-'*! Irving Stringham /O Digitized by tine Internet Arciiive in 2008 witii funding from IVIicrosoft Corporation littp://www.arcliive.org/details/collegealgebraOOwentricli WENTWORTH'S SERIES OF MATHEMATICS First Steps in Number. Primary Arithmetic. Grammar School Arithmetic. High School Arithmetic. Exercises in Arithmetic. Shorter Course in Algebra. Elements of Algebra. Complete Algebra. College Algebra. Exercises in Algebra. Plane Geometry. Plane and Solid Geometry. Exercises in Geometry. PI. and Sol. Geometry and PI. Trigonometry. Plane Trigonometry and Tables. Plane and Spherical Trigonometry. Surveying. PI. and Sph. Trigonometry, Surveying, and Tables. Trigonometry, Surveying, and Navigation. Trigonometry Formulas. Logarithmic and Trigonometric Tables (Seven). Log. and Trig. Tables {Complete Edition). Analytic Geometry. Special Terms and GircTilar on Application. COLLEGE ALGEBRA BY G. A. WENTWORTH, PROrBSSOK OF MATHEMATICS IN PHILLIPS EXETKR ACADEMY. DjOrJOO BOSTON, U.S.A.: PUBLISHED BY GINN & COMPANY. 1892. / Entered according to Act of Congress, in the year 1888, by G. A. WENTWORTH, in the Office of the Librarian of Congress, at Washington. Ttpoqrapht by J. S. CusHiNG & Co., Boston. Prbsswork by Ginn & Co., Boston. — i-J PREFACE. rpHIS work, as the name implies, is intended for Colleges and Scientific Schools. The first part is simply a review of the principles of Algebra preceding Quadratic Equations, with just enough examples to illustrate and enforce these principles. By this brief treatment of the first chapters, sufficient space is allowed, with- out making the book cumbersome, for a full discussion of Quadratic Equations, The Binomial Theorem, Choice, Chance, Series, Deter- minants, and The General Properties of Equations. Every effort has been made to present in the clearest light each subject discussed, and to give in matter and methods the best training in algebraic analysis at present attainable. The work is designed for a full-year course. Sections and problems marked with a star can be omitted, if necessary ; and for a half-year course many chapters must be omitted. The author gratefully acknowledges his obligation to Mr. G. W. Sawin of Harvard College, who has contributed the excellent chapter on Determinants, and been of invaluable assistance in revising every chapter of the book. Answers to the problems are bound separately in paper covers, and will be furnished free to pupils when teachers apply to the pub- lishers for them. Any corrections or suggestions relating to the work will be thank- fully received. G. A. WENTWORTH. Phillips Exeter Academy, September, 1888. 800565 TABLE OF CONTENTS. -♦- CHAPTER I. SECTION. PAGE. 1-18. Fundamental Ideas 1-8 CHAPTER II. 19-47. Fundamental Operations 9-26 CHAPTER III. 48-65. Factors 27-40 CHAPTER IV. Ir 66-76. Fractions 41-47 CHAPTER V. 77-90. Simple Equations 48-56 CHAPTER VI. 91-96. Simultaneous Equations of the First Degree 57-65 CHAPTER VII. 97-112. Involution and Evolution 66-76 CHAPTER VIII. 113-130. Exponents 77-86 CHAPTER IX. 131-143. Quadratic Equations 87-111 CONTENTS. V CHAPTER X. SECTION. PAGE. 144-146. Simultaneous Quadeatic Equations . . . 112-124 CHAPTER XI. 147-150. Equations solved like Quadratics . . . 125-131 CHAPTER XII. 151-160. Properties of Quadratic Equations . . . 132-142 CHAPTER XIII. 161-180. Surds and Imaginaries 143-152 CHAPTER XIV. 181-184. Inequalities 153-154 CHAPTER XV. 185r-215. Ratio and Proportion 155-175 CHAPTER XVI. 216-235. Progressions * 176-192 CHAPTER XVII. 236-238. Simple Indeterminate Equations .... 193-198 CHAPTER XVIII. 239-260. Binomial Theorem 199-214 CHAPTER XIX. 261-285. Logarithms 215-232 CHAPTER XX. 286-296. Interest and Annuities 233-242 CHAPTER XXI. 297-314. Choice 243-266 VI CONTENTS. CHAPTER XXII. SECTION. PAGB. 315-331. Chance 267-288 CHAPTER XXIII. 332-343. Continued Feactions 289-299 CHAPTER XXIV. 344^347. Scales of Notation 300-305 CHAPTER XXV. 348-355. Theory of Numbees 306-312 CHAPTER XXVI. 356-371. Variables and Limits 313-321 CHAPTER XXVII. 372-397. Seeies 322-358 CHAPTER XXVIII. 398-426. Determinants 359-383 CHAPTER XXIX. 427-482. General Properties of Equations . . . 384-435 CHAPTER XXX. 483-503. Numerical Equations 436-462 CHAPTER XXXI. 504-517. Geneeal Solution of Equations .... 463-479 CHAPTER XXXII. 518-535. Complex Numbees 480-494 COLLEGE ALGEBRA. CHAPTER I. FUNDAMENTAL IDEAS. 1. Quantity and Number. Whatever may be regarded as being made up of parts like the whole is called a quantity. In other words, whatever admits of division into parts all the same in hind as the whole is a quantity. To measure a quantity of any kind is to find how many times it contains another known quantity of the same kind. A known quantity which is adopted as a standard for measuring quantities of the same kind is called a unit. Thus, the foot, the pound, the dollar, the day, are units for meas- uring distance, weight, money, time. A number arises from the repetitions of the unit of meas- ure, and shows how many times the unit is contained in the quantity measured. 2. When a quantity is measured, the result obtained is expressed by prefixing to the name of the unit the number which shows how many times the unit is contained in the quantity measured. This result is called the measure of the quantity. The number which shows how many times the unit is taken is the numerical part of the measure. ALGEBRA. Thus, 7 feet, 8 pounds, 9 dollars, 14 days, are respectively meas- ures of a distance, a weight, an amount of money, and an interval of time ; the numerical parts being respectively the numbers 7, 8, 9, and 14. 3. For convenience, numbers are represented by symbols. In Arithmetic the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and combinations of these symbols, are employed to represent numbers. The series 0, 1, 2, 3, , obtained by counting, is called the natural series of numbers. Any figure or combination of figures represents one, and but one, particular number. 4. Numbers in General. Numbers possess many general properties, which are true, not only of a particular number, but of all numbers. For example, the sum of 12 and 8 is 20, and the differ- ence between 12 and 8 is 4. Their sum added to their difference is 24, which is twice the greater number. Their difference taken from their sum is 16, which is twice the smaller number. We shall see later on that these are general properties of numbers, namely : The sum of two numbers added to their diffefrence is twice the greater number ; the difference of two numbers taken from their sum is twice the smaller number. Or, (1) (greater number + smaller number) + (greater num- ber — smaller number) = twice greater number. (2) (greater number -f- smaller number) — (greater num- ber — smaller number) = twice smaller number. But these statements may be very much shortened ; for, as greater number and smaller number may mean any two numbers, two letters, as a and 5, may be used to represent FUNDAMENTAL IDEAS. them; then 2a will represent twice the greater number, and 2 h twice the smaller. Then these statements become : (1) {a-\-b) + {a-h) = 2a. (2) la + h)-{a-h) = 2h. In studying the general properties of numbers, letters used to represent numbers may represent any numerical values consistent with the conditions of the problem. 5. Algebra like Arithmetic is a science which treats of numbers. In any problem in which we are concerned with quantities, we use not the quantities themselves, but the numbers by which they are expressed. In Algebra as in Arithmetic we use the Arabic numerals to represent particular numbers. But in Algebra we also use other symbols, generally the letters of the alphabet, to represent numbers. Algebra is, then, a species oi generalized Arithmetic, and includes the ordinary Arithmetic. 6. Operations to be performed upon numbers are indicated in Algebra, as in Arithmetic, by signs. The chief signs of operation used in Arithmetic are the following : + (read, plus), the sign of addition. — (read, minus), the sign of subtraction. X (read, multiplied by), the sign of multiplication. -^- (read, divided by), the sign of division. 7. Positive and Negative Numbers. There are quantities which stand to each other in such opposite relations that, when we combine them, they cancel each other entirely or in part. Thus, six dollars gain and six dollars loss just cancel each other ; but ten dollars gain and six dollars loss cancel each other only in part. For the six dollars loss will cancel six dollars of the gain and leave four dollars gain. ALGEBRA. An opposition of this kind exists in assets and debts, in motion forwards and motion backwards, in motion to the right and motion to the left, in the degrees above and the degrees below zero on a thermometer. From this relation of quantities a question often arises which is not considered in Arithmetic ; namely, the sub- tracting of a greater number from a smaller. This cannot be done in Arithmetic, for the real nature of subtraction consists in counting backwards, along the natural series of numbers. If we wish to substract four from six, we start at six in the natural series, count four units backwards, and arrive at two, the difference sought. If we subtract six from six, we start at six in the natural series, count six units backwards, and arrive at zero. If we try to subtract nine from six, we cannot do it, because, when we have counted backwards as far as zero, the natural series of numbers comes to an end. 8. In order to subtract a greater number from a smaller, it is necessary to assume a new series of numbers, beginning at zero and extending to the left of zero. The series to the left of zero must ascend from zero by the repetitions of the unit, precisely like the natural series to the right of zero ; and the opposition between the right-hand series and the left-hand series must be clearly marked. This opposition is indicated by calling every number in the right-hand series a positive number, and prefixing to it, when written, the sign -f ; and by calling every number in the left-hand series a negative number, and prefixing to it the sign — •. The two series of numbers will be written thus : -4, -3, -2, -1, 0, +1. +2, +3, +4, I I t I I I I 1 I If, now, we wish to subtract 9 from 6, we begin at 6 in the positive series, count nine units in the negative direction FUNDAMENTAL IDEAS. (to the left), and arrive at — 3 in the negative series. That is, 6-9 = -3. The result obtained by subtracting a greater number from a less, when both are positive, is always a negative numher. If a and h represent any two numbers of the positive series, the expression a — h will denote a positive number when a is greater than h ; will be equal to zero when a is equal to b ; will denote a negative number when a is less than h. If we wish to add 9 to — 6, we begin at — 6, in the negative series, count nine units in the positive direction (to the right), and arrive at + 3, in the positive series. "We may illustrate the use of positive and negative numbers as follows : -5 8 20 It— 1 Q Suppose a person starting at A walks 20 feet to the right of u4, and then returns 12 feet, where will he be ? Answer : at C, a point 8 feet to the right of A. That is, 20 feet - 12 feet = 8 feet; or, 20 - 12 = 8. Again, suppose he walks from A to the right 20 feet, and then returns 25 feet, where will he now be ? Answer : at D, a point 5 feet to the left of A. That is, if we consider distance measured in feet to the left of A as forming a negative series of numbers, begin- ning at A, 20 — 25 =« — 5.' Hence, the phrase, 5 feet to the left of -4, is now expressed by the negative number — 5. 9. Numbers with the sign + or — are called algebraic numbers. They are unknown in Arithmetic, but play a very important part in Algebra. Numbers not affected by the signs + or — are called absolute numbers. Every algebraic number, as +4 or —4, consists of a sign + or — and the absolute value of the number; in this case 4. The sign shows whether the number belongs ALGEBRA. to the positive or negative series of numbers ; the absolute value shows what place the number has in the positive or negative series. When no sign stands before a number, the sign + is always understood; thus, 4 means the same as +4, a means the same as + a. But the sign — is never omitted. Two numbers which have, one the sign + and the other the sign — , are said to have unlike signs. Two numbers which have the same absolute values, but unlike signs, always cancel each other when combined; thus, +4-4 = 0, +a — a = 0. 10. Meaning of the Signs. The use of the signs -f and — , to indicate addition and subtraction, must be carefully dis- tinguished from their use to indicate in which series, the positive or the negative, a given number belongs. In the first sense, they are signs of operations, and are common to both Arithmetic and Algebra. In the second sense, they are signs of opposition, and are employed in Algebra alone. 11. Factors. "When a number consists of the product of two or more numbers, each of these numbers is called a factor of the product. When these numbers are denoted by letters, the sign X is often omitted; thus, instead of axh, we write ah \ instead of a X 5 X c, we write ahc. Factors expressed by letters are called literal factors ; factors expressed by figures are called numerical factors. 12. A known factor of a product which is prefixed to another factor to show how many times that factor is taken is called a coefficient. 13. Powers. A product consisting of two or more equal factors is called a power of that factor. FUNDAMENTAL IDEAS. The index or exponent of a power is a small figure or letter placed at the right of a number, to show how many times the number is taken as a factor. Thus, a? is written instead of aaa. a" is written instead of aaa io n factors. The second power of a number is generally called the square of that number ; the third power of a number, the cube of that number. 14. Signs. The principal signs used in Algebra in addi- tion to those of § 6 are the following : The signs of relation: =, >, <, which stand for is equal to, is greater than, and is less than, respectively. The signs of aggregation : the bar, | ; the vinculum, ; the parenthesis, ( ) ; the bracket, [ ] ; and the brace, \ ] . Thus, each of the expressions, ^ ^ x + y, {x + y), [x + y], {x + y], signifies that x + y is to be treated as a single number. The signs of continuation: dots, , or dashes, , read, and so on. The sign of deduction : .•., read, hence, or therefore. Remakk. AVhen a sign of operation is omitted between numerals it is the sign of addition; when between letters, or a numeral and a letter, it is the sign of multiplication. Thus, 423 means 400 + 20 + 3, but 2 abc means 2 X a X & X c 15. An algebraic expression is a number written with alge- braic symbols ; an algebraic expression consists of one sym- bol, or of several symbols connected by signs of operation. A term is an algebraic expression the parts of which are not separated by the sign of addition or subtraction. Thus, 3a6, bxi/, Sab -r- ^xy are terms. A monomial or simple expression is an expression with but one term. A polynomial or compound expression is an expression of 8 ALGEBRA. two or more terms. A binomial is a polynomial of two terms ; a trinomial, a polynomial of three terms. Like terms or similar terms are terms whicli have the same letters, and the corresponding letters affected by the same exponents. Thus, 1 o?cx^ and — ba^cx^ are like terms. 16. The degree of a term is the sum of the exponents of its literal factors. Thus, 2>xy is of the second degree, and bx^yz^ of the sixth degree. A polynomial is said to be homogeneous when all its terms are of the same degree. Thus, 1 x? — bx^y -^ xyz is homogeneous of the third degree. A polynomial is said to be arranged according to the powers of some letter when the exponents of that letter either descend or ascend in order of magnitude. 17. The value of an algebraic expression is the number which the expression represents. If the number represented by each symbol involved in an expression is known, the value of the expression can be found by putting for each symbol the number it represents and performing the indicated operations. The value of an expression evidently depends upon the values given to the several symbols involved. 18. Axioms, 1. Things which are equal to the same thing are equal to each other. 2. If equal numbers be added to equal numbers, the sums will be equal numbers. 3. If equal numbers be subtracted from equal numbers, the remainders will be equal numbers. 4. If equal numbers be multiplied into equal numbers, the products will be equal numbers. 5. If equal numbers be divided by equal numbers, the quotients will be equal numbers. CHAPTER II. FUNDAMENTAL OPERATIONS. — ADDITION. 19. An algebraic number which is to be added or sub- tracted is often inclosed in a parenthesis, in order that the signs -f and — which are used to distinguish positive and negative numbers may not be confounded with the -f and — signs that denote the operations of addition and subtrac- tion. Thus, -f 4 + (— 3) expresses the sum, and +4— (—3) expresses the difference, of the numbers -f- 4 and — 3. 20. Monomials. In order to add two algebraic numbers, we begin at the place in the series which the first number occupies, and count, in the direction indicated hy the sign of the second number, as many units as there are units in the absolute value of the second number. Thus, the sum of -f 4 + (+ 3) is found by counting from + 4 three units in the positive direction, and is, therefore, -f 7 ; the sum of -f 4 -f (— 3) is found by counting from + 4 three units in the negative direction, and is, therefore, + 1- In like manner, the sum of — 4 + (+ 3) is — 1, and the sum of — 4 + (- 3) is - 7. I. Therefore, to add two numbers with like signs, find the sum of their absolute values, and prefix the common sign to the sum. II. To add two numbers with unlike signs, find the dif- ference of their absolute values, and prefix the sign of the number absolutely greater to the difference. Thus, (1) +a + (+5) = a + ^,; (3) -« + (+ J) = _a + 5; (2) +a-\-(i-h) = a~h', (4) -a-\-{-h) = -a-h. 10 ALGEBRA. 21. It should be noticed that the order of the terms is iminaterial. Thus, +« — £== — h -\-a. This law is called the commutative law for addition. 22. By successive application of the above rules we readily obtain rules for adding any number of terms. Thus, • 4a + 5a + 3a + 2a=14a; -3a- 15a-7a + 14a-2a = 14a- 27a = -13a; 4a-36-9a + 76 = -5a + 46. 23. Polynomials. Two or more polynomials are added by adding their separate terms. It is convenient to arrange the terms in columns, so that like terms shall stand in the same column. Thus, Exercise 1. Add: 1. 9a2 + 3a + 45, 2a^-4a + 55, ba-2b-Qa\ 2. Ix'-^xy-^-y', ^xy-2y\ ^x' ~^xy -\-l2y\ 3. la'b-\-^ah^-lU\ Sa' + 2ab'-7b\ ab'-a'b~6a\ bb'-7a'-ab\ Ab' -2a' + a'b. 4. 5x* + 2x'-7, 4:x' + x-9, l-{-x-x\ x' + x'-x'-x'-7, 9x' + 9x'-12x-4:x' + 10. 5. 3m* + 2m'n + 5mV-9< 7 n' — 3 mn' - 8 m'n\ 11 mn' — 4 mV + 6 m\ 5 m* + 2 m'n — 15 mn' — 7 n\ 6. 2x^-{-Sx'y-4:xy, 2y^ - Sxy' -}- ia^y' ~10xy, bxY + AxY - 9y\ 8a;V - Ix^ + 6^y - Sx'y\ FUNDAMENTAL OPERATIONS. 11 SUBTRACTION. 24. Monomials. In order to find the difference between two algebraic numbers, we begin at the place in the series which the minuend occupies, and count in the direction opposite to that indicated hy the sign of the subtrahend as many units as there are units in the absolute value of the subtrahend. Thus, the difference between + 4 and + 3 is found by counting from +4 three units in the negative direction, and is, therefore, -f 1 ; the difference between + 4 and — 3 is found by counting from -f~ 4 three units in the posi- tive direction, and is, therefore, + 7. In like manner, the difference between — 4 and -f 3 is — 7 ; the difference between — 4 and — 3 is — 1. - . Compare these results with results obtained in addition ; it is evident that : Subtracting a positive number is equivalent to adding an equal negative number. Subtracting a negative number is equivalent to adding an equal positive number. To subtract, therefore, one algebraic number from another, change the sign of the subtrahend, and then add it to the minuend. Thus, -\-a — {+b) = a-b\ - a- (-{-b) = -a~b; -\-a-(-b)=^a + b; - a- (■- b) = - a + b. 25. Polynomials. When one polynomial is to be sub- tracted from another, place its terms under the like terms of the other, change the signs of the subtrahend, and add. From 4 a;' — 3 x'^y — xy^ + 23/^ take 2ar''- x^y ■\-hxy' -Zy" 12 ALGEBRA. Change the signs of the subtrahend and add : 4^x^-~5ar^y — xy^->r2y^ -2x^+ x'y — bxy'-^-^y^ 2x^-2x''y-'Qxy''-\-bf Instead of actually changing the signs of the subtrahend we need only conceive them to be changed. 26. Parentheses. From (§ 24), it appears that (1) a + (+5) = a + 5. (3) a-{^h) = a-h. (2) a + (- 6) = a - b. (4) a - (- ^») = a + h. The same laws respecting the removal of parenthesis hold true whether one or more terms are inclosed. Hence, when an expression within a parenthesis is preceded by a plus sign, the parenthesis may be removed. When an expression within a parenthesis is preceded by a minus sign, the parenthesis may be removed if the sign of every term within the parenthesis is changed. Thus, (1) a-^{h-c) = a + h-c. (2) a-{h-c)=-a-b-{-c. 27. Observe that the terms may be combined in any manner. Thus, a-\-b — c — d= {a-\-b) - {c -\- d) = (a-\-b — c) —d — a-\-{b — c—d). This is called the associative law for addition and sub- traction. 28. Expressions often occur with more than one paren- thesis. These parentheses may be removed in succession, by removing first, the innermost parenthesis ; next, the in- nermost of all that remain, and so on. FUNDAMENTAL OPERATIONS. 13 Thus, a-[h-\c + {d-e-f)\] ^a-[h~\c-\-{d-e-\-f)\] = a-[h~\c-\-d~e+f\] = a — \h — c — d-[-e — /] = a — h-\-c-\-d—e +/. 29. The rules for introducing parentheses follow directly from the rules for removing them : 1. Any number of terms of an expression may be put within a parenthesis, and the sign + placed before the whole. 2. Any number of terms of an expression may be put within a parenthesis, and the sign — placed before the whole ; provided the sign of every term within the paren- thesis he changed. Exercise 2. 1. From 4a+ 55 -3c take 2a + 95 -8c. 2. From 7a;^ — ar^ + 4a;-2 take 2a:^ + 8a;^ - 9a: + 8. 3. ^vom^a" -\-2>a^h-^ah''^W take 2a^ - 5 a'5 + 7 a^^ - 9 b\ 4. From-Ja5 + 4a'-|5HJa take a' — yV^' + i«- 5. From 4a;' — ^x^ + 8a; — 7 take the sum of 8a:' + 7-8a;^ + 7a: and - 9a:' - 8a;^ + 4a; + 4. Simplify : 6. 2-3a;-(4-6a;)- J7-(9-2a;)5. 7. 2>a-{a-h-c)--2[a-\-c-2{h~c)\. 8. 4a-[3a- {2a-(a-5)] + 55]. 9. [^a-2>{a-(h-a)\]-^[a-2\a-2{a-h)] + b\ 10. x{y + z)-\-y[x~{y -^ z)] — z[y — x{z - x)]. 11. 2a;'(a;-3a)-2[2a:*-aX^'-«')] - 3a[a;' - 2a:{a'* -^x{a-x)\ + a']. 14 ALGEBRA. MULTIPLICATION. 30. Let a and b be any two members. To obtain the product of a hy b we do to a what we do to unity to obtain b. Thus, to obtain 5 we take 1 five times, and to obtain the product 5 X 3 we take 3 five times. Similarly, to obtain —5 we take 1 five times, and then change the sign of the product ; hence, to obtain the product (— 5) x (— 3) we take — 3 five times, giving — 15, and then change the sign, giving + 15. In general, aX b^-j-ab] (—a)xb = — ab; aX(—b) = — ab; (— a) X (~ b) = -j- ab. From the preceding we obtain the rule : like signs give plus ; unlike signs give minus. The product of more than two factors, each preceded by the sign — , will be positive or negative, according as the number of such factors is even or odd. 31. Monomials. The product of numerical factors is a new number in which no trace of the original factors is found. Thus, 4x9 = 36. But the product of literal factors is expressed by writing them one after the other. Thus, the product of a and b is expressed by ab ; the product of ab and cd is expressed by abed. The product is evidently the same in whatever order the factors be written. This is the commutative law for multi- plication. 32. Index Law. The product of two or more powers of any number is that number with an exponent equal to the sum of the exponents of the several factors. FUNDAMENTAL OPERATIONS. 15 For, a"* X a" — (aaa to m factors) (aaa to w factors) = aaaaaa to (m + n) factors Similarly for more than two factors. This law is called the index law. 33. The product of three or mCre factors is evidently the same in whatever way the factors be combined. Thus, ahcde = (abc) X (de) = {ah) X (cde), etc. This is the asso- ciative law for multiplication. 34. Polynomials by Monomials. If we have to multiply a -\- h \>Y n, that is, to take {a~\-h) n times, we have, (a + 5) X n = (a + S) + (a + 5) + (a + 5) n times, = a-\-a-\- a n times -\-h-\-h-\-b n times, = aX7i + 6X7i, '= an-\- bn. As it is immaterial in what order the factors are taken, n X {a -\- h) = an -\- bn. In like manner, {a -\- b -\- c) X n -— an -\- bn -\- en, or, n{a-{-b -{- c) = an -{■ bn-\- en. The above is called the distributive law for multiplication. 35. Polynomials by Polynomials. If we have a-\-h-\- c to be multiplied hj m-\-n -{-p, we find, (a + 5 + c){m + n -{-p) = (a + 5-f-c)m-f(a + 5 + c)n + (a + 6 + e)p = am + bm -^em-\- an + bn -\- en -\- ap -\- hp -\- cp. 16 ALGEBRA. In multiplying polynomials, it is a convenient arrange- ment to write the multiplier under the multiplicand, and place like terms of the partial products in columns. (1) Multiply 5a — 6^ by 3a -46. 5a - 6 & 3a - 4 & \bc? -18a6 -20a5 + 24 62 15 a2 -38a6 + 2462 (2) Multiply a^ ^H c^ ~ ah — ho- -ac by a + 6 + c. Arrange according descending powers of a. a? — ah-ac^ 62- le + c2 a + 6 + c a? - o?h - c?c + a52_ a6c + ac^ + a26 ai2_ a6c + 53 - &2c + 6c2 H-a^c - a5c - ac^ + i^c -Z)C2 + (? a? — 3 a6c + &^ + c^ Observe that, with a view to bringing like terms of the partial products in columns, the terms of the multiplicand and multiplier are arranged in the same order. 36. Detached Coefficients. In multiplying two polyno- mials which involve but one letter, or are homogeneous (§ 16) and involve but two letters, we shall save much labor if we write only the coefficients. Thus, (1) Multiply 2:r^ + 4^7 + 7 by a;' - 3a; + 4. Since the a* term in the first expression is missing, we supply a zero coefficient. The work is as follows : 2+0+ 4+ 7 1-3+ 4 2+0+ 4+ 7 _6- 0-12-21 + 8 + + 16 + 28 2-6 + 12- 5- 5 + 28 FUNDAMENTAL OPERATIONS. 17 Writing in the powers of x, the product is 2a;5 - 6 x* + 12x' - Src^ _ 5a; + 28. (2) Multiply a'' + 2ax'-9x' + 4a''x hj x'' — 2ax — a\ Arranging by powers of x we have - 9a;3 + 2ax'^ + ia'^x + a^ and x^ - 2ax - a^. The work is as follows : -9+ 2+4+1 1- 2-1 -9+ 2+4+1 + 18-4-8-2 +9-2-4-1 -9 + 20 + 9-9-6-1 Hence, the product is 9a^ + 20ax* + 9a^a^ - 9a-V - 6a% - a\ 37. Special Oases. The following products are of great importance, and should be carefully remembered. (a + hy = a' + 2ab + h'; (a-by = a'-2ab + b'; (a + bXa -b) = a'-b'; (a-{-b + cy=a'i-b'-i-c'-{-2ab + 2ac-{-2be. The square of any polynomial may be immediately written down by the following rule : Add together the squares of the several terms and twice the product of each term into each of the terms that follow it. Also : (a =b by = a' ±Sa'b + Sab' ± J'; (a ± by = a*± 4a'b + 6a'b' ± 4ab^ + b*; and so on. 18 ALGEBRxi. 38. Again consider the product (x -\- a)(x -\- h) — x^ -\- (a -T h) X -\- ah. The coefficient of x is the algebraic sum of a and b ; the third term is i\iQ product of a and b. Thus, {x + Z){x + 7) = x2 + 10 re + 21 ; (a; - 3)(a; + 7) = a;2 + 4a; -21: (a; + 3)(x-7) = x2- 4rr-21 (a;_3)(a;_7) = a;2_i0a; + 21. Exercise 3. Find the product of : 1. ?>x-\-2y Q.ndi^x — by. 2. 2a;'-5 and 4:r + 3. 3. 2rr'4-4a:-3and2a;' + 3a7 — 4. 4. a;* + 2:c' + 4and:r*-2;r' + 4. 5. rr'^ + 2 0:3/ — 3 y^ and r?;^ — bxy-^-^if. 6. 9a--' + 3ri;y + 3/' — 6a; + 2y + 4 and 3:r-y + 2. 7. lla' + 45^-4a5(a-45)andaX5 + 3a)-45X« + ^)- 8. (a + 5y + (a - by and (a + 5)^ - (a - ^>y. 9. ^ — 2^ + 32 and ^— 2y + 32. 10. X^-\-2x^-4:X-l2.IidiX^-\-2a^-4:X-l. 11. 39c^^+^-^-54c^^-2^+^ + 60c^^+^^ and 30^^^-^+^^ 12. 24^"*+'"-^ — 42x''"-'"+' + 25:r'"+^'"-2 and 25:r'-'"-'". 13. a^-3a^-' + 4a^-=^-6a^-3 + 5a^-''and2a'-a2-f«- 14. a'"+* - a'*+^ - ft^ + a**-^ and «"+' — a' — a + 1. 15. a^ + 3a^-'^-2a^-^ and 2 a^+H a^+' - 3 a^. FUNDAMENTAL OPERATIONS. 19 DIVISION. 39. Division is the operation by which, when a product and one of its factors are given, the other factor is deter- mined. With reference to this operation the product is called the dividend; the given factor the divisor; and the required factor the quotient. The operation of division is indicated by the sign ~- ; by the colon : , or by writing the dividend over the divisor 12 with a line drawn between them. Thus, 12-^4, 12 : 4, — , each means that 12 is to be divided by 4. 40. Since a X b = -{- ab ; ■ {~a)Xb = — ab; aX {—b) = — ab; (— a) X(—b) = -\-ab; ,-, f. ab — ab tnerelore -r- — (^', ;- = — a ; b ' -f^> ' — ab . -{-ab __= + „; -zy=-«- Consequently, the quotient {^positive when the dividend and divisor have like signs. The quotient is negative when the dividend and divisor have unlike signs. » 41. Monomials. To divide one monomial by another. Write the dividend over the divisor with a line between theTYi ; if the expressions have common factors, re^nove the common factors. Thus 25a&a; ^ 5^ ^ 36&ca; ^ 6£^ \Olcx 2c ' 30abc 5a 20 ALGEBRA. A • (X" aaaaa , Again, — = — aaa = o? • or aa a^ _ aa _ 1 _ 1 a' aaaaa aaa a^ In general, ^ = ^^^ ^'Q ^ ^^^^^^^ a** aaa to n factors = aaa to m — w factors (if m'^n)] or = (if n > m). aaa to w — m factors Hence, if a power of a number be divided by a lower power of the same number, the quotient is that power of the number of which the exponent is the exponent of the dividend diminished by that of the divisor ; and if any power of a number be divided by a higher power of the same num- ber, the quotient is expressed by 1 divided by that power of the number of which the exponent is the exponent of the divisor diminished by that of the dividend. The term power has so far been restricted to positive in- tegral powers. The above is the index law for division. 42. Division of Polynomials by Monomials. The product {a -{- b -\- c) X p = ap -^ bp -\- cp. § 34 Therefore, {ap '\-bp-\- cp) -^p = a-^b -{- c. But a, b, and c are the quotients obtained by dividing each term, ap, bp, and cp, hj p. Therefore, to divide a polynomial by a monomial, divide each term of the polynomial by the monomial. This is the distributive law for division. FUNDAMENTAL OPERATIONS. 21 43. Division of Polynomials by Polynomials. If the divisor (one factor) is a-\-b-\-c, and the quotient (other factor) is n-\-p-}-q, {an -\- bn -\- en -\- ap -\- bp -\- cp -\-aq-\-bq-\-cq The first term of the dividend is an, the product of a the first term of the divisor, by n the first term of the quotient. The first term n of the quotient is therefore found by dividing an, the first term of the dividend, by a, the first term of the divisor. If the partial product formed by multiplying the entire divisor by n be subtracted from the dividend, ap the first term of the remainder is the product of a, the first term of the divisor, by p, the second term of the quotient. Hence, the second term of the quotient is obtained by dividing the first term of the remainder by the first term of the divisor. And so on. Therefore, to divide one polynomial by another : Divide the first term of the dividend by the first term of the divisor. Write the result as the first term of the quotient. Multiply all the terms of the divisor by the first term of the quotient. Subtract the product from the dividend. If there be a remainder, consider it as a new dividend and proceed as before. It is of great importance to arrange both dividend and divisor according to the ascending or descending powers of some common letter, and to heep this order throughout the operation. 22 ALGEBRA. (1) Divide 22a%^ + 15 6* + 3 a* - 10 a?h - 22 ab^ by a'^ + 3 6^ _ 2 ab. 3 a* - 10 a^b + 22 a^S^ _ 22 ab'^ + 15 6* | a^-2a5 + 3 6^ 3a*- 6a36+ Oa^S^ 3a2-4a& + 562 - 4a36 + 13a2&2_22a63 - 4a36+ 8a252_i2a63 5a2J2_i0a63^156* 5a262_l0a53 + 156* The operation of division may be shortened in some cases by the use of parentheses. (2) K^ + (a + 6 + c) a;2 + {ab + ac + bc)x + abc \ x + b x^ + { +b )x^ x'^ + {a + c)x + ac {a + c)x^ + {ab + ac + be) x {a + c)x'^ + {ab + be) x acx + abc acx + abc 44. Detached Coefficients. In Division as in Multiplica- tion, it is convenient to use only the coefficients when the dividend and divisor are expressions involving but one letter, or homogeneous expressions involving but two letters. Thus, the work of Ex. 1 of the last section may be arranged as follows : 3-10 + 22-22 + 15 |l-2 + 3 3- 6+ 9 3-4+5 - 4 + - 4 + 13- 8- -22 -12 5- 5- -10 + 15 -10 + 15 The quotient is 3 a^ _ 4 a6 + 5 b\ FUNDAMENTAL OPERATIONS. 23 45. Special Oases. There are some cases in Division which occur so often in algebraic operations that they should be carefully noticed and remembered. The student may easily verify the following results : a — 6 (2) ^'LZL^=.a'-{-a'b + a'b' + ab' + b'. a — b In general, it will be found that the difference of two like powers of any two numbers is divisible by the differ- ence of the numbers. n^ J- 7)3 (3) 9Ljt!L^a^-ab-^b\ (4) ^i!±i! =a'- a'b + a'b'' - ab^ + b', a-\-b In general, it will be found that the sum of two like odd powers of two numbers is divisible by the sum of the numbers. Compare the quotients in (3) and (4) with those in (1) and (2). (5) "^-f-x + y. X — y X — y -x'-\-x'y-\-xy'-\-y\ (6) t^-x y. (8)El^. = o^ — x^y-{-xy'' — y'. x-\ry x-\-y In general, it will be found that the difference of two like even powers of two numbers is divisible by the differ- ence and also by the sum of the numbers. The sum of two like even powers of two numbers is not divisible by either the sum or the difference of the numbers. But when the exponent of each of the two like powers 24 ALGEBRA. is composed of an odd and an even factor, the sum of the given powers is divisible by the sum of the powers expressed by the even factor. Thus, x^ + y^ is not divisible by a: + y, or by ^ — y, but is divisible by x^ + y^. The quotient may be found as in examples (3) and (4). It appears, then, that a factor of x"" — y** can always be found ; and that a factor of x"^ -j- y'^ can be found unless n is a power of 2. Thus, factors of x"^ + y^, x'^ + 3/*, x^ + 3/^, etc., cannot be found. Exercise 4. Divide : 1. (6 a^h^c X 35 a^5V) by (21 a^h\^ X 2 a^c"). 2 . 39 a V + 24 a V + 42 aV + 27 a V by 6 aV. 3. 35a;^ + 94a:z;' + 52a'a; + 8a^by 5a; + 2a. 4. :r^ — 5 ax^ — a^x + 14 c^ by a;^ — 3 ax — 7 c^. 5. 81a;* + 36a;y + 163/*by 9a;'-6:ry + 4y^ 6. a;* + 5* - aV + 2 5 V by x^ -\-h'' -\- ax. 7. a'-25'-3c'' + o5 + 2ac + 7Z'cby a-5 + 3c. 8. ^x'-^xY~^:i^-^i^-^^-\-y' hy y^-\-2x''-2-2>xy. 9. 2a"'+^-2a^+^-a'"+" + a''*by a"-2a. 10. 625a;* — 81 y* by 5^7 — 3y. 11. ;r'" + y'" by r?;« + y". ,^ 27 a^ P . 2>a h '^' T25"64^^T-4 13. (a + 2by-i-(b-Scyhja + S(b~c). 14. a*" - a'^+i + 37 «'"+' - 55 «"*+* + 50 a"^^ by l-3a+10a^ FUNDAMENTAL OPERATIONS. 25 15.4 A^+i - 30 }f + 19 h^~^ + 5 A^"-^ + 9 h"' by h^-^ - 7 h^-' + 2 A--^ - 3 h-~\ 16. 6 ;r"*-**+^ + -^m-n+i _ 22 a:*"-" + 19 a;"*""-^ — 4 a;"*-'*-^ by 3 a;'-" -4^7'-" + a;'-". 46. Extension of Meaning. The introduction of negative numbers requires an extension of the meanings of some terms common to arithmetic and algebra. But every such extension of meaning must be consistent with the sense previously attached to the term and with general laws already established. Addition in algebra does not necessarily imply augmen- tation^ as it does in arithmetic. Thus, 7 + (— 5) = 2. The word sum, however, is used to denote the result. Such a result is called the algebraic sum, when it is necessary to distinguish it from the arithnietical sum, which would be obtained by adding the absolute values of the numbers. The general definition of Addition is, the operation of uniting two or more expressions in a single expression writ- ten in its simplest form. The general definition of Subtraction is, the operation of finding from two given expressions, called minuend and subtrahend, a third expression, called difference, which added to the subtrahend will give the minuend. The general definition of Multiplication is, the operation of finding from two given expressions, called multiplicand and multiplier, a third expression, called product, which may be formed from the multiplicand as the multiplier is formed from unity. The general definition of Division is, the operation of finding the other factor when the product of two factors and one factor are given. 26 ALGEBRA. 47. Pundamental Laws. All the operations of algebra are performed subject to the following laws: I. The commutative law (§§ 21, 31). 11. The associative law (§§ 27, 33). III. The distributive law (§§ 34, 42). IV. The index law (§§ 32, 41). The various meaning of these laws as applied to the four fundamental operations have been explained as they oc- curred. We shall now simply formulate them, including Subtraction under Addition. I. The commutative law : For Addition a-\-b = h -\-a. For Multiplication ah = ha. II. The associative law : For Addition a-\-(b-\-c) ={a^h)-\-c. For Multiplication ahc = a (he) = {ah) c. III. The distributive law : For Multiplication n{a -\- h ^ c) = na -\- nh -\- 7t>g, For Division a + & + o^« j £. n n n n IV. The index law : For Multiplication a"* X a" = «'"+". For Division — = a"*"** (if m > w) ; a" ^ = J—(iin>m\ CHAPTER III. FACTORS. In multiplication we determine the product of two given factors ; it is often important to determine the factors of a given product. 48. The simplest case is that in which all the terms of an expression have one common factor. Thus, (1) x' + xy^x^x-^y). (2) 6a^ + 4a^ + 8a = 2a(3a^ + 2a + 4). Frequently the terms of an expression can be arrangetiL so as to show a common factor. Thus, (3) ac — ad — he -{- hd = {ac — ad) — (be — bd) = a(c— d) — b (c — d) = (a-b)(c-d). 49. The square root of a number is one of the two equal factors of that number. Thus, the square root of 25 is 5 ; for, 25 = 5 X 5. The square root of a* is a^ ; for, a*^a'x a\ In general, the square root of an even power of a number is expressed by writing the number with an exponent equal to one-half the exponent of the power. 50. Since a"Z>" X a^-^ = a^-^a^"^ = a^'a^'b^'b'' = o'"^'", a"^'* is the square root of a^"6'^'*. But a" is the square root 28 ALGEBRA. of tt'^", and Z)" of 5^". Therefore, the square root of the product of even powers may be found by taking the square root of each factor, and finding the product of the roots. The square root of a positive number may be either posi- tive or negative ; for, or, a^ = — a X — a ; but throughout this chapter only the positive value of the square root will be considered. 51. From § 38 it is seen that a trinomial is often the product of two binomials. Conversely, a trinomial may, in certain cases, be resolved into two binomial factors. (1) To find the factors of a;' + Ta; + 12. The first term of each binomial factor will obviously be x. The second terms of the two binomial factors must be two numbers of which the proc?wd is 12, and the s,um 7. These two numbers are 4 and 3. .-. a;2 + 7a; + 12 = (x + 4)(a; + 3). (2) To find the factors oi x^ — ^x — 36. The second terms of the two binomial factors must be two num- bers of which the product is — 36, and the sum — 9. These two numbers are — 12 and + 3. ... a;2-9a;-36 = (ar-12)(a; + 3). 52. Consider trinomials which are perfect squares. These are only particular forms of the trinomials of the last sec- tion, but from their importance demand special attention. (1) To find the factors of ^' + 18:r + 81. The second terms of the two binomial factors must be two num- bers of which i\iQ product is 81, and the sum 18. These two numbers are 9 and 9. .-. a;2 + 18a; + 81 = (a; + 9)(a; + 9) = (a; + 9)'. FACTORS. 29 (2) To find the factors of x^ - l^x + 81. The second terms of the two binomial factors must be two num- bers of which the product is 81, and the sum — 18. These two numbers are — 9 and — 9. .-. a;2 - 18a; + 81 = (a; - 9)(a; - 9) = (a; - 9)1 53. An expression in the form of two squares, with the negative sign between them, is the product of two factors which may be determined as follows : Take the square root of the first number, and the square root of the second number. The sum of these roots will form the first factor ; The difference of these roots will form the second factor. Thus, (1) a^-h^ = {a + b){a-b). (2) (a - by - (c - df = {(a -b) + (c - d)]{{a -b)-{c- d)} = {a — b + c — d}{a — b — c-{- d}. The terms of an expression may often be arranged so as to form two squares with the negative sign between them, and the expression can then be resolved into factors. (3) a^ + b''-c^-d^ + 2ab + 2cd = a2 + 2ab + b^-c^ + 2cd-d^ = (a2 + 2ab + 62) -{c''-2cd + d^) = (a + 6)2 _ (c - df = {(a + 6) + (c - c? )}{(a + 6) - (c - rf)} = {a-irh-\-c — d}{a -^b-c + d]. An expression may often be resolved into three or more factors. (4) Xl6_3/16 = (a;8 + y8)(a:8_y8) = (a;8 + 2/8)(a;* + 3/*)(a;*-y*) = (X8 + y8) (a;* + y^) {x^ + y^) (.^2 _ y2) - (a;8 + 3/«) {3^ + 2/*) {a? + y^) {x + y){x- y). 30 ALGEBRA. Any expression of the form x^ ± y'^ may be resolved into factors by the principles of § 45 with one exception ; viz., when the sign is + and n is a power of 2. 64. For a trinomial to be a perfect square, the middle term must be twice the product of the square roots of the first and last terms. The expression ^x'^ — 37:cy + 93/* will become a perfect square if 2bx^y'^ be added to the middle term. We must also subtract 25 ^y to keep the expression unchanged. This gives 4a;* - 37 ^y 4. 9^4 = (4 a;* - 12 xy + 9y*) - 25a;y = (2a;2-32/7^-25a;y = (2^2 - 3?/2 + ^xy){2x' -Sf-5xy) = (2x2 + 5x^ - 32/2)(2x2 -5xy- 3y^). 55. To find the factors of 6x' + x— 12. It is evident that the first terms of the two factors may be 6a; and a;, or 2a; and 3 a;, since the product of either of these pairs is Qx"^. Likewise, the last terms of the two factors may be 12 and 1, 6 and 2, or 4 and 3 (if we disregard the signs). From these it is necessary to select such as will produce the middle term of the trinomial. And they are found by trial to be 3a; and 2a;, and — 4 and + 3. .-. 6a:' + a;-12 = (3a;-4)(2a; + 3). 56. The factors, if any exist, of a polynomial of more than three terms can often be found by the application of principles already explained. Thus, it is seen that the expression a^ — 2xy +y^ + 2xz — 2yz + z^ FACTORS. 31 consists of three squares and three double products, and from ^ 37, is the square of a trinomial which has for terms x, y, z. It is also seen from the double product — ^.xy, that x and y have unlike signs ; and from the double product 2 xz, that x and z have like signs. Hence, x^ — 2xy -^-y"^ -\-2xz — 2yz + 2^ = (x — 3/ + zf. 57. Find the factors of ^x" -lxy — ?>y''-^x-{- 30y - 27. The factors of the first three terms are 2>x +y and 2x — Zy. Now — 27 must be resolved into two factors such that the sum of the products obtained by multiplying one of these factors by 3 x and the other by 2x shall be — 9a;. These two factors evidently are — 9 and + 3. Therefore, (6a^ - 7x2/ - 32/2 - 9a; + 30y - 27 = (3 x + 2/ - 9)(2a; -Zy^ 3). This result may be verified by actual multiplication. 58. The following method is often convenient for sepa- rating a polynomial into its factors : Find the factors of 2x''- bxy + 2y^ + 1 xz - byz + 822. (1) Reject the terms that contain 2. (2) Reject the terms that contain y. (3) Reject the terms that contain x. Factor the expression that remains in each case. (1) 2x'^-bxy ■V2y^ = {x-2y){2x-y). (2) 2x^ + 1 xz +322 =(x + 3 2)(2a; + 2). (3) 23/2-52/2+322=(-22/ + 32)(-3/ + 2). Arrange these three pairs of factors in two rows of three factors each, so that any two factors of each row may have a common term. Thus, x—2y, a; + 82, —2y + Zz\ 2x — y, 2x + z, ~y + z. 32 ALGEBRA. From the first row, select the terms common to two factors for one trinomial factor : From the second row, select the terms common to two factors for the other trinomial factor. 