u = 9, from which u = or or — 3. The four solutions of (3) and (4) are then (4, 1), (1, − 1), (— 3, i√2), (— 3, — i√2). From the first pair x = u + v 2 + 1 = 2, y = u V In the same way we find from the other pairs, = 142 1. X 1 x = 3+ i√2 y=2]' У 3 – i√2 } x = 3 - i√2 Y 3 + i√ē s * An equation is said to be symmetrical with respect to x and y whenever interchanging x and y leaves the equation unchanged. 82 SIMULTANEOUS QUADRATICS [CHAP. VII. 4. x² + y² + xy + x + y = 17, 2.x + y 2 xy = 5, - x² + 6 xy + y² — 4 x 4 y = 5. 3. 3x² + 3y² = 8(x + y) − 1, XY = x + y + 1. 6. x² + y² 13, x² + y² — 3 xy + 2 x + 2 y = 9. 5. 3x(1 + x)+3y(1+ y) = 54, 2 x + 2y + XY 16. x'y = 6. Many systems of equations of degree higher than two, and systems containing three or more unknowns may be solved by combinations and variations of the above methods, but these four cases do not by any means include all the simultaneous equations whose solution can be reduced to the solution of the quadratic. Usually the solution of such a system is in the nature of a puzzle for which no special rules can be given. Whatever method be used it must be kept in mind that the ultimate test of a solution is substitution in the given equations. Many equations coming under the cases mentioned here may be solved more easily by other special methods. For example, Exercise 6 under Case IV may be done as follows: Given x² + y² = 13, XY 6. Multiply the second equation by 2, add this result to and subtract it from the first equation, and thus obtain the system x² + 2xy + y² = 25, x² - 2 xy + y² — 1. Extracting the square root of each, we have x + y = ± 5, x − y = ±1. From these equations we find the same four results which were obtained by the general method for solving symmetrical equations. EXERCISES AND PROBLEMS 1. 3. 218 10 + x 3 4 x — 7 y — 5. 118 1 У = 5, = 1 2. 3x 4, Y 1 9x² + y/2 40. 4. x + y + 2 y² = 11, 3x-2y-2 y² + 9 = 0. 1 x2 + 1 233. y2 • ART. 53] 83 EXERCISES AND PROBLEMS 5. xy x = 12, xy + 3y = 35. 6. 2 y2 - 4 xy + 3x² 1/2 X2 : 16. 7. (x + 1)(y + 2) = 28, (x+3)(y+4)= 54. 8. x² + y² 8y+1 17, 2 xy, 9. x(y-4) = 14, y(x + 1) = 33. 11. (x − 1) (y + 5) = 80, (x + 4) (y − 2) = 39. 13. x² 4 xy + 3y² = 0, x² + y² = 5(y + 6). 15. (x4y-3)(x + y) = 0, x² + y² = 22 + 4 xy. 17. (x − y)² + (x − y) = 6, · x² = y(2x + 1) + 2 x − 18. HINT: Solve the first equation for (x − y). 19. x³ — y³ = 19, 2(x + y) 2 10x + 9 y + 3. 10. 3(x² y²) = 2 x + 17, x² - y² + x + y = 18. 12. (x + y)² + (x + y) = 12, 3x² + y² = x + y + 4. 14. 2(x + 1)(y − 4) = y + 1, (3 x − 1) (2 y − 9) = 5. 16. x² — xy — 2 y² = 0, x² + 2 y² - 3 x = 12. 18. x³ + y³ = 28, x + y = 4. HINT: Divide the first equation by the second. 20. x³ — y³ = 63, y3 x² + xy + y² = 21. уз X Y y -- 1. 21. x²+2xy+y²=(x+y)(3x+1), 22. √ x2 y² + 2xy + 1 = 0. - ay = 36. √y = f (x − y), 23. x2 y√xy = 14, 24. x² y = 20, y² — x √ xу x√xy = — - 7. 25. x² — xy + y² = 3 a², x + y = 3(x − y). 27. x² + y² = 13, y² + 2² = 25, 22 + x² = 20. 8, 29. x(y + z) y (z+x) z(x + y) = 20. = 18, 31. xy + zu = 17, xz + yu = 13, xu + yz = 11, x + y + z + u = 11. - 3+ √x² x4 — y² = 544. 26. x² + y² = α², √x + √y = √u. 28. x² + 2 y2 z² = 5, 2x + y + z = 6, x + 4 y z = 5. 30.x + y = = 11, z + u = 10, XY = ZU, x² + y² + z² + u² = 125. 32. x² + y² + z² + u² + v² = 19, x + y + z − u + v − 1, x + v = 2 + u, x + y + v = 5, y + z = v. 84 [CHAP. VII. SIMULTANEOUS QUADRATICS Eliminate x and y from the following equations. 33. x² + y² = a², x + y = :b, xy = c. 35. x² + xy = a², y² + xy = b², 34. x² + y² = a², x² — y² = b², xy = c². y2 x2 36. + a2 62 2, xy = ab, x² + y² = c². a²x² + b²y² = c¹. 37. A circular track is built so that the width of the track is of the inside diameter. After construction it was found that the area of the track was 2564 square yards. What are the inside and outside lengths of the track? 38. The fence around a rectangular field is 1400 feet long. The diagonal of the field is 500 feet long. What are the dimensions of the field? 39. The diagonal of a rectangular parallelopiped is 14 inches long. The sum of the three dimensions is 22. The reciprocal of one dimension is one half the sum of the reciprocals of the other two. What are the dimensions of the solid ? 40. A father divided $1000 between his two sons and kept it for them at simple interest until called for. At the end of 3 years, one son called for all the money due him and received $665.50. At the end of 4 years the other son received $576 as his share. How was the money originally divided and what rate of interest did the father pay? 41. A few days after the outbreak of the war a 25-pound bag of sugar cost the retailer 77 cents more than it did just before the outbreak. For $165 a grocer received 1550 pounds less sugar after the outbreak than he would have received before for the same amount. and after the outbreak of the war? What was the price before 42. A silver wire 1000 millimeters long is to be covered with a layer of gold until the diameter of the wire is increased one tenth. It is found that 4 cubic millimeters of gold is needed. What is the diameter of the wire? 43. A road between two towns is 33 miles long. At eight o'clock, from each of the two towns, a traveler starts toward the other and walks at a uni- form speed. At eleven o'clock they meet, but one traveler arrives at his destination 1 hour and 6 minutes earlier than the other. How many minutes. are needed for each to walk 1 mile ? 44. A man divides a tract of land into city lots. He sells the lots all at If the number of lots had been one the same price and realizes $4800. greater, and the price $8 per lot cheaper, he would have received the same amount of money. How many lots were there and what was the price per lot? ! ART. 53] 85 EXERCISES AND PROBLEMS √45. 45. A circular cylinder is inscribed in a sphere of radius 10. The total surface of the cylinder is half the surface of the sphere. What is the radius and what is the altitude of the cylinder ? J 46. The diagonals of the three faces of a rectangular parallelopiped which meet in a vertex of the solid are 10, 12, 14, respectively. What is the volume of the solid? 47. The floor of a rectangular room contains 273 square feet, one wall 189 square feet, and the adjacent wall 117 square feet. What are the dimen- sions of the room? 48. Psychologists assert that the rectangle most pleasing to the human eye is that in which the sum of the two dimensions is to the longer as the longer is to the shorter. If the area of a page of this algebra remains unchanged, what should its dimensions be? 49. An aeroplane, flying 75 miles per hour and following a long straight road, passed an automobile going in the opposite direction. One hour later it overtook a second automobile. The automobiles passed each other when the aëroplane was 100 miles away. If both automobiles travel with the same speed, how far apart were they when the aëroplane passed the second one and what was their speed? 50. Two students attempt to solve a problem that reduces to a quadratic equation. One in reducing has made a mistake only in the constant term of the equation, and finds 8 and 2 for the roots. The other makes a mistake only in the coefficient of the first degree term, and finds 9 and 1 for roots. What was the quadratic equation? 51. The radius of the front wheel of a carriage is 6 inches less than that of the rear wheel. If the front wheel makes 80 more revolutions than the rear wheel in going a mile, what is the circumference of each wheel to the nearest inch ? 52. Two polygons have together 12 sides and 19 diagonals. How many sides has each ? 53. A military ambulance traveling 12 miles per hour sent on ahead a motor cyclist messenger who could travel with twice the speed of the ambu- lance. A half-hour later it was found necessary to revise the message, and a second motorcyclist was sent to overtake the first and to give him the new message. The second messenger returned to the ambulance in forty-five minutes. What was his average speed during that time? “ - CHAPTER VIII INEQUALITIES 54. Definition. The expressions "a is greater than b" (a > b) and "e is less than d" (cd), when a, b, c, d, are real numbers, mean that a − b is a positive number and c d is a negative num- ber. Such expressions are called inequalities. Two inequalities a>b, c> d, which have the signs pointing in the same direction, are said to be alike in sense. If the signs point in opposite direc- The tions, as a > b, c 1, which is valid for all real values of a and b, is called an absolute inequality; while an in- equality such as 2-4 > 0, which holds only when x is greater than 4, is called a conditional inequality. In a conditional in- equality the letters cannot take all real values. 56. Elementary principles. The following elementary princi- ples, which follow at once from the definition of an inequality, must be observed in dealing with inequalities. I. The sense of an inequality is not changed if both sides are in- creased or decreased by the same number. In particular, the sense is not changed if we transpose a term, changing its sign. Let Then and ΟΙ Hence, a > b. a-b=n, where n is positive, a+k-b-k=n, (a + k) − (b + k) = n. a+kb+k. 86 ARTS. 54-56] 87 ELEMENTARY PRINCIPLES II. The sense of an inequality is not changed if both sides are multiplied or divided by the same positive number. III. The sense of an inequality is reversed if both sides are mul- tiplied or divided by the same negative number. The proofs of II and III are very similar to the proof of I. EXERCISES 1. If a and b are not equal, show that a² + b² > 2 ab. SOLUTION: (α — b)² > 0, since That is, By Principle I, a² a² — 2 ab + the square of any real number is positive 2 ab + b² > 0. b² + 2 ab > 0+2 ab, or a² + b² > 2 ab. 2. Show that a+b 2 > √ub, if a and b are positive and unequal. Show that the following inequalities subsist, the letters representing dis- tinct positive numbers. 3. a + b 2 2 ab a + b • a3 + 3 a + b 5. a242 ab 7. a² + 3 b² > 2 b(a + b). 9. If b, d, and ƒ are positive, and between and ( b e f α 4. a² + b² + c² > ab + ac + bc. 1 6. a + > 2, if a ‡ 1. 210 - α α 8. + > 2. α show that a + c + e lies b + d + ƒ 0123 10. Given a² + b² = 1, c² + 17² 1, show that ab + cd < 1. 11. Show that if we denote by | a |,* | b |, etc. the numerical values of a, b, etc., then 12. Given | a + b ≤ | a |+|b|, V [ a b ≤a+|b|, |ab|≥|a| - | b | · 101 a 0, ɑ < α»<▲, b₂>0, A, a0, show that a(b₁ + V₂+ ... + b₁) < a₁b₁ + agbe + ... + a„bn < A(b1 + b² + ... +(n). * The expression | a | is often read "absolute value of a," meaning the numerical value without regard to sign. 88 [CHAP. VIII. INEQUALITIES AY 57. Conditional inequalities. By transposing terms every in- equality may be reduced to an inequality of the form P>0, or P < 0. If one or both sides involves a variable, say x, it can be put in one of the two forms ƒ (x) > 0, or ƒ(x)<0. In this connection the most important prob- lem is to find the range of values of the variable for which the inequality holds. In the case of linear inequalities the solu- tion is easy. Thus, to find the values of x for which the inequality 31912 - x a holds, all the terms can be transposed to the left-hand side, and there results Hence the inequality in question holds only 4x+7 0. FIG. 15. for x> -7. Graphically, 3 x + 19 > 12 X for those values of x for which the graph of the function 3x+19 - 12 + x = 4 x + 7 lies above the X-axis (Fig. 15). The graph is of great service in determining the values of x for which one function of x is greater FIG. 16. X or less than another function. Thus, to find the range of values of x for which 2x² - 3x+8> x² + 2 x + 4, we transpose all terms to one side and have 22–5+4>0. ART. 57] 89 CONDITIONAL INEQUALITIES The graph of this function is shown in Fig. 16. It crosses the X-axis at 1 and 4, and for x > 4, or x < 1, the function x²-5 »+4 is positive; while for 4 >> 1, it is negative; hence, 2 x2 3x+8 > x² + 2 x + 4 for x >4, and x< 1, while for 4x1. 2x² - 3 x + 8 < x² + 2 x + 4 EXERCISES For what values of x do the following inequalities hold ? 1. 7x 3 > 32. x 3 4 3 3. 1 2 7 X 5. 3x2 11x > 4. x 3 2. >0. 1 2 4. x² <4. 1 6. x + >2. x 7. (x-5)(x-2)>(x-1)(x-3). 8. 2x² - 3 x 10 > x². 2x-5 α — x a2 x² 9. <0. 10. x+8 11. ax + b>0. a + x 12. x² - 4 x + 3 4x a²² + x² 0, 0, 0, 13. ax2 + bx + c = 0, <0. x + 5 14. <0; > 0. 8 x 3 - 2 3 1 2 8 16. 8 2 83 4 1 4 1 5 > 1 4 2 3 X بع من نن 1 3 - 3 2 NN 15. 2 1 - 2 >0. 17. Given a₁> A₂ > A3 > ... >a,, and a,>x> a,+1; show that (x — α1)(x — α2)(x — α3) (x an) represents a positive number when is even, and a negative number when r is an odd number. 18. Show that (x — α₁) (x — α)² is positive if x> a₁, and negative if x < .α1. CHAPTER IX MATHEMATICAL INDUCTION 58. General statement. Many important theorems in algebra can be proved by a method called mathematical induction. The method may be best explained after applying it to a simple example. Let it be required to show by mathematical induction that the sum of the first n odd numbers is the square of n. It should be noted that the nth odd number is 2n 1. We have then to prove 1+ 3+ 5+ ... + (2 n − 1) = n² for all positive integral values of n. We note that the theorem is true for n = 1 and n = 2. Assume it to be true for n = 7. We have then 1 + 3 + 5 + ··· + (2 n − 1) = p•². ... r (1) Adding the next odd number, that is, 2r+ 1 to both sides of (1), we obtain 1+3+5+ ··· +(2 r−1)+(2r+1)= r²+2r+1=(r+1)², (2) ... which states that the sum of the first +1 odd numbers is (1 + 1)². Hence the theorem is true for n = r + 1, if it is true for = 1. We know it is true for n = 2, hence it is true for N 2+1=3. n= Since it is true for n = 3, it is true for n = 4 Therefore, the sum of the first n odd numbers is the N and so on. square of n. In general, an argument by mathematical induction consists of two necessary parts. 90 ART. 58] METHOD OF MATHEMATICAL INDUCTION 91 First. To show by mere verification that the principle in question is true for some particular case, preferably for n = 1, or n = 2. Second. To show that if it is true for the case n = r, it is true =r+ 1. for the case n It is no proof simply to show that a theorem is true in a num- ber of cases. For example the above theorem is not proved by showing that it is true for n = 1, n 2, n = 4 and so on 3, n = for a definite number of cases. The second part of the proof is necessary. n = A celebrated example illustrating this point is the expression n² — n + 41. From the table we see that n² n + 41 is a prime +41 n 1 2 3 | 4 567 8 9 10 | 11 | 1 12 n² − n + 41 41 43 47 53 61 71 83 97 113 131 151 173 number for all integral values of n up to 12. N2 The table could be continued up to n = 40 and the lower row would still contain nothing but prime numbers. However we have no proof that n+41 is prime for all integral values of n. To prove this it is necessary to take the second step in the proof by mathe- matical induction, i.e. to prove that if №² r+41 is prime, then (~ — (~ (+1)² - ( + 1) + 41 is prime. But this is impossible. In fact, when n = 41, we have 2 122 n + 41 = 1681 = 412, a number which is not prime. Again, it is no proof simply to show that if a statement is true for n=r it is true for n=r+1. For example, assuming that the sum of the first even numbers is an odd number it follows that the sum of the first r + 1 even numbers is odd. Though the second part of the proof of this statement by mathematical in- duction can be correctly presented, we know the statement to be false. The first part of the proof is lacking. In a proof by mathematical induction both parts are necessary and are of equal importance. 92 [CHAP. IX. MATHEMATICAL INDUCTION EXERCISES Prove by mathematical induction. 1. 2 + 4 + 6 + SOLUTION: For n = ... +2n= n(n + 1). 1, we have 2 = 1. 2. Hence, the statement, 2 + 4 + 6 + + 2 n = n(n + 1), is true for n = = 1, and the first part of the proof is satisfied. Suppose it is true for r numbers. That is, assume that 2+4+6+ +2r=r(r + 1). (1) Adding the (r + 1)th term to both sides, 2+4+6+ ... + 2 r + (2 r + 2) = r(r + 1) + 2 r + 2 (r + 1) (r + 2). (2) By comparing (1) and (2), it is seen that (2) can be obtained from (1) by replacing r by r+ 1. That is, if the formula holds for r integers, it holds for we know the formula is true for n = 1, hence, it is true Hence, for any integer n, r + 1 integers. But for n = 2, and so on. V 2. 1+ 2+ 3+ 2+4+6+ 3. 3+ 6+ 9 + 12 + 4. 12 +22 + 32 + + 2n = n(n + 1). ገ (n + 1). 2 ... + 3 n = ... 3 n (n + 1). 2 + n² = {n(n + 1) (2 n + 1). 5. 2242 +62 + ... 3 +(2n) 2 = 2 n(n + 1) (2 n + 1). ... to n terms 6. 1+1+2+ • 2.3 3.4 N n + 1 7. X y is a factor of xn — yn if n is any positive integer. HINT: By verification we see that If x - V xr+1 — yr+1=x(x² — y¹) + y² (x − y). y′(x is a factor of x” — y', then it is a factor of xr+1 — yr+1. y æº \ 8. x²n — y2n is divisible by x + y if n is any positive integer. 9. 13 + 23 + 33 + ... + n³ = n² (n + 1)² = (1 + 2 + 3 + ... + n) 2 4 10. 2+22+ 23 + 11. xn Х an a ... — + 2n = 2(2n − 1). = xn−1 + axn--2 + + an−2x + un−1· ... 59. Meaning of r!. The symbol r!, read "factorial "," is used to indicate the product 1.2.3... 7. Thus, 3!1.2.3=6; 7! 1.2.3.4.5.6.7 5040. * The symbol [r is often used to represent factorial r. ARTS. 58-60] 93 BINOMIAL THEOREM EXERCISES Evaluate the following expressions : 9! 8! 3!5! 3!+5! 1. 2. 3. 4. 6! 7! 4! 4! 9! r! r! 5. 6. Prove = r. 7. = ? 316! (r− 1)! (r − 2) ! 60. Binomial theorem; positive integral exponents. By multi- plication, we find 2 (a + x)² = a² + 2 ax + x²- 3.2 (a (α + x)³· — α³ + 3 a²x + 3 ax² + x³ = a³ + 3a²x + ax² + 23. 2! (a + x)¹ = a¹ + 4 a³x + 6 a²x² + 4 ax³ + x¹ = α¹ + 4 α³x + a²x² + 4.3.2 3! ax³ + x¹. 4.3 2! If ʼn represents the exponent of the binomial in any one of the above three cases, we notice: (1) The first term is ɑ”. (2) The second term is na"-la. (3) The exponents of a decrease by unity from term to term while the exponents of a increase by unity. (4) If in any term the coefficient be multiplied by the expo- nent of a and divided by the exponent of a increased by unity, the result is the coefficient of the next term. For n < 5, we may then write (α + x)" —— an + nan−1x + n (n − 1) an-2x² + 2! + n(n−1) ... (n—r+2) (1)! an-r+¹µ−1 + ··· + 2". Here the question naturally occurs: Does the expansion hold for n≥5? It can be shown by mathematical induction that it holds for any positive integral value of n. Assume (a + x)m = am + mam-¹x + m(m 1) am-²x² + 2! + m(m — 1) ….. (m (~ — 1)! ~ + 2) am-r+1½r-1 + ... + xm. 94 [CHAP. IX. MATHEMATICAL INDUCTION Multiply both members of this assumed equality by a +x, and we obtain (a + x)m+1 am+1 + mamx + ... m(m 1)…….(m +2) pag am-r+2xr-1 + ... + axm (-1)! m(m + amx + + ... 1)···(m r + 3) (-2)! am-r+2xr-1 + ... + max" + xm+r = am+1+(m + 1) amx + (m + 1)m ... + (m ( − 1 ) ! 1)! • + 3) am--+2xr−1 + +(m + 1)axm + xm+1 This expansion is the same that would be obtained by substi- tuting m + 1 for m in the expansion of (a + x)". Hence, if the expansion is true for n = m, it is true for n m, it is true for n = m + 1. Since we know it is true for n = 2, it is true for n = when n is any positive integer, 3, and so on. Hence, (a + x)n = an + nan−1x + n(n − 1) an- 2x² + n(n + 2! 1)... (n r + 2) (λ − 1 ) ! an¬r+1xr−1 + +x". This expansion of a binomial is called the binomial theorem. In the expansion of (a + x)”, the rth term is n(n-1)(n-2)... (n − r + 2) a-+1-1 (1)! which may also be written n! (2° — 1)! (n − r + 1)! A”―r+1ær−1 The term involving a" is n (n − 1) (n − 2) (n 2! +1) a"-x"= n! r! (n − r)! а an-rx". Each of these terms is sometimes called the general term of the binomial expansion. In this chapter the exponent n in the binomial expansion is limited to positive integral values, but no assumption has been made with regard to a or x, so we are at liberty to use the expan- sion no matter what sort of numbers a and x may be. Thus, in ART. 60] 95 BINOMIAL THEOREM (2 b − 4 r²), a = 2b and x = 42. In a later chapter it will be shown that the expansion may be interpreted so as to hold when n is a negative or fractional number, but in that case the number x must lie between a and + a. EXERCISES Expand 1. (2x — 3 y³)4. SOLUTION: 4 (2 x − 3 y³) 4 = (2x) ¹ + 4 (2 x ) 2. (a + x)6. 4. (a+b)7. 6. (1 + 2 a) 4. 8. (x-3y2)5, 4 +(→ 3 y³) 4 ³ ( − 3 y³) + 6(2x)² ( — 3 y³)² + 4(2x) ( − 3 y³) 3 16 x¹ — 96 x³y³ + 216 x²y — 216 xy⁹ + 81 y¹². = 3. (a — x). - 5. (2-x)5. 7. (1 + a)5. 9. (a+√õ)ε. 6 10. (x + 1) " 11. ( 1 + 1 ) ° − ( 1 − 1 ) °·. e a 14 12. 2 b 1 14. (+√8) V2 16. (x−1 — y³)4. a b 6 18. vo √x 20. (√x + 2 + 1\3 x 13. (√x-vÿ)8. 15. (a³ + a³)6. 17. (3-1)5. 19. (a + b + r.)³. HINT: Consider (a + b) as repre- senting one number. x2 21. 3 X3 (~72 - 2323 + 1) 1. 4 22. Find the seventh term in the expansion of (x — 2 y²)11. SOLUTION: The rth term is given by the expression Here n(n − 1) (n (r− 1)! ... j + 2) an-r+1x-1. 11, r = 7, n = α = х x - · 2 y², n − r + 2 = 6. Substituting these in the expression for the rth term, we have 11 · 10 · 9 · 8 · 7 · 6 (x² )³ ( − 2 3²) 6 6! (x³)5( − 2 y²)6 = 29568 x‡y12. 23. Find the fourth term of (a — 4 b) 12. 24. Find the eleventh term of (2 x − y)17. 25. Find the middle term of (x² + 3 y²)8. 26. Find the fourteenth term of (a + b)18. 96 MATHEMATICAL INDUCTION X 27. Find the eighth term of √a y а 4|” 13 [CHAP. IX. 28. Find the sixth term of (x√y+y√x)°. 29. Find the middle terms of (1 — a³)7. 30. Number the terms of the expansion of (a + b)6, and with these num- bers as abscissas and the coefficients of the corresponding terms as ordinates, plot points. 31. On the same coordinate paper on which points described in Problem 30 are plotted, graph points for the expansion of (a + b)10. 32. Use the binomial theorem to find (102)5. HINT: 102 (100 + 2). = Use the binomial theorem to find 33. (99)6. 34. (51)5. 36. (1.1)10, correct to four significant figures. V 37. (1.1)15, correct to four significant figures. 35. (.98)6. 61. Direct variation. CHAPTER X VARIATION In Chapter III we have seen that if is Y a function of x, then in general y changes when a changes. We might say that y varies when a varies, but the word "varies" has come to have a more restricted meaning when used in this con- nection. Each of the statements "y varies as x,' y varies di- rectly as x," "y is proportional to x," "y is directly proportional to x," means that the ratio of y to x is constant. That is, Y = k, or y = kx. x 99.66 The constant k is called the constant of variation. The expres- sion "y varies as a" is often written Y xx. The area of a circle varies as the square of its radius. α That is, A ¤ r², or A = kr², if A represents the area and the radius. Here, k, the constant of variation, is equal to . If a train moves with a uniform speed, the distance s traversed varies as the time t. That is, sat, or s= kt. x 62. Inverse variation. Each of the statements "y varies in- versely as a," "y is inversely proportional to " means that y is equal to the product of the reciprocal of x and a constant. That is, k Y X The volume of a gas varies inversely as the pressure. k v Ρ if v represents volume and p pressure. 97 That is, 98 [CHAP. X. VARIATION 63. Joint variation. The statement "z varies jointly as x and y" means that z equals the product of xy and a constant. That is, z = kxy. The distance which a train moving with a uniform speed trav- erses varies jointly as the speed and the time, or d = kvt, where d is the distance covered, v the speed, and t the time. In this case k = 1, if v and d are measured with the same unit of length. 64. Combined variation. The statement "z varies directly as » and inversely as y" means that z varies jointly as a and the re- ciprocal of y. That is, 2= kx У The attraction F of any two masses m₁ and m₂ for each other varies as the product of the masses and inversely as the square of the distance r between the two bodies. That is, F __ km, m 2 2.2 elus EXERCISES AND PROBLEMS Write the following statements, 1 to 8, in the form of equations. 1. y varies as x, and y = 64, when x = 2. SOLUTION: y = kx. Substituting 64 for y and 2 for x, gives Hence, 64 = 2k, or k = 32. y = 32 x. 2. y is directly proportional to x, and y = 18 when x = 6. 3. s varies as t², and s = 64 when t = 2. 4. p varies inversely as v, and v = 128 when p 16. 5. z varies jointly as x and y. z = 120. When x = 2, and y = 3, it is found that When x = 2 and y = 3, it is 6. z varies directly as x and inversely as y. found that 2 = 120. z 7. The volume V of a sphere varies directly as the cube of its radius r. ARTS. 63-64] 99 EXERCISES AND PROBLEMS 8. The volume V of a circular cylinder varies jointly as its altitude h and the square of its radius r. 9. The number of feet a body falls varies directly as the square of the number of seconds occupied in falling. If the body falls 16.1 feet the first second, how many feet will it fall in 5 seconds? In 10 seconds? 10. The safe load of a horizontal beam supported at both ends varies jointly as the breadth and square of the depth, and inversely as the length between supports. If a 2 × 6 inches white pine joist, 10 feet long between supports, safely holds up 800 pounds, what is the safe load of a 4 x 8 beam of the same material 15 feet long? 11. The pressure of wind on a sail varies jointly as the area of the sail and the square of the wind's velocity. When the wind is 15 miles per hour, the pressure on a square foot is one pound. What is the velocity of the wind when the pressure on a square yard is 25 pounds? 12. The pressure of a gas in a tank varies jointly as its density and its absolute temperature. When the density is 1 and the temperature 300°, if the pressure is 15 pounds per square inch, what is the pressure when the density is 3 and the temperature 320 ? 13. Write in the form of an equation the following physical law: the bend b of a rod supported at both ends varies directly as the weight m hung at its middle point, directly as the cube of the length of the rod between supports, inversely as the width w of the rod and inversely as the cube of TO M 13 the depth d. b = 14. A beam 15 feet long, 3 inches wide, and 6 inches deep when sup- ported at each end can bear safely a maximum load of 1800 pounds. What is the greatest weight that can safely be placed on a beam of the same mate- rial 18 feet long, 4 inches wide, and 4 inches deep? (See Prob. 10.) 15. A plank 10 feet long, 10 inches wide, and 2 inches thick is supported at both ends. A weight of 180 pounds hung at the middle makes it bend 3 inches. How much will the plank bend if placed on edge? (See Prob. 13.) 16. The area of the top of a well-designed chimney varies as the quantity of coal used per hour and inversely as the square root of the height of the chimney. The top of a 150-foot chimney connected with a furnace using 11,000 pounds of coal per hour is 27 square feet. What should be the area of the top of a 100-foot chimney connected with a furnace using 2500 pounds of coal per hour? 17. If a heavier weight draws up a lighter one by means of a cord passed over a pulley, the number of feet passed over by each weight in a given time varies directly as the difference of the weights and inversely as their sum. If 5 pounds draw up 3 pounds 16.1 feet in 2 seconds, how high will 10 pounds draw 9 pounds in 2 seconds? 100 VARIATION [CHAP. X. 18. The time of a railway journey varies directly as the distance and inversely as the speed. The speed varies directly as the square root of the quantity of coal used per mile and inversely as the number of cars in the train. In a journey of 32 miles in hour with 12 cars, ton of coal is used. How much coal will be consumed in a journey of 64 miles in 2 hours with 10 cars ? 19. A paper disk is placed midway between two sources of light which are 12 feet apart. If the amount of light falling on the disk varies inversely as the square of the distance from the source of light, show that if the disk is moved parallel to itself a distance 2√3 feet, the whole amount of light falling on the disk is trebled. 20. How far must the disk in Problem 19 be moved from a point midway between the two lights so that the total amount of light on the disk is doubled ? 21. A solid spherical mass of glass 2 inches in diameter is blown into a hollow spherical shell whose outer diameter is 4 inches. If the volume of a sphere varies as the cube of the diameter, what is the thickness of the shell? 22. Kepler's third law states that the square of the number of years it takes a planet to revolve about the sun varies directly as the cube of the dis- tance of the planet from the sun. Let the distance from the earth to the sun be 1. How long would it take a planet whose distance from the sun is 100 to complete one revolution? 23. The number of inches a body falls in one second varies inversely as the square of the distance from the earth's center. At the surface of the earth a body falls 193 inches in one second. How far would it fall in one second if it were as far away as the moon? (The distance of the moon from the earth's center may be taken as sixty times the radius of the earth.) 24. The crushing load of a solid square oak pillar varies directly as the fourth power of its thickness and inversely as the square of its length. If a four-inch pillar 8 feet high is crushed by a weight of 196,320 pounds, what weight will crush a pillar of the same wood 6 inches thick and 12 feet high? CHAPTER XI PROGRESSIONS 65. Arithmetical progressions. An arithmetical progression is a sequence of numbers each of which differs from the next preceding one by a fixed number called the common difference. Thus, 2, 4, 6, 8, ... is an arithmetical progression with the common difference 2. the arithmetical progression. 10, 8, 6, 4, 2, ... the common difference is 2. In The numbers of the sequence are called the terms of the pro- gression. 66. Elements of an arithmetical progression. Let a represent the first term, d the common difference, n the number of terms considered, 7 the nth, or last term, and s the sum of the sequence. The five numbers a, d, n, l, and s are called the elements of the arithmetical progression. 67. Relations among the elements. Since a is the first term, we have, by definition of an arithmetical progression, That is, a + d = second term, a + 2 d third term, a + 3 d = fourth term, a +(n − 1)d = nth term. l= a + (n − 1)d. 101 (1) 102 [CHAP. XI PROGRESSIONS The sum of an arithmetical progression may be written in each of the following forms: s = a + (a + d) + (a + 2 d) + ··· + (1 − 2 d) + (l — d) + 1, s = l + (l — d) + (l − 2 d) + By addition ... ··· + (a + 2 d) + (a + d) + a. 2s=(a+1)+(a + 1) + (a + 1) + ··· + (a + 1) + (a + 1) + ( a + 1) ... = n(a + 1). Therefore, s=(a+1). (2) Whenever any three of the five elements are given, equations (1) and (2) make it possible to find the remaining two elements. Exercise. Establish formulas (1) and (2) by mathematical induction. 