\^ t Digitized by the Internet Archive in 2008 with funding from Microsoft Corporation http://www.archive.org/details/algebraschoolsOOnewcrich •MMOA M3N ^NVdWOO DNiaNiaXOOa QNV ONIXNIUd S.MOUX •OO 2y X10H Aa^THH *t88X '1881 ^xnoiUAcioo 3 >5 V "^ IV^''"'^ SMODM.KK isLOMIS - Aa eaoa^^oo a^v st:oohos \, r"^ v»-^a-tDT:v 'SaiEES pVOIXVM^BXVH S.SWOOAiSN \ PREFACE. The course of algebra embodied in the present work is substantially that pursued by students in our best preparatory and scientific schools and colleges, with such extensions as seemed necessary to afford an improved basis for more advanced studies. For the convenience of ;eacher3 the work is divided into two parts, the first adapted to well- prepared beginners and comprising about what is commonly required for admission to college ; and the second designed for the more advanced general student. As the work deviates in several points from the models most familiar to our teachers, a statement of the principles on which it is ' constructed may be deemed appropriate. One well-known principle underlying the acquisition of knowledge is that an idea cannot be fully grasped by the youthful mind unless it is presented under a concrete form. Whenever possible an abstract idea rnust be embodied in some visible representation, and all general theorems must be presented in a variety of special forms in which they may be seen inductively. In accordance with this principle, numerical exam- ples of nearly all algebraic operations and theorems have been presented. For the purpose of illustration, numbers have been preferred to literal symbols when they would serve the purpose equally well. The relations of positive and negative algebraic quantities have been represented by lines and directions from the beginning in order that the pupil might be able to give, not only a numerical, but a visible, meaning to all algebraic quantities. Should it appear to any one that we thus detract from the generality of algebraic quantities, it is sufficient to reply that the system is the same which mathematicians use to assist their conceptions of advanced algebra, and without which they would never have been able to grasp the complicated relations of imaginary quantities. Algebraic 183978 iv ^ PREFACE. operations with pure numbem are made to precede the use of symbols, and the latter are introduced only after the pupil has had a certain amount of familiarity with the distinction between algebraic and numer- ical operations. Another, but, unfortunately, a less familiar fact is, that all mathematical conceptions require time to become engrafted upon the mind, and the more time the greater their abstruseness. It is, the author conceives, from a failure to take account of this fact, rather than from any inherent defect in the minds of our youth, that we are to attribute the backward state of mathematical instruction in this country, as compared with the continent of Europe. Let us take for instance the case of the student, commencing the calculus. On the system which was almost universal among us a few years ago, and which is still widely prevalent, he is con- fronted at the outset with a number of entirely new conceptions, such as those of variables, functions, increments, infinitesimals and limits. In his first lesson he finds these all combined with a notation so entirely diJOferent from that to which he has been accustomed, that before the new ideas and forms of thought can take permanent root in his mind, he is through with the subject, and all that he has learned is apt to vanish from his memory in a few months. The author conceives that the true method of meeting this difficulty is to adopt the French and Grerman plan of teaching algebra in a broader way, and of introducing the more advanced conceptions at the earhest practicable period in the course. Accordingly, the attempt is made in the present work to introduce each advanced conception, disguised perhaps under some simple form, in advance of its general enunciation and at as early a period as the student can be expected to understand it. In doing this, logical order is frequently sacrificed to the exigencies of the case, because there are several subjects with which a certain amount of famil- iarity must be acquired before the pupil can even clearly comprehend general statements respecting them. A third feature of the work is that of subdividing each subject as minutely as possible, and exercising the pupil on the details preparatory to combining them into a whole. To cito one or two instances : a difficulty which not only the beginner but the expert mathematician frequently meets is that of stating his conceptions in algebraic language. Exercises in such statements have therefore been made to precede any solution of PREFACE, V problems. In general each principle whicli is to be presented or used is stated singly, and tbe pupil is practiced upon it before proceeding to another. Subjects have for the most part been omitted which do not find appli- cation either in the work itself or in subsequent parts of the usual course of mathematics, or which do not conduce to a mathematical training. Sturm's Theorem has been entirely omitted, and a more simple process substituted. The subject of the greatest common divisor of two polyno- mials has been postponed to what the author considers its proper place, in the genaral theory of equations. It has, however, been presented in Buch a form that it can be taught to pupils preparing for colleges where it is still required for admission. Thoroughness at each step has been aimed at rather than multiplicity of subjects. It is, the author conceives, a great and too common mistake to present a mathematical subject to the mind of the student wi^out sufficient fulness of explanation and variety of illustration to enable him to comprehend and apply it. If he has not time to master a complete course, it is better to omit entirely what is least necessary than to gain time by going rapidly over a great number of things. Some hints to those who may not have time to master the whole work may therefore be acceptable. Part I is essential to every one desiring to make use of algebra. Book YIII, especially the concluding sections on notation, is to be thoroughly mastered, before going farther, as forming the foundation of advanced algebra ; and affording a very easy and valuable discipline in the language of mathematics. Afterward, a selection may be made according to cir- cumstances. The student who is pursuing the subject for the sole purpose of liberal training, and without intending to advance beyond it, will find the theories of numbers and the combinatoiy analysis most worthy of study. The theory of probabilities and the method in which it is applied to such practical questions as those connected with insurance will be of especial value in training his judgment to the affairs of life. The student who intends to take a full course of mathematics with a view of its application to physics, engineering, or other subjects, may, if necessary, omit the book on the theory of numbers, and portions of the chapter on the summation of series. Functions and the functional notation, the doctrine of limits, and the general theory of equations will ' 'iim his VI {PREFACE. especial attention, while the theory of imaginary quantities will be studied mainly to secure thoroughness in subsequent parts of hiv course. As it has frequently been a part of the author's duty to ascertain what is really left of a course of mathematical study in the minds of those who have been through college, some hints on the best methods of study in connection with the present work may be excused. If asked to point out the greatest error in our usual system of mathematical instruction from the common school upward, he would reply that it con- sisted in expending too much of the mental power of the student upon problems and exercises above his capacity. With the exception of the fundamental routine- operations, problems and exercises should be confined to insuring a proper understanding of the principles involved : this once ascertained, it is better that the student should go on rather than expend time in doing what it is certain he can do. Problems of some difficulty are found among the exercises of the present work; they are inserted rather to give the teacher a good choice from which to select than to require that any student should do them all. It would, the author conceives, be found an improvement on our usual system of teaching algebra and geometry successively if the analytic and the geometric courses of mathematics were pursued simultaneously. The former would include algebra and the calculus, the latter elementary geometry, trigonometry, and analytic geometry. The analytic course would then furnish methods for the geometric one, and the latter would fiirnish applications and illustrations for the analytic one. The Key to the work contains not only the usual solutions, but the explanations and demonstrations of the less familiar theorems, and a number of additional problems. The author desires, in conclusion, to express his obligation to the many friends who have given him suggestions respecting the work, and espe- cially to Professor J. Howard Gore, of the Columbian University, who has furnished solutions to most of the problems, and given the benefit of his experience on many points of detail. Note. — Answers to exercises, requiring calculation or written worTc^ are published separately in pamphlet form, and will be supplied without cha1;ge when applied for by teachers. TABLE OF CONTENTS, PAET L— ELEMENTAKY COURSE. BOOK I.— THE ALGEBRAIC LANGUAGE, Chapter I.— Algebraic Numbers and Operations, 3, Genr ^l Definitions, 3. Algebraic Numbers, 4, Algebraic Addition, 6. Subtraction, 8. Multiplication, 9. Division, 11. Chapter II.— Algebraic Symbols, 12. Symbols of Quantity, 12. Signs of Operation, 13. Chapter III.— Formation of Compound Expressions, 17, Funda- mental Principles, 17. Definitions, 19. Chapter IV. — Construction of Algebraic Expressions, 22, Exer- cises in Algebraic Language, 25. BOOK IL— ALGEBRAIC OPERATIONS. General Remarks, 28. Definitions, 28. Chapter I. — Algebraic Addition and Subtraction, 30. Algebraic Addition, 30. Algebraic Subtraction, 33. Clearing of Parenthe- ses, 35. Compound Parentheses, 37. Chapter II,— Multiplication, 38. General Laws of Multiplication, 38. Multiplication of Positive Monomials, 40. Rule of Signs in Multiplication, 41. Products of Polynomials by Monomials, 44 Multiplication of Polynomials by Polynomials, 47. Chapter 'ill. — Division, 52. Division of Monomials by Monomials, 52. Rule of Signs in Division, 53. Division of Polynomials by Mono- Viii ' GDNTENT8. \ mials, 54. Factors and Multiples, 55. Factors of Binomials, 58. Least Common Multiple, 61. Division of one Polynomial by another, 62. Chapter IV.— Of Algebraic Fractions, 67. JSegative Exponents, 71. Dissection of Fractions, 73. Aggregation of Fractions, 74. Factor- ing Fractions, 78. Multiplication and Division of Fractions, 79. Division of one Fraction by another, 82. Reciprocal Relations ol Multiplication and Division, 83. BOOK III.— OF EQUATIONS. Chapter, I.— The Reduction of Equations, 85. Axioms, 87. Opera- tions of Addition and Subtraction — transposing Terms, 87. Operation of Multiplication, 89. Reduction to the Normal Form, 90. Degree of Equations, 93. Chapter II. — Equations of the First Degree with One Un- known Quantity, 94. Problems leading to Simple Equations, 99. Problem of the Couriers, 105. Problems of Circular Motion, 108. Chapter III.— Equations of the First Degree with Several Unknown Quantities, 109. Equations with Two Unknown Quantities, 109, Solution of a pair of Simultaneous Equations containing Two Unknown Quantities, 109. Elimination by Com- parison, 110. Elimination by Substitution, 111, Elimination by Addition or Subtraction, 112. Problem of the Sum and Differ- ences, 113. Equations of the First Degree with Three or More Unknown Quantities, 116. Elimination, 116. Equivalent and Inconsistent Equations, 121. Chapter IV. — Of Inequalities, 123. BOOK IV.— RATIO AND PROPORTION. Chapter I. — Nature of a Ratio, 128. Properties of Ratios, 132. Chapter II. — Proportion, 133. Theorems of Proportion, 134. The Mean Proportional, 138. Multiple Proportions, 139. ^ contents: IX BOOK v.— OF POWERS AND ROOTS. Chapter I. — Involution, 144. Involution of Products and Quotients, 144. Involution of Powers, 145. Case of Negative Exponents, 147. Algebraic Sign of Powers, 148. Involution of Binomials — tlie Binomial Theorem, 148. Square of a Polynomial, 153. Chapter II.— Evolution and Fractional Exponents, 155. Powers of Expressions with Fractional Exponents, 157. Chapter III. — Reduction of Irrational Expressions, 159. Defini- tions, 159. Aggregation of Similar Terms, 160. Factoring Surds, 161. Perfect Squares, 166. To Complete the Square, 167. Irra- tional Factors, 169. BOOK VI.— EQUATIONS REQUIRING IRRATIONAL OPERATIONS. Chapter I.— Equations with Two Terms only, 170. Solution of a Binomial Equation, 170. Special Forms of Binomial Equations, 171. Positive and Negative Roots, 172. Chapter II. — Quadratic Equations, 174. Solution of a Complete Quadratic Equation, 175. Equations which may be reduced to Quadratics, 179. Factoring a Quadratic Equation, 184. Equations having Imaginary Roots, 188. Chapter III. — Reduction of Irrational Equations to the Normal Form, 189. Clearing of Surds, 189. Chapter IV. — Simultaneous Quadratic Equations, 193. BOOK VII.— PROGRESSIONS. Chapter I. — Arithmetical Progression, 200. Problems in Pro- gression, 202. Chapter II.— Geometrical Progression, 207. Problems of Geo- metrical Progression, 208. Limit of the Sum of a Progression, ^M Compound Interest, 217. X CONTENTS. PART II.— ADVANCED COUESE. BOOK VIII.— RELATIONS BETWEEN ALGEBRAIC QUANTITIES. Functions and their Notation, 221. Equations of tlie First Degree between Two Variables, 224. Notation of Functions, 230. Func- tions of Several Variables, 282. Use of Indices, 233. Miscellaneous Functions of Numbers, 235. BOOK IX.— THE THEORY OF NUMBERS. Cttapter I.— The Divisibility of Numbers, 238. Division into Prime Factors, 239. Common Divisors of two numbers, 240. Relations of numbers to their Digits, 245. Divisibility of Numbers and their Digits-, 245. Prime Factors of Numbers, 248. Elementary Theorems, 251. Binomial Coefficients, 251. Divisors of a Number, 254. Chapter II.— Of Continued Fractions, 258. Relations of Converging Fractions, 267. Periodic Continued Fractions, 270. BOOK X.— THE COMBINATORY ANALYSIS. Chapter I. — Permutations, 273. Permutation of Sets, 275. Circular Permutations, 277. Permutations when Several of the Things are Identical, 279. The two classes of Permutations, 281. Sym- metric Functions, 284. Chapter II. — Combinations, 285. Combinations with Repetition, 287. Speeial Cases of Combinations, 289. The Binomial Theorem when the Power is a Whole Number, 296. Chapter III.— Theory of Probabilities, 299. Probabilities depend- ing upon Combinations, 300. Compound Events, 305. Cases of Unequal Probability, 310, Application to Life Insurance, 316. Table of Mortality, 318. BOOK XI.— OF SERIES AND THE DOCTRINE OF LIMITS. Chapter I.-— Nature of a Series, 321. Notation of Sums, 32- < CONTENTS. Xi Chapter IL— Development in Powers op a Variable, 326. Method of Indeterminate Coefficients, 327. Multiplication of Two Infinite Series, 333. Chapter III.— Summation op Series. Of Figurate Numbers, 336. Enumeration of Triangular Piles of Shot, 339. Sum of the Similar Powers of an Arithmetical Progression, 341. Other Series, 345. Of Differences, 350. Theorems of Differences, 355. Chapter IV. — The Doctrine of Limits, 358. Notation of the Method of Limits, 361. Properties of Limits, 364. y Chapter V. — The Binomial and Exponential Theorems. The Binomial Theorem for all values of the Exponent, 368. The Exponential Theorem, 373. Chapter VI.— Logarithms, 378. Properties of Logarithms, 378. Com- parison of Two Systems of Logarithms, 384. BOOK XII.— IMAGINARY QUANTITIES. Chapter I.— Operations with the Imaginary Unit, 391. Addi- tion of Imaginary Expressions, 393. Multiplication of Imaginary Quantities, 393. Reduction of Functions of i to the Normal Form, 396. Chapter II.— The Geometrical Representation op Imaginary Quantities, 404. BOOK XIIL— THE GENERAL THEORY OF EQUATIONS. Every Equation has a Root, 416. Number of Roots of General Equation, 418. Relations between Coefficients and Roots, 425. Derived Functions, 427. Significance of the Derived Function, 430. Forms of the Roots of Equation, 43 1. Decomposition of Rational Fractions, 433. Greatest Common Divisor of Two Functions, 438. Transformation of Equations, 442. Resolution of Numerical Equa- tions, 447. ^ or THE \ UNIVERSITY |. OF BOOK I. TUB ALGEBRAIC LANGUAGE. CHAPTER I. OF ALGEBRAIC NUMBERS AND OPERATIONS. General Definitions. 1. Definition. Mathematics is the science which treats of the relations of magnitudes. The magnitudes of mathematics are time, space, force, value, or other things which can be thought of as entirely made up of parts. 3. Def. A Quantity is a definite portion of any magnitude. Example. Any definite number of feet, miles, acres, bushels, years, pounds, or dollars, is a quantity. 3. Def. Algebra treats of those relations which are true of quantities of every kind of magnitude. 4. The relations treated of in Algebra are discovered by means of numbers. To measure a quantity by number, we take a certain por- >n of the magnitude to be measured as a unit, and express ow many of the units the quantity contains. Eemark. It is obviously essential that the quantity and its unit shall be the same kind of magnitude. o. Def, A Concrete Number is one in which the kind of quantity which it measures is expressed or understood ; as 7 miles ^ 3 days,, or 10 'pounds. 4 THE ALGEBRAIC LANGUAGE. 6. Def. An Abstract Number is one in wiiii^ix Uv/ particular kind of unit is expressed ; as 7, 3, or 10. Eemark. An abstract number may be considered as a concrete one expressing a certain number of units, without respect to the kind of units. Thus, 7 means 7 units. Algebraic Numbers. 7. In Arithmetic, the numbers begin at 0, and in- crease without limit, as 0, 1, 2, 3, 4, etc. But the quantities we usually measure by numbers, as time and space, do not really begin at any point, but extend without end in opposite directions. For example, time has no beginning and no end. An epoch of time 1000 years from Christ may be either 1000 years after Christ, or 1000 years before Christ. A heavy body tends to fall to the ground. A body which did not tend to move at all when unsupported would have no weight, or its weight would be 0. If it tended to rise upward, like a balloon, it would have the opposite of weight. If we have to measure a distance from any point on a straight line, we may measure out in either direction on the line. If the one direction is east, the other will be west. One who measures his wealth is poorer by all that he owes. If he owes more than he possesses, he is worth less than nothing, and there is no limit to the amount he may owe. 8. In order to measure such quantities on a uni- form system, the numbers of Algebra are considered as increasing from in two opposite directions. Those in one direction are called Positive; those in the other direction Negative. 9. Positive numbers are distinguished by the sign + , plus ; negative ones by the sign — , minus. If a positive number measures years after Christ, a negative one will mean years before Christ. If a positive number is used to measure toward the right, a negative one will measure toward the left. ALGEBRAIC NUMBERS. 5 If a positive number measures weight, the negative one will imply levity, or tendency to rise from the earth. If a positive number measures property, or credit, the nega- tive one will imply debt. 10, TBe series of algebraic numbers will therefore I be considered as arranged in the following way, the series going out to infinity in both directions. -=©91 NEGATIVE DIRECTION. POSITIVE DIRECTION. ^T Before. After. Downward, Upward. Debt. Credit, etc. etc. etc. -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, etc. Rem. It matters not which direction we take as the positive one, so long as we take the opposite one as negative. If we take time before as positive, time after will be nega- tive ; if we take west as the positive direction, east will be negative; if we take debt as positive, credit will be negative. 11. Positive and negative numbers may be conceived as measuring distances from a fixed point on a straight line, extending indefinitely in both directions, the dis- tances one way being positive, and the other way negative, as in the following scheme : * etc. - 7, -6. -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, +6, +7 . etc. I I I I I I I I I I I I I i I In this scale, the distance between any two consecu- tive numbers is considered a unit or unit step. 13. Def. The signs + and — are called the Alge- braic Signs, because they mark the direction in which the numbers following them are to be taken. * The student should copy this scale of numbers, and have it before ^im in studying tlio present chapter. 6 THE ALGEBRAIC LANGUAGE. The sign + may be omitted before positive numbers, when no ambiguity is thus produced. The numbers 2, 5, 12, taken alone, signify +2, +5, +12. But the negative sign must always be written when a negative number is intended. 13. Def, One number is said to be Algebraically- Greater than another when on the preceding scale it lies to the positive Cright hand) side. Thus, — 2 is algebraically greater than — 7 ; " " " " _2; 5 " " " " — 5. Alg^ebraic Addition. 14. Def. In Algebra, Addition means the combi- nation of quantities according to their algebraic signs, the positive quantities being counted one way or added, and negative ones the opposite way or subtracted. 15. Def. The Algebraic Sum of several quantities is the surplus of the positive quantities over the nega- tive ones, or of the negative quantities over the positive ones, according as the one or the other is the greater. The sum has the same algebraic sign as the prepon- derating quantity. Example. The sum of + ^ and — 7 is ; + 9 '' -^7 ^^ 4-2; 4-5 ^^ -_7 ^^ _2. The sum of several positive numbers may be represented on the line of numbers, § 11, by the length of the line formed by placing the lengths represented by the several numbers end to end. The total length will be the sum of the partial lengths. If any of the numbers are negative, the algebraic sum is represented by laying their lengths off in the opposite direction. Example 1. The algebraic sum of the four numbers 9, — 7, 1, —6, would be represented thus : algebbaIg addition. ET Here, starting from 0, we measure 9 to the right, then 7 to the left, then 1 to the right, then 6 to the left. The result would be 3 steps to the left from 0, that is, — 3. Thus, — 3 is the algebraic sum of +9, —7, +1, and —6. Ex. 2. If we imagine a person to walk back and forth along the line of numbers, his distance from the starting- point will always be the algebraic sum of the separate distances he has walked. Ex. 3. A man's wealth is the algebraic sum of his posses- sions and credits, the debts which he owes being negative credits*. If he has in money $1000, due from A $1200, due to X $500, due to Y $350, his possessions would, in the language of algebra, be summed up as follows : Cash, , . ... + $1000 Due from A, . . . _ +1200 Due from X, . - . . — 500 Due from Y, ,. . . . — 350 Sum total, . . . . + $1350 [In the language of Algebra, the fact that he owes X $500 may be expressed by saying that X owes him — $500.] 16. Bef. To distinguish between ordinary and algebraic addition, the former is called Numerical or Arithmetical addition. Hence, the numerical sum of several numbers means their sum as in arithmetic, without regard to their signs. 17. Rem. In Algebra, whenever the word sum is used without an adjective, the algebraic sum is understood. 8 THE ALOEBBAJg LANGUAGE. Algebraic Subtraction. 18. Memorandum of arithmetical defiyiitions and oijerations. The Subtrahend is the quantity to be subtracted. The Minuend is the quantity from which the subtrahend is taken. The Remainder or Difference is what is left. If we subtract 4 from 7, the remainder 3 is the number of anit steps on the scale of numbers (§ 11) from +4 to +7. This is true of any arithmetical difference of numbers. In Algebra, the operation is generalized as follows : 19. Def. The Algebraic Difference of two num- bers is represented by the distance from one to the other on the scale of numbers. The number from which we measure is the Subtra- hend. That to which, we measure is the Minuend. If the minuend is algebraically the greater (§ 13), the difference is positive. If the minuend is less than the subtrahend, the dif- ference is negative. In Arithmetic we cannot subtract a greater number from a less one. But there is no such restriction in Algebra, because algebraic subtraction does not mean taking away, but finding a difference. However the minuend and subtrahend may be situated on the scale, a certain number of spaces toward the right or toward the left will always carry us from the subtra- hend to the minuend, and these spaces make up the difference of the two numbers. 30. The general rule for algebraic subtraction may be deduced as follows : It is evident that if we pass from the subtrahend to on the scale, and then from to the minuend, the algebraic sum of these two motions will be the entire space between the subtrahend and minuend, and will therefore be the remainder required. But the first motion will be equal to the subtrahend, but positive if that quantity is negative, and vice versa, and the second motion will be equal to the minuend. ALGEBRAIC MULTIPLICATION. 9 Hence the remainder will be found by changing the algebraic sign of the subtrahend^ and then adding it algebraically to the minuend. EXAMPLES. Subtracting + 5 from + 8, the difference is 8 — 5 = 3. + 8 ^^ + 5, " " " 5 — 8 = — 3. + 8 ^^ — 5, ^^ " ^^_5_8 = — 13. — 8 ^^ h, " " ^^ 5 H 8 = + 13. + 13 " 0, " " -^ " ^ —13. _ 13 '' 0, " " ^ ^' _l_ 13. 31. By comparing algebraic addition and subtraction, it will be seen that to subtract a positive number ig* the same thing as to add its negative, and vice versa. Thus, To subtract 5 from 8 gives the same result as to add — 5 to 8, namely 3. To subtract — 5 from 8 gives 8 + 5, namely 13. Hence, algebraic subtraction is equivalent to the algebraic addition of a number with the opposite algebraic sign. Algebraists, therefore, do not consider subtraction as an operation distinct from addition. Algebraic Multiplication. 33, Memorandum of arithmetical definitions. The Multiplicand is the quantity to be multiplied. The Multiplier is the number by which it is multiplied. The result is called the Product. Factors of a number are the multiplicand and multiplier which produce it. 23. To multiply any algebraic quantity by a posi- tive whole number means, as in Arithmetic, to take it a number of times equal to the multiplier. Thus,' 4x3= 4 + 4 + 4=+ 12; — 4x3=— 4 — 4 — 4= — 12. The product of a negative multiplicand by a positive multiplier will therefore be negative. 10 THE ALGEBRAIC LANGVAOE. 34. If the multiplier is negative, the sign of the product will be the opposite of what it would be if the multiplier were positive. Thus, +4 X —3 = —12; -4 X -3 = + 12. The product of two negative factors is therefore positive. 35. The most simple way of mastering the use of algebraic signs in multiplication is to think of the sign — as meaning opposite in direction. Thus, in §11, — 4 is opposite in direction to + 4, the direction being that from 0. If we mul- tiply this negative factor by a negative multiplier, the direction will be the opposite of negative, that is, it will be positive, A third negative factor will make the product negative again, a fourth one positive, and so on. For example, -3 X -4= +12; —2 X —3 X —4 = — 2 X +12 = —24; —3 X —2 X -3 X -4 rrz -3 X -24 = + 72 ; etc. etc. Hence, 36. Theorem, The continued product of an even number of negative factors is positive ; of an odd num- ber, negative. Rem. Multiplying a number by —1 simply changes its sign. Thus, _|-4 X — 1 = — 4; -4 X -1 = + 4. EXERCISES. Find the algebraic sums of the following quantities : 1. 4 — 6 + 12 — 1—18. 2. — 6 — 3 — 8. 3. _ 6 — 10 — 9 + 34. 4. Subtract the sum in Ex. 3 from the sum in Ex. 2. 5. Subtract the sum 5 — 6 + 3—1 — 16, from the sum — 2 — 7-4 + 8. ALGEBRAIC DIVISION. 11 6. Subtract the sum 5 — 6+3-— 1 — 16, from the sum 7 _ 3 _ 8 + 4. 7. Form the product —7x8. 8. Form the product —8x7. 9. Form the product 6x— 5x7x — 4. 10. Form the product — 6x— llx8x— 2. 11. Form the product — Ix— -Ix— Ix— 1. 1 2. Subtract the sum in Ex. 1 from the sum in Ex. 3, and multiply the remainder by the sum in Ex. 2. 13. Subtract 8 from — 3, — 3 from -1,-1 from 8, and find the sum of the three remainders. 14. Subtract 7 from — 9 and the remainder from 2, and multiply the result by the product in Ex. 7. Algfebraic Division. 37. Memorandum of aritlimetical definitions. The Dividend is the quantity to be divided. The Divisor is the number by which it is divided. The Quotient is the result. 38. Bule of Signs in Division, The requirement of division in Algebra is the same as in Arithmetic ; namely. The product of the quotient hy the divisor must he equal to the dividend. In Algebra, two quantities are not equal unless they have the same algebraic sign. Therefore the product, quotient x divisor must have the same algebraic sign as the dividend. From this we can deduce the rule of signs in division. Let us divide 6 by 2, giving 6 and 2 both algebraic signs, and find the signs of the quotient 3 : + 3x+2=+6; therefore, +6 divided by +2 gives -i-3. + 3x— 2=— 6; '' —6 " ''—2 '' +3. — 3x+2=:— 6; " —6 '' '' -\-2 '' —3. — 3 X — 2 z= +6; " +6 " " —2 " —3. 12 THE ALGEBRAIC LANGUAGE. Hence, the rule of signs is tlie same in division as in mul- tiplication, namely : Like signs in dividend and divisor give + . Unlike signs give — . EXERCISES. Execute the following algebraic divisions, expressing each result as a whole number or vulgar fraction : 1. Dividend, — 7 + 10 — 11 + 25 ; divisor, 20 — 3. 2. Dividend, 12 — 3 + 15 — 10 ; divisor, 3 — 10. 3. Dividend, 25 — 36 + 6 — 20 ; divisor, —3 + 8. 4. Dividend, — 7 x — 8 ; divisor, —8 + 4. 5. Dividend, 56 + 8 x — 3 ; divisor, — 4 — 4. 6. Dividend, — 24 x — 1 ; divisor, — 3 x — 3. 7. Dividend, —13 x —10 x — 8 ; divisor, — 4x5x— 6. 8. Dividend, — 1 x — 1 ; divisor, — 3 x — 3. CHAPTER II. ALGEBRAIC SYM BOLS. Symbols of Quantity. 39. Algebraic quantities may be represented by letters of the alphabet, or other characters. 1 The characters of Algebra are called Symbols. 30. Def. The Value of an algebraic symbol is the quantity which it represents or to which it is equal. The value of a symbol may be any algebraic quan- tity whatever, positive or negative, which we choose to assign to the symbol. 31. The language of Algebra differs in one respect from ordinary language. In the latter, each special word or sign SIGJVS OF OPERATION. 13 has a definite and invariable meaning, which every one who uses the language must learn once for all. But in Algebra a symbol may stand for any quantity which the writer or speaker chooses, and his results must be interpreted according to this meaning. 33. The same character may be used to represent several quantities by applying accents or attaching numbers to it to distinguish the different quantities. Thus, the four symbols, a, a', a", a'", may represent four different quantities. The symbols ai, a.2, ^3, «4? «5? ^tc, maybe used to designate any number of quantities which are distinguished by the small number written after the letter a. Si^iis of Oi)eratioii. 33. In Algebra, the signs +, — , and x are used, as in Arithmetic, to represent addition, subtraction, and multiplication, these operations being algebraic, not numerical. 34. Signs of Addition and Subtraction. The com- bination a-\-h means the algebraic sum of the quantities a and 5, and a — h means their algebraic difference. If « = + 4 and ^> = + 3 If a = + 5 and d If « = — 6 and ^ = + 3, If a = — 6 and h The signs of addition and subtraction are the same as those used to indicate positive and negative quantities, but the two applications may be made without confusion, because the opposite positive and negative directions correspond to the opposite operations of adding and subtracting. 35. Sign of Multiplication. The sign of multipli- I cation, x , is generally omitted in Algebra, and when different symbols are to be multiplied, the multiplier is EXAMPLES. ■f 3, then a-{-h = -\-'1, a—h = + 1. — 7, then a-\-l = —'^, a—h ■= + 12. ■f 3, then a-\-b =1 —^, a—h = — 9. — 3, then «5 + ^ = — 9, «5— Z> = —3. 14 THE LANGUAGE OF ALGEBRA. written before the multiplicand without any sign be- tween them. Thus, 4« means « x 4. ax " X X a. Sahmy '* yxmxbxaxS. If numbers are used instead of symbols, some sign of mul- tiplication must be inserted between them to avoid confusion. Thus, 34 would be confounded with the number thirty-four, A simple dot is therefore inserted instead of the sign x . Thus, 3.4 = 4x3 = 12. 3.12-2 = 72. 1.2.3.4.5 = 120. 1.2.3.4.5.6 = 720. The only reason why the point is used instead of x , is that it is more easily written and takes up less space. 36. Division m Algebra is sometimes represented by the symbol -^, the dividend being placed to the left and the divisor to the right of this symbol. Ex. a -7-b means the quotient of a divided by 5. But division is more generally represented by writing the dividend as the numerator and the divisor as the denominator of a fraction. Ex. The quotient of a divided by i is written %• It is shown in Arithmetic that a fraction is equal to the quotient of its numerator divided by its denominator ; hence this expression for a quotient is a vulgar fraction. 37. Powers and Exponents, A Power of a quan- tity is the product obtained by taking that quantity a certain number of times as a factor. Def. The Degree of the power means the number of times the quantity is taken as a factor. If a quantity is to be raised to a power, the result mayf, in accordance with the rule for multiplication, be SIGI^S OF OPERATION. 15 expressed by writing the quantity the required number of times. Examples. The fifth power of a may be written axaxaxaxa or aaaaa ; and the fourth power of 7, 7.7.7.? = 2401. To save repetition, the symbol of which the power is to be expressed is written but once, and the number of times it is taken as a factor is written in small figures after and above it. Thus, aaaaa is written a^ ; 7.7.7.7 " " 74; XXX " " x^. Dsf. A figure written to indicate a power is called an Exponent. Def. The operation of forming a power is called /^ Involution. 38. Hoots. A Root is one of the equal factors into which a number can be divided. Def. The figure or letter showing the number of equal factors into which a quantity is to be divided is called the Index of the root. The square root of a symbol is expressed by writing the sign y" (called root) before it. Ex. I. V49 means the square root of 49, that is, 7. Ex. 2. \/x means the square root of x. Any other root than the square is represented by writing its index before the sign of the root. Ex. I. \^x means the cube root of ^. Ex. 2. Vo: means the fourth root of x, Def. The operation of extracting a root is called Evolution. 39. The operations of Addition, Subtraction. Multi- plication, Division, Involution, and Evolution, are the six fundamental operations of Algebra. 16 TEE ALGEBRAIC LANGUAGE. 40. Bef. An Algebraic Expression is any combi- nation of algebraic symbols made in accordance with the foregoing principles. EXERCISES. In the following expressions, suppose «j = — 7, ^ = — 5, c zz: 0, m = 3, ^ = 4, ^-^ = 9, and compute their numerical values. I. a + Z> + ^?z + 7J. 2. a + 7)1 + ?Z. 3. m — n — a — h. 4. 71 -{- p — m — a. 5- 3a — m + ^ — 271. 6. 2a — 7p-^2h — m 7. 3m np. 8. 7nncp. 9- hm7i. 10. hup. II. ahnp. 12. 2hih7ip. 13. am + In. 14. am — hi. 15- hp — «7Z. 16. 6p -\- an. 17. n^p + wi^Z^. - 18. 7)1^)1 — ap^. 19. a^ + R 20. a^ + h\ 21. a^ - ¥, 22. ahn —- h^n. 23. aW — mhi^. 24. aW - ¥m\ 25. aW- + a%. 26. ah^ — a%. 27. ah + ^^?^ 28. ac — hp hi — 7n2) 20. 2m%2 _ I0m3 ^0. ah — mp p — hc7}i m — n In the following expressions, suppose oj = 8, ^ = — 3, and X to haye in succession the fifteen values — 7, — 6, — 5, etc., to f 7, and compute the fifteen corresponding values of each expression : 31. x^-\-hx-^a. 32. • hx Arrange the results in a table, thus : x = -7; Expression 31 =: 78 ; Exp. 32 - -!!• X — —6; " = 62 ; etc. X — --6; " — 48. etc. etc. etc. COMPOUND EXPRESSIONS. 17 CHAPTER III. FORMATION OF COMPOUND EXPRESSIONS. Fundamental Principles. 41. The following are two fundamental principles of the algebraic language : First Principle. Every algebraic expression, how- ever complex, represents a quantity, and may be operated upon as if it were a single symbol of that quantity. Second Principle. A single symbol may be used to represent any algebraic expression whatever. 43. When an expression is to be operated upon as a single quantity, it is enclosed between parentheses, but the parentheses may be omitted, when no ambiguity or error will result from the omission. Example. Let us have to subtract h from a, and multiply the remainder by the factor m. The remainder will be ex- pressed hj a — h, and if we write the product of this quantity by m, in the way of § 35, the result will be ma — h. But this will mean h subtracted from ma, which is not what we want, because it is not a, but a — b which is to be multi- plied by m. To express the required operations, we enclose a — h in brackets or parentheses, and write m outside, thus : m {a — b). NUMERICAL EXAMPLES. 7(8 — 2) = 7-6 = 42; but 7-8 — 2 = 56 — 2 in 54 12(3 +4) = 12-7 = 84. (6 -f3)(2 + 6) = 9.8 - 72. (7 _ 4) (1 - 5) (2+ 7) = 3 X -4.9 = - 108. 2 Divided by m, 18 THE LANGUAGE OF ALGEBBA, Example 2. Suppose that- the expression a — h-\-c is to be added to 7n, subtracted from m, multiplied by m, divided by 7n, raised to the third power, or have the cube root extracted. The results will be written: Added to 7n, 7n + {a — b + c). Subtracted from m, m — {a — b + c). Multiplied by m, m (a — ^ + c), (a — b-i-c) m Cubed, {a--b-\-cY. Cube root extracted, ^/(a — b-\~c). There are two of these six cases in which the parentheses are unnecessary, although they do no harm, namely, addition and division, because in the case of addition, 771 -^ {a — b -^ c) is the same as m -\- a — b -\- c. [For example, 10 + (8 - 5 + 4) = 10 + 7 == 17, and 10 + 8 — 5 + 4 = 17 also.] Again, in the case of the fraction, it will be seen that it has exactly the same meaning with or without the parentheses. 43. An algebraic expression having parentheses as a part of it may be itself enclosed in parentheses with other expressions, and this may be repeated to any extent. Each order of parentheses must then be made larger or thicker, or different in shape to distinguish it. Examples, i. Suppose that we have to subtract a from ^, the remainder from c, that remainder from d, and so ouc We shall have. First remainder, h — a. Second, c — {b — a). Third, d-lc-ib- a)]. Fourth, e-\d---[c-^{b--ay\\. Fifth, f^le-^\d-^[c^{b^a)^\\ DEFINITIONS, 19 2. Suppose that we have to multiply the difference of the quantities a and ^ by ^ and subtract the product from m. The result or remainder will be m —p{a— b). Suppose now that we have to multiply this result by p + q. We must enclose both factors m parentheses, and the result will then be written : EXERCISES. In the following expressions, suppose a =z —1, b ^=2 3, m = 5. x =: — 3, — 1, + 1, -f 3, and calculate the four values of each expression which result from giving x the above four values in succession. x{x — a) {x — 2a) {x — 3a) [a{b — x) — b {a — x)f m (b — x)-\-b {m — x) 3. \ax -^ b(x -^ aY + m (x — a)^^ 4. [,y^{mx^ _f- ^) — ^(wx^ — b)] ^{fnb — a). Note, When the square root is not an integer, it will be sufficient to express it without computing it in full. Thus, for X = —'dj we shall have ^{mx^ + 6) - ^^{mx^' - &) = V^S - ^^42, This is a sufficient answer without extracting the roots. Defiiiitionso 44. Coefficient. Any number which multiplies a quantity is called a Coefficient of that quantity. A coefficient is therefore a multiplier. Example. In the expression 4,abx, 4 is the coefficient of abx, ia " " " bx, 4:ab '' " " X. 20 THE LANGUAGE OF ALGEBRA. Def. A Numerical Coefficient is a simple ntiinber, as 4, in the above example. Def. A Literal CoefEicient is one containing one or more letters used as algebraic symbols. Kem. Any quantity may be considered as having the coefficient 1, because \x is the same as x. Reciprocal, The ReciprocaL of a number is unity divided by that number. In the language of Algebra, Reciprocal of JV = -^^ Formula, A Formula is an expression used to show how a quantity is to be expressed or calculated. Term, When an expression is made up of several parts connected by the signs -h or — , each of these parts is called a Term. Example. — In the expression, a -^-Ix -\- 3ma^, there are three terms, a, bx, and dmx^. When several terms are enclosed between parentheses, so as to be operated on as a single symbol, they form a single term. Thus, the expression {a^ b x + Zmx^) {a + h) forms but a single term, though both numerator and denom- inator are each a product of several terms. Such expressions may be called compound terms. Aggregate, A sum of several terms enclosed be- tween parentheses in order to be operated upon as a single quantity is called an Aggregate. Algebraic expressions are divided into monomials and polynomials. A Monomial consists of a single term. DEFINITIONS. 21 A Polynomial consists of more than one term. A Binomial is a polynomial of two terms. A Trinomial is a polynomial of three terms. Note. The last three words are commonly applied only to sums of simple terms, formed of single symbols or products of single symbols. Entire. An Entire Quantity is one which is ex- pressed without any denominator or divisor, as 2, 3, 4, etc. ; a, &5 x^ etc. ; 2a6, 2mp, db {x — y\ etc. A Theorem is the statement of any general truth. 45. Other Algebraic Signs. Besides the signs al- ready defined, others are of occasional use in Algebra. >, the Sign of Inequality, shows when placed be- tween two quantities, that the one at the open end of the angle is the greater. Ex. I. a'yh means a is greater than 1). Ex. 2. m ), (« + &) — (m + n), (a + b) — (m -f n). 7. 7^3 — 2a2 ^ 3^^^ _ ^3 _ ^2 _ ^^^ _ 6^3 j_ 3^2 _ 2«i^;. 32 ALGEBRAIC OPERATIONS. 8. (m + nf + X, 2 (m + 7iY — y, 3 {7n + nf — 2a;, (m + 7i)^ -— y- 9. (i^ + ^Y - 6> (i^ + ^)' + ^^ (^ + ^)' + ^. (i^ + ^)H^. 10. 6a{x — y), ba(x — y), 2a{x — y), a{x — y), 11. 2 (m — 7^) a; 4- 2, 3 (m + t^) a; — 5, b{m + n)x — Q, 7 {m -{- n)x — 8. ^- ., ^^ 9^A.^y ^ y y ^ ^ ^ a a ha h' h 7 a 7 ir m ^x ^m ^x ^m .x ^ m 11. , 2 2 — , 3 3 — . 4 4 — y n y n y n y n m -\- n m + n m + n m + n 15. Of two farmers, the first had 2x — dy acres, and the second had x — y acres more than the first. How many acres had they both? 16. A had 2x dollars, B had y dollars less than A, and C had 2y dollars more than A and B together. How many had they all ? 17. A father gave his eldest son x dollars, his second 5 dol- lars less than the first, his third 5 dollars less than his second, and his fourth 5 dollars less than his third. How much did he give them all ? 55. Addition with Literal Coefficients, When dif- ferent terms contain the same symbol, multiplied by diiferent literal coefficients, these coeflScients may be added and the common symbol be aflBixed to their aggregate. EXAM PLE S. 1. As we reduce the polynomial ' 6x -{- 6X'—2x to the single term (6 + 5 — 2) ^ = 3a^, so we may reduce the polynomial ax ■\-hx -^ ex to the single term, (a -^-h — c)x. 2. The expression mx -i- ny — bx -{- dy -\- a + b may be expressed in the form {m —b) X -{- {?i 4- d) y + a -\-h SUBTRACTION. 33 EXERCISES. Collect the coefficients of x and y in the following ex- pressions : 1. ax -{-ly -\- mx + ny, 2. mnx + %by + lyqx — Aiy, 3. dx — 2y-\- 6bx -— 4^ + 7ax + m -{- n, 4. Sax -f 8^.-?; -}- by -{- 7x — 6y -{- x — by, 5. a:r + % ■+■ c;2 — r/7iC — ny — pz. 6. 2dx + 3ey + 4/2; — 2fx — 3cr — my + 2(:a/^ — dy -\- Vx. 16. hm^y — 6a; + 4a/^ — 3A/a; — y + Vy. 1 7. 4^/0; — 6«/ + aVy + ex — Vy — 4a V^/ + Va:. Algebraic Subtraction. 56. Def. Algebraic Subtraction consists in ex- pressing the difference of two algebraic quantities. jRule of Subtraction. It has been shown (§ 21) th^t to subtract a positive quantity, &, is the same as to add, algebraically, the negative quantity, —6. Also, that to subtract — & is equivalent to adding +6. Hence the rule : Change the algebraic sign of all the terms of the subtrahend, or conceive them to be changed, and then proceed as in addition. 34 . ALGEBRAIC OPERATIONS. NUMERICAL EXAMPLES. Min., 10 + 6 = 16 10+ 6 = 16 10+ 6 = 16 10+ 6 = 16 Subt., _9 ^=_9 9 — 4= 5 9— 8= 1 9 — 12 = -3 Eem., 1 + 6= 7 1 + 10=11 1 + 14=15 1 + 18= I9 ALGEBRAIC EXERCISES. I. From ^x — 4:ay + 5^ + c, Subtract x — 7a2/ — 8b -i- d. WORK. Minuend, 3x — 4:ay + 5^ + c Subtrahend with signs changed, — x -\- 7ay + Sb — d Difference, 2x + Say + 13b + c — d Next we may simply imagine the signs changed. 2. From 7x — Uxy — 12cy + 8& + 3ac Take 2x + Ibxy + 8cy — 6b — 2d Diff., 6x — 11^2:^ — 20cy + Idb + 3ac + 2d 3. From 8« + 9<^ — 12c — ISd — 4:X + 3cy Take I9a — 7^ — Sc — 2hd H- 3a: — A:y 4. From 257;? + 20W + 92?/ + 35aa; — 6 Take 140^ — 82^^ + 20z/ + %2ax + 14 5. From 8a + 14*^ subtract 6« + 20^. 6. From a — J + c — J take — a-\-b — G-^d. 7. From 8« — 2^ + 3c subtract 4:a — Q>b — c — 2d, 8. From 2x^ — 8a; — 1 subtract bx^ — 6.t + 3. 9. From 42;4 ^ 3:z;3 _ 2^2 — 7a; + 9 subtract a;^— 2a;3--2a> + 7a? - 9. ^10. From 2x^ — 2aa; + 3a^ subtract x^ — ax + ri^l 11. From «3 _ 3^2^ ^ 3^^2 _ ^3 subtract — «3 ^ 3^2^. 12. From 7a:8 _ 2:?;2 + 2a; + 2 subtract 4a;3..,2a.^— 2a;— 14. 13. From b{x — y) -\- 1 {x — z) + 9 (;>;— a;) take 9 (a; — y) + 7 (a; — 2;) + 5 (2; - a;). 14. From l2{a-'b) — 3(a^h)-{-'ia— 2b take 7 {a—b) -- 5 (a + ^>). 15. From 7^-11^-15^ take -- 5- + 6 ^ - 7 - + 8 ^ y z X y z X b SUBTRACTION, 35 Clearing of Parentheses. 57. In § 42, 2y it was shown that an aggregate of terms in- cluded between i)arentheses might be added or subtracted by simply writing + or — before the parentheses. When an aggregate not multiplied by a factor is to be added or subtracted, the parentheses may be removed by the rules for addition and subtraction, as follows: 58. Plus Sign he/ore Parentheses. If the paren- theses are preceded by the sign +, they may be removed, and all the terms added without change. Example i. 27+(8-~5— 4 + 7) = 27 + 8— 5-4 + 7 = 33. 2. m-{-{a — X— y + z)=i7n-\-a^x — y-^z, 3. 22; + (— 3a; — by) + m—^a) + (2y— 26?) — ^x — ^x — by-^^y — 4:a-\-2y — 2a — — X — 6a. The sign + which precedes the parentheses should also be considered as removed, but if the first term within the paren- thesis has no sign, the sign + is understood, and must be written after removing the parentheses, EXERCISES. Clear of parentheses and simplify 1. x — y+ (x + y). 2. X -i- y -i- {y — x). 3. 3ab — %mp + {ah — ox — 2mp). 4. 2ax — 3by + {7nx — 2ax — pz -\- Sby). '• 'f + (|-^3 + (5 + ^^)- 59. Minus Sign before Parentheses. If the paren- theses are preceded by the sign — , they may be removed and the algebraic sign of each of the included terms changed, according to the rule for subtraction in § 56. EXAMPLES. I. 27-(8-5-4 + 7)=:27 — 8 + 5 + 4 — 7 = 21; that is, 27 — 6 = 21. 36 ALGEBRAIC OPERATIONS, 2. m — {— a— p-\-y + x) =:m-\-a-\-p — y--x. 3. 3« 4- X — (2a — bx) — {9x — a) =1 3a -{- x — 2a + 6x — 9x -{- a. Simplifying as in § 54, this reduces to 2a — Sx, EXERCISES. Clear the following expressions of parentheses and reduce the results to the simplest form by the method of § 54. 1. ab — (m — Sab + 2ax) — 7a^. 2. X — {a — x) -{- {x — a). 3. 2b -i {b — 2c) — {b+ 2c). 4. ix — 3i/ -^ 2z — {— ^x -{■ 6y — dz) — (x — y). 5. lax — 2by — {^ax -f 3by) — {^ax — 3by), 6. {a — x) ^ {a -^ x) -\- 2x. 7. — {a — b) — (b — c) — {c — a). 8. — (3m + ^n) — {3m — 2n) -f 9m. 60. We may reverse the process of clearing of parentheses by collecting several terms into a single aggregate, and chang- ing their signs when we wish the parentheses to be preceded by the minus sign. The proof of the operation is to clear the parentheses introduced, and thus obtain the original expression. EXERCISES. Eeduce the following expressions to the form X — {an aggregate). 1. X — a — b. Ans. x — {a -\- b), 2. X — m — n. 3. a -\- X — 3x -{- 2y. Ans. x — { — a -\- 3x — 2y), 4. —3b-Yx-\-2c^ 6d. 5. 2x--2a -i- 2b. Ans, x — {— x -\- 2a — 2b). 6. 2x -^ a — b. 7. 3x — 2m + 2n. 8. 3x -\- ab — m — 3ab + 2m. 9. X — 2m — {3a — 2b). Ans. x — {2m -{- 3a — 2b)o 10. X -{- 3 — {a -}- b). 11. X -}- a — {b — c) -{■ {m — n). 12. X — {am + b) — {]) — q) — {am — n). 13. x— {a -\- b) — {p — q) — {m — n). SUBTMACTIOJV, * 37 Coini>ound Parentheses. 61. When parentheses of addition or subtraction are en- closed between others, they may be separately removed by the preceding rules. We may either begin with the outer ones and go inward^ or begin with the inner ones and go outward. It is common to begin with the inner ones. EXAMPLES. Clear of parentheses: 1. f-[e-\d-[c-ib-a)]\l Beginning with the inner parentheses, the expression takes, in succession, the following forms: f-[e- \d- [c-b + a]}'] = f—le— {d — c + b — afj = f-le-d+ c-b i-a^ = f — e-i-d — c + b — a, 2. x^[— {a -^ b) +• {m + ^) — {^ — y)]. Eemoving the inner parentheses, one by one, we have, X — [— a — b-\-m-]-n — x-\-y\ ^x-{-a-{-b— m — 7i-{-x — y. EXERCISES. Kemove the parentheses in the following expressions, and combine terms containing x and y, as in §§ 54 and 55. 1. m + [-(p-q) ^ (a-b) + {-c + d)l 2. m — [— {a — b) — {p + q) + {n — Jc)]. 3. 7 ax — [{2ax + by) — {3ax — by) + {— 7ax + 2by)]. 4. a — [a — \a — [a — {a — a)]\^. 5. j>j — [a — ^> — (5 + / + ^) + (— m — n)]. 6. 2ax — [Sax — by — {7ax + 2by) — {6ax — 3by)], 7. ax-{-by-\- cz + [2ax—3cz — {2cz-}-5ax) — C^by—Scz)], 8. x-^ \ 2x — y — [3x — 2y — {4x — 3y)] \. 9. ax — bz — \ ax -{- bz — [ax — bz — (ax + bz)] }. 10. 7ny — {x -^ 3y -\- [2my — 3 {x — y) — iab'] + 5 f. 38 ALGEBBAIG OPERATIONS, II 12 13 14. IS ax + 4icx — (mic + c^ — 2/) + [^^^ — (px 4- 2/)]. 3^a^ _ Ux — (— 3a«/ — 'daz + 3%) — Uz, Idax + 2:^y — d — [7ad f (a;?/ -f t?)] — 4:xy. m -\- 4:X — [— 4?/ H- 2^ 4- (^^ — ^) + p]- 2aVy — 3m — [^Va; — 6^ + {Vy — 2a/^)]. CHAPTER II. MULTI PLICATION. 63. The product of several factors can always be expressed Iby writing them after each other, and enclos- ing those which are aggregates within parentheses. E X A \ T p L E S . The product ot a + b hj c =^ c {a + b). The product of — ^r-^ by x — y = {x — y) — ^• The product of a \- l hj c -\- d z= i^c -\- d) (a -\- b). Such products may be transformed and simphfied by the operation of algebraic multiplication. General Laws of Multiplication. 63. Law of Commutation, Multiplier and multi- plicand may be interchanged without altering the product. This law is proved for whole numbers in the following way. Form several rows of quantities, each represented by the letter a, with an equal number in each row, thus, a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a MULTIPLICATION. 39 Let m be the number of rows, and n the number of a's in each row. Then, counting by rows there will be m X n quantities. Counting by columns, there will be n X m quantities. Therefore, m x n = n x m, or nm = mn, 64. Law of Association, When there are three factors, m, n, and a, 771 {no) = {7nn)a. Example. 3 x (5 x 8) = 3 x 40 = 120. {3x5)x8 = 15 + 8 = 120. Proof for Wliole Numbers. If a in the above scheme represents a number, the sum of each row will be na. Because there are m rows, the whole sum will be 7n (no). But the whole number of «'s is inn. Therefore, m (na) = {mn) a. 65. The Distributive Law. The product of an ag- gregate by a factor is equal to the sum of the products of each of the parts which form the aggregate, by the same factor. That is, 711 {jo -\- q -{- r) = mj) -\- jnq + mr. (1) Proof for Whole Numhers. Let us write each of the quan- tities p, q, r, etc., m times in a horizontal line, thus, p ^ p -\- p -^ etc., m times = mj)* q -^ q -{- q -\- etc., m. times = mq. r + r + r + etc., 7n times = 7nr, etc. etc. etc. If we add up each vertical column on the left-hand side, the sum of each will be ^ + §' + r + etc., the columns being all ahke. Therefore the sum of the m columns, or of all the quantt ties, will be , m{p -\- q -\- r, etc.). 40 ALGEBRAIC OPERATIONS. The first horizontal line of ^'s being mp, the second mq, etc., the sum of the right-hand column will be 7np + mq '\- mr, etc* Since these two expressions are the sums of the same quan- tities, they are equal, as asserted in the equation (1). Multiplication of Positive Monomials. 66. Rule of Exponents. Let us form the product X''^ X x\ By § 37, x"" means xxx, etc., taken m times as factor. X" means xxx, etc., taken 7i times as factor. The product is xxxxx, etc., taken {7n-\-n) times as factor. Therefore, x"^ x x'' = x'''+\ Hence, Theorem. The exponent of the product of like sym- bols is the sum of the exponents of the factors. 67. As a result of the laws of commutation and association, the factors of a product may be arranged and multiplied in such order as will give the product the simplest form. 68. Any product of monomials may be formed by combining these principles. Example. Multiply bniii^x^y^ by "Ihnx^y, By the rules of algebraic language, the product may be put into the form 6mn^x^y^7 bn x^ y. By interchanging the factors so as to bring identical sym- bols together, b'Umn^noi^x^y^y. | Multiplying the numerical factors and adding the exponents,, the product becomes 35 hmn^oi^ y^. \ MULTIPLICATION. 41 69. We thus derive the following Rule. Multiply the numerical coefficients of the factors, affix all the literal parts of the factors, and give to each the sum of its exponents in the separate factors. E X E RC I s E s.. 1. Multiply xy by oi^y. A7is, 'X^y\ 2. Multiply ^ax by 2ahx^. 3. Multiply hm^y hy^m^x, 4. Multiply 21m?/ by 2«2m. 5. Multiply 2«m by 2ma. 6. Multiply 5x'^yh by xh/h. 7. Multiply dxyz by dxyz. 8. Multiply 2abm by 2mba. 9. Multiply Sab^x^ by Samv. 10. Multiply 2'6mpqr by 2'6pqrs, 11. Multiply 12axy by 12xyz, 3 2 3 12. Multiply -m^^ by -m^/, 13. Multiply -r /^^^ by 4m^. 7 14. M.u\tiip\j -abcdhj Mefy. 70. When we have to find the product of three or more quantities, we multiply two of them, then that product by the third, that product again by the fourth, and so on. Ex. 2db X 2a^'b x '^aW' x ^bmxy = 36a^b^mxy, Exercises. Multiply 1 5 . mx X my x mz, 16. axxbxxcxx dx, 1 7. do?m X 4:Pn X m?i, 18. abx 2bc x ^ca, 19. %myi^ X hwp^ x 9;)m^ 20. abxacxadx am3 xyx 2yz x zx, 21. amx X anx x amxy x anxy x amxyz, 22. a^x X a^y x ax^ x ay'^ x «V x a^y^ x x^y^, 23. 2am X San xa^x m^ x ^mx x 2nx, Rule of Sig:ns in Multiplication. 71. It was shown in § 25 that a product of two factors is positive when the factors have like signs, and negative when they have unlike signs. Hence the rule of signs, + X + makes +? + X - " -, - X + " -, ~ X — " +. 42 ALGEBRAIC OPERATIONS, Examples. The quantity a Multiplied by 3 makes + 3a. 2 '' +2a. u iC 1 u + a. 6i u Ci 0. Ci a -1 66 — a. U a -2 U — 2a. The quantity — a Multiplied by 3 makes — 3a. u u 2 a — 2a. u u 1 u — a. (6 ii a 0. iC u -1 u + a. u u -2 6C + 2a. 72. Geometrical Illustration of the Rule of Signs, Suppose the quantity a to represent a length of one centimetre from the zero point toward the right on the scale of § 11. Then we shall have a = this line \ r The product of the line by the factors from -[-3 to —3 will be a X 3, p a X 2, a X 1, a X 0, a X — 1, a X — 2y a X — 3, ■ 1 T I We shall also have a z=z this line MULTIPLICATION. 43 The products by the same factors will be — a x^, I I I I — ax%, I I I — ri? X 0, — ax—1, I I — ax — 2, I I I — a X — 3, f I I I i These results are embodied in the following two theorems : 1. Multiplying a magnitude by a negative factor, multiplies it by the factor and turns it in the opposite direction. 2. Multiplying by —1 turns it in the opposite direc- tion without altering its length. Note. When more than two factors enter a product, the sign may be determined by the theorem, § 26. EXERCISES. I. am X ah x ac x ad. 2. ax x —bx x ex x dx. 3. X X —ax X —abx x —aicx. 4. dax X —^aW X — ba^mx, 5. —lm?y X —?tc^y^ x 6ax. 6. —2nzn x — 57i^x^ x —7i^yz — a^. 7. 2m X n X —a x —2b. 8. —3ax X —2km x —Hx x — 4ibmx» 9. —ny X gy X —2yx Sbm, 10. xy X 2y^ x y^x x 2ayxK 11. by'^ X —^gy X —2z^ X —axh. 12. bax X anx x ^z x b'^xy. 13. —4J)z X —xz X —yz X agz, 14. 2c^n X 2xh X —z^ X —bgz\ 15. —e^x X 3a: X eb^ x ay* 44 ALQEBRAIG OPERATIONS, i6. —2e X —2^ X a X Ix. 17. — 4a.r X ^ay x —2a?y x — i?^^. 18. a'^:r X — <^,V^ X ax^ x —x^y, 19. a:?;2 X _y2 X —1 X ^ax x — «V 20. wi^^ X —n^x X —mn^ x w.^ x — m^. 21. — a&2; X —ay'^ x ^^i-' x a^x^. 22. ^.^2 X ^^^ X xy X — flfiz:. 23. abc X — <:?^ x aa;^ x —1 X 'dax. 24. -ri^:?: x ^cx X — T^mx x — 4?/2 x 6m. 4 2 "^ 2K, —6mx X —2n^x x -^ac x — -m^. DO 26. —a X ^c X — 1 X T X 3a2 X ^xy x y. 27. —1 X ax X a^x x a^x^ x hx x d, 28. —an X 2am^ x —dmn x 6n^y x —m, 29. —mx X nx X —mn x —xy x —1. 30. —2px X —dqx X -m^x x -^y^ x — 1. bo Products of Polynomials by Monomials. 73. The rule for multiplying a polynomial is given by the disfcributive law (§ 65). Rule. Multiply each term of the polynomial hy the monomial, and talce the algebraic sum of the products. Exercises. Multiply 1. dx^ — 4:xy — 6y^ by — 4ax, Ans, — 12ax^ + IGax^y + 20axyK 2. 3.^2 — xy -\- y^ by 3.r. 3. x^ -\- xy -}- y^ by 3x, 4. ax -{- by + cz by axyz, 5. dax^—6ay^—7 by dabx, 6. 4m^ — 6nq by — 3mq. 7. o^y — 7«y — la'^y by Sab, 74. The products of aggregates by factors are formed in the same way, the parentheses being removed, and each term of the aggregate multiplied by the factor. MULTIPLICATION, 45 Example. Clear the following expression of parentheses : am {a — h + c) —p \a — {Ji ^ h) — m {a — b)]. By the rule of § 73, the first term will be reduced to ah7i — amd + amc. (1) The aggregate of the second term within the large paren- theses will be a — h -\- k — m {a — V) =: a — h -{- k — ma + mh^ (2) because, by the rule of signs in multiplication, — m{a—'b) = —m x a — m x —h = — ma + mb. Multiplying the sum (2) by —p and adding it to (1), we have for the result required: a^m — atnb + amc — pa -^ pli — pk -\- pma —pmb, EXERCISES. Clear the following expressions of parentheses : 1. p{a -{- m — p) -{- q{b ■— c) —-r {b + c). 2. {rn — an) x — {m + an) y + {a7i — m) z, 3. a{x — y)c — b{x — y) d+f{x + y) cd. Here note tliat the coefficient of ai — y in the first term is ac. 4. am [x — a{b -^ c)] — bn [ax -}- b{c + d)], 5. p [— a {m-\-n) + b{m—n)] — q[b {m — n)—a{m + n)], 6. dx (2q — 7ic) + 2y {6x — 3c)—z (2m + 7n), 7. am [m {a — b)c — 3h {2k — 4:d) + 4/^]. 8. 2pq [da --^b — ec—pq (2m — 3^z)]. 9. bn [—'7a — n{a-^c) — {3 — a — b)], 10. p{q — r)-{-q{r — p) + r{p — q). 75. The reverse operation, of summing several terms into one or more aggregates, each multiplied by a factor, is of fre- quent application. Thus, in § 65, having given mp -{• mq -\- mr, we express the sum in the form 7)2{p -i- q -{- r). 46 ALGEBRAIC OPERATIONS. The rule for the operation is If the sum of several terms having a commjon factor is to he formed, the coeffieients of this factor may he added, and their aggregate he multiplied hy the factor, Note. This operation is, in principle, identical with that of § 55. EXAMPLES. ahx — lex — ady -\-^dy— ohx ■i-4:ady-i- my — amy — dcmx + hinx. Collecting the coefficients of x and y as directed, we have {ab — hc — dh — dcm + bm) x -\- {—ad-\-3d+4rad-}-m—am) y. Applying the same rule to the terms within the parentheses, we find • ab — I}c — 3b = h{a — c — 3). — 3cm + bm = m.(b — 3c). — ad -\-M -\- 4:ad = Sad + 3d = {3a + 3)d = 3{a-}-l)d. m — am = m (1 — a). Substituting these expressions, the reduced expression becomes ^h{a — c — 3)+ 771 (b — 3c)'\ X -^ \3 {a -{- 1) d + m(l — «)] y. The student should now be able to reverse the process, and reduce this last expression to its original form by the method of § 74. EXERCISES. In the following exercises, the coefficients of y, z, and their products are to be aggregated, so that the results shall be expressed as entire functions of x, y, and z, as in § 55. 1. ax -\- bx — 3ax + 3bx + 6a: — 7:r. Ans. (— '^a -^ ^b — l)x. 2. my -\- py — my — 2py — - 3gy. 3. mx — ny -^ px '-' gy H- rx — sy. Ans, {7n -}- p -\- r) X -- (n + g + s) y, 4. 3az — y — 2az -\- z — az -\- y. MULTIPLICATION. 47 5- abxy — Icxy -f- J^iz/. 6. d^abxy — 24rc — a^^; — 7a;jf. 7. ^y — ^y — ^^^^ — ^^^^ + ^'^« 8. «m«/ — Z>/wy + a7iy — hny. 9- prz — "Iqrz — ^ppz + Sqhz. lO. cnx -{- b7ix — amy — 'Zhny. 76. An entire function of two quantities can be regarded as an entire function of either of them (§§ 49, 50), and when expressed as a function of one may be transformed into a func- tion of the other. Example. The expression {U + 3) a;3 _ (4^2 _ ^a) x^ + {d^ -^2a + l)x — a^ has the form of an entire function of x. It is required to express it as an entire function of a. Clearing of parentheses, it becomes 2ax!^ -\- dx^ — 4oa^x^ + 2ax^ + a^x — 2ax -\- x — a^. Now, collecting the coefficients of a^, a^, etc., separately, it becomes (_ 4:x^ ^x — 1) a^ + {2x^ + 2x^ — 2x) a + dx^ -}- x, which is the required form. EXERCISES. Express the following as entire functions of y : 1. {dy^—4.y)x^ + {y^—2y^-}-l)x^-{-{2y^-^6y^—7)x—y^—G. 2. {y^ — y^) x^ + {y^ — y)^ i- y^ — 1- 3. (\/5 — 2y^) x^ + (y4 _ 2y^) x^ + {y^ — 2y) x •{- y'^ — 2, 4. {y' + 3^/2) ^ + (2/4 + 3f/3) x^ + (^3 + 32/) x^ + (^/2 + 3) x. Multiplication of Polynomials by Polynomials. 77. Let us consider the product {a-{-l)){p + q^ r). This is of the same form as equation (1) of § 65, {a -{- h) taking the place of m. Therefore the product just written is equal to (« + ^) ^ + (« + ^) ^ + {ct + h) r. 48 ALGEBRAIC OPERATIONS, But {a + l))p :=z np -{- bp, (a -^ b) q = aq -{- bq. (a + b) r =^ a?' + br. Therefore the product is ap -\- bp -\- aq -\- bq -{- ar + br. It would have been still shorter to first clear the paren- theses from {a + b), putting the product into the form (^{P^ q -\-r) -\-b{p -\-q + r). Clearing the parentheses again, we should get the same result as before. We have therefore the following rule for multiplying aggre- gates : 78. Rule. Multiply each term of the inidtiplicancl hy each term of the multiplier, and add the products ivith their proper algebraic signs, EXERCISES. 1. (a + b) {2a — bn^ — 2bn^), 2. [a — b) {^m -f 271 — bahnn), 3- (y)i^ — ^2) {2mn + ^^m + qn), 4. {p'^ + q^ -\- r^) {pq + qr -f- rp). 5. (2a — db) {2a f 2b). 6. {)72x — ny) {mx + ny), 79. It is frequently necessary to multiply polynomials containing powers of the same letter. In this case the begin- ner may find it easier to arrange multiplicand, multiplier, and product under each other, as in arithmetical multiplication. Ex. I. Multiply Ix^ —62:2 + 5^ — 4 by ^x^ — Ax — 5. The first line under work. tlie multiplier contains 7.^'^ — 6x^-j-6x — 4 the products of the sev- 0^2 A^^ 5 eral terms of the multipli- candby3aj2 The second 21:^5 — 18.^4 _|_ 15.^3 _i2:z;2 contains the products by — 28^;^ + 24^;^ — 20rt^2_j_i(5;;p -4x, and the third by -5. _ 35^3 _^ 30^2_ 25^7; + 20 Like terms are placed — under each other to facil- 21x^—4:6x^+ 4:X^— 2x^-^ 92: + 20 itate the addition. MULTIPLICATION. 49 Ex. 2. Multiply m -\- nx -\- px^ hj a — hx, a — hx am + anx + a'px^ — hmx — hnx^ — hpoi^ am + {an — bm) x -f- {ap ~ hn) ^ — hpo? In the following exercises arrange the terms according to ^.ne powers and products of the leading letters, a^ h, x, y, or z Multiply 1. 3^2 + 5« + 7 by 2^2 _ 3a + 4. 2. a^ -\- ah + y^ hy a — K 3. a^ -\- a^ -\- ax^ + ^^ by ^ — x. 4. a^ — a^ '\- a — 1 by ^^ — a -\- 1, 5. x^ -\- ao^ -\- a^x^ + a^x ■\- a^ ]dj x — a. 6. a -\- hz -\- cz^ + f/2;^ by m — nz -\- pz\ 7. 3«2 f 5« + 7 by 2«2 + 3^ — 4. 8. a2 _ ^^ ^ ^2 by ^ ^ J. 9; a^ 4- a^o; + ax^ -\- a^ hj a — x, 10. a^ — a^ + a — 1 hj a^ -^ a — 1. 11. x/^ -{- ao(^ -^ a^x^ + aH -^ a^ hj x + a. 12. a -\- Iz -{- cz^ -{- dz^ by m + ^^2; —pz\ 13- (^ + ^''^) (^ + ^^)- 14. (a + Z^:c + c:?:2) i^'jfyi j^ fix ^ px^), 15- (y« - 3^ + 2) (*/2 - 2). 16. (2/8 + 2/3 + y + 1) (2/3 + y + 1) 17- («/' - 2«/3 + 3?/ - 4) (^3 + 22/2 + 3?/ + 4). 18. Sft^^a; — 3a2y + 2a2» by a^ — a" 19. a? + 6«5 + -5 by « — ~i. 20. (« 4- ^) 4- (a — i) by (« 4- ^) — (a — ^). 21. a2 — ^2 _j. (^ „ ^) by «2 ^ Z>2 ^ (a 4- Z^). 22. « 4- ^ -f- ^ by ^ — ^ 4- c. 23. ^2 + ^2 _ (3^2 _^ j2) by 2a + 2^ — 2 (aj — ^). 24. 2{a ^h) -\- X ~ y hy a '\- h — {x -{- y), 25. «:^"* 4- hx"* — «^^:c by aa;2 4- ba^, 26. «'" — b'' by a"* 4- ^". 4 50 ALGEBRAIC OPERATIONS. 27. — Ibx^y + ^xy'^ — 12^^ by — hxy. 2 7 1 28. -x^ + 3«.?; — ~a^ by 20;^ — aj:?; — -al 3 D ^ 4 ISToTE. Aggregates entering into either factor should be simplified before multiplying. Special Fornis of Multiplication. 80. 1. To find the square of a binomial, as a -{-l. We multiply « + ^ by a + Z>. a {a -\- d) = a^ -\- ab b{a + b) = ah + l^ {a + h){a^l)) = o? + "^ah + W' Hence, {a + Vf = «2 + %ab + ^2 (1) 2. We find, in the same way, {a — ly = a^ — 2ab + P, (2) These forms may be expressed in words thus: Theorem. The square of a binomial is equal to the sum of the squares of its two terms, plus or minus twice their product. 3. To find the product of a -f- ^ by a — b, a{a -{- b) = a^ -{- ab -^b(a + b) = -ab-W Adding, {a + b) {a — b) = a^ — b\ (3) That is . Theorem. The product of the sum and difference of two numbers is equal to the diflference of their squares. The forms (1), (2), and (3) should be memorized by the student, owing j to their constant occurrence. ^ When Z> = 1, the form (3) becomes (a + 1) (a - 1) = 6^2 — 1. The student should test these formulae by examples like the following : (9 + ^f = 92 + 2.9.4 H- 42 = 81 + 72 -f 16 rrr 169. (9 _ 4)2 =rr 92 — 2.9.4 -I- 42 =: 81 - 72 -f 16 = 25. 2. (m — - 2w)2. 4- (ix- -5yy. 6. (3x + 1) (3x - ■!)• 8. (5a^5 - 3) (5a;3 + 3) MULTIPLICATION. 51 (9 + 4) (9 — 4) r= 92 — 4- = 65. Prove these three equations by computing the left-hand member directly. EXERCISES. Write on sight the values of I. {m -\- 2^)2. 3. {da - 2b)\ 5. {%x + y) {%x - y). 7. {4ic2 + 1) {4rX'^ — 1). 81. Because the product of two negative factors is positive, it follows that the square of a negative quantity is positive. Examples. {— of — a^ =. {+ a)\ {b - ay = a^ — 2ab + b^ = (a - by. Hence, 27ie ejopression a^ — 2ab + b^ is the square hoth of a — b and of b — a, 83. We have ^ a x a := — a^. Hence, The pj^oduct of equal factors with opposite signs is a negative square. Example, — {a — b){a — b) = — a^ + 2ab — b% which is the negative of (2). Because — {a — b) = b — a, this equation may be written in the form, {b — a){a — b) = — a^ + 2ab — i^, which is readily obtained by direct multiplication. EXERCISES. Write on sight the values of I. — {a -i- b) X — {a + b). 2. (^ — y){y — ^)- 3- {^ -\-y){-^ — y)- 4. {2a — 3b) (3b — 2a). 5. (3^ — 2a) {— 3b + 2a). 6. (am — bn) {bn — am). 7. {xy — 2) (2 — xy). 52 ALGEBRAIC OPERATIONS, CHAPTER III. DIVISION. 83. The problem of algebraic diyision is to find such an axpression that, when multiplied by the divisor, the product shall be the dividend. This expression is called the quotient. In Algebra, the quotient of two quantities may always be indicated by a fraction, of which the numerator is the divi- dend and the denominator the divisor. Sometimes the numerator cannot be exactly divided by the denominator. The expression must then be treated as a frac- tion, by methods to be explained in the next chapter. Sometimes the divisor will exactly divide the dividend. Such cases form the subject of the present chapter. Division of Monomials by Monomials. 84. In order that a dividend may be exactly divisi- ble by a divisor, it is necessary that it shall contain the divisor as a factor. Ex. I. 15 is exactly divisible by 3, because 3-5 =r 15. 2. The product a¥G is exactly divisible by ac, because ac is a factor of it. To divide one expression by another which is an exact divisor of it: EuLE. Remove from the dividend those factors the product of ivhich is equal to the divisor. The remain- ing factors luill be the quotient, 85. Rule of Exponents. If both dividend and divisor contain the same symbol, with different exponents, say m and fly then, because the dividend contains this symbol m times as a factor, and the divisor yi times, the quotient will contain it m — n times. Hence, DIVISION. 63 In dividing, exponents of like symbols are to he sub^ traeted. EXERCISES. 1. Divide 26.t«/ by 2^. Ans, l^x. 2. Divide 21^2^,^ by Ihc. 3. Divide o(^ by x^, Ans. x, 4. Divide ISa^ by 6«. Ans, 3a. 5. Divide 15a^m by 3^^. Afis. bam. 6. Divide 14«3m2 by 7fi^m. 7. Divide' 16^^^^ by Sahn^. 8. Divide 36a:i/22;3 by Qxyz. 9. Divide 40a2a;8^5 by lOa^:^^^. 10. Divide 3ba}p by Ta^'. Rule of Signs in Division. 86. The rule of signs in division corresponds to that in multiplication, namely : // dividend and divisor have the same sign, the quo- tient is positive. If they have opposite signs, the quotient is negative. Proof. ■\-mx -V- ( + m) = +x, because +x x {-\-m) = -\rmx. . -\-mx -~ (— '/^O = —^9 " —^ X (—^^0 = +^?^^- ^mx -=- ( + m) = —X, " —X X ( + W) = —mx. —mx -^ {—m) = -\-x, " -\-x X (— m) = —7nx. The condition to be fulfilled in all four of these cases is that the product, quotient x divisor, shall have the same alge- braic sign as the dividend. EXERCISES. Divide I.. -\-a by -f- a. Ans. + 1. 2. + «5 by — a. Ans. — 1. 3. — a by + a. Ans. — 1. 4. —ah\ —a. Ans. + 1. 5. _ 33ahnx by llax. Ans. — dam. 6. — 24:X^yz by 12xyz. Ans. — 2x. 7. 21ani^x'" by - - 7amx"', A ns. — 3m.T"'~". 54 ALGEBRAIC OPERATIONS. 8. 9 lO II 12 13 14, 16 17 — ISa^p^ by — 6^^^. ^/is. 3a'^~'*jy'^-i. — 16a^xiny^ by 4aa;2?/'*. 14^^^ by — 7by^, — lUH^'lC^ by — ^lH^h\ V^Ka — Vfd' by 3(^-Z>)2c. 42 (2; — 2/)^ by — 7 (x — yy. — 44a* {x — yY by lla^ {x — yy. — 45^^ {a — Z>)^ by 9Z/^ {a — Z>)^. — 48 {m + 7z)^ by — 8 (^7? + oi)^, 64 (a + ly {x — ij)^ by 4 (a + I) {x — y). Ans. 4z{a — i) c^. Division of Polynomials by Monomials. 8*7. By the distributive law in multiplication^, whatever quantities the symbols m, a, b, c, etc., may represent, we have : {a + b -^ c -{• etc.) x m = 7na -\- mb -\- mc + etc. Therefore, by the condition of division, {ma + 7nb + mc + etc.) -^m=za-\-b-\-c-\- etc. We therefore conclude, 1. In order that a polynomial may be exactly divisi- ble by a monomial, each of its terms mnst be so divisible. 2. The quotient will be the algebraic sum of the separate quotients found by dividing the different terms of the polynomial. EXERCISES. Divide 1. 2fl^2 _|. 6^3^ _ 8^5^2 }3y 2a\ Ans. 1 4- ^ax — 4a% 2. Qm^n — 12m^/^2 — ISmn^ by Qmn, 3. ^a%^ — 16^4^,4 + 8a5J3 by 4.aW. 4. ^xy^ — Sa^y^ + 4x^y by -— 4:xy. 5. 12abx — 24:abx^ by — Uabx, 6. 21am^x:^ — Ua^m^x^ + 2Sa^7n^x^ hy — 7amx!^, 7. '72a^x + 24^:^ + A8ax^ by 24^:?;. 8. a{b — c) -}- b (c — a) -{- c {a — b) ■\- abc by a^^c. 9. 27 (« - Z')5 _ 18 (« - by + 9{a~ by by 9 (a - b). 10. a^ {a — by — ci^ (« — by' by a'' {a — Z>)'^. Divisioif, 55 11. {a + by {a - hf + {a-^-hf {a-hy by {a + l) {a-h). 12. 10 (x + y)^{x -.yY-.^[x ^ yy (x — yf by 6{x -\- y) (x — y). 13. (« + Z^) (a - Z^) by a^ - ^l Factors and Multiples. 88. As in Arithmetic some numbers are composite and others prime, so in Algebra some expressions admit of being divided into algebraic factors, while others do not. The latter are by analogy called Prime and the former Composite. A single symbol, as a or x, is necessarily prime. A product of several symbols is of course composite, and can be divided into factors at sight. A binomial or polynomial is sometimes prime and some- times composite, but no universal rule can be given for dis- tinguishing the two cases. 89. When the same symbol 01 expression is a factor of all the terms of a polynomial, the latter is divisible by it. EXAMPLES. I. ax -{- dbx^ + c^GX^ ^=1 a (x -t- Ix^ -\- «c.t^). c^h^x + c^Wx^ =r aWx {h + ax^. 2 EXERCISES. Factor I. ax^ -f c^x, 2. a^'^cy + c^hc^y + ahh^y. 3. a^^ h^ + a'^ h'^'^. 4. a^^ x^ — a^^ x^^ + a^ x^^. 5. a^i^^c^^ -(- a^^b^^c^ + a^^b^c^^, 90. There are certain forms of composite expressions which should be memorized, so as to be easily recognized. The following are the inverse of those derived in § 80. 1. a^ + 2ab -^ b^ = {a -\- b)\ 2. a} — 2ab -{- b^ =: (a — b)\ ' 3. a'-b'^ = {a + b) {a - b). The form (3) can be applied to any difference of even powers ; thus, 56 ALQEBBAIG OPERATIONS, a^-b^=. {a^ + h^) (^2 _ ^2) . and, in general, a^'^ — W^ — {a'' + l'^) (a'' — b""). If the exponent is a multiple of 4, the second factor can be again divided. EXAM PLES. a^-h^ ^ (a? + h^) («2 - ^2) ^ {d^j^W) (a-{-h) (a-b). a?^¥ = {a' + ¥) {a^ - ¥) = {a' + ¥) (a^ + b^) (a-^b) (a^b). When b is equal to 1 or 2, the forms become a^ — l =z {a + l)(a — 1). a^-.4c = {a + 2) {a — 2). a^-^2a + l = {a + 1)K a^-2a + l = {a-lY = {1- a^. a^ — 4^. + 4 = {a — 2)2 = (2 — ay. By putting 2b for b, they give a2 _ 4^2 = (a + 2b) (a — 2Z>). ^2 + 4a^ + 4Z>2 = (^ ^ 2b)\ EXERCISES. Divide the following expressions into as many factors as possible : 1. x^ — 16. Ans. (^2 4- 4) (a: + 2) (x — 2). 2. y^ — 16a;4. 3. a;2 + 6:r + 9. Ans. (:r + 3)2. 4. a;2 _ 6:r -f 9. 5. 4.a^x^ — Wy\ 6. lea^a;^ __ 1, 7. 9:^2 __ i2:r^ -f- 4^2. 8. ^2:^2 + 2«a;y + y\ 9. 4fl^2^2 ^ ^^^bxy + 62i^^l 10. «4 + 4«2^ ^ 4J2. 11. x!^ — 2.Ty _|. ^4^ 1 2. a;4 _ 4^2^2 ^ 4^4^ 13. ^4 _ 4^2^2 ^ 4^4; 14. a^--a%\ 15. ^2/1 _ 2^w 4. 1. 16. x^ — 4aa;^ + 4:a\ 17. 1 — 2/*- 18, 2:6^ + 2x^yh + t/^js;. • ^725. Z{X^ + 2xhf + y^) = ^ (aj3 4. ^8)2. DIVISION, 19. 21. 23- 25- 20. 22. 24. 26, ^2m _ ^» 4^^ — 9itY. 0^ _ i^;2^6^ 2:4m „ 2:6^ ^ 1. 27. a;2 + a; + -• 28. ^2m ^ ^ ^ 4 57 91. By combining the preceding forms, yet other forms may be found. For example, the factors («2 + ah-\- b^) (a2 - aJ + b^), (1) . are respectively the sum and difference of the quantities Or + W- and ab. Hence the product (1) is equal to the difference of the squares of these quantities, or to («2 ^ 52)2 _ aW = a^ + aW + ¥, Hence the latter quantity can be factored as follows : a* + aW ■\-¥ — {a^ + ab + b^) {a^ -^ ab ■\- b'^). EXERCISES. Factor I. x^ + xY -\- y^' 2. a^ + 8«2^2 ^ iej4, 3. «4 + 9a2a;2 + Sla;^. 4. a^"" ^ d^^ b^^ -\- ¥^, 8. m^ — a^ _|_ 2a^ — ¥, Ans, {m — a -\- b) {m -\- a — b)* Here the last tliree terms are a negative square. Compare § 82. 9. a^ — 4^2 _|_ 4^^ — ^^ 10^ ^8 — 4^j2 _|_ 4^j^ — fjf.2^ 93. The following expression occurs in investigating the area of a triangle of which the sides are given : (a ^ b -\- c) {a ^ b — c) {a — b -\- c) {(I — b — c), (1) By § 80, 3, the product of the first pair of factors is (a + Z>)2 — c2 =r a2 4- 2«& + Z>2 _ c2 ; and that of the second pair, 68 ALOEBRAIG OPERATIONS. By the same principle, the product of these products is (^2 4- 2>2 _ ^2)2 _ ^aW, which we readily find to be ^4 _|_ ^4 _!_ ^4 _ 2^2^ _ 2Z»%2 _ 2c^a\ (2) Hence this expression (2) can be divided into the four factors (1). Factors of Binomials. 93. Let us multiply x""-^ + ax""-^ 4- a^x^~^ + + a'^'^^x + a^-i by x — a. OPERATION. a; — a it'« _j_ ax^-^ + «2^w-2 + a%^-3 4_ 4_ ^/^-i ^ — ax'^-'^ — a^a;'^-^ — ahf'-^ — .... — fl^^^-i^. — a^ Prod., ii;^ — a^ The intermediate terms all cancel each other in the product, leaving only the two extreme terms. The product of the multiplicand by .^; — a is therefore x^ _ a'^. Hence, if we divide x'^ — a'^ by x — a, the quotient will be the above expression. Hence the binomial x'^ — a^ may be factored as follows : x'>^ -^aP' = {x — a) {x'^-^ + ax'^-^-i-a^x^-^+ -\-a'^~^x-\-a^-^). Therefore we have, Theorem. The difference of any power of two num- Tbers is divisible by the difference of the numbers themselves. Illustration. The difference between any power of 7 and the same power of 2 is divisible by 7 — 2 = 5. For instance, 72 _ 22 ^ 45 ^ 5.9. 73 — 23 = 335 = 5.67. 74 -2^ = 2385 = 5.477. etc. etc. etc DIVISION. 59 94. Let us multiply ^71-1 _ axr^-"- + a^x^-^ — + (— «)'^-2^ + (— a)^-i bj a; + a = i^; — (— ^). Eem. This expression is exactly like the preceding, except that — a is substituted for a. It will be noticed that the coefficients of the powers of x in the multiplicand are the powers of — a, because {—ay = —a, (_a)3 ^ ^a% {-ay =: ^aS etc. etc. The sign of the last term will be positive or negative, according as n — 1 is an even or odd number. OPERATION. xn-^—ax'^'^ + a^x'^-^—aH^'^-^...,-\-{-'aY-^X'\-{—ay-^ X -{- a := X — {— a) xn — ax""-^ -h o?x^-^ — aH'^-^ . . . + {—ay-^ + ax'^-^ — a^x^-^ + a^x^-^.,.. — {—a)^-^x — (— r/)^ Prod., :r^ — {—aY' Ik f^' The multiplier x -[- a is the same as x — {— a) (§ 59). In multiplying the first terms, we use + a, and in the last ones — (— a), because the latter shows the form better. Hence, reasoning as in (I), the expression x^ — (— a)^ admits of being factored thus : a^^ — (— a)"" = {x + a) [x"^'^ — ax"^'^ + a^x^'^ — .... + (— a)^-^x 4- (~ a)^-i]. If n is an even number, then (— a)^ = a% and xn _ (— a)^ — x^ — a^. If n is an' odd number, then {— a^) =z — a% and xn — (_ aY = x^ + a^ Therefore, Theorem 1. When n is odd, the binomial af'-^-a^ is divisible by riJ+ a. 60 ALGEBRAIC OPERATIONS. Theorem 2. When n is even, tlie binomial x^—a"^ is divisible 'by x-\-a. Note. These theorems could have been deduced imme- diately from that of § 93, by changing a infco — a, because X — a would then have been changed to x -{- a, and x'^ — a^ to x^ -f- a'^ or x'^ — a^, according as 7i was odd or even. The forms of the factors in the two cases are : When n is odd, • x'^ -\- a"" = {x + a) {x""-^ — ax'^-'^ + a^x^~^ — . . . . -f a^-i). When n is even, r^n __an — (x -{- a) {x^-^ — ax^-^ + a^x^'^ — . . . . -a^-^), (a) In the latter case^ the last factor can still be divided, be- cause x^ — a^ is divisible hjx — « as well as by x + a. We find, by multiplication, {x — a) {x""-^ + a^x^-^ -f «4^-6 + .... + «7i-2) — x''-^ — ax^-^ + a^x""-^ — aH""-^ + .... + oP'-'^x — oP'-K Therefore, from the last equation {a) we have : When n is even, x^ — oP z=L (x -\-a) (x—a) {x"^'^ -f a^x^-^-{- d^x^-^ — .... + op-'^). EXERCISES. Factor the following expressions, and when they are purely numerical, prove the results. Ans, (5 + 2) (5 — 2). 52 — 22 =r 25 — 4 = 21 ; (5 4_2)(5-2) = 7.3 rr: 21.] I. 5^ — Jd^. VP roof. (5 2. 53 - 23. 4. 55 — 25. 6. 73 + 23. 8. 7* — 2*. 10. o^-aK 12. x^ -—aK 14. x^ + ct\ 16 %a^ __ 2753. 18. ^ + %y\ 20. Sa^ + 27¥. 3. 5^ — 2*. 5- 56-26. 7. 73 — 23. 9- x^ — a\ II. x^ - a\ IS- x^ -\-a^ IS. a^ - 8^3. 17. 16^4 — ¥. 19. xi _ 162/4. 21. x^ — G4a6„ DIVISION. 61 Least Common Multiple. 95. Def. A Common Multiple of several quanti- ties is any expression of which all the quantities are factors. Example. The expression am/^n^ is a common multiple of the quantities a, m, n^ arriy amn, am^, m^n^, etc., and finally of the expression itself, mn^nK But it is not a multiple of a^, nor of X, nor of any other symbol which does not enter into it as a factor. Def, The Least Common Multiple of several quantities is the common multiple which is of lowest degree. It is written for shortness L. C. M. KuLE FOR eii^di:n^g the L. C. M. Factor the several quantities as far as possible. If the quantities have no common factor, the least common multiple is their product. If several of the quantities have a common factor, the multiple required is the product of all the factors, each of them heing raised to the highest power which it has in any of the given quantities, Ex. I. Let the given quantities be 2ab, U\, 6ac. The factors are 2, 3, a, h, and c. The highest power of h is l^, while a and c only enter to the first power. Hence, L. C. M. =: 6ab^c. Ex. 2. a^ — b% a^ + 2ab + b^ a^ — 2ab + IP, a^ — ¥. Factoring, we find the expressions to be, (^ + b){a-^ b\ {a + bf, {a - bf, {a^ + b^' (a + b){a^ h). By the rule, the L. C. M. required is (a + byia-^-bfia^ + V'). 62 ALGEBRAIC OPERATIONS, EXERCISES. Find the L. CM. of I. xy, xz, yz, 2. d^i, ¥c, cH, d^CL 3. a, ah, abc, abed. 4. a^, a¥, Id^. 5. x^ — y\ X -^y, X — y. 6. ^2 _ 4^ ^ _ 4^ _l_ 4^ x^ + 4rX -{- 4. 7. 16a^x^ — 4:771% 2ax + m, 2ax — m, 8. ^2 _ 1^ ^2 ^ 1^ ics — 2a; + 1, x'^ -^-^x -\- 1. 9. 4a {h + c), h(a — c), 2aK 10. 2(^ — ^)2, 2(^ + ^)2, 2 (a — Z>)(« + Z>). 11. 3(:r + ^), ^{x-y), d{x^ + y% 12. a — ^>, ^2 __ ^2^ ^3 _ ^3^ ^4 _ ^4. 13- X -{-y, x — y, a -^1), a — K 14. o:^ — a^, ic^ + a^, rr^ — c^^ X -\- a, 15. 2^ — 64^6, x^ — 16aS x^ — 4^2. 16. a- + b, a^ + 2«^ + b% a^ — ¥, Division of one Polynomial by another. If the dividend and divisor are both polynomials, and entire functions of the same symbol, and if the degree of the numer- ator is not less than that of the denominator, a division may be performed and a remainder obtained. The method of dividing is similar to long division in Arithmetic. 96, Case I. When there is only one algebraic sym^ hoi in the dividend and divisor. Let us perform the division, 3a;4 _ 4^3 + 2ir2 4- 3a: — 1 -f- x^ — x-\- 1. We first find the quotient of the highest term of the divi- sor x^, into the highest term of the dividend ^x^, multiply the whole divisor by the quotient dx^, and subtract the product from the dividend. We repeat the process on the remainder, and continue doing so until the remainder has no power of x so high as the highest term of the divisor. The work is most conveniently arranged as follows : DIVISION. Dividend. Divisor. 3^^ _ /^x^ _f 2.^'2 4- 3a; — 1 \x^ — x-i-l 3a;« X Divisor, '^X^ — Zx^ + dx^ 3a;2 _ ^ _ 2 First Remainder, — CC^ — X^ -{- 3x — 1 —X X Divisor, — iK^ + i^:^ — X Second Remainder, — 2^;^ + 4:X — 1 -2 X Divisor, —2x^-{-2x — 2> Third and last Remainder, 2^ + 1 63 Quotient. The division can be carried no farther without fractions, because x^ will not go into x. We now apply the same rule as in Arithmetic, by adding to the quotient a fraction of which the numerator is the remainder and the denominator the divisor. The result is, x^ — x-{-l x^—x-\-l ^ ' This result may now be proved by multiplying the quotient by the divisor and adding the remainder. There is an analogy between the result (a) and the cor- responding one of Arithmetic. An algebraic fraction like (a), in which the degree of the numerator is greater than that of the denominator may be called an improijer fraction. As in Arithmetic an improper fraction may be reduced to an entire numler plus a proper fraction, so in Algebra an improper frac- tion may be reduced to an entire function of a symbol plus a proper fraction. EXERCISE^. Execute the following divisions, and reduce the quotients to the form {a) when there is any remainder. 