2aj — 2/ + 2. Then, 2x^-bxy + 2y'^ ^Ixz-byz + ?,z^ = {x-2y + ?,z){2x -y -\- z). When a factor obtained from the first three terms is also a factor of the remaining terms, the expression is easily- resolved. Thus, x2 - 3a;y + 23/2 _ 3a; + 6^/ = (cc - 2y){x - y) -2>{x-2y) =^{x-2y){x-y-3). Exercise 5. Resolve into factors : 1. 9x'-{-6x'-{-Sx^-\-2x. 2. 2a'-3a'b~Ua'i-21ab. 3. 5x'-\-15x'i/-4:xf-12y\ 4. a^ctf — b^xy"^ — c^cx^ + b'^cy^. 5. :r^ + 8a;+7. 11. :r' + :^-72. 6. x^-Vlx^m. 12. ^^-14:r-176. 7. x'^lx-V^. 13. 81a;*-196^y. 8. o(?-2x-2^. 9. 9:r2 + 30a; + 25. 15. 64a;^ + ^/. 10. 16a;2-56a; + 49. 16. {p^-y''y-y\ 17. {a^-\-2b''y-a^h\ 18. {2x-Zyy-{x~2yy. 19. {2x^-^x-{-iy — x\x-\-^y. HIGHEST COMMON FACTOR. 33 20. X* - 2(b' - c'y + b'-2 b'c' + c\ 21. lbx''-7x-2. 22. lla^-54:X + QS. 23. 21a;^ + 26a;-15. 24. 70a;2-27a;-9. 25 . x*-2 abx' -a*- a'b' - b*. 26. 5a;^ + 4:i;^-20a;-125. 27. 2a;* -5:?;' -a;' -2. 28. 6a;* — aa;'-2aV + 3A — 2a*. 29. 12a;*+10a;V-12a:y — 6a:y-4y*. HIGHEST COMMON FACTOR. 59. A common factor of two or more expressions is an expression which is contained in each of them without a remainder. Two expressions which have no common factor except 1, are said to be prime to each other. The highest common factor of two or more expressions is the product of all the factors common to the expressions. For brevity, H. 0. F. will be used for highest common factor. Ex. Find the H. C. F. of 8aV — 24a'a; + 16a'' and Uax"^ - 12axy -24:mj. 8aV - 24a^x + IQa^ = 8a'^{x^ -3x + 2) ^2^a\x-l){x-2)', 12ax^y - I2axy - 24 ay - 12 ay {x^-x- 2) -2»x3ay(a; + l)(a;-2). .-. theH.C.F. = 22a(a:-2) = 4a(a;-2). 34 ALGEBRA. Hence, to find the H. C. F. of two or more expressions : Resolve each expression into its lowest factors. Select from these the lowest power of each common factor, and find the product of these powers. 60. When it is required to find the H. C. F. of two or more expressions which cannot readily be resolved into their factors, the method to be employed is similar to that of the corresponding case in arithmetic. And as that method consists in obtaining pairs of continually decreasing numbers which contain as a factor the H. C. F. required ; so in algebra, pairs of expressions of continually decreasing degrees are obtained, which contain as a factor the H. 0. F. required. The method depends upon two principles : I. Any factor of an expression is a factor also of any multiple of that expression. Thus, if i^ represent a factor of an expression A, so that A = nF, then rn^A = mnF. That is, mA contains the factor F. II. Any common factor of two expressions is a factor of the sum or difference of any multiples of the expressions. Thus, if F represent a common factor of the expressions A and B so that A — mF, and B = nF; then pA =pmF, and qB = qnF. Hence, pA ± qB =pmF± qnF, = {pm ± qn)F. That is, pA ± qB contains the factor F. HIGHEST COMMON FACTOR. 35 61. The general proof of this method as applied to num- bers is as follows : Let a and h be two numbers, of which a is the greater. The operation may be represented by : h)a{p p2 42)154(3 126 nF)mF(p pnF c)h{q qc_ 28)42(1 28 cF)nF(q qcF d)c{r rd 14)28(2 28 F)cF(c cF p, q, and r represent the several quotients, c and d represent the remainders, and d is supposed to be contained exactly in c. The numbers represented are all integral. Then c = rd, h = qc-\- d= qrd -\- d— {qr + l)d, a=ph-\-c = pqrd -\- pd -\- rd = {pqr -\-p + r) d. .'. d is a common factor of a and h. It remains to show that d is the highest common factor of a and h. Let/ represent the highest common factor of a and h. Now c = a —ph, and/ is a common factor of a and h. .'. by (II.) /is a factor of c. Also, d=b — qc, and/ is a common factor of h and c. .'. by (II.) /is a factor of d. That is, d contains the highest common factor of a and b. But it has been shown that c? is a common factor of a and b. .'. d is the highest common factor of a and b. 36 ALGEBRA. Note. The second operation represents the application of the method to a particular case. The third operation is intended to rep- resent clearly that every remainder in the course of the operation contains as a factor the H. C. F. sought, and that this is the highest factor common to that remainder and the preceding divisor. 62. This method is only needed to determine the com- pound factor of the H. 0. F. Simple factors of the given expressions should, be separated, and the highest common factor of these factors reserved to be multiplied into the compound factor obtained. Modifications of this method are sometimes needed. (1) Find the H. C. F. of 4:x' ~Sx-b and 12:r^ - 4:X - 65. 4a;2_8a;_5)12a;2- 4a! -65(3 12x^-243; -15 20a; -50 The first division ends here, for 20 x is of lower degree than 4 x^. But if 20 a; — 50 be made the divisor, ix^ will not contain 20a; an in- tegral number of times. Now, it is to be remembered that the H. C. F. sought is contained in the remainder 20 a; — 50, and that it is a compound factor. Hence if the simple factor 10 be removed, the H. C. F. must still be con- tained in 2 a; — 5, and therefore the process may be continued with 2 a; — 5 for a divisor. 2a; -5)4x2- 8a;-5(2a; + l 4a;'' -10a; 2a;-5 .-. the H. C. F. = 2a; - 5. 2a;-5 (2) Find the H. C. F. of 21r^--4:r^-15a;-2 and 21ar»- 32rr2- 54^;- 7. Writing only the coefficients (§ 44), the work is as follows : 21 - 4 - 15 - 2) 21 - 32 - 54 - 7 (1 21- 4-15-2 _ 28 - 39 - 5 HIGHEST COMMON FACTOR. 37 The difficulty here cannot be obviated by removing a simple factor from the remainder, for — 28 a;^ — 39 a; — 5 has no simple factor. In this case, the expression 21 a^ - 4 x'* — 15 a; — 2 must be multiplied by the simple factor 4 to make its first term divisible by — 28 x^. The introduction of such a factor can in no way affect the H. C. F. sought ; for the H. C. F. contains only factors common to the remain- der and the last divisor, and 4 is not a factor of the remainder. The signs of all the terms of the remainder may be changed ; for if an expression A is divisible by — F, it is divisible by + F. The process then is continued by changing the signs of the remain- der and multiplying the divisor by 4. 28 + 39 + 5)84 _ 16- 60- 8(3 84 + 117+ 15 _133- 75- 8 Multiply by — 4, - 4 532 + 300 + 32(19 532 + 741 + 95 Divide by - 63, - 63) -441 -63 7+ 1 7 + 1)28 + 39 + 5(4 + 5 28_ + 4 35 + 5 .-. thell.C.F. is7a; + l. 35 + 5 In practice the work is most conveniently arranged as follows 21- 4- 15- 2 21 -32-54-7 1 4 21 _ 4-15-2 84- 16- 60- 8 _l)_28-39-5 84 + 117+ 15 28 + 39 + 5 28+4 3 + 19 -133- 75- 8 - 4 35 + 5 35 + 5 532 + 300 + 32 532 + 741 + 95 -63)- 441 -63 7+ 1 4 + 5 .-. the H.C.F. is 7a; + 1. 38 ALGEBRA. In the preceding work each quotient is placed opposite the corre- sponding divisor ; but the position of the quotients is evidently a matter of indifference. 63. From the foregoing examples it will be seen that, in the algebraic process of finding the highest common factor, the following steps, in the order here given, must be care- fully observed : I. Simple factors of the given expressions are to be re- moved from them, and the highest common factor of these is to be reserved as a factor of the H. C. F. sought. II. The resulting compound expressions are to be ar- ranged according to the descending powers of a common letter ; and that expression which is of the lower degree is to be taken for the divisor; or, if both are of the same degree, that whose first term has the smaller coefficient. III. Each division is to be continued until the remainder is of lower degree than the divisor. IV. If the final remainder of any division is found to contain a factor that is not a common factor of the given expressions, this factor is to he removed; and the resulting expression is to be used as the next divisor. V. A dividend whose first term is not exactly divisible by the first term of the divisor, is to be multiplied by such an expression as will make it thus divisible. The H. 0. F. of three expressions will be obtained by finding the H. C. F. of two of them, and then of that and the third expression. LOWEST COMMON MULTIPLE. 64. A common multiple of two or more expressions is an expression which is exactly divisible by each of them. The lowest common multiple of two or more expressions LOWEST COMMON MULTIPLE. 39 is the product of all the factors of the expressions, each factor being written with its highest exponent. The lowest common multiple of two expressions which have no common factor will be their product. For brevity L. 0. M. will be used for lowest common multiple. Find the L. C. M. of 12a'c, Ubc\ S6ah\ 12a'c = 22x3aV, I4:bc'=2 Xlhc\ S6ab' = 2'xS'ab\ .-. the L. C. M. = 2' X S' X la'b'c' = 252a'b'c\ 65. When the expressions cannot be readily resolved into their factors, the expressions may be resolved by find- ing their H. G. F. Find the L. C. M. of Qx' ■\\x^y-\-2if 6-11+0 + 2 6- 8-4 and ^x'-22xy' 9+ 0-22- 8 2 - 3+4+2 - 3+4+2 18+ 0-44-16 18-33+ 0+ 6 11)33-44-22 3- 4- 2 8/. 2-1 Hence, ^^ -Wx^y ^2y^ ^{2,x-y) {^x'' -^xy -2y% and 9a;3 _ 22xy'^ -^f = {Zx + 42/)(3a;2 - 4a?y - 2y^. .'. the L. C. M. = (2 aj - y)(3 x + 4 2/)(3 x^-Axy- 2y'^\ In this example we find the H. 0. F. of the given expres- sion, and divide each of them by the H. C. F. Instead of dividing both expressions by their H. 0. F., we might have divided only one expression, and have mul- tiplied the quotient by the other expression. 40 ALGEBRA. The object of finding the H. G. F. is to obtain some means oi factoring the given expressions. Exercise 6. Find the H. 0. F. of: 1. 12a;' -17^ + 6, ^x' + ^x-S. 2. x'~a\ x^ + Zax — 4:a\ x^ ~ 5ax-\-4:a\ 3. x'-6x'+lSx' — 12x + 4:, x'-4:X^-i~Sx''-16x-\-16. 4. Sx* — x'-2x'' + 2x — S, 6a;' +13^;^ 4- 3a;' -1-20 37. 5. 96a;* + 8a;^-2ar, 32a;^- 24a;'- 8:f f 3. 6. a;* + 5a;^-7a;'-9a;-10, 2a;'- 4a;' + 4a;- 8. 7. 2a;' -16a; + 6, 5a;«+ 15a;^ + 5a; + 15. 8. 2a* + 3a'a;-9aV, 6a'a;- 3aa;'- 17aV+ 14aV. 9. 2a'-4a* + 8a'-12a' + 6a, 3a«-3a^-6a' + 9a'-3a'. 10. Sa^-lx'^ — y' + bxf, a; ^ + 3 a;?/' - 3 a;' - ^, 3a^ -\- 5x'^^-{-x7/^ — 7/^. 11. 36a;'-28a;^ + 32a;* + 8a;'-16a;', 12a;^ - 14a;* - 20a;' -f- lOo;' + 4a;. Find the L. 0. M. of: 12. a;' — 3a;-4, a;' - a; - 12, ai-' + 5a;4-4. 13. 6a;'- 13 a; -f- 6, 6a;' -f 5a; -6, 9a;' -4. 14. 3a;*-a;'-2a;2_^2a;-8, 6a;'+ 13a;' + 3 a; + 20. 15. 15aV+10(2V + 4aV + 6a''a;-3a^ 12 X* + 38 ao;' + 16 a'a;' - 10 a^x. 16. 2a;* + a;'-8a;'-a; + 6, 4a;* + 12 a;' -a:' — 27a;- 18, 4a;* + 40;' -17a;' -9a; +18. CHAPTER IV. FRACTIONS. 66. An algebraic expression is integral when it consists of a number of terms connected by + and — signs, each term being the product of a coefficient into powers of the letters involved. In an integral algebraic expression the coefficients may be frac- tional. Thus, ic' — f aa;2 + fa is an integral algebraic expression. 67. An algebraic fraction is the quotient of two integral expressions, and is generally written in the form -• h The dividend, a, is called the nnmerator; the divisor, h, the denominator. Tje numerator and denominator are called the terms of the fraction. 68. Since the quotient is unchanged if the dividend and divisor are both multiplied (or divided) by the same factor, the value of a fraction is unchanged if the numerator and denominator are multiplied (or divided) by the same factor. 69. To reduce a fraction to lower terms, Divide the numerator and denominator hy any c&mmon factor. A fraction is expressed in its lowest terms when both numerator and denominator are divided by their H. C. F. 42 ALGEBRA. (1) Reduce to lowest terms •D a KK 6 a;2 — 5 a; — 6 By I 55, — - — — 8 x'^ — 2 ic — 15 (2) Reduce to lowest terms (2 a' -3)(3a; + 2) 15 3a; + 2 (2 a. ■m a -3)(4a; + 5) 4x + 5 3a^-14a2+16a Since no common factor can be determined by inspection, it is necessary to find the H. C. F. of the numerator and denominator by the method of division. We find the H. C. F. to be a - 2. Now, if a^ — 7a2 + 16 a — 12 be divided by a — 2, the result is a2 _ 5^ ^ g . and if 3 a^ — 14a2 + 16a be divided by a - 2, the re- sult is 3 a'^ — 8 a. a3_7^2 + 16(j_12 a2-5a + 6 3a=»-14a2 + 16a Zo? 70. Mixed Expressions. If the degree of the numerator of a fraction equals or exceeds that of the denominator, the fraction may be changed to the form of a mixed or integral expression hy dividing the numerator hy the denominator. The quotient will be the integral expression ; the remain- der (if any) will be the numerator, and the divisor the denominator, of the fractional expression. To reduce a mixed expression to a fractional form. Multiply the integral expression hy the denominator, to the product annex the numerator, and under the result write the denominator. The dividing line has the force of a vinculum or paren- thesis affecting the numerator ; therefore, if a minus sign precede the dividing line, and this line be removed, the sign of every term of the numerator must be changed. -_,, a — 6 en — (a — h) cn — a-i-b Thus, n = = FRACTIONS. 43 71. Lowest Oommoii Denominator. To reduce fractions to equivalent fractions having the lowest common denominator. Find the L. 0. M. of the denominators. Divide the L. 0. M. by the denominator of each fraction. Multiply the first numerator by the first quotient, the sec- ond numerator by the second quotient, and so on. The products will be the numerators of the equivalent fractions. The L.C.M.. of the given denominator's will be the denom- inator of each of the equivalent fractions. Thus. l£, 2^, A 1 , 9 ax 8 a^v 10 , • i are equal to , ^. , respectively, ^ 12a^ 12a' 12a' ^ ^ The multipliers 3 a, ia"^, and 2, being obtained by dividing 12 a^ the L. C. M. of the denominators, by the respective denominators of the given fractions. 72. Addition and Subtraction of Practions. To add fractions : JReduce the fractions to equivalent fractions having (he lowest common denominator. Add the numerators of the equivalent fractions. Write the result over the lowest comm.on denominator. To subtract one fraction from another we proceed as in addition, except that we subtract the numerator of the sub- trahend from that of the minuend. (1) Simplify. ^a~^h _ 2a-b + c 13a-4£^ The L. CD. is 84. The multipliers are 12, 28, and 7 respectively. 36 a — 48 6 = 1st numerator, - 56 a + 28 6 - 28 c = 2d numerator, 91 a — 28 c = 3d numerator. 71 a — 20 6 — 56 c == sum of numerators. 44 ALGEBRA. . 3a-45 2a-6 + c 13a-4c ^ 71a-206-56c "7 3 12 "^ 84 Since the minus sign precedes the second fraction, the signs of all the terms of the numerator of this fraction are changed after being multiplied by 28. (2) Simplify J!_^-^+1 2a:y x^ — lf- x-\-y ' ' x^-\-y^ The L. C. D. is (a; + y){x - y^x" + y^). The multipliers are x^ + y^, {x — y) {x^ + 2/^), {x + y){x — y), (o^ + 2/2), {x + y)(x — y), respectively. ^■2y2 + 2/* = 1st numerator, — x* + 2a^y--2 x^y^ + 2 xy^ — y^ = 2d numerator, x^ — 2/* = 3d numerator, 2x^2/ ~2xy^ = 4th numerator. 4 cc^y — x'^y'^ — 2/* = sum of numerators. .-. Sum of fractions - ^^y-^Y-f , a* — y* 73. Since ^ = «, and ~ ^, = a, it is evident that if the signs of both numerator and de- nominator be changed, the value of the fraction is not altered. Since changing the sign before the fraction is equivalent to changing the sign before every term of the numerator or denominator, therefore the sign before every term of the numerator or denominator may he changed^ provided the sign before the fraction is changed. Since, also, the product of + « multiplied by + ^ is ah, and the product of — a multiplied by — J is ah, the signs of two factors, or of any even number of factors, of the numerator or denominator of a fraction may be changed without altering the value of the fraction. FRACTIONS. 45 By the application of these principles, fractions may often be changed to a form more convenient for addition or subtraction. Ex. Simplify ^~-l— + ^£zi^. Change the signs before the terms of the denominator of the third fraction, and change the sign before the fraction. The result is, 2 3 2x-S a;2x — 14a;'^ — 1 in which the several denominators are written in similar form. The L. C. D. is x{2x - l)(2a; + 1). 8 a;'' — 2 = lat numerator, — 6x'^ — 3x = 2d numerator, -- 2 a;2 + 3 ic = 3d numerator. 2 = sum of numerators. -2 .'. Sum of the fractions = a;(2a;-l)(2a; + l) 74. Multiplication of Practions. Let it be required to find the product of the two fractions y and -• a If we multiply the dividend a by c, we multiply the quotient - by c ; if we multiply the divisor b by d, we divide the quotient - by d. Hence, the product of the two frac- tions - and - is -— • Therefore, to find the product of two b d od fractions, Find the product of the numerators for the numerator of the product, and the product of the denominators for ihe denominates of the product. « 46 ALGEBRA. 75. Division of Fractions. Multiplying by the reciprocal of a number is equivalent to dividing by the number. Thus, multiplying by ^ is equivalent to dividing by 4. The reciprocal of a fraction is the fraction with its terms interchanged. Therefore, to divide by a fraction, Interchange the terms of the fraction and multiply hy the resulting fraction. If the divisor be an integral expression, it may be changed to the fractional form. 76. A complex fraction is one which has a fraction in the numerator, or in the denominator, or in both. To simplify a complex fraction, Divide the numerator hy the denominator. It is often shorter to multiply both terms of the fraction by the L. C. D. of the fractions contained in the numerator and denominator. Exercise 7. Reduce to lowest terms : ^ 42a^ — 3Qa^:r ^ 6aV- 2a^ + 18g^- 6a' 35aa;'-25/ ' 4a* + 2aV + 12a' + 6c''' 2 2x^-\-bx'-\2x 4 x'-{-{2h''~a^)x''-\-h' l3^-\-2bx'-l2x ' x'+2aa^-\-a'x'-b''' ^ 6:r^-9:i;*+ll:r^ + 6:<;'-10:r 4:x^ + lOa;^ + 10:^;* + 4^^ + QOx' Simplify : Sx — 2y 4:y + 2x . 22y-9x 3 5 "^ 15 * „ _2 1 2a + 3 , 1 ■ Sa-2h ; 3a 2b 6a' '^ 2x' Qab FRACTIONS. 47 8, 3 , 4a ^a" 11. x~a (x — of {x — of 9 a-\-h . h-\-c a — c {b~c){c-ay {c-a){a-h) (a-b){b-c) 10. . } , + . 1 • 1 a{a — b){a — c) b(b — c)(b~a) c(c — a)(c — b) 6.2:'' -17:^+12 27x'+l8x-24: 25x'-25x + 6 12^:2 _ 25^+ 12"*" Ux'i-lx-U 20x^-2Sx-{-6 , ^ 2 aV ^ 5 a^5^ ^ 15 5^c^ 25 a*:r 36« 4cV 4a9:c 13. f ^' — y* . ^ + y\ .f ^' + y' . ^ + y\ V^;"' — y"' x^-xy) ' \x -y ' xy - yV ^g : r''— 7a;+12 ^ or'' + a: - 2 ^ 2a:^ + 5a;-3 > + 5a; + 6 x''-6^ + 4: Sx'-7x-6 ^g 6a^ - a - 2 ,, Sa' - lOa + 3 ,, 12a^ + 17a + 6 8a' -2a -3 12a*' + a- 6 6a' + a-2 2x±y 17. -^ ?^. 19. («^+A.)(«'+*') a; + 2/ a; a-\-b a — b 1 + x l-\-x^ 64 a^- 96 g^g^H- 36 a:r ^g 1 + x^ l + x' 2^ 36a'-729:r' 1+a;' 1+.'^' 48a' -27^'^ l + r' 1 + a;* 8a'-72aa; + 162a:» CHAPTER V. SIMPLE EQUATIONS. 77. Two different expressions which involve the same symbols will generally have different values for assumed values of the several symbols ; for certain values of the symbols involved the two expressions may have the same value. 78. An equation is a statement that two expressions have the same value ; that is, a statement that two expressions represent the same number. Every equation consists of two expressions connected by the sign of equality ; the two expressions are called the sides or members of the equation. An equation will in general not hold true for all values of the symbols involved ; it will hold true for only those values which give to the two members the same value. Thus, the equation, 4a;2-3.'c + 5 = 3a;2 + 4a;-5, holds true when for x we put 2, since both members then have the value 15 ; also when for x we put 5, since both members then have the value 90. If we give to x any other value, the two members will be found to have different values, and the equation will not hold true. 79. An equation of condition is an equation which holds true for only certain particular values of the symbols involved. An identical equation, or an identity, is an equation which holds true for all values of the symbols involved. The two members of an identical equation are identical expressions. SIMPLE EQUATIONS. 49 In identical equations it is customary to use the sign =, called the sign of identity, instead of the sign of equality. Thus, the two expressions {x-\-yf and x^ ^- 2xy -\- y^ have the same value for all values of x and y, and we accordingly write the identity, {x + 2/)2 = x^ + 2x3/ + y^- This is read, {x + yY is identically equal to x^ ^^ 2xy + y^ \ or, (re + yf is identical with x"^ -\-2xy + y"^. Wherever the term equation is used, it is to be under- stood that an equation of condition is meant, unless the contrary is expressly stated. 80. In any particular problem we have two kinds of numbers to consider : (1) Numbers which are either given, or supposed to be given, in the problem under consideration. Such numbers are called known numbers ; if given, they are generally rep- resented by figures ; if only supposed to be given, by the first letters of the alphabet. (2) Numbers which are not given in the problem under consideration, but are to be found from certain given rela- tions to the given numbers. Such numbers are called unknown numbers, and are generally represented by the last letters of the alphabet. The relations between the known and unknown numbers are generally expressed by means of equations. To be able to determine all the unknown numbers, we must have as many equations as there are unknown num- bers. If there are two or more equations, we have a system of simultaneous equations. 81. To solve an equation, or a system of simultaneous equations, is to find the unknown numbers involved. 50 ALGEBRA. 82. The degree of an equation is the same as the sum, in the term in which that sum is greatest, of the exponents of the several unknown numbers involved in the equation. If the equation involves but one unknown number, the degree is the same as the exponent of the highest power of the unknown number involved in the equation. Equations of the first, second, third, and fourth degrees are called, respectively, simple equations, quadratic equa- tions, cubic equations, and biquadratic equations. 83. Literal equations are equations in which some or all of the given numbers are represented by letters. 84. An equation which involves but one unknown num- ber, represented for example by x, will hold true for those values of x which give to the two members the same value (§ 78), and for no other values of x. The values of x for which the equation holds true are called the roots of the equation. Thus, the roots of the equation 4.'c2 — 3 x + 5 = 3 x^ + 4a; — 5 are 2 and 5. To solve an equation which involves one unknown num- ber is therefore to find its roots. 85. The various methods of solving equations are based mainly upon the following general principle : If similar operations be performed upon equal numbers, the results ivill be equal numbers. Thus, the two members of a given equation are equal numbers. If the two members be increased by, diminished by, multiplied by, or divided by, equal numbers, the results will be equal numbers. Similarly, if the two members be raised to like powers, or if like roots of the two members be taken, the results will be equal numbers (§ 18). SIMPLE EQUATIONS. 61 86. Any term inay he transposed from one side of an equation to the other, provided its sign be changed. Suppose we have x -\- a=' b. Now, a = a. Subtract, x =b~a. The a which appeared in the left member with the posi- tive sign, now appears in the right member with the negative sign. Similarly for any other equation. 87. The signs of all the terms on each side of an equa- tion may be changed ; for this is in effect transposing every term. 88. To solve an equation with one unknown number, Transpose all the terms involving the unknown number to the left side, and all the other terms to the right side: combine the like terms, and divide both sides by the coefficient of the unknown number. To verify the result, substitute the value of the unknown number in the original equation. Ex. Solve (x~2)(x + ^) = (x.-\-l)(x-{-2). Multiply out, a;2 + 2 ic - 8 - a'2 + 3 a; + 2, or 2x-^ = 2>x + 2. Transpose, 2a; — 3x = 2+8, - X = 10, a =: — 10, Ans. 89. Fractional Equations. To clear an equation of fraction, Multiply each term by the L. C. M. of the denomjinators. If a fraction is preceded by a minu^ sign, the sign of every term of the numerator must be changed when the denominator is removed (§ 73). 52 ALGEBRA. « 1-^ = ^-^- Multiply by 33, the L. C. M. of the denominators. Then, 11a; - 3a; + 3 = 33a; - 297. Transpose and combine, — 25 a; = — 300. .-. a; = 12. Since the minus sign precedes the second fraction, in removing the denominator, the + (understood) before x, the first term of the numerator, is changed to — ; and the — before 1, the second term of the numerator, is changed to +. If the denominators contain both simple and compound expressions, it is best to remove the simple expressions first, and then each compound expression in turn. ^ ^ 14 "^"6:^ + 2 7 Multiply both sides by 14. Then, 8a; + 5 + ^^^^-^ = 8a; + 12. 3a;+l Transpose and combine, — ^^-^ — = 7. ^ 3x + l Multiply by 3 a; + 1, 49 a; - 21 = 21 a; + 7. .-. a; = 1. Exercise 8. Solve : 1. 8 (10 -a;) = 5 (^+3). 2. 2x-S(2x-S) = l~4:(x-2). 3. (x-bXx + 6) =(x-lXx~2). 4. (2x-{-S){Sx — 2') = x' + x(5x-i-S). 5. (x-SXx + b) = (x + lX2x-S)-x\ SIMPLE EQUATIONS. 53 6. (a; + 4)(a7-2) = (a;+3)(3a;-f 4)-(2a; + l)(^--6). 7. {x - ^){2x i-b) = x{x-\-^)-^{x-\- l){x + 3). 8. {x-\-2y~r^x = {x-2y^b{l%-x). 9. {x-^Y + {x-^y = {x-2y^{x-^^f. -rt 3a; X 26 .. 5^ — 6 3a; a; — 9 12. 3a;- 5 2 ^^^ ""^ -^' + "" ~"" zz= 7. 13. 2a; + 10 3 3(5a;-3) _6 ^^ 2(4a; + 3) 5 * 2a;+l ' 4a;-3 5 4 10 12-3a; 3a;- -11 4 3 4a;+17 3a;- -10 a; + 3 1 X — -4 .-3 2x- i=:l. 18. 4a; + 3 3a;-4 ^ 7 3a; + 4 4a;-3 '12 19. 6^__^^2a; + l 3 a;-f2 2 20. 2^±i + 2£_5, a+1 ^ a rt, ax — h bx 4- c ■, 21. ■ — = abc. c a 22 ^• + « I x-\-h __b 3(a; + 6j 2(a; + a) 6* a; -2a 13ft''-2a;' _g a;-f3a a;'' -9a' 23. 24. «|^| «(^ — Q^) a;(a; + a) ^ aa; ^ a; a a; (a; + a) a (a; — a) d^ — x^ 54 ALGEBRA. 90. Problems. In the statement of problems it is to be remembered that the unit of the quantity sought is always given, and it is only the number of such units that is to be found. We have nothing to do with the quantities them- selves; it is only numbers with which we have to deal. Thus, X must never be put for a distance, time, weight, etc., but for a number of miles, days, pounds, etc. (1) A and B had equal sums of money ; B gave A $5, and then 3 times A's money was equal to 11 times B's money. "What had each at first ? Let X = numher of dollars each had. Then a; + 5 = numher of dollars A had after receiving 1 5, and X — 5 = number of dollars B had after giving A $5. .-. 3(a; + 5) = ll(a;-5), 3a; + 15 = 11a; -55, -8a; = -70, a; = 8|. Therefore each had |8,75. (2) A can do a piece of work in 5 days, and B can do it in 4 days. How long will it take A and B to do the work together ? Let X = the number of days it will take A and B together. Then - = the part they can do together in one day. Now, I = the part A can do in one day, and ^ = the part B can do in one day, .'. -^ + $ = the part A and B can do together in one day. 5 4 a; 4a; + 5x = 20, 9a; = 20, a; = 2|. Therefore they can do the work together in 2f days. SIMPLE EQUATIONS. 55 Exercise 9. l; The difference of two numbers is 3 ; and three times the greater number exceeds twice the less by 18. Find the numbers. 2. If a certain number be increased by 16, the result is seven times the third part of the number. Find the given number. 3. A boy was asked how many marbles he had. He replied, " If you take away 8 from twice the number I have, and divide the remainder by 3, the result is just one- half the number." How many marbles had he ? 4. The sum of the denominator and twice the numerator of a certain fraction is 26. If 3 be added to both numer- ator and denominator, the resulting fraction is |. Find the given fraction. 5. A courier sent away with a despatch travels uni- formly at the rate of 12 miles per hour ; 2 hours after his departure a second courier starts to overtake the first, trav- elling uniformly at the rate of 13-|- miles per hour. In how many hours will the second courier overtake the first ? 6. Solve the above problem when the respective rates of the first and second couriers are a and h miles per hour, and the interval between their departures is c hours. 7. A certain railroad train travels at a uniform rate. If the rate were 6 miles per hour faster, the distance travelled in 8 hours would exceed by 50 miles the distance travelled in 11 hours at a rate 7 miles per hour less than the actual rate. Find the actual rate of the train. 8. A can do a piece of work in 10 days ; A and B to- gether can do it in 7 days. In how many days can B do it alone ? 56 ALGEBRA. 9. A can do a piece of work in a days ; A and B to- gether can do it in b days. In how many days can B do it alone ? 10. If A can do a piece of work in 2 m days, B and A together in n days, and A and C in m + ^ days, how long will it take them to do the work together ? 11. A boatman moves 5 miles in f of an hour, rowing with the tide ; to return it takes him \\ hours, rowing against a tide one-half as strong. "What is the velocity of the stronger tide ? 12. A boatman, rowing with the tide, moves a miles in h hours. Eeturning, it takes him c hours to accomplish the same distance, rowing against a tide m times as strong as the first. What is the velocity of the stronger tide? 13. If A, who is travelling, makes -J of a mile more per hour, he will be on the road only |- of the time ; but if he makes -J- of a mile less per hour, he will be on the road 2J hours more. Find the distance and the rate. 14. The circumference of a fore wheel of a carriage is a feet ; that of a hind wheel, h feet. What distance will the carriage have passed over, when a fore wheel has made n more revolutions than a hind wheel ? 15. A wine merchant has two kinds of wine which he sells, one at a dollars, and the other at h dollars per gallon. He wishes to make a mixture of / gallons, which shall cost him on the average m dollars a gallon. How many gallons must he take of each ? Discuss the question (i.) when a = h\ (ii.) when a ov b = m ; (iii.) when a=^b^=m\ (iv.) when a > 5 and < in ; (v.) when a > & and b'p-m. CHAPTER VI. SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 91. Equations that express different relations between the unknown numbers are called independent equations. Thus, a; + 2/ = 10 and x — y = 2 are independent equations ; they express different relations between x and y. But x + y = 10 and 3a; + 3y = 30 are not independent equations; one is derived imme- diately from the other, and both express the same relation between the unknown numbers. 92. Equations that are satisfied by the same values of the unknown numbers are called simnltaneons equations. 93. Simultaneous equations are solved by combining the equations so as to obtain a single equation with one unknown number ; this process is called elimination. There are three methods of elimination in general use : I. By Addition or Subtraction. II. By Substitution. III. By Comparison. We shall give one example of each method. (1) Solve: 2x-2>y= 4") (1) 3a; + 2y = 32j (2) Multiply (1) by 2 and (2) by 3, 4 a; - 6 y = 8 (3) 9a; + 63/- 96 (4) Add (3) and (4), 13 a; =104 .-. a; = 8. Substitute the value of x in (2), 24 + 2 3/ = 32. .-. 3/ = 4. In this solution y is eliminated by addition. 58 ALGEBRA. (2) Solve: 2:r + 3y = 8| (1) 3a;+7y=7j (2) Transpose 3 2/ in (1), 2x = 8 — 3y. Divide by coefficient of x, x = —^ — ^ (4) Substitute the value of x in (2), 2if^.r, = 7. 24 -92/ + 142/ = 14. 52/ = -10. .••3/-- 2. Substitute the value of y in (4), .-. a; = 7. In this solution y is eliminated by substitution. (3) Solve: 2:r-9y=ll) 3:i;-4y= 7j (1) (2) transpose 9y in (1) and 42/ in (2), 2a;=ll + 9y, (3) 3a; = 7+*42/. (4) Divide (3) by 2 and (4) by 3, x~ 11±_^, (5) . 7 + 4^. 3 (6) Equate the values of a;, 11 + ^ I±i^. ^ '23 (7) Reduce (7) 33 + 27y = U + 8y. .-. 2/ = -l. Substitute the value of?/ in (5), .-. a; = 1. In this solution x is eliminated by comparison. Each equation must be simplified, if necessary, before the elimination is performed. SIMULTANEOUS EQUATIONS. 59 (4) Solve: (.^ - l)(y + 2) = (:^; - 3)(y - 1) + 8^ (i) 2a;-l 3(y-2) _.^ C ^^^ Simplify (1), xy + 2x-y -2-= xy ~x -Sy + 3 + S. Transpose and combine, 3 a; + 2y = 13. (3) Simplify (2), 8 cc - 4 - 15 y + 30 = 20. Transpose and combine, 8 a; — 15 y = — 6. (4) Multiply (3) by 8, 24 a; + 16y = 104. (5) Multiply (4) by 3, 24 x - 45 3/ = - 18. (6) Subtract (6) from (5), 61 y = 122. .-. 2/ = 2. Substitute the value of y in (3), 3 a; + 4 = 13. .-. a; = 3. Fractional simultaneous equations, with denominators which are simple expressions containing the unknown numbers, may be solved as follows : (5) Solve: A-lA=7 m 3a7 5y ^ ^ Multiply (2) by 4, li_A=12. (3) o X oy Add (1) and (3), ^^=19. Divide both sides by 19, — = 1. 3aj ■ 0, _ 1 . . x — ^. Substitute the value of x in (1), 5y 2 Transpose, — = 2. 53/ Divide both sides by 2, — = 1. 5y 60 ALGEBRA. 94. Literal Simultaneous Equations. The method of solv- ing literal simultaneous equations is as follows i Ex. Solve: ax-{-hy = m\ cx-{-dy = n } (1) (2) To find the value of y : Multiply (1) by c. Multiply (2) by a, acx + bey = cm acx + ady = an (3) (4) Subtract (4) from (3), (be — ad) y = cm- - an Divide by coefficient of - -ff -an -ad To find the value of x : Multiply (1) by d, Multiply (2) by h, adx + bdy = d7n bcx -f bdy = bn (5) (6) Subtract (6) from (5), {ad —bc)x = dm - -bn Divide by coefficient of 0: = ^^- -bn 1. 95. If three simultaneous equations are given, involving three unknown numbers, one of the unknown numbers must be eliminated between two pairs of the equations; then a second between the resulting equations. Likewise, if four or more equations are given, involving four or more unknown numbers, one of the unknown numbers must be eliminated between three or more pairs of the equations ; then a second between pairs of the re- sulting equations ; and so on. Solve: 2x—Sy-{-4:Z= 4) (1) (2) (3) 2x -3y + Az=-- = 4 Sx + 5y- 72 = ^2 bx - 3/- 82 = -- 5 SIMULTANEOUS EQUATIONS. 61 Eliminate z between Multiply (1) by 2, (3) is two Dfa; of a pairs of these equations. 4a;-6y + 82= 8 5a;- y-82= 5 (4) Add. Multiply (1) by 7, Multiply (2) by 4, 9>x-ny = 13 14a;-2l3/ + 282= 28 12a; + 203/-282= 48 (5) Add, Multiply (6) by 7, (5) is 26a;- 3/ = 76 182a; -7t/ = 532 9a;-72/= 13 (6) (7) Subtract (5) from (7), Substitute the value Substitute the values 173 a; =519 .-. a; = 3. in (6), 78 - y = 76. .••2/ = 2. and y in (1), 6-6+4z = 4. .-. 2 = 1. Exercise 10. Solve the following sets of equations : 1. 6a;+ 5y 10^7+ Sy 4. 46 66 2^+ 7y-52| ^x- 5?/=16j 4.r+ 9y=79| 7a;-17v = 40j = 19 3 2a; 4y Vly 9^7 5. a:=16 — 4y| y = 34-4a;) 6. 5:i;--23/+ 781 33/= a:+104r 7. 2f - ^. 3 ^2 -10 y_5a; -7 4 19 8. 4 + y = 3a: ^ 4 a:-8 = 4^ 6 62 ALGEBRA. 9. 10. x±y_ 15 8 2^ + 2.y-5 21 13. 11 12. 5 + ?^3 15_4_ _4_ _5_^86^ 5a; 6y 15 I 10- 4a; 5y 20^ a;_ 3/ — 10 " 3 14. a; 2y--^ 23 4 + .y+13 8 20 4- ^^~^^ ' 2 18 = 30 73-3.y V. 15. 2:r — 33/ = 5a — 5 3a;-2y = 5a + 5 16. a 6 c 1 h ' a ' c ^ 19. 8:i; + 4y-32 = 6- 4:r-5y + 42 = 8 2a; 20. Sx-^ + z =7^ 17. 18. x-y x-\- c h-c y y + h a + c j X — a _ a — b '\ y — a a-\-b 21. 2x-% + 4:z 2 2£_1 1 3^-7 2 5i 11 a^ + 6^ bz~x 2y~?>z ^y-2x 1 = 1 SIMULTANEOUS EQUATIONS. 63 22. -+^+-=3 X y z X y z 23. xy x + y xz 2a X y z x-\- z = h 96. Problems. It is often necessary in the solution of problems to employ two or more letters to represent the numbers to be found. In all cases the conditions must be sufficient to give just as many equations as there are unknown numbers employed. If there are more equations than unknown numbers, some of them are superfluous or inconsistent ; if there are less equations than unknown numbers, the problem is in- determinate. Ex. If A gives B f 10, B will have three times as much money as A. If B gives A $10, A will have twice as much money as B. How much has each ? Let X = number of dollars A has, and y = number of dollars B has. Then y + 10 = number of dollars B has, and rr — 10 = number ol dollars A has after A gives $ 10 to B. .-. 2/ + 10 = 3(a; - 10), and a; + 10 = 2(3/ - 10). From the solution of these equations, a; = 22 and y Therefore A has |22 and B has $26. 26. Exercise 11. 1, Three times the greater of two numbers exceeds twice the less by 27 ; and the sum of twice the greater and five times the less is 94. Find the numbers. 2. A fraction is such that if 3 be added to each of its terms, the resulting fraction is equal to \ ; and if 3 be sub- tracted from each of its terms, the result is equal to i. Find the fraction. 64 ALGEBRA. 3. Two women buy velvet and silk. One buys 3^ yards of velvet and 12| yards of silk ; the other buys 4 J yards of velvet and 5 yards of silk. Each woman pays $ 63.80. Find the price per yard of the velvet and of the silk. 4. Each of two persons owes $1200. The first said to the second, " If you give me f of what you have, I shall have enough to pay my debt." The second replied, " If you give me |- of what your purse contains, I can pay my debt." How much does each have ? 5. Two passengers have together 400 pounds of baggage. One pays $1.20, the other $1.80, for excess above the weight allowed. If all the baggage had belonged to one person he would have had to pay $4.50. How much bag- gage is allowed free ? 6. A number is formed by two digits. The sum of the digits is 6 times their difference. The number itself ex- ceeds 6 times the sum of its digits by 3. Find the number. 7. A number is formed by two digits of which the sum is 8. If the digits be interchanged, 4 times the new num- ber exceeds the original number by 2 more than 5 times the sum of the digits. Find the original number. 8. Three brothers, A, B, C, have together bought a house for $32,000. A could pay the whole sum if B would give him f of what he has ; B could pay it if would give him f of what he has ; and C could pay the whole sum if he had J of what A has together with -^^ of what B has. How much does each have ? 9. A and B entered into partnership with a joint capital of $3400. A put in his money for 12 months; B put in his money for 16 months. In closing the business, B's share of the profits was greater than A's by -^ of the total profit. Find the sum put in by each. SIMULTANEOUS EQUATIONS. 65 10. A capitalist makes two investments ; the first at 3 per cent, the second at 3j per cent. His total income from the two investments is $427. If $1400 were taken from the second investment and added to the first, the incomes from the two investments would be equal. Find the amount of each investment. 11. A cask contains 12 gallons of wine and 18 gallons of water ; a second cask contains 9 gallons of wine and 3 gallons of water. How many gallons must be taken from each cask, so that, when mixed, there may be 14 gallons consisting half of water and half of wine ? 12. A and B ran a race to a post and back. A return- ing meets B 30 yards from the post and beats him by 1 minute. If on arriving at the starting place A had imme- diately returned to meet B, he would have run ^ the dis- tance to the post before meeting him. Find the distance run, and the time A and B each makes. 13. A and B together can do a piece of w^ork in 15 days. After working together for 6 days, A leaves off and B finishes the work in 30 days more. In how many days can each do the work ? 14. A and B together can do a piece of work in 12 days. After working together 9 days, however, they call in C to aid them, and the three finish the work in 2 days. finds that he can do as much work in 5 days as A does in 6 days. In how many days can each do the work? 15. A pedestrian has a certain distance to walk. After having passed over 20 miles, he increases his speed by 1 mile per hour. If he had walked the entire journey with this speed, he would have accomplished his walk in 40 minutes less time ; but, by keeping his first pace, he would have arrived 20 minutes later than he did. What distance had he to walk ? CHAPTER VII. INVOLUTION AND EVOLUTION. 97. The operation of raising an expression to any re- quired power is called Involution. Every case of involution is merely an example of multi- plication, in which the factors are equal. 98. Index Law. If m is a positive integer, by definition a"* = a X a X a to m factors. § 13 Consequently, if m and n are both positive integers, (a")"* = a** X a" X a" to m factors = (a X a to n factors)(a X a ••••• to n factors) taken m times = aXaX a to mn factors The above is the index law for involution. Also, (pry = a-»' = (a")"* ; (ahy ~ah X ab to n factors = (aX a to n factors)(5 X b to 7i factors) 99. If the exponent of the required power is a composite number, the exponent may be resolved into prime factors, the power denoted by one of these factors found, and the result raised to a power denoted by another factor of the exponent. Thus, the fourth power may be obtained by taking the second power of the second power; the sixth by taking the second power of the third power ; and so on. INVOLUTION AND EVOLUTION. 67 100. From the Law of Signs in multiplication it is evi- dent that all even powers of a number sue positive ; all odd powers of a number have the same sign as the number itself. Hence, no even power of ani/ number can be negative; and the even powers of two compound expressions which have the same terms with opposite signs are identical. Thus, (5 - a)^ = J - (a - b)^ = (a- h)\ 101. Binomials. By actual multiplication we obtain, {a-\-hy = a^-\-2ah-^h'') la-\-bJ = a^ + Sa'^S + 3a5^ + 5'; (a + hy = a* + 4a'6 + 6a^5^ + 4a6H ¥. In these results it will be observed that : I. The number of terms is greater by one than the ex- ponent of the power to which the binomial is raised. II. In the first term, the exponent of a is the same as the exponent of the power to which the binomial is raised ; and it decreases by one in each succeeding term. III. h appears in the second term with 1 for an exponent, and its exponent increases by 1 in each succeeding term. IV. The coefficient of the first term is 1. V. The coefficient of the second term is the same as the exponent of the power to which the binomial is raised. VI. The coefficient of each succeeding term is found from the next preceding term by multiplying the coefficient of that term by the exponent of a, and dividing the product by a number greater by one than the exponent of h. If h is negative, the terms in which the odd powers of b occur are negative. Thus, (a _ hy = a* - 4a^5 -\-^a'b^ - 4a5» + b\ By the above rules any power of a binomial of the form a±b may be written at once. 68 ALGEBRA. 102o The same method may be employed when the terms of a binomial have coefficients or exponents. (1) (a - hf = a' - 3a26 + Saft* _ h\ (2) {bx^-2y^f, = (5x2)3 _ 3(5a;2)2(2y3) + 3(5 a;2)(2 2/3)2 _ (2^3)3^ = 125 a;« - 150 xy + 60a;y - 8 2/9. In like manner, a polynomial of three or more terms may be raised to any power by enclosing its term in paren- theses, so as to give the expression the form of a binomial. (3) (x3-2a:2 + 3a: + 4)2, = {(a;3-2a;2) + (3a; + 4)}2, = (V - 2 x^Y + 2(^3-2 a;2)(3 a; + 4) + (3 a; + 4)^, = ^6 _ 43j5 + 4a;4 + 6a;4 _ 4a^ - IGrc^ + 9a;2 + 24a; + 16, - jb6 - 4a^ + 10a;* - 4x3 - 7a;2 + 24a; + 16. Exercise 12. In the following expressions perform the indicated oper- ations : 1. i2ay. 4. {-Wcf. 7. {-ba'b'xj. 2. (3aV)». 5. (-a^5^c)*. 8. {Qa'b'cJ. 3. /2aW g {^a'h'Y g (-3aVy V3cV/ ' {^a'hj ' (Qa'bxj' (3a'a^y(Wxy ^j (4:x*yy . «y^)^ * (6 6V)Xa^Z>7* * (9xy)^ * (3 3/)^' 12. (:.+ 3)^ 15. (1-4.:)^ 18. (^-yj 13. {l-2xy. 16. (l-^y 19. (l + 3a:7. 14. (3-^y. 17. (l+?|-^: 20. g-^-J INVOLUTION AND EVOLUTION. 69 21. f2a'b'^-~X 23. (1 + Sx-x'y. 103. The nth root of a number is one of the n equal factors of that number. The operation of finding any required root of an expres- sion is called Evolution. Every case of evolution is merely an example of factor- ing, in which the required factors are all equal. Thus, the square, cube, fourth roots of an expression are found by taking one of the two, three, four equal factors of the expression. The symbol which denotes that a square root is to be extracted is ■>/ ; and for other roots the same symbol is used, but with a figure written above to indicate the root ; thus, -y/, -y/, etc., signifies the third root, fourth root, etc. 104. Index Law. If m and n are positive integers we have (§ 98), {cry = a»"». CoHsequently Va™^ = a*^. Thus, the cube root a^ is a^ ; the fourth root of 81 a^^ is 3 a' ; and 80 on. The above is the index law for evolution. Also, since (aby — a^'b'', therefore, Va*^ = ab = Vc^ Vb^. Hence, the root of a simple expression is found by divid- ing the exponent of each factor by the index of the root, and talcing the product of the resulting factors. 