68. Arithmetical means. The first and last terms of an arith- metical progression are called the extremes; while the remaining terms are called the arithmetical means. To insert a given number of arithmetical means between two numbers it is only necessary to determined by the use of equation (1) and to write down the terms by the repeated addition of d. EXERCISES Find and s for the following seven sequences: 1. 2, 11, 20, to 10 terms. ... SOLUTION: Here, l = a + (n − 1)d. - a = 2, d 9, n = 10. = 2 2+9.9 9 · 9 = 83. n S = (a + 1) 2 2. 2, −5, —8,. to 20 terms. ... 4. 5, 1, 3, to 20 terms. - 6.7, 29, 15, to 16 terms. 8. Given a = 19, d d= 9. Given d 10. Given l 5(2 + 83) = 425. 3. 3, 7, 11, to 15 terms. ... 5. 1, 12, 1, to 8 terms. 1. 2, s=91; 7. ༢༡ 11, - £1, 웃​, find,n and 1. 4, n = 15, 759; find a and s. - }, n = 14, d = §; find a and s. V 11. Given a = 12, l — 64, s —— 520; find n and d. 12. Given a = 7, 1 = 43, d = 4; find n and s. ... to 17 terms. ARTS. 67-71] RELATIONS AMONG THE ELEMENTS 103 13. Given aɑ = 4, l = — 64, n = 21; find d and s. 14. Given d = 去​っ​て ​It, 8 = 1633 1633; find a and n. 15. Given l 50, n = 50, s = 1275; find a and d. 16. Given d= 10, n = 10, s = 17. Given α = 7, n = =7 = 10; find a and 1. 7, s = 7; find d and 1. 18. Insert 6 arithmetical means between 3 and 8. SOLUTION: We have to find d, when a = 3, 18, and n = 3, 1 = 8, and n = 6 + 2 = 8. Since we have l = a + (n − 1)d, 8 www-dand 3+ 7 d, or d = $. Hence, the 6 arithmetical means between 3 and 8 are 26, 31, 36, 47, 46, 57. 51 W 19. Insert 3 arithmetical means between 2 and 14. 20. Find the arithmetical mean between 10% and 16. 21. Insert 9 arithmetical means between 1 and — 1. 69. Geometrical progressions. A geometrical progression is a sequence of numbers in which the same quotient is obtained by dividing any term by the preceding term. This quotient is called the ratio. Thus, 3, 6, 12, 24, ... is a geometrical progression with a ratio 2. 70. Elements of a geometrical progression. The elements are the same as those for an arithmetical progression with one exception. Instead of the common difference of an arithmetical progression, we have here a ratio represented by r. 71. Relations among the elements. If a represents the first term, then αν second term, ar² = third term, ar.3 fourth term, ajn-1 nth term. That is, 1 = arm-1. (1) By definition, s = a + ar + ar² + ar³ 4 ... +arn-1, (2) Then, sr = ar + ar² + ar³ + ... + arn-1 +ar”. (3) 104 PROGRESSIONS [CHAP. XI. Subtracting members of (2) from members of (3), we have sr s = arn a. Hence, arm a (1 S = a r-1 pn) 1 γ Since arn-1, (4) may be written in the form ri-a S r — 1 (4) (5) Here, as in an arithmetical progression, whenever any three of the five elements are given, relations (1) and (5) make it possible. to find the other two. Exercise. Establish formulas (1) and (4) by mathematical induction. 72. Geometrical means. The first and last terms of a geo- metrical progression are called the extremes, while the remaining terms are called the geometrical means. To insert n geometrical means between two given numbers is to find a geometrical pro- gression of n+2 terms having the two given numbers for extremes. /1 1. Given a = 1, r = 3, n = = 2. Given a = — 2, d EXERCISES 9; find 7 and s. 3, n = 8; find and s. 3. Given a = 1, r = 21 1, n = 10; find 7 and s. 4. Given a = 6, r 2, n = 8; find l and s. - 5. Given s = 242, a = 2, n = 5; find r and l. 6. The 3d term of a geometrical progression is 3, and the 6th term is 81. What is the 10th term ? % 7. What is the 7th term of a geometrical progression whose first term is 2 and third term 3? 8. What is the sum of the first 5 terms of a geometrical progression whose first term is 2 and third term 8? 9. The first term of a geometrical progression is 3, and the last term 81. If there are four terms in the geometrical progression, find the common ratio the sum of the series. and 10. Insert one geometrical mean between 7 and 343. 11. Insert two geometrical means between 2 and 1024. 12. Insert four geometrical means between 12 and §. 13. What is the eighth term of the progression a², ab, b², 14. If each term of a geometrical progression is multiplied by the same number, show that the products form a geometrical progression. ... .? Arts. 71-73] NUMBER OF TERMS INFINITE 73. Number of terms infinite. gression 105 Consider the geometrical pro- b, b, b, A, **. 1 1 21 4 1 169 It may at first thought appear that the sum of the first n terms of this progression could be made to exceed any finite number previously assigned by making n large enough. That this is not. the case and that the sum can never exceed unity, will be seen from the following illustration. Conceive a particle moving in a straight line towards a point one unit distant in such a way as to describe the distance in the 1st second, the remaining distance in the 2d second, the remaining distance in the 3d second, and so on indefinitely. This is represented in Fig. 17. 1224 The distance AB represents one unit of distance. In the first second the particle moves from A to P₁. In the 2d second it moves from P₁ to P2, and so on. The total distance traversed by the particle in n seconds is given by the sum } + 1 + 1 + ... to n terms, which sum cannot exceed nor equal 1, no matter how many terms we take, but can be made to differ from 1 by as small a positive number as we please by making the number of terms large enough. A B P₁ FIG. 17. P P3 PPP 102 Thus, when n=10, the sum is 1833 (Prob. 3, Art. 72). In this illustration, 1 is said to be the limiting * value of the sum of the first n terms of the series. If s, represents the sum of the first n terms of the series, we write lim S = 8个​& s₁ = 1, n which reads, "the limit of s, as n increases beyond bound is 1.”† * For definition of limit, see Art. 150. + The symbol (6 "stands for "n increases beyond bound," or its equivalent "n becomes infinite.” 106 [CHAP. XI. PROGRESSIONS The limit s is called the sum of the geometrical progression with infinitely many terms. For any geometrical progression in which the ratio is less than 1, the above argument can be repeated, and it can be shown that there is a limiting value to the sum of the first n terms of such a series. In Art. 71, we have shown that the sum of the geometrical progression +arn-1 a + ar + ar² + is given by a(1 − 2•”) a arn Sn 1 — r 1 1 1- r. We may then write lim 8个 ​lim а S " n → ∞ 1 — r lim Ap n∞ 1 Τ (See Art. 152.) It will be proved in the chapter on Limits (Art. 154) that Hence, lim aron n→ ∞ 1 — j =0 when < 1. lim a Sn 1-r S = 8个​U 74. Series. A series is an expression which consists of the sum of a sequence of terms. Thus, the indicated sum of the terms of a progression is often called a series. A finite series is one which has a limited number of terms. An infinite series is one in which the number of terms is infinite; that is, the number of terms has no bound. 75. Repeating decimals. Repeating decimals furnish good illustrations of infinite series which are at the same time the sum of the terms of a geometrical progression with infinitely many terms. For example, .33333 may be written as the series where a = .3+.03+ .003 + .0003 + ..., .3 and r = .1. The limit of the sum of n terms of this series as the number n increases indefinitely is. Again, .9828282 ... may be written .9 + .082 + .00082 + ARTS. 73-77] 107 HARMONICAL MEANS where the terms after the first form a geometrical progression in which a = .082 and r = .01. The expression "limit of the sum of n terms of the infinite series as ʼn increases beyond bound" is often abbreviated by say- ing merely "sum of the infinite series." EXERCISES Find the sum of the following infinite series : 1. 3 + 1 + 1 + } 2. 1 5. 4 } + { − 2 + · · 3. // — 1 } + 1 3+ 4 6. 4. 24 12 + 6 3+ Find the limiting value of the following repeating decimals : 7. .636363 ….. 10. .363636 ·· 8. .44444... 11. 40.909090 9. .83333 12. .54123123 .. ” 76. Harmonical progressions. Three or more numbers are said to form a harmonical progression if their reciprocals form an arith- metical progression. The term "harmonical as here used comes from a property of musical sounds. If a set of strings of uniform tension whose lengths are proportional to 1,,,,, be 1 sounded together, the effect is harmonious to the ear. The sequence 1 1 1, 1, 1, 4, }, 6 is a harmonical progression since the reciprocals form the arith- metical progression 77. Harmonical means. 1, 2, 3, 4, 5, ..., To find n harmonical means between two numbers, find n arithmetical means between the reciprocals of these numbers. The reciprocals of the arithmetical means are the harmonical means. EXERCISES 1. Insert two harmonical means between 3 and 12. 2. Insert two harmonical means between 2 and §. 3. Insert four harmonical means between and 1. 4. What is the harmonical mean between a and b ? 5. Show that 4, 6, 12 are in harmonical progression, and continue the series for two terms in each direction. 108 PROGRESSIONS [CHAP. XI. 6. If A, G, and I stand respectively for the arithmetical, geometrical, and harmonical means between two numbers a and b, show that G² = AH. 7. The numbers a, b, c are in arithmetical progression, and b, c, d are in harmonical progression; prove that ad = bc. PROBLEMS 1. What distance will an elastic ball traverse before coming to rest if it be dropped from a height of 20 feet and if after each fall it rebounds one third of the height from which it falls? Σ 2. If a falling body descends 16 feet the first second, 3 times this dis- tance the next, 5 times the next, and so on, how far will it fall the 30th sec- ond, and how far altogether in 30 seconds? 3. Assume that a baseball will fall 16 feet the first second, 48 the next, 80 the next, and so on. A baseball was dropped from the top of Washington Monument, 550 feet high, and caught by an American League catcher. About how fast was the ball falling when caught? About 4. A swinging pendulum is brought gradually to rest by friction of the air. If the length of the first swing of the pendulum bob is 30 centimeters, and the length of each succeeding swing is less than the preceding one, what is the distance passed over in the fifth swing? ΤΟ 5. What is the total distance passed over by the pendulum bob de- scribed in Problem 4 in 5 swings? 6. A ball rolls down an incline of 20 degrees, 5.47 feet the first second, and in each succeeding second, 10.94 feet more than in the preceding second. How far will it roll in 10 seconds ? From a point in l₁ 7. Two straight lines l1 and l2 meet at the point 0. drop a perpendicular to l2. From the foot of this perpendicular drop a per- pendicular to l, and so on indefinitely. If the lengths of the first and second perpendiculars are 6 and 5 respectively, what is the sum of the lengths of all the perpendiculars ? 8. In a raffle, tickets marked 1, 2, 3, 4, etc., are shaken up in a hat and drawn by the purchasers one at a time. The price of a ticket is the num- If the raffled article ber of cents corresponding to the number on the ticket. is worth $10, what is the least number of tickets which will insure no loss to the vendors of the tickets? - 9. A person contributes one cent and sends letters to three friends ask- ing each to contribute one cent to a certain charity and to write a similar letter to each of three friends, each of whom is to write three letters, — and so on until ten sets of letters have been written. If all respond, how much money will the charity receive ? ART. 77] V.. PROBLEMS 109 10. Twenty-five stones are placed in a straight line on the ground at in- tervals of 4 feet. 10 feet from the end of the row is a basket. A runner starts from the basket and picks up the stones and carries them, one at a time, to the basket. How far does he run altogether? 11. An employer hires a clerk for five years at a beginning salary of $500 per year with either a rise of $100 each year after the first, or a rise of $25 every six months after the first half year. Which is the better propo- sition for the clerk ? 12. The population of a town is 10,000. It loses annually 2 per cent of its population by deaths and gains 3 per cent by births, and at the end of each year, it has gained 100 people as a result of movings into and away from the What will the population be in 10 years? town. 13. Find the limit of the sum of the series V: 1 + 1 n + 1 (n + 1) 2 1 + ... + where n > 1). (n + 1) 3 14. What is the sum of the first n odd numbers ? 15. What is the equation whose roots are the arithmetical and the har- monical means between the roots of x² - 16 x + 48 = 0 ? 1 1 1 16. If form an arithmetical progression, show that x, — У x 2Y Y 2 y, and z form a geometrical progression. ✓17. 17. What is the number which added to each of the numbers 1, 3, 2, produces a geometrical progression ? 18. The fourth term of a geometrical progression is 108, the seventh term is 864. What is the 10th term ? 19. There are four numbers in arithmetical progression. When these numbers are increased by 2, 4, 8, 15 respectively, the sums are in geometri- cal progression; find the four numbers in arithmetical progression. 20. Three numbers whose sum is 18 are in arithmetical progression. If you multiply the first by 2, the second by 3, and the third by 6, the resulting products form a geometrical progression. Find the three numbers. 21. Find the present value of ten annual payments of $1000 each made at the ends of the next ten years, if money is worth 5 per cent compounded annually. CHAPTER XII COMPLEX NUMBERS 78. Number systems. If our number system consisted of zero and positive integers only, the solution of an equation such as 3x-2=0 would be impossible; for no number in the system considered satisfies this equation. We can extend the number system so as to include the class of numbers to which the solution belongs. These new numbers are the rational fractions. x While the solution of 3 x 2 = 0 is possible in a number system composed of zero, positive integers, and rational fractions, the solution of an equation such as a + 4 = 0 is impossible. To meet the demands of such equations, we find it expedient again to ex- tend the number system so as to include the negative numbers. In a number system thus extended an equation ax + b = 0, a and b are any integers or fractions, has a solution. 0, where The solution of an equation such as x² 2 demands a further extension of our number system. It must be made to include irrational numbers, that is, numbers which cannot be represented by the quotient of two integers (see p. 23). But the number system thus extended is not sufficient to meet all the demands of the equations met in algebra. In this number system it is im- possible to solve certain quadratic equations, for example, the equations a² + 1 = 0 and x² 0 and x² - 6x + 13 = 0. It is necessary again to extend the system so as to include numbers of the form a+bi, where a and b are real numbers, discussed in Art. 1; and where i is a symbol whose square is -1, that is, i = √1 (see Art. 22). These numbers are usually called complex numbers and sometimes imaginary numbers. When a = O they are called pure imaginary numbers. The term "imaginary number" is here used in a technical sense. The numbers are imaginary in the same sense that a 110 ARTS. 78, 79] 111 GRAPHICAL REPRESENTATION fraction, a negative number, or an irrational number is imaginary for a number system consisting of positive integers. At this point the question may be asked,-In working with this new system which includes complex numbers, inay we not find it necessary to add new numbers, at present unknown, just as we found it necessary to add fractions, negative numbers, and irrational numbers to our system of positive integers? The answer to this question is that the system of complex numbers is sufficient to meet the demands of the equation. While we have seen that the solution of equations with integral coefficients demands fractions, negative numbers, irrational num- bers, and complex numbers, it is not to be inferred that all numbers are roots of equations with integral or rational coeffi- cients. For example, the irrational number # cannot be the root of an equation with rational coefficients. The proof is beyond the scope of this book.* -1+4i I 2+3i 21 79. Graphical representation of complex numbers. We have seen that all real numbers may be represented by points on a straight line. The complex number x + iy depends on two real numbers x and y, and may be represented graphically by a point in a plane. Two lines, X'X, Y'Y, are drawn perpen- dicular to each other and in- tersecting at O, Fig. 18. To represent the number 2 + 3 i, measure off on X'X to the right the distance 2, and up the distance 3. In general, the graph of the number x + iy is the point whose coordinates are (x, y). The line X'X is often called the axis of →X 2 2-3i 3-4i 5 i Ÿ' FIG. 18. "reals" and the line Y'Y the axis of imaginaries. * See Klein, Famous Problems in Elementary Geometry, translation by Beman and Smith, p. 68. 112 [CHAP. XII. COMPLEX NUMBERS *It is often convenient to represent complex numbers by another method. Connect the point AY Xxx & FIG. 19 x+iy リ ​which represents x + iy with the origin as in Fig. 19. Let the length of this line be r. The point can then be repre- sented by giving the length r and the angle . From the figure x = r cos 0, y = r sin 0, 2.2 x² + y² = 7¹². Hence, the number x + iy may be written in the form x + iy = r (cos 0 + i sin 0). This form is called the polar form of a complex number. The angle is called the argument or amplitude, the length r the modulus or absolute value of the complex number. It should be noted that the complex numbers include all real numbers. In Fig. 18, the real numbers are represented by points on the line X'X. The pure imaginary numbers are represented by points on the line Y'Y. 80. Equal complex numbers. If two complex numbers a + bi and c+di are equal, then a = c and b d. For, if a + bi = c + di, (1) by transposing, a c = (d — b)i. (2) Unless a c = d d — b = 0, we should have a c, a real num- ber, equal to (db)i, an imaginary number. Conversely, if a = c, and bd, a + bi= = c + di. Hence, when any two expressions containing imaginary and real terms are equal to each other, we may equate the real parts and the imaginary parts separately. In particular, if a + bi = 0, a = 0 and b = 0. * * The remainder of this article and the articles marked may be omitted by those who have not studied trigonometry. ARTS. 79-81] ADDITION AND SUBTRACTION 113 EXERCISES Represent graphically the following numbers and in each case find the argument and the modulus. 1. 2-3 i. SOLUTION: The number is represented in Fig. 18. The modulus r is given by r = √x² + y² = √4 + 9 = √13. To find the argument we have 312 Y tan 0 = X 3 0 arc tan 2. - 2 - 3 i. 3. 2 + 3 i. 5. 1 — i. 6. 54 i. -8. 4 i. 9. - 6 i. 11. 4+0 i. 12. .5. 14. 0.7 + 1.1 i. 3 sin @= √13 4. 1 + i. 7. i. 10. 3 i. 13. 3 2 Write the following complex numbers in the form + iy. 15. 3(cos 30° + i sin 30°). 17. 2(cos 150° + i sin 150º). 19. (cos 225° + i sin 225°). 21. (cos 270° + i sin 270°). 16. 2(cos 60° + i sin 60°). 18. 4(cos 90° + i sin 90°). 20. 6(cos 0° + i sin 0°). − 22. If (x + 1) + i(y − 1) = 0, what are the values of x and y ? What must be the value of x and y in order that the following equations may be true ? 23. x + y + i( x − y) = 2 + 4 i. 24. 2x+7y+i (3 x − 2 y) = — 3 — i. 26. 3x + xyi + 2 y — ix — 3 y—5—0. 25. 2x² + y² + i(x − y) = 1 + i. 27. x² + ix + iy + y² = 130 + 8 i(x − y). We assume 81. Addition and subtraction of complex numbers. that the number i like other numbers obeys all the laws of algebra. Given two complex numbers a+bi, c+di, we may write the sum and difference. Thus, (a + bi)+(c + di) = (a + c) + (b + d)i, (a + bi)− (c + di) = (a — c) + (b − d)i. Hence, to add (or subtract) complex numbers, add (or subtract) the real and imaginary parts separately. The result is a complex number. 114 [CHAP. XII. COMPLEX NUMBERS To add two complex numbers, a + bi and c + di, graphically, we represent the numbers as points A and B in Fig. 20. Connect I B A R C D FIG. 20. X each point with the origin 0. Complete the parallelogram, having OA and OB for adja- cent sides. The vertex P represents the sum of the two given complex numbers. For, from the figure the coordinates of the fourth vertex are OQ and QP. But OQ OD + DQ = a + c, QPQR + RP = b + d. Hence, P represents the point (a+c)+(b + d)i which is the sum of a + bi and c + di. To subtract one complex number from another graphically, say c + di from a + bi, we graph the points which represent and a + bi, and proceed as for addition. EXERCISES Perform the following operations algebraically and graphically. 1. (1 + i) + (2 + 3 i). 3. (2+)-(1 + 4 i). 5. (3-5)-(3+5i). 7. (0 + 3 i) + (1 − 4 i). 9. (4+ i)− (3 + 3 i) + (1 − i). 2. (2+2 i) + (1 − 3 i). 4. (3-5)+(3 + 5 i). 6. (3-4) + (5 − i). i) C di 8. (6 +0 i) + (−3 +7 i) − (4+2 i). 10. (4+3 i) — (4 + 3 i). *82. Multiplication of complex numbers. Let a + ib and c id be any to complex numbers. Since i obeys all the laws of alge- bra, we have (a + ib)(c+id) ac + ibc + iad ibd (acbd)+ i(bc + ad). = + = The result is a complex number. To multiply two complex num- bers graphically, let the two numbers a + ib, c + id be represented by the points P1 and P₂ (Fig. 21). Reducing to the polar form, we have 0₁ a+bir₁(cos + i sin 01), c+dir₂(cos ½ + i sin 02). ARTS. 82-83] CONJUGATE COMPLEX NUMBERS 115 By actual multiplication, (a + bi) (c + di) 1. 712[COS 01 cos 02 + ¿(sin 0, cos 02 + cos 01 sin 02) — sin 0₁ sin 02] =rir₂[cos(0₁+02) + i sin (01 +02)]. 1 Hence, the modulus of the product of two complex numbers is the product of their moduli and the argument is the sum of their arguments. The point P, which represents (a + bi)(c + di) may then be con- structed by drawing through O a line making an angle 0 = 0 + 02 with the line OX, and constructing on this length a segment OP, whose length is 172. 1 83. Conjugate complex numbers. Numbers which differ only in the sign of the imaginary parts are called AY P 1 2 A 1 A. FIG. 21. conjugate numbers. Thus, 3+2i and 3-2 i are conjugate. Since and = (a + bi)+(abi) 2 a, (a + bi)(a — bi) = a² + b², (a + bi) — (a — bi)= 2 bi, we see that the sum and the product of two conjugate complex numbers are real numbers, but the difference of two conjugate complex numbers is an imaginary number. EXERCISES Multiply both analytically and graphically, finding the arguments and moduli of the products. 1. (3+ √3 i) (2 + 2 i). SOLUTION. (3 + √3 i)(2+2 i) = 6+ 6 i + 2√3 i + 2√3 i² = 6-2√3+i(6 +2√3) Putting the numbers in the polar form, we have, 3+√3 i=2√3(cos 30° + i sin 30°), 2 + 2 i = 2√2 (cos 45° + i sin 45°). 116 [CHAP. XII. COMPLEX NUMBERS P • 45 30 75 FIG. 22. X 6. (1+i)²(22√3 i). 8. (2-2) (2 + 2 i). i) 10. (0+3i) (2 + i). * Hence, r1 r₁ = 2√3, r₂ = 2√2, 01 = 30°, 02 = 45°. The modulus of the product is, then, rir2 = 4√6, and the argument is 75°. Let Pi and P₂ in Fig. 22 represent the two given numbers. Through O draw a line mak- ing an angle of 75° with the line OX. On this line measure off the distance OP = 4√б. The point P then represents the product of the two numbers. 2. (3+√3 i) (2 + {√3 i). г 3. (1+√3i) (4 + √3 i). 4. (√2+√-2)². 5. (1+i)¹. 7. (-2+2i) (2 + 2 i). 9. (1 + i)(1 + 2 i) (1 + 3 i). 11. (0+2)(0 – 2 i). 84. De Moivre's theorem. If two complex numbers are equal, then as a special case of Art. 82, we have Ө r(cos 6 + i sin 0) r (cos ✪ + i sin 0) = r² (cos 0 + i sin 0)² 72 (cos 20+ i sin 2 0). Multiplying both sides of this identity by r(cos +i sin 6), we have r³ (cos 0 + i sin 0)³ = 7³ (cos 3 0 + i sin 3 0), and it can easily be proved by mathematical induction that [r(cos + i sin 0)]” Ꮎ y” (cos n 0 + i sin n 0), where n is any positive integer. This relation is known as De Moivre's theorem, and holds also for fractional values of the exponent. To prove the theorem when the exponent is the reciprocal of a 1 positive integer, consider the expression (cos + i sin 0)" in which n is a positive integer. ARTS. 83-85] ROOTS OF COMPLEX NUMBERS 117 Let θ = ηφ, 1 1 then (cos + i sin 0)" = (cos no + i sin no)” 1 =[(cos + i sin p)"]" = cos & + i sin & Ө = COS +isin N Ө η De Moivre's theorem thus gives an easy method of finding any power or any root of a complex number. The proof can be readily extended to the case of an exponent which is any rational fraction. * 85. Roots of complex numbers. From Art. 84, the nth root. of x + iy is 1 0 (x+iy)" =[r(cos + i sin 0)]" If m be any integer, cos (0+ m 360°) = cos 0, We may then write 1 sin (0 + m 360°)= sin 0. 1 Ꮎ cos - + i sin n n +isin (0+ m 360°)}]" 360°)}] 0+ m 360° COS +i sin 0+m 360°7 n N 1 (x + iy)" =[r(cos + i sin 0)]" =[r {cos (0+m 360°) If now we let m take the values 0, 1, 2, 3, ……., n 1, we find n results, all different numbers whose nth powers are x + iy. We may then state the following THEOREM. Any number has n distinct nth roots. EXERCISES Using De Moivre's theorem, find the indicated powers and roots. 1. (3+√3i)4. SOLUTION: Writing 3 + √3 i in the polar form, 3+√3i=2√3(cos 30° + i sin 30°). By De Moivre's theorem, 4 (3 + √3 ¿)¹ =[2√3(cos 30° + i sin 30°) ]4 = 144(cos 120° + i sin 120°) 144 ( − 1 + ¿√3 i) =− 72 + 72√3 i. 118 [CHAP. XII. COMPLEX NUMBERS -2. (3+ √3 i)5. 4. (− } - {√3 i)³. 3. (3+√3 i)3. -5. (+1√3i). 6. [2(cos 10° + i sin 10°)]º. -7. (2 + 2 i)5. 8. (1+i). 9. V-2+2 i. SOLUTION: Writing 2+2 i in the polar form, we have − 2 + 2 i = 2 √2 (cos 135° + i sin 135°). By De Moivre's theorem, 3 V — 2+2 i=( − 2+2 i)3 = [2√2 {cos (135°+m 360°) + i sin (135°+m 360°)}]} P÷2+2 2 + 2 i = √2 '2 [cos (45° + m 120°) + i sin (45° + m 120°)]. P₁ P FIG. 23. 12. √3+√3 i. 14. V64(cos 60 + i sin 60°). 10 16. Vcos 300° + i sin 300°. 18. Vi HINT: Write in the form VO + i. For m = 0, 1, and 2, this expression reduces to 1+ i, √2(cos 165° + i sin 165°), and √2 (cos 285° + i sin 285°) respectively. Any one of these three X numbers is a cube root of 2 + 2 i. The points P1, P2, P3 representing these three numbers lie at equal intervals on a circle of radius √2 (Fig. 23). 10. 2+2 i. 11. √ −8 +8√3 i. 13. √64(cos 60° + i sin 60°). 15. 5 64(cos 60 + i sin 60º). 17. V-4-4i. 19. 27 i. 20. √4. 21. VI. 22. $8. 23. 81 ¿. Find all the roots of the following equations and represent them graphi- cally. -24. x3 27 = 0. HINT: x3 = 27. of 27 +0 i. The roots of the equation are then the three cube roots 25. x5 — 1 = 0. 28. x4 - 16 = 0. 26. x5 - 32 = 0. 27. x³- 1 = 0. 29. x6 - 1 = 0. 30. x8-1=0. Arts. 85, 86] 119 DIVISION * 86. Division of complex numbers. The quotient of two com- plex numbers may be obtained as follows. a + ib a + ib C id c + id c + id с id ac + bd — i(ad — bc) c² + d2 ac + bd ad bc i c² + d² c² + d² This is a complex number. Writing the two given complex numbers in the polar form, we have r₁(cos 0, + i sin ₁) 1 72(cos 02 + i sin ₂) 2 r₁₂(cos r2(cos 1 + i sin 0₁) (cos 0₂- i sin (2) + i sin 02) (cos 02 - i sin 02) rir₂[cos (0,0) + i sin (01-02)] "½²(cos² 0₂ + sin² 0₂) 12 [cos (0,0)+ i sin (0, — ½)]. 1 Hence, the modulus of the quotient of two complex numbers is the quotient of their moduli, and the argument is the difference of their arguments. If, in Fig. 24, Pi and P₂ represent the points ail and c+id respectively, the point P which represents the quotient a+ib may be constructed by drawing c + id through a line making an angle 0=0,- 02 with the line OX, and constructing on this line a segment OP, whose length is νι P₁ 12 P X FIG. 24. 120 [CHAP. XII. COMPLEX NUMBERS EXERCISES Find the quotients of the following pairs of numbers, analytically and graphically. 1. (4+4 i) ÷ (2 + {√ši). SOLUTION: 4 + 4 i 4 + 4 i 2 − 3 √ ³ i _ 3 + √3, 3 − √? + 2+3√3 i 2+}√ši 2-3√3i Writing both numbers in the polar form, we obtain 2 4 + 4 i = 4√2(cos 45° + i sin 45°), 2+}√3i = √3 (cos 30+ i sin 30°). r₁ = 4√2, r₂ = 4√3, 01 = 45°, 02 = 30°. Hence, Y P1 P 45 780 15° FIG. 25. -X 2 The modulus of the quotient is then M 71 vő, 7·2 15°. Let and the argument is = 01 02 = 15°. P1 and P2 in Fig. 25 represent the two given numbers. Through O draw a line, making an angle of 15° with the line OX. Measure off on this line the distance OP = √6. The point P then represents the quotient of the two numbers. 2. (−2+2√3i) ÷ (1 +√3i). 3. (√3 i) + ( − 1 + 1 √³ i). 2 5. (1 + i)² ÷ (2 – 2√3 i). 7. 2 ÷ (1 + i). · 9. (1 + i)÷(1 – i). 4. (4+4)(1 − i). 6.4i (2+2i). 8. (−2+2i)÷ (— 4 i). 10. (3+√3 i) ÷ (√3 + 3 i). Graph the following complex numbers and their reciprocals. 11. 1 i. 12. 3+ √3 i. 13. 2-2 i. 14. 3+ 7 i. 15. −3+5 i. CHAPTER XIII THEORY OF EQUATIONS 87. The polynomial of the nth degree. (Art. 27) of a polynomial of the nth degree is -1 -2 ɑox² + α₁xn−1 + A₁₂xn−2+ a ... The general form + an where n is a positive integer, ao, ɑ, ……., α, are independent of x, and a。0. The polynomial of the second degree has been dis- cussed in connection with the quadratic equation. When f(x) is used in this chapter, it is to be understood to mean a polynomial in x. EXERCISE By comparing the following polynomials with the general form, deter- mine n, ɑo, ɑ1, •••, ɑn. (1) (2) (8) f(x) = x² + 3 x¹ +8. ƒ(x) = 4 x8 + 3 x¹ + 10 x² + ix². f(x)=(i +√3)x³ + 5 x² + 10. 88. Remainder theorem. If f(x) is divided by x der is f(r). f(x)= α¸Ã² + ɑ¸Ñ”−¹+ ... + ɑ₂-12 + α „ a ... + α-11' + an -1 — ... 2°, the remain- (1) Given Then, ƒ(r) = αr + α₁−1 + (2) and f(x)-ƒ(r) = α。(2” — 7") +α₁ (x”−1 — p•”−¹) + ··· +α„-1(x−r). (3) But since x r is a factor of each of the expressions an -1 ― xn−1 — pn−1, ..., X J 2012 r (Ex! 11, Art. 58, p. 92), it is a factor of the right-hand member of (3). If we should inclose the whole right-hand member of (3) in a parenthesis, we could remove the factor x r, and call what re- mains inside the parenthesis Q(x). Then we have from (3), f(x)-ƒ(r)=(x − r) Q(x). (4) 121 122 [CHAP. XIII. THEORY OF EQUATIONS Transposing f(r) and dividing by ar, we have f(x) Q (x) + £ (r), f X 2° Xx which is what was to be proved. COROLLARY. If f(x) vanishes when xr, then f(x) is exactly divisible by x 1'. EXERCISES 1. Show, by the remainder theorem, that x" an is exactly divisible by xa if n is even. 2 x 2. Show that an+an is divisible by x + a if n is odd. 3. Without performing the division, find the remainder when x³-3x² + 1 is divided by x — 2. 4. By means of the remainder theorem, find a value for k such that x³ + 3x² + 6x + 2 is divisible by x 1. 