1. Divide x^ — 2x — \ by x -\-\, 2. Divide x? -\- 2x^ — %x — \ by i?:; — 1. 3. Divide x^ — Zx^ + 2a; — 1 by a;^ — ir. 2^:4 __ 2^:3 _|_ ^2 _ ^ _ 5 4. Keduce x^ — x — \ 5. Divide 24^3 — 38«2 — 32a + 50 by 2a — 3. Ans. Qaot. = 12^2 — a — —\ Rem. — —-> 64 ALGEBRAIC OPERATIONS, 6. Divide :r* — 1 by a: — 1. When terms are wanting in the dividend, they may be considered rs zero. In this last exercise, the terms in x^, x-, and x are wanting. But the beginner may write the dividend and perform the operation thus : a^ + Ox^ + Ox^ + Oaj - 1 | x- 1 ^ — ^ aj3 + a;2 + a; + 1 x^ + Ox' x^ - X' aj-^ + Ox of' — X x-1 Xj-1 The operation is thus assimilated to that in which the expression is complete ; but the actual writing of the zero terms in this way is un- necessary, and should be dispensed with as soon as the student is able to do it. 7. Divide a^ — 2« + 1 by a — 1. 8. Divide x^ -{- 1 hj x -{- 1, 9. Divide Sa^ + 125 by 2a -+ 5. 10. Divide* ^^^ + 1 by a + 1. 1 1. Divide a^ + 2^2 + 9 by a^ -]- 2a -^ 3. 12. Divide a^ — 1 hj a^ -{- 2a^ + 2a + 1. 13. Divide x^ — 12a;4 + 36x^ — 32 by x^ — 2. 14. Divide {x^ — 2x-^ 1) {x^ — 12^' — 16) by x"^ — IG. For some purposes, we may equally well perform the operation by beginning with the term containing the lowest power of the quantity, or not containing it at all. Take, for instance, Example g : 125 + Sa^ \ 5 + 2a ^^^ + 50a 25 - 10a + AcC' -60a - 50a - 20a:' 20a'^ + 8a3 20a^ + Sa^ 15. Divide 1 + 3:?; + 3x^ + x^ by 1 + x. 16. Divide 1 — 4:X -{- 4x^ — x^ hj 1 — x. 17. Divide 15 + 2« ~ da^ + a^ + 2a^ — a^ by 5 + 4<^ — a\ 18. Divide 1—f by 1 -f- 2?/ -f 2^2 ^ y\ 19. Divide 64— 64:^: + 16:z:2_8^_l_4^_^ by — 4 + 2:?; + ^^. 20. Divide 64 — 16^^:^ j^ ofi by 4 — 4a; + x^. DIVI8I0JS'. 65 97. Case II. When there are several algebraic sym- hols in the divisor and dividend. Let us suppose the dividend and divisor arranged accordiug to powers of some one of the symbols, which we may suppose to be X, as in § 76. Let us call A the coefficient of the highest power of x in the dividend, and ^ the term independent of x^ so that the dividend is of the form Ax'^ + (terms with lower powers of x) + H. Let us call a the coefficient of the highest power of x in the divisor, and h the term of the divisor independent of x, so ,that the divisor is of the form ax'^ + (terms with lower powers of x) + li. Then we have the following Theorem. In order that the dividend may be exact- ly divisible by the divisor, it is necessary : 1. That the term containing the highest power of x in the dividend shall be exactly divisible by the cor- responding term of the divisor. 2. That the term independent of x in the dividend shall be exactly divisible by the corresponding term of the divisor. Reason, The reason of this theorem is that if we suppose the quotient also arranged according to the powers of x, then, 1. The highest term of the dividend, Ax'^, will be given by multiplying the highest term of the divisor, aaP', by the high- est term of the quotient. Hence we must have, Ax^ Highest term of quotient = — ^• 2. The lowest term of the dividend will be given by multi- plying the lowest term of the dividend by the lowest term of the quotient. Hence, we must have, TT Lowest term of quotient = -j- • Eem. 1. Since we may arrange the dividend and divisor according to the powers of any one of the symbols, the above 66 ALGEBRAIC OPERATIONS. theorem must be true whatever symbol we take in the place of X. Eem. 2. It does not follow that when the conditions of the theorem are fulfilled, the division can always be performed. This question can be decided only by trial. We now reach the following rule : I. Arrange both dividend and divisor according to the ascending or descending powers of some coimnon symbol. II. Form the first term of the quotient by dividing the first terjn of the dividend by the first term of the divisor, III. Multiply the whole divisor by the term thus found, and subti^act the product from the dividend. IV. Treat the remainder as a new dividend in the same way, anfid repeat the process until a remainder is found which is not divisible by the quotient. Ex. I. Divide x^ + 3ax^ + Sa^x + a^ by x + a. OPERATION. x^ + dax^ + 3a^x + a^ \ x -{- a ^+ ^^^ x^ + '^ax + a^ 2ax^ 4- da^x 2ax^ + 2 a^x d^x + a^ cfix + a^ "O O" Ex. 2. Divide x^ — ax^ -\- a (h -\- c) x — ahc—bx^—cx^ + hex by X — a. Arrang x^ — (a-\-h-\-c)x^ -^ {al)-\-'bc-^ca)x — abc \ x — a I Arranging according to § 76, we have the dividend as follows : | ^— O^ x^—{b^c)x-\-ic — {b-}-c)x^-{- {ah-\-bc-\-ca)x — {b-\-c)x^-\- (ah + ac) x hex — ahc hex — ahc ~0 FRACTIONS. 67 EXERCISES, 1. Divide the dividend of Ex. 2 above by x — h. 2. Divide the dividend of Ex. 2 above by x — c. 3. Divide a^ + ¥— (^ + dabc hj a + b — c. 4. Divide a^ -^¥ + 3ab — l hj a + b — 1. 5. Divide u^I^ + 2abx^ — {a^ -f I^)x^ by ab + {a — b) x. 6. Divide (a^ — bcf -f ^bV by o? + be, 7. Divide (a + 5 + c) (a2> -{• be -\- ea) — abe hj a + b. 8. Divide {a + b — e){b -\- c — a) {e -{- a — b) by a^ — 52 — c2 H- 25c. 9. Divide a^ + 5^ + c^ — 3a5c })j a -\-b + c. 10. Divide a;^ -f- 4a^ by a;^ — 2^a; + "Ha^ 1 1. Divide a^{b + x) — bf^ix — a) + {a — b)oo^ + abx by ic + a + i 12. Divide 01^ — ax^ — 5^:^; -\- aW by (ic — t^) (a; + b), 13. Divide 12a%9 — 14^5^ + 6^^^ — a' by %a^7? — «3. ■♦♦♦- CHAPTER IV. OF ALGEBRAIC FRACTIONS. 98. Bef. An Algebraic Fraction is the expression of an indicated quotient when the divisor will not ex- actly divide the dividend. Example. The quotient of ;? -7- ^ is the fraction — • Def. . The numerator and denominator of a frac- tion are called its two Terms. Transformation of Single Fractions, 99. Reduction to Lowest Terms, If the two terms of a fraction are multiplied or divided Iby the same quantity, the value of the fraction will not be altered. 68 ALGEBRAIG OPERATIONS, CLCC Example. Consider the fraction — If we divide both ay terms by a, the fraction will become -• J y ax X ^'~ y Corollary, If the numerator and denominator con- tain common factors, they may be cancelled. Def, When all the factors common to the two terms of a fraction are cancelled, the fraction is said to be reduced to its Lowest Terms. To reduce a fraction to its lowest terms, factor hoth terms, ivhen necessary, and cancel all the common factors. Ex. I. — ^ = — . acny'^ en The factor ay^ common to both terms is cancelled. -^ . aW a^ The factor aW common to both terms is cancelled. Ex. 3. Keduce -1=-- aH Here a^x is a divisor of both terms of the fraction. Di- 1 o^x 1 vidinff by it, the result is -^» Hence -^- = -* ° "^ (T (vx a^ -r, mu — nu (m — n)u u Ex. 5. = ) ^ = -. mx — nx (m -— n)x x EXERCISES. Reduce the following fractions to their lowest terms : aWp^ am I. — -^—* 2. • a^¥p ahnx lOpqr^ 12axy FRACTIONS. 69 5- 7. 9- II. IS- 17. 72(6g — :r)(^-g) o^y — 'b y _ ax — hx a"- W •a^ — ^ah + l^ x^ + y^ ~a{x-\-'y) a^ U a^ - V'^ f x^ — y^ axm — axn hym — hyn 10 12 14 16 20 {a + x){m — n) 24 {a^—'ilax + x^){m—n) ahf — b^y^ ay — by ' a^ + 4:ax + 4:X^ a^ — 4:X^ a^ + 8Z^ 3 ay + 2J?/ V mx — nx {a -\- b) {m — n) 100. Rule of Sifpis in Fractions, Since a fraction is an indicated quotient, the rule of signs corresponds to that for division. The following theorems follow from the laws of multiplication and division : 1. If the terms are of the same sign, the fraction is positive ; if of opposite signs, it is negative. 2. Changing the sign of either term changes the sign of the fraction. 3. Changing the signs of both terms leaves the frac- tion with, its original sign. 4. The sign of the fraction may be changed by changing the sign written before it. 5. To these may be added the general principle that an even number of changes of sign restores the fraction to its original sign. 1'~~irb- F"""~ ^=^' a — a — a a b b — a a — b h — a Ex. Ex.2. -% = Ex. 3, m — n n — m m m — n 70 ALGEBRAIC OPERATIONS, EXERCISES. Express the following fractions in four different ways with respect to signs : X — y I. -* a m X -y a a a -^ + c a + m — 'X p-q m — n p -\- q — r "a — m + x Write the following fractions so that the symbols x and y shall be positive in both terms : X — h ^ m — X 7. H 8. c-y a -\- X — b 9' + I n - -y a — ■ X J- ■X a + ^- — X a — X -\- b X — a -\-b II. ^ • *«. , b — X a — b -\- y 101. When the numerator is a product, any one or more of its factors can be removed from the numerator and made a multiplier. ^ abmx ^ mx 7 ^ 7 1 Ex. = ab == abm =: abmx- p -vq p + q p -\-q p -\- q EXERCISES. Express the following fractions in as many forms as possi- ble with respect to factors : pax ab abc I. -— • 2. — . 3. — -T« mn c a -\- X^ —iP- «4 _ J4 r^_ 16^4 ^* a —b ^' X ' X + 2a 103. Reduction to Gimn Denominator. A quan- tity may be expressed as a fraction with any required denominator, i>, by supposing it to have the denomi- nator 1, and then multiplying both terms by D. For, if we call a the quantity, we have ^ = - = ^ • FRACTIONS. 71 Ex. If we wish to express the quantity ab as a fraction having xy for its denominator^ we write ahxy If the quantity is fractional, both terms of the fraction must be multiplied by that factor which will produce the required denominator. Ex. To express t with the denominator nh^, we multiply both members by n¥ -r-b := nb^. Thus, a _ anV^ b ~~ nb^ This process is the reverse of reduction to lowest terms. EXERCISES. Express the quantity 1. a with the denominator b, 2. ax " " " 3. ab '' " 4. — " ^^ " 6 - 1 an u a ■\-b H ^ '^ y u cc a ^ — y * X -\-l a — l Negative Exponents. 103. By the principle of § 85, we have % = a^-K a^ If we have hy n, the exponent of the second member of the equation will be negative, and the first member, by can-' n{x -y)' X. «2- ■l\ a2- ■f- x^ + 3a; + 1. a*- •1. 72 ALOEBRAIG OPERATIONS. celling n factors from each term of the fraction, will become -^r— -• Hence -j~- = a^'K By putting for shortness k — n=:s, the equation will be 1 — = a'^ a' Hence, A ne£ativ6 exponent indicates the reciprocal of the corresponding quantity with a positive exponent. If in the formula a^'^ =r -^ we suppose h := n, it will become a^ =z ~, or a^ -—1. Hence, because a may be any quantity whatever, Any quantity with the exponent is equal to unity, Tliis result may be made more clear by suc- cessive divisions of a power of a by a. Every time we eftect this division, we diminish the ex- ponent by 1, and we may suppose this diminution to continue algebraically to negative values of the exponent. On the left-hand side of the equations in the margin, the division is effected symbolically by diminishing the exponents ; on the right the result is written out in the usual way, EXERCISES. In the following exercises, write the quotients which are fractional both as fractions reduced to their lowest terms, and as entire quantities with negative exponents, on the principle. d^ = aaa a' = aa a} = a a^ r= 1 a-i = 1 a a-2 = 1 aa etc. etc. a ah-\ a:^ =: a%-\ etc. Divide . I. x} by X. Ans. X. 2. x by x^, ' 3. _ 2^^3 by l\ Ajis. - or x~'^. X -'^ov -%a-W. a . 4. ^aW' by — "^a^. Ans. FRACTIONS. 73 5. — ^a% by 4:ad\ 6. \%a%^xy by ^abx. 7. 14«^4^V by — naWd^. 8, Uapqxy by 18a2»c. 9. — dQ>a^p^x^y by — 2^a^xy. 10. 486^2 (a; — 2/)2 by 36 (iz: — y). 11. 42^2^4'bj^Op-i^^y. \x — yf '' \x — yf 12. 22 (a — V) (m — ^) by 15 [a -\-h)(m ■\- n). 13. 25 («2 — Z>2) ^^2 _ ^2) by 15 (^ __ 5) (^^ ^ ^^), 14. (r^ - 1) (a2 - 4^) by (ir2 -!)(«+ 2^^). 15. i^:^ — 1 by a^ + 1. 16. a^¥x^y^ by a^b^x^y\ 17. mhi'^yH by mn^y^A 18. 77i(m + l) (m + 2) (m + 3) by m (m— 1) (m— 2) (m— 3). 19. a^ by a**. 20. a^^c^ by ^J'^c'^. Dissection of Fractions. 104. If the numerator is a polynomial, each of its terms may be divided separately Tby the denominator, and the several fractions connected by the signs + or — . The principle is that on which the division of polynomials is founded (§ 87). The general form is m mm m The separate fractions may then be reduced to their lowest terms. Example. Dissect the fraction d^aWx — l^amy + l^inz — I21^n^u IQabx The general form (1) gives for the separate fractions, d2aWx __ 18am^ Ubnz _ 12Pn^u IQabx IQabx 16abx 16abx Eeducing each fraction to its lowest terms, the sum becomes dmy Ibnz dbnho "" Sbx 16ax 4:ax 74 ALGEBRAIC OPERATIONS, EXKRCISES. Separate into sums of fractions, abc + hcd + cda + dab abed — xyzu + x^yzu^ + xyh'^u — x^yhV x^yh^v? (m — n){n -^ q) — (m -\- n)(p — q) {m — n) {]) — q) (x — a){y — b)-j- {x ^y){a — b) + {x — b){y — a) ^ x^ — y'^ (a 4- b) {m — n) — {a — b){m -{• n) Agg^regation of Fractions. 105. When several fractions have equal denomina- tors, their sum may be expressed as a single fraction by aggregating their numerators and writing the com- mon denominator under them. m m m m ^ a — b b — c , c — a Ex. 2. ^ — y y — ^ ^ — y _ a — b c—b c — a_2c — 2b_2{c—'b) ~ X — y X — y X — y ~ x — y ~~ x — y Hem. This process is the reverse of that of dissecting a fraction, EXERCISES. Aggregate a ab abc a b abc abc abc ' {a — by {a — b)^ ^ — a , y— b a-{- b x — y a^x a^x a^x a^x c d + a—b b — a a — b b — a a — b a — c c — b c A- a 5. 1 ±- — m — n m — n n — 7n n — m FRACTIONS. 75 106. When all the fractions have not the same denomina- tor, they must be reduced to a common denominator by the process of § 102. Any common multiple of the denominators may be taken as the common denominator, but the least common multiple is the simplest. To KEDUCE TO A COMMOi^ DENOMINATOR. CJlOOSe a common multiple of the denominators. Multiply both terms of each fraction by the multi- plier necessary to change its denominator to the chosen multiple. Note 1. The required multipliers will be the quotients of the chosen multiple by the denominator of each separate fraction. ISToTE 2. When the denominators have no common fac- tors, the multiplier for each fraction will be the product of the denominators of all the other fractions. Note 3. An entire quantity must be regarded as having the denominator 1. (§ 102.) EXAMPLES. I. Aggregate the sum 1 1^ _ J^ J\_ a ab cibc abed in a single fraction. The least common multiple of the denominators is abccL The separate multipliers necessary to reduce to this com- mon denominator are abed, bed, cd, d, 1. The fractions reduced to the common denominator abed are abed — bed -\- ed — d +1 abed^ abed ' abed ' abed^ abed ^, . aled — bed -\- ed — d -\- 1 The sum is ^r-i » abed By dissecting this fraction as in § 104, it may be reduced to its original form. 76 ALGEBRAIC OPERATIONS. 2. Eeduce the sum 1 ah c a T) c'~'^ to a single fraction. The multipliers are. by Note 2, bed, acd, aid, ale. Using these multipliers, the fractions become led — a^cd a¥d — al(^ abcd^ alcd ' alcd^ alcd ' from which the required sum is readily formed. 3. Eeduce the sum 1 X ^ 01? "^ x — 1 "^ 2^ + 1 ■^" ^-^* The least common multiple of the denominators is x^ — 1. The multipliers are, by Note 1, x^ — 1, X -\-l, X — 1, 1. The sum of the fractions is found to be x'^ — 1 -{- X -{ ■ 1 -{- x^ — X + x^ __ 3x^^ x^ — 1 ~~ x^ — 1 EXERCISES. Eeduce to a single fraction the sums, 3 5 7 9 10. II 11 11 4. -, — z + 1 — i?;l + a; 1 —X 1 + X ax x^ . a I 6. a + X a + X 'a — I a -\- I a X ^ 2x — 6 6 3 x(a — x) a{a — x) ' 4:X^ — 1 2a; — 1 x 1 I gy 1 X -}- y x^ — y^ X — y 1 1 + tT— - + a— I I — c c — a fi. a + - — -; 12 a ■\-l a— I X ■\- y ' x — y *"' a—'l^ a-Vl FRACTIONS. 77 «2 -f. J2 I a 1 1_ 1 ^4- ^{x-1) ^{x -{-1) x^ _a L ^\ ^^' a — b \ a — b/' m -\- n ^ — y y m -\- y m — n X -\- y m^ m {m — y) ^ ^ a x^ 20. \a-b^ b-al a — X c^ — x^ a — b 1) — c c — a {a — b) (b ■— c) {c — a) '^' a~+~b "^ b~+~c "*" c -{- a "^ {a + b) {b + c) {c~^aj' a b m — (x — a) m — (x -}- a) 21. ^ ^^ -» X -{- y x — y c ^ a h 22. -^ + — -^ ab be ac a b c ^3- (^ _ ^) (^ _ c) + (^ _ a) {b-c)'^ {c-a) (c-b) X -\-l X — 1 , 24. + ^x. X — 1 X -\-l ab a^ a {a^ + P) 26. 1 27. 1 — X -\- a X — a x^ — '^xy + y^ x^ + y'^ 28 1 — ?— ±-^^-Zl^l 1^ 1_ Jl ^9- (^ _!_ j^y + (« - Z>)2 + a2 - 6« a2 _ <^ab + ^>2 3^- ^ + 4^1 78 ALGEBRAIC OPERATIONS. Factoring Fractions. 107. If several terms of the numerator contain a common factor, the coefficients of this factor may be added, and their aggregate multiplied by the factor for a new form of the numerator. EXAMPLES. ax — - h x + ex + dx _ {a — b + c -\- d) x m ~ m =^{a-h + c + d)^^. (§101.) dbx + hex + acy — dby __ {ab -\- he) x {ac — ah) y ahn ~~ ahn ahn = (a + c)^^ + ic-l)l- EXERCISES. Eeduce ahy — hey — acy mnu + 7npu + pnu 3 4 5 6 8, 9 10 ahe ' mn ahq + hcq + ahr 4- her ahe ax — hy — dhx — 4:ay 2ma 4:7nx -i- 2y — dax — 6ex -f ay xyz a^ + ^a^h + a l^ a^x — ^ahe — (3^/ — 4g) o? xy ' '* i? + S' a;2y — [4a; -f a; {2h -- 4g) + 3^cg] a-^h ax^ — 4ca; — 3 \mx -\- m {a — x) — am] 2a — dh 4:aVx — - 2eV^ + 2hVx — 2 iif)in\/x — 4:Vx). '~' da — ^b FRACTIONS. 70 Multiplication and Division of Fractions. 108. Fundamental Theorems in the Multiplication and Division of Fractions : Theorem L A fraction may be multiplied by any quantity by either multiplying its numerator or dividing its denominator by that quantity. Cor, 1. A fraction may be multiplied by its denominator by simply cancelling it. Cor. 2. If the denominator of the fraction is a factor in the multiplier, cancel the denominator to multiply by this factor, and then multiply the numerator by the other factors. m Ex. — -; TT X a^ {x^ — IP) == mn {x + h), Ct \pC — 0) because the multiplier «2 (^2 _ ^2^ = a{x — b)a {x + b). Theorem 11. A fraction may be divided by either dividing its numerator or multiplying its denominator. Theorem IIL To multiply by a fraction, the multi- plicand must be multiplied by the numerator of the fraction, and this product must be divided by its de- nominator. Let us multiply t by — We multiply by m by multiplying the numerator (Th. I), and we divide by n by multiplying the denominator (Th. II). Hence the product is y— • That is, tJxe product of the numerators is the numer- ator of the required fraction, and the product of the denominators is its denominator, EXERCISES. Multiply ab ■\- y , ah ^ X I. m X — a, 2, — by -• X —' a '^ X ^ a ab . «^ u 9 9 3. -zTx ^y ^y- . 4. --::z^ by x^ - a\ 80 ALGEBBAIO OPERATIONS. ^ xhj -^ ^ x^ -^ x — m 7. ^Zl_*by^ + i. 8. « + !^by« + ^. iz; , v—ab m -\- n . n — m g. ab by ay + 10. by , ^ y X 7n — n ^ 7n -\- n Ti^ ii- 1 Ix ^ a h X 11. Multiply «+- by 3 + - + -- ^ ^ / mn \ I mn \ 12. Eeduce [m -\ ]\m ■ — )• v 711 — nJ \ m -\- n1 13. Eeduce \a - "f) (^ " y)* 14. Multiply ^ - ^ by ^. Ans* — 15. Divide — by p. 16. Divide 7 by a + J. a — 17. Divide — — ■ by x — 1. iC + 1 18. Divide ^^ + J by 1 + xK 19. ^. ., — 2a — 3m , ,^ DiYide — T-~ by J'* — a^. 109. Reciprocal of a Fraction. The reciprocal of a fraction is formed by simply inverting its terms. Eor, let T be the fraction. By definition, its reciprocal will be 1^ a b Multiplying both terms by b^ the numerator will be b and the denominator z x b, that is, a. Hence the reciprocal required will be -, or, in algebraic language, I FRACTIONS. 81 a ^ a b 110. Def. A Complex Fraction is one of which either of the terms is itself fractional. a b Example. m A — y a OR is a complex fraction, of which ^ is the numerator, and m + - the denominator. The terms of the lesser fractions which enter into the numerator and denominator of the main fraction may be called Minor Terms. Thus, b and y are minor denominators, and a and x are minor numerators. To reduce a complex fraction to a simple one, multi- ply both terms hy a multiple of the minor denominators. am Example. Eeduce y2 Multiplying both terms by xy^, the result will be amx bxy + Jiy^* which is a simple fraction. EXERCISES. Reduce to simple fractions : X b a ^ - X I. ^* 2. 1 -^i 1 a X — X a -{-X b a X ab mil a -\- X ^' ~bd' a-^x km 6 82 ALGEBBAIG OPERATIONS, 1 + ^"' n + 1 am -\ h an n 2ab a^ + l + 2 a a + 2b a a -{- 2h a b a -\-b 8. II. 12. ^3. ^ , o. 1 14. l-\-x 1 — X 1—x + 1 + X 1 -{- X 1 — X 1—x 1 + X 2x- 3 ' y a -{- b — x 1 + a I + a 1 — a 1 a 1-a r+ a 1 a a 1 1 b b + a x-y + y + X x^y f- ■ x^ x-\-y x^- f X — y x^ — y^ Division of one Fraction by Another. a 711/ 111. Let us divide t by — The result will be expressed by the complex fraction a b m n Reducing this fraction by the rule of § 110, it becomes an bm^ which is equal to t x — • That is, b m ' To divide hy a fraction, we have only to Tnultiply by its reciprocal. Divide FRACTIONS. 83 EXERCISES. ab , a X -i- 1 , 2x X ^ X a^ — ¥ , a^-[-db 5. r + 1 , a; 4- 1 ^ « m , h n —— T by -^ T- 6. y H by X---1 "^ x^ — 1 n '' a m a h c , m n p 7. -4--4.-by h--i--- * x y z ' x y z a—- b a -\- b ^ a—b a ■\- b Reciprocal Relations of Multiplication and Division. 113. The fundamental principles of the operations upon fractions are included in the following summary, the under- standing of which will afford the student a test of his grasp of the subject. 1. The reciprocal of the reciprocal of a number is equal to the number itself. In the language of Algebra, 1 T = ^- a 2. The reciprocal of a monomial may be expressed by changing the algebraic sign of its exponent. 3. To multiply by a number is equivalent to dividing by its reciprocal, and vice versa. That is, jy X a or — = aN, a and vice versa, iv x - = — a a 84 ALGEBRAIC 0PEBAT10N8. 4. When the mimerator or denominator of a fraction is a product of several factors, any of these factors may be transferred from one term of the fraction to the other by changing it to its reciprocal. That is. , 7 - abc abc DC p par ~~ 1 ~~ qr ^^ - par ^ etc. ^ abc Ic p~^dbc Or, = -— j = , etc. pqr a ^pqr qr 5. Mxdtiplicaticm by a factor greater than nnity increases^ less than unity diminisJies. Bivision by a divisor greater than unity diminishes^ less than unity increases. 6. («) When a factor becomes zero, the product also becomes zero. {^) When a denominator becomes zero, the quotient becomes infinite. That is, Oxa==:«xO = 0. - =: infinity. Note. The following way of expressing what is meant by this last statement is less simple, but is logically more correct : If a fraction has a fixed numerator, no matter how small, we can make the denominator so much smaller that the fraction shall be greater than any quantity we choose to assign. EXERCISE. If the numerator of a fraction is 2, how small must the denominator be in order that the fraction may exceed one thousand? That it may exceed one million? That it may exceed one thousand millions ? BOOK III. OF E QUA TIONS. CHAPTER I. THE REDUCTION OF EQUATIONS. Definitions. 113. Def. An Equation is a statement, in the lan- guage of Algebra, that two expressions are equal. 114. Def. The two equal expressions are called Members of the equation. 115. Def. An Identical Equation is one which is true for all values of the algebraic symbols which enter into it, or which has numbers only for its members. Examples. The equations 14 + 9 r= 29 — 6, 5 _l_ 13 _ 3 X 4 — 6 = 0, which contain no algebraic symbols, arc identical equations. So also are the equations X ■= X, X — X ^=1 0, (x -{- a) {x — a) — x^ — a^, because they are necessarily true, whatever values we assign to X, a, and y. Eem. All the equations used in the preceding two books to express the relations of algebraic quantities are identical ones, because they are true for all values of these quantities. 86 EQUATIONS. 116. Def, An Equation of Condition is one whicli can Ibe trne only when the algebraic symbols are equal to certain quantities, or have certain relations among themselves. Examples. The equation :c + 6 = 22 can be true only when x is equal to 16, and is therefore an equation of condition. The equation X -\- h =z a can be true only when x is equal to the difference of the two quantities a and h Rem. In an equation of condition, some of the quantities may be supposed to be known and others to be unknown. 117. Def. To Solve an equation means to find some number or algebraic expression which, being sub- stituted for the unknown quantity, will render the equation identically true. This value of the unknown quantity is called a Root of the equation. EXAMPLES. 1. The number 3 is a root of the equation 2a;2 _ 18 m 0, because when we put 3 in place of x, the equation is satisfied identically. 2. The expression is a root of the equation "zcx — 4:a + 2b = 0, when x is the unknown quantity, because when we substitute this expression in place of x, we have 2c(^^^)^^a + 2b = 0, ^ or ^a — 2b — 4ca-{-2b = 0, which is identically true. AXI0M8. 87 Eem. It is common in Elementary Algebra to represent unknown quantities by the last letters of the alphabet, and quantities supposed to be known by the first letters. But this is not at all necessary, and the student should accustom him- self to regard any one symbol as an unknown quantity. Axioms. 118. Def. An Axiom is a proposition wMch is taken for granted, without proof. Equations are solved by operations founded upon the fol- lowing axioms, which are self-evident, and so need no proof. Ax. I. If equal quantities be added to the two members of an equation, the members will still be equal. Ax. II. If equal quantities be subtracted from the two members of an equation, they will still be equal. Ax. III. If the two members be multiplied by equal factors, they will still be equal. Ax. IV. If the two members be divided by equal divisors (the divisors being different from zero), they will still be equal. Ax. V. Similar roots of the two members are equal. These axioms may be summed up in the single one. Similar operations upon equal quantities give equal results, 119. An algebraic equation is solved by performing such similar operations upon its two members that the unknown quantity shall finally stand alone as one member of an equation. Operations of Addition and Subtraction— Trans- posing Terms. 130. Theorem. Any term may be transposed from one member of an equation to the other member, if its sign be changed. 88 EQUATIONS. Proof. Let us put, in accordance with § 41, 2d Prin., t, any term of either member of the equation. a, all the other terms of the same member. h, the opposite member. The equation is then Now subtract t from both sides (Axiom II), or by reduction, a ^= h — t. This equation is the same as the one from which we started, except that t has been transposed to the second member, with ifcs sign changed irom + to — . If the equation is b — t =^ a, we may add t to both members, which would give b = a + L NUMERICAL EXAMPLE. The learner will test each side of the following equations : 194.3_94-4 rr: 7 + 10. Transposing 4, 194-3 — 9 = 7 + 10— 4. 9, 19 + 3 = 7 + 10—4 + 9. 19, 3 = 7 + 10-4 + 9-19. "3, = 7 + 10—4 + 9-19—3. 131. Rem. All the terms of either member of an equation may be transposed to the other member, leaving only on one side. Example. If in the equation b = a + t, we transpose b, we have = a + t — b. By transposing a and t, we have b-^a^t = 0. 123. Changing Signs of Members. If we change the signs of all the terms in hoth memhers of an equation, it will still be true. The result will be the same as multiplying both REDUCTION. 89 members by — 1, or transposing all the terms of each member to the other side, and then exchanging the terms. Example. The equation 17 + 8 = 11 + 14 may be transformed into = 11 + 14 — 17 — 8, or, = — 11 — 14 + 17 -f 8, or, — 17 - 8 =z — 11 _ 14. Operation of Multiplication. 133. Clearing of Fractions, The operation of multipli- cation is usually performed upon the two sides of an equation, in order to clear the equation of fractions. To clear an equation of fractions : First Method. Multiply its members by the least common multiple of all its denominators. Secoi^d Method. Multiply its members by each of the denominators in succession, Eem. 1. Sometimes the one and sometimes the other of these methods is the more convenient. Rem. 2. The operation of clearing of fractions is similar to that of reducing fractions to a common denominator. Example of First Method. Clear from fractions the equation Here 24 is the least common multiple of the denominators. Multiplying each term by it, we have, Q,x-\-Ax + ^x = 624, or 13:2; r= 624. Example of Secokd Method. Clear the equation X — a X '\- a X Multiplying by x — a, we find ax — a^ ex — ca a A 1 = 0. X ■\- a X 90 EQUATIONS. Multiplying by rr -f- «, ax -\- a^ -{- ax — a^ -\ = 0. X Keducing and multiplying by Xy 2ax^ + cx^ — ca^ = 0. EXERCISES. Clear the following equations of fractions : /iX rt r\ XX ^_ --6 = 0. .. 5-7 = ^^- ^^^_K X x^ _ b 2+3~i-^- 4- « + ««-«■ 1 .y>l - J-. 6 ?.^-? ab^ a^ b~ aW " 3 "^ 4 ~ o' II 13 cc — a X -\- a X ■\- a X — a x^ 4- 2ax X — a ^_y y ^ a X . y - 10. 14. a — ^ h — a "" Here the second term is the same as X -{- a X —h X ^ a ^ x+ b x — 2 x-^2 x — 6 ~' X + b X — a X -f a X -\- a X — a -y a a — X X — a Reduction to the Normal Form. ( 124. Def, An equation is in its Normal Form when its terms are reduced and arranged according to the powers of the unknown quantity. In the normal form one member of the equation is expressed as an entire function of the unknown quantity, and the other is zero. (Compare §§ 50, ?6.) To reduce an equation to the normal form : I. Transpose all the terms to one mewiber of the equa- tion, so as to leave as the other mernber. REDUCTION, 9] II. Clear the equation of fractions. III. Clear the equation of -parentheses hy performing all the operations indicated. lY. Collect each set of tains containing lihe powers of the unhnown quantity into a single one. V. Divide hy any common factor which does not con- tain the unhnown quantity. Eem. This order of operations may be deviated from according to circumstances. After a little practice, the student may take the shortest way of reaching the result, without re- spect to rules. EXAMPLES. 1. Eeduce to the normal form (a; — 2) {x — 3) _ (x + 2) (a: + 4) x — h ~^ i?; + 5 1 Clearing of fractions, {x + 5) {x — 2) {x — Z)^(x — 5) (x + 2) {x + 4). 2. Performing the indicated operations, ^ _ 19^' _{_ 30 = ^3 _|_ ^2 _ 222; — 40. 3. Transposing all the terms to the second member and reducing, ^ z=.x^ — ?>x— 70, which is the normal form of the equation. Kem. Had we transposed the terms of the second member to the first one, the result would have been ~ a;2 + 3:^; + '^0 = 0. Either form of the equation is correct, but, for the sake of uniformity, it is customary to transpose the terms so that the coefiicient of the highest power of x shall be positive. If it comes out negative, it is only necessary to change the signs of all the terms of the equation. Ex. 2. Reduce to the normal form, hmx^ 2ax ^mx^ ^ -„ z=^ 2mx — D«. X — a X -{- a X? — ci'^ 92 EQUATIONS. 1. Transposing to the first member, ^mx^ 2ax 3ma^ . __ 2mx -i- 6a ~ 0. X — a X -\- a x" — a'^ 2. To clear of fractions, we notice that the least common multiple of the denominators is x'^ — a\ Multiplying each term by this factor, we have, 6mx^{x+a)—2ax{x—a)—3mx^—2?nx{x^—a^) + 6a(x^—a^) = 0. 3. Performing the indicated operations, 6ma^+6amx^—2ax^+2a^x—dma^—2mx^-\-2a^mx+6ax^—6a^=0. 4. Collecting like powers of x, as in § 76, {da + 6am) x^ + {2a^ + 2a^m) x — 6a^ = 0, 6, Every terra of the equation contains the factor a. By Axiom IV, § 118, if both members of the equation be divided by a, the equation will still be true. The second member being zero, will remain zero when divided by a. Dividing both members, we have (3 + 6771) x^ + 2a{l+m)x^ 6a^ = 0, which is the normal form. EXERCISES. Eeduce the following equations to the normal form, x, y, or z being the unknown quantity : dy^ -\-2y __y -—1 x — a__^x-{-a 7 "~"2 ' X -j- a ~~ X 3- a; — 7 2x-{-6 2x -h 10 4a; — 2 a^ ^ Za^x + 2^3 ^ ^ n^ — 6ax^ 2x -\- a 2x — a 6. '^-^- + -±- = 0. a-\-bb-{'Z^a + z 7? z^ ah a —z a^ — x^ ^ — c^ REDUCTION. 93 7 + ^^S-^+i3=0. y y^ y^ a a? ^^ _ 1 X — a x^ — a^ x^ — a^ I)^ ¥ h^ + c — z c^ — z^ d^ — z^ c^ — Z"" a b mm II. = 12. = -. 1 X — a n .1 nx X -^ — X X tX/ 13. T + 1 ' 1 ^ a^ a a' -2 X x 14. 15. 16. _3z__ _ 6z^ _ 1 I ^ 3 " z' ax hx 1+ ' X -{- a X — a a h X a — x a b "" i a X X Degree of Equations. 135. Def. An equation is said to be of the n^^ de- gree when n is the highest power of the unknown quantity which appears in the equation after it is re- duced to the normal form. EXAMPLES. The equation Ax -{- B ^ is of the first degree. Ax'^ + B =z " " second " Ax^ + Bx-\- C =:: " " third etc. etc. An equation of the second degree is also called a Quadratic Equation. 94 EQUATIONS OF THE FIRST DEGREE, An equation of the third degree is also called a Oubic Equation. Example. The equation ax^ + hx^y^ -\- y^ -\- ah z:^ is a quadratic equation in x, because x^ is of tlie highest power of x which enters into it. It is a cubic equation in y. It is of the first degree in z. CHAPTER II. EQUATIONS OF THE FIRST DEGREE WITH ONE UNKNOWN QUANTITY. 136. Eemark. By thQ preceding definition of the degree of an equation, it will be seen that an equation of the first degree, with x as the quantity supposed to be unknown, is one which can be reduced to the form Ax-\- B = 0, {a) A and B being any numbers or algebraic expressions which do not contain x. Such an equation is frequently called a Simple Equation. Solution of Equations of the First Degree. 137. If, in the above equation, we transpose the term B to the second member, we have Ax = ^B. If we divide both members by A (§ 118, Ax. IV), we have^ B Here we have attained our object of so transforming the equation that one member shall consist of x alone, and the other member shall not contain x. ONE UNKNOWN QUANTITY. 95 To prove that j is the required value of x, we sabsti- tute it for x in the equation {a). The equation then becomes, or, by reducing, — ^ + ^ = 0, an equation which is identically true. Therefore, — -^ is the required root of the equation (a), (§ 117, Def.) 138, In an equation of the first degree, it will be unneces- sary to reduce the equation entirely to the normal form by transposing all the terms to one member. It will generally be more convenient to place the terms which do not contain x in the opposite member from those which are multiplied by it. Example. Let the equation be mx -\- a =1 nx -\- l, (I) We may begin by transposing a to the second member and nx to the first, giving at once, mx — nx ^= h — a, or (m — n)x =z b -— a, without reducing to the normal form. The final result is the same, whatever course we adopt, and the division of both members hjm — n gives h — a 7n — 71 139. The rule which may be followed in solving equations of the first degree with one unknown quantity is this : I. Clear the equation of fractions, II. Transpose the terms which are multiplied hy the unhnown quantity to one member ; those which do not contain it to the other, III. Divide hy the total coefficient of the unhnown quantity. 96 EQUATIONS OF THE FIRST DEGBEE. Note. Rules in Algebra are given only to enable the beginner to go to work in a way which will always be sure, though it may not always be the shortest. In solving equations, he should emancipate liimself from the rules as soon as possible, and be prepared to solve each equa- tion presented by such process as appears most concise and elegant. No operation upon the two members in accordance with the axioms (§ 118) can lead to incorrect results (provided that no quantity which becomes zero is used as a multiplier or divisor), and the student is therefore free to operate at his own pleasure on every equation presented. EXAMPLES. ^. ax . 1. Given ^ = 1. hy It is required to find the value of each of the quantities a, h, X, and y, in terms of the others. Clearing of fractions, we have ax = hy. To find a, we divide by x, which gives III a — —' x To find ^, we divide by y, which gives ax _ ~y ~ To find X, we divide by a, which gives by a To find y, we divide by d, which gives ax Thus, when any three of the four quantities a, h, x, and y, are given, the fourth can be found. 2. Let us take the equation, x — 'i — ^^ + ^ ^X-\- 10 "~ A:X — % Clearing of fractions, we have 4t2 — 30.^ -j- 14 = 4:«;2 _|_ 32^^ 4. 60. ONE UNKNOWN QUANTITY 97 » Transposing and reducing, — %^x — 46. Dividing both members by •— 62, _ 46 _ _ 46 ""■ ~ - 62 ~ ~ 62 23 31' This result should now be proved by computing the value of both 23 members of the original equation when — — is substituted for a*. ol X X __ax 1 ^* m n~ h m Proceeding in the regular way, we clear of fractions by multiplying by mnh This gives nbx + 7nl)X = amnx — nh. Transposing and reducing, {nd -f- md — amn) x =z — nh. Dividing by the coefficient of x, __ nh _ nb nb -^ mb — amn aynn — mb — nb These two values are equivalent forms (§ 100). But we can obtain a solution without clearing of fractions. ax Transposing -y, we have X X m n ax T which may be expressed in the form /I 1 a\ \m n b! 1^ m Dividing by the coefficient of x, m X ^=z — 1 ^_? m 71 b This expression can be redi ced to the other by § 110. 7 98 EQUATIONS OF THE FIRST DEGREE. EXERCISES. Find the values of x, y, or u in the following equations : b — ^x Sx — % X. 3-3 3- •^ 4. ^' ^ ^ - 22 i + 2 + 3 - ^^• S- y ^y y -1 a^ b c~ 7- U U tl- 3-4 + 5=='^- 9- u u 1 1 2. — X := a. 4. a; + 23 _ x-^1 -^• 6. 36 _ 45 u — b~ u u — 2Q. 8. a— dx =: b -}- ax, 10. 3a; H — = X. a c c — x~~ a — X 12. x — 1 x — 2 X — 6 X ■ x^2 x — d x — Q x — n 13- —y = a — h. 1 11 1 14. x — 2 X — 4: x — Q X — 1 ^5- l(^-i)-|("--I)+i(^-i) = <>- tl u a 16. - + a h — a h -\- a X 1 17. ax -{- b = - + y a b ■ u — a u — b u — c _ u — {a ■\- b -\- c) b c a ~~ abc 711 (x -\- a) n (x -\- b) X -\-b X -{- a 20. (.^— ^)3 4-(.T_Z>)3 4. {x—cf =z 3 (:^;— «) (x—b) {x—c). Find the values of each of the four quantities, a, b, c, and d, in terms of the other three, from the equations 21. 7 h T -7 =0. 22. — + 1 =r: 0. b — c b — d cd ONE UNKNOWN QUANTITY. 99 Problems leading to Simple Equations. 130. The first difficulty which the beginner meets with in the solution of an algebraic problem is to state it in the form of an equation. This is a process in which the student must depend upon his own powers. The following is the general plan of proceeding : 1. Study the problem, to ascertain what quantities in it are unknown. There may be several such quantities, but the problems of the present chapter are such that all these quan-^ titles can be expressed in terms of some one of them. Select that one by which this can be most easily done as the unknown quantity. 2. Represent this unknown quantity by any algebraic sym- bol whatever. It is common to select one of the last letters of the alpha- bet for the symbol, but the student should accustom himself to work equally well with any symbol. 3. Perform on and with these symbols the operations re- quired by the problem. These operations are the same that would be necessary to verify the adopted value of the unknown quantity. 4. Express the conditions stated or implied in the problem by means of an equation. 5. The solution of this equation by the methods already explained will give the value of the unknown quantity. It is always best to verify the value found for the unknown quan- tity by operating upon it as described in the equation. EXAMPLES. I. A sum of 440 dollars is to be divided among three people so that the share of the second shall be 30 dollars more than that of the first, and the share of the third 80 dollars less than those of the first and second together. What is the share of each? Solution. 1. Here there are really three unknown quantities, but it is only necessary to represent the share of the first by an unknown symbol. 100 EQUATIONS OF THE FIB8T DEGREE, 2 Therefore let us put X = share of the iSrst. 3. Then, by the terms of the statement, the share of the second will he X + 30. To find the share of the third we add these two together, which makes 2x + 30. Subtracting 80, we have 2x — 50 as the share of the third. We now add the three shares together, thus, Share of first, x " " second, x-^-Z^ " " third, 2^--_50 Shares of all, ^x — 20 4. By the conditions of the problem, these three shares must together make up 440 dollars. Expressing this in the form of an equation, we have ^x — 20 z= 440. 5. Solving, we find x :=! 115 — share of first. Whence, 115 -f 30 =r 145 = share of second. 115 + 145 — 80 = J^O =: share of third. Sum = 440. Proof. Ex. 2. Divide the number 90 into four parts, such that the first increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, shall all be equal to the same quantity. Here there are really five unknown quantities, namely, the four parts and the quantity to which they are all to be equal when the operation of adding to, subtracting, etc., is performed upon them. It will be most convenient to take this last as the unknown quantity. Let us therefore put it equal to u. Then, Since the first part increased by 2 must be eqaal to ?/, its value will be u — 3. Since the second part diminished by 2 must be equal to u, its value will be \i + 2. Since the third part multiplied by 2 must be \i, its value will be ^ • Since the fourth part divided by 2 must make u, its value will be 2?i. ONE UNKNOWN QUANTITY, 101 Adding these four parts up, their sum is tbund i;o oe -- • By the conditions of the problem, this sum must make up the num- Iber 90. Therefore we have ? = - Solving this equation, we find U = 20. Therefore 1st part = u — 2 = IS. 2d '' =:u + 2 = 22. ^ dd '' =u-T-2 = 10. 4th " =2u = 40. The sum of the four equals 90 as required, and the first part increased by 2, the second diminished by 2, etc., all make the number 20, as re- quired. r PROBLEMS FOR EXERCISE. 1. What number is that from which we obtain the same result whether we multiply it by 4 or subtract it from 100? 2. What number is that which gives the same result when we diyide it by 8 as when we subtract it from 81 ? 3. Divide 284 dollars among two people so that the share of the first shall be three times that of the second and $1G more. 4. Find a number such that ^ of it shall exceed \ of it by 12. 5. A shepherd describes the number of his sheep by saying that if he had 10 sheep more, and sold them for 5 dollars each, he would have 6 times as many dollars as he now has sheep. How many sheep has he ? 6. An iapplewoman bought a number of apples, of which 60 proved to be rotten. She sold the remainder at the rate of 2 for 3 cents, and found that they averaged her one cent each for the whole. How many had she at first ? 7. If you divide my age 10 years hence by my age 20 years ago, you will get the same quotient as if you should divide my present age by my age 26 years ago. What is my present age ? 8. Divide $500 among A, B, and 0, so that B shall have $20 less than A, and C $20 more than A and B together. 102 EQUATlOm OF THE FIRST DEGREE. 9. A- father -left $10000 to be diyidecl among his five chil- dren, directing thai each -should receive $500 more than the next younger one. What was the share of each ? 10. A man is 6 years older than his wife. After they have been married 12 years, 8 times her age would make 7 times his age. What was their age when married ? 11. Of three brothers, the youngest is 8 years younger than the second, and the eldest is as old as the other two together. In 10 years the sum of their ages will be 120. What are their present ages ? 12. The head of a fish is 9 inches long, the tail is as long as the head and half the body, and the body is as long as the head and tail together. What is the whole length of the fish ? 13. In dividing a year's profits between three partners, A, B, and 0, A got one-fourth and $150 more, B got one-third and $300 more, and C got one-fifth and $60 more. What was the sum divided ? 14. A traveller inquiring the distance to a city, was told that after he had gone one-third the distance and one-third the remaining distance, he would still have 36 miles more to go. What was the distance of the city ? 15. In making a journey, a traveller went on the first day one-fifth of the distance and 8 miles more ; on the Second day he went one-fifth the distance that remained and 15 miles more ; on the third day he went one-third the distance that remained and 12 miles more ; on the fourth he went 35 miles and finished his journey. What was the whole distance travelled ? 16. When two partners divided their profits, A had twice as much as B. If he paid B $300, he would only have half as much again as B had. What was the share of each ? 17. At noon a ship of war sees an enemy's merchant vessel 15 miles away sailing at the rate of 6 miles an hour. How fast must the ship of war sail in order to get within a mile of the vessel by 6 o'clock ? 18. A train moves away from a station at the rate of h miles an hour. Half an hour afterward another train follows it, running m miles an hour. How long will it take the latter to overtake it ? 19. What two numbers are they of which the difference is 9, and the difference of their squares 351 ? 20. A man bought 25 horses for $2500, giving $80 a piece ONE UNKNOWN QUANTITY 103 for poor horses and $130 each for good ones. How many of each kind did he buy ? 21. A man is 5 years older than his wife. In 15 years the snms of their ages will be three times the present age of the wife. What is the age of each ? 22. How far can a person who has 8 hours to spare ride in a coach at the rate of 6 miles an hour, so that he can return at the rate of 4 miles ah hour and arrive home in time ? 23. A working alone can do a piece of work in 15 days, and B alone can perform it in 12 days. In what time can they perform it if both work together ? Method of Solution. In one day A can do ^^ of the whole work and B can do yi-. Hence, both together can do (xV + tV) of it. If both together can do it in x days, then they can do - of it in 1 day. X XT 111 Hence, - = — + — X 13 15 is the equation to be solved. 24. A cistern can be filled in 12 minutes by two pipes which run into it. One of them alone will fill it in 20 minutes. In what time wotild the other one alone fill it ? 25. A cistern can be emptied by three pipes. The second pipe runs twice as much as the first, and the third as much as the first and second together. All three together can empty the cistern in one hour. In what time would each one sepa- rately empty it ? 26. A marketwoman bought apples at the rate of 5 for two cents, and sold half of them at 2 for a cent and the other half at 3 for a cent. Her profits were 50 cents. How many did she buy ? 27. A grocer having 50 pounds of tea worth 90 cents a pound, mixed with it so much tea at 60 cents a pound that the combined mixture was worth 70 cents. How much did he add ? 28. A laborer was hired for 40 days, on the condition that every day he worked he should receive $1.50, but should for- feit 50 cents for every day he was idle. At the end of the time $52 were due him. How many days was he idle ? 29. A father left an estate to his three children, on the condition that the eldest should be paid $1200 and the second $800 for services they had rendered. The remainder was to be equally divided among all three. Under this arrangement. 104 EQUATIONS OF THE FIRST DEGREE. the youngest got one-fourth of the estate. What was the amount divided ? 30. A person having a sum of money to divide among three people gave the first one-third and $20 more, the second one-third of what was left and $20 more, and the third one- third of what was then left and $20 more, which exhausted the amount. How much had they to divide ? 31. One shepherd spent $720 in sheep, and another got the same number of sheep for $480, paying $2 a piece less. What price did each pay? 32. A crew which can pull at the rate of 9 miles an hour, finds that it takes twice as long to go up the river as to go down. At what rate does the river flow ? 33. A person who possesses $12000 employs a portion of the money in building a house. Of the money which remains, he invests one-third at four jier cent, and the other two-thirds at five per cent., and obtains from these two investments an annual income of $392. What was the cost of the house ? 34. An income tax is levied on the condition that the first $600 of every income shall be untaxed, the next $3000 shall be taxed at two per cent., and all incomes in excess of $3600 shall be taxed three per cent, on the excess. A person finds that by a uniform tax of two per cent, on all incomes he would save $200. What was his income ? 35. At what time between 3 and 4 o'clock is the minute- hand 5 minutes ahead of the hour hand ? 2,6. One vase, holding a gallons, is full of water ; a second, holding h gallons, is full ot brandy. Find the capacity of a dipper such that whether it is filled from the first vase and the water removed replaced by brandy, or filled from the second vase and the latter then filled with water, the strength of the mixture will be the same. 37. Divide a number m into four such parts that the first part increased by a, the second diminished by a, the third multiplied by a, and the fourth divided by a shall all be equal. 38. Divide a dollars among five brothers, so that each shall have n dollars more than the next younger. 39. A courier starts out from his station riding 8 miles an hour. Four hours afterwards he is followed by another riding 10 miles an hour. How long will it require for the second to overtake the first, and what will be the distance travelled ? If X be the number of hours required, the second will have travelled X hours and the first (aj + 4) hours when they meet. At this time they must have travelled equal distances. ONE UNKNOWN QUANTITY. 105 Problem of the Couriers. Let us generalize the preceding problem thus : 131. A courier starts out from his station riding c miles an hour ; h hours later, he is followed by another riding a miles an hour. How long will the latter he in overtaking the first, and wha/t ivill he the distance from the point of departure. Let us put t for the time required. Then the first courier will have travelled {t-\-1i) hours, and the second t hours. Since the first travelled c miles an hour, his whole distance at the end of ^ + ^ hours will be (t + Ji) c. In the same way, the distance travelled by the other will be at. When fhe latter overtakes the former, the distances will be equal ; hence, at = c{t^ h). (1) Solving this equation with respect to t, we find t = -^. (2) a — c . \ / Multiplying by a gives us the whole distance travelled, which is Distance = a — c This equation solves every problem of this kind by substi- tuting for a, c, and li their values in numbers supposed in the problem. For example, in Problem 39, we supposed a = 10, c=zS, h=^4:. Substituting these values in equation (2), we find t = 16, which is the number of hours required. To illustrate the generality of an algebraic problem, we shall now inquire what values f shall have when we make dif- ferent suppositions respecting a, c, and h. (1.) Let us suppose a = c, or a — c=zO^ that is, the rates of travelling equal. Then equation (2) will become . _ ch 106 EQUATIONS OF THE FIRST DEOBEE. an expression for infinity (§ 112, 6), showing that the one conriei would never overtake the other. This is plain enough. But, (2. ) Let us supipose that the second courier does not ride so fast as the first, that is, a less than c, and a — c negative. ch Then the fraction will not be infinite, but will be nega- it — c tive, because it has a positive numerator and a negative denom- inator. It is plain that the second courier would never overtake the first in this case either, because the latter would gain on him all the time ; yet the fraction is not infinite. What does this mean ? It means that the problem solved by Algebra is more gen- eral, that is, involves more particular problems than were implied in the statement. If we count the hours after the second courier set out as positive, then a negative time will mean so many hours before he set out, and this will bring out a time when, according to our idea of the problem, the horses were still in the stable. The explanation of the difficulty is this. Suppose S to be the point from which the couriers started, and AB the road alonfif which they travelled from A S B S toward B. Suppose also that i——....-^— the first courier started out from S at 8 o'clock and the second at 12 o'clock. By the rule of positive and negative quantities, distances towards A are negative. Now, because algebraic quantities do not commence at 0, but extend in both the negative and positive directions, the algebraic problem does not suppose the couriers to have really commenced their journey at S, but to have come from the direction of A, so that the first one passes S, without stop- ping, at 8 o'clock, and the second at 12. It is plain that if the first courier is travelling the faster, he must have passed the other before reaching S, that is, the time and distance are both negative, just as the problem gives them. The general principle here involved may be expressed thus : In Algebra, roads and journeys, like time, have no begin- ning and no end. ONE UNKNOWN QUANTITY. /lO? (3.) Let us suppose that the couriers start out at the same time and ride with the same speed. Then li and a — c are both zero, and the expression for t assumes the form, '=1- This is an expression which may have one vahie as well as another, and is therefore indeterminate. The result is correct, because the couriers are always together, so that all values of t are equally correct. The equation (1) can be used to solve the problem iu other forms. In this equation are four quantities, a^ c, h, and t, and when any three of these are given, the fourth can be found. There are therefore four problems, all of which can be solved from this equation. First Problem, that already given, in which the time required for one courier to overtake the other is the unknown quantity. Secokd Problem. A courier sets out from a station, riding e miles an hour. After h hours another follows him from the same station, intending to overtake him ill t hours. How fast must he ride ? The problem can be put into the form of an equation in the same way as before, and we shall have the equation (1), only a will now be the unknown quantity. If we use the numbers of Prob. 39 instead of the letters, we shall have, in- stead of equation (1), the following : 16a =: 8 (16 + 4) = 8-20 = 160, whence a = 10. If we use letters, we find from (1), .(^ + 70 and the problem is solved in either case. Third Problem. TJxe second courier ca,n ride just a miles an hour, and the first courier starts out h hours 108 EQUATIONS OF THE FIRST DEGREE, before him. How fast must the latter ride in order that the other maij take t hours to overtake hiin ? Here c, the rate of the first courier, is the unknown quan- tity, and by solving equation (1), we find at Fourth Problem. The swiftest of two couriers can ride a miles an hour, and the slower c miles an hour. How long a start must the latter have in order that the other m^ay require t hours to overtake him ? Here, in equation (1), h is the unknown quantity. By solving the equation with respect to /?, we find, , at — ct 11 = —-, which solves the problem. PROBLEMS OF CIRCULAR MOTION. 40. Two men start from the same point to run repeatedly round a circle one mile in circumference. If A runs 7 miles an hour and B 5, it is required to know : 1. At what intervals of time will A pass B ? 2. At how many different points on the circle will they be together ? We reason thus : since A runs 2 miles an hour faster than B, he c:ets away from him at the rate of 2 miles an hour. When he overtakes him, he will have gained upon him one circumference, that is, 1 mile. This will require 30 minutes, which is therefore the required interval. In this interval A will have gone round 8^ and B 2^ times, so that they will be together at the point opposite that where they were together 30 minutes previous. Hence, they are together at two opposite points of the circle. 41. What would be the answer to the preceding ques- tion if A should run 8 miles an hour, and B 5 ? 42. Two race-horses run round and round a course, the one making tlie circuit in 30, the other in 35 seconds. If they start out together, how long before they will be together again ? Note. In x seconds one will make j^ circuit and the other jr^. o\j 00 43. If one planet revolves round the sun in T and the other in T' years, what will be the interval between their coni unctions? TWO UNKNOWN QUANTITIES, 109 CHAPTER III. EQUATIONS OF THE FIRST DEGREE WITH SEVERAL UNKNOWN QUANTITIES. Case I. Equations with Two Unknown Qiian- titles. 133. Def. An equation of the first degree with two unknown quantities is one which admits of being re- duced to the form ax -{-by = 6*5 in which x and y are the unknown quantities and a, &, and c represent any numbers or algebraic equations which do not contain either of the unknown quantities. Def. A set of several equations containing the same unknown quantities is called a System of Simulta- neous Equations. Solution of a Pair of Simultaneous Equations containing^ Two Unknown Quantities. 133. To solve two or more simultaneous equations, it is necessary to combine them in such a way as to form an equation containing only one unknown quan- tity. 134. Def. The process of combining equations so that one or more of the unknown quantities shall dis- appear is called Elimination. The term "elimination" is used because the unknown quantities which disappear are eliminated. There are three methods of ehminating an unknown quan- tity from two simultaneous equations. 110 EQUATIONS OF THE FIRST DEGREE. Elimination by Comparison. 135. Rule. Solve each of the equations with respect to one of the unhnown quantities and put the two values of the unhnown quantity thus obtained equal to each other. This will give an equation with only one unhnown quantity, of which the value can he found from the equation. The value of the other unhnown quantity is then found hy substitution. Example. Let the equations be ax -{- hy ^^ c, \ ax + h'y = c\ \ From the first equation we obtain, (1) .'. = ^-^. (2) (3) a From the second we obtain, c' — h'y a Putting these two values equal, we have c — hy _ c' — h'y Eeducing and solving this equation as in Chapter II, we find, _ ac' — a'c which is the required value of y. Substituting this value of y in either of the equations (1), (2), or (3), and solving, we shall find _ Vc - he' ~~ ah' — a'h If the work is correct, the result will be the same in which- ever of the equations we make the substitution. TWO UNKNOWN QUANTITIES. Ill Numerical Example. Let the equations be ^ + ^ = 28,) 3x -2ij ^ 29. j ^^ From the first equation we find and from the second x = — ^, o from which we have 28 — y = — -, ^ = 11. Substituting this value in the first equation in x, it becomes a: = 28 - 11 = 17. If we substitute it in the second, it becomes 29 + 22 51 X = -3— - y - n, the same value, thus proving the correctness of the work. Elimination by Substitution. 136. Rule. Fijtd the value of one of the unknoivn quantities in terms of the other from either equation, and substitute it in the other equation. The latter will have hut one unhnown quantity. Example. Taking the same equations as before, ax -^llj r:^ c, a'x ■\- y ij = c', the first "equation gives x = — — -* Substituting this value instead of x in the second equation, it becomes a'c — a'biJ 1, , a -^ Solving this equation with respect to y, W^ get the same result as before. 112 EQUATIONS OF THE FIRST DEOBEE. Numerical Example. To solve in this way the last nu- merical example, we have from the first equation (4), X = 2% — y. Substituting this value in the second equation, it becomes 84 — 3y — 2y = 29, from which we obtain as before, 84-^29 y = — ^- := 11. This method may be applied to any pair of equations in four ways : 1. Find X from the first equation and substitute its value in the second. 2. Find x from the second equation and substitute its value in the first. 3. Find y from the first equation and substitute its value in the second. 4. Find y from the second equation and substitute its value in the first. Elimination by Addition or Subtraction. 137. EuLE. Multiply each equation hy such a factor that the coefficients of one of the unknown quantities shall become numerically equal in the two equations. Then, by adding or subtracting the equations, we shall have an equation luith but one unknown quantity. Rem. We may always take for the factor of each equation the coefficient of the unknown quantity to be eliminated in the other equation. Example. Let us tako once more the general equation ax -}- by = c, a'x + b'y =z c\ Multiplying the first equation by «', it becomes aa!x + a!hy = a!c. Multiplying tho second by a, it becomes aa'x + aVy = ad. TWO UNKNOWN QUANTITIES. 113 The unknown quantity x has the same coefficient in the last two equations. Subtracting them from each other, we obtain {ah — ah') y z=. a'c — ac\ a'c — ac' y ah — ah' Rem. We shall always obtain the same result, whichever the above three methods we use. But as last method is the most simple and elegant. of the above three methods we use. But as a general rule the Problem of the Sum and Difference. The following simple problem is of such wide application that it should be well understood. 138. Problem. The sum and difference of two num- bers hein^ given, to find the numbers. Let the numbers be x and y. Let ,s* be their sum and d their difference. Then, by the conditions of the problem, x + y = s, X — y =i d. Adding the two equations, we have '^x =1 s + d. Subtracting the second from the first, ^y z=z s — d. Dividing these equations by 2, _ s -\- d __ s d ^ -* ~2"~ -2 + 2 __ s — d __ s d y "" ~"2~ "^ 2 ■" 2* We therefore conclude : The greater number is found by adding half the dif- ference to half the sum. The lesser nuynber is found by subtracting half the difference front' half the sum. 114 EQUATIONS OF THE FIMST DEGREE. This result can be illustrated geometrically. Let AB and BC be two lines placed end to end, so that AC is their sum. To find their difference, we cut off from AB a length ^ ^\ ^\ ^ AC r= BC ; then C'B is the i P difference of the two lines. If P is half way between C and B, it is the middle point of the whole line, so that AP = PC = iAC = i sum of lines. CT r= PB = iC'B = i difference of lines. If to the half sum AP we add the half difference PB, we have AB, the greater line. If from the half sum AP we take the half difference C'P, we have left AC, the lesser line. EXERCISES. Solve the following equations : I. Zx — 2y = 33, 2x — 3y = 18. 2. 3x — 6y -=13, 2x + 7y = 81. 3- 7x-{- (jy = a, Qx-\-6y = b. 4- 2x + dy = m, 2x-'3y = n. 5- ax + by := p, ax — by = 5'- 6. 6"^ 7 = 26, X y 6 ""7 = 2. 7. = 18, ^ .y 8 "^2 = 29. 8. 2^3 == a, X y 2 3 = b. 9- 'i (*• + «/) + 3 {x - y) = 103, Note. Solve this equation first as if aj+^ and x—y were single sym- bols, of which the values are to be found. Then find x and y hy % 138 preceding. lo. x-^y -\- {x-^y) = U, x + y ^ ^c - y) = 10. ^^' x'^ y^ 12' x ^ ~ 12* TWO UNKNOWN QUANTITIES. 115 Note. Equations in this form can be best solved as if - and - were the unknown quantities. See next exercise. ^ ^ ' X y 1^^ X y Solution. If we multiply the first equation by 4, and the second by 8, we have 12_ 8 _ 44 _ 22 X y ~ 10 "" 5 ' ^1 + ^A = , = % X y 5 Subtracting the first from^he second, we have 23 __ 23 y - b' whence, y = b. Again, to eliminate - , we multiply the first equation by 5 a»d the second by 3 and add. Thus, 15 __ 1 _ 11 X y ~ 2' 8 ^ 10 „ 12 whence, 23 _ 23^ X ~ 2' X = 2. 2 3_^ ?_3___L ^^' x'^ y~ 12' X y~ 12 1 2 _ b_ 2_^_^ ^^' x'^ y'' 12' X y~ 24' 5_3__1 3_1__1^ X y~ Q' X y ~ ^O' 5 _ 3 _ _ 1 _3 1_ _ 1^ x + 1 y — l~~ 6' x-i-l y — l'^SO' 2 3 _ 7 2 3 _ 1 ^7- x + 2'^ y — d~' 12' x-i'2~y--d'~' 12 15. 16. 116 EQUATIONS OF THE FIRST DEGBEE. ^ a I a h ^ X y ^ y Case II. Equatio7is of the First JDegree with Three or More Unknown Quantities. 139. When the values of several unknown quantities are to be found, it is necessary to have as many equations as un- known quantities. If there are more unknown quantities than equations, it will be impossible to determine the values of all of them from the equations. All that can be done is to determine the value of some in terms of the others. If the number of equations exceeds that of unknown quan- tities, the excess of equations will be superfluous. If there are n unknown quantities, their values can be found from any n of the equations. If any selection of n equations we choose to make gives the same values of the unknown quantities, the equations, though superfluous, will be consistent. If different values are obtained, it will be impossible to satisfy them all. Elimination. 140. When the number of unknown quantities exceeds two, the most convenient method of elimination is generally that by addition or subtraction. The unknown quantities are to be eliminated one at a time by the following method : I. Select an unknown quantity to he first eliminated. It is best to begin ivith the quantity which appears in the fewest equations or has the simplest coejflcients. II. Select one of the equations containing this un^ Icnown quantity as an eliminating equation. III. Eliminate the quantitxj between this equation and each of the others in succession. THREE OB MORE UNKNOWJS^ QUANTITIES. 117 We shall then have a second system of equations less bj one in number than the original system and containing a num- ber of unknown quantities one less. IV. Repeat the process on the new system of equations, and continue the repetition until only one equation with one unhnown quantity is left. Y. Having found the value of this last unhnown quantity, the values of the others can he found by suc- cessive substitution in one equation of each system. Example. Solve the equations (1) 42: — 3y — 2; + ^ — 7 = 0, (2) X— y -^2z-\-2u — 10 =zO (3) 2a; + 2y — z-^^ti— 2 = 0,^ ^^^ (4) a; + 2y + ;$; + 2t — 19 = 0. We shall select x as tlie first quantity to be eliminated, and take the last equation as the eliminating one. We first multiply this equation by three such factors that the coefficient of x shall become equal to the co- efficient of X in each of the other equations. These factors are 4, 1, and 2. We write the products under each of the other equations, thus : Eq. (1), ^x-^^y — z-\- u— 1 = 0, (4) X 4, 4:r 4- 8y + ^^ + ^u — 76 ^ 0. Eq. (2), X— y -\-'^z-\-2ii — 10 = 0, (4) X 1, :g -f 2y + ^ + ^^ — 19 rzz 0. Eq. (3), 2x-^2y — z — 2u~- 2 = 0, (4) X 2, 2a; + 4^/ + 2^ + ^u — 38 = 0. By subtracting the one of each pair from the other, we obtain the equations, lly -f 5;^ + 3i^ — 69 = 0, ^ 2y -{-dz-\- 4:11 — 36 = 0. ) The unknown quantity x is here eliminated, and we have three equa- tions with only three unknown quantities. Now eliminating y by means of the last equation, in the same way, and clearing of fractions, we find the two equations, 232; + 38?^ — 258 = 0, llz + 14?^ — 90 = 0, : } (^) 118 EQUATIONS OF THE FIRST DEGREE, The problem is now reduced to two equations with two unknown quantities, which we have already shown how to solve. We find by solving them, u — %, We next find the value of y by substituting these values of z and u in either of the equations (b). The first of them thus becomes: lly _ 10 + 24 — 69 = 0, from which we find, 2^ = 5. We now substitute the values of y, 2, and u in either of equations {a). The second of the latter becomes iT — 5 — 4 + 16 — 10 = 0, and the fourth becomes, o; + 10 — 2 + 8 — 19 = 0, either of which gives o: = 3. We can now prove the results by substituting the values of a*, y, z, and u in all four of equations {a)^ and seeing whether they are all satisfied. EXERCISES. 1. One of the best exercises for the student will be that of resolving the previous equations {a) by taking the last equa- tion as the eliminating one, and performing the elimination in different orders; that is, begin by eliminating u, then repeat the whole process beginning with z, etc. The final results will always be the same. 2. Find the values of x^, x^^ x^, and x^, from the equa- tions, i^l + ^2 + ^3 + ^4 = ^^> X^ "T ^2 ' *^3 *^4 "~~ ^ iV -j ^"^ U^ n ~|~ %fj q "~~ »^ A ——— V, This example requires no multiplication, but only addition and sub traction of the different equations. 3a 2x+6y + 3z = 13, 2a; + 2^ — z = 12, 6x + 5y--2z = 29. PROBLEMS. 119 dz + 2u- •hy = 18, 3x -\-y — Au - 9, X -\-lz- .6y =z 33, bz- -2x- -8y + 2u = 15. X -{- y + z y + z + u 6. 1 1 X y 1 1 = m z -}- U + X 1^ + ^ + y =z C, y~ z 1 1 -z^x PROBLEMS FOR SOLUTION. 1. A man had a saddle worth $75 and two horses. If the saddle be put on horse A he will be double the value of B, but if it be put on B his value will be equal to that of A. What is the value of each horse ? 2. What number of two digits is equal to 7 times the sum of its digits, and to 9 times the difference of its digits increased by 4? Let X be the first digit, or the number of tens, and y the units. Then the number itself will be lOar+y. Seven times the sum of the digits are 7aJ4-7^, and 9 times the difference is 9(aj— ^ + 4). 3. A number of two digits is equal to 6 times the sum of its digits, and if 9 be subtracted from the number the digits are reversed. What is the number? 4. Find a number of two digits such that it shall be equal to 6 times the sum, of its digits increased by 1, while if 18 be subtracted from the number the digits will be reversed. 5. Find a number which is greater by 2 than 5 times the sum of its digits, and if 9 be added to it the digits will be reversed. 6. What number is that which is equal to 9 times the sum of its digits and is 4 greater than 11 times their diifereYice ? 7. What fraction is that which becomes equal to f when the numerator is increased by 2, and equal to f when the de- nominator is increased by 4. 8. Two drovers A and B went to market with cattle. A sold 50 and then had left half- as many as B, who had sold none. Then B sold 54 and had remaining half as many as A. How many did each have? 120 EQUATIONS OF TEE FIRST DEGREE. 9. A boy bought 42 apples for a dollar, giving 3 cents each for the good ones and 2 cents each for the poor ones. How many of each kind did he buy ? 10. Find a fraction which becomes equal to ^ when its denominator is increased by 13, and to f when 4 is subtracted from its numerator. 11. Find a fraction which will become equal to f by adding 2 to its numerator, or by adding to its denominator 3, will be- come ^. 12. A huckster bought a certain number of chickens at 32 cents each and of turkeys at 75 cents each, paying $14 for the whole. He sold the chickens at 48 cents each, and the turkeys at $1 each, realizing $20 for the whole. How many chickens and how many turkeys had he ? 13. An apple woman bought a lot of apples at 1 cent each, and a lot of pears at 2 cents each, paying $1.70 for the whole. 11 of the apples and 7 of the pears were bad, but she sold the good apples at 2 cents each and the good pears at 3 cents each, realizing $2.60. How many of each fruit did she buy? 14. When Mr. Smith was married he was \ older than his wife ; twelve years afterward he was | older. What were their ages when married.? 15. A and B together can do a piece of work in 6 days, but A working alone can do it 9 days sooner than B working alone. In what time could each of them do it singly ? 16. A husband being asked the age of himself and wife, replied: ^^If you divide my age 6 years hence by her age 6 years ago, the quotient will be 2. But if you divide her age 12 years hence by mine 21 years ago, the quotient will be 5. 17. The sum of two ages is 9 times their difference, but seven years ago it was only seven times their difference. What are the ages now ? 18. Two trains set out at the same moment, the one to go from Boston to Springfield, the other from Springfield to Bos- ton. The distance between the two cities is 98 miles. They meet each other at the end of 1 hr. 24 min., and the train from Boston travels as far in 4 lirs. as the other in 3. What was the speed of each train ? 19. A grocer bought 50 lbs. of tea and 100 lbs. of coffee for $60. He sold the tea at an advance of J on his price, and the coffee at an advance of \, realizing $77 from both. At what price per pound did he buy and sell each article ? Note. If x and y are the prices at which he bought, then \x and |y are the prices at which he sold. INCONSISTENT EQUATIONS. 121 20. For p dollars I can purchase either a pounds of tea and h pounds of coffee, or m pounds of tea and n pounds of coffee. What is the price per pound of each ? 21. A goldsmith had two ingots. The first is composed of equal parts of gold and silver, while the second contains 5 parts of gold to 1 of silver. He wants to take from them a watch- case having 4 ounces of gold and 1 ounce of silver. How much must he take from each ingot ? 22. A banker has two kinds of coin, such that a pieces of the first kind or h pieces of the second will make a dollar. If he wants to select c pieces which shall be worth a dollar, how many of each kind must he take ? 23. A has a sum of money invested at a certain rate of interest. B has $1000 more invested, at a rate 1 per cent, higher, and thus gains $80 more interest than A. C has in- vested 1500 more than B, at a rate still higher by 1 per cent., and thus gains $70 more than B. What is the amount each person has invested and the rate of interest ? 24. A grocer had three casks of wine, containing in all 344 gallons. He sells 50 gallons from the first cask; then pours into the first one-third of what is in the second, and then into the second one-fifth of what is in the third, after which the first contains 10 gallons more than the second, and the second 10 more than the third. How much wine did each cask contain at first ? Equivalent and Inconsistent Equations. 141. It is not always the case that values of two unknown quantities can be found from two equations. If, for example, we have the equations a; 4- 2^ = 3, 2x -f 4^ r= 6, we see that the second can be derived from the first by multi- plying both members by 2. Hence every pair of values of x and y which satisfy the one will satisfy the other also, so that the two are equivalent to a single one. If the equations were a; + 2y = 5, 2x-\-4.y = 6, there would be no values of x and y which would satisfy both equation.s. , 122 EQUATIONS OF THE FIRST DEGREE, For, if we multiply the first by 2 and subtract the second from the product, we shall have, 1st eq. X 2, 2:^ 4- % = 10 2d eq., %x_^rj^y_j=_^ Remainder, 0=4, an impossible result, which shows that the equations are incon- sistent. This will be evident from the equations themselves, because every pair of values of x and y which gives 2:c + % = 6, must also give ic + 2^ == 3, and therefore cannot give a; + 2^ = 5. 143. Generalization of the preceding result. If we take any two equations of the first degree between x and y which we may represent in the form ax-\-ly — c, ) , . dx + Vy =z c', \ ^^ and eliminate x by addition or subtraction, as in § 137, we have for the equation in y, {ad — ab') y =: a'c — ac', Now it may happen that we have, a'b — ab' = identically. (2) In this case y will disappear as well as x, and the result will be a'c — ac' = 0. If this equation is identically true, the two equations (1) will be equivalent ; if not true, they will be inconsistent. In neither case can we derive any value of y or x. If we divide the above equation, (2), by aa' we shall have b _ b^^ a '~ a' Hence, Theorem. If the quotient of the coefficients of the unknown quantities is the same in the tv^o equations, they will be either equivalent or inconsistent. INEQUALITIES, 123 This theorem can be expressed in the following form : If the terms containing the unknown quantity in the one equation can he multiplied by such a factor that they shall both become equal to the corresponding terms of the other equation, the two equations will be either equivalent or inconsistent. Proof, If there be such a factor m that multiplying the first equation (1) by it, we shall have ma = a\ mb =1 V. Eliminating m, we find a!b — aV = 0, the criterion of inconsistency or equivalence. 143. When two equations are inconsistent, there are no values of the unknown quantities which will satisfy both equa- tions. When they are equivalent, it is the same as if we had a single equation ; that is, we may assign any value we please to one of the unknown quantities, and find a corresponding value of the other. CHAPTER IV. OF INEQUALITIES. 144. Def. An Inequality is a statement, in the language of Algebra, that one quantity is algebraically greater or less than another. Def. The quantities declared unequal are called Members of the inequality. The statement that A is greater than B, or that A — B is positive, is expressed by A > B. 124 INEQUALITIES. That A is less than B, or that A — B i^ negative is expressed by ^ ^ A G indicates that the quantity^ is less than A but greater than Co The form A^ B indicates that A may be either equal to or greater than B, but cannot be less than B, Properties of Inequalities. 145. Theorem I. An inequality will still subsist after the same quantity has been added to or subtracted from each member. Proof, If the inequality hQ Ay B, A — B must be posi- tive. If we add the same quantity H to A and B, or subtract it from them, we shall have A ±^ H — {B ± H), which is equal to A — B, and therefore positive. Hence, if j A> B, then A±H> B ±E. Cor. If any term of an inequality be transposed and its sign changed, the inequality will remain true. Theorem IL An inequality will still subsist after its members have been multiplied or divided by the same positive number. Proof. If ^ — i? is positive, then (m or n being positive) m {A — B) or mA — 7nB will be positive, and so will A -^ A B or n 71 n Hence, if A> B, then mA > mB, and n n INEQUALITIES. 125 It may be shown in the same way that if m or n is negative, A B mA — niB or will be negative. Hence, n 01 ^ Theorem III, If botli members of an inequality Tbe multiplied or divided by the same negative number, the direction of the inequality will be reversed. That is, if A > B, then — mA < — mB, and < n n Theorem IV. If the corresponding members of several inequalities be added, the sura of the greater members will exceed the sum of the lesser members. Theorem F. If the members of one inequality be subtracted from the non-corresponding members of another, the inequality will still subsist in the direction of the latter. That is, if Ay B, then A — y y B — X. The proof of the last three theorems is so simple that it may be sup- plied by the student. Theorem VI If two positive members of an in- equality be raised to any power, the inequality will still subsist in the same direction. Proof, Let the inequality be Ay B. {a) Because A is positive, we shall have, by multiplying by A (Th. II), A^ > AB, (1) Also, because B is positive, we liave, by multiplying (a) AB y &. (2) 126 INEQUALITIES. Therefore, from (1) and (2), A^ > B\ (3) Multiplying the last inequality by Ay A^ > AB\ (4) Multiplying (2) by ^, AB'^ > B\ (5) Whence, A^ > B\ The process may be continued to auy extent. Examples of the Use of Inequalities. 146. Ex. I. If a and h be two positive quantities, such that «2 + ^2 ^ 1^ Tve must have « + Z> > 1. Proof, If a-\~h^l, we should have, by squaring the members (Th. VI), ^2 _^ 2a& + ^>2 = 1 . and by transposing the product 'Zah (Th. I, Cor.), «2 + Z>2 = 1 _ <^ad. Because a and ^ are positive, "Hah is positive, and 1 _ 2aZ> < 1. Therefore we should have a^ + y^ < 1, and could not have a^ + 1)^ —1, as was originally supposed. Ex. 2. If flj, Z>, w, and ^^ are positive quantities, such that a ^ m , V b > ^' ^«> then the value of the fraction will be contained between a + n ct fH the values of ^ and — ; that is, n (^) (3) WEQUALITIES. 127 a a + m m b ^ b-\-n ^ n' ^^ To prove the first inequality, we must show that a a + m b b -\- n is positive. Keducing this expression by § 106, it becomes an — hn b~{b~+n)' From the original inequality (a) we have, by multiplying by the positive factor bn, an > bm. That is, an — bm is positive ; therefore the fraction (3) with this positive numerator is also positive, and (2) is positive as asserted. The second inequality (1) may be proved in the same way. EXERCISES. I. Prove that if a and b be any quantities different from zero, and 1 > ic > — 1, we must have «2 _ 2abx + b^> 0. (a + bV — - — ) > ab, 3. If 3a; — 5 > 13, then x > 6. 4. If 6a; > y + 18, then a; > 4. 5. If ^-^>|-3, thena;>5. 6. li m — nx y p — qx, then x > 7. If " < 1 '■, and m and y of like sign : x < y. 'my 8. If «2 _{_ J2 _^ ^2 _ 1^ and a, b, and c are not all equal, then ab -\- be + ca I I I I I (7 is I of D, We express this relation by saying, 9 The ratio of A to D is 2^ or -; '' ^ to i> is ^; ) (1) " '' C to D is ^' 4 148. The ratio of one quantity to another is expressed by writing the unit of measure after the quantity measured, and inserting a colon between them. The statements (1) will then be expressed thus : A:D = 2i=:l; ^'•^ = h ^ ' ^ = l Def. The two quantities compared to form a ratio are called its Terms. RATIO. 129 Def. The quantity measured, or the first term of the ratio, is called the Antecedent. The unit of measure, or the second term of the ratio, is called the Consequent. Eem. When the antecedent is greater than the consequent, the ratio is greater than unity. When the antecedent is less than the consequent, the ratio is less than unity. 149. To find the ratio of a quantity ^ to a standard U, we imagine ourselves as measuring off the quantity A with Z7as a carpenter measures a board with his foot-rule. There are then three cases to be considered, according to the way the measures come out. Case I. We may find that, at the end, A comes out an exact number of times U. The ratio is then a whole number, and we say that U exactly measures A, or that ^ is a multiple of U, Case IL We may find that, at the end, the measure does not come out exact, but a piece of A less than U is left over. Or, A may itself be less than U. We must then find what fraction of U the piece left over is equal to. This is done by dividing U up into such a number of equal parts that one of these parts shall exactly measure A or the piece of A which is left over. The ratio will then be a fraction of which the num- ber of parts into which U is divided will be the denominator, and the number of these parts in A the numerator. Example. If we find that i i ; i _. jj by dividing U into 7 parts, 4 of p^—^-— ^-. these parts will exactly make A, \ \ \ \ \ = ^ then ^ = 4 of U, and we have for the ratio of A to Z7, A: U=^. If we find that A contains U 3 times, and that there is then a piece equal to 4- of Z7 left over, we have 130 RATIO. The 3 U^B, are equal to ^i}- of Z7, so that we may also say A = ^j-oiU, or A:U=^. which is simply the result of reducing the ratio 3f to an im- proper fraction. In general, if we find that by dividing U into n parts, A will be exactly m of these parts, then A : IJ = —, n whether m is greater or less than n. When the magnitude of A measured by U can be exactly expressed by a vulgar fraction, A and U are said to be com- mensurable. Case III. It may happen that there is no number or frac- tion which will exactly express the ratio of the two magnitudes. The latter are then said to be incommensurable. 