70 ALGEBRA. 105. It is evident from § 100 that I. Any even root of a positive number will have the double sign, ±. II. There can be no even root of a negative number. Thus, V— a;^ is neither + x nor — x, for (+ xy = + x"^, and (— xy = + x^. The indicated even root of a negative number is called an impossible, or imaginary, number. III. Any odd root of a number will have the same sign as the number. 106. Square Boots of Compound Expressions. Since the square of a -\- b is d^ -{- 2 ab -\- b"^, the square root of a'' + 2ab + b'isa-{-b. It is required to devise a method of extracting the square root a-{-b when d^ -\~2ab -{- b"^ is given. Ex. The first term, a, of the root is obviously the square root of the first term, a^, in the expression. a' + 2ab + b'' \a + b jf the a? be subtracted from the given ^ expression, the remainder is 2ab + b^. 2a + b 2ab + b'^ Therefore the second term, b, of the root 2ab + b'^ is obtained by dividing the first term of this remainder by 2 a, that is, by double the part of the root already found. Also, since 2ab + b'^ = (2 a + b)b, the divisor is completed by adding to the trial-divisor the new term of the root. The same method will apply to longer expressions, if care be isi^en to obtain the trial-divisor at each stage of the process, by doubling the part of the root already found, and to obtain the complete divisor by annexing the new term of the root to the trial-divisor. INVOLUTION AND EVOLUTION. 71 Find the square root of 1 + lOo;'^ + 25 a;* + 16a;« - 24a;^ - 20a? - 4a;. 16a«-24a;5 + 25a;^-20a^ + 10a:^~4a; + l |4a;^-3a;' + 2a;-l 16 »6 8x3 -"3^- 24x5 + 25 a;* 24x5+ 9 ^^4 8x3-6x2 + 2x 16x*- 20x3 + 10x3 16x*-12x3+ 4x2 8x3-6x2 + 4x-l - 8x3+ 6x2-4x + l - 8x3+ 6x2-4x + l The expression is arranged according to descending powers of x. It will be noticed that each successive trial-divisor may be obtained by taking the preceding complete divisor with its last term doubled. 107. Arithmetical Square Koots. In the general method of extracting the square root of a number expressed by figures, the first step is to mark ofi" the figures in periods. Since 1 = l\ 100 = 10^, 10,000 = 100^, and so on, it is evident that the square root of any number between 1 and 100 lies between 1 and 10 ; the square root of any number between 100 and 10,000 lies be- tween 10 and 100. In other words, the square root of any number expressed by one or two figures is a number of one figure ; the square root of any number expressed by three or four figures is a number of two figures ; and so on. If, therefore, a dot be placed over the units figure of a square num- ber, and over every alternate figure, the number of dots will be equal to the number of figures in its square root. Find the square root of 3249. 3249 (57 In this case, a in the typical form a^ -\-2ah -{-h^ 25 represents 5 tens, that is, 50, and h represents 7. 107) 749 The 25 subtracted is really 2500, that is a^, and the 749 complete divisor 2a + &is2x50 + 7 = 107. The same method will apply to numbers of more than two periods by considering a in the typical form to repre- sent at each step the part of the root already found. 72 ALGEBRA. 108. If the square root of a number has decimal places, the number itself will have twice as many. Thus, if 0.21 be the square root of some number, this number will be (0.21)2 _ 0.21 X 0.21 = 0.0441 ; and if 0.111 be the root, the number will be (0,111)2 _ 0.111 X 0.111 = 0.012321. Therefore, the number of decimal places in every square decimal will be even, and the number of decimal places in the root will be half as many as in the given number itself. Hence, if the given square number contains a decimal, and a dot be placed over the units' figure, and then over every alternate figure on both sides of it, the number of dots to the left of the decimal point will show the number of integral places in the root, and the number of dots to the right will show the number of decimal places. Ex. Find the square roots of 41.2164 and 965.9664. 4i.2i64(6.42 965.9664(31.08 36 9 124)521 61)65 496 61 1282)2564 6208)49664 2564 49664 It is seen from the dotting that the root of the first example will have one integral and two decimal places, and that the root of the second example will have two integral and two decimal places. 109. If a number contains an odd number of decimal places, or gives a remainder when as many figures in the root have been obtained as the given number has periods, then its exact square root cannot be found. We may, how- ever, approximate to the root as near as we please by annexing ciphers and continuing the operation. When a number contains an odd number of decimal places, we separate the decimal part into periods by placing a dot over the second decimal figure from the decimal point; another over the fourth figure from the decimal point ; and so on. INVOLUTION AND EVOLUTION. 73 Ex. Find the square roots of 3 and 357.357. 3.(1.732.... 1 27)200 189 357.3570(18.903 1 28)257 224 343)1100 1029 869)3335 3321 3462)7100 6924 37803)147000 113409 Exercise 43. Simplify : 1. -VUa'b*. 3. ah -\- h"^ to represent at each stage of the process the part of the root already found. 111. Arithmetical Cube Eoot. In extracting the cube root of a number expressed by figures, the first step is to mark it off into periods. Since 1 = l^, 1000 = lO^, 1,000,000 = 100^, and so on, it follows that the cube root of any number between 1 and 1000, that is, of any number which has one, two, or three figures, is a number of one figure; and that the cube root of any number between 1000 and 1,000,000, that is, of any number which has four, five, or six figures, is a number of two figures ; and so on. Hence, if a dot be placed over every third figure of a cube num- ber, beginning with the units' figure, the number of dots will be equal to the number of figures in its cube root. INVOLUTION AND EVOLUTION. 76 If the cube root of a number contains any decimal figures, the number itself will contain three times as many. * Hence, if the given cube number have decimal places, and a dot be placed over the units figure and over every third figure on both sides of it, the number of dots to the left of the decimal point will show the number of integral figures in the root ; and the number of dots to the right will show the number of decimal figures in the root. If the given number is not a perfect cube, zeros may be an- nexed, and a value of the root may be found as near to the true value as we please. 112. It is to be observed that if a denotes the first term of the root, and h the second term of the root, the first com- plete divisor is, and the second trial-divisor is 3 (a + b)^, that is, Sa'^ + eaS + SS^ which may be obtained from the preceding complete divisor by adding to it its second term and twice its third term. Ex. Extract the cube root of 5 to five places of decimals. 5.000(1.70997 1 3x102 = 300 3(10x7) = 210 72= 49 >! 559 \ 4000 3913 259 J ) ) L [ 87000000 3 X 17002 = 867000( 3(1700x9)= 4590( 92= 81 871598] 78443829 45981 ^ 3 x 17092 = 8762043 85561710 78858387 67033230 61334301 76 ALGEBRA. After the first two figures of the root are found, the next trial divi- sor is obtained by bringing down the sum of the 210 and 49 obtained in completing the preceding divisor ; then adding the three lines con- nected by the brace, and annexing two ciphers to the result. The last two figures of the root are found by division. The rule in such cases is, that two less than the number of figures already obtained may be found without error by division, the divisor to be employed being three times the square of the part of the root already found. Since the fourtli power is the square of the square, and the sixth power the square of the cube, the fourth root is the square root of the square root, and the sixth root is the cube root of the square root. In like manner, the eighth, ninth, twelfth roots may be found. Exercise 14. Extract the cube root of : 1. 21-\mx-\-U^x^-^^a?. 2. x^-2>3(f'-^bx^-2>x-l. 4. 1 - 6 a; -f- 21 rr'^ - 44 a:' + 63 x' - 54 :rH 27 x\ 5. 27 + 296a;^ - 125:r« - 108^ + 9a;^ - 15:r* - 2>00x^. 6. 12a;^-i?5_54a;-59 + ^ + 8a;^ + ^. X^ X x^ 7. 8a:'-36aa;^ + -' + ^^' + 66A-^'-63a'. of X ^ Extract to three places of decimals the cube roots of : 8. 517. 9. 1637. 10. 3.25. 11. 20.911. CHAPTEK VIII. EXPONENTS. 113. Positive Integral Exponents. If ti is a positive integer, we have, by definition, a"* — aX aX a to n factors. From this definition we have obtained the following laws, which hold true for positive integral exponents : I. a'" X a" =- a'"+". §32 11. >n, §41 «^- 1 if. >m. III. (a'-y .= a««. §98 IV. J^^mn ^ ^m^ §104 V. (ahy ■= a^"^. §98 From law I., and the commutative and associative prin- ciples, laws II.-V. may readily be obtained. 114. In the case of fractional and negative exponents, we proceed as follows : We assume laws I. and V. to hold for such exponents, and then proceed to investigate what meaning of fractional and negative exponents is consistent with these laws. It being assumed that laws I. and V. hold true, it is easily proved that laws II., III., and IV. must hold true. 78 ALGEBRA. 115. Positive Practional Exponents. If w is a positive in- teger, _ is a positive fraction. n "We have, by III., 1 1 any- = an =a^ = a', but (Va)" = a. .'. an = -\/a. Again, if m and n are both positive integers, by III., but (Va^f = or. :. a« = Va^. Hence, in a fractional exponent, the numerato'r indicates a power, and the denominator a root. 116. Negative Integral Exponents. Dividing d? successively by a in the ordinary manner, v^e have the series a\ a\ a, 1, I \ i- (1) a a^ a^ Dividing again by a by subtracting 1 from the exponent of the dividend, we have, since II. holds true, the series a^ a\ a\ a\ a-\ a-\ a-\ (2) Comparing (1) and (2), we see that n 1 1 -1- 9 i. 1 J- a" = l, a —-1 a ^^=—, a ^ = — a a a 117. Negative Fractional Exponents. If w is a positive in- teger, — - is a negative fraction, and we have n ^ ^ a but fxy= 1 =1. :.a-i=^=L EXPONENTS. 79 Again, if m and n are both positive integers, by III. >ut (^\=^-^ = l 1 . ^-^__ 1 _ 1 Hence, whether the exponent be integral or fractional, we have alwavs a~"'= — 118. From the application of these laws, we obtain p r pr P P P ^ 1^ 1 i „ - „*- "s/ab = (aby == a^b»= -\/a '\/b ; and so on. 119. Compound expressions are multiplied and divided as follows : (1) Multiply x^ + rr* yi + y^ ^J ^^ — x^y^ -\- y^. X + x^y'^ + a'y* ^2/* — x^y^ — x*y* + x'^y^ + x^y^ + y + x^y^ +y. (2) Divide ■(.»,. ... (^7(=)--. Find equivalent fractions with rational denominators for the following, and find their approximate values : 3 7 4-V2 6 15. 16. V7 + V5' 2V5-V6' I + V2' 5-2V6 2 . 1 . 7V5 . 7-2V3 + 3V2 V3' V5-V2' V7 + V3' 3 + 3V3-2V2 CHAPTER IX. QUADRATIC EQUATIONS. We now resume the subject of equations where we left it at the end of Chapter VI. Having considered equations of the first degree with one or more unknowns, we come next to the consideration of quadratic equations. 131. A quadratic equation which involves but one un- known number can contain only : (1) Terms involving the square of the unknown number. (2) Terms involving the first power of the unknown number. (3) Terms which do not involve the unknown number. Collecting similar terms, every quadratic equation can be made to assume the form ax^ -\- bx -j- c = 0, where a, h, and c are known numbers, and x the unknown number. If a, 5, c are given numbers, the equation is a mimerical quadratic. If a, b, c are numbers represented wholly or in part by letters, the equation is a literal quadratic. Thus, a;2 — 6a; + 5 = 0isa numerical quadratic, and ax^ + 2bx + S c — ab = is a, literal quadratic. 132. In the equation ax^ -}-bx-{-c = 0, a, b, and c are called the coefficients of the equation. The third term c is called the constant term. 88 ALGEBRA. If the first power of x is wanting, the equation is a pure quadratic ; in this case, & = 0. If the first power of x is present, the equation is an affected or complete quadratic. 133. Solution of Pure Quadratic Equations : (1) Solve the equation bx^ — ^S — 2x^. We have 5 a;^ - 48 = 2 a;^. Collect the terms, 3a;2 = 48. Divide by 3, x^ = 16. Extract the root, a; = ± 4. Observe that the roots are numerically equal, but one is positive and the other negative. There are but two roots, since there are but two square roots of any number. It may seem as though we ought to write the sign ± before the x as well as before the 4. If we do this, we have + a;= + 4, — «= — 4, +a; = — 4, — a;= + 4. From the first and second, a; = 4; from the third and fourth, » = — 4; these values of x are both given by x = ±4. Hence it is unnecessary, although perfectly correct, to write the ± sign on hoth sides of the reduced equation. (2) Solve the equation 3a;^— 15 = 0. We have 3a;2 = 15, or ic^ = 5. Extract the root, x = ±\/5, The roots cannot be found exactly, since the square root of 5 can- not be found exactly ; it can, however, be found as accurately as we please ; for example, it lies between 2.23606 and 2.23607. (3) Solve the equation ^x^ -\-lb = 0. We have 3 x^ = — 15, or x"^ = - 5. Extract the root, x = ±V— 5, There is no square root of a negative number, since any number, positive or negative, multiplied by itself, gives a positive result. QUADRATIC EQUATIONS. 89 The square root of — 5 differs from the square root of + 5 in that the latter can be found as accurately as we please, while the former cannot be found at all. 134. A root which can be found exactly is called an exact or rational root. Such roots are either whole numbers or fractions. A root which is indicated but can be found only approx- imately is called a surd or irrational root. Such roots involve the roots of imperfect powers. Exact and surd roots are together called real roots. A root which is indicated but cannot be found, either exactly or approximately, is called an imaginary root. Such roots involve the even roots of negative numbers. Exercise 18. Solve : 3 6 2"- ' 4:x 3 . 3 l-{-x 1 — x 8. 4. bx'-9 = 2x'-{-2A. x^ a:'^ - 10 _ >. 50 + x'' * 5 15 25 * g 3a;^-27 90 + 4a;^ _^ x' + S x'-{-9 7. 8. 4a;' + 5 2x'-5 _ 7x'-25 10 15 20 103:^ + 17 12a;' + 2 _ 5a;'^-4 18 liar' -8 9 9. x^ -j- bx -{- a = bx (I — bx). 10. ax'-^-h^c. 90 ALGEBRA. 11. x^ — ax-\-h=^ ax {x — 1). ah —X b — ex 12 13. 14. 15. h — ax be — X ?){x-\-a) 2x-j-a _ -. 4a; — a 2a-\-x 3a , x-}-4:a ^ 1 a''-\-2ax- x'' x — ba x-{-3a (x — ba)(x-{-Sa) 2(a + 2b) a-2x ^ b^ a-\-2x a-i-b (a+b)(a-{-2x) 135. Solution of Affected Quadratic Equations : Since (x =b bf is identical with x'^ ±2bx-{- b^, it is evi- dent that the expression x"^ ±2bx lacks only the third term 6^ of being a perfect square. This third term is the square of half the coefficient of x. Every affected quadratic may be made to assume the form x'^it2bx = e, by dividing the equation through by the co- efficient of x^ (§ 131). To solve such an equation : The first step is to add to both members the square of half the coeffieient of x. This is called completing the square. The second step is to extraet the square root of each mem- ber of the resulting equation. The third step is to reduee the two resulting simple equations. (1) Solve the equation x? — %x=^2^. We have a;^ - 8 a; = 20. Complete the square, x'^ — 8a; + 16 = 36. Extract the root, a; — 4 = ± 6. Reduce, a; = 4 + 6 = 10, or a; = 4-6 = -2. QUADRATIC EQUATIONS. 91 The roots are 10 and — 2. We write the ± sign on only one side of the equation, for the rea- son given after the first example of § 133, Verify by putting these numbers for x in the given equation: a;=10, x = -2, 102 - 8 (10) = 20, (-2)2 -8 (-2) = 20, 100 - 80 = 20. 4 + 16 = 20. (2) Solve the equation ^±i = i^^. Free from fractions, (x + l)(rr + 9) = (x-l)(4a;-3). Simplify, 3a;2-17a; = 6. Divide by 3, a;2_.y.x = 2. Complete the square, a;^ -^v-(fM- Extract the root. ^-f-f Reduce, a,_.17 19_36_g 6 6 6' _ ^ 17 19 2 1 '^" = -6-¥ = -6 = -3 The roots are 6 and Verify by putting these numbers for x in the original equation : x = f5. 6 + 1 ^ 24 6-1 6- 7_21 5 15 x = 1^ 3* 1 3 + 1 -1- 1 3 -1 -h« 2 3 3 13 3" 26 • 3 2 4 13 26* 92 ALGEBRA. 136. When the coefficient of x^ is not unity, we may pro- ceed as in the preceding section, or we may complete the square by another method. Since {ax db hy is identical with aV ±i 2 ahx + ^^ it is evident that the expression aV dz 2 ahx lacks only the third term h^ of being a perfect square. This third term is the square of the quotient obtained by dividing the second term by twice the square root of the first term. Every afiiected quadratic may be made to assume the form aV ± 2ahx-=c (§ 131). To solve such an equation : The first step is to complete the square ; to do this, we divide the second term hy twice the square root of the first term, square the quotient, and add the result to both mem- bers of the equation. The second step is to extract the square root of each mem- ber of the resulting equation. The third step is to reduce the two resulting simple equations. 137. Numerical Quadratics are solved as follows : (1) Solve the equation 1^ x^ -{- b x — ^ = 1 x^ — x -\- Ab. 16a;2 + 5a; - 3 - 7a;2 - a; + 45. Simplify, 9a;2 + 6x = 48. Complete the square, ^x^ -\-Qx + 1 = ^^. Extract the root, ' 3a; + l = ±7. Reduce, 3a; = _l + 7or— 1 — 7; 3a; = 6 or -8. .-. a; = 2 or - 2f . Verify by substituting 2 for x in the equation 16a;2 + 5a;-3 = 7a;2-a;+45. 16(2)2 + 5(2) - 3 = 7(2)2- (2) + 45^ 64 + 10 - 3 = 28 ~ 2 + 45, 71 = 71. QUADRATIC EQUATIONS. Verify by substituting — 2f for x in the equation 16x^ + 5x-3 = 7x'^-x + 4:5. 1024 40 o 448 , 8 , .. o = h - + 45, 9 3 9 3' 1024 - 120 - 27 = 448 + 24 + 405, 877 = 877. (2) Solve the equation Sx^ — 4:X = S2. Since the exact root of 3, the coefficient of x^, cannot be found, it is necessary to multiply or divide each term of the equation by 3 to make the coefficient of x^ a square number. Multiply by 3, 9 a;^ - 12 a; = 96. Complete the square, 90;^ — 12 a; + 4 = 100. Extract the root, 3 a; — 2 = ± 10. Eeduce, 3a; = 2 + 10 or 2 - 10; 3a! = 12or-8. .-. a; = 4 or — 2f . Or, divide by 3, ^2__4a;_32 3 3 Complete the square. 2 4a; 4 32 4 100 3 9 3 9 9 Extract the root. -h^f .... = 4L0. = 4or-2f. Verify by substituting 4 for X in the original equation. 48 - 16 = 32, 32 = 32. Verify by substituting — 2f for x in the original equation, 21i + (10i) = 32, 32 = 32. 94 ALGEBRA. (3) Solve the equation — 3a;^ + 5a; = — 2. Since the even root of a negative number is impossible, it is necessary to change the sign of each term. The resulting equation is 3a;2_5,T = 2. Multiply by 3, 9 a;^ - 15 a? = 6. Complete the square, 9x^ Extract the root, Reduce, Or, divide by 3, Complete the square. Extract the root, 15.+ 25^49. 4 4 2 2 3a. 2 . 3a; = 6 or - -1. .'. a; = 2 or - 1 3' x^ 5a;_2 3 3 5a; 25 49 " 3 36 36 5 7 ^-6 = "6- . ^ 5±7 = 2or- 1 "3* If the equation 3 a;'' — 5 a; = 2 be multiplied hj four times the coeffi- cient of x^, fractions will be avoided : 36a;2-60a; = 24. Complete the square, 36; r2-60a; + 25 = 49. Extract the root. 6a;-5 = ±7. 6a; = 5±7, 6a; =12 or -2. .•.. = 2or-|. It will be observed that the number added to complete the square by this last method is the square of the coefficient of x in the original equation 3 a;'' — 5 a; = 2, QUADRATIC EQUATIONS. 95 3 1 (4) Solve the equation = 2. ^ b — x 2x — b Simplify, 4aj2 _ 23 a; = - 30. Multiply by four times the coeflSicient of x^, and add to each side the square of the coefficient of x, 64 a;2 - ( ) + (23)^ = 529 - 480 = 49. Extract the root, 8 a; - 23 = ± 7. Reduce, 8a; = 23±7; So; = 30 or 16. .-. a; = 3| or 2. If a trinomial is a perfect square, its root is found by taking the roots of the first and third terms and connecting them by the sign of the middle term. It is not necessary, therefore, in completing the square, to write the middle term, but its place may be indicated as in this example, (5) Solve the equation 72a;^ - 30ar = - 7. Since 72 = 2' X 3^*, if the equation be multiplied by 2, the coeffi- cient of x^ in the resulting equation, 144 x'^ — 60 a; = — 14, will be a square number, and the term required to complete the square will be I ^ j = I _ I = _. Hence, if the original equation be multiplied by 4x2, the coefficient of x^ in the result will be a square number, and fractions will be avoided in the work. Multiply the given equation by 8, 576a;2-240a; = -56. Complete the square, 576 a;^ - () + 25 = — 3L Extract the root, 24 a; - 5 = ± V^^ . Reduce, 24 a; = 5 ± V^^^. .■.x = M5±V:^l). Note. In solving the following equations, care must be taken to select the method best adapted to the example under consideration. -, - Exercise 19. Solve : 1. x^-2x=lb. 3. x^-x = l2. 2. a^-Ux = -^S. 4. x'-Bx = 28, 96 ALGEBRA. 6. a;^- 13a; + 42 = 0. 9. 3a;^ - 19a; + 28 = 0. 6. a;^- 21a; + 108 = 0. 10. 4a;^ + 17a; - 15 = 0. 7. 2a;''' + a; = 6. 11. 6a;^-a;=12. 8. 4a;H7a; = 15. 12. 5a;'-^a^ + 4 = 0. 13. 6a;'^-7a; + 5 = 0. 14. ^!±i + (a;+l)(a; + 2) = 0. 15. (a;-57 + a;^-5 = 16(a; + 3). 3a;-19_ll + a; 16. ^ + 17. 6 ' 3 3 11 .T+1 2.r + 3 2 18 ^+ 1 , a;^ 11 X 6 2x -Q a;' — 4 , 2a; , 1 — 2a7 20. a; + ^i| = 2(a;-2). X ~ o 21. ^ ■ - 8 22. 2a; — 6 3 — a; x x + 2 4-a; _7 a;-l 2a; 3' 23. ^ §4-_^dlA=l- a;-2 2a; + l 24 ^ — ^ _L ^ — 4 ^ 1 a; + 4~^2(a;-l) 2 25 ^ + 1 , l-^ _ 2 a;'-4'^a; + 2 5(a;-2) QUADRATIC EQUATIONS. 97 26 ^-5 :r-8__ 80 1 27. -i^+ ^ 1^ ^ 28 3^ + 5 I x + ^_ x-\ x-\-2> x-2, x^-^ 29. ^+1 I :r + 2_ 2a; + 13 .r — 1 a; — 2 a:+l 30. 2^ + 3^ + 1^ = 4. a;-fl a; + 2 x—1 31. 32. 33. 3:^ + 2 ■ x-1 ^ix''-x-\-l) ._Q x-\-l l~x _ 4 9-4:^2 22: + 3 2a; -3 2^+1 2(£+l) _ 2-1 a; + 3"^a: + 2 ^^' 138. Literal Quadratics are solved as follows : (1) Solve the equation aoi^ -\-hx-\-c — 0. Transpose c, ax^ + hx = — c. Multiply the equation by 4 a and add the square of 6, 4aV + ( ) + 62 = 62 - 4ac. Extract the root, 2aa; + 6 = ± V62 - 4 ac. Reduce, 2ax = — h± Vb^ — 4 ac. — 6 ± V62 — 4 ac 2a (2) Solve the equation adx — acx^ = hex — bd. Transpose hex and change the signs, o^x'^ + hex — acZa; = 6(^. Express the left member in two terms, acx^ + (6c — ad) x = hd. 98 ALGEBRA. Multiply by 4ac, 4 a^cH"^ + 4:ac (be — ad) a; = 4 abed. Complete the square, 4a2cV + ( ) + (6c - adf = bH^ + 2abcd + a^^. Extract the root, 2aex + (6c — acZ) = ± (6c + acZ) Reduce, 2aca;-i — (6c — a^) ± (6c + ac?) = 2ac? or — 26c. d b .•. a; = - or c a (3) Solve the equation px^ —px-{- qx^ + qx Express the left member in two terms, pq {p + q)x-' - (p - q)x =^-^. Multiply by four times the coefficient of x"^, 4 (jp + qY rc^ — 4 (p^ — q^)x = 4:pq. Complete the square, 4:{p + qfx' -{) + {p-qf =p^ + 2pq + f. Extract the root, 2{p + q) x — (p — q) = ± {p + q). Reduce, ^P -^ 9)^ = {p- 9)-{p + 9), = 2p or —2q. - ^ or- ' p+q p+q Observe that the left member of the simplified equation must be expressed in two terms, simple or compound, the first term involving x^, the second involving x. ^ , Exercise 20. iSolve : x'-2ax---=Sa\ (a? + «)' _ (a? - «)' ,2 X S 2. x'' + 7a' = 8ax. 3. 4:x(x-a)-\-a':=b\ '^' ^'~a~4^^' 4. ^~~^ = 2aix+2a). ^' ^' - (^ + b)^ = -^b. , \ \, 9. x^-'^^±^x^l = 0. 5. x^ — ax-\-b. mn QUADRATIC EQUATIONS. 99 ^x{a — x) _a X a + 6_ a ?>a — 2x 4 * a — x x a — x 11. 2x^-{-^ = (a-\-b)x. ' 16. t.ZlA = ^±^. 2 x — b 2 12. (x-^mf-\-{x—my=bmx. a-\-b 2a-\-b_x Q^2 ' x — 2a a a 13. oaj2 + 5a^a; + ^-0. 14. b(a-xy = {b~l)x'. ' b''^ X 2b ' 19. ^^ =a + b-(a-b)x. ax — bx bab ~2>¥ — ax ^ 2a-\-x 2a-x 3 21. .--..^ (3^ + 2.)6 3(.^ + 5- X 2 4 22 3a ■ 2a _4a . . a x-\-a x-\-2a x x-{-Sa 22 a — b-{-x ■ a + 5 _^2 a + 6 + a; ic4-5 24 o^ + 4Z> g — 45 _46 a;4-26 £c — 25 a a-{- b X -\- a 26. (^a'-m(^' + l) = 2x. 4:a' + 9b' 27. (Sa' + b')(x'-x+l) = (a'+3b')(x' + x + l). ^a' b' 4:0" -b' 28 x + 2 x — 2 a;(4-ar') ft + 26 _ g'' 'W a -2b (a — 2b)x x^' 100 30. ALGEBRA. x+1 2 x+2 c ex ax — bx a — c X — a ?>h{x- -c) X — a a — c [a — c){x — a) 32. x{x^h''-h)^ax{a^V)-{a^ hf (a - b). ' 2~^ X ~ 2 ' + z = 0. m-^n \ mnj m n 35 2^5 (3a;-l)5^ ^ (2a;+l)a^ 3:r+l 2a;+l 3a; + l 36 ^ + 2a — 45 8b — la , rg — 4a _ 25a; ' a^-25a; ' 2(ab-2b') o^ 1 ^ ■ x~bb _ x-{-19b — 2a ' a-j-2b a'-U^ (a + 2b)x~ 2bx-ax 3g a-2b 2(:r + 4a + 35) _Q x-{-2b x-ba + Sb 39 ^ + 35 35 _ a + Sb 8a'-12ab 4:a'-9b'' (2a + 35) (rr- 35)' 40 1 I 1 ^ Q^ , 2bx-\-b 2x^-\-x~\ 23?-%x^\ 2bx-b a-ax^' 4j 1 4aa;'^4-35(2-^) _o a; 2aa:' + 2a+35 42 ^ — Q^ I 2 (a5 — g^ + 2 5^) _ 1 25 (.2; + a) a(a; + a)^ a 2ax + b 2aa;-5 _^ 95V+(4a^-65^)a;-(a^ + 5^) aa: + 5 ax — b a V — 5^ a: + a + ^ . 3(a + c) ^o a; — 3a + 5 a7 + 5 + c 43. 44. QUADRATIC EQUATIONS. 101 139. Solutions by a Formula. Every affected quadratic may be reduced to the form x^ -\-px + g' = 0, in which p and q represent numbers, positive or negative, integral or frac- tional. Solve : x^ -{- px -\- q = 0. ^x'^()-\-p^=p^-4:q, 2x -{-p = d= 'Vp'^ — 4:q. By this formula, the values of x in an equation of the form a;^ +^a; -f 5- = 0, may be written at once. Thus, take the equation Divide by 3, a;2-|a;+| = 0. o Here, p = - 1 and ^ = | 5 + l-x/25 6 2>9 3 6 6 = 1 or — 3 140. Solutions by Factoring. A quadratic which has been reduced to its simplest form, and has all its terms written on one side, may often have that side resolved hy inspection into factors. In this case the roots are seen at once without complet- ing the square. (1) Solve x''-\-1x-m = 0. Since a;^ + 7a; -60 = (a; + 12) (a; -5), the equation x'^ + 7 a; — 60 = may be written (x + 12) (x — 5) = 0. 102 ALGEBRA. It will be observed that if either of the factors a; + 12 or a; — 5 is 0, the product of the two factors is 0, and the equation is satisfied. Hence, a; + 12 = 0, or ic — 5 = 0. .'. « = — 12, or a; = 5, (2) Solve x''-\-1x = 0. The equation a;^ + 7 a; = becomes a;(a; + 7) = 0, and is satisfied if a; = 0, or if a; + 7 = 0, .'. the roots are and — 7. It will be observed that this method is easily applied to an equa- tion all the terms of which contain x. (3) Solve 2x^-x^-Qx^0. The equation 2x^ -x^-Qx = becomes a; (2 a;^ - a; - 6) = 0, and is satisfied if a; = 0, or if 2x'^ — x — Q> = 0. q By solving 2 a;^ - a; - 6 = the two roots 2 and — - are found. q .-. the equation has three roots, 0, 2, — -• (4) Solve x^ + x^-4:X-4. = 0. The equation «^ + a;^ — 4a; — 4 = becomes x'^{x + 1) - 4(a; +' 1) = 0, (a;2 - 4) [x + 1) = 0. .•. the roots of the equation are — 1, 2, — 2. (5) Solve a;' -2:^2- 11a; + 12 = 0. By trial we find that 1 satisfies the equation, and is therefore a root (§ 84). Divide by a; — 1 ; the given equation may be written (a; _ 1) (a;2 _ a; _ 12) = 0, and is satisfied if a; — 1 = 0, or if a;'^ — a; — 12 = 0. The roots are found to be 1, 4, — 3. (6) Solve the equation x{x'^ — 9) — a(a'^ — 9). If we put a for x, the equation is satisfied; therefore a is a root (g 84). QUADRATIC EQUATIONS. 103 Transpose all the terms to the left member, and divide hj x — a. The given equation may be written {x - a) {x^ + aa; + a2 - 9) = 0, and is satisfied if re — a = 0, or if o;^ + ax + a'^ — 9 = 0. The roots are found to be -a + V36-3a'^ -a- V36-3a^ ''• 2 ' 2 ' Exercise 21. Find all the roots of : ,1. {x-l){x~2){x'-^x-{-^) = 0. 2. {x^-2x-\-2){o(^-^x + 1) = Q, 3. a;' + 27 = 0. 4. a;* -81=0. 5. a;^-27 + 4(a;^-9) = 0. 6. a;* + 9a;2— 16(a;H9) = 0. 7. 2a;' + 3a;^-2a;-3 = 0. 8. x'-^c(^-\-M-2>2x-=0. 9. x^-x-Q = 0. 10. a^-e>x^-\-llx-^ = 0. 11. a;*-3a;^ — 8a;' + 6a; + 4=:0. 12. a;' + a;'-r4a;-24 = 0. 13. a;*-6a;' + 9a;^ + 4a;-12 = 0. 14. a;(a;-3)(a;+l) = a(a-3)(a + l). 15. a;(a;-3)(a; + l) = 20. 16. (a:-l)(a;-2)(a;-3) = 24. 17. (a; + 2) (a: -3) (a; + 4) = 240. 18. (a;+l)(a; + 5)(a;-6) = 96. 104 ALGEBRA. 141. Character of the Eoots. Every quadratic equation can be made to assume the form ax^ -{-hx-\- c^=0. Solving this equation (§ 138, Ex. 1), we obtain for its two roots h + V^' - 4ac —h — V^' - 4 ac 2a 2a There are two roots, and but two roots, since there are two, and but two, square roots of the expression b'^ — 4a 0, roots real and different ; II. b^ — 4:ac = 0, roots real and equal ; III. 6^ — 4ac < 0, roots imaginary. 142. By calculating the value of i^ — 4 ac we can deter- mine the character of the roots of a given equation without solving the equation. Examples : (1) x'-5x + 6==0. Here a = l, b = — 5, c = 6. 62-4ac = 25-24 = l. The roots are real and different, and exact. (2) Sx'-{-1x-l = 0. Here a = 3, 6 = 7, c = — 1. 62-4ac = 49 + 12 = 61. The roots are real and different, and are both surds. (3) 42:^-12a; + 9 = 0. Here a = 4, 6 = -12, c = 9. 62 _4ac = 144 -144 = 0. The roots are real and equal. (4) 2x'-,Sx-{-4: = 0. Here a = 2, 6 = - 3, c = 4. 62_4ac = 9-32 = -23. The roots are both imaginary. 106 ALGEBRA. (5) Find the values of m for which the equation 2 wa;"^ + (5m + 2)a; + (4m + 1) = has its two roots equal. Here a=2m, 6 = 5m + 2, c = 4m+l. If the roots are to be equal, we must have ¥ — 'iac-=0, or (5m + 2)2-8m(4m + l) = 0. 2 This gives m = 2 or For these values of m the equation becomes 4a;2 + 12a; + 9 = 0, and 4a;2 - 4a; + 1 = 0, each of which has its roots equal. Exercise 22. Determine, without solving, the character of the roots of each of the following equations : 1. .2r^-6a; + 8 = 0. 6. 16:c^-56:r + 49 = 0. 2. x'-4:X-\-2 = 0. 7. 2>x^~2x-\-l2~-=0. 3. a;2 + 6:r+13 = 0. 8. 2x'' -I9x +11 = 0. 4. ^x''-\2x-{-1 = 0. 9. 9a;' + 30a: + 25 = 0. 5. 5a;^-9a; + 6 = 0. 10. 17a;'- 12a: + 5^ = 0. Determine the values of m for which the two roots of each of the following equations are equal : 11. (3m + l)a;' + (2m + 2)a; + m = 0. 12. (m-2)a;' + (m-5)a;4-2m-5 = 0. 13. 2ma;'-fa;'-6ma;-6a; + 6m + l = 0. 14. ma;' + 2a;' + 2m = 3ma;- 9a; +10. QUADRATIC EQUATIONS. ' 107 143. Problems involving Quadratics. Problems which in- volve quadratic equations apparently have two solutions, as a quadratic equation has two roots. When both roots are positive integers, they will give two solutions. Fractional and negative roots will in some problems give solutions ; in other problems they will not give solutions. No difficulty will be found in selecting the result which belongs to the particular problem we are solving. Some- times, by a change in the statement of the problem, we may form a new problem which corresponds to the result that was inapplicable to the original problem. Imaginary roots indicate that the problem is impossible. (1) The sum of the squares of two consecutive numbers is 481. Find the numbers. Let X = one number, and a; + 1 = the other. Then, aj" + (a; + l)^ = 481, or 2a;2 + 2a; + 1 = 481. The solution of which gives a; = 15 or — 16. The positive root 15 gives for the numbers, 15 and 16. The negative root —16 is inapplicable to the problem, as consecu- tive numbers are understood to be integers which follow one another in the common scale, 1, 2, 3, 4.... (2) What is the price of eggs per dozen when 2 more in a shilling's worth lowers the price 1 penny per dozen ? Let X = number of eggs for a shilling. Then, - = cost of 1 egg in shillings, 12 and — = cost of 1 dozen in shillings. But if a; + 2 = number of eggs for a shilling, 12 = cost of 1 dozen in shillings. a; + 2 ^ 12 12 1 .-. = — (1 penny being -^ of a shilling). X a? -f- ^ IZ 108 * ALGEBRA. The solution of which gives a; = 16 or — 18. And, if 16 eggs cost a shilling, 1 dozen will cost || of a shilling, or 9 pence. Therefore the price of the eggs is 9 pence per dozen. If the problem be changed so as to read : What is the price of eggs per dozen when 2 less in a shilling's worth raises the price 1 penny per dozen ? the algebraic statement will be 12 _ 12 _ _1_ a; -2 a? 12" . The solution of which gives re = 18 or — 16. Hence, the number 18, which had a negative sign and was inap- plicable in the original problem, is here the true result. Exercise 23. 1. The product of two consecutive numbers exceeds their sum by 181. Find the numbers. 2. The square of the sum of two consecutive numbers ex- ceeds the sum of their squares by 220. Find the numbers. 3. The difference of the cubes of two consecutive num- bers is 817. Find the numbers. 4. The difference of two numbers is 5 times the less, and the square of the less is twice the greater. Find the numbers. 5. The numerator of a certain fraction exceeds the de- nominator by 1. If the numerator and denominator be interchanged, the sum of the resulting fraction and the original fraction is 2-^. What was the original fraction ? 6. The denominator of a certain fraction exceeds twice the numerator by 3. If S^^j be added to the fraction, the resulting fraction is the reciprocal of the original fraction. Find the original fraction. QUADRATIC EQUATIONS. 109 7. A farmer bought a number of geese for $24. Had he bought 2 more geese for the same money, he would have paid f of a dollar less for each. How many geese did he buy, and what did he pay for each? State the problem to which the negative solution applies. 8. A laborer worked a number of days, and received for his labor $36. Had his wages been 20 cents more per day, he would have received the same amount for 2 days' less labor. What were his daily wages, and how many days did he work ? State the problem to which the negative solution applies. 9. For a journey of 336 miles, 4 days less would have sufficed had I travelled 2 miles more per day. How many days did the journey take ? State the problem to which the negative solution applies. 10. A farmer hires a number of acres for $420. He lets all but 4 for $420, and receives for each acre $2.50 more than he pays for it. How many acres does he hire ? 11. A broker sells a number of railway shares for $3240. A few days later, the price having fallen $9 per share, he buys, for the same sum, 5 more shares than he had sold. Find the number of shares transferred on each day, and the price paid. 12. A man bought a number of sheep for $300. He kept 15, and sold the remainder for $270, gaining half a dollar on each sheep sold. How many sheep did he buy, and what did he pay for each ? 13. The length of a rectangular lot exceeds its breadth by 20 yards. If each dimension be increased by 20 yards, the area of the lot will be doubled. Find the dimensions of the lot. 110 ALGEBRA. 14. Twice the breadth of a rectangular lot exceeds the length by 2 yards ; the area of the lot is 1200 square yards. Find the length and breadth. 15. Three times the breadth of a rectangular field, of which the area is 2 acres, exceeds twice the length by 8 rods. At $5 per rod, what will it cost to fence the ^eld? 16. Two pipes running together fill a cistern in 10-f- hours ; the larger will fill the cistern in 6 hours less time than the smaller. How long will it take each, running alone, to fill the cistern ? 17. Three workmen, A, B, and C, dig a ditch. A can dig it alone in 6 days more time, B in 30 days more time, than the time it takes the three to dig the ditch together ; C can dig the ditch in 3 times the time the three dig it in. How many days does it take the three, working together, to dig the ditch ? 18. A cistern holding 900 gallons can be filled by two pipes running together in as many hours as the larger pipe brings in gallons per i^inute ; the smaller pipe brings in per minute one gallon less than the larger pipe. How long will it take each pipe by itself to fill the cistern ? 19. A number is formed by two digits, the second being less by 3 than one-half the square of the first. If 9 be added to the number, the order of the digits will be re- versed. Find the number. 20. A number is formed by two digits ; 5 times the second digit exceeds the square of the first digit by 4. If 3 times the first digit be added to the number, the order of the digits will be reversed. Find the number. QUADRATIC EQUATIONS. Ill 21. A boat's crew row 3 miles down a river and back again in 1 hour and 15 minutes. Their rate in still water is 3 miles per hour faster than twice the rate of the current. Find the rates of the crew and the rate of the current. 22. A jeweller sold a watch for $22.75, and lost on the cost of the watch as many per cent as the watch cost dollars. What was the cost of the watch ? 23. A farmer sold a horse for $138, and gained on the cost "I" as many per cent as the horse cost dollars. Find the cost of the horse. 24. A broker bought a number of $100 shares, when they were a certain per cent below par, for $8500. He afterwards sold all but 20, when they were the same per cent above par, for $9200. How many shares did he buy, and what did he pay for each share ? 25. A drover bought a number of sheep for $110; 4 having died, he sold the remainder for $7.33i^ a head, and made on his investment four times as many per cent as he paid dollars for each sheep bought. How many sheep did he buy, and how many dollars did he make? 26. A certain train leaves A for B, distant 216 miles ; 3 hours later another train leaves A to travel over the same route ; the second train travels 8 miles per hour faster than the first, and arrives at B 45 minutes behind the first. Find the time each train takes to travel over the route. 27. A coach, due at B twelve hours after it leaves A, after travelling from A as many hours as it travels miles per hour, breaks down ; it then proceeds at a rate 1 mile per hour less than half its former rate, and arrives at B three hours late. Find the distance from A to B. CHAPTER X. SIMULTANEOUS QUADRATIC EQUATIONS. Quadratic equations involving two unknown numbers require different methods for their solution, according to the form of the equations. 144. Case I. When from one of the equations the value of one of the unknown numbers can be found in terms of the other, and this value substituted in the other equation. Ex. Solve: ,_yU \ (2) Transpose x in (2), y = x — 2. Substitute in (1), 3 a;^ - 2 x {x-2) = 5. The solution of which gives x =1 or — 5. .-.y = — 1 or — 7. Special methods often give more elegant solutions than the general method by substitution. I. When equatiojis have the form, x±:y = a, and xy = b ; a;2 -t y2 — ^^ Q^.y^d rf-y ~h ; or, x±y = a, and x^ -\-y'^^= h. (i) Solve: " + ^,^.''1 S Square (1 ), x' + 2xy^y'' = 1600. (3) Multiply (2) by 4, ^xy =1200. (4) Subtract (4) from (3), x^-2xy-vy'^ = 400. (5) Extract root of each side, x — y = ±20. (6) Add (6) and (1), 2 a! = 60 or 20, Subtract (6) from (1), 2y = 20 or 60. .;. a; = 30 > or a; = 10 1 2/ = 10 J 3/ = 30/' SIMULTANEOUS QUADRATICS. 113 t (2) Solve: 1~^ = !J ,^ol Square (1), Subtract (2) from (3), Subtract (4) from (2), Extract the root, -2xy=-2\. x^ ^2xy +2/2 = 64. a; + 2/ = ± 8. (3) (4) (5) By combining (5) and (1), cc = 6> ora; = -2J (3) Solve : x^ y 20 1,1 41 x^'^ f 400>' (1) (2) Square (1), 1 + 2 1^81. x' xy 2/2 400 (3) Subtract (2) from (3), 2 40 xy 400 (4) Subtract (4) from (2), 1 2 1 1 a;2 xy 3/^ 400" • Extract the root,^ x y 20 (5) By combining (1) and : (5), a; = 4") or a; = 5') II. When one equation may he simplified by dividing it by the other. (4) Solve: ^ + 2/ = 9lT 0) ^ ' x+ v=7 i (2) Divide (1) by (2), «2 -xy + y'^ = 13. (3) Square (2), x^ + . 2rc2/+y2 = 49. (4) Subtract (3) from (4), Sxy = 36. Divide by - 3, -xy = -12. (5) Add (5) and (3), x'' - 2xy + y^ = l. Extract the root, x — y = ±l. ' (6) By combining (6) and (2), a; = 4 > or y = si x = = 3 = 4 y = 114 ALGEBRA. 145. Case II. When each of the two equations is homo geneous and of the second degree. Solve 3/2-^:^=16 Let y = vx, and substitute vx for y in both equations. From (1 ), 2 V V - 4 va;2 + 3 a;2 = 1 7, . ..^ 17 2i,2_4v + 3 From (2), ^2^2 -a;2 = 16, Equate the values of x"^, 17 16 2v^ _4v + 3 v2_i 32i-2- Giv + 48 = 17^2_i7^ 15t;2- 64?; = -65. The solution gives, 5 13 ^ = -or-. 5 v = -> 3 5' 5a; 13a; y-vx^-. y = vx- ^. Substitute in (2), Substitute in (2), 25^' x^ = 16, 169-^ .-16. 9 25 a;2 = 9, .^=25, a; = ± 3, 9 5a; f, a; = ±5, 2/ = Y = ^5. 3 13a; 13 •^5 3 (1) (2) 146. Case III. When the two equations are symmetrical with respect to x and y ; that is, when x and y are simi- larly involved. Thus, the expressions 2a;^ + Sx'^y^ + 2y^, 2xy — 3x — Sy + I, a;* — Sx^y — 3 xy^ + y*, are symmetrical expressions. In this case the general rule is to combine the equations in such a manner as to remove the highest powers of x and y. SIMULTANEOUS QUADRATICS. 115 (1) Solve: X -\-y =12 j (1) (2) Divide (1) by (2), a;2 rry+Z- /• (3) To remove x^ and /, square (2), x2 + 2a;2/ + 2/2=144. (4) Subtract (4) from (3), 3.3/ = ^f 144, which gives We now have, a;y = 32. a;.+ 2/ = 12l (ry = 32 » Solving as in Case I., we find, a; = 8') or a; = 4) (2) Solve: x' + y' = 2>Z1 \ (1) a;+y=7 j (2) To remove aj* and 3/*, raise (2) to the fourth power, x^ + 4x^2/ + 6icV + 4ic?/3 + 2/* = 2401, Subtract (1), x^ + y' = 337, ^x^y + Q x^y'^ + 4 0:3/3 = 2064. Divide by 2, 2 x^^ + 3 x^y'^ + 2 0:3/3 = 1032. (3) Square (2) and multiply the result by 2xy, 2 0:33/ + 4 .23/2 ^2xf = 98 xy. (4) Subtract (4) from (3), - xy = 1032 - 98 xy, or xV- 98x3/ = -1032. This is a quadratic equation, with xy for the unknown number. Solving, we find, xi/ = 12 or 86. We now have to solve the two pairs of equations, x + y= 7) X + y = 7 ) X3/ = 12 / ' X3/ = 86 ) From the first, x = 4l or x = 3"» y = 3i y = 4i From the second, _ 7 ± V- 295 ^ y = 7t\/^^295 116 ALGEBRA. The preceding cases are general methods for the solution of equa- tions which belong to the kinds referred to ; often, however, in the solution of these and other kinds of simultaneous equations involv- ing quadratics, a little ingenuity will suggest some step by which the roots may be found more easily than by the general method. Exercise 24. 1. a; + y= 8) 13. x"-^ xy^^O')^ xy-=lb)' 2^ -32/= 1 3 2. :r + y = 6) 14. :?;'-- y' = 13 j xy-\-21 = 0)' 307 - 2y = 9 J ^-^= H. 15. i-fl^-^ a;y = 24 ) x y IS 4. x-y = ie>\ xy=b^ X y m xy -\- 60 xy = 18 8. 9. 16. 1-1 = 1 17. x^ + 4:y + ll = 6. 2x + Sy = l} Sx +2y+ 1 = ^ 18. x + Sy + l = -\ 7. y = 9 — 3a:) v Ay-l-l (■ . ^y -> ^ ' :r-f 2y a; + 2y=12) 19. ^r^ + y^^lOe) xy -f- 3/2 = 35 j a:?/ = 45 3 a;-3y-f9 = 01 20. x' +y' =52} ^y~ y' + 4 = 0r :ry + 24= 03 10. a;'^ + y'^ = 100 ) 21. x^ — xy = 3) X -\-y = 14: ) y^ ~{-xy= 10 ) 11. x' + y^ = 11)^ 22. xy +y'= 4) 3 2a;^-v' = 17'3 4a7 +y =15 3 2x'-y'' 2^:2 - y2 + 8 = ) 23. a;^ + 3:ry Bx —y — 2 = 0J ^y— y^ SIMULTANEOUS QUADRATICS. 117 = 60) = 40 j ;. X'' — 4:X^ = 4:6) v^ — XV = 6 J 24. x^ -]- xy 7/ + X1/ 25. x^-{-2xi/- y' = 28 Sx^+2xi/+2f = 12 26. 30. 8:^" — Srcy 27. a;' + 3a;y = 55) 2y2_|_ rry^lSJ 28. a;'— xyi-y'' = S7) a;2 + 2a;y + 8 = Oj 29. x^-}-xy + 2y' = U 2x^- xy+ 3/2=16 - 2/^ = 401 }■ 9:^2+ :ry + 2y'^ = 60 31. Sx'-{-3xy+ 7/ bx^ + 7xy + 4:y^ 52 140 !^ 34. x' + y^ = Q5} X +y = 53 35. a^-y^==98} X -y = 23 36. a,'^ + y^ = 2791 X -j-y = 33 37. a;^-.3/' = 218) a;-y= 23 38. x''-xy + 7/= 19 3 39. a;' -3/^ = 1304) :i;' + :ry + 2/'= 163 3 40. x' + f =91} xy(xi-y) = 8^\ 32. 4a;' + 3:ry + 52/'' = 27) 7a;'' + 5a:y + 93/' = 47r 33. 5a:' + 3a;3/ + 2y'^ = 188) x'^ — xy ■\- y'^= 19 3 41. x^-y^ = 9S- 42. 43. 44. 118 ALGEBRA. 45. x'-y'=1xy\ x-y= 2 i 57. :^'^ + y^ = ^3/ + 19) X +y =xy- 7 3 46. 58. x-y ' ^ + y 3 [• 47. bxy 59. a;^ + a;y,+ 2/* = 133) .^^-a:y +y^=- 19 3 a;* + a;y + y^ = 931 ) ;r^ + ^y +y^= 49 3 60. 48. xY-l^ xy-{-m^ ;i x-\-y = 61. x"^ a;y + y2=84), a; + -\Jxy + 2/ = 6 3 49. xy = ^xy + \2- ^ xy = x-\-y-\-l . 62. :r +y =21 + V^3 " 60. -^ + /= 36 1 63. a;^ + y^ = 49-:ry 3 -+^=^' J 51. x^^f = m-xy\ ^^' 2x^+Sxy+12 = Sf ,+y = xy~5\' 3.+5y+ 1 = 52. :.^ + y^ = l-3.;yj 65. - + |=1 x' + f = xy+S7 ) ^ ^ 1 +y =-. 23 53. a;* + y*=706) - + " = 4 X +y ah X y QQ. x-\-y = a ' } 54. x^-y-211) Jy=.a?-w\- X —y = 1 j 55. x^ + y' = 3368-) ''• ^; = «^+*y|. 1 1 ^3 I 69. a;^ + 2/= + 2; + y = 18 a; y 4 '' a;y = 6 SIMULTANEOUS QUADRATICS. 119 71. a^-\-f = 2xY-lb^ X +2/ =^xy-\-\ 3 72. ay'^-\-hxy- 74. 75. 76. rc'^ + y^ = ^^3/ 1 77. 2(a;'^ + 2/')-=5:py-9a5 ) ah)] ^{a^V){x^y)^Z{xy-ah) 78. a;^+ y' + 2' = 49 2a; +3y -4z = 79._ xy -\- yz -\- xz = ^^ ) * 4a; = 32/ = 2z + 43 80. a;^ + y' + z^ = 84- = 49-) = 11 = 63 X -\-y +z y"^ — xz 81. 2a:y + a; + y = 22 22/z+y 2a:2: + »4-\ 14 . a;z J a; + y = 22^ y 4- z -^ 58 [■ . a; + z-=323 \ah\ ^ 3 82. a;^ + ^y + ^2 = d^ y''-\-yz + xy=2ah z^ -\- xz -\- yz — h^ 120 ALGEBRA. Exercise 25. 1. If the length and breadth of a rectangle were each increased 1 foot, the area would be 48 square feet ; if the length and breadth were each diminished 1 foot, the area would be 24 square feet. Find the length and breadth of the rectangle. 