89. Synthetic division. The operation of dividing a polyno- mial by a binomial x r can be performed rapidly by means of a process called synthetic division. This rapid division, com- bined with the remainder theorem, gives a convenient method of evaluating f(x) for different values of x. For example, divide 5x46x38x224 x 6 by x-2. By the ordinary method C 5 x¹ 6 x³ + 8 x² - 24 x 6 x 2 5 x4 - 10 x³ 23 5x³+4x²+16x+8 4x³ + 8 x² 4 203 8x2 16 x2 24 x 16 x2 32 x 8 x 6 8x 16 +10 ARTS. 88, 89] 123 SYNTHETIC DIVISION Manifestly, the work can be abridged by writing only the coef ficients, thus, 56+8-24-611-2 " 5 10 +4+8 +4-8 5 + 4 + 16 + 8 + 16 — 24 +16-32 Since the coefficient of x in x +8 - 6 +8-16 +10 is unity, the coefficient of the first term of each remainder is the coefficient of the next term to be obtained in the quotient. Further, it is not necessary to write the terms of the dividend as part of the remainder, nor the first term of the partial products. The work thus becomes : 5682461-2 10 +4 8 +16 32 + 8 16 +10 We may omit the first term of the divisor and write the work in the following more compact form: .5 6+ 8-24 — 6 | 2 10 8-32 16 5+4+16+ 8 +10 If we replace 2 by +2, we may add the partial products to the numbers in the dividend. Then, we have: 5- 6+ 8 - 24 8-24 6 12 +10 + 8 +32 + 16 5+ 4+16+ 8 +10 The quotient is 5 x³ + 4 x² + 16x +8, and the remainder is 10. 124 [CHAP. XIII. THEORY OF EQUATIONS 90. Rule for Rule for synthetic division. To divide f(x) by x 1, arrange f(x) in descending powers of x, supplying all missing powers by putting in zeros as coefficients. Detach the coefficients, write them in a horizontal line and in the order α, a1, a2, ···, An⋅ Bring down the first coefficient a。; multiply a。 by r, and add the product to a; multiply this sum by r, and add the product to u2. Continue this process; the last sum is the remainder, and the pre- ceding sums are the coefficients of the powers of x in the quotient, arranged in descending order. • PROOF OF RULE. This rule may be established by mathematical in- duction. By long division, αox² + α₁xn−1 +α₂xn−2+ Aoxn --- аorxn−1 -2 ... - + α¸xn−s + As+1X”—s−1 + -2 |αoxn−1 + (α₁+αor)xn−2+ (α1 + αor)xn−1 + A2xn−2 (a1 + αor)xn−1 — (α₁r + αo r²)xn−2 ... ... + anx +(as-1 + ra¸-2 + + rs-1α0) xπ-8 We note that the coefficient of x-2 in the quotient is formed according to the rule. Assume that the coefficients in the quotient down to that of x²- are formed according to the rule. On this hypothesis, proceed by long division to find the coefficient of x-s-1 in the quotient. This may be ex- hibited as a continuation of the division above as follows: (as−1 + ras−2 + (as−1 + ras-2 + -8-1 + rs−1αo)xn−s+1 + α„xn−s + ɑ;+1xn−s−1 + + rs−1αo)xn−s+1 — (raç-1 + r²ɑs-2 + (as + ras−1 + r²ɑs-2 + + rsαo)xn−s + αs+1xn−s−1 + ... ... ... +an +rsαo)xn-s + an This shows that if the coefficients in the quotient down to that of xn−s are formed according to the rule, the coefficient of the next lower power is formed according to the rule. Hence, the rule is established. EXERCISES 1. Divide x4 + 3 x³ − 5 x + 3 by x SOLUTION: 1+ 3 + 0 4 by synthetic division. 5+ 34 + 4 + 28 + 112 + 428 1 + 7 + 28 + 107 + 431 The quotient is x3 + 7 x2 + 28 x + 107 and the remainder is 431. 2. Divide 3x³ + 5 x² + 2 x + 1 by x + 5. 3. Divide 5 x³- 6x + 3 by x + 2. 4. Divide 7x4 3x2 2 by x+ - 5.) (In this case › — — 5. Divide x3 125 by x-5. ARTS. 90, 91] 125 GRAPHS OF POLYNOMIALS 91. Graphs of polynomials. When the coefficients of f(x) are real numbers, the march of the function for different values of a can be clearly presented by the use of the graphic methods. explained in Arts. 27-28. To any value assigned to x, there cor- responds one and only one value of the polynomial f(x). This is sometimes expressed by saying that f(x) is single valued. The fact that the graph of ƒ (x) is a continuous curve (see Art. 27) makes it of much service in the theory of equations. EXERCISES Construct the graphs of the following functions and locate their real zeros approxi- mately (to within 0.5). 1. f(x) = x³ x3 6 x² + 11 x 6. As pointed out in Art. 89, synthetic divi- sion furnishes a convenient method of evaluating f(x) for different values of x. Thus ƒ (0.5) is obtained as follows: 1611 6 + 11 – 6 0.5 0.5 2.75 +4.125 1 5.5 + 8.25 1.875 Y Hence, ƒ (0.5)=-1.875. In this way the following values are obtained: ƒ ( − 2) = — 60. ƒ (−1) = — 24. ƒ (— 0.5) = — 13.125. f(0)=-6. ƒ (0.5)=— 1.875. f f(1)=0. ƒ (1.5) = 0.375. ƒ (2) = 0. ƒ (2.5) = — 0.375. ƒ (3) = 0. ƒ (4) = 6. ƒ (5) = 24. The graph is shown in Fig. 26; it presents to the eye the following facts : (1) f(x) has zeros at 1, 2, and 3. (2) f(x) is positive when 2 > x > 1, and when x > 3. (3) f(x) is negative when x < 1 and when 3 > x>2. 2. x3 6x2+8x. 3. x3 7x2 +14x-8. FIG. 26. 4. 235 +4. 5. x42x3 - 7 x² + 8 x + 12. 6. 7. 3x4 - 4x³ — 12 x² + 3. x4 - 3x² + 5 x − 6. 8. x³- 2x – 4. I 126 [CHAP. XIII. THEORY OF EQUATIONS 92. General equation of degree n. By equating to zero the gen- eral polynomial of the nth degree, we obtain what is known as the general equation of the nth degree in one unknown. That is to say, 1 α。x" + α₁ж"-¹ + A²x²-2 + is the general equation of the nth degree. + an = 0 The principal object of this chapter is to present methods which aid in determining exactly or approximately the real roots of special numerical equations included under this type. It is largely for this purpose that we discuss the graphs of poly- nomials. The zeros of the polynomial are the roots of the equa- tion formed by equating the polynomial to zero. The real roots of the equation may then be looked upon geometrically as the abscissas of the points of the X-axis where the graph of the poly- nomial cuts this axis. 93. Factor theorem. If r is a root of the equation f(x) = 0, then XC r is a factor of ƒ (x). Since a zero of f(x) is a root of the equation f(x) = 0, this theorem follows directly from the corollary to the remainder theorem (Art. 88). Exercise. State and prove the converse of the factor theorem. 94. Number of roots of an equation. Every equation, f(x) = 0, of the nth degree has n roots and no more. To prove this proposition we assume the fundamental theorem More ex- that every equation, f(x) = 0, has at least one root. plicitly, we assume that There always exists at least one number, real or complex, which will satisfy an algebraic equation of the nth degree, whose coefficients are any real or complex numbers. † *The term "numerical equations" is here used to indicate that the coeffi- cients are not literal. This fundamental theorem was first proved by Gauss in 1797. For proof, see Fine's College Algebra, p. 588. ARTS. 92-94] 127 FUNDAMENTAL THEOREM Let r be a root, then (Art. 93), er, is a factor of f(a) and 1 x ƒ(x)=(x— r₁)ƒı(x), n - (1) where f(x) is a polynomial of degree 1, beginning with the term a-1. By the theorem assumed, f(x)= 0 has at least one root. Let r½ be a root; then and f(x)=(x — 12)₤2(x), f(x)=(x − r₁)(x — 12).ƒ2(x), (2) in which f(x) is a polynomial of degree n-2, beginning with the term ɑx”-². Continuing this process, we separate out n linear factors with a quotient a。, so that f(x)=α。(x-r₁)(x − r₂) ··· (x − r) — ••• — where 71, 72, ……., "', are n roots of f(x)= 0. (3) If ƒ(x)=0 has another root different from any of these, let » denote such a root. Then, from (3), ... — ɑ。(1 (1° a。(r — 1'1) (1 — 1°2) ··· (†' — 1",) = 0. But here we should have the product of factors equal to zero when Lo one of the factors is zero. As this is impossible (III, Art. 5), there are not more than n roots of ƒ (x) = 0. Fence, every equation of the nth degree has n roots and no more Furthermore, every polynomial of the nth degree is the product of linear factors. It is not, however, possible, in general, to determine these factors if n exceeds 4 (see Art. 106). Two or more of the roots of n f(x) = 0 may be equal to each other, in which case the equation is said to have multiple roots. If (x − r) then f(x)=0 is said to have m roots equal to has five equal roots, and (x 1)²(x − 3)(x — 4)3 equal to 1, one equal to 3, and three equal to 4. — m. r. is a factor of f(x), Thus, (x-2)=0 5 O has two roots COROLLARY I. If two polynomials of degrees not greater than n are equal to each other for more than n distinct values of the variable, the coefficients of like powers of the variable are equal. ... + an box" + b₁xn−1 + ... · + b₂ (4) Let α。x¹· + α₁xn−1 + for more than n values of x. From (4), (α- b₁)x+(α₁ — b₁)x−1 + ... +(a,b) = 0. (5) 128 [CHAP. XIII THEORY OF EQUATIONS Then ao - α。 — b₁ = 0, αι b₁ = 0, (པེ - b₁ = 0. For, if any coefficient in (5) were not equal to zero, we should have an equation of degree equal to or less than n with more than n roots, which is contrary to the theorem just proved. α Hence, α = b。, a₁ = b₁, •••, a„ = b„• ❝o COROLLARY II. If two polynomials of degrees not greater than n are equal for more than n distinct values of the variable, they are equal for all values, and the equality is an identity. + an 95. Graphs of аox² + а₁xn−1 + = 0( ao(x − ri) ( − 1)…( − 1). r1)(x • We assume that a, a, ..., a, are real numbers, and further for convenience of expression that a is positive, although this is not a necessary limitation. In Art. 91 the graphs of a few poly- nomials are plotted. Some important properties of these graphs appear when the polynomial is separated into linear factors. We cannot at this point make the treatment so complete as later, but we can well consider two important cases: 1. When the factors x -21, x 12,···, x — r₁ are all real and distinct. · ... n r„ Arrange the factors so that r₁ > r₂ > ••• > "'-1>7. When a>" all the factors are positive and the graph is above the X-axis. r 4 16 r 07 1 FIG. 27. When r₁> x>r2, one factor is negative and the graph is below the X-axis. When r₂ > x>r, two factors are negative, and the graph is again above the X-axis. Continuing this process, we see that the graph crosses the X-axis at the n points, x=r1, x=r2, ARTS. 94-96] 129 COMPLEX ROOTS X 7, and we obtain a general notion of the nature of the curve. See Fig. 27. 2. When the factors are real but some of them repeated. To discuss the graph in this case, take for example, and let − ƒ(x)= αo(x — 1°1)² (x — 1½) (x — 1′3)5, 71 72 73. Since the factors x- r₂ and x 13 occur to powers with odd ex- ponents, it follows as above that the curve crosses the X-axis at x = r₂ and x = 1‍3. But it does not cross at x = 71, since the sign of f(x) is the same when x > r₁ as when 71 > > 12, and the curve touches the axis at x = √1. In general, if a factor x - r occurs to a power with an odd exponent, the curve crosses the X-axis at x=1, while if it oc- curs to a power with 1 all even exponent, it merely touches the X 12 X-axis at x=1. See Fig. 28. FIG. 28. Another case is discussed in Art. 97, where imaginary factors occur. 96. Complex roots. If a complex number a + bi is a root of an equation f(x) = 0 with real coefficients, the conjugate complex number bi is also a root. By the hypothesis, a+bi satisfies the equation ɑox” + ɑ12”-¹+ ... + an 0. (1) Put a+bi for a in this equation and expand the powers of a+bi by the binomial theorem. Represent the real part of this ex- pansion by P and the imaginary part by iQ. Then Whence, P+ iQ=0. P = 0 and Q =0 (Art. 80). (2) (3) In the binomial expansion of a + bi with any exponent, imagi- nary terms occur when and only when a term involves an odd power of bi as a factor. The result of substituting a bi instead of abi is clearly obtained by replacing i byi in (2). We obtain thus P-iQ. 130 [CHAP. XIII. THEORY OF EQUATIONS But, by (3), hence, P=0 and Q=0; PiQ = 0, and a -bi is a root of (1). 97. Graphs of f(x) when some linear factors are imaginary. Since imaginary factors of f(x) occur in conjugate pairs when the coefficients in f(x) are real, it follows that in this case f(x) may be regarded as the product of a,, of real linear factors of the type x, and quadratic factors of the type 2 (x − a)² + b² = (x a — bi)(x a+bi), where a, b, and rare real numbers. When all the roots of ƒ (2)=0 are real, the polynomial f(x) is the product of real linear factors, but if f(x)=0 has imaginary or complex roots, f(x) contains quadratic factors of the type (x − a)² + b² which cannot be sepa- rated into real linear factors. In Art. 95 the graph of ƒ(x) is discussed when the polynomial is the product of real linear factors, and it is shown that, corre- sponding to each linear factor x r, the graph meets the X-axis at x = r. It should now and x=- x2 FIG. 29. X be noted that (x — α)² + b² > 0, for all real values of x, and there is, therefore, corresponding to quad- ratic factors of f(x), no intersection of the graph with the X-axis. EXAMPLE: Graph x4 ƒ (x) = x¹ — 7 x3 — 4x²+78x = x(x+3)(x²-10x+26) = x(x+3)[(x-5)²+1]. Corresponding to the linear factors x and x + 3, the graph intersects the X-axis at x 0 3 respectively (Fig. 29). Corresponding to the quadratic factor x² - 10x + 26 there is no intersection with the X-axis. (In Fig. 29 one horizontal space represents one unit, while one vertical space represents twenty units.) ARTS. 96-98] TRANSFORMATIONS OF EQUATIONS 131 EXERCISES 1. If "'1, 12, ••, î½ are roots of an equation, show that n (x — r₁) (x — r2) (X — 1°3) ··· (x — în) = 0 is the equation or an equivalent equation. ... - 2. Form equations which have the following roots and no others. (a) 2, 3, 5. (b) 1 + 2 i, 1 — 2 i, when i 2 =— 1. (c) 1 + √2, 1 − √2, 3. (d) √2, − √2, √3, − √3. (e) 1, − 2, 3, 0. _____ − (f) 2 + √3, 2−√3, −2+√3, 2-v3. 3. By means of the theorem concerning the number of roots of ƒ (x) = 0, show: (1) if f (x) = 0 be multiplied by a polynomial in x, the resulting equa- tion has more roots than ƒ (x) = 0; (2) if ƒ (x) = 0 be divided by a polyno- mial in x, which is a factor of ƒ (x), the resulting equation will have fewer roots than ƒ (x) = 0. 4. Plot the graphs of the following: (a) f(x)=(x − 1)²(x − 3)². (b) f(x) = x(x − 1)(x − 4). (c) f(x) = (x − 1) (x + 2) (x + 7). (d) f(x) = (x + 5) (x − 6)². (e) f(x) = (x − 2)²(x+2). 5. Show that 3 and are double roots of 9x5 and find the other root. 51x4+58x³ + 58 x² 51x+9= 0, M Separate the following polynomials into real linear and quadratic factors and plot the graphs. x3 6. x³-1. 9. 261. 7. x3 + 1. 10. x¹ + 4x³ + 3 x² - 4 x 4. 8. x¹ — 1. 11. Show that an equation f(x) = 0 of odd degree and with real coeffi- cients has an odd number of real roots. 98. Transformations of equations. Frequently an equation can be solved more readily, or its properties can be discussed better, by transforming it into an equation whose roots are related to those of the given equation in some specified manner, For ex- ample, in solving numerical equations for their real roots (Arts. 132 [CHAP. XIII. THEORY OF EQUATIONS 102–105), we shall have use for the following transformations of f(x)=0: x = x¹ X x', and x = x' + h. ԴՈՆ x' If we make a = (or x' = mx) in f(x) = 0, we obtain an equa- m X tion in a' whose roots are m times those of f(x) = 0. If we make x′ in ƒ(x)= 0, we obtain an equation in x' whose roots are equal in absolute value but opposite in sign to those of f(x)= 0. If we make x = x' + h in ƒ(x) = 0, we obtain an equation in x' each of whose roots is less by h than the corresponding root of f(x)= 0. These transformations can be performed rapidly by means of the following rules: 1. To obtain an equation each of whose roots is m times a cor- responding root of f(x)=0: multiply the successive coefficients be- ginning with that of x-1 by m, m2, m³, ..., respectively. * For example, to find the equation each of whose roots is double the roots of the equation a¹ 4 x³ + 3 x² + 1 = 0, we make m 23 and obtain 24-2(4x³)+2²(3 x²) + 2³(0 · x) + 24 = 0, x¹ 8 x¹³ + 12 x² + 16 = 0. — To establish this rule, substitute x = -2 ... x' M in f(x)= αox" +α₁x²-1 + α²x² + + α- 1 x + α, The result of this substitution is = 0. (1) 2, n-1 1-2 x' dol m + all + αz + +α-1 + a₂ = 0, (2) m m M -1 ... + m²¬¹a,-1x' + m”a„ α 0, (3) oг ɑ。x'n + ma₁x'n−1 + m²α‚x'"-2 + a after multiplying by the constant m". The rule is thus estab- lished. 2. To obtain an equation each of whose roots is equal in absolute value to a root of ƒ (x) = 0, but opposite in sign: change the signs of the odd degree terms in ƒ (x)= 0. *In carrying out this rule any missing power of x should be supplied with zero as a coefficient. ART. 98] 133 TRANSFORMATIONS OF EQUATIONS For example, the roots of the equation a4 — 2 x³ 13 x² + 14 x + 24 = 0 >>> are 2, 4, — 1, 3, and the equation with roots -2, 4, 1, 3 is x4 + 2x3 13 x² 14x+24 0. The rule follows at once from rule 1, by making m =— 1. 3. To obtain an equation each of whose roots is less by h than a corresponding root of a given equation f(x)=0: divide f(x) by x - h and indicate the remainder by R. Divide the quotient by æ R„ h, and indicate the remainder by R-1 Continue this process to n divisions. The last quotient, ao, and the remainders, R1, R2, ···, R are the coefficients of the transformed equation. The new equa- tion is then, 18.- -1 Ax!" + R₁x¹¹ + Rox¹² + ... R-1' + R₁ = 0. The division should be performed by the method of synthetic division. For example, find the equation each of whose roots is less by 2 than the roots of the equation 23 x³- 4x² - 3 x + 2 = 0. The work is as follows: 1 − 4 − 3 + 2] 212 +2 4 14 1 - 2 - 7 12 R3 - 12, +2 0 1+0-7 R2 7, 2 1+2 R₁ Alo 2- 2, 1. The required equation is 23 + 2x² - 7x - 12 = 0. x' To establish the rule, substitute a = ' + h in ɑox n + A₁xn−1 + • This gives the equation in a' 0. (1) ... =0 +α, (2). a。(x' + h)" +ɑ1(x' + h) n−¹ + ··· + a„-1(x' + h) + a„ = 0 134 [CHAP. XIII. THEORY OF EQUATIONS whose roots are less by h than those of (1). Expanding the bi- nomial powers and arranging in powers of x', we may present the result in the form 2-1 ɑqï'" + µx¹»-¹ + Aga!"-2 + ··· + 4-12' + A₁ = 0. cuốn đế t A„-12' A„ If in (3), we make x' = x • (3) h, we obtain α。(x − h)" + A1(x − h)”~¹ + A₂2(x − h)”−² + n-2 ... + A„-1(x − h) +1 n 0 (4) h. in powers of x n- n which is the same as equation (1) arranged From the form of equation (4), it follows that 4, is the re- mainder when f(x) is divided by x h; 4-1 is the remainder when the quotient of the last-named division is divided by a h; continuing this process to n divisions, A, is the last remainder, and a。 is the last quotient. That is, which establishes the rule. An A₁ = R₁ A-1 R₂-19 A₁ = R₁₂ R1, EXERCISES Obtain equations whose roots are equal to the roots of the following equa. tions multiplied by the number opposite. x2 8 1. x³ + 2x² — 7 x − 1 = 0. (5) 2. x³ 0. (5) 5 25 3. x410 x2 3x-2=0. 2 = 0. (-1) 4. x3 3 x² + 10 0. (-2) Obtain equations whose roots are equal to the roots of the following equa- tions multiplied by the smallest number which will make all the coefficients integers and that of the highest power unity. 5. x³- 2x² + x 10 = 0. x3 7. 3 x3 + 4 = 0. 3 6. 2 x¹ + 3 x³ 6x25x-1= 0. 8. x3 + x² + Ꮖ 83 = 0. 6 36 9. Obtain equations whose roots are equal in absolute value but opposite in sign to the roots of equations given in Exs. 1-4. Obtain equations whose roots are equal to the roots of the following equa- tions diminished by the number opposite. ARTS. 98, 99] DESCARTES'S RULE OF SIGNS 135 10. 2x¹ — 3x² + 4 x − 5 = 0. (2) SOLUTION: We apply transformation 3, Art. 98. By synthetic division, this gives 2 + 0 3+ 4 512 + 4+ 8 + 10 + 28 2 + 4+ 5+14 +23 R₁ = 23, + 4 + 16 + 42 2 + 8 +21 + 56 R3 = 56, + 4 + 24 2 + 12 + 45 R₂ 45, + 4 2 + 16 R₁ = 16, ao 2. Hence, 2x4 + 16 x³ + 45 x² + 56 z + 23 = 0 is the required equation. 1 11. x² 7 x + 6 = 0. (33) 12. x³ 7x+7 = 0. (1) 13. x3 27 x 36 0. (3) 14. x56x4 + 7.4 x³ +7.92 2 x² x 17.872 -0.79232 = 0. (1.2) 15. 2x4 + 16 x³ + 45 x² + 56 x + 23 = 0. (-2) 99. Descartes's rule of signs. In a polynomial arranged in descending powers of a, if two successive terms differ in sign there is said to be a variation in sign. For example, 204 4x³ + 3 x² + 4 x − 5 has 3 variations of sign as is shown more clearly by writing down the signs + -+ + Multiply this polynomial by x-1. There results 25-524+7 x3 + x²-9x+5 2:3 with 4 variations of sign. This last polynomial has one more positive zero (see Art. 95) than the first. If increasing the num- ber of positive zeros of a polynomial always increases the number of variations in sign by at least one, then the number of positive roots is never greater than the number of variations of sign. THEOREM. The number of positive roots of an equation f(x) = 0 does not exceed the number of variations of sign of f(x), nor does the number of negative roots exceed the number of variations of sign of ƒ(− x). This is Descartes's rule of signs. 136 [CHAP. XIII. THEORY OF EQUATIONS The part which relates to positive roots will be established by showing that whenever a positive root is introduced into an equation, there is added at least one variation of sign. Let f(x)=0 be an equation of degree m. It is only necessary to show that (x − r) f(x) has at least one more variation of sign than f(x). Group the terms of f(x) between consecutive variations of sign in brackets. In general, for a function of degree m, we have -1 ƒ(x) = [b。xm + b₁æm−1+ - [b₁p+1xm-p-1 + b - - ... + bxxm-p] p+ 2 xm-p-2 + +bqxm−⁹] +1 1 ± [bm-vx" +bm-v+120-1+ ... where bo, b1, b2, b are positive. +b] Multiplying this function first by x, then by — r, and adding, we obtain (x − r) f(x) = [b。xm+1 + (b₁ — bor)xm ± …..] − [(bp+1+rbp)x¹−r ± ...] + [(bq+1 + rbq) xm¬¶±···] ± [(bm-v + rVm-v-1) xv+1 ± ...] From: It will be noticed that the coefficients of the first terms in the several brackets, that is, b。, (b+1+rbp), (bq+1 +rbq), ···, are all posi- (bq+1+rbq), tive. Hence, the signs between the brackets remain unchanged. The signs within the brackets are uncertain, but however they may occur there is at least one variation between the first term of one bracket and the first term of the next bracket. Hence, as far as the terms in the brackets are concerned the number of variations remains the same or is increased. But there is added the variation caused by the term rbm whose sign differs from the sign of the last bracket. Therefore, there is at least one more variation in (x − r)f(x) than in f(x). The part of Descartes's rule which relates to negative roots follows from the fact that the roots of ƒ (— x) = 0 are equal in absolute value but opposite in sign to the roots of f(x) = 0. ARTS. 99, 100] 137 LOCATION OF ROOTS BY GRAPH EXERCISES Find the maximum number of positive and of negative roots, and any other information about the nature of the roots of variation in sign, hence, there is not more than 1. x35x-7= 0. SOLUTION: There is one one positive root. f(x) = there are no negative roots. are imaginary and one positive. 2. 3 x¹ + x² + 2 = 0. 3. x5 + 1 = 0. 4. x3 + 3x² + 1 = 0. 5. x6 + 4x² + x = 0. x³ — 5 x — 7 with no variation in sign, hence, X3 Since there are three roots of the equation, two 6. xn 1 = 0 (n odd). 7. xn 1 = 0 (n even). 8. Given that the roots of 5 x3 3x² - 4 x + 11 mine the signs of the roots. 4 x + 11 = 0 are all real, deter- 9. Show that the equation 7 x6 - 2 x² − 2 x + 4 imaginary roots. 0 has at least two 10. Show that the equation x² + 5 x³ + 4 x — 10 = 0 has six and only six imaginary roots. 100. Location of roots by graph. If the real roots of an equa- tion ƒ (x) = 0 are greater than a and less than b, these roots are said to be contained in the interval a to b along the X-axis. The number a is a lower limit and the number b is an upper limit of the interval. We are concerned with the graph of f(x) chiefly throughout an interval along the X-axis which contains the roots of ƒ (x) = : 0. To avoid the labor of plotting the graph outside of this interval, it is desirable, in evaluating f(x) for x = b by synthetic division, to know whether b is greater than any root. The following eri- terion will be found helpful. If all the sums are positive in the synthetic division by x – b (b positive), then b is greater than any root. For, a greater number than b would make the sums still greater. For example, to show that 6 is greater than any real root of x4 - 5 x³ + 3 x3 - 42 x – 50 = 0, 138 [CHAP. XIII. THEORY OF EQUATIONS we divide by x 6 by synthetic division, 1-53-42-50 6 50|6 6 +6 +54 +72 1 + 1 + 9 + 12 + 22 and we observe that a number greater than 6 would increase each sum. To find a lower limit of the negative roots of ƒ (x)=0, it is only necessary to find as above, by synthetic division, an upper limit of the positive roots of f(x)= 0. After the graph of ƒ (x) is plotted throughout an interval which contains the roots, at least the approximate values of the real roots of ƒ (x)=0 are presented geometrically. The following ΑΥ a P₁ 2 X principle aids in locating roots during the process of plotting the graph. If f(a) and f(b) have contrary signs, the equation f(x) = 0 has at least one real root between a and b. 1 2 1 For the points P₁ and P₂ (Fig. 30) which correspond to x = a and ∞ = b are on opposite sides of the X-axis, and any continuous curve connecting P₁ and P₂ crosses the X-axis at least once between Since, to every intersection of the graph with the X-axis there corresponds a real root of the equation (Art. 95), we assume this principle. a and b. FIG. 30. EXERCISES 1. Show that x3 + 8 x 70 has a real root between 0 and 1, and that the other roots are imaginary. 2. Show that 1 is a lower limit of the roots of C X4 5 x3 + 3x² 42 x 50 0. — = 3. Show that 2 is an upper limit of the roots of x4-2 x³ +3 x²-5x+1=0. Find the integral part of each real root of 4. x3 + x² 2 x 10. — 5. x³ + 2x + 5 = 0. 7. x³ + 2x 5 = 0. 8. x4 12 x3 + 12 x − 3 = 0. - 9. 8x3 36 x²+46 x 150. 6. x3 2 x 5 - 0. 10. x³-3x² - 4x + 11 = 0. 11. x³- 2 x − 5 = 0. ARTS. 100-102] 139 RATIONAL ROOTS 101. Equation in the p-form. An equation -1 x² + Þ₁x²¬¹ + P»xn−2+ ... +P n + P₂ = 0, (1) where P1, P2, …**, P₁ are any numbers, may be called the p-form of the equation of the nth degree. The general equation, n-2 A ïť" + A₁x" - 1 + α₂x²² + + a„= 0, ... (2) can clearly be reduced to the p-form by dividing its members. The p-form is more convenient for the statement of certain theorems (Art. 102) than the form (2). by do. Exercise. Reduce 6x³-3x² + 2 x − 50 to the p-form. 102. Rational roots. Any rational root of an equation f(x) = 0) in the p-form with integral coefficients is an integer and an exact divisor of P a To prove this theorem, suppose, if possible, that is a root of f(x) = 0, a b where is a fraction in its lowest terms. Then, a b n-1 α\n-2 + +P-1 ()" + (1) + (1) Multiplying (1) by b″-1, we obtain ( + P₁ = 0. (1) +P₁а"~¹ +P₂а"~28++P-1ab"-2+p,b"-1 = 0, an b or An b (P₁α"~1 + P₂"-2b+ »−² +P„-1ab”−2 +P₁-1ab" "+p',,b"−1). (2) All the terms of the right-hand member of (2) are integers, while the left-hand member is a fraction in its lowest terms. Hence, the hypothesis that is a root leads to an absurdity. α b Next, suppose that c is a root, where c is an integer. Then, c² + P₁c"−1 +P₂c²¬² + ... +Pn-1c+P₂ = 0. Transposing p₁ and dividing through by c, we obtain (3) cn−1 +P₁cn−² + ?2c”−3+ ... +Pn-1= — Pn. (4) C 140 [CHAP. XIII. THEORY OF EQUATIONS Each term of the left-hand member of (4) is an integer. Pn. Pn Hence, " is an integer; that is, c is an exact divisor of p„ С To obtain the rational roots of an equation in p-form with integral coefficients, it is only necessary to test whether the integers which are the exact divisors of p₂ satisfy the equation.* EXERCISES Find the rational roots of the following equations. 1. x³ — 9 x² + 23 x − 15 = 0. SOLUTION: By Descartes's rule of signs, this equation has no negative Hence, we need try only 1, 3, 5, and 15. By synthetic division, roots. 1 9 +23 − 15|1 +1 8 +15 — 1 − 8 + 15 + 0 The depressed equation is x2 − 8 x + 15 = (x — and 5 are the roots. 2. 108 x3 54x2 + 45 x 13 = 0. - SOLUTION: In the p-form this equation is X3 5 T2 5) (x 5) (x − 3) = 0. Hence, 1, 3, 1 x² + + ½ x − 783³½ = 0. - (1) Transform (1) into an equation whose roots are 6 times those of (1). This gives X3 x³-3x² + 15 x — 26 = 0. The rational roots of (2) divided by 6 give the rational roots of (1). Descartes's rule, (2) has no negative roots. 13, 26. Depressing the equation, (2) By Hence, we need try only 1, 2, 1 − 3 +15 — 26|1 p +1 2 + 13 1 Hence, 1 is not a root. The depressed equation x2 the only rational root of (2) 2 + 13 13 1-3+15-2612 1 — +2 2 + 26 1 +13 + 0 x + 13 and 3. x4 15 x² + 10 x + 24 = 0. 5. x³ + 3x² - 4 x 12 = 0. =0 has no rational roots. Hence, 2 is is the only rational root of (1). 4. x3 4x² + 2x 1 = 0. - 6. 2x3 + x² + 2 x + 1 = 0. * If p, is a large number, this method is too long to be practical. ARTS. 102, 103] 141 IRRATIONAL ROOTS 7. 3x3 + 8 x² + x − 2 = 0. X3 9. x4 — x³ — 8 x² - 14 x + 60 = 0. 11. 4x³- 12x² + 27 x - 19 = 0. 13. x4 - 45 x² + 40 x + 84 8. 4x³ 8x² 5 x − 1 822 +5 – 1 = 0. 10. 10 x4 + 17 x³ 16 x² + 2x=0. 4 x2 17 x + 60 = 0. 0. 12. x3 14. 8x3 13x 1 15. 24 x4 4x³ — 2 x2 + 0. 24 4 16. 324 x 54x2 + 45 x − 13 – 0. ― 2x² - 4x + 1 = 0. x5 17. x³ — 8x4 + 15 x³ + 20 x² − 76 x + 48 = 0. 103. Irrational roots. Horner's method. The irrational roots of a numerical equation can be obtained to any desired number of decimal places by a method of approximation called Horner's method. The method can be best explained by first applying it to an example. In case an equation has some rational roots, it should always be depressed by removing such roots before con- sidering irrational roots. EXAMPLE: Find the real roots of x4 — 2 x³ + 4 x² - 15 x + 14 = 0. 1. Test for rational roots as in Art. 102. It results that 2 is the only rational root. ΑΥ (1) 1 − 2 + 4 15 + 14|2 + 2 + 0 + 8-14 1+0+4 7+0 X 1 The depressed equation is x³+4x-7=0. (2) 2. Test for the interval which contains the real roots. From Descartes's rule, equation (2) has not more than one positive root, and it has no negative root. Furthermore, 2 is greater than any root (Art. 100). 0 to x 2. FIG. 31. 3. Plot x³ + 4x - 7 from x The graph (Fig. 31) shows that 1 is the first figure of the root. 4. Transform to diminish roots by 1; or graphically, change the origin to the point marked 1. The numerical work is as follows: 142 [CHAP. XIII. THEORY OF EQUATIONS 71 1 + 0 + 4 - 7 | +1 +1 + 5 1 +1 + 5 2 +1+2 1+2+7 +1 1+ 3 The first transformed equation is then 3 x₁³ +3x²+7x₁-2=0. 2 (3) This equation has a root between 0 and 1, since (2) has a root between 1 and 2. By evaluating ƒ(x) = x³ + 3x² + 7 x − 2 for successive tenths (0.0, 0.1, 0.2, ..., 0.9), we find that this function is negative when ₁ = 0.2 and positive when x₁ = 0.3. Hence, (3) has a root between 0.2 and 0.3. An approximation to this root is given by neglecting the second and third degree terms in (3) and solving 7x1 2 = 0. The root of this equation between 0 and 1 is a₁ = 0.2.... It is important to observe from the graph of ƒ (x) that the sign of the known term in each transformed equation is to be the same as that of the original equation after the rational roots have been removed. Transforming (3) into an equation whose roots are less by 0.2, we have 7 1 3 0.2 0.64 - 2 0.2 1.528 1 3.2 7.64 0.2 0.68 - 0.472 or 1 3.4 8.32 0.2 1 36 x²+3.6xy²+ 8.32 x2 -0.4720 (4) as the second transformed equation. The root of equation (4) which we seek lies between 0 and 0.1. Neglecting powers of x2 higher than the first, it appears from the equation 8.32 x2 -0.472 = 0 ART. 103] 143 IRRATIONAL ROOTS that a₂ lies between 0.05 and 0.06. That the root is in this in- terval may be tested by evaluating for a2 = 0.05 and 0.06. 