150. Theorem. The ratio of two incommensurable magnitudes may always be expressed as near the true value as we please by means of a fraction, if we only make the denominator large enough. Examples. Let us divide the unit of measur-e into 20 parts, and suppose that the antecedent contains more than 28 but less than 29 of these parts. Then, by supposing it to con- tain 28 parts, the limit of error will be one part, or ^V of the standard unit. In general, if we wish to express the ratio within 1 n^^ of the unit, we can certainly do it by dividing the unit into n or more parts, or by taking as the denominator of the fraction a number not less than n, Ilhcstration hy Decimal Fractions, The square root of 2 cannot be rigorously expressed as a vulgar or decimal fraction. But, if we suppose a/2 = 1.4 = \^, the error will be < yV ; a/2 = 1.41 =\U, " " |. f - S- = A- ; whence, f > f . I - f == A ; whence, | > ^ etc. etc. General Proof, Let « : 5 be the original ratio, and let both terms be increased by the quantity u, making the new ratio a-\-u : & + u. The new ratio mmus the old one will be {h — a)u If h is greater than «, this quantity will be positive, show- ing that the ratio is increased by adding u. If ^ is less than a^ the quantity will be negative, showing that the ratio is dimin- ished by adding u. CHAPTER II. PROPORTION. 155. Def. Proportion is an equality of two or more ratios. Since each ratio has two terms, a proportion must have at least four terms. • Def. The terms which enter into two equal ratios are called Terms of the proportion. \i a\h be one of the ratios, and p \ q the other, the pro- poiiion will be, a:h^p\q. • (1) 134 PROPORTION, A proportion is sometimes written, a \ h : : p : q^ which is read, " As <^ is to & so is p to g." The first form is to be pre- ferred, because no other sign than that of equality is necessary, but the equation may be read, " As a is to & so is ^ to g," whenever that expres- sion is the clearer. Def. The first and fourth terms of a proportion are called the Extremes, the second and third are called the Means. Theorems of Proportion. 156. Theorem I. In a proportion the product of the extremes is equal to the product of the means. Proof. Let us write the ratios in the proportion (1) in the form of fractions. It will give the equation. Multiplying both sides of this equation by hq, we shall have aq = b]), (3) Cor. If there are two unknown terms in a propor- tion, they may be expressed by a single unknown symbol. Example. If it be required that one quantity shall be to another as jj to q, we may call the first px and the second qx, because px \ qx z=i p : q (identically). 15*7. Theorem IL If the means in a proportion be Interchanged, the proportion will still be true. Proof, Divide the equation (3) by pq. We shall then have, instead of the proportion (1), V'^ q or a : p =1 h \ q. PROPORTION. 135 Def. The proportion in which the means are inter- changed is called the Alternate of the original pro- portion. The following examples of alternate proportions should be studied, and the truth of the equations proved by calculation : 1 : 2 rn 4:8; alternate, 1:4 =2:8. 2:3=6:9; " 2:6=3:9. 5 : 2 = 25 : 10 ; " 5 : 25 = 2 : 10. 158. Theorem IIL If, in a proportion, we increase or diminish each antecedent by its consequent, or each consequent by its own antecedent, the proportion will still be true. Example. In the proportion, 5 : 2 = 25 : 10, the antecedents are 5 and 25, the consequents 2 and 10 (§ 148). Increasing each antecedent by its own consequent, the proportion will be 5 + 2 : 2 = 25 + 10 : 10, or 7 : 2 = 35 : 10. Diminishing each antecedent by its consequent, the proportion will become, 5 - 2 : 2 = 25 - 10 : 10, or 3 : 2 = 15 : 10. Increasing each consequent by its antecedent, the proportion will be 5 : 2 + 5 ^ 25 : 10 + 25, or 5 : 7 = 25 : 35. These equations are all to be proved numerically. General Proof, Let us put the proportion in the form h q ^ ^ If we add 1 to each side of this equation and reduce each side, it will give ^+ ^ _ p + q ~ I - ~q ' that is, a -^1) : h =: p -]- q \ q, (5) In the same way, by subtracting 1 from each side, it will be a — h \ h ^^ p — q : q. (6) 136 PROPORTION. If we invert the fractions in equation (4), the latter will become a I) By adding or subtracting 1 from each side of this equation, and then again inverting the terms of the reduced fractions, we shall find, Tlie form (5) was formerly designated as formed "by composition," and (6) as formed " by division." But these terms are now useless, be- cause all the above forms are only special cases of a more general one to bL^ now explained. 159. Theorem IV, If four quantities form the pro- portion a : b = c : d^ {a) and if m, n^ p^ and q be any multipliers whatever, we shall have ma + nb : pa -\- qb — mc -\- nd \ pc-\- qd. Proof, The proportion {a) gives the equation, a h ~ c d' Multiplyii member, ig this equation by - and adding 1 to each qh qd Keducing each member to a fraction and inverting the terms. qh _ qd pa -^ qh pc -{- qd Dividing both members by q, h__ d___ pa -\- qh'~ pc + qd The original proportion {a) also gives, by inversion, (7) PROPORTION, 137 from^which we obtain, by multiplying by -, adding 1, etc., qh -\- pa _ qcl + pc pa ~ pc ^_ ^ __J_ /g) pa + qh pc -\- qd (8) xm + (7) X w gives the equation, ma -\- nh __ mc + 7id pa -{- qh '" pc -\- qd ^ or ma + nh : pa -\- qb := mc + nd \ pc + qd, (9) which is the result to be demonstrated. 160. Theorem V. If each term of a proportion be raised to the same power, the proportion will still subsist. Proof. If a : b = p : q, a p or T = y b q' then, by multiplying each member by itself repeatedly, we shall have Hence, in l^" q^' a^ p^ b^ "" q^'' etc. etc. general. a^ : b"' =z p^ : (p. Cor. If a : b =^ p ' q^ then a^ \ a'' ±b^ — pi^ \ p'' ±(f' \ and a^±b^ \ If' = p^±q^ I ^». Theorem VI. When three terms of a proportion are given, the fourth can always be found from the theorem that the product of the means is equal to that of the extremes. 138 PROPORTION, We have shown that whenever a : h =z 2^ : q, then aq == hp. Considering the different terms in succession as unknown quantities, we find, a J. ~ q' I _aq - P' P aq Q ~" a Cor. 1. If, in the general equation of the first degree ax + l)y =^ c?5 the term c vanishes, the equation determines the ratio of the unknown quantities. Proof. If ax + by = 0, then ax = — %, X b and - = , y a or X : y =z — b : a. Cor. 2. Conversely, if the ratio of two unknown quantities is given, the relation between them may be expressed by an equation of the first degree. The Mean Proportional. 161. Def. When the middle terms of a proportion are equal, either of them is called the Mean Propor- tional between the extremes. Tlie fact that b is the mean proportional between a and c is expressed in the form, a : b =^ b : c. PROPORTION, 139 Theorem I then gives, b'^ =z ae. Extracting the square root of both members, we have b = Vc^c, Hence, Theorem VIL The mean proportional of two quan* tities is equal to the square root of their product. Multiple Proportions. 163. We may have any number of ratios equal to each other, as a \ b ^=i c : d =: e : f, etc. 6 : 4 = 9 : 6 = 3 : 2 = 21 : 14. (a) Such proportions are sometimes written in the form 6 : 9 : 3 : 21 = 4 : 6 : 2 : 14. {b) In the form {b) tlie antecedents are all written on one side of the equation, and the consequents on the other. Any two numbers on one side then have the same ratio as the cor- responding two on the other, and the proportions expressed by this equality of ratios are the alternates of the original propor- tions {a). For instance, in the proportion {b) we have, 6:9 =4:6, which is the alternate of 6 : 4 = 9:6. 6:3=4:2, '' " " 6:4=3:2. 6 : 21 = 4 : 14, " " " 6 : 4 = 21 : 14. 9 : 21 = 6 : 14, " " " 9 : 6 = 21 : 14. 163. A multiple proportion may also be expressed by a number of equations equal to that of the ratios. Since a \ b := c \ d z=z e : f, etc., let us call r the common value of these ratios, so that a c , -^ = r, -^ = r, etc. Then a = rb, c = rd, (c) c = rf, 140 PBOPORTION. will express the same relations between the quantities a, 2>, c, dy e, f, etc., that is expressed by a I i ^=^ c : d =z e : f, etc., {a) or a : c : e : etc. =z h \ d : f : etc. {h) It will be seen that where r enters in the form (c) there is one more equation than in the first form {a). [In this form each = represents an equation.] This is because the additional quantity r is introduced, by eliminating which we diminish the number of equations by one, as in eliminating an unknown quantity. 164. Theorem, In a multiple proportion, the sum of any number of the antecedents is to the sum of the corresponding consequents as any one antecedent is to its consequent. Ex. We have - — --=-- = -—' Then 5 15 25 30 2 + 6 + 10 + 12 30 5 + 15 + 25 + 30 75' which has the same value as the other four functions. General Proof, Let A, B, C, etc., be the antecedents, and a, I, c, etc., the corresponding consequents, so that B : 1= \ : c, etc. (1) ion ratio A : a, B: b, etc.. so that A = ra, B = rh, C = re. etc. etc. Adding these equations, we have A + B -{- C+ etc. = r{a + b + c-^ etc.), . A + B -{- C -\- efcc. a -\- h -\- c -\- etc. ' that is, the ratio A -\- B -\- C -\- eta, : a -{- d -{- c -\- etc. is equal to r, the common value of the ratios A : a, B : b, etc. PROBLEMS. I . A map of a country is made on a scale of 5 miles to 3 inches. PBOPOETION. 141 (1.) What will be the length of 8, 12, 17, 20, 33 miles on the map ? (2.) How many miles will be represented by 6, 8, 16, 20, 29 inches on the map ? Rem. 1. If Xy y, z, u, v be the required spaces on the map, we shall have 5 : 3 == 8 : aj = 12 : y, etc. If «, b, c, etc., be the required number of miles, we shall have S : 6 = 6 : a = S : b rzilQ : c, etc. Rem. 2. When there are several ratios compared, as in this problem, it will be more convenient to take the inverse of the common ratio, and multiply the antecedent of each following ratio by it to obtain the conse- quent: In the first of the above proportions the inverse ratio is f , and X = iotS, y = f of 12, etc. In the second, a =: f of 6, & = f of 8, etc. 2. To divide a given quantity A into three parts whicli shall be proportional to the given quantities a, h, c, that is, into the parts x, y, and z, such that X \ a =1 y : t =^ z : c, or X \ y : z ^=^ a '. h \ c. Solution. By Theorem IV, X y z X -\- y -\- z A a~h~c~~a+h-\-c'~'a-\-h + c Therefore, _ aA _ ^^ _ ^^ ^ "~ a + ^ + c' ^ "" a + b -\- c' ^ ~ a -\- h + c 3. Divide 102 into three parts which shall be proportional to the numbers 2, 4, 11.- 4. Divide 1000 into five parts which shall be proportional to the numbers 1, 2, 3, 4, 5. 5. Find two fractions whose ratio shall be that of a:b, and whose sum shall be 1. 6. What two numbers are those whose ratio is that of 7 : 3 and whose difference is 24. 7. What two numbers are those whose ratio is m : n, and whose difference is unity ? 8. Find x and y from the conditions, X : y = a : b, ax — by = a -{- b. 142 PROPORTION. 9. Show that if a : b = A : B, c : d=i C : D, we must also have ac : bd = AC : BD, 10. Having giveu x = ay, find the value of ^ "^ ^^. x-2y II. Having given find the value of x — 'Zy x + y '^ = 5, x — y 12. If a : b ^= p : q, a^ 2j3 prove a^ + b^ : 7 ^^ p^ ■\- q^ \ —^- — , a-\- b ^^'^ i^+S' 13. If a -\- b -{- c -}- d a — b -{- c — d a + b — c — d a — b — c + d^ show that a \ b =: c \ d. 14. A year's profits were divided among three partners, A, B, and C, proportional to the numbers 2, 3, and 7. If should pay B $1256, their shares would be equal. What was the amount divided ? 15. In a first year's partnership between A and B, A had 2 shares and B had 5. In the second year, A had 3 and B had 4. In the second year, A's profits were $3200 greater and B's were $1700 greater than they were the first. What was each year's profits ? 16. In a poultry yard there are 7 chickens to every 2 ducks, and 3 ducks to every 2 geese. How many geese were there to every 42 chickens ? 17. A drover started with a herd containing 4 horses to every 9 cattle. He sold 148 horses and 108 cattle, and then had 1 horse to every 3 cattle. How many horses and cattle had he at first ? 18. If a bowl of punch contains a parts of water and b parts of wine, what is the ratio of the wine to the whole punch ? What is the ratio of the water? What are the sums of these ratios ? PROPORTION. 143 19. One ingot consists of equal parts of gold and silver, while another has two parts of gold to one of silver. If I combine equal weights from these ingots, what proportion of the compound will be gold and what proportion silver ? 20. What will be the proportions if, in the preceding prob- lem, I combine one ounce from the first ingot with three from the second ? 21. One cask contains a gallons of water and h gallons of alcohol, while another contains m gallons of water and n of alcohol. If I draw one gallon from each cask and mix them, what will be the quantities of alcohol and water ? 22. What will be the ratio of the liquors in the last case, if I mix two parts from the first cask with one from the second ? 23. What will it be if I mix p parts from the first with q parts from the second ? 24. A goldsmith has two ingots, each consisting of an alloy of gold and silver. If he combines two parts from the first ingot with one from the second, he will have equal parts of gold and silver. If he combines one part from the first with two from the second, he will have 3 parts of gold to 5 of silver. What is the composition of each ingot ? Suggestion. Call r the ratio of the weight of gold in the first ingot to the whole weight of the ingot ; then 1 — r will be the ratio of the sil- * ver in the first to the whole weight of the ingot. See the following question. Note. Problems 18-24 form a graduated series, introductory to the processes of Problem 24. 25. Point out the mistake which would be made if the solution of the preceding problem were commenced in the fol- lowing way : If the first ingot contains p parts of gold to q parts of silver, and the second contains r parts of gold to 8 of silver, then Two parts from the first ingot will have 2p of gold and 2q of silver. One part from the second ingot will have r of gold and « of silver. Therefore, the combination will contain 2p + r parts of gold, and ^q-\-s parts of silver. Show also that if we subject p, q, r, and s to the condition p-V-q = r + s, the process would be correct. 26. Show that if the second term of a proportion be a mean proportional between the third and fourth, the third will be a mean proportional between the first and second. BOOK V. OF POWERS AND ROOTS. CHAPTER I. INVOLUTION. Case L Involution of Products and Quotients. 165. Def. The result of taking a quantity, A^ n times as a factor is called the n*'*^ power of A^ and as already known may be written either ^ A J., etc., 72, times, or A^. Bef. The number n is called the Index of the power. Bef, Involution is the operation of finding the powers of algebraic expressions. The operation of involution may always be expressed by the application of the proper exponent^ the expression to be involved being inclosed in parentheses. Example. The n^^ power of « + J is (aj + lY. The iif'^ power of abo is {abcY, 166. Involution of Products. The n^^ power of the product of several factors a^ h^ c, may be expressed without exponents as follows : dbcahcahC) etc., each factor being repeated n times. INVOLUTION, 145 Here there will be altogether n a's, n Vs, and n (fs, so that, using exponents, the whole power will be a'^b^c^ (§ 6^, 67). Hence, {abc)^ = aP'hH^ That is, Theorem. The power of a product is equal to the product of the powers of the several factors. 167. Involution of Quotients. Applying the same methods to fractions, we find that the n^^ power of - is — • For (xY' XXX (-) = , etc., n times: V yyy _ XXX, etc., n times .^ . ^ ~" yyy^ ^tc, n times ^^ ' ' __ x^ EXERCISES. Express the cubes of I. ahc, 2. — • 3. ahcK c mn a -\- h mn {a + 5) pq a — h ' pq {a — b)' Express the n^^ powers of the same quantities, the quanti- ties between parentheses being treated as single symbols. Case II. Involution of Powers. 168. Problem. It is required to raise the quantity a^ to the n^^ power. Solution. The n*^ power of a^ is, by definition, a'^ X a^ X a^y etc., n times. By § 66, the exponents of a are all to be added, and as the exponent m is repeated n times, the sum m -f m -h m 4- etc., n times, is m7i. Hence the result is a^^, or, in the language of Algebra, 10 146 IJS VOLUTION, Hence, ^eorem. If any power of a quantity is itself to be raised to a power, the indices of the powers must be multiplied together. EXAM P LE S. Note. It will be seen that this theorem coincides with that of Case I when any of the factors have the exponent unity understood. EXERCISES. Write the cubes of the following quantities: 4.a I. dxyl 2. 3- aP", 4. bx'. 5. 2ahn\ 6. 6a^ b Write the 71^^ powers oi 7. a. 8. a^. 9- aWc, 10. a'^a^. II. 2p^q\ 12. {a-\-b) {c-^d). ^3. 0'^ + «/)(^- -y)- 14. 7 {a + b — c){a- -by. Ans, 7^ '{a-\-b- ■ c)^ (a — b^P. a '5- r 16. a2 17. x-\-y x-y 10. „ • xy^ ^'''' ~xnfn • ab {c - df "9- (^ _ ^) c3 • Reduce : 20. {^ab^n^f. 21. (— ^mnx^f. 22. 2a{—3b^m7i^)K 23. {IpqVy. 24. {ab^y^ 25- {2ah^)\ 26. {m^)\ Note 1. If the student find any of these exponential expressions difficult of expression, he may first express them by writing each quantity a number of times indicated by its exponent. Note 3. The student is expected to treat the quantities in paren- theses as single symbols. INVOLUTION. 147 Eem. The preceding theorem finds a practical application when it is necessary to raise a small number to a high power. If, for example, we have to raise 2 to the 30th power, we should, without this theorem, have to multiply by 2 no less thf».n 29 times. But we may also proceed thus : 2^ = 4, 2^ = 22.22 = 4-4 ^ 16, 28 = 24.2^ =16.16 = 256, 216 := 28.28 = 2562 ^ 65536^ 224 ^ 216.28 = 216.256 = 16777216, 230 ^ 221.26 = 224.64 = 1073741824. Case of IVegative Exponents, 169. The preceding theorem may be applied to negative ^^xponents. By the definition of such exponents, g = aP5-.. (1) Eaising the first member to the n^^ power, we have. This is the same result we should get by applying the theorem to the second member of (1), and proves the proposi- tion. EXERCISES. Express the 6th powers of I. alrK 3. a7np~K Eeduce : II. {ab-^c-^)-^. 13. {x^y-i)-\ After forming the expressions, write them all with positive exponents, in the form of fractions. 2. a^h'\ 4. a-^h-\ 6. {x-^yY{x-\-z) Q {a + h)-^ 0. {a - by 10. {ah-^c-'^y. 12. {m^n-J)-^, 14. \a%^c-''Y. 148 INVOLUTION. Algebraic Signs of Powers. 110. Since the continued product of any number of posi- tive factors is positive, all the powers of a positive quantity are positive. By § 26, the product of an odd number of negative fac- tors is negative, and the product of an even number is positive. Hence, Theorem. The even powers of negative quantities are positive, and the odd powers are negative. EXAM PLES. ' (— of = a^ ; (— ay =: —a^; (— aY = a*, etc. EXERCI SES. Find the value of I. (-3)^. 2. 1 [— 3)'- 3- 4*. 4. (-5)^. 5- < [- 6)3. 6. {-by. 7. {-a -by. 8. ( [—mny. . 9- i-pqy. lO. (— «)2». II. [- 5)2»+>. 12. (_a_J)2«-i 13. (-1)^. 14. [_i)2n+t. 15- (- l)^->. Case III. Involution of Binomials— the Bino- mial Theorem. 1*71. It is required to find the n^^ power of a hinomial. 1. Let a + ^ be the binomial ; its n^^ power may be written Let us now transform this expression by dividing it by «^, and then multiplying it by «% which will reduce it to its orig- inal value. We have (§ 167), Multiplying this last expression by V, by writing this power outside the parentheses, it becomes «^^ (' - 'J. « INVOLUTION. ' 151 preceding one, until we come to the s^^ or last, which will be 71 — 8 -{- 1. Such a ijroduct is written, n(?i — l){n — 2) (n — s + l). The dots stand for any number of omitted factors, because s may be any number. We have written 4 of the s factors, so that s — 4 are left to be represented by the dots. The denominator of the fraction is the product of the s factors, 1.2.3.... 5, each factor being greater by 1 than the preceding one, and the dots standing for any number of omitted factors, according to the value of s. Thus, the s^^ coefficient in the 7i^^ line will be n{n- l )(n-2) (n - s + 1) 1.2.3 s " ^"^^ If 5 is greater than ^n, the last factors will cancel some of the preceding ones, so that as s increases from ^n to n, the values of the preceding coefficients will be repeated in the reverse order. Thus, suppose n = 6, Then, by cancelling common factors, = ^J = -• 1.2.3.4.5.6 ~ ^' If we should add one more factor to the nul^erator, it would be 0, and the whole coefficient would be 0. The conclusion we have reached is embodied in the follow- ing equation, which should be perfectly memorized : /-I , \*, -I . , n hi — 1) ^ ii(n — l) (n —2) . (1 + x)n = 1 + iix 4- --^372 — ^ + -^ iT^ ^ ^ 1.2.3.4 ^ ^^*** ^-^'^ 6.-5.4.3 1-2.-3.4 6.5.4.3.2 1.2.3.4.5 6.5.4.3.2.1 152 INVOLUTION, EXERCISES. 1. Compute from the formula {d) all the binomial coeffi- cients for ^ =: 6, and from them express the development of (1 + xf. 2. Do the same thing for n = 8, and for n = 10. 173. To find the development of (a -f- b)^, we replace x by - , and then multiply each term by a^, (t [See equations (1) and (2).] We thus have ^7 i/ll 1^ {a + hY z=i oP' ^ na^'-'^h + V -^ a'^-W + etc. to l^ The terms of the development are subject to the following rules : I. The exponents of b, or the seeoncl term of the hiiio- mial, are 0, 1, 2, etc., to n. Because W is simply 1, a" is the same as W^V^. II. The sum of the exponents of a and b is n in each term. Hence the exponents of a are n, n ~ 1, n — 2, etc., to 0. III. The coefficient of the first term is unity, and of the second n, the index of the power. Each following coefficient may he found from; the next preceding one hy multiplying by the successive factors, n — 1 n — 2 n — d ""2~^ —3—, — J-. etc. IV. // b or a is negative, the sign of its odd powers will he changed, hut that of its even powers will remain the same. (Compare § 170.) Hence, {a — hY — a^ — na^-^b the terms being alternately positive and negative. {a - hY =:a^- nan-^b + ^^.-Jil «7i-2^ _ etc., ('-r^ INVOLUTION. 153 EXERCISE S — Continued. 3. Compute all the terms of {a + by, using the binomial coefficients. 4. What is the coefficient of W in the development of {a + ^)^«. 5. What are the first foui terms in the development of {2am + 3^)8. 6. What are the first three terms in the development of Vl8 ? What are the last two terms ? 7. What are the first three and the last three terms of 8. What is the development of (a + -) • 9. What are the first four terms in the development of the following binomials : (1 + a^)n ; (1 + 2x^Y ; (1 - "^^^Y ; 10. What are the sum and difference of the two develop ments, (1 + xf and (1 — a;)^? Case IV. Square of a JPolynoinial. 173. 1. Square of any Polynomial Let a-}-b + c-{-d + etc., be any polynomial. We may form its square thus : a-{-b-i-c + d-{- etc. a + b-\-c-\-d-{- etc^^ a^ -\- ab -\- ac -{- ad + etc. ab -{-b^ -{-be -hbd + etc. ac -}- be -{- €^ -}- cd -^ etc. ad 4- bd -\- cd -{- d^ + etc. a^ -j-b^ -\- c^ + ^^ + etc. + 2ab + 2ac + 2ad + etc. + 2bc + 2bd + etc. -f- 2cd + etc. 164 INVOLUTION, We thus reach the following conclusion : Theorem, The square of a polynomial is eqnal to the sum of the squares of all its terms plus twice the product of every two terms. 2. Square of an Entire Function, Sometimes we wish to arrange the polynomial and its square as an entire function of some quantity, for example, of x. Let us form the square oi a -\-1jx -{- cx^ + d^ + etc. a -\- dx ^ cx*^ -\- dx^ + etc. a -\- hx -^ cx^ -\- dx^ -f etc. «2 -f adx + acx^ + adx^ + etc. abx -f- ¥x^ + hcx^ -\- bdx^ + etc. acx^ -h bcx^ + c^x^ + etc. ada^ -\- hdx^ + etc. a^ + 2al)x + {2ac + b^) x^ -^-J^ad + 2I)c) x^ + etc. We see that : The coefficient of x^ is ac -\- hh -{- ca, " " " c^ is ad-^lc -^ cb -\- da, " " " ^ is ae -\- M -{- gg -\- dh -\- ea, etc. etc. The law of the products ae, hd, gg, etc., is that the first factor of each product is composed successively of all the co- efficients in regular order up to that of the power of x to which the coefficient belongs, while the second factor is composed successively of the same coefficients in reverse order. EXERCISES. Form the squares of I. 1 + 2:i; + ^x\ ■ 2, 1 -\- 2x + 3x^ + 4^. 3. l-\-2x-{- 3x^ + 4:0^ + 5a;5. 4. 1 + 2x + Sx^ + ^x^ + 5a;5 + (jx^ 5. 1 — 2x + dx'^ — 4:X^ 6. a—b -\- c — d, a evolution: 155 CHAPTER 11. EVOLUTION AND FRACTIONAL EXPONENTS. 174. Def. The it*'* Root of a quantity q is snch a number as, being raised to the ii}^ power, will produce q. When n = 2^ the root is called the Square Root. When 72, = 3, the root is called the Cube Root. ' Examples. 3 is the 4th root of 81, because 3.3.3.3 ■= 34 = 81. As the student already knows, we use the notation, 71^^ root oiq z=z ^^. There is another way of expressing roots which we shall now describe. 175. Division of Exponents. Let us extract the square root of a^. We must find sach a quantity as, being multiplied by itself, will produce a^. It is evident that the required quan- tity is a^, because, by the rule for multiplication (§§ ^Q, 166), ■a^ X a^ = a', n The square root of a'^ will he a^, because n n IL^Vl a^ X 0^ — a^ ^ = a^. n In the same way, the cube root of a^ is a^, because n n^ n^ 0^ X a^ X a^ = a^. The following theorem will now be evident : Theorem. The square root of a power may be ex- pressed by dividing its exponent by 2, the cube root by dividing it by 3, and the n^^ root by dividing it by n, 176. Fractional Exponents. Considering only the origi- nal definition of exponents, such an expression as n^ would 156 EVOLUTION. have no meaning, because we cannot write a 1^ times. But by what has just been said, we see that cfi may be interpreted to mean the square root of a^, because 3 3 a'^ X a'^ — a\ Hence, A fractional exponent indicates the extraction of a root. If the denominator is 2, a square root is indi- cated ; if 3, a cube root ; if n^ an n^^^ root. A fractional exponent has therefore the same meaning as the radical sign \/, and may be used in place of it. EXERCISES. Express the following roots by exponents only : I. "s/m, 2. ^/(w^ + 7i). 3. a/(« + ^)^. 4. ^/{a + hY. 5. ^/mK 6. ^^^ 7. \/a\ 8. ^/{a-\-bY, 9. ^/{a + h)^. 177. Since the even powers of negative quantities are positive, it follows that an even root of a positive quantity may be either positive or negative. This is expressed by the double sign ± . EXERCISES. Express the square roots and also the cube roots and the nf'^ roots of the following: I. (a + If. 2. {a + hf. 3. a-^h 4. {x-\-yf. 5. (^ + #. 6. (:x + y)K 178. If the quantity of which the root is to be ex- tracted is a product of several factors, we extract the root of each factor, and take the product of these roots. Example. The n^^ root of am^p is a^ni^p^, because {Jirn^p^Y = arri^p, by §§ 168 and 176. If the quantity is a fraction, we extract the root of both members. EVOLVTIOJSr. 157 Proof, (§§167,168.) Because ~7 taken 7i times as a factor makes t 9 therefore, ^n ct by definition, it is the n*^ root of t- EXERCISES. Express the square roots of I. 4:X\ 2. -— Express the cube roots of 4. 27-64. 5. 27«3. 7. ab^(^d\ 6. 64.27«35«' 8a^ 125a;^^ Express the ^^^^ roots of 9. 7. 10. 4.7. 12. 6a^b^^, II. 4.7.10. 6«2|^^ 14. 15- /^7W+l ^/J >^— 2 C^cln dTnn J^ 16. 35^ a-'^"" (a + Z>)4'^ (r?; — ^jY 4^ (5 — c + cZ)-^. Eeduce to exponential expressions : 17. v^^r=^. 18. ^^^^. 19. '^^^O^. + bY 20. 21 m/(a bY Powers of Expressions with Fractional Expo- nents. 179. Theorem. The p^^ power of the n^^ root ig equal to the n^^ root of the p^^ power. 158 FRACTIONAL EXPONENTS. Ill algebraic language, or (fl^-)" — (a^)", Example. (v^s)^ =: 2^ == 4, or, in words, the square of the cube root of 8 (that is, the square of 2) is the cube root of the square of 8 (that is, of G4). Gejieral Proof. Let us put :^; = the n^^ root of a, so that x^ = a. (1) The p^^ power of this root x will then be xp. (2) Eaising both sides of the equation (1) to the 2^^^ power, we have af'P = aP = pi^ power of a. The 71^^ root of the first member is found by dividing the exponent by n, which gives n*^ root ofp^^ power = x^, the same expression (2) just found for the p^^ power of the 71^^ root. This theorem leads to the following conclusion : 1. The expression p a/" 1 may mean indifferently the p^^ power of a^, or the ?ztli root of 6i^, these quantities being identical. 2. The pov^ers of expressions having fractional ex- ponents may be formed by multiplying the exponents by the index of the power. EXE RCISES. Express the squares, the cubes, and the 7i^^ powers of the following expressions : I. a^. 2. ak 3. ai 4. n'^i 5. ab^. 6. ab^c^. IRRATIONAL EXPRESSIONS. 159 m m P Q 7. a^h^, 8. a^h v. m 9. {a + hf{a — b)~^. 10. a-^h"". II. a^h\ 12. ^ ^ ^-^ ^ Reduce to simple products and fractions : ^y ^) . 14. {cfih^c'i)^, 15. {aH^y^. 16. \a ^1 9, (^:) '^- \-A} • '^- .^^ • .-R- CHAPTER III. REDUCTION OF IRRATIONAL EXPRESSIONS. Definitions. 180. Def. A Rational Expression is one in which the symbols are only added, subtracted, multiplied, or divided. All the operations we have hitherto considered, except the extraction of roots, have led to rational expressions. Def. An expression which involves the extraction Oi a root is called Irrational. Example. Irrational expressions are Va, ^/{a + h), a/27; or, in the language of exponents, ^4, . {a + ^)i, 27*. In order that expressions may be really irrational. 160 IRRATIONAL EXPRESSIONS. they must be Irreducible, that is, incapable of being expressed without the radical sign. Example. The expressions ^'cf^2ah + h^, a/36, ' are not properly irrational, because they are equal to a -\- h and 6 respectively, which are rational. Def. A Surd is the root which enters into an irrational expression. Example. The expression a + hVx is irrational, and the surd is Vx. Def. Irrational terms are Similar when they con- tain the same surds. Examples. The terms ^30, 7^/30, {x + y) a/30, are similar, because the quantity under the radical sign is 30 in each. The terms {a + h) ^/x -f- y, dVx + ^, fnVx + y are similar. Ag^gregation of Similar Terms. 181. Irrational terms may be aggregated by the rules of §§ 54—56, the surds being treated as if they were single sym- bols. Hence : WTien similar irrational terms are conneetecl hy the signs -{-or — , the eoeffieients of the similar surds may he added, and the surd itself affixed to their sum. Example. The sum aV{x + y) — W{x + ij) + dV(x + y) may be transformed into (a — b + 3) V{x + y). EXERCISES. Keduce the following expressions to the smallest number of terms : I. 7a/3 — 5a/2 + 6a/3 + 7a^A/2. IRRATIONAL EXPRESSIONS, 161 2. W{x + y) + W{x -y) + 2{a + b) V{x + y) — 3{a + b) V{x — y)' 4. {fl + y) a/^2/ + (a — Z>) Viz^^. 5. Vx {a — b)-\-(b — c) ^/x ■\- {c — d) Vx. 6. aVx — ^/x + "la^/x — {a -\- b) Vx. 7. - Vi?^ — aVx + QV^ — cVx + ^ v''^. 8. ^ "r Va: — GcV^ — "4~~ "^^ + ^^' 9. -Vx — Vx+ {a — h) Vx + ^^~ ^ Vi«^. 10. a/^ — hVa — a/o; H 4-" '^^ ~ o '^^" 11. jv^— V^H — ^— o — -yx, 12. 4A/i?^ — • Q 'V'^^ + («^ — ^) Vi^. o Factoring Surds. 183. Irrational expressions may sometimes be transformed so as to have different expressions under the radical sign, by the method of § 178, applying the following theorem : Theorem. A root of the product of several factors is equal to the product of their roots. In the language of Algebra, ^/abcd, etc. = "s/a ^/h "s/c \^d, etc. Proof, By raising the members of this equation to the n^^ power, we shall get the same result, namely, a X I? X c X d, etc. Example. VsO = V^ a/S. 11 162 IRRATIONAL EXPRESSIONS. EXERCISES. Prove the following equations by computing both sides : a/4 V'49 = V4:.49 = Vl96. Pm/. a/4: a/49 r:^ 2-7 = 14, and a/196 =: 14. V4 a/9 = a/36. a/4 ^25 rr a/4725. a/9 a/16 = A/90L6. V25 Vse = a/25^. Express with a single surd the products : I. V{a + d) V{a — b). SoLUTioi;r. V{a + b) V(a —b) = V{a + b){a — b) V7V6. 3. V^Va. Va V{a + y), 5. Va Vb V{a + b). V{x + I) V{x - 1). V{x^ + 1) V{x + 1) \/{x - 1). l(a + ^)^ {a - b)ij. 183. If we can separate the quantity under the radical sign into two factors, one of which is a perfect square, we may extract its root and affix the surd root of the remaining factor to it. EXAMPLES. Va^b = Va^ Vb — aVb. Vab Vac = Vcf^bc = aVbc, a/12 a/6 =: a/72 = a/36 a/2 = 6 a/2. V{^a^ + 8a^J — \^ah^ = V^d^ {a ^2f^~iac) — 2aV{ct + 25 — 4ac). (x^— Ax^y -h Ax}/)^ = (a; — 2^) x^. IRRATIONAL EXPRESSIONS. 163 EXERCISES. Eeduce, when possible : I. a/8. 2. a/32. 3. a/128. 4. a/3 a/27. 5. ^'ab ^/'c~a ^/~hc. 6. a/2 a/72. 7. a/4 a/72. • 8. a/(:z: + 1) a/(^ + 1). 9. a/175. ro. a/150c II. a/108. 12. a/^^^T^T+T). 13. a/(«^^ + '^cbbx + ^2^). Here the quantity under the radical sign is equal to In questions of this class, the beginner is apt to divide an expression like ^a + 6 H- c into /y/(X + ^Jh + y'c, which is wrong. The square root of the sum of several quantities cannot be reduced in this way. 14. \^dhj At ^CLy •\- ^y- 15. ^/^mh + "^mz -f 4;$;. Eeduce and add the following surds : 16. 4a/2 — 6a/8 + 10a/32. 17. a/12 + a/27 -f- a/75. 18. A/4a — 2a/«. 19. 125^ — 45^ — 80*. 20. 'v/81 — A^192. 21. {aWf^ + (aSc^ji Mviltiplication of Irrational Expressions. 184. Irrational polynomials may be multiplied by com- bining the foregoing principles with the rule of § 78. The following are the forms : To multiply a + h^/x by m + n^/y, a (m + nVy) = (i^n x an^/y. hVx {m -f nVy) = hm^/x + In^/xy, The product is am + mWy -}- hnVx + In^/xy, EXERCI SES. Perform the following multiplications and reduce the results to the simplest form (compare § 80) : I. (2 + 3 a/5) (5 — 3 a/2). 2. (7 + 2 a/32) (9 — 5 a/2). 164 IRRATIONAL EXPRESSIONS, 3. {a + ^/h) {a — V^). 4. {^/a-\-^/h-^Vc-{- \^d)\ 5. (m + n^) (m + 2/1*). 6. («i — c^s) (a^ -f- a^), 7. (^ + «-^)^. 8. (a* -a-*)'. 9. [« + &\/(^ + y)'\ [a - 2>a/(^ + y)l 10. [m -|- /^V(«5 -f ^)] [w — /^a/(« — h)\ 11. [a; + VCt'^ — 1)] \x — V(^2 _ 1)]. 12. [(^2 + 1)4 + ^J [(^H- l)i - ^]. Expressions may often be transformed and factored by combining the foregoing processes. Example. To factor axi + bxi + cxi + dx~2, we notice that 1 X a 5 i ^ ; x-^ = x'^x^, x-^ =z x-^x% etc. so that the expression may be written, ax^xi 4- bx^x^ + cxxi + dxi = (ax^ + hx^ -{- ex + J) x^, EXE RCISES. Keduce the following expressions to products : 13. 2 4-V2. 14. 3^ + 2-3*. 15. {a 4- Z>)2. 16. V^ + ay^ — %3, 17. X — y — ^/x — y. Eeduce to the lowest terms : i». — - 2 ^/a'^-h 20. 22. ax^ + ho^ axi — bx^ a — X + Vcc — x Va^ - b^ 21, *^. a — .T — Vet — X a -jr b 185. Rationalizing Fractions. The quotient of two surds may be expressed as a fraction with a rational numerator or a rational denominator, by multiplying both terms by the proper multiplier. Example. Consider the fraction -— • IRRATIONAL EXPRESSIONS. 165 Multiplying both terms by a/7, the fraction becomes — — , and has the rational denominator 7. 5 Multiplying by y'S, it becomes — — , and has the rational numerator 5. ^ ^^ The numerator or denominator may also be made rational when they both consist of two terms, one or both of which are irrational. Let us have a fraction of the* form A -^dVB P + Qy/R' in which the letters A, D, P, Q, and R stand for any algebraic or numerical expressions whatever. If we multiply both nu- merator and denominator by P — Q^ R, the denominator will become p2 _ Q^n, The numerator will become AP + PB^/B - A QVR - DQ^BR. so that the value of the fraction is AP + pdVb -^ AQVR - dqVbr P' - Qm EXE RCI SES. Reduce the following fractions to others having rational denominators : 7a/3 2V18 . 5V24 9V5 3\/6 2a/^ 166 PEMFEGT SQUARES, a + 2V(^ + y) 2a/3 + 7a/5 lO. : • II. a + V(x -\-y) * V5 _ V3 12. V^ - V(>^' + ^) ^^^ ^ 1 V^ + «^ + '\/x — a «i + (a + 1)3 "s/x 4- a — V^ — a Perfect Squares. 186. Def. A Perfect Square is an expression of whicli the square root can -be formed without any surds, except such as are already found in the expression. Examples. 4/72^, 4^^ + 4^ + 1 are perfect squares^ be- cause their square roots are 2m\ 2a + 1, expressions without the radical sign. The expression a + 2Vcib -\- b, of which the root is Vet + Vb, may also be regarded as a perfect square, because the surds Vcc aad \^b are in the product 2\/ab, Criterion of a Perfect Square, The question whether a trinomial is a perfect square can always be decided by compar- ing it with the forms of § 80, namely : a^ + 2ab -^ b'^ = {a -^ b)% or «2 _ 2ab + ^2 — (^ _ ^)2. We see that to be a perfect square, a trinomial must fulfil the following conditions : (1.) Two of its three terms must be perfect squares. (2.) The remaining term must be equal to twice the product of the square roots of the other two terms. When these conditions are fulfilled, the square root of the trinomial will be the sum or difference of the square roots of the terms, according as the product is positive or negative. The root may have either sign, because the squares of posi- tive and negative quantities have the same sign. COMPLETING THE SQUARE, 167 If the terms which are perfect squares are loth negative, the trinomial will be the negative of a perfect square. EXAMPLES. ^/W~^ab~^^ = a -\- b or — (a -{- b). b OY b ~a. ^/^2 _2ab -\- b^ = a- — a^ + 2ab — b^= — {a ■ of, EXERCISES. Find which of the following expressions are perfect squares, and extract their square roots : I. 9 + 12 + 4. 2. r?:^ + 4^ + 4. 3. 4^ _l_ 2x^ + h 4. a2 ^ab — b\ 5- 4:a^^ + 12a^Z»^^+ 9b'^^. 6. a^ + %ab — W. 7. x^ — ax^y + - a^y\ 8. aW — ^abccl + c^d^. 9- 771 + 2m^n^ + n. 10. a^ — 2ax + y\ II. a + 4a4z>i + 42>. 12. f^ — 2 + «-!. IS- 25y + 9g2 — ^Op\ 14. ejim^n _^ 7^2 ^ 9^^^4;i, IS- ^9xY + 9^2 _ 42:r?/^. 16. 9m^^ — 2m>g + ^ To Complete the Square. 187. If one term of a binomial is a perfect square, such a term can always be added to the binomial that the trinomial thus formed shall be a perfect square. This operation is called Completing the Square. Proof, Call «^ the root of the term which is a perfect square, which term we suppose the Jirsi, and call m the other term, so that the giyen binomial shall be a^ + m. Add to this binomial the term — ^ , and it will become a^ + 7)7 + 4^2 168 COMPLETING THE SQUARE. This is a perfect square, namely, the square of m thatis, a^ + m + ^, = [a + -). Hence the following EuLE. Add to the hinomial the square of the second term divided hy four times the first tei^m. Example. What term must be added to the expression x^ — 4:ax (1) to make it a perfect square ? The rule gives for the term to be added, 4:X^ Therefore the required perfect square is x^ — ^ax + 4r^2 ^ {-x — 2af. We may now transpose 4:0?, so that the left-hand member of the equation shall be the original binomial (1). Thus, . x'^ — 4tax z=i {x — 2ay — 4^^. The original binomial is now expressed as the difference of two squares. Therefore, the above process is a solution of the problem : Having a binomial of which one term is a ^perfect square, to express it as a difference of two squares, EXERCISES. Express the following binomials as differences of two squares : I 3 5 7 9 IT 13 x'^ + 2xy, 2. x^ -\- 4:xy, x^ + ^ax. 4. 4a;2 -f ^xy, 4:X^ + 4:xy. 6. dx^ -\- ax, IGx^ + S2mx, 8. x^ + 4.x, a\r? + 2a^x, 10. Wx^ + 2. m%2 + 1. 12. ^p^x^ + Ix, 4:X^ ' 9a^x^ IRRATIONAL FACTORS. 169 Irrational Factors. 188. When we introduce surds, many expressions can be factored which have no rational factors. The following theorem may be applied for this purpose : Theorem. The difference of any two quantities is equal to the product of sum and difference of thek square roots. In the language of algebra, if a and h be the quantities, we shall have which can be proved by multiplying and by § 80, (3). EXERCISES. Factor I. m — n. 2. m — 1. 3. am — hn. 4- 4«2m _ 9, 5. x^ — m. 6. x^ — {m -f n\ 7. {x — af - 1 , --{m- -n). 8. x^ — (in — n). 9. {a + hf — (4;/ __ qy jq^ -^2 _j_ ^xy + y^ — {m -f w)i Find the irrational square roots of the following expressions by the principles of § 186 : 11. a — 2 -f a~K Ans, a^ — a~^. 12. X — 2Vxy + y, 13. 4 -{- 4a/3 -f 3. 14. 9 + 5 — 6a/5. 15. 4a 4- ^ — 4tah^. 16. fl + ^» + 2(a4-Z^)ir(;-f 2:1 17. 3 + 2a/15 + 5. 18. 3 + 5 — 2a/15. 20. a — 2\/a + 1. ^ i 1 22. a + 2^3 4- 1. a a a 24. 4 + 3 + 9- 26. ^5 _|. 2 -f a -5. 28. a -\- b — 4: -\- 19. X y Vxy 4 "^4 2 ■ 21. « — 2a^ 4- ai 23. 7 . ai 25- - + - + -. 16 ^ 4 ^ 4 27. 4^:3 _ 8 -t- 4a;-8 a -\-h BOOK VI. EQUATIONS REQUIRING IRRA TIONAL OPERATIONS. CHAPTER I. EQUATIONS WITH TWO TERMS ONLY. 189. In the present chapter we consider equations which contain only a single power or root of the unknown quantity. Such an equation, when reduced to the normal form, will be of the form Ax^ + ^ = 0. By transposing B, dividing by A, and putting B the equation may be written, x^ — a = 0. or x^ =i a, (1) Here n maybe an integer, or it may represent some fraction. Such an equation is called a Binomial Equation, because the expression x'^ — a is a binomial. Solution of a Binomial Equation. 190. 1. When the exponent ofx is a ivhole numher. If we extract the n^^ root of both members of the equation (1), these roots will, by Axiom Y, still be equal. The n^^ ri8f6t of x'^ being X, and that of a being a^, we have ^ X = a% and the equation is solved. BINOMIAL EQUATIONS, 171 2. When the exjjonent is fractional. Let the equation be m ^ — a. Raising both members to the n^^ power, we have x^ = aP', Extracting the m^^ root, X — a^. If the numerator of the exponent is unity, we only have to suppose m = 1, which will give X = a^. Hence the binomial equation always admits of solution by forming powers, extracting roots, or both. Special Forms of Binomial Equations. Bef, When the exponent n is an integer, the equa- tion is called a Pure Equation of the degree n. When 71 = 2^ the equation is a Pure Quadratic Equation. When n = 3, the equation is a Pure Cubic Equa- tion. EXERCISES. Find the values of x in the following equations : I. P x^ Ans. X = -' 2. a -^ h ^ =c. ■ x^ 3. a h xi — b xi — a 4. 9 x^ a; "" 24' 5- x — 2a 2x— b X — a ~~ x—b 6 x^ — na nx^ — h 7- 9- a At xa. x^ — a ~ x^ — l yi xi ^ "«-* 8. \^x -\- a^ b — a a + b ^x — a^ 172 POSITIVE AND NEGATIVE BOOTS. In the last example, clearing the equation of fractions, we shall have or {x^ - a'^)^ = 62 _ a\ We square both sides of this equation, which gives another in which a*^ only appears. lo. {x — a)i = h^. II. {x^ — a^p = mx. 12. (V^ — 'v/^)i = nx^. Positive and Neg^ative Roots. 191. Since the square root of a quantity may be either positive or negative, it follows that when we have an equation such as x^ z=z a, and extract the square root, we may have either X = + a^, or X =z — a^. Hence there are two roots to every such equation, the one positive and the other negative. We express this pair of roots by writing X z=z ±a^, the expression ± ^^ meanr.ig either + a^ or — a^. It might seem that since the square root of x^ is either +ir or —x, we should write having the four equations, ± X - ±aK X z= ai X = — aK — X = + ai, — X = — aK But the first and fourth of these equations give identical values of x by simply changing the sign, and so do the second and third. PROBLEMS LEADINO TO PURE EQUATIONS. 1. Find three numbers, such that the second shall be double the first, the third one-third the second, and the sum of their squares 196. 2. The sum of the squares of two numbers is 369, and the difference of their squares 81. What are the numbers? I PROBLEMS. 173 3. A lot of land contains 1645 square feet, and its length exceeds its breadth by 12 feet. What are the length and breadth ? To solve this equation as a binomial, take the mean of the length and breadth as the unknown quantity, so that the length shall be as much greater than the mean as the breadth is less. 4. Find a number such that if 9 be added to and subtracted from it, the product of the sum and difference shall be 175. 5. Find a number such that if a be added to it and sub- tracted from it the product of the sum and difference shall be 2a + 1. 6. One number is double another, and the difference of their squares is 192. What are the numbers ? 7. One number is 8 times another, and the sum of their cube roots is 12. What are the numbers ? 8. Find two numbers of which one is 3 times the other, and the square root of their sum, multiplied by the lesser, is equal to 128. 9. What two numbers are to each other as 2:3, and the sum of their squares = 325 .^ Note. If we represent one of the numbers by 2x, the other will be ^x. 10. What two numbers are to each other as m\ny and the square of their difference equal to their sum ? 11. What two numbers are to each other as 9 to 7, and the cube root of their difference multiplied by the square root of their sum equal to 16 ? 12. Find X and y from the equations ax^ -f hy'^ = c, a!x^ 4- Vy'^ = c'. 13. The hypothenuse of a right-angled triangle is 26 feet in length, and the sum of the sides is 34 feet. Find each side. Note. It is shown in Geometry that the square of the hypothenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. In the present problem, take for the unknown quantity the amount by which each unknown side differs from half their sum. 14., Two points start out together from the vertex of a right angle along its respective sides, the one moving m feet per second and the other 7i feet per second. How long will they require to be c feet apart ? 15. By the law of falling bodies, the distance fallen is pro- portional to the square of the time, and a body falls 16 feet the first second. How long will it require to fall h feet ? 174 QUADBATIG EQUATIONS. CHAPTER II. QUADRATIC EQUATIONS. 193. Def. A Quadratic Equation is one which, when reduced to the normal form, contains the second and no higher power of the unknown quantity. A quadratic equation is the same as an equation of the second degree. Def. A Pure quadratic equation is one which con- tains the second power only of the unknown quantity. The treatment of a pure quadratic equation is given in the preceding chapter. Def. A Complete quadratic equation is one which contains both the first and second powers of the un- known quantity. The normal form of a complete quadratic equation is ax^ -^ bx -\- c = 0. (1) If we divide this equation by a, we obtain (2) x^ + ^-x a a Putting, for brevity, - = --P^ c _ a ~ -<1, equation will be written in the form, x^ -\-px + q =1 0. (3) Def. The equation x^ -^ px ■\- q -^^ is called the General Equation of the Second Degree, or the General Quadratic Equation, because it is the form to which all such equations can be reduced. QUADRATIC EQUAT10N8, 175 Solution of a Complete Quadratic Equation. 193. A quadratic equation is solved hy adding such a quantity to its two members that the inember contain- ing the unhnoiun quantity shall he a perfect square, (§187.) We first transpose q in the general equation, obtaining x^ + px =z — g. We then add j- to both members, making x^ -\-px+^ ='- — q. The first member of the equation is now a perfect square. Extracting the square roots of both sides, we have = Wi ,+l-^./«= From this equation we obtain a value of x which may be put in either of the several forms, X =z -iwr- X = p Vp^ — 4:q 2^ 2 1 If instead of substituting jt) and q, we treat the equation in the form (2) precisely as we have treated it in the form (3), we sliall obtain the several results, _ - -b± V{I^ -^ 4:ac) "~ 2a 176 QUADRATIC EQUATIONS. 194. The equation in the normal form, (1), may also be solved by the following process, which is sometimes more con- venient. Transposing ^/and multiplying the equation by a, we obtain the result a^x^ + dbx = — ac. To make the first member a perfect square, we add — to each member, giving a^x^ + ahx + — n^ ac, 4 4 Extracting the square root of both sides, ive have ^^ + o = ^ V(^^ — ^ac), from which we obtain the same value of x as before. 195. Since the square root in the expression for x may be either positive or negative, there will be two roots to every quadratic equation, the one formed from the positive and the other from the negative surds. If we distinguish these roots with x^ and x^^ their values will be _ — ^ — V{b^ — ia c) ' ^^ - 2a We can always find the roots of a gi ven quadratic equation by sub- stituting the coefficients in the preceding expression for x. But the stu- dent is advised to solve each separate equation by the process just given, which is embodied in the following rule : I. Reduce the equation to its normal or its general fornh, as may he inmost convenient, II. Transpose the terms which do not contain x to the second member, III. // the coefficient of x^ is unity, add one-fourth the square of the coefficient of x to both manbers of the equation and exti^act the square root, IV. // the coefficient of x^ is not unity, either divide by it so as to reduce it to unity ^ or multiply all the terin^ QUADBATIG EQUATIONS. 177 hy such a factor that it shall become a perfect square, and complete the square hy the rule 0/ § 187. EXAMPLE. Solve the equation • x-l _^ Clearing of fractions and transposing, we find the equation to become 2x^--4:lx + 1 = 0, (5) , 41:?; 1 ^4 — , 2 - 2 Adding J the square of the coefficient of x to each side, we have 2 11 1681 _ 1681 1 _ 1673 ^ ■" 2 '"^ "^ 16 "^ 16 2 ~" 16 ' Extracting the square root and reducing, we find the values of x to he X, = ^(41 + V1673), and x^ = j(41-\/l673). Using the other method, in order to avoid fractions, we multiply the equation (5) by 3, making the equation, 4^2 _ 822; = -> 2. Adding -j- ~ —7— to each side of the equation, we have . o.. 4:12 1681 1673 4a;2 — 82^; + -— ^ — j 2 = — ; — 4 4 4 Extracting the square root, 41 /1673 ^1673 2a; /1673 2 "■ r 4 2 ' whence we find 41 ± \/l673 the same result as before. EXERCISES. Reduce and solve the following equations . -^ + 2 _ 2; — 2 _ 6 ^ + 4 y — i _ 10^ ^' ,T-.2 :^+~2 "^ 6' ^* i^--.4 ^ ?/ + 4 *"* 3' 12 178 QUADBATIG EQUATIONS, 1 2 _ 4 4. y^ — %ay + ^2 _ ^2 _ q. 1 1 1 1 X + a 1 + ^-'^ =3. X — a X -{- a y 2 + 2/ 2/'-4'2-^v y -{- a y — a _ 1 1 8- ^rfT-.-;r-- + ^r— . = 4. « -f- iz; a — i?; + 3 = 0. y — a y -\- ci y — ci' y^ — a^ V — (^ X X PROBLEMS. 1. Find two numbers such that their difference shall be 6 and their product 567. 2. The difference of two numbers is 6, and the difference of their cubes is 936. What are the numbers ? 3. Divide the number 34 into two such parts that the sum of their squares shall be double their product ? 4. The sum of two numbers is 60, and the sum of their squares 1872. What are the numbers ? 5. Find three numbers such that the second shall be 5 greater than the first, the third double the second, and the sum of their squares 1225. 6. Find four numbers such that each shall be 4 greater than the one next smaller, and the product of the two lesser ones added to the product of the two greater shall be 312. 7. A shoe dealer bought a box of boots for $210. If there had been 5 pair of boots less in the box, tlfey would have cost him %\ per pair more, if he had still paid $210 for the whole. How many pair of boots were in the box ? Rem. If we call x the number of pairs, the price paid for each pair must have been QUADRA TIG EQUATIONS. 179 8. A huckster bought a certain number of chickens for $10, and a number of turkeys for $15.75. There were 4 more chickens than turkeys, but they each cost him 35 cents a piece less. How many of each did he buy? 9. A farmer sold a certain number of sheep for $240. If he had sold a number of sheep 3 greater for the same sum, he would have received $4 a piece less. How many sheep did he sell? 10. A party having dined together at a hotel, found the bill to be $9.60. Two of the number having left before pay- ing, each of tlie remainder had to pay 24 cents more to make up the loss. What was the number of the party? 11. A pedler bought $10 worth of apples. 30 of them proved to be rotten, but he sold the remainder at an advance of 2 cents each, and made a profit of $3.20. How many did he buy ? 12. In a certain number of hours a man traveled 48 miles ; if he had traveled one mile more per hour, it would have taken him 4 hours less to perform his journey ; how many miles did he travel per hour ? 13. The perimeter of a rectangular field is 160 metres, and its area is 1575 square metres. What are its length and breadth ? 14. The length of a lot of land exceeds its breadth by a feet, and it contains m^ square feet. What are its dimen- sions ? 15. A stage leaves town A for town B, driving 8 miles an hour. Three hours afterward a stage leaves B for A at such a speed as to reach A in 18 hours. They meet when the second has driven as many hours as it drives miles per hour. What is the distance between A and B ? Note. The solution is very simple when the proper quantity is taken as unknown. Equations which may be Reduced to Quad- ratics. 196. Whenever an equation contains only two powers of the unknown quantity, and the index of one power is double that of the other, the equation can be solved as a quadratic. 180 QTTADBATIG EQdATIONS. Special Example. Let us take the equation 7^ + M + c =z 0. (1) Transposing c and adding -Vi to each side of the equation, it becomes 4 4 The first member of this equation is a perfect square, namely, the square of a:^ + - J. Extracting the square roots of both members, we have ^ + |S = /(jS^ - c) = ± J V(52 - 4c). Hence, ^ = \[-i± V{V - 4c)]. Extracting the cube root, we have x=z -\-h±V(P-4:c)'\^. General Form, We now generalize this solution in the following way. Suppose we can reduce an equation to the form ax^^ + hx^ + c = 0, in which the exponent n may be any quantity whatever, entire or fractional. By dividing by a, transposing, and adding J 1 7»2 - -7, to both sides of the equation, we find 4a^ ^ ' a 4«2 4«^ a The first side of this equation is the square of Hence, by extracting the square root, and reducing as in the general equation, we find QUADRATIC EQUATIONS. 181 Extracting the n^^ root of both sides, we have X = -j— L [—b± V(b^ — 4:ac)]K "" \ 2a / If the exponent ^ is a fraction, the same course may be followed. Suppose, for example, axi + ix^ + c = 0. Dividing by a and transposing, we have 4 . J 2 c xt J x'^ = a a Adding —^ to both sides, 4 . & 2 . J^ W- c a Aa^ 4:0^ a The left-hand member of this equation is the square of Extracting the square root of both members, I _^ _ /^ _ c\* __ { b^ — ^ac)^ , whence, a;^ =:_ Eaising both sides of this equation to the f power, w^have 2a X = -]^ EXERCISES. 1. Eind a number which, added to twice its square root, will make 99. 2. What number will leave a remainder of 99 when twice its square root is subtracted from it. 182 QUADBATIG EQUATIONS. 3. One-fifth of a certain number exceeds its square root by 30. What is the number ? 4. What number added to its square root makes 306 ? 5. If from 3 times a certain number we subtract 10 times its square root and 96 more, and divide the remainder by the number, the quotient will be 2. What is the number ? Solve the equations : 6. \y^ — 2^2 _ 15. 7. 3^ _ 7^2 _ 25. 8. 5«/4 — ^y^ = 13, mm-. 9. {x^ + a^Y" — 4:{x^ + a^)^ — «2 _ 2 + - • €1 197. When the unknown quantity appears in the form x^ -\ — 2, the square may be completed by simply adding 2 to ^ . 1 . this expression, because x^ -\- 2 -\- -^ is a perfect square, 1 ^ namely, the square of x + -* The value of x may then be X deduced from it by solving another quadratic equation. 3 Example. Zx^ + "2 "= ^^• We first divide by 3 and add 2 to each side of the equation, obtaining , , ^ 1 ^^ . o ^^ ^+^ + ^2 = y + ^ = y Extracting the square root of both sides, , 1 2V7 2^/21 2 .., X ^^ 3 3 By multiplying by x, this equation becomes a quadratic, and can be solved in the usual way. Let us now take this equation in the more general form, 1 x-\-- = e, (a) 2 which reduces to the foregoing by putting e = -^ ^21. Clear* ing of fractions and transposing, x^ — ex -}- 1 =z ; QUADRATIC EQUATIONS. 183 which being solved in the usual way, gives e ± V(e^ — 4) X = -^ The two roots are therefore ""^ - 2 ~ ' _ e — V{e^ — 4) ^3 - 2 If in the first of these equations we rationalize the numer- ator by multiplying it by e — \/{e^ — 4) (§ 185), we shall find 2 1 it to reduce to , that is, to — • Therefore, e - V(e2 - 4) ^2 iCj r= — identically. Vice versa, x^ is identically the same as — This must be the case whenever we solve an equation of the form {a), that is, one in which the value oi x + - is given. X 50 Let us suppose first that e = — , so that the equation is 1 50 It is evident that a; = 7 is a root of this equation, because ; when we put 7 for x, the left-hand member becomes 7 + ^r, 50 1 which is equal to —• If we put ^ for x, the left-hand mem- ber will become 7^1 7 ^ ' Hence x and - exchange values by putting - instead of 7, X i so that their sum x -\- - remains unaltered by the change. 184 QUADRATIC EQUATIONS. The general result may be expressed thus : Because the value of the expression x + - remains un- altered when we change x into , therefore the reciprocal of any root of the equation 1 X + - = e X is also a root of the same equation. EXERCISES. Find all the roots of the following equations without clear- ing the given equations from denominators : 1 17 1 I. iz;2 _|_ — . 2. a^x^ + --^ = 7)1^ — 2. x^ 4 a^x^ 3. 16^/2 + ^3 = 28. 4. j,-{-y' = 2m^ 5. Show, without solving, that if r be any roofc of the equation • 1 ^' + ^2 = «> then — r, - , and will also be roots. Factorings a Quadratic Equation. 198. 1. Special Case, Let us consider the equation a;2_22;— 15 = 0, or a;2 — 2a; + 1 — 16 = 0, or {x - 1)2 _ 42 = 0. Factoring, it becomes (§ 90), (ic — 1 + 4) {x — 1 — 4) = 0, or {x + 3) {x — 5) =: 0. Therefore the original equation can be transformed into (x + 3) (a; -~ 5) = 0, a result which can be proved by simply performing the multi- plications. qUJiDBATIG EQUATIONS, 185 This last equation may be satisfied by putting either of its factors equal to zero ; that is, by supposing a; + 3 = 0, whence a; = — 3 ; or ;c — 5 = 0, whence :r = + 5. These are the same roots which we should obtain by solving the original equation. 2. Factoring the General Quadratic Equation, Let us con- sider the general quadratic equation, x^ -\- px -\- q z= 0. {a) !N'ow, instead of thinking of a; as a root of this equation, let us suppose x to have any value whatever, and let us con- sider the expression x^ + 2^x -f qy (1) which for shortness we shall call X, Let tis also inquire how it can be transformed without changing its value. First we add and subtract -^p^, so as to make part of it a perfect square. It thus becomes, X = x^ 4- 2^x + -;;2 _ _^2 + ^ . . or, which is the same tiling, Factoring this expression as in § 188, it becomes X = ^ + 2P + The student should now prove that this expression is really equal to x'^ + px + q, by performing the multiplication. Let us next put, for brevity, 1 "2 -Ip-i^f-gf. 186 QUADRATIC EQUATIONS. The preceding value of X will then become, X= {x-a){x-. (3), (3) an expression identically equal to (1), when we put for a and P their values in (2). Let us return to the supposition that this expression is to be equal to zero, and that x is a root of the equation. The equation (a) will then be {x -a){x-(3) z= 0. (4) But no product can be equal to zero unless one of the fac- tors is zero. Hence we must have either X — cc = 0, whence x = cc-, or X — P = 0, whence x = fi Hence, a and ^ are the two roots of the equation (a). The above is another way of solving tlio quadratic equation. To compare the expressions (1) and (3), let us perform the multiplication in the latter. It will become, X = x^— {a + P)x + ap. . Since this expression is identically the same as x^-^px-^q, the coefficients of the like powers of x must be the same. That is, cc + P = -p,) . • , . which can be readily proved by adding and multiplying the equations (2). This result may be expressed as follows : Theorem, When a quadratic equation is reduced to the general form x^ ■\- px ■\- q — ^^ the coefficient of x will be equal to the sum of the roots with the sign changed. The term independent of x will be equal to the product of the roots. The student may ask why can we not determine the roots of the quadratic equation from equations (5), regarding a and /3 as the unknown quantities ? QUADBATIG EQUATIONS. 187 We can do so, but let us see what tlie result will be. We eliminate either « or /3 by substitution or by comparison. From the second equation (5) we have, a = ^-- a Substituting this in the first equation, we have a + P- Clearing of fractions and transposing, «2 + ^e« + ^ = 0. We have now the same equation with which we started, only a takes the place of x. If we had eliminated cc, we should have had the same equation in /3, namely, /32+^/3 + g^ = 0. So the equations (5), when we try to solve them, only lead us to the original equation. 199. To form a Quadratic Equation when the Roots are given. The foregoing principles will enable us to form a quad- ratic equation which shall have any given roots. We have only to substitute the roots for a and /3 in equation (4), and perform the multiplications. EXERCISES. Form equations of which the roots shall be : I, +1 and — 1. 3. — 3 and — 2. 5. 7 + 2a/3 and 7— 2^3. 7. ~ 1 and + 2. 9. +1 and — 2. II. - and -• 4 5 13. 2 + ^/2 and 2 — ^/2. 14. 15- 5 + 7V5 and 5 — 7^/5. 16 3 and 2. 3 + 2Vl0and 3-2^10. + 1 and + 2. — 1 and — 2. 2 + a/S and 2 — ^/b, 7,9 2 "^^ 2- 9 + 2\/2 and 9 — 2a/2 a -\'h and a — h. 17. a + A/a^ — h^ and a — ^/a^ — l\ 188 IMA OINAR Y MOOTS. Equations having Imaginary Roots. 300. When we complete the square in order to solve a quadratic equation, the quantity on the right-hand side of the equation to which that square is equal must be positive^ else there can be no real root. For if we square either a positive or negative quantity, the result will be positive. Hence, if the square of the first member comes out equal to a negative quantity, there is no answer, either positive or negative, which will fulfil the conditions. Such a result shows that impossible conditions have been introduced into the problem. E X AMPLES. 1. To divide the number 10 into two such parts that their product shall be 34. If we proceed with this equation in the usual way, we shall have, on completing the square, x^ — lOiT + 25 = — 9, or {x — 5)2 = — 9. The square being negative, there is no answer. On con- sidering the question, we shall see that the greatest possible product which the two parts of 10 can have is when they are each 5. It is therefore impossible to divide the number 10 into two parts of which the product shall be more than 25 ; and because the question supposes the product to be 34, it is im- 'possible in ordinary numbers. 2. Suppose a person to travel on the surface of the earth to any distance ; how far must he go in order that the straight line through the round earth from the point whence he started to the point at which he arrives shall be 8000 miles? It is evident that the greatest possible length of this line is a. diameter of the earth, namely, 7,912 miles. Hence he can never get 8,000 miles away, and the answer is impossible. In such cases the square root of the negative quantity is considered to be part of a root of the equation, and because it is not equal to any positive or negative algebraic quantity, it is called an imaginary root. The theory of such roots will be explained in a subsequent book. I IRBATIONAL EQUATIONS. 189 CHAPTER III. REDUCTION OF IRRATIONAL EQUATIONS TO THE NORMAL FORM. 201. An Irrational Equation is one in which tlio unknown qnantity appears under the radical sign. An irrational equation may be cleared of fractions in the same way as if it were rational. Example. Clear from fractions the equation Vx + 6? -f Vx — a _ 2a Vx -\- a — Vx — a Vx^ — a^ Multiplying both members by Vx^ — a^ = \/x + aVx—a, we have {x 4- a) Vx — a -\- {x — a) Vx -^ (^ __ ^ Vx -i- a — Vx — a Next, multiplying by Vx -{- a ^ Vx — a, we have {x-\-a) Vx — a -\- {x — a) Vx -\- a =: 2aVx-\-a — 2aVx — a. Transposing and reducing, we have {x + da) Vx — a -i- {x — 3a) Vx + « = 0, and the equation is cleared of denominators. Clearing of Surds. 303. In order that an irrational equation may be solved, it must also be cleared of surds which contain the unknown quantity. In showing how this is done, we shall suppose the equation to be cleared of denominators, and to be composed of terms some or all of which are multiplied by the square roots of given functions of x. Let us take, as a first example, the equation just found. Since a surd may be either positive or negative, the equation in question may mean any one of the following four: 190 IRRATIONAL EQUATIONS. {x -f 3«) ^/x — a+ {x — 3«) Vx -\- a = 0, (1) {x + da) "^x — a — {x — oa) ^/x + « = 0, (3) — (x -\- ?id) ^/x — a -\- (x — ?>a) \/x -{- a = 0, (3) — {x -\- 3a) "s/ X — a — {x — 3a) ^/x + a = 0. (4) But the third equation is merely the negative of the second, and the fourth the negative of the first, so that only two have different roots. Let us put, for brevity, P — (x \3a)^/x — a-\- {x — 3a) \/x~^\^, ) Q = (x -\- 3a) Vx ^ a— {x — 3a) "s/x + r?, ) and let us consider the equation, PQ = 0. (6) Since . this equation is satisfied when, and only when, we have either P = or § =: 0, it follows that every value of x .which satisfies either of the equations (1) or (2) will satisfy (G). Also, every root of (6) must be a root either of (1) or (2). If we substitute in (6) the values of F and Q in (5), we shall then have {x 4- 3aY{x — a) — {x — 3a)^ {x -{- a) = 0, which reduces to 6x^ — 9a^ = 0, , . 3a and gives x =: ± — — • V5 It will be remarked that the process by which we free the equation from surds is similar to that for rationalizing the terms of a fraction employed in § 185. As a second example, let us take the equation, V^ -h 11 + Vx — 4: — 5 = 0. (a) We write the three additional equations formed by combin- ing the positive and negative values of the surds in every way: •— a/S~+Ti + Vx — 4: — 5 r= 0, Vx + 11 — a/S'^^^ — 5 = 0, — Vx + 11 — Vx — 4: — 5 = 0. The product of the first two equations is IRRATIONAL EQUATION'S. 191 (^^/^Z^l^ 5)2 _ {x + 11) = 0, or 10 — lOVi"^^ = 0. (1) The product of the last two is 10 + l0\/^^"4 z= 0. (2) The product of these two products is 100 — 100 (^ — 4) = 0, which gives x =i b. It will be remarked that (2) differs from (1) only in having the sign of the surd different. This must be the case, because the second pair of equations formed from [a) differ from the first pair only in having the sign of the surd '^/x — 4 different. Hence it is not necessary to write more than one pair of the equations at each step. The general process is as follows : I. Change the sign of one of the surds in the given equation, and multiply the equation thus formed hy the original equation. II. Reduce this product, in it change the sign of an- other of the surds, and form a neiu product of the two equations thus formed, III. Continue the process until an equation without surds is reached. Example. Solve A/8a; + 9 + V^x -{- 6 + Va? + 4 = 0, Changing the sign of ^/x + 4, ^/Sx + 9 + ^/^x + 6 — v'o; + 4 = 0. The product is (VSx + 9 + \/%x + 6/ — {x + 4) = 0, or, after reduction, 9:^ + 11 -!_ 2^y^x + 9 \/'lx + 6 = 0. Changing the sign of ^/'ilx -f- 6, we have ^x 4- 11 — 2^/8^ + 9 V^x + G — 0. 192 IRRATIONAL EQUATIONS. The product of the last two equations reduces to 17^2 _ QQx — 95 = 0, which being solved gives x = — ~ Remark. Equations containing surds may often reduce to the form treated in § 196. In this case, the methods of that section may be fol- lowed. EXERCISES. Solve the equations : 1 1 2Va — Wx '\/x + ^a Vx — ^/a X — a Vx''-\-a X "s/a^ —X ^ 4. Vx + i-i + a/uj^^iT= 14. 5. (3 — x)i — (3 + x'^)i = 0. 6. Vet + ^x + Vet — ^x ■=. 2Vx + — • Vx-\-2 ^ - 4: Vx-2 V'Ex + 3 2 ' Vx -\- 3 — Vx--4: =:1, 9. Va^ — 2X + Va^ — 2x X + Vx X (x — 1) 10. — = . X — Vx ^ Vr+a 1 31. ; = , Vx — a + yctx — 1 Vx — 1 SIMVLTANE0U8 QUADEATIG EQUATIONS. 193 CHAPTER IV. SIMULTANEOUS QUADRATIC EQUATIONS. Between a pair of simultaneous general quadratic equations one of the unknown quantities can always be eliminated. The resulting equation, when reduced, will be of the fourth degree with respect to the other -unknown quantity, and cannot be solved like a quadratic equation. But there are several cases in which a solution of two equa- tions, one of which is of the second or some higher degree, may be effected, owing to some of the terms being wanting in one or both equations. 303. Case I. When one of the equations is of the first degree only. This case may be solved thus : KuLE. Find the value of one of the unhnown quan- tities in terms of the other from the cquaMon of the first decree. This value being substituted in the other equa- tion, we shall have a quadratic equation from which the other unhnown quantity may he found. Example. Solve 2iz:2 + dxy ^ 6y^ — x -^ 6y i= 26, ) 2x^^ =z 5. ) ^^^ From the second equation we find Whence, .^ = V + g^+jg. Substituting this value in the first equation and reducing, we find 4?/2 + 16y + 10 =r 26. Solving this quadratic equation, 13 194 SIMULTANEOUS QUADBATIG EQUATIONS ^=:— 2±V'8=— 2± 2\/2. This value of y being substituted in the equation {h) gives, 1 -{- sv^s — 1 ± 6a/2 X =z 2 ~ 2 The same problem may be solved in tbe reverse order by eliminating 4/ instead of x. Tlie second equation (a) gives 2x — 5 If we substitute this value oiy in the first equation, we shall have a quadratic equation in x, from which the value of the latter quantity can DC found. Solve EXERCISES. I. x^ — 2xy + 4?/2 = 21. 2x -\- y = 12. 2. 3^2 _ 2^2 _|_ 5^ _ 2^ = 28. 2; + 2/ + 4 = 0. 3- X -{-2y = 0, 4. 3x^ + 2^^ = 813, 7a; — 4^ = 17. 5. ^ + y = 7, a: y _ '^ y~~x~ 12' 304. Case II. When each equation contains only one term of the second deffvee^ and that term ''his the sayne 2>^oduct or square of the unknown quantities in the two equations. Such equations are ax^ -{• dx -{- ey -{- f = 0, | , . a'x^ + d'x + e'y +/ = 0, f ^^^ where the only term of the second degree is that in a^. If we eliminate x'^ from these equations by multiplying the first by a' and the second by a, and subtracting, we have SIMULTANEOUS QUADRATIC EQUATIONS, 195 {a'd — ad') x + {a!e — ae') y -\- a'f — af = 0. Bui iving this equation with respect to x, we find _ {ae'^a'e )y^ af - a'f ^~ d'd-ad' ^^> By substituting this value of x in either of the equations (a), we shall have a quadratic equation in y. Solving the latter, we shall obtain two values of y. Substituting these in (b), we shall have the two corresponding values of x, and tlie solution will be complete. Hence the rule, Eliminate the term of the second decree by addition or subtraction, and use the resulting equation of the first degree with either of the original equations, as in Case I, Example. Solve 2xy — 4:x-^6y-jzz 23, ) 3xy + 7x + y = 41,) ^^^ Multiplying the first equation by 3 and the second by 2, and subtracting, we have — 26x + 13y = — 13 ; (b) whence, x = -y -\- -- (c) Substituting this value in tl>e first equation, we find a quadratic equation, which, being solved, gives y = -2± V29. Substituting these values in (c), the result is The two sets of values of the unknown quantities are therefore ''i^-l + l^^^' ^, = -l-lV29. y^ = -2 + V2d, y^=: -2- a/29. We might have obtained the same result by solving the equation (c) with respect to y, and substituting in (a). The student should practice both methods. 196 SIMULTANEOUS QUADRATIC EQUATIONS. EXKRCISES. 1. %x^ - dx — 4y = 25, "' ^' ' ''''' x^ -\-2x — 3y = 18. 2. 2y^ + y := 28, ^2 _|. ^r^ '-4:y = 18. dxy ^2x -i- 6y = 70. 305, Case III. When neitliei^ equation €on° tains a term of the first degree in oc or y. EuLE. Eliminate the constant terms by multiplying each equation by the constant term of the other, and adding or subtracting the two products. The result ivill be a quadratic equation, from luhich either unhnoivn quantity can be determined in terms of the other. TJxen substitute as in Case L (1) Example. Solve x^ •{- xy — «/^ = 5, 2x2 _ 3^^ ^ 2^2 -- 14. 14 X 1st eq., 14a:2 H- 14.xy — 14?/^ = 70. 5 X 2d eq. , 10:^8 -^ Uxy -f 10^^ - . 70. Subtracting, ix^ + 29xy — 24/ = 0. This is a quadratic equation, by which one unknown quan- tity can be expressed in terms of the other without the latter being under the radical sign. Transposing, 4:X^ + 2dxy — 24«/l (2) 841 1225 Completing square, 4,0(^ 4- 2^xy + 'T'^-y^ = ~T^^^* Extracting root, 2x -\- -j-y :=^ :t-ry* wi, — 29 ± 35 3 ^ Whence, x — ~=- — y = jy or -— 8y. Substituting the first of these values of x in either of the original equations, we shall have f = 16 ; SIMULTANEOUS QUADBATIG EQUATIONS, 197 whence, ?/irr±4; 2;=±3. SubstiiTuting the second value of x^ we have • II'. ^2 ^ Whence, ^-±:^; "* ^ ^ 7lT Therefore the four possible values of the unknown quanti- ties are. Each of these four pairs of values satisfies the original equation. A slight change in the mode of proceeding is to divide the equation (2) by either x^ or y"^, and to find the value of the quotient. Dividing by y^ and putting X y the equation will become 4z^2 j^ 2du ^ 24 = 0. This quadratic equation, being solved, gives 29 i: 35 _ 3 8 *~ 4 IC =:: —^ — ^ - or — 5, X Putting - for ic, and multiplying by y, 3 X =: -y ov — Sy, as before. EXERCISES. Solve 1. x^ — xy + y^ — 3 = 0, x^ — 2xy -f- 4y^ — 4: = 0. 2. 2x^ + 3xy — 1^2 _ 2 = 0, 198 SIMULTANEOUS QUADRATIG EQUATIONS. 306. Case IV. When the expressUms contain- inn tJte itn known quantities in the two equations have comnion factors. Rule. Divide one of the equoMons which can he fac- tored by the othe7% and cancel the common factors. Then clear of fractions, if necessary, and we shall have an equation of a lower degree, EXAMPLES. 1. x^-\-y^ = 91, q: + y — 7. We have seen (§ 94, Th. 1) that x^-\-y^ is divisible hy x-{-y. So dividing the first equation by the second, we have x^ — xy -\- y'^ =: 13. This is an equation of the second degree only, and when combined with the second of the original equations, the solu- tion may be effected by Case I. The result is, :r = 3 or 4, y = 4 or 3. 2. xy + 2/2 = 133, x^ — y^ z=z 95. Factoring the first member of each equation, the equations become y{x + y) :=. 133, (x ^ y) {x - y) = 95. Dividing one equation by the other, and clearing of fractions, 7 I2y = "ix, or y = —x. The problem is now reduced to Case I, this value of y being combined with either of the original equations. 307. There are many other devices by which simultaneous equations may be solved or brought under one of the above cases, for which no general rule can be given, and in which the solution must be left to the ingenuity of the student. Sometimes, also, an equation which comes under one of the cases can be solved much more expeditiously than by the rule. Let us take, for instance, the equations, x% ^ y2 — 65, xy =z 28. These equations can be solved by Case III, but the work would be long and cumbrous. We see that by adding and ^ SIMULTANEOUS QUADBATIC EQUATIONS. 199 subtracting twice the second equation to and from the first, we can form two perfect squares. Extracting the roots of these squares, we shall have two simple equations, which shall give the solution at once. Each unknown quantity will have four values, namely, ± 7 ± 4. PROBLEMS AND EXERCISES. The following equations can all be solved by some sliort and expe- ditious combination of the equations, or by factoring, without going through the complex process of Case III. The student is recommended not to work upon the equations at random, but to study each pair until he sees how it can be reduced to a simpler equation by addition, multi- plication, or factoring, and then to go through the operations thus sug- gested. 1. y^ + xy = 14, x^ + xy = 35. 2. 4:X^ — 2xy = 208, 2xy — y^ =r 39. 3. x^ '{- y = 4^, y^ -j- X = 4:y. If we subtract one of these equations from the other, the difference will be divisible by x — y. 4. .^3 _!_ y^ _|. 3^ _|. 3^ ^ 378^ :^ j^y^^Zx — Zy z=z 324. 5. x^ + ^2 ^ 74, x^y ^ 12. 6. x^ -^ xy ■=: G3, ,x^ — y'^ = 77. ^/x — yy x^ -\r xy =: a, y'^ -{- xy ^ h, x^ + xy"^ = 10, y^ + x^y =2 5. X =z aV^ -\- y, y -=^ 'b\/x -f y, x'^x + y r= 12. y^/x + 3/ = 15. 2x^ + 2y^ = X -{- y, x^ i- y^ = x — y. 5x^ — 5y^ = X -\- y, 3x^ — Sy^ = x — y. x^ + y^-i-z^ = 30, xy-\-yz-\-zx = 17, x — y ^ z = 2. 7 8 9 10, II 12 13 14 15 /x -\- y x + y — 2\/ — '-^ = 8 -y 200 PROBLEMS. 1 6. A principal of $5000 amounts, with simple interest, to $7100 after a certain number of years. Had the rate of inter- est been 1 per cent, higher and the time 1 year longer, it would have amounted to $7800. What was the time and rate? 17. A courier left a station riding at a uniform rate. Fiye hours afterward, a second followed him, riding 3 miles an hour faster. Two hours after the second, a third started at the rate of 10 miles an hour. They all reach their destination at the same time. What was its distance and the rate of riding ? 18. In a right-angled triangle there is given the hypothe- nuse = a, and the area = Z>^; find the sides. 19. Find two numbers such that their product, sum, and difference of squares shall be equal to each other. 20. Find two numbers whose product is 216; and if the greater be diminished by 4, and the less increased by 3, the product of this sum and difference may be 240. 21. There are two numbers whose sum is 74, and the sum of their square roots is 12. What are the numbers ? 22. Find two numbers whose sum is 72, and the sum of their cube roots 6. 23. The sides of a given rectangle are m and n. Find the sides of another which shall have twice the perimeter and twice the area of the given one. 24. A certain number of workmen require 3 days to com- plete a work. A number 4 less, working 3 hours less per day, will do it In 6 days. A number 6 greater than the original number, working 6 hours less per day, will complete the work in 4 days. What was the original number of workmen, and how long did they work per day ? 25. Find two numbers whose sum is 18 and the sum of their fourth powers 14096. Note. Since the smn of the two numbers is 18, it is evident that the one must be as much less than 9 as the other is greater. The equa- tions will assume the simplest form when we take, as the unknown quan- tity, the common amount by which the numbers differ from 9. 26. Find two numbers, x and y, such that C(^^y^ : x^ — y^ :: 35 : 19, xy = 24. 27. Find two numbers whose sum is 14 and the sum of their fifth powers 161294. BOOK VII. PROGRESS/OA^S, CHAPTER I. ARITHMETICAL PROGRESSION. 308. Def. When we have a series of numbers each of which is greater or less than the preceding by a con- stant quantity, the series is said to form an Arithmet- ical Progression. Example. The series 7, 12, 17, 22, 27, 32, etc. ; 7, 5, 3, 1, —1, —3, etc.; a -\- b, a, a — h, a — 2^, a — 3^, etc., are each in arithmetical progression, because, in the first, each number is greater than the preceding by 5 ; in the second, each is less than the preceding by 2 ; in the third, each is less than the preceding by h. Def, The amount by which each term of an arith- metical progression is greater than the preceding one is called the Common Difference. Def. The Arithmetical Mean of two quantities is half their sum. AH the terms of an arithmetical progression except the first and last are called so many arithmetical means between the first and last as extremes. Example. The four numbers, 5, 8, 11, 14, form the four arithmetical means between 2 and 17. 202 ABITHMETIGAL PROGRESSION. EXERCISES. 1. Form four terms of the arithmetical progression of which the first term is 7 and common difference 3. 2. Write the first seven terms of the progression of which the first term is 11 and the common difference — 3. 3. Write five terms of the progression of which the first term is a — ^n and the common difference 27^. Problems in Progression, 209. Let us put a, the first term of a progression. d, the common difference. Uy the number of terms. I, the last term. 2, the sum of all the terms. The series is then ay a-\-d, a-\-2d, ....?. Any three of the above five quantities being given, the other two may be found. Pkoblem I. Given the first term, the cormnon differ- ence, and the number of terms, to find the last term. The 1st term is here a, M " " a + d, 3d " " a + 2d. The coefficient of d is, in each case, 1 less than the number of the term. Since this coefficient increases by unity for every term we add, it must remain less by unity than the number of the term. Hence, The i^^ term is a + {i — 1) d, whatever be i. Hence, when i = n, I = a + {n'^l)d. (1) From this equation we can solve the further problems : Problem II. Given the last term I, the common dif- ference d, and the number of terms ^i, to find the first term. ARITHMETICAL PR0ORES8I0N. 203 The solution is found by solving (1) with respect to a, which gives a = l-{n^ 1) d. (2) Problem III. Given the first and last terms, a and I, and the number of terms n, to find the common differ- ence. Solution from (1), d being the unknown quantity, d = ^. (3) 71—1 ^ ' Problem IV. Given the first and last terms and the common difference, to find the niimher of terms. Solution, also from (1), I — ct ^ ^ I — a + d ... « = -^- + l = ^— (4) Problem V. To find the sum of all the terms of an arithmetical progression. We have, by the definition of S, 2 = a -f («5 + cO -^ (« + ^^) + {l — d) + l, the parentheses being used only to distinguish the terms. Now let us write the terms in reverse order. The term before the last is ? — d, the second one before it I — 2dy etc. We therefore have, ^ =,1 j^ (l — d) -\- {l — '^d) -{■ {a -\- d) -\- a. Adding these two values of 2 together, term by term, we find 22 = (« + ?) + {a + l) + {a + l) + + (« + + («4-0, the quantity {a + 1) being written as often as there are terms, that is, 71 times. Hence, 22 = 72 {a + l\ j: = 7i—-' (5) Eemark. The expression — -— , that is, half the sum of the extreme terms, is the mean value of all the terms. The 204 ARITHMETICAL PROGRESSION. sum of the n terms is therefore the same as if each of them had this value. 