2. A farmer laid out a rectangular lot containing 1200 square yards. He afterwards increased the width Ij yards and diminished the length 3 yards, thereby increasing the area by 60 square yards. Find the dimensions of the original lot. 3. The diagonal of a rectangle is 89 inches ; if each side were 3 inches less, the diagonal would be 85 inches. Find the area of the rectangle. 4. The diagonal of a rectangle is 65 inches ; if the rectangle were 3 inches shorter and 9 inches wider, the diagonal would still be 65 inches. Find the area of the rectangle. 5. The difference of two numbers is f of the greater, and the sum of their squares is 356. Find the numbers. 6. The sum, the product, and the difference of the squares of two numbers are all equal. Find the numbers. Hint. Represent the numbers hj x +y and x — y. 7. The sum of two numbers is 5, and the sum of their cubes is 335. Find the numbers. 8. The sum of two numbers is 11, and the cube of their sum exceeds the sum of their cubes by 792. Find the numbers. 9. A number is formed by two digits. The second digit, is less by 8 than the square of the first digit ; if 9 times SIMULTANEOUS QUADRATICS. 121 the first digit be added to the number, the order of the digits will be reversed. Find the number. 10. A number is formed by three digits, the third digit being the sum of the other two ; the product of the first and third digits exceeds the square of the second by 5. If 396 be added to the number, the order of the digits will be reversed. Find the number. 11. The numerator and denominator of a certain fraction are each greater by 1 than those of a second fraction ; the sum of the two fractions is ^. If the numerators were interchanged, the sum of the fractions would be f . Find the fractions. 12. There are two fractions. The numerator of the first is the square of the denominator of the second, and the numerator of the second is the square of the denominator of the first ; the sum of the fractions is •^, and the sum of their denominators 5. Find the fractions. 13. The sum of two numbers which are formed by the same two digits is -f|- of their difference ; the difierence of the squares of the numbers is 3960. Find the numbers. 14. The fore wheel of a carriage turns in a mile 132 times more than the hind wheel ; if the circumference of each were increased 2 feet, the fore wheel would turn only 88 times more. Find the circumference of each wheel. 15. Two travellers, A and B, set out from two distant towns, A to go from the first town to the second, and B from the second town to the first, and both travel at uni- form rates. When they meet, A has travelled 30 miles farther than B. A finishes his journey 4 days, and B 9 days, after they meet. Find the distance between the towns, and the number of miles A and B each travel per day. 122 ALGEBRA. 16. Two boys run in opposite directions around a rec- tangular field, of which the area is one acre ; they start from one corner, and meet 13 yards from the opposite corner. One boy runs only f as fast as the other. Find the length and breadth of the field. 17. A man walks from the base of a mountain to the summit, reaching the summit in 5j hours; during the last half of the distance he walks 5 mile less per hour than during the first half. He descends in 3 1 hours, walking 1 mile per hour faster than during the first half of the ascent. Find the distance from the base to the summit and the rates of walking. 18. A besieged garrison had bread for 11 days. If there had been 400 more men, each man's daily share would have been 2 ounces less ; if there had been 600 less men, each man's daily share could have been increased by 2 ounces, and the bread would then have lasted 12 days. How many pounds of bread did the garrison have, and what was each man's daily share ? 19. Three students. A, B, and 0, agree to work out a set of problems in preparation for an examination ; each is to do all the problems. A solves 9 problems per day, and finishes the set 4 days before B ; B solves 2 more problems per day than 0, and finishes the set 6 days before C. Find the number of problems in the set. 20. A cistern can be filled by two pipes ; one of these pipes can fill the cistern in 2 hours less time than the other ; the cistern can be filled by both pipes running to- gether in 1|- hours. Find the time in which each pipe wiU fill the cistern. SIMULTANEOUS QUADRATICS. 123 21. A and B have a certain manuscript to copy between them. At A's rate of work he would copy the whole man- uscript in 18 hours ; B copies 9 pages per hour. A finishes his portion in as many hours as he copies pages per hour ; B is occupied with his portion 2 hours longer than A is with his. Find the number of pages copied by each. 22. A and B have 4800 circulars to stamp, and intend to finish them in two days, 2400 each day. The first day A, working alone, stamps 800, and then A and B stamp the remaining 1600, A working altogether 3 hours. The second .day A works 3 hours and B 1 hour, and they ac- complish only y^^ of their task for that day. Find the number of circulars each stamps per minute, and the num- ber of hours B works on the first day. 23. A, in running a race with B, to a post and back, meets him 10 yards from the post. To come in even with A, B must increase his pace from this point 41-f- yards per minute. If, without changing his pace, he turns back on meeting A, he will come in 4 seconds behind A. Find the distance to the post. 24. A boat's crew, rowing at half their usual speed, row 3 miles down stream and back again, accomplishing the distance in 2 hours and 40 minutes. At full speed they can go over the same course in 1 hour and 4 minutes. Find the rate of the crew and of the current. 25. A farmer sold a number of sheep for $286. He received for each sheep $2 more than he paid for it, and gained thereby on the cost of the sheep ^ as many per cent as each sheep cost dollars. Find the number of sheep. 26. A person has $1300, which he divides into two parts and loans at diff'erent rates of interest, in such a 124 ALGEBRA. manner that the two portions produce equal returns. If the first portion had been loaned at the second rate of interest it would have yielded annually $36; if the second portion had been loaned at the first rate of interest it would have yielded annually $49. Find the two rates of interest. 27. A person has $5000, which he divides into two portions and loans at different rates of interest in such a manner that the return from the first portion is double the return from the second portion. If the first portion had been loaned at the second rate of interest it would have yielded annually $ 245 ; if the second portion had been loaned at the first rate of interest it would have yielded annually $90. Find the two amounts and the two rates of interest. 28. A number is formed by three digits ; 10 times the middle digit exceeds the square of half the sum of the three digits by 21 ; if 99 be added to the number, the digits will be in reverse order ; the number is 11 times the number formed by the first and third digit. Find the number. 29. A number is formed by three digits ; ^ the sum of the last two digits is the square of the first digit ; the last digit is greater by 2 than the sum of the first and second ; if 396 be added to the number, the digits will be in re- verse order. Find the number. 30. A railroad train, after travelling 1 hour from A, meets with an accident which delays it 1 hour ; it then proceeds at a Tate 8 miles per hour less than its former rate, and arrives at B 5 hours late. If the accident had happened 50 miles further on, the train would have been only 3|- hours late. Find the distance from A to B. CHAPTER XI. EQUATIONS SOLVED LIKE QUADRATICS. 147. Some equations not of the second degree may be solved by completing the square. (1) Solve: 8x'-i-6Sx'=8. This equation is in the quadratic form !f we regard x^ as the un- known number. We have, 8x« + 63a;3=8. Multiply by 32 and complete the square, 256a;«+() + (63)2=4225. Extract the square root, 16ar^ + 63 = ± 65. Hence, a^ = ^ or — 8. Extracting the cube root, two values of x are ^ and — 2. To find the remaining roots, it remains to solve completely the two equations We have, 8a;3_i_o, or, (2a;-l)(4ic2 + 2a; + l) = 0. .'. either 2 a; — 1 = 0, or, 4a;2 + 2a; + l =0. Solving these, we find for three values of x, We have, a^ + 8 = 0, or, {x + 2){x^-2x + 4:) = 0. .-. either a; + 2 = 0, or, a;2-2a; + 4 = 0. Solving these, we find for three values of x, - 2, 1 + V^, 1 - V^. 4 4 These six values of x are the six roots of the given equation. (2) Solve: •Vx'-S\^=4.0. Using fractional exponents, we have x^ — 3x^= 40. This equation is in the quadratic form if we regard x^ as the un- known number. 126 ALGEBRA. Complete the square, 4 x^ — 12 x^ + 9 = 169, Extract the root, 2 x* - 3 = ± 13. .-. 2a;*=16or-10, «* = 8 or — 5, » = 16 or — 5 Vs. There are other values of x which we shall not at present attempt to find. ™ , Exercise 26. bolve : 1. a;« + 7:r« = 8. 17. ^^x^' ^ ^x'"" =b. 2. a;*-5a7'^ + 4 = 0. 18. 4:r^- 3a:i= 10. 3. :.« + 4^^ = 96. 19, 2:.*- 3a;* = 9. 4. 37a;' -9 = 4^;*. 5. 16:r«=17:i;'-l. 6. 32a;^° = 33:^^-1. 7. ^« + 14^^:^ + 24 = 0. 8. 19:^^216:^;^ = a;. 1 13 9. a;«- 22a;* + 21 = 0. 24. — ^+— z = -. 20. -\/^^=V?+12. 21. a; = 9V^ + 22. 22. -v^?- 4^/^ = 32. 23. 2V?-3-^ 35. 4a:^ - 3(a;^ + 1) (x^ - 2) = a;^(10 - 3a;^). 36. (a;f - 2) (a;t - 4) = a;t (a;f - 1)^ - 12. 37. SVar^ + 17 + V^^+1 + 2V5ar\+ 41 = 0. EQUATIONS SOLVED LIKE QUADRATICS. 131 38. 2 a; ^4: x\ 36 X 40. 41. 42. X + V2-X' 1 + V2 1 = rp. 2a? 1 + Vl 1-vr 9 1 + -Vax — b ^ax + 6 — ■\Jax 1 — ■yo.x — h ^a — x-\- ■\/b — x_ -Vx + -y/b Va ^b — x -y/x—^b^ 43. V^ + '\a — Vaa; + a;^ = Va. 44. a7' + y'^ + a; + y = 481 a;y = 12 J 45. X -{- y -\- -y/ X -\- y x — y-\- Va; — y 46. x'^-\- a:y + 2/^ = ^ + Va;y + y = n n 47. 3V^ + 2Vy 4V^ — 2V^ 6. ^2_j_1^^2__64 16 a^ 48. V^— V3/ = a;^(V^+ V3/)]. 49. 4 3a: 2 54 CHAPTEK XII. PROPERTIES OF QUADRATIC EQUATIONS. 151. Representing the roots of the quadratic equation ax^ -\- bx -^ c — by a and jS, we have (§ 141), — b-{--Vb''-4:ac a = > 2a -h--Vb'-4cac ^= 2a Adding, a+p^^l; multiplying, ^-l: If we divide the equation ax^ -\-bx-\-c=^0 through by b c a, we have the equation x"^ -{- - x -{- - =^ \ this may be b e written x^ -\- px -{- q =^ where p ^= —, a It appears, then, that if any quadratic equation be made to assume the form x^^px^ q=^^, the following relations hold between the coefficients and roots of the equation : (1) The sum of the two roots is equal to the coefficient of X with its sign changed. (2) The product of the two roots is equal to the constant term. Thus the sum of the two roots of the equation a;'' — 7a; + 8 = is 7, and the product of the roots 8. PROPERTIES OF QUADRATIC EQUATIONS. 133 152.* The expressions a + ^, a^, are examples of symmetric functions of the roots. Any expression which involves both roots, the two roots entering to similar powers and with similar coefficients, is a symmetric function of the roots. From the relations a-\- /3 = —p, ap = q, the value of any symmetric function of the roots of a given quadratic may be found in terms of the coefficients. Given that a and )8 are the roots of the quadratic x'^ — 7 a; + 8 = 0, we may find the values of symmetric functions of the roots as follows : (1) a? + 0^. We have o + )8 = 7, oj8 = 8. Square the first, . o^ + 2a)8 + iS^ = 49. Subtract, 2aj8 =16, and we have (2) 0? + j8^ a^ + &' = 33. a3 + 3a2)8 + 3 «i82 + fi' = 343. 3 a^B + 3 aB,-' = 168. 3 a)3 (o + )3) or Subtract, o^ + jS^ = 175. )8 a This is which is 175 8 * 153. Eesolntion into Factors. By § 151, if a and /3 are the roots of the equation x"^ -\-px -\- q = 0, the equation may be written 2 r \ o\ \ o c\ The left member is the product of a; — a and x — P, so that the equation may be also written (x -.a)(x-P) = 0. It appears, then, that the factors of the quadratic expres- sion x^ ■\-px + q are x — a and x — ^, where a and /8 are the roots of the quadratic equation x^ -{■px-\- q = 0. 134 ALGEBRA. The factors are real and different, real and alike, or imaginary, according as a and P are real and unequal, real and equal, or imaginary. If /? = a, the equation becomes (x — a){x — a) — 0, or {x~af = 0] if, then, the two roots of a quadratic equation be equal, the left member, when all the terms are transposed to that member, will be a perfect square as regards x. If the equation be in the form ax"^ + ^:r + c = 0, the left member may be written a{x'^-{--x~\--\ or (§ 151) a(x — a) (x ^ P). 154. If the roots of a quadratic equation be given, we can readily form the equation. Form the equation of which the roots are 3 and The equation is {X or {X or 3) 3) {2x + 5) = 0, a;2 _ a; - 15 = 0. 155. Quadratic expressions may be factored by the prin- ciples of § 153. (1) Resolve into two factors x^ — 5x -{-S. Write the equation a;"^ — 5 a; + 3 = 0. rru t t J ^ 1. 5 + Vis 5 - Vl3 The roots are found to be , 2 2 The factors of a;^ — 5 a; + 3 are 5 + VfS . 5 - \/l3 X and X 2 2 (2) Resolve into factors 2>x^ — 4:X -\-b. Write the equation Sa;^ — 4a;-f5 = 0. m, . t A ^ x. 2+V^ni 2-\/-ll The roots are found to be > o o Therefore the expression Sa:^ — 4 a; + 5 may be written (§ 153) 2 4-\/^ni\/ 2-\/^Tl\ X — —)v PROPERTIES OF QUADRATIC EQUATIONS. 135 Exercise 28. Form the equations of which the roots are : 6. a + 35, a — Sb. a + 2b 2a + b 3 ' 3 ■ 8. 2 + V3, 2-V3. 9. -1+V5, ~1-V5. 10.1+41-^1 il or imaginary : 15. x''-2>x-\-4:. 16. x'-^-x-^-l. 17. 4:r2-28a; + 49. 18. 4a;^ + 12a; + 13. Note. The remainder of this chapter may be omitted if it is desired to abridge the course. In examples 19-27, a and p are to be taken as the roots of the equation a;^ — 7 :?; + 8 = 0. Find the values of : 1. 3,2. 2. 4,-5 ». 3. -6, - -8. 4. 2 1 3' 2 5. 1 3' 3 4' Resolve into f actors 11. 2>x'- 15a;- -42. 12. ^x-"- ■27 a;- -70. 13. 492;'^ + 49a: + 6. 14. 169 o;^ ^-52, 2; + 4, 19. (a-^)^ 20. a'/3 + a;8^ 21. 22. «+^. ^ 0. 23. 24. a' + ^ 25. i+i. 26. (a^-py. 27. 136 ALGEBRA. In examples 28-33 a and fi are to be taken as the roots of the equation x^ -{- px-^ q = 0. The results are to be found in terms oi p and q. Find the values of: 28. 1 + 1. a ^ 29. a'y8 + a)Sl 30. a^ + /3l 31. a'^ + af3\ 32. a' + l3\ 33 ^V'^' 34. When will the roots of the equation ax^ -\- bx -^ c = be both positive? Both negative? One positive and one negative ? 35. When will one root be the square of the other ? 36. When will the sum of the reciprocals of the roots be unity ? 37. Show that the roots of the equation x'-{-2(a + b)x + 2(a' + h') = are imaginary if a and b are real and unequal. 38.* Show that the roots of the equation — x"" + (x~b) (x-c) + (x—c) (x-a) + (x-a) (x-b) = are real if a, b, and c are real. 39.* Show that the equations ax"^ -\-bx-[- c — 0, a'x + c' = 0, will have a common root if — r + -7 = —n' a'^ c'^ a'c' 40.* Show that the equations ax" -\- bx -{- c -= 0, a'x' -\'h'x-{-c' = 0, will have a common root if (a'c - ac'y = (b'c - bc'Xa'b - ab'). PROPERTIES OF QUADRATIC EQUATIONS. 137 156.* The Roots in Special Oases. The values of the roots of the equation ax^ + ^a; + c = are (§ 141) 2a ' 2a ' Multiplying both numerator and denominator of the first expression by — 5 — V^'^ — 4 ac, and both numerator and denominator of the second expression by — 5 -f V^^ — 4ac, we obtain these new forms for the values of the roots : 2g 2c -g We proceed to consider the following special cases : I. Suppose a to be very small compared with b and c. In this case b'^ — Aac differs but little from b^, and its square root but little from b. The denominator of the first root in B will be very nearly —2b, and the root itself very nearly — j; the denominator of the second root in B will be very b small, and the root itself numerically very large. The smaller a is, the larger will the second root be, and the less will the first root differ from — -• The first root may be found approximately by neglecting the a^ term and solving the simple equation bx-{- c = 0. In fact, the quadratic equation itself approximates to the form Ox'^-]-bx-\-c = 0. II. Suppose both a and b to be very small compared with c. In this case the first root, which dififers but little from — |, also becomes very large, so that both roots are very large. The smaller a and b are, the larger will the roots be. 138 ALGEBRA. The quadratic equation in this case approximates to the form Ox'^-{-Ox-{-c = 0. III. Suppose c = while a and b are not zero. In this case the first root in A becomes zero, the second root becomes _b a The quadratic equation becomes ax^ -\-bx = or x (ax -f 5) = ; one root is 0, the other, obtained by solving the equation axi-b = 0, is (§ 140) ; these are the values just found. IV. Suppose b — and c = while a is not zero. In this case both roots in A become zero. The equation reduces to ax"^ ^ 0, of which both roots are zero (§ 140). V. Suppose b = while a and c are not zero. In this case the two roots become +\/ and — \ Ma Ma The equation becomes the pure quadratic ax'^-}-c = 0. Solving this, we obtain for x the values just found. 157.* Collecting results, we have the following : I. a very small compared with b and c ; one root very large. II. a and b both very small compared with c ; both roots very large. III. c = 0, a and b not zero ; one root zero. IV. b = 0, c = 0, a not zero ; both roots zero. V. b = 0, a and c not zero ; a pure quadratic ; roots numerically equal but opposite in sign. PROPERTIES OF QUADRATIC EQUATIONS. 139 158.* Variable Coefficients. When the coefficients of an equation involve an undetermined number, the character of the roots may depend on the value given to the unknown number. For what values of m will the equation 2ma;' + (5m -j- 2)^ + (4m 4- 1) = have its roots real and equal, real and unequal, imaginary? We find 62_4^c = (5m + 2)2-8m(4m + l) = 4 + 12m-7m2 = (2-m)(2 + 7m). Boots equal. In this case 6^ — 4 ac is to be zero. We must have either 2-m = 0, or 2 + 7m = 0. 2 .•. m = 2, or m = 7 Roots real and unequal. In this case S'^ — 4 ac is to be positive. The factors 2 — m, 2 + 7w, are to be both positive or both negative. 2 If m lies between 2 and — -, both factors are positive ; both fac- tors cannot be negative. Roots imaginary. In this case S'^ — 4 ac is to be negative. Of the two factors 2 — m, 2 + 7m, one is to be positive and the other negative. If m is algebraically greater than 2, 2 — m is negative and 2 + 7m 2 positive ; if m is algebraically less than — , 2 + 7m is negative and 2 — m positive. 159.* By a method similar to that of the last section, we can often obtain the maximum or minimum value of a quad- ratic expression for real values of x. (1) Find the maximum or minimum value of 1 -J- ^ — ^ for real values of x. Let 1 + a; — x^ = m ; then x^ — x=\ — m. 140 ALGEBRA. Solving, ^^ lW5-4m , Since X is real, we must have 5>'im or 5 = 4 m. .'. 4m is not greater than 5, 5 m is not greater than — 5 1 The maximum value oi \ -\- x — x^ is - ; for this value a; = — 4 2 (2) Find the maximum or minimum value oi x^-\-2>x-\-^ for real values of x. Let a;2 ^ 3 a; + 4 = m ; then a;'^ + 3 a; = ?/i — 4. o 1 • - 3 ± V4m-7 Solving, X = A Since x is real, we must have 4m > 7 or 4m = 7. .-. 4 w is not less than 7, 7 m is not less than — 4 7 3 The minimum value of a;^ + 3 a; + 4 is - ; for this value a; = 4 2 Note. Instead of solving for x, we might have used the condition for real roots, viz., 6^ — 4ac greater than or equal to zero. 160.* The existence of a maximum or minimum value may also be shown as follows : Take the first expression of the last article, 1 +x — x^. This is l~i\~^'^^^\ is positive for all real values of x ; its least value is zero, c and in this case the given expression has its greatest value, -• Similarly for any other expression. PROPERTIES OF QUADRATIC EQUATIONS. 141 Exercise 29.* For what values of m are the two roots of each of the following equations (1) equal, (2) real and unequal, (3) im- aginary ? 1. (3m+l)a;'^ + 2(m + l)a;-f ^ = 0. 2. (m — 2)a;' + (m-5)rp + 2m — 5 = 0. 3. 2ma;^ + a;^ — 6ma; — 6a;+ 6m + 1 = 0. 4. m2;'^ + 2:c'^ + 2m — 3ma; + 9:r-10 = 0. 5. 6ma;^4-8m:i; + 2m = 2a; — a;^ — 1. Find the maximum or minimum value of each of the following expressions, and determine which : 6. a;*'' -6a: +13. 7. 4a;2 — 12.T + 16. 8. Z-\-Vlx-^x\ 9. a;' + 8a; + 20. 10. 4a;^-12a; + 25. 11. 25a;^-40a;-16. 12. 0^ 13. {x-\-l2){x-^) x' 14. 4:X 1 f> x'-x-l x^-x+1 1 R x' + 2x-S x'-2x-{-S 17. 1 1 2 + a; 2-x 18. x' + Sxi-^ x' + l 1Q (^+iy x'-x-i-l on 2x'-2x + 6 (x + 2y x'-2x + B 21. Divide a line 2 a inches long into two parts such that the rectangle of these parts shall be the greatest pos- sible. 142 ALGEBRA. 22. Divide a line 20 inches long into two parts such that the hypotenuse of the right triangle of which the two parts are the legs shall be the least possible. 23. Divide 2a into two parts such that the sum of their square roots shall be a maximum. 24. Find the greatest rectangle that can be inscribed in a given triangle. 25. Find the greatest rectangle that can be inscribed in a given circle. 26. Find the rectangle of greatest perimeter that can be inscribed in a given circle. CHAPTER XIII. SURDS AND IMAGINARIES. 161. Quadratic Surds. The 'product or quotient of two dis- similar quadratic surds will be a quadratic surd. For every quadratic surd, when simplified, will have under the radical sign one or more factors raised only to the first power ; and two surds which are dissimilar can- not have all these factors alike. 162. The sum or difference of two dissim.ilar quadratic surds cannot he a rational number, nor can it be expressed as a single surd. For if Va ± V6 could equal a rational number c, we should have, by squaring, and transposing, rfc 2 -yfab — & — a — b. Now, as the right side of this equation is rational, the left side would be rational ; but, by § 161, Va6 cannot be rational. Therefore V« =b V^ cannot be rational. In like manner, it may be shown that Va =b Vo cannot be expressed as a single surd V^. 163. A quadratic surd cannot equal the sum of a rational number and a surd. For, if Va could equal c + V^, we should have, squar- ing, and transposing, 2c^b^a-b-c\ That is, a surd equal to a rational number, which is impossible. 144 ALGEBRA. 164. If a-\- V^ — x-{- Vy, then a will equal x, and b will equal y. For, transposing, -\fb ~ Vy = a; — a ; and if h were not equal to y, the difference of two unequal surds would be rational, which by § 162 is impossible. .•. h = y and a^=x. In like manner, if a — V^ = a: — Vy, a will equal x, and h will equal y. Expressions of the form «+ V^, where V^ is a surd, are called binomial surds. 165. Square Eoot of a Binomial Surd. Let V a + V^ = Vic + Vy. Squaring, a + V6 = a; + 2 Vxy + y. .'. X + y = a, and 2\/a;2/ = Vb. (§ 164) From these two equations the values of x and y may be found. This method may be shortened by observing that, since ■\/b = 2Vxy, we have a — Vb = x — 2Vxy + y. By taking the root, v a — Vb = Vx — Vy. .-. (Va + \/6) (Va - V6) = (Vx + Vy) (V^ - Vy). .-. Va^ — b = x — y. And, as a = x + y, the values of x and y may be found by addition and subtraction. (1) Extract the square root of 7 -f 4 VS. Let V^ + Vy= V 7 + 4\/3. Then Vx - Vy = V7-4\/ 3. Multiplying, x — y= V49 — 48, .-.x-y-^l. But < x + y = 7, .'. X = 4, and y = 3. SURDS. 145 .-. Vx + Vy = 2+ Vs. .-. V? + 4 V3 = 2 + Vs. A root may often be obtained by inspection. For this purpose, write the given expression in the form a + 2 V6, and determine what two numbers have their sum equal to a, and their product equal to h. (2) Find by inspection the square root of 18 + 2 VtY. It is required to find two numbers whose sum is 18 and whose product is 77 ; these are evidently 11 and 7. Then 18 + 2 V77 = 11 + 7 + 2 yTTx^, = (Vll + V7)2. That is, VTl -f- V7 = square root of 18 + 2 V77. (3) Find by inspection the square root of 75 — 12V21. It is necessary that the coefficient of the surd be 2 ; therefore, 75 — 12 V2I must be put in the form 75-2V756. The two numbers whose sum is 75 and whose product is 756 are 6S and 12. Then 75-2 V756 = 6S + 12 - 2y6S x 12, = (V63- Vl2)2. That is, V63 - Vl2 = square root of 75 - 12 V2I ; or, 3 V7- 2 V3 = square root of 75-12 V2I. Exercise 30. Extract the square roots of : 1. 14 + 6V5. 6. 20-8V6. 11. 14-4V6. 2. 17 + 4V15. 7. 9-6V2. 12. 38-12V10. 3. IO + 2V2I. 8. 94-42V5. 13. 103-12VlT. 4. I6 + 2V55. 9. 13-2V30. 14. 57-I2VI5. 5. 9-2V14. 10. II-6V2. 15. 3i-VT0. 16. 2a + 2V^^^^. 18. 87-12V42. 17. a'-25V^^^=^'. 19. {a-\-hf-^{a-h)^ab. 14^ ALGEBRA. IMAGINARY EXPRESSIONS. 166. An imaginary expression is any expression which in- volves the indicated even root of a negative number. It will be shown hereafter that any indicated even root of a negative number may be made to assume a form which involves only indicated square roots of negative numbers. In considering imaginary expressions, we accordingly need consider only expressions which involve the indicated square roots of negative numbers. Imaginary expressions are also called imaginary numbers and complex numbers. 167. Imaginary Square Eoots. If a and h are both posi- tive, we have (§ 118) I. Vab=--^-K/b. 11. (V^y = a. If one of the two numbers a and h is positive and the other negative, law 1. is assumed still to apply ; we have accordingly : V^ = V5(=l) = V5 V=^ ; V— a = Va(-- 1) = Va V— 1 ; and so on. It appears, then, that every imaginary square root can be made to assume the form aV— 1, where a is a real number. 168. If a and h are both negative, both laws cannot apply, for they give different results : I. gives IMAGINARY EXPRESSIONS. 147 II. gives . (+ V=^) (+ V^ = (+ V=^)^ - - a. We therefore assume that II. holds true ; hence I. does not hold true. This assumption gives us : v^ X v=n: --= ( v=i)' = - 1 ; V^^ X V^ = Va V^ X -^/h V^ = V^(-l) The law V^ X V^ = ( V^)^ = - 1 is very im- portant. Observe that the law Va V^ = Va6 holds true unless both a and h are negative. 169. It will be useful to form the successive powers of (V-iy=(V^)'V^ =(--l)V-^l = -V:^; (V^)' = (V-iy (V^y = (- 1) (- 1) = + 1 ; and so on. It appears that the successive powers of V— 1 form the repeating series -\- V— 1, — 1, — V— 1, + 1, and so on. 170. An imaginary expression will generally consist of two parts : a real part and an imaginary part. Thus, the roots of the quadratic equation a;'* — 6a:+13 = are 3 + V=l, 3--v^l; that is, 3 + 2V=n[, 3-2V=l; each of these imaginary expressions consists of a real part and an imaginary part. 148 ALGEBRA. 171. Every imaginary expression may be made to assume the form a + 5V~ 1, where a and h are real numbers, and may be integers, fractions, or surds. If 5 = 0, the expression consists of only the real part a, and is therefore real. If a = 0, the expression consists of only the imaginary part 5V— 1, and is a pure imaginary. 172. The form a + ^V— 1 is the typical form of imaginary expressions. Eeduce to the typical form 6 + V— 8. This may be written 6 + V8 V^, or 6 + 2V2 V^ ; here a = 6, and 6 = 2V2. 173. The sum of two imaginary expressions is generally an imaginary expression. Add a+^V=l, and c + c?V— 1. The sum is (« + c) + (^ + d) V^. This is an imaginary expression unless b-\-d = 0; in which case the expression is real. 174. The product of two imaginary expressions is generally an imaginary expression. Multiply a-\-h V— 1, by c + d V^ . ac -\- he V— 1 -f ad^— 1 — hd The product is {ac — hd) + (he + ad)^— 1, an imaginary expression unless hc-^-ad^^ 0. IMAGINARY EXPRESSIONS. 149 f 175. The quotient of two imaginary expressions is gener- ally an imaginary expression. Divide a + ^ V— 1 by 2a5. For (a — hf must be positive, whatever the values of a and h. That is, a2-2a6 + 62>0; .-. a^ + J^ > 2a6. 184. It can be easily shown that the principles applied to the solution of equations may be applied to inequalities, except that if each side of an equality have its sign changed, the inequality will be reversed. Thus, if a > &, then - a will be < -h. 154 ALGEBRA. (1) If a and h are positive, show that a^-\-b^> a^b -\- alP", We shall have a^ + 6^ > a% + aS^, if (dividing each side by a + h), a?~ah^h'^> ah, if a2 + 62>2a6. But this is true (§ 183). .-. a? + b^> a^h + ab\ (2) Show that o? ^ h" -\- c^ > ah -\- ac -\- he. Now, a2 + 62>2a6, a2 + c2>2ac, (^83) Adding, 2a2 + 26^ + 2c2 > 2ah + 2ac + 26c, .-. a^ _(- 52 _,. g2 -v^ ^5 ^ (jg _,. 5g_ Exercise 32. Show that, the letters being unequal and positive : 1. o?-\-W>2h{a^h). 2. oJ'b-^ah^>2a^h\ 3. {p?-{-h'){a' + h')>{a'-^hy. 4. a^h + aV + ah'' + 5'c + ac' + ^c' > 6 a5c. 5. The sum of any fraction and its reciprocal > 2. 6. If a;^ = a^ + 5^ and y^ = c^-\-d^, xyi^ac-\-hd, or ac?+^c. 7. a5 + ac+6c<(a4-& — c)'H-(a + c-5)'+(5 + c — ay. 8. Which is the greater, (a' + 5=^) {c' + ^') or (ac + hdj ? 9. Which is the greater, a^'—h^ or 4a^(a — ^) when a'>hl 10. Which is the greater, J^ + J- or Va -f V^ ? 11. Which is the ejreater, or ? ^ 2 a + ^ 12. Which is the greater, 7-2 + ^ °^ 7 + ~ CHAPTEB XV. RATIO, PROPORTION, AND VARIATION. 185. Ratio of Numbers. The relative magnitude of two numbers is called their ratio, when expressed by the indicated quotient of the first by the second. Thus the ratio of a to 5 is -» or a ^ 5, or a : h ; the quotient is generally written in the last form when it is intended to express a ratio. The first term of a ratio is called the antecedent, and the second term the consequent. When the antecedent is equal to the consequent, the ratio is called a ratio of equality ; when the antecedent is greater than the consequent, the ratio is called a ratio of greater inequality ; when less, a ratio of less inequality. When the antecedent and consequent are interchanged, the resulting ratio is called the inverse of the given ratio. Thus, the ratio 3 : 6 is the inverse of the ratio 6 : 3. 186. A ratio will not be altered if both its terms be mul- tiplied by the same number. For the ratio a : 6 is represented by ~> the ratio ma : mb is repre- sented by -— • ; and since t?^ = -. we have ma:mb = aib. mo mb b A ratio will be altered if dififerent multipliers of its terms be taken ; and will be increased or diminished according as the multiplier of the antecedent is greater than or less than that of the consequent. J.«JU .a.j_jVTX JI3S\,I\.. If m'>n, ma > na, If ma:h. or ma :nh<,a:b. 187. Katios are coinpounded by taking the product of the fractions that represent them. Thus, the ratio compounded of a : 6 and c : c? is ac : hd. The ratio compounded oi a-.h and a : & is the duplicate ratio a?:h^\ the ratio compounded of a : 6, a : 6, and a : 6 is the triplicate ratio a' : 6^ ; and so on. 188. Katios are compared by comparing the fFactions that represent them. Thus, a : & > or < c : c? according as ^ > or < - . d ad ^ ^ be 843 — >■ or •< — . hd'^ bd as acZ > or < 6c. 189. Proportion of Numbers. Four numbers, a, b, c, d, are said to be in proportion when the ratio a : 5 is equal to the ratio c : d. We then write a'.b=^c\d, and read this either, the ratio of a to 5 equals the ratio of c to d, or a is to 5 as c is to d. A proportion is also written a\h : : c \ d. The four numbers a, b, c, d are called proportionals ; a and d are called the extremes, b and c the means. RATIO AND PROPORTION. 157 190. When four numbers are in proportion, the product of the extremes is equal to the product of the means. For, if a:b- = c: d, then a b" c = — . d Multiplying by hd, ad- = bc. The equation ad = be gives he a h = ad c so that an extreme may be found by dividing the product of the means by the other extreme ; and a mean may be found by dividing the product of the extremes by the other mean. If three terms of a proportion are given, it appears from the above that the fourth term can have one, and but one, value. 191. If the product of two numbers is equal to the prod- uct of two others, either two may be made the extremes of a proportion and the other two the means. For, if ad = be, then, dividing by bd, ~ = ^, bd bd .'. a:b==c:d. 192. Transformations of a Proportion. If four numbers, a, b, e, d, be in proportion, they will be in proportion by : I. Inversion : b will be to a as c? is to c. For, if - a:b = c:d, then a c b^d 158 ALGEBRA. and h d a c b : a = d: c. II. Composition : a-\-b will be to 6 as c + c? is to d. For, if a:b -= c : d, then and a c b~ d f + -5-^- a + b c + d b d a-\-b:b^c-\- d: d. III. Division \ a — h will be to ^ as c — c? is to d. For, if then and a:b = c: d. a e b d b d a-b c-d b d •.a — b:b = c — d: d. IV. Composition and Division \ a-\-h will be to a — h as c + c? is to c — c?. For, from II., a + b ^c_±d^ b d and from III., a^^c-d^ b d Dividing, q±b c_±d a—bc-d .'. a + b ; a — b = c + d : c — d. RATIO AND PROPORTION. 159 V. Alternation : a will be to c £ For, if a:b = c:d, then a c b~d Multiplying by -, ab _ be be cd or a_b c d .-. a:c=b:d. 193. In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. r may be put for each of these ratios. Then - = r, j=r, ^=r, f = r. a J h .•. a = br, c = dr, e =fr, g = hr. .-. a + c + e + g = {b + d +f + h)r. , a+c+e+g_ a " b+d+f + h^'^^b .'. a + c + e + g : b + d +f + h = a : b. In like manner it may be shown that ma + nc +pe + qg-.mb + nd +pf + qh'^ a:b. 194. If a, b, c, d be in continued proportion, that is, if a\h = b\c = c:d, then will a:c = o?:b'^ and a:d—a^:b^. For, Hence, Also a b c b c d Ov. b a. a -X ■■ - y b c a _ c b b .'. a : C = -a': &». a h r. a a. a x-x = T X T X b c d b b b 160 ALGEBRA. 195. If a, h, c are proportionals, so that a : b — b : c, then b is called a mean proportional between a and c, and c is called a third proportional to a and b. If a : :b = = b : c, then b = Vac. For, if a : & "= 6 : c, then a 6 b~ c and &2 = ac. /. b = Vac. 196. The products of the corresponding terms of two or more proportions are in proportion. For, if a:b = e:d, e:f=g:h, k : I = m : n, .1 ■ a c e q h m then y ^_ a J h I n Taking the product of the left members, and also of the right members of these equations, aek _ cgm bfl dhn .'. aek : bfl = cgm : dhn. 197. Like powers, or like roots, of the terms of a propor- tion are in proportion. For, if a:b = c:d, ,v a c then T = T o d Raising both sides to the nth power, RATIO AND PROPORTION. 161 Extracting the nih root, i_ £ 1 i i ^ 198. If two numbers be increased or diminished by like parts of each, the results will be in the same ratio as the numbers themselves. I ^ m\ m 1± - a a+—a a \ n n For (-^) h±-b n .'. a : = a ± — a : ± — 0. n n 199. The laws that have been established for ratios should be remembered when ratios are expressed in fractional form. (1) Solve : t±^±l ^ ^^-^ + 2 By composition and division, 2a?2 2a;2 2{x + l) -2(a;-2) and this equation is satisfied, when a; = ; or, dividing by — , when = ; 2 X + 1 2 — a? that is, when a? =» J. (2) If a : h = c : d, show that a^ -\- ab :b^ — ab = c^-^-cdid^ — cd. If then a±h^c±d^ a—b c—d ' and a c ~b -d . a ^^a + b c ,^c + d. A A - » — 6 a — b — d c — d that is, a^ + ab c2 + cd b^ -ab d^- cd or a? + ab -.b"^ - ab = c^ + cd : d"^ - cd. (3) li a \ h ^= c \ d, and a is the greatest term, show that a + c? is greater than b -\- c. Since ^ = ^, and a':> c, b d .'. b>d. Also, a-b^c_-d^ b d and b "> d. .'.a — b^c — d. Adding b + d=b + d, we have a ^d^b + c. Exercise 33. 1. Write down the ratio compounded of 3 : 5 and 8 : 7. Which of these ratios is increased, and which is diminished by the composition ? 2. Compound the duplicate ratio of 4 : 15 with the trip- licate of 5 : 2. 3. Show that a duplicate ratio is greater or less than its simple ratio according as it is a ratio of greater inequality or a ratio of less inequality. 4. Arrange in order of magnitude the ratios 3 : 4, 23 : 25, 10 : 11. 5. If a > 5, which is the greater ratio, a-\-b:a-boxa''-\-b'':a^-b'"? RATIO AND PROPORTION. 163 Find the ratios compounded of: 6. 3 : 5, 10 : 21, 14 : 15. 7. 7:9, 102 : 105, 15 : 17. 8. a^ — x"^ : a^ -{• S ax -}- 2x^ and a -{- x : a ~ x. 9. a;'-4:2a;'^ — 5a; + 3anda;-l:a;-2. 10. Prove that a ratio of greater inequality is diminished, and a ratio of less inequality increased, by adding the same number to both its terms. 11. Prove that a ratio of greater inequality is increased, and a ratio of less inequality diminished, by subtracting the same number from both its terms. 12. Show that the ratio a:b is the duplicate of the ratio a -\~ c : b -\- c, if e^ = ab. 13. Two numbers are in the ratio 2 : 5, and if 6 be added to each, they are in the ratio 4 : 7. Find the num- bers. 14. What must be added to each of the terms of the ratio m : n, that it may become equal to the ratio p : q'? 15. If X and y be such that, when they are added to the antecedent and consequent respectively of the ratio a : b, its value is unaltered, show that x\y=^a\b. Find X from the proportions : 16. 27: 90 = 45: a:. 17. \\\-A\ = Z\'.x. 54' 7c 166 • ■ Find a third proportional to : xagandA. 20. «'-*\nd«-* c c 164 ALGEBRA. Find a mean proportional between : 21. 3 and 16i. 22. i^^^'and ^^1+^. If a : b = c : d, prove that : 23. 2a + b:h = 2c + d:d. 24. Sa — b:a = 3c—d:c. 25. 4a + 35 :4a. — 35 = 4c + 3c^:4c-3d 26. 2a' + Sb':2a'-Sb' =2c' + 3d': 2c' - Sd\ If a : b = b : c, prove that : 27. a' + ab-.b' + bc-.'.a-.c. 28. a: c : : (a+bf : (b-\-c)\ 29. If a : 5 = 5 : c, and a is the greatest of the three numbers, show that a-\~ c'>2b. 30. If - — ^ = ^ = , and x, y, z be unequal, show I m n that I -{- m -{- n — 0. Find X from the proportions : 31. x + l:x—l = x-{-2:x-2. 32. a; + a : 2a; — Z> = 3a; + ^ : 4^ — a. 33. x' — 4:x + 2 : x^ — 2x — 1 = x^ — 4:x : x'' — 2x - 2. 34. 3 + 0;: 4 + 07 = 9 + 0;: 13 + a;. 35. a-{-x:b-}-x = c-\-x:d-{-x. 36. If a : 5 = c : c?, show that a-f- c-f- a 37. When a, b, c, d are proportional and all unequal, show that no number x can be found such that a-\-x,b-{-x, c-\-x, d-\- x shall be proportionals. RATIO AND PROPORTION. 165 200. Eatio of Quantities. To measure a quantity of any- kind is to find out how many times it contains another known quantity of the same kind, called the unit of measure. The number which expresses the number of times that a quantity contains the unit of measure is called the nu- merical measure of that quantity. Thus, if a line contains the linear unit of measure, one yard, 5 times, the measure of the length of the line is 5 yards, and the numerical measure of the line is 5. 201. Commensurable Quantities. If two quantities of the same kind are so related that a unit of measure can be found which is contained in each of the quantities an in- tegral number of times, this unit of measure is a common measure of the two quantities, and the two quantities are said to be commensurable. If two commensurable quantities be measured by the same unit, their ratio is simply the ratio of their numerical measures. Thus, ^ of a foot is a common measure o#2^ feet and 3f feet, being contained in the first 15 times and in the second 22 times. The ratio of 2J feet to 3f feet is therefore the ratio of 15 : 22. Evidently two quantities different in kind can have no ratio. 202. Incommensurable Quantities. We cannot expect, how- ever, that two quantities of the same kind chosen at random will have a common measure. Thus, the side and diagonal of a square have no common measure ; for, if the side be a inches long, the diagonal will be aV2 inches long, and no measure can be found which will be contained in each an integral number of times. Again, the diameter and circumference of a circle have no common measure, and are therefore incommensurable. 166 ALGEBRA. In this case, as there is no common measure of the two quantities, we cannot find their ratio by the method of § 201. We therefore proceed as follows : Suppose a and b to be two incommensurable quantities of the same kind. Divide b into any integral number, n, equal parts, and suppose one of these parts is contained in a more than m times and less than w + 1 times. Then y lies between — and ^"^ , and cannot differ from eithei n n of these by so much as -. But by increasing n indefinitely, - can be made to de- n crease indefinitely, and to become less than any assigned value, however small, though it cannot be made absolutely equal to zero. Hence, the ratio of two incommensurable quantities can- not be expressed exactly by numbers, but it may be expressed approximately to any desired degree of accuracy. Thus, if h represent the*side of a sq^^are, and a the diagonal, Now y/2 = 1.41421356 , a value greater than 1.414213, but less than 1.414214. If, then, a millionth part of h be taken as the unit, the value of the ratio ? lies between 1M4213 ^^^ 1414214^ ^^^ therefore diflfers from b 1000000 1000000 either of these fractions by less than 1000000 By carrying the decimal farther, a fraction may be found that will differ from the true value of the ratio by less than a billionth, a trillionth, or by less than any other assigned value whatever. Hence the ratio -' while it cannot be expressed by numbers ex- actly , may be expressed by numbers as accurately as we please. RATIO AND PROPORTION. 167 203. The ratio of two incommensurable quantities is an incommensurable ratio ; and is a fixed value toward which its successive approximate values constantly tend as the error is made less and less. 204. Equal Incommensurable Eatios. As the treatment of Proportion in Algebra depends upon the assumption that it is possible to find fractions which will represent ratios, and as it appears that no fraction can be found to represent the exact value of an incommensurable ratio, it is necessary to show that two incommensurable ratios are equal if their approximate values remain equal when the unit of measure is indefinitely diminished. Let a : b and a' : 5' be two incommensurable ratios of which the true values lie between the approximate values — and 4-1 '^ ^^ , when the unit of measure is indefinitely diminished. n 1 Then they cannot differ by so much as -• Let d denote the difi'erence (if any) between a : b and a':b'; then d< — n Suppose the fixed value d is not zero ; now n can be made as larse as we please, and - as small as we please ; 1 V' . hence - can be made less than dii dia not zero. n Therefore d = 0, and there is no difference between the ratios a : b and a' : b'. Therefore a:b = a':b'. 205. Proportion of Quantities. In order for four quanti- ties, A, £, C, D, to be in proportion, A and B must be of the same End, and C and B of the same kind (but C and I) need not necessarily be of the same kind as A and B), 168 ALGEBRA. and in addition the ratio of J. to ^ must be tlie same as the ratio of C to D. If this be true, we have the proportion When four quantities are in proportion, their numerical measures are four abstract numbers in proportion. 206. The laws of § 192, which apply to proportion of abstract numbers, apply to the proportion of concrete quantities, except that alternation will apply only when the four quantities in proportion are all of the same kind. Exercise 34. 1. A rectangular field contains 5270 acres, and its length is to its breadth in the ratio of 31 : 17. Find its dimensions. 2. If five gold coins and four silver ones be worth as much as three gold coins and twelve silver ones, find the ratio of the value of a gold coin to that of a silver one. 3. The lengths of two rectangular fields are in the ratio of 2:3, and the breadths in the ratio of 5:6. Find the ratio of their areas. 4. Two workmen are paid in proportion to the work they do. A can do in 20 days the work that it takes B 24 days to do. Compare their wages. 5. In a mile race between a bicycle and a tricycle their rates were as 5 : 4. The tricycle had half a minute start, but was beaten by 176 yards. Find the rate of each. 6. A railway passenger observes that a train passes him, moving in the opposite direction, in 2 seconds; but moving in the same direction with him, it passes him in 30 seconds. Compare the rates of the two trains. RATIO AND PROPORTION. 169 7. A vessel is half full of a mixture of wine and water. If filled up with wine, the ratio of the quantity of wine to that of water is ten times what it would be if the vessel were filled up with water. Find the ratio of the original quantity of wine to that of water. 8. A quantity of milk is increased by watering in the ratio 4:5, and then 3 gallons are sold ; the remainder is increased in the ratio 6 : 7 by mixing it with 3 quarts of water. How many gallons of milk were there at first ? 9. Each of two vessels, A and B, contains a mixture of wine and water ; A in the ratio of 7:3, and B in the ratio of 3:1. How many gallons from B must be put with 5 gallons from A to give a mixture of wine and water in the ratio of 11 : 4 ? 