3+ 3.6 x² + 8.32 x 0.472 Transforming (4) by synthetic division into an equation whose roots are less by 0.05, we obtain xz³ +3.75x3² + 8.6875 x3 — 0.046875 = 0. (5) Neglecting powers of x, higher than the first, it appears from the equation 8.6875 x3 0.046875 = 0 that x, lies between 0.005 and 0.006. Transforming (5) by synthetic division into an equation whose roots are less by 0.005, we have 3 x¸³ + 3.765x² + 8.725075 x¸. 0.003343625 — 0. - (6) The root of this equation between 0 and 0.001 can be obtained at least as far as the first figure by neglecting powers of 24 above the first. This gives X4 0.0003+. Transforming (6) into an equation whose roots are less by 0.0003, we obtain 3 2 x5³ +3.7659 x¿² + 8.72733427 x5 0.000725763623 = 0. The root of this equation between 0 and 0.0001 can be obtained at least so far as the first significant figure by neglecting powers of a above the first. This gives X5 = 0.00008+. Taking the sum of successive diminutions of the roots of (2), we obtain as the approximate value of the root sought x = 1.25538+. The preceding work of transformation may be compactly ar- ranged as follows: 144 [CHAP. XIII THEORY OF EQUATIONS 1 0 4 7 1. 1 1 5 1 1 1 1 21 627 5 2 1 3 7 2 0.2 0.2 0.64 1.528 1 3.2 7.64 - 0.472 0.2 0.68 1 3.4 8.32 0.2 1 3.6 8.32 - 0.472 0.05 0.05 0.1825 0.425125 1 3.65 8.5025 - 0.046875 0.05 0.1850 1 3.70 8.6875 0.05 1 3.75 0.005 1 3.755 8.6875 – 0.046875 0.005 0.018775 0.043531375 J 8.706275 -0.003343625 0.005 0.01880 1 3.760 8.725075 0.005 1 3.765 8.725075 – 0.003343625 0.0003 0.0003 0.00112959 1 3.7653 8.72620459 .002617861377 -0.000725763623 0.0003 0.00112968 1 3.7656 0.0003 1 3.7659 8.72733427 8.72733427 - 0.000725763623 0.00008 0.0003012784 1 0.00008 0.000698210843872 3.76598 8.7276355484 -0.000027552779128 The heavy type indicates the successive transformed equations. The process can evidently be continued to find the root to any required number of decimal places. If a root of an equation is known to be small, one important point to note is that such a root can, in general, be well estimated by dividing the known term, with its sign changed, by the coeffi- cient of the first degree term. The coefficient of the first degree term is, for this reason, sometimes called the trial divisor in ob- taining approximate roots. A still better estimate of a root can, ARTS. 103-105] 145 SUMMARY in general, be obtained by dropping terms of degree higher than the second, and solving the quadratic. When an equation has more than one irrational root, each is treated separately as we have treated the single irrational root in this example. If two roots of an equation f(x) = 0 are nearly equal, their separation may become laborious, but the separation may be accomplished by assigning values to a sufficiently near each other in plotting the graph of f(x). For example, 4 x³ 24 x² + 44 x 23 = 0, M has two roots between 2 and 3. By assigning successively the values a= 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3, in plotting the graph we find that one of these roots is between 2.2 and 2.3, while the other is between 2.8 and 2.9. 104. Negative roots. The negative roots of f(x)=0 are obtained by finding the positive roots of ƒ(−x)=0, and changing their signs. It is therefore sufficient to discuss the method of obtaining positive roots. 105. Summary. In solving a numerical equation f(x) = 0 for all its real roots, the following rules may be found helpful in systematizing the work: 1. Test for rational roots; and if any exist, depress the equation by removing the corresponding factors. 2. Determine an interval which contains the irrational roots. 3. Plot the depressed polynomial throughout this interval to locate the roots. 4. Apply Horner's method to find the irrational roots. To do this, fix the attention upon some positive root whose location is known to be between two consecutive integers. Obtain by synthetic division (Art. 89) an equation whose roots are less than those of the given equation by the smaller of these two integers. The new equa- tion has a root between 0 and 1. Locate this root between two suc- cessive tenths; and decrease the roots by the smaller of these tenths. The equation thus obtained has a root between 0 and 0.1. Locate this root between two successive hundredths, and again decrease the 146 [CHAP. XIII. THEORY OF EQUATIONS roots by the smaller of these hundredths. Continue this process to any required number of decimal places. Add together all the diminutions of the roots to obtain the re- quired root. If more than one root is contained between two consecutive integers, separate them by means of the location principle. 5. Treat negative roots in the same manner as positive roots after changing f(x) = 0 into ƒ( — x) = 0. EXERCISES AND PROBLEMS Find a positive root of each of the following equations to two decimal places. 1. x³-3x² - 2 x + 5 = 0. 2. x48x³ + 14 x² + 4 x 8 0. = 3. x5 + 12 xª + 59 x³ + 150 x² + 201 x 207 : 0. 4. x3 · 100 0. = 5. x5 1000 0. Find the rational roots, and the value of irrational real roots to two decimal places, of the following equations. * 6. x³- 3 x² - 2 x + 6 = 0. 8. x3 3 x2 4x + 13 = 0. 10. x³ + 4x² - 5 x 12. 2x4. 12 x3 + 12 x − 3 0. 20 — 0. - = 14. x4 3 x³ + 3 − 0. 7. x3 + 30 x 420 = - 0. • 9. x³ + 3 x² + 4 x + 5 = 0. 11. 3 x¹ - 2 x³- 21 x² - 4 x + 11 = 0. x3 13. x3 3 x 10. 15. x3 + 4x² + 4 x + 3 = 0. 16. An open box is made of a rectangular piece of tin 10 inches by 20 inches by cutting equal squares from the corners and turning up the sides. Find (to two decimal places) the side of a square cut out if the volume of the box is 187 cubic inches. 17. A piece of property can be bought for $7550 cash or $8000, payable in four equal annual instalments of $2000 each, the first instalment being paid at once, and the remaining instalments at the ends of 1, 2, and 3 years. What yearly rate of interest compounded annually gives the two offers equal present values? 18. A sphere of yellow pine 1 foot in diameter floating in water sinks to a depth x given by 2 x3 3x²+0.657 : 0. Find the depth to 3 significant figures. 19. A sphere of ice 1 foot in diameter floating in water sinks to a depth x given by the equation 2 x3 3x²+0.93 = 0. Find the depth to 3 significant figures. ! ART. 105] 147 EXERCISES AND PROBLEMS 20. A cork sphere 1 foot in diameter floating on water sinks to a depth x given by the equation 2 x³-3x² + 0.24 = 0. If the sphere is 2 feet in diameter, the immersed depth is given by x3 2 x³ - 6 x² + 1.92 0. Find the depths to 2 significant figures. 21. The width of the strongest beam which can be cut from a log 12 inches in diameter is given by the positive irrational root of the equation X3 x³- 144 x + 665 = 0. Find the width to 3 significant figures. 22. The speed in feet per second of a 1-inch manila rope transmitting 4 horse power, under a tension of 300 pounds on the tight side, is given by the equation v3 - 19200 v + 211200 = 0. Find the velocity to 3 significant figures. 23. The diameter of a water pipe whose length is 200 feet, and which is to discharge 100 cubic feet per second under a head of 10 feet, is given by the real root of the equation x5 - 38 x 101 = 0. Find the diameter to 3 significant figures. (Merriman and Woodward, Higher Mathematics, p. 13.) 24. The algebraic treatment of the trisection of an angle whose sine is a involves the solution of the cubic equation 4x3 = 3 x (t. a= }√2, The unknown, x, is the sine of one third the given angle. When a= find x to 3 significant figures. If 25. A vat in the form of a rectangular parallelopiped is 8 × 10 × 12 feet. the volume is increased 500 cubic feet by equal elongations of the dimensions, find elongations in feet to two decimal places. 26. In Problem 25, if the volume is increased by elongations proportional to the dimensions, find each elongation. 27. From the American Report on Wholesale Prices, Wages, and Trans- portation, for 1891, the median wage is given in dollars by of a value of x in the equation 2561/1 = αo + α1X + α2x² + A3X³ + ɑ4X4, where do = 69721928, α1 =— 6578, α2 =— 3314, a3 = 197, as == the median wage correct to mills. 128. Find 148 [CHAP. XIII. THEORY OF EQUATIONS 106. Algebraic solution of equations. In Arts. 102-105, methods are discussed by which we obtain approximately the real roots of numerical equations. We turn now to a brief consideration of equations with literal coefficients. Solving such an equation consists in obtaining an expression in terms of the coefficients which satisfies the equation. In other words, it consists in finding a formula which gives the roots in terms of the coefficients. For example, the roots of the typical quadratic are − ax² + bx + c = 0, √b² b ± √ b² − 4 ac 2 a The roots of an equation are functions of the coefficients, and it is important to inquire into the character of these functions. The solution is said to be an algebraic solution, if these functions of the coefficients involve no operations except a finite number of additions, subtractions, multiplications, divisions, and extractions of roots. The algebraic solution of an equation is often called the solution by radicals. In Arts. 107, 108, the general cubic. α¸Ã³ +α₁x² + ɑx + α; = 0, and the general biquadratic A024 + α₁2¹³ + A2x² + α3x + A4 0, are solved by radicals. The algebraic solution of the general fifth degree equation αx + α₁α + Ɑ»x³ + ɑzx² + α4x + α5 0 engaged the attention of mathematicians during the eighteenth and the first quarter of the nineteenth century. In 1826 Abel proved that the typical fifth degree equation has no algebraic solution. Since that time a branch of mathematics, known as the theory of substitution groups, has been much developed. While a treat- ment of substitution groups is beyond the scope of this book, it ARTS. 106, 107] THE CUBIC EQUATION 149 may be stated that, by means of this theory, it is shown that no typical equation ɑox² + A₁x²-1 + ··· + α-1x+Ɑ„= 0 has an algebraic solution if n exceeds 4; and necessary and suffi- cient conditions that an equation has an algebraic solution are established. 107. The cubic equation. The general cubic equation is α₁²²³ + α₁x² + A2X + Az = 0. a1 (1) (2) By making x = Y 3 Ag equation (1) is transformed into 2 ɑo. doy³+(a₂ 2 a13 Y+ ɑ1ɑ² + a3 = 0, 3 do. 27 a62 3 ao 2 3 α,α2 2 a,3 Az α z or' y³ + y + + a3 = 0, (3); 3 a62 27 a03 3 a02 ao which has no term of the second degree. Let 3 H Задаг a₁² 2 (4) 3 a² 2 2 a,3 Aj Al and G + a3. (5) 27 a3 3 a62 do Then (3) takes the form, y³ + 3 Hy + G = 0. (6) Now assume , 3 y = u³ + v³, (7) and H³ = uv. (8) From (6), (7), and (8), G = u + v. (9) Eliminating v from (8) and (9), we have and solving this quadratic in u, we find for a solution, u² + Gu — H³ = 0, (10) V H3 − G+√G² + 4 H³ И 2 From (8) and (11) we have U 2 (11) G-VG2+4H3 (12) 150 [CHAP. XIII. THEORY OF EQUATIONS The double sign before the radical in the solution of the quad- ratic in u is omitted because taking the negative sign before the radical would simply interchange the values of u and v. Since 7 ? = u$ + v³, the three values of y are : H Yı 23 Y₂ = wus Y2 H гоге wus (13) Y3 = w²u} 23 H W² u š where u is any one of the three cube roots of u, and w is a com- plex cube root of unity (Art. 85). Exercise. Test the solution by substitution of these values of y in (6). By means of (2) and (13), the roots of equation (1) are * x₁ = u} H α1 Xg = wu 1 u 23 1 3 do H а α1 " ww} Зад X3 = w²u} H ພ2,3 1 3 do (14) When the coefficients of the equation are real numbers, the numerical character of the roots depends upon the number under the radical sign in (11) and (12). When G² + 4 H³ is negative, u is a complex number. In this case, to obtain y from (7) would involve the extraction of the cube root of complex numbers. As we have no general algebraic rule for extracting such a cube root, the case in which G² +4 H³ is negative is called the irreducible case. These roots may, how- ever, be obtained by a method involving trigonometry (see Art. 85). Even when G² + 4 H³ is positive, the solution presented above is not, in general, so well adapted to obtaining real roots of numeri cal equations as the methods of Arts. 102-105. * This solution of the cubic is due to Tartaglia, but was first published by Cardan (1545). ARTS. 107, 108] THE BIQUADRATIC EQUATION 151 108. The biquadratic equation. The general biquadratic may be written in the p-form (Art. 101) as 2 a024 + a1223 + A2x² + Azx + A4 = = 0 x²+P₁x³ +P²x² + P3x + P4 = 0. Adding (mx + 1)² to both members of (1), we have (1) 2:4 + P₁20³ + (P₂+ m²)x² + (P3 + 2 mb)x + P4 + b² = (mx + b)². (2) Assume the identity, 2º¹ + P¹Ñ³ + (P²+m²)x² +(P3+2 mb)x+Ps+b²= (x²+1}x+q)². (3) Equating coefficients of like powers of x, we have 2 P₁2 P₂ + m² 4 + 29, P3 + 2 mb = P19, P₁ + b² = q². Eliminating m and b from (4), (5), and (6), we obtain 2 (p₁² + 8q4 p₂) (q² — P₁)=(P14 — P3)², (4) (5) (6) or 893-4 p₂q² +(2 P1P3-8 P₁) + 4 P2P4 P1P4 P320. (8) This is a cubic in q. cals in Art. 107, we P4)I +4 - Since the general cubic is solved by radi- may assume a value of q known. When q is known, the values of m and b are obtained from (4) and (6). From (2) and (3), we have (x² + 1 x + q)² = (mx + b)², 2 which is equivalent to the two quadratic equations (9) and P1 202 + x + q mx — b = 0, 2 P1 x² + x + q + mx + b = 0. 2 The solutions of these two quadratics give the four roots of (1). *This solution is due to Ferrari and was first published by Cardan (1545). 152 [CHAP. XIII. THEORY OF EQUATIONS EXERCISES 1. By Tartaglia's method, solve x³ — 4 x² + 6 x − 4 = 0 and verify the results by substitution. SOLUTION: Here do = 1, ai 4, a2 = 6, α3 = − 4. From (4) and (5), Art. 107, G = 34, H = 3. From (11), 10+ 6√3 U 27 やる ​1 + √3 3 From (14) the roots of the given equation are 2, 1 + i, 1 i. - Substitution for x shows that each of these numbers satisfies the equation to be solved. 2. Solve a¹ — 6 x³ + 12 x2 — 20 x 12 = 0. · (1) SOLUTION: Adding (mx + b)2 to both members of this equation gives x4 — 6 x³ + (12 + m²) x² + (2 mb — 20)x + b² 12 = (mx + b)². Assume the identity (2) X4 M 6.3+(12+ m2)2 + (2 mb 2 + - — 20) + 02 12 = (x² - 3x+q)². (3) Equating coefficients, we obtain 12 + m² 2 mb - 20 9+29, (4) 64, (5) 12 = q². Eliminating m and b from these three relations, we have the cubic b2 - 12 (6) q³ — 6 q² + 42 q — 68 = 0. (7) This cubic has a root q values of m2, b2, and mb are m² 2. From (4), (5), and (6), the corresponding 1,62 16, mb= 4. (8) From (2), (3), and (8), (x² This equation is equivalent to the two quadratic equations 3 x + 2)² = (x + 4)². (9) x2 3x+2 2 − (x + 4) = 0, and x2 3 x + 2 + x + 4 = 0. €€ (10) (11) The roots of (10) are 2 ± √б, and those of (11) are 1±i√5. These four values satisfy the given biquadratic. Solve the following equations, and verify the results by substitution. 3. x³ + 4x² + 4 x + 3 = 0, 5. 3 x32 x2 6x + 4 0. 7. x4 + x³ x2 7 x 6 = 0. 9. x3 + 12 x² + 57 x + 74 = 0. 4. 2 x³-3x² + 2 x 30. 6. 8x3 28 x² + 30 x − 9 = 0. 36x3+ 45 x²+54 x −81=0. 8. x³- 2x² + 3 x3 10. 4 x4 0. ARTS. 108-110] VARIABLE COEFFICIENTS AND ROOTS 153 of 109. Coefficients in terms of roots. Let 71, 72, ,, be the roots (1) Pn x" +P₁x"¹ + P22-2 + ... + P₁ = 0, Then, from Art. 94, -2 − 1°1)(x x² + P₁x²-¹ + P₂x² + ··· + P₁ = (x — T₁)(x — 1°2) ··· (x — 1',), = x" — (~1 + r½ + + r'„)xn−1 + (1172 + 7173 + + 1'n-1¹'n)X”−2 • — (r1r2l3 + ··· + "'n-2?"-1"'n)x−³ + ··· + (− 1)"7′17″27′3 ··· Tu • • (2) 119 by actual multiplication of the binomial factors of the second member. Equating coefficients in (2) (Art. 94, Cor. I), we have − P₁ = 21+ 12+ P1 ... +1 + "'n-17n9 + Tn-27n-1*n P² = 7172 + r1^3 + 17ο - P3 = 717273 + ... (− 1)”P„ = ''1²°27′3 ··· În· (4) That is, P₁ = sum of the roots, P2: = sum of products of roots taken two at a time, - P3 = sum of products of roots taken three at a time, (-1)p, product of the roots. If certain relations among the roots are given, the expressions (A) of the coefficients in terms of the roots may aid in solving the equation. 110. Variable coefficients and roots. are the roots of αox” + α₁ 支¹ + ɑ22n−2+ Given that T1, T2, •, In ... · +a, = 0, (1) relations (4), Art. 109, may be written in the form α1 21+ 12+ ... for yo + do = ?12 + 713 + ... + "'n-1¹µ9 ао аз Clo ུ ~1~27°3 + ?°17°27'4 + ··· + 1'n-2”n−17n9 a = 71273 • (−1)" αo (B) 154 [CHAP. XIII. THEORY OF EQUATIONS If a remains fixed, it follows from (B) that ང་ n (ཅ an -*0, if one root approaches zero ; -0 O and a-10, if two roots approach zero; a„ → 0, α„-1 → 0, a .. a₁−7+1 →→→ 0, if r roots approach zero. In certain problems of analytic geometry, it is desirable to know the character of the coefficients of (1), if some of the roots become indefinitely large. In (1), put x = = 1 This gives X1 Clo a1 Ar Cb-1 + + + + + an = 0, X₁n X1 n-1 -2 X1" X1 or, αo + α₁x1 + α₂α₁² + For our purposes, we may nite) as n-1 + A₁₂-121" - ¹ + α₂x₁n 0. (2) define x = ∞ (read, a becomes infi- = 1 when 21 →0. The conditions which make x∞ are 21 then precisely the conditions which make a₁→0. Hence, from (2), if a remains fixed, it follows that N. if a。 → 0, one root of equation (1) becomes infinite; do 0, and a₁ →0, two roots of equation (1) become infinite; if a。 → 0, do if α → 0, α₁ → 0,... a,-10, roots of equation (1) become in- do finite. 1 1 EXERCISES 1. Solve x³- 2x² - 4x + 8 x3 — 4x+8= 0, the sum of two of the roots being 4. 2. Solve 2 x3 3x² + 2x 3 = 0, the sum of two of the roots being zero. 3. The roots of x³ – 6 x² + 3x + 10 are in arithmetical progression. Find them. - 4. Solve x³- 8 x² + 5 x + 50 = 0, two of the roots being equal. 5. Obtain the roots of ax² 13 x + 1 = 0, to three decimal places, when a = 10, 1, 0.1, 0.01, and 0.001. x² 6. Obtain the roots of x2 - 13 x + a = 0, to three decimal places, when a = 10, 1, 0.1, 0.01, 0.001. * The symbol → is read "approaches." ART. 110] 155 EXERCISES 7. Determine m and b so that the quadratic 3x² - (mx + b ) 2 − 4 x + 2 = 0 b)? 4x shall have two infinite roots. 8. Determine m and b so that the cubic 3x² x3 + x(mx + b) 2 − 3 x² + 5 = 0 x³ shall have two infinite roots. 9. Determine m and b so that the quadratic 9x2 shall have two infinite roots. 16(mx + b)² = 25 2 10. Determine m and b so that 3x² + 34x(mx + b) + 11 (mx + b)² − x + 21(mx + b) = 0 shall have two infinite roots. CHAPTER XIV LOGARITHMS 111, Generalization of exponents. In Art. 6, a* is defined when x is a positive integer; and a meaning is obtained (Arts. 7–9) from the laws of exponents for a when x is any rational number. Thus, 45 = 4 · 4 · 4 · 4 · 4, and 83 is the square of the cube root of 8. But no meaning has been obtained thus far for a when x is ir- rational; for example, 4 is thus far undefined. We have, however, defined √2 as the limit of a sequence of rational num- bers 1, 1.4, 1.41, 1.414, 1.4142,... When a variable, z, taking this sequence of rational values, ap- proaches √2 as a limit, it can be proved that a (a <0) has a limit, and we define a as this limit. In accordance with this illustration, a*(a > 0), if x is irrational, is defined as the limit of a² when z approaches x.* 112. Definition of a logarithm. said to be the logarithm of y to x = loga Y. If a* − y(a >0, a ‡1), then x is the base a, and this is written The two equations and ax y X log ay (1) (2) thus mean exactly the same thing; and the terms logarithm and exponent are equivalent. We shall assume in what follows that: 1. Corresponding to any two positive numbers y and a (a ‡ 1) there exists one and only one real number x such that a³ Y. This assumption is sometimes expressed by saying that any *a* can also be defined, consistently with the laws of exponents, as the limiting value of an infinite series of positive integral powers of x (Art. 170). 156 ARTS. 111-113] PROPERTIES OF LOGARITHMS 157 positive number has one and only one logarithm, whatever positive number is the base (unity excepted). 2. The laws of exponents (Art. 6) which apply to rational expo- nents are also valid when irrational exponents are involved. EXERCISES 1. log2 8 = ? logɑ a ? loga α = ? log4 2 = ? log4 256 = ? log16 4 ? 2. Fill out the following table. BASE NUMBER LOGARITHM 3 81 4 10 160000 5 123 23 13 2 32 3 2- 32 - 5 113, Derived properties of logarithms. 1. The logarithm of a product equals the sum of the logarithms of its factors. Let then, and loga u = x and log。 v = =y, (1) (Definition of logarithm.) (Art. 6 and Art. 112, Assumption 2.) ax = u, a" = v', uv = arty. Hence, log, uv = x + Y, that is, log₁ uv = loga u + log₁v. Similarly, loga(uvw) = loga u + loga v + loga W, and so on for any number of factors. EXAMPLE: log10 255 = log10 3 + log10 5 + log10 17. 158 [CHAP. XIV. LOGARITHMS 2. The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. As above, let then, and Hence, that is, EXAMPLE: log, u = x and log, v = y, ax u, a" 2, ૫ = a*~". ย ル ​loga a y, U loga V. V ՂՆ 10g a 10% a v 25 logio = log10 625 - log10 133. 133 3. The logarithm of u is equal to v times the logarithm of u. To prove this, let Then, from (1), Hence, x log。 u or a² = U. αυπ. U" = αvx. (1) (Law of Indices, and Art. 112, 2.) log₁ u” = vx = v logu (2) EXAMPLE: logio (257) log10 257. 1 Making vn and v respectively, we have N (a) The logarithm of the nth power of a number is n times the logarithm of the number. (b) The logarithm of the real positive nth root of a number is the logarithm of the number divided by n. EXERCISES Express the logarithms of the following expressions in terms of the loga- rithms of integers. 1. * log V8 9 36 3 SOLUTION: 1/8 log log V8-log 98 log 63 (1 and 2, Art. 113.) = – 96 = = log 8 — log 9 - 3 log 6. (3, Art. 113.) 1 22 2. log 33 3. log √13 1048 253 4. log 113234 *When in a problem the same base is used throughout, it is customary not to write the base. ARTS. 113, 114] 159 COMMON LOGARITHMS Express the logarithms of the following in terms of the logarithms of prime numbers. 5. log (35) (36)²(72) 6. log (88) (75) 4 (12) 2 7. log(100)2 (20) 1 (75) 3 9. log (√2√66 √25). Given log10 2 0.3010, log10 3 = 8. log (√2V7²√6). 10. Prove that log, 1 = 0. 0.4771, log10 7 = 0.8451, find the logarithms of the following numbers to the base 10. 11. 12. 12. 30. 13. 42. 14. 420. 15. 189. 16. 900. 17. 343. 18. 343. 19. 3. 3 35 23. 294. 20. z. 24. VI. 21. 1029. 22. √504. 25. V1715. 26. $43218. 114. Common logarithms. While any positive number can be used as the base of some system of logarithms, there are two systems in general use. These are the common or Briggs's system and the natural or Naperian system. In the common system the base is 10, while in the natural system the base is a certain irra- tional number e= 2.71828 .. It may be stated that the com- mon system is adapted to numerical computation, while the natural system is adapted to analytical work.* In the following discussion of common logarithms, log x is written as an abbreviation of log10 x. Since, 10-1 = 0.1 10-2 = 0.01 10⁰ = 1 10¹ = 10 102 100 10-3 0.001 103 1000 10-40.0001 it follows that log 1 0 log 0.1 1 log 10 1 log 0.01 2 log 100 2 log 0.001 3 log 1000 = 3 log 0.0001 — 4 * The notation ln x for log, ≈ and log x for log10 x is frequently used when both kinds of logarithmns appear in the same problem. 160 [CHAP. XIV. LOGARITHMS So far as these powers of 10 are concerned, it may be observed that the logarithm of the number becomes greater as the number increases. In accordance with this observation, we may assume, if a < x < b, that For example, ΟΙ log a < log x < log b. log 100 < log 765 < log 1000, 2 < log 765 < 3. (1) When the logarithm of a number is not an integer, it may be represented at least approximately by means of decimal fractions. Thus, log 765 = 2.8837 correct to four decimal places. The integral part of a logarithm is called the characteristic and the decimal part is called the mantissa. In log 765, the charac- teristic is 2 and the mantissa is 0.8837. For convenience in constructing tables, it is desirable to select the mantissa as posi- tive even if the logarithm is a negative number. For example, log 0.3010; but since 0.30109.6990 — 10, 10, this may be written log 9.699010 with a positive mantissa. The fol- lowing illustration shows the method of writing the characteristic and mantissa : 2 3.8564 log 7185 log 718.5 - 2.8564 log 71.85 1.8564 log 7.185 0.8564 log 0.7185 9.8564 — 10 log 0.07185 = 8.8564 10 - 115. Characteristic. With our decimal system of notation, the characteristic in the case of the base 10 is very easy to deter- mine by a simple rule. Herein lies the advantage of this base. If y is a number which has n digits in the integral part, then 10″−1 = y < 10", and by Art. 114, (1), n-1 log y < n. log y = n − 1 +f, (1) Hence, where fis positive and less than 1. Hence, to find the characteristic of the common logarithm of a number which has an integral part, subtract 1 from the number of digits in the integral part. ARTS. 114-116] 161 USE OF TABLES If y represents a decimal fraction, we may move the decimal point ten places to the right, and apply the rule just stated to the integral part of the number so formed, provided we subtract 10 from the resulting logarithm. That is, У 1010 y 1010 Y1 1010' where log y = log y₁ — log 10¹º, · Yı log y₁ — 10. · The result so obtained could manifestly also be obtained by the following rule: To find the characteristic of the common logarithm of a decimal fraction, subtract from 9 the number of ciphers between the decimal point and the first significant figure. From the number so obtained subtract 10. If two numbers contain the same sequence of figures, and therefore differ only in the position of the decimal point, the one number is the product of the other and an integral power of 10, and hence, by Art. 113, the logarithms of the numbers. differ only by an integer. Thus, log 3722 = log 37.22 + log 100 log 37.22 +2. Hence, the mantissa of the common logarithm of a number is in- dependent of the position of the decimal point. In other words, the common logarithms of two numbers which contain the same sequence of figures differ only in their characteristics. Hence, tables of logarithms contain only the mantissas, and the computer must find the characteristics by the foregoing rules. 116. Use of tables. On pp. 162, 163, a "four-place" table of logarithms is given. In this table, the mantissas of the loga- rithms of all integers from 1 to 999 are recorded correct to four decimal places. "Five-place," "six-place," and "seven-place" tables are in common use, but this four-place table will serve for our present purposes. Methods by which such a table can be made will be discussed after applying the logarithms found in the table to purposes of 162 [CHAP. XIV. LOGARITHMS N 1 2 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 12 13 14 1234 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 16 17 18 19 5678✪ 1761 1790 1818 1847 1875 1903 1931 1959 1987 · 2014 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 3 222 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 36 5563 5575 5587 37 5682 38 5798 5809 5453 5465 5478 5490 5502 5514 5527 5539 5551 5599 5611 5623 5635 5647 5658 5670 5694 5705 5717 5729 5740 5752 5763 5775 5786 5821 5832 5843 5855 5866 5877 5888 5899 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 46 6628 6637 6646 6656 6665 6675 47 6721 6730 6739 6749 6758 6767 48 6812 6821 6830 6839 49 6902 6911 6920 6928 6848 6937 6590 6599 6609 6618 6684 6693 6702 6712 6776 6785 6794 6803 6857 6866 6875 6884 6893 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 54 7324 7332 7340 7348 7356 7364 7372 7380 7300 7308 7316 7388 7396 ART. 116] 163 LOGARITHMS N 1 2 3 4 10 5 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 55 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 88839 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 77777 1234 70 8451 8457 8463 8470 8176 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 73 74 8591 8597 8633 8639 8645 8651 8692 8698 8704 8710 8603 8609 8615 8621 8627 8657 8716 8663 8669 8675 8681 8686 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 7777 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 84 9243 9248 9206 9212 9217 9222 9227 9232 9238 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 86 9345 9350 9355 9299 9304 9309 9315 9320 9325 9330 9335 9340 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 6888888 9445 9450 9455 9460 9465 89 9494 9499 9504 9509 9513 9420 9425 9430 9435 9440 9469 9474 9479 9484 9489 9518 9523 9528 9533 9538 91 92 93 9685 9689 90 9542 9547 9562 9590 9595 9600 9605 9609 9638 9643 9647 9652 9552 9557 9566 9614 9657 9661 9694 9699 9703 94 9731 9736 9741 9745 9750 9754 9571 9576 9581 9586 9619 9624 9628 9633 9666 9671 9675 9680 9708 9713 9717 9722 9727 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 96 9823 9827 9832 9836 9841 9845 97 98 99 9868 9872 9877 9881 9912 9917 9921 9926 9930 9956 9961 9965 9969 9974 9886 9809 9814 9818 9850 9854 9859 9863 9890 9894 9899 9903 9908 9934 9939 9943 9948 9952 9978 9983 9987 9991 9996 164 [CHAP. XIV. LOGARITHMS arithmetical calculation. In order to use the tables we must know how to take from the tables the logarithm of a given num- ber, and how to take from the tables the number which has a given logarithm. 117. To find from the table the logarithm of a given number. EXAMPLES 1. Find the logarithm of 821. Glance down the column headed N for the first two significant figures, then at the top of the table for the third figure. In the row with 82 and the column with 1 is found 9143. Hence, log 821 = 2.9143. 2. Find the logarithm of 68.42. This number has more than three significant figures, so that its logarithm is not recorded in the table. It may, however, be obtained approximately from logarithms recorded in the table by a process of interpolation. In this process, it is assumed that to a small change in the number, there corre- sponds a change in the logarithm which is proportional to the change in the number. This assumption is called the principle of proportional parts. As in Ex. 1, we find that the mantissas of 6840 and 6850 are 8351 and 8357, respectively. The difference between these two mantissas is 6. Since 6842 is two tenths of the interval from 6840 to 6850, by the principle of proportional parts, we add to 8351, Hence, 0.2 × 6 = = 1+. log 68.42 = 1.8352. 118. To find from the table the number which corresponds to a given logarithm. EXAMPLES 1. Find the number whose logarithm is 2.4675. The mantissa 4675 is not recorded in the table, but it lies between the two adjacent mantissas 4669 and 4683 of the table. The mantissa 4669 corresponds to the number 293 and 4683 corresponds to 294. The number 4675 is of the interval from 4669 to 4683. By the principle of proportional parts, the number whose mantissa is 4675 is 2930 + 1º × 10: = 2934+. Hence, log 293.4 = 2.4675. 17 2. Find the number whose logarithm is 9.3025 — 10. From the table, (9.3025 log 0.2000 9.3010 — 10 log 0.2010 = 9.3032 10 Difference = 0.0022 10) — (9.3010 — 10)=0.0015. By the principle of proportional parts, the number is 0.2000 + 15× 0.0010 0.2007. 2 ARTS. 116-119] COMPUTATION WITH LOGARITHMS 165 EXERCISES Obtain, from the table, the common logarithms of the following. 