310. In the equation (5) we are supposed to know the first and last terms and the number of terms. If other quan- tities are taken as the known ones, we have to substitute for some one of the quantities in (5) its expression in one of the equations (1), (2), (3), or (4). Suppose, for example, that we have given only the last term, the common difference, and the number of terms, that is, /, d, and n. We must then in (5) substitute for a its value in (2). This will give, ^ =z nil — - d\ =z nl ^—- d. (6) EXERCISES. In arithmetical progression there are 1. Given, common difference, + 3; third term = 10. Find first term. Ayis, First term — 4. 2. Given 4th term = d, common difference = — c. Find first 7 terms, their sum and product. 3. Given 3d term z= a ^ h, 4th term = a + 2^. Find first 5 terms. 4. Given 1st term ■=: a — h, 9th term =: 9^ + Ih. Find 2d tprm and common difference. 5. Given, sum of 9 terms = 108. Find middle term and sum of 1st and 9th terms. 6. Given 5th term =r 7^^ — by, 7th term = 9.r — ^y. Find first 7 terms and common difference. 7. Given 1st term = 12, 50th term = 551. Find sum of all 50 terms. 8. To find the sum of the first 100 numbers, namely, 1 + 2+3 +99 + 100. Here the first term a is 1, the last term I 100, and the number of terms 100. The solution is by Problem V. 9. Find the sum of the first n entire numbers, namely, 1+2 + 3 + n. ARITHMETICAL PROGRESSION. 205 10. Find the sum of the first n odd numbers, namely, 1 + 3 + 5 +27^ — 1. Here the number of terms is n. 11. Find the sum of the first n even numbers, namely, 2 + 4 + 6 + 2/^. 12. In a school of m scholars, the highest received 134 merit marks, and each succeeding one 6 less than the one next above him. How many did the lowest scholar receive ? How many did they all receive ? 13. The first term of a series is m, the last term 2m, and the common difiference d. What is the number of terms? 14. The first term is Ic, the last term lOh — 1, and the number of terms 9. What is the common difference ? 15. The middle term of a progression is s, the number of terms 5, and the common difference — K What are the first and last terms and the sum of the 5 terms ? 16. The sum of 5 numbers in arithmetical progression is 20 and the sum of their squares 120. What are the numbers ? Note. In questions like tins it is better to take the middle term for one of the unknown quantities. The other unknown quantity will be the common difference. 17. Find a number consisting of three digits in arithmeti- cal progression, of which the sum is 15. If the number be diminished by 792, the digits will be reversed. 18. The continued product of three numbers in arithmet- ical progression is 640, and the third is four times the first. What are the numbers ? 19. A traveller has a journey of 132 miles to perform. He goes 27 miles the first day, 24 the second, and so on, travelling 3 miles less each day than the day before. In how many days will he complete the journey ? Here we have given the first term 27, the common difference —3, and the sum of the terms 132. To solve this, we take equation (5), and sub- stitute for 1 its value in (1). This makes (5) reduced to a -{- a -^ (n — 1) ^ n(n — 1) d ^ — n ^ '— = na-\ ^ 2, a, and d are given by the problem, and n is the unknown quan- tity. Substituting the numerical value of the unknown quantities, the equation becomes 206 ARITHMETICAL PROGRESSION This reduced to a quadratic equation in n, the solution of which gives two values of n. The student should explain this double answer by continuing the progression to 11 terms, and showing what the negative terms indicate. 20. Taking the same question as the last, only suppose the distance to be 140 miles instead of 132. Show that the answer will be imaginary, and explain this result. 21. A debtor owing $160 arranged to pay 25 dollars the first month, 23 the second, and so on, 2 dollars less each month, until his debt should be discharged. How many pay- ments must he make, and what is the explanation of the two answers ? 22. A hogshead holding 135 gallons has 3 gallons poured into it the first day, 6 the second, and so on, 3 gallons more every day. How long before it will be filled ? 1 23. The continued product of 5 consecutive terms is 12320 and their sum 40. What is the progression ? 24. Show that the condition that three numbers, p, q, and r, are in arithmetical progression may be expressed in the form q — r 25. In a progression consisting of 10 terms, the sum of the 1st, 3d, 5th, 7th, and 9th terms is 90, and the sum of the re- maining terms is 110. What is the progression ? 26. In a progression of an odd number of terms there is given the sum of the odd terms (the first, third, fifth, etc.)^ and the sum of the even terms (the second, fourth, etc.). Show that we can find the middle term and the number of terms, but not the common difference. 27. In a progression of an even number of terms is given the sum of the even terms = 119, the sum of the odd terms = 105, and the excess of the last term over the first = 26. What is the progression ? 28. Given a and I, the first and last terms, it is required to insert i arithmetical means between them. Find the expres- sion for the i terms required. GEOMETRICAL PROGRESSION. 207 CHAPTER II. GEOMETRICAL PROGRESSION. 311. Def. A Geometrical Progression consists of a series of terms of which each is fornied by multiply- ing the term preceding by a constant factor. All arithmetical progression is formed by continual addi- tion or subtraction; a geometrical progression by repeated multiplication or division. Def. The factor by which each term is multiplied to form the next one is called the Common Ratio. The common ratio is analogous to the common difference in an arithmetical progression. In other respects the same definitions apply to both. EXAM PLE s. 2, 6, 18, 54, etc., is a progression in which the first term is 2 and the common ratio 3. 2 1 ^ ^ ^ etc 2' 4' 8^ is a progression in which the ratio is -• + 3, — 6, + 12, — 24, etc., is a progression in which the ratio is — 2. Note. A progression like the second one above, formed by dividing eadi term by. the same divisor to obtain the next term, is included in the general definition, because dividing by any number is the same as multi- plying by the reciprocal. Geometrical progressions may therefore be divided into two classes, increasing and decreasing. In the increasing progression the common ratio is greater than 1 and the terms go on in- creasing ; in a diminishing progression the ratio is less than unity and the terms go on diminishing. Eem. In a progression in which the ratio is negative, the terms will be alternately positive and negative. 208 OEOMETRIGAL PROGRESSION'. Def. A Geometrical Mean between two quantities is tlie square root of their product. EXERCISES. Form five terms of each of the following geometrical pro- gressions : 1. First term, 1 ; common ratio, 5. 2. First term, 7 ; common ratio, — 3. 3. First term, 1 ; common ratio, — 1. 2 3 4. First term, - ; common ratio, -• 4 1 5. First term, - ; common ratio, -• Problems of Geometrical Progression. 313. In a geometrical progression, as in an arithmetical one, there are five quantities!, any three of which determine the progression, and enable the other two to be found. They are a, the first term. r, the common ratio. Uy the number of terms. I, the last term. 2, the sum of the n terms. The general expression for the geometrical progression will be a J ar, ar^^ a7% etc., because each of these terms is formed by multiplying the pre- ceding one by r. The same problems present themselves in the two progres- sions. Those for the geometrical one are as follows : Problem I. Given the first term, the eommorv ratio, and the nunider of terms, to find the last term. The progression will be a, ar, ar^, etc. We see that the exponent of r is less by 1 than the number of the term, and since it increases by 1 for each term added, it m^LIFORg^ GEOMETRICAL PB0OBE88I0N, 209 must remain less by 1, how many terms so ever we take. Hence the n^^ term is I =: ar'^-K (1) Problem II. Given the last term, the cormnon ratio, and the number of terms, to find the first term. The sokition is found by dividing both members of (1) by ?'^~S which gives Problem III. Given the first term, the last term, and the number of terms, to find the common ratio. From (1) we find r'^~^ = -• Extracting the (n -^ ly^ root of each member, we have 1 =©" [The sohition of Problem IV requires us to find n from equation (1), and belongs to a higher department of Algebra.] Problem V. To find the sum of all n terms of a geo- metrical progression. We have !> = a -{- ar + ar^ -\- etc. + ar^~K Multiply both sides of this equation by r. We then have rl = ai^ + ar"^ -\- ar^ -\- etc + ar^. Now subtract the first of these equations from the second. It is evident that, in the second equation, each term of the second member is equal to the term of the second member of the first equation which is one place farther to the right. Hence, when we subtract, all the terms will cancel each other except the first of the first equation and the last of the second. Illustration. The following is a case in which a = 2,T = d,n — Q', fe' 2 = 2 + 6 + 18 + 54+162 + 486. 32 r= 6 + 18 + 54 + 162 + 486 + 1458. Subtracting, 32-2 = 1458 - 2 = 1456, or 22 = 1456, and 2 r= 728. 14 210 GEOMETRICAL PROGRESSION. Returning to the general problem, we have (r-l)l. — ar^' — a —a {r^ — 1) ; whence, 2 = a — = a- (4)- r — 1 1 — r ^ \' It will be most convenient to use the first form when r > 1, and the second when r < 1. By this formula we are enabled to compute the sum of the terms of a geometrical progression without actually forming all the terms and adding them. EXERCISES 3 1. Given 3d term =z 9, common ratio = -• Find first 5 terms. 32 2 2. Given 5th term = — , common ratio = — -- Find first 5 terms. 3. Given 5th term = x^jj'^, 1st term z= yK Find common ratio. 4. Given 1st term = 1, 4th term =: a^. Find common ratio and first 3 terms. 5. Given 2d term = m, common ratio :=. — m. Find first 4 terms. 6. A farrier having told a coachman that he would charge him $3 for shoeing his horse, the latter objected to thq price. The farrier then offered to take 1 cent for the first nail, 2 for the second, 4 for the third, and so on, doubling the amount for each nail, which offer the coachman accepted. There were 32 nails. Find how much the coachman had to pay for the last nail, and how much in all. (Compare § 168, Rem.) 7. Find the sum of 11 terms of the series 2 + 6 + 18 + etc., in which the first term is 2 and the common ratio 3. 8. If the common ratio of a progression is r, what will be the common ratio of the progression formed by taking I. Every alternate term of the given progression ? II. Every 71^^^ term ? GEOMETRICAL PROGRESSION. 211 9. The same thing being supposed, what will be the com- mon ratio of the progression of which every alternate term is equal to every third term of the given progression ? 10. Show that if, in a geometrical progression, each term be added to or subtracted from that next following, the sums or remainders will form a geometrical progression. 11. Show that if the arithmetical and geometrical means of two quantities be given, the quantities themselves may be ' found, and give the expressions for them. 1 2. The sum of the first and fourth terms of a progression is to the sum of the second and third as 21 : 5. What is the common ratio? 13. Express the continued product of all the terms of a geometrical progression in terms of a, r, and 71 ? Limit of the Sum of a Progression. 313. Theorem. If the common ratio in a geometri- cal progression is less than unity (more exactly, if it is contained between the limits —1 and +1), then there will be a certain quantity which the sum of all the terms can never exceed, no matter how many terms we take. For example, the sum of the progression 111^ ^+j + g + etc., in which the common ratio is ~, can never amount to 1, no matter how many terms we take. To show this, suppose that one person owed another a dollar, and proceeded to pay him a series of fractions of a dollar in geometrical progression, namely, 1111 r 4' 8' 16' ^^^* When he paid him the - he would still owe another -, when he paid the j he would still owe another j? and so on. 212 GEOMETRICAL PBOGBESSION. That is, at every payment he would discharge one-half the re- maining debt. Now there are two propositions to be under- stood in reference to this subject. I. The entire debt ean never he discharged hy such payments. For^ since the debt is halved at every payment, if there was any payment which discharged the whole remaining debt, the half of a thing would be equal to the whole of it, which is impossible. II. The debt can he reduced helow any assignable limit by continuing to pay half of it. For, however small the debt may be made, another pay- ment will make it smaller by one-half; hence there is no smallest amount below which it cannot be reduced. These two propositions, wliicli seem to oppose each other, hold the truth between them, as it were. They constantly enter into the higher mathematics, and should be well understood. We therefore present another illustration of the same subject. A B I - I I I I I Suppose AB to be a line of given length. Let us go one- half the distance from A to B at one step, one-fourth at the second, one-eighth at the third, etc. It is evident that, at each step, we go half the distance which remains. Hence the two principles just cited apply to this case. That is, 1. We can never reach B by a series of «uch steps, because we shall always have a distance equal to the last step left. 2. But we can come as near B as we please, because every step carries us over half the remaining distance. This result is often expressed by saying that we should reach B by taking an infinite number of steps. This is a convenient form of expres- sion, and we may sometimes use it, but it is not logically exact, because no conceivable number can be really infinite. The assumption that in- finity is an algebraic quantity often leads to ambiguities and difficulties in the application of mathematics. GEOMETBIGAL PBOGRESSION. 213 Def. The Limit of the sum 2 of a geometrical progression is a quantity which s may approach so that its difference shall be less than any quantity we choose to assign, but which s can never reach. . EXAMPLES. 1. Unity is the limit of the sum 2. The point B in tlie preceding figure is the limit of all the steps that can be taken in the manner described. The following principle will enable us to find the limit of the sum of a progression : 314. Principle. If r < 1, the power r^ can be made as small as we please by increasing the value of n^ but can never be made equal to 0. Suppose, for instance, that _ 3 _ 1 ^ ~ 4 "~ 4' Then every time we multiply by r we diminish r^ by - of its former value ; that is, 4 4 4 3 — ^a — — A»3 4 4^ etc. etc. etc. Now let us again take the expression for the sum of a series of n terms, namely, 1-r^ 2 = a- , 1 — r which we may put into the form 1 — r 1 — r 214 GEOMETRICAL PR0OBE8S10N, If r is less than unity, we can, by the principle just cited, make the quantity r'^ as small as we please by increasing n indefinitely. From this it follows that we can also make the term -^ r^ as small as we please. 1 — r Proof. Let us put, for brevity, h 1-r' so that the term under consideration is If we cannot make hr'^ as small as we please, suppose 5 to be its smallest possible value. Let us divide s by k, and put ^- k No matter how small s may be, and how large k may be, T, or t, will always be greater than zero. Hence, by the pre- fc ceding principle, we can find a value of n so great that y^ shall be less than L That is, ,» < _. Multiplying both sides of this inequality by k, kr'^ < s. That is, however small we take s, we can take n so large that kr"^ shall be less than s, and therefore s cannot be the smallest value. Since 2 =: kr^, I 1 — r and since we can make kr^ as small as we please, it follows Limit of 2 = 1 — r This is sometimes expressed by saying that when r < 1, a -\- ar -{- ar^ + ar^ + etc., ad infinitum = , and this is a convenient form of expression, which will not lead us into error in this case. OEOMETBICAL PROGRESSION. 215 EXERCISES. Haying giyen the progression 1.1 .1.1 .etc of which the limit is 1, find how many terms we must take in order that the sum may differ from 1 by less than the follow- ing quantities, namely : Firstly, .001 ; secondly, .000 001 ; thirdly, .000 000 001. To do this, we must find what power of ^ will be less than .001, what power less than .000 001, etc. What are the limits of the sums of the following series : 1. Q + To + Ts + ^t^v ctd infi^iitum. o o o 2. q + Q + 97 + ^^C'j ^^ mfinitum, 3. Q — 02 + Q3 — etc., ad infinitum. 4 42 43 4. Q + Q2 + 93 + ^^^'^ ^^ infinitum. 5. j-p^ + ^ ^ p + (Y+~^p + ^^^-^ ^^ infinitum. 6. ^-^ - W^f "^ W^Y' ~ ^^^'' ^^ infinitum. 7. 1 1 5 ^H ;: — etc., ad infinitum. m ni? m?' m* "^ 8. What is that progression of which the first term is 12 and the limit of the sum 8. 9. On the line AB a man starts from A and goes to the point c, half way to B ; then he re- turns to d^ half way back to A ; then 1 1 1 I turns again and goes half way to c, then back half way to dollars every year, letting the whole sum constantly accumulate at com- pound interest, WTiat will the amount be at the end of n years? Solution. The first investment will have been out at interest n years, the second n — 1 years, the third n — 2 years, and so on to the 7i^^, which will have been out 1 year. Hence, from the last formula, the amounts will be : 218 COMPOUND INTEREST. Amount of 1st payment^ i^ (1 + pY- 2d a p(i + pY-\ 3d Si p{i + pY-\ 4th i( p(l+ p)»-K 5th a p(l + p)'^-*. etc. etc. The sum of the amounts is : p{l-\'P) + p{l + py + P{1 + pY -h , . , .p(l+p)^. This is a geometrical progression, of which the first term is p (1 + p), the common ratio 1 + p, and the number of terms w. So in the formula (4), § 212, we put J)(l+p) for a, 1 + p for r, and thus find, 2 =p{i4.o) (l+i^-i =^(l±£)!!L-^(i±p). ^"^'^l + p — 1 ^ p EXERCISES. I. A man insures his hfe for $5000 at the age of 30, pays for his insurance a premium of 80 dollars a year for 32 years, and dies at the age of 62, immediately before the 33d payment would have been due. If the company gains 4 per cent, inter- est on all its money, how much does it gain or lose by the insurance p Note. Computations of this class can be made with great facility by the aid of a table of logarithms. 2. What is the present value of a dollars due n years hence, interest being reckoned at c per cent. ? Note. If p be the present value, Problem I gives the equation, 3. What is the present value of 3 annual payments, of a dollars each, to be made in one, two, and three years, interest being reckoned at 5 per cent. ? 4. What is the present value of n annual payments, of a dollars each, the first being due in one year, if the rate of in- terest is c per cent. ? What would it be if the first payment were due immediately ? SECOND PART. ADVANCED COURSE. f BOOK VIII. RELATIONS BETWEEN ALGEBRAIC QUANTITIES. Of Algebraic Functions. 317. Bef. When one quantity depends npon an- other in such a way that a change in the value of the one produces a change in the value of the other, the latter is called a Function of the former. This is a more general definition of the word " function " than that given in § 49. Examples. The time required to perform a journey is a function of the distance because, other things being equal, it varies with the distance. The cost of a package of tea is a function of its weight, be- cause the greater the weight the greater the cost. An algebraic expression containing any symbol is a func- tion of that symbol, because by giving different values to the symbol we shall obtain different values for the expression. Bef. An Algebraic Function is one in which the relations of the quantities is expressed by means of an algebraic equation. Example. If in a journey we call t the time, 8 the average speed, and d the distance to be travelled, the relation between these quantities may be expressed by the equation, d = st. Any one of these quantities is a function of the other two, defined by means of this equation. An algebraic function generally contains more than one 222 FUNCTIONS. letter, and therefore depends upon several quantities. But we may consider it a function of any one of these quantities, se- lected at pleasure, by supposing all the other quantities to remain constant and only this one to vary. For example, the time required for a train to run between two points is a func- tion not only of their distance apart, but of the speed of the train. The speed being supposed constant, the time will be greater the greater the distance. The distance being constant, the time will be greater the less the speed. Def. The quantities between which the relation ex- pressed by a function exists are called Variables. This term is used because such quantities may vary in value, as in the preceding examples. Def. An Independent Variable is one to which we may assign values at pleasure. The function is a dependent variable, the value of which is determined by the value assigned to the independent variable. Def, A Constant is a quantity which we suppose not to vary. Eem. This division of quantities into constant and varia- ble is merely a supposed, not a real one ; we can, in an algebraic expression, suppose any quantities we please to remain constant and any we please to vary. The former are then, for the time being, constants, and the latter variables. iLLUSTRATIOiq". If We put d, the distance from New York to Chicago ; s, the average speed of a train between the two cities ; t, the time required for the train to perform the jour- ney, then, if a manager computes the different values of the time t corresponding to all values of the speed 5, he regards 6? as a constant, s as an independent variable, and ^ as a function of 5. If he computes how fast the train must run to perform the journey in different given times, he regards t as the independ- ent variable, and 5 as a function of t. FUNCTIONS. 223 When we have any equation between two variables, wo may regard either of them as an independent variable and the other as a function. Example. From the equation ax -\-hy =1 c, we derive by c X a ax + a' c y — ■""T + v in one of which x is expressed as a function of y^ and in the other ^ as a function of x, 218. Names are given to particular classes of functions, among which the following are the most common. 1. Def, A Linear Function of several variables is one in which each term contains one of the variables, and one only, as a simple factor. Example. The expression Ax ^ By -\- Cz is a linear function of x^ y, and z, when A, B, and C are quan- tities which do not contain these variables. A linear function differs from a function of the first degree (§ 52) in having no term not multiplied by one of the varia- bles. For example, the expression Ax + By + C is a function of x and y of the first degree, but not a linear function. The fundamental property of a linear function is this: If all the variables he -multiplied hy a coimnoii fac- tor, the function will he multiplied hy the same factor. Proof. Let Ax -}- By -\- Oz be the linear function, and r the factor. Multiplying each of the variables x, y, and z by this factor, the function will become Arx -{- Bry + Orz, which is equal to r {Ax + By + Cz). 224 FUNCTIONS. Moreover, a linear function is the only one which pos- sesses this property. 2. Def. A Homogeneous Function of several va- riables is one in which each term is of the same degree in the variables. (Compare § 52.) Example. The expression ax^+'bx^y-\-cyh-\-dz^ is homo- geneous and of the third degree in the variables x, y, and z. Kem. a linear function is a homogeneous function of the first degree. Fui^DAMEKTAL PkOPERTY OF HOMOGENEOUS FUKCTIOi^S. // all the variables he multiplied hy a common factor, any homogeneous function of the ii*'^ degree in those va- riables will be multiplied by the n*^ power of that factor. Proof, If we take a homogeneous function and put rx for .T, ry for y, rz for z, etc., then, because each term contains x, y, or z, etc., n times in all as a factor, it will contain r n times after the substitution is made, and so will be multiplied by r^. 3. Def. A Rational Fraction is the quotient of tvro entire functions of the same variable. A rational fraction is of the form, a -\-'bx -^ cx^ -\- etc. m -\- nx -\- px^ 4- etc. Any rational function of a variable may be expressed as a rational fraction. Compare § 180. Equations of the First T^e^ee between Two Variables. 319. Since we may assign to an independent variable any values we please, we may suppose it to increase or decrease by regular steps. The difference between two values is then called an increment. That is, Def. An Increment is a quantity added to one value of a variable to obtain another value. INCREMENTS, 225 Eem. If we diminish the variable, the increment is negative. Theorem, In a function of the first degree, eqnal in- crements of the independent variable cause equal incre- ments of the function. Example. Let x be an independent variable, and call u the function -^x'^ 11, so that we have u = -x + 11. If we give x the successive values —2, — 1, 0, 1, 2, etc., and find the corresponding values of the function u, they will be Values of ic, — 2, — 1, 0, 1, 2, 3, 4, etc. " « u, 8, 9^, 11, 12|, 14, 154, ^"^y etc. We see that, the increments of x being all unity, those of y are all 1^, General Proof. Let au -\- hx = c he any equation of the first degree between the variable x and the function u. Solving this equation we shall have c — bx c b u = = x. a a a Let us assign to x the successive values, r, r -\- h, r -\- 27^, etc., the increment being h in each case. The correspondnig values of the function u will be c d c h h ^ c I %b -. , r, r n, r /^, etc., a a a a a a a a of which each is less than the preceding by the same amount, ~h. Hence the increment of u is always h, which proves the theorem. 230. Geometric Construction of a Relation of the First Degree. The relation between a variable x and a function u of this variable may be Shown to the eye in the following way .• 15 226 aEOMETRIO C0N8TBUCTI0M. \ el'*] -X -3 -8 ^^gi y Take a base line AX, mark a zero point upon it, and from this zero point lay off any values of x we please. Then at each point of the line corresponding to a value of x erect a vertical line equal to the corresponding value of %i. If u is positive, the value is measured upward ; if negative, downward. The line drawn through the ends of these values of u will show, by the distance of each of its points from the base line AX, the values of u corresponding to all values of x. Let us take, as an example, the equation hu ■\-dx = 10, 3 the solution of which gives u = 2 — -x. o Computing the values of u corresponding to values of x from —3 to +6, we find : X = -3, -2, -1, 0, +1, +2, +3, +4, +5, +6. u= +3f, +3i 2|, 2, If, f, i, -I, -1, ^1|. Laying off these values in the way just described, we have the above figure. Wherever we choose to erect a value of u, it will end in the dotted line. We note that by the property of functions of the first de- gree just proved, each value of u is less (shorter) than the pre- ceding one by the same amount ; in the present case by -• It is known from geometry that in this case the dotted line through the ends of u will be a straight line. We call this line through the ends of u the equation line. OF EQUATIONS OF THE FIB8T DEGBEE, 227 331, When we can once draw this straight line, we can find the value of u corresponding to every value of x without using the equation. We have only to take the point in the base line corresponding to any value of x^ and by measuring the distance to the line, we shall have the corresponding value of u, Now it is an axiom of geometry that one straight line, and only one, can be drawn between any two points. Therefore, to form any relation of the first degree we please between x and u, we may take any two values of x, assign to them any two values of u we please, plot these two pair of values of u in a diagram, draw the equation line through them, and then measure off, by this line, as many more values of u as we please. Example. Let it be required that for x=. -\-l we shall have u^= -\-ly and for a; = +5, «^ = + 3. What will be the values of w corresponding to a; =r — 3, —2, —1, 0, etc. Drawing the base line AX below, we lay off from 1 the ver- tical line +1 in length, and from the point 5 the vertical line + 3. Then drawing the dotted line through the ends, we measure off different values of u, as follows: X = -3, ^2, -1, 0, +1, +2, +3, +4, +5, +6, etc. u = -1, -}, 0, +1 1, +li, +2, +21-, +3, +3i, etc. EXE RCISES. 1. Plot the equation 2u -{- dx — 6. 2. Plot a line such that for X = — 6 we shall have u = -f 4, for a; rr: + 6 " " u = —4:, and find the values of u for ^ == 1, 2, 3, 4, and 5, 228 GEOMETBIG G0N8TBUGT10N 333. The algebraic problem corresponding to the con- struction of § 220 is the following: Having given two values of y corresponding to two given values of x, it is required to construct an equation of the first degree such that these two pairs of values shall satisfy it. Example of Solution, Let the requirement be that of the equation plotted in the preceding example, namely, for X ^=1 1 we must have t^ — 1, for X — b " " u — Z, The problem then is to find such values of a, d, and c, that in the equation ax 4- hit = 6', (1) we shall have u =^1 for x '=: 1, and u = 3 for x =z 5. Sub- stituting these two pairs of values, we find that we must have axl + bxl = c, ax5 + bxS = c; or a -{- b =: c, 5a -\- dh = c. Here a, ^, and c are the unknown quantities whose values are to be found, and as we have only two equations, we cannot find them all. Let us therefore find a and h in terms of c. Multiplying the first equation by 3, and subtracting the product from the second, we have 2« = — 2c or a =i — c. Multiplying the first equation by 5, and subtracting the second from the product, we have 2Z> = 4c or l = 2c. Substituting these vahies of a and b in (1), we find the re- quired equation to be 2c2i — ex = c. We may divide all the terms of this equation by c (§ 120, Ax. Ill), giving 2u — X = 1, i OF EQUATIONS OF THE FIRST DEGBEE. 221) thus showing that there is no need of using c. The solution of this equation gives from which, for x = —3, —2, —1, etc., we shall find the same values of u which we found from the diagram. EXERCISES. Write equations between x and y which shall be satisfied by the following pairs of values of x and y. 1. For X = 2, y =r 1 ; and for x == 5, ?/ =: — 1, 2. For X =^ — 2, y =: — 1 ; and for x = -\-2, y =z +1. 3. For X =z — 5, y =: + 2 ; and for x =: -{-5, y = — 2. 4. For X .= 0, y = — 7 ; and for x = lb, y = 0. 5. For X z= 26, y = 2 ; and for x = 30, y = 3. 333. Geometric Solution of Ttvo Equations with Two Un- hnown Quantities, The solution of two equations with two unknown quantities consists in finding that one pair of values which will satisfy both equations. If we lay off on the base line the required value of x, the two values of y corresponding to this value of a: in the two equations must be the same ; that is, the two equation lines must cross each other at the point thus found. Hence the following geometric solution : I. Plot the two equations froin the same hase line and zero point. II. Continue the equation lines, if necessary, until they intersect. III. The distance of the point of intersection froin the hase line is the value of y which satisfies both equations. IV. The distance of the foot of the y line from, the zero point is the required value of oc. EXERCISES. Solve the following equations by geometric construction : 1. X — 2u = 3, 2x -\- u = 6. 2. 2u 4- '^.'^ = 4, Su -f X — 1. 230 NOTATION OF FUNCTIONS. 334. Geometric Explanation of Equivalent and Inconsist- ent Equations, If we have two equivalent equations (§ 200), each value of a: will give the same value of the other quantity u or y. Hence the two lines representing the equation will coincide and no definite point of intersection can be fixed. If the two equations au ■{• hx =: c, a'u + i'x =z c\ are inconsistent we shall have (§ 142), I V \ a a' ] If h be any increment of x, the increments of u in the two equations (§219) will be A and ,h Therefore these ^ ^*^ ^ a a increments will be equal, and the two equation lines will be parallel. Hence, To inconsistent equations correspond parallel lines, ivhich have no point of intersection. If the two equations are equivalent (§ 141, 143), their lines will coincide. I Notation of Functions. ^ 335. In Algebra we use symbols to express any numbers whatever. In the higher Algebra, this system is extended thus : We may use any synvbol, having a letter attached' to it, to express a function of the quantity represented hy that letter. Example. If we have an algebraic expression containing a quantity x, which we consider as a function of x, but do not wish to write in full, we may call it F{x), or {x), or \x\ or Ax, or, in fine, any expression we please which shall contain the symbol x, and shall not be mistaken for any other expression. In the first two of the above expressions, the letter x is enclosed in parentheses, in order that the expression may not be mistaken for x mul- tiplied by F, or 0. The parentheses may be omitted when the reader knows that multiplication is not meant. NOTATION OF FUNCTIONS, 231 The fundamental principle of the functional notation is this: When a syiyihol with a letter attached represents a function, then, if ive substitute any other quantity for the letter attached, the combination will represent the function found by substituting that other quantity. Example. Let us consider the expression ax^ + 5 as a function of Xy and let us call it (j)(x), so that {x) = ax^ + Z>. Then, to form (p (y), we write y in place of x^ obtaining (y) = af + ^. To form (p {x -}- y), we write x-\-y in place of x, obtaining (t){x-\-y) = a{x -\- yY + h. To form (p (a), we write a instead of x, obtaining (a) =: a^ -{- b» To form {ay^)-, we put ay^ in place of x, obtaining {ay^) = a {ay^f + & = ay + b. The equation (p {z) z=zO will mean az^ +d = 0. EXERCISES. Suppose {x) = ax^ — a^x, and thence form the values of I. 0(^). 2. 0(^). 3. 0(%). 4. (a; + ^). 5. (.^' + 05). 6. (.T — «). 7. (ic + «?/). 8. (j){x — ay). 9. (:i^2)^ Suppose F (.^') = 0;^=^, and thence form the values of 10. FQj). II. F{^), 12. Fm). 13. i^(:^ + ^). 14. F{x-y). is^ F {1). Suppose / {x) =z x% and thence form the values of 16. /(I). 17. f{x^). 18. /(.T«). 19. /(a^). 20. /(a,-=). 21. /(a;«). 232 FUNCTIONS OF SEVERAL VARIABLES. 2 2. Prove that if we put (x) = a^, we shall have (a; + ^) = {x) X (/> Cv), (:r/y) = [0 {x)\y = [0 (y)]^. Let lis put (m) =. m{m — 1) (m — 2) (m — 3) ; thence form the values of 23. 0(6). 24. 0(5). 25, 0(4). 26. 0(3). 27. 0(2). 28. 0(1). 29. 0(0). 30. 0(-l). 31. 0(-^)' Functions of Several Variables. 336. An algebraic expression containing several quantities may be represented by any symbol having tlie letters v^hich represent the quantities attached. Examples. AYe may put {x, y) =z ax— hy, the comma heing inserted between x and y, so that their product shall not be understood. We shall then have, {m, 7i) = am — hi, {y, '^) =^ cty — bx, the letters being simply interchanged. (f>{x-{-y, x—y) z=i a{x + y)—l{x — y) — (a— b) X + («5 4- Z>) y. (P{a,b) = a^-P. {d, a) = ab — da = 0. ^{a -\- b, ab) = a {a -\- b) — aP. (a, a) = a^ — ba, etc. etc. If we put (^, b, c) = 2^ + 3Z> — 5c, we shall have (:r, z, y) =^ ^x -\- ?fZ — by, {z, y, x) := 2z + 3y — 6x. (f) (?/?., m, — m) — 2m -i- dm -{- 6m = 10m. 0(3,8, 6) = 2'd + 3.8 — 5-6 = 0. EXERCISES. Let us put (x, y) z= 3x — 4:y, fix, y) = ax -^ by, f{x, y, z) = ax -i- by — abz. USE OF INDICES, 233 Thence form the expressions : I.

{a, h). 3- «/>(3, 4). 4. S3z=l + 2 + 3i=:6, etc., etc. Using the preceding notation, find the values of the ex- pressions : 3- S^-\- S^ + Sq + 8^. 4. ct^ -\- a^ + a^ -\- a^, 5. 2/S'5 — ag. 6. 2/S'g-^6. 337. Sometimes the relations between quantities distin- guished by indices are represented by equations of the first degree. The following are examples : Let us have a series of quantities, Aq, A^, Aq} A^, A^, etc., connected by the general relation, Am = ^i + ^i-i' W It is required to express them in terms of ^^ and A^, We put, in succession, ^ = 1, i = 2, i = 3, etc. Then, when i = 1, we have from (a), A^ — A^ +Aq. When i = 2, A.^ — A^ + A^ = 2A^ + ^o- i = 3, A^ = A^-{- Aq = 3A^ + 2^0- i = 4, A^ =: A^ + A^ ^ 6A, + 3^0- I = 5, A^ = A^ + A^ z= SA^ + 6Aq, and so on indefinitely. MISCELLANEOUS FUNCTIONS, 236 EXERCISES. 1. If ^ifi = Ai — Ai^i, what will be the values of ^3 . . . . ^lo, and in what way may all subsequent values be determined ? 2. If Ai^i = 2Ai — Aq, find A2 to Aq in terms of ^^ and A^, 3. If Ai^t = iAi H- Ai_\, find A^ to ^5. 4. If Ai = Ai_\ + Ji, find the sum A^ -{- A^ -\- A^ + , , . . + An, in terms of A^^ n and n. (Comp. § 209, Prob. V.) 5. If ^zHi = rAi, find ^1 + ^3 + ^3 + . . . . + ^n? in terms of J^, and n 6. If ^e+i = ^*^-4i + Ai^\, find ^3, ^3, .... ^g, in terms of ^^ and A^, Miscellaneous Functions of Numbers. 238. We present, as interesting exercises, certain elemen- tary forms of algebraic notation much used in Mathematics, and which will be employed in the present work. 1. When we have a series of synabols the number of which is either indeterminate or too great to be all written out, w^e may write only the first tw^o or three and the last, the omitted ones being represented by a row of dots. Examples. a, h, c, . . . . t, J., /^, (5, . . • • /vO, 1, 2, . . . . 7^, n being in the last case any number greater than 2. The number of omitted symbols is entirely arbitrary. EXERCISES. How many omitted expressions are represented by the dots in the following series: I. 1, 2, 3, n. 2. 1, 2, 3, n — 2. 236 MISCELLANEOUS FUNCTIONS 3. 1, 2, 3, n + 2. 4. n, n — 1, 71 — 2, , . , , n — s, 5. n, n — I5 7t — 2, , , , , ?i — s — 1. 6. n, n — 1, n — 2, , , . . n — s -{-1. What will be the last term in the series : 7. 2, 3, 4, etc., to 71 terms. 8. n, n — 1, n — 2, etc., to s terms. 9. 2, 4, 6, etc., to Jc terms. 2. Product of the First n Numbers. The symbol n\ is used to express the product of the first n numberSj 1-2.3 n. Thus, 1 ! = 1. 2! = 1-2 =3 2. 3! = 1.2.3 = 6. 4! = 1.2.3.4 = 24. etc. etc. It will he seen that 2 ! = 2. 1 ! 3! = 3.2! And, in general, ?^ ! = ^ (^ — 1) ! whatever number n may represent. EXERCISES. Compute the values of I. 5! 2. 6! Z' ^^- _7j_ 8! ^* 3! 4! ^* 3! 5! 6. Prove the equation 2.4.6.8 .... 27z :=; 2'»w! 7 . Prove that, when n is even, ^ ^ _ ^ (^ — 2) (7^ — 4) . . . .• 4. 2 3. Binomial Coefficients, The binomial coefficient n(ri — 1)(7^ — 2) . . . . to 5 terms 123 s is expressed in the abbreviated form, MISCELLANEOUS FUNCTIONS, 237 ©■ the parentheses being used to show that what is meant is not the fraction -• s ( EXAMPLES. (?) = ? = s. © 7-6.5.4.3 ■~ 1.2.3.4.5 ~ 21. (f) n (1) n (n — l){n- ~ 1-2-3 -2) © n{n — l).. ..2.1_j ~ 1.3-3... . ^ '» + 4\ _ (re + 4) {71 + 3) (w + 2) 1.2.3 EXERCISES. Compute the vaUies of the expressions : - ©+(^(i)-©-©+(i)-(^©- '■ (l)-e)-(^©+(D- Prove the formulae : /5\ _ 5! (n\ __ nl ^* \2/ ~" 2TF! "^^ \s/ ~ si {n - s) ! 5. (•^^D = !^:(;)- ^- © + (l) = m' '■ (i)+(i)=m- «• (i)+(i)=m- BOOK IX. THE THEORY OF NUMBERS. CHAPTER I. THE DIVISIBILITY OF NUMBERS. 329. Def, The Theory of Numbers is a branch of mathematics which treats of the properties of integers. Def. An Integer is any whole number, positive or negative. In the theory of numbers the word nwnber is used to ex- press an integer. Def. A Prime Number is one which has no divi- sor except itself and unity. The series of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc. Def A Composite Number is one which may be expressed as a product of two or more factors, all greater than unity. Kem. Every number greater than 1 must be either prime or composite. Def. Two numbers are prime to each other when they have no common divisor greater than unity. Example. The numbers 24 and 35 are prime to each other, though neither of them is a prime number. Eem. a vulgar fraction is reduced to its lowest terms when numerator and denominator are prime to each other. DIVISIBILITY OF NUMBERS, 239 Division into Prime Factors. 330. Every composite number may by definition be di- vided into two or more factors. If any of these factors are composite, they may be again divided into other factors. When none of the factors can be farther divided, they will all be prime. Hence, Theokem. Every composite number may he divided into prime factors. Example. 180 rr 9-20, 9 z= 3.3, 20 = 4.5 = 2.2.5. Whence, 180 = 2-2.3.3.5 = 22.32.5. Cor, 1. Because every number, not prime is composite, and because every composite number may be divided into prime factors, we conclude: Every nitrnber is either prime or divisible hy a prime. Cor, 2. Every number, prime or composite, may be ex- pressed in the form p°-q^ry etc., (a) where ^, q, r, etc., are different prime numbers ; S ft y^ ^tc, the exponents, are positive integers. Rem. If the number is prime there will be but one factor, namely, the number itself, and the exponent will be unity. EXERCISES. Divide the following numbers or products into their prime factors, if any, and thus express the numbers in the form {a) : I. 24. 2. 72. 3. 260. 4. 169. 5. 225. 6. 256. 7. 91. 8. 143. 9. 360. 10. 217. II. 3072. 12. 1.2.3.4.5.6.7.8.9. kEEM. In seeking for the prime factors of a number, it is lever necessary to try divisors greater than its square root, for f a number is divisible into two factors, one of these factors 240 DIVISIBILITY OF NUMBERS, Coinmon Divisors of Two Numbers, 331. Theoeem I. // two mnnhers have a common factor, tJieir sum will have that same factor. Proof, Let a be the common factor ; m, the product of all the other factors in the one number; n^ the corresponding product in the other number. Then the two numbers will be am. and an. Their sum will be a (m + ^). Because m and n are whole numbers, m-^n will also be a whole number. Therefore a will be a factor of am + an. Theoeem II. // two numhers have a comm>on factor, their difference will have the same factor. Proof. Almost the same as in the last theorem. Cor, If a number is divisible by a factor, all multiples will be divisible by that factor. Eem. The preceding theorems may be expressed as follows : // two nioinhers are divisible hy the saiiie divisor, their sum, difference, and multiples are all divisihle hy that divisor, Eem. If one number is not exactly divisible by another, a remainder less than the divisor will be left over. If we put Z), the dividend; d, the divisor; q, the quotient; r, the remainder; we shall have, D := dq -\- r, or D — dq =: r. Example. 7 goes into QQ 9 times and 3 over. Heno this means 66 = 7.9 + 3, or 66 --7.9 = 3. I DIVISIBILITY OF NUMBERS. 241 333. Peoblem. To find the greatest common divisor of two numbers. Let m and n be the numbers, and let m be the greater. 1. Divide m by n. If the remainder is zero, n will be the divisor required, because every number divides itself. If there IS a remainder, let q be the quotient and r the remainder. Then m — nq = r. Let d be the common divisor required. Because m and n, are both divisible by d, m — nq must also be divisible by d (Theorem II). Therefore, r is divisible by d. Hence every common divisor of m and n is also a common divisor of n and r. Conversely, because m =z nq + r, every common divisor of n and r is also a divisor of m. There- fore, the greatest common divisor of m and n is the same as the greatest common divisor of /^ and r, and we proceed with these last two numbers as we did with 7)1 and n, 2. Let r go into n q' times with the remainder r'. Then n = rq -\- r\ or n ^rq' ^= r'. Then it can be shown as before that 6? is a divisor of r\ and therefore the greatest common divisor of r and r'. 3. Dividing r by r\ and continuing the process, one of two results must follow. Either, cc. We at length reach a remainder 1, in which case the two numbers are prime ; or, P, We have a remainder which exactly divides the pre- ceding divisor, in which case this remainder is the divisor required. To clearly exhibit the process, we express the numbers m, n, and the successive remainders in the following form : 16 242 GREATEST COMMON DIVISOR, m = n-q + r, {r < n); n = r-q' + r', (r' < r); r = r'-q" -f->", (/' < r'); r' = r"'q"' + r"\ {r'" < r") -, etc. etc. etc., iinfcil we reach a remainder equal to 1 or 0, when the series terminates. EXERCISES. I. Find the G. C. D.* of 240 and 155. ividend . Div. Quo. Rem. 240 r= 155 .1 + 85. 155 Z=i 85 •1 + 70. 85 Z=l 70 1 + 15. 70 =: 15 4 + 10. 15 = 10 1+ 5. 10 — 5 2. Therefore 5 is the greatest common divisor. Note. Let the student arranf^e all the following exercises in the above form, first dividing in the usual way, if he finds it necessary. Find the greatest common divisor of 2. 399 and 427. 3. 91 and 131. 4. 8 and 13. 5. 1000 and 212. 6. 799 and 1232. 7. 800 and 1729. 8. 250 and 625. 9. 1000 and 370. 10. If p be a number less than n and prime to n, show that n —p is also prime to n, 11. It p be any number less than ?i, the greatest common divisor between 7i and p is the same as that between n and n —-p. 12. If 7z is any odd number, — ~ — and — 7^ — are both prime to it. Corollaries, 1. When two numbers are divided by their greatest common divisor, their quotients will be prime to each other. * The letters G. C. D. are an abbreviation for Greatest Common Divisor. GEARING OF WHEELS. 243 I 2. Conversely, if two numbers, n and n\ prime to each other, are each multiplied by any number d, then d will be the G. C. D. of dn and dn', 233. Gearing of Wheels. An interesting problem con- nected with the greatest com- mon divisor is afforded by a common pair of gear wheels. Let there be two wheels, the one having m teeth and the other n teeth, gearing into each other. If we start the wheels with a certain tooth of the one against a certain tooth of the other, then we have the questions : (1. ) How many revolutions must each wheel make before the same teeth will again come together ? (2. ) With how many teeth of the one will each tooth of the other have geared ? Let q be the required number of turns of the first wheel, having m teeth. Let^ be the required number of turns of the second, hav- ing n teeth. Then, because the first wheel has m teeth, qni teeth will have geared into the other wheel during the q turns. In the same way, pn teeth of the second wheel will have geared into the first. But these numbers must be equal. Therefore, when the two teeth again meet, pn = qm. Conversely, for every pair of numbers of revolutions 2^ and q, which fulfil the conditions, pn =. qm, the same teeth will come together, because each wheel will have made an entire number of revolutions. This equation gives p m q~ n 244 QEABING OF WHEELS, Hence, if we reduce the fraction — to its lowest terms, we n ' ^ shall have the smallest number of revolutions of the respective wheels which will bring the teeth together again. To answer the second question : After the first wheel has made q revolutions, qm of its teeth have passed a fixed point. Any one tooth of tlie other wheel gears into every n^ passing tooth of the first wheel. Therefore QTYl any such tooth has geared into - — teeth of the first wheel, that is, into p teeth, because, from the last equation, qm ^— —p. n If d be the G. 0. D. of m and n, then m = dp^ n := dq\ m n Therefore each tooth of the one wheel has geared into only every d^ tooth of the other. In the figure on the preceding page, m = 21 and /^ = 6. Hence, ^ = 3, and each tooth of the one will gear into every third tooth of the other. The numbers on the large wheel show the order in which the gearing occurs. How long soever the wheels run, the same contacts will be repeated in regular order. Hence, if each tooth of the one wheel must gear with every tooth of the other, the numbers m and n must he prime to each other. EXERCI SES. 1. If one wheel has 40 teeth and the other 10, show how they will run together. Show the same thing for the following cases: 2. w? r= 72, w = 15. 3. m =1 24:, n = 18. 4. m = 36, 71 = 25. 5; m = 24, 7i = 7. 1 NUMBERS AND THEIR DIGITS. 245 Relations of Numbers to their Digits. 334. In our ordinary method of expressing numbers, the second digit toward the right expresses lO's, the third lOO's, etc. That is, each digit expresses a power of 10 correspond- ing to its position. Def. The number 10 is the Base of our scale of numeration. ISToTE. The base 10 is entirely arbitrary, and is supposed to have originated from the number of the thumbs and fingers, these being used by primitive people in counting. Any other number might equally well have been chosen as a base, but in any case we should need a number of separate characters (digits) equal to the base, and no more. Had 8 been the base, we should have needed only the digits 0, 1, 2, etc., to 7, and different combinations of the digits would have represented numbers as follows : 1 = 1, 7 = 7, 10 = 1-8 + = eight. 17 = 1-8 + 7 = fifteen. 20 = 2-8 + = sixteen. 56 = 5*8 -f 6 r= forty-six. 234 = 2-82 + 3-8 + 4 = one hundred fifty-six, etc. Let us take the arbitrary number z as the base of the scale. As in our scale of lO's we have 234 = 2.102 + 3.10 + 4, so in the scale of ^'s the digits 234 would mean 2^2 + 3;^ + 4. In general, the combination of digits abed would mean az^ + dz^ + cz -\- d. Divisibility of Numbers and tlieir Digits, 235. Theorem. If the sum of the digits of any num her he subtracted from the nurriber itself, the remainder will he divisible by z — 1. 246 DIVISIBILITY OF NUMBERS. J Proof. Let the digits be a, b, c, d. The number expressed wiU be az^ j^ M j^ cz + d Sum of digits =: a -\- h -{- c -\- d Subtracting, rem. = a{z^—l) + h(f—l) + c{z—l). The factors z^ — 1, z^ — 1, and z — 1 are all divisible by 2; — 1 (§ 93). Hence the theorem is proved. (§ 231.) Theorem. Ijv any scale having z as its base, the sum of the digits of any number, when divided by z — 1, ivill leave the same remainder as will the number itself when so divided. If we put : n, the number ; s, the sum of the digits ; r, r', the remainders from dividing by z—lj q, q' , the quotients ; we shall have, ^Number, n =1 q{z — 1) -\- r Sum of digits, s = q' {z — 1) + r' Eemainder, {q — q') {z — 1) -\- r — r. Because 7i — s and {q — q') {z — 1) are both divisible by z — 1, their difference r — r' must be so divisible. Since r and r' are both less than z — 1, this remainder can be divided by ;2; — 1 only when r = r', which proves the theorem. Zero is considered divisible by all numbers, because a re- mainder is always left. If a be any factor oi z — 1, the same reasoning will apply to it, and therefore the theorem will be true of it. In our system of notation, where z = 10, the above theo- rems may be put in the following well-known form : If the sunt' of the digits of any number be divisible by 3 or 9, the number itself ivill be so divisible. These are the only numbers of which the theorem is true, because 3 is the only divisor of 9. Theorem. If from any number we subtract the digits of the even powers of z, and add those of the alternate -powers, the result will be divisible by z -\- 1. Proof. To az^ + bz^ + cz + d Add a — b -\- c — d Result, a (f^-\) ^b{z^—l) + c(z^\). NUMBERS AND THEIR DIGITS, 247 The factors of a^ h, and c are all divisible by 2; + l (§§ 93, 94), whence the result itself is so divisible. Applying this result to the case of ^ = 10, we conclude: // on suhtracting the sum of the digits in the place of units, hundreds^ tens of thousands, etc., froin the sum of the alternate ones, the remainder is divisible hy 11, the number itself is divisible by 11. If 771 be any factor of ^, it will divide all the terms of the number az^ + bz^ -\- c^ + d, except the last.- Hence, if it divide this last also, it will di- vide the number itself. Applying this result to the case of z = 10, we conclude : If the last digit of any number is divisible by a fac- tor of 10, the number itself is divisible by that factor. The factors of 10 being 2 and 5, this rule is true of these numbers only. It will be remarked that if the base of the system had been an odd number, we could not have distinguished even and odd numbers by their last figure, as we habitually do. For example, if the base had been 9, the figures 72 would have represented what we call sixty-five, which is odd, and 73 would have represented what we call sixty-six, which is even. The use of the base 10 makes it easy to detect when a num- ber is divisible by either of the first three prime numbers, 2, 3, and 5. If the last figure is divisible by 2 or 5, the whole num- ber is so divisible. To ascertain whether 3 is a factor, we find whether the sum of the digits is divisible by 3. In taking the sum, it is not necessary to include all tlie digits, but in adding we may omit all 3's and 9*s, and drop 3, 6, or 9 from the sum as often as convenient. Thus, if the number were 921642712, we should perform the operation mentally, thus : Drop 9 ; 3 + 1 = 3, which drop ; 6, drop ; 4 + 2 = 6, which drop; 7 + 1 = 8 + 2 = 10, which leaves a remainder 1. EXERCISES. T. Prove that if an even number leaves a remainder 1 when divided by 3, its half will leave a remainder 2 when so divided. 248 DIVISIBILITY OF NUMBERS. 2. If from any number we subtract the sum of units' digit plus the product of the tens' digit by ^, plus the product of the hundreds' digit by i% etc., the remainder will be diyisible by 10 — i. (^ may be any integer, positive or negative.) Note. When i — 1, this gives the rule of 9's and when ^ = — 1, the rule of ll's. Prime Factors of Numbers. 336. First Fui^damental Theorem. A product can- not he divided hy a prime numher unless one of the fac- tors is divisible by that prime number. Note. This theorem is not true of composite divisors. For exam- ple, neither 8 nor 9 is divisible by 6, but the product 8 • 9 = 72 is so divisible. But if we take as many numbers as we please not divisible by 7, we shall always find their product to leave a remainder when we try to divide it by 7. To make the demonstration better understood, we shall first take a special case : The product Q^a is not divisible by 7, unless a is divisible byl. Proof. Suppose . . . . 66^ di v. by 7 7 goes into 66 9 times and 3 over, because 7 • 9 = 63, 63a di v. by 7 Therefore, by Theorem II, § 231, 3a div. byl' 2 3 goes into 7 2 times and 1 over. Multiply by 2, 6a div. by 7 Subtracting, 7a div. by 7 We have left, a div. by 7 Hence, if 66a is divisible by 7, then a is divisible by 7. Gauss's Demonstration, If it be possible, let am be the smallest multiple of m which is divisible by p, when neither a nor m is so divisible. If a is greater than p, then let p go into a b times and r over, so that a := bp + r, or a — bp z=z r. Then, am div. by p. Subtract bpm " " Eemainder, (a — bp) m " " Or r77i " " PRIME FACTORS OF NUMBERS. 249 That is, if am. is divisible by p, so is ririy where r is less than p. Therefore the smallest multiple of m which fulfils the con- ditions must be less than pm. Therefore, let a (2) = 1 0(3) = 2 0(6) ^ 2 0(9) = 6 (18) = _6 Sum, 18. To show how this comes about, write down the numbers 1 to 18, and under each write the greatest common divisor of that number and 18. Thus, Num., 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18. G.C.D., 12321612921612321 18. Necessarily the numbers in the second line are all divisors of 18 as well as of the numbers over them. The divisor 1 is under all the numbers prime to 18, so that there are (18) = divisors 1. If n be any number over the divisor 2, then - and -^, or 9, must be prime to each other. (§ 232, Cor. 1.) That is, the 256 DIVISORS OF A NUMBER. numbers n are all those which, when divided by 2, are prime to 9. So there are (f> (9) divisors 2. The divisor 3 marks all numbers which, when divided by 3, 18 are prime to — = 6. Hence, there are o (6) divisors 3. In the same way there are (3) divisors G, (2) divisors 9, and (j) (1) divisor 18. The total number of these divisors is both 18 and (p (18) + etc. etc. If for each x we substitute its expression, the fraction — will take the form , * 1 11 X. 1 1 '"^ «3.-j , etc., etc. x^ If the substitutions are continued indefinitely, the form will be 1 1 «i +- 1 «2 +- 1 «3 +- Such an expression is called a continued fraction. Def. A Continued Fraction is one of which the denominator is a whole number plus a fraction ; the denominator of this last fraction a whole number plus a fraction, etc. CONTINUED FRACTIONS. 259 A continued fraction may either terminate with one of its denominators or it may extend indefinitely. Bef. When the number of quotients a is finite, the fraction is said to be Terminating, 344. Problem. To find the value of a continued fraction. We first find the value when we stop at the first denomina- tor, then at the second, then at the third, etc. * Using only two denominators, the fraction will be 11 Xc> F z=. 1 a.x^ + 1' Xn F being put for the true value of the fraction. To find the expression with three terms, we put, in the preceding expression, a^ ■\ in place of x^. This gives x^ 1 rp ^ 3 __ ^2^3 + -*- . " n n O-^l J- 1 ~ (^1^2 +1)^3 +«1* ^1^3 + — +-•- To find the result with the fourth denominator, we substi- tute x^i=lcu A The fraction becomes: 3 ^ ' x^ ^ ^ (^2^3 +1)^:4 +^a , X \{a^ac^ + 1) «3 + ^1] ^4 + ^1^2 +1 To investigate the general law according to which the successive expressions proceed, we put P, the coefficient of :c in any numerator; P', the coefficient of x in the denominator ; Q^ the terms not multiplied by x in the numerator ; §', the terms not multiplied by x in the denominator ; and we distinguish the various expressions by giving each P and Q the same index as the x to which it belongs. 260 CONTINUED FRACTIONS, Then we may represent each value of F in the form, where i may take any yalue necessary to distinguish the frac- tion. Comparing with the fractions as written, we see that : P, =0, Q, =1, P^, =1, Q\ =0; F, =1, Q, = 0, P; = a,, (?; - 1 ; (6) To show that this form will continue, how far soever w^e carry the computation, we put in the expression (b) the general value of Xi, 1 Xi =z ai -i , I.- 1, • jjy {ctiPi-\- Qi) n+1 + Pi f.. wh.chg..es, F = ^^p-^-^^^-^r (,) To show the general law of succession of the terms, let us compare the general equation (b) with (d). Putting ^ + l for i in (b), it becomes, Comparing this with (d), we find Fi+t =z aiFi -\- Qi, Qibl = Pi\ whence, Qi = P^^i. Substituting this value of Qi in the equation previous, it becomes P,+i = ^^P^ + Pi_i. (/) Working in the same w^ay with the denominators, we find Q'i.t = P'v By supposing i to take in succession the values 1, 2, 3, etc., CONTINUED FRACTIONS, 26i these formulae show that the successive values of P may be computed thus : pUiJ ^'""'^' Pg = a,P, + P„ P, = a,P, + P^, etc., to any extent. Also, P; = 1, p; = «!, ^ . -P 3 =^ tt^Ps + Pi, P^ = ^S-^S + -f*2» etc. etc. Since each value of Q is equal to the value of P having the next smaller index, it is not necessary to compute the $'s sep- arately. If the fraction terminates at the n^^ value of a, we shall have Xn = an, exactly. If it does not terminate, we have to neglect all the denom- inators after a certain point ; and calUng the last denominator we use the n*^, we must suppose Xji "=. dn* In either case, the expression (Z>) will give the value of the fraction with which we stop by putting i =:n and Xn = an- Therefore, F = "^^r ^ ^f , Ctn Pn + Qn or, substituting for Q'^ and Q'^ their values in (^), ■p ^n I^n "T" I^n—1 But the general expressions (/) and (g) give 262 CONTINUED FRACTIONS. ' (^n 't^n "f" ^n—\ "^^^ -^n+h (^n Pn + Pn-\ = Pn+1* Therefore, F=^' Pn+l Therefore, to find the value of the fraction to the n^^ term, we have only to comjjute the values of Pn+i and Pn+h without tahing any account of Q, Example. Take the fraction, 1 i + i 2 4-i 3 + i 4 + ^ 5 etc. Here, a^ =1, CI2 = 2, a^ = 3, , . , . ai z= i. We now have, by continuing the formulae (c) and (/), and using those values of a^, a 2, etc. : Pi =0, P2 = 1. Ps ^^ ^3-^2 + Pi =^2=2^ P4 = «3^3 + Pg = 3.2 + 1 = 7, P, =a^P^ + Pg =4.7 + 2 = 30, Pe = a,P, +Pi = 5.30 + 7 = 157, etc. etc. etc. p; = i, P'a = «i = 1, p; = a,P', + P\ = 3-1 + 1 = 3, p; = (jjp; + p; = 3.3 + 1 = 10, p', = «4^4 + K = 4-10 + 3 = 43,. p; = a,P', + P; = 5.43 + 10 = 225. Therefore, supposing in succession, n = l, n — 2, n = d, etc., we have, for the successive approximate values of tlie fraction, •. CONTINUED FRACTIONS. For n = l, 2 For ^ = 2, ^« P\ - 3 For n = 6, ^ ~ P'~ 225' 263 These successive approximate values of the continued frac- tion are called Converging Fractions, or Convergents. 345. The forms (/) and {g) may be expressed in words as follows : The numerator of each convergent is formed by mul- tiplying the preceding numerator by the corresponding a, and adding the second numerator preceding to the product. The successive denominators are formed in the same way. Example. The ratio of the motions of the sun and moon relative to the moon's node is given by the continued fraction : 1 12 + -^ i + i 2 + i 1+^ 4 + ^ 3 + etc. Let us find the successive convergents. We put the de- nominators «>t = 12^ ^2 = 1, etc., in a line, thus: a = 12, 1, 2, 1, 4, 3. P = J^ J^ 3 4 19 61 P' = 1' 12' 13' 38' 51' 242' 777* Under a^ we write the fraction -, whicli is always the one with which to start, because P^ = and P\ = 1 (§ 244, c). Next to the right is — , because Pg = 1 and Pg = a. After this, we multiply each term by the multiplier a above it, and 264 CONTINUED FRACTIONS. add the term to the left to obtain the term on the right. Thus, 2.1 + 1 = 3, 2.13 + 12 = 38, etc. Ex. 2. To compute the convergents of 1 ^+ 1 ^ + 4 etc. a = 3, 4, 2, 4, 2, 4, Numerators, Denominators, 1 4 9 40 1' 3' 9' 20' 89' 89 198' etc. etc. EXERCISES. Reduce the following continued fractions to vulgar frac- tions : 3 + ^. 3 + ^ ^ + 4- 2 + ^ 3 + 1 3 4- 1 ^1 3 + —— 3 + a + 1 .1 '1 1+ 5 + -. b + -. 3 + ^ "^ '■ 1 + r 346. Problem. To express a fraetional quantity as n eontinued fraetion. Let R be the given fraction, less than unity. Compute x^ from the formula, _ 1 ^1 - ^• Let a^ be the whole number and R' the fraction of ^^j. Then compute __ 1 ^2 -^- CONTINUED FRACTIONS. 265 Let a^ be the whole number and R" the fraction of x^. We continue this process to any extent, unless some value ^f X comes out a whole number, when we stop. 26 Example. Express ^ as a continued fraction. 1 73 ^'^ ^ :^ "^ 26 ^ ^ + 26 ' R' Jl_ R" 26 21 21 1 + 21 21^ 5 ~^ + 5^ So the continued fraction is 1_ 2 + 1- ^1 = 2 ttg = 1 «3 = 4 ^4 = 5 R' R" 21 26* 21* R'" — -. R" = 0. It will be seen that the process is the same as that of find- ing the greatest common divisor of two numbers. EXERCISES. Develop the following quotients as continued fractions : 113 1049 628 ^' 925* 355 2. 3326 347. The most simple continued fraction is that arising from the geometric problem of cutting a line in extreme and mean ratio. The corresponding numerical problem is : To divide unity into two such fractions that the less shall he to the greater as the greater is to unity. Let r be the greater fraction. Then 1 —r will be the lesser one. We must then have 1 — r I r : : r : 1, 266 CONTINUED FRACTIONS. which gives r^ = 1 -- r, or r^ 4- r = 1, or r (r + 1) = 1, 1 or r z=z 1 + r Now, let us put for r in the last denominator the expression , and repeat the process indefinitely. We shall have. 1 + ^ 1 r z=z - i + i i + i 1 + ^ 1 etc., ad infiniUim, Now we may form the successive convergents which approximate to the true value by the rule. As all the denom- inators a are 1, we have no multiplying, but only add each term to the preceding one to obtain the following one. Thus we find: 011235^132134 1^ 1' 2' 3'' 5' 8' 13' 21' 34' 55' ^*''' The true value of r may be found by solving the quadratic, y2 -f r = 1, . . , . -1± a/5 which gives r = ^r • The positive root, with which alone we are cOYiCemed, is r =: "-^j" ^^ = 0.61803399. The values of the first nine convergents, with their errors. are: 1:1= 1.0, error = + 0.382. 1:2= 0.5, " —0.118. 2:3= 0.666...., " +0.0486. 3:5= 0.600, '' -0.0180. 5:8= 0.625, '' +-0.00697. CONTINUED FRACTIONS. 267 8 : 13 = 0.61538.... , error = — 0.00265. 13 : 21 = 0.61904.... , " + 0.00101. 21 : 34 = 0.617647.... , " - 0.000397. 34 : 55 = 0.618182...., " + 0.000148. etc. etc. etc. Relations of Successive Convergents. 348. Theorem I. The successive conver gents are alternately too large OMid too small. Proof, The first convergent is — The true denom- inator being a^ -\ , the denominator a^ is too small, and therefore the fraction is too large. In forming the second fraction, we put — instead of — a.2 x^ Because ^2 < i?:^3, this fraction is too large, which makes the denominator a^ -\ too great. The third denominator a^ is too small, which will make the preceding one too large, the next preceding too small, and so on alternately. 171 7Yi! Theorem II. // — and —, he any two consecutive convergents, then mn! — m'n = ± 1. Proof, We show : («) Thatthe theorem is true of the first pair of con vergents. {p) That if true of any pair, it will be true of the pair next following. {a) The first pair of convergents are J_ 1 _ ^2 ""^■^^ which gives 7nn' — m'n = 1, thus proving («). 268 CONTINUED FRACTIONS, \p) Let — , — 7, —77, ^ ^ n n n be three consecutive convergents, in which 7nn' — m'n =. ±1, (1) By (/) and {g) we shall have ^" = an' + ^. Multiplying the second equation by m' and subtracting the product of the first by n', we have * m'n" — m"n' r= m'71 — mn', Avhich is the negative of (1), showing that the result is =F 1- The theorem being true of the first and second fractions, must therefore be true of the second and third ; therefore of the third and fourth, and so on indefinitely. Corollaries, Dividing (1) by n7i\ we have m w! . 1 T_r ; = ± — -,- Hence, n n nn I. The difference between the two successive converge ents is equal to unity divided hy the product of the denominators. Because the denominator of each fraction is greater than that of the preceding one, we conclude : II. T]^e difference between two consecutive convergents constantly diminishes. Combining these conclusions with Th. T, we conclude : III. Each value of a convergent always lies between the values of the two preceding convergents. For if R^, R^, R^ be three such fractions, and if R^ is greater than R^, then R^ will be less than R^. But it must be greater than R^, else we should not have R^ — Rq numer- ically less than R^ — R^. Hence, if we arrange the successive convergents in a line in the order of magnitude, their order will be as follows : -"'4^ -^^e^ -*^8' • • • • ^9? Ri9 R^9 each convergent coming nearer a true central value. Hence, CONTINUED FRACTIONS. 269 IV. The true value of the continued fraction al- ways lies between the values of two consecutive con- vergents. Comparing with (I), we conclude : V. The error which we mahe hy stopping at any con- vergent can never he greater than unity divided hy the product of the denominators of that convergent and the one next following. EXAM PLE. Referring to the table of values of |-(\/5 — 1) in § 247, we see that : Error of 2 : ^^<3-5' (for .0486 < 1). Error of 3 ; ^^<5'8' (for .018 's among themselves. We shall then have 5 ! r ! X Xn permutations. From each of these may be found t ! permutations by in- terchanging the d^ among themselves. Proceeding in the same way, we shall have Xn X r\ X s\ X t\ X etc. possible permutations of all 7i things. But this number has been shown to be n\ Therefore, Xn X r\ X s\ X t\ X etc. = n ! By division, X„ = ^^j^^----, (3) which is the required expression. We remark that if any symbols are not repeated, the for- mula (3) will still be true by supposing the number correspond- ing to r, s, or t to be 1. EXAMPLES. I. The number of possible permutations of aabl are z=z — — =r 6, as already found. 2! 2! ~ 2-2 2. The possible permutations of aaahhcd are 7 ! 5040 3! 2! 6-2 420. EXERCI SES. Write all the permutations of the letters: I. aaah, 2. aabc, 3. aaabc. 4. How many different numbers of seven digits each can be formed by permuting the figures 1112225 ? PERMUTATIONS, 281 5. If every different permutation of letters made a word, how many words of 13 letters each could be formed from the word Massachusetts. The Two Classes of Permutations. 255. The n\ possible permutations of n things are divisi ble into two classes, commonly distinguished as even permu- tations and odd permutations in the following way: We suppose the 7i things first arranged in alphabetical or numerical order, a,d,Cyd,,,., or 1, 2, 3, 4, ... . n, and we call this arrangement an evmi permutation. Then, having any other permutation, we count for each thing how many other things of lower order come after it, and take the sum. If this sum is even, the permutation is an even one ; if odd, an odd one. EXAMPLES. 1. Consider the permutation 265143. Here 2 is followed by 1 number of lower order, namely, 1. " Q " " 4 " " " " 5,1,4,3. " 5 " " 3 " " '' " 1,4,3. a \ " ^* Q '' '*• ^< a 4 ^^ " 1 '' '' '^ ^^ 3 Then 1+4 + 3 + + 1 = 9. Hence the permutation is odd. 2. Consider cdbea. Here c is followed by 2 letters before it in order, namely, ba, '' d ," " 2 " " " " la. 6i Jj ii 6i 1 (< 66 66 66 ^ 66 Q " '' 1 ^^ ^^ ^^ ^^ n Then 2 + 2 + 1 + 1 = 6. Hence the permutation is even. Def, The total number of times which a thing less in order follows one greater in order is called the Number of Inversions in a permutation. 282 PERMUTATIONS. Example. In the preceding permutation^ 265143, the number of inversions is 9. In cdhed it is 6. Eem. It will be seen that the class of a permutation is even or odd, according as the number of inversions is even or odd. Theorem I. If, in a perinutaUon, two things are interchanged, the class will he changed from even to odd, or from odd to even. Proof. Consider first the case in which a pair of adjoining things are interchanged. Let us call : ilc, the two things interchanged. A, the collection of things which precede i and Jc. C, the collection of things which follow them. The first permutation will then be AikC.'^ {a) After interchanging i and k, ifc will be AUG. \b) Because the order of things in A remains undisturbed, each thing in A is followed by the same things as before. In the same way, each thing in G is preceded by the same things as before. Hence, the number of times that each thing in A or G is followed by a thing less in order remains unchanged, and, leaving out the pair of things, i, k, the number of inversions is unchanged. But, by interchanging i and k, the new inversion ki is in- troduced. Therefore the number of inversions is increased by 1. * This form of algebraic notation differs from those already used in that the symbols A and C do not stand for quantities, but mere collec- tions of letters. It is an application of the general principle that a single symbol may be used to represent any set of symbols, but must represent the same set throughout the same question. A and C are here used to show to the eye that in forming the permutations of (5) from {a), all the letters on each side of ik preserve their relative positions unchanged. PERMUTATIONS. 283 If the first arrangement is Id, this one inversion is removed. Hence^ in either case the number of inversions is changed by 1, and is therefore changed from odd to even, or vice versa. Illustration, In the permutation 265143, the inversions, as already found, are the following nine : 21, 65, 61, 64, 63, 51, 54, 53, 43. Let us now interchange 5 and 1, making the permutation 261543. The inversions now are ^ 21, 61, 65, 64, 63, 54, 53, 43, the same as before, except that 51 has been removed. Next consider the case in which the things interchanged do not adjoin each other. Suppose that in the permutation h a d e h c f we interchange a and h. We may do this by successively in- terchanging a with d, with e, and with h, making three inter- changes, producing I d e h a c f . Then we interchange h with e ^and with d, making two interchanges, and producing h h d e a c f , which effects the required interchange of a with h. The number of the neighboring interchanges is 3-|-2 r= 5, an odd number. Because the number of inversions is changed from odd to even this same odd number of times, it will end in the opposite class with which it commenced. Theorem II. I7^e possible permutations of n things are one-half even and one-half odd. Proof, Write the n ! possible permutations of the n things. Then interchange some one pair of things {e.g., the first two things) in each permutation. We shall have the same permutations as before, only differen ]y arranged. 1 3 odd. 1 -i even. 2 3 even. 3 1 odd. 3 2 odd. 3 1 even. 284 PERMUTATIONS. By the change, every even permutation will be changed to odd, and every odd one to even. Because every odd one thus corresponds to an even one, and vice versa, their numbers must be equal. Illustration. The permutations in the second column fol- lowing are formed from those in the first by interchanging the first two figures : 12 3 even, 2 13 2 odd, 3 2 13 odd, 1 2 3 1 even, 3 3 12 even, 1 3 2 1 odd, 2 EXERCISES. Count the number of inversions in each of the following permutations ; I. hcdagef. 2. Magdef. 3. 325941. 4. 5432. 5. 82917364. 6. 82971364. S56 Def. A Symmetric Function is one which is not changed by permuting the symbols which enter into it. An Alternating Fuhction is one which, when any two of its symbols are interchanged, changes its sign without changing its absolute value. EXERCISES Show which of the following functions are symmetric and which are alternating : I, a + 6 + c. 2. ahc. 3. ^ (6 + c) -f 5 (c + a) + c (a + 6), 4. a^ (b - c) + h' (c-a) + c' {a - b). 5. a' {b + c) + b' (c+a) -he' (a + 5). 6. (a-b) {b - c) (c - a), 7. ab -\- be -{- oa. COMBINATIONS. 285 CHAPTER II. COMBINATIONS. 357. Def, The number of ways in which it is pos- sible to select a set of .^ things out of a collection of n things is called the Number of Combinations of s things in n. Ex. I. From the three symbols a, h, c, may be formed the couplets, ah, acy he. Hence there are three combinations of 2 things in 3. Ex. 2. From a stud of four horses may be formed six dif- ferent span. If we call the horses A, B, C, J), the dijBFerent span will be AB, AC, AD, BC, BD, CD. Rem. 1. A set is regarded as different when any one of its separate things is different. Eem. 2. Combinations differ from permutations in that, in forming a combination, no account is taken of the order of arrangement of things in a set. For instance, ah and ha are the same combination. Hence, we may always suppose the letters or numbers of a combination to be written in alpha- betical or numerical order. Notation, The number of combinations of s things in 7i is sometimes designated by the symbol, c» Problem. To find the ninnher of combinations of s things in n. If we form every possible set of s things out of n things, and then permute the s things of each set in every possible way, we shall have all the permutations of n things taken s at a time (§ 252). That is, Cg X Is 286 GOMBmATIONS. express the number of permutations of n things taken 5 at a time. But we have found this number to be n (n — 1) (m — 2) . . . . (m — s + 1). We have also found P, = .s! = 1.2-3-4 s. Hence, Cf xsl = n{n — l) {iu— 2) .... (re - - s + 1), , n _ M (« - 1) (w - 2) {n - -s + 1) ^' - 1.2.3. 4.... s = g) (§ 228, 3) ; C^= . ,^' ... which is the required expression. Rem. For every combination of s things which we can take from n things, a combination ot n— s things will be left. Hence, Cf = Cls. This formulae may be readily derived from the expression for the number of combinations. For, if we take the equation pn _ ^ ^^ - s\ {n-s)V this formula remains unaltered when we substitute n — s for s, and therefore also represents the combinations of 7i — s things in n. Def. Two combinations which together contain all the things to be combined are called two Complement- ary combinations. EXERCISES. 1. Write all combinations of two symbols in the five sym- bols, a^ h, Cy d, e. 2. Write all combinations of three symbols in the same letters, and show why the number is the same as in Ex. i. GOMBINxiTIONS. 287 3. A span of horses being different when either horse is changed, how many different span may be formed from a stud of 3 ? Of 7 ? Of 9 ? 4. If four points are marked on a piece of paper, how many distinct lines can be formed by joining them, two and two ? How many in the case of n points ? From each one of the points can be drawn 7i -—1 lines to other points; then why are there not 71 {n — 1) Hnes? 5. If five lines, no two of which are parallel, intersect each other, how many points of intersection will there be ? How^ many in the case of n lines ? 6. If n straight lines all intersect each other, how many different triangles can be found in the figure ? 7. In how many different ways may a set of four things be divided into two pairs ? 8. In how many ways can a party of four form partners at whist? 9. In how many ways can the following numbers be thrown with three dice : {a) 1,1,1; {d) 1,2,2; {c) 1,2,3. 10. A school of 15 young ladies have the privilege of send- ing a party of 5 every day to a picture gallery, provided they do not send the same party twice. How many visits can they make? Combinations with Repetition. 358. Sometimes combinations are formed with the liberty to repeat the same symbol as often as we please in any set. Example. From the three things a, h, c, are formed the six combinations of two things with repetition, aa, ah, ac, hh, he, cc, Pkoblem. To -find the ninnhcr of comhinations of s things in n, wheiz repetition is allowed. Solution. Let the n things be the first 71 numbers, 1, 2, 3, 4, ... . u. 288 COMBINATIONS. Form all possible sets of s of these numbers with repetition, the numbers of each set being arranged in numerical order. Let Rs be the required number of sets. Then, in each set, Let the first number stand unchanged. Increase the 2d number by 1. " " 3d " " 2. '' '' 4th " " 3. We shall then have Bs sets of s numbers, each without rep- etition. Example. From the numbers 1, 2, 3 are formed with repetition, 11, 12, 13, 22, 23, 33. Then, increasing the second numbers by 1, we have 12, 13, 14, 23, 24, 34. The greatest possible number in any set after tlie increase will be n -{- s — 1, because the greatest number from which the selection is made is n, and the greatest quantity added is 5—1. Hence all the new sets will consist of combinations of s numbers each from the n -\- s — 1 numbers, 1, 2, 3, 4, .... :?^ .... :^ + 5 — 1. {a) No two of these combinations can be the same, because then two of the original combinations would have to be the same. Hence the new sets are all different combinations of s numbers from the n -[- s — 1 numbers {a). Therefore the number of combinations cannot exceed the quantity (7/^. Conversely, if we take all possible combinations of s differ- ent numbers in n -{- s — 1, arrange each in numerical order, and subtract 1 from the second, 2 from the third, etc., we shall have different combinations from the first n numbers with repetitions. Hence the number of combinations in the second class cannot exceed those of the first class. Hence we conclude that the number of combinations of s things in n with repetition is the same as the combinations of s things in n -{- s — 1 without repetition, or COMBINATIONS. 289 »7i j-iTh^s-i In -{- s — 1\ __ n{n + l){n -\-2) {n-\- s — 1) "" 1.2.3.4 s EXERCISES. 1. Write all possible combinations of 3 numbers with repe- tition out of the three numbers 1, 2, 3 ; then increase the second of each combination by 1 and the third by 2, and show that we haye all the combinations of three different numbers out of 1,2,3,4,5. 2. How many combinations of 4 things in 4 with repeti- tion ? Oi n things in tz ? In the last question and in the following, reduce the result to its lowest terms. 3. How many combinations of 7^ + 1 things m. n—l with repetition ? Special Cases of Combinations. 359. It is plain that because each of these combinations consist simply of one of the n things. Hence, also, Cl-i = n, because in every such combination one letter is omitted. It is also plain that because the only combination of 7i letters is that comprising the n letters themselyes. Hence we write, by analogy, G^ = 1, although a combination of nothing does not fall within the original definition of a combination. 360. The formulae of combinations sometimes enable us to discover curious relations of numbers. 1. Let us inquire how we may form the combinations of *290 GOMBINAriONS. s -\- 1 things when we have those of 5 things. Let the n things from which the combinations are to be formed be the letters a, d, c, d, e,f, g, etc {n\n number). Let all the combinations of 5 + 1 of these n letters be writ- ten in alphabetical order. Then : 1. In the combinations beginning with a, the letter a will be followed by all possible combinations of s letters out of the 71 — 1 letters d, c, d, etc., of which the number is Cf'^. 2. In the combinations beginning with h, the letter h is followed by all combinations of s letters out of the n — 2 let- ters c, d, e, f, etc. Therefore there are G'^'^ combinations beginning with h, 3. In the same way it may be shown that there are C'^~^ combinations beginning with c, C^~^ beginning with d, etc. The series will terminate with a single combination of the last s + 1 letters. Since we thus have all combinations of s + 1 letters, we find, by summing up those beginning with the several letters a, b, c, etc., or' + cr' + or' + .... + ci = c»+i. («) Substituting for the combinations their values, we find By the notation (§ 228, 3), all the terms of the first member have the common denominator s ! , while the numerators are each composed of the factors of s consecutive numbers. Mul- tiplying both sides by s ! and reversing the order of terms in the first member, we have 1.2.3 5 + 2-3.4 5 + 1 + etc. etc. etc. + (n — s — l) (^ _ 3) (^ — 2) + {n — s) {71 — 2) {71 — 1) _ {n — s) . , , , {n — 2 ) (n — 1) ^ _ ____ \ COMBINATIONS. 291 The student is now recommended to go over the preceding process with special simple numerical values of n and s which he may select for himself. EXAMPLES. If ^ = 5 and s = 2, we have 1.2 + 2.3 + 3.4 = ^. 