10. The time which an express train takes to travel 180 miles is to that taken by an ordinary train as 9 : 14. The ordinary train loses as much time from stopping as it would take to travel 30 miles ; the express train loses only half as much time as the other by stopping, and travels 15 miles an hour faster. What are their respective rates ? 11. A and B trade with different sums. A gains $200 and B loses $50, and now A's stock is to B's as 2:i. But if A had gained $100 and B lost $85, their stocks would have been as 15 : 31. Find the original stock of each. 12. A line is divided into two parts in the ratio 2 : 3, and into two parts in the ratio 3:4; the distance between the points of section is 2. Find the length of the line. 13. A railway consists of two sections ; the annual ex- penditure on one is increased this year 5%, and on the other 4%, producing on the whole an increase of 4^%- Compare the amounts expended on the two sections last year, and also the amounts expended this year. 170 ALGEBRA. VARIATION. 207. A quantity which in any particular problem has a fixed value is called a constant quantity, or simply a con- stant; a quantity which may change its value is called a variable quantity, or simply a variable. Variable numbers, like unknown numbers, are generally represented by x, y, z, etc. ; constant numbers, like known numbers, by a, 5, 3. .-. TToc Z)3. I 214, 1 Put W = 'mL^; then, since 20 and 4 are corresponding volumes of W and D, 20 = m X 64. . -«, 20 5 :. when i> = 5, W^ ^-^ of 125 = 39xV. 174 ALGEBRA. Exercise 35. 1. li y ccx^ and y = 4 when a; = 5, find y when x = 12. 2. If y X a;, and when ^ = J, y = i, find y when 3. If z vary jointly as x and y, and 3, 4, 5 be simulta- neous values of x, y, z, find 2 when a; = y = 10. 4. If y X -, and when y = 10, a; = 2, find the value of X X when y = 4. 5. If 2; cc -, and when z = 6, :p = 4, and y = 3, find y the value of z when a; = 5 and y = 7. 6. If the square of x vary as the cube of y, and a; = 3, when y = 4, find the equation between x and y. 7. If the square of x vary inversely as the cube of y, and x — 2 when y = 3, find the equation between x and y. 8. If z vary as a; directly and y inversely, and if when 2; = 2, a; = 3, and y = 4, find the value of z when a; = 15 and y = ^. 9. If y oc rr + c where c is constant, and if y — 2 when 2: = 1, and if y = 5 when a; = 2, find y when a; = 3. 10. The velocity acquired by a stone falling from rest varies as the time of falling ; and the distance fallen varies as the square of the time. If it be found that in 3 seconds a stone has fallen 145 feet, and acquired a velocity of 96|- feet per second, find the velocity and distance fallen at the end of 5 seconds. VARIATION. 175 11. If a heavier weight draw up a lighter one by means of a string passing over a fixed wheel, the space described in a given time will vary directly as the difference between the weights, and inversely as their sum. If 9 ounces draw 7 ounces through 8 feet in 2 seconds, how high will 12 ounces draw 9 ounces in the same time ? 12. The space will also vary as the square of the time. Find the space in Example 11, if the time in the latter case be 3 seconds. 13. Equal volumes of iron and copper are found to weigh 77 and 89 ounces respectively. Find the weight of lO-J- feet of round copper rod when 9 inches of iron rod of the same diameter weigh Sl^ ounces. 14. The square of the time of a planet's revolution about the sun varies as the cube of its distance from the sun. The distances of the Earth and Mercury from the sun being 91 and 35 millions of miles, find in days the time of Mercury's revolution. 15. A spherical iron shell 1 foot in diameter weighs -^^ of what it would weigh if solid. Find the thickness of the metal, it being known that the volume of a sphere varies as the cube of its diameter. 16. The volume of a sphere varies as the cube of its diameter. Compare the volume of a sphere 6 inches in diameter with the sum of the volumes of three spheres whose diameters are 3, 4, 5 inches respectively. / 17. Two circular gold plates, each an inch thick, the diameters of which are 6 inches and 8 inches respectively, are melted and formed into a single circular plate 1 inch thick. Find its diameter, having given that the area of a circle varies as the square of its diameter. CHAPTER XVI. PROGRESSIONS. 216. A succession of numbers that proceed according to some fixed law is called a series ; the successive numbers are called the terms of the series. A series that ends at some particular term is a finite series; a series that continues without end is an infinite series. 217. The number of different forms of series is unlimited ; in this chapter we shall consider only Arithmetical Series, Geometrical Series, and Harmonical Series. ARITHMETICAL PROGRESSION. 218. A series is called an arithmetical series or an arith- metical progression when each succeeding term is obtained by adding to the preceding term a constant difference. The general representative of such a series will be a, a-{-d, a-\-2d, a-{-Sd...... in which a is the first term and d the common difference ; the series will be increasing or decreasing according as d is positive or negative. 219. The nth Term. Since each succeeding term of the series is obtained by adding d to the preceding term, the coefficient of d will always be one less than the number of the term, so that the nth term is a-\-{n— 1) d. ARITHMETICAL PROGRESSION. 177 If the nth term be represented by I, we have l-=a^{n-l)d. I. 220. Sum of the Series. If I denote the nth term, a the first term, n the number of terms, d the common difference, and s the sum of n terms, it is evident that s^ a +(a + c?)-j-(a + 2c?) + + (^-cf)+ I, or s= I J^ll-d)-^{l-2d)^ + (a+c?)+ a. ■■• 2s = (a+0+(a+0+(«+0 + -{-(a+l) +(a+0 — 71 (a + I). .•.. = |(a + 0. II. 221. From the two equations, l=a-\-(n-l)d, L s = l(a + ll 11. any two of the five numbers a, d, I, n, s may be found when the other three are given. (1) Find the sum of ten terms of the series, 2, 5, 8, 11, Here a = 2, d=S, n = 10. From I., 1 = 2 + 27 = 29. Substituting in II., s = — (2 + 29) = 155. Ans. (2) The first term of an arithmetical series is 3, the last term 31, and the sum of the series 136. Find the series. From I. and II., 31 = 3 + (n - 1) d, (1) 136 = ^(3 + 31). (2) From (2), n = 8. Substituting in (1), c? = 4. The series is 3, 7, 11, 15, 19, 23, 27, 31. 178 ALGEBRA. (3) How many terms of the series, 5, 9, 13, , must be taken in order that their sum may be 275 ? From I., ? = 5 + (n-l)4; .\l = 4:n + l. (1) From II., 275 = - (5 + I). (2) Substituting in (2) the value of I found in (1), 275 = |(4n + 6), or 2n2 + 3n=275. We now have to solve this quadratic. Complete the square, 16n2 + () + 9 = 2209. Extract the root, 4n + 3 = ±47. .-. n=ll, or -12^. * We use only the positive result. (4) Find n when d, I, s are ) given. From I., a = l-{n- 1) d. From II., a = 2s -In n 2 s — In Therefore, l-{n-l) n .'. In — dv? -\- dn=^2s — In, .\dn'^-{2l-\-d)n = -2s. This is a quadratic with n for the unknown number. Complete the square, 4^2^2 -{) + (2l + dy = {2l + df - 8ds. Extract the root, 2dn-{2l + d) = ± V{2l + df-8ds. 2l + d± V(2 l + df-Sds .". n = ^^ ^ • 2d Note. The table on the following page contains the results of the general solution of all possible problems in arithmetical series, in which three of the numbers a, I, d, n, s are given and two required. The student is advised to work these out, both for the results obtained and for the practice gained in solving literal equations in which the unknown quantities are represented by letters other than x, y, z. ARITHMETICAL PROGRESSION. 179 No. 1 2 Given. Required Results. a d n ads I l = a + {n-l)d. l = -^d±V[2ds + {a~^df] 3 a n s J 2s 1 = a. n 4 d n 8 ^_s ^{n-l)d n 2 5 a d n s=^n[2a^{n-\)d\ G a d I ' 2 ' 2cZ • 7 a n I s = {l^a)l 8 9 d n I s^ln[2l-{n-l)d\ dn I a = l-{n-l)d. 10 d n s a-..' (^-1)^ 11 d I s a n 2 a = ^d± V{l + ^df-2ds. 12 13 n I s a = ^-^-l. n a n I aJ-. n-1 14 a n s d a d 2(s-an) n (n - 1) 15 a I s d- ''--' . 2s-l-a 16 17 18 19 20 n I s 2{nl-s) n{n-l) a d I ads a I s d I s n ^-'-=f^- d-2a±^/{2a-df+^ds " 2d l + a 2l + d± V(2^ + c/)«-8(& "- 2d 180 ALGEBRA. 222. The aritlimetical mean between two numbers is the number which stands between them, and makes with them an arithmetical series. If a and h represent two numbers, and A their arithmet- ical mean, then, by the definition of an arithmetical series, A — a = h — A. a + h .'.A 223. Sometimes it is required to insert several arithmeti- cal means between two numbers. Ex. Insert six arithmetical means between 3 and 17. Here the whole number of terms is eight ; 3 is the first term and 17 the eighth. By I., 17 = 3 + 7^, The series is 3, [5, 7, 9, 11, 13, 15,] - 17, the terms in brackets being the means required. 224. When the sum of a number of terms in arithmet- ical progression is given, it is convenient to represent the terms as follows : Three terms by ^ — y, oc, x -\- y; four terms by a? — 3y, x — y, x-\-y, x-\-^y] and so on. Ex. The sum of three numbers in arithmetical progres- sion is 36, and the square of the mean exceeds the product of the two extremes by 49. Find the numbers. Let x — y, X, X + y represent the numbers. Then, adding, 3 a; = 36. .-. a = 12. Putting for x its value, the numbers are 12 -y, 12, 12+3/. ARITHMETICAL PROGRESSION. 181 By the conditions of the problem we have (12)2 ==(12 -2/) (12 + 3/) + 49. 144 = 144-3/2 + 49, 3/ = ±7. The numbers are 5, 12, 19; or 19, 12, 5. Exercise 36. Find : 1. The 10th term of 3, 8, 13 2. The 8th term of 12, 9, 6 3. The 12th term of - 4, - 9, - 14 4. The 11th term of 2|, If, 1^ 5. The 14th term of U, i, - 1 ^4 6 Find the sum of: 6. 8 terms of 4, 7, 10 7. 10 terms of 8, 5, 2 8. 12 terms of -3, 1, 5 9. n terms of 2, 1^, - 10. n terms of 2^, 1|, l^ 11. Given a = 3, Z=: 55, 71 == 13. Find elands. 12. Given a == 3i, ^ = 64, n = 82. Find d and s. 13. Given a = l, n = 20, s = 305. Find c? and <^. 14. Given I = 105, n = l^,s = 840. Find a and d, 15. Given c?:= 7, 71=12, s = 594. Find a and /. 16. Given a = 9, c^ = 4, s = 624. Find n and /. 17. Given c? = 5, ^ = 77, 5 = 623. Find a and n. 182 ALGEBRA. 18. When a train arrives at the top of a long slope, tne last car is detached and begins to descend, passing over 3 feet in the first second, three times 3 feet in the second second, five times 3 feet in the third second, etc. At the end of 2 minutes it reaches the bottom of the slope. What was its velocity in the last second ? 19. Insert eleven arithmetical means between 1 and 12. 20. The first term of an arithmetical series is 3, and the sum of six terms is 28. What term will be 9 ? 21. How many terms of the series — 5, — 2, + 1, + must be taken in order that their sum may be 63 ? 22. The arithmetical mean between two numbers is 10, and the mean between the double of the first and the triple of the second is 27. Find the numbers. 23. The first term of an arithmetical progression is 3, the third term is 11. Find the sum of seven terms. 24. Arithmetical means are inserted between 8 and 32, so that the sum of the first two is to the sum of the last two as 7 is to 25. How many means are inserted ? 25. In an arithmetical series the common difference is 2, and the square roots of the first, third, and sixth terms form a new arithmetical series. Find the series. 26. Find three numbers in arithmetical progression of which the sum is 21, and the sum of the first and second f of the sum of the second and third. 27. The sum of three numbers in arithmetical progres- sion is 33, and the sum of their squares is 461. Find the numbers. 28. The sum of four numbers in arithmetical progres- sion is 12, and the sum of their squares 116. What are these numbers? ARITHMETICAL PROGRESSION. 183 29. How many terms of the series 1, 4, 7 must be taken, in order that the sum of the first half may bear to the sum of the second half the ratio 7 : 22 ? 30. The sum of the squares of the extremes of four num- bers in arithmetical progression is 200, and the sum of the squares of the means is 136. What are the numbers ? 31. A man wishes to have his horse shod. The black- smith asks him $2 a shoe, or 1 cent for the first nail, 3 for the second, 5 for the third, etc. Each shoe has 8 nails. Ought the man to accept the second proposition ? 32. A number consists of three digits which are in arithmetical progression ; and this number divided by the sum of its digits is equal to 26 ; if 198 be added to the number, the digits in the units' and hundreds' places will be interchanged. Required the number. 33. There are placed in a straight line upon a lawn 50 eggs 3 feet distant from each other. A person is required to pick them up one by one and carry them to a basket in the line of the eggs and 3 feet from the first egg, while a runner, starting from the basket, touches a goal and re- turns. At what distance ought the goal to be placed that both men may have the same distance to pass over ? 34. Starting from a box, there are placed upon a straight line 40 stones, at the distances 1 foot, 3 feet, 5 feet, etc. A man placed at the box is required to take them and carry them back one by one. What is the total distance that he has to accomplish ? • 35. The sum of five numbers in arithmetical progres- sion is 45, and the product of the first and fifth is | of the product of the second and fourth. Find the numbers. 184 ALGEBRA. GEOMETRICAL PROGRESSION. 225. A series is called a geometrical series or a geometrical progression when each succeeding term is obtained by mul- tiplying the preceding term by a constant multiplier. The general representative of such a series will be in which a is the first term and r the constant multiplier or ratio. The terms increase or decrease in numerical magnitude according as r is numerically greater than or numerically less than unity. 226. The nth Term. Since the exponent of r increases by one for each succeeding term after the first, the ex- ponent will always be one less than the number of the term, so that the nth term is ar''~\ If the nth term is represented by I, we have l=ar^-\ I. 227. Sum of the Series. If I represent the nth term, a the first term, n the number of terms, r the common ratio, and s the sum of n terms, then s ^= a -\- ar -\- ar^ -\- ar"~\ Multiply by r, rs= ar -{- ar^ -\- ar^ -\- ar""'^ -{- ar^. Subtracting the first equation from the second, rs — s^= ar'^ — a, or (r — 1) s = a (r" — 1). .■.«=^':^^. II. r — 1 GEOMETRICAL PROGRESSION. 185 Since l = ar''-^, aT^ = rl, and II. may be written rl—a III 228. From the two equations I. and II., or the two equations I. and III., any two of the five numbers a, r, I, n, s, may be found when the other three are given. (1) The first term* of a geometrical series is 3, the last term 192, and the sum of the series 381. Find the num- ber of terms and the ratio. (1) (2) From I. and III., 192 = = 3r«-», From (2), 381 = r- 192r-3 r-1 = 2. Substituting in (1), 2"-i = = 64. .•. n = = 7. The series is 3, 6, U 5, 24, 48, 96, 19 (2) Find I when r, , n. s are given. From I., a = I rn-i Substituting in III., s = - r-1 ' (r- -l)s- r«-i .\l- (r-l)r«-is r«-l Note. The table on page 186 contains the results of all possible problems in geometrical series in which three of the numbers a, r, I, n, s are given and the other two required, with the exception of those in which n is required ; these last require the use of logarithms with which the student is supposed to be not yet acquainted. 186 ALGEBRA. No. 1 2 3 4 Given. Required. Results. a r n a r s ans r n s I l = ar^-\ ^_a + {r-l)s r ;(s_Z)n-i_a(s-a)"-i = 0. 5 6 7 8 a r n arl a n I ml ' 3 a(r--lX r-l r** - r«-i 9 10 11 12 ml r n s rls n I s a r"-l a = W-(r-l)s. 13 14 15 16 a n I ans als nls r s , s — a A a a s-l s-l 229. The geometrical mean, between two numbers is the number which stands between them, and makes with them a geometrical series. GEOMETRICAL PROGRESSION. 187 If a and b denote two numbers, and G their geometrical mean, then, by the definition of a geometrical series, a G .'.G=V'^. 230. Sometimes it is required to insert several geometri- cal means between two numbers. Insert three geometrical means between 3 and 48. Here the whole number of terms is five ; 3 is the first term and 48 the fifth. By I., 48 = 3r*, r* = 16, r = ±2. The series is one of the following . 3, [ 6, 12, 24,] 48; 3, [-6, 12, -24,] 48. The terms in brackets are the means required. 231. Infinite Geometrical Series. When r is less than 1, the successive terms become numerically smaller and smaller ; by taking n large enough we can make the nth term, ar""\ as small as we please, although we cannot make it absolutely zero. The sum of n terms, , may be written ; r— 1 1 — rl — r this sum differs from by the fraction ; by taking 1 — r 1 — r enough terms we can make I, and consequently this dif- ference, as small as we please ; the greater the number of terms taken the nearer does their sum approach- . 1 — r Hence — — is called the sum of an infinite number of 1 — r terms of the series. 188 ALGEBRA. (1) Find the sum of the infinite series 1-1+1-1+ 2 4 8^ Here, a = l, r = 2 1 2 The sum of the series is or — Ans. 1+^ 3 "We find for the sum of n terms ( — ] ,or- + - 3 3 V 2/ '33 2 this sum evidently approaches - as n is increased. {~ir (2) Find the value of the recurring decimal .12135135. Consider first the part that recurs; this may be written 135 135 135 J ,1 f i-u- • • 100000 — — 1 \- and the sum of this series is , 100000 100000000 ^ 1_ 1000 which reduces to Adding .12, the part that does not recur, we 449 obtain for the value of the decimal -— ^- Ans. 3700 ■n- 1 . Exercise 37. 1. The eighth term of 3, 6, 12, 2. The twelfth term of 2, -4, 8, .... 3. The twentieth term of 1, — -. -» o y 4. The eighteenth term of 3, 2, 1^, 5. The nth term of 1, — IJ, 1|, Find the sum of : 6. Eleven terms of 4, 8, 16, 7. Nineteen terms of 9, 3, 1, GEOMETRICAL PROGRESSION. 189 8. Twelve terms of 5, —3, If, 9. n terms of li, -. — , ^ 8 40 Sum to infinity : 10. 4-2+1- 12. l-? + i--.. 5^25 "• 1+1+1+ ''■ l+i+h+ Find the value of the recurring decimals : 14. 0.153153 16. 3.17272 15. 0.123535 17. 4.2561561. 18. Given a = 36, 1=2^, n — 5. Find r and s. 19. Given I = 128, r = 2,n='J. Find a and s. 20. Given r = 2,n=7,s = 635. Find a and I. 21. Given ^=1296, r = 6, s = 1555. Find a and w. 22. Insert three geometrical means between 14 and 224. 23. Insert five geometrical means between 2 and 1458. 24. If the first term is 2 and the ratio 3, what term will be 162 ? 25. The fifth term of a geometrical series is 48, and the ratio 2. Find the first and seventh terms. 26. Four numbers are in geometrical progression ; the sum of the first and fourth is 195, and the sum of the second and third is 60. Find the numbers. 27. The sum of four numbers in geometrical progression is 105 ; the difference between the first and last is to the difference between the second and third in the ratio of 7 : 2. Find the numbers. 190 ALGEBRA. 28. The first term of an arithmetical progression is 2, and the first, second, and fifth terms are in geometrical progression. Find the sum of 11 terms of the arithmetical progression. 29. The sum of three numbers in arithmetical progres- sion is 6. If 1, 2, 5 be added to the numbers, the three resulting numbers are in geometrical progression. Find the numbers. 30. The sum of three numbers in arithmetical progres- sion is 15; if 1, 4, 19 be added to the numbers, the results are in geometrical progression. Find the numbers. 31. There are four numbers of which the sum is 84; the first three are in geometrical progression and the last three in arithmetical progression ; the sum of the second and third is 18. Find the numbers. 32. There are four numbers of which the sum is 13, the fourth being 3 times the second ; the first three are in geometrical progression and the last three in arithmetical progression. Find the numbers. 33. The sum of the squares of two numbers exceeds twice their product by 576 ; the arithmetical mean of the two numbers exceeds the geometrical by 6. Find the numbers. 34. A number consists of three digits in geometrical progression. The sum of the digits is 13 ; and if 792 be added to the number, the digits in the units' and hundreds' places will be interchanged. Find the number. 35. Find an infinite geometrical series in which each term is 5 times the sum of all the terms that follow it. 36. If a, h, c, d are four numbers in geometrical pro- gression, show that (a^ + b' + c') {b' -i-c'-j- d') = (ab + be -{- cd)\ HARMONICAL PROGRESSION. 191 HARMONICAL PROGRESSION. 232. A series is called a harmonical series, or a hannonical progression, when the reciprocals of its terms form an arith- metical series. The general representative of such a series will be 1^ 1 1 ^ 1 a a-\-d a -{-2d a-\-(n — l)d Questions relating to harmonical series are best solved by writing the reciprocals of its terms, and thus forming an arithmetical series. 233. If a and b denote two numbers, and ff their har- monical mean, then, by the definition of a harmonical series, JI a b H' . 2 ^1 l_ a-hb ' H a b ab 2ab H a-^b 234. Sometimes it is required to insert several harmoni- cal means between two numbers. Ex. Insert three harmonical means between 3 and 18. Find the three arithmetical means between - and — • These are found to be i^, — , — ; therefore, the harmonical means 235. Since ^ = ^i^ and = Vab, G' H='^ or 0=y/AH, 192 ALGEBRA. Th^ is, the geometrical mean between two numbers is also the geometrical mean between the arithmetical and harmonical means of the numbers, or Hence O lies in numerical value between A and H. Exercise 38. 1. Insert four harmonical means between 2 and 12. 2. Find two numbers whose difference is 8 and the har- monical mean between them If. 3. Find the seventh term of the harmonical series 3, 3f, 4 4. Continue to two terms each way the harmonical series of which two consecutive terms are 15, 16. 5. The first two terms of a harmonical series are 5 and 6. What term will be 30? 6. The fifth and ninth terms of a harmonical series are 8 and 12. Find the first four terms. 7. The difference between the arithmetical and harmon- ical means between two numbers is If, and one of the numbers is four times the other. Find the numbers. 8. The arithmetical mean between two numbers exceeds the geometrical by 13, and the geometrical exceeds the har- monical by 12. What are the numbers? 9. The sum of three terms of a harmonical series is 39, and the third is the product of the other two. Find the terms. 10. When a, 5, c are in harmonical progression, show that a: c = a — b : b — c. 11. If a and b are positive, which is the greater, A or £[? CHAPTER XVII. SIMPLE INDETERMINATE EQUATIONS. 236. If a single equation involving two unknown numbers be given, and no other condition be imposed, the number of solutions of the equation is unlimited ; for if one of the unknown numbers be assumed to have any value, a corre- sponding value of the other may be found. Such an equation is called an indeterminate equation. Although the number of solutions of an indeterminate equation is unlimited, the values of the unknown numbers are confined to a particular range ; this range may be fur- ther limited by requiring that the unknown numbers shall be 'positive integers. 237. Every indeterminate equation of the first degree, in which X and y are the unknown numbers, may be made to assume the form 7 ax ± by =^ ± c, where a, 5, and c are positive integers and have no common factor. 238. The method of solving an indeterminate equation in positive integers is as follows : (1) Solve 3^ -f- 4y = 22, in positive integers. Transpose, 3 a; = 22 — 43/. the quotient being written as a mixed expression. .-. a; + 3/ - 7 = — 3^- 194 ALGEBRA. Since the values of x and y are to be integral, x +y — 1 will be integral, and hence ~ -^ will be integral, though written in the form of a fraction. -r , 1 — V i^et — ~- = m, an integer. Then l-y = Sm. .'. y = 1 — 3m. Substitute this value of y m the original equation, 3a; + 4- 12m = 22. .*. « = 6 + 4to. The equation y = 1 — 3 m shows that m may be 0, or have any negative integral value, but cannot have a positive integral value. The equation a; = 6 + 4m further shows that m may be 0, but can- not have a negative integral value greater than 1. .•. m may be or —1, and then a;=6| x =2 ) y = lj' y = 4) (2) Solve 5a; — 14?/ = 11, in positive integers. Transpose, 5 a; = 11 + 14 3/, , = 2 + 2y + iiii^. (1) .•..-2y-2 = l±l^. Since x and y are to be integral, a; — 2^ — 2 will be integral, and hence — — ^ will be integral. o Let — — ^ = m, an integer. o Then V ^^-1 4 2/-^ + ^- (2) _j -1 Now — - — must be integral. Let — ^ = n, an integer. Then m =4?i + 1 SIMPLE INDETERMINATE EQUATIONS. 195 Substituting in (2), y = 5n+l. Substituting in (1), a; = 14n + 5. Obviously x and y will both be positive integers if n have any positive integral value. Hence, a; = 5, 19, 33, 47, y = l, 6, 11, 16, Another method of solution is the following : * From the given equation we have x = — — 2^. 5 Here y must be so taken that 11 -f 14 y is a multiple of 5 ; take 2/ = 1, then x = 5, and we have one solution. Now 5a;— 142/ = 11, and 5 (5) -14(1) = 11. Subtract, 5(a;-5)-14(y-l) = 0, 2/.-1 5 Since x—5 and y — 1 are integers, a; — 5 must be the same mul- tiple of 14 that y — 1 is of 5. Hence, if a; — 5 = 14m, then y — \ = 5m. .-, «= 14m + 5, and y = 5m + l. Therefore a; = 5, 19, 33, 47 y = l, 6, 11, 16, It will be seen from (1) and (2) that when only positive integers are required, the number of solutions will be limited or unlimited according as the sign connecting a; and y is positive or negative. (3) Find the least number that when divided by 14 and 5 will give remainders 1 and 3 respectively. If N represent the number, then = a?, and = y. 14 5 -^ .-. iV= 14a; + 1, and iV= 5y + 3. .-. 14ar + 1 = 53/ + 3. 5y = 14x-2, 53/= 15a;- 2 -a;. Q 2 + a; Let = m, an integer. 5 /. a; = 5m — 2. 196 ALGEBRA. y == -J (14 a; — 2), from original equation. .-.2/ = 14m — 6. Ifw=l, a; = 3, and 2/ = 8, .-. iV= 14x + 1 = 52/ + 3 = 43. Ans. (4) Solve 5a: -f- 6y = 30, so that a; may be a multiple of y, and both x and y positive. Let Then and X = = my. (5m + Q)y- = 30. • ■y = 30 5m + 6 x = 30 m 5m + 6 x = ~-H,y = n- X = = ^,y = If. Ifm = 2, Ifm = 3, (5) Solve 1457 + 223/ — 71, in positive integers. a? = 5 — V H — — — ^• ^ 14 If we multiply the fraction by 7 and reduce, the result is -4y + i a form which shows that there can be no integral solution. There can be no integral solution of ax ±by = ± c if a and b have a common factor not common also to c ; for, if cZ be a factor of a and also of h, but not of c, the equation may be written mdx ± ndy = ± c, or nx ± ny = ± - ; d which is impossible, since - is a fraction, and mx ± ny is an integer, if X and y are integers. Exercise 39. Solve in positive integers: 1. x + y=l2. 4. 8a; + 5y = 74. 2. 2a; + lly = 83. 5. 5a: + 3y = 105. 3. 4a7 + 9y = 53. 6. fa; + 5y = 92. SIMPLE INDETERMINATE EQUATIONS. 197 7. ix+\y = 21. 8. 2^ + iy = 53. Solve in least possible integers : 9. 1x-2y^l2. 12. lla;-62/=73. '10. 92:-5y = 21. 13. 15a;-47y = ll. 11. 7:c-4y--45. 14. 23a;-14y = 99. 15. Find two numbers which, multiplied respectively by 7 and 17, have for the sum of their products 1135. 16. If two numbers are multiplied respectively by 8 and 17, the difference of their products is 10. What are the numbers ? 17. If two numbers are multiplied respectively by 7 and 15, the first product is greater by 12 than the second. Find the numbers. 18. Divide 89 in two parts, one of which is divisible by 3, and the other by 8. 19. Divide 314 in two parts, one of which is a multiple of 11, and the other a multiple of 13. 20. What is the smallest number which, divided by 5 and by 7, gives each time 4 for a remainder ? 21. The difference of two numbers is 151. The first divided by 8 has 5 for a remainder, and 4 must be added to the second to make it divisible by 11. What are the numbers ? 22. Find pairs of fractions whose denominators are 24 and 16, and whose sum is -^f . 23. How can one pay a sum of $87, giving only bills of $5 and $2? 198 ALGEBRA. 24. A man buys calves at $5 apiece, and pigs for $3 apiece. He spends in all $114. How many did he buy of each ? 25. A person bought 40 animals, consisting of pigs, geese, and chickens, for $40. The pigs cost $5 apiece, the geese $ 1, and the chickens 25 cents each. Find the num- ber he bought of each. 26. Solve 18a; — 5y= 70 so that y may be a multiple of X, and both positive. 27. Solve Sx-\- \2y = 23 so that x and y may be posi- tive, and their sum an integer. 28. Divide 70 into three parts which shall give integral quotients when divided by 6, 7, 8, respectively, and the sum of the quotients shall be 10. 29. In how many ways can $3.60 be paid with dollars and twenty-cent pieces ? 30. In how many ways can 300 pounds be weighed with 7 and 9 pound weights ? 31. Find the general form of the numbers that, divided by 2, 3, 7, have for remainders 1, 2, 5, respectively. 32. Find the general form of the numbers that, divided by 7, 8, 9, have for remainders 6, 7, 8, respectively. 33. A farmer buys oxen, sheep, and hens. The whole number bought is 100, and the total cost £100. If the oxen cost £5, the sheep £1, and the hens Is. each, how many of each did he buy ? 34. A farmer sells 15 calves, 14 lambs, and 13 pigs, and receives $200. Some days after, at the same price, he sells 7 calves, 11 lambs, and 16 pigs, for which he receives $141. What is the price of each ? CHAPTER XVIII. BINOMIAL THEOREM. 239. Binomial Theorem, Positive Integral Exponent. By suc- cessive multiplication we obtain the following identities : {a-\-hf = d + 2>a'h^^ah'' +5^; {a + by = a'-i-4:a'b + Qa'b' + ^ab' + b\ The expressions on the right may be written in a form better adapted to show the law of their formation : {a + by = a' + 2ab +f^^^ ^a-^by^a^ + Sa^b + ^^ab^ +rS^'' Note, The dot between the Arabic figures means the same as the sign X. 240. Let n represent the exponent of (a -\- b) in any one of these identities ; then, in the expressions on the right, we observe that the following laws hold true : I. The number of terms is w + 1. II. The first term is a", and the exponent of a is one less in each succeeding term. The first power of b occurs in the second term, the second power in the third term, and the exponent of h is one greater in each succeeding term. The sum of the exponents of a and i in any term is n. 200 ALGEBRA. III. The coefficient of the first term is 1 ; of the second term, n; of the third term, '—^~—-^\ of the fourth term, n{n-V) {n-2) . ^'^ i.' A' o 241. Consider the coefficient of any term : the number of factors in the numerator is the same as the number of factors in the denominator, and the number of factors in each is the same as the exponent of h in that term ; this exponent is one less than the number of the term. 242. Proof of the Theorem. That the laws of § 240 hold true when the exponent is any positive integer, is shown as follows : We know that the laws hold for the fourth power; suppose, for the moment, that they hold for the kth power. We shall then have (a + hf = a^ + ha^'-'b + ^(^~^) a^-^&^ 1 * A + ^(^-IK^- 2) ^.-3^3 _f_ (1) i .iJ ' O Multiply both members of (1) by a + ^ ; the result is {a + bf-^' = a*+i+ {h + 1) a^5 +i^±^ a^-^6^ X ' A ^ (^ + l)^-(/^- 1) ^^-233 _|_ (2) In the right member of (1) for h put ^+ 1 ; this gives ^ r2^3 "^ ^"^ BINOMIAL THEOREM. 201 This last expression, simplified, is seen to be identical with the right member of (2), and this in turn by (2) is identical with (a -\- 5)*+\ Hence (1) holds when for h we put ^ + 1 ; that is, if the laws of § 240 hold for the Tcth power, they must hold for the (Jc + 1) th power. But the laws hold for the fourth power ; therefore they must hold for the fifth power. Holding for the fifth power, they must hold for the sixth power ; and so on for any positive integral power. Therefore they must hold for the nth power, if 7i is a positive integer ; and we have (a + hf = a" + na^-'h + ""^""""^ a"-^6' n(7i-l)(n-2) 353 , ^ ^ 1-2-3 Note. The proof of § 242 is an example of a proof by mathematical induction. 243. This formula is known as the binomial theorem. The expression on. the right is known as the expansion of (a + by-, this expansion is definite series when n is a positive integer. That the series is finite may be seen as follows : In writing out the successive coefficients we shall finally arrive at a coefficient which contains the factor n — n\ the corresponding term will vanish. The coefficients of the succeeding terms likewise all contain the factor n — n, and all these terms will vanish. 244. If a and b be interchanged, the identity A may be written (a + br = {b + ar = b- + nb--'a + '^^^f^b^-W I ^rn-l)(n-2)^n-3^3^ ^ 1-2-3 202 ALGEBRA. This last expansion is the expansion of A written in reverse order. Comparing the two expansions we see that : the coefficient of the last term is the same as the co- efficient of the first term ; the coefficient of the last term but one is the same as the coefficient of the first term but one ; and so on. In general, the coefficient of the rth term from the end is the same as the coefficient of the rth term from the beginning. In writing out an expansion by the binomial theorem, after arriving at the middle term, we can shorten the work by observing that the remaining coefficients are those already found written in reverse order. 245. If h be negative, the terms which involve even powers of h will be positive, and those which involve odd powers of h negative. Hence, (a - by ~a^- na^-'' b + !L(!L_J0 ^^-^ h" 1 ' Ji («-l)( «-2)^,._,y^ 3 L' 2t' o Also, putting 1 for a and x for b, in A and B, 1-2-3 ^ ^ (\-xf=l-nx^ ^^'l J^ x — -- ./, 1-2 n{n~l){n-2) ^ [ .... j) X.' 2i' o BINOMIAL THEOREM. 203 246. Examples; (1) Expand (1 + 2:z:y. In for X put 2x, and for n put 5. The result is (1 + 2xf = 1 + 5(2a;) + ^4x2 ^^-^^^^ ' ^ ' \-l 1-2-3 + f^:3:|l6.^ + 5-4-3>2-l3,^ l-2-3'4 1-2-3-4-5 = 1 + lOo; + 40^2 + 80r» + 80 x* + 32 a^. (2) Expand to three terms {^ - — Y- Put a for 1. and 6 for — ; then, by B, X 3 -^ (a-&)6 = a6-6a56 + 15a*62 + Replacing a and 6 by their values, ^J^_ J ^20_ a;^ «^ 3 247. Any Eequired Term. From A it is evident (§ 241) that the (r -|- 1) th term in the expansion of {a + 6)" is n{n—V) (n — 2) to r factors ,^7^ l'2-3 r " The (r + 1) ^A term in the expansion of (a — by is the same as the above if r be even, and the negative of the above if r be odd. Ex. Find the eighth term of U - ^T' Here a = 4, 6 = -, n = 10, r = 7. '2 The term required is 10-9-8 7•6•5•4 /^^3/a?'V 1-2-3-4-5-6-7 which reduces to — 60 a;". (4)'(f) 204 ALGEBRA. 248. The G-reatest Coefficient. Suppose that the coefficient of the (r -j- 1) th term is the numerically greatest coefficient. This coefficient, the preceding and following coefficients, are the following : r ih term, ^ ^^^\ \ rf^-^ ; (, + l),^term, -A \.2^3 (T-l)r ' (r + 2)^Aterm, n — r; r < — ^' and r > — - — If n is even, r = -, and r + 1 = — ^— ; in this case there is one middle term and its coefficient is the greatest coefficient. If n is odd, we can only have r= . T" , or r = — - — ; A A in this case there are two middle terms ; their coefficients are alike, and are the two greatest coefficients. 249. A trinomial may be expanded by the binomial theorem as follows : Expand {^ -^ "Ix - x^J , Put 2a;-aj2 =2; then (1 + 2)3 =» 1 + 32 + 3^2 + 23 .-. (1 + 2 a; - a;2)3 = 1 + 3 (2 a; - a;^) + 3 (2 a; - x^J + (2 ir - x^f =- 1 + 6a; + 9a;2 - 4a^ - 9a;* + 6a^ - a;8. BINOMIAL THEOREM. 205 ^ . Exercise 40. Expand : 1. (l + 3:r)^ 4. {2 + xJ. 7. {?>x- 2yy. H'^t)' "(MT '^(f- -f)' »^('-f)' «■ (f -^.j •■ (^i 10. (1+Ax-^Sx'y. 11. (a' -ax -2: vj. Find: 12. The fourth term of /^a; + -iY- V 2a;y 13. The eighth term of ( 2 --^J . 14. The twelfth term of (" -^)"- 15. The twentieth term of {'"wJ- 16. The fourteenth term of f ^^ ^ ) . 17. The (r + 1) ^A term of fV^ + -^^Y- 18. The (r+l)th term of ^^^ - ^T. 19. The (r + 3) th term of (^ - -^Y'- ^ ^ V2y V3^/ 20. Find the middle term of (- ^^-^ j . 206 ALGEBRA. 21. Find the two middle terms of f — ^+ \^] 22. Find the r th term from the end of f aI— ] . 23. In the expansion of {a + by show that the sum of the coefficients is 2". 24. In the expansion of {a -\- by show that the sum of the even coefficients is equal to the sum of the odd co- efficients. ^-f 25. Expand 26. If A is the sum of the odd terms, and B the sum of the even terms, in the expansion of {a + by, show that A' -B' = {a' - by. 250. Convergent and Divergent Series. By performing the indicated division, we obtain from the fraction the 1 —X infinite series 1 -}- x -\- x"^ -\- x^ -{- This series, however, is not equal to the fraction for all values of x. 251. Suppose X numerically less than 1. In this case we can obtain an approximate value for the sum of the series by taking the sum of a number of terms ; the greater the number of terms taken, the nearer will this approximate sum approach the value of the fraction. The approximate sum will never be exactly equal to the fraction, however great the number of terms taken ; but by taking enough terms, it can be made to differ from the fraction as little as we please. BINOMIAL THEOREM. 207 Thus, if a; = ;5» the fraction is r = 2, and the series is ^^hhl The sum of four terms of this series is 1^ ; the sum of five terms, l^f ; the sum of six terms, 1|| ; and so on. The successive approxi- mate sums approach, but never reach, the finite value 2. When X is numerically less than 1, the series is equal to the fraction, and we have the equation 1 l~x = l + :r + a;' + a;' + 252. A series is said to be convergent when the sum of a number of terms, as the number of terms is indefinitely- increased, approaches some fixed finite value ; this finite value is called the sum of the series. 253. In the series 1 -\- x -\- x^ -\- x^ -{- suppose x nu- merically greater than 1. In this case, the greater the number of terms taken, the greater will their sum be ; by taking enough terms we can make their sum as large as we please. The fraction, on the other hand, has a definite value. Hence, when x is numerically greater than 1, the series is not equal to the fraction, and we c?o not have the equation =1 + x -\- x^ + ce^ + 1 — x Thus, if a; = 2, the fraction is —^ = - 1 ; the series is 1 — Z 1+2 + 4 + 8 + The greater the number of terms taken, the larger will their sum be. Evidently the fraction and the series are not equal. 208 ALGEBRA. 254. In the same series suppose x=l. In this case the fraction is = -» and the series 1 + 1 + 1 + 1 + The more terms we take, the greater will the sum of the series be. "We do i? jt know whether or not the fraction is equal to the series. If X, however, is not exactly 1, but is a little less than 1, the value of the fraction will be very great, and the sum of the series also very great ; and the fraction will be equal to the series. Suppose x = — l. In this case the fraction is =-, ^^ 1 + 12 and the series 1 — 1 + 1 — 1 + If we take an even number of terms, their sum is ; if an odd number, their sum is 1. Hence, when x = — l, the fraction is not equal to the series. 255. A series is said to be divergent when the sum of a number of terms, as the number of terms is indefinitely increased, either increases without end, or oscillates in value without approaching any finite value. No reasoning can be based on a divergent series ; hence, in using an infinite series it is necessary to make such restrictions as will cause the series to be convergent. Thus we can use the infinite series 1 + x + x"^ + xi^ + when, and only when, x lies between + 1 and — 1. Observe that any series of the form A + Bx + Cx"^ + is convergent when x = 0, since in this case the series reduces to the first term. 256. Identical Series. If two seizes which are arranged by powers of x be equal for all values of x which make both series convergent, the corresponding coefficients are equal each to each. BINOMIAL THEOREM. 209 Let the equation a-^-hx^ ex" -f- da^ + -=:^A + Bx+Cx''-\- Bx"-}- (1) hold true for all values of x which make both series con- vergent. Since this equation holds true for all values of x which make both series convergent, it will hold true when x = 0. Let 37 = 0; then a = A. (2) Subtract (2) from (1) and divide both members by x; then b + cx + dx' + = B+Cx-{-Dx^ + (3) Let ar = ; then b = B. (4) Subtract (4) from (3) and divide both members by x ; then c-i-dx + = C-i-Dx + Let x~0; then c= C. And so on. 257. Binomial Theorem, Any Exponent, We have seen (§ 245) that when w is a positive integer we have the identity /I I N» II I n(n — l) 2 , ^(^ — l)(w — 2) , , (l + xf-l + nx-i- -y-^^ + -^.^.3 ^^ + We proceed to the case of fractional and negative expo- nents. I. Suppose n is a positive fraction, -• We may assume that (1 4- xy =(A + Bx + Cx' + Dx' + y, (1) provided x be so taken that the series A + Bx+Cx' + Bx'i- is convergent (§ 252). That this assumption is allowable may be seen as follows : Expand both members of (1). 210 ALGEBRA. We obtain and 1 Z 1. A o In the first k coefficients of the second series there enter only the first k' of the coefficients A, B, C, D, If, then, we equate the coefficients of corresponding terms in the two series (| 256) as far as the kth term, we shall have just k equations to find k unknown num- bers A, B, C, D, Hence the assumption made in (1) is allowable. Comparing the two first terms and the two second terms, we obtain ^3 = 1, .'.A = l; qAi-^B^p, .\B = ~- Extracting the qth root of both members of (1), we have (1 + xf = 1 + -X+ Cx' + Da^ + (2) where x is to be so taken that the series on the right is convergent. II. Suppose w is a negative number, integral or frac- tional. Let n = — 7n, so that m is positive ; then 1 (l+:r)'^ = (l + ^)- (1 + ^)^ From (2), whether w is integral or fractional, we may assume 1 1 (1 -]-xY 1 + wrr + ex'' -f da? + * By actual division this gives an equation in the form (l-\- x)-"^ = l-mx-{-Cx' -{- Da^ -{- (3) BINOMIAL THEOREM. 211 258. It appears from (2) and (3) that whether n be inte- gral or fractional, positive or negative, we may assume (1 + xf = l-\-nx+Cx'-\- Dx^ + , provided the series on the right is convergent. Squaring both members, Also, since a;' + 2Z> b^ + -\-2nC\ (1) we have, putting 2x-\-x^ for y, (1 + 2a; + x^f = 1 + w(2a; + x^) + (7(2a; + x'Y -{- D{2x -\- x'J + 8i) 3(? + = 1 -\-2nx-\-n +4e Comparing corresponding coefficients in (1) and (2), 4(7+8i) = 2i)+27i(7. n{n — 1) (2) \2C =n' C 1-2 3i) = (^-2)(7, ^^'/L (^-1)(^-2X 1. A o and so on. Hence, whether n be integral or fractional, positive or negative, we have {\-{-xf=l + nx-[- n{n-l) ^, _^ n{n-l){n-2)^ _^ 1-2 ■ l'2-3 provided, always, x be so taken that the series on the right is convergent. The series obtained will be an infinite series unless n is a positive integer (§ 243). 212 ALGEBRA. 259. If X is negative, Also, if X a, (a + ^)« = (rr + ay = x''fl + H „ A , a , ?z(n — 1) a^ , \ = a;" + nax^-' + ^ifc^nHaV-^ + 1-2 ^ 260. Examples. (1) Expand (1 + x)^. ^ 3-6 3-6-9 The above equation is only true for those values of x which make the series convergent. (2) Expand — Vl—x _1._6 _JL._5._9 1 - (- i)- + -Iry-^-^ -- VsV^^ + l + lx + —x^ + ^'^'^ a^ + ^ 4-8 4-8-12 if X is so taken that the series is convergent. BINOMIAL THEOREM. 213 A root may often be extracted by means of an expansion. (3) Extract the cube root of 344 to six decimal places. 344-343(l+A) = 73(l+J-3} - 7 (1 + .000971815 - .000000944), = 7.006796. (4) Find the eighth term of (a; — -r—^j • Here (§ 147) a = ^^ b=.-^^J-, n = -i, r=7. Wx 4 a'* ^ ^-(A)' The term is 1-2-3-4-5-6-7 1-3-5-7-9-11-13-37 or . — — . 2 • 4 • 6 • 8 • 10 • 12 • 14 • 4^ • a" Exercise 41. Expand to four terms : 1. (l + x)^. 4. (l-x)-\ 7. iS'+ 8R, third year =S+8R^8I^, n th year =8+8R + 8R''\- + SR'-K That is, the amount A = 8+8R-^ 8R' + + ^^"'• .'. AR^--8R + 8R' + 8I^-{- + 8R^. .-. AR-A^8R''-8 . ,_ 8(R^-1) " R — 1 ' 8(R^-1) or, A= (1) If $10,000 be set apart annually, and put at 6 per cent compound interest for 10 years, what will be the amount ? A SjE^ - 1) . $ 10.000(1.06^0 _ 1) r 0.06 By logarithms the amount is found to be 1 131,740 (nearly). 236 ALGEBRA. (2) A county owes $60,000. What sum must be set apart annually, as a sinking fund, to cancel the debt in 10 years, provided money is worth 6 per cent ? o Ar $60,000x0.06 m.