1. 163. 4. 1.41. 7. 7.854. 2. 89. 3. 999. 5. 0.00785. 6. 6563. 8. 3.142. 9. 0.5236. 12. 0.0298. 10. 1.732. 11. 0.8665. Obtain, by means of the table, the numbers whose common logarithms are the following. 13. 2.7182. 14. 9.8532 - 10. 15. 3.1416. 16. 0.5236. 17. 7.8321 - 10. 18. 4.2631. 19. 8.5432 — 10. 20. 1.4142. 21. 0.4343. 119. Computation by means of logarithms. The application of logarithms to shorten calculations depends upon the properties of logarithms given in Art. 113. By means of logarithms labo- rious multiplications and divisions may be replaced by additions and subtractions; and involution and evolution may be replaced by multiplication and division. EXAMPLES 1. Find the value of N: 6.320 × 8.674 2.851 to four significant figures. 0.8007 log 6.320 log 8.674 = 0.9382 log (6.320) (8.674) = 1.7389 log 2.851 0.4550 log N = 1.2839 -- N = 19.23. In using logarithms, much time is saved and the liability of error is decreased by making a so-called form for all the work before using the table at all. Thus, in Example 1, the "form" is log 6.320 = log 8.674= log (6.320) (8.674)= log 2.851 = log N N 166 [CHAP. XIV. LOGARITHMS 2 2 2. Make a form for evaluating N (6.85) ½ V8.542 log 6.85 log (6.85): log 8.542 = log V8.542 = log (6.85) 8.542 log 65.27 log √65.27 log N N= √65.27 3. Evaluate N 58.61.* log 58.61 1.7680 n log V58.61 = 0.5893 n N 3.885. EXERCISES AND PROBLEMS Evaluate to four significant figures by logarithms. 2. 0.0631 × 7.208 x 0.5127. 1. (0.2386)³. 0.3384 3. 4. 2563 × (— 3.442). 8.659 714.8 x 0.5110 5. 645.3. 6. (0.9323) 5. 763.2 × 2.163 7. 986.7 8. V0.08244. V0.0001289 3 10. 9. 11. 0.62305. $8.193 × √(0.06285)3 0.9834 13. V0.7684. 3 4 15. (0.4754) (0.6782)}. 17. (3.142)(0.5236) (85%) 19. √3.1416 × (16)³. $0.0008276 12. √2 × V10 × V0.01. X 14. (0.008543). √4300 16. (1.06) 0.03296 7.962 3 18. 20. 1852 1122. — = HINT: 1852 1122 (185 + 112) (185 — 112). * When a number is negative, find its logarithm without regard to sign, writing n after a logarithm that corresponds to a negative number so as to keep the negative sign in mind. ART. 119] 167 PROBLEMS 21. 2102 1672. - 23. (-0.03574) &. 25. 12(0.52363 (-52.36) 22. V100. 24. (3) (√5) (37). 26. (V8) (19) 27. The time t of oscillation of a simple pendulum of length l feet is given in seconds by the formula π = し ​t = π 32.16 Find the time of oscillation of a pendulum 3.326 feet long. (Take 3.142.) 28. What is the weight in tons of a solid cast-iron sphere whose radius is 5.343 feet, if the weight of a cubic foot of water is 62.355 pounds and the specific gravity of cast-iron is 7.154 ? 29. Find the volume and surface of a sphere of radius 14.71. 30. The stretch of a brass wire when a weight is hung at its free end is given by the relation S = mgl πr²k" = where m is the weight applied, g 980, is the length of the wire, r is its radius, and k is a constant. Find k for the following values: m = m = 944.2 grams, 219.2 centimeters, r = 0.32 centimeter, and S 0.060 centimeter. l = = 31. Find the length of a wire which stretches 5.9 centimeters for a weight of 1826.5 grams hanging at its free end, the diameter of the wire being 0.064 centimeter, and k = 1.1 x 1012. 32. The weight P in pounds which will crush a solid cylindrical cast-iron column is given by the formula P = 98,920 d3.55 11.7' What weight will where d is the diameter in inches and the length in feet. crush a cast-iron column 6 feet long and 4.3 inches in diameter ? 33. For wrought-iron columns the crushing weight is given by P = 299,600 d3.55 12 What weight will crush a wrought-iron column of the same dimensions as that in Problem 32 ? 34. The weight W of one cubic foot of saturated steam depends upon the pressure in the boiler according to the formula W = P0.941 330.36' where P is the pressure in pounds per square inch. sure is 280 pounds per square inch? What is Wif the pres- 168 [CHAP. XIV. LOGARITHMS 35. By using the formula given in Problem 34, find the pressure in a boiler when a cubic foot of steam weighs just one pound. 36. The diameter in inches of a connecting rod depends upon the diam- eter D of the engine cylinder, the length of the connecting rod, and P the maximum steam pressure in pounds per square inch, according to Mark's formula d = 0.02758 √ D. 1. VT. What is d when D = 30, 37. For D= 10, l = formula in Problem 36, steam pressure used? 75, and P = 150 ? 60, a table in Kent's Pocket Book, based upon the gives d = 2.14 inches. What was the maximum 38. The discharge of water from a triangular weir is given by 5 8c 11² √29, 15 where c is a constant 0.592, g is the acceleration due to gravity 32.2 feet per second, and H is the water head. Find 39. Given pv1.41 Find q when H = 0.3 foot. 400 as the relation between pressure and volume of air expanding under certain conditions. Find p when v = 24. 40. The number, n, of vibrations per second made by a stretched string is given by the relation 1 Mg N 21 m where is the length of the string, M the weight used to stretch the string, m the weight of one centimeter of the string, and y = 980. Find n, when し ​= 0.00670 gram. M 6213.6 grams, l = 84.9 centimeters, and m = 41. What must be the weight per centimeter length of a wire which is 70.95 centimeters long and is stretched by a weight of 4406.5 grams, in order that it may vibrate 178 times per second ? 42. The work in foot pounds done during the adiabatic expansion of a gas from pressure p₁ to pressure p2 is k--1 W = 144 P1v1 K J 1 [1-(1)] where v is the original volume of the gas and k is a constant. Find W for k = 1.41, P1 = 3.5. 60, P2 = 15, vi 43. If $1500 is placed at 3 per cent interest, converted annually, what will it amount to in 10 years ? HINT: Inn years $1 will amount to $ (1.03)". 44. What will $10,000 amount to if left at interest for 10 years at 4 per cent: (a) converted annually? (b) converted semiannually? (c) con- verted quarterly? ARTS. 119, 120] 169 PROBLEMS 45. What sum of money left for 20 years at 5 per cent, converted annu- ally, will amount to $10,000 at the end of that time? 46. If $1 had been kept at interest 5 per cent, compounded annually, from the beginning of the Christian era to the present time, how many digits would occur in the integral part of the accumulated amount when expressed in dollars? 47. The formula y = 0.0263 11 gives the relation between y and x when x stands for the stress in kilograms per square centimeter of cross section of a hollow cast-iron tube subject to tensile stress, and y for the elongation of the tube in terms of cm. as a unit. Compute y when x = 101.8. 1 600 48. The formula y=ks*gez, where log k=5.03370116, log s=−0.003296862, log y = 0.00013205, log c = 0.04579609, gives the number living at age x in Hunter's Makehamized American Experience Table of Mortality. Find, to such a degree of accuracy as you can secure with a four-place table of loga- rithms, the number living (1) at age 10, (2) at age 30. 49. Given that one kilometer is equal to 0.6214 mile. Find the number of miles in 2489 kilometers. 50. Given that one kilometer equals 0.6214 mile, and that the area of Illinois is 56,625 square miles. Express the area of Illinois in square kilo- meters (to four significant figures). 120. Change of base. The logarithm of a number y to the base bis equal to the product of its logarithm to the base a and the loga- rithm of a to the base b. That is, Let Then, and log, y = log, y · log, a. U = logy and v = logy. तय y, b y, a" b". a - b", 21 logь a, и (1) (2) (3) (4) (5) From (2) and (6), v = u log, a. log₁ y = loga y logr a. (6) (7) EXAMPLE: 10 logo 128 log, 128 log10 2. By making y = b in (7), we obtain 1 = log, b log, a. That is, log, a (8) 1og a 170 [CHAP. XIV. LOGARITHMS The number log, a is often called the modulus of the system of base b with respect to the system of base a. In Art. 115, attention is called to the advantages of 10 for the base of a system of logarithms to be used in numerical calculations. For analytical purposes, as will appear in the calculus, it is con- venient to use natural logarithms. This system has for its base. an irrational number e 2.71828…... In the chapter on Infinite Series, there will be given a series from which this approximation to e is obtained, and another series from which the logarithm of a number to the base e can be obtained to any number of decimal places. It turns out that log, 10 = 2.3026, 1 and log10 e = 0.4343. log, 10 By (1), logio y = log, y log10 e, 0.4343 log, y. = The number logo e 0.4343 is the modulus (to four significant figures) of common logarithms with respect to natural logarithms. 1. Given log, 2 value in table, p. 162. EXERCISES 0.6931, find log10 2 and compare the result with the 2. Given log10 2 3. Given log10 3, find logs 3. HINT: By Art. 120, 0.3010, find log, 4. log, 3 • logio 3 log 10, log10 3 logio 5 4. Given log10 3, find log, 81. Find the logarithms of the following numbers, 5. 10 to the base 2. 7. 10 to the base 3. 9. 75 to the base 3. 11. 130 to the base 20. 6. 10 to the base 4. 8. 10 to the base 5. 10. 13 to the base 20. 12. 1300 to the base 20. ARTS. 120, 121] 171 EXPONENTIAL EQUATIONS 121. Graph of y = log(a > 1). A general notion of the value of the logarithm of any number can be easily fixed by reference to the graph of y = log. x. This graph is also the graph In the graph (Fig. 32) we take a = e = 2.718 · but of x a". · "" IY FIG. 32. the general form of the curve is not changed if a be given any other positive value greater than 1. If the student retains this picture, he should find it easy to keep in mind the following facts when the base is greater than unity. 1. A negative number does not have a real number for its loga- rithm. 2. The logarithm of a positive number is positive or negative. according as the number is greater than or less than 1. 3. If a approaches zero, log a decreases without limit. x 4. If a increases indefinitely, log a increases without limit. EXERCISES 1. Plot the graph of y = logo by using tables to find log10 x. 2. Plot the graph of y HINT: log, x. log5 x = logio log 10 5 3. Plot the graph of x = logɩ y. 4. Plot the graph of x = logy. 172 [CHAP. XIV. LOGARITHMS 122. Exponential and logarithmic equations. An equation which involves the unknown or unknowns in the exponents is often called an exponential equation. Thus, 2º = 16 is an expo- nential equation in a. In this simple example, the value of a can be obtained by inspection; but a table of logarithms is, in general, of value in solving exponential equations. Such equations arise in a variety of problems. For instance, at compound interest, the amount of one dollar at a nominal rate of 0.05 per annum is (see Problem 43, p. 168) A 1+ 0.05\nt =(1 0.05) dollars, in which t is the time in years and n is the number of times per year that interest is converted. We may also write A = [(1 + 0.05 n 0.05t 1+ 10.05. ጎ When n is increased beyond bound, the interest is said to be con- verted continuously. It turns out that, in this case (see Art. 170), A e0.056 where e is the base of natural logarithms. EXAMPLE: What will $1000 amount to in one year at 5% interest con- verted continuously? SOLUTION: Let S be the amount of $1000 at the end of a year, then · S = 1000 €0.05 log S = log 1000 + 0.05 log e = 3.0217, S = $1051. An equation which involves the logarithm of an expression that contains an unknown is sometimes called a logarithmic equa- tion. Thus, log10 2 x = 3 is a logarithmic equation. To solve this equation, we may write, from the definition of a logarithm, 2 x = 10³ 1000. Hence, x = 500. ART. 122] 173 PROBLEMS EXERCISES AND PROBLEMS Solve the following equations for x. 1. 5² = 10. SOLUTION: Since 5² = 10, log 10 5* = log 10 10 = 1. ≈ log10 5 = 1. 1 X log10 5 1 1.431. .6990 2. 23x 52x-1 = 45≈ 3x+1. SOLUTION: log10 23x 52x-1 = log10 45x 3x+1, 3 x log10 2 + (2 x − 1) log10 5 = 5 x log10 4 + (x + 1) log10 3 = 10x log10 2 + (x + 1) log103. Transposing and collecting terms, we have x(2 log10 5 – 7 log10 2 — log10 3) = log10 3 + log10 5. X logio 3+ log10 5 2 logio 5 - 7 log10 2 0.4771 +0.6990 log10 3 1.3980 2.1070 0.4771 0.9916. 3. 16 log10 x2. SOLUTION: From (1), 4. 117.. 16= logio x², x2 x² = 1016, x=108. 5. (0.3) 0.8. = 7. 5(x²-3x) — 25. 8. 212-2x 9261. (1) (2) (3) 6. 3x² = 532. 9. In a geometrical progression, l and r. 10. In a geometrical progression, s = r, and s. arn-1, solve for n in terms of a, 1, αγη α solve for n in terms of a, 1 Solve for x and y the following systems of equations. 11. SOLUTION: From (1) and (2), (1) 5x+v = 82, (2) 3x-1= 4. (3) (x + y) log 5 = log 82, (4) (x − y) log 3 = log 4. 174 [CHAP. XIV. LOGARITHMS Solving the linear equations (3) and (4) for x and y, we get x = log 82 2 log 5 log 4 + 2 log 3' log 82 log 4 y = 2 log 5 2 log 5 (5) (6) Complete by computing the value of (5) and (6) to three decimal places by the use of tables. 12. 2x+y=18, 3 x = 29. 13. 4*+ = 62x, log (x + 1) = log (y — 3). 14. In how many years will $1000 amount to $2000 at a nominal rate of 0.06 per annum, (1) when interest is converted annually, (2) when it is con- verted quarterly, (3) when it is converted continuously? 15. Solve for x the equation ex+e-*= y; (a) when y = 2, (b) when et y = 4. 16. If fluid friction be used to retard the motion of a flywheel making Vo revolutions per minute, the formula V Voe-kt gives the number of revo- lutions per minute, after the friction has been applied t seconds. If the con- stant k = 0.35, how long must the friction be applied to reduce the number of revolutions from 200 to 50 per minute? 17. The pressure, P, of the atmosphere in pounds per square inch, at a height of z feet, is given approximately by the relation P = Poe-kz, where Po is the pressure at sea level and k is a constant. Observations at sea level give Po 14.72, and at a height of 1122 feet, P = 14.11. What is the value of k? 18. Assuming the law in Problem 17 to hold, at what height will the pressure be half as great as at sea level? 19. If a body of temperature Ti° be surrounded by cooler air of tempera- ture To°, the body will gradually become cooler and its temperature, T°, after a certain time, say t minutes, is given by Newton's law of cooling, that is, T = To + (T1— To) e-kt, where k is a constant. In an experiment a body of temperature 55° C. was left to itself in air whose temperature was 15° C. After 11 minutes the tem- perature was found to be 25°. What is the value of k? 20. Assuming the value of k found in Problem 19, what time will elapse before the temperature of the body drops from 25° to 20° ? 21. Solve the equation log, (3 x + 1) = 2 for x. 22. Solve the equation log10 (x² - 21 x) = 2 for x. ARTS. 122, 123] CALCULATION OF LOGARITHMS 175 23. In solving an important problem in the elements of mechanics, it turns out that 1 ks + √ k²x² + 20² t loge k 20 (1) where s is the distance traversed by a moving point in time t. general, more useful to have s in terms of t than t in terms of s. express s in terms of t in equation (1). It is, in Hence, 123. Calculation of logarithms. At this point the inquiring student will naturally bring up the question as to how the loga- rithms of numbers are computed so as to make a table of logarithms. Logarithms were invented by Napier about the year 1600 and common logarithms by Briggs a little later. The invention grew out of the comparison of two series of numbers - the one in arithmetical progression and the other in geometrical progression. The following theorem lies at the foundation of the early methods of computing logarithms: If a series of numbers are in geometrical progression, their corre- sponding logarithms are in arithmetical progression. Let the numbers in geometrical progression be a, ar, ar², ar³, ..., ar”-1. Then, log a, log ar, log ar², log ar³, ..., log arn-1 are in arithmetical progression. (1) (2) In this arithmetical progression, the first term is log a, and the common difference is log r. The following example illustrates the use of this principle in calculating logarithms. Given log 10, log10 1000 = 3, the geometrical mean between 1 and 1000 is ✓1000 31.62. Then 1, 31.62, 1000 is a geometrical progression, and 0, 1.5, 3 is the corresponding arithmetical progression, so that 1.5 log10 31.62. Next, insert a geometric mean between 1 and 31.62, also between 31.62 and 1000. This gives = 1, 5.624, 31.62, 177.8, 1000 as the geometrical series, and 0, 0.75, 1.5, 2.25, 3 as the corresponding logarithms. We could next insert between any two of these numbers a geo- metrical mean, and find its logarithm. By continuing this process, we could insert means until the numbers would differ by as little 176 [CHAP. XIV. LOGARITHMS as we please. This method of calculating logarithms has the dis- advantage of giving the logarithms of numbers spaced unequally, since the numbers are in geometrical progression. Another method of obtaining logarithms, which has many ad- vantages over the one just given, is discussed briefly in Art. 171 of the chapter on Infinite Series; but its more complete treatment belongs to the calculus. CHAPTER XV PARTIAL FRACTIONS 124. Introduction. Early in the study of algebra we added together algebraic fractions and found the sum to be a single fraction whose denominator is the lowest common multiple of the denominators. Thus, 6 3 + x + 1 x+2 11 9x+15 x²+3x+2 1 1 It is often necessary to perform the inverse operation, that is, to decompose a given fraction into a sum of other fractions (called "partial" fractions) having denominators of lower degree. Thus 2 x it is easily shown that can be decomposed into + x²-1 x+1 x — 1 An algebraic fraction is said to be proper when its numerator is of lower degree than its denominator. In this chapter it is necessary to consider only proper fractions; for if the degree of the numerator is not lower than that of the denominator, the fraction may be reduced by division to the sum of an integral part and a proper fraction. Thus, 24 3x² 3 x¹ − 3 x² + 2 x 2 x = 3x² + x² - 1 22 x² - 1 We shall assume the possibility of decomposing any proper fraction whose denominator contains factors prime to each other into the partial fractions of the types А B X а (x — α)»' Cx + D x²+mx + n Ex + F (x² + mx + n) i where A, B, C, D, E, F are constants, p, q positive integers, and x² + mx + n an expression without real linear factors.* With this assumption we shall show how to decompose certain classes of fractions. * See Chrystal's Algebra, Fifth edition, Part I, Chapter VIII. 177 1.78 [CHAP. XV. PARTIAL FRACTIONS 125. CASE I. When the denominator can be resolved into factors of the first degree, all of which are real and different. EXAMPLE: Resolve 1 x + 6 x² into its simplest partial 2 X3 fractions. The sum of three fractions A B + + C x 1. x 1 + x will give a fraction whose denominator is æ 23. We, therefore, try to determine A, B, and C so that 1 − x + 6 x² x - x³ A B + + C 2 1. x 1 + x A(1-x)(1 + x) + Bx (1 + x)+ Cx (1 − x). x(1 + x)(1-x) Then, 1 − x + 6 x² = A (1 − x) (1 + x) + Bx (1 + x)+Cx (1−x). (1) The two members of (1) are equal for all values of x except pos- Hence, by Art. 94, Corollary II, In (1), making sibly for x 0, x = 1, x=-1. x= they are equal for these values. making making Therefore, x = 0, we obtain A=1; x = 1, x we obtain B = 3; 1, we obtain C 4. 1 x+6x2 1 3 4 + x x3 x 1. 1 + x The values of A, B, and C could also have been obtained by arranging the right-hand member of (1) in powers of a and equat- ing coefficients of like powers (Art. 94, Corollary I); thus, x 1−x+6x²= A + (B+ C) ≈ +(− A + B − C') x². Λ A = 1, B+C=-1, - A+B C = 6. 1 These equations when solved yield A = 1, B = 3, C = −4. In resolving a fraction into partial fractions, for every factor (x - a) occurring in the denominator there is a single partial frac- tion of the form A where A is a constant. x α Exercise. Resolve 5x+1 x²-2x-35 into partial fractions. ARTS. 125, 126] 179 LINEAR FACTORS 126. CASE II. When the denominator can be resolved into real linear factors, some of which are repeated. 6 x³- 8 x²-4x+1 EXAMPLE: Resolve into its simplest partial x²(x − 1)² fractions. The sum of four fractions A + B C D X + x² + x — 1' (x − 1)² - will give a fraction whose denominator is x²(x — 1)2; we therefore try to determine A, B, C, D so that Then, 6 x³ — 8 x² - 4 x + 1 A B x²(x-1)² = + + C D X 2.2 + x − 1 ' (x − 1)² · 6 x³ — 8 x² - 4 x + 1 = Ax(x − 1)² + B(x − 1)² + Сx²(x − 1) + Dx² =(A+C)x³ +(-2A+B-C + D)x² +(A− 2 B)x + B. Equating coefficients of like powers (Art. 94, Corollary I) we have A+ C = 6, − 2 A + B − C + D = − 8, - A-2B=-4, B=1. Solving these equations for A, B, C, D, we find A=-2, B=1, C=8, D= −5. 623 8x2 -4x+1 Hence, x²(x-1)² 2 8 LO 5 + + ! x x2 2 - 1 (x − 1)2 In this case, for every factor (x − a) which occurs times there are r partial fractions of the form A₁ > x A2 a (x — α)²³ A, (x − a)¹³ where A1, A2, …, A, are constants. xercise. Separate 3 x³ + 4 x² + 8 x + 2 into partial fractions. x(x+3)3 180 [CHAP. XV. PARTIAL FRACTIONS 127. CASE III. When the denominator contains quadratic factors which are not repeated and which cannot be separated into real linear factors. EXAMPLE: Resolve fractions. 3x2 2 (x² + x + 1)(x + 1) into a sum of partial Ax+ B Ꮯ Let + x + 1 Then, 3x2 2 (x² + x + 1)(x + 1) x² + x + 1 3x²-2=(Ax + B)(x + 1) + C(x²+x+1), =(A+C)x²+(A + B + C)x + B + C. Equating coefficients of like powers, we have whence, and A+ C = 3, A+B+C= 0, B+C=-2, A=2, B=-3, C=1; 3x² - 2 2x-3 1 + (x² + x + 1)(x+1) x² + x + 1 x+1 In this case, for every factor x² + mx + n occurring once, there is a single partial fraction of the form B are constants. Exercise. Resolve 3x3 - 4x² + 6x − 1 (x² + x + 1)(x² − x + 1) Ax + B x² + mx + n' where A and into partial fractions. 128. CASE IV. When the denominator contains quadratic factors which are repeated. EXAMPLE: Resolve x¹ + x³ − 2 x² – 5 x − 4 (x − 1)(x² + x + 1)² into partial fractions. x1+x³-2x²-5x-4 A Let + Bx + C x²+x+1 + Dx + E (x² + x + 1)²° then, (x − 1)(x² + x + 1)² X 1 x²+x³-2x²-5x-4=4(x²+ x + 1)² + (Bx+C)(x-1)(x²+x+1) +(Dx + E)(x − 1) - =(A+B)x² +(2 A+ C)µ³ +(3 A+ D)x² + (2 A - B - D+ E)x + (4 −C- E). ARTS. 127-128] 181 QUADRATIC FACTORS Equating coefficients, we have: A + B = 1, 2A+ C = 1, 3A+D=2, 2A-B-D + E = − 5, 1 A-C-E-4. Solving these equations for A, B, C, D, E, we find 4-1, B=2, C3, D= 1, E = 0. Hence, x²+x³-2x²-5x-4 x x3 1 2x + 3 + + (x − 1) (x² + x + 1)² X - - 1 x² + x + 1 (x² + x + 1)? In this case, for every factor (x² + mx + n) occurring r times, there are › partial fractions of the form, A₁x + B₁ A2x + B2 x²+mx+n' (x² + mx + n)²² A‚x + B, (x² + mx + n)r' where A1, A2, …**, A, B1, B2, …, B, are constants. 2x4 Exercise. Resolve 5x3 + 15 x² + 21 x + 16 x(x² − 2 x + 4)² into partial fractions. EXERCISES Resolve into the simplest partial fractions: 5 x + 4 x 1. 2. 3. x2 X - 6 M 2x x2 - x3 — x2 2 x 3 5x2 6 x 5 2x-5 4. 5. 6. (x − 1)³(x + 2) (x − 1)(x − 2) x² + 1 x(x − 1)(x − 2) +1 x² + 1 x² - 4x + 5 7. • 8. 9. x2 1 x(x − 1)² 2 x4 + x³ + 2 x² х 1 10. 11. x3 x 5x2+8x+11 12. 13. (x² + 1)(x + 1) (x − 3) (x − 3)(x² 3x+2) 6(2x- - 1) 14. 15. x(x+2)(x − 3) (x² - 4x + 3) (x 2) 4 4 x² + 1 16. 17. ic³ + 4x X3 1 19. 1 + 7 x x2 (1 + 3x)²(1 − 10x) 20. 18. 2x+5 x³ (x-1) (x − 1)³(x − 3) (x − 1)²(x² + 1) x + 4 ( − 1)(x2 – 5x+6) 2x² -5x+7 x²+ x - 3 182 [CHAP. XV. PARTIAL FRACTIONS 9 .2 x 5-9x 21. 22. (1 − 3 x)³(1 + x) (x + 2) (x² - 2x + 5) x + 2 x³ + 3x² Ꮖ 6 x³ + 2x² + 2x 2 -23. 24. X (x + 1)(x² −x + 1)2 X4 1 2 x³ — 8 x² - 7 x + 1 4x³- 18x2 25. 26. x4 + x3 x-1 (2 x − 3) 4 27. 3x 1 20 x2 - 2 x³ 28. x²(x + 1) 2 X4 16 X3 x³- 2x²-6x-21 5x2 4x+16 29. 30. x2 x² - 4x 5 (x-3) (x² x + 1)2 6x38x2 4x + 1 3 x³ + 19 x² + 35 x 31. 32. x²(x − 1)² (x+2)³ 3 45 + 36 x x2 x² + x + 2 33. 34. x4 - 6 x² - 27 (x-1)2(x2-x+1) x3 2 x³ + 2x² + 2 35. 36. (x² + x + 1)(x² + x + 2)² (x² + 1)² 2 17 - 11 x + 7 x² 37. 38. (1 + x + x²) (1 − x)³ 4x4 +8x³ + 6 x² + 6x + 5 (3 x + 2)(x² + 1)² CHAPTER XVI PERMUTATIONS AND COMBINATIONS 129. Introduction. Two positions are to be filled in an office one that of stenographer and the other that of messenger. There are 12 applicants for the position of stenographer, and 3 for that of messenger. In how many ways can the two positions together be filled? The position of stenographer can be filled in 12 ways, and with each of these there is a choice of 3 messengers. Hence, the two positions can be filled in 12 × 3 = 36 ways. This example illustrates the following FUNDAMENTAL PRINCIPLE. If one thing can be done in m dif- ferent ways; and if, after this is done in one of these ways, a second thing can be done in n ways, then the two together can be done in the order stated in mn ways. For, corresponding to each of m ways of doing the first thing, there are n ways of doing the second thing. In other words, there are n ways of doing the two together for each way of doing the first thing. Hence, there are in all mn ways of doing the two things together. A convenient and evident extension of the fundamental princi- ple may be stated in the following form: If one thing can be done in m₁ ways, a second in m2 ways, a third in m3 ways, and so on, the number of different ways in which they can be done when taken all together in the order stated is m¿m¿m3 130. Meaning of a permutation. Each different arrangement which can be made of all or part of a number of things is called a permutation. By the expression "number of permutations of n things taken r at a time” is meant the number of permutations consisting of r 183 184 [CHAP. XVI. PERMUTATIONS AND COMBINATIONS things each which can be formed from n different things. Thus, the permutations of the letters abc taken all at a time are abc, acb, bac, bea, cab, cba The permutations of the four letters a b c d taken three at a time are a b c b a c c a b d a b ась b c a c b a d b a a c d b c d c b d d b c ad c b d c c d b d c b a b d b a d c a d d a c a d b b d a cda d ca The special cases 131. Permutations of things all different. just considered lead us to the problem of deriving a formula for the number of permutations of n things taken at a time. The symbol P, is used to represent this number. n The number of permutations of n different things taken rat a time is " P₁ = n (n − 1) ... (n r + 1). The number „P required is the same as the number of ways of filling different positions with n different things. We may represent the n things by a1, a2, …, a, and ask how many permu- tations of letters can be formed from them. 2° For the first place there is a choice of n letters, for the second a choice of n 1, for the third of n - 2, and so on. For the rth place there is then a choice of n r+1 letters. It follows (Art. 129) that nPr = n(n − 1) ··· (n (n − r + 1). When r = n, (1) becomes (1) nPn = n(n − 1) 1)... 2. 1 = n!. (2) That is, the number of permutations of n things taken n at a time is n! 132. Permutations of n things not all different. number of permutations of the letters in the gives no new permutation to interchange the o's. Consider the word book. It Let P be the ARTS. 131, 132] 185 EXERCISES AND PROBLEMS number of permutations. If we should replace oo by dissimilar characters 0102, there would be 2! permutations of 0102 corre- sponding to each of the P permutations. But if the letters were all different the number of permutations would be 4!. Hence, 4! 4! = 2! P, 2! P, P 12. 2! This example illustrates the THEOREM. If P is the number of permutations of n things taken all at a time, of which n₁ are alike, n₂ others alike, n, others alike, and so on, then n! P= nj! ng! nz! To establish the theorem, suppose we should replace n, like things by n₁ unlike things, there would be P.n₁! permutations obtained from the original P permutations. In each of these permutations there would be n₂ things alike, and n, others alike. Similarly, replacing the n like things by n dissimilar things, we get Pning! permutations in each of which there would be ", alike. Continuing this argument, we find that the number of permutations of n things taken all at a time, when ₁ are alike, ??2 others alike, n, others alike, and so on, is given by n! P n EXERCISES AND PROBLEMS 1. How often can 6 ball players take seats together on a bench without sitting twice in the same order? 2. In how many different orders can the colors violet, indigo, blue, green, yellow, orange, and red be arranged when taken all together? 3. How many different permutations can be made of the letters of the word "stone" when taken 3 at a time? 4. Five different positions are to be filled, and there are 20 applicants each applying for any one of the positions. In how many ways can the positions be filled? 5. In how many ways can ten books be arranged on a shelf if the places of two of them are fixed ? 6. Given P₁ = 5,P3, find n. 186 [CHAP. XVI. PERMUTATIONS AND COMBINATIONS 7. How many permutations can be made of the letters of the word "Illinois"? ? Of the word "Mississippi"? 8. In how many ways may a party of 8 people take their places at a round table ? 9. How many different combinations may be struck from 8 bells if only 3 are struck at one time? 10. Four persons enter a railway carriage in which there are 6 seats. In how many ways can they take their places ? 11. Find the number of permutations of letters in the word "level." 12. How many different numbers of six figures each can be formed by permuting the figures 233455 ? 13. Write all the permutations of the letters abcd, when taken (1) two at a time, (2) four at a time. 133. Combinations. A set of things or elements without refer- ence to the order of individuals within the set is called a com- bination. Thus, abc, acb, bac, bca, cab, cba are the same combination. By the "number of combinations of n things taken r at a time" is meant the number of combinations of r individuals which can be formed from n things. Thus, the combinations of a b c taken two at a time are ab, ac, bc. N 134. Combinations of things all different. Let C, denote the number of combinations of n things taken r at a time. Then a formula can be derived for C, by establishing the relation be- tween C, and „P„. Take one combination of r things; with this r! permutations can be made. Take a second combination; with this r! permu- tations can be made. There are thus r! permutations for each combination. Hence, there are in all C, r! permutations of î „C,r things taken r at a time. That is, whence „C₁r!