6 If II = 7 and 5 = 3. 1.2.3 + 2.3.4 -f 3.4.5 -f- 4.5.6 = i:^!^-. If iz =2 7 and s = 4, 1.2.3.4 + 2.3.4.5 -f 3.4.5.6 = ^-— ^. 5 If w =: 9 and 5 = 3, 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + 5.6.7 + 6.7.8 =. ?1^-. Prove these equations by computing both members. 261. Another curious example is the following: Let us have p + q things divided into two sets, the one containing p and the other q things. Then, to form all possi- ble combinations of s things out of the whole p -{- q, we may take : Any s things in set p ; Or any combination of 5 — 1 things in set^ with any one thing of set ^ ; Or any combination of s — 2 things in set p with any com- bination of 2 things in q ; Or any combination of 5 — 3 things in p with any 3 out of g, etc. We shall at length come to the combinations of all 5 things out of q alone. Adding up these separate classes, we shall have : c? + cu ci+cua + .... + C1 CU + Gl This sum makes up all combinations of s things in the whole p-\-q, and is therefore equal to 6'^^^. Putting the numerical expressions for the combinations, we have the theorem : 292 COMBINATIONS. p-^')=©-(A)(f)-(A)©-- If, as an example, we put 5 = 3, ^ = 4, g — 5, tliis theo rem will give 9 ^7 _ 4.3»2 4^5 4 5 -^ 5-4.3 1-2.3 ~ 1.2.3 "^ 1-2*1 "^ 1*1.2 "^ 1-2^^ the correctness of which is easily proved by computation. EXERCISES. 1. Write all the combinations of three letters out of the five, a, Z>, c, d, e, and show that C^ of them begin with a, CI with h, and 0% with c, according to the reasoning of § 260. 2. Prove that C| = 6^t + C^, Cl = Cl + CI, and in general, C^ = C^ -{- C^s-i- In the following two ways : (1.) Let all combinations of s letters in the n letters a, b, c, , . . . n be formed, their number being (7^. Then suppose one letter added, making the number n -\- 1. The combinations of s letters out of these n -\- 1 will include the Cg formed from the ^ letters, plus each combination of the additional (w + 1)*^ letter with the combinations of ^ — 1 out of the first n letters. (2.) Prove the same general result from the formula, c^ = (g. 3. If we form all combinations of 3 things out of 7, how many of these combinations will contain a 7, and how many will not ? 4. If we form all the combinations of s letters out of the n letters a, h, c, . , , , n, COMBINATIONS. 293 how many of these combinations will contain a, and how many will not ? 5. In the preceding case, how many of the combinations will contain the three letters a, d, c ? 363. Theoeem I. The total number of comhination$ ivhich ean he formed from n things, including 1 zero combination, is 2^. . In the language of Algebra, c^ + c^+c^ + .... + Cl-t -\-cl = t. Proof, Let ns begin with 3 things, a, h, c, and let us call the formal zero combination, 1 = C^, Then we have d, blank, Number = 1 (7i, a,d,c. " =3 C\, ah, ac, be, " =3 Cl, abc, " =J_ Sum m 8 = 23. Now introduce a fourth letter d. The combinations out of the four things, a, h, c, d, will consist of the above 8, plus the 8 additional ones formed by writing d after each of the above eight. Their number will therefore be 16. In the same way, it may be shown that we double the pos- sible number of combinations for every thing we add to the set from which they are taken. We have found, for n = 3, Sum of combinations = 8 = 2^; 71 =z 4., " " =2-8 = 24; n — ^, " " = 2-24= 25; etc. etc. which shows the theorem to be general. Theorem II. If the signs of the alternate combina- tions of n things be changed, the algebraic sum will be zero. In algebraic language, Cl _ Cl -^Ct-Cli- etc. ± Cl ^ 0. {a) 294 COMBINATIONS. Proof. If in the formula of § 261, Ex. 2, Dumely, ^ S ^ S -T ^ 8-ly we put n — 1 for n, it becomes pn pn~\ , /^^~} L/ s — ^ S "T" ^ s—1* Putting s successiyely equal to 0, 1, 2, ... . n, we have ri^ p^'' 1 . — f 7i — i ? Cl— Co +Ci =1+Gl , C 3=02+03; O w-1 — O 7i-2 + O/^-i — O 7i_3 + 1. Substituting these values in the expression (a), it becomes 1 _ (1 + or') + {cr' + en - (er' + or') + . . . . ^ 1 _ 1 _ cr' + cr' + c^r' - or' - cr' + etc. How far soever we carry this process, all the terms cancel each other except the last. Therefore, if we con tin ae the addi- tions and subtractions until we come to Cn-i ? the sum will be Cl -C'ii-Cl- etc ± CU = ± OVi =±1. The last term will be + C'JI; === =F 1? and will therefore just cancel the sum of the preceding terms. Note. Theorem I may be demonstrated by these same formulae, since the sum of all the terms taken positively will be duplicated every time we increase ti by 1. 363. Independent Comlinations. There is a system of combinations formed in the following Avay : It is required to form a comhinatioTi of s things, by taking one out of each of s different collections. How many combinations can be formed' ? Let the 1st collection contain a things, u 2d '' " b " " 3d " '' c' " etc. etc. C0MBmATI0N8. 295 Then we may take any one of a things from the first col- lection. With each of these we may combine any one of the h things in the second collection. With each of these we may combine any one of the c things of the third collection. Continuing the reasoning, we see that the total number of combinations is the continued product ahc .... to cS factors. If the number in each collection is equal, and we call it a, the number of combinations will be a^. This form of combinations is that which corresponds most nearly to the events of life, and is applicable to many questions concerning probabilities. For example, if any one of live dif- ferent events might occur to a person every day, the number of diiferent ways in which his history during a year might turn out is 52^, a number so enormous that 255 digits would be re- quired to express it. EXERCISES. 1. A man driving a span of horses can choose one from a stud of 10 horses, and the other from a stud of 12. How many different span can he form ? 2. It is said that in a general examination of the public schools of a county, the pupils spelt the word scliolar in 230 different ways. If in spelling they might replace ch hj c ov Tc\ by au, mu, or 00 \ 1 by II', a by e, 0, u, or ou; r hj re; in how many different ways might the word be spelt ? 3. If a coin is thrown n times in succession, in how many different ways may the throws turn out ? 4. If there are three routes between each successive two of the five cities, Boston, New York, Philadelphia, Baltimore, Washington, by how many routes could we travel from. Boston to Washington ? 296 COMBmATIONS. The Binomial Theorem when the Power is a Whole Number. 364. The binomial theorem (§ 172), when the power is a positive integer, can be demonstrated by the doctrine of com- binations, as follows : Let it first he required to form the product of the n hinomial factors, To understand the form of the product, let us first study the special case when /?. = 3. Performing the multiplication of the first three fac- tors, the product will consist of eight terms : ' ^2'^'i^3 ~T (t^X^X-2 -f- X^X^X^ This product is the expression {a) developed when 7i = 3. IS''?* We conclude, by induction, that the entire product (a) when developed in this same way, will be composed of a sum of terms, each term being a product of several literal factors. When (a) is thus multiplied out, we shall call the result the developed expression. The developed expression has the following properties : I. Each term contains n literal factors, a's and x's, and no more. For, suppose x^r=i a^, x^ ^ a^, to Xn — cfn- Then the expression (a) will reduce to %^a^a^a^ , , , , an, {V) and the developed expression must assume the same value ; that is, it must consist of terms each of which reduces to the expression a^a^a^ .... an, {c) when we change x into a. Now if it contained any term with either more or less than n factors, it could not assume this form. THE BINOMIAL THEOREM. 297 II. The factors of each term have all the n indices 1, 2, 3, n. For, the index figure of no term is altered by changing x into a^ as in I. Hence, if in any term any index figure were missing or repeated, that term would not reduce to the form (c), whence there can be neither omission nor repetition of any index. III. Because each term has n factors^ it Tivust either have n factors a; n — 1 factors a and one factor x; n — 2 factors a and two factors x ; In general, a term may ha^ve the factor a repeated n — i times, and x repeated i times. IV. In a term which contains i factors x, these / factors must be affected with some combination of i indices out of the whole number 1, 2, 3, .... ^ ; and the n — i «'s must be affected by the complementary combination of n — i indices. We next inquire whether there is a term corresponding to every such combination. Let 1, 3, 4, 7, ... . be any combination of i indices, and 2, 5, 6, 8, ... , the complementary combination of ^ — / indices. Since the developed expression must be true for all values of a and x, let us put in (^), a^ ■-— 0, ccg = ; «3 = % ^'6 = ^; «4 = 0, x^ —^\ {d) a^ =0, x^ z=^ Q'y etc. etc. The product {a) will then reduce to the single term, x^a^x^x^a^a^x^a^ {e) By the same change the developed expression must reduce to this same value, and it cannot do this unless the expression (c) is one of its terms. 298 GOMBmATIONS. Hence the developed expression must contain a term corresponding to every combination. V. Since every combination of i figures out of 1, 2, 3, .... :^ will, in this way, give rise to a term like (e), containing the symbol a i times, and the symbol x n — i times, there will be C'i such terms. Now suppose a^ =1 a^ := a^ =1 , , , , cin =^ ci' /y — /y /y — /v» /y *t/ -| -^— «^2 " '^ ^ ~~~~ • • • • *^n «^» The expression (a) will then reduce to {a -f x)^. In the developed expression, all the Cf terms containing x i times and a n — i times will now be equal and their sum will reduce to 0^ a'~^ x. Hence, putting in succession i = 0, ^ = 1, etc., to i = n, we shall have {a-\-xY = a^+ Cia'^-^x+C^a'^'^x^ + +Cl tax^-^-^:^. Substituting for Ct its value, we shall have {a + xY = a« + naP'-'^x + (^L^-2^2 + . . . . + ( — ^ Xix'^-'^ + l-jx'^, which is the Binomial Theorem, enunciated, but not demon- strated, in Book V, Chapter I. Note. If the studeAt has any difficulty in understanding the steps of the preceding demonstration, he should suppose 71 = 3, and refer the demonstration to the developed expression (aO. PM0BABILITIE8, - 299 CHAPTER III. THEORY OF PROBABILITIES. 365. Bef, The Theory of Probabilities treats of the chances of the occurrence of events which cannot be foreseen with certainty. Notation, Let a bag contahi 4 balls, of which 1 is white- and 3 black. If a ball be drawn at random from the bag, we should, in ordinary language, say that the chances were 1 to 3 in favor of the ball being white, or 3 to 1 in favor of its being black. In the language of probabilities we say that the probability 1 3 of a white ball is -r. and that of a black one -• 4^ 4 In general, if there are m chances in favor of an event, and n chances against it, its probability is Hence, Bef, The Probability of an event is the ratio of the chances which favor it to the whole number of chances for and against it. Illustrations, If an event is certain, its probability is 1. If the chances for and against an event are even, its prob- ability is --• If an event is impossible, its probability is 0. Cor, 1. If the probability that an event will occur is^, the probability that it will fail is 1—p, Cor, 2. A probability is always a positive fraction, greater than and less than 1. 266. Method of ProbaMUties, To find the probability of an event, we must be able to do two things : 300 PnOBABILlTIES. 1. Enui7ierate all possible ways in which the event may occur or fail, it being supposed that these zuays are all equally probable. 2, Determine how many of these zuays will lead to the event. If n be the total number of ways, and m the number which lead to the event, the probability required is — • EXERCISES. 1. A die has 2 white and 4 black sides. What is the prob- ability of throwing a whifce side ? 2. A bag contains n balls numbered from 1 to n, the even numbers being white and the odd ones black. What is the probability of drawing a black ball when n is an odd number? What, when n is an even number ? 3. A bag contains 3w-f-2 balls, of which numbers 1, 4, 7, etc., are white; 2, 5, 8, etc., are red; 3, 6, 9, etc., are black. What are the respective probabilities of drawing a white, red, and black ball ? , . Rem. In tlie last example tlie probabilities are all less than ^ ; tli^re- fore, should one attempt to guess the color of the ball to be drawn, he would be more likely to be wrong than right, no matter what color he guessed. This exemplifies a lesson in practical judgment to be drawn from the theory of probabilities. If there are three or more possible re- sults of any cause, it may happen that the best judgment would be more likely to be wrong than right in attempting to predict the result. Thus, if there are three presidential candidates with nearly equal chances, the chances would be against the election of any one that might be named. Gamblers of the turf are nearly always found betting odds against every horse that may be entered for a race, though it is certain that one of them will win. Hence, if a naturaj. event may arise from a number of causes with nearly equal facility, it is unphilosophical to have any theory whatever of the cause, because the chances may be against the most probable cause being the true one. Probabilities depending: upoi?! Combinations. 267. Problem i. Two coins are thrown. What are the respective probabilities that the result will be : Both heads ? head and tail ? both tails ? PnOBABlLITIES, 301 At first sight it might appear that the chances in favor of these three results were equal, and that therefore the probabil- ity of each was ^- But this would be a mistake. To find the probabilities, we must combine the possible throws of the first coin (which call A) with the possible throws of the second (which call B), thus : A, head ; B, head. . A; head ; B, tail. A, tail ; B, head. A, tail ; B, tail. These combinations are all equally probable, and while there are only one each for both heads and both tailsVthere are 1^1 1 two for head and tail. Hence the probabilities are 7, -, ^^ The sum of these three probabilities is 1, as it ought always to be when all possible results are considered. Proh. 2. Five coins are thrown. What are the respective probabilities: q heads, 5 tails? 1 head, 4 tails? .^ m- ^,^^ 2 heads, 3 tails? etc. etc. Let the several coins be marked a, 1), c, d, e. Coin a may be either head or tail, making two cases. Each of these two cases of coin a may be combined with either case of b (as in the last example), making 4 cases. Each of these 4 cases may be combined with either case of coin c, making 8 cases. Continuing the process, the total number of cases for five coins is 5^ ±= 32. Of the^e 32 cases, only one gives no head and 5 tails. There fare 5 cases of 1 head, namely: a alone head, I? alone head, etc.,' to e. 2 heads may be th ^wn by coins r/, b; a, c, etc. ; b, c ; b, d, etc. ; c, d, etc. ; that is, by any combination of two letters out of the five, a, b, c, d, e. Hence the number of cases is Ol = 10. 302 PBOBABILiriES. In the same way the number of cases corresponding to 3, 4, and 5 heads are, respectively, Gl = 10, Cl = 6, Cl = 1. Dividing by the whole number of cases, we find the respec- tive probabilities to be 32' 32' 32' 32' 32' 32* The following general proposition is now to be proved by the student : Theorem. If there are n coins, the prohahility of throwing s heads and n — s tails is From this result we may prove the theorem in combina- tions of § 262. If we suppose, in succession, s = 0, 8 = 1, s =: 2, etc., to s =: H, the respective probabilities of iiead, 1 head, 2 heads, etc., will be ' Cl G^ Cl ' - Cl 2^' 2^' 2^*' 2^' Because the sum of all these probabilities must be unity, we find C^ + ^5l + C'l + .... + a;: =:: 2^ Prol, 3. Two dice are thrown at backgammon. What are the respective probabilities of throwing 5 and 6 and two 6's ? If we call the dice a and h, any number from 1 to 6 on « may be combined with any number from 1 to 6 on Z>. There- fore, there are in all 36 possible combinations. In order to throw two 6's, a must come 6 and h also. Therefore there is only one case for this result, so that its probability is --• To bring 5 and 6, a may be 5 and Z> 6, or ^ 5 and a 6. So there are two cases leading to this result, and its probability is A _ _L •' 36 ""^ 18' PROBABILITIES. 303 Note. That 5 and 6 are twice as probable as a double 6 may be clearly seen by supposing that the two dice are thrown in succession. If the first throw is. either 5 or 6, there is a chance for the combination 5, 6, but there is no chance for a double 6 unless the first throw is 6. Proh 4. If three dice are thrown, what are the respective probabihties that the numbers will be : 1, 1, 1? 1, 1, 2? 1, 2, 3? The solution of this case is left as an exercise for the student. Proh. 5. From a bag containing 3 white and 2 black balls, 2 balls are drawn. What are the respective probabilities of Both balls white? 1 white and 1 black ? Both black? Since any 2 balls out of 5 may be drawn, the total number of cases is Cg. Only one of these combinations consists of two white balls. C\ of the cases bring both balls black. A w^hite and black are formed by combining any one of the three white with any one of the two black. The respective probabilities can now be deduced by the student. EXERCISES. 1. It takes two keys to unlock a safe. They are on a bunch with two others. The clerk takes three keys at random from the bunch. What is the probability that he has both the safe keys? 2. A party of three persons, of whom two are brothers, seat themselves at random on a bench. What are the probabilities {a) that the brothers will sit together, {h) tliat they will have the third man between them ? 3. If two dice are thrown at backgammon, what are the probabilities {a) Of two aces ? {h) Of one ace and no more ? 4. In order that a player at backgammon may strike a cer« 304 PROBABILITIES. tain point, the sum of the numbers thrown must be 8. What are his cliances of succeeding in one throw of his two dice ? 5. A purty of 13 persons sit at a round table. What is the probability that Mr. Taylor and Mr. Williams will be next to each other? (See § 253.) 6. An illiterate servant puts two works of 2 volumes each upon a shelf at random. What is the probability that both pair of companion volumes are together? 7. A gentleman having three pair of boots in a closet^ sent a blind valet to bring him a pair. The valet took two boots at random. What are the chances that one was right and tlie other left ? What is the probability that they were one pair ? 8. If the volumes of a 3- volume book are placed at random on a shelf, what is the probability that they will be in regular order in either direction ? 9. A man wants a particular span of horses from a stud of 8. His groom brings him 5 horses taken at random. What is the probability that both horses of the span are amongst them ? 10. From a box containing 5 tickets, numbered 1 to 5, 3 are drawn at random. What is the probability that numbers 2 and 5 are both amongst them ? 1 1. The same thing being supposed, what is the probability that the sum of the two numbers remaining in the box is 6 ? 12. Of two purses, one contains 5 eagles and another 10 dollar-pieces. If one of the purses is selected at random, and a coin taken from it, what is the probability that it is an eagle ? 13. From a bag containing 3 white and 4 black balls 2 balls are drawn. What is tlie probability that they are of the same color ? 14. The better of two chess players is twice as likely to win as to be beaten in any one game. What chance has his weaker opponent of winning 2 games in a mateli of 3 ? 15. From a bag containing m white and n black balls, two balls are drawn at random. What is the probability that oii'.' is white and the other black ? PROBABILITIES. • 305 1 6. From a bag containing 1 white, 2 red, and 3 black balls, 3 balls are drawn. What is the probability that they are all of different colors ? 17. If ^ coins are thrown, what is the chance that there will be one head and no more ? 18. From a Congressional committee of 6 Republicans and 5 Democrats, a sub-committee of 3 is chosen by lot. What is the probability that it will be composed of two Eepublicans and one Democrat ? Compound Events. 368. Theorem I. The probability that tiuo independ- ent events will both happen is equal to the product of their separate probabilities. Proof, For the first event let there be m cases, of which p are favorable; and for the second ?2 cases, of which ^ are favorable. Then, by definition, the respective probabilities will be — and - • m n When both events are tried, any one of the in cases may be combined with any one of the n cases, making in all in x n combinations of equal probability. The combinations favorable to both events will be those only in which one of the p cases favorable to the first is com- bined with one of the q cases favorable to the second. The number of these combinations is p x q. Therefore the probability that both events will happen is P >< q ^ Z X 2 m X n m n^ which is the product of the individual probabilities. If there are three events of which the probabilities are ^, q^ and r, and we wish to find the probability that all three will happen, we may by what precedes regard the concumng of the first two events as a single event, of which the probitbility is pq. Then the probability that the third event will also con- cur is the product of this probability into r, or pqr. 20 306 PROBABILITIES. Proceeding in the same way with 4, 5, 6, ... . events, we reach the general Theorem II. The -probability that any number of in^ dependent events will all occur is equal to the continued product of their individual probabilities. Eem. This theorem is of great practical use as a guide tcj our expectations. It teaches that if success in an enterpr requires the concurrence of a great number of favorable cil cumstances, the chances may be greatly against it, although each circumstance is more likely than not to occur. This is illustrated by the following Example i. A traveller on a journey by rail has 8 connec- tions to make, in order that he may go through on time. There are two chances to one in favor of each connection. What is the probability of his keeping on time ? The probability of each connection being - , the probabil- o ity of successfully making the first two connections will, by the preceding theorems, be ( - 1 , the first three I - 1 , and all eight /2\8 _ 28 \3/ ~" 38 28 256 1 3-8== 6561 ^2:6'^^"^^^- Therefore there are 25 chances to 1 against his going- through on time. On the other hand, if, instead of any one accident being fatal to success, success can be prevented only by the concur- rence of a series of accidents, the probability of failure may become very small. Ex. 2. A ship starts on a voyage. It is an even chance that she will encounter a heavy gale. The probability that 9 she will not spring a leak in the gale is ^n* ^^ ^ ^^^^ occurs, 9 there is a probability of — that the engine will be able to 3 pump her out. If they fail, the probability is j that the com- PROBABILITIES, 307 partments will keep the ship afloat. If she sinks, it is an even chance that any one passenger will be saved by the boats What is the probability that any individual passenger will be lost at sea ? The probability that the ship will meet a heavy gale is ^ the ship will spring a leak in the gale is To the engines cannot pump her out is . . — the compartments cannot keep her afloat is - the boats cannot save the passenger is . : - The continued product of these probabilities is ^Tuio^ which is the probability that the passenger will be lost. 369. The preceding theorem as enunciated supposes that the several events are independent, that is, that the probability of the occurrence of any one is not affected by the occurrence or non-occurrence of the others. To investigate what modifi- cation is required when the occurrence of one of the events alters the probabihty of another of the events, let us distinguish the two events as i\\Q first and second. We then reason thus : Let the total number of equally possible cases be m, and let p of these cases favor the first event. Its probability will then be — • ra It is certain that the events cannot both happen unless the first one happens. Hence the cases which favor both events can be found only among the p cases which favor the first. Let q of these p cases favor the second event. Then the prob- ability of both events will be — • In case the first event happens, one of the p cases which 308 PROBABILITIES. favor it must occur, aud the probability of the second event will then be ^. Then P Probability of both events = — = — x -• Hence, •^ m 7)1 p Theokem. The probability that two events will both oceur is equal to the probability of the first event mitlti- pliecl by the probability of the seeond, in case the first occurs. By continuing the reasoning to more events, we reach the general Theokem. The probability that a nuT)%ber of events will all occur is equal to the product {X Prob. of second in case first occurs. X Prob. of third in- case first two occur. X Prob. of fourth in case first three occur, etc. etc. etc. Example. From a bag containing 2 white and 3 black balls, 2 balls are drawn. What are the probabilities (1) that both balls are white, (2) that both are black ? This problem has already been solved, but we are now to see how the answers may be reached by the last theorem. It is evident that we may suppose the two balls drawn out one after the other, and the probabilities of their being white or black will be the same as if. both were drawn together. I. Both balls white. The probability that the first ball 2 drawn is white is -• If it really proves to be white, there will o be left 1 white and 3 black balls. In this event, the probabiliti^ that the second also will be white is - 4 Hence the probability that both are white is 2 1 _ 2. 5 ^ 4 "" 10* PROBABILITIES, 309 II. Both halls hlach. Applying the same reasoning, we find for the probability of this case, 3 1 _ ^ 5 '"^ 2 ~" 10* EXERCISES. I. Two men embark in separate commercial enterprises. The odds in favor of one are 3 to 2 ; in favor of the other, 2 to 1. What are the probabilities (1) that both will succeed ? (2) that both will fail? -2. The probability that a man will die within ten years is -, and that his wife will die is —• What are the respective probabilities that at the end of ten years, (a) Both are living ? (/3) Both are dead ? (y) Husband living, but wife dead? {6) Husband dead, but wife living ? 2 3. The probability that a certain door is locked is - • The o key is on a bunch of 4. A man takes 2 of the four keys, and goes to the door. What are the chances that he will be able or unable to go through it ? 4. Two bags contain each 4 black and 3 white balls. A jDerson draws a ball at random from the first bag, and if it be white he puts it into the second bag, mixes the balls, and then draws a ball at random. What is the probability of drawing a white ball from each of the bags ? 5. If a Senate consists of m Democrats and 71 Eepublicans, what is the probability that a committee of three will include 2 Democrats and 1 Eepublican? 6. A bag contains 2 white balls and 5 black ones. Six people, A, ,B, 0, D, E, F, are allowed to go to the bag in alpha- betical order and each take one ball out and keep it. The first one who draws a white ball is to receive a prize. What are their respective chances of winning? Note. A's chance is easily calculated, because lie has the draw from all 7 balls. 310 PROBABILITIES In order that B may win, A must first fail. Therefore, to find B's probability we find (1) the probability that A fails, (2) the probability that if A fails then B will win. We then take the product of these probabili- ties. In order that C may gain the prize, (1) A must fail, (2) B must fail, (3) C himself must gain. So we find the successive probabilities of these occurrences. Continuing to F, we find that he cannot win unless the 5 men before him all miss. He is then certain to gain, because only the two white balls would be left. 7. Two men have one throw each of a coin. X offers a prize if A throws head, and if he fails, but not otherwise, B may try for the prize. If both fail, X keeps the prize himself. What are the respective chances of the three men having the prize ? 8. A and B are alternately to throw a coin until one of them throws a head and becomes the winner. If A has the first throw, what are their respective chances of winning ? 9. A crowd of 71 men are allowed to throw in the same way for a prize, in alphabetical order, the game ceasing as soon as a head is thrown. What are the respective chances of the con- testants? 10. Three men take turns in throwing a die, and he who first throws a 6 wins. What are their respective chances ? 11. If 4 cards are drawn from a pack of 52, show that the probability that there will be one of each of the four suits is 39 26 13 51*50*49' 12. One purse contains 5 dimes and 1 dollar, and another contains 6 dimes. 5 pieces are taken from the first purse and put into the second, and after being mixed 5 are taken from the second and put into the first. Which purse is now most likely to contain the dollar ? 13. Of two purses, one contains 4 eagles and 2 dollars, the other 4 eagles and 6 dollars. One being taken at random, and a coin drawn from it, what are the respective probabilities that it is an eagle or a dollar ? PROBABILITIES. 311 Cases of Unequal Probability. 370. Def. If two or more possible events are so related that only one of them can happen, they are called Mutually Exclusive Events. Theorem. The prdbahility that some one of several exclusive events, we care not which, ivill occur, is equal to the sum of their separate probahilities. Proof, Let there be m possible and equally probable eases in all; let p of these cases be favorable to one event, q to the P Q T second, r to the third, etc., so that — , — ., — , are the re- ,,.,.,. m m m spective probabilities. Since only one of the events is possible, the p cases which favor one must be entirely different from the q cases which favor the second, and these cases p-\-q must be entirely differ- ent from the r which favor the third, etc. Hence there will be j9 + ^^ + r + etc. , cases which favor some one or another of the events. Hence the probability that some one of these events will occur is p -\- q -^^ r -\- etc. m ' which is equal to the sum of the probabilities, par, — + — H h etc. m m m Eem. If the concurrence of some two events, say the first and second, had been possible, some one or more of the p cases which favor the first would have been found among the q cases which favor the second. Then the whole number of cases which favored either event would have been less than p-{-q, and the probability that one of the two events would happen less than the sum of their respective probabilities. 3*71. General Problem. To find the probahility that an event of ivhich the proiahility on any one trial is p, will happen exactly s tiines in n trials. 312 PROBABILITIES. This problem is at the basis of some of the widest applica- tions of the theory of probability to practical questions, espe- cially those associated with life and fire insurance. The con- ditions which it implies are therefore to be fully comprehended. We may conceive a trial to mean giving the event an opj^or- hmity to happen. The simplest kind of trial is that of throw- ing a coin or die. At each throw, any side has an opportunity to come up. Then, if we throw 50 pieces, or which amounts to the same thing, throw the same piece 50 times, there will be 50 trials; and we may inquire into the probability that a given side will be thrown exactly 9 times in these trials. The same conception occurs in another form if we have 50 men, each of whom has an equal chance of dying within 5 years. Waiting to see if any one man will die in the course of the 5 years is a trial, so that there are 50 trials in all, and we may inquire into the probability that 9 of the men will die during the trials, just as in the case of 50 throws of a die. Let us distinguish the several trials by the letters a, h, c, d, e, , , , . ii, which must be ?^ in number. 1. In order that the event may not happen at all, it must fail on every one of the n trials. The probability of this (§ 268, Th. II) is (1 —pY. This is therefore the probability that it will not happen at all. Because the probability of tlie event happening on any one trial is }), the probability of its failing is 1 — p. We now compare the possible results. 2. The event may happen once on any one of the n trials, «, h^ c, etc. In order that it may happen only once, it must fail on the other /z. — 1 trials. The probability that it will happen on any one trial, say e, and also fail on the remaining n — 1 trials is, by the same theorem, p (1 —py-'^. Because there are 7i trials on which it may equally happen, the probability that it will happen once and only once is np (1 — pY-\ PROBABILITIES, 313 3. The event may happen twice on any two trials out of the n trials. In order that it may happen twice only, it must fail on the other ^ — 2 trials. Taking any one combination, say Happen on d, d\ Fail on a, c, e, , , . , n, the probability is p^ (1 — pY~'^, But it may happen twice on any combination of two trials out of the n trials, a, h, c, , , , . n. Because these combina- tions are mutually exclusive (§ 270), the total probability, of happening twice is 4. In general, in order that the event may happen just s times, it must happen on some combination of s trials, and fail on the complementary combination of n — 5 trials. The probability on any one combination is p^ (1 — pY~^ and there are (7? such combinations. Hence the general probability of happening s times is C^,ps{l-p)n-s, {a) If there is on each trial an equal chance for and against the event, then p = -. and 1 —p = -- The probability of the event happening s times then becomes 2n' This case corresponds to that already treated in § 267, Problem 2, and the result is the same there found. HXERCISES. I. A die having two sides white and four sides black is thrown 5 times. What are the respective probabilities of a white side being thrown 1, 2, 3, 4, and 5 times? Note. Here p, tlie probability of a white side on one throw, is ^ , and 2 1 — p = - • The number n of trials is 5. o 314 PROBABILITIES. 2. Of 6 healthy men aged 50, the probability that any one will live to 80 is J- What is the probability that three or more of them will live to that age ? 3. A chess-player whose chances of winning any one game from his opponent are as 2 to 1, undertakes to win 3 games out of 4. What is the probability that he will be able to do it? Note. It would be a fallacy to suppose that the probability required is that of winning exactly 3 games, because he will equally win if he wins all four games. 3*73. Events of Maximum ProhaUlity. Keturning to the general expression {a), let us inquire what number of times the event is most likely to occur on n trials. The required ' number is that value of s for Avhich the probability is the greatest. If we call Ps the probability that the event will happen exactly s times, and if s is to be the number for which the probability is greatest, we must have Pa > Ps-U Ps> Ps^^l^ Substituting for these quantities the corresponding forms of the expression (a), which is equal to Ps, we have C'sP'ii -pY-' > otip'-'ii -pY-'^\ (7?iJ^(i -pY-' > c^+tp'^'ii-pY-'-'^ The general formula for (7? in § 257 gives (b) Og — ^ s-h cS + 1 w Hence we have, by dividing both terms of the first in- equality (Jb) by C^-ip'-^ (1 —pY~'^ 1 p > 1 — /?. PROBABILITIES. 315 Multiplying by s, tiiis becomes np — sjj -\- 2^ > ^ ~ "^i^- Interchanging the members and reducing, we have s < p (n + 1). (d) Now divide the second inequality (b) by C^ p^ (1 — ^j)^~^^ and reducing by the second equation (c), we have -I ^ n — s Multiplying by 5 + 1 and reducing, we find s > p {n + 1) - 1. {e) Comparing the inequalities (d) and (e), we see that s lies between the two quantities p {n -{- 1) and p {n -|- 1) — 1 ; that is, s is the greatest whole number in p (n -\- \), If the number of trials 7i is a large number, and ^ is a small fraction, p{n -\- 1) and pn will differ only by the fraction p. We shall then have, very nearly, 6^ z=i pn. That is : Theorem L The most probable number of times thai an event will happen on a great number of trials is the produet of the number of tibials by the probability on eaeh trial. Example. If a life insurance company has GOOO members, and the probability that each member will live one year is on tlie average — , then the most probable number of deaths during the year is 100. Eem. It must not be supposed that in this case the num- ber of deaths is likely to be exactly 100, but only that they will fall somewhere near it. There is a practical rule for determining what deviation must be guarded against, the demonstration of which requires more advanced mathematical methods than those employed in this chapter. It is: 316 PE0BAB1LITIE8. Theokem II. Deviations from the most probable num- ber of deaths, equal to the square root of that number, will be of frequent oeeurrenee. Deviations mueh greater than this square root will be of infrequent occurrence, and deviations more than twice as great zvill be rare. Examples. In a company of which the probable annual number of deaths is 10, the actual number will commonly fall between 10 — VlO and 10 + V 10, or between 7 and 13. It will very rarely happen that the number of deaths is. as small as 4 or as large as 16. If the company is so large that the most probable number of deaths is 100, the actual number will commonly fall between 100 — VlOO and 100 + VlOO, or between 90 and 110. If the most probable number of deaths is 1000, the actual number will commonly range between 968 and 1032. We now see the following result of this theorem: The greater the number of deaths to be expected, the greater will be the probable deviation, but the less ivill be the ratio of this deviation to the whole number of deaths. Examples. The reductions of the cases just cited are shown as follows : Expected number Probable Ratio of deviation of deaths. deviation. to expected number. 10 3 0.33 100 10 0.10 1000 32 ' 0.03 Application to Life Insurance. 373. At each age of human life there is a certain proba- bility that a person will live one year. This probability di- minishes as the person advances in age. It is learned from observation, on the principle described in the preceding section, that events in a vast number of trials are likely to happen a number of times equal to the product of their probability on each trial, multiplied by the number of trials. PROBABILITIES. 317 Therefore, by dividing the whole number of times the event has happened by the whole number of trials, the quotient is the most probable value of the probability on one trial. Example. If we take 50,000 people at the age of 25, and record how many of them are alive at the end of one year, this is making 50,000 trials whether a person of that age will live one year. If 49,650 of them are alive at the end of the year, and 350 are dead, we would conclude : ProbabiHty of living one year, .... 0.993 Probability of dying within the year, . , 0.007 The probability for all ages may be determined by taking a great number of infants, say 100,000, and counting how many die in each year until all are dead. If n are living at the age y, and n' at the age ^ + 1, then the probability of dying within one year after the age y will be , and that of livinsr will be — • ^ n It is not, however, necessary to wait through a lifetime to reach this conclusion. It is sufficient to find from observation what proportion of the people of each age die during any one year. Suppose, for instance, that the census of a city is taken, and it is found that there are 2500 persons aged 30, and 2000 aged 50. At the end of a year another inquiry is made to ascertain how many are dead. It is found that 20 of the 30 year old people, and 30 of the 50 year old people have died. This would show : At age 30, probability of dying within 1 year = 0.008. " 50, " " " " — 0.015. This saine probability being obtained for every year of life, the probability of living 1 year at all ages would be known. Then a table of mortality could be formed. A table of mortality starts out with any arbitrary number of people, generally 100,000, at a certain age, frequently 10 years. It then shows how many of these people will be living at the end of each subsequent year until all are dead. The following is a specimen of such a table. 318 PROBABILITIES. Table of Mortality. Ages. Living. Dying. Prob. of surviving Prob. of within Ages. Living. Dying. Prob. of surviving Prob. of dying within 10 442 a year. the year. a year. the year. 1 00000 .99558 .00442 60 58373 1677 .97127 .02872 II 99558 407 .99591 .00408 61 56696 1760 .96895 .o3io4 12 99i5i 385 .99611 .00388 62 54936 1849 .96634 .03365 i3 98766 376 .99619 .oo38o 63 53087 1936 .96353 .03646 14 98390 379 .99614 .00385 64 5ii5i 2014 .96062 .03937 i5 98011 396 .99595 .00404 65 49 '37 ^080 .95766 .04233 i6 97615 426 .99563 .00436 66 47057 2i38 .95456 .04543 n 97189 469 .99517 .00482 67 • 44919 2186 .95133 .04866 i8 96720 .525 .99457 .00542 68 42733 2224 .94795 .o52o4 19 96195 58 1 .99396 .oo6o3 69 4o5o9 2268 .94401 .05598 20 95614 621 .99350 .00649 70 38241 233i .93904 .06095 21 94993 645 .99321 .00679 71 35910 2401 .93313 .06686 22 94348 653 .99307 .00692 72 33509 2469 .92631 .07368 23 93695 65i .9930 5 .00694 73 3 1 040 253i .91846 .08154 24 93044 647 .99304 .00695 74 28509 2567 .90995 .09004 25 92397 647 .99299 .00700 75 25942 2542 .90201 .09798 26 91750 65 1 .99290 .00709 76 23400 2476 .89418 .io58i ^^ 91099 668 .99266 .00733 77 20924 2369 .88678 .11321 90431 686 .99241 .00758 78 18555 2247 .87890 .12109 29 89745 703 .99216 .00783 79 i63o8 21 10 .87061 .12938 3o 89042 718 .99193 .00806 80 14198 1969 .86i3i .13868 3i 88324 726 .99178 .00821 81 12229 1823 .85092 .83932 .82573 .14907 32 87598 733 .99163 .00836 82 10406 1672 .16067 33 86865 743 .99144 .00855 83 8734 l522 .17426 34 86122 754 .99124 .00875 84 7212 i36o .81142 .18857 35 85368 768 .99100 .00899 .00932 85 5852 1186 ,79733 .20266 36 84600 789 .99067 86 4666 1014 .78268 .21731 ll 838ii 811 .99032 .00967 87 3652 849 .76752 .23247 38 83000 83o .99000 .01000 88 2803 689 .75419 .24580 39 82170 844 .98972 .01027 89 2114 548 •74077 .25922 40 8i326 854 .98949 .oio5o 90 1 566 435 .72222 .27777 41 80472 860 .98931 .01068 91 ii3i 336 .70291 .29708 42 79612 869 .98908 .01091 92 It 247 .68930 .3 1 069 43 78743 888 .98872 .01127 93 181 .66970 .643o5 .33029 44 77855 913 .98827 .01172 94 36-1 i3i .35694 45 76942 948 198698 .01232 95 236 86 .63559 .36440 46 75994 989 .oi3oi 96 i5o 56 .62666 .37333 H 75oo5 1029 .98628 .01371 97 fo 44 .53191 .46808 48 73970 1067 .98557 .01442 98 33 .34000 .66000 49 72909 1102 .98488 .oi5i 1 99 17 II Vs % DO 71807 ii33 .98422 .01577 100 6 4 H % 5i 70674 1167 .98348 .oi65i 101 2 2 52 69507 1204 .98267 .01732 102 .... 53 54 683o3 67052 I25l i3o4 .98168 .98055 .oi83i .01944 Note. The abo\ e table is that of 55 65748 1358 .97934 .97804 .02065 the English Institu te of Act iiaries, 56 57 58 59 64390 1414 .02195 pre pared between 18 62andl86S , from 62976 6i5o5 59974 147 1 i53i 1601 .97664 .97510 .97330 .02335 .02489 .02669 the lea continued expei ding life insuranc 'ience of t le compan wenty ies. PROBABILITIES. 819 Pkoblem. To find the probability that a person of age a will live to age y. Solution. We take from the table the number living at age y, and divide it by the number living at age a. The quo- tient is the probabiUty. 374:, The principle on which the value of a contingent payment is determined is the following : Theorem. The value of a -prohdble payiiieut is equal to the sum to he paid, multiplied hy the pi^ohahility that it will he paid. Proof, Let there be n men, for each of whom there is a probability p that he will receive the sum s. Then by § 272, Th. I, pn of the men will probably receive the payment, so that the total sum which all will receive will probably be /j?k^. Now, before they know who is to get the money, the value of each one's share is equal. Therefore, to find this value, we divide the whole amount to be received, namely, pns, by the number of men, n. This gives ps as the value of each one's chance, which proves the theorem. Note. In this proof it is tacitly supposed that the pns dollars are as valuable divided among the pn men as divided among all n men. But this, though supposed in mathematical theory, is not morally true. Morally, the money will do more good when divided among all the men than when divided among a portion selected by chance. All gambling, whether by lotteries or games of chance, is in its total efiects upon the pecuniary interests of all parties a source of positive disadvan- tage. This disadvantage is treated mathematically by more advanced methods in special treatises. EXERCISES. . Find from the table th e probabilities 5 that a person a. Aged 30 will live to 70. I, i6 30 6i (i 80. c. ii 50 (i i( 60. d. (C 60 (( ii 70. . 320 PROBABILITIES. e. Aged . 70 will live to 80. /. a 80 " " 90. 9- 66 90 " " 95. h. a 95 " " 100. 2. What age is that at which it is an even chance whether a person aged 40 will be living or dead ? 3. Show that the probability that a person aged 30 will live to 70 is equal to the product of the probability that he will live to 60 multiplied by the probability that a man aged 60 will live to 70. (Apply the theorem of § 269.) 4. What premium ought a man of 65 to pay for insuring his life for $7000 for 1 year ? 5. Ten young men of 25 form a club. What is the proba- bility that it will be unbroken by death for ten years ? 6. The probability that a planing mill will burn down within any one year is -• What ought an insurance company o to charge to insure it to the amount of $3000 for 1 year, for 2 years, for 3 years, and for 4 years, respectively ? 7. If the probability that a house will burn down in any one year is ^, what ought to be the premium for insuring it for s years to the amount of a dollars ? Note. In cases like the last two, it is assumed that only one loss will be paid for. 8. What is the probability that if a man aged 25 marry a wife of 20, they will live to celebrate their golden wedding? 9. A company insures the joint lives of a husband aged 70 and a wife aged 50 for $5000 for 5 years, the stipulation being that if either of them die within that time the other shall be paid the money. What ought to be the premium, no allow- ance being made for interest ? 10. A man aged 50 insures the life of his wife, aged 35, for $10,000 for 20 years, with the promise that the money is not to be paid unless he himself lives to the age of 70. What ought to be the premium ? Note. In computations relating to the management of life insurance, it is always necessary to allow compound interest on all paym(;nts. But the above exercises are intended only to illustrate the application of the theory of probabilities to the subject, and therefore no allowance for in- terest is expected to bu made in the answers. BOOK XI. OF SERIES AND THE DOCTRINE OF LIMITS. CHAPTER I. NATU RE OF A SERIES. 275. Def. A Series is a succession of terms follow- ing each other according to some general law. Examples. An arithmetical progression is a series deter- mined by the law that each term shall be greater than the preceding one by the same amount. A geometrical progression is a series subject to the law that the ratio of every two consecutive terms is the same. These two progressions are the simplest form of series. A series may terminate at some term, or it may continue indefinitely. Def, A series which continues indefinitely is called an Infinite Series. Def. The Sum of a series is the algebraic sum of all its terms. Hence the sum of an infinite series will consist of the sum of an infinite number of terms. 276. The law of a series is generally such that the n^^ term may be expressed as a function of n. For example, in the series 1.1.1.1 2+3+4 + 5 + '*"- the n^^ term is r« /J + 1 21 322 SERIES. In the series --^ + ^-^ +-3. "4 + ^^^'^ the 'nP^^ term is n {n + 1) _Z>(g/; The expression for the n^^ term of a series as a function of 7^ is called the General Term of the series. EXERCISES. Express the n^^ term of each of the following series : '• 3^+4^ + 5^ + ^*"- 2. 1-3 + 3-4 + 5-6 + etc. 3- 1 + ra + r?. 3 + '^^^^ IX ft. i)u ^ Write four terms of each of the series having the following- general terms : 4^^2 _ X 5. The n^^ term to be — -r, 7- ^ 47i^ + 1 6. The i^^' term to be «: (i. -j- 1) (^ + 2) ^. (^ + 3) (/^ + 4) 2:^+1 7. The (^ + ly^ term to be 8. The {n — ly^ term to be {n + 5) (^ + 6) 1-2 ^ 1 277. The most common nse of a series is to enable us to j compute, by approximation, the values of expressions which it is difficult or impossible to compute directly. Suppose, for 1 -{- X example, that we have to compute the value of when x is a small fraction, say — , and to have the result accurate to eight decimals. We shall see hereafter that when x is less than 1, we have CONVERGENCE OF 8EIUES. 82c = 1 -\- 2x -}- 2x^ -i- 2x^ + etc., ad infaiitum. 1—x 50 Suppose X z= — =z ,02. We compute this series thus: 1 2 X .02 =: .04 Multiplying by .02, .0008 a (e .000016 i( <£ .00000032 ^ 1.02 Sum = -^ = 1.04081632 .098 which IS much more expeditious than dividing 1 02 by .98. It will be seen that every term we add makes the quotient accurate to one or two more decimals, so that there is no limit to the precision which may be attained by the use of the series. If, however, x had been greater than unity, the series would give no result, because the terms 2x, 2x% 2x% would have gone on increasing indefinitely, whereas the true value of the frac- 1 -{- X tion would have been negative. This example illustrates the following two cases of series : I. There may he a certain limit to ivhieh the sum of the series shall approach, as we increase the numher of terms, hut which it can never reach, how great soever the number of terms added. For example, the series we have just tried, 2 2 2 2 ^ + 50 + 50-^ + 50-3 + 505 + "*''-' 1 02 approaches the limit tt-^t., but never absolutely reaches it. II. As we increase the numher of terms, the sum may increase without limit, or may vihrate hach and forth in consequence of some terms heing positive and others negative. These two classes of series are distinguished as convergent and divergent. 324 SERIES. Def, A Convergent Series is one of whicli the sum approaches a limit as the number of terms is increased. Refer to § 213 for an example of infinite series in geometrical pro- gressions whicli have limits. Def. A Divergent Series is one of which the sum does not approach a limit. Examples. The series 1-f 2-f 3 + 4-f etc., ad iJifaiihirn, IS divergent, because there is no limit to the sum of its terms. The series 1 — 1+ 1 — 1 + 1— etc., is divergent, because its sum continually fluctuates between +1 and 0. Rem. When we consider only a limited number of terms, the question of convergence or divergence is not important. But when the sum of the whole series to infinity is to be con- sidered, only convergent series can be used. Notation of Stims. 378. The sum of a series of terms represented by common symbols may be expressed by the symbol 2, followed by one of the terms. Example. The expression ^a means "the sum of several terms, each represented by «." When it is necessary to distinguish the different terms, different accents or indices are affixed to them, and represented by some common symbol. Example. The expression means the sum of several terms represented by the symbol a with indices attached ; that is, the sum of several of the quan- tities a^y ^3, «3, a^, etc. When the particular indices included in the summa- tion are to be expressed, the greatest and least of them are written above and below the symbol 2. SIGJS' OF SUMMATION, 325 Examples. The expression i=15 i=5 means : " Sum of all the symbols ai formed by giving i all in- tegral values from i = 5 to i= 15." That is, e=15 lai=zas + ^6 -r ^7 + <^8 + «9 + ^10 + «^ii + ^12 + ^13+^14 + ^15' i=5 i=5 l^irn means + m +- 2m +- 3/>2 + 4//z +- 5/^.. i=0 's (i,y) means (1,/) + (2,7) + (3,/) + (4,^). %-\ '~Hhj) = {t, 2) + {h 3) + (^, 4) + (^, 5) + {i, 6). ''i?^! — l! + 2!+-3! + 4! = 1 + 2 + 6 + 24 = 33. '~2^ = 7 + 8 + 9 +- 10 +■ 11 = 45. ' 2^2 = 22 + 32 + 42 + 5^ = 54. i=2 EXERCISES. Write out the following summations, and compute theii values when they are purely numerical : i=7 n=Q n=6 I. ^j\ 2. ^nin-^l), 3. 2^(^ + 1). i=8 n=7 «=6 4. 2wi. 5. 2n>fc. 6. l:(7^+l)(y— 1). i=4 ?i=5 «=5 ^ 2 7. ^imi, 8. 2/^%2^ 9. 2 — — -• i=2 w=2 71=0 ^ + 1 Express the following sums by the sign 2 : 10. h^+lH+h^+h^+h^. lu 13 + 23 + 33 + 48. 12. 1.2 + 2.3 + 3.4 + 4.5. 13. I + I + I + J + I 326 8EUIES. CHAPTER II. DEVELOPMENT IN POWERS OF A VARIABLE. 379. Among the most common series employed in math- ematics are those of which the terms are multiplied by the successive powers of some one quantity. An example of such a series is 1 -\- ^z -\- dz^ -{- 4^3 + 5;2^ + etc., in which each coefficient is greater by unity than the power of z which it multiplies. A geometrical progression, it will be remarked, is a series of this kind, in which the terms contain the successive powers of the common ratio. The general form of such a series is j in which the successive coefficients a^, a^, a^, etc., are formed according to some law, but do not contain z. Such a series as this is said to proceed according to the ascending powers of the variable z, Eem. The sum of a series is often equal to some algebraic expression containing the variable. Conversely, we may find a series the sum of all the terms of which shall be equal to a . given expression. Def. A series equal to a given expression is call< the Development of that expression. e^ To Develop an expression means to find a series the sum of all the terms of which are equal to the ex- pression. The most extensively used method of development is that of indeterminate coefficients. INDETEBMINATE COEFFICIENTS. 327 Method of Indeterminate Coefficients. 380. The method of indeterminate coefficients is based upon the following principles : Let us have two equal expressions, each containing a varia- ble z, and one or both containing also certain indeterminate quantities^ that is, quantities introduced hypothetically, and not given by the original problem, the values of which are to be subsequently assigned so as to fulfil a certain condition. The condition to be fulfilled by the values of the inde- terminate quantities is that the two expressions containing z and these quantities shall be made identically equal. Then, because the equations are to be identically equal, wo can assign any values we please to z, and thus form as many equations as we please between the indeterminate quantities. If these equations can be all satisfied by one set of values of these quantities, then by assigning these values to them in the original equation, the latter will be an identical one, as required. The student should trace the above general method in the following examples of its application. 281. Theorem I. // a series proceeding according to the ascending -powers of a quantity is equal to zero for all values of that quantity, the coefficient of each sepa- rate term must he zero. Proof, Let the several coefficients be a^, a^, ^g, etc., and z the quantity, so that the series, put equal to zero, is a^ -f- a^z + (^2^2 -f a^z^ + etc. = 0. Because the equation is true for all values of z, it must be true when z=zO. Putting z = 0, it becomes a^ = 0. Dropping a^, the equation becomes a^z + a2Z^ + a^z^ -f- etc. = 0. Dividing by z, a^ + a2Z + a^z^ + etc. = 0. From this we derive, by a repetition of the same reasoning, a^ = 0. 328 SERIES. Continuing the process, we find «2 = 0, ^3 = 0, etc., indefinitely. Theoeem II. // two series proeeeding hy ascending powers of a quantity are equal for all values of that quantity, the coefficients of the equal powers must he equal. Proof, Let the two equal series be aQ-\-a^z-\-a^z^-\-eiG, = Z>o+^i^+^2^^ + ^^c. {a) Transposing the second member to the left-hand side and collecting the equal powers of z, the equation becomes ^0 — ^0 + (^1 — ^i) ^ + (^2 — ^3) ^^ + etc. = 0. Since this equation is to be satisfied for all values of z, the coefficients of the separate powers of z must all be zero. Hence, ^0 ■"" ^0 = ^9 a^ —b^ = 0, a^ — lc^ = 0, etc. or a^ =: h^, a^ = b^, a^ = h^, etc. ExEECiSE. Let the student demonstrate these last equa- tions independently from (a), by supposing 2; = 0, then sub- tracting from both sides of {a) the quantities found to be equal ; then dividing by z ; then supposing ^ = 0, etc. Rem. The hypothesis that {a) is satisfied for all values of z is equivalent to the supposition that it is an identical equa- tion. In general, when we find different expressions for the same functions of a variable quantity, these expressions ought to be identically equal, because they are expected to be true for all values of the variable. Theorem III. A function of a variable can only he developed in a single way in ascending powers of the variahle. For if we should have Fz = A^-\- A^z + A^z^ + A^z^ + etc., and also Fz = B^ -{- B^z + B^z^ + B^z^ + etc., INDETERMINATE COEFFICIENTS. 329 these two series, being each identically equal to Fz, must be identically equal to each other. But, by Th. II, this cannot be the case unless we have Aq =z Bq, J-i = B^, A^ = B^, etc. The coefficients being equal, the two series are really one and the same. 383. Expansion ly Indeterminate Coefficients, The above principle is applied to the development of functions in powers of the variable. The method of doing this will be best seen by an example. 1. Develop in powers of x, JL "Y" X Let us call the coefficients of the powers of x a^, a^, etc. The series will be known as soon as these coefficients are known. Let us then suppose — - — = a^ + a^x + a^oG^ + a^o(^ + etc. J- -J— X Here we remark that, so far as we have shown, this equa- tion is purely hypothetical. We have not proved that any such equation is possible, and the question whether it is possi- ble must remain open for the present. We must find whether we can assign such values to the indeterminate coefficients, a^, a^, a 2, etc., that the equation shall be identically true. Assuming the equation to be true, we multiply both sides by 1 -^ X. It then becomes I = a^ -^ {Gq + a^) X -\- (a^ -}- a^) x^ + etc. ; or transposing 1, = t?o -r- 1 + {aQ-\-a^)x + {a^-\-a2)x^ + {a2+a^)a^ -\- etc. By Theorem I, the coefficients must be identically zero. Hence, a^ — 1 =0, which gives a^-\.a, = 0, a^ -I- a, = 0, a^^a^ = 0, etc. «o = 1; ^1 =: -«« := — 1; «2 := — «i = 1; «3 etc. = — 1; 330 SERIES. Substituting these values of the coefficients in the original equation, it becomes = 1 — x-\-x^^a^-\-o(^--- etc. 1 + ^ This same method can be applied to the development of any rational fraction of which the terms are entire functions of some one quantity. Let us, for instance, suppose m -\- nx -\- im^ o i ^ . Multiplying by the denominator of the fraction, this equa- tion gives a + 1)X ^^ mA^ + (nA^-^mA^ x + {^^pA^-^nA^ +mA^)x^ + (pA 1 + /^^ 2 + mA s)x^ -{■ etc. We now see that when i > 1, the coefficient of x^ in this equation is mAi + nAi_t + pAi-2' Equating the coefficients of like powers of x, mAr. = a, whence Ar. = ~; mA^ + tiAq = h, " mAc^ + nA^ +P^o = ^^ " VI A ^ 4- nA^ +pA^ = 0, ^^ We have from the general coefficient above written, when Ai= Ai-t — ^Ai-2. m m That is, each coefficient after the second is the same linear function of the two coefficients next preceding. Such a series is called a Recurring Series. EXERCISES. Develop by indeterminate coefficients : 1 _1 ^' 1 — x ^' 1 — 2^' ^1 — m — ""a ? ^2 = — P. m ^0- m ^^ ^3 = — L m ^1- m ^ UNDETERMINED MULTIPLIERS, 331 1—X 1 -\- X 1 + X 1 + x 4. 6. 8. 1-x 1-x 1 4_ 2a: -f 3^:2 1 — 2^ + S^^ l — 2x-\-x^ 1-x 1 + 2:z; + dx^ I ^x — x^ 383. The development of a rational fraction may also be effected by division, after the manner of §§ 96, 97, the opera- tion being carried forward to any extent. Example. Develop j-^ — -L — X 1+X \l — x 1—X 2x 2x — 2x^ 1 + ^^ + 2x^ + 2x^ + etc. 2a;2 + 2x^ — 2x^ 2x^, etc. EXERCISES. Develop by division the expressions : 1 — 2^ 1 -\-x I. 1 -\- X 1 —X -\- x^ 384. Elimination hy Undetermined Multipliers, There is an application of the method of undetermined coefficients to the problem of eliminating unknown quantities, which merits special attention on account of its instructiveness. Let any system of. simultaneous equations between three unknown quantities be ax -\- ly -\- cz =^ Ji, (1) a'x -h h'y + cz = 7i', (2) a"x + b^'y + c'^z = It", (3) Can we find two such factors that, if we multiply two of the equations by them, and add the results to the third, two of the three unknown quantities shall be eliminated ? 332 SERIES. This question is answered in the following way : If there be such factors, let us call them m and n. If we multiply the first equation by m, the second by 7i, and add the product to the third equation, we shall have {am + a'ji + a") x ) + {b7n + b'n + b") y ^ = hn + h'71 + h'\ (b) + {cm 4- en + c") z ) In order that the quantities ^ and ^ may disappear from this equation, we must have hn + b'n + b" = 0, cm + c'n -j- c" = 0. Since we have these two equations between the quantities m and n, we can determine their values. Solving the equations, we find : h'c" - b"c' m n = Id -Vc ' Vc - h e" be' - b'c ' These are the required values of the multipliers. Substi- tuting them in the equation {b), we find that the coefficients of 2/ and 2; vanish, and that the equation becomes [' a{b'e" — b"e') -i-a' {b"e - be") bd^ b'c "^ "^ k {b'c" - b"e') + It' {b"e - be") ^ ^„ Clearing of denominators and dividing by the coefficient of X, we find __ h {b'c" - b"e') + h' {b"c - be") + li" {be' - b'e) ^ - a {b'e" - b"c') + a' {b"c - be") + a" {be' - b'e) EXERCISES. I. Find the values of y and z by the above process for finding x. MULTIPLICATION OF SERIES. 333 For this purpose we may begin with the equation (&) and find values of m and n such that the coefficients of x and z in (p) shall vanish. These values will be different from those given in {c). By substituting them in (6), X and z will be eliminated, and we shall obtain the value ofy. We then find a third set of values of m and n^ such that the coeffi- cients of X and y shall vanish, and thus obtain the value of z. 2. Solve by the method of indeterminate multipliers the exercise 3 of § 140. Multiplication of Two Infinite Series. 284a. Problem. To express the product of the two series and Iq + d^x + Ic^x^ + h^x^ + etc. The method is similar to that by which the square of an entire function is formed (§ 173, 2). We readily find the first two terms of the product to be The combinations which produce terms in x'^ are a^b^x^ + a^l^x^ -f a^h^x^. Those which produce terms in x^ are ^0^3^ + ct'J)^x^ + ^2^1^ + a^l^x^. In general, to find the terms in x'^ we begin by multiplying Gq into the term InX^ of the lower series, and then multiplying each succeeding of the first series by each preceding term of the second, until we end with anb^x'^. Hence, if we suppose Product = Aq -\- A^x -\- A^x^ -]-.,, , + AnX"^ + etc., we shall have, for all values of n, An = a^bn + a^bn-i + a^bn-% + ....+ anb^. By giving /I all integral values, we shall form as many values as we choose of J^, and so as many terms as we choose of the series. 334 SERIES. EXERCISES. 1. Form the product of the two series: ^ x^ oc^ ofi , ^ "~ 2! + 4! ~ 6l + ^^""'^ /y»3 /y^ '^-1 tAj tlU >Aj . ^ ~~ 3~! "^ 5l "~ 7l "^ 2. Form the square of each of these series. 3. Can you, by adding the squares together, show that their sum is equal to unity, whatever be the value of ir? To effect tins, multiply each coefficient of x'^ in the sum of the squares by n\ , substitute for each term its value CT given in § 257, and apply § 262, Th. II. 385, Series proceeding according to the Powers of Two VariaUes. Such a series is of the form in which the products of all powers of x and y are combined. By collecting the coefScients of each power of x, the series will become + (^0 + (^\y + ^2^^ + <^3^^ + — )^^ + etc.^ etc., etc., etc. Hence, the series is one proceeding according to the powers of one variable, in which the coefficients are themselves series, proceeding according to the ascending powers of another variable. Let us have the identically equal series proceeding accord- ing to the ascending powers of the same variables, + etc., etc., etc. Since these series are to be equal for all values of x, the coefficients of like powers of x must be equal. Hence, 8EBIE8. 335 «o + f^iV + «s^^ + etc. = ^0 + ^iV + -^32/^ + e^c- ^0 + ^1^ + ^3^^ 4- etc. = ^0 + ^1^ + B^y^ + etc. etc. etc. Again, since these series are to be equal for all values of y, we must have «o — ^0^ «i == A^, (^2 = ^3^ etc, d, = B,, h = ^1, ^, = B,, etc. etc. etc. etc. Hence, in order that two series proceeding according to the ascending powers of two variables may he identi- cally equal, the coefficients of every liJce product of the powers must he equal. 336 SERIES. CHAPTER III. SUMMATION OF SERIES. Of Figiirate Numbers. 386. The numbers in the following columns are formed according to these rules : 1. The first column is composed of the natural numbers, 1, 2, 3, etc. 2. In every succeeding column each number is the sum of all the numbers above it in the column next preceding. Thus, in the second column, the successive numbers are : 1, 1 + 2 = 3, 1 + 2 + 3 = 6, 1 + 2 + 3 + 4 = 10, etc. In the third column we have j 1, 1 + 3=4, 1 + 3 + 6 = 10, etc. ' 1 (^) 1 3 3 1 1 3 4 1 G 5 1 4 10 6 10 15 7 5 15 20 35 21 6 21 35 7 etc. etc. etc. It is evident from the mode of formation that each number is the difference of the two numbers « next above and below it in the col- ^ ^ umn next following. # • • The numbers 1, 3, 6, 10, etc., in • • • • the second column are called trian- • • • • • gular numbers, because they repre- iv'^ 1+2+3+4+5. SERIES, 337 sent numbers of points which can be regularly arranged over triangular surfaces. The numbers 1, 4, 10, etc., in the third columns are called pyramidal numbers, because each one is composed of a sum of triangular numbers, which being arranged in layers over each other, will form a triangular pyramid. All the numbers of the scheme are called figurate num- bers. The numbers in the i^^ column are called figurate numbers of the i^^^ order. 387. If we suppose a column of I's to the left of the first column, and take each line of numbers from left to right in- cUned upward, we shall have the successive lines 1, 1 ; 1, 2, 1 ; 1, 3,3, 1, etc. These numbers are formed by addition in the same way as the binomial coefficients in § 171, 2. We may therefore conclude that all the numbers obtained by the pre- ceding process are binomial coefficients, or combinatory expres- sions. This we shall now prove. Theorem. Tlie n^^ nuinher in the i^^ column is equal to CT'~' or to y^(7^4-l)^^ + 2) (^ + / - 1) 1.2-3.... I * ^^ Proof. Because the com.binations of 1 in any number arc equal to that number, we have, when i — 1, n^^ number in 1st column =: 7^ = Ci, which agrees with the theorem. When i = 2, we have, by the law of formation of the numbers, n^^ number in 2d column = 6'i + Ci + (7i + . . . . + Ci, which, by equation (a) (§260, 3), is equal to Cg^ . Therefore the successive numbers in the second column, found by supposing 7^ = 1, 7^ = 2, etc., are Cl G%, G%,.... (fl'\ 33 338 SERIES. Since the n*^ number in the third column is equal to the sum of all above it in the second, we have n^^ number in 3d column ^ Cl^-Cl+G\+ CV^ = C^^ which still corresponds to the theorem, because, ^hen i =: 3^ n -\- i — 1 z=z n -^ 2- To prove that the theorem is true as far as we choose to carry it, we must show that if it is true for any value of t, it is also true for a value 1 greater. Let us then suppose that, in the r^^ column the first n numbers are j^r ryr+1 ryr+2 pT+n—l Since the n^^ number in the next column is the sum of these numbers, it will be equal to ^ I which is the expression given by the theorem when we suppose i = r -^ 1. Now we have proved the theorem true when i = 3; there- •fore (supposing r = 3) it is true for i = 4. Therefore (sup- posing r = 4) it is true for i =: 5, and so on indefinitely. If in the general expression (1) we put i = 2, we shall have the values of the triangular numbers ; by putting i = 3, ^ we shall have the pyramidal numbers, etc. Therefore, 71 {n + 1) The n^^ triangular number The n^ pyramidal number 1.2 Qi {n + 1) {n + 2) 1.2-3 By supposing ^ = 1, 2, 3, 4, etc., in succession, we find the succession of triangular numbers to be 1-2 2.3 4.5 1.2^ 1.2' 1.2' ^^^'' and the pyramidal numbers, 1.2.3 2.3.4 3.4.5 1.2.3' 1.2-3' 1.2.3' ' which we readily see correspond to the values in the scheme (A). 8EBIE8. 339 Enumeration of Triangular Piles of Shot. 388. An iuteresting application of the preceding theory is that of finding the number of cannon-shot in a pile. There are two cases in which a pile will con- tain a figurate n umber \ I. Elongated projectiles, in which each rests on two projectiles below it. II. Spherical projectiles, each rest- ing on three below it, and the whole forming a pyramid. Case I. Elongated Projectiles, Here the vertex of a pile of one vertical layer will be formed of one shot, the next layer below of two, the third of three, etc. Hence the sum of n layers from the vertex down will be the n^^ triangular number. It is evident that the number of shot in the bottom row is equal to the number of rows. Hence, if 7i be this number, and N the entire number of shot in the pile, we shall have, _ n {n + 1) 2 ' If the pile is incomplete, in consequence of all the layers above a certain one being absent, we first compute how many there would be if the pile were complete, and subtract the number in that part of the pile which is absent. Example. The bottom layer has 25 shot, but there are only 11 layers in all. How many shot are there? 25*26 K the pile were complete, the number would be — - — There being 14 layers wanting from the top, the total number 14*15 of shot wanting is —^ — Hence the number in the pile is til _ 25-26 — 14-15 _ (14 + 11)(15 + 11) — 14-15 2 ~ 2 = ^m^ + i^ + n) ^ ^^^^ 340 SERIES. Note. This particular problem coald have been solved more briefly by considering the number of shot in the several layers as an arithmetical progression, but we have preferred to apply a general method. EXERCISES. 1. A pile of cylindrical shot has 7i in its bottom row, and r rows. How many shot are there ? 2. From a complete pile having h layers, s layers are re- moved. How many shot are left ? 3. A pile has n shot in its bottom row, and m in its top row. How many rows and how many shot are there? 4. A pile has p rows and Tc shot in its top row. How many shot are there ? 5. Explain the law of succession of even and odd numbers in the se- ries of triangular numbers. 6. How many balls are necessary to fill a hexagon, having n balls in each side ? Note. In the adjoining figure, 389. Ca^se II. Pyramid of Balls, If a course of balls be laid upon the ground so as to fill an equilateral triangle^ having n balls on each side, a second course can be laid upon these having n — \ balls on each side, and so on until we come to a single ball at the vertex. Commencing at the top, the first course will consist of 1 ball, the next of 3, the third of 6, and so on through the tri- angular numbers. Because each pyramidal number is the sum of all the preceding triangular numbers, the whole num- ber of balls in the n courses will be the n^ pyramidal number, or n{n + \)(n + ^) ^ ~" 1.2.3 EXERCISES. I. How many balls in a triangular pyramid having 9 balls on each side ? 8ERIE8. 341 2. If from a triangular pyramid of n courses Jc courses be removed from the top, how many balls will be left ? 3. How many balls in the frustum of a triangular pyramid having n balls on each side of the base and m on each side of the upper course ? Sum of the Similar Powers of an Arithmetical Progression. 390. Put «!, the first term of the progression ; di the common difference; n, the number of terms ; m, the index of the power. It is required to find an expression for the sum, < + («i + d)"^ + (^1 + ^^)'^ + + [«^i + (^ - 1) clY", which sum we call Sm- Let us put, for brevity, a^, a^, a^, a^, , . , . an for the sev- eral terms of the progression. Then «2 = ^1 + ^^ a^ =: a^ + 2d =.^2 + ^^ an = a^ + (^^ — l)d=z an-i + d. Raising these equations to the (m + iy^ power, and adding the equation an^ =z an + d, we have am+i ^ «m+i -I- (m + 1) a^d + ^H^+^lHl af-H^ + etc ^^+1 =: a^+i + (m + 1) afd + ^^^^^^ af-^d'^ + etc. ^m+i- ^ ^m+i + (^ 4. 1) a^d 4- ^^+^)^ ^m-1^2 4. etc. «m^j ^ ^m+i ^{m + 1) ay + l^i+il^ ^m-i^ + etc. If we add these equations together, and cancel the common termSj a^+i + a^+^ + . . . . -\- a^+^, which appear in both members, we shall* have 342 SERIE8. . (^ + 1 ) m (m - 1) ,30. . i 172^^ '(^^m-2, etc. From this we obtain, by solving with respect to Sm^ C:i-«r' m m{m-l) which will enable us to find Sin when we knoAv Si, S2, » * - - Sm-h that is, to find the snm of the n^^ powers when we know the sum of all the lower powers. It will be noted that S^ means the sum of the arithmetical series itself, as found in Book VII, Chap. I ; and that Sq = n, because there are 71 terras and the zero power of each is 1. By § 209, Prob. V, q). The bottom course will contain jjq balls, the next course {p — 1) (^ — 1)> etc. The total number of balls in the pile will be i\rz:.;,^ + (j,->l)(g-l)+(^-2)(^-2) + ....+(p-^ + l). (6) To find the sum of this series, let us first suppose p^=^q, and the base therefore a square. We shall then have N' = q^ -^ {q -If + {q -2f + . . . . -{-1, which is the sum of the squares of the first q numbers. Therefore, by § 291, (4), ^, ^ ^(^-fl)(2^+ l) ^^^ Next let us put r for the number by which p exceeds q in the general expression (6). This expression will then become ]V= q{q + r) + {q-l){q-l-^r) + {q-2){q-^2-{-r) + + (1 + r) == ^' -f (^ - 1)' + (^ -2)2 + . . . . + 22 + 1 + [^ + (^-l)+(^-^) + .... +1]^ = l(lill)^.±i) + iil±l) , (§ 291, 4.) _ q{q-{-l) (3r -{- 2q + 1) ~ 6 EXERCISES. 1. Find the sum of the first 20 numbers, 1 + 2 + 3+ ... . + 20, then the sum of their squares, and the sum of their cubes, by successive substitutions in the general equation (2). 2. Express the sum and the sum of the squares of the first r odd numbers, namely, 1 + 3 + 5 +.... + (2/- - 1), and 12 -f 32 + 52 + + (2r - 1)2. 3. Express the sum of the first r even numbers and the sum of their squares, namely, 2 + 4 + 6 + . . . , + 2r, and 22 + 42+62 + +(2r)2. 8ERIE8, 345 4. A rectangular pile of balls is started with a base of p balls on one side and q on the other. How many balls will there be in the pile after 3 courses have been laid ? How many after s courses ? 5. Find the value of the expression S (a + Z>a; + cx% 6. Find the value of 2 (a + te + cx^), x=l 294. To find the sum of n terms of the series 111 1 1-2 ^ 2.3 ^ 3.4 ^ ^ n{n H- 1) Each term of this series may be divided into two parts, thus : J___l_l 1 _ 1 1 1-2 "" 1 2' 2.3 ""2 3' 1 11 n{n -{- 1) n n -\-\ Therefore the sum of the series is in which the second part of every term except the last is can- celled by the first part of the term next following. Therefore the sum of the n terms is If we suppose the number of terms n to increase without limit, the fraction will reduce to zero, and we shall have ^^ + 1 jT^ + 273 + 3T4 + ^^^"> ^^ i'y^finU'iim — 1. This is the same as tlie sum of the geometrical progression, s + 7 + o i, and put, for brevity, h =zj — i, the terms may be put into the form ij h \i jr ij ic \l J/ P_ P / JL L\ - jc \i + 1 _/ + ir {^ + 1) U + 1) ^ Vi* + 1 y + etc. etc. P ^li—1 1_). k\i 4- n — 1 1 -^n—V (i -\- n — 1) (y + /^ + 1) k\i -\- n — 1 j -{- n When we add these quantities, the second part of each term will be cancelled by the first part of the ¥^ term next follow- ing, leaving only the first part of the first k terms and the second part of the last k terms. Hence the sum will be PJl + J_ + +^_ _ J: ^!__.... L__). Ic\i i-^1^ ' y+1 i-hn i + n—1 j + n—1/ SERIES. 347 Example. To find the sum of n terms of the series 1 1 1.1. . 1 2. 1 1/1 3-6 2-5 ' 3.6 ' 4.7 • 5-8 (n-\-l){n + 4:) Each term may be expressed in the form 2-5 3\2 5/' !\3 6/' 4.7 3 \4 7/' __!__= 1(1 1_) TC (w + 3) 3 Vw w + 3/' 1 = 1 (_i ^v (?z + 1) (^ + 4) 3 \/^ + 1 ?^ + 4/ Therefore, separating the positive and negative terms, we find the sum of the series to be 1/1111 1 1 3\2 3 45 n n -\-l 11 111 _ __1 l\ 5 6 n n + 1 n-{-2 n + 3 ?^-f4/' or, omitting the terms which cancel each other. 1(1 + 1 + 1 1 i_ 3\2^3^4 2^ + 2 u -\- 3 n -]- y- When n is infinite, the sum becomes 3\2 ^3^4/ 3 12 36 EXERCISES. What is the sum of n terms of the series : '- 3^ + 4^ + 5^ + ^*'^- j_ j_ j_ 1 3.5 ' 5.7^7-9^ ^ {2n + l)(2n + 3) 348 8EBIE8- 2 3- 4- 1-3 ' 2-4 ' 3-5^ 5- Sum the series 1 1 ■•■ (w + 1) (;^ + 4)' 7^ (?i + 2) ^J^T) + (« + 1) (a + 2) + (a + 2) (a 4- 3) + ^^'' ""^ ^^^' 396. To sum the series ^ = 1 + 2r + 3r2 + 4r3 + etc. Let us first find the sum of n terms, which we shall call 8n^ Then ^^ == 1 4- 2r + 3r2 + A.r^ + /^r~-l. Multiplying by r, we have r8n = r -^^ 2r2 + Sr^ + 4r4 + . . . . + ^ir'*. By subtraction, {l — r)8n = l-]-r-\-r^-\-7^ + r^^-i — nr^ = 3 :^ir'* (§ 212, Prob. V). 1 — r ^Q y / Therefore, Sn = j. (1 ^ry 1 — r Now suppose n to increase without limit. If r > 1, the sum of the series will evidently increase without limit. If r < 1, both 7'^ and nr^ will converge toward zero as n increases (as we shall show hereafter), and we shall have 1 S = (1 - ry EXERCISES. Find in the above way the sum of the following series to n terms and to infinity, supposing r < 1 : 1. G5 -f Sar -h 5ar^ -f 7ar^ .... + (2/i — 1) ar^-~^» 2. 2fl5 4- 4ar -f- 6ar^ + Sar^ -f 2nar'^--\ 3. {a + J) r -f (^ -f 2^) r2 + . . . . + (a + /^&) r^. SERIES. 349 397. Sum the series r^ + ark + ra + '*''•' <«) of which the general term is _^_j-J^— ^-. Let us find whether we can express this series as the sum of two series. Assume n{n-\-l){n -\- 2) ^ n{n-\-l) "^ (n -}- 1) {n + 2) ' where, if possible, the values of the indeterminate coefficients A and B are to be so chosen that this equation shall be true identically. Keducing the second member to a common denominator, we have 1 {A + B) n-\-2A n {n + 1) {n -f- 2) ~ n {n -\- 1) {n -{- 2)' In order that these fractions may be identically equal, we must have {A -]- B) n + 2A =z 1, identically y which requires that we have (§ 281), A -{- B = 0, 2A = 1. This gives A = -, i5 == — -• Therefore, 1 1111 n{n + l){n-\-2) 2n{n-\-l) 2 {n + 1) {n + 2)' 60 that each term of the series (a) may be divided into two terms. The whole series will then be We see on sight, that by cancelling equal terms, the sum of n terms is ^^ 1 1 On = -J 4 2{n + l){7i + 2)' and the sum to infinity is j* 350 SERIES. 398. Consider the harmonic series 1 1 1 X 1 + ^ + 3+j + etc., of which the oi^^ term is — This series is divergent, because we may divide it into an unlimited number of parts, each equal to or greater than -, as follows: 1st term = !,>-; 2d term = ^; 3d and 4th terms > ^ ; etc. etc. In general, if we consider the n consecutive terms, the smallest will be ^ , and therefore their sum will be greater than :r- X ^, that is, greater than -• Zn Z ]^ow if in {a) we suppose n to take the successive values, i, 2, 4, 8, 16, etc., we shall divide the series into an unlimited number of parts of the form {a), each greater than ^- There- fore, the sum has no limit and so is divergent. Of Differences. 299. When we have a series of quantities proceeding ac- cording to any law, we may take the difference of every two consecutive quantities, and thus form a series of differences. The terms of this series are called First Differences. Taking the difference of every two consecutive differences, we shall have another series, the terms of which are called Second Differences. The process may be continued so long as there are any dif- ferences to write. SERIES. 351 Example. In the second column of the following table are given the seven values of the expression ^ _ 10a;3 ^ 30^2 _ 40:c + 25 = (px, for X = 0,1, 2, 3, 4, 5, 6. In the third column a' are given the differences, 6 — 25 = — 19, 1 — 6 :^ ~ 5, — 14 — 1 = — 15, etc. In column a" are given the differences of these differences, namely, _ 5 _ (_ 19) r= + 14, _ 15 — (- 5) = — 10, etc. X (j>X A' A'' A"' Aiv Av +25 — 19 1+6 +14 — 5 —24 2+1 —10 +24 — 15 3—14 —10 +24 — 25 +24 4 _ 39 +14 +24 — 11 + 48 5 _ 50 +62 + 51 6 '+ 1 The process is continued to the fourth order of differences, * which are all equal, whence those of the fifth and following orders are all zero. It will be noted that the sign of each difference is taken so that it shall express each quantity 7ninus the quantity next preceding. We have therefore the following definitions : 300. Def. The First Difference of a function of any variable is the increment of the function caused by an increment of unity in the variable. The Second Difference is the difference between two consecutive first differences. ■ In general, the n*^' Difference is the diflference be- tween two consecutive {n — ly diflferences. 352 8EBIE8. To investigate the relation among the differences, let ua represent the successive numbers in each column by the indices 1, 'Z, 3, etc., and let us put A^, Ag, A3, etc., for the values of ^x. We shall then have the following scheme of differences, in which a; = Aj — Ao, a; = Ag - Ai, a; = A3 - Ag ; a; = a;-a;, a';=:.a;-a;, a; = a;-a;; a: = a;-a;, a- = a;-a';, a;:=a;-a- etc. etc. etc. the w'* order of differences being represented by the symbol A ■with n accents. Ao Ai A'; ^3 • a; A' A"' ^2 ^1 ^0 /// 1 An-1 Let us now consider the following problem : To express Ai in teriJis 0/ Aq, Aq? Aq? etc. We have, by the mode of forming the differences, Ai =z Ao + a;, a; = a'o + Ao, Ai = A^ + a'o, etc. {a) Ag = Ai + a;, Ag =: Ai + Ai, Ag = Aj + a''' etc. If in this last system of equations, we substitute the values of A^, Ai, etc., from the system (a), we have Ag = Ao + 2a; 4- a;, a; = a; + 2a; + a';, etc. (h) Again, A3 ::=. Ag -h a;, a; = a; + a';, a; = a; + Ag, etc. DIFFERENCES. 