^f-f- / j^ S= — = ^ — '- — = 1 4555 (nearly). i2«-l 1.061" -1 ^ ^ ^^ Note. The amount of tax required yearly is $3600 for the interest and 1 4555 for the sinking fund ; that is, $8155. 291. Annuities. A sum of money that is payable yearly, or in parts at fixed periods in the year, is called an annuity. To find the aviount of an unpaid annuity when the inter- est, time, and rate per cent are given. The sum due at the end of the first year = S, second year = 8+ 8R, third year = S -\- 8R -{- 8R\ n th year = 8 ■\- 8R + 8^" + + 8R''-\ That is, A= ^^\~^^ §290 An annuity of $ 1200 was unpaid for 6 years. What was the amount due if interest be reckoned at 6 per cent ? ^ _ S{R''-\) _ $1200(1.06e-l) _ ^ gg^^ r 0.06 292. To find the present worth of an annuity when the time it is to continue and the rate per cent are given. Let P denote the present worth. Then the amount of P for n years will be equal to A the amount of the annuity for n years. But the amount of P for n years = P(l + r)« = Pi^, §289 and ^=^?(f^i). §291 R — 1 . p^r,^ 8{Rr~l) : *■ R-l INTEREST AND ANNUITIES. 237 P=.^x^-l Br R-\ This equation may be written B-l B' R-l\ B^J As n increases, the expression ('-*) approaches 1. Therefore if the annuity he perpetual, B-1 r (1) Find the present worth of an annual pension of $ 105, for 5 years, at 4 per cent interest. P- ^ w -?^"-l j|105 1.0#-1 ^ 1.0# 1.04-1 ^ ^ ^^ (2) Find the present worth of a perpetual scholarship that pays $300 annually, at 6 per cent interest. p=^=i^ = $5000. r 0.06 * 293. To find the present worth of an annuity that begins in a given number of years, when the time it is to continue and the rate per cent are given. Let p denote the number of years before the annuity begins, and q the number of years the annuity is to continue. Then the present worth of the annuity to the time it terminates is i?'+« B-l ' 238 ALGEBRA. and the present worth of the annuity to the time it begins is R^ R-l Hence, \B'+' R~\ 1 \R' R i} If the annuity is to begin at the end of p years, and to be perpetual, the formula becomes P = _ , _ — — - X - R^+' R - 1 R\R - 1) R' And since — - — approaches 1 (§ 292), R^ R'(R - 1) (1) Find the present worth of an annuity of $5000, to begin in 6 years, and to continue 12 years, at 6 per cent interest. i^^+e E - 1 1.0618 ^ 0.06 * ' (2) Find the present worth of a perpetual annuity of $ 1000, to begin in 3 years, at 4 per cent interest. P = ^ = ^^^'^'^ = $22,225. Rp{B-l) 1.043x0.04 INTEREST AND ANNUITIES. 239 294. To find the annuity when the present worth, the time, and the rate per cent are given. i^'^ - 1 Br~l What annuity for 5 years will $4675 give when interest is reckoned at 4 per cent ? S=FrX -^—- = $4675 X 0.04 X ^'^^^ = $ 1050. R^ - 1 1.04^ - 1 295. Life Insurance. In order that a certain sum may be secured, to be payable at the death of a person, he pays yearly a fixed premium. If P denote the premium to be paid for n years to insure an amount ^, to be paid immediately after the last pre- mium, then R-l .^ p_ A{E-V) _ Ar If A is to be paid a year after the last premium, then p^ A{R-l) _ R{Br-l) B{I^-l) Note. In the calculation of life insurances it is necessary to em- ploy tables which show for any age the probable duration of life. 296. Bonds. If F denote the price of a bond that has n years to run, and bears r per cent interest, S the face of the bond, and q the current rate of interest, what interest on his investment will a purchaser of such a bond receive ? Let X denote the rate of interest on the investment. 240 ALGEBRA, Then P(l -\- x^ is the value of the purchase money at the end of n years. jSr(l + qf-'' + jSr(l + qY'' + + Sr+S is the amount of money received on the bond if the interest received from the bond is put immediately at compound interest at q per cent. But Sr{l + qy-'' + jSr(l-\- qf-' + + Sr + S ••'+" U+ Fq \ ^q ^ r (1) What interest will a person receive on his invest- ment if he buys at 114 a 4 per cent bond that has 26 years to run, money being worth 3i per cent? 1 ■:. / 3.5 + 4(1.035)^«-4 y^ ^^ 3.99 ; ■ By logarithms, 1 + x = 1.033. That is, the purchaser will receive 3J per cent for his money. (2) At what price must 7 per cent bonds, running 12 years, with the interest payable semi-annually, be bought, in order that the purchaser may receive on his investment 5 per cent, interest semi-annual, which is the current rate of interest ? P- <7(1 ^xf In this case B= 100; and, as the interest is semi-annual, g = 0.025, r = 0.035, w-24, a; = 0.025. Hence. p^ 2.5 + 3.5(1.025^- 3.5. 0.025(1.025)24 By logarithms, P= 118. INTEREST AND ANNUITIES. 241 Exercise 46. 1. In how many years will $100 amount to $1050, at 5 per cent compound interest? 2. In how many years will $A amount to $5 (1) at simple interest, (2) at compound interest, r and H being used in their usual sense ? 3. Find the difference (to five places of decimals) be- tween the amount of $ 1 in 2 years, at 6 per cent compound interest, according as the interest is due yearly or monthly. 4. At 5 per cent, find the amount of an annuity of f A which has been left unpaid for 4 years. 5. Find the present value of an annuity of $100 for 5 years, reckoning interest at 4 per cent. 6. A perpetual annuity of $1000 is to be purchased, to begin at the end of 10 years. If interest is reckoned at 3 J per cent, what should be paid for it ? 7. A debt of $1850 is discharged by two payments of $ 1000 each, at the end of one and two years. Find the rate of interest paid. 8. Reckoning interest at 4 per cent, what annual pre- mium should be paid for 30 years, in order to secure $2000 to be paid at the end of that time, the premium being due at the beginning of each year ? 9. An annual premium of $ 150 is paid to a life-insurance company for insuring $5000. If money is worth 4 per cent, for how many years must the premium be paid in order that the company may sustain no loss? 242 ALGEBRA. 10. What may be paid for bonds due in 10 years, and bearing semi-annual coupons of 4 per cent each, in order to realize 3 per cent semi-annually, if money is worth 3 per cent semi-annually ? 11. When money is worth 2 per cent semi-annually, if bonds having 12 years to run, and bearing semi-annual coupons of 3J per cent each, are bought at 114i, what per cent is realized on the investment ? 12. If f 126 is paid for bonds due in 12 years, and yield- ing 3^ per cent semi-annually, what per cent is realized on the investment, provided money is worth 2 per cent semi- annually ? 13. A person borrows $600.25, How much must he pay annually that the whole debt may be discharged in 35 years, allowing simple interest at 4 per cent? 14. A perpetual annuity of $100 a year is sold for $2500. At what rate is the interest reckoned? 15. A perpetual annuity of $320, to begin 10 years hence, is to be purchased. If interest is reckoned at 3-^ per cent, what should be paid for it ? 16. A sum of $10,000 is loaned at 4 per cent. At the end of the first year a payment of $400 is made; and at the end of each following year a payment is made greater by 30 per cent than the preceding payment. Find in how many years the debt will be paid. 17. A man with a capital of $100,000 spends every year $9000. If the current rate of interest is 5 per cent, in how many years will he be ruined ? 18. Find the amount of $365 at compound interest for 20 years, at 5 per cent. CHAPTER XXI. CHOICE. 297. Fundamental Principle. If one thing can he done in a different ways, and, when it has been done, a second thing can he done in b different ways, then the two things can he done together m a X b different ways. For, corresponding to the first way of doing the first thing, there are h different ways of doing the second thing ; corresponding to the second W3ij of doing the first thing, there are h different ways of doing the second thing ; and so on for each of the a different ways of doing the first thing. Therefore there are aX b different ways of doing the two things together. (1) If a box contains four capital letters. A, B, C, D, and three small letters, x, y, z, in how many different ways may two letters, one a capital letter and one a small letter, be selected ? A capital letter may be selected in four different ways, since any one of the letters A, B, C, D, may be selected A small letter may be selected in three different ways, since any one of the letters x, y, z, may be selected. Any small letter may be put with any capital letter. Thus, with A we may put x, or 3/, or 2 ; with B we may put x, or y, or 2 ; with we may put x, or y, or z; with D we may put x, or y, or z. Hence the number of ways in which a selection may be made is 4 X 3, or 12. These ways are : Ax Bx Cx Dx Ay By Oy Dy Az Bz Cz Dz 244 ALGEBRA. (2) On a shelf are* 7 English, 5 French, and 9 German books. In how many ways may two books, not in the same language, be selected ? An English book and a French book can be selected in 7x5, or 35, ways. A French book and a German book in 5 x 9, or 45, ways. An English book and a German book in 7 X 9, or 63, ways. Hence, there is a choice of 35 + 45 + 63, or 143, ways. Ans. (3) Out of the ten figures, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, how many numbers, each consisting of two figures, can be formed? Since has no value in the left-hand place, the left-hand place can be filled in 9 ways. The right-hand place can be filled in 10 ways, since repetitions of the digits are allowed (as 22, 33, etc.). Hence, the whole number is 9 x 10, or 90. Ans. 298. By successive application of the principle of § 297 it may be shown that, If one thing can be done in a different ways, and then a second thing can be done in b different ways, then a third thing in c different ways, then a fourth thing in d different ways, etc., the number of different ways of doing all the things together will Je a X b X c X d, etc. For, the first and second things can be done together in axb different ways (§ 297), and the third thing in c differ- ent ways; hence, by § 297, the first and second things and the third thing can be done together m{aXb)X c different ways. Therefore, the first three things can be done in aX b X c different ways. And so on for any number of things. Ex. In how many ways can four Christmas presents be given to four boys, one to each boy ? The first present may be given to any one of the boys ; hence there are 4 ways of disposing of it. CHOICE. 245 The second present may be given to any one of the other three boys ; hence there are 3 ways of disposing of it. The third present may be given to either of the other two boys ; hence there are 2 ways of disposing of it. The fourth present must be given to the last boy ; hence there is only 1 way of disposing of it. There are, then, 4 X 3 x 2 x 1, or 24, ways. Arts. 299. Selections and Arrangements. (1) In how many ways can a vowel and a consonant be chosen out of the alphabet ? Since there are in the alphabet 6 vowels and 20 consonants, a vowel can be chosen in 6 ways and a consonant in 20 ways, and both (g 297) in 6 X 20, or 120, ways. (2) In how many ways can a two-lettered word be made, containing one vowel and one consonant ? The vowel can be chosen in 6 ways and the consonant in 20 ways ; and then each combination of a vowel and a consonant can be written in 2 ways ; as ac, ca. Hence, the whole number of ways is 6 x 20 x 2, or 240. These two examples show the difference between a selec- tion or combination of different things, and an arrangement or permutation of the same things. Thus, ac form a selection of a vowel and a consonant, and ac and ca form two different arrangements of this selection. From (1) it is seen that 120 different selections can be made with a vowel and a consonant ; and from (2) it is seen that 240 diflferent arrangements can be made with the same. Again, a, 6, c is a selection of three letters from the alphabet. This selection admits of 6 different arrangements, as follows : ahc bca cab acb bac cba A selection or combination of any number of things is a group of that number of things put together without regard to their order. 246 ALGEBRA. An arrangement or permutation of any number of things is a group of that number of things put together, regard being paid to their order. 300. Arrangements, Things all different. The number of different arrangements (or permutations) of n different things taken all together is n{n-l)(n~2){n-2>) 3x2x 1. For, the first place can be filled in n ways, then the second place in n — 1 ways, then the third place in n — 2 ways, and so on to the last place, which can be filled in only 1 way. Hence (§ 298) the whole number of arrangements is the continued product of all these numbers, n{n- l)(n - 2){n - 3) 3x2x1. For the sake of brevity this product is written \n, and is read factorial n. Observe that 1x2 (w — 1) n =|n. Ex. How many different arrangements of nine letters each can be formed with the letters in Cambridge ? There are nine letters. In making any arrangement any one of the letters can be put in the first place. Hence, the first place can be filled in 9 ways. Then the second place can be filled with any one of the remain- ing eight letters ; that is, in 8 ways. In like manner, the third place can be filled in 7 ways, the fourth place in 6 ways, and so on ; and, lastly, the ninth place in 1 way. If the nine places be indicated by Roman numerals, the result is (§ 298) as follows : I. II. III. IV. V. VI. VII. VIII. IX. 9x8x7x6x5x4x3 x 2x1= 362,880 ways. Hence, there are 362,880 different arrangements possible. CHOICE. 247 301. The number of different arrangements of n different things taken x at a time is n(n~ l)(n — 2) to r factors, that is, n{n~ l){n — 2) [n — (r - 1)], or n{n~-l){n~2) (n — r + l). For, the first place can be filled in n ways, the second place in n — 1 ways, the third place in n — 2 ways, and the rth place in n — (r — 1) ways. Let P„_ y represent the number of arrangements of n dif- ferent things taken r at a time. Then Pn^r = n(n — l)(n — 2) to r factors. = n(n- l)(n — 2) (n - r -f 1). Ex. How many different arrangements of four letters each can be formed from the letters in Cambridge ? There are nine letters and four places to be filled. The first place can be filled in 9 ways. Then the second place can be filled in 8 ways. Then the third place in 7 ways, and the fourth place in 6 ways. If the places be indicated by I., II., III., IV., the result is (§ 298) I. II. III. IV. 9x8x7x 6 = 3024 ways. Hence, there are 3024 different arrangements possible. 302. Selections, Things all different. The number of dif- ferent selections (or combinations) of n different things taken I at a time is n(n-l)(n-2) (n-r + 1) \r To prove this, let C„, r represent the number of different selections (or combinations) of n different things taken r at a time. Take one selection of r things ; from this selection \r arrangements can be made (§ 300). 248 ALGEBEA. Take a second selection ; from this selection [r arrange- ments can be made. And so on for each of the Cn, r selec- tions. Hence, (7„_ ^ X |^ is the number of arrangements of n dif- ferent things taken r at a time ; or ^n, r X [^== Jr^, r- n -t^n, r [r _n{n-l){n-2) (n - r + 1) Ex. In how many different ways can three vowels be selected from the five vowels a, e, i, o, u. The number of different ways in which we can arrange 3 vowels out of 5 is (§ 301) 5 X 4 X 3, or 60. These 60 arrangements might be obtained by first forming all the possible selections of 3 vowels out of 5, and then arranging the 3 vowels in each selection in as many ways as possible. Since each selection can be arranged in [3, or 6, ways (§ 300), the number of selections is -\°- or 10. Ans. The formula applied to this problem gives C. 3 = ^^1X^ = 10. ' 1x2x3 303. Selections, Second Formula. Multiplying both numer- ator and denominator of the expression for the number of selections in the last example by 2 X 1, we have C -^ 5x4x3x2xl _ [j ''' 1x2x3x2x1 [3|^' In general, multiplying both numerator and denomina- tor of the expression for C^^r in § 302 by \n — r, we have n(n — l) (w — r+l)(n — r) 1 ^"•*- ^ \r_x(n-r) 1 """ \r \n — r CHOICE. 249 This second form is more compact than the first, and is more easily remembered. Note. In reducing a result expressed in the above form, it is to be observed that \n — r cancels all the factors of the numerator from 112 1 up to and including n-r. Thus, in -^=-. LZ cancels all the fac- tors of [12 from 1 up to and including 7 ; so that Ijj _ 12x11 X 10x9x8 [5[7 1x2x3x4x5 = 792. 304. Theorem. The number of selections of n things taken T at a time is the same as the number of selections of n things taken n — i at a time. \n \n ror, ^n,n-r- |^_^ , |^^_^^_^) \n - r \r ^''•'^ This is also evident from the fact that for every selection of r things taken, a selection of n — r things is left. Thus, out of 8 things, 3 things can be selected in the same number of ways as 5 things ; namely, JL = 8x7x6^^e 1315 \3 ^ |1 1 Note. Evidently Ci, i = 1 ; also Ci, i = ~- = - ; .-. 1 = 1, and 10 = 1. [0 305. Examples in Selections and Arrangements. Of the arrangements possible with the letters of the word Cam- bridge, taken all together. (1) How many will begin with a vowel ? In filling the nine places of any arrangement the first place can be filled in only 3 ways, the other places in 1 8^ ways. Hence, the answer is (g 298) 3 X [8 = 120,960. 250 ALGEBRA. (2) How many will both begin and end with a vowel ? The first place can be filled in 3 ways, the last place in 2 ways (one vowel having been used), and the remaining seven places in [7 ways. Hence, the answer is (§ 298) 3 X 2 X [7 = 30,240. (3) How many will begin with Cam? The answer is evidently [6 ; since our only choice lies in arrang- ing the remaining six letters of the word. (4) How many will have the letters cam standing together ? This may be resolved into arranging the group c a m and the last six letters, regarded as seven distinct elements, and then arranging the letters cam. The first can be done in [7 ways, and the second in l_3_way8. Hence both can be done in [7 xi^= 30,240 ways. Ans. In how many ways can the letters of the word Cam- bridge be written : (5) Without changing the place of any vowel ? The second, sixth, and ninth places can be filled each in only 1 way ; the other places in [6 ways. Therefore, the whole number of ways is [6 = 720. Ans. (6) Without changing the oi^der of the three vowels ? The vowels in the different arrangements are to be kept in the order a, i, e. One of the six consonants can be placed in 4 ways : before a, be- tween a and i, between i and e, and after e. Then a second consonant can be placed in 5 ways, a third conso- nant in 6 ways, a fourth consonant in 7 ways, a fifth consonant in 8 ways, and the last consonant in 9 ways. Hence the whole number of ways is 4x5x6x7x8x9, or 60,480. Ans. CHOICE. 251 (7) Out of 20 consonants, in how many ways can 18 be selected ? The 18 can be selected in the same number of ways as 2; and the number of ways in which 2 can be selected is ^><^ = 190. Ans. 2 (8) In how many ways can the same choice be made so as always to include the letter b ? Taking b first, we must then select 17 out of the remaining 19 consonants. This can be done in 1^^^ = 171 ways. Ans. (9) In how many ways can the same choice be made so as to include b and not to include c ? Taking b first, we have then to choose 17 out of 18, c being excluded. This can be done in 18 ways. Ans. (10) From 20 Republicans and 6 Democrats, in how many ways can 5 different offices be filled, of which three particular offices must be filled by Republicans, and the other two offices by Democrats? The first three offices can be assigned to 3 Repubhcans in 20x19x18 = 6840 ways; and the other two offices can be assigned to 2 Democrats in 6 X 5 = 30 ways. There is, then, a choice of 6840 X 30 = 205,200 different ways. (11) Out of 20 consonants and 6 vowels, in how many ways can we make a word consisting of 3 different conso- nants and 2 different vowels? Three consonants can be selected in ^^.'^^^^^^ = H^O ways. i X ij X ; that is, ->x. ^ "^ r + etc. Again, ?<^ + -. + etc. ^^r + etc. 1 < 1 P + - pi- ^ ^ r + etc. ; that is, < X ; and so on. P + :: 336. If — , — , — 0,7-6 any three consecutive convergents, Vl v., Vg and if nii, nia, ma be the quotients that produced them, then Us^msU^-fUi V3 niaVa + Vi For, if the first three quotients are p, q, r, the first three convergents are (§ 334), I g , r±l_ (1) p pq + l {pq+iy+p From (§ 334) it is seen that the second convergent is formed from the first by writing in it ^ + - for ^ ; and the third from the second by writing ^^ + - for q. In this way, any convergent may be formed from the preceding con- vergent. 292 ALGEBRA. Therefore, — will be formed from — by writing m, + — for m2. In (1) it is seen that the third convergent has its numer- ator = r X (second numerator) + (first numerator) ; and its denominator = r X (second denominator) -f- (first denomi- nator). Assume that this law holds true for the third of the three consecutive convergents ^^ ^, !i^ so that, !fg^ ^^^i + ^'o . (2) ^0 Vi V2 ' Vi m./Vi -f Vo Then, since — is formed from — by using m, H for m.^, V3 V2 -^ ^ ^3 Us _ V mj _ m^ (m^Ui -j- Uq) -f Ui Substitute u^ and Va for their values maWi + Uq and ^2'i^i + Vo; then Uz _ nizUi -\- Ui Therefore the law still holds true ; and as it has been shown to be true for the third convergent, the law is gen- eral. (Note on p. 201.) 337. The difference between two consecutive convergents, ^and^is±. Vi V2 V1V2 The difference between the first two convergents is 1 g _ 1 p pg + l p{pq + l) CONTINUED FRACTIONS. 293 Let the sign '^ mean the difference between, and assume the proposition true for — and — ; then Wo^Wi^Wo?v;-WiVo_J^ Vo Vi VqVi VqVi But ■i^2 ^ Wi _ U2V1 ^ U1V2 _ {m^Ui + Uq) v^ ^ Ux (mjVi + Vp) V2 Vi V1V2 ViV-i (substituting for W2 and V2 their values, mjWi + Wo and ^2^1 + 1^0). Eeducing, U2 Ui UqVi ^ UiVq '^ ; V2 Vi ViV.2 = — (by assumption). V1V2 Hence, if the proposition be true for one pair of consecu- tive convergents, it will be true for the next pair ; but it has been shown to be true for the j?rs^ pair, therefore it is true for ever^/ pair. (Note on p. 201.) Since by § 335 the true value of x lies between two con- secutive convergents, — and — , the convergent — will Vi V2 Vi differ from a; by a number less than _ ^ ~ _^ ; that is, by a number less than — ; so that the error in taking — for x is less than — , and therefore less than — , as V2 > Vi since V2 = W2V1 + Vo. Any convergent, — , is in its lowest terms ; for, if Wi and Vi had any common factor, it would also be a factor of U1V2 '^ Wa^i ; that is, a factor of 1. 294 ALGEBRA. 338. The successive convergents approach more and more nearly to the true value of the continued fraction. Let — , — , — be consecutive convergents. Vq Vi V2 Now — differs from x, the true value of the fraction, only because ma is used instead of m^ ~{ •' W3 + etc. Let this complete quotient, which is always greater than unity, be represented by M. Then, since ^' = "'''^' + "° , ^ = :Mfl±^«. t'2 m.iVi + Vo ■ Mvi 4- Vq Ui _ Mu^ + Up Ml _ UqVi ^ UiVq __ 1 Vi Mvi + Vq Vx Vi{Mvy^VQ) v^{Mvi-^Vq) And Uq __ Wo -M^i 4- Mq __ M{UqVx'^UxVq) __ M Vo Vo Mvi + Vo Vo{Mvi-{-Vq) Vo(Mvi-{-Vo) Now 1 < M and Vi > Vq, and for both these reasons Ul ^Uq Vi Vo That is, — is nearer to x than — is. Vi Vo 339. Any convergent — is nearer the true value x than any other fraction with smaller denominator. Let - be a fraction in which b < Vi. If - is one of the convergents, a^ ~ - > — ~ x. § 338 h Vi If - is not one of the convergents, and is nearer to x CONTINUED FRACTIONS. 295 than — is, then, since x lies between — and — (§ 335), ■5- must be nearer to — than — is ; that is, ^ _2 ^ _i ^ _? , or - — j-^ < — ; V.2 Vi V2 V20 yiVg and since b < Vi, this would require that v^a ~ Uzb < 1 ; but ViCt'^Uib cannot be less than 1, for a, b, U2, v^ are all integers. Hence, — is nearer to x than - is. Vy_ ■ 340. Find the continued fraction equal to -f-J-, and also the successive convergents. Following the process of finding the H. C. F. of 31 and 75, the successive quotients are found to be 2, 2, 2, 1, 1, 2. Hence the con- tinued fraction is 2 + 2 + -1- 2 + ■'^ To find the successive convergents : Write the successive quotients in line, ^ under the first quotient, \ under the second quotient, and then (^ 336) multiply each terra by the quotient above it, and add the term to the left to obtain the corresponding term to the right. Thus, Quotients = 2, 2, 2, 1, 1, 2. Convergents = f , ^ f , ^^, ^7, ^f , f ^. It is convenient to begin to reckon with f , but the next conver- gent, in this case \, is called i]iQ first convergent. Note. Continued fractions are often written in a more compact form. Thus, the above fraction may be written 111111 2 + 2-h2 + l + 1+2 296 ALGEBRA. 341. A quadratic surd may be expressed in the form of a non-terminating continued fraction. To express V3 in the form of a continued fraction. Suppose then \/3 = i+; 1=V3 X - (as 1 is the : c -1. .-. X = 1 V3 + 1 V3-1 2 Suppose Vs + i 2 -K" 1 is the great then 1 — = y Vs + 1 2 ^ V3 - 1 2 ••• y = 2 Vs + i V3-1 1 Suppose V3 + 1 1 =..!(. 2 is the great then 1 — = z . z = V3 + 1 1 1 - 2 = V3 - 1. \/3 4-l' V3 + 1 n V3-1 This is the same as x above ; hence, the quotients 1, 2, will be continually repeated. .-. V3 = l+-1- ^ 2 + etc. of which will be continually repeated, and the whole expres- sion may be written, The convergents will be 1, 2, |, |, if, ff, |i, etc. CONTINUED FRACTIONS. 297 342. A continued fraction in which the denominators recur is called a periodic continued fraction. The value of a periodic continued fraction can be ex- pressed as the root of a quadratic equation. Find the surd value of r- 4_ o' Let X be the value ; then X = = - — - ; 2 + x .'. a;2 + 2a;=2, a; = -l + V3. "We take the + sign since x is evidently positive. 343. An exponential equation can be solved by con- tinued fractions. Solve by continued fractions 10* = 2. Suppose a: = o + |; then 10i^ = 2. or 10 = 2!'. .-. y =. 3 + i (as 10 lies between 2^ and 2*). Then 10 =-2 * = 23x2'; or 2^ = Y = f. and 2 -(f)'. ... 2 _ 3 + 1 fas 2 lies between ( - ) and / Then 2 = (f)'^« = (f)»X(f)«; or (f)»-m and 1 f-(lM)«. 298 ALGEBRA. The greatest integer in u will be found to be 1 Hence, a; = + 3 + 1 9 + etc. The successive convergents will be ^, -^q, ||-, etc. The last gives a^ = f f = 0.3010, nearly. Observe that by the above process we have calculated the common logarithm of 2. By § 337 the error, when 0.3010 is taken for the common logarithm of 2, is considerably less than ; that is con- siderably less than 0.00011 ; so that 0.3010 is certainly correct to three places of decimals, and probably correct to four places. Logarithms are, however, much more easily calculated by the use of series, as will be shown in a following chapter. Exercise 52. 1. Find the values of: 111. 111. 11111 4 + 3 + 2' 2 + 3 + 7' 1 + 2+1 + 4 + 5' 2 . Find continued fractions for f|^ ; J^ ; J^ ; -^ ; V5 ; Vll ; 4V6 ; and find the fourth convergent to each. 3. Find continued fractions for ^Vt ; «i ; fMI ; "^V- ) and find the third convergent to each. 4. Find continued fractions for V21; V22; V33; V66. 5. Obtain convergents, with only two figures in the de- nominator, that approach nearest to the values of : V7 ; VIO; Vl5; VTZ; Vl8; V20; 3 - V5 ; 2 + VIT. 6. Find the proper fraction which, if converted into a continued fraction, will have quotients 1, 7, 5, 2. 7. Find the next convergent when the two preceding convergents are ^j and If, and the next quotient is 5. CONTINUED FRACTIONS. 299 8.' If the pound troy is the weight of 22.8157 cubic inches of water, and the pound avoirdupois of 27.7274 cubic inches of water, find a fraction with denominator less than 100 which shall differ from their ratio by less than 0.0001. 9. The ratio of the diagonal to a side 6f a square being V2, find a fraction with denominator less than 100 which shall differ from their ratio by less than 0.0001. 10. The ratio of the circumference of a circle to its diam- eter being approximately the ratio of 3.14159265 : 1, find the first three convergents to this ratio, and determine to how many decimal places each may be depended upon as agreeing with the true value. 11. In two scales of which the zero-points coincide the distances between consecutive divisions of the one are to the corresponding distances of the other as 1 : 1.06577. Find what division-points most nearly coincide. 12. Find the surd values of : , i i. 3 i i. i 1 i. , i i i. ^■^4 + 2' "^1 + 6' 3 + 1 + 6' ^2 + 3 + 4 13. Prove that (" + J + a)(i + a)~f' 14. Show that the ratio of the diagonal of a cube to its edge may be nearly expressed by 97 : 56. Find the greatest possible value of the error made in taking this ratio for the true ratio. 15. Find a series of fractions converging to the ratio of 5 hours 48 minutes 51 seconds to 24 hours. 16. Find a series of fractions converging to the ratio of a cubic yard to a cubic meter, if a cubic yard is -^i^j^ of a cubic meter. CHAPTER XXIV. SCALES OF NOTATION. 344. Definitions. Let any positive integer be selected as a radix or base ; then any number may be expressed as a polynomial of which the terms are multiples of powers of the radix. Any positive integer may be selected as the radix ; and to each radix corresponds a scale of notation. In writing the polynomials they are arranged by descend- ing powers of the radix, and the powers of the radix are omitted, the place of each digit indicating of what power of the radix it is the coefficient. Thus, in the scale of ten, 2356 stands for 2x103 + 3x102 + 5x10 + 6; in the scale of seven for 2x7=^ + 3x72 + 5x7 + 6; in the scale of r for 2r3 + 3r2 + 5r + 6. 345. Computation. Computations are made with numbers in any scale, by observing that one unit of any order is equal to the radix-number of units of the next lower order ; and that the radix-number of units of any order is equal to one unit of the next higher order. (1) Add 56,432 and 15,646 (scale of seven). 56432 "^^^ process differs from that in the decimal scale 15646 ^^^y ^^ ^^^^ when a sum greater than seven is reached, we divide by seven (not ten), set down the remainder, lOolll ^^^ carry the quotient to the next column. SCALES OF NOTATION. 301 (2) Subtract 34,561 from 61,235 (scale of eight). 61235 34561 ^® ^^^ ^ight, instead of ten as in the common 24454 '*'^^^' (3) Multiply 5732 by 428 (scale of nine). 5732 428 51477 We divide each time by nine, set down the remain- 12564 der, and carry the quotient. 24238 2612127 (4) Divide 2,612,127 by 5732 (scale of nine). 5732)2612127(428 24238 17732 12564 51477 51477 346. Integers in Any Scale. If i be any 'positive integer ^ any positive integer N may he expressed in the form JSr= c?r" + 5r"-^ + +pr'' -j-qr + s, in which the coefficients a, b, c, , are positive integers, each less than r. For, divide iVby r**, the highest power of r contained in iV, and let the quotient be a with the remainder JVi. Then, JSf = ar"" + JS^,. In like manner, JV; = ^r""^ + iV; ; JV, = cr^-' + JV^ ; and so on. By continuing this process, a remainder s will at length be reached which is less than r. So that, ]^= ar"" + &r**-^ + -{-pr"^ -\r qr-{-s. 302 ALGEBRA. Some of the coefficients s, q,p, may vanish, and each will be less than r ; that is, their values may range from zero to r — 1. Hence, including zero, r digits will be required to express numbers in the scale of r. To express any positive integer N in the scale of r. It is required to express N in the form and to show how the digits a, h, may be found. If N= ar"" + br""-' + -\-pr' + qr + s, N s then — = ar~-' + hr'^-'' + -\-pr + g' + - That is, the remainder on dividing iV by r is s, the last digit. Let iVi = ar""-^ + W-'' + -^pr + q ; N a then — = ar"-' + W'^ + -\-p + ^^ That is, the remainder is q, the last but one of the digits. Hence, to express an integral number in a proposed scale, Divide the number by the radix, then tKe quotient by the radix, and so on; the successive remainders will be the successive digits beginning with the units' place. (1) Express 42,897 (scale of ten) in the scale of six. 6 )42897 6 )7149 . ... 3 6 )1191 3 6 )198 3 6 )33 5 3 Am. 530,333. SCALES OF NOTATION. 303 (2) Change 87,214 from the scale of eight to the scale of nine. The radix is 8 ; and hence the two digits on the 9 )37214 left, 37, do not mean thirty-seven, but 3x8 + 7, 9)3363 ... 1 or thirty-one, which contains 9 three times, with 9]305. ... 6 the remainder 4. 9 )25 8 The next partial dividend is 4 X 8 + 2 = 34, 2. ... 3 which contains 9 three times, with the remainder Ans. 23,861. 7. and so on. (3) In what scale is 140 (scale of ten) expressed by 352 ? Let r be the radix ; then, in the scale of ten, 140 = 3r2 + 5r + 2 or 3r2 + 5r=138. Solving, we find r = 6. The other value of r is fractional, and therefore inadmissible, since the radix is always a positive integer. 347, Eadix-Practions. As in the decimal scale decimal fractions are used, so in any scale radix-fractions are used. Thus, in the decimal scale, 0.2341 stands for 10 102 IQS 10* • and in the scale of r it stands for M m2 A«0 A«« (1) Express fff (scale of ten) by a radix-fraction in the scale of eight. Assume 245^a 6 £ d 256 8 82 8' 8* Multiply by 8. 7f i = a + | + ^ + | + .-. a = 7. and 21^6 £ rf 32 8 8=» 8» 304 ALGEBRA. Multiply by 8, 5^ = & + ^ + ^ + 8 8^ .•.* = 6,and 1 = 1 + 1 + Multiply by 8, 2 = c + f + 8 .-. c = 2, and = d, etc. The answer is 0.752. (2) Change 35.14 from the scale of eight to the scale of six. 3 6 We take the integral and fractional parts separately. 16)18(1 16 2 Integral part: 6)35 4 5. 6 16)12(0 Fractional part : 14 12 3 8 82 64 16 6 16)72(4 64 8 This is reduced to a radix-fraction in the scale 6 16)48(3 48 of six as follows : 45.1043. Arts, Exercise 53. 1. Add together 435, 624, 737 (scale of eight). 2. From 32,413 subtract 15,542 (scale of six). 3. Multiply 6431 by 35 (scale of seven). 4. Multiply 4685 by 3483 (scale of nine). 5. Divide 102,432 by 36 (scale of seven). 6. Find H. 0. F. of 2541 and 3102 (scale of seven). 7. Extract the square root of 33,224 (scale of six). SCALES OF NOTATION. 305 8. Extract the square root of 300,114 (scale of five). 9. Change 624 from the scale of ten to the scale of five. 10. Change 3516 from the scale of seven to the scale of ten. 11. Change 3721 from the scale of eight to the scale of six. 12. Change 4535 from the scale of seven to the scale of nine. 13. Change 32.15 from the scale of six to the scale of nine. 14. Express y^-S^ (scale of ten) by a radix-fraction in the scale of four. 16. Express ^^% (scale of ten) by a radix-fraction in the scale of six. 16. Multiply 31.24 by 0.31 (scale of five). 17. In what scale is this true ? 21 X 36 = 746. 18. In what scale is the square of 23 expressed by 540 ? 19. In what scale are 212, 1101, 1220 in arithmetical progression ? 20. Show that 1,234,321 is a perfect square in any scale (radix greater than four). 21. Which of the weights 1, 2, 4, 8, pounds must be selected to weigh 345 pounds, only one weight of each kind being used ? 22. If two numbers are formed by the same digits in different orders, prove that the difference of the numbers is divisible by r — 1. CHAPTER XXV. J THEORY OF NUMBERS. 348. Definitions. In the present chapter, by number will be meant positive integer. The terms prime, composite, will be used in the ordinary arithmetical sense. A multiple of a is a number which contains the factor a, and may be written ma. An even number, since it contains the factor 2, may be written 2m; an odd number may be written 2m + 1, 2m-l, 2m + 3, 2m-3, etc. A number a is said to divide another number h when h . - is an integer. a 349. Eesolution into Prime Factors. A number can be resolved into prime factors in only one way. Let iV be the number; suppose N=ahc , where a,b,c, are prime numbers; suppose also iV— a^y where a, j8, y, are prime numbers. Then, abc =o.Py Hence, a must divide the product abc ; but a, b, c, are all prime numbers ; hence a must be equal to some one of them, a suppose. Dividing by a, bc'---;=py , and so on. Hence, the factors in ajSy are equal to those in abc , and the theorem is proved. 350. Divisibility of a Product. 1. If a number a divides a vroduct bo, and is prime to b| it must divide c. THEORY OF NUMBERS. 307 For, since a divides he, every prime factor of a must be found in he ; but since a is prime to h, no factor of a will be found in h; hence 'all the prime factors of a are found in c ; that is, a divides c. From this theorem it follows that : II. If a prime number a divides a product bcde...., it mtost divide some factor of that product ; and conversely. III. If a prime numher divides b**, it must divide b. IV. If a is prime to b and c, it is prime to be. Y. If B, is prime to b, every power of a is prime to every power ofh. 351. Theorem. ^ - , a fraction in its lowest terms, is equal b to another fraction —, then c and d ao'c equitnultiples of a and\i, If- = -, then — - = c. Since h will not divide a, it o d h must divide d ; hence c? is a multiple of h. Let d = mh, m being an integer ; since 7 = -, and b d d = mh, - = — - ; therefore c = ma. mo Hence, c and d are equimultiples of a and h. From the above theorem, it follows that in the decimal scale of notation a common fraction in its lowest terms will produce a non-terminating decimal if its denominator con- tains any prime factor except 2 and 5. For a terminating decimal is equivalent to a fraction with a denominator 10". Therefore, a fraction - in its lowest terms cannot be equal to such a fraction, unless 10" is a multiple of h. But 10", that is, 2" X 5", contains no factors 308 ALGEBRA. besides 2 and 5, and hence cannot he a multiple of b, if 6. contains any factors except these. 352. Square Numbers. If a square number is resolved into its prime factors, the exponent of each factor will be even. For, if JSr = aP XM Xc"- , ]V'=a''^Xb'^Xc'' Conversely : A number which has the exponents of all its prime factors even will be a perfect square ; therefore, to change any number to a perfect square, Eesolve the number into its prime factors, select the fac- tors which have odd exponents, and multiply the given number by the product of these factors. Thus, to find the least number by which 250 must be multiplied to make it a perfect square. 250 = 2x5^, in which 2 and 5 are the factors which have odd exponents. Hence the multiplier required is 2 x 5 = 10. 353. Divisibility of Numbers. I. J7* ^'^0 numbers N and N', when divided by a, have the same rem^ainder, their difference is divisible by a. For, if iVwhen divided by a have a quotient q and a remainder r, then N —- qa-{-r. And if iV' when divided by a have a quotient q^ and a remainder r, then iV^'z= q^a + r. Therefore, N-W={q- q^) a. II. If the difference of two numbers N and W is divisible by a, then N and W when divided by a will have the same remainder. THEORY OF NUMBERS. 309 For, if N-N^ = {q-q')a, then Therefore, q = — ~q'. • That is, N-aq =]V'-aq'. III. If two numbers N and N', when divided hy a given number a, have remainders r and r', then NN' and rr' when divided by a will have the same remainder. For, if N = qa-\-r, and N'=q'a-\-r\ then NW = qq^a^-\- qar'-\- q'ar-{-rr* ' — (qq'a -f qr'-\- q'r) a + rr\ Therefore, by II., NN' and rr' when divided by a will have the same remainder. As a particular case, 37 and 47 when divided by 7 have remainders 2 and 5 respectively. Now 37 X 47 = 1739 and 2x5 = 10. The remainder, when each of these two numbers is divided by 7, is 3. Note. From II. it follows that, in the scale often . (1) A number is divisible by 2, 4, 8, if the numbers denoted by its last digit, last two digits, last three digits are divisible respectively by 2, 4, 8, (2) A number is divisible by 5, 25, 125 if the numbers denoted by its last digit, last two digits, last three digits, are divisible respectively by 5, 25, 125, (3) If from a number the sum of its digits is subtracted, the remainder will be divisible by 9. 310 ALGEBRA. For, if from a number expressed in the form a + 10b + Wc + 10'd+ a + b + e+ d + is subtracted, the remainder will be (10 -1)6 + (10^ - 1) c + (10^ -l)d + and 10 - 1, 102 _ 1, 10^ - 1, will be a series of 9's. Th,erefore, the remainder is divisible by 9. (4j Hence, a number N may be expressed in the form 9 n + s (if s denotes the sum of its digits) ; and iV will be divisible by 3 if s is divisible by 3 ; and also by 9 if s is divisible by 9. (5) A number will be divisible by 11 if the difference between the sum of its digits in the even places and the sum of its digits in the odd places is or a multiple of 11. For, a number N expressed by digits (beginning from the right) a, b, c, d may be put in the form of i\^= a + 10 6 + 102c + 103 (^ + ... ]:^^a+b-c + d- = {10 + l)b + {W-l)c + {l(fi + i)d + But 10 + 1 is a factor of 10 + 1, 10^ - 1, 10^ + 1, Therefore, iV- a + b — c + d — is divisible by 10 + 1 = 11. Hence, the number N may be expressed in the form 11 n + (a + c + ) -{b + d + ), and will be a multiple of 11 if (a + c + ) — {b + d + ) is or a multiple of 11. 354. Theorem. The joroduct of r consecutive integers is divisible hy |r. Kepresent by P„^ ^ the product of Ic consecutive integers beginning with n. Then, P„, , = n(n + 1) ••-'(^ + ^ - 1) ; Pn+i. .+1 = (^ + 1)(^ + 2) (71 + lc){n + ^ + 1) = w(7i+l)(w + 2) {n^k) + (^ + l)(n + l)(n + 2) (n + ^). •'• -fn+l, *+l ^^ ^n, A+1 "h ("^ H~ 1) -fn+l,&- THEORY OF NUMBERS. 311 Assume, for the moment, that the product of any k con- secutive integers is divisible by \k. Then, P„+i. ^, = P^,^, + {Jc-\-V)M\h- or, P„+i, ^, = P„, ,+1 + M\h±\ ; where M is an integer. From this it is seen that if P„_ j+i is divisible by [^-f- 1 , -^n+i,t+i is also divisible by |^+1 ; but Pi,t+i is divisible by 1^+1 since Pi,ic+x = \k-\- 1 . .'. Pz.^+i is divisible by 1^+1 ; •*• -fs.t+i is divisible by | ^-f 1 ; and so on. Hence, the product of any ^ + 1 consecutive integers is divisible by | ^+ 1 , if it is known that the product of any h consecutive integers is divisible by \]c. But the product of any 2 consecutive integers is divisible by |_2 ; therefore, the product of any 3 consecutive integers is divisible by [3 ; therefore, the product of any 4 consecutive integers is divisible by [£ ; and so on. Therefore, the product of any r consecutive integers is divisible by |r. 355. Examples. (1) Show that every square number is of one of the forms 5w, 5n— 1, bn-\-\. Every number is of one of the forms ; 5m — 2, 5w — 1, 5m, 5m + 1, 5m + 2. (5m ± 2)2 = 25m2 ± 20m + 4 = 5(5m2 ± 4m + 1) - 1 ; I (5m±l)2 = 25m2dzl0m + l = 5(5m2±2m) + l; (5m)2 = 25m2 =5(5m2). .*. every square number is of one of the three forms ; 5n, 5n — 1, 5n + l. Hence, in the scale of ten, every square number must end in 0, 1, 4, 5, 6, or 9. (2) Show that ?2^ — n is divisible by 30 if n is even. n* — n = n(7i - l)(n + l)(n2 + 1) = 7i(n-l)(n + l)(n2-4 + 5) = n(n - l)(n + l)[(n - 2)(n + 2) + 5]. 312 ALGEBRA. n{n- l){n + 1) is divisible by |_3 (^ 354). One of the five consecutive numbers n — 2,n — l,n,n+l,n + 2, ^ is divisible by 5, and n^ — n is therefore divisible by 5. Hence n^ — n is divisible by 5 [3, that is by 30. Exercise 54.* Find the least number by which each of the following numbers must be multiplied in order that the product may be a square number. I. 2625. 2. 3675. 3. 4374. 4. 74088. 5. If m and n are positive integers, both odd or both even, show that m^ — m? is divisible by 4. 6. Show that n^ — n\B always even. ■ 7. Show that w^ — n is divisible by 6 if n is even ; and by 24 if n is odd. 8. Show that ■rv' — n is divisible by 240 if n is odd. 9. Show that n} — n is divisible by 42 if n is even ; and by 168 if n is odd. 10. Show that n(n-\- l)(n -\- 5) is divisible by 6. II. Show that every cube number is of one of the forms, 9n, 971-1, 9n+l. 12. Show that every cube number is of one of the forms, In, 771—1, 771+1. 13. Show that every number which is both a square and a cube is of the form 77i or 77i + 1. 14. Show that in the scale of ten every perfect fourth power ends in one of the figures 0, 1, 5, 6. CHAPTER XXVI. VARIABLES AND LIMITS. 