= P₁₂ C .c, = 2 P. r! r" ARTS. 132-136] 187 COMBINATIONS Since „P, = n (n − 1)... (πr + 1), (Art. 131) we have n(n 1) (n r + 1). n.Cr r! Multiplying numerator and denominator by (n − r)!, we get Since nCr n! r! (n − r) ! C n n(n − 1) ... (r + 1) (n − r)! n! (n − r)!r!' it follows that the number of combinations of n things taken r at a time is the same as the number taken n N r at a time. 135. Binomial coefficients. It may be noted that the formula for C, is the coefficient of the (r+1)st term of the binomial expansion (a+a)". The binomial theorem for positive integral exponents may therefore be written in the form (α + x)" = a" + C₁α-¹x + C₂α-²x²+ a +C-1αxn-¹+C₁. If we 136. Total number of combinations. The total number of com- binations of n things taken 1, 2, 3, ..., n at a time is 2º 1. write the binomial theorem as in the last section, we obtain (1 + x)” = 1 + „С₁x + „С²x² + Putting a = 1, we get ... + C₁-12-1+ „C₁₂". n 2n 1 = nC1 + nC2 + · ... N + nCn−1 +nCn• EXERCISES AND PROBLEMS 1. A woman with 10 friends to invite can have how many dinner parties with 6 guests without having the same company of 6 twice? In how many 2. A man and his wife wish to invite 4 men and 5 women to dinner, but find they can entertain only 6 guests at one dinner. ways can they invite 3 men and 3 women out of this group? 3. A man has 5 friends. In how many ways can he invite one or more of them to dinner? 4. How many different assemblages of 1000 persons can be selected from an assemblage of 1002 persons? 5. In a certain town, there are 4 aldermen to be elected. and there are 8 candidates. How many different tickets can be made up? 188 [CHAP. XVI. PERMUTATIONS AND COMBINATIONS 6. Find 20 C17; 12C9. 7. Given „C₁ = 4 210, find n. 8. Given „P, = 272, and „Cr 272, and „C, = 136, find n and r. 9. How many different sums of money can be formed with a penny, a nickel, a dime, a quarter, and a half dollar? 10. How many different sums may be formed with a penny, a nickel, a dime, a quarter, a half dollar, and a dollar ? 11. There are five letter boxes in a town. In how many ways can a person post two letters ? 12. In how many ways can 5 books be selected from a set of 11 ? 13. A committee of 6 is to be chosen from 7 Englishmen and 4 Americans. If the committee is to contain at least 2 Americans, in how many ways may the committee be chosen? 14. Prove that Cr=nCn-r• N. 15. Make use of the theorem n Cr=nCn-r to evaluate 100 C98. 16. Out of 15 consonants and 4 vowels how many words can be formed each containing 3 consonants and 2 vowels ? 17. How many straight lines can be drawn through pairs of points selected from 10 points no three of which are in the same straight line? 18. In how many ways can a pack of 52 playing cards be divided into 4 hands, the order of the hands, but not the cards in the hands, to be regarded? In how many ways may 19. There are 5 trails to the top of a mountain. a person go up and return by a different trail ? 20. In how many ways can 8 books be arranged on a shelf so that two particular books will not be together? 21. How many baseball teams of 9 men each can be chosen from 15 players of whom 8 are qualified to play in the infield only, 5 in the outfield only, and 2 in any position (battery included in infield) ? 22. How many different combinations can be formed with the following weights ? 1 decigram, 1 2-decigram, 1 3-decigram, 1 5-decigram, 1 gram, 1 2-gram, 1 3-gram, 1 5-gram, 1 10-gram, 1 20-gram, 1 30-gram, 1 50-gram. 23. On how many nights may a different guard of 4 men be posted out of 16 soldiers? On how many of these nights will any particular man be on guard ? ART. 136] 189 EXERCISES AND PROBLEMS 24. With 4 white balls, 6 black balls, and 9 red balls, how many different combinations can be formed each consisting of 1 white ball, 3 black balls, and 5 red balls? 25. A town which has 11 physicians, 13 teachers, and 8 lawyers can form how many committees each consisting of 3 physicians, 4 teachers, and 2 lawyers? 26. A company consists of 100 soldiers. In how many ways is it possible to leave 60 men to garrison a fort and to divide the remainder into two scouting parties of 20 men each? CHAPTER XVII PROBABILITY 137. Meaning of probability. If a bag contains three white and five black balls and one ball is drawn out at random, what is the probability that this ball is white? The event in question is said to happen if a white ball is drawn, and to fail if a black ball is drawn. The number of ways in which the event may happen is 3, and the total number of pos- sible ways in which it may happen and fail is 8. For this reason, is said to be the probability of drawing a white ball. This illustrates the following definition of probability: If all the happenings and failings of an event can be analyzed into r+s possible ways each of which is equally likely; and if in r of these ways the event will happen, and in s of them fail, the proba- bility that the event will happen is and the probability that it will fail is S r+s COROLLARY. 2° r + s The sum of the probability that an event will hap- pen and the probability that it will fail is 1, which is the symbol for certainty. In applying the definition of probability, the fact should not be overlooked that the ways are assumed to be "equally likely." To illustrate the need of precaution in this matter, consider the following EXAMPLE: What is the probability that a man, A, in good health will die within the next 24 hours? We might argue that the event can happen in only one way and fail in only one way, and that the probability that A will die in the next 24 hours is therefore. What is the flaw in this argument? 190 ARTS. 137-139] 191 EXPECTATION OF MONEY The expression "equally likely" indicates that we have no more reason to expect the event to take place in one way than in any other. 138. Probability derived from observation. If it be observed that an event has happened m times in n possible cases (n a large num- ber); then, in the absence of further knowledge, it is assumed that the best estimate of the probability that the event will happen on a m given occasion in question is and that confidence in this estimate increases as n increases. n Such estimates of probability are of much practical value in insurance and statistics. For example, according to the American Experience Table of Mortality, of 85,441 men living at the age of 30, the number living ten years later is 78,106. The probability that a man aged 30 will live ten years is taken to be 18106 85441 139. Expectation of money. If p is the probability that a person will win a sum of money m, we may define his expectation as pm. PROBLEMS 1. A bag contains ten times as many white balls as black balls, and one ball is to be drawn out at random. What is the probability that the ball drawn is white ? 2. Five coins are tossed. them are heads? What is the probability that exactly two of SOLUTION: Since each coin can fall in two ways, the five can fall in 25 32 ways. The two coins can be selected from the five in 5С₂ = 10 ways. Hence, the probability is 19. 3. From a bag containing 5 black and 4 white balls, 3 are drawn at random. Find the probability that 2 are black and 1 is white. 4. If from a suit of 13 cards, 2 cards are drawn, what is the probability that an ace and a king are drawn? 5. A gambler is to win $ 30 if an ace is thrown with a single die; what is the value of his expectation ? 6. According to a mortality table, it appears that of 100,000 persons at the age of ten years, only 69,804 reach the age of fifty. Find the proba- bility that a child aged ten will reach the age of fifty years. 192 [CHAP. XVII. PROBABILITY 7. From a committee of 3 sophomores, 4 juniors, and 5 seniors, a sub- committee of 4 is selected by lot. Find the probability that it will consist : (1) of 2 juniors and 2 seniors; (2) of 1 sophomore, 1 junior, and 2 seniors; (3) of 4 seniors. 8. From a bag containing 10 five-dollar bills have the privilege of drawing a bill at random. expectation? and 20 two-dollar bills, I What is the value of my 140. Events of a set are said to be independent or dependent according as the occurrence of any one of them does not or does affect the occurrence of others in the set. They are said to be mutually exclusive when the occurrence of any one of them on a particular occasion excludes the occurrence of any other on that occasion. Independent events. The probability that all of a set of independ ent events will happen on a given occasion when all of them are in question is the product of their separate probabilities. • Let P be the probability that all events of the set will happen and P1, P2, p, be their separate probabilities. It is to be proved that P-P1PP, Suppose the event corresponding to P₁ can happen in a₁ ways and fail in b₁ ways: the event corre- sponding to p₂ can happen in a ways and fail in by ways; and So on. P2 Then, P1 a₁ a₁ + b₁ , P2 ar A 2 + b 2 , • Pr a, + b, the separate events By the fundamental principle (Art. 129) all can happen together in a₁α, a, ways out of ... (α1 + b₁)(ɑ2 + b₂) ….. (α, + b,) possible ways of happening and failing. Hence, P= a1a2 ... ar (α₁ + b₁) (α₂ + b₂). (α, + b,) ... = P1 P2 ••• Pr• Dependent events. If the probability of a first event is pi, and if, after this has happened, the probability of a second event is p₂; then the probability that both events will happen in the order specified is P1 P2. The extension to any number of events is obvious. ARTS. 139, 140] 193 EVENTS Exclusive events. If the separate probabilities of r mutually ex- clusive events be P1, P2, ***, Pr, the probability that one of these events will happen on a particular occasion when all of them are in ques- tion is p₁ + P₂ + + Pr. ... This proposition may be regarded as an immediate consequence of the definition of probability for mutually exclusive events. To illustrate, the probability of throwing an ace or a deuce in single throw is clearly += }}· 1 6 1 • PROBLEMS 1. If the probability is that the age of a man selected at random from a group of men is between 20 and 25 years, and that it is between 25 and 35, what is the probability that his age is between 20 and 35 years? 2. What is the probability of throwing an ace with a single die in two trials ? 3. The probability that A will live ten years is and the probability that B will live ten years is 4. What is the probability that they will both live ten years? 4. Find the probability of drawing 2 white balls in succession from a bag containing 5 white and 6 black balls if the first ball drawn is not replaced before the second drawing is made. 5. One purse contains 9 coins consisting of 2 dimes, 3 quarters, and 4 half dollars. If one coin is drawn at random from the purse, what is the probability of its being either a quarter or a half dollar? 6. In a bag are 4 white and 6 black balls; find the chance that out of 5 drawn, 2 and only 2 are white. 7. Find the probability of throwing at least 8 in a single throw with two dice. 8. A traveler has three railroad connections to make. If the probability is that he would make any particular connection taken alone, what is the probability of his making all three connections? 9. The probability that a man of a certain age will die within 20 years is 0.2, and that his wife will die within that time is 0.15. What is the proba- bility that at the end of 20 years (1) both will be dead? living? (3) the man will be living and his wife dead? be dead and his wife living? (2) both will be (4) the man will 194 [CHAP. XVII. PROBABILITY 141. Repeated trials. 71 If p is the probability that an event will happen in any single trial, then „C.p'q" is the probability that this event will happen exactly r times in n trials, where q probability that the event will fail in any single trial. = 1 p is the For, the probability that it will happen in specified trials and fail in the remaining » › is p'q”-” (Art. 140), and v trials can be selected from n trials in C, ways. These ways being mutually exclusive, we have, by Art. 140, that the probability in question is n nC₁p'q"-". It will be observed that „C,p'qn-r is the (n − r+1)th term of the binomial expansion of (p + q)". We next inquire into the probability that an event such as is. described above happens at least r times in n trials. The event happens at least r times if it happens exactly n, n − 1, n — 2, ….., or 7 times in n trials. Hence, we have the following ... THEOREM: The probability that an event will happen at least r times in n trials is p" +C-1p"-¹q + C-2 p"-2q² + +С-r P'q"-". This expression is the first n − r +1 terms of the binomial ex- pansion of (p+q)". PROBLEMS 1. In tossing a coin, what is the probability that in six tosses (1) exactly three result in heads; (2) at least three result in heads? 2. According to the American Experience Table of Mortality, out of 100,000 persons living at the age of 10 years, 91,914 are living at the age of 21 years. Each of five boys is now 10 years old; what is the probability that exactly four of them will live to be 21? That at least four of them will live to be 21 ? 3. A's chance (probability) of winning any single game against B is 4. Find the chance of his winning at least three games out of seven. 4. In tossing ten coins, what is the probability that at least four of them will be heads? 5. Find the expectation of a man who buys a lottery ticket in a lottery of 100 tickets where there are four prizes of $100, ten of $50, and twenty of $5. 6. If, in the long run, one vessel out of every 50 is wrecked, find the probability that of 6 vessels expected (1) exactly 5 will arrive safely, (2) at least 5 will arrive safely. ART. 141] 195 PROBLEMS 7. Which is the greater, the probability of throwing at least one ace in six trials of throwing a die, or the probability of throwing at least one head of a coin in two trials ? 8. According to the American Experience Table of Mortality, out of 89,032 persons living at the age of 25 years, 26,237 will be living at the age of 75. A husband and wife are 25 each at the date of marriage; what is the probability that they will live to celebrate their golden wedding? What is the probability that at least one of them will be living 50 years after the marriage? 9. A machinist works 300 days in a year. If the probability of his meeting with an accident on any particular work day is 1000, show that the probability of his entirely escaping injury for a year is approximately 2. (Use logarithms.) 3 10. A card is to be drawn from a whist deck and replaced by a joker, and then a second card is to be drawn. What is the probability that both cards drawn will be aces ? 11. A bag contains 6 balls. A part or all of the balls are drawn. Under the condition that the ways in which balls can be drawn from 6 balls are equally likely for r 1, 2, 3, r= 1, 2, 3, ……., 6, what is the probability of drawing an even number of balls? 12: An Italian nobleman, interested in gambling, had, by continued ob- servation of a game with three dice, noticed that the sum 10 appeared more often than the sum 9. He expressed his surprise at this to Galileo and asked for an explanation. Find the probability of (1) the sum 10, (2) the sum 9, and explain the difficulty of the nobleman. 13. A and B take turns in throwing with a single die, A throwing first. The one who throws an ace first is to receive a prize of $66. What are the values of their expectations? 14. In the population of continental United States as given in the census of 1900 there were 75,994,575 persons of whom 64,763 were blind and 89,287 were deaf. What is the best estimate from these figures of the proba- bility that a person chosen at random from such a population would be (1) blind? (2) deaf? (3) both blind and deaf if the two were independent? Estimate to the nearest integer the number in the total population that would be both blind and deaf if it were correct to assume the two defects independent. CHAPTER XVIII DETERMINANTS 142. Extension of the determinant notation. Determinants of the second and third orders were used in Chapter V in the solution of systems of linear equations in two and three unknowns; and a determinant of the second order was so defined that the pair of values 1 b C₂ X = У (1) a₁ bi a₁ bi a b az bz satisfies the system of equations, provided a α₁x + b₁y = c1, A₁x + b₂y = c29 საყ α V₁ 0. Az br (2) Analogously, a determinant of the third order was so defined that the set of values dr b₁ c₁ a₁ di c₁ a₁ b₁ d dy b₂ Ca V 2 az d2 C2 A2 ხა d2 dg b3 C3 d3 az dz C3 Az bz d (3) X= " Y , C1 a ₂ br C a ₂ by co 2 2 Az bz C3 As by C2 az b x Co Az bz b3 C3 Az Cs 196 ART. 142] 197 EXTENSION satisfies the system of equations ά₁x + b₁y + c₁% = d₁, = Aqx + b₂Y + CqZ = d₂, ძვე AzX+bzY+CgZ = №39 provided аг b 2 C₂+0. (13 Vz C3 (4) The determinant notation is extended in the present chapter to the solution of systems of linear equations containing more than three unknowns, and to certain problems of elimination. It will be observed that each term (e.g. ab₂ and ɑ,b,c) in the expansions, ar br cr a₁ b az b₂ 19 (5) = а₁b₂С3 + а₂b½º1 + аžb¸Ñ½ — ɑ‚b‚¤¸ — аžb₁cз — α¸b¸с₂, (6) Az bz Cz Ay b₂ C2 13 of determinants of orders 2 and 3 respectively, consists (except for sign) of the product formed by taking one and only one ele- ment from each row and column. This fact suggests the exten- sion of determinants to represent certain expressions in n² elements by means of an array, α ₁ bi ci di ... aq b₂ cz d₂ ··· l½ ... as bz Cz dz 13 as b₁ c₁ ds... ls an V n C n Cn dn (7) where the expansion is to consist of terms which are products formed by taking one and only one element from each row and column, and where the signs of terms are to be consistent with the special cases of n = 2 and n = 3. A determinant such as (7) which has n rows and n columns is called a determinant of the nth order. The diagonal abcg called the principal diagonal. ... In is 198 [CHAP. XVIII. DETERMINANTS To fix the signs of terms in the expansion of a determinant of any order, the notion of an inversion is introduced. If, in an arrangement of positive integers, a greater precedes a less, there is said to be an inversion. Thus, in the order 12543, there are three inversions: 5 before 4, 5 before 3, 4 before 3. In 2341576, there are four inversions. When applied to any term in the expansion of a determinant such as (7), we say there is an inver- sion if the order of the subscripts presents an inversion when the letters (apart from subscripts) have the order abcd of the 7 principal diagonal. With respect to determinants of orders 2 and 3, it may be observed that the number of inversions is even when the term is positive, and that the number of inversions is odd when the term is negative. ... Consistently with these conditions, we lay down the following DEFINITION. A square array of n² elements, such as has been considered in the cases n=2 and n=3, is called a determinant of the nth order. It is an abbreviation for the algebraic sum of all the different products that can be formed by taking as factors one and only one element from each column and each row of the array, and giving to each term a positive or a negative sign according as the number of inversions of the subscripts of the term is even or odd, when the letters have the same order as in the principal diagonal. It may be added that if in any case the number of inversions in the principal diagonal is different from zero, the sign of a term is or according as the number of inversions in its subscripts + differs from the number in the principal diagonal by an even or odd number. Since the subscripts fix the signs of terms, it may appear necessary to carry subscripts along in any numerical case, but we shall derive other modes of expansion (Art. 144) which make this unnecessary. We shall, in general, use the Greek let- ter ▲ to represent a determinant. 143. Properties of determinants. The following theorems em- body the most important properties of determinants. I. The expansion of a determinant ▲ of order n contains n! terms. Since the number of terms is the same as the number of permutations of the subscripts 1, 2, 3, ..., n, the number is n! (Art. 131). * ART. 143] 199 DEFINITION II. If in a determinant ▲ corresponding rows and columns are interchanged, the expansion is unchanged. Thus, a₁ b₁ C1 A1 A2 аз A2 b2 C2 b₁ bg bз V 2 3 Az bz C3 C1 C2 C3 III. If two rows (or columns) of a determinant A are interchanged, the sign of the determinant is changed. Let us take for simplicity a determinant of the third order, but the argument used will clearly apply to any determinant. Thus, Az bz b3 C3 a2 b₂ Cz Az bz b3 C3 Az b₂ c₂ G b, c, In the first place, interchanging two adjacent rows will simply interchange two adjacent subscripts in each term of the expan- sion. This will change the sign of every term of the expansion, Consider next the effect of interchanging any two rows (or columns) separated by m intermediate rows. The lower row can be brought just below the upper one by m interchanges of adja- cent rows. To bring likewise the upper row into the original position of the lower row, m +1 further interchanges are neces- sary. Hence, interchanging the two rows in question is equiva- lent to 2m+1 interchanges of adjacent rows. Since 2m+1 is an odd number, this process changes the sign of the determinant. IV. If a determinant ▲ has two rows (or columns) identical, its value is zero. - If we interchange two rows, we obtain, by III, A. But since. the interchange of two identical rows does not alter the deter minant we have that is, or A = — A, 2A=0, A = 0. V. If all the elements of a row (or column) of ▲ are multiplied by the same number m the determinant is multiplied by m. For, one element from the column multiplied by m must enter into each term of the expansion of A. 200 [CHAP. XVIII. DETERMINANTS VI. If one row (or column) of ▲ has as elements the sum of two or more numbers, ▲ can be written as the sum of two or more deter- minants. That is, "1 a₁ + a₁' + α₁" b₁ C1 a₁ b₁ c₁ a b₁ c1 a₁" b₁ c₁ A ▲ = a₂+ a + ag" by C₂ = Ag by Cy+ a by c₂+ a ay b₂ b₂ c₂ + a2" b₂ c₂ 11 Az bz Cz ag' by C3 Az" bz C 3 As + az + Az″ bg C3 This theorem is evident for this special case, since each term in the expansion of ▲ is evidently equal to the sum of the corre- sponding terms of the three determinants. Similarly, we can prove the general case. VII. The value of any determinant ▲ is not changed if each element of any row (or column), or each element multiplied by any given number, m, be added to the corresponding element of any other row (or column). By V and VI, A2 a 3 a₁+maz Az аз α Ay Az 1 b₁ + mb3 b b z Ն, Նջ V₂ bz + m * b 2 B3 3 C1 C2 C3 C3 C2 C3 G₁ + MC3 C2 αι Ɑ 2 аз b₁ bg bg+0, by IV. C1 C2 C3 Likewise, the theorem can be proved for a determinant of any order. 144. Development by minors. If we suppress both the row and column to which any element, say C, of the determinant belongs, the unsuppressed elements form a determinant called the first minor of c, and which we shall denote by the capital letter C. Thus, in a, b, c, the minor of b₂ is Cl2 V 2 C2 аз bz из Сз C1 A3 C3 * This notation means that the determinant is multiplied by m. ART. 144] 201 DEVELOPMENT BY MINORS A determinant A may be expressed in terms of the elements C1, C2, ***, c₂ of a column (or row) and their first minors as follows: Form the product of each element such as c in the column by the corresponding minor C. Give each of the products thus formed a positive or a negative sign according as the sum of the number of the row and the number of the column containing c is even or odd, and take the algebraic sum of these results. This sum is equal to ▲. ai bi ci b2 C2 b, C1 b₁ bi c₁ Thus, a₂ by c₂ = = a1 A2 + az Vz C3 b3 V x C 2 A z V3 C3 If we can establish this theorem, we have a systematic method for expanding any determinant, since the first minors of A are again determinants which can be expressed in terms of their own minors. This process can be continued until we have the expan- sion of A. The proof of the theorem involves two steps: 12 (1) The coefficient of the leading element a, in the expansion of A is the minor, Д₁, of a₁. For, A, is a determinant of order n-1 in elements b₂, b3, ….. b„,... and its expansion therefore con- tains a term for each permutation of 2 3 4 N. As to the signs of terms, the number of inversions is not changed by pre- fixing a₁. K (2) The coefficient of any element c, in the expansion of A is its minor C with a + or a sign, according as the sum of the number of the row and the number of the column containing c is even or odd. If c is in the hth column and kth row, we can bring it to the leading position (column 1, row 1) without disturb ing the relative positions of elements not found in column h or row k. This is done by interchanging the column in which cz stands with each preceding column in turn until c, is in column 1, and the row in which c, stands with each preceding row in turn until c is in row 1. In making these changes, the sign of the determinant is, changed, h − 1 + k −1 = h+k − 2 times (Art. 143, III). Hence, if ▲' denotes this determinant with c as the leading letter, -2 ▲' = ( — 1) + k − 2 ▲ = ( − 1)*+*A. k 202 [CHAP. XVIII. DETERMINANTS k Let C' be the minor of c in A'. By (1), the sum of the terms in the expansion of A' which contain c is CC Since the minor of c in A' is the same as in A, the coefficient of c in the expansion of ▲ is (-1)+C. This establishes the second step. 1 0 1. Develop 3 1 SOLUTION: 1 1 2 1 0 2 3 4 A = =1 ∞ 3 2 1 0 1 1 1 1 EXERCISES 2 1 2 3 4 NMH1 1 2 2 1 0 - 1 1 1 k. 2 3 4 1 2 1 1 2 1 2 1 0 2 1 0+3 2 3 4 1 1 1 1 1 1 1 1 1 | 1 1 2 1 - 1 2 3 4 = 48. 2 1 1 0 2 1 2 1 3 3 1 2. Develop A 142 1 4 4 3 2 5 5 2 HINT: Subtracting column 2 from column 1, we have by VII, Art. 143, 1 1 2 1 3 1 4 A = 03 1 4 0 4 3 2 4 3 2 5 2 1 05 2 1 3. How many inversions are there in the arrangement 1 4 5 2 3 6 8 7 ? 1 2 0 1 2 7 6 5 3 4 4. Develop 1 1 2 1 1 2 3 2 1 1 7 3 5. Develop 1 1 5 3 4 3 1 4 7 5 6 1 1 1 6. Show that A α b c = (a — b) (b − c)(c — a). a2 b2 c2 HINT: When ab, two columns are identical so that A vanishes, and by the factor theorem, Art. 93, ab is a factor of A. ARTS. 145, 146] 203 SOLUTION OF EQUATIONS 1 a a² a³ 1 V b2 b3 7. Factor A into linear factors. 1 C c² C3 1 دل دل d 1 a Ն 8. Show that A = 1 a² b² = ab(a — b) (a − 1) (1 — b). 1 a3 b3 а ს C а ( a! 9. Factor A= a² b² c² 10. Factor A a , a³ f³ C3 a с 145. We shall now establish a theorem of determinants which is useful in performing the eliminations required in the solution of equations in two or more unknowns. THEOREM. In developing a determinant by minors with respect to a certain column (or row), if the elements of this column (or row) are replaced by the corresponding elements of some other column (or row), the resulting expression vanishes. For example, we have, by Art. 144, a₁ b₁ c₁ d₁ 1 A 2 by C2 do A: аз V3 C3 d3 A4 b4 C4 We are to prove that b¡A₁ — b₂A? + ¿3A3 — ¿4A4 = 0. (1) The left member of (1) is equal to the expression of the de- terminant derived from A by replacing the column of a's by the l's with corresponding subscripts. But this gives a determinant with two columns identical, which therefore vanishes (Art. 143, IV). The same method of proof can manifestly be applied to a determinant of any order. 146. Systems of linear equations containing the same number of equations as unknowns. In Chapter V, we used determinants to express the solution of simultaneous equations containing two and three unknowns. We are now in a position to make use of deter minants to solve a system of n linear equations in n unknowns. 204 [CHAP. XVIII. DETERMINANTS For simplicity of notation, take n=4, and consider the system of equations ɑ₁x + b₁y + c₁²+d₁w = k₁, AqX + b₂y + C₂² + d₂w = k₂, A3X + b¿Y + C3² + d¸w = k», α4x + b4Y + C4² + d4w = k4, (1) (2) (3) (4) to be solved for x, y, z, and w if a solution exists. It is conven- ient to call the determinant of the coefficients of the unknowns, αι a₁ b₁ c₁ A2 by C₂ de A = Свіз V z cz dz A 4 b4 C4 d4] the determinant of the system of equations. CASE I. When A÷0. ... As above, let A1, A2, ..., B1, B2, be the minors of a, a, ., respectively. Multiplying both members of (1), (2), (3), and (4) by A1, A2, A3, and 4, respectively, we obtain V1, V2, ... А₁₁ж+ А‚b‚у + Д‚ç‚½ + А‚d₂w= — — -Аα- Abыy Acaz AdwДk AzɑzX + AzhzY + A3€32 + A3dzw Ak₁, (5) (6) Agk3, (7) (8) - Â4Ɑ4X — Â4b4у — А4€42 — ¸øw — — Akş · Adding (5), (6), (7), (8), we obtain A for the coefficient of x (Art. 144), and zero for coefficients of the other unknowns (Art. 145). That is, Similarly, and Ak₁ A. Ak₁₂+ Agkg Ask4. ▲⋅ y = B₁k₁ + Byk - Bhg + B₁ks, A.2 A⋅ w = - (9) (10) 49 (11) (12) C₁k₁ — C₂ką + Czliz — C4K49 Ciki D₁₁ + Dyk - Dylez + D44. If, in A, we replace the right-hand member of (9). a's by k's and expand, we have the Similarly, replacing the b's, c's, and d's respectively by k's, we have the right-hand members of (10), (11), and (12). It follows that ART. 146] SOLUTION OF EQUATIONS 205 k₁ b₁ c₁ dı a L Ką by c₂ dy bz cz dz аз K3 K s V ± C+ d4 A 4 " У X A α₁ b₁ k₁d, di k₁ c₁ dı ką c₂ dą k3 C3 d3 K₁ CĄ dĄ A 4 b2 az b₂ ky d₂ V K3 dз Cl3 V 3 C2 C3 だって ​kiz аз ая A s b + ks ds a4 V₁ C4 Tit го 2 A is a solution, and the only solution, of (1), (2), (3), (4). The following rule may then be applied to obtain the solution of any system of n linear equations containing n unknowns when A, the determinant of the system, is not zero: Any unknown is equal to a fraction whose denominator is the determinant of the system, and whose numerator is the determinant formed from the determinant of the system by substituting for the coefficients of the unknown sought the corresponding known terms with that sign attached to each known term which it has when on the side of the equation opposite the unknowns. CASE II. When A 0. If a solution exists when A 0, it cannot take the preceding form, since division by zero is excluded from algebraic operations. While the theory becomes too complicated in this case to be presented in full here, certain particular cases may well be considered. As a rule (subject to certain exceptions), a system of equations has no solution when ▲ = 0. For example, the system A 3x+4y=5, 6x+8y=9 has no solution. Likewise the system x+y-z=5, 4x+y-2z=9, 5x+2y-3z=1 has no solution. 206 [CHAP. XVIII. DETERMINANTS A system may, however, have an infinite number of solutions when A=0. For instance, the equations − x + y 2 =0, 4x+y=2z=0, 5x+2y-3%=0 z (13) (14) (15) constitute such a system. These equations are manifestly satis- =z=0. To obtain other solutions solve (13) and fied by xy: (14) for x and y in terms of z. This gives x = }} %, y = z z. (16) These values of x and y satisfy (15) as well as (13) and (14). Hence any value assigned to z with the corresponding values x and y obtained from xz, y=z satisfies (13), (14), and (15). Since z may have any value, there is an infinite number of solutions of the system in question. Systems with an infinite number of solutions may be more generally illustrated by the homogeneous * equations A₁x + b₁y + c₁2 = 0), (17) ax + by + c₂² = 0, (18) A3X+b₂Y + C3² = 0, (19) when A: (12 b₂ c₂ =0, (20) A3 b3 C3 but some minor of A is not zero, say 0. (21) A2 Az b₂ To prove that (17), (18), (19) have an infinite number of solu- tions, substitute in (19) the values C₁z b₁ a1 C₁2 Co2 by A2 — C2Z XC У α1 b₁ α1 b₁ Alis ხა A₂ b2 * A homogeneous equation is one in which all the terms are of the same degree in the unknowns. ARTS. 146, 147] 207 SOLUTION OF EQUATIONS which satisfy (17) and (18) when condition (21) is fulfilled. This substitution gives, after clearing of fractions, ZA3 Cr Vi C2 C₂ by C1 a b zb3 + 203 C, b, c, = 2 A2 V₂ C2 аг Ɑ 2 V₂ Az bz C3 Hence, 13 which, by (20), vanishes whatever value be assigned to z. z can take an infinite number of values, each of which with the corresponding x and y satisfies (17), (18), and (19). 