353 Substituting the values of Ag, Ag, etc., from (5), we have A3 = A„ + 3a'o + a; + a; + 2a; + a; or A3 = A„ + 3a; + 3a; + a;' (c) As = A„ + 3a; + 3a; + a;' A's = a; + 3a:+ a; + a; + 2a;'+a'j a; = a; + 3a; + 3a7 + aj Forming A^ = A3 + A3, etc., we see that the coefficients of Aq, Ao, etc., which we add, are the same as the coefficients of the successive powers of x in raising 1 + a; to the rf^ power by successive multiplication, as in § 171. That is, to form A^, A'^, etc., the coefficients to be added are 13 3 1 1 3 3 1 14 6 4 1 and these are to be added in the same way to form A 5, and so on indefinitely. Hence we conclude that if i be any index, the law will be the same as in the binomial theorem, namely, Ai = Ao + ^'a'o + (I) a; + (I) a;' + etc. ) A;-A;+tA; + (|)A: + (|)A- + etc.) To show rigorously that this result is true for all values of 1, we have to prove that if true for any one value, it must be true for a value one greater. Now we have, by definition, whatever be ^, Ai+i = Ai + Ai, Ai^i = A^ + Ai', etc. Hence, substituting the above value of A^ and Ai, Ai+i = Ao + (i + 1) a; + [(I) + ^j a; 23 354 8EB1E8. We readily prove that (i)--=m. (i)+(i)=m. etc. etc. Substituting these values in {e), the result is the same given by the equation {d) when we put ^ + 1 for L The form {c) shows the formula to be true for i = 3. Therefore ifc is true for i = 4. Therefore it is true for i =z 5, etc., indefinitely. EXAMPLES AND EXERCISES. I. Having given Aq = 7, A^ = 5, Aq = — 2, and A'", A'", etc. =: 0, it is required to find the values of A^, A^, A3, etc., indefinitely, both by direct computation and by the formula (d). We start the work thus: The numbers in column A'' are all equal to — 2, because A''' = 0. Each number in column A' after the first is found by adding A'' or — 2 to the one next above it. Each value of Aj is then obtained from the one next above it by adding the appropriate value of A^ . This process of addition can be carried to any extent. Continuing it to i = 10, we shall find Ajo = —38. Next, the general formula (d) gives, by putting A^ = 7, A'jj = 5, A'^ = — 2, and all following values = 0, A, = 7 + 5^-3^1), and the student is now to show that by putting i — 1, i = 2, etc., in this expression, we obtain the same values of Ai, Ag, A3, .... Aio, that we get by addition in the above scheme. It is moreover to be remarked that we can reduce the last equation to an entire function of i, thus : Ai = 7 + 6^* - i\ % Ai Ai At 7 + 5 1 + 12 + 3 -% 2 + 15 + 1 — 2 3 etc. — 1 — 2 4 etc. -2 etc. etc. DIFFERENCES. 355 2. Having given A^ — 5, A'^ = —20, A'^ = —30, A'^' = + 9, it is required to find in the same way the values of Aj to A5, and to express Ai as an entire function of i by formula {d). 3. On March 1, 1881, at Greenwich noon, the sun's longi- tude was 341° 5' 10". 9 ; on March 2 it was greater by 1° 0' 9".0, but this daily increase was diminishing by 2" each day. It is required to compute the longitude for the first seven days of the month, and to find an expression for its value on the n^^^ day of March. 4. A family had a reservoir containing, on the morning of May 5, 495 gallons of water, to which the city added regularly 50 gallons per day. The family used 35 gallons oh May 5, and 5 gallons more each subsequent day than it did on the day preceding. Find a general expression for the quantity of water on the n*^ day of May ; and by equating this expression to zero, find at what time the water will all be gone. Also ex- plain the two answers given by the equation. Theorems of Differences. 301. To investigate the general properties of differences, we use a notation slightly difibrent from that just employed. If u be any function of x, which we may call ^Xy so that we put U = (j)X, then A^ = (a; -f 1) — 1,000, X > 100,000, X > 10,000,000, etc., to make — less than a, X In the language of limits, the above theorem is expressed thus : The limit of - , when x is indefinitely increased, is zero. Theorem II. If a fraction have any finite numerator, and an independent variable for its denominator^ zve may assign to this denominator a value so small that the fraction shall exceed any quantity, hoivever great, which we may assign. Proof. Put as before - for the fraction, and let A be any number however great, which we choose to assign. Let 71 be a number greater than A. Divide a into n parts, and let a be one of these parts ; then a = na. Consequently, a Therefore, if we take for x a quantity less than a, we shall have ^>n> A, X or i>'- Eem. If we have two independent variables, x and y: We may make x any number of times greater than y. 360 LIMITS, Then we may make y any number of times greater than this value of x. Then we may make x any number of times greater than this value of y» And we can thus continue, making each variable outstrip the other to any extent in a race toward infinity, without either ever reaching the goal. Theorem III. // h he any fixed quantity, however great, and a a quantity ivhieh zve may mahe as small as lue please, lue may mahe the product ha less than any assignable quantity. Proof, If there is any smallest value of ha, let it be s. Because we may make a as small as we please, let us put Multiplying by h, we find ha < s. So that ha may be made less than s, and s cannot be the smallest value. Def, The Limit of a variable quantity is a value which it can never reach, but to which it may approach so nearly that the difference shall be less than any assignable quantity. Rem. In order that a variable X may have a limit, it must be a function of some other variable, and there must be certain values of this other variable for which the value of X cannot be directly computed. EXAMPLES. I. The value of the expression ^ x^ — a^ X — a can be computed directly for any pair of numerical values of x and a, except those values which are equal. If we suppose x^=ia, the expression becomes LIMITS. 361 a — a 0' which, considered by itself, has no meaning. 2. The sum of any finite number of terms of a geometrical progression may be computed by adding them. But if the number of terms is infinite, an infinite time would be required for the direct calculation, which is therefore impossible. 3. The area of a polygon of any number of sides, and hav- ing a given apothegm, may be computed. But if the number of sides becom.es infinite, and the polygon is thus changed into a circle, the direct computation is not practicable. EXERCISE. ^x 8 If we have the fraction, X = ~ , show that we may 7 1 make x so great that X shall differ from - by less than -— , less than _-^, lesss than j^-^-, and so on indefinitely. Notation of the Method of Limits. 304. Put X, the quantity of which the value is to be found ; X, the independent variable on which X de- pends, so that Xis a function of ic; «, the particular value of x for which we can- not compute X; X, the limit of X, or the value to which it approaches as x approaches to a. Then the limit L must be a quantity fulfilling these two conditions : 1st. Supposing X to approach as near as we please to a, we must always be able to find a value of x so near to a that the difference L -— X shall become less than any assignable quan- tity. 2d. X must not become absolutely equal to L, however near x may be to a. 362 ' LIMITS. Eem. The quantity a, toward which x approaches, may be either zero, infinity, or some finite quantity. Example i. Suppose Y — ^ '^ ^ ^ X — a By § 93, this* expression is equal to x^ + ax + dj^, {a) except when x:=za. But suppose ^ to be the difference be- tween x and a^ so that a; = « + (J. Substituting this value in the expression (a)^ the equation becomes /TflS __ /y3 z=z 3a2 + Sad + ^2. x — a Now we may suppose 6 so small that daS -\- 6^ shall be less than any quantity we choose to assign. Hence we may choose ry'S /vS a yalue of x so near to a that the value of shall differ X — a from 3^2 by less than any assignable quantity. Hence, if ^ _ x^ — a^ x — a ^ then L =z 3a% oj3 __ ^3 or Sa^ is the limit of the expression as x approaches a. X — • Ctr X Ex. 2. The limit of -, when x becomes indefinitely great, is unity. For, subtracting this expression from unity, we find the difference to be 1 X + 1 By taking x sufficiently great, we may make this expression less than any assignable quantity. (§ 303, Th. I.) Therefore, X approaches to unity as x increases, whence unity is its X -p J. limit. LIMITS. 363 Notation, The statement that L is the limit of X as x approaches a is expressed in the form Lim. X(a..=.a) = L. The conclusions of the last two examples may be ex- pressed thus : ^3 qZ rt* Lim. ix=a) =z Sa^, Lim. (x=^) = 1. . X — a X -\- 1 Rem. This form of notation is often used for the follow- ing purpose. Having a function of x which we may call X, the form X(x=a) means, " the value of X when x:= a." EXAMPLES. (x^ + a)(x=:a) = a^ -\- a. {x^ — a^)(x=^a) = 0. If we require the limit of a fraction when both terms be- come zero or infinite, divide both terms hy some commorb factor which hecomes zero or infinity. Rem. If the beginner has any difficulty in understanding the pre- - ceding exposition, it will be sufficient for him to think of the limit as simply the value of the? expression when the quantity on which it de- pends becomes zero or infinity. For instance, Lim. ~ (a; = oo ), the value of which we have found to be unity, may be regarded as simply the value of the expression, oo GO + 1* Although this way of thinking is convenient, and generally leads to correct results, it is not mathematically rigorous, because neither zero nor infinity are, properly speaking, mathematical quantities, and people are often led into paradoxes by treating them as such. EXERCISES. Find the limit of I. when X approaches infinity. Divide both terms by x. 2. - — -— when X approaches infinity. uix? 3. — ^ when X approaches infinity. 364 LIMITS, \ X 4. z. — when X approaches infinity. /v2 r^ 5. ' when X approaches a, X — €t 6. when x approaches infinity. a — x ^^ Properties of Limits. 305. Theorem I. // tivo functions are equal, they must hajve the same limit. Proyf. If possible^ let Z and L be two different limits for the respective functions. Put SO that L and V differ by 2^;. Because Z is the limit of the one function, the latter may approach this dmifc so nearly as to differ from it by less than z. In the same way, the other function may differ from L by less than z. Then, because L and L differ by %z, the func- tions would differ, which is contrary to the hypothesis. Theorem II. The limit of the sum of several func- tions is equal to the sum of their separate limits. Proof. Let the functions be X, X', X", etc. Let their limits be L, L\ L'\ etc. Let their differences from their limits be a, a', a'\ etc. Then X = L — a, X' = L' -- a\ X" = Z" - a", etc. etc. Adding, we have .X+X' + X" + etc. = L-{-L' + r'-{-etc.-{a + a' + a'' + etc.) The theorem asserts that we may take the functions so near their limits that the sums of the differences a -\- a' -\- a" -^ etc, shall be less than any quantity we can assign. LIMITS. 365 Let h be this quantity, which may be ever so small ; n, the number of the quantities a, a', a", etc. ; a, the largest of them. Because we can bring the functions as near their limits as we please, we may bring them so near as to make k a < - y or 7ia < h n Then «+«'-j-a"4-etc. < na (because a is the largest) ; whence, « + cc' + «" 4- etc. < h. Therefore the sum X+X' + JT^ + etc. will approach to the sum L -\- L' + L" -^ etc., so as to differ from it by less than L Because this quantity Jc may be as small as we please, L-{-L'-^L"-{- etc. is the limit of X+ X' + X" + etc. Theorem III. The limit of the product of two func- tions is equal to the product of their limits. Proof, Adopting the same notation as in Th. II, we shall have XX' = LL' -aU -a'L + aa'. Because L and L' are finite quantities, we may take a and «' so small that aL' -^a' L — aa shall be less than any quan- tity we can assign. Hence XX' may approach as near as we please to LL', whence the latter is its limit. Cor. 1. The limit of the product of any number of functions is equal to the product of their limits. Cor. 2. The limit of any power of a function is equal to the power of its limit. Theorem IY. TJ%e limit of the quotient of tiuo func- tions is equal to the quotient of their limits. Proof, Using the same notation as before, we have for the quotient of the functions, X' L' — a u while the quotient of their limits is -^-^ 366 LIMITS. The difference between the two quotients is L' U — ex! La — La L L — a L{L — a) If L is different from zero, we may make the quantities a and a' so small that this expression shall be less than any quantity we choose to assign. Therefore, - is the limit of L — a . ^ A. —p , that IS, OT ^^' L — a X jQn qti 306. Problem. To find the limit of as x 7 X — a approacries a. Case I. When n is a positive whole number. We have from § 93, when x is different from a, A x^ a"' ^ x — a Now suppose X to approach the limit a. Then x'^-'^ will approach the limit a^"'^, x^~^ the limit a^~% etc. Multiplying hy a,a^, etc., we see that each term of the second member approaches the limit a^~K Because there are n such terms, we have ^. x^ — a^ Lim. — — : — (x=a) = na^~\ X — a Case II. WJien n is a positive fraction. Suppose n =i~, p and q being whole numbers. Then x^ — a'^ x^ — a^ x — a X — a Let us put, for convenience in writing, x"^ =z y, (f =z h'^ then X ■= 'f, a — 1)^\ x^ — a^ _y^ — If _ y — b 1^^^^ ~ '"f^^^^ ~ "FEZ ' y-b LIMITS. 367 As X approaches indefinitely near to a, and consequently y to b, the numerator of this fraction (Case I) approaches to pl)p~^ as its limit and the denominator to ql)Q~^. Hence, the fraction itself approaches to phP-^ _ p qb^-^ q bP-^, Substituting for b its value a^, we have Lim. — ^^ ix^a) = ^ bP-Q = ^-a? = ^ J~' X — a q q q Hence the same formulae holds when n is a positive fraction. Case III. When n is negative. Suppose n = —p,p itself (without the minus sign) being supposed positive. Then \ X ^ a / = = x-P arP — — x~P a'P - X — a When X approaches a, then x~p approaches arP, and a^ — a^ approaches ^a^~^ Substituting these limiting values, X d we have X^ nj^ Lim. {x=a) z=z — a'^PpaT^^ = — parP^K X — a ^ ^ Substituting for — p its value /^, we have /jfjfl /77l Lim. (cc=a) = na^~^. X ^ a Hence, Theorem. The formulce x^ — oP- ^ . Lim. ix=d) = naP''^ X -- a is true for all values of n, whether entire or fractional, positive or negative. 368 BINOMIAL TUEOUEM CHAPTER V. THE BINOMIAL AND EXPONENTIAL THEOREMS. i The Binomial Theorem for all Values of the Exponent. 307. We have shown in §§ 171, 264, how to develop {l-\-xY when ^ is a positive whole number. We have now to find the development when n is negative or fractional. Assume (1 + xY = -^0 + B^x + B^x^ + B^a^ + etc., {a) Bq, B^y etc., being indeterminate coefficients. Because this equation is by hypothesis true for all values of x, it will remain true when we put another quantity a in place of x. Hence, (1 -{-aY = B^ -i- B^a + B^a? + B^a^ + etc. {b) Subtracting {V) from (a), and putting for convenience X =Z 1 -{- X, A = 1 -\- CCy the difference of 'the two equations (a) and (b) will be X''-A'' = B^ {x - a) + B., {x^ - a^) + B^ {x^ - a^) + etc. The values we have assumed for Xand A give X— A =: X — a. Dividing the left-hand member by X -- A, and the right- hand member by the equal quantity x — a, we have -X^TX- = ^1 + ^. -^ZTa + ^3 ^— ^ + etc. Now suppose X to approach a. The limit of the left-hand member will be nA^'K Taking the sum of the corresponding* limits of fche right-hand member, we shall have nA%-^ = B^ + 2B^a + dB^a^ -f 4.B^a^ + etc. Eeplace A by its value, 1 + a, and multiply by 1 + a^ We then have BINOMIAL THEOREM. 369 ^(1 + aY = ^1 (1 + «) + 2^3«(1 + «) + 3Z?3a2(l + a) + 4i?4a3 (1 _|. ^) + etc. = J5i + (^, + 2i?2) a + i2B, + 3^3) a^ + (3^3 +4i?Ja3 4-eto. Multiplying the equation (b) by ^, we have /2 (1 + a)^ = Ti^o + ^-^1^ + ^^2^2 + iiB^a^ Equating the coeflBcients of the like powers of a in these equations (§ 281), we have, first, B^ = 7iB^. By putting a = in equation (&), we find Bq = 1, whence Then we find successively, 2B^ = (n-l) Bj, whence B^ = -^ B^ = !i(^ll. 3B, = in-2)B,, " S3='^'^.=^^^|='-^. Substituting these values of Bq, B^, B^, etc., in the equa- tion {a) and using the abbreviated notation, w^e obtain tho equation (1 + xY = l + nx-\-(^'^)x^ + (I) x^ + etc., (c) which equation is true for all values of n, 308. There is an important relation between the form of this development when niQ a, positive integer, as in §§ 171 and 264, and when it is negative or fractional. In the former case, when we form the successive factors n — 1, n — 2, u — 3, etc., the ?i^^ factor will vanish, and therefore all the coefficients after that of x^ will vanish. But if n is negative or fractional, none of the factors n — 1, 71 — 2, etc., can become zero, and, in consequence, the series will go on to infinity. It therefore becomes necessary, in this case, to investigate the convergence of the development. If i«; > 1, the successive powers of x will go on increasing indefinitely, while the coefficients (-), (^), etc., will not go 24 370 BINOMIAL THEOEEM. on diminisliing indefinitely in the same ratio. For, let us consider two successive terms of the development^ the (^ + 1)*^, and the {i 4- 2)"^, namely. The quotient of the second by the first is (n \ /n\ n — i As I increases indefinitely, tliis coefficient of x will approach the limit — 1 (§ 304), while x is by hypothesis as great as 1. Therefore, by continuing the series, a point will be reached from which the terms will no longer diminish. Therefore, | The development of {I + xy^ in powers of x is not con- vergent unless a; < 1. j In consequence, if we develop {a -f b)^ when n is negative or fractional, we must do so in ascending powers of the lesser of the two quantities, a or b, EXAM PLES. I. Develop (1 + x)^, or the square root of 1 + a;. Putting 7Z = ^, we have © = r (1) = 2V2-') M 1/1 2_ 1.2 ^ ""2.4' 1.2.3 ~" 2.4.6' - ^- 1.1.3.5 in\ __ 2 /^\ _ _ ll W "" 4 \3/ "" 2T4.6.8 etc. etc. etc. BINOMIAL THEOREM. 371 Whence, /■. , ^i 1,1 1-1^ . 1-1-3 , 1-1-3.5 , , ^ If iz; is a small fraction, the terms in x^, a^, etc., will be much smaller than - x itself, and the first two terms of the series will give a result very near the truth. We therefore conclude : Tl%e square root of 1 plus a small fraction is approxi- mately equal to 1 plus half that fraction. 2. To develop a/10. We see at once that VlO is between 3 and 4. We put 10 in the form 3^ + 1=^32(1 + 1), when VlO = 3 (l + ^^ Then, by the development just performed, l28""9^ + ^^''- (l+J) -l + 2'9- . 1-4- 1 8.92 16-93 We now sum the terms : 1st term. 1.0000000 2d " = 1st ~ 18, ... . + .0555556 3d " :::3 2d ^ - 36, . . . - .0015432 4th " := 3d -^ 18, . . . + .0000857 5th " = 4th X — 5-r-72, . — .0000060 6th " = 5th X —7-^90, . + .0000005 Sum = (1 + ^ = 1.0540926 lence, VlO r= 3 X sum = 3.1622778 which may be in error by a few units in the last place, owing to the omission of the decimals past the seventh. 372 EXPONENTIAL THEOREM, 3. To develop VS. We see that 3 is the nearest whole number of the root. So we put V8 = V(3^-l)=^3^(l-J-)=3(l-J)*, from which the development may be effected as before. EXERCISES. 1. Compute the square root of 8 to 6 decimals, and from, it find the square root of 2 by § 183. 2. Develop (1 — x)^. 3. Develop (1 — x)~^y and express the term in xK ^ 1 1-3 , 1.3.5 , • ^+2^ + 2:4^+2:4:6'^^ +^^^- Term in cc* = 2.4.6 M 4. Develop y and express the general term. (1 + xy 5. Develop (1 + -) and express the general term. 6. Develop (1 — xf, and express the general term. 7. Develop the m^^ root oi 1 -{- m, 8. Develop (a — b)~% when a 1, we transform thus : !-.= _. (l-l), / IV and so put (1 — x)-^ = (— x)~^ (1 1 xt 10. Develop the wP^ power of 1 H 11. Compute the cube root of 1610 to six decimals. EXPONENTIAL THEOREM, 373 12. Develop {Va + Viy, 13. Using the functional notation, (m) = 1 + (^):, + (1)^ + g),,3 + etc, multiply the two series, (m) and (^), and show by the for- mulas of § 261 that the product is equal to (m + n). The Exponential Theorem. 309. Let it be required, if possible, to develop a^ in powers of a;, a being any quantity whatever. Assume a"" = Cq + C^x + C^x^ + C^x^ + etc. (1) to be true for all values of x. Putting any other quantity y in place of x^ we shall have ay = G, + C^y + C,y^ + G^y^ + etc. (2) By the law of exponents we must always have a"" X av =: tt-^+y. Now the value of a^'^v is found by writing x -{- y for x in (1), which gives a^^y= C, + G^{x+y) + C^{x + yf+C^{x^yY^QtQ. (3) On the other hand, by multiplying equations (1) and (2), we find a^av = G,^ + G^G^y + G^G^y"^ + G.G^f + etc. + (7o 6> + G^^xy + 6\ (72^^ _|_ etc + (7o^2^' + C^i^2^^^ + etc. ^ ^ + ^0^3^^ + etc. By § 285, the coefficients of all the products of like powers of ir and y must be equal. By equating them, we shall have more equations than there are quantities to be determined, and, unless these equations are all consistent, the development is impossible. To facilitate the process of comparison, we have in equation (4) arranged all terms which are homogeneous in x and y under each other. 374 BINOMIAL THEOBEM. By putting x — in (1), we find aO ::::, Q^^ wlienCG C^ = 1. (§ 103.) Comparing the terms of the first degree in x and y in (3) and (4), we find Coefficient of x, C^ — Cq C-^ ; " y, C, - C,G,. These two equations are the same, and agree with C^ = 1 ; but neither of them gives a value for C\, which must therefore remain undetermined. Comparing the terms of the second degree, we find, by de- veloping [x + yY, C^ {x^ + 2xy + 2/^) == C,x^ + G^'xy + C^y\ which gives ^^2 = ^V^ whence (7o =z -— (7.^. ^ 1-2 ^ Comparing the terms of the third order in the same way, we have C^{x^^^x^y + Zxy^ + y^) = C,a^-\-C,C^x^y + C^C^xy^+C,y^ which gives ^0^ =: 0^0^ z= -C^^; whence by definition, ( lo2r -„ = — 2 ; 1 I 111 Hence, V. The logarithm of a nujnher between mid 1 is negative. Again, as we increase ?^, the value of a'^ increases without limit, and that of — approaches zero as its limit. Hence, VI. The logarithm of is negative infinity. VII. Theorem. T1^e logarithm of a product is equal to the sum of the logarithm's of its factors. Proof. Let p and q be two factors, and suppose h = log p, k = log q. Then a^ =^ P^ a^ =: q. Multiplying, a^a^ = a^^^ = pq. Whence, by definition, h + k = log (pq), or log p + \ogq z= log (pq). The proof may be extended to any number of factors. VIII. ' Theorem. The logarithm of a quotient is found by subtracting the logarithm of the divisor from that of the dividend. Proof. Dividing instead of multiplying the equations in the last theorem, we have -, — a^-^ = -• a* q 380 LOGARITHMS. Hence, by definition, h — k=z\og^y or logj;— log^ = log^. IX. Theorem. The logarithm of any power of a num- her is equal to the logarithm of the numher multiplied by the exponent of the power. Proof. Let h = log ij, and let n be the exponent. Then ofi' =: p. Raising both sides to the n^^ power. Whence nh = log^'^, or n log p •=. log p^, X. Theorem. The logarithm of a root of a numher is equal to the logarithm of the numher divided hy the index of the root. Proof. Let s be the number, and let p be its n^^ root, so that p = \^s and s = p^. Hence, log s = log p^ = 7i log p. (IX.) logs Therefore, log^ = n , n/ lo2r s or log Vs = — -- — Note. We may also apply Th. IX, since p = A Con- sidering - as a power, the theorem gives ^ogp = -logs. EXERCISES. Express the following logarithms in terms ol hgp, log g^ log X, and logy, a being the base of the system : 1 LOGARITHMS. 381 1. Log p% Arts. 2 log^ + log q, 2. Log 2^^^' 3. Log^Y- 4- l^og pqVy^. 5. Log - = log xp~\ and explain the identity. 6. Log|| = logxyp-^q-\ Ans. Log X + log y — logp — log ^. 7- Log^,. 8- Log ^^3- 9. Log \/ic. 10. Log Vx Vy* II. Logy|. 12. Log V«. 13. Logaa;. 14. Log-. T ^ IS. Log-. |yj. I^^^g («^ + ^) + log (« — aj) 17. Log '\/a^ — x\ 18. LogVl-«^. 19. Log (a^ — ^-2), Logarithmic Series. 313. Rem. The logarithm of a number cannot be deyel- 2 »3 {n — l)x _ X 1.2.3 n ~^ ^n Hence, (l + y)^ =il+x\^y — I + I — |- + etc.j + terms in x\ a^, etcJ On the other hand, the exponential development, § 309, (8), gives, by putting 1 + ^ for c^. , (1 + y)^ = 1 + C^x -\- terms in x% a^, etc. ^ Equating the coefficients of x in these two identical series we have ^i=2/-f + |'-^+etc. (1) The value of C^ is given by the theorem of § 310, putting 1 -^'y ioY a] that is, C^ is here defined by the equation 6^- =il -\- y. Hence, if we take the number e (§ 310) as the base of a system of logarithms, we shall have G, = log (1 + y). Comparing with (1), we reach the conclusion: Theorem. Assuming the J^aperian base e as a base, we have log (1 + 2/) = 2/ - f- + I - |- + etc., ad inf. (2) LOGARITHMS. 383 Def. Logarithms to the base e are called Naperian Logarithms, or Natural Logarithms. The appellation " natural " is used, because this is the simplest system of logarithms. Kem. The series (2) is not convergent when y "> 1, and therefore must be transformed for use. Putting — y for y in (2), we have log(l--y) = -^-| -| -^tc- Subtracting this from (2), and noticing that log (l+y)- log (1 - 2/) = log ^ (Th. VIII), we have log ^^ = 2y + ^ + ^ + etc. (3) Now n being any number of which we wish to investigate ippose y = ^_ 1-y n ' whence log ip-^- = log = log (^^ + 1) — log n. Substituting these values in (3), we have log {n + 1) -log.. ^ ^ + 3^-iy3 + ^p^-iy5 + etc. (4) This series enables us to find log {n + 1) when we know log n. To find log 2, we put ^ = 1, which, because log 1 = 0, gives log2 = 3g + 3^3 + A3 + ^ + etc.) Summing five terms of this series, we find log 2 =. 0.693147 the logarithm, let us suppose y = - — — — • This will give 384 L00ARITHM8, Putting ^^ =r 2 in (4), we have log3=:log2 + 2g + 3l3 + J^ + J^ + etc.), ^ which gives log 3 — 1.098612. Because 9 = 32, log 9 := 2 log 3 = 2.197224. Putting ^ = 9 in (4), we have log 10 = log 9 + 2 (1 + 3i-3 + ^~ + etc.), whence log 10 = 2.302585. In this way the [N^aperian logarithms of all numbers may be computed. It is only necessary to compute the logarithms of the prime numbers from the series, because those of the com- posite numbers can be formed by adding the logarithms of their prime factors. (§ 312, VII.) 314, Definitive Form of the Exponential Series, We are now prepared to give the exponential series (§ 309, 8) its defi- nite form. Since the coefficient 0^ is defined by the equation e^i = a, the quantity G is the Naperian logarithm of a. Hence, the exponential series is «. = 1 + ^l2g^ + (^«1 + (^M^V etc., which is a fundamental development in Algebra. By putting « = e, we have log a = 1, and the series be- comes that for e^ already found. By putting x=il, we have an expression for any number in terms of its natural logarithm, namely, Comparison of Two Systems of Logarithms. 315. Put e, the base of one system ; a, the base of another; riy a number of which we take the logarithm in both systems. LOGARITHMS. 385 Putting I and V for the logarithms in the two systems, we have ^ =1 n, d' = n^ and therefore e^ = d', (1) Now put h for the logarithm of a to the base e. Then e^ — a, and raising both members to the V*^ power, e^^' = d'. Comparing with (1), I = kV, or ^' = ^ X i- (2) This equation is entirely independent of 7i, and is therefore the same for all values of ii. Hence, Theoeem. // ive multiply the logarithm of any nuTThber to the base a by the logarithin of a to the base e, we shall have the logarithin of the number to the base e. 316. Although there may be any nnmber of systems of logarithms, only two are used in practice, namely : 1. The natural or Naperian system, base = ^ = 2.718282 2. The common system, base — 10. The natural system is used for purely algebraic purposes. The common system is used to facilitate numerical calculations. Assigning these values to e and a in the preceding section, the constant k is the natural logarithm of 10, which we have found to be 2.302585. Therefore, by (2), for any number, nat. log = common log x 2.302585, , , nat. lo2r. and common log - ^^^^i;;;^ =: nat. log X 0.4342944..,. Hence, 25 386 LOGARITHMS, Theoeem. The comnion logarithm of any ninnher vianj he found by multiplying its natural logarithm by 0,4342944 .... or by the reciprocal of the JS^aperian loga- rithm of 10. Def, The number 0.4342944 is called the Modulus of the common system of logarithms. EXERCISES. 1. Show that if a and h be any two bases, the logarithm of a to the base b and the logarithm of b to the base a are the re- ciprocals of each other. 2. What does this theorem express in the case of the natu- ral and common systems of logarithms ? Common Logarithms. 317. Because 102 -- 100, 101 ^ 10^ 100 ^ 1^ _ 1 ) we have to base 10, ^^ - 10^ 10-2 _ __ ^^ - 100' etc. etc. The following conclusions respecting common logarithms will be evident from an inspection of the above examples : I. The logarithm of any number between 1 and 10 is a fraction between and 1. II. The logarithm of a number ivith two digits is 1 plus some fraction. III. In general, the logarithm of a number of i digits is i — 1, plus some fraction. IV. The logarithmj of a fraction less than unity is negative. V. Tlie logarithms of two numbers, the reciprocal of each other, are equal and of opposite signs. \ log 100 = 3, log 10 = 1, log 1 = 0, log 1 To ~ -1, log 1 100 ~ -2, LOGARITHMS. 387 VI. // one number is 10 times another, its logarithm^ will he greater by unity. Proof. If 10^ = n, then 101+^ = 10 X l(y = lOn, Hence, if I = log n, then ? + 1 = log 10^. 318. To give an idea of the progression of logarithms^ the following table of logarithms of the first 11 numbers should be studied. The logarithms are not exact, because all logarithms, ex- cept those of powers of 10, are irrational numbers, and there- fore when expressed as decimals extend out indefinitely. We give only the first two decimals. log 1 = 0.00, log 7 = 0.85, log 2 =: 0.30, logs =: 0.90, log 3 = 0.48, log 9 = 0.95, log 4 = 0.60, log 10 = 1.00, log 5 = 0.70, log 11 = 1.04. log 6 =r 0.78, It will be noticed that the difference between two consecu- tive logarithms continually diminishes as the numbers increase. For instance, the difference between log 20 and log 10 must by § 312, YIII, be the same as between log 1 and log 2. 319. Gomjmtation of Logarithms, Since the logarithms of all composite numbers may be found by adding the loga- rithms of their factors, it is only necessary to show how the logarithms of prime numbers are computed. We have already shown (§ 313) how the natural logarithms may be computed, and (§ 316) how the common ones may be derived from them by multiplying by the modulus 0.4342944.... It is not how- ever necessary to multiply the whole logarithm by this factor, but we may proceed thus: We have, putting M for the modulus, com. log n =z Mneii. log n, com. log {n i- 1) zzz M nat. log {n -j- 1) ; 388 L0GABITHM8, whence, by subtraction, com. log {n + 1) — com. log n =:^ M [nat. log {n -f 1) — nat. log n] ; and, by substituting for nat. log (^ + 1) -— nat. log n its yalue, § 313, com. log {n + 1) = com. log n + 2m[-^^ + 372]^-VI? 1 + T-JK ^Tr + etc. By means of this series, the computations of the successive logarithms may be carried to any extent. Tables of the logarithms of numbers up 100,000, to seven places of decimals, are in common use for astronomical and mathematical calcula- tions. One table to ten decimals was published about the beginning of the present century. The most extended tables ever undertaken were constructed under the auspices of the French government about 1795, and have been known under the name of Les Grandes Tables du Cadadre. Many of the logarithms were carried to nineteen places of decimals. They were never published, but are preserved in manuscript. 330. It may interest the student who is fond of computa- tion to show how the common logarithms of small numbers may be computed by a method based immediately on first principles. Let 7Z be a number, and let - be an approximate value of its logarithm. We shall then have, n = 10^, or, raising to the q^^ power, flQ — lOP. Hence, could we find a power of the number which is also a power of 10, the ratio of the exponents would at once give the logarithm. This can never be exactly done with whole numbers, but, if the condition be approximately fulfilled, we shall have an approximate value of the logarithm. Let us seek log 2 in this way. Forming the successive powers of 2, we find 210 _ 1024 = 103 (1.024). (1) LOGARITHMS. 389 Hence, 3 : 10 = 0.3 is an approximation to log 2. To find a second approximation, we form the powers of 1.024 until we reach a number nearly equal to 2 or 10, or the quo- tient of any power of 2 by a power of 10. Suppose, for instance, that we find 1.024^^ = 2. Because 1.024 = 2^^-r- 10^, this equation will give ' 2^^\^ = )i, or Z''^^^ = ^.lu dx )x = 2, or 210^^ = 2. 103^, or 2io^-i = lO^a^, which will give log 2 = lOic- If we form the powers of 1.024 by the binomial theorem, or in any other way, we shall find that x is between 29 and 30, from which we conclude that log 2 = 0.301 nearly. To obtain a yet more exact value, we form the 30th power of 1.024 to six or seven decimals, and put it in the form 1.02430 = 2 (1 + a), where a will be a small fraction. Then we find what power of 1 -\- cc' will make 2. Let y be this power. Kaising the last equation to the yth power, we have 1.02430y = 2^(1 + a)y = 2y+K Putting for 1.024 its value, 2^^ -r- 10^, this equation becomes 2^1/ 10902/ 2^+1, or 2299y-i = 1090y, whence,. log 2 = g^|^. By a little care, tlie value of y can be obtained so accurately that the value of log 2 shall be correct to 8, 9, or 10 places of decimals. The power to which we must raise 1 + cc to produce 2 will be approximately — — — ~ — , when a is very small. 390 LOGARITHMS. EXERCISES. 1. In the common system {a = 10) we have log 2 = 0.30103, log 3 =z 0.47712. Hence find the logarithms of 4, 5, 6, 8, 9, 12, 12^, 15, 16, 16|, 18, 20, 250, 6250. Note that 5 = ^ , 12J = if ^, 16| = if^, and apply Th. VIII. 2. How many digits are there in the hundredth power of 2? 3. Given log 49 = 1.690196; find log 7. 4. Given log 1331 = 3.124178 ; find log 11. 5. Find the logarithm of 105 and 1.05 from the above data ? 6. Find the logarithm of 1,06^^, 7. If $1 is put out at 5 per cent, per annum compound interest for 1000 years, how many digits will be required to express the amount? (Compare § 216.) 8. Prove the equation log X = ^ log (:?; -f- 1) + ^ log {x--l) + M + o /o^9 '-1 \.q + K /o^5! iTs + ^^^' _2x^ _ 1 ^ 3 (2:r2 — 1)3 ^ 5 {2x^ — 1)' 9. lty = log 71, of what numbers will y -i- ly y + 2, y — 1 and «/ — 2 be the logarithms ? 10. Find X from the equation cP^ =:lu Solution. Taking the logarithms of both members, we have X log c = log h ; whence, x = :r^— • logo 1 11. c^ = n. 12. (^ = —> m 13. b^ = -. 14. h-^ = p. Show that the answers to (13) and (14) are and ought to be identical. 15. a^^ z=z m, 16. bc^ = h 17. Find X and y from the equations ] BOOK XII. IMA GIN A RY QUANTI TIE S. CHAPTER 1. OPERATIONS WITH THE IMAGINARY UNIT.* 331. Since the square of either a negative or a positive quantity is always positive, it follows that if we have to extract the square root of a negative quantity, no answer is possible, in ordinary positive or negative numbers (§§ 170, 200). In order to deal with such cases, mathematicians have been led to suppose or imagine a kind of numbers of which the squares shall be negative. These numbers are called Imagi- nary Quantities, and their units are called Imaginary Units, to distinguish them from the ordinary positive and negative quantities, which are called real. 332, The Imaginary Unit, Let us have to extract iJie square root of — 9. It cannot be equal to + 3 nor to — 3, because the square of each of these quantities is + 9. We may therefore call the root V— 9, just as we put the sign \/ before any other quantity of which the root cannot be extracted. But the root may be transformed in this way : Since — 9 = +9 x —1, it follows from § 183 that • * It is not to be expected that a beginner will fully understand tliis subject at once. But he should be drilled in the mechanical process of operating with imaginaries, even though he does not at first understand their significance, until the subject becomes clear through familiarity. 392 IMAOmART QUANTITIES. Def, The surd V— 1 is the Imaginary Unit. Tlie imaginary unit is commonly expressed by the symbol 1 This symbol is used because it is easier to write i than The unit Hs a supposed quantity such that, when squared, the result is — 1. That is, i is defined by the equation %^ = — 1. Theoeem. Tl%e square root of any negative quantity may he expressed as a number of imaginary units. For let — ^ be the number of which the root is required. Then V — n =z V -\- n V— 1 = VnL Hence, To extract the square root of a negative quantity , extract the root as if the quantity were positive, and affix the symbol i to it. 333. Complex Quantities, In ordinary algebra, any num- ber may be supposed to mean so many units. 7 or a, for example, is made up of 7 units or a units, and might be writ- ten 7-1 or al. When we introduce imaginary quantities, we consider them as made up of a certain number of imaginary units, each repre- sented by the sign i, just as the real unit is represented by the sign 1. A number h of imaginary units is therefore written hi, A sum of a real units and h imaginary units is written a + hi, and is called a complex quantity. Hence, Def. A Complex Quantity consists of the sum of a certain number of real units plus a certain number of imaginary units. Def. When any expression containing the symbol of the imaginary unit is reduced to the form of a com- plex quantity, it is said to be expressed in its Normal Form. IM AGIN ART QUANTITIES. 393 Addition of Complex Expressions. 334. The algebraic operations of addition and subtraction are performed on imaginary quantities according to nearly the same rules which govern the casQ of surds (§ 181), the surd being replaced by L Thus, aV— 1 + b^/— I = ai -\- U = {a -\- b) i. Hence the following rule for the addition and subtraction of imaginary quantities : Add or subtract all the real terms, as in ordinaj^j algebra. Then add the coefficients of the imaginary unit, and affix the symbol i to their sum. Example. Add a -t- bi, 6 + li, 5 — 10/, and subtract ^a — 2bi + z from the sum. We may arrange the work as follows: a + bi 6 4- W 5 — lOi — z — da -{- 2bi (sign changed). Sum, —z — 2a-{-ll + {3b — 3) i. EXE RC I S E S. 1. Add dx + ^yi + m, 2m + 6ni, 6m — 62/i. 2. Add 4a^, 17 i, 3a + 6bi, x + yi. 3. From the sum a -{- bi -\- m — ni — ;^ + qi subtra<)t the sum ■\- yi — z — ui. Reduce to the normal form : 4. ' a -{-bi — {m — ni) — {x + yi), 5. 7n{a — bi) — n{x — yi). Multiplication of Complex Quantities. 325. Theorem. All the even powers of the imagi- nary unit are real units, and all its odd powers are imaginary units, positive or negative. 394 IM AGIN ART QUANTITIES, Proof. The imaginary unit is by definition such a symbol as when squared will make — 1. Hence, ^2 = - 1. Now multiply both sides of this equation by i a number of times in succession, and substitute for each power of i its value ■ given by the preceding equation. We then have i^ = — ^2 =rr -f 1 (because v^ =. — 1), i^ — _«4 — -f ^2 — — 1, etc. etc. etc. It is evident that the successive powers of i will always have one of the four values, % — 1, — i, or + 1. iy i^, ^^, etc., will be equal to i; t\ ^^ i^o, etc., '' " —1; ^, P, P-\ etc., '' " - ^; ^^ ^8, ^l2^ etc., " " ■\- 1. We may express this result thus : > If n is any integer, then: To multiply or divide imaginary quantities, we proceed as if they were real and substitute for each power of % its value as a real or imaginary, positive or negative unit. Ex. I. Multiply ai by xi. By the ordinary method, we should have the product, axi'^. But i^ = — 1. The product is therefore — ax. That is, ai x xi =: — ax, Ex. 2. Multiply a -f bi by m + ?ii, ni (a + hi) = ani — dn (because ni x M = —hn) m {a -f M) = hmi + rt??z (77i + ni) {a + ^i) = «77z — ^^^ + {an + Z>w) i, which is the product required. IMAOINART QUANTITIES, 395 EXERCISES Multiply I. X -\- yi hy a — b. 2, m -\- 7ii by ai, 3. m — ni by M. 4. 1 + ^ by 1 — i. 5. a; — y^ by ^ + M. 6. a; — «/* hy x -\- yi. 7. a — «^^ — M by a + ai + M. Develop 8. (flj + My. 9. (7?z + 7^^)3. 10. (1 + ^)2. II. (l — iy. 336. Imaginary Factors. The introduction of imaginary units enables us to factor expressions which are prime when only real factors are admitted. The following are the princi- pal forms : a^^b^ = {a + M) {a — bi), a^-W± Ubi = {a± bif. The first form shows that the sum of two squares can always be expressed as a product of two complex factors. For example, 17 = 4^ + 12 = (4 + i) (4 — i). EXERCISES. Factor the expressions : I. x^ -\- 4.. 2. x^-\-2. 3. ^^^^x + b =1 (x — 1)2 + 4. 4. x^ — 4:X -}- 13. 5. a -\- b. 6. a2 ^ 2an + 5^^2. 7. x^ + 2xy + 2y\ 337. Fu:n^damental Prin'ciple. A complex quantity A -\- Bi cannot he equal to zero unless zve have hoth A :=zO and B = 0. Proof. If ^ and ^ were not zero, the equation A-\-Biz=zO would give A that is, the imaginary unit equal to a real fraction, which is impossible. Cor. If both members of an equation containing imagi- 396 IMAGINARY QUANTITIES. nary units are reduced to the normal form, so that the equation shall be in the form A-\-Bi = M-^Ni, we must haye the two equations, A = M, B =z N, For, by transposition, we obtain A^M^{B-N)i = 0, whence the theorem giyes A — M=iO^ B — JSF=0. Hence, Every equation hetween complejo quantities involves two equations hetween real quantities, formed by equating the numhers of real and imaginary units. Reduction of Functions of i to the Normal Form. 338. 1. If we have an entire function of i, a -\-M -\- ci^ + di^ -f ei^ -{-fi^ -f etc., we reduce it by putting i^ = — 1, i^ = — i, i^ = 1, etc., etc., and the expression will become {a — c -^ e — etc.) + {b — d +/— etc.)^; which, when we put X = a — c -\- e — etc, y = h — d -i-f — etc, becomes x + yi, as required. 2. To reduce a rational fraction of i to the normal form, we reduce both numerator and denominator. The fraction will then take the form a -\- bi m + 7ii Since this is to be reduced to the form x -f yiy let us put a -^-bi — - — -. = X -\- yi, m -}- m ^ X and y being indeterminate coefficients. Clearing of fractions, a -\- bi = mx — 7iy + (my -f nx) i. IMA OINAB 7 Q UANTITIES, 397 Comparing the number of real and imaginary units on each side of the equation, we have the two equations 7nx — ny^ztty nx + my = 5. Solving them, we find ma -\- nb _ mb — na *,, ^ a -\- hi ma + nh ml) — na . Therefore, — ^ — . = — ^— — ^ -\ ^— — ^ t, 771 4- ni m^ + w- 7)1^ + 11? which is the normal form. EXERCISES. Eeduce to the normal form : 2. 1 + i — i^ + i^ — i^ — 1*5 + i\ ''' i^i 6 + 5i 1 + ^* ^ ^2^* (^ — c^i) A, — - — :• q. — — ;• 6. ^^ -' ^ (j — 6i ^ 1 — ^ X -\- ai 1 — i a -\-U (a-\-M) {a— hi) 7- 2~T5' • 'a^Yi ^' (^^i)2~-" 10. What is the value of the exponential series which gives the development of e^? We put x = i in § 310, Eq. 10. 11. Develop (1 + xiY hj the binomial theorem. 12. What are the developed values of (1 + UY + (1 — uy and (1 -i^biY — {l — UY^ 13. Write eight terms of the geometrical progression of which the first term is a and the common ratio t. 14. Find the limit of the sum of the geometrical progres- sion of which the first term is a and the common ratio -• z 339. To reduce the square root of an imaginary expres- sion to the normal form. Let the square root be ^a + bi. We put X + yi z= ^J a + bi. Squaring, x^ — y'^ -\- %xyi = «5 + bi. 398 IM AGIN ART QUANTITIES. Comparing units, x^ -— ?/ = a, Solving this pair of quadratic equations, we find _ V{Va ^-\-^ + a) V2 ' \/(V^3+ I)' -a) Therefore, '= V2 EXERCISES. Eeduce the square roots of the following expressions to the normal form : I. 3 + 4^. 2. 4 + di. 3. 12 + 5^. 4. Find the square roots of the imaginary unit i, and of — i, and prove the results by squaring them. Note that this comes under the preceding fomi when a = and 6 = ±1. 5. Find the fourth roots of the same quantities by extract- ing the square roots of these roots. 330. Quadratic Equations loith Imaginary Roots, The combination of the preceding operations will enable us to solve any quadratic equation, whether it does or does not contain imaginary quantities. Example i. Find x from the equation x^ -^ 43: + 13 = 0. Completing the square and proceeding as usual, we find x^ -^ 4cx-^ 4. = — 9, whence a; + 2 = a/— 9 — ±3/, and X =z _ 2 ± 3^. Ex. 2. x^ + Ixi — c = 0. Completing the square, X^ + OXl — — := C — —' 4 4 IMAGINABT QUANTITIES, 399 Extracting the root, hi _ VJg^^g ^■^ 2 " 2 ' 1 hi whence x = ±,- a/(4c — h^) — — • EXERCISES. Solve the quadratic equations: I. a;2 -f- a; + 1 = 0. 2. 2;2 — a; + 1 — 0. 3. a;2 + 3^2; -f 10 r= 0. 4. :zj2 _|_ iqo; + 34 = 0. Form quadratic equations (§ 199) of which the roots shall be 5. a + hi and a — hi. 6. ai + h and ai — h, 331. Exponential Functions. When in the exponential function a^ we suppose z to represent an imaginary expression X + «/i, it becomes This expression could have no meaning in any of our pre- vious definitions of an exponent, because we have not shown what an imaginary exponent could mean. But if we suppose the effect of the exponent to be defined by the exponential theorem (§§ 309, 314), we can develop the above expression. First we have, by the fundamental law of exponents, f^x+yi — a^avK Next, if we put c = Nap. log a, we have a — e<^; whence, ay^ = ePVK K we put, for brevity, cy = u, we shall now have The value of a^ being already perfectly understood, we may leave it out of consideration for the present, and investi- gate the development of e^*. By the exponential theorem (§ 310, 10), . ^ . . ii^ij^ u^i? tiH^ uH^ e«i = 1 + m + -p + -3 J + -J- + -J- + etc. 400 IMAQINABT QUANTITIES. Substituting for the powers of i their values (§ 325), .» = 1 - -J + jj - gj + etc. + ^M - ^ + gj - etc.) I. These two series are each functions of u, to which special names have been given, namely : Def. The series 1 — ^, + ji — ^. + 51 — etc., is called the cosine of if, and is written cos u, Def. The series '^ — 07+^7 — 7"^ + ^^ — ^^c-? is called the sine of u^ and is written sin u. Using this notation, the above development becomes, QUI — COS u + i sin u, {a) which is a fundamental equation of Algebra, and should be memorized. Remarks. These functions, qo& u and sin u, have an ex- tensive use in both Trigonometry and Algebra. To familiarize himself with them, it will be well for the student to compute their values from the above series for u = 0.25, u = 0.50, u=zl, u=:2, to three or four places of decimals. This can be done by a process similar to that employed in computing e in § 310. If the work is done correctly, he will find : For u = ~, 4 " - = l' " U =zl, " u = 2, 333. Let us now investigate the properties of the functions cos u and sin u, which are defined by the equations, u^ ii'^ u^ cos^. = l-^ + jj--^ + etc. ^ ^ 7 / (^) U^ VP w , ^ miu — u — — Ar -^^— -^ + etc. 1 COS -7 = 4 0.969, sin \ =. 0.247 4 cos| = 0.878, sin 1 = 0.479. COS 1 = 0.540, sin 1 = 0.841 COS 2 = - 0.416, sin 2 — 0.909. IM AGIN ART QUANTITIES, 401 Since cos ti includes only even powers of u, its value will remain unchanged when we change the sign of u from + to — 5 or vice versa. Hence, cos {^u) =z cos ti. (1) Since sin u contains only odd powers of u, its sign will change with that of u. Hence, sin (— -w) = — sin u, (2) If in the equation {a) we change the sign of Uy we have, by (1) and (2), or e""^* = cos u — I sin u, Now multiply this equation by {a). Since 1 ^m X e-'^^ =: e^* X -rr = 1, we have 1 = (cos ?^)^ — i^ (sin -2^)2, or 1 = (cos uj^ + (sin uy. It is customary to write cos^ u and sin^ u instead of (cos uf and (sin uY, to express the square of the cosine and of the sine of u. The last equation will then be written cos^ u + sin^ u == 1. {c) Although we have deduced this equation with entire rigor, it will be interesting to test it by squaring the equations {V), First squaring cos u, we find (§ 284), C0S2 ^ = 1 -^ ,,2 + ,,4 (i. ^ _i__ + _) « etc. The coefficient of u"^ is found to be 1 1 1^ 1 n\ +2! (?z — 2)! "^4!(7^— 4)! "^ ^ 7i\ when ^ is double an even number, and to the negative of this expression when n is double an odd number. Again, taking the square of sin u, we find sm'* u 26 ^. + ,,4(__l^j^-J-^--) + etc. .402 IMAGINARY QUANTITIES. the coefficient of u'^ being 1 1 1 1 ! 0* — 1) ! 3 ! (/i — 3) ! 5 ! (^ — 5) ! (^-1)! 1!' or the negative of this expression, according as ;^ ^ is even or odd. ^ Adding sin^ i^ and cos^ u, we see that the terms u^ cancel each other, and that the sum of the coefficients of w* can be arranged in the form ^ 1_ _1 1_ 1 4! 1! 3! "^2! 2! 3! l! "^4!* Let us call this sum A, If we multiply all the terms by 4 ! , and note that by the general form of the binomial coeffi- cients, n\ _ ln\ wefind 4!^ = l^g)+g)-(3^)+g), which sum is zero, by § 2G2, Th. 11. Therefore the coefficients of u^ cancel each other. Taking the sum of the coefficients of u'^, we arrange them in the form n\ 1! 0^-1)! "^2! (?^-2)! 3! (^^ - 3)! + ^ ^'^ which call A, Then multiplying by n\, we have »^^=i-©+(i)-(i)+--+e). which sum is zero. Therefore all the coefficients of u'^ cancel each other in the sum sin^ u + cos^ u, leaving only the first term 1 in cos^ u, thus proving the equation (6) independently. This example illustrates the consistency which pervades all branches of mathematics when the reasoning is correct. The conclusion (c) was reached by a very long process, resting on many of the fundamental principles of Algebra ; and on reach- IMAOINAUY QUANTITIES. 403 ing a simple conclusion of this kind in such a way, the mathe- matician always likes to test its correctness by a direct process, when possible. Let us now resume the fundamental equation {a). Since u may here be any quantity whatever, let us put nu for u. The equation then becomes, ^nui _ cos nu + i sin nu. But by raising the equation {a) to the n*^ power, we have Hence we have the remarkable relation, (cos u -\- i sin u)^ := cos nu + i sin nu. Supposing n=z2, and developing the first member, we have cos^ u — sin^ u + 2i sin u cos tc =z cos 2u -f i sin 2u. Equating the real and imaginary parts (§327, Cor.), we have cos^ u — sin2 to = cos 2u, 2 sin u cos i^ = sin 2u, relations which can be verified from the series representing cos u and sin u, in a way similar to that by which we verified sin^ 2c + cos^ u = 1. EXERCISES. 1. Find the values of cos^ u, sin^ u, cos* u, and sin* u by the preceding process. 2. Write the three equations which we obtain by putting u =. a, u =z b, and u =i a -\- h in equation [a). Then equate the product of the first two to the third, and show that cos {a -{- h) =. cos a cos h — sin a sin Z>, sin (a 4- ^) = sin a cos h + cos a sin ^. 3. Eeduce to the normal form, {x — i) {x — 2i) {x — 30 {x — U). 4. Develop (a^ + U)^ by the binomial theorem, and reduci the result to the normal form. " 404 QEOMETBIG REPRESENTATION. CHAPTER II. THE GEOMETRIC REPRESENTATION OF IMAGINARY QUANTITIES. 333. In Algebra and allied branches of the higher raaUie- maties^ the fundamental operations of Arithmetic are extended and generalized. In Elementary Algebra we have already had several instances of this extension, and as we are now to have a much wider extension of the operations of addition and mul- tiplication, attention should be directed to the principles involved. In the beginning of Algebra, we have seen the operation of addition, which in Arithmetic necessarily implies increase, so used as to produce diminution. The reason of this is that Arithmetic does not recognize negative quantities as Algebra does, and therefore in employ- ing the latter we have to extend the meaning of addition, so as to apply it to negative quantities. When thus applied, we have seen that it should mean to subtract the quantity which is negative. In its primitive sense, as used in the third operation of Arithmetic, the word JnuUiplij means to add a quantity to itself a certain number of times. In this sense, there would be no meaning to the words " multiply by a fraction.^' But we ex- tend the meaning of the word multiply to this case by defining it to mean taking a fraction of the quantity to be multiplied. We then find that the rules of multiplication will all apply to this extended operation. This extension of multiplication to fractions does not take account of negative multipliers. In the latter case we can extend the meaning of the operation by providing that the algebraic sign of the quantity shall be changed when the mul- tiplier is negative. We thus have a result for multiplication by every positive or negative algebraic number. Now that we have to use imaginary quantities as multi- OEO METRIC REPRESENTATION. 405 pliers, a still further extension is necessary. Hitherto our operations with imaginary units have been purely symbolic ; that is, we have used our symbols and performed our operations without assigning any definite meaning to them. We shall now assign a geometric signification to operations with imagi- nary units, subject to these three necessary conditions : 1. The operations must be subject to the same rules as those of real quantities. 2. The result of operating with an imaginary quantity must be totally different from that of operating with a real one, and the imaginary quantity must signify something which a real quantity does not take account of. 3. If the imaginary quantity changes into a real one, the operation must change into the corresponding one with real quantities. 334. Geometric Represenfatmi of Imaginary Units. Cer- tain propositions respecting the geometric representation of multiplication have been fully elucidated in Part I, and are now repeated, to introduce the corresponding representations of complex quantities. I. All real numbers, positive and negative, may be arranged along a line, the positive numbers increasing in one direction, the negative ones in the opposite direction from a fixed zero point. Any number may then be represented in magnitude by a line extending from to the place it occupies. We call this line a Vector. II. If a number a be multiplied by a positive multiplier (for simplicity, suppose +1), the direction of its vector will remain unaltered. If it be multiplied by a negative multiplier (suppose — 1), its vector will be turned in the opposite direc- tion (from — a to -\- a, or vice versa). Compare § 72, where the coarse lines are the vectors of the several quantities. —a +« 1 I I III. If the number be multiphed twice by — 1, that is, by (—1)2, its vector will be restored to its first position, being twice turned, and if it be multiplied twice by + 1, that is, by (-[- 1)% its vector will not be changed at all. Its vector will 406 IMAOINABT QUANTITIES. +ia therefore be found in its first position, whether we multiply it by the square of a positive or of a negative unit; in other words, both squares are positive. IV. To multiply the hne + a twice by the imaginary unit i, is the same as multiplying it by i^ or — 1. Hence, Multiplying by the iiivaginary unit i must give the vector such a motion as, if repeated, will change it from -\- a to — a. Such a motion is given by turn- ing the vector through a right angle, into the position + ia, A second motion brings it to the position __^ — a, the opposite of -h «. A third motion brings it to — ia, a position the opposite of -f ia, A fourth motion restores it to the original position + a. If we call each of these motions multiplying hy i, we have. -ia la, i^a a, i^a ta. _4 -3 —2 —1 ! \ [ L_ +4i from the diagram, a = a, ia ikt = a, which corresponds exactly to the law governing the powers of i (§ 325). Hence : // a quantity is represented hy a vector extending from a zero point, the multiplication of this quantity hy the imaginary unit may he represented hy turning the vector through 90°. V. In order that multiplier and multiplicand may in this op- eration be interchanged without affecting the product, we must suppose that the vertical line which we have called ia is the same as ai, that is, that this line represents a imaginary units. We have therefore to count the imaginary units along a vertical line on the saine system that we count the real units on a horizontal line. + i 1 2 ^-2i 3i U OEOMETRIG HEP RESENT ATION. 407 '«+6i U U 7>^ bi \ ^bi 835. Qeoinetric Representation of a Complex Quantity, We have shown (§ 15) that algebraic addition may be represented by putting lines end to end, the zero point of each line added be- ing at the end of the line next preceding. The distance of the end of the last line from the zero point is the algebraic sum. On the same system, to repre- sent the algebraic sum of the real and imaginary quantities a + Ji, we lay off a units on the real (horizontal) line, and then h units from the end of this line in a vertical direction. The - end of the vertical line will then be the position corresponding to flj -f hi. It is evident that we should reach the same point if we first laid off h units from on the imaginary line, and then a units horizontally. Hence this system gives })% j^ a =^ a -\- hi, as it ought to, to represent addition. If « or J is negative, it is to be laid off in the opposite di- rection from the positive one. We then have the points cor- responding to — a -\-l)i, — a — hi, and aj — hi, shown in the diagram, which should be carefully studied by the pupil. The result we have reached is the following : Every complex quantity a -f hi is considered as be- longing to a certain point on the plane, namely, that point which is reached by laying off from the zero point a units in the horizontal direction and h units in the vertical direction, R 336. Addition of Com- plex Quantities, If we have several complex terms to add, as a + hi, m — ni, p + qi, we may lay them off separately in their ap- propriate magnitude and di- 408 IMAGINABT QUANTITIES, rection, as in the figure, the last line terminating in a point K. If we first add the quantities a -\- U, etc., algebraically (§ 324), the result will be a + m -\- p -\- {b — n -\- q)i. We may lay off this sum in one operation. The sum a-hm- -\-p will carry us from to M, and the sum {b ^ n -\- q) i from M to R, because MU=b — n + q. Therefore we shall reach the same point R whether we lay the quantities off sepa- rately, or take their sum and lay off its real and imaginary parts separately. 337. Vectors of Complex Quantities, The question now arises by what straight line or vector shall we represent a sum of complex quantities ? The answer is : T]^e vector of a sum of sev- eral vectors is the straight line from the heginning of the first to the end of the last vector added. For example, the sum of the quantities OX == a and XP = bi is the vector OP. It might seem to the student that the length of the vector represent- ing the sum should be equal to the combined lengths of all the separate vectors. This difficulty is of the same kind as that encountered by the beginner in finding the sum of a positive and negative quantity less than either of them. The solution of the difficulty is simply that by addition we now mean something different from both arithmetical and algebraic addition. But the operation reduces to arithmetical addition when the quantities are all real and positive, because the vectors are then all placed end to end in the same straight line. Therefore there is no inconsistency between the two operations. Two imaginary quantities are not equal, unless both their real and imaginary parts are equal, so that their sum shall ter- minate at the same point P. Their vectors will then coincide with each other. Hence : Tivo vectors are not considered equal unless they agree in direction as well as length. GEOMETRIC REPRESENTATION, 409 In other words, in order to determine a vector com- pletely, we must know its direction as well as its length. This result embodies the theorem of the preceding chapter (§ 327), that two complex quantities are not equal unless both their real and imaginary parts are equal. It is only in case of this double equality that the two complex quantities will belong to the same point on the plane. Because OXP is a right angle, we have by the Pythagorean theorem of Geometry, (length of veetor)2 = a^ _|_ j2^ or length of vector = ^/d^ -j- b\ We are careful to say length of vector, and not merely vec- tor, because the vector has direction as well as length, and the direction is as important an element as length. To avoid repeating the words " length of," we shall put a dash over the letters representing a vector when we consider only its length. Then OX will mean length of the line OX. Def. The length of the vector, or the expression Va^ -f b% is called the Modulus of the complex ex- pression a 4- M. The modulus is the absolute value of the expression, con- sidered without respect to its being positive or negative, real or imaginary. Thus the different expressions, — 5, +5, 3 + 4/, 4 — 3i, 5^, all have the modulus 5 (because Vs^ 4- 4^ =: 5). The points which represent them are all 5 units distant from the zero point, and so lie on a circle, and their vectors are all 5 units in length. The German mathematicians therefore call the modulus the absohite value of the complex quantity, and this is really a better term than the English expression modulus, Def. The Angle of the vector is the angle which it makes with the line along which the real units are measured. If OA is this line, and OB the vector, the angle is AOB. 2. 4 - 31 3- — 4 + Si. 5- 3 + 4^. 6. 3 — 4^. 8. — 3 — 4:1. 9- 5 + 7i. II. 5 + 6i, 12. 5 + 4i. 14. 3 + i- 15. S^i. nta" I and vertical line ; mark several 410 IM AGIN ART QUANTITIES. EXERCISES. Lay off the following complex quantities, draw the vectors corresponding to them, and find the modulus both by measure- ment and calculation : I. 4 + U. 4. — 4 — 3^. 7. — 3 -f U. 10. 5 + 6i. 13. 3 + 2i. 16. 3 — 2^. 17. Draw a he points on the plane of these lines, and find by measurement the complex expressions for each point. Also, draw the sev- eral vectors and measure their length. Continue this exercise until the relation between the complex expressions and their •points is well apprehended. I^OTE. The student may adopt any scale he pleases, but a scale of millimeters will be found convenient. 338. Geometric Multiplication. The question next arises whether the results we obtain for multiplication of complex quantities follow, in all respects, the usual laws of multiplica- tion, especially the commutative and distributive laws. I. To multiply a vector hy a real factor. Let the vector he a -\- hi and the jf factor m. The product will be ^ ,-^'1 ma + mhi. In the geometric construction, let OK = a and Al^^U. We shall O- then have, by the rule of addition. Vector OB =: a + U. "When we multiply a by m, let OA' be the product ma, and A'B' the product mhL Because the lines OA and AB are both multiplied by the same real factor m to form OA' and A'B', we shall have OA : AB : OB = OA' : A'B' : OB'. GEOMETRIC REPRESENTATION. 411 Therefore the triangles OAB and OA'B' are similar and equiangular, so that angle A'OB' = angle AOB. This shows that the lines OB and OB' coincide, so that BB' is the continuation of OB in the same straight line. More- over, the above proportion gives OB' = 7??0B, or, from (1), vector OB' = m vector OB. Therefore, multiplying a vector hy a real factor changes its length without altering its direction. II. To multiply a vector hy the imaginary unit. Multiplying a + hi by i, the ^q result is — b + ai. The construction of the two '^' vectors being made as in the fig- ure, we have p _y OB = a + M, OQ. = — /^ + ai. Because the triangles OPQ and OAB are right-angled at P and B, and have the sides containing the right angle equal in length, they are identically equal, and angle POQ = angle OBA =z 90^ — angle BOA. Hence the sum of the angles POQ and BOA is a right angle, and because POA is a straight line, therefore, angle BOQ .-= 90°. Therefore, the result of multiplying the vector OB by the imaginary unit is to turn it 90° ivithout changing its length. We have assumed this to be the case when the vector represents a real quantity, or lies along the line OB ; we now see that the same thing holds true when the vector represents a complex quantity. If instead of the multiplier being simply the imaginary unit, it is of the form ni, then, by (I), in addition to turning the vector through 90°, we multiply it by n. 412 IMAGINARY QUANTITIES, III. To multiply a vector by a complex quantity, m + ni. This will consist in multiplying separately by m and ni^ and adding the two products. Put OB = a + 6^, the yector to be multiplied ; ON = m + ni, the multiplier. To multiply OB by m^ we take a length OC, deter- mined by the proportion, OC : OB = m : 1, (I) whence by (I), OC = m-OB = w (a + hi). To multiply OB by ni, we take a length CD determined by the condition, length GY) — n length OB, CD : OB or nil; and to multiply by i, we place it perpendicular to OB. (II) We then have, CD = OB X ni. In order to add it to OC, the other product, we place it as in the diagram, and thus find a point D which corresponds to the sum OC + CD = OBxm + OBxm'; that is, to the product {m + ni) (a + hi). Now because OC == OB x m and CD = OB x n, we have OC" : CD = m:n=zOM : MN, and because the angles at M and C are right angles, the tri- angles OCD and OMN are similar. Therefore, angle COD = angle MON. Hence the angle AOD of the product- vector is equal to the sum of the angles of the multiplier and multiplicand. For the length OD of the product-vector we have, GEOMETBIG REPRESENTATION^, 413 length 0& = 00^ + CJ? = (m2 + n^) OBl Extracting the square root, length OD = Vni^+n^ • OB Therefore the length of the product-vector is equal to the products of the lengths of the vectors of the factors. Combining these two results, we reach the conclusion: The modulus of the produet of two coinplex factors is equal to the product of their moduli. The angle of the product is equal to the sum; of the angles of the factors. 339. TJie Roots of Unity. We f? have the following curious problem : | Ny> Given, a vector A, which call « ; i \ it is required to find a complex factor i \ X, such that when we multiply a n ¥ 70 A times by x, the last product shall be a / itself. That is, we must have / o^a = a. /C The required factor must be one which will turn the vector round without changing its length. Let us begin with the case of n=:3. Since three equal motions must restore OA to its original position, the condition will be satisfied by letting x indicate a motion through 120°, so that OA shall take the position OB when angle AOB = 120°. Then, P being the foot of the per- pendicular from B upon AO produced, we shall have angle POB = 60°, and angle PBO = 30°. Therefore, FO = la, PB = ^^a, and vector OB =: xa =z — -a + --;-aL 414 IMAOINART QUANTITIES. Because the factor x has not changed the length of the line, the modulus of x is unity, and because it has turned the line through 120°, its angle is 120°. Therefore its value is -OP-fPBi on a scale of numbers in which OB = 1 ; that is, 1 , a/3. ^==--2 +-2-^- Eeasoning in the same way with respect to the product x^a, which, produces the vector OC, we find ^ - ~ 2 ~ 2 '^ an equation which we readily prove by squaring the preceding value of X and reducing. Multiplying these values of a: and x^, v/e find x^ = 1, which ought to be the case, because x^a = a. Hence, 1 a/S - The complex quantity — ^ H — \y- i ^s a cube root of unity. But the vector 00, of which the angle is 240°, also repre- sents a cube root of unity, if we suppose 00 = 1, because three motions of 240° each turn a vector through 720°, or two revolutions, and thus restore it to its original position. This also agrees with the algebraic process, because, by squaring the above value of x^, we have ^^ ^ -^ a/3. 1 . a/3 2 ^=-2 + -2-^ = ^^ V2 2 7~4 4+^ and by repeating the process we find Since 1 itself is a cube root of unity, because 1^ = 1, we conclude : • There are three cube roots of unity. GEOMETRIC REPRESENTATION. 415 We readily find, by the process of § 334, IV, that i, — 1, — i, and 1, are all fourth roots of unity. By a course of reasoning similar to the above for any value of n, we conclude : The n^^ roots of unity are 7i in number. EXERCISES. 1. Form the first eight powers of the expression V2 + V3 ' show that the eighth power is 1, and lay off the vector corre- sponding to each power. 2. Form the first twelve powers of a/3 1. and show that, the twelfth power is +1. 3. Find the fifth and sixth roots of unity by dividing the cir- cle into five and six parts, and either computing or measuring the lengths of the Hnes which determine the expression. Note. The student will remark the similarity of the gen- eral problem of the n*^ roots of unity to that of dividing the circle into n equal parts (Geom., Book VI). BOOK XIII. 777^ GENERAL THEORY OF EQUA^ TIONS. Every Equation has a Root. 340. In Book III, equations containing one unknown quantity were reduced to the normal form Ax'' + Bx"^-^ + Gx^-'^ + . . . . + i^ = 0. If we divide all the terms of this equation by the coefficient A^ and put, for brevity, B Pi = J. C etc. etc. F Pn = 3, the equation will become X^ + p^X^-^ +i?2^"^ + -i-Pn-l^ -{-Pn = 0. (a) This equation is called the General Equation of the fith Degree, because it is the form to which every algebraic equation can be reduced by assigning the proper values to n, and to j^i, P2, Ps7 etc. The n quantities Pi, P29 * » » * Pn are called the Coeffi- cients of the equation. We may consider pn as the coefficient of xf^ = 1. 341. Theoeem I. Every algebraic equation has a root, real or imaginary. That is, whatever numbers we may put in place o^ p^, P2, ^3, . . . . Pn, there is always some expression, real or imaginary, which, being substituted for x in the equation, will satisfy it. GENERAL THEORY OF EQUATIONS. ^Yl Rem. The theorem that every equation has a root is demonstrated in special treatises on the theory of equations, but the demonstration is too long to be inserted here. If we suppose the values of the coefficients p^p^} etc., to vary, the roots will vary also. Hence, Theorem II. ITie roots of an algebraic equation are functions of its coefficients. Example. In Chapter YI we have shown that the roots of a quadratic equation are functions of the coefficients, because if the equation is x^ + px -^ q ^= 0, the root is x = — - — -— -, which is a function of p and q. 342, Equations tvliich can de solved. If the degree of the equation is not higher than the fourth, it is always possible to express the root algebraically as a function of the coefficients. But if the equation is of the fifth or any higher degree, it is not possible to express the value of the root of the general equation by any algebraic formulae whatever. This important theorem was first demonstrated by Abel in 1825. Previous to that time, mathematicians frequently at- tempted to solve the general equation of the fifth degree, but of course never succeeded. This restriction applies only to the general equation, in which the coefficients p^, p^, p^, etc., are all represented by separate algebraic symbols. Such special values may be assigned to these coefficients that equations of any degree shall be soluble. 343. The problem of finding a root of an equation of the higher degrees is generally a very complex one. If, however, the equation has the roots — 1, 0, or + 1, they can easily be discovered by the following rules : I. If the algebraic sum of the coefficients in the equa- tion vanishes, then -\-lis a root. 27 418 GENERAL THEORY OF EQUATIONS. II. // the sum of the coeffieients of the even powers of X is equal to that of the eoeffieients of the odd powers, then — 1 is a root III. // the absolute term p^ is wanting, then is a root. These rules are readily proved by putting «= +1, then x— —1, then a; — in the general equation {a) and noticing what it then reduces to. The demonstration of II will be a good exercise for the student. Number of Roots of General Equation. 344. In the equation (a), the left-hand number is an en- tire function of x, which is equal to zero when the equation is satisfied. Instead of supposing an equation, let us suppose x to be a variable quantity, which may have any value whatever, and let us study the function of x, ^ -{-p^x!>^-^+PzX^-^ + +Pn^lOC +pn, which for brevity we may call Fx, Whatever value we assign to x, there will be a correspond- ing value of Fx, Example. Consider the expression Fx = x^ — 7^2 -I- 36. Let us suppose x to have in succession the values — 4, — 3,-2, — 1, 0, 1, 2, etc., and let us compute the corre- sponding values of Fx, We thus find, X = — 4, — 3, — 2, — 1, 0, Fx=z — 140, — 54, 0, + 28, + 36, xz=z 1, 2, 3, 4, 5, 6, 7, S.'^ Fx= -\- 30, + 16, 0, - 12, — 14, 0, + 36, + 100. We see that while x varies from — 4 to +8, the value of Fx fluctuates, being first negative, then changing to positive, then back to negative again, and finally becoming positive once more. We also see that there are three special values of x, namely, — 2, +3, and + 6, which satisfy the equation Fx = 0, and which are therefore roots of this equation. GENERAL THEORY OF EQUATIONS, 419 345. Representation of Fx ly a Curve. In Book VIII ifc was shown how a function of a variable of the first degree might be represented to the eye by a straight line. The relation between a variable and any function of it may be represented to the eye in the same way by a curve, as shown in Geometry, Book VIL We take a base line, mark a zero point upon it, and lay off any number of equidistant values of x. At each point we erect a perpendicular proportional to the corresponding value of Fx at that point, and draw a curve through the ends. The fluctuations of the vertical ordinates of the curve now show to the eye the corre- sponding fluctuations of Fx, When Fx is negative, the curve is below the base line. When Fx is positive, the curve is above the base line. The roots of the equation Fx^=.^ are shown by the points at which the curve crosses the base line. In the present case these points are — 2*, 4-3, -f- 6. In order to distinguish the roots from the variable quantity Xj we may call them «, (3, y, d, etc., or x^^ x^, x^, etc., or a^, ^2, ^3, etc., the symbol x being reserved for the variable. The distinction between x and the roots will then be this: X is an independent variable, which may have any value whatever. Fx is a function of x of which the value is fixed by that of x, a, ft y, etc., or a; 1, a; 2, 3^3, etc., are special values of x which, being substituted for x, satisfy the equation Fx = 0. Theorem. An equation' mith real coefficients, of which the degree is an odd number, must have at least one real root. Fx 430 GENERAL THEORY OF EQUATIONS. Proof. 1. When n is odd, x'^ will have the same sign (4- or — ) as X. 2. So large a value, positive or negative, may be assigned to X that the term x'^ shall be greater in absolute magnitude than all the other terms of the expression Fx, For, let us put the expression Fx in the form = ^(i+^+5+....+|). 1) If we suppose x to increase indefinitely either in the posi- tive or negative direction, the terms — , ^, etc., will all ° X x^ approach as their limit (§ 303, Th. I). Therefore the expression \ ^-tl ^(^ j^ etc. will approach unity as its limit, and will therefore be positive for large values of x, both positive and negative. The whole expression will then have the same sign as the factor x'^, and, n being odd, will have the same sign as x, 3. Therefore, between the value of x for which Fx is negative and that for which it is positive there must be some value of x for which Fx = 0, that is, some root of the equation Fx = 0. For illustration, take the preceding cubic equation. Cor. An equation of odd degree has an odd numher of real roots. For, as Fx changes from negative to positive infinity, it must cross zero an odd number of times. 346. Theorem I. // we divide the expression Fx by X — a, the remainder will he Fa, or Kemainder = a^ -^ p^a^~^ + p^oP''^ + . . . . + ptf Special Illustration, Let the student divide ^-^bx^ -{- dx H- 1 by x — a, according to the method of § 96. He will find the remainder to come out a^-^ba^ -\-Za + 1. GENERAL THEORY OF EQUATIONS. 421 General Proof, When we divide Fxhy x^ a, let us put Qy the quotient ; R, the remainder. Then, because the dividend is equal to the product, Divi- sor X Quotient + Remainder, {x — a)Q + R= Fx. Two things are here supposed: 1. That this equation is an identical one, true for all values of X, This must be true, because we have made no supposition respecting the value of x, 2, That we have carried the division so far that the remain- der R does not contain x. Because it is true for all values of x, it will remain true when we put x = a on both sides. It thus reduces to R = F{a), which is the theorem enunciated. The value of x being still unrestricted, let us in dividing take for a a root a of the general equation Fx = 0. Then, by supposing x = a, the equation (a) will be satisfied, or Fa z=z 0. Therefore if we divide the general expression Fx hy x -— a, the remainder Fa will be zero. Hence. Theorem II. If we denote by a a root of the eqitation Fx = 0, the expression Fx will be exaetly divisible by X— a. Illustration. One root of the equation a;3 _ ^2 _ 11^ _|. 15 _. is 3. If. we divide the expression 7^ — x^ — ll:r + 15 by X — 3, we shall find the remainder to be zero. 347. When we divide Fxhj x — a, the highest power of X in the quotient will be x^~K Therefore the quotient will be an entire function of x of the degree n — 1. 422 GENERAL THEORY OF EQUATIONS, Illustration, The quotient from the last division was x^ + 2x- 5, which is of the second degree, while the original expression was of the third degree. If we call this quotient F^x, we shall have, by multiplying divisor and quotient, Fx =z (x -— a) F^x, Now suppose (i a root of the equation i^i« = ; then F^x will, by the preceding theorem, be exactly divisible hj x — (i. The quotient from this division will be an entire function of X of the degree n — 2. This function may again be divided by x — y, representing by y the root of the equation obtained by putting the function equal to zero, and so on. The results of these successive divisions may therefore be expressed in the form Fx :=: {x — a) F^x .... (Degree ^ — 1), ^ F^x — {x — P) F^x (Degree ti — 2), >- (1) F^x = {x — y) F^x .... (Degree n — d),) etc, etc. etc. Since the degree is diminished by unity with every division, we shall at length have a quotient of the first degree in x, of the form x-^e, ^ e being a constant. I Then, by substituting in the equations (1) for each func- tion of x its value in the equation next below, we shall have ^ I Fx = (x — a){x — P){x-'y) {x -- e), the number of factors being equal to the degree of the original equation. Hence, Theorem I. Every entire function of x of the nth degree may he divided into n factors, each of the first decree in x. NVMBER OF ROOTS. 423 Since a product of several factors becomes zero whenever any of the factors is zero, it follows that the equation Fx — will be satisfied by putting x equal to any one of the quantities €c, P, y, , . » . e, because in either case the product {x — a)(x — P) [x — y) . , , , {x — e) will vanish. Therefore the quantities «, /3, y, £, are all roots of the original equation Fx = 0. Hence, Theorem IL An algebraic equation of the n^^ degree has n roots. We have seen (§ 195) that a quadratic equation has two roots. In the same way, a cubic equation has three roots, one of the fourth degree four roots, etc. Moreover, a product cannot vanish unless one of the factors vanishes. Hence the product Fx or (x — a)(x ^P){x — y^ , . , , (x^ e) cannot vanish unless x is equal to some one of the quantities, «, i3, y, . . . . e. Hence, An equation of the rfi^ degree can have no more than n roots. 348. We may form an equation of which the roots shall be any given quantities, a, Z>, c, etc., by forming the product, (x — a)(x — h) {x — c), etc. Example. Form an equation of which the roots shall be — 1, -f 1, 1 + 2i, 1 — 2/. Solution, We form the product {x+l)(x^l){x-^l- 2i) {x-^l + 20, which we find to be x^ — 2a^ -i- ^x^ -{- 2x -- 5. Therefore the required equation is x^ — 2:i^-\- 4t2 -^2x — 6 = 0. 424 GENERAL THEORY OF EQUATIONS. EXERCISES. Form equations with the roots : 1. 2 + V3, 2 ~ \/3, — 2, +1. 2. 3 + V5, 3 - ^5, - 3. 3. 2, -2, 4 + V7, 4-V7. 4. 1 + V3, 1 - V3, 1 + V5, 1 - V5. 349. When we can find one root of an equation, then, hy dividing the equation by x minus that root, we shall have an equation of lower degree, the roots of which will be the remain- ing roots of the given equation. Example. One root of the equation x^ — x^ — 11.T + 15 =: is 3. Find the other two roots. Dividing the given equation by a: — 3, the quotient is x^ j^^x — 5. Equating this to zero, we have a quadratic equation of which the roots are — 1 4- V6 and — 1 — a/6. Hence the three roots of the original equation are 3, — l+VO, — 1 — V6. EXERCISES. 1. One root of the equation x^ _ 32;2 — Ux + 12 = is— 3. Find the other two roots. 2. Find the five roots of the equation a:5 _ 4^ _^ I2x^ ^ 4^2 __ 13^ _- 0. (Compare § 343.) 350. Equal Roots, Sometimes, in solving an equation, several of the roots may be identical. For example, the equation 2;3 _ 6^2 _|. 12^ _ 8 = COEFFICIENTS AND ROOTS. 425 has no root except 2. If we divide it by x — 2, and solve the resulting quadratic, its roots will also be 2. Hence, when we factor it the result is {x -2)(x — 2) (x-2) = 0. In this case the equation is said to have three equal roots. Hence, in general, TJte n roots of an equation of the n^^ degree are not all necessarily different from each other, hut two or more of them may he equal. Relations between Coefficients and Roots. 351. Let us suppose the roots of the general equation of the n^^ degree x^ + p^x^-^ + 2h^~^ + .,..