356. Constants and Variables. A number that, under the conditions of the problem into which it enters, may be made to assume any one of an unlimited number of values is called a variable. A number that, under the conditions of the problem into which it enters, has a fixed value is called a constant. Variables are generally represented by x, y, 2, etc. ; con- stants, by the Arabic numerals, and by a, h, c, etc. 357. Punctions. Two variables may be so related that a change in the value of one produces a change in the value of the other. In this case one variable is said to be a function of the other. Thus, if a man walks on a straight road at a uniform rate of a miles per hour, the number of miles he walks and the number of hours he walks are both variables, and the first is a function of the second. If y be the number of miles he has walked at the end of x hours, y and X are connected by the relation y = ax, and y is a function of x. Also X = - ; hence, x is also a function of v- When one of two variables is a function of the other, the relation between them is generally expressed by an equa- tion. If a value of one variable is assumed, the corre- sponding value of the other variable can be found from the given equation of relation between the two variables. The variable of which the value is assumed is called the independent variable ; the variable of which the value is 314 ALGEBRA. found from the given relation of the two variables is called the dependent variable. In the last example we may assume values of x, and find the cor- responding values of y from the relation y = ax\ or assume values of y y, and find the corresponding values of x from the relation x= -. In the first case x is the independent variable, and y the dependent ; in the second case y is the independent variable, and x the dependent. 358. Limits. As a variable changes its value, it may- approach some constant; if the variable can be made to approach the constant as near as we please, but cannot be made absolutely equal to the constant, the variable is said to approach the constant as a limit, and the constant is called the limit of the variable. Let X represent the sum ofn terms of the infinite series 1 + i + i + i + ; ien {I 227), . a)"- -1_ 1 = 2^-1 = 2- 2n-l 1 2n-l Suppose n to increase ; then, — — decreases, and x approaches 2. Since we can take as many terms of the series as we please, n can be made as large as we please ; therefore, — — • can be made as small as we please, and x can be made to approach 2 as near as we please. We cannot, however, make x absolutely equal to 2. If we take any assigned value, as i-^-q-q, we can make the dif- ference between 2 and x less than this assigned value ; for we have only to take n so large that is less than rxr^oiF ! ^^^ ^^> ^1^^^ 2»»-i is greater than 10,000 : this will be accomplished by taking n as large as 15. Similarly, by taking n large enough, we can make the difference between 2 and x less than any assigned value. Since 2 — x can be made as small as we please, it follows that the sum of n terms of the series 1 + ^ + ^ + 1 + , as n is constantly increased, approaches 2 as a limit. VARIABLES AND LIMITS. 315 359. Test for a Limit. In order to prove that a variable approaches a constant as a limit, it is necessary and suffi- cient to prove that the difference between the variable and the constant can be made as near to zero as we 'please^ but cannot be made absolutely equal to zero. A variable may approach a constant without approaching it as a limit. Thus, in the last example x approaches 3, but not as a limit ; for 3 — ic cannot be made as near to as we please, since it cannot be made less than 1. 360. Infinites. As a variable changes its value, it may constantly increase in numerical value; if the variable can become numerically greater than any assigned value, however great this assigned value may be, the variable is said to increase without limit, or to increase indefinitely. When a variable is conceived to have a value greater than any assigned value, however great this assigned value may be, the variable is said to become infinite; such a variable is called an infinite number, or simply an infinite. 361. Infinitesimals, As a variable changes its value, it may constantly decrease in numerical value ; if the vari- able can become numerically less than any assigned value, however small this assigned value may be, the variable is said to decrease without limit, or to decrease indefinitely. In this case the variable approaches as a limit. When a variable which approaches as a limit is con- ceived to have a value less than any assigned value, how- ever small this assigned value may be, the variable is said to become infinitesimal; such a variable is called an infini- tesimal number, or simply an infinitesimal. 362. Infinites and infinitesimals are variables, not con- stants. There is no idea of fixed value implied in either an infinite or an infinitesimal. 316 ALGEBRA. An infinitesimal is not 0. An infinitesimal is a variable arising from the division of a quantity into a constantly increasing number of parts ; is a constant arising from taking tbe difference of two equal quantities. A number which cannot become infinite is said to be finite. 363. Eolations between Infinites and Infinitesimals. I. If X is infinitesimal, and a is finite and not 0, then ax is infinitesimal. For, ax can be made as small as we please since x can be made as small as we please. II. If X is infinite, and a is finite and not 0, then aX is infinite. For aX can be made as large as we please since X can be made as large as we please. III. If X is infinitesimal, and a is finite and not 0, then is infinite. For - can be made as large as we please a X " X since x can be made as small as we please. IV. 7)^ X is infinite, and a is finite and not 0, then — is infinitesimal. For — can be made as small as we please since X can be made as large as we please. In the above theorems a may be a constant or a variable ; the only restriction on the value of a is that it shall not become either infinite or zero. 364. It appears from § 157 that one root of the quadratic equation ax"^ -\~ bx -\- c = is infinite when a is infinites- imal ; and that both roots are infinite when a and b are both infinitesimal. , 365. Abbreviated Notation. An infinite is often repre- sented by 00. In § 363, III. and IV. are sometimes written : - = 00, — = 0. ' 00 • VARIABLES AND LIMITS. 317 The expression - cannot be interpreted literally, since we cannot divide by ; neither can — = be interpreted literally, since we can find no number such that the quotient obtained by dividing a by that number is zero. - = 00 is simply an abbreviated way of writing : if -= X, and x approaches as a limit, X increases without limit. — = is simply an abbreviated way of writing : if — = x, and X increases without Ji. limit, X approaches as a limit. 366. Approach to a Limit. "When a variable approaches a limit, it may approach its limit in one of three ways : (1) The variable may be always less than its limit. (2) The variable may be always greater than its limit. (3) The variable may be sometimes less and sometimes greater than its limit. If X represent the sum of n terms of the series 1 + i + ^ + | + , X is always less than its limit 2. If X represent the sum of n terms of the series ^ — ^ — \ — j — , X is always greater than its limit 2. If X represent the sum of n terms of the series 3 — f + | — 1 + , we have (§ 227) x = 3^itd£ = 2-2(-H». As n is indefinitely increased, x evidently approaches 2 as a limit. If n is even, x is less than 2 ; if n is odd, x is greater than 2. Hence, if n be increased by taking each time one more term, x will be alternately less than and greater than 2. If, for example, n- 2, 3, 4, 5, 6, 7, x=ii 2i, H, 2tV. im. m- In whatever way a variable approaches its limit, the test of § 359 always applies. 318 ALGEBRA. 367. Equal Variables. If two variables are equal and are so related that a change in the one produces such a change in the other that they continue equal, and each ap- proaches a limit, then their limits are equal. Let X and y be the variables, a and b their respective limits. To prove a — b. We have (§ 359) a — x-\- x\ b = y-\-y\ where x^ and ?/' are variables which approach as a limit. Then, since the equation x = y always holds, we have, by subtraction, a~b = x^ — y\ x^ — y' can be made less than any assigned value since x^ and 2/' can each be made less than any assigned value. Since x^ — y' is always equal to the constant a — b, x^ — y must be a constant. But the only constant which is less than any assigned value is 0. Therefore a;' — y' = 0, and hence a — 6 = 0. :. a^b. 368. Limit of a Sum. The limit of the algebraic sum of any finite number of variables is the algebraic sum of their respective limits. Let x,y, z, , be variables ; a, b, c, , their respective limits. Then a — x, b—y, c — z, , are variables which can each be made less than any assigned value (§ 359). Then {a — x) -\- {b — y) -\- {c — z) -{■ can be made less than any assigned value. For, let V be the numerically greatest of the variables a — x, b ~y, c — z, , and n the number of variables. Then, {a - x) + {b-y) + {c-z) + — + — + ton terms; therefore >nx-—\ thatis, >-• 2n 2n 2n 2 Now, the first term is 1, the second term is J, the sum of the next two terms is greater than ^, the sum of the succeeding four terms is greater than J ; and so on. So that, by increasing n indefinitely, the sum will become greater than any finite multiple of ^. Therefore, the series is divergent. Ex. To determine whether the following series is con- vergent (§ 267). 1+1+1+1+ 1,1,1, 1 ' [2 ' [3 \ n-l \n \n-\-l The nth term is . The sum of the remaining terms is 1+_JL_+__1_+ _ Wi 1 1 , 1 In |n + l |n + 2 " \n\ n + 1 (n + l)(n + 2) CONVERGENCY OF SERIES. 823 This is<-/'l+i + - + ] ; therefore [n V n r? j [nl,_l) [n\n— 1/ (n — 1) |?». — 1 n But as w increases indefinitely, this last approaches as a limit. Hence, the series is convergent. 373. Test for Oonvergenoy of a Series. If the terms of an infinite series are all positive, and the limit of the nth term is 0, then if the limit of the ratio of the (n + 1) th term to the nth term, as n is indefinitely increased, is less than 1, the series is convergent. Let Ui, Ui, U3, Un, w„4.,, Un+2 be ail infinite series. Let r represent the limit of the ratio — ^^ as n increases indefinitely, and suppose r to be positive and less than 1. Let Jc be some fixed number between r and 1, and take k so near 1 that ^ ^^, , shall each be < k Then, ""-+'< Jc, Un+l ""^^Kh .'. Un+i < hun, Un+2 < ^Wn+i, Un+3 < JcUn+2, .-. U„^i 1, there must be in the series some term from which the succeeding term is greater than the next preced- ing term ; so that the remaining terms will form an in- creasing series, and therefore the series is not convergent. If r = ± 1, this value gives no information as to whether the series is convergent or not ; and in such cases other tests must be applied. If r < 1, but approaches 1, or — 1, as a limit, then no fixed value k can be found which will always lie between r and zfc 1, and other tests of convergency must be applied. Thus, in the infinite series i+i-+^+ +1+ 1 ■ im 2'» S** n"" (n + 1)«* r, the ratio of the {n + l)th term to the nth term, is n + lj \ n + lj ' which approaches 1 as a limit as n increases. Suppose m positive and greater than 1 ; then the first term of the 2 series is 1. The sum of the next two terms is less than — . The sum 2™ of the next four terms is less than — . The sum of the next eight Q terms is less than — ; and so on. Hence, the sum of the series is less 3m than 1+A + 1+A + , or <1 + J_ + J_ + J_ + , 2m 4m 8*" 2"'~^ 4m-l 8"*~^ which is evidently convergent when m is positive and greater than 1. CONVERGENCY OF SERIES. 325 If m is positive and equal to 1, the given series becomes l+i + i + J + , which is the harmonical series shown in § 372 to be divergent. If m is negative, or less than 1, each term of the series is then greater than the corresponding term in the harmonical series, and hence the series is divergent. 374. Special Case. If the terms of an infinite series are alternately positive and negative; if, also, the terms contin- ually decrease, and the limit of the nth term is zero, then the series is convergent. Consider the infinite series, Ui — U2~\- Uz — W4 + ^ Wn ± w„+i =F w^2 ± The sum of the terms after the nth term is which may be written Since the terms are continually diminishing, each of the groups in either form of expression is positive, and there- fore the absolute value of the required sum is seen, from the first form of expression, to be less than u^.^^ ; and from the second form of expression, to be greater than w„+i — Wn+2- But both Un^x and u^^^ approach zero as n increases indefi- nitely ; therefore the sum of the series after the nth term approaches zero, and the series is convergent. In finding the sum of an infinite decreasing series of which the terms are alternately positive and negative, if we stop at any term, the error will be less than the next succeeding term. The series 1 — + — - + ± - t ± is convergent. 2 3 4 n n+ 1 326 ALGEBRA. For, we may write the series l-^ + a-i) + (i-i) + .or l-(|-i)-(i-^)- , which shows that its sum is greater than ^, and less than 1. Observe that the present test applies to series in which — ^^ approaches 1, or —1, as a limit; to these series the test of § 373 will not apply. 375. Oonvergency of the Binomial Series. In the expan- sion of (1 + ^Yi the ratio of the (r + l)th term to the rth term is (§ 247) n — r -\-l /w + l • X, or If X is positive, and r greater than n-\-l, — — 1 is negative ; hence the terms in which r is greater than n + 1 are alternately positive and negative. If X is negative, the terms in which r is greater than n-\-l are all positive. In either case we have Ur \ r j as r is indefinitely increased, this approaches the limit — x. Hence (§ 373), the series is convergent if x is numerically less than 1. If n is fractional or negative, the expansion of (a + 6)" must be in the form a^il-^ -Y if a > & ; and in the form 6« (\ + ^ if & > a (^ 259). 376. Examples. (1) For what values of x is the infinite series X }-' rb — =F convergent? 2^3 n ^ CONVERGENCY OF SERIES. 327 Here, r = -* = ( — -rr ) a; = ( 1 — r- j x. As n is indefinitely increased, r approaches a; as a limit. Hence, the series is convergent when x is numerically less than 1 ; and divergent when x is numerically greater than 1. When x = l, the series is convergent by ^ 374. When x = — l, the series becomes -M-l ^b-)' the harmonical series already shown to be divergent (^ 372). (2) For what values of x is the infinite series X , of , x^ J 1- — — convergent? 1x22x33x4 w(w-fl) Wn-fl Here, Un As n is indefinitely increased, r approaches a; as a limit. If a; is numerically less than 1, the series is convergent. If a; is numerically greater than 1, the series is divergent. If a; = 1, every term of the series 1 +^ + J_ + 1X2 2x3 3x4 is less than the corresponding term of the series ^^¥h This last series is a special case of the series -l+^+l^ Im 2'* 3'» and is therefore convergent (^ 373). Hence, -^ + --^ + - — - + is convergent. 1X2 2x3 3x4 If a; = — 1, the series becomes ~r>r2"^2l<3~3xl and is convergent by § 374. 328 ALGEBRA. Exercise 56. Determine whether the following infinite series are con- vergent or divergent : 1 l+l-j-l-J-l-l- 4 2,345 2 3' 4^ * 12~*~2' 3' 4' l!_i_--f-- ' . -, , 1 , 2=^ . 3=^ [2 [3 [4' 2. 1+^ + — + — + 5. 1 + — + — + — -!- ^12^13^14^ ^12-t-32-i-42-t- 92 03 ^4 1 I*" 9»» ^»* ^[2 [3 [4 x«^2'" 3"* 4"* SERIES OF DIFFERENCES. 377. Definitions. If, in any series, we subtract from each term the preceding term, we obtain a first series of differ- ences ; in like manner from this last series we may obtain a second series of differences ; and so on. In an arithmetical series the second difierences all vanish. There are series, allied to arithmetical series, in which not the first, but the second, or third, etc., differences vanish. Thus take the series 1 5 12 24 43 71 110 1st differences, 4 7 12 19 28 39 2d differences, 3 5 7 9 11 3d differences, 2 2 2 2 4th differences, In general, if ai, a^, a^, be such a series, we have «! a-i a^ ^4 a^ a^ a-i 1st differences, hi h^ h^ b^ h^ h^ 2d differences, Ci c^ C3 C4 c^ 3d differences, d^ d^ d^ d^ 4th differences, ex e^ e^ and finally arrive at difierences which all vanish. SERIES OF DIFFERENCES. 329 378. Any Eequired Term. For simplicity let us take a series in which the fifth series of differences vanishes. Any other case can be treated in a manner precisely similar. From the manner in which the successive series are formed we shall have : a2 = ai-}-bi 03 = a2-\-b2 = ai-\-2bi-\- Ci ^2=^1+^1 53 = ^2 +x ~1 (1) Resolve — into partial fractions. (x — • AjiX — o ) The denominators will be a; — 2 and a; — 3. Assume 3x-7 ^.A_ + ^,; (x - 2){x -3)x-2 a;-3 then Zx-1=:A{x-^) \- B{x-2). . /. ^ + ^ = 3 and 3^ + 25= 7; g 256 whence, A = \ and 5 = 2. Therefore, ^ ^l'~^ ^^ = -^ + -^' {x-2){x-Z) x-2 x-Z This identity may be verified by actual multiplication. 340 ALGEBRA. 3 (2) Resolve — into partial fractions. ^ -f- 1 Since x^ + 1 — {x + l){x^ — x + 1), the denominators will be x -f 1 and x^ — X + 1. A 3 _ A , Bx+ Assume — = h x^-j-l x + 1 x"^ — X + 1' then 3 = A{x^-x + l) + {Bx + C){x + 1) = {A + B)x^ + {B + C-A)x + {A + C); whence, 3 = A + C, B+C-A = 0, A + B = 0;^25Q and ^ = 1, S = -l, C=2. 3 _ 1 x-2 Therefore, a;^ + l x + 1 x"^ — X + 1 (3) Resolve — ^ into partial fractions. The denominators may be x, x"^, x + 1, {x + If. Assume 4.3-..-3.-2^i j^^ _^, x^x + iy X a;2 x + 1 {x + lf .-. 4a;3 - »2 - 3 a; - 2 = .4a;(a; + 1)^ + B{x + If + Cx^x + 1) + Dx^ = {A + C)^ + {2A + B + C+D)x' + {A+2B)x + B■ v{\ielice, ^ + (7=4, ^256 2A + B + C+D = -1, A + 2B = -3, or, A = l, B = -2, 0=3, D = -^. Therefore, i^i^^!^^^^^l-l+ ' ^ x\x + lf X x'^ x + 1 {x + iy Exercise 60. Resolve into partial fractions : 7a: +1 - bx—l (x + 4:)(x-b) {2x-lXx-5) x'-l „ x"^ — X 2. 6 . .-2 (x + SXx + 'i) ;r'-3a;-10 x(x''-4) MISCELLANEOUS PROPERTIES OF SERIES. 341 ^x" -4 x\x + 5) Jx'-x (X- 2x' 10. 11. 9. "•" ••:'" • 12. 7x 6x'-bx + l 13a7 + 46 12a:2-lla;-15* 2a:^-lla: + 5 :r^-l :r^ 384. Expansion in Series. A series which is obtained from a given expression is called the expansion of that expression (§ 243). The given expression is called the generating func- tion of the series. Thus (^ 250), the expression is the generating function of 1 — X the infinite series l+x + x^ + a^ + When the expression is a finite series, the generating function is equal to the expansion for all values of the symbols involved. Thus, (l±l^Y = \ + e + i2x + &^. \ X J or X When the expansion is an infinite series, the generating function is equal to the expansion for only such values of the symbols involved as make the expansion a convergent series. Thus, is equal to the series l+x + x^-\-oi^ + when, and 1 — a; only when, x is numerically less than 1 (§ 251). 385. The expansion of a given expression may be found : By division, By the binomial theorem. By the method of undetermined coefficients. By other methods, which involve a knowledge of the Differential Calculus. 342 ALGEBRA. (1) Expand -— — - in ascending powers of x. Divide a? by 1 + x"^, then l+x" provided x is so taken that the series is convergent. By § 373 x must be numerically less than 1. (2) Expand :^ in descending powers of x. Divide a; by a;^ + 1, then X 1_1^1_ 1 + x'^ X X^ 0^ provided x is so taken that the series is convergent. By ^ 373 x must be numerically greater than 1. In the two preceding examples we have found an expansion of — - — for all values of x except ± 1. (3) Expand ^ in ascending powers of x by the X -J- X binomial theorem. 1 = (1 + a;2)-i = 1 - a;2 + a* l+x"" X — x^ + x^ — 1+x^ provided x is so taken that the series is convergent. (4) Expand - — — in ascending powers of x. Assume 2 + 3x _ ^ + ^3. + cfc2 ^ jr>^ 1 + X + x^ then, by clearing of fractions, 2 + 3x = A + Bx + Cx^ + Da^ + + Ax + Bx^ + Ca? + + Ax^ + j5x3 ^ MISCELLANEOUS PROPERTIES OF SERIES. 343 By §256, A^2, B + A = S, C+B + A = 0, I) + C+B = 0; whence ^=1, C= — 3, D = 2, and so on. ... A±A^^2 + x-3x'' + 2r^ + x*-33^ + 1+x + x'^ The series is of course equal to the fraction for only such values of X as make the series convergent. Remaek. In employing the method of Undetermined Coefficients, the form of the given expression must determine what powers of the variable x must be assumed. It is necessary and sufficient that the assumed equation, when simplified, shall have in the right member all the powers of x that are found in the left member. If any powers of x occur in the right member that are not in the left member, the coefficients of these powers in the right member will vanish, so that in this case the method still applies ; but if any powers of x occur in the left member that are not in the right mem- ber, then the coefficients of these powers of x must be put equal to in equating the coefficients of like powers of x ; and this leads to absurd results. Thus, if it were assumed in problem (4) that ^•^^■'' = Ax -f J5x2 + Cj^ + , 1 + a? + x'^ there would be in the simplified equation no term on the right cor- responding to 2 on the leftr so that, in equating the coefficients of like powers of x, 2, which is 2x^, would have to be put equal to Ox" ; that is, 2 = 0, an absurdity. (5) Expand (a — xy in a series of ascending powers of x. Assume {a- xf = A-{- Bx + Cx^ + D:^ + Square, a-x = A^\2ABx^{2AC^W)x'^{2AD^2BQ)x^-\- Therefore, by \ 256, A^ = a, 2AB = -\, 2^(7+52 = 0, 2^2) + 25(7= 0, etc., and ^ = a*, 5=---, G=-\. ^ = --^i 2a2 8a^ 16a^ Hence, (a-x)^ = a* - -^ --^ - -^- Cf. g 258 2a* 8 a* 16 a^ 344 ALGEBRA. (6) Expand powers of x. Assume 1 + x (l + :r)(l + ^^) in a series of ascending 1 + x _ A Bx + C (1 + x){l + a;2) 1 + a; 1 + x^ .: 1 + x = {A + B)x'' + (5 + C)x + (^ + C). :.A^B = 0, B + C=l, A + 0=7. Whence, A = 3, B = -3, C-4. 7 + x _ 3 4-3x " {l+x){l+x^)~l +x 1 +x^' -^ = 3 (-^\ - 3(1 - a? + a;2 -x3 + a;*- ) 1+a; Vl+^y == 3 - 3a; + 3x' -^x^ + Sx^- , But and izil^=(4-3») l + a;2 Adding the two series, 7 + x 4-3a;-4a;2 + 3a;3 + 4a;* = 7-6x-x'' + 7x* (1 + a;)(l + a;2) 386. Reversion of a Series. Given y = ax -\- bx"^ -\- cx^ + dx^ + , where the series is convergent, to find x in terms of y. Assume x = Ay + By"^ + Cy^ + Dy^ + In this series for y put ax + hx^ + cx^ + dx^ + result is x^-\- the x = a Ax -\- hA x'+cA 3?-\-dA -\-a'B + 2abB + h'B -{-a'C + 2acB + 3a^^C -\-a'D MISCELLANEOUS PROPERTIES OF SERIES. 345 Comparing coefficients (§ 256), aA = l] bA + a'B = 0; cA + 2abJB -\- a'C =0; dA + h'B + 2acB + 3a^5(7+ a'D = 0. a 6? a* (1) Given y^=^x^x^-\-o?-\- ; find x in terms of y. Here, a = l, 6 = 1, c = l, d=\ A = \, B = -l, C=l, D = -l, Hence, x^y ^y^ + y^ —y* + /y»* /v»" /y»^ (2) Reverty = a;--+--- + Here, a = l, & = — J, c = J, d = — ^,... y2 y3 y4: Hence, ^ = 2/ + ^ + % + g + Exercise 61. Expand to four terms in ascending powers of x : 1 4 1-^ 7 x(x-l) l-2a; ' l + ^ + :i;^ * (a:+l)(a:'+l) 1. 2. _A_. 5. ^-^^ . 8. ^1Z1^±1. 2-3a; l-}-x-x' x'(x'-l) ^ 1+x ^ ix-Qx" ^ 2r'-l 2 + 3a; l-2a; + 3a;^ a:(r'+l) 346 ALGEBRA. Expand to four terms in descending powers of x: 10. -A_. 12. -Izi2^. 14. 3^-2 2 + x 1 + 3:^-07'' x(x-iy 11 ^ — ^ 12, x^ — x-i-l ^^ x'-x-{-l 3-[-x x(x-2) (x-lXx'+l) Revert : 16. y = x--2x'' + Sc(^-4:x' + ' ^ f\ 7 17. y = ^--+---+ /v»* /v»** /y»^ 18. , = . + ^+^ + ^ + 387. The following series have been already studied : (1) Arithmetical series (§§ 218-224). (2) Geometrical series (§§ 225-231). (3) Harmonical series (§§ 232-235). (4) Expansions obtained by the binomial theorem (§§ 239-260). (5) Series of Differences (§§ 377-380). We shall now consider series obtained by division, or by the method of undetermined coefficients (§ 385). \ A- X 388. Kecurring Series. From the expression ^ we obtain by actual division, or by the method of unde- termined coefficients, the infinite series 1 + 3a; + 7^' + 17:f' + 41 a;* + ^x^ + In this series any required term after the second is found by multiplying the term before the required term by 2x, the term before that by x"^, and adding the products. MISCELLANEOUS PROPERTIES OF SERIES. 347 Thus, take the fifth term : In general, if w„ represent the nth term, W„ = 2 XUn-i + x^u,^_2. A series in which a relation of this character exists is called a recurring series. Recurring series are of the first, second, third, order, according as each term is dependent upon one, two, three, preceding terms. A recurring series of the first order is evidently an ordi- nary geometrical series. In an arithmetical, or geometrical, series any required term can be found when the term immediately preceding is given. In a series of differences, or a recurring series, several preceding terms must be given if any required term is to be found. The relation which exists between the successive terms is called the identical relation of the series ; the coefficients of this relation, when all the terms are transposed to the left member, is called the scale of relation of the series. Thus, in the series 1 + 3a; + 7a;2 + 17a;3 + 41.x* + 99^5 + the identical relation is Un = 2 XUn-l + x'^Un-2 ; and the scale of relation is l-2a;-x2. 389. If the identical relation of the series is given, any required term can be found when a sufficient number of preceding terms are given. Conversely, the identical relation can be found when a sufficient number of terms are given. 348 ALGEBRA. (1) Find the identical relation of the recurring series 1 + 4a; + Ux' + 49 r' + 171 x* + 597^^ + 2084:i:« + •- Try first, a relation of the second order. Putting n = 3, and, then, w = 4, 14 = 4/) + q, 49 = 14:p +4:q; whence, P = h q = ^- This gives a relation which does not hold true for the fifth and following terms. Try next a relation of the third order. Assume Un =pxun-i + qx^Un-2 + ra;^Wn-8- Putting n = 4, then n = 5, then n = 6. 49= lip + 4:q+ r, 171- 49p + 14^+ 4r, 597 = 171|) + 49g + 14r; whence, p = 3, q = 2, r -= — 1. This gives the relation Un=S XUn- 1 + 2 x'^Un-2 — X^Un-3 which is found to hold true for the seventh term. The scale of relation is 1 — 3x — 2x'^ + x^. (2) Find the eighth term of the above series. Here, Wg = 3 xu.j + 2x^Uq — x^u^ = Sx (2084 x^) + 2x\d97a^)-x*(l1l x*) = 1-21bx\ Am. . 390. Sum of an Infinite Series. By the sum of an infinite convergent numerical series is meant the limit which the sum of n terms of the series approaches as n is indefinitely increased; a divergent numerical series has no true sum. By the sum of an infinite series of which the successive terms involve one or more variables is meant the generating function of the series (§ 384) ; that is, the expression of which the series is the expansion. MISCELLANEOUS PROPERTIES OF SERIES. 349 The generating function is a true sum when, and only when, the series is convergent. The process of finding the generating function is called summation of the series. 391. Sum of a Eecurring Series. The sum of a recurring series can be found by a method analogous to that by which the sum of a geometrical series is found (§ 227). Take, for example, a recurring series of the second order in which the identical relation is Uu=puu-i-^quu-2, or Uj, —puu-x — quj,_^ = 0. Let s represent the sum of the series ; then s = w, + W2 + W3 + w^.i + w„, — pS = —pUi —pU2 — —pUn-2 —pUn-l—pUn, - qs= — qUi — — qUn-3 — qu^-i — qu^-i — qu^. Now, by the identical relation, u^—pu^— qui = 0, Ui—pu^—qu2 = 0, Un—pUn-i—qUn-i. = 0. Therefore, adding the above series, „ ^ -^1 + (^2 —pui) pu„ i- q(un + u^.i) l-p-q l-p-q Observe that the denominator is the scale of relation. If the series is infinite and convergent, u^ and w„_i each approaches as a limit, and s approaches as a limit the fraction ^i + K-J^^O. l-p-q If the series is infinite, whether convergent or not, this fraction is the generating function of the series. 350 ALGEBRA. For a recurring series of the third order of which the identical relation is Ui-\-(u2— pu^ + (ws — pu2 — qui) we find s = :, 1—p—q—r l—p~q—r Similarly for any recurring series. (1) Find the generating function of the infinite recurring series l + 4:x+lSx'-{-^Sa^ + 14:2x' + By § 389 the identical relation is found to be Ufe = 3 XUjc-l + X^U}c-2. Hence, s = 1 + 4 oj + 13 a;^ + 43 a;^ + 142 a;* + ..... -3a;s= -3a;-12a;2-39a;3-129a;*- - a^8= - a;2- 4x3- i3a;4_ Adding, {1 — 3x — x^)s = l + x, 1 +x 1-3 X — X- (2) Find the generating function and the general term of the infinite recurring series 1 _ 7a; _ a;2 _ 43^ ._ 49^4 _ 397^5 _ Here Uk = xuk-i + 6 x'^Uk-2. s = l_7a;_ x2-43a;3_49.x*- _ X8=- — x + 1x'^+ a;3 + 43a;* + _6aj2s= -6a;2 + 42a;3+ 6x^ + l-8rp __ l-8a; ^ l-a;-6a;2 (1 + 2a;)(l - 3x)' MISCELLANEOUS PROPERTIES OF SERIES. 351 By g 383 we find l-8a; ^ 2 1 (l + 2a;)(l-3a;) 1 + 2a; l-3a;" By the binomial theorem or by actual division, — - — = 1 - 2a; + 22a;2 - 2^a^ + + 2^(- lYx^ + , 1 +2a; — - — = 1 +3x + 3'^x^ + S^a^ + Z^x^ + 1-Sx Hence the general term of the given series is [2r+l(_l)r_3rja;r. (3) Find the identical relation in the series P + 2^ + 3^ + 4^ + 5^^+6^4-7^-}- The identical relation is found from the equations 16= 9p + iq+ r, 25 = 16j9+ dq + ir, SG = 25p + 16q + 9r, . to be Uk = 3 ujc-i — 3 u^^i + ujcs. Exercise 62. Find the identical relation and generating function of: 1. l + 2x+7x'' + 2Ss^+76x*+ 2. S + 2x + Sx'-i-7x'-{-18x*-i- Find the generating function of : 3. 2-\-Sx + 5x' + 9x' + 11x*-{-SSx^-{- 4. 7 - 6^7 + 9x' + 27a:'+ 54a;*+ 189:r' + 5. l-{-bx-{-9x'+13x'+nx*-}-21c(^-\- 6. l + a:-7a;' + 33^*- 130a;^ + 499^'+ 7. S-{-6x+14:x' + S63^-{-9Sx* + 2763^-\- Find the sum of n terms of : 8. 2 + 5 + 10 + 17 + 26 + 37 + 50 + 9. P + 2' + 3H4H5' + 352 ALGEBRA. EXPONENTIAL AND LOGARITHMIC SERIES. 392. Exponential Series. By the binomial theorem \ nj n 1x2 'nr . nx {nx — l)(yi:y — 2) v. J- . "^ 1x2x3 n^"^ = l+x f i\ ( ly 2\ l;^ ^ + (1) This equation is true for all real values of x, since the binomial theorem may readily be extended to the case of incommensurable exponents by the method of § 264 ; it is, however, only true for values of n numerically greater than 1, since - must be numerically less than 1 (§ 375). n As (1) is true for all values of x, it is true when x=^l. [(-9">(-y But Hence, from (1) and (2) •• (2) §264 1 + 1+^+ ^ "A n) + = l + ^-f x{x~^^ ^(^^-_g(^_?) [2 li EXPONENTIAL AND LOGARITHMIC SERIES. 353 This last equation is true for all values of n numerically- greater than 1. Take the limits of the two members as n increases without limit. Then (§ 367) {'^'■"tt J and this is true for all values of x. It is easily seen by § 373 that the second series is convergent for all values oi X] the first series was proved convergent in § 372. The sum of the infinite series in parenthesis is called the natural base (§ 267), and is generally represented by e ; hence, by (3), ^=1+^+1+1+ ^ To calculate the value of e we proceed as follows : 1.000000 • 2 3 4 5 6 7 8 9 1.000000 0.500000 0.166667 0.041667 0.008333 0.001388 0.000198 0.000025 ding, ten places. e = £ = 0.000003 -. 2.71828. = 2.7182818284. 393. In A put ex in place of x ; then Put €f = a\ then c = \og^a, and e'* = a'. .-. a-=l + ^log,a + ^iM + ^^i^'+ B L£ l£ The series in B is known as the exponential series ; B re- duces to A when we put e for a. 354 ALGEBRA. 394. Logarithmic Series. In A put e* = 1 + y ; then a7 = loge(l+y), and by A, x^ , x^ , x^ 2'="+ll+[l+g+ Revert the series (§ 386), and we obtain ^^y^yl+t^t^ ^ y 2^3 4 ' But a; = loge(l+y). .-. log,(l+y) = y-5 + ^-{ + Similarly from B, The series in D is known as the logarithmic series ; D re- duces to when we put e for a. In and D y must be between — 1 and + 1, or be equal to + 1, in order to have the series convergent (§ 376, Ex. 1). 395. Modulus. Comparing and D we obtain log„(H-y)=-l-log,(l + y); or, putting iV for 1 + y, iog«i\^=-J-iogeiv: log«a Hence, to change logarithms from the base e to the base a, multiply by =logae; and conversely (§283). loge^ The number by which natural logarithms must be multi- plied to obtain logarithms to the base a is called the modu- lus of the system of logarithms of which a is the base. Thus, the modulus of the common system is logjo/! -{-nx approaches as n ap- proaches as a limit. 3. Provethat l = 2(i + | + i + ) 4. Calculate to four places, loge4, log^S, loge6, loge?. 5. Find to four places the moduli of the systems of which the bases are : 2, 3, 4, 5, 6, 7. 6. Show that 1 /8\_ 5 7 9 ^^\ej Ix2x3"^3x4x55x6x 7 7. Show that log.a-log.5 = — - + -^_-J + -^— j + 8. Show that, if x is positive, 93 03 A3 9. Show that l + ^ + | + l=:5.. [^ 12 [^ 10. Show that 6^^^ = X+ YV^^ where /y«2 /y,4 ^6 -,,3 ^y^ , rt^T 11. Expand ^^^^ in ascending powers of a;. 12. Expand :== in ascending powers of a;. we obtain CHAPTER XXVIII. DETERMINANTS. Origin. Solving the two simultaneous equations 0-2^ + b-ii/ = ^2, Similarly, from the three simultaneous equations a^x + b,y + c^z = di, a2(^3 + di^zCi + asbiCi — aib^Ci — aihiC^ — a^iCx, which agrees with § 398. There is, no simple rule for expanding determinants of orders higher than the third. 364 ALGEBRA. Exercise 64. Prove the following relations by expanding : 1. 2. a^ ^ ^2 a h i cti = a, ai ' ai a^ a. 0,3 a^ ^1 bi c, ai h h. h = (-3 Ci Ci = — h = c. Hence, a — 6, 6 — c, and c — a are factors. A is of the third degree in a, 6, c, and these are easily seen to be all the factors. It remains to determine the sign before the product. In A as given c^h is + ; in the product (a — 6)(6 — c){c — a) the term (j?h is — . Hence, A = - (a - 6)(6 - c){c - a). 370 ALGEBRA. (2) Resolve into factors d^ a h -{-0 h'' b c + a c^ c a-\-b As in the last example a — h, b — c, c — a are found to be factors. There is one other factor of the first degree. To the third column add the second ; the result may be written a" a 1 (a + b + c) or, by Ex. 1, 62 6 1 c2 c 1 {a + b + c){a - b){b — c){e -a). Show that : a b a c b c 3. Exercise 65. 2abc. 2. a' 1 b' b' b' 1 c' c' c' 1 d' d' d' b -\-c a b e-{- a a b a + b = 4:abc. bed a d^ o? cda b b' b' dab c c^ c^ abc d d^ d^ 4. 1 1 1 1 c' b' 1 c^ a^ — 1 b' a' Find the value of 20 15 25 17 12 22 19 20 16 a b c a c b b c a c b a 6. 3 23 13 7 53 30 7. 9 70 39 Resolve into simplest factors 8. a a" b ¥ a a" b b' be ca ab 10. 22 29 27 25 23 30 28 26 24 a' be 1 b^ ca 1 c^ ab 1 DETERMINANTS. 371 11. a h c h c a . 12. a b 1 a a? a' 1 1 h c h' c' . 13. 1 d 6? d> a b c d h a d c c' d a b d e b a 14. If all the elements on one side of a diagonal term are zeros, show that the expansion reduces to this term. Show that : 15. 16. o? — be a 1 b-'-ca b 1 c^ ~ ab c 1 = 0. a + 25 a + 46 a + 6^» a + 3S a + bb a-{-7b a + 4:b a-\-6b a + 8b 17. b^ + c^ ba ca ab & + c^ cb ac be a^ + b'^ = 4:a'b'c\ 18. (a-hbf (b + ef (e + af 2abe(a-^b + cy 19. 1+^ 2 3 4 1 2-{-x 3 4 1 2 3 + ^ 4 1 2 3 4 + a; = x' + lOx'. 20. a^ -f 1 ba ea da ab b' + l eb db ae be c^ + 1 de ad bd cd d^-{-l -i-d'+l. 372 ALGEBRA. 417. Minors. If one row and one column of a determi- nant be erased, a new determinant of order one lower than the given determinant is obtained. This determinant is called a first minor of the given determinant. Similarly, by erasing two rows and two columns we ob- tain a second minor ; and so on. Thus, in the determinant {a^ b^ c^\, erasing the second row and third column, we obtain the first / 7^ a \ ^ ' 2 This minor is said to correspond to q Cg ^2 I the element 63, and is generally represented by Aj ; so that, in this case, Ai =\ ^ % ai br h ^3 b. Ci c, C3 Ci di di ds d. In general, to every element corresponds a first minor obtained by erasing the row and column in which the given element stands. 418. Theorem. If all the elements of the first row after the first element are zeros, the determinant reduces to a^^a^. Consider the determinant Ci d. Every term of A contains one, and only one, element from the first row ; and all the terms that do not contain (Xi contain one of the zeros, and therefore vanish. The terms that contain a^ contain no other element from the first row or column, and, consequently, contain one, and only one, element from each row and column of the determinant ^2 ^^3 ^4 ^2 <^3 ^4 C?2 C?3 0?4 Hence, disregarding the sign, each term of A consists of «! multiplied into a term of A^^. or A, DETERMINANTS. 373 Take any particular term of A, as ai h^, Cg d^ ; the sign is fixed (§ 404, Rule I.) by the number of inversions in the series 14 3 2; the sign of the term i^ c^ d^ of A^^ is fixed by the number of inversions in the series 4 3 2. Adding <2i makes no new inversions among either the letters or the subscripts. Consequently the sign of the term in A is the same as the sign of the term in aiA^^. Since this is true of every term of A, we have A = a^Ha,- Similarly for any determinant of like form. 419. Terms containing an Element. From § 418 it appears that the sum of the terms which contain ax may be written ttiAfl^. For, no one of the terms which contain ai can con- tain any one of the elements aj, <^3> «4, » *ri(x) be the quotient obtained by dividing /(^) by a; — A, we have f{x) = {x-h)4>{x\ and the equation f(x) — may be written {x -A)<^(a;) = 0, of which A is evidently a root (§ 84). GENERAL PROPERTIES OF EQUATIONS. 387 435. Synthetic Division. Let the quantic 3x' - Ax* + x^~ 12:r^ + 3a; + 6 be divided hj x — 2. The work is as follows : 3a^-4a;* + a;3_ 12x2 + 3a; + 6 Sc^-ex* + 2x^ + ^ + 2x^ ~^a^ + 5x3 _ .12x'» + 5x3- 10x2 . 2x2 + 3x • 2x2 + 4x - x + 6 - x + 2 3a;* + 2x3 + 5x2-2x-l The work may be abridged by omitting the powers of x, and writ- ing only the coefficients. We now have 3-4 + 1-12 + 3 + 61 1-2 3-6 3+2+5-2-1 + 2 + 1 + 2-4 + 5-12 + 5-10 - 2 + 3 - 2 + 4 -1 + 6 -1 + 2 But the operation may be still further abridged. As the first terra of the divisor is unity, the first term of each remainder is the next term of the quotient, and we need not write the quotient. Second, we need not bring down the several terms of the dividend Third, we need not write the first terms of the partial products. 388 ALGEBRA. The work is now as follows : 3_4 + l_12 + 3 + 6|l + 2 -4 + 5 -10 2 + 4 -1 +2 + 4 Omitting the first term of the divisor, which is now useless, chang- ing — 2 to + 2, and adding, instead of subtracting, we have, raising the terms and bringing down the first coefiicient, 3-4 + 1-12 + 3 + 6 |_2 •+6 + 4 + 10 -4-2 3+2+5- 2-1+4 The last term below the line gives us the remainder, the preceding terms the coefficients of the quotient. In this particular problem the quotient is 3a;* + 2a;^ + 5a;2 — 2a;— 1, and the remainder is 4. This method is called the method of Synthetic Division. For the application of this method to the division of any quantic by x~h we have the following rule : Write the coefficients a, b, c, etc., in a horizontal line. Bring down the first coefficient a. Multiply a hy h, and add the product to b. Multiply the sum so obtained by h, and add the product to c. Continuing this process, the last sum will be the remainder, and the preceding sums the coeffiicients of the quotient. Remark. If there are any powers of x missing, their places are to be supplied by zero coefficients. GENERAL PROPERTIES OF EQUATIONS. 389 Ex. Divide 2x' — Qx"" -{- 6x-2 by x~S. 2 + 0- 6+ 5- 2U ^ + 6 + 18 + 36 + 123 2 + 6 + 12 + 41 + 121 The quotient is 2a^ + 6a;2 + 12 a; + 41, and the remainder 121, 436. Value of a Quantic. By the principles of division it is evident that the operation of dividing a given quantic f(x) hj X — h can be carried on until the remainder does not involve x. Kepresent the quotient by ^(a;), and the remainder by H. Then, we have /(x)=(x-h)<}>(x)-\-Ii. Putting h for x, /(A)=0 + i?. Hence, the value which a quantic f (x) assumes when we put hfor X is equal to the last remainder obtained in the operation of dividing f (x) by x — h. This remainder, and, consequently, the value of the quantic, may be easily calculated by the method of syn- thetic division. The truth of the above theorem may also be shown by another method, which has the advantage of showing the form of the quotient and remainder. Take, for example, the quantic ax*" -f bx^ + co(^ -{- dx-\-e. Divide the quantic by x — h. The work is as follows : a b c d e \ h ah Bh Ch Dh 390 ALGEBRA. where B~—ah -\-h, C=Bh -{-€-= ah^ -{-hh-^c, D=Ch-^d=ah^-^bh?-\- eh + d, B = Dh-\- e = ah' -{-hh^ -{- cK" + dh-\-e. The remainder M is evidently the value which the quan- tic assumes when we put h for x. The quotient is ax^ + {ah + h)x'' + {ah^ ^hh-\- c)x-]- (ah^' + hh^-^-ch^ d). Similarly for any quantic. Exercise 68. Find the quotient and remainder obtained by dividing each of the following quantics by the divisor opposite it. 1. x'-?>x'' — x^-^2x-~l x — 2. 2. x'-?>x'' + 2x-l x — 2>, 3. 2x'-\-2>x^-^x'-lx-l0 x~2. 4. 3a;' + 2a:' — 6:17 + 50 x + ^. ^, ax^-{-2>hx^^?>cx-\-d x -\- h. Are the following numbers roots of the equations oppo- site them (§ 434) ? 6. (3) x'']-x'-^x-\-2 = 0. 7. (-7) a;* + 7;r^ + 21a;+147 = 0. 8. (0.3) a;* - 2.3 a:' + 3.6a;^ + 4.9a; +1.2 = 0. Find the value of the following expressions when for x we put the number in parentheses : 9. 3a;' + 2a;'- 6a: +1 (-3). 10. 2a;* + 6a:' — 9a7-5 (6). 11. a:5 + 7a:'-2a:'-49 (-4). 12. a;* + 6a:' -7a;'- 3a; +1 (-0.2). GENERAL PROPERTIES OF EQUATIONS. 391 437. Number of Eoots. We shall assume that every rational integral equation has at least one root. The proof of this truth is beyond the scope of the present chapter.* Let/(^) = be a rational integral equation of the wth degree. This equation has, by assumption, at least one root. Let aj be a root. Then, by § 433, J(x) ~{x- a^Mx), where f I (x) is a quantic of degree n—1. The equation /i (a:) = must, by assumption, have a root. Let og be a root. Then, by § 433, f, (x) = (:r -^ a,) /, (x), where /a (a;) is a quantic of degree n — 2. Continuing this process, we see that at each step the de- gree of the quotient is diminished by one. Hence, we can find n factors a? — ai, x — Oi x — a„. The last quotient will not involve x, and is readily seen to be ao, the coeffi- cient of a;" mf(x). Now, f(x) = (x- ai)/i (x) = (x — ai)(x — a2)f2X = ao(x — ai)(x — a2) (x — On), so that the equation f(x) = may be written aQ(x — ai)(a; — a^) (x — a„) = 0, which evidently holds true if x has any one of the n val- ues tti, Oa a„. It follows, then, that if every rational integral equation has one root, an equation of the nth degree has n roots. * See Burnside and Panton, Theory of Equations, 2d ed., Art. 195 ; Briot et Bouquet, Fonctiona Elliptiqite, Art. 23. 392 ALGEBRA. 438. Linear Factors. The factors x — ai, x — a^ x—On are linear functions of x (§ 429). When f{x) is written in the form ao{x — ai)(a; — a,) {x — a^), it is said to be resolved into its linear factors. From § 437 it follows that a quantic can be resolved into linear factors in only one way. To resolve a quantic /(:r) into linear factors is evidently equivalent to solving the equation f{x) = 0. 439. Multiple Eoots. The n roots of an equation of the nth. degree are not necessarily all different. Thus, the equation ^^ — 7:^;^ + 15^ — 9 = may be written (x — 1)(^ — 3)(a; — 3) = 0, and- the roots are seen to be 1, 3, 3. The root 3, and the corresponding factor a; — 3, occurs twice ; hence, 3 is said to be a double root. When a root occurs three times, it is called a triple root; four times, a quadruple root; and so on. Any root which occurs more than once is a multiple root. 440. Eoots Griven. When all the roots of an equation are given, the equation can at once be written. Ex. Write the equation of which the roots are 1, 2, 4, — 5. The equation is {x — 1) (a; - 2) (x - 4) (a; + 5) == 0, or a;*-2a;3-21a;2 + 62a;-40 = 0. 441. Solutions by Trial. When all the roots of an equa- tion but two can be found by trial, the equation can be readily solved by the process of § 437. The work can be much abbreviated by employing the method of synthetic division (§ 435). Of. § 140. GENERAL PROPERTIES OF EQUATIONS. 393 Ex. Solve the equation Try + 1 and — 1. Substituting these values for x, we obtain 1 - 3 ~ 6 + 14 + 12 = 0, 1 + 3-6-14 + 12 = 0, which are both false, so that neither + 1 nor — 1 is a root. Try 2. Dividing by a; - 2, 1 _3 _6 +14 +12L2 + 2 -2 -16 -4 1 _1 _8 - 2 + 8 we see that 2 is not a root. Try 3. Dividing by a; - 3, 1 _3 -6 +14 +12|_3 + 3 +0 -18 -12 1+0-6-4 we see that 3 is a root. The quotient is a^ — 6x — 4:. In this quotient try 3 again. Dividing by x — Z, 1 +0 -6 -4L3 +3 +9 +9 1 +3 +3 +5 we see that 3 is not again a root. Try - 2. Dividing by a; + 2, 1+0-6 -4 | -2 -2 +4 +4 1-2-2 we see that — 2 is a root. The quotient is a:* — 2 a; — 2. Hence the given equation may be written (a;- 3)(a; + 2)(a:»- 2a;- 2) = 0. Therefore one of the three factors must vanish. If a;-3 = 0, a; = 3; if a; + 2 = 0, a;=_- 2; if a;*- 2a; - 2-0, solving this quadratic, we find a; =» 1 + V3 or a; =. 