147. Systems of equations containing more unknowns than equa- tions. Consider first the single equation 3x+5y-60 (1) with two unknowns. It is manifest from our work on loci of equations (Art. 38) that there are an infinite number of pairs of values of x and y which satisfy this equation. Consider next the two equations, 3x-4y-2z+1=0, 4x+3y-2-6=0 with three unknowns. We may solve (2) and (3) for x and y in terms of z. gives X 2z-1 · 4. 2+6 3 10 z+21, CO 4 25 4 3 3 2z-1 4 У 2+6 5 z+22 3 4 25 4 3 (2) (3) This (4) (5) Any value assigned to z and the corresponding x and y obtained from (4) and (5) satisfy (2) and (3). Hence, the system has an infinite number of solutions. The main point to be brought out by these illustrations is that, in general, from n equations containing more than n unknowns, we may solve (Art. 146) for some selected of the unknowns in 208 [CHAP. XVIII. DETERMINANTS terms of the remaining unknowns. We are then at liberty to assign any values to these remaining unknowns, and thus obtain an infinite number of solutions. The problem in the exceptional cases in which it is impossible to solve for a selected set of n un- knowns is too complicated to be treated here. 148. Systems of equations containing fewer unknowns than equations. Consider the equations α₁x + b₁y + c₁ = 0, A₂x + b₂y + c₂ = 0, Azx+b3y+C3=0. (1) (2) (3) In order that these three equations may be consistent,* it is neces- sary that X = آن آن bi b У а₁ b₁ α 2 ხა | ая ar C1 A C₂ a₁ b₁ +0, ar br az b₂ az b₂ which satisfy (1) and (2) shall also satisfy (3). This requires that b₁ а1 C Co სა A.2 C₂ 2 V3 + C₁₂ = 0. a1 V₁ A₂ ხ.ა Az b₂ ხა cr br Clearing of fractions, and interchanging columns in we C2 ba obtain by ci C₁ a₁ b₁ аз b3 + C3 = 0, b2 C2 Az Cz d₂ b₂ αι δι A₂ Ag b₂ C₂ =0, (Art. 144) (4) or, As bs Vz C3 * Two or more equations are consistent (Art. 39) when they have a com- mon solution. ARTS. 148, 149] 209 SOLUTION OF EQUATIONS as a condition to be satisfied in order that equations (1), (2), and (3) be consistent. Stated in words, in order that three linear equations in two unknowns have a common solution, it is neces- sary that the determinant formed of the coefficients of the un- knowns and of the known terms vanish. The method used for three linear equations in two unknowns can clearly be extended to any number n of linear equations in N -1 unknowns. It results that the determinant formed of the coefficients of the unknowns and of the known terms must vanish in order that the n equations in n-1 unknowns have a common solution. While the vanishing of this determinant is a necessary condi- tion for the existence of a common root, it is not a sufficient con- dition as is shown by the following example. Take the system of equations x+y-4=0, 2x+2y+5=0, xy-6=0. (5) (6) (7) Here, 1 1 - 4 2 2 5 = 0, 1 1 -61 but any two of the equations are inconsistent. This condition is In establishing the above necessary condition, we assumed that two of the equations have a solution. satisfied by no two of equations (5), (6), (7). 149. Common roots of quadratic and higher degree equations in one unknown. Consider first the system, ax+b=0, a 0, a'x² + b'x + c'=0, a' 0, (1) (2) consisting of one linear and one quadratic equation. In order that (1) and (2) have a common root, it is necessary and suffi- cient that the solution x= isfy (2). This requires that b α which satisfies (1) shall also sat- 210 [CHAP. XVIII. DETERMINANTS or a'b² b'b +c' = 0, a² (3) α (4) a'b² — abb' + a²c' = 0. The relation (4) among the coefficients is the condition that (1) and (2) have a common root. The left-hand member of (4) may be put into determinant form as follows: Multiply (1) by 2, and the resulting equation in combination with (1) and (2) gives the system ax+b=0, ax² + bx = 0, a'x²+ b'x + c'=0, (1) (5) (2) which should be thought of as linear equations in two unknowns, From Art. 148, it is necessary that a and x². O a b al V 0 = 0, a' b' c' (6) But (6) is in order that (1), (2), and (5) have a common root. merely (4) written in determinant form, as can be easily verified. Consider next the equations ax² + bx + c = 0, a 0, a'a¨³ +b'x² + c'x + d' = 0, a'+0. (7) (8) If we multiply (7) by a and by a², and (8) by x, to form additional equations, we obtain ax² + bx + c = 0, ax² + bx² + cx = 0, ax¹ + bx³ + cx² =·0, a'x³ + b'x² + c'x + d' = 0, a'x¹ + b'x³ + c'x² + d'x = 0, which can be treated as linear equations in four unknowns, x, x², x³, and x¹. From Art. 148, it is necessary that ART. 149] COMMON ROOTS OF EQUATIONS 211 0 0 a b c 0 αυ c 0 a b c 0 0 = 0, (9) O a' b' c' d' a' b' c' d' 0 in order that (7) and (8) have a common root. Moreover, when condition (9) is satisfied, (7) and (8) have a common root, but the proof* is beyond the scope of this book. A relation such as (6) or (9) which results from eliminating the unknowns from systems of equations is often called the eliminant of the equations. The method of elimination employed in the last two examples is called Sylvester's method of elimination. It consists in form- ing from the given equations additional equations by multiplying the given equations by successive powers of x until we have one more equation than powers of x. These powers of a are then treated as distinct unknowns and the eliminant is obtained as in Art. 148. EXERCISES Solve by using determinants. 1. 4x-3y = 5, 8x + y = 17. 1 1 1 3. 1, i Y 2 1 1 1 + X Y 2 2. 3x+4y 2 z = 5, 40 - 3y+ 8 z = - 4, 2x+8 y 32-5=0. 4. 3x + 2y + 4 z 5x+y 20 - 13 = 0, 2 + 2 w 9 = 0, 7z+3 w- 140, 2 x + 3 y − 7 z + 3 w 4 x − 4 y + 3 z 5 w - 4 = 0. - 1 1 1 + + = 1. 4 y 22 4 x 5. Find a value of k such that kx-3y-5= 0, 8x + y = 17 = 0, kx+2y-10 = 0 are consistent equations. Can k take more than one value? * For proof, see Bocher's Introduction to Higher Algebra, pp. 200-202. 212 [CHAP. XVIII. DETERMINANTS 6. Discuss the number of values of x, y, z which satisfy x + 3y 2 = 0, 2y+z 0, 5x + y + 2 z = 0, and find the ratios x y z of corresponding values (apart from x = y=z=0). 7. Eliminate k from the equations k²x - 2 kx² + 1 = 0, by Sylvester's method. x² + k − 3 kx 0 8. Find the eliminant of ax² + bx + 9. Are the equations 2 + 3x + 2 sistent ? 10. Determine b so that x2 c = 0 and 2³ = 1. 0 and 2+8x² + 9 x + 2 = 0 con- 3x²- 8 x 3 = 0, − x³- bx² x p 6 have a common root. ai bi 11. By Art. 148, it is shown that O is a necessary condition az be that the two equations а₁x + b₁ = 0, 1x (α10) α 2 x + bx = 0 (α20) be consistent. Show that this condition is also sufficient for this special case. CHAPTER XIX LIMITS 150. Definition. If a is a constant and x is a variable which assumes in order a given sequence of values in such a manner that | a x | * becomes and remains less than any assigned number d (d>0), then x is said to approach a as a limit. We have had many illustrations of limits in elementary mathe- matics. Thus, in geometry the area of a circle is considered as the limiting value of the area of the inscribed regular polygon as the number of sides is indefinitely increased. Here the terms of the "given sequence of values" are the areas of the inscribed polygons as the number of sides is increased. Again, as we annex 3's to the decimal .3333 its value runs through the sequence of numbers .3, .33, .333, etc., which can be made to approach as near to as we please. In the geometrical pro- gression 1 + 1 + 1 + 1 + ···, .... 8 S₁, the sum of the first n terms, runs through the sequence 1, 4, 7, 15, ···, and approaches the limiting value 2. The essence of the definition of a limit lies in the words "be- comes and remains less." For example, if a runs through the sequence of values,-,,-, †, − 4, ···, 3 3, the difference |1 — a❘ takes on the values 1 4 1 5 į, į, į, j, k, f ... }, 4 2 and becomes less than any assigned number but it does not re- main so. limit. In this case we cannot say that x approaches 1 as a To indicate that a approaches a as a limit, we use the notation xa, or lim x = a. *The notation | a x means the absolute value of a x. (Cf. footnote, p. 87.) 213 214 [CHAP. XIX. LIMITS 151. Infinitesimals. A very important class of variables which are assumed to run through a sequence of values consists of those which have the limit zero. They are called infinitesimals. The area between a circle and the inscribed regular polygon as the number of sides increases, the weight of the air in the re- ceiver of a perfectly working air pump, and the difference 2-S where S is the sum of the first n terms of the series 1+ 2+ +···, are examples of infinitesimals. n THEOREM. If u→ 0 and v→0, and X and Y are always nu- merically less than some positive constant k, then Xu + Yv→0). In other words, if u and v are infinitesimals, then Xu + Yv is an infinitesimal. Let d be any positive number however small. Since lim u = 0, and lim v = 0, │u| and | v | will ultimately become and remain less For these values of u and v, we have than d 2 k | Xu│< |X|d 2 k Yd | Yv│< 2 k |Xu|+| Yv│< (|X|+| Y|) d¸ 2 k whence, By hypothesis, hence, |X| + | Y│< 2 k, | Xu |+|Yv| nwhere d 1 d 1 a is any number greater than 1. Then 2 1 3+ x x2 2 1 and since x2' x' x² are infinitesimals by Theorem I, Art. 154, we 1 have 3 ∞ The symbols, for the limit of the fraction. are called indeterminate forms. Among • other such forms which may arise arè 0 ∞ and ∞ ∞, but the expressions which give rise to these forms may be reduced to the form o as shown in the following examples. 0 220 [CHAP. XIX. LIMITS 1 x 1 takes the form 0.∞ when x = 1. For EXAMPLE 1: (x² + x − 2) · any other value of x we may write 1 x² + x 2 (x² + x − 2) · X 1 х 1 Hence, lim x1 1 lim (x² + x − 2). - х 1 X 1 EXAMPLE 2: x² + x 6 - (22 + 2 = 2) == x1 X 1 Kontak takes the form ∞∞ x² x = 3. X2 9 x(x² - 9) For any other value of x, we may write x 1 x² + x 6 2(x 3) 2 x2 9 x(x² — 9) x(x² — 9) x(x+3) Hence, lim x x→3 1 x² + x 6 lim 2 = X2 — 9 x(x² - 9) x→3x(x+3) 9 when x = 3. EXERCISES What values should be given to the following expressions in order to make them continuous for values of x indicated? X3 1 X2 22 – 5x +6 x 6 1. when x 1. 2. when x = 2. X - 1 x2 4 x 3 3. when x = 3. 4. 1 X5 when x = 1. √x √3 1 x2 x3 + as 1 5. when x = a. 6. (x² — 4) · when x = 2. x2 a² X 1 2 7. when x=0. 8. X x(x + 2) x2 (x − 1)² - - 2 2x 1 when x = 1. (x − 1)2 x² + x + 1 9. when x=- - 1. x + 1 As x increases without limit find the limits of the following fractions. 10. x + 1. XC x 11. x² + 1 x³-3x х 12. 2x3+5x2 CHAPTER XX INFINITE SERIES 157. Definition. Let u₁, u2, ., Un be ... any unending sequence of real numbers positive or negative. The expression ... гиз + глз + гиз + + Un + Un + 1 + ··, when the terms are formed according to some law of succession, is called an infinite series. In the discussion of geometrical and harmonical progressions. we have met such series, for example, 1 + 1 + 1 + 3 + ···, and 1 + 1 + 1 + 1+ 158. Convergence and divergence. In the series U1 + U2 + U3 + ... + Un let S, represent the sum of the first n terms, that is, S1 2112 S₂ = U1 + U2, S3 = U₁ + U² + U3, S₂ = U₁ + U2 + 1 For example, in the series. 1 ... + 2119 1 + } + { + } + ··· given in Art. 157, we have S₁ = 1, S₂ = 1 + 1 = 3/3, S3 √3 = 1 + 1 + 1 = 4, 7 1 1 1 1 S₁ = 1 +=+=+ ... + =2 - 2 4 2n-1 2n-1 221 222 [CHAP. XX. INFINITE SERIES In the series 1+ 2+ 3+ 4+ we have S₁ = 1, S₂ = 1+2=3, S3 = 1 + 2 + 3 = 6, S N =1+2+3+ N ... + n 2 5 (1+n). we have In the series 1 − 1+1-1+1 n - S₁ = 1, S₂ = 0, S2 S3 = 1, S₁ = 1 or 0. 11. These three examples illustrate three cases which may occur. I. S approaches a limit as n increases without limit. In the first example above, S, is never greater than 2, no matter how large a number n represents and approaches 2 as a limit, when n increases without limit. II. S is numerically larger than any assigned number for a sufficiently large value of n. This case is illustrated in the second problem. III. S remains finite but does not approach a limit as n in- creases without limit. This case is illustrated in the third series, where S may have either of the values 0 or 1, according as n is even or odd. Series which come under Case I are called convergent series and are by far the most important. Series which are included in Cases II and III are called divergent series. We have then the DEFINITION. When in an infinite series the sum of the first n terms approaches a limit as n increases without limit, the series is said to be convergent. Otherwise it is divergent. The limit of the sum of n terms of a convergent series, written lim S, is often called the sum of the series. In connection *The word " * It is not " is here used in a purely conventional sense. sum to be understood as the sum of an infinite number of terms, but as the limit of the sum of n terms as n increases without limit. * 1 ARTS. 158, 159] FUNDAMENTAL ASSUMPTION 223 with convergent series we shall also use the expression "limiting value of the series" to mean lim S. Many important mathematical investigations and the solution of many practical problems require the use of infinite series and depend upon the question of convergence. The central problem of this chapter will be the problem of deciding whether a given series is convergent or divergent. It will be found convenient to take up first those series in which the terms are positive numbers. SERIES WITH POSITIVE TERMS 159. Fundamental assumption. An infinite series of positive terms is convergent if S is always less than some definite number, however great n may be.* 71 Let K be a number such that S,< K for all values of n. 77. Since the series contains positive terms only, S, is a variable which in- creases as n increases. Since it can never attain so great a value as K, we assume that there is some number less than K which S approaches as a limit. n To illustrate this assumption, consider the series O 1 1 1 1+ + 十二​十 ​22 33 44 S3 ㅜㅜ ​S1 S2 S4 2 FIG. 33. I and take points on the line OX (Fig. 33), to represent S1, S2, S3, *** so that the measure of OS1, is S1, of OS2, is S2, etc. S₁ = 1, 1 S₂ = 1 + 1.2500, 22 1 1 S3 = 1 + + 1.2870, 22 33 1 1 S₁ = 1 + + + = 1.2909, 22 33 44 * For proof see Pierpont's Theory of Functions of a Real Variable, Art. 109. Fine's College Algebra, p. 59, Art. 192. 224 [CHAP. XIX. INFINITE SERIES We can show that the sum of n terms of this series is less than 2. (See Art. 160, Example 1.) Hence, according to the assumption of this section, there is some point not farther to the right than 2 which S, approaches as a limit when n→∞. An analogous assumption exists for a series all of whose terms are negative. An infinite series of negative terms is convergent, if S₁ is always algebraically greater than some definite number, however great ʼn may be. EXERCISES 1. It can be shown that the sum of the series, 1 + + + + + + 1 1 1 1 1 ! 2! 3! 4! is always less than 3. by means of this series. 1 (n + 1)! Illustrate the assumption of this section graphically 2. Illustrate graphically the assumption for a series of negative terms by means of the series, - 1 1 1 1 22 33 44 160. Comparison test for convergence. Consider the series of positive terms U1 + Uq + Uz + ... + W₂+ If from some term on, the terms of this series are equal to or less than the corresponding terms of a known convergent series, V₁ + V2 + V z + ... +vn +.. of positive terms, then the u series is convergent. Let and Sn U₁ + U₂+ ... + Wπ S₁₂ = v₁ + v₂ + ... + Un° Suppose that after the kth term Uk+1Z Vx+19 Uk + 2 Ž Vk+29 ***, Un Ž Vn, ***, then, if and we have or Sk ... S₁₂ = V1 + V2 + ··· + 1½³ S - S S - St k Sn ≤ En' - Sk' + Sk < lim S,' — Sk' + S'‰· * For meaning of 1 !, 2, 3 !, etc., see Art. 59. Sx = U₁ + Uz + ... ··· + U xr Ика Vk9 ARTS. 159, 160] 225 COMPARISON TEST Since by hypothesis lim S,' exists, it follows that S, is always less than a definite number, and by the assumption of Art. 159 the series + U₂ + Uz + ... EXAMPLE 1: Prove the series to be convergent. +un+ ... 1 1 1 2+1+ + + + 22 33 44 ... + is convergent. 1 (n − 1)n−1 +. SOLUTION: For purposes of comparison take as the v series the geometri- cal progression 1 + 1 1 2n-1 + + + + 2 22 which we have shown to be convergent (Art. 157). Write the given series under the comparison series : 1 1 1 1 + + + + 2 22 + + 23 27-1 1 ... + + 22 33 (n − 1)n−1 1 1 2+1+ + + After the third term, each term in the second series is less than the corre- sponding term just above it. That this is true for every term after the third is shown by examining the two nth terms. If n>3, then 1 1 < (n − 1)n-1 2n-1 Beginning with the fourth term, the sum of n terms of the first series is Hence, the sum of the second series can never exceed always less than 1. 1 1 2+1+ + 3. 22 4 In comparing two series it is not sufficient to compare a few terms at the beginning of the series. The nth terms should be com- pared. EXAMPLE 2: Test for convergence the series 3.33 2.32 3 1 1 1 + 1 1 + +1+ + 1 + + 3 2.32 3.33 If we neglect the first three terms of this series and prove the remainder to be convergent, the given series must converge. For, if the series beginning with the fourth term has a sum, the sum of the entire series will be the sum of terms after the third plus 3.33 +232 + 3 = 102. Beginning then with the fourth term and comparing with the series, 1 + } + { + (Art. 73), we have which is known to converge to 1 1 1 1 1 + 1+ + + it ... + + 32 33 37-1 and 1 + 13 1 1 1 + + + ... + 2.32 3.33 - (n − 1).3n-1 + 226 [CHAP. XX. INFINITE SERIES Each term in the second row is equal to or less than the corresponding term in the first. Hence, the second series converges to some number not greater than and the sum of the entire series in question is not greater than 103.5. In testing for convergence it is often convenient to omit a finite number of terms as in the above example. That this is always justified is shown by the following THEOREM. The convergence of a series is not affected by neglect- ing a finite number of terms. For the sum of the terms neglected is a definite number which added to the sum of the new series gives a definite number for the sum of the entire series. EXERCISES Test the following series for convergence : 1 1. 1+ 1 1 + + + 2.2 3.22 4.28 1 1 2. 1+ + + + 2 32 43 1 3. + 1 1 1 + + + 1.2 2.3 3.4 4.5 SOLUTION: Write S, in the form Sn 2 3 8 = (1 − 1 ) + ( − 1 ) + ( − 1 ) + + ( 11 ) = 1 - 2 1 1 • n + n + 1 lim Hence, 4. 1+ 5. 1++ + + 个 ​n∞ Sn 1 1 + + 3! 4! S₂ = 1, + 2 ! 1 1 1 22 42 1 1 1 6. 1++ + + 23 33 43 1 1 7. 1 + + 2P 3P T X2 8. + + 1.2 2.3 3.4 1 1 9. 1+ + +, where p≥2. X3 +, where x < 1. where p > 1. 2p 3P ARTS. 160, 161] 227 COMPARISON TEST SOLUTION: Write down the inequalities, 1 1 2 1 + 2P 3r 2p 2p-1' 1 1 1 4 1 + + + 4P 5P 6p 7P 4p 4p-1 1 1 1 8 1 + + ... 8p Эр + < 15P 8p 8p-1 Add the members of the inequalities, thus 1 1 1 1 1 1 + + + ... + < 2p-1 + + 4p-1 8x-1 2p 3r 4p The right-hand side of this inequality is a geometrical progression whose 1 ratio is 2p-1 which is <1 when p>1. Hence, the series is convergent. This is a useful series for testing other series. 10. 1+ x x2 + X3 + where x1. 21 31 41 22 32 161. Comparison test for divergence. tive terms, ... Given the series of posi- Ալ — ՛ջ — + u₁ + .... Uπ If from some term on, the terms of the series are equal to or greater than the corresponding terms of a known divergent series of positive terms, V₁ + V2 + ... +~+ then the given series is divergent. Sn S₁ = U₂ + U₂ + + U₂ ... Un S₁ = V1 + V2 + ··· + v₂. Let and After the kth term, suppose vn Uk+15 Vx+19 Ux+2 = Vx+2, •··, Un > Un S-SS-S', then ก S„ > Sn' — S'x' + Sx. or By hypothesis S' increases without limit as n increases. Hence, if n is made large enough, S, will exceed any given number, and the u series is divergent. 228 [CHAP. XX. INFINITE SERIES A useful comparison series for divergence is the harmonical series 1 2 + 1 + 1 + 1 + ···, 3 which can be shown to be divergent by means of the inequalities: 1 + 1 > 1, 1 + 1 > 1 + 1 = 1, 4 } + { + } + { >}+↓↓↓+ } = {}, } + 10 + 16 + 1/10 > 1/1/ Adding members of the inequalities, we have ... 1 + 1 + 1 + 1 + · >1+} + { + } + ···· 1+1+1+1+ ar² ... But the right-hand member of this inequality can be made as large as we please. The series in question is therefore divergent. Any geometrical progression a + ar+ar2+ in which the ratio is greater than 1 can be shown to be divergent by compari- son with the series a + a + a + ...; and such a geometrical pro- gression is often useful as a comparison series in testing for divergence. EXERCISES Prove the following series divergent. 1. 1+ 2+ 3+ 4+ 1 1 1 2. 1+ + + + √2 √3 √4 1 1 1 3. 1+ 2p + + + 3P 4p where p is positive and less than 1. 4. 112 + 14 1 1 + + + 5. 1+ 3 +計​計 ​162. Summary of standard test series. When any new series has been shown to be convergent or divergent, we evidently in- crease our supply of series for comparison purposes, but a few standard series are so useful as to deserve special mention. For convergence: 1. a + ar + ar² + 1 2. 1 1 ... + aron-1 + ... (r < 1). 1 + + + 1.2 2.3 3.4 ... + n (n + 1) + ARTS. 161-163] 229 RATIO TEST 1 1 3. 1+ + + 2p 3p For divergence: 1. a + ar + ar² + 1 1 1 2. 1+ +=+=+ 2 3 4 1 + ... + ··· (p > 1). ... +ar"-1 + ... (r ≥1). ... 1 +-+ n 163. Ratio test for convergence and divergence. Another im- portant test for the convergence or divergence of an infinite series is the so-called ratio test. THEOREM. If, as n increases beyond bound, the ratio proaches a limit λ, the series of positive terms ... U1 + U₂+ + u₁ + is convergent if λ < 1 and divergent if λ > 1. Un ap- If λ=1, this test fails. 1. λ<1. Since lim 'n.. λ, we can make Un Un+1 И. n differ from λ X 2 î 1 FIG. 34. by as small a number as we please. Hence, if we plot values of 3. n+1 Un Un+1 Un on the line OX, as n increases the points representing will concentrate about the point λ. If n is taken large enough, they will lie to the left of the point r, where λ 1. In this case the points representing "+ will ulti- mately concentrate about the point A on the line OX, and if n is X 1 FIG. 35. λ large enough, they will lie to the right of the point r, where 1 1, Un + 1 > run i Un + ? > p² Un n Therefore, since the series. is divergent for > 1, ru„ + p²u„ + 2º³µ‚„ + ·· ... the series UĮ + U₂ + Uz + ... + u₁ + .. Un ... is divergent (Art. 161). 3. λ=1. If lim U₁+1 1, this test fails. This is illustrated. in the two series, 1 1 1 + + and 1.2 2.3 3.4 1 1 1 1++ +=+ 2 3 4 The first has been shown to be convergent (Art. 160), the sec- ond divergent (Art. 161), but for each lim Un+1 'n+1 1. И ル ​ART. 163] 231 RATIO TEST EXAMPLE 1: Consider the series 1:2 + 2 3 4 + + 22 23 24 + U Here, Un+1 22 + 1 2n+1 Un = 2n Un+1 n + 1 2n n+1 22 السبع 1 + Un 2n+1 n 2 n 2 n 2 n lim Un+1 lim + Un 2 n 2 Hence, the series is convergent. EXAMPLE 2: Consider the series 2 22 23 24 + + + + 22 32 42 52 2n+1 112 + 1 2 n 2n Here, Un+1 = Un = n + 2) 2' (n + 1) ² ² Un+1 2n+1 (n + 1)2 n + 1 1 2 2 Ите ( n + 2)² 2n n + 2 lim Hence, the series is divergent. Un+1 2. Un 1 1 1 2! 3! 4! 22 32 42 Apply the ratio test to the following series: 1. 1+ + + + 2. EXERCISES 112 + 2/2 + 3. 1+ + + + 2p 6. 1+ མས མིས 2! 3! 4! 22 23 + 24 2! 3! 4! 4. + + + 100 1002 1003 + 1 5. 2+ 22 + 3 4 3p Ap + + 2! 3! 4! 7. 1 + + + 3! 5! 22 32 42 9. 1+ + + + 2! 3! 4! 23 24 + + where p may have any value. 8. 1+2 + 4 + 8 + · 232 [CHAP. XX. INFINITE SERIES SERIES WITH BOTH POSITIVE and neGATIVE TERMS 164. Thus far we have considered only series whose terms are all positive or all negative. The following theorem will throw light on the convergence of series whose terms are not all of the same sign. THEOREM. An infinite series of real terms which are not all of the same sign is convergent if the series formed by making all the terms positive is convergent. After all the minus signs have been changed to plus signs, let the series be U₁ + U2 + Uz + ••·. By hypothesis, this series is convergent and therefore has a limiting value S. The sum of the first n terms of this series is then less than S. Hence, the sum, S, of the first n terms of the original series is numerically less than S. Let these n terms con- sist of p positive and q negative terms. If P, be the sum of the positive terms and N, the sum of the negative terms, then S₁ = P₂ — N. But P, and N, are always less than S. Hence, by Art. 159, P, and N, approach fixed numbers P and N respectively as n increases without limit. Then lim 22∞ S₁ = P― N, a definite number, and the series is convergent. 165. Ratio test extended. The ratio test can readily be ex- tended to series whose terms are not all of the same sign. Since a series of positive terms is convergent if lim Un+1 u <1, n → ∞ Un it follows, from the theorem just proved in Art. 164, that any series is convergent if the numerical value of lim n+1 is less N→ ∞ than 1. That is, if lim 'n+1 < 1. 12∞0 འ༦ ༦ To extend the ratio test for divergence we need the following important ARTS. 164, 165] RATIO TEST EXTENDED 233 THEOREM. In any convergent series, the limit of the nth term as n increases without limit is zero. That is, lim n→ ∞ n U₂ = 0. S-S-1- For, we may always write u, As n increases both S, and S-1 approach the same limit S; hence (Art. 152), lim 22∞ ab i༦ lim 8个 ​(SS-1) = S― S=0. n While it is necessary that lim u₂ = 0 in order that a series be convergent, it is not sufficient. That is, lim u may equal zero and the series be divergent. Exercises 2, 3, 4, 5, p. 228, furnish examples of such series. If lim+11, the nth term cannot approach zero as a 8个 ​Un limit, hence, the series is divergent. We may then write the ratio test for any infinite series as follows: U1 + U2 + Uz + · If lim +1 22∞ < 1, the series converges. Un lim If If 22 →∞ lim 8个 ​འཆ +1 U₂ 'n+1 Un > 1, the series diverges. = 1, the test fails. EXAMPLE: Test for convergence and divergence the series 1 − 2 x + 3 x² − 4x³ + ― Here, Un+1 (n + 1)xn Ить 17-1 lim n + 1 N )이 ​Un+1 and = |x|. 8个 ​Un Hence, if x lies between 1 and - - 1, the series is convergent. For |x|> 1, the series is divergent. The interval between + 1 and 1 is called the interval of convergence of the series, and is represented graphically by the heavy part of the line in Fig. 36. For the points 1 and 1 the test tells us nothing. DIVERGENT CONVERGENT DIVERGENT -1 1 FIG. 36. 234 [CHAP. XX. INFINITE SERIES 166. Alternating series. A series whose terms are alternately positive and negative is convergent if each term is less than the pre- ceding term, and if the nth term approaches zero as a limit when n increases without limit. Let the series be U1 U2 + Uz 14 + n- +(-1)π-¹u, ± ''', where U1, U2, 1:39 are positive, and U2 < U19 U3 < 22, ***, and where lim n∞ n U₂ = = 0. Let n be an even number. We may then write S, in the form S₁ = (U1 − U2) + (U3 − U4) + ... + (Un-1 — Un). Since each parenthesis contains a positive number, S, is positive and increases as n increases. We may also write S, in the form ル ​S = U1 — (Uq — Uz)· — U π- -2 一 ​U-1) — U₂• Since the parentheses are again positive, S₁ 0, a >0). x + a x + 2 a x + 3 a 22 32 42 6. 1+ + + + 2! 3! 41 7. 1.1 11 +1.02 1 1 1 + + + 1 10. 2+x(x+1) + 1.2 8. +++ ·· 1.003 1.0004 x(x + 1)¸ x(x + 1)(x + 2) Compute, correct to four decimal places. 9. 1.3 1.3.5 + + + 1 1.4 1.4.7 + •••, (x>0). 1.2.3 1 1 1 11. 1 + 32 53 74 1 1 1 1 1 1 12. • 2 23 13. 1 1 1 12 -12 - + + 3! 25 5! 27 7! 1 1 + + 22 33 44 1 1 1 1 1 1 14. • + + 1 3 2 32 3 33 4 34 Write down the first five terms of the series in which u, has the following values, and investigate the series for convergence or divergence. 1 15. Ить 3n 2n 17. u₂ = Ить 1 + 2n+1 1 16. Un 222 1 18. Un = 1 + n √ ñ 19. 1 1+x2 1 + + For what values of x are the following series convergent ? 1 2+2x2 3+3x² + X x2 x3 20. + + 2 4 + 1 1 1 1 1 21. + + + 2x + 1 3 (2 x + 1)3 5 (2x+1)5 22. 2x + 3. 2x² + 4.3 x³ + 23. 1 x+2 + 1 + 1 (x + 2)² ' (x + 2)³ 168. Power series. + The series аo + α₁x + α²x² + + α₁x² + ···, ... ... in which ao, A1, A2, … are independent of a, is called a power series. Such a series may converge for all values of x, may diverge for all values of x except for x = 0, or it may converge for some å values of x and diverge for others. ARTS. 167, 168] 237 POWER SERIES THEOREM. If in the series the ratio An+1 an аo + α₂ x + A2x² + · approaches a limit r, then the series converges for | x | < and diverges for |x|> r H This result follows directly from Art. 165. Applying the ratio test lim n+1 lim a,+12+1 rx. 8个​袋 ​a༦༦ 8个 ​апост From Art. 165, the series converges if |rx|<1, that is, for |2|< 2° H For x> the series diverges. If r = 0, the series converges 1 for all values of x. For x the test fails; the series may or グ ​may not be convergent for this value of x. COROLLARY. If a power series is convergent for x = b, it is con- vergent for every value of x numerically less than b. EXERCISES Find the interval of convergence of the following series. Exhibit the results graphically. 1. X 22x2 32x3 + + + 22 23 n2 SOLUTION: lim An an+1 = 2n (n + 1)² 2n+1 An+1 an (n + 1)² 2 n² 8个 ​8个 ​An+1 Un lim (n + 1)² lim 2 n2 ( + 2 n2 2 n² The series is therefore convergent for |x|<2 (Fig. 37). CONVERGENT n2 2n + 1 8个 ​2 n² 2 DIVERGENT -2 DIVERGENT 0 FIG. 37. 2 2. 1+2x+(2 x)² + (2x) ³ + 3. 1++ x8 + + 3 32 33 4. 1 + x + 2 ! x² + 3 ! x3 + 2 x2 ... (x0). 3 x2 5. x + + + 2! 3! 238 [CHAP. XX. INFINITE SERIES 52x2 53x3 6. 1+2+ + + 2! 3! 7. 1-2x + 2.3 2! x2 2.3.4 3! 23 + 2.3.4.5 4 ! x4 + By division, expand the following fractions into series and test for con- vergence. 1 1 1 8. 9. 10. 1 M x 1 4. Ꮖ 2 Bir 1 11. (1 + x)² The expansions of sin a, cos x, are sin x, and are tan x are given below Find the interval of convergence of each of the series. 12. sin x = x x3 x5 + x7 + 3! 51 7! 13. cos x = 1 x2 x1 + x6 + 2! 4! 6! 1 x3 1 3 14. are sin x = x + + 2 3 2 4 x5 1 + 5 2 • 15. are tan x = x x3 X5 + XC7 + 3 5 7 304 5 x7 4 6 815 7 + 16. From the series in Exercise 12, find the value of sin 5° to four decimal places. HINT: Express 5º in radians. 169. Binomial series. The power series 1 + mx + m (m 2! c.com 1) m (m − 1) (m x² + 2)203 + 3! is called the binomial series. If m is a positive integer, the series ends with the (m + 1)th term and has been shown to be the ex- pansion of (1+x)m. If m is not a positive integer, the series is an infinite series, but it can be shown that it converges towards (1 + x)" when x has any value which makes the series convergent. In other words, it can be shown that for these values of a, the binomial expansion holds for any exponent, integral or fractional, positive or negative.