+ pn-\ ic + jt?» = to be «, ^, y, . . . . €. We have shown (§ 341) that these roots are functions of the coefficients p-^, p,^, . . . . pn-. To find these functions is to solve the equation, which is generally a very difficult problem. But the coefficients can also be expressed as functions of the roots, and this is a very simple process which we have already performed in some special cases by forming equations having given roots (§ 348). If we form an equation with the two roots, a and i3, the result will be . = {x — a){x-^(i) = x^—{a + p)x + af). Comparing this with the general form, x^ -{- p^x 4-i?2 = ^> we see that p^ = — [a ^ (3)^ P2 = ^P, a result already reached (§§ 198, 199). Next form an equation with the three roots, a, P, y. Multiplying {x — a) (x — (3) by x — y, we find the equa- tion to be a^— {a + 13 + y)x^ + {ap -\- Py + ya)x — aPy = 0. 426 GENERAL THEORY OF EQUATIONS. So in this case, p^ = — (« + /^ + y), ;?2 = «3 + Py + ya, Vz = — «/^r- Adding another root (5, we find the result to be Pi = - (« + ^ + y + ^), ;?2 = «/3 + c«y + «(5 + i3y + /3(y + yd, (2) Ps = — ccpy — apd — ayd — jSyd, p^ = €cpy6. Generalizing this process, we reach the following conclu- sions : The coefficient p^ of the second term of the general equa- tion is equal to the sum of the roots taken negatively. The coefficient i^g of the third term is equal to the sum of the products of every combination of two roots. The coefficient p^ of the fourth term is equal to the sum of the products of every combination of three roots taken negatively. The last term is equal to the continued product of the neg- atives of the roots. 353. Symmetric Functions, It will be remarked that the preceding expressions for the coefficients p^, pc^, etc., are all symmetric functions of the roots a, p, y, etc, (§ 256.) The following more extended theorem is true : Theorem. Every rational symmetric function of the roots of an equation may he expressed as a rational function of the coefficients. Example. From the equations (2) we find p,^ - 2p^ = C.2 4. /32 + ^2 + ^^ ^PiP2 -Pi^ _ 3j93 = «3 + i33 + y3 + ^. We thus reach the curious conclusion that although we may not be able to find any individual root of an equation, yet there is no difficulty in finding the continued product of the roots, their sum, the sum of their squares, of their cubes, etc. The general demonstration of this theorem, and the methods by which any rational symmetrical function of the roots may be determined, are found in more advanced treatises. DERIVED FUNCTIONS. 427 Derived Functions. 353. Def. If in the expression we substitute x^Ti for x^ and then develop in powers of Ti^ the coefficient of the first power of Ti is called the First Derived Function of oc. To find the First Derived Function. Putting x -{-h for x, the result is F{x-\-Ji) = {x+h)^-{-p^{x + h)^-^ + ,...+pn-i{^+Ji)+Pn' (a) Developing the several terms of the second member by the binomial theorem, we have n in —— 1 1 {x + hy = ^'^ + nx^'-^n H ^-- -x^-'^W + etc., {x + Tif-^ = x"^-^ + {n — 1) x'^-^h + etc., {x + ny-'^ = x^-^ + (n — 'Z) x^-^h H- etc., etc. etc. etc. Substituting these expressions in the equation {a) and leaving out the terms in h^, h% etc. (because we do not want them), we have F{x + h) = x^ + p^x'^-^ + p^^x^''^ + + Pn-i ^ + Pn + [nx^'-^^{7i—l) p^x'^-^-]-{n—'Z)p^x^-^-\- -\-pn-i\ h + omitted terms 7nultiplied ly li^, W, etc, (b) We see that the first line is here the original Fx, while the coefficient of A in the second line is by definition the derived function. So, if we put F'x, the derived function of Fx, we have F{x + h) = Fx + 7i F'x + terms x h^, h% etc. Let the student, as an exercise, now find the derived function of ^4 + 30^3 _ 5:^2 + 7^. _ 9 by the process just followed, commencing with equation (a). Examining the coefficient of h in (Z>), we see that the de- rived function is formed by the following rule : 428 GENERAL THEORY OF EQUATIONS. Multiply ecwli term by the exponent of the variable in that term, and diminish the exponent by unity. The last or constant term disappears entirely from the ex- pression. EXERCISES. Form the derived function of the following expressions : 1. ^5 _!_ 5^ _l_ 8^ _ 2a;2 _ ^. _l_ 1. Ans. bx^ + 20.^•3 + 24:^:2 _ 4^. __ 1, 2. x^ — 2x^ — 2.^3 _ %x. 3. i^ + 12:^:5 _ 242;3 -f x^ + 7. 4. r^ — 2ax^ + 3¥x^ + a^x. 5. x^ — ^mx^ + lOma^ — 16m.x\ Rem. The student shoald obtain the result by substituting x-^h toT X in each equation and developing, until he is master of the process. 3o4, Second Form of the Derived Fmiction, If, as be- fore, we put a, f3, y, 6, etc., for the roots of the equation Fx =z 0, we shall have Fx = {x — a) (x — P) {x — y) . , . , {x — e). (c) Let us form the derived function from this expression. Putting X -\- h for x, it will become {h -i- X — a) {h + X — p) {h + X — y) {h -^ x — e). Studying this expression, and forming the products which contain h when three or four factors only are included, we see that the coefficient of the h in the first factor is (x—p) {x — y) . . . . , in the second factor (x—a) (x—y) . . . . , etc. That is, the total coefficient of h will be (x — p) {x — y) . . . , {x — e), omitting first term ; ^ i^x — cc) (x — y) . . . . {x — e), omitting second term ; etc. etc. etc. + {x — a) (x — P) {x — y) . . . , omitting last term. But comparing with (c), we see that the first of these Fx Fx products is , the second is — — ^ , etc., to the last, X — cc X — p Fx which is Hence, X — e DERIVED FUNCTIONS, 429 X— ax — [3x — y X — E ^ ' Illustration, Let us take once more the expression of § 344, Fxz=zQ^^nx^ + 36, of which the three roots are —2/3, and 6. Its derived func- tion, by method (1), is Zx^ — Ux. Expressing Fx as a product of factors, it is Fx = {x -i- 2) {x — 3) {x — 6). By (d) the derived function is (x -3){x^6) + (x + 2){x-6) + {x + 2) {X - 3), which reduces to dx^ — Ux, the same value as by the first method. 355. Theorem I. When the derived function is posi- tive, the original function increases with x; when it is negative, the function decreases as x increases. Proof, When we increase x by the quantity h, Fx is changed to F{x -{- 1i), and is increased by the difference F{x -\-h) — Fx. But, by {h) and (&'), we have F {x -\- h) — Fx =z h F'x + ¥ x other terms =z h {F'x + A X other terms). {e) Now we may take the increment h so small that h x other terms shall be less than F'x, and then F'x -^ hx other terms will have the same sign (+ or — ) as F'x. Then, supposing h positive, the increment F{x -\- h) — Fx will be positive when F'x is positive, and negative when it is negative. Theorem II. If an equation has equal roots, such root will also he a root of the derived function. 430 GENERAL THEORY OF EQUATIONS. Proof. Let be the root which Fx =:0 has in duplicate. Then when Fx is factored, it will be of the form Fx =: {x — a) {x - P){x--P){x — y) {x - e). Now when we form F'x by method (2), the factor {x — |3) will be left in all the terms. Therefore x — fi will be a factor of F'x, Therefore, when x = 13, then F'x = 0, so that P is a root of the equation F'x = 0. 356. If the equation Fx = contains no equal roots, and if we suppose x = a in equation (d), all the terms except the first will vanish, because the common numerators Fx contain X — a as a factor. In the case of the first term, both numerator and denomi- nator vanish when x = a; therefore we must find the limit of when X approaches «. This is easy, because Fx -— - = {x-(i){x-y) (z- e). Therefore, by supposing x to approach a, we shall have Fx Lim. (x=a) = (a — p) {a ^ y) , , . , [a — e). Therefore, by changing x into a in (d), we find F'a = {a-ti){a-y) (a - e). Hence The derived function of a root which has no other root equal to it is the continued product of its difference from all the other roots. Significance of the Derived Function. 357. Theokem. The derived function expresses the rate of increase of the function as compared with that of the variable. Proof, The equation {e) may be expressed in the form F(x -f li) = Fx-\-n {F'x + Bh), FORM OF BOOTS, 431 where Bli^ is the sum of the remaining terms of the develop- ment in powers of h. We then have Increment oix =z h. Corresponding increment of Fx = F{x -\- h) ^ Fx =: h {F'x + Bh). Eatio of these increments, — '^ -—■ = F'x + Bh. If we suppose the increment h to approach zero as its limit, the product Bh will also approach zero, and the ratio will approach F'x as its limit. But this ratio of the increments may be considered as the ratio of the average rate of increase of the function F to that of the variable x. Hence, when we plot the values of Fx by a curve, as in § 345, the derived function shows the slope of the curve at each point. When the derived function is positive, the curve is running upward in the positive direction, as from x=i—d to ^ = 0, and from x=i -f5 to x=z +oo. When the derived function is negative, the curve slopes downward, as from a; = to x =z +4. When the derived function is zero, the curve at the corre- sponding point runs parallel to the base line, as at and +4f. If this point corresponds to a root of the equation, the curve will coincide with the base line at this point, and will there- fore be tangent to it. Hence, from § 356, Th. II, A pair of equal roots of an equation are indicated hy the curve touching the base line without intersecting it Forms of the Roots of Equation. 358. Theorem I. Imaginary roots enter an equation with real coefficients in pairs. That is, \i a + hi be a root of such an equation, then a — bi will also be a root. 432 GENERAL THEORY OF EQUATIONS. Proof. Let X^ -^p^X^-^ + PzX^''^ + + Pn-\X -{- pn = ^ (1) be the equation with real coefficients, and let us suppose that a + hi is a root of this equation. If we substitute a + hi for X, we shall have a;« = a« + na^-^ hi — !Ll!^lJ ^/i-2 ^,2 _ ^ ^n-3 ^,3^-4. etc. p^x^-^ = p^a'^~^ + p^ (n — l)a'*""% — etc. If we substitute all the terms thus formed in equation (1), and collect the real and imaginary terms separately, we shall have a result A + Bi = (§ 324), A signifying the sum of all the real terms, a^ ^^ ^ «^-2 yi^ p ^ (^n-i^ etc., and Bi the sum of all the imaginary ones. In order that this equation may be satisfied, we must have identically A =zO, B z=0 (§ 327). Next let us substitute a — hi for x. Since the even powers of hi are all real, and the odd powers all imaginary, this change of sign will leave all the real terms in (1) unchanged, but will change the signs of all the imaginary terms. Hence the result of the substitution will be A - Bi. But if « + hi is a root, then, as already shown, ^ = and ^ = ; whence A- Bi=zO also, and therefore a — hi is also a root. Def, A pair of imaginary roots v^hich diflfer only in the sign of the coefficients of the imaginary unit are called a pair of Conjugate Imaginary Roots. Theorem II. In the expression Fx every pair of conju- gate imaginary factors form a real product of the second decree in x. DECOMPOSITION OF RATIONAL FRACTIONS. 433 Proof. If in the expression Fx — {x — a) {x — (^){x ^y) . . , . {x — s), we suppose a and i3 to be a pair of conjugate imaginary roots, which we may represent in the form a =1 a -^ hi, (3 = a — hi, then the product of the terms {x — a) {x •— h) or of {x — a — hi) {x — a + hi), will be {x — of 4- b\ or x^ — '^ax -\- a^ + }^, a real expression of the second degree in x. Cor, Since Fx can always be separated into factors of the first degree, either real or imaginary (§ 347, Th. I), and since all the imaginary factors enter in pairs of which the product is real, we conclude : Every entire function of x with real coefficients may he divided into real factors of the first or second degree. Decomposition of Rational Fractions. 359. Def. A Rational Fraction is one whicli may Ibe reduced to tlie form ax>^ + hxi^-^ + cx!^-^ + ....+/ X^ + p^X^-^ + p^X^-^ + . . . . -{- pn If the exponent m of the numerator is equal to or greater than the exponent n of the denominator, we may divide the numerator by the denominator, obtaining a quotient, and a remainder of which the highest exponent will not exceed 71-^1. If we put fx, the numerator of the above fraction ; Fx, its denominator ; Q, the quotient; (px, the remainder ; "^x d)X we shall have, Rational fraction r= 4t- — (^ -f -^* f§ 96.) Fx ^ ^ Fx ^^ ' 28 434 GENERAL THEORY OF EQUATIONS, Q will be an entire function of x, with which we need not now farther concern ourselves. The problem now is, if possible, to reduce the fraction •^ to the sum of a series of fractions of the form Fx ABC E x — a X ^ 13 X — y ' ' * ' ~ x — e' A, By C, etc., being constants to be determined, and a, p, y, etc., being the roots of the equation Fx — 0. Let us then suppose Fx'~x — a'^X'--li'^x — y'^ '^ x -• e ^^ Multiplying both sides by Fx, we have AFx BFx CFx EFx X — a X — P x — y x — E ^ ' We require that this equation shall be an identical one, true for all values of x. Let us then suppose a; = «. Then because by hypothesis cc is a root of the equation Fx = 0, we have Fa = 0, and the terms in the second member will all vanish except the first. If there is only one root «, we have (§ 357), ^. Fx ^, Lim. {x=a) = Fa. X — a Therefore, changing x to a, we have (pa = AF'a, which gives A = ~j — In the same way we may find B--^ (c) etc. etc. Sttbstittttiiig these values of A, B, etc., in tlie equation {b), it becomes DECOMPOSITION OF RATIONAL FRACTIONS. 435 _0X _ (Pa (t>? _ (t>y , i^o; ~ \x - a) Fa ^ {x ~ fi) F'[3 '^ {x - y) F'y "^ '^• Note. The critical student should remark that in the preceding analysis we have not proved that the expression of the rational fraction in the form ih) is always possible, but have only proved that if\i be possible, tlien the coefficients A^ B, G must have the values (c). To prove that the form is possible, the second member of Q)) may be reduced to a com- mon denominator, which common denominator will be Fx, and the sum of the numerators equated to (px. By equating the coefficients of the separate powers of x, we shall have n equations to determine the n unknown quantities A, B, G, etc. Since n quantities can, in general, be made to satisfy n equations, values of J, B, G, etc., will in general be possible. It will be instructive to solve the following exercises, both directly and by the comm.on denominator. I. Decompose EXAMPLES. 2a;2 _ 3a; + 5 a^^^x^j^ 36 We have already found the roots of the denominator to be - 2, 3, and 6. Using the formulas (c), we find (t>x = 2x^ — dx-\-5, Fx = x^— 7x^ H- 36 = {x-{- 2) {x — 3){x^ 6), F'x = Sx^ — Ux; (pa = 19, (pp = 14, 0y = 59 ; F'a = 40, F'P = - 15, F'y = 24. 2x^ — 3x + 5 19^ 14 59 x^ — W + 36 "" 40 (x + 2) '~ \h{x — 3) "^ 24 (a; — 6)' _ 2^3 — 7^ + 3 2x^ — lx-^3 2. Decompose — ' x^ — 2x^ — X -^2 (^+1) (^—1) {x—2) Here the roots of the denominator are — 1, 1, and 2. Let us effect the decomposition by the following method. Assume 436 GENERAL THEORY OF EQUATIONS, 2:^2-7^ + 3 A ^ B G {x + l){x — \){x-'2) x-\-l^ x — 1^ x — ^ Reducing the second member to a common denominator, it becomes A {x^ — 3a; + 2) + ^ (^^ — ^ — 2) + { x^ — 1) {x J^l)(x — l){x — ^)~ Since both members now have the same denominator, their numerators must also be equal. Equating them, after arrange ing the last one according to powers of x, we have {A^B-^C)x^- {dA-\-B)x-^^A—'^B--C— 2.7:2.-7^ + 3. Since this must be true for all values of x, we equate the coeflBcients of a; in each member, giving A -\- B -^ C = ^, 'dA-\- B = 1, 2A-2B- C= 3. These equations being solved give A = 2, B z=l, (7 =: — 1. Substituting in (d), 2a;2 — 7a: 4- 3 2 1 1 (a: + 1) (a; — 1) (a; — 2) a; + 1 "^ a: — 1 x — 2 EXERCISES. Decompose : a^jMO 2a^ — 12a^ — 8ar + 12 ^* x^-^ox^^fl 2a x^ 4- Sx -\-4: x^ -{. x^ — 4.x — 4: X x^-^a^ aW (a;2 _ «2) {^ - h^) x^ — w^ 360. When the equation Fx =zO has two or more equal roots, the preceding form fails, because all the terms of the second member of {h') will then vanish when we suppose x equal to one of the multiple roats. In this case we must pro- ceed as follows : DECOMPOSITION OF RATIONAL FRACTIONS. 437 If Fx = {x — a)^ {x — 0)^ (x — y)P, we suppose iric (:^ _ a)^ '^ {x— cc)^-^ '^ (x — «)'^-2 "^ "^ o; — €« + (^ _ /3)^ + J^'^yi^i + + ^z:^ + ^ + ^-1 4- 4. _^^. ^ {x — y)P ^ {x-^ y)P-i ^ * * ' ^ ^ — y etc. etc. etc. In the case of m, n, or p = 1, this form will be the same as (b), as it should. By reducing the second member to a common denominator, and equating the sum of the numerators to (px, we shall have, as before, a number of equations the same as the degree of x in Fx. EXAMPLE. ^ Sx^ — 9x^ — 2:r — 1 Decompose x^ _ 2x^ — 2x^ -^ 4:X^ -^ X — 2' of which the roots of the denominator are — 1, — 1, 1, 1, 2. Solution. Because of the roots just given, the expression to which the fraction is to be equal is A A, B B. G (x^lf^x--l^(x-{-lf^x-\-l^x — 2 Eeducing to a common denominator, and equating the co- efficients of the powers of x to the coefficients of the corre- sponding powers in the numerator 8::^^ — 4cX^ — 2a: — 1, we have A^-{-B^-{-C= 0, — Ji + ^ -3^1 + ^ = 8, — 3J[i + ^1 - 4^ — 2C = — 9, 2A^ —2A^2B^-\-2B+C=—l. Solving these equations, we find, A = 1, B = 2, 0=3. A, =-2, B,=.-l, 438 GENERAL THEORY OF EQUATIONS. The given fraction is therefore equal to {x — iy x — l'{x-\-lY x-j-1 ' x^ 2 EXERCISES. Decompose ^^^- ^ns. ^ + j;J-jy,. x-1 x^-2 3- • {X + 1)2 '^' x^ — X^-X + 1 x -^2 '^' cx^ -\-~x^ — x — 1 Greatest Common Divisor of Two Functions. 361. When we have two equations, some values of the unknown quantity may satisfy them both. They are then said to have one or more common roots. Such equations, when factored as in § 347, will have a common factor or divisor for each common root. Hence, Theorem. The common roots of two equations majj be found frovv their greatest comjmon divisor. Problem. To find the greatest cowymon divisor of two equations. This problem is solved by dividing the two polynomials by the methods of §§ 96, 97, and 232. Example i. To find the greatest common divisor of the two polynomials, and x^—2x^ + 4:X^ + 2:^; — 5. FIRST DIVISION. a;5 _ 4^ ^ 12a;3 ^ 4^2 _ 13^ I x^ — 2a^ + ^x^ + 2x — 5 a;5 _■ 2x^ + 4a^ + 2x^ -- 6x \ x —2 — 2a;4+ 8.^3+ 2x^ — Sx — 2x^ 4- 4:X^ — Sx^ — 4a; + 10 4:X^ + 10^:^ — 4:X — 10 = first remainder. GREATEST COMMON DIVISOR. 439 SECOND DIVISION. - 4t -10 x^ + I- ^^ — i?:2 _ 1^ i^ -1 -^x^+ 6x^-{-%x-6 ^:^;2 «_^ — second remainder; or, 5^(:r2 — 1) = second remainder. In the next division, we may omit the fractional factor ^-, because every value of x which satisfies the equation x^ — 1 =z will also make -^ {x^ — 1) ~ 0, so that these two equations have the same roots. In this process we may always multiply or divide the terms of each remainder by any factor which will make their coefficients entire. THIKD DIVISION. 4:X^ + 10:^2 — 4:X — 10 x^-1 4,0? ^4,x 4.x +10 10a;2 — 10 10a;2 — 10 Hence, the G.C.D. of the two functions is x^ — 1, and their common roots are +1 and —1. This result may also be reached by factoring the given equations, and multiplying the common factors, thus : a^ -_ 4^4 _|_ i22;3 + 4a;2 — 13a; = X {x — 1) {x + 1) {x -' 2 — di) {x — 2 + 30, x^^2x^-\- 4a;2 + 2:^; — 5 = {x — l){x + l){x^l-- 2i) (a; - 1 + 2i), We see that the common factors are (x -~ 1) (r^; + 1) = a;2 - 1. 440 GENERAL THEORY OF EQUATIONS. The rules for throwing out factors from divisor or dividend are as follows : I. If both given polynoTnials contain the same factor in all their terms, re7)%ove this factor, and after the G. C. D. of the remaining factors of the two polynomials is found, multiply it by this factor. Proof. If a be such a factor^ and X and Y the quotients after this factor is removed from the two polynomials, the lat- ter, as given, will be aX and a Y. Since a is now a common divisor of both given polynomials, if we call D the G.O.D. of Xand Z, it is evident that aD will be the G.C.D. of aX and aY. II. Any factor common to all the terms of any divi- sor, and not contained in the dividend, ma;y be thrown out. Proof. If this factor were any part of the G.C.D. sought, it would, by § 232, be a factor of each dividend. Since the only factors we require are those of the G.C.D, factors In a divisor only may be rejected. EXERCISES. Find the G.C.D. of the following polynomials: I. x^ — 1 and ^ — 1. 2 3 x^ — 1 and x^ — 1. «5 _ 2a^ — ^3 + 3«3 _ 2a— 15 and a^^a^—^a^—a + 6. 26x^ -j-6x^ — x — l and 20a;4 -\-x^ — l. a^ +2^2 + 9 and a^ + 2a^ _ 6a — 9. m^ -\- 3m^ + 3m + 1 and m^ — 1. xi^Sa^ + 21.^2 _ 20:r + 4 and 2c(^ — 12a;2+21i2^~10. a^ -{- a^ — a — 1 and a^ + a^ — a — 1. 363. The given polynomials may be functions of two or more symbols, as in § 97. We then arrange them accord- ing to the powers of one of the symbols, and perform the divi- sions by the precepts of § 97. I GREATEST COMMON DIVISOR. 441 Ex. Find the greatest common divisor of q:^ —- ax^ + a {b -{- c) X — ahc — hx^ — cx^ + hex and 01? — ax^ — a{b + c)x — abc + hx^ -\- cx^ + hex. The quotient of the first division will be unity, so we write the two functions under each other, thus : a? — {a + h -{- c)x^ -^ {ah -\- he -^ ca)x — ahc a? + {— a -^h -^ e)x^ — {ah — he -\- ca) x — ahe — 2{h + e)x^+ 2 (a^ + ae) x = 1st rem. Dividing this remainder by —2{h+ e), we have the next divisor. We then perform the next division as follows : a? + {—a-^h + e)x^ — {ah—he+ea) x — ahe x^ — ax^ x^ — ax X -V^b^rc) {b-\-G) x^ — {ah—hC'\-ca) x — ahe {h -\-e)x^ — {ah + ca) x hex — ahe = 2d rem. Dividing this by the factor he, which is contained in all its terms, we have x — a for the next divisor, which we find to divide the last divisor, and therefore to be the G.C.D. required. 'EXERCISES. Find the G.C.D. of a^ + Shcx-h ¥ — €^ and a? 4-{c-^h) x^-i-{P'i-he+(?) x, a? + Sax -\- a^ — 1 and x^ — {a^ — 2a) x -{- a — 1. {a-{-h-{- c) {ah + ^c + ca) — ahe and a^ -j- ah — ae — he, x^ + 4a^ and x^ — 2a^x -f- 4:aK x^ — ax^ — ¥x + ah^ and x^ — a^, x^ -\- a^ -^ ¥ — dahx and x^ + 2ax + a^ — 5^. ^4 _ 2a;2 + 2 - -, + ^ and 0:4 __ 2^:2 + -o - 1^. X^ X^ 0^ 7^ x^ — x^y + xy^ — y^ and x^ -\- x^y^ + y^. 442 GENERAL THEORY OF EQUATIONS. Traiisformation of Equations. 363. Def. An equation is said to Tbe Transformed when a second equation is found whose roots bear a known relation to those of the given equation. Rem. Sometimes we may be able to find a root of the transformed equation, and thence the corresponding root of the original equation, more easily than by a direct solution. Problem I. To change the signs of all the roots of an equation. Solution, By changing x into — ic in a given equation, the signs of the terms containing odd powers of x will be changed, while those of the even powers will be unchanged. Hence, if a be any root of the original equation, — a will be a root of the equation after the signs of the alternate terms are changed. Hence the rule : Change the signs of the alternate terms, of odd and even degree, in the equation. Problem II. To diminish all the roots of an equa- tion hy the same quantity h. Solution, If the giyen equation is . x^ + p^x^-^ + p^xP'-^ + -\-pn = 0, and if y is the unknown quantity of the required equation, we must have y =. X -- h. Therefore, x :=: y -{- h. Substituting this value of x in the equation, it will become ?/''-i-(i?i+^/02/''"H ^,^ + (^_l)^^7, + g)/,2 l^^-2 + etc. {a) When h, n, and the ^'s are all given quantities, the coeffi- cients of y become known quantities. I GENERAL THEOET OF EQUATIONS. 443 EXERCISES. 1. Transform the equation x^ — ^x — ^ z=z into one in which the roots shall be less by 1. 2. Transform x^ — dx^ -\- blx — 1 = into one in which the roots shall be greater by 5. 364:, Removing Terms from, Equations, The quantity li may be so chosen that any required term after the first in the transformed equation shall vanish. For, if we wish the second term of the equation {a) to vanish, we have to suppose j)^ + nil =z 0, which gives ^ = — ^. We then substitute this value of Ji in the equation {a), which gives an equation in which the second term is wanting. If we wish the third term to vanish, we must determine h by the condition which requires the solution of a quadratic equation. Each consecutive term is one degree higher in the unknown quan- tity h, and the last term is of the same degree as the original equation. This method is principally applied to make the second term disappear, which requires that we put n Example. Make the second term disappear from the fol- lowing equation, x^ ■\- px -^ q •=: 0, Solution, Hence, w = 2 and p^ = p, so that ;. = -|. P 444 GENERAL THEORY OF EQUATIONS. f-P +g = 0, Making this substitution, the equation becomes 4 which i« the required equation. Eem. This process affords an additional elegant method of solving the quadratic equation. The last equation gives /p2 1 The value of x, being equal to i/ + h, then becomes which is the correct solution. EXERCISES. Eemove the second term from the following equations : 1. x^—%X^-\-^X—l=: 0. 2. ii;4 _ 4^3 _|_ 3^2 _ 8 zz: 0. 3. ^ -^ 5^4 + 2a;3 _!_ 2^2 _ 3^ __ 0, 4. x^ — 12^5 4- 2:^2 — X =z 0. Eem. The theory of the above process will be readily com- prehended by recalling that the coefficients of the second term is equal to the sum of the roots taken negatively, or if a, P, y, etc., be the roots, It is evident that if we subtract the arithmetical mean of all the roots, that is, — — , from each of them, their sum will vanish, because ec4-^ + /J+^ + y + ^ + etc. = ~i?i +n^ = 0. n n ' n ^^ n Hence, when we put «/ — — for ic in the equation, the sum of the roots, and therefore the second term, vanish. GENERAL THEORY OF EQUATIONS. 445 365. Problem. To transform an equation so that the roots shall he mivltiplied by a given factor m. Solution, Since the roots are to be multiplied by m, the new unknown quantity must be equal to mx. So if we call this quantity y, we have y z= mXy which 2jives x ^=l —* Substituting this in the general equation, it becomes ^ + Pi —^1 + P2 —^2 + • • • • +2Jn = 0, Multiplying all the terms by m% the equation becomes yn _^ 7np^y^~^ + m^p^y^-^ + .... + yn'^pn = 0. Hence the rule,^ Multiply the coefficient of the second term by m, that of the thii^d by m^, and so on to the last term, which will be multiplied by m^. If the roots are to be divided, we divide the terms in the same order. EXERCISES. 1. Make the roots of ri;^ — 2a; + 3 = four times as great. 2. Divide the same roots by 2. 366. Problem. To transform an equation so that its roots shall be squared. Solution. Let the given equation be a;4 + PxX^ + p^x'^ + l^x + p^ =: 0. If ^ be the unknown quantity of the new equation, we must have y — x\ which gives x =: ±2/^. If we substitute a; = ^/^ in the given equation, It may be reduced to the form y'^ 4- P%y + ^4 + {PiV + Pz) r - 0. 446 GENERAL THEORY OF EQUATIONS. If we substitute a; = — y^, the result will be y'^ + ihv -^-v^ — {Piy + p-i) r = 0- Since the value of y must satisfy one or the other of these equations, it must reduce their product to zero ; we therefore multiply them together. Considering them as the sum and difference of a pair of expressions, the product will be {y^ -^p^y + V4I - {Piy ^ p^Yy = 0, or y'+{^P2-Pi')y'+{P2'+^P4-^PiP3)f+(^:P2P^'-Pz')y+p,' = 0. EXERCISES. 1. Transform the quadratic, x^ — 5x -\- 6, of which the roots are 2 and 3, into an equation in which the roots shall be the squares of 2 and 3, using the above process. 2. Transform in tlie same way ^3 ^ 12:^2 _!_ 44^. _!_ 43 _ 0, 3. Transform x^ — 4:X^ — 10ir3 + 40a;2 -{. dx — 36 = 0. Generalization of the Preceding Problems. 367. Problem. Given, an equation of any degree in an unhnoivn quantity x ; Required, to transform this equation into another of which the root shall he a given function of x, Solution. Let ?/ be a root of the required equation, and fx the given function. We must then have f^ = y' Solve this equation so as to obtain x as a function of y. Substitute this value of x in the original equation, and form as many equations as there are values of y. The product of these equations will be the required equa- tion in «/. GENERAL TUEOBT OF EQUATIONS. 447 EXERCISES. 1. Transform iC2 _ 7x + 10 =: SO that the roots of the new equation shall be '^x\ 2. Transform x^ — ^x^ + 2cc = so that the roots shall be ax + h. 3. Transform x^ — ^x -\- 1^ =: so that the roots shall be ^x^' — 3. o Resolution of Nvimerical Equations. 368. Convenient method of computing the numerical value of an entire function of x for an assumed value of x. If we have the entire function of x, Fx = ax^ + bx^ + cxP -^ dx + e, we may put it in the form Fx =1 "^[^(^ax -\- l))x -{■ c'\x + d\ X -\- e. Therefore^ if we put ax -\-h ^ V , h'x -\- c =: c\ c'x + c? =: d', d'x -{- e =^ e\ we shall liave Fx = e\ Numerical Example. Compute the values of Fx = 2qP — ^x!^ — 6a;3 + 82) — 9 f or a; =: 3 and a; = — 2. We arrange the work thus : Coefficients, Prod, by {x=^), 2 -3 -6 + 6 +9 ~+3 +3 + 9 + 9 + 8 + 27 + 35 - 9 + 105 + 96 Hence, F^ = 96. For 2: = — 2, 2 -3 - 6 —4 +14 -7 +8 —16 —16 + 8 + 32 + 40 — 9 —80 -89 Hence, F(~2)^ - 89. 448 GENERAL THEORY OF EQUATIONS, This, it will be noticed, is a more convenient process than that of forming the powers of x and multiplying and adding. 369. Having an entire function of x, and jMtting x:=.r-\-li^ it IS required to develop the function in poiuers of lu It will be remarked ihat this problem is substantially identical with that of § 362, and the solution of this will be the solution of the formero But in the former case h was supposed to be a given quantity, whereas it is now the unknown quantity corresponding to y in the former problem. Example of the Problem. If we have the expression Fx — 2x^ + 3a;2 + 4, and put X = 2 -\- h, it will become, by developing the sepa- rate terms, F {2 -\-h) = 2¥ + loh^ + 367^ + 32. Ge:n"eral Rule for the Process. First compute the value of Fr by the process employed in § 366. Then repeat the process, using the successive sums ob- tained in the first process instead of the corresponding coefficients, and stopping one term before the last. The result will be the coefficient of li. Repeat the process with the new sums, stopping yet one terin sooner. The result luill be the coefficient of W. Continue the repetition until ive have the first term only to operate upon, which will itself be the coefficient of the highest power of h, Ex. I. The example above given is performed as follows: Coefficients, + 2 + 3 + 4 Product by r, 4 14 28 First sums, 7 14 32 Second products, 4 22 Second sums, 11 36 Tliird product, 4 15 Result, F{2 + h) 1 ^ 2h^+inm+mh+ 32. Ex. 2. In the function^ Fx = 2x^ —7x^ + 52^ — 2x^ ^ Qx — 8, let us put X =1 3 -{- Ji, and express the result in powers of y^. GENERAL THEORY OF EQUATIONS. 449 CoefEcients, 2 Products by 3, First sums. Second products, -7 6 -1 + 6 + 5 -3 + 2 + 15 -2 + 6 + 4 + 51 + 6 + 12 + 18 + 165 -8 + 54 + 46 Second sums, Third products, + 5 6 + 17 33 + 55 150 183 Third sums. 11 6 17 6 23 50 51 101 205 Result, F{^-]-h) = 2h' + 237^^ + lOl/i^ + 205^^ + 183^ + 46. EXERCISES. 1. Compute 2A5 _^ 23/^4 _|_ 101/^3 _^ 205/^2 -f 183^ + 46, when h = x — d. 2. Compute a^ — "^/x + 7 f or i^; = — 4 + 7^, — 3 + k, etc., to +3 4- h. Proof of the Preceding Process. If we develop the ex- pression a{h + r)'' -\-'b{n-\-rY-^ + c(7i + r)^-2 _^ d{7i-^rY-^ + etc., and collect the coefficients of like powers of h, we shall find Coef. of 7^^ — a, li^-^ r=r nar + h, nn-'i = {^\ ar'^ H- {n -\)lr -\- c, {A) in-^ = [|] ar^ + (--^) br^ + (^^ - ^) cr + d, Now examining Ex. 2 preceding, it will be seen that we can make the computation by columns, first computing the whole left-hand column and thus obtaining the coefficient of h^~^, then computing the next column, thus obtaining the coeffi- cient of h^~% and so on. Commencing in this way, and using the literal coefficients, a, b, c, etc., and the literal factor r, we shall have the results : 450 GENERAL THEORY OF EQUATIONS. ah c j ar ar^ + br ar -\- b ar^ -{- hr -{- c ar 2ar^ + br 2ar + b dar^ + 2br + c ar 3ar^ + b r 3ar + b 6ar^ + 3br + c nar -^ b (9) ^^^ + {n —l)br-[- c. If n is the degree of the equation, then, by the preceding process, we shall add the product ar to b n times, the n sepa- rate sums being ar-\-b, 2ar-\-b, 3ar-\-b, .... nar-{-b. To form the second column, we multiply each of these sums except the last by r, and add them to the coefficient c. The terms in ar added being ar^, 2ar^, 3ar% etc., the sum will be (1 + 2+3 + -\-7i — l) ar^. The coefficient is a figu- ni (^ __ '^\ rate number equal to — ^^-^r (§§ 286, 287). The sum of 2 the coefficients of br is n — 1, because there are n — 1 of them used, each equal to unity. Therefore the final result is ( - j ar'^' + {n — l)br -\- c, - which we have found to be the coefficient of /^^~^. In this second column the partial sums or coefficients of ar'^ are ^ 1, 1 + 2 = 3, 1 + 2 + 3=: 6, etc., to 1 + 2 + 3 + .... +(/^— 2)= Therefore the numbers successively added to form the co- efficients of ar^ in the third column are 1, 1 + 3 = 4, 1 + 3 + 6 = 10, etc. The coefficients of br'^ will be the same as those of ai^ in the column next preceding. Continuing the process, we see that the coefficients uro formed by successive addition, as in the following table, whero each number is the sum of the one above it plus the one on its GENERAL THEOBT OF EQUATIONS. 451 ^ y. |.2 ^ y.4 ^ y.8 q\^q^ w 1 1 1 1 1 1 etc. h 2 3 4 5 6 etc. A2 3 6 10 15 etc. h^ 4 10 20 etc. U 5 15 etc. ¥ 6 etc. h^ etc. etc. etc. left. We have carried the table as far as nz=z% and the ex- pressions at the bottom of each column will, when 7^ = 6, be formed from the numbers in this table, taken in reverse order, thus : Column under hy Qar + b ; ^' " c, 15ar^+ hh* -f- c; « '' d, 20ar^ + lObr^ -{-4:cr + d; " '' e, 15ar^ + 10br^-\-dcr^+3dr + e; '' '' f, 6ar5+ 5Z»r4 + 4cr3 4-3^r2 + 26r+/; ■" " g, ar^-j- br^-i- cr^-\- dr^+ er^ +//•+?/. IS^ow the numbers of the above scheme are the figurate numbers treated in § 287, where it is shown that the n^^ num- ber in the i^ column after the column of units is n(n + 1) (^^ + 2) (n -{- i — 1) _ / nj-i-^n 1.2-3 i "~ \ i /' Comparing with the coefficients in the equations (A), we see that the two are identical, which proves the correctness of the method. Sao. Application of the Preceding Operation to the Ex- traction of the Roots of Numerical Equations* Let the equa- tion whose root is to be found be a^ + Ix^-'^ + cx'^-^ 4- . . . . -^ g — 0. We find, by trial or otherwise, the greatest whole number in the root x. Let r be this number. We substitute r-]-h for 452 GENERAL THEOllT OF EQUATIONS. X in the above expression, and, by the preceding process, get an equation in /?, which we may put in the form ali^ -f Vli^-^ + cVi^-^ + d!li'^-^ + 4- ^' = 0. Let /be the first decimal of li. We put r' + /^' for 7^ in this equation, and, by repeating the process, get an equation to determine h\ which will be less than 0.1. If r" be the greatest number of hundredths in li\ we put li! = r" -\-h", and thus get an equation for the thousandths, etc. 371. The first operation is to find the number and approx- imate values of the real roots. There are several ways of doing this, among which Sturm's Theorem is the most celebrated, but all are so laborious in application that in ordinary cases it will be found easiest to proceed by trial, substituting all entire • numbers for x in the equation, until we find two consecutive numbers between which one or more roots must lie, and in difficult cases plotting the results by § 345. It is, however, necessary to be able to set some limits be- tween which the roots must be found, and this may be done by the following rules : I. An equation in which all the coefficients, including the ahsoliite term, are positive, can have no positive real root. For no sum of positive quantities can be zero. II. If in computing the value of Fx for any assumed positive value of x, hy the process o/ § 366, we find all the sums positive, there can he no root so great as that assumed. For the substitution of any greater number will make all the sums still greater, and so wdll carry the last sum, or i^^*, still further from zero. III. If the sums are alternately positive and nega- tive, the value of x we employ is less than any root. IV. // two values of x give different signs to Fx, there must he one or some odd numher of roots hetween these values (compare § 345). GENERAL THEORY OF EQUATIONS, 453 V. Two values of x which lead to the same sign of Fx iiicliule either no roots or an even number of roots be- tween them. Let us take as a first example the equation 2;3 _ 72: -h 7 = 0. Let us first assume 2; = 4. We compute as follows : Coefficients, 1 —7 +7 Products, 4 16 36 Sums, +1 T^ T43 So F {i) rrz +43, and as all the coefficients are positive, there can be no root as great as 4. Putting 2; = — 4, the sums, including the first coefficient 1, are 1, —4, +9, —29. These being alternately positive and negative, there is no root so small as — 4. Substituting all integers between —4 and -j-4, we find i^(-4) = -29, F{0) :=: + 7, F{-^)= + 1, F{1) = + 1, F{-2) = 4-13, F{2) = + 1, F{—1) = +13, i^(3) = +13. If we draw the curve corresponding to these values (§ 345), we shall find one root between —3 and —4, and very near —3.05, and the curve will dip below the base line between +1 and +2, showing that there are two roots between these num- bers ; that is, there are two roots of the form l-\-h, li being a positive fraction. Transforming the equation to one in A, by putting 1 + A for .r, we find the equation in h to be 7^3 ^ 3/^2 _ 47^ ^ 1 :3:: 0. (1) Substituting 7^ = 0.2, 0.4, 0.6, 0.8, we find that there is one root between 0.3 and 0.4, and one between 0.6 and 0.7. Let us begin with the latter. If in the last equation we put 7^==0.6 + ^', we find the transformed equation in li^ to be Fli! = h'^ + 4.87/2 ^ 0.687/ - 0.104 = 0. (2) If we substitute different values of h' hi this equation, wo 29 451 GENtJUAL THEORY OF EQUATIONS. shall find that it must exceed .09, and as it must be less than 0.1, we conclude that 9 is the figure sought, and put h' =z .09 + h". Transforming the equation (2), we find the equation in h" to be h"^ + o.07A"2 + 1.5683//' - 0.003191 = 0. (3) Since h" is necessarily less than 0.01, its first digit, which is all we want, is easily found, because the two first terms of the equation are very small compared with the third. So we simply divide .003191 by 1.5683, and find that .002 is the re- quired digit of h". We now put h" = .002 + h'", and transform again. The resulting equation for h'" is jr^ + 5.0767z'"2 + 1.588592y^'" - 0.000034112 = 0. (4) The digits of x, h, Ji', and h" which we have found show the true value of a; to be , X = 1.692 + h'". By continuing this process, as many figures as we please may be found. But, after a certain point, the operation may be abbreviated by cutting off the last figures in the coefficients of the powers of h. The work, so far as we have performed it, may be arranged in the following form (see next page). The numbers under the double lines are the coefficients of the powers of h, h', h", etc. It will be seen that for each digit we add to the root, we add one digit to the coefficient of li^, two to that of h, and three to the absolute term. We have thus extended the latter to nine places of decimals, which, in most cases, will give nine figures of the root correctly. If this is all we need, we add no more decimals, but cut off one from the coefficient of h, two from that of h^, and so on for each decimal we add to the root. We shall find the next figure after 1.692 to be zero ; so we cut off the figures without making any change in the coeffi- cients. The next following is 2, so we cut off again for it, and multiply as shown in the following continuation of the process : GENERAL THEORY OF EQUATIONS. 455 -7 +1 + 1 +1 -6 +1 +2 + 2 -4.00 + 1 + 2.16 + 3.0 -1.84 + .6 + 2.52 + 3.6 + 0.6800 .6 + 0.4401 4.2 + 1.1201 6 + .4482 + 4.80 + 1.568300 9 10144 4.89 + 1.578444 9 10148 4.98 9 + 1.58859^ + 5.070 2 5.072 2 5.074 2 + 7 -6 1.692 + 1.000 -1.104 - ,104000 + .100809 - .003191000 + .003156 888 -34112 + 5.076 CONTINUATION OF FROCESS. + 15.076 + 1.5885|9!2 1 -34112 31774 1.5887 1 -2338 1589 1.5|8|8|8 — 749 636 —113 111 -2 I 021471 It will be seen that from this point we make no use of the coefficient 1 of /^^, and only with the second decimal do we use the coefficient of JiK After that, the remaining four figures are obtained by pure division. There is one thing, however, which a computer should always attend to in multiplying a number from which he has cut off figures in this way, namely : Always carry to the product the ninnher ichich would have been carried if the figures had not hceri cut off, and 456 GENERAL THEORY OF EQUATIONS. increase it hjj 1 if the figure following the one carried would have been 5 or greater. For instance, we had to multiply by 7 the number 151888. If we entirely omit the figures cut oif, the result would be 105. But the correct result is 1111216; we therefore take 111 in- stead of 105. Again, in the operation preceding, we had to multiply 158|88 by 4. The true product is 635|52. But, instead of using the figures 635, we use 636, because the former is too small by |52, and the latter too great by |48, and therefore the nearer the truth. For the same reason, in multiplying 1.58818 by 1, we called the result 1589. Joining all the figures computed, we find the root sought to be 1.692021471. Let us now find the negative root, which we have found to lie between — 3 and — 4. Owing to the inconvenience of using negative digits, and thus having to change the sign of every number we multiply, we transform the equation into one having an equal positive root by changing the signs of the alternate terms. The equation then is 2;^ — Ix — 7 =: 0. The work, so far as it is necessary to carry it, is now ar= rammed as follows: 3^ 3 3 6 o o 4 9.04 _4 9.08 _^ 9120 8 9.128 8 9.136 ^ (9l|44 -7 2 18 -7 'I 3.0489173395 6 20.0000 .8616 20.3616 .3632 20?724800 73024 20.7978-M 730 88 20.87091 12 823 1 20.87914|2 823 20T8873J7 9 20|.8|8|7|5 -1.000000 814464 -0.185536000 .166882592 - .19153408 18791228 -862180 2088': 5 ^53305 146213 -7092 6266 627 -199 188 -11 OENEUAL THEORY OF EQUATIONS. 4^7 The negative root of the equation is therefore — 3.0489173395. EXE RCI SES. Find the roots of the following equations: 1. a;s _ 3^2 ^ 1 _ (3 real roots). 2. x^ — Zx -\-l — (3 real roots). 3. x^ — 4^^ -^ 2 =2 {% positive roots). 4. ^;2 + ,^; _ 1 — 0. 5. Prove that when we change the algebraic signs of the alternate coefficients of an equation, the sign of the root will be changed. 372. The preceding method may be applied without change to the solution of numerical quadratic equations, and to the extraction of square and cube roots. In fact, the square root of a number ^ is a root of the equation x^ — ^ = 0, or x^ -^ Ox — n =z 0, and the cube root is a root of the equation a^ -\-0x^ -\-()x — n = 0, Ex. I. To compute a/2. 10 -2 1 1.4142135a 1 1 ■ 1 -1.00 1_ .96 2:0 0.4 -.0400 281 24 -11900 4 - - 11296 = -60400 2.80 56564 2.81 — J- -1008 2.820 849 1 -159 2.824 JM -18 2.8280 IZ 2 1 2.8282 2 2|.8|2|8|4 458 GENERAL THEORY OF EQUATIONS. Ex. 2. To compute the cube root of 9842036. 2 4 2 4 2 8 4 1200 2 61 60 1261 1 62 .— -- 61 132300 1 2536 62 134836 1 2552 630 137388.00 4 192.69 634 137580.69 4 638 192.78 137773.47 1.93 642.0 .3 1317|7175|4 642.3 3 642.6 3 -9842036 I 214^30303242 8 ■ -1842 1261 -581036 539344 41692000 41274207 -417793 413326 4467 4133 276 58 55 8 :64^i^ APPENDIX, SUPPLEMENTARY EXERCISES. Note. The following additional exercises and problems are of the same general character with those in the body of the book. They are partly original, and partly selected from the best recent German col- lections of problems. They are arranged under the section numbers to which they pertain, so that the teacher, on arriving at those sections, will be able to select as many of them as he deems necessary for the drill of his class. SUPPLEMENTARY EXERCISES. Algebraic Addition and Subtraction. §15. Supposing one to start from a certain point on the scale of numbers, and then move over positive and negative spaces as follows, it is required to find his stopping-point in each of the following cases: 1. Starts from + 4, and moves through + 2 — 3 + 9 — 7 — 2 units. 2. Starts from + 9, and moves through — 1 — 6 — 9 + 5 + 8 units. 3. Starts from — 1, and moves through + 2 — 3 + 4 — 5 + 6 units. 4. Starts from — 8, and moves through — 1 + 3 — 5 + 7 — 9 units. 5. Starts from — 12, and moves through —9-6 + 8 + 5 + 8 units. §31. I. How far is A from B (positively or negatively) when they have severally made the following motions from the same point on the scale of numbers: , A . , B . a. -2-3-5 + 7. +1 + 2 + 3 + 4 + 5. 2>. -5 + 5-6 + 6. +5 + 6-2-4 + 12. 0, -2 + 7 + 8 + 9 + 10. -7-3 + 4-5-6. a, _i-2 + 6-2-l. +3 + 4 + 5-8-3. Ans.-l. 462 ALGEBRAIC ADDITION 2. What is the meaning of the following expressions : That man is — 6 years older than his wife ? Kichmond is — 70 miles north of Washington ? You are — 3 inches taller than your brother ? 3. The Autocrat of the Breakfast Table tells of a Parson Turrel who, dying in the last century, bequeathed a noted chair to the oldest member of the Senior class in Harvard College, which was to be passed down from class to class indefinitely. The first Senior who got it was to pay 5 crowns, but each succeeding one was to get it at a price 1 crown less than that paid by his predecessor. How would the require- ment of the will work at the end of 7 and of 100 years? §34. I. Find the value oi a — h and of b — a when a and b have the following sets of values : (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) a= +2, +7, -9, -5, -17, + 8, -33, -18, +12, +22 b= -3, -9, -3, +8, -29, +14, +13, -19, -12, -22 a-b= +5 b—a— —5 2. Compute the values of 1 + 3^ and of 1 — 3:^; for the following 11 values of x : :^ =- 5, - 4, - 3, - 2, - 1, 0, + 1, + 2, + 3, + 4, + 5. 3. Compute the values of a -\-2b and ot a — 2b for each of the 10 sets of values of a and b in Ex. 1. §56. 1. How much is a -\- 2x greater than a — 3x, and vice versa 9 2. How much \s a — b greater than b — a? 3. How much is greater than a — 2b? 4. How much is greater than — x? Than -\- x? AND SUBTliACTION, 463 5. A party of 9 boys were formed into a solid square of 3 rows, with 3 boys in each row. The rear left- hand boy B was t inches tall. Every other boy was X inches taller than the boy next behind him, and 2/ inches shorter than the boy on his left. Ex- press the height of each boy, and the sum of the heights of all the boys. 6. During six successive days a man earned m cents more every day than he did the day before, and paid out n cents less. On the first day his earnings were h cents, and his pay- ments h cents. How much had he left at the end of the sixth day? 7. Of two travellers, X went east h miles and then returned h miles toward the west ; Y went west x miles and then returned y miles toward the east. If they started together, how far was X east of Y when they stopped ? How far was Y east of X ? 8. There were three travellers on the same road, B being X miles west of C, and C y miles west of A. A went m miles toward the east ; B went twice as far as that toward the east; and C went 4??i miles toward the west. How far was each west of the two others when they stopped ? 9. Of two men, A and B, A had a dollars and B had x dollars on Monday morning. On Monday evening A paid B d dollars, and B returned y dollars of this to A. Each fol- lowing evening during the week A paid B g dollars less than before, and B returned A z dollars less than he did the evening before. How much had each on each morning from Tuesday to Saturday? 10. Four casks, marked A, B, C and D, each containing r gallons of water, stood at the corners of a square. Then m gallons were poured out of A into B, 71 gallons out of B into C, p gallons out of into D, and q gallons out of D into A. How much was then in each cask ? Prove the result by showing that the sum of the quantities in all the casks is 4r. II. The same four casks at first contained a, h, c and d gallons respectively. Then x gallons were poured out of B into A. Then a quantity equal to what was left in B was 464 ALGEBRAIC ADDITION. poured from C into B ; a quantity equal to wliat was left in C was poured from D into ; and, finally, a quantity equal to what was left in D was poured from A into D. How much was then left in each cask ? Prove as before. 12. Three traders, A, B and 0, had a, 1) and c dollars respectively. A bought c dollars' worth of goods from B ; B, a dollars' worth from C ; and C bought I dollars' worth from A. When each paid the other for the goods, how much money had each left ? What was the sum-total of money possessed by the three ? 13. Given a quadrangle the lengths of whose sides are a, l, c and d respectively. Enough of the side b is cut off and added to a to double the latter; the re- mainder of I is then doubled by cutting off from c ; and the re- mainder of c is doubled by cutting off from d. How long will each side then be ? 14. Of two men starting out from the same point, A walked m miles west the first day, and h miles more each fol- lowing day than he did the day before ; B walked p miles west the first day, and x miles less each day than ho did the da^ before. How far was A west of B, and how far was B west of A, at the end of the first, second, third and fourth days respectively ? 15. If, on this line, we suppose the point B to be at the East. ^ ? ^^West. dislance h west of A, and C to be at the distance c west of A, then, ill alirebraic language: How fjir is A west of B ? How far is A west of ? How fjir is west of B ? How for is B west of C ? How far is the middle point between B and west of A ? How far is the middle point between C and A west of B ? CLEABING OF PARENTHESES, 465 How far is the middle point between A and B wesb of ? What is the algebraic sum of these last three distances ? Note. Should the student find any difficulty in this or the next question, he should begin by expressing the distances a and h in num- bers, and noticing the processes by which the measures are found. 1 6. The three points A, B and C are at the respective M A B C I ill distances a, h and c west of a fourth point, M. Express alge- braically the three distances B west of A ; A west of C ; west of B, and take their sum. Express also the distances A west of the middle point between M and B, B " " " M and C, C '^ " " MandA, and find the sum of these three distances. Then express A west of the middle point between B and 0, B " " " C and A, C " " " A and B, and find the algebraic sum of tlie three distances. Express also the three mutual distances !)ct\veeu tbe middle points of lines AB, BO and CA respectively — that is : Mid. point betw. A and B west of mid. point betw. B and 0, etc. etc. etc. §61. Clear the following expressions of parentheses, and com- bine the terms by addition: 1. 3 m — [A — 2m — {Ji -\- m) — (h — m)\ 2. (ct — y) ~ (a '\- V) -^ (a — m) — {a — m). 3. (a + h)-{a-^h)^ \_{a ^ h) ^{a + ^)]. 4. Vi — {37^ — [4A — {bh — m) + m] + 2m}. 5. 3c- 'M-(2d-'dc) + [- [c- d) - (3c + 2^?)]' 6. A:h - In - (4.h + In) - [3A + (4m - a) - (5m + h)]. 466 MULTIPLICATION, 7. x-\- {x — a— {2a — 2x) + [a — {a ~ x)] }, 8. 6a+ {6a -~ 2x+[4a - Sx - {3a - 4.x)]\. 9. {a + b - c) + {a- b + c)+{- a + b + c)-{a + b + c). 10. a + dx— {2a + 2x) — {da + x) — [a — {a — x)]. 11. b + [-2b-dc-{dc-b)-3b]. 12. -[- (3w-27i) + (2m-3/i)] + [{5m-A7i)-{3?i~4:m)]. Multiplication and Addition. §74. ■ Clear the following expressions of parentheses: 1. b { a{c — x) -\-b{c-\-x) -{- ax[b ~ c{x — a)]}, 2. m [x — n {b — y) -\- b {n -\- y) -\- y {n + b)\ . 3. a?i [an (1 — a7i) -\- a'^n'' (1 — an)\, 4. h{l-Vh[l + h{l + li)^\. 5. x{l-x\l-x{l-x)']}. 6. X {2) -{- x\_q~\-x (r + ^)] }, 7. a; {;? — ir [^ — a; (?' — a;)]}. 9. a I [(a:c — Z>) ic — c] 2: — 6?}. 10. i>i[(;?^+/).'?^+/]a:+y}. 11. { [{7nx — nt") X — m^] .t — ^m^ } mx, 12. [a' (^ - c) + b' {c - «) +c=^ {a - b)] abc, 13. m [a' {x + y) + b' {x-y)- x{a' + ^'^J + {a' - b') ym. 14. a {« — ^ [rt — c («^ — d)]}, .15. « |aZ> — c [a^^— c^ (a''^ — (i')]}. 16. (^ + a;)(^-.v) + («^-a;)(^+2/)- 17. {m -\- n) {x -\- y) — {m — n) {x — y), (§ 76.) Arrange the following expressions according to powers of x: 1. {x'-x'+l)a' + {x'--'X+l)a'+a\ 2. 1 -{-X — x^ "-x^ — a{l -{-X — X*) -\-a'^{l — x) — a^» 3. 7na^—-na^{x — l)--ma''(^'^— a: + 1)"~ ^i<:i^{^'— ^''+i^ — 1). 4. « — a; { ^ — :?; [c — c?; (^ — a:)] } . DIVISION, 467 5. { {ax — h) {ax -\-h) — {bx — a) {hx + a) ] {a^ + h'')x. 6. I {mx -\- ay -[- (mx — aY] (mV — a^), 7. { {inx + ccY — [mx — ay \ (m — nx) {m + fix), 8. a + X {b + X [c + X {d + x)]\ ~ a \x -^^ I \x -\- c {d -^ x)\\. §80. Write out the results of the following powers and pro- duets on sight: I. {ax-\-lyy. 2. {ax — hyy. 3. {ay + lxy. 4. {ay — bxy. 5. {ax-^2byy. 6. {ax-2byy. 7. {ax + dbyy. 8. {ax-dbyy. 9. (m + ny X. 10. (771 -j- ^^)'^ ^. II. ir(a;+.v)*- 12. ic (:?; — ?/)'. 13. a(a: — ,?/) (:r + ^). 14. ax'' {a ~ x) {a-}-x). 15. 7??/^ (2m + n) {2m — 7^). 16. mW (3m + ^?.) (3m — n). 17. (^ + by + {a- by. 18. (6j + by -{a- by. Form the values of the following quantities, and arrange according to powers of x, y and z\ 19. {ax-\-byy -\-{bx — ayy. 20. {ax -\- byy ~ {bx — ayy 21. {2mx — nyy -\- {mx — 2nyy. 22. {2mx — nyy — {mx — 2nyy, 23. {x -\- ny) {x — 7iy) {x^ — n'y^) — (y + nx) {y — nx) {y^ — n^x^). 24. {ax -\- by -{- cz) {ax + by — cz) {ax — by -\- cz) {ax — by — cz). Division. §(85.) 1. Mx^'bc^ -^2ax^b. 2. 24«':r"?/ -V- 6fl^'^V- 3. 12a^icy -^ 4a'icV. 4. a'^o; (c + ^ ) -^ ax{c-\-d). 5. a C^-y) -^a(ic~-2/). 468 FACTORING, 6. 'bc{x + y)-^h{x-^y). 7. x^'y {a — h) -^ X {a — Z>), 8. 10.T^ (a-V)-^ ^x^ {a'\'V)-^x. 10. 10 {a + Vf - 15 {a + ^)^ ~ 10 (^ + Z>) ~ 5 (a + Z>). 12. (a+a;r(a+2/r-(«+^r(^+2/r. 13. —V^oTl'^-^^a^h'^, Factoring. §89. Factor the following expressions: 2, X — ax^ + ^^^- 3- — ^' + m^?i — m^n^. 4. «m — «'^/i^ + a%^ 5. c"Z>'" — c^'^Jf'. 6. — abc-\-if)fabc, 7. ft"^^^** — a^:c" " \ 8. ahVy"^ — a'lfx^y, 9. m/z-^jf?^ — S7n^7i^p. 10. 2a"ic'" - 7a'":c^ 11. 8a^c - 12aWc\ §91. In the following exercises, first take out all monomial factors common to the several terms, as in § 89, and factor the remaining terms by the rules: 1. a^ — aV^. 2. mW — w*. 3. ^ififx — ^yi^x. 4. mx^ — m. 5. aV - a\ 6. ax^ — ax. 7. rrfx^ — m*x. 8. 9X' - 4:X\ 9. aV - 4:a'x\ xo. 7n''y^ — 7/iy. II. 16my - 2bmY' 12. 4.9aV - ma'x. 13. am^ -a^ + ^a^h - ab\ 14. a.V - 4bV + 4hcx^ — c'x\ 15. «ic'" - 4a'2:" + ^a\ 16. a'h - 4.a'F + 4.ah\ 17. 4yx' - 12xY + 9x'y\ 18. ^xY + 12^:^' + ^^Y' §93. 1. 2 {x' +y* + z*) -4: {xY + y'z' + zV). 2. a' + 16b' + c'- Sa'F - SFc' - 2cV. FAGTOUING. 469 3. 2 (xY + y'^z' + z'x') -x' - f - z\ 4. 8a^Z>V+ Z%Ve + 8cV - ^* - 16^^ - 16c*. §94. In the following, begin by removing all factors common to the two terms as in § 89: 1. «^ + 4a V + 4a V. A71S, a' {a + 2^;')'. 2. a^ — aV. Ans, a'' {a -\- x) {a — x) {a* + ^^^^ + ^*)- 3. «V — x". 4. ay' — ay . 5. (a - S)^ - c\ 6. (« + ^)^ - c^\ 7. a;' (a; - yY - a;^ 8. :c* + 8a:^/^ 9. {a + by + a\ 10. a{a + ly + a\ II. a;' + /. 12. a'+ aZ>'. 13. a" + 64m V. 14. m' + Um'x'. IS. a;= + ^'. 16. a' + 8. 17. «' + 216. 18. 64:^'' + 125cl 19. a;^ + /- 20. ic' — a'. 21. Sa' - 27b\ 22. 64m' - M\ 23. .-K' + l. 24. 64a' + Z^^ 25. a' + ab'. 26. a - 27a\ 27. ab'-b\ 28. a' - 243. 29. 32a'° + l. 30. 16a' — a. 31. 3a;' + 16. *32. 27a^ + 8a2:^ 33- (2^ + 2/)'- (a;- -yy 34. {x + f/)' + (^ - 2^)'. 35- {x + yy-{x~ -yy 2,6, 1 — a^ + 2aa; — x". Factoring Trinomials. A trinomial of the form x' -\- ax -\-h can always be factored when we can find two numbers whose sum is a and whose product is Z>. For if m and n are these numbers, the trinomial is x^ -\- {m -\- n) X -{- mn, which is equal to (x -\- m) (x -\- n). ♦ 470 FACTO BIN Q. Factor: 1. x" -{- {a-\- h) X + ah. Arts, {x -{- a) {x -{- h), 2. y'+^ + 2. Ans. {^ + l)(y + 2). 3. 2/^ + 4«^ + 3. 4. cr^ + 5.T + 4. 5. n" + 5yi + 6. 6. t^'^ + i5n + 8. 7. a' + 7«^ + 10. 8. a' + 8a + 12. 9. m' + 7m + 12. 10. m'* + 8m + 15. II. x' + Ix" + 10a;. 12. y' -\- 6y' + 8y. 13. x' + 7x' + 12a;^ 14. a' + Sa' + 15a'. 15. x' -f 19^;^ + 88. 16. a' + 12a' + 35. 17. x'"" + 9^:^ + 20. 18. y'"" + by'"" + 6. 19. ir'' + (^ "" ^^) ^ "" ^^- ^^s. {x + m) (a; — 7i), From this last example it is seen that when the quantities m and n have opposite signs the last term of the trinomial will be negative^ while the middle term will have the sign of the greater of those quantities, being equal to their algebraic sum or numerical difference. 20. x^ — X — 6, Ans. {x — 3) (x -{- 2). 21. x' -x'- 12. 22. y* - 2y' - 15. 23. a' + a — 30. 24. a^ —a — 30. 25. m* + 2m — 8. 26. m"^ — 2m — 8. 27. 7i' - dn' - 40. 28. m' + 3m' - 40. 29. a;^ + {2a — db) x — 6ab, 30. x'' — dax — 4td'. 31. a;* + ao;' - 6a\ 32. a:''* - Ux'' - 12b\ If the quantities m and 7i are both negative, the sum m + n will be negative and the product positive, because {x — m) (x — n) = x^ — {m -\- n) x -\- mn. ZZ- x" - {a-^b)x + ab. 34. y' - oy + 6. 35. f-^ + ^' 36. x-" - nx + 10. 37. x" - IZx + 40. 38. x" - 8.T + 15. 39. ax" — dax + 2a, 40. ax^ — Qax"" -{- Sa, 41. 7n^x^ — bmx -j- 4. 42. m.Y — 5ma: + 4. 43. ml'?;^ — dmx + 2. 44- ^^^^ — 4:ma: + 4. 45. aV — la^'x -\- 12a, 46. mV — 7m V + 12m*. 47. Tiy - 7/^y + 10n\ 48. ry - 7r'?/' + 12ry. FACTORING. 471 In the following exercises trinomials of all the preceding classes are contained: I. x^ + 10^ + 24. • 2. x^ -Qx + 8. 3. x''-\-^x- 20. 4. x' + 3:r' + 2. 5. a;' - 7:?:" + 12. 6. :cy - ^Ixy + 26. 7. a'c' - Uabc + 396'^ 8. a'b'' - Ma'hx + 143:r^ 9. a' - 12a + 20. 10. x' + 5O2; + 49. II. a;' + 4a: -32. 12. a^- 7a -18. 13. a' + a' - 132. 14. a'^^V + Wb^d" - 22. 15. a* + 17a' - 390. 16. a" - 7a -^ 12. 17. x' + x- 72. 18. a;^ - 12a; + 27. 19. x" — 39x + 108. 20. a;' — a; — 12. 21. a;'* - 7a; - 60. 22. 15a;* - 17a;^ + 4. 23. (a + by - lie (a + b) + dOc\ 24. a;^ + 4a; - 77. 25. a;' + 6a; - 135. 26. a;' — 14a; + 48. 27. a;' + 12a; + 35. Miscellaneous Exercises in Factoring. I. ax' — 2bx -\- ex. 2. ax — 2^a; + dcy — ex -\- 2y. 3. a^x — 2eay — 4:X — y + x. 4. a^x + a^^'o;^ — 3a^"'a;'. 5. ax'— 2bx^~\- ex\ 6. tx' — pqx^ -{-px"". 7. ex' — abx^ — 2y -\- Say'. 8. aexy + 20-^ — dx'y*. 9. 2c'x'y' — x^y"^ + dx'y'. 10. 4:x'y — 3a;^?/'^+ 2x^y'. II. (a;* - 4). 13. {x' - 81a;''). 15. o;^ -a; + i. 17. a^Z^'^ - a^Z>^ + ^. 19. ia' -\-l — 4:a. 21. a;' + 2a;'y + xy'. 23. 25aV - 30aa;^.V + 9a;y . 28. a' x' ^5.-4- 3y ^ 9^"- x' a; 6y' 18,«/' 12. (o;' - 9a;). 14. 4.x' -12x'y + 9y'. 16. a'x' - y'. 18. 9a' - 1. 20. a;^^ + a;*" + i. 22. 16a*^ - 1. 24. 12a* - 36a'xy + 27x'y\ 26. a' « , 1 8 6^ "^ ISb'' 472 FAGTOEING. 29. %4.a'h - na'V + b4.ah\ 30. x'* - y'\ 31. (4 — gij- 3-- mV + 2mV + mV. 33. 289a'^Z^^c' + 102a;>^c'^J + 9^V^. 34. 121a' - 2^Qa'¥ + 169Z>\ 35.. 98a''Z>^ - b^ahx + 8:c\ 36. 16^;^ + ^x' + ^^ 37. fK - 2/". 38. d^a'h^e - bVc\ 39. j-« + l. 40. 16a^ {^aF - lObc) + 20c {6c - 4:aFy 41. — — Sax''. 42. 2^xy — Id^xyz". 43. (a - by - (a + by. 44. (3« - 5)" - (« + 2*)'- 45- ^^^^- - ^^y. 46. (« + 5)= - K -2«& + J^). 47. (a - 5)' - (4a'' — 12ab + 9b'}. 48. {a + if - (4a' + 12ab + 9b'). 49- i - 2 + J. SO- «' - 2««/" + 2/^ X tl> 51. ^'^ -4a;^^^ + 4^^. 52. a'^*'^ - 2^c& + c\ 53. a^'^—y^^ 54. 16a' — 4c'. 57. 4a;' - 4.x' + i?;\ 58. 4.aV - {a' + b' - cy, 59. {2x + yy-^{x + yy. Products of Two Binomials. We have {a + i) {x + y)=ax+bx + ay + by. Hence a polynomial of four terms may sometimes be ex- pressed as a product of two binomial factors. We can do this when, two terms of the polynomial {ax -}- bx for example) being divided by a common factor {x), and the two remain- ing terms by a common factor {y), the quotients are equal. We can thus factor the following: 1. ax — bx -\-ay ■— by, Ans, {a — b) {x -}- y). 2. ax -{-bx — ay — by, 3. ax -- bx — ay -}- by. DIVISION BY POLYNOMIALS. 473 4. nv" + mn -{-71^ + n, 5. mn — m^ -\- if — tfm, 6. l + a + a" + a\ t. l-x-x' + x\ 8. l-\-x — x'' — x\ 9. a' + fl^^c?: + ax" + a:'. 10. (a — ?i:r) {a + ^za;) — (w — - ax) {n + a^). 11. h^ - Wx + hx' - dx\ 12. a^ + «^ - a^ - a\ 13. m^ — 3m' + m' — 3m'. 14. m' + 3m' — m' — 3m. Division by Polynomials. §97. 1. a" + 4:ax + 4a;' ~ a + 2x. 2. 6a' - 6b' -f-2a'^ - 2b\ 3. «^ - da'b' + 3a'b' - b' -^ a' - da'b + Sab' - b\ 4. a' - 9«' + 27a - 27 -f- r? - 3. 5. 48a'' - Hea'b - 64:ab^ + 106b' ~ 2a - 3b. 6. ^a' + a'+^a + i^ia+l. 7. dSa'b' - 77a^b' + 121a'b' ~- Sa'b - 7a¥ + llab\ 8. 100«* - UOa'b + 2d5aW - dOa'b' ^ ba' - 2a''Z^. 9. 37a'Z>' - 26a''Z> + 3a' - Uab' -- 3«'* - 6ab'+ 2b\ 10. 0^" + ^ + x'^y + a:?/"* + ,^"* + ^ -h :z;"* + y*". 11. a^ + a^'^'W'' + Z>*» -^ a^^ + a"Z>" + b^, 12. 10a' - 27a'^ + 3Wb' - 18«Z>' - 8^' -^ 2a' - 3fl^ + UK 13. 4a:;l — 3o(^y^ — y -7- x^ — «/i 14. 8al -— 6a^ -\- a^ --■ 2a^ — a^. 15. 9a-2 + 12a-^^-4-^3a-^+2. 16. 4a; - 10.Tt - 62:«;i - 30^^:1 -f- 2:^:* + 5. 17. a;^y~^+ ^~y ^ ^~V + ^^~^' 18. x^"" — f"" -^ x"" - «/". 19. 4:?;^ + 6 - 3bx' + 58a;* - 70a;' - 23a; -f- 6a;' - 5a; + 2 - 7a;'. ^ 19 2 2a^x a' . a' 20. a;* — — a a;' + -^ + ^ -^ ^' — 2aa; + - . 21. {a' - 2ab + b' - c') {a + b + c) -^ a - b - c. 22. {ax + byY + {ay - bxy -^ a' + b\ 23. 12a;' - 14a;' - llo:' + 19a; - 6 — 3a;' - 5a; + 2. 24. 4:a'F - 3b\ + llaF + 12a' -^ 34 a'^ -^ ^' + 6a' - hah. 25. 6aZ>c' - 9Z>'6'' + ^a'Z^' - a'c' -=- 3^c - ac + 2ab. 26. 2Z>c; - 1 + a' + 2c - Z^' - 2^ - c' -V- a + c - 1 + ^. 4,74 DIVISION BY POLYNOMIALS. 29. 12 {x - yf -3x{y -z)- 2y {x + z) - 20z {y + 3z) -r-6{x + 2z)-3y, 30. {4.x' - 9y') {Sx' - 27y') -~ {2x - Zy)\ 31. 12 + 82^' + 106^' - 70^' - 112a' - 38 -r- 3 - 5« + 7a'. 32. a^ (1) + c) + y' (^ - c) + c" {a — I) -\- ale -^ a -^l -^ c. 33. a' + V-\-c'-Mhc-^a + l)-\-c. 34. x^ — (a +^) i^"" + (5^ + (^P) X — aq -^ X — a, \ 35. a* - 13a' + 36 -^ a' + 5a + 6. 36. x'' + x'y"" -\-y^ -T-x"^ — Q:y + ^/^ 37. 3a^ - 8a'^' + 3aV + 61' - W& -v- a' - l\ Z'^. y' - 3^V + 3z/V - 2;' -^ ,^' - 3y'cc + 3yx' - x\ 39. 16aV— 7 ate — c'— ^^a^lx — ^a'h'' -^^acx -^ Sax — 6ab — c. 40. x' + (a' - 2b') x' - {a' - h') x'-a'- 2a'b' - a'b' -■x' -a' - y"' 41. 6 {x^-^y^) + {V^xy - 4) (x-^y) -%{x'-\- y') - 16xy - 120 -^'*+^' + 2.T(l+^) + 2,y + 6. 42. a' -b' ^a^- U. 43. a'b'"^ + 2acZ?^ + " + 2a:z:Z>^ + c'V'' + 2^;^^;'* + .t' -V- aZ^'" + cZ>" + X. 44. (^"-^^)(^^-^^)-^^ + /. 45. 20a'^Z>^ - ^OSa'b'' - 121a''¥' + 132a'"^' + 245a^^Z>^' -^ 9a'^^ - 16a^' + lla^Z^'. 46. 1 + 34 a;*^ - 20x' + 20.t' - 4a;' + 12:?;' - 31^;^ -f- 2a; + 4a;' - 3a:' + 1. 47. 2a;3^ — 6a;2^^" + Qx^y^"" — 2^/^'^ -v- a:" — y"". 48. a (a - 1) a;' + (a' + 2a - 2) a:' + (3a' - a') x - a' -i- aa;' — 2a; — a'. 49. a'^' — ^ (a' + Z>) 2/ + ^^^ -^ ay — b. 50. (a + J) (a + 6') — (a + Z>) (^ + c) -^- a — d 51. a;' + (4a^ - J)'^) x ^ {a - 2b) (a' + db') ^ x - a + 2h 52. a;' — ^~^^ o:^ + j^"^' 53. * - 6^' + 27^^ ^ i + 2^ + 3z\ 54. a^'^-^^'W^c — a2'^ + '^-^^^"^6'^ + a-''^"^c"* + a^^-'*Z>^^ + ^c'* FRACTIONS. 475 Fractions. § 108. Execute the following multiplications of fractions by entire quantities by dividing the denominators: X a—h. 2. 3 7^ X « + ^. a-h'^ a' - ¥ m^ jf -\- Q^ a^ — ¥ , , „ 7)1" -\-n'' a -\-ab m — m n ^ ' 9. z \ ^3 4 X 1 — m'. 10. — - — 3 x\-\-x. Execute the following multiplications by dividing the denominator by one factor of the multiplier, when denomi- nator and multiplicator have a common divisor, and then multiplying the numerator by the other factor of the multi- plier: Here the denominator is {a — by, and the multiplier is (a — b) {a + b). We multiply by (a — b) by dividing the denominator, and by a-{-b by multiplying the numerator, m ( « 4- b) Hence the product is a — b -5 X m^ — n^. 3. — ^ — -; —7-^ X m^ — 4:n^. m' — 4:7nn -\- 4:n 6. — ^ — \-T-nr X «' - *'• ax + a?/ -\- ox -\- by 476 Fll ACTION 8. 1 Xa' + h\ *' ax — ay -\- Ix — ly 1 8. X mx + mil — nx — nil. mx — nx — my -\-ny ^ ^ Execute the following divisions by dividing the numerator by as many factors of the divisor as possible, and multiplying the denominator by the remaining factors: ax . X my ~ ar. Ans, . 2. -~- -r- mq, mn mnr Zpr ^ 3 5 7 9 II ah ^ a" -\-al) ^ ^^ mn a — m -{- p -^ X ~ 2xy -\- y cx-\- cy 2 2 o ^^ ~~ ^y t \2 ax — ay ^ ox-\-oy^ ^ ' — '--rr- -^ (« + ^hy. 10. '—- -^ a" — h\ a — zb ^ ' a — ac — he ax -\- Ix h\ Execute the following indicated multiplications or divi- sions, and aggregate each product or quotient into a single fraction: la ,'b\la_h\ 2 fl __ i") 1. wi nl \m nj' ' \a h ) c' 3 5 7 9 II m \ m -\- X \' l — mJl-\-m \a-{-b a — hi a ^ ^ \a — h J \m nJ \m nJ \ ml f 1 4. Z _ V!}:\ f 1 _ Z _ !?_^ \ m pJ \ m pj' FRACTIONS, 477 13 14, 16 21 \ ^ ml \7n / ' J \^^' my (m + n) ( h -) — (^^ — ^0 • ^ ^ ^ \m nJ ^ \m nl (-^ + -^ + -^) {x + y^z). \x — yy — zz — xj^ ^- 771 + ^. w n 17. m — 71, a + l a + b c' + h:' a — b ' a 19 a b , a' + b' ' a+ b a — b ' a — b '^x'^ x'~^ x' ' x'^ x'' Factor th^ following fractional expressions: c X ' x^ c^' ^' ¥ ~^ xy ~^ y'' 1 + a ' ^' 2c' Sx'' 9. S--3I + 2. h' h 2. a' V 4. x' /■ /; m ip' + q') n{f-^n^ c X -\- a X 4- a^ , X -\- a^ x -\- a^ a a^ ' a^ a^ 21. [ax + ^) (Z>.^' — a) — {ax — Z>) (Z^.-r + «) = (i^ -f Z>, « + a; Z^ + a; 22. ^''^ = — ■ — ah . a 1 ,7)1 — 71 1 , 771 -\- n 23. = ■ — . m — 71 x 711 -{-n X 1 p 24. - {771 -\- Olf 771-^71 2 {7)1 + 71)' o 1 p X p 480 25. I = 26. m — 27 SIMPLE EQUATIONS. nx 1 -- 7nx P_±J^^ q + x q-\-x' 1 : + m. ah — ax he — hx ac — ax' 28. (m + nV = 3m' + ?i' — -'^ — . X 30. ^^ = ^^~g^ ^6?(^ + c) 771 m 1 + :^ l-x ■ 31. = 1. n n m-\ — X 32. = 2/w. 1 + ic 1 — a; 1 + ^-^ ZZ- X -\- a m 1 + = ^. 34. X -\- a 1 + = 2^. Equations of the First Degree witli Two Unknown Quantities. §§137-140. j mx — ny — 0. ^ + y = a. I fl^o; + a^y = ap, [ hx -j- h'^y = ^^. \px + qy = a. \ X — y=-h. 4. < y h -{- y 3a -\- x' ax + '^hy — d. 1 x-\-y^3' -^=3. y ^ \hx^ y-a. \ X -\-hy •= 2a. TWO UNKNOWN QUANTITIES. 481 X _ y x-\-y:=s. a , h - + - = c. X y --{-- z=zd. ^x y 12. a h __ X y \,x y b {x — a) -\- a {y — h) X — a : y — b = b : a. 13. iix-{-b = my -{- d = c. 14. mx = ny —p = x -\- qy. 15. m{x + y) = 71 {x — y) = r. X ^ _ 1 16. la + b a — b a — b a-\-b' y _ 1 6' *7- 1 x-\-y 1 + 1 1-x- 1 L 1 — ic + .V 2 3* 4 3* Equations with Three or More Unknown Quantities. [x-{-y-^z = a. mx = ny. [py =qz. ( X -\- y = a. }y-\-z = 2a. [z~i-x = da-\-b. i ¥ = iy- iiz=ix + l. (x — z = am. )y-^z = bm. (^x — y= cm. (x = t/ — 2z. \y = '3z- 2x. (z=y + l. 6. x+y+z=i X _y _z a b c' m n - - - _ ^. X y z' ^ + y + ^ = S' a; + ^ + ^ == 30. 8a; + iy + 2z = 50. 272; + '9y + 3z= 64. ''x-\-y -\- z = au. x-}-y = bu. X — = m. 7ix + y -{- z = a. X -|- ny -j- z =^ b. X -{- y -{- nz = c. 482 8IMFLE EQUATIONS. II. < IS- IS- f 2; + 2?/ = 8. ^ + 2^ == 12. z -{-u = 8. ( «^ + ^^ + cz — d. \a^x-\-¥y-^ez=d\ 14. 17. i 19. i 2a; + 3^/ + 52; =: 67. 2a; + 3^ 4- 4^ = 35. 16. 2^ - 3«/ + 5^ = 13. a; 4- ^ = fl^. y^z^l, ] z^ — X =^ d, -hx -\-ay _ a — l {b—c){a—cy h — c cy c + hz az a + ex {c--a){b—ay c — a [a-b){c-b)' a a — r a — s ^ "^ ^ - r "^ Z^ - ;^ c c — r c — Si X y a .- + - = -• ^y z c 3^ + 2/ + ^ = 3. X -\- 4:y -\- z = 4t. X -\- y + 6z = 5, = 1. = 1. b+ c y + + ■ y c ~ a z a + b. a -\- a ' a — b z X a + b :=b + C, — c -\- a. PROBLEMS LEADING TO EQUATIONS WITH ONE UNKNOWN QUANTITY.* 1. A capitalist earned 4 per cent interest fi*om f of his in- vestment, and 5 per cent from the remaining \, making a total annual interest of $2940. What was tlie amount invested? 2, What quantities must be added to each term of the TYl fraction — that it may take the following series of values: * Although only one unknown quantity is really necessary in these problems, the student may often find it convenient to use two or more. niOBLEMS. 483 AVhat qiifintities musfc be subtracted from each term to pro- duce tlie same results ? Explain the relation between the answers in the two cases. 3. A man is 40 years old, and his wife is 36. In how many- years will the sum of their ages be 5? Explain the results when Ave put, in succession, s = 100; s = 76; and s = 50. 4. A railway train passed a station at" the speed of m miles an hour. Then k hours later another passed in the same direction, going n miles an hour. Supposing the speeds uniform, at what distance and at what time did they meet? Explain the relation of the answers when m > n and when 7)1 < 71, 5. If, in the preceding problem, the second train went in the opposite direction, what would the answer be? Explain the relation between the answers. 6. A ship sailed from port with a speed Ic knots per houi*. In 7i hours after sailing she was followed by a steamer, who overtook her in 7i hours. What was the speed of the steamer? 7. An oarsman who pulls 6 miles an hour rows from his house down a river whose current is 2 mi.les an hour, and re- turning gets back 3 hours after he started. How far did he go? 8. On the same stream one rower pulling 6 miles an hour going down stream, and another pulling 7 miles an hour going up stream, started out at the same moment; but the starting-point of the second was 5 miles below that of the first. At what point and in what time did they meet? 9. A steamer goes down the Rliine from Mayence to Co- logne, 117 miles, in 8^ hours, but requires 14 hours f6r the return journey. What is the speed of the current? 10. On an ocean the crests of the waves are y^ of a mile apart, and are moving at the rate of 40 miles an hour. If a ship steams 15 miles an hour, how many times an hour will she pitch when going with the waves, and how many times when goinsr aarainst them? 484 SIMPLE EQUATIONS. 11. A number is divided into three parts, of which one is 30 less than a half, a second 10 less than a third, and the re- maining part 8 greater than a fourth. Find the number and the three parts. 12. From a line was taken J its length and 2 feet more, and from what was left ^ its length and 2 feet more, leaving ^ the whole line and 2 feet more. What was the length of the line? 13. A team performed a journey in 8 hours, going one third the way at the rate of 25 miles an hour, and the remain- ing two thirds at the rate of 40 miles an hour. What was the disitance? 14. A grocer has 60 pounds of tea worth 75 cents a pound, formed by mixing one kind worth 80 cents a pound with another worth 50 cents a pound. How many pounds of each kind were in the mixture? 15. Divide a line of length / so that | of one part shall be equal to f of the other part. 16. A man is 6 years older than his wife. Ten years hence the sum of their ages will be 7 times the age of the wife 14 years ago. What are their ages? 17. A man who must be back in 1 hour starts in a coach going m miles an hour, and walks back at the rate of n miles an hour. How far can he go and be back in time? 18. The earth performs a revolution round the sun in 1 year; Mars, in 1| years. What is the mean interval between conjunctions; ihafc is, between the times at which the earth passes Mars? 19. The periodic time of Jupiter is 11-f years; of Saturn, 29|- years. At what intervals will the earth be in conjunction with each of them, and at what intervals will they be in con- junction with each other? 20. Two persons, A and B, were mounting a tower, B be- ing always 24 steps behind A. When A was half way up he said to B, " When I reach the top, you will be 8 times as high as you are now." What was tlie height of the tower? 21. The circumference of the front wheels of a carriage is 9 feet; of the hind wheels, 12 feet. How far has the carriage PBOBLEMS. 485 driven when the front wheels have made m turns more than the hind ^wheels? 2 2. The members of a club have to raise a certain sum of money. If each member contributes $2, there will be $28 too much; if $1.25, there will be $32 too little. How many mem- bers are there, and what is the amount to be raised? 23. If a dealer sells a piece of cloth at 7n cents a yard, he gains d dollars; if at 7i cents a yard, he loses c dollars. What is the length of the piece, and the purchasing price per yard? 24. A merchant by the profits of trade increases his capi- tal each year by 20 per cent of the amount at the beginning, but takes out $1000 at tlie end of each year for his board. At the end of the third year he has increased his capital by $200 more than -| of its original amount. With what amount did be start? 25. A boat which steams 12 miles an honr makes her (rip in 3 kours going down stream, and in 5 houi-s going up stream. W^hat is the speed of the current and the length of the trip? 26. A number is increased by n, and the sum multiplied hyn\ this product is then increased by 71, and the sum multi- plied by n, with the result 2?^.^ What is the numbei'? 27. A number is diminished by n, and the remainder multi- plied by n ; the same operation is repeated on the product, and again repeated on the second product, with the result — w^ What is the number? 28. What number is that whose fourth part exceeds its sixth part by 2? 29. If you add 4 to a certain number, the sum is 2 less than twice the number. What is it? 30. Divide $520 among three people so that the first may have $20 less than the second, and tlie second $10 more than one fourth the share of the third. What must each receive? 31. Divide c dollars among three people so that the first may have a dollars less than the second, and the second m dollars more than one fourth the share of the third. What must each receive? 32. A left a certain town at G miles an hour, and in 8 486 SIMPLE EQUATIONS. hours after was followed by C at 8 miles per hour. In how many hours did overtake him? 33. A left a certain town at b miles an hour, aud in n hours after was followed by D at c miles per hour. In how many hours did D overtake him? 34. A farmer said, if he had 5 more sheep, and sold them at $4 each, he would have 5 times as many dollars as he now has sheep. How many sheep has he? 35. A farmer said, if he had a more sheep, and sold them all at n dollars each, he would have c times as many dollars as he now has sheep. How many sheep has he? 36. If you divide my age 10 years hence by my age 10 years ago, you will get the same quotient as if you should divide my present age by my age 15 years ago. What is my present age? 37. If you divide my age c years hence by my age a years ago, you will get the same quotient as if you should divide my present age by my age d years ago. What is my present age? 2,^, Divide $415 among A, B and C so that A shall have $40 less than B, and C $20 more than half as much as A and B together. 39. Divide %a among C, D and E so that C shall have $m less than D, and E %7i more than one third the share of and D together. 40. A can do a piece of work in 20 days, B in 24 days, and C in 30 days. In what time can they together do the work? 41. A, B and can do a piece of work in 4 days, A alone in 12 days, and B alone in 10 days. How long would it take C to do it? 42. A, B and can do a piece of work in 6 days, A alone in 9 days, and B alone in 12 days. How long would it take C to do it? 43. A can do a piece of work in a days, B in Z> days, and C in c days. In what time can they together do it? 44. A man is 12 years older than his wife; four years ago 8 times her age was 5 times his. What arc their present PROBLEMS. 487 45. A man is a years older than his wife; b years ago c times her age was m times liis. What are their present ages? 46. Divide $1200 profit so that A may have one fourth and $100 more, B $5*0 less than one third, and $'^50 moro than one sixth. 47. The interest on -^^ of a certain capital at 5 i)er cent added to the interest on the remainder at 6 per cent is equal to $1680. What is the capital? 48. A person, asking the distance to a certain city, was told that after he had gone one fourth the distance and two thirds the remaining distance, he would still have 20 miles to travel. What was the distance? 49. How far can a person who has 5 hours to spare ride at 6 miles per hour so as to walk back in time at 4 miles per hour? 50. How far can a person who has n hours to spare ride at 1) miles per hour so as to walk back in time at c miles per hour? 51. A man bought 15 horses for $1665, paying $120 for each good horse, and $75 each for the poor ones. How many of each did he buy? 52. The difference of the squares of two consecutive num- bers is 15. What are the numbers? 53. The difference of two numbers is 2, and the difference of their squares is 28. What are the numbers? 54. The sum of two numbers is 12 ; the square of the greater is 48 more than the square of the less. What are the numbers? 55. The product of two consecutive numbers is 4 more than the square of the less. What are the numbers? 56. Divide 60 into three such parts that one third of the first, one fourth of the second, and one fifth of the third shall be equal to each other? 57. Divide 80 into four such parts that if the first be in- creased by 3, the second diminished by 3, the third multiplied by 3, the results shall be equal. 58. The greater of two numbers is 4 times the less; if each 488 SIMPLE EQUATIONS. be increased by 3, the greater will be 3 times the less. What are the numbers? 59. A man is 10 years older than his wife; in 10 years twice the sum of their ages will be 6 times her present age. What is the age of each? 60. A man bought a certain number of sheep for $1200; he reserved 80, and sold the remainder for $960. How many did he buy? 61. A father aged 48 years has a son aged 12. In how many years will the age of the father be three times that of the son? 62. A merchant has two kinds of tea; one cost $1.50 a pound, and the other $2. He wishes to mix them so as to have 50 pounds worth $1.80 a pound. How much of each must he use? 67,, In a certain quantity of mortar the sand was 15 pounds more than f of the whole, the lime 9 pounds less than ^ of the whole, and the plaster- of -paris 6 pounds less than \ the sand. What was the amount of the mortar? 64. A laborer agreed to work 50 days on the condition that he should receive $1.50 for everyday he worked, and for- feit $0.75 for every day he was idle. At the end of the time he received $48. How many days did he work? 65. A grocer having 60 pounds of coffee worth 15 cents a pound mixed it with so much coffee at 18 cents a pound that the mixture was worth 16 cents. How much did he use? 66. The interest on a certain capital at 5 per cent is $20 less than the interest on $900 more at 1 joer cent less. What is the capital? 67. A woman bought 200 apples at 5 for 3 cents, and sold part at 2 for a cent, and part at 5 for 4 cents, thereby making 10 cents. How many of each kind did she buy? 68. A and B play at cards. A begins with $120, and 13 with $180 ; when they stop playing B has four times as much as A. How much did B win? 69. From a cask of wine one fourth leaked out, then 20 gallons were drawn, when it was found to be 10 gallons less than half full. How much did it hold? PROBLEMS. 489 70. An estate of $4680 is to be divided among 4 sons and 3 daughters. Each son is to receive $40 more than the next younger; the eldest daughter is to have $20 less than the eldest son. and each of her sisters $20 less than the next older. What did each child get? 71. A sum of $2880 is to be divided among A, B and C. Five times A's share is to be equal to three times C's, and B is to have twice as much as A and C. What does each receive? 72. Six plasterers, 8 journeymen and 12 apprentices re- ceive at the end of a certain time $387.50. The plasterers receive $2 a day, the journeymen $1.25, and the apprentices 75 cents. How many days did they work? 73. In the above problem, what should each class of work- men receive if each plasterer worked 3 days more than the journeymen, and the apprentices 6 days less? 74. A man wished to give 10 cents each to some beggars, but found he had not enough of money by 14 cents; he then gave each one 8 cents, and found that he had 10 cents re- maining. How many beggars were there? 75. A post is 6 feet more than J in the mud, 2 feet less than ^ in the water, and 4 feet in the air. What is the length of the pole? 76." A and B begin trade. A has $1000, and B $1210. The former gains a certain per cent on his investment, and the latter loses the same per cent, when their capitals are found to be equal. What was the amount lost and gained? 77. A person in play lost ^ of his money, then won $60, after which he lost \ of what he then had, when he found he had but $350 remaining. Wliat had he at first? 78. In a camp of 3294 soldiers there were 3 cavalry to every 26 infantry, and half as many artillery as cavalry. What was the number of each? 79. The right-hand digit of a certain number is 2 less than the second; and if the number be divided by the sum of the digits, the quotient will be 7. What is the number? 80. The length of a town lot exceeds its width by 12 feet. If each were 3 feet greater, there would be an increase of 645 square feet in its dimensions. What is the length? 490 • SIMPLE EQUATIONS. 8 1. A house was sold for $0800, by which there was a cer- tain gain. If it had been sold for $1000 less, 3 times the resulting loss would liave been twice the present gain. What was the cost of the house? 82. A can do a piece of work in 12 days, and B in 15. After A has worked 4 days B comes to help him. In what time can they both finish it? St,, a tank has two filling and one emptying pipe. One can fill it in 12 hours, the other in 24 hours ; and the third can empty it in 18 hours. If tliey are started at the same time, how long will it take to fill the tank? 84. In the preceding problem, suppose the third can empty it in 8 hours. How long will it take to fill it? 85. Suppose it is full already, and the third can empty it in 6 hours. How long will it take to empty it? S6, A person travelled 168 miles, of which he went 3 by boat and 4 by coach to every 6 by rail, and walked one third as far as he went by boat. How many miles did he travel by each? 87. The sum of two numbers is 42. If the less be divided by the greater, the quotient will be less by ^ than when the less is divided by half the greater. What are the num- bers? 2>S, A and B are of the same age. Three times A's age 6 years ago is equal to twice B's age 9 years hence. What is the age? 89. In tossing pennies, A threw heads 3 times out of 5, and B 4 times out of 7. In all they get 41 heads. How many times did they toss? 90. What two numbers are those whose sum is 13, and whose product added to the square of the less makes 50? 91. A tank has five pipes. No. 1 can fill it in 6 hours, No. 2 in 8 hours, and No. 3 in 12. No. 4 can empty it in 9 hours, and No. 5 in 18. If they begin at the same time, how long will it take to fill the tank? 92. A starts from a certain place, and travels at the rate of 17 miles in 5 hours. One hour and 53 minutes after, B PROBLEMS. 491 starts at the rate of 19 miles in 4 hours. How far will thoj travel before B overtakes A? 93. Two persons start from the same place at the same time^ going in the same direction. One travels 2 miles an hour faster than the other. After they had gone as many hours as tlie slower goes miles per hour, their distance apart was equal to half the distance travelled by the faster. How long did they travel? 94. Two men travel in opposite directions; the rate of one is 1 mile ntore than two thirds the rate of the other. When they had gone 4 hours the distance apart was equal to 44 miles. What were their rates? 95. An officer in arranging his men in the form of a square found that he needed 5 men to complete the square, and by increasing the file by 6 and diminishing the rank by 5 he had 5 men too many. How many men had he? 96. A coach that travels 6 miles an hour starts 50 minutes after another that goes 5 miles an hour. How far will the first-named travel in order to be 11 miles ahead of the other? 97. A merchant withdrew from his capital $500 at the end of each year for current expenses; his profits each year were 33^ per cent of his unexpended capital. In 3 years his original stock was doubled. What was his original stock? 98. What fraction is that whose denominator is 2 more than the numerator, and if 3 be subtracted from both numer- ator and denominator the fraction will be f ? 99. Divide 40 into two such parts that the greater dimin- ished by 4 and divided by the less increased by 6 shall be 1| ? 100. On a note interest is paid at 6 per cent. At the end of the first year $200 is credited on the principal, and the rate of interest is reduced to 5 per cent, when the annual interest is diminished by one fifth. What was the face of the note? loi. The difference between the simple and compound in- interest of a certain principal during the second year at 5 per cent is $10. What is the principal? 102. The fore and hind wheels of a carriage have circum- ferences of 12 and 16 feet. How far will the carriage have 492 SIMPLE EQUATIONS. gone when the sum of the revolutions made by the wheels is 287? 103. During the first 3'ear a broker gains 20 per cent on his capital, the second year he gains 30 per cent on his in- creased capital, and the third 25 per cent on his re-increased capital, when he finds that his capital is $4910 more than what he began with. What was his first capital? 104. A man sold a house and furniture for $6400; f of the price of the house was $200 less than f the price of the furni- ture. What was the value of each? 105. A purse contains 65 coins, i)art cents and part dimes. How many of each are there if the total value is $2? 106. Each member of a base-ball club subscribes as many cents as there are members. If there had been 10 more members, each subscription would have been 9 cents less. How many members were there? 107. A man purchased a number of lemons at 2 cents each, and f as many at 3 cents each; he sold them all at the rate of 2 for 5 cents, and gained 25 cents. How many of each kind did he purchase? 108. A boy in flying his kite lost | of his string, then added 65 feet, and found that it was just f of its original length. What was the length at first? 109. A and B start from two towns that are 133 miles apart and travel towards each other. They meet at the end of 10 hours, and find that A has travelled 1^ miles an hour more than B. How many miles had each travelled? no. A man owning a cow and horse found that 4 loads of hay would keep them both 6 months. Having disposed of his horse, ho found that the same quantity of h-ay would last the cow 14 months. How long would 1 load last each? 111. A has $647, which is $33 loss than 4 times what B has; is worth twice as much as A and B together, lacking $72. How much have B and C? 112. A boat which could move 14 miles in still water was accelerated 2^ miles per hour going down stream, and retarded the same returning; it was 10 hours longer coming up a cer- tain distance than going down. What was the distance? PROBLEMS. 493 113. A and B have i\\% same income. A spends |- of liis, and B by spending $200 a year more than A finds himself at the end of 5 years $450 in debt. What was their income? 114. A farmer bought 22 cows at a certain price; had he paid 8 per cent less he could have purchased 1 more cow and had $21 left. What was the price of each cow? 115. A son is -|- the age of his father, and 11 years ago lie was f of his age. How old is each ? 116. A man rows 5 miles an hour in still water. How far can he row up a stream and back in 3 hours, the stream flowing a mile an hour? 1 17. A man bought some sheep for $94. Having lost 7 of them, he sold J of the remainder at first cost for»$20. How many did he buy? 118. The pei-imeter of a rectangle is 28 feet; if 2 feet be taken from its length and added to its breadth, its area is in- creased by 12 square feet. Find its original breadth? 119. A man can row 9 miles an hour with the stream, and 3 against it. How far can he go so as to be back in 6 hours? 1 20. The first digit of a certain number exceeds the second by 5, and if the digits be* inverted the new number will be f of the original number. What is the number? 121. Divide $900 in two such parts that the interest on one part at 4|- per cent may exceed that on the other at 3|^ per cent by 50 cents. 122. How much foreign brandy at $8 a gallon and whisky at $3 a gallon must be mixed together so that the compound may be sold for $9, and the merchant thereby gain 30 per cent. 123. A person has two kinds of coins. Four pieces of one make a dollar, or 10 pieces of the other. How many of each must be taken so as to have 7 pieces equal a dollar? 124. Find two numbers whose product is 72, and whose difference multiplied by the greater is found by subtracting the product from 18 times the greater. 125. A person after spending $200 more than ^ of his in- come had remaining $75 less than ^ of it. What was his in- come? 494 • SIMPLE EQUATIONS, 126. Divide 77 into two such parts that the quotient of the first divided by 8 added to tlie quotient of the second divided by 9 shall be 9? 127. The sum of three numbers is 155. If the second be divided by the first, the quotient is 2, and 2 for a remainder, and the third divided by the second gives 3 for a quotient and 3 for a remainder. What are the numbers? 128. At a ball there were twice as many gentlemen as ladies. When 8 couples danced there were remaining three times as many gentlemen as ladies. What was the number of each? 129. A can build 7 cubic yards of wall in 4 days, B 12 yards in 5 days, and C 9 yards in 2 days. How long will it take all three to build 850 yards? 130. Each of the three digits of a certain number is greater than the next folloAving by 1; when the digits are inverted, the new number will be 18 more than ^ the first number. What is the number? 131. A farmer bought 30 sheep and 10 calves for the same sum. If the sheep had cost 25 per cent more and the calves 35 per cent less, 7 sheep would have cost $3 more than 4 calves. What did each sheep cost? 132. Upon withdrawing from the business A takes J of the capital and $100 more, B ^ of the new remainder and $100 more; gets $300. What was the capital? 133. What number is that which gives the same continued product when divided into 3 equal parts as when divided into 4 equal parts? 134. Find a number of two digits, the first of which is 4 times the second, and the number is 2 less than 3 times the number formed by inverting the digits. 135. In going from one town to another a traveller found at a certain place that the distance travelled was -| the whole distance, and when he had gone 11 miles further he had f of the whole distance yet to go. What was the distance? 136. A wine-merchant has wine in casks of two sizes. One containing 2|^ gallons he charges $8.50 for; the other, 3^ PROBLEMS. 496 gallons, is priced at $10.90. Wliat is the price of the casks, supposing them to cost the same? 137. A miin's income was $800 the first year, and increased $50 each succeeding year. Afc the end of 3 years he had L.ived $15.75. What were his annual expenses? 138. If A gives B $10 he will liave twice as much as B; but if B gives A $10 he will have '^ as much as A. How much had each? (§ 140.) PROBLEMS INVOLVING EQUATIONS WITH TWO OR MORE UNKNOWN QUANTITIES. 1. It is found that when a ship steams 12 knots (sea-miles) an hour with the waves she pitches 1 in 15 seconds, and steaming at the same speed against them she pitches 1 in G seconds. What is the speed of the waves, and how many w Jives are there in a sea-mile? 2. Two men start at the same time to miike the same journey. The first goes 10 miles the first day, and goes a cer- tain fixed distance more every following day than he did the day before. He overtakes the second at the end of the 8th day,, and finishes his journey at the end of the 11th, while the second finished at the end of the 12th. What is the length of the journey, and how far did the second go each day? 3. A cannon being fired while a heavy wind was blowing, it was found that the sound required 4|- sec(>nds to go a mile with the wind, and 4|- seconds to go a mile against the wind. What was the velocity of the wind, and what time would have been required for the sound to go a mile in still air? 4. The greatest distance between Venus and the earth ij 160 millions of miles; the least, 22 millions. What is the dis- tance of each from the sun, supposing that each moves around the sun in a circular orbit having the sun in its centre? 5. A brother and sister being asked how large the family was, the brother replied, "1 have as many brothers as sisters." The sister replied, ^' I have twice as many brothers as sisters." How many boys and girls were in the family? 496 SIMPLE EQUATIONS. 6. Find that fraction whose value becomes ^ when n is sub« tracted from each of its terms, and ^ when m is added to each of its terms. 7. Find two nunit)ers such that their difference is 153, and the lesser goes into the greater 9 times and 1 over. 8. One number divided by another gives the quotient 4, with 3 as a remainder. Increasing divisor and dividend by 10, the quotient is 2 and the remainder 23. Find the numbers. 9. Find two quantities such that half their sum added to half their difference shall make a, and half their difference subtracted from half their sum shall leave the remainder h, 10. Find two quantities whose sum and quotient are each equal to m. 11. Find two numbers of three digits of which one is formed by simply reversing the order of digits in the other, and which fulfil the following conditions: (1) the sum of the digits in each is 15; (2) the sum of the first and last digits is 3 greater than the second one; (3) the difference of the num- bers is 99. 12. Each of two vessels, A and B, was partly filled with water. A man poured from A into B as much ms was already in B, then from B into A as much as was left in A, then from A into B as much as was left in B, when each vessel con- tained 8 quarts of water. How much did each contain at first? 13. Find two quantities the sum of whose reciprocals is 5, and ^ til J one added to \ the other is equal to twice their product.* 14. For $6.60 one can buy either 20 pounds of coffee and 25 of sugar or 14 of coffee and 34 of sugar. What is the price of ejich per pound? 15. A river stermier can run 90 miles down stream and back again in 15 hours; but if she runs 120 miles down, she can only get back 70 miles on her return journey at the end of 15 hours. What is her speed and the flow of the river? * 16. Cn a river were two steamers, the speed of the swift * Compare with Exercises 11 to 20, § 138. FEOBLBiMS. 497 one being 3 miles an hour greater than that of tlie slow one. A man who went 58 miles down the river on the slow boat and 30 miles back on the swift one found that he had been 9 hours on the water. But when lie went 87 miles down on the slow boat and 90 miles back on the swift one, he found that it took 18 hours. What was the speed of each boat and the flow of the river? 17. A quadrilateral has four sides, a, I, c and d. If \ of a be added to h, then \ of the extended J) be added to c, and then ^ of the extended c to d^ the four sides will each be equal to m. What was the length of each side at first? 18. Three pedestrians started on a journey. The first performed it in a certain time; the second, going 1 mile an hour slower, took 12 hours longer; the third, going 2 miles an hour slower than the first, took 33 hours longer. What was the distance, and the speed of each? 19. The perimeter of a triangle whose sides are a, h, c, is 771 feet. If I" the side a be added to b, then ^ of the prolonged h be added to c, and then \ of the prolonged c be added to a, the sides will be equal. What is the length of each side? 20. Divide 232 into three parts. A, B and C, such that, whether we subtract A from the sum of B and C, B from ^ the sum of A and C, or C from J the sum of A and B, the remainders shall all be equal. 21. Find two quantities whose difference and product are each equal to n, 22. The quotient of two numbers is 2, and 2 times their sum is equal to 6 times their difference. What are the num- bers? 23. A man has a saddle, worth $50, and two horses. If the c:idule be put on horse A, he will equal B in value; but if put on B, his value will be double that of A. What is the value of each horse? 24. What number of two digits is equal to 4 times their sum and 12 times their difference? 25. What number of two digits is equal to 4 times their sum, and when the digits are reversed equal to 7 times their Gum? 498 SIMPLE EQUATIONS. 26. Find a number of two digits that is equal to 4 times the Slim of its digits increased by 3, and if 9 be added to the number the digits will be reversed. 27. Find a number which is greater by 2 than 6 times the sum of its digits, and if 9 be subtracted from the number the digits will be reversed. 28. What number is that which is 4 times the sum of its digits, and is 3 greater than 11 times their difference? 29. What fraction is that which becomes \ when 2 is added to the denominator, and ^ if 5 be subtracted from the numer- ator? 30. Two drovers went to market with sheep. A sold 90 and then had left ^ as many as B. Then B sold 72, and had f as many as A remaining. How many did each have? 31. A woman bought 60 apples for a dollar, giving 3 cents for every 2 bad ones and 2 cents each for the good ones. How many of each did she buy? 32. Find a fraction that becomes ^ when 4 is added to its denominator, or 2 subtracted from its numerator. T,T^, A marketman had 4 more ducks than chickens. He sold the chickens for 30 cents apiece and the ducks for 40 cents apiece, gaining 40 cents more than if the prices had been reversed. How many of each had he? 34. A boy bouglit a number of apples at 2 cents each and peaches at 3 cents each, paying $4.36 for the wliole; 12 of the apples were bad and 9 peaches were rotten. He sold the good apples at 2 for 5 cents and the peaches 3 for 10 cents, receiv- ing $4.50 for the whole. How many of each fruit did he buy? 35. When I was married I was ^ older than my wife; 10 years after her age was f of mine. What were our ages when we were married? :^6. A and B can do a piece of work in 12 days; but if A worked twice as fast they could do it in 8^- days. In what time could eacli of thorn do it singly? 37. B and C can do a piece of work in 12 days; with the assistance of A they can do it in 9 days. In what time can A do it alone? 38. A farmer sold 60 fowls, a part tujkcys and a part PROBLEMS. 499 chickens; for tke turkeys he received $1.10 apiece, and for the chickens 50 cents apiece, receiving for the whole $51.60. How many were there of each? 39. A tank has 4 pipes, A, B, C and D. A, B and C can fill it in 6 hours; B, C and D, in 8 hours; C, D and A, in 10 hours; D, A and B, in 12 hours. How long will it take each and all to fill it? Explain the negative result for D. 40. A tank has two pipes, of which one may be made to run either in or out. If both run in the tank is filled in 2 hours; if one in and the other out, in 5 hours. In what times would the separate pipes fill it? 41. A grocer bought 50 pounds of sugar and 100 pounds of coffee for $26. He sold the sugar at an advance of 25 per cent and the coffee at a discount of 10 per cent, receiving $25.50 for the whole. What was the buying and selling price of each? 42. Find the sum of two numbers the difference of whose squares is equal to the difference of the numbers. 43. Divide 168 into three such parts that the second divi- ded by the first gives 5 as a quotient and 10 for a remainder, and the difference between the third and second multiplied by 3 is equal to 4 times the first. 44. A father is 5 times as old as his son. Six years hence he will be only 3 times as old. What are their present ages? 45. The sum of the ages of two persons is f of what it will be 12 years hence. The difference between their ages is ^ of w^hat it will be 24 years hence. What are their ages? 46. A farmer sold to one person 40 bushels of oats and 30 bushels of wheat for $44.50, and to another the same amount of oats, at 10 cents a bushel more, and wheat, at 5 cents a bushel less, for $57. What was the price per bushel of each? 47. There is a number of 3 digits whose sum is 10. The first and second is 4 times the third, and if 29? be added the digits will be reversed. What is the number? 48. There is a number of 3 digits whose first and third digits are 6 more than the second. Four times the first is 14 more than the difference between the second and third; and 500 SIMPLE EQUATIONS. if 97 be added to the number the digits will be reversed. What is the number? 49. A certain number of 3 digits is 34 times the sum of its digits, and also 102 times the difference between the first and second; and if 36 be added to the number the second and third will exchange places. What is the number? 50. An oarsman who can row 20 miles and back in 7 hours finds that he can row 10 miles with the current in the same time that it takes him to go 4 miles in the contrary direction. Find the rate of the current. 51. A merchant has two kinds of sugar; one cost 8 cents a pound, and the other 11 cents. How much of each must be taken to make 120 pounds worth 9 cents per pound? 52. A grocer mixed tea that cost $1.10 a pound with tea that cost 95 cents per pound. The cost of the mixture is $101. He sells it at $1 a pound and gains $2. How many pounds of each did he use? 53. A, B and C can earn $25 in 5 days; B and C, $28 in 7 days; A and C, $22 in 8 days. What does each man earn in 1 day? ^4. A and B can do a piece of work in 2 days; A and C, 4 times as much in 9 days; A, B and C, 11 times as much in 18 days. In how many days could each do it alone? 55. A sum of money at simple interest amounted in 5 yeais to $1500, and in 8 years to $1680. What was the principal and rate? 56. A ]icrson has $1200 invested at a certain rate and for a certain time; had the rate been 1 per cent less and the time 2 years more, he would have had $24 more interest; while with a rate 2 per cent less and a time 1 year more he would have had $144 less interest. Eind the rate and time. 57. A sum of money at simple interest for c years amounted to t dollars, and the same for h years amounted to a dollars. What was the principal and rate? 58. In a race over a course 4000 feet long A gives B 300 feet start, and wins by 1 minute and 20 seconds. In a second trial A gives him 40 seconds start, and wins by 900 feet. What was the rate of each? FllOBLEMS. 501 59. A, B and C promised to give $1000 to a church. A gave one third less than he agreed to, so B increased his by one fourth, which left $55 more for C. Now if B had given one fifth less than promised, and C $70 more, A's share would have been his original subscription. What was the amount of the first pledge? 60. The fore wheels of a carriage are 10^ feet in circum- tVreiice, and the hind wheels 13. in going a journey tlie fore wheels make 2500 more revolutions than the hind wheels. What was the distance? 61. A coach has 2 more outside passengers than inside. Six outsiders could travel at an expense of $1 more than 4 in- siders. The fare of all amounted to $20.50. At the end of half the journey 2 were added to the outside and 1 inside, Avhich increased the total fare by $2.50. What was the num- ber and fare of each class? 62. A person has two creditors; at one time he ])ays them $680, giving to one f of the sum due him, and to the other $40 more than ^ of his debt; at another time he pays them $580, giving to the first f of what remains due to him, and to the other -f- of what remains due to him. Wiiat was the amount of each debt?- 63. If a certain croquet-ground were 5 feet longer and 3 feet broader it would contain 320 more feet; but if it were 3 feet longer and 5 feet broader it would contain 310 more feet. What is its present area? 64. The sum of two numbers is 12, and the difference of their squares is 24. What are the numbers? 65. Two boats, 320 and 360 feet long respectively, are moving with uniform speed. If they go in opposite directions it requires 10 seconds to pass each other; but if they go in the same direction it takes 90 seconds for them to pass. What is the speed of each boat? 66. A train runs a certain distance at a uniform rate. If the rate be increased by 5 miles an hour the distance would be travelled in f of the time; but if the rate be diminished by 5 miles an hour the time Avould be increased by 3 hours. Wliat is the rate and distance? 502 SIMPLE EQUATIONS. 67. What number of 3 digits is greater by 99 when its digits are reversed; greater by 270 than the sum of its digits; and greater by 45 than when the second and third are trans- posed ? 6^, A and B could have completed a certain piece of work in 12 days; but after both had worked 4 days B was left to finish it alone, which he did in 24 days more. How long would it have taken each to do it alone? 69. A number consists of 2 digits whose sum is 12, and if 15 be subtracted from the number, and the remainder be divided by 2, the digits will be inverted. What is the number? 70. A boy spent his money in oranges. If he had bouglit 5 more, each orange would have cost a half-cent less; if 3 less, a half-cent more. How much did he spend, and how many did he buy? 71. A person bought apples at 4 cents a dozen, and 1| times as many peaches at 12 cents a dozen; after mixing them he sold them at 8 cents a dozen, losing 4 cents on the whole. How many dozen of each did he buy? 72. Fi»nd a fraction that becomes f when 2 is added to its numerator, and ^ when 4 is added to the denominator. 73. Five pounds of tea and 12 pounds of sugar cost $7.44. If tea were to rise 10 per cent and sugar fall 25 pe: cent, 8 pounds of tea and 6 pounds of sugar would cost $11.10. What is the price per pound of each? 74. A^s income is half as much again as B% while his ex- penses are twice as great as B's. A spends $60 more than his income, and B $60 less than his. What is the income of each? 75. A invested some money at 5 per cent, and B at 6 per cent, both receiving the same amount of income. If A had invested $1000 more than he did, his income would have been 11 per cent on B's investment. What did each invest? 76. An oarsman can row 9 miles up stream and 13 miles down in 4 hours, or 13 miles up and "9 miles down in 5 hours. What is the rate of the stream and of the rowing? 77. Six years hence the product of two people's ages will be greater by 348 than it is now. What will then be the sum of their ages? PROBLEMS, 503 78. A invests money at 4 per cent, B at 5 per cent, and C at 6 per cent. A and B together receive ^560, B and C $520, and A and C $360. How much does eacli invest? 79. Find the quotient of two numbers whose sum is n times their difference. 80. A and B can finish a job in 12 days. A worked 2 days, and B 3. How long will it take C to finish it if he could have done the whole in 15 days with B^s assistance, and in 10 days with A's? 81. A cai'penter and apprentice received $16.80 for 7 days' wages, the carpenter getting 20 cents more for 2 days' work than the boy for 3 days'. What was the daily wages of each? 82. A man paid $50 for 7 photographs and 12 prints; if he had paid $1 more he could have had 7 prints and 15 photographs. What was the price of each? Ratio and Proportion. §164. 1. Divide 126 into three parts that shall be proportional to the numbers 3, 4, 7. 2. Find two fractions that shall be to each other as 3 : 4, and whose sum shall be f. • 3. Divide .0444 into three parts that shall be to each other as i : i : |. 4. Find two numbers which are to each other as 4 : 3, and whose difference is ^ of the less ? 5 If 0- : ^ :: 6 : 8 and 4a; — 3^ = 7, what is the value of X and y'^ 6. A year's profits were divided among two partners in the proportion of 3 : 4. If the second should give $425 to the first, their shares would be equal. What was the amount divided? 7. In a first yearns partnership A had 3 shares, and B 4. In the second, A had 1, and B 2. In the first year A gained $300 more than he did the second, and B gained $200 less than he did the second. What were the profits each year ? 504 RATIO AND PROPORTION. 8. In a farm-yard there are 4 sheep to every 3 cattle, and 5 cattle to 6 hogs. How many hogs are there to every 20 sheep? 9. A drover started to market with a herd of 7 horses to every 5 mules. He sold 27 horses and bought 3 mules, and then had 3 horses to every 4 mules. How many of each had he at first? 10. Find two quantities whose sum, difference and product are proportional to 5, 1 and 12. 11. What number is that to which if 2, 6 and 12 be sever- ally added, the first sum shall be to the second as the second is to the third? 12. What two numbers are to each other as 3 to 4, and if 4 be added to each the sums will be as 4 to 5? 13. What quantity must be taken from each term of the ratio m : n that it may equal the ratio c \ d? 14. If a : Z> be the square of the ratio of a -\- c : b -\- c, show that c is a mean proportional between a and b, 15. If a : Z> == Z>. : c, show that a:a-{-b=^a — b\a — c, 16. And under the same conditions show that {a' + ¥) {V + c') = {ab + bc)\ 17. li a : b — c : d, show that a {a -{- b -{- c -{- d) = {a'\-b) {a -\- c). 18. In a milk-cUn, the quantity of milk is to the entire contents (milk and water) as 5 : 6. Five gallons are sold, and 1 gallon of water is added; then the ratio of the milk to the whole is 4 : 5. How many gallons of each were there at first? 19. In a two-mile race between a bicycle and a horse, their rates were as 5 to 6. The bicycle had 1 minute start, but was beaten by 312 yards. What was the rate of each? 20. A line is divided by one point into two parts in the ratio of 3 : 5, and by another point into two parts in the ratio of 1 : 3. The distance between the points of division is 1 inch. What is the length of the line? 21. The sum of the two digits of a number is 6, and the number is to the number expressed by the same digits reversed as 4 : 7. What is the number? 22. One ingot contains two parts of gold and one of silver, PROBLEMS. 505 and anotlier two parts of gold and three of silver. If equal parts are taken from each ingot, what will be the proportion of the gold to the silver in the alloy? 23. If two ounces be taken from the first and three from the second, what will be the ratio of the gold to the silver? 24. A cask contains 4 gallons of water and 18 gallons of alcohol. How many gallons of a mixture containing 2 parts water and 5 parts alcohol must be put in the cask so that there may be 2 parts of water to 7 of alcohol? 25. Which is the greater I'atio, 1 -\- a \ 1 — a or 1 -\- a^ : 1 — a^, a being positive and less ihan 1? 26. AVhicli is the greater ratio, a^— ah -{-If : a^ -\- ab -\- F or a* — a^'b'' + ^* : a* + a^b^ + ^\ ^ '^^^ ^ having like signs? 27. What number must be taken from the second term of the ratio 2 : 34 and added to the first that it may equal 5:6? 28. What number must be taken from each term of the ratio 19 : 30 that it may equal the ratio 1:2? 29. It a :b = c : d, show that a^ : b"" = a" -\- c" : F + d\ 30. A bankrupt owed two creditors $1800. The sum of their credits is to the less as 3 : 1. What did he owe each? 31. Discuss the general problem: To divide a given quantity iVinto parts proportional to the given numbers m, n, p, etc. 32. Divide the number iV^into three parts, x, y and z, such that X shall be to t/ as 2 : 3, and z to the difference between x and y as 3 : 2. 2,2,' The speed of the steamship Servia is to that of the Both- nia as 13 to 10, and the first steams 5 miles farther in 8 hours than the second does in 10 hours. What is the speed of each? 34. The speed of two pedestrians was as 4:3, and the slower was 5 hours longer in going 36 miles than the faster was in going 24. What was the rate of each? 35. A chemist had two vessels, A, containing acid, and B, an equal quantity of water. He poured one third the acid into tlie water,. and then poured one third of this mixture back into the acid. What was then the ratio of acid to water in A? 2)6. If 24 grains of gold and 400 grains of silver are each worth one dollar, what will be the weight of a coin containing equal parts of gold and silver and Avorth a dollar? 506 llATIO AND PROPORTION 37. AVluit common quantity must be subtracted from the four quantities m, n, x and y that the remainders may form a proportion? 2,^. A cliemist has two mixtures of alcohol and water, the one containing 90 per cent, of alcohol, the other 50 per cent. How much of the first must he add to 1 litre of the second to make a mixture containing 80 per cent, of alcohol? 39. It is a law of mechanics that the distances through which heavy bodies will fall in a vacuum in different times are 23roportional to the squares of the times. If a body fall 48 feet fai'ther in 2 seconds than in 1 second, how far will it fall in 1 second? How far in t seconds? 40. Find an expression snch that if yon subtract m + n and 711 — 71, the ratio of the remainders shall be n : m. 41. On a line are two points whose distance is a. The first point divides the line into parts whose ratio is 2 : 3; the second into parts whose ratio is 5 : 7. What is the length of the line? 42. If a line is divided into two ])arts whose ratio is m : n, what is the ratio of the length of the whole line to the distance of the point of division from the middle point? 43. A line is divided into three segments propoitional to the numbers m, p and q. What is the ratio of the parts into which the middle point of the line divides the middle segment? 44. Divide $285 among tlii'ee persons, A, B and C, so that the share of A shall be to that of B as 6 . 11, and that of C shall be $30 more than those of A and B together. 45. A sailing-ship leaves port, and 12 hours later is fol- lowed by a steamship. If the ratio of the speeds is 3 : 8 how long will it take the steamer to overtake the ship? 46. A courier started from his post, going 7 miles in 3 hours. Two hours later another follow'ed, going 7 miles in 2 hours. How long will the second be overtaking the first? 47. The areas of the openings of two water-faucets are in the ratio 3 : 5; the speeds, of flow of the water through the openings are in the ratio 3:4. At the end of an hour 1221 gallons more have flowed through the second than through the first. What was the flow from each? PROBLEMS, 507 48. Tlie flows from two faucets into two equal vessels is in the ratio 4 : 7, and both vessels were placed under them at the same moment. When the vessel under the larger faucet was full, it was removed and the other put into its place. In 80 seconds from the time of beginning both vessels were filled. How long would it take each f;iucet to fill one of the vessels? 49. Three numbers, a, b and c, are so related that a : h -\- c — m : n, b : c -\- a = p : q. Find the ratio c : a -{- b. Find a, h, and then a -\- h, in terms of c, 50. If, in the preceding problem, the &\\m a -{- h -\- c =^ X, express each of the numbers a, h and c in terms of N. 51. The speeds of two trains, A and B, are as m : ti, and the journeys they have to make as ^ : 5'. It took train B t hours longer to make its journey than it did train A. What was the time required by each train for each journey? 52. A street-railway runs along a regular incline, in conse- quence of which the s})eeds of the cars going in the two direc- tions are as 2:3. The cars leave each terminus at regular intervals of 5 minutes. At what intervals of time will a car going up hill meet the successive cars coming down, awdivicG versa'^ 53. The same thing being supposed, two cars starting out simultaneously from the termini meet at the end of 30 minutes. How long in time is the journey for each car? 54. The same thing being again supposed, a rider gallops up hill at such a rate that he passes the successive cars going up hill at the same time that they meet the successive cars coming down, so that every time he passes a car going up he meets one coming down. What is the ratio of his speed to that of each of the cars? 55. Give the algebraic answers to the three preceding questions when the ratio of the speeds is m : w. 56. Three given points. A, B and X, lie in a straight line. A and B are taken as base- ^ y b x points from which distances ' ill 508 RATIO AND PUOPOIITION. are measured. Having given Distance AB = b, Distance AX = x, it is required to find the jDositioii of a fourth point, Y, between A and B, such that we shall have AX :YB = AX :BX = X : X - h. Do this by finding the distance of Y from A in terms of h and X. 57. Show that in the preceding construction we have AY ' AX AB* 58. Show that, in the preceding problem, the product of the distances of X and Y from the middle point of the line AB is \h\ 59. If, instead of the point X, the point Y is given, find the distances AX corresponding to the following values of AY, in order that the same proportion may hold true, and explain the results when negative: («) AY = i J. Ans. X = !»• {e) KY = \ A. (^)AY = |s. (,?) AY = ^ A. (;/) AY = i 6. (V) AY =. 1 A. {d) AY = (i + «)^>. (e)AY = ^A. I1p:mark. When four points on a strnight line fulfil the preceding proportion, they arc called four harmonic points, and the line AB is said to be divided harmonically. 60. It is a theorem of mechanics that, in order that two masses, V and AV, at the ends of a lever, AB, may be In equi- librium, the distances of their points of suspension, A and B, from the fulcrum, F, must be inversely j^voportional to their weights; that is, we must have Weight V : weight W = FB : FA. PROBLEMS. C09 iNow, if the leiigtli AB of the lever is I, and the weights of A F B V and W are respectively m and n, express the lengths AF and FB of the arms of the lever. 6 1. The weights at the end« of a lever are 8 and 13 kilo- grammes, and the fulcrum is 3 inches from the middle of the lever. What is the length of the lever? 62. The sum of the two weights is 25 pounds, and the ratio of the distance of the fulcrum from the middle point to the length of the lever is 2 : 9. What are the weights? 6:^, The weights are m and n {in > 71), and one arm of the lever is h longer than the other. Express the length of the lever. 64. A lever was balanced with weights of 7 and 9 kilo- grammes at its ends. One kilogramme being taken from the lesser and added to the greater (making the weights 6 and 10 kilogrammes), the fulcrum had to be moved 2 inches. What was the length of the lever? 65. A line is divided into three parts proportional to the numbers 3, 4 and 5. What is the ratio of the parts in which the middle point of the line divides the middle segment? 66. To 300 pounds of a mixture containing 2 parts of zinc, 3 of copper and 4 of tin was added 200 pounds of another mixture of the same metals, when it was found that the pro- portions were now as 3, 4 and 5. What Avere the proportions in the mixture added? 67. Find two numbers whose sum, difference and product are to each other as the numbers 5 : 1 : 18. 6S, Find two numbers in the ratio 7 : 3, the ratio of whoso difference to their product is 1 : 21. 510 • RATIO AND PROPORTION, 69. Find fcAvo numbers such that the first sliall be to the second as their sum is to 3^, and as their difference is to 2f . 70. Find three numbers whose sum is 73, and such tliat if 2 be subtracted from the first and second their differences will be to each other as 1 : 2, and if 2 be added to the second and third their sums Avill be to each otlier as 4 : 5. 71. Two boats start in a race. The second boat rows 25 strokes to the first's 28, but 10 strokes of the second are equal to 12 of the first. If the distance between the boats at starting is 30 strokes of the second boat, how many strokes will it: make before reaching the first? 72. One cask contains 18 gallons of wine and 6 gallons of water; another contains 12 gallons of wine and 18 gallons of water. How much must be taken from each to form a mix- ture containing 8 gallons of wine and 8 gallons of water? 73. Two mixtures of wine and water contain respectively \ and I wine. How much of each must be taken to form 44 gallons of a mixture of which the wine is to the water as 5 : 6? 74. A and B ran a race in 6 minutes. B had a start of 20 yards; but A ran 5 yards while B ran 4, and Avon by 10 yards. What was the length of the race, and the rate o'f running? 75 . A jeweller has three ingots of metal. A pound of the first contains 7 ounces of gold, 3 ounces of silver and 6 ounces of copper; a pound of the second contains 12 ounces of gold, 3 ounces of silver and 1 ounce of copper; a i^ound of the third contains 4 ounces of gold, 7 ounces of silver and 5 ounces of copper. He wishes to form an alloy weighing 1 pound, which shall have 8 ounces of gold, 3f ounces of silver and ^\ ounces of copper. How much must be taken from each ingot? 76. The king of Syracuse gave a goldsmiih 10 pounds of gold with which to make a crown. When it was finished the king gave (lie crown to Archimedes to ascertain if it was pure gold. The philosopher knew that gold weighs .948 as much in water as in air, and silver .901. When the crown was weighed in water he found it lost 10 ounces. What was the quaniity of gold and silver in the crown? IRRATIONAL EXPRESSIONS. gH Irrational ExpressionSc §179, Execute tlie following divisions: 1 . x^^ -f- or^. 2. )llx^h^^ci ~ 9a^Wc. 3. 12^Wi -^4aW. 4. x^^^-yz^ -V- x'^y^z~l. 5. a-'b-'-^a-'h-K 6. x-\i~^z~'^ -, x^yz^. 7. a-^"^-^c-^^-«w. 9. 24:cy^ + 15:c-y;s' - '^xy-' H- B.^-y. 10. ^hx'yz-' - ^Wy-^z' + ly-'z' ~ - IxY^-'. 1 1 . 20a; V - 'z' - 4.y'z' -12x-'y-'-^4:X-'y- 'z\ 12. 28%^-' + l%xy-''z^ — 12x-^y-'z-' -^ 4x-hj-'z-\ 13. ai —a^U -^ «i 14. X^ — xW -\-X^ -^ Xh. 15. 12at -36at -^ l-^ftl m 16. 2a;« - 60;-^-^ 2a;'. 17. 80;^ —4:xi -^ 2.T~i 183. Express the following ])rodacts of irrational qnantities with a single fractional exponent by reducing the fractional exponents to a common denominator, and then reversing the process of §182: X y z 1 I. Prove the equation" «w^«c« = {d^b^c^Y, 2/ aWck Ans. a^bk'^ = {a^b'c)i. 3- m^n^pK 4. 2m. All IS. 12i 5. 2i3i 6. am. 7. MrisK 8. 1 1 x^y^^. 9- d2iam. 10. vrvvvi II. ami. 12. cm. 13. 7^ . 5i 14. h-m 15- 2-^3^. 16. 8^12- i 512 IRRATIONAL EXPRESSIONS. §183. Keduce the following expressions to monomials: 1. Vbi+ V2i+ '/K Alls. ¥6{Vd+ V4t-]-i)= oVif, 2. VT2 + V27 + i^i^. 3. V6- V80 + VlTo. 4. V2- Vis + V32, 5. V76 + 4/48 - VS, 6. 1^2 + 2 ^27"+ 3 fT5 - 9 1^48. 7. V4^ - Vda + V2da. 8. V'^- + Vb^ - V7^. Eeduce the following to their simplest form and factor: 9. Vl^^F + VbUa^\ 10. {4:a'by - {a'dy. 11. {2''a''b'c)^ - {4..b'a'b'c')i + {4..6'ah'c)l 12. {64.a'^ + 'F)h - {Ua'^-Wy^ + {2a*"* + ^)i + {2a'^cy. 13. Kc^ + ^W. 14. [{ci + ^y{x + y)]K ,5. iL+i l/^^ZJ. x6. 7)1 — 11 f mp - - q 2^ -{- q ' 7ri-\- h\ni' — 2m)i-{- if J ' §184o Multiply: I. {c + b V7) {c - a V7). 2. (m + Vji) {vi - VJi). 3. {am -\- n Va — z) {n — m Va — z). 4. (4 _[_ 3 |/2) (4-3 V2), 5. (5 - Qn V2) (5 + ijn ^2). 6. Mi-(^ - \m^ + 2(i^iw. 7. ( i^i? + ^ + '^/^ - ^) ( ^/^ + ^ - ^^^ - ^)- 8. {a + x^ + yi){a~'0^-\-y^.^ I Vvi Vn UVm _ Vn\ ^' ^Vn VmJ^Vn VnJ' {x+a) ^ \ \ {x-aY {x - ay \\{x+ ay ^ Aggregate the following fractional expressions and sim- plify when possible: ri . a^ {c-\-xy (c-x)^ a r {c — xp [c -\- xy ml , m^ , m , . ^' + ^ ' ^3* —T -^ r -^ — • 14. 7- — t -\ ; . QUADRATIG EQUATIONS. 513 Equations of the Second Degree. §§ 195-302. I. ^x' + nx = 1 1.0. 2. 2.r + 8a; = 64. 3. a;^ - llx + 6 == 57. 4. ^' - 37a; = - 320. 5. a;' + 6.T ==: 7. 6. a;' - %x = - 12. 7. x — mx — — n, Z. X ^ — — --. 2 2^/' 15 72 - 6a; ^ , , ^ Q. --^ — = 2. 10. a; — ax — bx := — ao. X 2x II. V'V — 1 = a; — 1. 12. a;^ — of a; = 18. 13. 3a;^ + a; = 7. 14. 4.i' ^^ == 46. 40 , 27 ,,, ^48 165 19. (a; — fl^) (x — h) = 0, 20. (a; + 4) (a; + 1) = 6 (x' + 1) - 8a;^ 21. 3 (a;' - 1) - 24 = 4 (a: + 5) {x - 3). 22. {x - 2) (3a; + 1) = 10 - (2a; + 1) (^ _ 3). X a; — 3 a; — 8 x — 1 2 (a; - 3) x - I' ^' a; + 2 2a; + 10* 5 29 _ _ 3 2 — X 4 — 5a; 2a;* a; + 2 _ 3 16a; 32- (2a: - 1) ~" ^ "^ 4a;' - 1* X — 1 2a; — 3 a; — 8 ^ a x + b 2*7, = , 28. = ' — . 'x a; — 1 a; — 9' ' x — b %x — a 3 2 1 ^5- 26. 29. ?>a — a; 2« — a; a — 3a;* 30. 6 |/.i; + -^ == 37. 31. |/3i^; - 5 + |/a; + 6 3^ 9. 32. \^x — 2 + 1/4 — a; = 4/0 — a;. 614 >l'^2'^^ OA'E UNKNOWN QVANTJTT. a-x 9 --3.7; , , 33. 10 -^- = ---- + 00;. X + 3 ,16 --- -^x __ 26 35. 14 + 3.T - ^;^^---- = yx H ^ — . X H- 19 _ 4 _ 9.7; - 8 ^ * ' 3 a; ~ ;^/6^+78. 99. 3 Va;' + 5 - V9^~+4^~+~5 ~ 2. 100. 4/0:' + 3a: + /a;' + a: + 2 = 4. § 198. Factor the following expressions by adding such a quan- tity as will make the trinomial a perfect square, and subtract- ing the same quantity. a"^ ■— 2ah — ZW. Add 4Z>^ and subtract it; then a^ - 2ah + Z>' - ^y = {a- by - W = [{a -b)- 2b] [{a -'b) + 2h]. WITH TWO UNKNOWN QUANTITIES. 517 ' I. a' + '^a'h' - %\ 2. x' - 2ax - 3a^ 3. x" - 4cX + 3. 4. 2:' + 82:?/ -- 9y\ 5. a' - 20abc + 64^V. 6. 2;^;' - ^x + 3. 7. ^'^ + 2% - 8b\ 8. 4^^ - 4.ab - Ub\ 9. :?;' + 6x + 5. 10. 6a' + bab - 6b\ II. - x' + x' + 12. 12. a' + 9«^<^^ + S\b\ 13. 2a' - 2ab + b\ 14. a' + 4.ab + 3b\ 15. x' - 6x' - 16. 16. x' + a^y. 17. x' — Qx'y' + Dx'f, 18. 12^' -f 24a:y + %'. 19. ia' - 37 a' b' + 9>. (§3O70 Simultaneous Quadratics. Two Unknovvk Quantities. I. X -\- y = 7, 2. X — y = 5. x' +y'= 25. x' - 2xy = 21. 3. 22; — 3?/ = 1. 4. 2a: — 2?/ = 5. 3x^— 4cxy = 15. 5a;' — 3a:^ — v'' — 161- 5. a;' + ^' = a'. 6. a; + y = 28. x" - y'' = b\ xy = 147. 7. x' + If/' = 169. 8. o;^ + y^' =: 224. ^^ = 60. xy = 12. 9.x + y= 8. 10. -- = --. x'+ y'= 224. 3a;y + 2x + y = 485. II. a; + y + 4/^y = 19. 12. lOa; + ^ = 3^:?/. x' + y' = 97. .V = ^ + '^^' 13. ^' = ia;y. 14. X + 2y =: 30. cc - 7/ = 15. .V' - 10^ == 1^^ + ^6- 15. 2a; + 3?/ = 17. 16. 3a; + by = 31. xy = 12. xy + y'= 18. 17. 5a; — 3y = 1. 2^' - a;' - 3a:f/ + 10a; - 5^ = 1. 18. 4a; — 5^ = 1. lly' - 6x' - 9xy + 22a; - 7y' = 20. 19. 7a:^ - 13xy + 5/ = - 5. 20. 3a;' - lla;y + 7y' = 7. 6x^ - 9xy + 4:t/ = 6. 5a;' - 17a^^ +11/ - 1*^- 518 QUADRATIC EQUATIONS. 21. S^r'' - nxy + 7^/' = 5. 22. Zx" - 2^xy + ty' = 15. Qx' - Ihxy + 9^/^ = 15. 7^' - 48^'^ +19/= 11. 23- X + y = ij. 24. a; + ^ = 8. X + ;?/ = 72. 2:^ + y' = 3368. 25- X + y = 5. 26. .t' — 2/' = 56. ^' + .' = '7- ^^ (^ -- ^) = 16. 27. X J^ y ^xy. 28. xy + a.y = 18. xy = x' - y\ X -{- xy^ =: 27. 29. x' +y' + x + y^l^. 30. :z;^ + ^^ - a; - ?/ = 78. xy = 6. ^y + ^' + .V = ^^' 31- 1 + i = ^. ^ ^ 6 32. 3y^ - 2a;' = 19. r?; + ^ = 5. / + a;?/ = 15. 33- x\-\-y\ =z 6. 34. 2.r'' + ?^xy = 26. :rf + ^1 .= 126. 3?/' + )lxy = 39. •^ 36. ^'^^^ - 'ixy = 24. 35- "^^y -x-y 820 a;^ 4- ^'^ = . a:i/ - 2.?/' =4. 37. , a; 5 ^•^ + .^-3- 38. ^ + .y = ^+2. 1 4. ^ - io ~T~ "~~ • ^ + ^ = ^5: xy X 3 y ^ 39- ^x' + 3/ = 27. 40. 4.x' - 57/' = 16. ^x" - .^'^ = 15. ^x' + 2^^^ = 35. 41. x' -^y' =45. 42. Zx - 2y = 6. a: =2y. :r?/ ~ .1; =8. 43- xy =12. 44. 4cX^ — 5y = 4:xy. 3a; — 2.y = 1. 5x + 3y = 37. 45. X -\- xy = 24. 46. ^' + ^' + ^ + ^ =- 36. y 4- 2:?/ = 21. ^^ - 2/' + ^ - 2/ = ^4. 47. x' + xy=zd5. 48. x' + xy + f = 7. y'' -f o;^^ = 14. x'-xy + y' = 3. 49. x' ~ 2i:«/ + ^' =7. 50. ^ + ^y + y = 11- x' - Zxy + 2^^ = - 2. ^' + ^y + y" = 19. WITH THREE UNKNOWN QUANTITIES. gig 52. x"" + y =. ^{x-y). X +y=4(a;-,v). ^y. 54. x^ — xy^^l^, x^ J^y' = 74. 56. x'' - ^xy + 2y^ = 1. x"^ + 2.T^ — 4y'^ = 5. 58. i^+ Vy =z 5. ic V.'c -f- ^ '^.V = 35. 60. 4:x' - 9y' = 7. 2x +^ =7. 62. a:' + y^ = 25. X +y = 7. 64. X — 2y = 2. x' + 4y' = 100. Three or more Unk:n^ovvn Quantities. I. xy = 24. 2. x"" + .T2; = 24. {7 -y)z = 8. ,-z' + XZ = 12. (3 -x) {z- 11) = 3. y' + yz + z' = 28 - 3?/. 3. :/;?/2; = 3 (.t^ + 4) = 12 (2; + ^) = 4 (o;^ + ;^ - 10). 4. :z:' + .y' + ^' = 84. 5. ^ + ^Z + ^ = 14. •^ + .V + ^ = 14. a:' + y' + z' = 84. :r7/ = 8. xz = y\ X -\- y _b 6. a: + f/ + 2; =r 12. 7- — ^ - g . 5^. x + 2 _ ^ x'+ 2 19 53. 55. ^+2" 3* 2-{-y = 40 — X :. + 2^ =- 7. x^ — xy = 35. ^^_ ^^ = 10. 57. .T^' + ^:y = 18. 2;?/^ + a; = 27. 59- X -y = 8. rr^ - ?/^ = 80. 61. 4 (.r + 2^) = 12. rr'^ - 4//*^ := 33. (>z- X — y = 2. ic^ 4- 2/^ = 34. ^y -\- yz -}- zx = 47. a;+ ^ 3 xz 4' ^•' + 2/' - ^' = 0. y + ^ ^ _7 y^ 12* 8. 2x' + 2.T^y + y' = 49. 9. :x^ - ^^y + 2^ = 2. x' — .T2; + ;a^' = 28. x' + f + / = 49. // + 2yz + i^' = 25. xy — z-\-y — 3. :^ + ?/ + 2; = 9. II. a; + ^'zy + 2; = 10. " + y/'^ + ;2;^ = 29. x^ +'^« + ^' =- 38. ^' = 4^ + 1. ^^ _|_ ^2; = .t'. 10. .7; 520 QUADRATIC EQUATIONS, 12. x-{- y =1, 1^, X -\- y = 9. II J^ V = 1. u -\- V = ^, X + u' ^ S, x' + u' = 52. y + v' := 4. y' + v' = 41. 14. xii = yv. 15. xy = 35. a; -[- y =14. tcv = 18. ^^4-^=7. x + u = 13. ~+^= 4. ^ + i; =9. Problems Leading to Quadratic Equations. 1. A principal of $6000 amounts with simple interest to 17800 after a certain number of years. Had the rate been 1 per cent, higher and the time 1 year longer, it would have amounted to $720 more. What was the time and rate? 2. A courier left a town riding at a uniform rate. Three hours afterwards another followed, going 1 mile an hour faster. Two hours after the second another started, going 6 miles an hour. They arrive at their destination at the same time. What was the distance and rate of riding? Ans. Dist. = 60 or 6. Speeds, 4, 5 and 6 or 1, 2 and 6. 3. In aright-angled triangle the hypothenuse is 5 and tlie area 6. What are the sides? 4. Find two numbers whose product is 180, and if the greater be diminished by 5 and the less increased by 3, the product of the sum and difference will be 150. 5. Find two numbers whose sum is 100 and the sum of their square roots 14. 6. Find two numbers whose sum is 35 and the sum of their cube roots 5. 7. By selling a horse for $130 I gain as much per cent, as the horse cost me. What did I pay for him? 8. What is the price of apples a dozen when four less in 20 cents' worth raises the price 5 cents per dozen? 9. The sum of the squares of three consecutive numbers is 149. What are the numbers? 10. If twice the product of two consecutive numbers be divided by three times their sum the quotient will be f . What are the numbers? PROBLEMS. 521 11. A womnn bought a number of oranges for 3G cents. If she had bought 4 more for the same money she would have paid i of a cent less for each orange. How many did she buy? 12. In mowing 60 acres of grass, 5 days less would have been sufficient if 2 acres more a day had been mown. How many acres were mown per day? 13. A broker bought a certain number of shares (par $100 each) at a discount for $6400. When they were at the same per cent, premium, he sold all but 20 for $7200. How many shares did he buy, and at what price? 14. If the length and breadth of a rectangle were each in- creased by 2, the area would be 238; if both were each dimin- ished by 2, the area would be 130. Find the length and breadth. 15. Twice the product of two digits is equal to the number itself; and 7 times the sum of the digits is equal to the number formed by the same digits reversed. What is the number? 16. The sum of two numbers is ^ of the greater, and the difference of their squares is 45. What are the numbers? 17. The numerator and denominator of two fractions are each greater by 2 than those of another, and the sum of the two fractions is 2-|; if the denominators were interchanged, tlie sum of the two fractions would be 3. What arc the frac- tions? 18. A man starts from A to go to B. During the first half of the journey he drives \ mile an hour faster than the other half, and arrives in 5f hours. On his return he travels a mile slower during the first half than when he went in going over the same portion, and returned in 6f hours. What was the distance and rate of driving? 19. .A person who has $8800 invests a part of it in one enterprise and the rest in another; the dividends differ in rate, but are equal in amount. If the sums invested liad exchanged rates of dividends, the first would have yielded $200 and the other $288. What were the rates? 20. Divide 50 into two such parts that their product may be to the sum of their squares as 6 to 13. 21. A company at a hotel had $12 to pa}^ but before set- 522 QUADRATIC EQUATIONS. tling 2 left, when those remaiiiiiig had 30 cents apiece more to pay than before. How many were there? 2 2. A drover bought a nnmber of sheep for $180; after keeping 10 he sold the rest for $200, and gained 33^ cents apiece. How many did he buy? 23. Two partners, A and B, gained $140 in speculation; A's money was 3 months in trade, and his gain was $60 less than his capital; B's money, which was $50 more than A's, was in 5 months. What was each man's capital? 24. Divide 30 into two such parts that their product may be 36 times tlieir difference. 25. A aiid B set out from two towns which are 126 miles apart, and travelled until they met. A went 8 miles an hour, and the number of hours they travelled was 3 times greater than the number of miles B travelled an hour. What were their hourly rates? A^is.y in part. B's rate, VoS — 4. 26. In a purse containing 28 pieces of silver and nickel, each silver coin is worth as many cents as there are nickel coins, each nickel is worth as many cents as there are silver coins, and the whole are worth $1.50. How many are there of each? 27. Find two such numbers that the product of their sum and difference may be 7, and the product of the sum and dif- ference of their squares may be 144. 28. A grocer received an order for 12 pounds of sugar at 12 cents a pound. If he should have none for that price, he was to send as many pounds more or less than 12 as the sugar cost less or more than 12 cents a pound. The bill amounted to $1.35. How many pounds had he sent, and what was the price per ]>ound? 29. A grocer sold 50 pounds of pepper and 80 pounds of ginger for $26; but he sold 25 ])ounds more of pepper for $10 than he did of ginger for $4. What was the price per pound of each? 30. A and B's shares in speculations together amounted to $675. A had his money invested 5 months and B 4^ months, and each receives in capital and profits $455. What did each be2:in with? PROBLEMS. 523 ^ 31. A person rents a certain number of acres of land for $120; he retains 10 acres, and sublets the rest at 20 cents an acre more than he gave, and receives $12 more than he pays for the whole. How many acres were there, and how much per acre? 32. A person bought a certain number of shares for as many dollars per share as the number he buys; after they rose as many cents per share as he had shares, he sold them and gained $4. How many shares did he buy? 2,2,. The income of a certain railway company would justify a dividend of 5 per cent, of the whole stock; but as $150,000 of the stock is prefei-red^ guaranteeing 6 per cent., the divi- dend for the remaining stock is reduced to 4f per cent. What is the whole amount of stock? 34. The length of a rectangular farm is to its width as 4 to 3; f is in grass, and the remaining 45 acres is cultivated. What are the dimensions of the field? 35. If a straight line be divided into two such parts that the rectangle contained by the whole line and one part is equal to 6 times the square of the other part, what will be the ratio of these two parts? 2,6, Out of a sphere of clay whose diameter is 16 inches, two spheres are formed with radii of 3 and 5 inches respec- tively. If the volumes of spheres vary as the cubes of their radii, what will be the radius of the sphere that can be made of the clay that remains? 37. The two digits of a certain number differ by 1, and their product is i of the noxt higher number, what is the number? 38. Find five numbers having equal differences, and such that their sum shall be 40, and the sum of their cubes 3520. 39. A merchant bought a barrel of wine for $60; he re- tained 12 gallons for his own use and sold the remainder at an advance of 80 per cent, on each gallon and gained 20 per cent, on the whole. At what price per gallon did he sell it? 40. Find two numbers that are to each other as 9 to 7; and the square of their sum is equal to the cube of their dif- ference. 524 QUADRATIC EQUATIONS. 41. The panel in a door is 12 by 18 inches, and it is to be SRiTouiided by a margin of uniform width and equal surface to the panel. How wide must the margin be? 42. The fore wheel of a coach makes 6 more revolutions than the hind wheel in going 160 yards; but if the circumfer- ence of each wheel be increased by 4 feet, the fore wheel will make only 4 more revolutions in 160 yards. What is the cir- cumference of eacli wheel? 43. The sum of three numbers is 15; the difference between tlic first and third is 3 more than the difference between the second and third, and the sum of their squares is 93. What are the numbers? 44. Tiie product of two numbers is 15, and if their differ- ence be added to the difference of their squares the sum will be 18. What are the numbers? 45. A certain number consists of two digits; the number is 4 times the sum of its digits; and 3 times the number is equal to twice the square of the sum of its digits. What is the number? 46. Find two numbers whose sum is 14, and if their prod- uct be added to the sum of their squares the result will be 148. 47. Two brokci's begin business with a joint capital of $10,000. A withdraws at the end of 12 months and receives $4960 in capital and profits. B remains 3 months longer and receives $7800 stock and gain. What was the original capital of each? 48. Find five equal numbers whose sum is equal to their continued product. 49. A jockey bought a horse and sold it at a certain per cent, profit; with the money he bought another horse and sold it at the same per cent, profit, and with the proceeds he was able to buy 2 horses each costing 2 per cent, less than the first. What per cent, did he make on each transaction? 50. Two travellers start from the same place at the same time, one goes due north 16 miles a day, and the other due east 21^ miles a day. How long must they travel in order to be 160 miles apart? PROBLEMS. 525 51. What is the length of a side of a square whose area is increased by |of its amount when 4 feet is added to each side? 52. Find the length of the side of a square such that the number of square feet in its area exceeds the number of linear feet in its perimeter by 12. 53. The perimeter of a rectangle is 34 feet; if its length were increased by 4 feet, while its perimeter remained the same, the former area would exceed the double of the second by 6 feet. What were the original dimensions? 54. If 3 feet be taken from one side of a rectangle whose perimeter is 14 feet and added to the other side, the area would be doubled. What were the first dimensions? 55. A man invests his money at a certain rate of interest for two years, and finds that he will get 1 per cent, more for it if he reckon by compound interest compounded annually than by simple interest. What is the rate of interest? 56. A person bought a certain number of shares when they were at a discount and sold them when they rose to a premium of the same rate per cent. His profit on the first investment was A percent, more than the common value of the premium and discount. What was the latter and the rate of profit? 57. A regiment of 2196 soldiers is formed into two squares, one having 6 more men on a side than the other. How many men are there on a side of each square? 58. Find two numbers whose product is twice their sum, and the sum of their squares 45. 59. Find two numbers whose product is 8 times their dif- ference, and the difference of their squares 48. 60. Find two numbers whose difference is 6, and | of their product is equal to the square of the less. 61. Find two numbers such that their product added to twice the square of the greater is 65, and the product added to the square of the less is 24. 62. Find two numbers such that their sum multiplied by the sum of their squares is 715, and the difference multiplied by the difference of their squares is 99. 6^, Two trains start at the same time from two towns and run at a uniform rate towards the other town. When Ihcy 526 PROGRESSIONS. meet it is found that one train has travelled 90 miles more than the other, and that if they continue at the same rates they will finish the journey in 6 and 13|- hours. What are the distance and rates? 64. A man receives $2200 a year interest. If he had in- vested his capital at ^ per cent higher, he could have lessened his investment by $4000 and received the same income as be- fore. How much had he invested? Progressions. Note.— Tlie abbreviations A. P., G. P., C. D., and C. R. are but for Arithmetical Progression, Geometrical Progression, Common Difference, and Common Ratio, respectively. 1. If the first and last terms of an arithmetical progression are a and I and the number of terms n, express the sum of all the intermediate terms. 2. If the first and last terms of an A. P. are 4 and 28 respectively, what possible values may the sum of the inter- mediate take? 3. Sum to n terms distinguishing the cases when n is even and odd, when necessary: 1-3+5-7+ 4. 2-4+6-8+ ■ 5. p,p + n,p+2n, 6. If the square of the fourth term of an A. P. is equal to the product of the first and sixth, show that the tenth term must vanish. 7. If the square of the second term of an A. P. is equal to the product of the first and fourth, show that tlie square of the sixth is equal to the product of the fourth and ninth. 8. Generalize the preceding result by showing that, in order that the square of tlie nth. term may be equal to the product of the first and n'ih, and the square of the mth to the product of the ^^'th and m'th, it is necessary and sufficient that m, m', n and n^ fulfil the conditions m' = 2 {m — 7i) +1; 2u^ = ni + n 9. Find three quantities in A. P. whose sum shall be 3« and the sum of whose squares shall be ll«^ PUOGRESSIONS, 527 10. Find 7 terms of an A. P. such that their sum shall be 14 and the sum of their squares 84. 11. In an A. P. the product of the first and eighth terms is less by h than the product of the second and seventh. How much less is the product of the third and sixth than that of the fourth and fifth? 12. Express the sum of n terms of an A. P. in terms of the first term and the CD. 13. If a and h are the first two terms of an A. P., express the last term and the sum of n terms. 14. Prove that if the sum of m terms of an A. P. be n, and the sum of n terms be m, we shall have 2 (m + ^0 + ^^^ — ^• 15. If a^, If, & be ill A. P., then, a-\-V c -\-a' h -\- c will also be in A. P. 16. The sum of the first three terms of an A. P. is 15 and the sum of their squares is 83. What is tlie sum of n terms? 17. In a progression of 9 terms, the third term is 10 and the sum 153. Find the first term and common difference. 18. In an A. P. a certain term is h\ there are 2n terms before k and n terms after it, and the sum of all the terms is 3^ + 1. Find the 0. D. 19. Two men start simultaneously from the same point in the same direction. The one walks m miles the first day, and diminishes his walk by h miles each day; the other walks n miles the first day, and increases his walk h miles each day. How far will the latter be ahead at the end of i days? 20. Express the sum of the G. P.'s: . ^n _(_ ^2n _|. ^3n _|_ _ _ _^ ^lOn. ^n _(_ ^2n ^ ^3n _|_ _ _ _|_ ^mn.^ 1 + |/3 + 3 + + 3^ 21. The sum of the first and seventh terms of a G. P. is A, and the sum of the second and eighth is h. Find the first term and the C. K. 22. The sum of the first and fifth terms of a G. P. being 528 PROGRESSIONS. added to twice the third term gives a sum which is 9 times the first term. Find the C. R. 23. The fifth term of a G. P. exceeds the first by 16, and the fourth exceeds the second by 4 1^3. Find the first term and 0. R. 24. In a G. P. the sum of n terms is S and the sv> WjI 2n terms is 6S, Express the 0. R. and first term. 25. In a G. P. of 271 -f 1 terms, whose first term is 5, the sum of the first and last terms is 125 greater than twice (he middle term. Find the C. R. 26. The first term of a G. P. is 2, and the continued product of the first 5 terms is 128. What is the 0. R. ? 27. Find that G. P. of which the product of the first and second terms is 3, and that of the third and fourth terms is 48. 28. A person who each year gained half as much again as he did the year before, gained $2059 in 7 years. What was his gain the first year? 29. A man who had a principal out at 5 per cent, per annum compound interest for 4 years found that the interest gained during the second and fourth years was greater by $84.10 than that gained during the first and third years. What was the principal? 30. Show that \ia,h,c,d, , . . ^, ? be in G. P. we shall have (a + ^ + c + -{-k) {h -^ c^ d-\- + /) = ^(^ + ^ + ^ + .... + 0^ 31. If a, h, c, d be in G. P. prove that (^« J^V" -\- c') {V + c' + d') = [ah + ic + cd)\ {h - cy + (c - ay -\- {d- by =: (a- dy. 32. Generalize the first of the preceding results by show- ing that if we multiply the sum of the squares of the first n terms of a G. P. by the sum of the squares of the n terms following the first term, the product will be equal to the square of the sum of all the products formed by multiplying each term from the first to the ^th by the term following it. 33. Sum to 71 terms (^_Ly+(,«»_i,y+(,«^-i,)V.... PIWGRESSI0N8. 629 34- I^^ *i Cr- P- <^f ^ terms are given: The sum of all the terms except the first = 33; The sum of all the terms except the last = — 22. Find the series. 35. Find two quantities of which the arithmetical mean is a and th. geometrical mean is g, and prove the result. 2,6. In a G. P. of 8 terms the product of the four alternate terms beginning with the first is 1, and the product of tlie four alternate terms from the second to the eighth is 16. Find the progression. 37. A party of m persons have s dollars unequally divided among them. Each simultaneously divides his money equally among his m — 1 fellows. If one of the party had a dollars in the beginning, how much will he have after 1.2, and p such divisions? ' 7)1 m — l\m J m (m — l)\m / m {m — 1)^ \m ) Find the limits of the sums of the progressions: 38.1 + 1 + 1+.... 39. - + ! + «+.... n m 7171^' 43..1 + (r + i)"'+(r+3-)"%.... 43. l-(r + 9"V(r + i 44. r+{l+ay+{l+a+a'y+{l+a+a''+a'y+ r and ar being each less than unity. 45. r+{l-ay+{l-a+a'y+{l~a+a''~ay+ 530 FUNCTIONAL NOTATION 46. r-{l-aY+(l-a+a'y-{l-a+a'-a'y-\- . X[. ^''' n + l {n+ ly "^ (^n + ly Functional Notation. Prove: 1. {2n)l = 2^{l.d.5. . . . 2/^-l) .nl 2. (2^) ! = 2^^ (1.3.5 15) (1.3.5.?) (1.3). Using the notation [m] = 1.3.5.7 . . . . m y^ = 2« Show that we have 3. ^! = 2''-[^-l][|-l][|-l]. ... [3]. [2u - ly ,j 5. If S{)i) represent the sum of the first n terms of a -- geometrical progression whose 0. R. is r, show that S{2n) = (r^ + l)S{?i). 6. What will be the last factors in the numerators and denominators of the following expressions: m- i"^)-- ^"^y- (^D^ c^i)^ (^')- «-e)+c-ii)=(:^>. FUNCTIONAL NOTATION. 531 Sn represent the sum of the first n natural numbers, ^hat s, /y,= l + 2 + 3 + +n, aow that: 10. /S'n : Sn + 1 = n : n + 2. 11. ^2X /S', X /5i = 3! [7]. 12. .^2 X ^S', X >S; X >^2n = nl [2n + 1]. 13. S,XS,XS, X .%n+i - (^ + 1)! [2n + 1]. 14. S,X S,X S, X S,n = {2n + 1) (t^!)^ [2^ - l]^ 15. /S'i+.?,+ ^8+ +^2. = 4.{V + 2' + d' + 4:'+ + 7i'). 6. S,+ S,+ S,+ + S,r^ + ^ == 1^ 4_ 3« + 5« + . . . . 4. (27^ + ly. 17. IlCi=^h + sCi^i find the values of 0^, C3, Ci, and Of terms of A, 5 and Co, and find the value toward which 6^ oaches as i increases indefinitely, assuming 5 < 1. 1 8. Apply the preceding notation to the following problem : A. person having a full and an empty cask pours half the con- tents of the full one into the other; then half of this last one back again. He repeats this double operation an indefinite number of times. Find what fraction of the liquid will re- main in the first cask after 1, 2, 3, 4, and i such double opera- tions. To do this assume that d and 1 — d represent the fractions of the . aid in the two casks after the ^th operation, and then find the fractions after the {t + l)st operation. 19. A vintner has one cask containing a gallons of wine and another containing b gallons of water. He pours half the wine into the water, then half that mixture back into the wine, and so on indefinitely. Find an expression- for the quantities and proportions of wine and water in each cask after 2n and also after 2n -\-l such operations. 532 PERMUTATIONS AND COMBINATIONS. Permutations and Combinations. 1. A regular cube is to have its sides numbered 1, 2 .... G. In how many ways may the numbering be done? 2. In how many ways might the numbering be done in the last problem if only three of the six sides were to be numbered ? 3. A party of 3 boys and 4 girls has to walk in single file, the boys ahead. In how many ways can they be arranged? 4. What would be the number of aiTangements in the last problem if the only condition were that the boys must be to- gether in one group and the girls in another? 5. If the combination of auy three different letters in any , order made a word, how many words of three letters could be formed from the 26 letters of the alphabet? 6. If in the last problem the words thus formed were divided into sets such that the different words of a set should be formed of the same letters, how many sets would there be, and how many letters in a set? 7. Six men with their wives are to stand in a row. In how many ways may they be arranged subject to the condition that each man must remain alongside his wife? 8. What would be the answer to the last problem in case each man had to keep his wife on his right? 9. A boy has the letter blocks which form the words you are mad. In how many of the arrangements will all three words be recognized, supposing that any word may be recog, nizedwhen its first letter stands first, and its other letters fol- low it in any order? 10. If every permutation of two or more letters made a word, how many words could be formed from 10 letters? 11. In how many permutations of n letters will the first letter retain its place? The second letters retain their second places? The last letter retain the last place? 12. If we write under each other all possible permutations of the first n numbers 1, 2, 3 .... ^^ what will be the sum of each column? Ans, ^{n -\-l)\ ) PERMUTATIONS AND COMBINATIONS. 533 13. What will be the sum of each column if the possible permutations of the figures 1 2 2 3 3 3 4 are all written under each other? 14. From a collection of 5 capital letters and 7 small ones how many combinations of 1 capital with 2 small ones can be formed? 15. The driver of a four-horse coach can choose his horses from a stable of 6 white and 8 black horses, but he must not pair 2 horses of different colors. In how many different ways may he choose his 4 horses? 16. How many of the possible combinations of 3 letters in the first 10 will contain the letter c? How many will con- tain both the letters c and e?? 17. Of the possible combinations of s things in n, how many will contain a designated thing? How many 2 desig- nated things? How many h designated things? 18. A party of 6 meet for whist, 2 waiting while the other 4 play. Each 4 must play one game with each possible ar- rangement of partners. How many games will be played in all; how many will each person play, and how many times will any two designated persons have met as partners? 19. From a collection of 5 letters and 6 numbers how many combinations, each consisting of 1 letter and 2 num- bers, can be formed? How many consisting of 2 letters and 3 numbers? Of 5 letters and 4 numbers? 20. From a collection of m letters and n numbers how many combinations of r letters with s numbers can be formed? 21. In how many ways may a pile of 20 balls be divided into two piles, the one having 15 balls and the other 5 ? 22. How many different signals may be made with 4 flags of different colors, it being assumed that each different order of each combination forms a different signal, but that the signal remains the same when the order is reversed? 23. What would be the answer to the preceding problem if each combination of several flags could be arranged either horizontally or vertically, and an inversion of each vertical arrangement made a different sig,nal? 634 PERMUTATIONS AND COMBINATIONS. 24. How many different signals can be made with 10 flags, of which 2 are white, 3 red, and 5 blue, all hoisted together in a vertical row? 25. How many different arrangements can be made of a base-ball " nine," supposing that only one man can pitch, and only two can catch ? 26. Supposing that, in a game of chess, the first player always has a choice of two good moves and the second player of three, how many games of 20 moves each are possible? 27. If the 8 pieces at chess could be arranged in any order on the 8 squares of the first rank, how many different arrange- ments would be possible? 28. In how many different ways can 4 pawns be arranged upon the 64 squares of a chess-board? How many different arrangements can be made with a king, queen, knight, and rook? Explain the relation of the two answers. 29. In how many ways may 12 balls be divided into three piles, containing, the one 3 balls, the second 4, and the thirds? 30. In how many ways may n balls be divided into 3 piles, containing, the one p, the second q, and the third r balls {p+q + r^^n)? 31. What must be the value of r in order thai (1'"' /7W ? 32. The ratio of the number of combinations of 2n things in 4:71 to that of the combinations of 71 things in 2/1 is {271 + 1){271 + S) , . . . {4:71 ~ 3) {4:71 - 1) 1.3.5 . . . . (2/^-l) 33. Show that the sum of the 7i\ different numbers that can be formed by permuting any 71 different digits is divis- ible by {71 — 1) times the sum of the digits, and that the quotient is 111 .... 34. If we define a magic square as an arrange- 6 18 ment of ?i' numbers in a square such that the sum of every line and every column is equal to the same '2' 5 3 quantity; show that if one such arrangement is pos- 004 sible with given numbers, then {niy are possible. SERIES. 635 See margin for example of square when ti = 3, and note that we leave out of consideration the diagonal lines of numbers. 35. Given m different letters and n different numbers, find the number of different permutations each containing r letters and s numbers. 2t(i, Given n unequal straight lines; how many non-identi- cal rectangular parallelopipeds may be formed, each of whose edges must be equal to some one of these lines in the two cases ; (1) When the same line cannot be repeated in a figure and (2) When it can be repeated without restriction. 37. The same thing being supposed and case (1) taken; how many different parallelopipedons may be built upon the same horizontal plane as a base, with their vertical faces toward the four points of the compass ; two figures being regarded as different when they cannot be brought into coin- cidence without turning them around or over. 38. Given n—1 sets containing respectively 2^^, 3« . . . . na different things; show that the number of combinations com- prising a of the first set, 2a of the second, etc., is ^ ^>1' . Series. Indeteeminate Coefficients. Develop the following expressions in powers of x by the method of indeterminate coefficients: I. 1 + nx 1-x' 2. 1 + x 1 — nx' 3- 1 +mx 1 +nx' 4. x + a C — X 5- a{a + x) a' + x'' 6. a' + x' a-\-x' •7. x' + a' x" + a'* 8. (i^x){i^hxy 9. X 10. x^ (1 ^ ex) (1 - CJ l + ic*" 1 12. 1 a" -\- ax-\' x^* 0^ — ax-^ x^' 536 SERIES. Products of Series. Form the products: 1. (1 - a; + x'- x'+ . . . .){l + x+x'+x'+ . . . .)• 2. (l+2^+3^'^+4a;^+ . . . .) {l-^x+^x'-^x^+ ....). , (,j y\y' y\ \{^j^y\y\y' jl. \ 4. {i+ax+ av + — ) (i+|+|; +|;+ . . . .). 5.(i-..+av-....)(i-i+i;-:-;+....). 6. (1 + 2a; + 3a:'' + 4a;' + . . . .)\ 7. (1 - 2a; + 3a;'' - 4a:' + . . . ,)\ Carry the products as far as x^ and express the 7i^^ term of the product in terms of n in each case for which you can form it. FiGURATE Numbers. 1. Enumerate an incomplete pile of cylindrical shot (§ 288) haying n shot in its bottom row, and as many in its top row as there are rows. Show that in this problem the number n must be odd. 2. The top and bottom rows of an incomplete pile of cylin- drical shot, having 8 rows in all, contain 9 shot less than one third the pile. How many shot are in the pile? 3. In an incomplete pile of 63 cylindrical shot 35 are in the interior of the pile, so as to be completely surrounded by others, and 28 form the top, bottom and sides. Describe the pile, and show that two piles may be formed which fulfil the conditions. 4. In a triangular pyramid of balls the ratio of the whole number of balls to the number in the bottom layer is 14 : 3. How many balls form the pile? 5. In a triangular pyramid having n balls on each edge, how many balls form the four faces? 6. If 20 balls in a triangular pyramid are completely sur- rounded by others, how many form the entire pyramid? SERIES. 537 ^ 7. A rectangular pile has 15 balls in its top row and its lesser side has 10 balls. Enumerate the balls in the pile. 8. If one side of the base contains m balls and the other n (m > n), how many balls will the pile contain; how many layers, and how many balls in the top row? 9. If 495 balls form a complete rectangular pile, having 10 balls on one side of the base, how many will the other side comprise? 10. How many balls in a square pyramid having 12 balls on each side of the base? 11. A rectangular pile has 84 shot in its bottom layer and 66 in the next layer. How many in the whole pile? Prove: 12. 1.2 + 2.3 + 3.4+ .... +7i{n + l) _ n{n + l){n + 2) ~ 3 • 13. 17^ + 2(72,-1) + 3 (^-2)+ , . , . + n[n-{7i-l)] __ n{n + l) {n + 2) " 3! 14. 1.2 + 2.4 + 3.6+ .... +n.2n _ n{n + l) (27 1 + 1) "" 3 15. 1 (2 - ^) + 2 (4 - w) + 3 (6 - 7^) + . . . . + n' _ n{n + l) {n + 2) ~ 3! 16. If we multiply the corresponding terms of the two progressions: a, a + Ji, a + 2hy , . , , a + ih, i, 1) — h, h — 2h, , , , , h — ill, the s.um of the products will be (^•+1) \ai> +'^^^y -i^i±^\ . 17. Find the sum of the products when, in the second series, the C. D. is + li instead ot — h. 538 SERIES. Express the values of i8. ai + ia^ h) {h + k) + (a + 2A) {h + 2k) -\- . . . . to n terms. 19. 1.3 + 3.5 + 5.7+ .... +^(^ + 2). 20. 1^ + 3 (a — 3) + 5 («5 — 6) + . . . .to n terms. 21. l.tt + 3 (a + 3) + 5 (nj + 6) + . . . ,ton terms. 22. Prove the equations: 4 5 6 7 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 = 1^1.1 by subtracting from the second member the successive terms of the first member, beginning with the last. 23. Generalize the preceding result by proving in the same way the general equation: _ (n + 1\ Note that the first operation will be" to deduce By means of the preceding formulae write, on sight, the values of: 24. 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + 4.5.6.7 1.2.3 2.3.4 3.4.5 4.5.6 5.6.7 ^^* 1. 2. 3 "*" 1.2. 3 + 1.2. 3 "^1.2. 3 "^1.2. 3* 26. 1.2. 3. 4 + 2. 3. 4.5 + . . . . + ?^(^ + 1) (^+2) (w+3). 27. Show that the sum of the products of the first n natu- ral numbers taken by 2's is ^^ ^- — ^ 7^ -^ ^ — \ 28. In the following scheme we start with a column of ar's on the left, and with the top line a, ft, y, d, etc. Then, each number following, in each column, is formed by adding the number above it to the number on the left of the latter. It 8ERIE8. 539 is now required to write the general expression for the ni\i number in the 2d, 3d, 4th, and iih. columns. 1 2 3 4 a P 7 d a P^- a r^ V ^ ^^ h r a /5 + 2a yj -215- - a 6 - -2y- - ^ a y5 — 3a: yj -3/i- -3a 6- -3r- -3/i+ a a y^ + 4a r - h4/^H h6a 8 + ir- \- 6/i + 4a 29. A trader starts with a capital of a dollars; he gains, and adds to his capital, h dollars the first year, and c dollars more each year than he did the year before. Express his accumulated capital at the end of n years in terms of n. SuMMATioiT OF Series. Sum to infinity: 1. 1 + 7^ + (1 + 2?z) a: + (1 + Zn) x^-\- (1 + ^n) 2:'+ .... 2. 1 + 3a; + ^x^ + 10^'+ . 3- 4. 5- 6. 7. 8. 1 + 4a; + ^x^ + 16a;' + . +n'x'"'^-i- . . . . 1.3 1 2.4 3.5 1 2.5 + 5.8"'"8.11 + 2.4^4.6^6.8^ 2.3.4 1 1.3.5 1 + + + 3.4.5 1 2.4.6 1 + + + 4.5.6 1 3.5.7 1 + ^ 1.4.7 ' 2.5.8 ' 3.6.9 10. l-i^{a+l+a-^) r + {a'+ a + 1 + a-^-{-a-y''+ 11. {71 + lyx +{n + 2)V + {71 + 3)V + . . . . 12. - + -i + -:^ + -4 + . . . . 540 LIMITS. Sum to n terms: 13. a'J^{a + iy + {a + %Y+ 14. 3 + 5 + 9 + 14 + . . . . + M!i+i). z 15. 3 + 8 + 15 + 24 + . ,,,-]- 71 {n + 2). 16. 1 + /^ + 2 (2 + y?;) + 3 (3 + ^) + . . . , J[-n{n + r 16a, Show that the series: 2^3 4^5 may be transformed into eitiier of the three forms: Jl_ 4. _1 ■ _1 . 1.2^3.4^5.6^ * * • • . _ J ^ ___^ _ ^^ 2.3 4.5 6.7 • • • • .. 1 1 , 111 ^^ 2 "^1.2.3"^ 3.4.5 "^5.6.7"^7.8.9"^**** 17. How do two of the preceding results enable us to sum r~2 "^ 2~3 "^ 3~4 + * • * * ^^ wfinitum? 18. What number is equal to the continued product: 2^. 4i.8T's. 16^^.32^5 .... ad infinitum? 19. To what limit approaches the indefinitely continued product: 1 ?_ ?_ i_ a^.a'^^a^^a»** . . . . ? Limits. Find the limits of {x - by x" as X increases indefinitely. 2. ax ax 3- -.- LIMITS. 541 4. '^ — ~ as X approaches a indefinitely. X — a X a x' -a' x' -a' a X X a 12. X — a (x + a)'' — (x — ay . • -, ^ .^ 1 7. -^^ n^ ^s X increases indefinitely. (1 + axY • (1 + bxy (1 - axr ,, ,, ,, ^* (1 - bxy 10. ^ as n increases indefinitely. „ r + r + 3'+.... + n' ,,,, ,. in^ 4- 2^ + 3"^ + 4^ + . . . . + ^^ ^^ 13. The first term of a series is — , the second — tt, and o 6 each succeeding term one half the sum of the two which pre- cede it. To what limits will the nth term and the sum of the series approach as n increases indefinitely? 14. Find the limit toward which the nth term approaches when First term = a-{-2b; second term = a — b; cc u ^ 1. .. rod- uct reduces identically to unity. 5. Show by what relations the development of e^^ becomes identical with the square of that of e^. Logarithms and Logarithmic Series. I. Express the logarithm of the continued product of all tlio terms of a geometrical progression. Calling h the arbitrary base of the system of logarithms, ve the following equations so as to express x in terms of y: 2: 4. 6. 8. 10. log X = y, 3. log X = ay, log 2x = y, 5. log 7}ix — a -\- y, log ax = my, 7. log x^ = y, logx"" := my. g, y — V''«'^. Eec iuce to their simplest form the expressions: 12. Prove the identities: ^log aj^log y ? £ 14. m^^^ = ^yloemx - _^^_ "^ _. ^losmy^ogn. / LOGARITHMS AND LOGARITIIMIG SERIES. 545 16. If a, h, and c be the mth, ^th and ^tli terms of a geo- metrical progression show that {P — q) log a-\-{q— m) log h + (m — p) log c = 0. 17. Prove that the value of the expression is independent of n and equal to log a, 18. Prove the equation: 2 log X — log {x -\- a) — log {x — a) _^.^j a' , 1 ^^ ,1 «" , ) -'^^]2a;^ _ ^^+3 ' (2x'-a'y~^ 6 (2x'-ay^ ' ' ' ' ]' 19. If a, J and c are three consecutive numbers show that 21og*-log«-log.= 3Jf]^-+3^J-p^, + ....[. 20. Prove: Nap. log4 = 1 + j-|^ + ^+^+ .... 21. If a, I, Cy d, etc.y are in geometrical progression, then, in order the equations 1111^ a-^ =M = cp = d^= . . . . may be satisfied, the quantities m, n, p, q, etc., must be in arithmetical progression. 1 1 22. If 2/ = IQi-iog^, and z^K^^-'^^sv^ show that 23. Prove the development and by making the development in another form and com- paring the coefficients of x^ prove the identity 2"-g_o.-. «-3g„_, I (w-4)( «-5) the series terminating with tlie last exponent which is not negative. HINTS ON A COURSE IN ADVANCED ALGEBRA. For the benefit of students wlio may contemplate a course of reading in the various branches of Advanced Algebra, the following list of sub- jects and books has been prepared. As a general rule, the most extended and thorough treatises are in the German Language, while the French works are noted for elegance and simplicity in treatment. To pursue any of these subjects to advantage, the student should be familiar with the Differential Calculus. I. THE GENERAL THEORY OF EQUATIONS.— In English, Tod- hunter's is the work most read. Serret, Algehre Superieurey 2 vols., 8vo, is the standard French work, covering all the collateral subjects. oToT?T»A^^ Theorie des Substitutions et des Equations Algefyriques, 1 vol., 4to, s the largest and most exhaustive treatise, but is too abstruse for my but experts. _-. -.i^PERMINANTS.— Baltzer, Theorie der Beterminanten, is the standard treatise. There is a French but no English translation. A recent English work is Robert F. Scott, The Theory of Deter- minants and their Applications in Analysis and Geometry. III. THE MODERN HIGHER ALGEBRA, resting on the theories of Invariants and Covariants. Salmon, Lessons introductory to the Modern Higher Algebra^ is the standard English work, and is especially adapted for instruction. CLEB8CH, Theorie der lindren Algebraischen Formen^ is more exhaustive in its special branch and requires more familiarity with advanced systems of notation. IV. THE THEOKi. OF NUMBERS. There is no recent treatise in English. Gauss, Disquisitiones Arithmeticm, and Legendre, Theorie des Norribres, are the old standards, but the latter is rare and costly. Lejeune Dirichlet, Vorlesungen uber Zahlentheorie, is a good German Work. There is also a chapter on the subject in Serret, Algehre Super ieure. V. SERIES. — This subject belongs for the most part to the Calculus, but Catalan, Traite eUmentaire des Series, is a very convenient little French work on those Series which can be treated by Elementary Algebra. VI. Q JATERNIONS.— Tait, Elementary Treatise on Quaternions^ is prepared especially for students, and contains mfiny exercises. The original works of Hamilton, Lectures on Quaternions and Elements of Quaternions, are more extended, and the latter will be found valuable for both reading and reference. UNIVERSITY OB CALIFORNIA LIBRARY DUE on the last date stamped below. OCT 21 1947 0cll4'48J§ LD 21-lOOw .l2,'46(A2012sl6)4120 Wl^t?5'? ' Xf^* » 4 UNIVERSITY OF CAUFORNIA UBRARY lt-M.^1^