* *For a proof of the binomial expansion for any exponent, see Fine's College Algebra, p. 553. ARTS. 168, 169] 239 BINOMIAL SERIES The binomial series is convergent for < 1, and divergent for x>1. For, we have an+1 (n = m(m − 1)(m − 2) ... (m — 11)¸ (n + 1)! m(m − 1) (m − 2) ... (m − n + 1) IN 17 n! 1), n "+1 m 1 · 1. lim n lim an+1 lim m 12 n → ∞ l l n n→ ∞ n + 1 22 → ∞ 1+ 1 n Hence, from Art. 168, the series converges for - 1<<1. In expanding (b+x)" for fractional or negative values of m, we may write it in the form b(1+ ʊm (1 + 23")". then proceed in powers of 01+ 772 X The expansion will thus: bm (1 x m(m ጎ 1)(x 2! b 2 ¿ (1 + 1 ) = ( 1 + 2 + (n = 1) (+)" + ). X This series converges when <1, that is, the interval of con- vergence for the expansion of (b+x)" is the interval between –b and +b (Fig. 38). DIVERGENT CONVERGENT DIVERGENT -b b FIG. 38. EXERCISES Expand the following binomials to five terms and indicate the interval for which the expansion holds. 1. (1 + 3x)−2. 3. V1+2. 5. (2+x). 7. (5 — 4x²)¯¾. 9. (1 + 4x²)−1. 2. (1+2x) ¹³. 4. (2-3x)-4. 6. 1 √1 - 8x 8. (1 − x)˜³. 10. (3a+2x)§. 240 [CHAP. XX. INFINITE SERIES Extract the following roots to four places of decimals by the binomial expansion. 11. $65. SOLUTION: √65 = (4³ + 1)³ = 4(1 + ; 4(1 + ا 43 1 1 1 1 1 3 43 3 =4(1+0.00521 : 4.0207+. + ..) 3 46 0.00003 + ...) 12. √5. 13. √15. 14. v1.02. 15. 998. 6 16. 65. 17. $130. 18. 731. 19. $258. 20. 0.515. 170. Exponential series. The power series 1+ +1+1 x2 23 xn-1 + + + 2! 3! + (n − 1)! is called the exponential series. It is convergent for all values of X. For, we have 1 Ant! n! , an ༦ 1 antl 1 (n − 1)!' a„ N hence, It can be proved that lim m∞ m-> lim an+1 -0. n→ ∞ a n (1 1\mx 1+ = m x x². 23 1+ + + + 1 2! 3! where ex, 1 1 e=1+1+2+ + 2! 3! = 2.71828, correct to five places of decimals. This number e is the base of the natural system of logarithms ах Let a² = eh, then x log, a= h. Hence, a² = exlog.a =1+x log, a + x² (log, a)2, x³ (log, d)3 a)² + 2! 3! + ARTS. 169-171] 241 LOGARITHMIC SERIES 171. Logarithmic series. The power series x2 XC 2 + 203 3 + +(-1) n-1 xn 12 土 ​is called the logarithmic series. Since we have 1 an+1 = ± a n + 12 +1' 8个 ​lim an+1 An – 1, and the series is convergent for x < 1. 72 F 1 n DIVERGENT CONVERGENT DIVERGENT -1 1 FIG. 39. It will be shown in the calculus * that this series converges to log. (1+x) for any value of x for which the series is convergent. The series can then be used to find logarithms of numbers to the base e. Thus, log. (3)= log. (1 + }) ( 1 ) 2 4 127 + + 2 3 The logarithmic series can be used to calculate logarithms of positive numbers less than 2. However, it converges so slowly that it is not well adapted to numerical computation. To derive a more convenient series for the calculation of natural logarithms, we proceed as follows: † x3 log,(1 + x) = x + + 2 3 4 x2 x3 X4 Hence, loge (1 − x) - -= X + 2 3 4 1 + x By subtraction, loge (1 + x) — log, (1 − x) = log. 1 — Xx =2(+1+1+..) 3 * Townsend and Goodenough's First Course in Calculus, p. 326. For a more detailed discussion, see Osgood's Introduction to Infinite Series, pp. 23 and 44. 242 [CHAP. XX. INFINITE SERIES Let whence We have then 1 1 + x 1 -x x m + 1 m 1 2 m + 1 + 1 3(2 m + 1)³ · 5(2 m + 1) 5 loge m + 1 m 1 1 = 2 + 2 m + 1 or log(m + 1) = log, m + 2 m = 1, we have 1 + 2 m + 1 1 + 1 + ·), +...). 3 3(2 m + 1)³ ' 5(2 m +1) 5 If 1 1 loge 2 = 0 + 2 + 34 5.35 + ...) = 0 = 0.6931 + Letting m = 2, = log, 3 log, 2+2 5 2+ 1 + 3.53 ...) 1.0986 + = 0.6931 + 0.4055 + In this way the logarithm of any number to the base e may be computed. From Art. 120, if a is any positive number, we have log10 α = loge a loge 10 1 loge 10 • log, a. Hence, if we have computed the logarithm of a number to the base e, we can find its logarithm to the base 10 by multiplying by 1 nificant figures, = 0.43429. loge 10 1 To five sig- log, 10 In computing a table of logarithms we need compute the logarithms of prime numbers only. The logarithms of composite numbers may then be found by means of the theorems on logarithms. EXERCISES 1. By computing the logarithms of 2, 3, 5, 7, construct a table of logarithms to the base 10 for the numbers 1 to 10. 2. Using the series for loge (1 + x), compute log, to three places of decimals. 3. Find log, 16 and log, 17 to four significant figures. ANSWE [The answers to some exere are intentionally omitted.] Artic Page 11 1. 5. 2. 9. 3. 6. 4. 3, 2, 3. 5. 5, 4. 6. 7. a. 8. 1. x5 68 9. 0. m¹n². 11. y5 16 a²x6 12. 8 a³ + 12 a2b2 + 6 ab4 + b6. 13. a3n. 14. a³n. 15. an. 16. a2n+2. 17. a2n-2. 18. (4)*- 19. (-3)nx2n 4nyn gnx4n ท n 20. 16ny2n 22. x²y²+xy + 1. 23. a2n b2n. 21. 2 + n m 24. x2n−2 — y2n−2. 25. p2n-2-p2n−5 + p2n−1 — p2n−4. 26. a6n-9.x12n-3. y9n-3. 27. x³n. Pôn. yn². 7 a4-m . xn-3 30. 15 28. x6. 31. a³". 29. 1. 32. xn² + yn². 33. 4 a2n+2 anfn + b²n. 35. (a² + ab + b²)² (x + y)². 34. x2n 36. 27 p³n x² + 1. 893n. Article 12. Pages 15-17 5x3 1. 2. 5 a3x3. 3. . 4. a3 5. 3. 9. 22. 6. . 7. 2. 8. 25. 34 10. 0.7. 11. 12. 12. 2.54 13. Ꮖ . 10 b7 14. 15. a. 16. a b¹c. a a7c3 x2 a0.06 14 उ 17. 2xy³. 18. 19. x 20. Y ย 20.14 α 21. - x14у7z-21. 22. p-0.14. 25. aš – bš af_b. 7 26. x³yš + x³y³. 27. x−1 — y-¹. — 24. b-b- y−1. 28. 1+x². 23. x y. 243 244 ANSWERS 1 29. a α -2 30. a+2ab+b. 31. a+3a³b³ +3a}{b} 32. x-4+x−² + 1. 34. 2 + x + x 4. 2+x‡ 36. x-18. + 38. x-3 2 x−²y−1 + 4 x−¹y-2 — 8 y−3. 39. a-8a-7 40. x + x²². α-4 +3 a³b³ + b. 33. p-3-3p-2 + 3p-1 — 1. 35. 2xy-5уx-1. 37. m-2 - m−1 + 1. a-5 — a-¹ — a−3 + a−1 + 1. 41. a³b¯½ +2a³ — 3 61. 42. x¯}+ a‡ +a³xt. 45. (a+b)5. 43. √x Vy². 3 √ 44. Va 1 46. 47. m/xn 62 48. √(a - b) 5 Y Valo 1 49. 7. 50. y√x³z. 51. √x Vaz³ 52. xlyzi 53. a-4x 55. x³z². a36 33 b3 56. x 58. 24. 2 61. an²-1. 1 2 59. anang. p²-q²+2 62. a 2p 60. X 22 1 54. art 57. 4x²; 1 1 • n. yn 778 63. x³ 64. a on². b. 5 65. x28 67. 170.4. 68. 10-6, 3. 10-9, 15. 10-7. 72. 10-6. 69. 3, -1, 0, 34. 71. 666.10-10. 73. 0.6867, 0.5896, 0.4861, 0.3934 microns 74. 1 preceded by twenty-four zeros after the decimal point. Article 14. Page 19 2. a b+c+d-e. 1. 5 a + 12 b. 3. 6x+2. 4. x+2y. 5. 0. 6. 4c + 3b − (5 c² — 4 a − 4 ď³). 7. a-d― (c — b). 8. 20. 10. (x² + y²)2. 9. 6a+5b-(a+b)-(4a+3b). Article 15. Pages 20-21 Z א x + 1 2 ab 2 1. 2. 3. 4. 2 У ४ +62 2 m - 1 1 a X 1 x²+ x + 1 XZ 72 5. 6. 7. 8. α 2+1 x સે xz + y² ANSWERS 245 m + n 1 9. 10. 11. x + y. 12. x + 1. m n x + 1 x²y x 13. x + y. 14. 15. x + y - 2 x+2 16. 2. 2αx 17. x2 x + 1. 18. 19. x - 1. 20. n. a² + x² 1 1 2x + h 21. 22. 23. x(x + h) (x − 1)(x+h−1) x²(x + h) 2 1 E 24. 25. 26. rs(r+s) 3r+ Rt 6p + 14 R R 27.2 a + b 28. a2b. 29. x. 30. a • 31. a4b²xy³ ( 33. (a² — x²) ž. ab 1 2 ყ32 32. a + b. y3 2x3 a2 x2 34. x Article 17. Pages 24-25 2. /99. 3 3. 40. 4. V112. 3 5. 104. 6. 2058. 3 7. 192. 10. 25. 11. √250. 12. √28. 13. 81. 14. 314. 15. √42. 18. 24. 19. 3√ž. 20. 5√5. 21. 2√3. 22. 33. 23. 20√2. 24. 3√3. 27. √20. 28. 9. 29. √13. 30. vi. 31. 8. √5 34. 5 35. VA. 36. 1+ √3 2 37. 15 5 √s. 3√5-3√3 38. 2 39. (√20+ √2). 40. − 4 (v2+ √3). 1 = 0.4142. 41. 1+ √2 1 √5 1 1 – √3 4 √2 − √3 = 0.4472. - 1.3660. 12.585. Article 18. Pages 25-26 8.(5-2√3). *³ (5 − 2 √/3). 13 2. √300. 3.√35. 7. 26-7√3 23 11. 5 + 21/6. 12. 33 11 4. V121. 5. 9√15. 6. √2. 9. √6+ √15 3 10. - 5. 14. x³ +2. α 36+7√15 13. Va+1+Va. 246 ANSWERS 2. 3V3. 8. √ √ 12. 16(3) 2. 3. V6. √2. Article 19. Page 26 4. x√3x. 5. 9√3. 6. √2. 9. 15. va 7. (b+x 7. (b+x-ay) √α. 11. 7√ā-2vā. 10. 92x. 14. 3(5). 15. 5. 1 α 2. √675. 3. 12. 5. 16. 2n/a7. $/72 3. a 72 a. 4. 100. 5. 245. 8. 6. 9. 12 + 7 √6. 10. 2√3-V10. 13. 3 − 5 α + 2√a(1 + a). 17. 54. 13. 15(3). Article 20. Page 27 6. 77. 7. 12u¹b³. 11. m + 3√m²n + 3√mn²+n. 15. x4y6z3. 12 12 14. x4 +1. 18. V3. 12 19. 2108. 20. 23. ᎥᏆ 21. V У 22. √5. 23. Vab. 27. 7' = 4√/75. $75 28. √a = 1 √57a. 8 24. }V15122. 1 29. V18 a. 30. 96 ax². 2a Article 21. Pages 27-30 25. 6. 26. 78 10 1. 2288 feet per sec. 6. 0.0205. 7. 1.9 feet. 13. 1077. 14. 30. 2. 25.30. 3. 500.7. 4. 10.71. 5. 0.740. 8. 9.39. 9. 26.95. 10. 29.83. 11. 154. 15. 50.9. Article 22. Page 30 2. 5. 8. 625. 14. 0. 3. 1. 4. - 2 − 2 i√3. 5. x² + α². 6. i. 9. i. 10. 0. 11. 14. 12. 2 a(a² - 3 b²). 7. 7. 13.-1054. 15. (x − a)² + b². α Article 25. Pages 32-33 1. Aπr2. = 2. V=ƒ(x). 3. y = √100 — x². 4. P = C, C = f(t). 5. a² + a + 1, 1, 3, 1, 91, a² + 3 a + 3. 1 + x² x + 1 7. 1, C 1 x2 x 1 6. 3, — 1, 0, 1– √2, 1, 0, 1− √2, a + b + 1 a + b 1 9. x++ x² + 1, x² + 3 x + 3, x4 + 2 x³ + 4x² + 3x + 3. 5. (— 12, − 3). 11. 1+2 t 3+ 4t 7 t + 10 12. 15 t + 22 − Article 26. Page 35 6. (1, 4), (2, 4), (5, 4), (5, 1), (5, 2), (2, − 2), (— 1, 2). (— 1, 1). 8. (a) (— 1, 5), (3, 5), (3, 1), (— 1, 1). (b) (1, 5.83), (3.83, 3), (1, 0.17), (— 2.83, 3). ANSWERS 247 Article 29. Page 40 2. 6. 4, -3. 3. - 2. 1 7. 1, 5. 4. 1, 5. 8. 1, 1, 5. 0, - 4. - 3. 9. 0.5. - 5. 10. 0 and 1; - 1 and 0; 6 and - 5. Article 36. Pages 47-48 (b) 2x(x − 3) = 5(x − 3) 4) with the additional 1. (a) 2x2 = 5c with the additional root 0. with the additional root 3. (c) 2x(x² — 4) = 5(x² roots 2 and 2. 2. (x²+2)(x − 3) = 3 x (x − 3). This equation is re- 3. 0. 4. (a) x² - 6 = 3 or x² = 9; dundant with respect to a root 3. (b) 2. 5. 8x2 - 24 x + 16 0. 9. 7x= 12. 13. x64 = 0. 16. 7x75 0. = 10. 3x + 8 = 0. 14. x2-5 x 14 = 0. 17. x = 1. 6. a. 11. x 7. a. 5 = 0. 8. 7x-1= 0. 12. x · 250. 15. (x-1)(7 x² - 16x — 15)=0. Article 38. Page 51 2. x = ¹², y = 7. 13 21 y = 2. 6. x=7, y = 3. x = 16, y : 11. = 15. 4. x = 1, y = 2. 5. x=7, 7. x = 6, y = 12. ་ 2. x = 1, y = 1. 3. x = 3, y y = — 1. 9. x = 18, y 6. x = 2, y = 3. = 10. solution. 12. x = · }, y = 3. 13. x = 0, y = 7. Article 39. Pages 52-53 4. 4. x = 1, y = 3. 7. x = 4, y = 2. 5. x = 1, 8. x = 1, y = 2. 11. No 10. An infinitely large number of solutions. 4. x = 5, y = 2. Article 40. Page 54 2. 154. 8. 0. 3. 41. 4. 39. 5. 15. 6. axv+ byu, 7. x² −30x + 40. Article 41. Page 56 2. x = 1, y = 2, z = 3. z = 3. 3. y = 2, z 3. x = = 10, 5. x = 2, y = − 1, z = 3. 6. 4. x = 10, y = 2, 7. x = — 4, 5. 8. No solution. 3. 0. y = 8, z = 6. x = 2, y = 2, z = 2. -9. x = 7, y = 4, z = 3. z= Miscellaneous Exercises and Problems. 4. 2y-2x. 5. 6. Pages 56-58 6. x = 3. 8. 30 gallons of 25 per cent alcohol and 20 gallons of 30 9. 3 oz. and 5 oz. 10. 18 and 72°. 11. 512,000 sq. ft. 13. 428,571. 17. 7. x = 2, y = 5. per cent alcohol. 12. 27 and 72. 5 min. 29 sec.; 20. $20,000, 14. 5 mo. 15. 6900. 6 min. 45 sec. 19. c = 0.036, b = 999.5. $15,000. 21. Analytic geometry 85, algebra 95, trigonometry 90. 248 ANSWERS 2, 7, 12, 17, 22. 0, 6, 12, 18, 24. 22. 2, 3, and 6 hr. 25. First 8, second 15. 27. Going 9 miles uphill, 134 miles level, 4½ miles 23. 12, 9, 3, 21. 26. 1020 bonds. 4, 8, 12, 16, 20. 24. downhill. 6, 9, 12, 15, 18. 8, 10, 12, 14, 16. 10, 11, 12, 13, 14. Article 44. Pages 62-63 1. 7, 2. 2. 1, -7. 3. 5, 4. 4. I Gallag 1. 5. 8 6. 5, ૐ. 7. 27 4. 8. 1, 47 9.2, 4. 10. 6, 11. 7, 2. 12. 2 ± 3 i. 13. 3 ± i. 14. 2, 2. 15. 14, 16. α, a - 1. 1 m 272 17. 5, 0. 18. 3 ± √5. 19. √2. 20. 100 21. 9. 22. 10. 23. 5. 25. 5. 26. 50. 29. 9, 33. 33. 16 or 6.2. 30. No solution. 34. 5.4 or -0.51. 27. 16, 23. 31. 1.6- or 0.75. 1+e' 1-e 24. 2. 28. 1, 3. 32. 3.1 or -2.2. Article 45. Pages 64-65 2. ± 2, ±3. 3. ±1√7, ± i 2 N! 5. 1, -5+i√3 6, 2 7. 0. 8. 10. 11. 12. - 1, 3. 4. 2,2 i. 6. 1, 1, −3 ± 2√2. 9. 92, 1. 13. — 2 a, a, (−1 ± i√³), - a(1 + i√³). 15.2 ± 2 i, 2 ± √5. 10. 25, 9. α 2 - 1±√13 14. 2 - 1± i√3 2 16. 1. 17. 1, 1. 18. 1. 19. (1√21)3. ก 20. 2, -1±i√3. 21. V b±√b² - 4 ac 2 a 22. 2, 23. 0, 49. 24. 1. 25. 0, -5, 5 + √− 15 2 2. x2 x-6=0. Article 46. Pages 65-66 4. x²+(5−√5)x – 5√5 = 0. 6. x2 - x ⋅ 2 √5 + 4 = 0. 3. x27x=0. 5. x2 5 = 0. 7. x²+1=0. ANSWERS 249 8. x2 - 8x+25= 0. 10. x² + x + 1 = 0. 9. x² - 4x — 71 = 0. x2 11. x2 · ½ x + 1 = 0. 12. x² (a² + 1)x + 1 = 0. 13. x2 a · (m² + m² ) x x + 1 = 0. 1. k = ±√5. 3. cr 3 - k 3, or 1. 5. k Article 48. Page 68 2. k 8. 4. m 122, k = - 8. 6. k = 0, or 1 p ጎ 一 ​7. k = 9. k = 1. 11. k 4. - 1, m = 4. 1. Real and unequal. 4. Real and unequal. 8. k = 9. 11. k=-2, or ㄨ​ˇ 14. k =±r√1 + m². 8. k =± 2. 10. k = 3. 12. k 3, or — 1. = Article 50. Pages 69-70 2. Imaginary. 5. Real and equal. 9. k = 9. ་་ 12. k = . 3. Real and equal. 7. k±4. 10. k5. h 13. 1. 15. kva²m² + b². 16. 17. - ½, — 1}. 11 18. 2 m 6 m 2 m m² 20. 22. k = 3. 1 e2' 1 e2 24. k = - 4. 25. k = 1. 一 ​2. 51 per cent. 4. 33.89, 81.89. 6. 3.1+ sec. 8. 38.8+ ft. Problems. Pages 71-74 3. 50 x 75. 4, - 3. 19.1. 23. k = 20. 26. k = ± 13. 5. 0.8+ sec. or 2.3- sec. 7. 1.5+ sec. 9. One half the whole time. 10. 4.9 ft. Ball begins to fall before the end of the second second. 11. 3.7 sec. 14. 2 in. 18. 0.625. 23. 14 × 7. 26. 8 ft. per sec. 12. 80.5 ft. per sec. 15. 5.828+. 21. 1 1 • 24. 1 ft. 4 in. 27. 33.47. 18. 144.9 ft. 16. 12+7√3 = 24.125. 22. 25. 25. 1 ft. 2 in. 28. 238 ft. if time is measured to nearest tenth of a second. 250 ANSWERS 31. R=1±√1 – 4 g² X 2 2 g rPX±√P²X²+4 R²W(P—W) 2 RW 29. 198³-. 30. 5. 221= 32. n- n¹ α ± √a² a 4 T2b 33. x= 2 Tb 34. 1 part of corn to 4 parts rye. Article 53. Case I, Page 77 2. (§√7, 4), (&√7, − 1), (— §√7, 2), (— §√7, − 2). 3. (0, 3), (0, -3). 4 3 3 4 4 i i 4. (5 √5, & √21), (5¹ √5, -2 √ž), (-5 √, & vπ), i 4 (− 5 1 √5, − 2 √21). 3 4 5. (5, 0), (— 5, 0). 15 6. (11√34, 1√34), (}§√34, — }§√34), (— }{√34, 14√34), (— }{√34, — }{√34). 7. (†√65, † √91), (†√65, — †√91), (—†√65, f√91), - 6 - 語 ​(— 1º½ √65, — f§√91). ∞ oi 8. (0.19, 3.1), ( - 0.19, · 3.1), (0.19, ― 3.1), ( — 0.19, 3.1). 9. (1.1, 2.3), ( — 1.1, 2.3), (1.1, — 2.3), ( — 1.1, — 2.3). · Case II, Pages 78-79 2. (1, 2). 3. に ​1-8 i√6 18+ 6 i√6 5 5 (- 1+8 i√6 5 — 18 – 6 i√ē). 5 4. (3,7), (7, 3). 6. (8, 5), (− 5, — 8). 8. (4, 10), (100, — 110). 10. (1, 1), (- }, — }). 5. (4, 9), (9, 4). 7. (1, 6), (— 4, — 2). 9. (1, 2). 号​,一​子​). 11. (2, 3), (— }, }). 12. (0, 0), (4, 5). 13. (0,0), (— 1, 1, — 2.4). - 14. (1.4, -0.17), (0.034, — 7.1). - 15. (1.1, 2.2), (0.03, — 0.71). 17. r>5√2, r = 5√2, r<5√2. 18. c<25, c = 25, c> 25. 19. Two points for any real value 20. Two points for any real value of a. 21. b=±r√1 + m². of b. 22. k = +Va²m² + b². Case III, Pages 80-81 (— 2. (4, 1), (− 4, − 1), (14, – 4), ( − 14, 4). 4. (3, 5), (— 3, — 5), (§, ¹³), (— §, — ¹³). 5. (4, 7), (− 4, — 7), (7, 4), (− 7, — 4). - 3. (– VŨ, VŨ), (V6, – V6) (−√á, – √6) ANSWERS 251 6. (3, 5), (3, 5), (8, 5), (-8, 5). — 7. (§√21, §√21), (— § √21, — ‡√21). 8. (6,3), (6,3), (6,3), (— 6, 3). 9. (4, 5), (4, 5), (3√3, √3), (— 3√3, −√3). 10. (2, 4), (-2, -4), (181 i √101, - 16 i 101 Bi V101), ( 18 i 101 16 i √101, 101 V101) 11. (√5, √5), (— §√5, — {√/5). 12. (2.7, 0.85), (1.09, 2.11), ( — 1.09, 13. (10, 1.4), ( — 10, — 1.4), (11, 17), ( — 11, - 17). 1.09, 2.11), ( — 2.7, — 0.85)., - Case IV. Page 82 2. (5, 0), (0, 5), (1, — 4), (— 4, 1). 3. (3, 2), (2, 3), 1+i√23 6 1-i√23 6 √23), に ​-1-i√23 6 1+i√23 6 4. (3, 1), (1, 3), に ​19 + √309 19√309 8 8 19 √309 19+ √309 8 8 5. (2, 3), (3, 2), (− 5 + ¿√II, 5-i√II), (—5— i√11, −5+ ¿VII). - 6. (3,2), (2, 3), (3,2), (2, 3). ― Exercises and Problems. 1. (3, 1), (— 17, — 15). 5 3. (1, 1), (— }, — 1³). 5. (2, 7), (18, ). Pages 82-85 2. (2, 1), (-, - b). 4. (1, 2), (— 32, − 24). 6. (3, 5), (3, 5), (§, 1), (— §, — f²). 7. (6, 2), (3, 5). J 8. (2,3), (— 89, 52 - 233). 9. (2, 11), (4, 12). 11. (9, 5), (- 43, — 8.1). 10. (5, 4), (4, — §9). - 12. (1, 2), (1, 4), (− 1 + i√³, — 3 — ¿√³), (— 1 — i√3, − 3 + i√³). 9 13. (6, 2), (−2, −3), (5+√265 5+ √265), (5-√265 5 √265 (- 4 4 14. (2,5), (, V). 4 4 15. (7, 1), (— 49, — 13), (} √/33, — } √/33), ( − 1 √/33, § √/33). 16. (4, 2), (— 2, − 1), 17. (− 4, − 6), (5, 3), — √17 (1 − VI, −1 + √π7), (1 + VI7, -1-VI). (1 — V1 2 2 - (− 9 + √141, − 3 +√141 2 2 2 2 9 - √141 - 3 — √/141 2 2 252 ANSWERS 19. (3,2), (2, 3). C 18. (3, 1), (1, 3). 21. (0, 1), (2, 5), (į √2, − ↓ √2), ( — {√2, {√2). 22. (9, 4), (4, 9), (6, 6), (− 6, — 6). 23. (4, 1). 25. (2 a, a), (— 2 a, — a). · 26. (0, α), 27. (2, 3, 4), (2, — 3, 4), (2, — 3, — 4), (2, 3, (-2, 3, 4), 20. (4, 1), (1, 4). 24. (5, 9), (− 5, 9). (α, 0). 4), (— 2, 3, 4), (— 2, 3, 4), (— 2, — 3, — 4). · — — 29. (1, 3, 5), (— 1, — 3, — 5). (3, 8, 4, 6), (3, 8, 6, 4). 28. (2, 1, 1), (§, — 10, — 74). 30. (8, 3, 6, 4),* (8, 3, 4, 6), 31. (1, 2, 3, 5), (2, 1, 5, 3), 32. (1, 1, 2, 2, 3),† 34. a4 — b4 = 4 c4. 37. 628.5, 653.3. 5 (1, ½, 1, 1), (1, 1, †, 1), (3, 5, 1, 2), (1, 1, 1, 1), (5, 3, 2, 1), 7 (1, 1, 1, 1). 33. b² + c² + 2 a². 36. a4b4 — c4. 39. 12, 6, and 4. (2, 1, 1, 3, 2). 35. aª + b¹ = c²(a² + b²). 38. 300 and 400. 40. $550, $450; 7 %. 42. 0.1556. 44. 24 lots, $200 per lot. 46. 467.8. 50. x² - 10x+9= 0. 52. 7 and 5. 41. 4.4 cents, 7.5 cents. 43. 10 and 12. 45. Radius, 2√5, altitude, 8√5. 47. 9 × 13 × 21 ft. 49. 100 miles, 25 miles per hour. 51. 12 ft. 11 in., 16 ft. 1 in. 53. 36 miles per hour. Article 57. Page 89 1. x 5. 2. x > 3. 3. x < 9. 4. - 2 < x < 2. 5. 7. x < 3. 10. 0 < x < a, if a > 0; a < x < 0, if a <0. x>4 and x < − }. 8. x5 and x < − 2. 6. x positive but ‡ 1. 9. - 8 < x < 35/2. 11. x> " if a is positive; x < if a is negative. a α 12. x2 4x + 3 > 0, if x > 3 or < 1. = 0, if x = 1 or 3. < 0, if 1 < x < 3. 13. ax2 + bx + c> 0, if a is positive and x x‡ tween these values. b + √b² - 4 ac ax² + bx + c = 0, if x = 2 a 6 ±√b2 4 ac nor be- 2 a - ax² + bx+c<0, if a is a positive and x lies between − b +√b² − 4 ac b = √b2 G √b² 4 ac · and 2 a *(8, 3, 6, 4) means o=8, y = 3, z= 6, u = 4. † (8, 1, 2, 2, 3) means = 1, y = 1, 2 = 2, u = 2, v = 3. 2 a ANSWERS 253 14. 25>0, if x > 3 or <-5; 50, if - 5 < x < 3. x 3 15. x <- 4, and x > 4. x< · x 3 16. For all real values of x. Article 59. Page 93 1. 504. 2. 8. 3. 30. 4. 2. 5. 84. 7. r2 — r. Article 60. Pages 95-96 6. 16+ a + 6 a² + 16 a³ + 16 aª. 8. x5 15x4y² + 90 x³y¹ — 270 x²y6 + 405 xy³ — 243 y¹0. 9. aº + 6 а³ √õ + 15 ab + 20 a³b √b + 15 a²b² + 6 ab² √õ + 6³. 10. x4 + 4x² + 6 + + 4 1 x2 x4 12 40 12 + + €3 11. e 12. a4b4 - 8 a³b³ + 24 a²b2 — 32 ab + 16. 3 еб 13. x¹ — 8 x³µ³ + 28 x³y — 56 x³y + 70 x²y²— 56 x²y³ + 28 xу³ — 8 x³y² + ya — 14. 911. 1 4 + 16. x−1 — 4x-³y'¹³+6x−²y³ — 4 x−¹y + y‡. 17 29 a²² 15. a⁹+6a² +15 a 20 a 13 2 + 15 a²² 3 + 6a²²⁹ + a¹. ૫ +15a4 20a30 3 x +15 a2b3 ab 3 6 16 x2 x2 X3 x-4 3 ασ 17. 32. 18. 1,3 6 a5 3 1 b² x · x2 19. a³ +3a²b + 3 ab² + b³ + 3 a²c + 6 abc + 3 b²c + 3 ac² + 3 bc² + c³. 20. x² + 6x + 11 + 12 x² + y¹ 12 3 12 1 + + + + x } X x2 x3 x8 8 21. + x6 x3 + 81 27 27 32 16 + 3x7 x12 $15 N + = x² x + 1 3 8100 8 + + 3x² 8 16 + 24 x3 X4 32 x9 23. - 14,080 a9b³. 24. 2489344 x7y10. 13 26. 18! 5! 13! a5b13 27. 13! 7!6! 1 хозба 25. 5670 x8ys. 28. 126 x 2 y7. 29. - 35 a³, +35 a³, 32. 11,040,808,032. 33. 941,480,149,401. 36. 2.594. 34. 345,025,251. 35. 0.885842380864. Article 64. Pages 98-100 37. 4.177. 2. y = 3x. 3. s 16 t2. 4. p = 2048 v 5. z = 20xy. 6. 2 = 180 x Y 7. V = kr³ = 1 π³. 254 ANSWERS 8. V khr2 =mhr2 9. 402.5 ft. 1610 ft. 10. 1896+ lb. kml3 11. 25 mi. per hour. 12. 48 lb. 13. b = wd3 14. 888 lb. 15. 0.12 in. 16. 7.5 sq ft. 20. 3√5±√17. 23. 0.05+ in.. 17. 3.4-. 21. 0.087- in. 24. 441,720 lb. 18. T. 25 128 22. 1000 yr. Article 68. Pages 102-103 2. 1=-59, 8=-610. 3. l = 59, s = 59, s = 465. 5. 1 = √2, s= 1 121 3. 6. 1 = 48, 8 = 142. 8. n = 7 or 13, l = 7 or — 5. 4. l 71, s - 660. = 7. 1 = 22 {, s = 9 5 891. 9. α = 3, s = 465. - 105 10. α = 7, 8 2 12. n = 16, s = 142. 14. α=- a $, n = = 23. 16. a 44, l = 46. 19. 5, 8, 11. 20. 131. l 1. 6561, s = 9841. 3. 1=1024, 8= 6. 6561. 9. r = 3, s = 120. 12. . 11. n 20, d 4. 13. d 3, s 714. 1, d 1. - 2. 15. a = 1, 17. = し ​5, d 21. 1, §, k, †, 0, − 1, − 3, − 4, — 4. Article 72. Page 104 2. l- 4374, s=- 6560. • 4. l=-768, s=—510. –768, 5. r = 3, l = 162. 7. 27. 8. 62. 10. 49. 11. 16, 128. 67 13. a5 Article 75. Page Page 107 བཤ 1. . 2. . 3. 4. 7. II. 8. 4. 9. . 4. 16. 10. . 5. 16. 11. 4019. 6. — 37. 12. 18028 33300 Article 77. Page 107 2 ab 1. 4, 6. 2. 1, 4. 3. 16, 1, 1, 1. 4. a+b Problems. Pages 108-109 1. 40 feet. 2. 94811 and 14475 feet. 3. Between 185 and 190 feet per second. ་ 4. 19.683. 5. 122.853. 6. 547. 7. 36. 8. 45. 9. $885.73. 10. 2900 feet. ANSWERS 255 11. The latter proposition is worth $25 per year more to the clerk. 1 12. 12,092. 13. n 15. x2 14x+48 0. 17. 3. 19. 6, 8, 10, 12. 20. 3, 6, 9. Article 80. Page 113 2. r = √13, 0 = arc tan 2. 4. r = √2, 0 ㅠ ​Ө 4 6. r = √41, 0 = arc tan 8. r = 4, 0 π = 2 14. n². 18. 6912. 21. $7721.73. 3. " = √13, 0 √13, 6 = arc tan 5. r = √2, 0 7. r = 1, Ө 9. r = 6,0 π 2 4 4 π 4 • #. π - 2 10. - Ө 12. 1 = 5, 0 = π. 2 ㅠ ​2 11. = 4, 0 = 0. 13. = V13, 0 arc tan - 14. "' =√Ï.7, 0= arc tan . 4. 15. 3√3 + 3 i. 16. 1+ i√3. 17. −√3 + i. 18. 4 i. 19. √2-ži. 20. 6. 21. i. 22. x 1, y = 1. 23. x = 3, y = — 1. 24. x=- ૐ, જી 13, y=-15 25. x = 0, y = ~ 1, or x = }, y - 3. 26. x = 0, y = 5, or x = 2, y y = = 1. 27. x = 9, y = 7, or x 9, Y =- — 7. Article 81. Page 114 1. 3 + 4i. 2. 3 i. 3. 13i. 4. 6. 5. Ga 6. 2-5 i. 7. 1 i 8. - 1 + 5 i. 9. 2-3 i. 10. 0. Article 83. Page 116 2. 4+4 i√3, 0 4. 4 i, 0 = 1 = 4. 6. 4√3 + 4 i, 0 = π = r' = 8. 16 3. 10 v3 1,0 = 1, r = 10 √3. 16 3 3 5. - 4, 0 = π, r = 4. π r = 8. 6 7. - 8,0 = – 8, 0 = π, r = 8. 8. - 3 п 8 i, 0 = r = 8. 9. 2 - 10, 0 = π, r = 10. 10. 6, 0 = = π r = 6. 8 ? 11. 4, 0, r' — 4. 2 - 10 i. 256 ANSWERS Article 85. Page 118 2. 432 + 144 i√3. 3. 24 i√3. 7. - 128 - 128 i. 6. 512 i. 10. 4. 1. 5. 1. 8. 32 ¿. − 1 + i, √Ž(cos 15° + i sin 15°), VŽ(cos 255° + i sin 255°). 11. 2+2 i√3, – 2 – 2 i√3. 12. †12(cos 15° + i sin 15”), V12(cos 195° + i sin 195°). 13. 4√3+4 i, — 4√3 — 4 i. 14. 4(cos 20° + i sin 20°), 4(cos 140° + i sin 140°), 4(cos 260° + i sin 260³). 15. 2(cos 10° + i sin 10°), 2(cos 70° + i sin 70°), 2(cos 130° + i sin 130°), 2(cos 190° + i sin 190°), 2(cos 250° + i sin 250°), 2(cos 310° + i sin 310°). These results are represented graphically by the six points at equal inter vals on a circle of radius 2, beginning with 0 = 10º. 16. ¿√3 + ¦ i, cos 66° + i sin 66°, cos 102° + i sin 102°, cos 138° + i sin 138°, cos 174° + i sin 174°, – į√3 — § i, cos 246° + i sin 246°, cos 282° + i sin 282º, cos 318+ i sin 318°, cos 354° + i sin 354°. These results are represented graphically by the ten points at equal inter vals on a circle of radius 1, beginning with 0 ≈≈ 30°. 17. 1 + i, √2(cos 117° + i sin 117°), √2(cos 189° + i sin 189°), V(cos 261° + i sin 261°), √2(cos 333° + i sin 333°). 18. cos 18° + i sin 18°, i, cos 162° + i sin 162°, cos 234° + i sin 234°, cos 306° + i sin 306°. 3 19. {√3 + ¦ i, — {√3 + ¦ i, − 3 i. 20. ± 2. 22. 2, 1 + i√3, − 1 − i√3. 21. 1, − ¦ + ¦↓√ši, – ↓ — §√ši. 23. 3(cos 221° + i sin 221°), 3(cos 1121+ i sin 1124°), 3(cos 2021 + i sin 2021°), 3(cos 2921° + i sin 2921°). 24. 3, + i√3, — — $i√3. § { 25. 1, cos 72° + i sin 72°, cos 144° + i sin 144°, cos 216° + i sin 216”, cos 288° + i sin 288°. 26. 2, 2(cos 72° + i sin 72°), 2(cos 144° + i sin 144°), 2(cos 216° + i sin 216°), 2(cos 288° + i sin 288°). 27. See answer to 21. į 1 { រ 28. ±2, ±2 i. 29. 1, § + ¿¿√³, − † + į i√3, − 1, − ¦ — { ¿√ā, ↓ — ¦ i√3. 30. ±1,±i, ± {√2, ± {√2 i. ANSWERS 257 Article 86. Page 120 7. 1 2. 1 + i√3. i. 3. 12 3. − 1 + ½ i√3. 8.-1-i. 4. 4 i. 5. - {√³+i. 6. 1+i. 9. i. 10. Vi. - Article 88. Page 122 3. - 1. 4. k = −3. Article 91. Page 125 2. 0, 2, 4. 3. 1, 2, 4. 4. Zero at 1, between 1.5 and 2, and between 3 and 2.5. 5. Zeros at 2, 1, 2, 3. 6. Zero between 1.5 and 2, and between 2 and M 2.5. 7. Zero between - 1.5 and - 1, between 8. Zero at 2. 1 and 0.5, between 0 and 0.5, and between 2.5 and 3. Article 97. Page 131 30= 0. 2. (a) x3 - 10 x² + 31 x (c) x³ — 5 x² + 5 x + 3 = 0. (e) x¹ — 2 x³ — 5 x² + 6 x = 0. (b) x² - 2 x + 5 = 0. (d) x¹ — 5 x² + 6 = 0. (ƒ) SA 14 c² + 1 = 0. Article 98. Pages 134-135 1. x3 + 10 x2 175 x 125 = 0. 3. x4. 10x² + 3x 2 = 0. 2. x3 x240 0. = 4. x³+6x² - 80 = 0. 2700. 5. x3 6 x² + 3 x 7. x3 + 36 0. 11. x2 - x-6=0. 13. x39x² — 90 — 0. 15. 2x43x² + 4 x 5 = 0. 6. x¹ + 3 x³ – 12 x² + 20 x − 8 = 0. 498 = 0. 8. x³+6x² + 6 x 12. x3 + 3x² 4x + 1 = 0. 14. x5 7x3 + 2x 8 = 0. 2. 4 imaginary roots. Article 99. Page 137 3. One negative and 4 imaginary roots. 5. 4 imaginary roots, one zero root, 1 imaginary roots. 7. 1 posi- 8. 2 positive and 1 negative 4. 1 negative and 2 imaginary roots. one negative root. 6. 1 positive and n tive, 1 negative, and n roots. S 2 imaginary roots. 4. 1..... 8. 1. ···, 1. ……., 3. 1..... ... 0. 11. ... 1. Article 100. Page 138 ... 0..... 11. 2. 0. ... 5. - 1..... 9. 0...., 1. .•*, 6. 2. 2... 2. .. 10. 7. 1. ..., 1..., 258 ANSWERS Article 102. Pages 140-141 ci co oi - 1, 2, 3, 4. 4. 1. 5. 2, -3, 2. 6. - . (. - 3. 8. 1,, 11. 1. 10. 0. 9. 3. 12. - 4, 5, 3. 13. 2, 6, — 7, — 1. 14. . 15. No rational roots. 16. 17. 1, 2, — 2, 3, 4. · Article 105. Pages 146–147 1. 3.13-, 1.20. 2. 1.41. 7. 6.17. 5.24-. 3. 0.64. 4. 4.64. 8. 2.36, 2.69, - 2.05-. 9. - 2.21. - 0.95. 12. 5.83, 0.27, 0.93,1.03. 5. 3.98. 6. 3, 1.41, 10. 2.24-. 13. 1.88, 11. 3.01, 0.63, – 2.02, - 0.35, - 1.53. 14. 1.18, 2.87. 15. 3. 16. 1.73. 17. 4.00 per cent. 18. 0.606. 19. 0.860. 20. 0.32, 0.64. 21. 6.86. 22. 11.07. 23 2.92. 24. 0.259. 26. 2√232-81.20, 25. 1.46. 3√29² – 12 = 1.80. 27. $1.536. 2 92 10 = 1.50, 9. có có ô 3. 3, w, w2. 6. 1, 2, 3. 2, 5+2 i√3. Article 108. Page 152 4., i, i. M 7. 2, 1, − 1 ± i√2. 10., 1, 33√2. Article 110. Pages 154-155 5. }, + √2, −√. 3 8.1, i√2. § ± 1. 2, 2, 2. 2. 3, ± i. 士 ​5. For a 10, {1.218, 0.082}; a = 1, {12.923, 0.077}. 3. - 1, 2, 5. 4. 5, 5, 2. 129.923, a = 0.1, a = 0.01, { 1299.923, 0.077}, \ 0.077, a = 0.001, { 12999.923, 0.077}. 6. For a = 10, {12.179, 0.821 }, a = 1, { 12.923, 0.077 }, Ɑ= 0.1, {12.992, 0.0077}, a = 0.01, {12.999, 0.0008}, a = 0.001, { 13.000¯, 0.000+}. 7. m ± √3, b = = 3√3. 10. m -- 3, 1) = 2, or m = 3 i 8. m = ±i, b = F 9. m = ± 2, b = 0 2 fr, 6 = fr• I' Article 113. Pages 158-159 2. 2 log 2 4. 3 log 3. log 25 & log 11 log 23. 3. log 13 - log 10 - log 48. C 5. log 5+ log 7 4 log 2 4 log 3-log 2 — log 3. 6. log 2 log 3 M log 5-log 11. ANSWERS 259 7. 3 log 2 + 9. log 2+ log 11 + log 3 + log 5. 11. 1.0791. 13 12 log 5 — 4 log 3. 6 2 12. 1.14771. 8. log 2 + log 3 + ½ log 7. 13. 1.6232. 14. 2.6232. 15. 2.2764. 19.0.1370. 23. 0.8228. 16. 2.9542. 17. 2.5353. 18. 1.3313. 20. - 2.4013. 24. - 0.1505. 21.3.0124. 22. 1.3512. 25. 0.6469. 26. 0.7726. Article 119. Pages 166–169 1. 0.01359. 2. 0.2332. 3. 0.03908. 4. 24.15. 5. 8.641. 9. 6. 0.7038. 7. 1.673. 8. 0.4352. 0.9097. 10. 0.2979. 11. 0.03229. 12. 0.7295. 13. 0.9630. 14. 0.3857. 15. 0.4420. 17. 0.2917. 18.0.1606. 19. 6.636. 21. 25.31. 22. 3.162. 23. 0.0696. 16. 64.32 20. 27.89. 24. 0.9047. 25.0.8646. 26. 1.639. 27. 1.01 sec. 28. 142.5 tons. 29. Volume = 13,330, Surface = 2719. 31. 11,660. 32. 834,200. 33. 1,476,000. 30. 1051 107. • 34. 0.608. 35. 476. 36. 4.578. 37. 100 pounds. 38. 0.125 cubic foot. 39. p = 4.529. 41. 0.0068. 42. 24,470. 44. (a) 14,790. (b) 14,860. 40. 177.5. 43. $2014. With a seven-place table the following more accurate results are obtained : (c) 14,860. 45. $3767. (a) 14,802.4. (b) 14,859.4. (c) 14,888.7. 46. 41 digits until the year 1935, when it will require 42 digits. 47. 4.251. 48. (1) 100,100; more accurate value 100,081; (2) 85,450; more accurate value, 85,442. 49. 1547 miles. 50. 146,700 sq. kilometers. Article 120. Page 170 2. 1.3862. 6. 1.661. 3. 0.6825. 4. 2.7302. 5. 3.322. 7. 2.096. 8. 1.431. 9. 3.930. 10. 0.856. 11. 1.625. 12. 2.393. 260 ANSWERS Article 122. Pages 173-175 4. ≈ 0.8115. 7. x = 1 or 2. 9. n = log rl - log a log r 5. 0.1853. 6. x = ± 2.390. 8. x=3 or 1. 10. log [(r− 1)s + a] — log a log r 12. x = 1.61, y = 2.56. 14. (1) 11.90, (2) 11.58, (3) 11.55. 13. x = = 6.84, y = 10.84. With a seven-place table of logarithms, the following more accurate results may be obtained: (1) 11.89, (2) 11.64, (3) 11.55. 15. 0,1.32. 16. 3.96. 17. 0.00003776. 18. 18,360. e2 1 19. k 0.126. 20. 5.5 minutes. 21. x = 3 22. * = 25 and C 4. 23. s = v0 2 k (ekt e-kt). 1 1. 8! :00 1 x+2 3 1 3. + Article 128. Pages 181-182 2. 218 4. 5. 4(x-3) ' 4(x + 1) 3 1 Ꮖ 1 x 2 يم 5 + x x 2 3 7. + x + 1 x 1 9. + 1 11. х 2 5 (x − 1)² 2(x-1) X 1 2(x 132218-22 1 2x+2 + x² + 1 6 7 + 12. - 3) يع 1 14. HI8 + 5 3(1 - x) 1 3(x+2) 1 2 2 + 1 (x − 1)² (x − 1)3 1 2 5 6. + 2 x 1 8. + x 1 2(x-2) *2 x (x 1)2 10. x + 1 + 1 + 1 + 1 1 x + 2 X X 1. x + 1 2 x+2 1 M 3 x² + 1 x + 1 2 3 + х XC 3 x + 2 1 3 1 x 15. + 16. 2(x-1) X 2 2(x-3) x x² + 4 4 17. 3(x-1) 1 19. 1 + 4x+8 3(x² + x + 1) 1 - 10x 3(13x) 3(1 + 3 x) 2 11 20. 8(x-3) 7 11 11 2(x − 1)³ 4(x − 1)² 8(x − 1) 18. Į 2!8 1 1 2 小 ​x2 x3 X 1 1 ANSWERS 261 3 21. + 21 2(1–3)3' 8(1 − 3 x)² · 21 7 + + 32(1 − 3 x) 32(1 + x) 1 2 x 1 1 22. + 23. + x + 2 x2 2x+5 1 (x2 x + 1) 2 2 2 2x + 2 1 2 24. 3 x 4 + + 25. + x + 1 Ꮖ 1 x²+1 ic + 1 Ꮖ. 1 x² + x + 1 1 27 27 5 1 5 4 26. 27. 2(2z-3) 2(2x-3)3 (2 x − 3) 4* C x2 ic + 1 (i + 1)² 2 3 C 10 3 4 28. ic 29. +2 + + C 2 x+2 x² + 4 x + 1 Ꮧ. 5 1 x + 2 2x + 3 1 30. 31. C 3 x2 x + 1 (x2 x + 1)2 L2 28 + 8 1 5 18 1 2 32. 3 + 33. + x + 2 (x+2) 2 (x + 2) 3 * x + 3 8 ここ ​8 ― 1 (x − 1)2 3 x + 4 x² + 3 5 4 1 X 3 34. + - (x − 1)2 1 X 1 x2 x + 1 1 I 35. x + 2 I + 36. x² + x + 2 x² + x + 1 (x² + c + 2)² x² + 1 (x² + 1)² 2 1 2x + 4 1 37. x³ + 2x² + 2 + + 38. 1 x (1 − x)³ X)3 + 1 + x + x² 3x + 2 (x² + 1)² Article 132. Pages 185-186 1. 720. 2. 5040. 3. 60. 4. 1,860,480. 5. 40,320. 9. 336. 6. 8. 10. 360. 7. 3360; 34,650. 11. 30. 8. 5040. 12. 180. Article 136. Pages 187-189 1. 210. 4. 501,501. 2. 40. 3. 31. 5. 70. 6. 1140; 220. 7. 10. 8. n = 17, r = 2. 9. 31. 10. 63. 11. 25. 12. 462. 13. 371. 15. 4950. 16. 327,600. 52!4! 17. 45. 18. (13!)4 20. 30,240. 26. 23. 1820; 455. 100! 60! 20! 20! 21. 3360. 24. 10,080. 19. 20. 22. 4095. 25. 3,303,300. 262 ANSWERS Article 139. Pages 191-192 1. 19. (3). 3. 10. 8. $3. 4. 78. 5. $5. 6. 0.69804. 7. (1); (2) i 8 33 Article 140. Page 193 1. 7. 2. 11. 9. (1) 0.03; (2) 3. 25. 4. fr. 5. 3. 6. 11. 10 7. 12. 5 TE 8. 27. 8 0.68; (3) 0.12; (4) 0.17. rico 1. (1)%; (2) 31. 21 3. 16173 6384 5. $10. 7. The latter. 10. g. 3 676 1 12. (1) }; (2) 2%. 25 14. (1) 64,763 75,994,575 (4) 76. Article 141. Pages 194-195 0.000852; (2) 2. (1) 0.2886; (2) 0.9446. 3 4. 584. 6. (1) 0.108; (2) 0.994. 8. (1) 0.087; (2) 0.503. 11. 3. 13. A's expectation is $36; B's is $30. 89,827 75,994,575 = 0.00117; (3) 0.00000100; Article 144. Page 202 2. 25. 3. 5. 4. 2. 5. - 99. 1. 2, 1. 4. 2, 4, 1, − 3. - Article 149. Pages 211–212 ·- 2. 1, 0, -1. 3. ∞0, 1. 5. k = 4. 6. xyz=-1:1:2. 7. 5 x6 — 2x4 — 9 x² + 6 x − 1 = 0. 8. 9. Yes. a³ +63 + c3 - 3 abc 10. 2, or 3 abc = 0. 154. Article 156. Page 220 1. 3. 2. 1. 5. 3a 3 a 2 6. 4. 9. Increases without limit when x→→ 10. 1. 11. 0. 3. 2√2. 7. 1. - 1. 12. Article 163. Page 231 4. §. 8. 1. 1. Convergent. 2. Convergent. 3. Convergent. 4. Divergent. 5. Divergent. 6. Convergent. 7. Convergent. 8. Divergent. 9. Convergent. ANSWERS 263 1. Convergent. 4. Convergent. 7. Divergent. 10. Divergent. 13. 0.7834. 16. Convergent. Article 167. Pages 235-236 2. Convergent. 5. Convergent. 8. Divergent. 11. 0.8965. 14. 0.2877. 17. Divergent. 19. Divergent for all values of x. 20. Convergent for |x|< 1, also for x =— 1. 21. Convergent for x > 0, and x <- 1. 23. Convergent for x > − 1 and x < − 3. 3. Convergent. 6. Convergent. 9. Convergent. 12. 0.4794. 15. Convergent. 18. Convergent. 22. Convergent for |x| < 1. Article 168. Pages 237–238 2. Convergent for |x|< }. 3. Convergent for |x| < 3. 4. Divergent for all values of x, for which the series is defined. 5. Convergent for ∞