1 - VS. Hence the four roots of the given equation are 3, -2, 1 + V3, 1-V5. 394 ALGEBRA. Exercise 69. Solve the equations : 1. x'- 7:r^+ 16a; -12 = 0. 2. x'+ 9x''+ 2a; -48 = 0. 3. x^— ^x^~ 8:r+ 8 = 0. 4. x^~ 5a;^— 2a; + 24 = 0. 5. a;'+ 2^2+ 4a;+ 3 = 0. 6. a^— 6a;' + 6a; + 99 = 0. 7. 6a;' - 29a;' + 14a; + 24 = 0. 8. 2a;^+ 3a;'— 13a; — 12 = 0. 9. x' — 15a;' — lOo; + 24 = 0. 10. a;*+ 6a;'- 5a;'- 45a;- 36 = 0. 11. x'+ 4a;' — 29 a;' -106 a; +130 = 0. 12. x'— 5a;'— 2a;' + 12a; + 8 = 0. 13. 6a;*- 5a;'- 30a;' + 20a; + 24 = 0. 14. 4a;* + 8a;' -23a;'- 7a; + 78 = 0. Form the equations which have the following roots : 15. 2, 6, -7. 19. 5, 3+V^, 3-V=I 16. 2, 4, -3. 20. 2, ^, 2, -f 17. 2, 0, -2. 21. 2, 3, -2, -3, -6. 18. 2, 1,-2,-1. 22. 1, I, -I, -|. 23. 3 + V2, 3-V2, 2 + V3, 2- V3. 24. 0.2, 0.125, -0.4. 25. 0.3, -0.2, --^,-1 26. 2 + V^, 2- V^, 1 + 2^^=1, 1_2V=^. GENERAL PROPERTIES OF EQUATIONS. 395 442. Eolations between the Boots and the Ooeffioients. The quadratic equation of which the roots are a and yS is (§ 153) or, multiplying out, The cubic equation of which the roots are a, )8, y is (x-a){x-p)(x-y) = 0, or x'~(a + l3-}-y)x' + (al3 + ayi-Py)x-aPy = 0. The biquadratic equation of which the roots are a, /S, y, Sis (^ - a)(x - PXx - y)(x - 8) = 0, or x'-~(a-}-P + y + B)x'+(api-ay + aS + py + p8-{-yS)x' - (a^y + a^8 + ayS + fiyS) x + ajSyS = 0. And so on. Take any equation in which the highest power of x has the coefficient unity. From, the above we have the follow- ing relations between the roots and the coefficients : The coefficient of the second term, with its sign changed, is equal to the sum of the roots. The coefficient of the third term is equal to the sum of all the products that can be formed by taking the roots two at a time. The coefficient of the fourth term, with its sign changed, is equal to the sum of all the products that can be formed by taking the roots three at a time. The coefficient of the fifth term is equal to the sum of all the products that can be formed by taking the roots four at a time ; and so on. If the number of roots is even, the last term is equal to the product of all the roots. If the number of roots is 396 ALGEBRA. odd, the last term, with its sign changed, is equal to the product of all the roots. Observe that the sign of the coefficient is changed when an odd number of roots are taken to form a product ; that the sign is unchanged when an even number of roots are taken to form a product. 443. By dividing the equation through by the coefficient of the highest power of x, any rational integral equation whatever can be reduced to a form in which the coefficient of the highest power of x is unity. "We shall write an equation reduced to this form, called the "^/orm," as follows : Let a, p, y, etc., be the roots of this equation. Represent by Sa the sum of the roots, by %aP the sum of all the products that can be formed by taking the roots two at a time ; and so on. From § 442 we now have 2a =—pi, pi = — Sa, ^a^y = —^3, Pz = — ^o.Py, a^yS = (- 1)%. pr, = (- Ifa^yh Ex. Let a, jS, 7 be the roots of the equation Then, ' 2o=a+)3 + 7= 7, So)8 = /37 + 70 + a)3 = — 9, 0J87 = — 4. The relations between the roots and the coefficients of an equation do not assist us to solve the equation. In every case we are brought at last to the original equation. GENERAL PROPERTIES OF EQUATIONS. 397 Thus, in the equation we have a + i3 + 7 =- 7, iSy + 7a + oj8 = - 9, afiy = - 4. Eliminating ^ and 7, we have to solve the equation a3-7a2-9a + 4 = 0; that is, we have to solve the given equation. 444. Symmetric Punctions of the Eoots. The expressions 2 a, 2a^, Sa^y, , are examples of symmetric functions of the roots (§ 152). Any expression which involves all the roots, the roots all entering to similar powers and with similar coefficients, is a symmetric function of the roots. From the relations %a = —pr, ^ap = -\-p2, 2ay8y = — ^3, , the value of any symmetric function of the roots of a given equation may be found in terms of the coefficients. If a, p, y are the roots of the equation we may calculate the values of symmetric functions of the roots as follows : We have o + /3 + 7 = 4, (1) /37 + 7a + a)8 = 6, (2) aj87 = 5. (3) (1) 2o2 = a2 + i82 + 72. Square (1), a"" + fi^ + 7^ + 2)87 + 2ya + 2a$ = 16 But, by (2), 2)87 + 27a + 2a)8==12 .-. a" + )8'» + 7' - ■* (2) Sa'/S = a'fi + a?y + ff'y + &^a + y^a + y'$. Multiply (1) by (2), Sa^iS + Za$y = 24 But, by (3), 3a$y^l5 .'. 5a2)3 - 9 398 ALaEBRA. (3) :Za' = a' + fi' + y\ Multiply o2 + j82 + 72 ^y a + fi+y; the result is a^ + fi^ + y^ + Sa^jS = 16 But Sa'^iS- 9 .'. a^ + &' + -/ =-7 And so on. Cf. § 152. 445. By the aid of tlie preceding sections we can find the condition that a given relation should exist among the roots of an equation. Find the condition that the roots of the equation x^ -\- ^^x"^ -{- qx-\-r-- = shall be in geometrical progression. . Let )8 be the mean root. Then, + ^ + 7 = -J), i87 + 7a + 0)8 - J, a)87=--r. and j82 = 7a. From (2) and (4), i87 + a;8 + j82 = q. or, by (1), -pfi = q. .-.--I- Substituting in (3), ( ~ I ) = ~ ^' (1) (2) (3) (4) or (f=p^r, the required condition. 446. Imaginary Eoots. If an imaginary number is a root of an equation with real coefficients, the conjugate imagi- nary (§ 176) is also a root. GENERAL PROPERTIES OF EQUATIONS. 399 Let a + P^', where i = V— 1, be a root of the equation tto^" + a^a;"-' + an = 0, the coefficients being real. Put a + pi for X in the left member of this equation, and expand the powers of a + Pi by the binomial theorem. All the terms which do not contain i, and all the terms which contain even powers of ^, will be real ; all the terms which contain odd powers of i will be imaginary. Eepresenting the real part of the result by P, and the imaginary part of the result by Qi, we have (§ 432), since a + pi is a root, and therefore P=: and Q = (§ 179). Now put a — /3i for x in the given equation. The result may be obtained from the former result by changing i to — i. The even powers of i will be unchanged while the odd powers will have their signs changed. The real part will therefore be unchanged, and the imaginary part changed only in sign. The result is P- Qi, which vanishes, since by the preceding P = and Q = 0. Therefore a — pi is a root of the given equation (§ 432). This theorem is generally stated as follows : Imaginary roots enter equations in pairs. The above proof will be more readily understood if applied to an equation of the third or fourth degree. Corresponding to a pair of imaginary roots, we shall have the factors x — a — ^i, x — a-{- fii. The product of these, is positive, provided x is real. Hence, corresponding to a pair of imaginary roots, we have a factor of the second de- gree, which for real values oix does not change sign (§ 180). 400 ALGEBRA. Exercise 70. 1. Form the equations of which the roots are : 2,4,-3; 3,-2,-4. If a, y8, y are the roots of x^ — bx^ + 4a; — 3 = 0, find the value of: 2. %a\ 5. :Sa^/?y. 8. Sa*. 3. %a^p. 6. %o?p\ 9. :Sa«^y. 4. ^a\ 7. :^a% 10. :Sa-^^V. If a, p, y are the roots of x^ -\-]px^ ■\- qx-\-r=-^, find in terms of the coefiicients the values of : 11. 2a^ 16. (i8 + y)(y + a)(a + /J). 12. %o^^, 17. fe+:)^+^. a /? y ^^' ^"''' 18. y8^ + / , Z + g^ , a^ + /8^ 14. :§a^^^ ^y y"- "^ 15. 2g% ■'•^* o , ^ ^ To" In the equation a;^ +^^7^ -f g-o; + r = 0, find the condi- tion that : 20. One root is the negative of one of the other two roots. 21. One root is double another. 22. The three roots are in arithmetical progression. 23. The three roots are in harmonical progression. GRAPHICAL REPRESENTATION OF FUNCTIONS. 401 GRAPHICAL REPRESENTATION OF FUNCTIONS. The investigation of the changes in the value oif{x) cor- responding to changes in the value of x is much facilitated by using the system of graphical representation explained in the following sections. E AA F' 447. Oo-ordinates. Let JT'JT and Y' Y be two perpen- dicular straight lines drawn in a plane, intersecting at 0. The lines X'X and Y'Y are called axes of reference ; ^ the point O is called the origin. Distances measured from , along JT'X, as OA, OC, ^~ OE, and OG, are called abscissas j distances meas- ured from X^X parallel to TY, as AB, CD, EF, and OS, are called ordinates. Abscissas are considered positive if measured to the right; negative, if measured to the left. Ordinates are considered positive if measured upwards ; negative, if meas- ured downwards. Thus, OA, 00, CD, and EF Q,re positive ; OE. 00, AB, and GE are negative. An abscissa is generally represented by x, an ordinate is generally represented by y. The abscissa and ordinate of any point are called the co-ordinates of that point. Thus the co-ordinates of £ are OA and A£, 402 ALGEBRA. The co-ordinates of a point are written thus : (x, y). Thus, (7, 4) is the point of which the abscissa is 7 and the ordi- nate 4. The axis JT'JT is called the axis of abscissas, or the axis of X ; the axis F' Y, the axis of ordinates, or the axis of y. 448. It is evident that if a point B is given, its co-ordi- nates referred to given axes may be found by drawing the ordinate and measuring the distances OA and AB. Conversely, if the co-ordinates of a point are given, the point may be readily constructed. Thus, to construct the point (7, —4), a convenient length is taken as a unit of length. A distance of 7 units is laid off on OX to the right from Oto A. At J. a perpendicular to X^X is drawn downwards, of length 4 units, to B. Then B is the required point. Ex. Construct the points (3, 2); (5, 4); (6, -3); (-4,-3); (-4,2); (-3,-5); (4,-3). 449. Graph of a Punction. Let/(:^) be any function of x, where ^ is a variable. Put y =f{oc) ; then y is a new variable connected with x by the relation y—f{pc). If f{x) is a rational integral function of x, it is evident that to every value of x corresponds one, and only one, value ofy. If different values of x be laid off as abscissas, and the corresponding values of f{x) as ordinates, the points thus obtained will all lie on a line ; this line will generally be a curved line, or, as it is briefly called, a curve. This curve is called the graph of the function /(a;) ; it is also called the locus of the equation y =/(^). We proceed to construct the graphs of several functions. Eemaek. In constructing, or plotting, as it is called, the graph of a function, the student will find it convenient to use the paper called plotting, or co-ordinate, paper. This is ruled in small squares, and therefore saves much labor. GRAPHICAL REPRESENTATION OF FUNCTIONS. 403 (1) Construct the graph of 3 — 2a;. Put y =» 3 — 2x. The following table is readily computed : If a; = ^l, y= 5. 2. 2/= 7. li x = l, y=- 1. '« x = 2, y = -l. " x = S, y = -3. " a; = 4, y = -5. " a; = 5, 3/ = - 7. a; = — a; = -3, y= 9. a; = — 4, y = 11. a; = _ 5, y = 13. Constructing the above points, it appears that the graph of the function 3 — 2 x is the straight line MN. X' M \ "4—1- \-- :\ ;: \ \ \ N In general, where the equation y=f{x) contains only the first powers of x and y, the locus will be a straight line. 404 ALGEBRA. (2) Plot the graph of J^ - 4. Putting y = ^a^~4:, we readily compute the following table Y X' / 't I-- 1 / J. Y' 4- I / X a;= 0, 2/ = -4. x = ± 1, 2/ = - 3.5 x = ±2, 3/ = -2. a; =±3, 2/ = + 0.5 x = ±A, 2/= + 4. a; = ± 5, 7/ = + 8.5 a; = ±6, 2/ = + 14. Plotting these points, we obtain 3 curve here given. (3) Plot the graph of x^ ~ x^ -{■ X — b. Putting 2/ = a^ — a;'^ + a? — 5, we compute the following table : .1 X 18 y IS 0.5, - 4.625. 1.0, - 4.000. 1.5, - 2.375. 2.0, + 1.000. 2.5, + 6.875. 0.0, -5.000. 0.5, -5.875. 1.5, - 12.125. Interpolation (§ 381) shows that if 2/ = 0, x = 1.88+ . Does the re- sult agree with the figure ? 450. Consider any rational integral function of x, for example oc^ -{-x — ^^. Put y = x'^ + x~ -^^. GRAPHICAL REPRESENTATION OF FUNCTIONS. 405 Assuming values of x^ we compute the corresponding values of y, and construct the graph. Now, any value of X which makes y — satisfies the equation x^ -\-x — ^ = 0, and is a root of that equation ; hence, any abscissa whose corresponding ordinate is zero represents a root of this equa- tion. The roots may be found, y approximately, by measuring the abscissas of the points where the graph meets XX\ for at these points y = 0. xr From the given equation the fol- lowing table may be formed : a; is 2/ is If a; is y is 0, -15.75. - 1, - 15.75. 1, -13.75. -2, -13.75. 2, - 9.75. -3. - 9.75. 3, - 3.75. -4, - 3.75. 4, + 4.25. -5, + 4.25. The table shows that one root is between 3 and 4 (since y changes from — to +, and therefore passes through zero); and, for a like rea- son, the other is between — 4 and -5. ^X Y' 451. An equation of any de- gree may be thus plotted, and T V^ the graph will be found to cross the axis X'X as many times as there are real roots in the equa- x^-\- tion. When an equation has no real roots, the graph does not meet XX. In the equation x^ — 6 x + 13 = 0, both of whose roots are imagi- nary, the graph, at its nearest approach, is 4 units distant from X^X. \ I Y' 406 ALGEBRA. X'- If an equation has a dou- ble root, its graph touches JT'X, but does not intersect it. The equation a;2 + 4a; + 4 = has the roots — 2 and — 2, and the graph is as shown in the figure. Exercise 71. Construct the graphs of the following functions : 1. x^-\-Zx-Vd. 4. x^-^x-^\^. 2. x^ — 1x^-\-\. 5. :?;*-5^' + 4. 3. :r* — 20:^2 + 64. 6. x"" -\x^ -^ x-\. DEBIVATIVES. 452. Definition. Ijet :r be a variable, and J{pc) any func- tion of a;. Suppose X to have a particular value a ; the correspond- ing value of /(;r) is /(a) (§ 431). Now suppose X to increase to a-\-h\ the corresponding value oif{x) is /(a + h). The increase in the value oi f{x), called the increment of f{x), isf(a + h) —f(a) ; the increase in the value of x is h. Dividing the increment of f(x) by the increment of x, we obtain f(a + h)-f(a) h In the same manner for any particular value of x ; that is, for any value of x. DERIVATIVES. 407 Hence, in general, we shall obtain the expression h where x may have any value ; that is, is variable. The limit which this expression approaches, as h ap- proaches zero as a limit, is called the derivative of the function /(:i') with respect to x. The derivative of a func- tion of X is, in general, a new function of x. The above may be written : Derivative with respect to x oif{x) _Umit pincrement of /(a;)! A = [_ increment of x y{x- \-h)~f{x) - h Note. A = is read " as h approaches zero." limit j J The particular value of the derivative corresponding to x = a h limit A: it r/(a + A)-/(a)- | An increment may be either positive or negative. In general, the derivative with respect to u of -y, where V is a function of u, is the limit, as the increment of u ap- proaches 0, of increment of v increment of u The derivative with respect to x of f{x) is represented by DJ(x) ; that of /(y) with respect to y by I)yf{y) ; that of V with respect to u by D^v ; and so on. The derivative of f(x) with respect to x is also repre- sented hj /(x). Thus Dj(x)=f'(x)', i),/(y)=/'(y); and so on. 408 ALGEBRA. 453. Eule for finding a Derivative. In the given function change x to x + h. From the new value of the function subtract the old, and divide the remainder by h. Take the limit of the quotient as h. approaches zero as a limit. The derivative of a constant is 0, since the increment of a constant will always be 0. (1) Find Z),(aa;). The function is ax. Change x to x + h, a{x + h). From the new value subtract the old, ah. Divide by h, a. Take the limit as h approaches as a limit f .-. Dx{ax) = a. If a = 1, DxX = l. (2) Find D,{x^-\-^x+l). The function is cc^ + 4 a; + 1. Change a; to a; + A, {x + Kf + 4(a; + h) + l, or a;3 + 3 hx^ -V ^K^x + ¥ + 4:X + ^h + 1. From the new value subtract the old, 3 Aa;2 + 3 K'x + h^ + 4:k Divide by A, Sx"^ + 3hx + h"^ + 4. Take the limit as h approaches as a limit ; .-. i)x(a^ + 4a; + l) = 3a;2 + 4. 454, Derivative of x". The function is x"". Changing x to X-]- h, we obtain (x -\- hy. Now, whatever the value of n, (x + hy can be expanded by the binomial theorem, and we obtain (x + hy = x^ + nx^'-'h + ^'('^~'^) x^-'h:' + I-?. From this new value of the function subtract a;", the old value, and divide by h. DERIVATIVES. 409 We now have = nx^~^ ; the sum of the terms after the first approaches as a limit by § 375. Hence, to find the derivative with respect to x of any power of X, multiply hy the exponent^ and diminish the exponent of x hy one. Thus, X>x (a;*) = 4 ar» ; D^ {x'^) = - 3 a'* ; Exercise 72. Find the derivatives with respect to a; of : 1. x\ 5. x\ 9. x^-\-2x'. 2. r-. '■-> 10. (x + a)\ 3.1. X 7. x-\ "■ ^s 4. x-\ 8. x' + x. 12. (x+iy\ 455. Derivative of a Sum. Let f(x) and <^(a;) be two functions of x ; their sum f(x) + x(/+ <^ + ) = A/+ i>x<^ + Here /is an abbreviation iov f{x), for <^(a:), etc. By means of the above and §§ 453, 454, we can find the derivative with respect to x of any rational integral func- tion of X. Ex. Find D^(2x' + 4:x' -Sx + S). D42a^ + 4a;2-8a; + 3) = D^{2x^) + D^i^x") - D^{^x) + Z)a,(3) = 2D^x^ + 4i)^a;2 -^D^x + D^3 = 2(3a;2) + 4(2a;)-8(l) + = 6x^ + Sx-S. 456. Derivative of a Product. Let f(x) and <^(^) be two functions of x ; their product f{x) (x)] = ^'"^'^ r /(a; + h)(x-{-h) -f(x) cf> (x) l limit 'f(x -\-h)(x + h) -f(x i-h) (x) +f(xi-h)ct>(x)-f(x) (x) ^ limit fy^^ _^ ^N <^(a7+/0-<^(:r) "| since ^^^^^ [f(x + A)] =/(a;). DERIVATIVES. 411 The above may be formulated Similarly for three or more functions. Thus, 457. Derivative of (x — a)*". A {x - af = 1^^^^ nx-a + hr-{x-ay- \ A=OL h J ^ limit [ {x -af-\-n{x- af-'h + -{x- af \ A = o|_ h J limit r , N., 1 . 1 = A=o[^(^-«) + ] = n{x-ay-\ Ex. Z),(a;-3)* = 4(x-3)«. Exercise 73. Write the derivatives with respect to a; of : 1. a;' + 4. 4. x^-Zx^-]-x^. 2. a;^ + 3a;^-l. 5. 4a:* + 6a;' + 2. 3. x^-\-x' + 2. 6. 6a;^-7a;'4-7a:. 7. 3a;5 + 4a;* + a;'-a:'-6a; + 5. 8. 4ar'-2a;*-a;' + 6a;' — 7. 9. {x-2){x^Z). 12. (a;-47(a;-2)(a: + l). 10. (2:-l)(ar-2)(a:-3). 13. {x-a)\x-p)\ 11. (a;-3)X:^ + 4). 14. {x-a){x-P){x-y). 15. (a;-2)(a;-3)(a; + 5)(a; + 4). 16. (a;^ + 2)(a;'-4a; + 8). 412 ALGEBRA. 458. Successive Derivatives. The derivative of a function of X is itself a function of x, and has itself a derivative with respect to x. The derivative of the derivative is called the second de- rivative ; the derivative of the second derivative, the third derivative ; and so on. By derivative is meant the first derivative, unless the contrary is expressly stated. The second derivative with respect to x of f(x) is repre- sented by D^f{x), or by /" {x) ; the third derivative by D^f{x), or by /'"(a;) ; and so on. Evidently, f\x) = D:^f{x) = D,D,f{x) ; f\x) = D:f{x) = D.DJf(x) = A AAA ; and so on. 459. Values of the Derivatives. The value which f(x) assumes when for x we put a is represented by /(a). Similarly, the value which f'(x) assumes when for x we put a is represented by f'(a) ; the value of f"(x) by /"(a) ; and so on. Thus, if f{x) =a^-2a;2 + a; + 4, we obtain f^{x) = 3 x^ _ 4 a; + 1, f^{x) =6x-4, /i^(«), /""(x), etc., all vanish. Putting 2 for x we obtain /(2) = 6; /(2) = 5; //(2) = 8; /^/(2) = 6. Similarly for any function. 460. Sign of the Derivative. In the function f{x) let x increase by the successive addition of very small incre- ments. As X increases, the value of /(x) will change, sometimes increasing, sometimes decreasing. DERIVATIVES. 413 Suppose X to have reached a fixed value a ; the corre- sponding values oif{x) and/'(ar) will be /(a) and /'(a). Let X increase by an increment h from aio a-{- h. By § 452, /'(a) = limit [ /(« + A) -/m If f{x) is increasing as a; passes through the value a, f{a + A) >f{a) and /'(a) is positive. li f{x) is decreasing as a; passes through the value a, /(a + h) - 8. 414 ALGEBRA. Exercise 74. Write the successive derivatives with respect to a: of : 1. x^-4:X^ + 2. 3. 2x^-{-2x^-4:X+l. 2. x'-^4:x^~bx. 4. Sx' + ^x^-x^ + x. 5. ax^ + Shx^ + Scx + d. 6. ax' + 4:bx^ + 6cx' + 4:dx-\-e. 7. (x-a)\x-l3). 8. (x — a)(x — PXx — y). 9. (a; - a)^:^ - ;8)^ Find whether the following functions are increasing or decreasing as x increases through the value set opposite each of them : 10. x^-x^-^1 2. 12. 2x'-[-3x''-6x 1. 11. x'-x' + Qx—l 4. 13. 4:x' — Sx''-{-4:X-6 -3. 461. Derivative in Terms of the Eoots. Take the cubic f(x) = a(x — a)(x - P)(x — y), since I),(x-a) = l, D,(x-P) = l, I),(x-y) = l, (§ 457) we have, by § 456, fXx) = a(x-l3)(x-y) + a(x-aXx — y)-{'a(x—a)(x-l3) x—a x—p X—y Similarly, for any quantic, f.(,. ^ zw_+ /(^ + /M^^ ym. '' ^ ■' a; — tti x — Oi x — On ^x — a DERIVATIVES. 415 462. Multiple Boots. In the quantic f{x) let a be a triple root. Then we can write (§ 439), where the degree of <^(^) is less by 3 than that oif(x). By § 456, f\x) = (x- af ' (x) + S(x- af x = (x- ay [{x — a)'x + 3 <^a:]. Hence, if f(x) has a triple root a, the factor (x ~ a)' occurs in the H.C. F. off(x) and /'(a:). Similarly for a multiple root of any order. To find the multiple roots oif(x). Find the H.O.F. oi f{x) and f\x), and resolve it into factors. Each root will occur once more in f{x) than the corresponding factor occurs in the H. 0. F. Ex. Find the multiple roots of Here f{x) = o,-^ - x* — ^or* + ar^ + 8 x + 4. /(x) = 5x* - 4aj3 - 15x2 + 2a; + 8. Find the H.C. F. of/(x) and/(a;) as follows : 5_4_15+ 2 + 8 5 + 0-15-10 -4+ + 12 + 8 -4+ + 12 + 8 5-5- 5-4- -25 + -15 + 5 + 2 + 40 + 8 20 20 100 8 -1- -5- -5-f -10+ 3+ 32 + -50 + 15 + 160 + 4 + 15 _ 2 - 54)- -54 + + 162 + 108 ■ 1 + + 3 + 2 1-1 -5 + 4 Hence, x» - 3aj - 2 is the H. C. F. We find, by trial, that — 1 is a root of the equation aJ-3x-2 = 0. The other roots are found to be — 1 and 2 (§ 441). Hence, a;»-3a;-2 = (a; + Vf{x - 2). 416 ALGEBRA. Therefore, — 1 is a triple root, and 2 is a double root, of the given equation. As the given equation is of the fifth degree, these are all the roots, and the equation may be written (a; + l)3(a:-2)2 = 0. Having found the multiple roots of an equation, we may- divide by the corresponding factors, and find the remaining roots, if any, from the reduced equation. Exercise 75. The following equations have multiple roots. Find all the roots of each equation : 1. x^ — Sx' + lSx- 6 = 0. 2. a;' — 7a;' + 16a:- 12 = 0. 3. x* — 6x^- 8x- 3 = 0. 4. x' -■ 1 a^ + dx'-^ 27 X — 54:^0. 5. x' + Qa^+ a;'^- 2407 +16 = 0. 6. a;^ - 11a;* + 19^ +115a;^- 200.1; -500 = 0. 7. Resolve into linear factors x^-5a^ + 6x'-\-9x'-Ux'-4:X-{-8. 8. Show that an equation of the form x'^ = a" can have no multiple root. 9. Show that the condition that the equation x'' + Bqxi-r = shall have a double root is 4 g'" + r* = 0. 10. Show that the condition that the equation a^ + Spx^ + r = shall have a double root is r (4^^ + r) = 0. DERIVATIVES. 417 463. Expansion of f(x-f h). Consider a quantic of the fourth degree, viz. : f(x) = ax*' -\-h3^-\- cx^ -\-dx-\-e. Put x-\-hm place of x, then fix -\- h) = a{x -\- hy ^h{x -^ hj + c{x -{- Kf + d{x + A) + e. Expanding the powers of a; -f A, and arranging the terms by descending powers of x, the above identity becomes f{x -\-h)~a x' + Aah a^ + ^ah? x' + ^ah' + h + 35A + Sbh' + c + 2ch + d x-\-ah*' -i-bh' + ch' + dh + e But f(h) = ah'+ hK'-\- ch?-\-dh-\-e, f\h) = 4:ah' + Sbh'-}-2ch +d f\K) =12aA^ + 65A +2c f"Xh)=24:ah +Qb /X^)=24a, rw =0, and we have f(x+ h)^f{h)+xf\h) + ,?£^ + ^-^ + r."^. If we had arranged the expansion of f{x + A) by powers of A, we should have found f{x + h) =f{x) + hfix) + h-CM + A'-^ + h'^. Similarly for any quantic 418 ALGEBRA. 464. Calculation of the Coefficients. The coefficients in the expansion of fix + K) may be conveniently calculated as follows : Take f{x) = ax^ -f- hx^ -\- cx^ + dx + e. Put fix + A) = Ax" + Bx"" + Or' -\-Dx-^E, where A, B, C, D, E are to be found. In the last identity put ^ — A for x. Then, since f{x — h-\-h) =f(x'), we obtain f(x) = Aix-hy + B(x-- hf + C(x - Kf + B(x-h) + E. From the last identity we derive the following rule for finding the coefficients of the powers of x in the expansion off{x+h). Divide fi^x) by x— h; the remainder will be E, that is /(A) ; and the quotient Aiix-ky + B(x- Kf + Cix -h)-\-D. Divide this quotient by {x— h); the remainder will be Z), that is /'(A) ; and the quotient Aix~Kf-\-B{x-h)+a Continuing the division the last quotient will be A or a. The above division is best arranged as follows (§ 436) : \_h a h c d € ah b'h c'h d'h a V c' d' E ah b"h c"h a ah c" b"'h n a iw ah B TRANSFORMATION OF EQUATIONS. 419 and we Lave fix + A) = ax'' + Bx'^ + Cx"^ -\-Dx-\-E. The above method is easily extended to equations of any degree. Exercise 76. In the following quantics put for x the expression oppo- site, and reduce. 1. x'-2>x'-\-^x -^ x-\-2. 2. x'-2x^-^^x -3 a: + 4. 3. ^x'-23(^^2x''- x — ^ a; + 3. 4. 2x'-^x^-\-^x^-1x-^ x-2. 5. 2a:*-2:r' + 4a;'^-5a;-4 x-^. TRANSFORMATION OF EQUATIONS. 465. The solution of an equation, and the investigation of its properties, is often facilitated by a change in the form of the equation. Such a change of form is called a trans- formation of the equation. 466. Roots with Signs changed. The roots of the equation f(_x) = are those of the equation f(x) = 0, each with its sign changed. For, let a be any root of equation f{x) = 0. Then, we must have /(a) = 0. In the quantic /(— x) put — a for a; ; that is, a for — x. The result is /(a). But we have just seen that /(a) vanishes, since a is a root of the equation fix) = 0. Hence, /(- x) vanishes when we put - a for x, and (§ 432) -a is therefore a root of the equation /(- x) = 0. 420 ALGEBRA. Similarly, the negative of each of the roots of f{x) = is a root of /(— x) = 0; and, since the two equations are evidently of the same degree, these are all the roots of the equation /(— x) = 0. To obtain the /(— x) we change the sign of all the odd powers of x in the quantic/(^). Thus, the roots of the equation aj* _ 2 a:3 - 13 a;2 + 14 a; + 24 = are 2, 4, — 1, — 3 ; and those of the equation a;* + 2x3 - 13 a;2 - 14a; + 24 = are - 2, - 4, + 1, + 3. 467. Boots multiplied by a G-iven Number. Consider the equation ax' + bx' + cx'' + dx-\-e = 0. (1) Put y = mx, then ^ = — ; and the equation becomes UMiM^M^y-'- (2) The left-member of (2) differs from the left-member of (1) y only in that — is put in place of x. Let a be any root of (1) ; the left-member of (1) vanishes when we put a for x, and we obtain aa' + ba^ + ca^ + da-i-e = 0. In the left-member of (2) put ma for y ; we obtain aa* -f ha^ -{- ca^ -\- da -\- e, which, as we have just seen, vanishes. Hence, if a is a root of (1), ma is a root of (2). Since the above is true for each of the roots of (1), and the two equations are evidently of the same degree, the roots thus obtained are all the roots of (2). TRANSFORMATION OF EQUATIONS. 421 Similarly, for an equation of any degree. Equation (2) may be written in the form ay* + mbi^ -\- m^cy^ + Tn^dy -\- m^e = 0. The above form, if written with x in place of y, gives the following rule : Multiply the second term hy m ; the third term by m^ ; and so on. Zero coefficients are to be supplied for missing powers of x. Ex. "Write the equation of which the roots are the doubles of the roots of the equation 3a;* - 2x3 + 4a;2 - 6a? - 5 = 0. Here ?n = 2, and the result is 3a;* - 2(2)a;3 + 4(2)2a;2 - 6(2)3a; - 5(2*) = 0. or 3a;*-4aj3 + 16a;2-48a:-80 = 0. 468. Eemoval of Practional Coefficients. If any of the co- efficients of an equation in the form x^ +p,x^-' +i?2^""' + Pn = are fractions, we can remove fractions as follows : Multiply the roots by m ; then take m so that all of the coefficients will be integers. Ex. Reduce to an equation, in the p form, with integral coefficients 2x3_ia;2 + 5a; + i = 0. Dividing by 2, a^ -^^x^ + ^\x + 1 = 0. Multiplying the roots by m (^ 467), The least value of m that will render the coefficients all integral is seen to be 6. Putting 6 for m, we obtain ar'-a;2 + 15a; + 27 = 0, the equation required. Any multiple of 6 might have been used instead of 6, but the smaller the number the easier the work. 422 ALGEBRA. 469. Eeciprocal Eoots. Consider the equation ax^ + hx^ ^cx^-\-dx-\-e^O. (1) Put 2/ = - ; then :r = - ; and the equation becomes The left-member of (2) differs from the left-member of (1) only in that - is put in place of x. Let a be any root of (1) ; then we must have aa* + i>o^ -\- Co? -{- da-{- e = 0. In the left-member of (2) put a for - ; that is, - for y ; y a we obtain aa* + ^a^ + Col^ -\-da-\- e, which, as we have just seen, vanishes. Hence, - is a root of (2). Since the above is true for a each of the roots of (1), and the two equations are evidently of the same degree, the reciprocals of the roots of (1) are all the roots of (2). Similarly for an equation of any degree. Equation (2) may be written « + ^y + cy' + dy^ + ey* = 0, or, writing x in place of y, ex^ -\- dx^ -{- cx^ -\- hx -\- a = Qi ) so that the coefficients are those of the given equation in reversed order. Ex. Write the equation of which the roots are the reciprocals of the roots of 2rc* - Sx^ + 4a;2 _ 5a; - 7 = 0. The result is 2 -Zo^ + Ax^ -bx^ -1 x^ = 0, or 7a;* + 5a;3-4a;2 + 3x-2 = 0. TRANSFORMATION OF EQUATIONS. 423 470. Eeciprocal Equations. The coefficients of an equation may be such that reversing their order does not change the equation. In this case the reciprocal of a root is another root of the equation. That is, one-half the roots are recip- rocals of the other half. An equation in which the above is true is called a recip- rocal equation. Thus, the roots of the equation 6x5 - 29a;* + 27x3 + 27.r2 _ 29a; + 6 = are — 1, 2, 3, |, |. Here, — 1 is the reciprocal of itself; ^ of 2 ; | of 3. 471. Eoots diminished by a Given Number. Consider the equation ax' + ha^ -{-cx^ + dx + e^Q. (1) To obtain the equation which has for its roots the roots of the above equation each diminished by h, we proceed as follows : Put y = x — h] then x^=y -\~h\ and the equation be- comes a{y + hy + h{y + Kf + c{y-{-hy^-d(:y-\-h) + e^O. (2) The left-member of (2) differs from the left-member of (1) only in that y -j- A is put in place of x. Let a be any root of (1) ; then we must have aa* + 5a' + ca' -f c?a -f e = 0. In the left-member of (2) put a for y + A ; that is, a — h for y ; we obtain aa* -{- ha? -\- Co? -\- da -\- e ] which, as we have just seen, vanishes. Hence, a — A is a root of (2). Since the above is true for each of the roots of (1), and the two equations are evidently of the same degree, the roots thus found are all the roots of equation (2). Similarly for an equation of any degree. 424 ALGEBRA. Putting X in place of y in equation (2), that equation may be written f(x + A) = 0, equation (1) being /(a;) = 0. Equation (2) may be also written (§ 463) in the form ax' + Bx' + Cx' + I)x + E= 0, where E=f{hl D^f'(h), C=^-^, B^-^-^. \± \± The coefficients are most easily calculated by the method explained in § 464. To increase the roots by a given number h, we diminish the roots by — h. Ex. Obtain the equation which has for its roots the roots of the equation 2a;* - 3a;3 - 4x2 + 2a; + 9 = 0, each diminished by 2. The work (§ 464) is as follows : 2 - 3 + 4 ~ 4 + 2 + 2 +9 - 4 -4 2 + 1 + 4 - 2 + 10 - 2 +5 + 16 2 + 5 + 4 + 8 + 18 + 14 2 + 9 + 4 a is 13 x' + 26 ' + 26a;2 2 +13 The required equatioi 2a;4 + + 14aj + 5 = 0. 472. Transformation in General. In the general problem of transformation we have given an equation in x, as f(x) = 0, and we have to form a new equation in y where y is a given function of ^, such as <^(^). When from the equation y = {x) we can find an expres- sion for X, the transformation can be readily accomplished TRANSFORMATION OF EQUATIONS. 425 by substituting this expression for x in the given equation, and reducing the result. (1) Given the equation x^ — 2,x-\-\ = 0, to find the equation in y where y = ?>x~2. We find X = ^ ^ . Substituting this expression for x in the given equation, that becomes which reduces to y3^ 62/2 -152/ -19 = 0. (2) Given the equation of which a, ^, y are the roots. Find the equation of which the roots are ^ + y-a, y + a-^, a + ^ - y. We have y^^^r-^-a. = o + i3 + 7-2a = 2-2«. H42 2-2/ But, since o is a root of the given equation, a3-2a2 + 3a-5 = 0. Putting =^^ for o, and reducing, we obtain 3/J_2y« + 8y + 24-0, the equation required. 426 ALGEBRA. Exercise 77. Multiply the roots of each of the following equations by the number placed opposite the equation. 1. x^-Sx'' + 2x-4. = -1. 2. x' + Sx^-2x~l = —2. 3. 2x'-Sx^-{- x'-e>x-4: = -3. 4. 2x'-Sx' + Qx-8 = -2. 5. ^x'-4:x^-2x+1 = -2. Transform to equations with integral coefficients in the p form the equations 6. 12a:^- 4:X^+6x + l =0. S. 10^*+ bx'~4:x^-{-25x-S0 = 0. 9. 6x'+ Sx' + 4:x'-2a^ -{-6x-18 = 0. Write the equations which have for their roots the recip- rocals of the roots of the following equations : 10. Sx'-2a^-}-bx'-6x+7 = 0. 11. 2x^-4:x^-5x''-7x-8 = 0. 12. x^- x^ + 2x''-{-4:X-l = 0. Diminish the roots of each of the following equations by the number opposite the equation : 13. a;'-lla;' + 31a;-12 = 1. 14. x'-6c(^ + 4:X^ + lSx-b = 2. 15. x' + 10x' + 13x-24: = -2. SITUATION OF THE ROOTS. 427 16. a;* + a;' -16a;'' -4a; + 48-0 4. 17. a;* + a;2-3a; + 4 = 0.3. 18. a;*-3a;'~a;'' + 4a;-5=0 -0.4. 19. Form the equation which has for its roots the squares of the roots of the equation a;^-2a:^ + 3a;-5-=0. 20. Form the equation which has for its roots the squares of the differences of the roots of ^3_4^2_^2a;-3 = 0. 21. Given the equation a;'-2a;' + 4a;-4--=.0; find the equation in y where y = ^x^ — Z. SITUATION OF THE ROOTS. 473. Pinite Value of a Quantic. Any positive integral power of X is finite as long as x is finite. The product of a positive integral power of a; by a finite number will be finite when x is finite. A quantic consists of the sum of a definite number of such products, and will, consequently, have a finite value as long as x is finite. The derivatives of a quantic are new quantics, and will, consequently, have finite values as long as a; is finite. 474. Sign of a Quantic. When x is taken numerically large enough, the sign of a quantic is the same as the sign of its first term. Write the quantic affif 4- aix^-'^ + aiX"^"^ + a„ 428 ALGEBRA. in the form aoX-(l+^+ -^^-{- -^„\ By taking x large enough, each of the terms in paren- theses after the first can be made as small as we please. If a^ is numerically the greatest of the coefficients «!, ttj, ^n. the sum of the terms in parenthesis after the first will be numerically less than that is (§ 231), less than ^\ £ . The value of this expression can be made less than 1, or indeed less than am/ assigned value, by taking x large enough. Hence, even in the most unfavorable case, that in which all the terms in parenthesis after the first are negative, the sum of these terms can still be made less than 1 ; the sum of all the terms in parenthesis will then be positive. The sign of the quantic will be the same as the sign of aoX^, its first term. 475. When x is taken num>erically small enough, the sign of a quantic is the same as the sign of its last term. Write the quantic in the form f a^yC ■ a^x . an-\X , i y The proof follows the method of the last section. 476. Continuity of a Kational Integral Function. A func- tion of x, f{x), is continuous when an infinitesimal (§ 361) change in x always produces an infinitesimal change in /(a;), whatever the value of x. SITUATION OF THE ROOTS. 429 We proceed to show that if f{x) is a rational integral function of x, it is a continuous function. Give to X any particular finite value a ; the correspond- ing value oif{x) is /(a). Increase a; to a-\- h\ the corresponding value of /{x) is f{a -j- li), and the increment in the value off(x) is or (§ 463), f(a+h)-f{al {na)+^na) + |->(a)). The derivatives /'(a), /"(a), /"(«) all have finite values (§ 473) ; and it is easily seen from § 475 that when h is very small the expression in parenthesis is numerically less than 2f\d). Since 2hf\a) approaches as a limit (§ 363, I.) when h approaches as a limit, the increment oi f(x), which is less than 2hf\a), will approach as a limit when h approaches as a limit. Since the above is true for any particular finite value of X, we see that an infinitesimal change in x always produces an infinitesimal change mf(x). It follows that as J(x) gradually changes from f(a) to f(b), it must pass through all intermediate values. The derivatives of a rational integral function of a: are them- selves rational integral functions of X, and are therefore continuous. The changes in the value of a quantic f{x) are well illustrated by the graph of the function. Since f{x) is continuous, we can never have a graph in which there are breaks in the curve, as in the curve here given. In this curve there are breaks, or discontinuities, at a — — 2,/ and a; =- -h 2. 430 ALGEBRA. 477. Tteorem on Change of Sign, Let two real numbers a and b be put for x in f (x). If the resulting values of f (x) have contrary signs, an odd number of roots of the equa- tion f (x) = lie between a and b. As X changes from a to b, passing through all interme- diate values, f(x) will change from f(a) to f{b), passing through all intermediate values. Now, in changing from f(a) to f(b), f(x) changes sign. Hence, /(^) must pass through the value "zero. That is, there is some value oi x between a and b which causes /(:r) to vanish ; that is, some root of the equation f(x) = lies between a and b. But f(x) may pass through zero more than once. To change sign, f(x) must pass through zero an odd number of times ; and an odd number of roots must lie between a and b. Applied to the graph of the equation, since to a root cor- responds a point in which the graph meets the axis of x (§ 450), the above simply means that to pass from a point below the axis of a: to a point above that axis, we must cross the axis an odd number of times. Thus, in a^ -2x'' + 3x -7 ^0. If we put 2 for x, the value of the left-membei' is — 1 ; if we put 3 for X, the value is +11. Hence, certainly one root, and possibly three roots, of the equation lie between 2 and 3. 478. An equation of odd degree has at least one real root For, if the first coefficient is not positive, change signs so as to make it positive. If the last term is negative, make X positive and very large ; the sign of the left-member is -f (§ 474). Put a; = ; the sign of the left-member is — . Hence, there is at least one real positive root. Similarly, if the last term is positive, there is at least one real negative root. SITUATION OP THE ROOTS. 431 479. Descartes' Eule of Signs. An equation in which all the powers of x from x^ to x^ are present is said to be com- plete ; if any powers of x are missing, the equation is said to be incomplete. An incomplete equation can be made complete by writing the missing powers of x with zero coefficients. A permanence of sign occurs when + follows +, or — follows — ; a variation of sign when — follows +, or -f fol- lows — . Thus, in the complete equation ic6-3a5 + 2a;* + a^-2a;2-a;-3, writing only the signs + - + + ---, we see that there are three variations of sign and three permanences. For positive roots, Descartes' rule is as follows : The number of positive roots of the equation f (x) = can- not exceed the number of variations of sign in the quantic f(x). To prove this it is only necessary to prove that for every positive root introduced into an equation there is one varia- tion of sign added. Suppose the signs of a quantic to be + - + + + --+, and introduce a new positive root. We multiply by a; — A, or, writing only the signs, by -| . The result is + - + + + -- + -f - + - + + + -- + - + --- + + - 432 ALGEBRA. The ambiguous signs =fc, =F indicate that there is doubt whether the term is positive or negative. Examining the product we see that to permanences in the multiplicand correspond ambiguities in the product. Hence, we cannot have a greater number of permanences in the product than in the multiplicand, and may have a less number. But there is one more term in the product than in the multipli- cand. Hence we have at least one more variation in the product than in the multiplicand. For each positive root introduced we have at least one more variation of sign. Hence the number of positive roots cannot exceed the number of variations of sign. Negative Boots. Change :?; to — :?;. The negative roots of the given equation will be positive roots of this latter equation (§ 466), and the preceding rule may then be applied. 480. From Descartes' rule we obtain the following : If the signs of the terms of an equation are all positive, the equation has no positive root. If the signs of the terms of a complete equation are alter- nately positive and negative, the equation has no negative root. If the roots of a complete equation are all real, the number of positive roots is the same as the number of variations of sign, and the number of negative roots is the same as the number of permanences of sign. 481. Existence of Imaginary Eoots. In an incomplete equation Descartes' rule sometimes enables us to detect the presence of imaginary roots. Thus, the equation a;^ + 5a; + 7 = may be written o(?±Ox^-\-bx + *1^0. We are at liberty to assume that the second term is positive, or that it is negative. SITUATION OF THE ROOTS. 433 Taking it positive, we have the signs + + + +; there is no variation, and the equation has no positive root. Taking it negative, we have the signs + - + +; there is but one permanence, and therefore not more than one negative root. As there are three roots, and as imaginary roots enter in pairs, the given equation has one real negative root and two imaginary- roots. Exercise 78. All the roots of the equations given below are real; determine their signs. 1. .T* + 4.x^ - 43^-2 - 5Sx + 240 = 0. 2. x^ - 22ic2 + 155 cc - 350 = 0. 3. X* + 4.x^ - 3ox^ - 78£t; + 360 = 0. 4. a;«-12£c2-43.T-30 = 0. 5. x^ -3x^-5x^ + 15ic2 + 4.T - 12 = 0. 6. x^- 12a;2 + 47.r-60 =0. 7. x^ - 2x^ - 13a;2 + 38cc - 24 = 0. 8. x^-x'^- 187 cc^ - 359ic2 + ^^q^ + 3(50 ^ q^ 9. ic«-10x'«+ 19ic*+ 1100^8- 5360^2+ 800a;- 384 = 0. 10. If an equation involves only even powers of x, and the signs are all positive, the equation has no real root. 11. If an equation involves only odd powers of x, and the signs are all positive, the equation has the root 0, and no other real root. 434 ALGEBRA. 12. Show that the equation has at least two imaginary roots. 13. Show that the equation x'+l5x'+1x~n = has two imaginary roots, and determine the signs of the real roots. 14. Show that the equation x^ -\- qx-\-r = has one negative and two imaginary roots when q and r are both positive ; and determine the character of the roots when q is negative and r positive. 15. Show that the equation :r" — 1 = has but two real roots, + 1 and — 1, when n is even ; and but one real root, + 1, when n is odd. 16. Show that the equation x"" -}-l = has no real root when n is even ; and but one real root, — 1, when n is odd. 482. Limits of the Eoots. In solving numerical equations it is often desirable to obtain numbers between which the roots lie. Such numbers are called limits of the roots. A superior limit of the positive roots of an equation is a number greater than any positive root. An inferior limit to the positive roots of an equation is a positive number less than any positive root. General methods for finding limits to the roots are given in most text-books ; but in practice close limits are more easily found as follows : (1) x'~6x' + 4:0x' - 8a; + 23 = 0. Writing this a^{x-5) + 8x{5x-l) + 23 =0. we see that the left-member is positive for all values of x as great as 5 ; consequently, it cannot become for any value as great as 5, and there is no root as great as 5. SlTUATtOl^ OF tflE ROOTS. 435 (2) x' + Sx^ + x'-Sx'-blx+lS^O. Writing this x^{a^ - 8) + 3x{x^ -17) + a^ + 1S = 0, we see that the left-member is positive for all values of x as great as 3 ; consequently there is no positive root as great as 3. Sometimes we can find close limits by distributing the highest positive powers of x among the negative terms. (3) x* + sf' — 2x^ — 4:X — 24: = 0. Multiplying by 2, 2 ic* + 2 a;^ - 4 a;^ - 8 a; - 48 = 0. Writing this x'^ {x'~i) + 2x {x^ - 4) + x* - 48 = 0, we see that there is no positive root as great as 3. An inferior limit to the positive roots is found by putting a; = - (§ 469), and finding a superior limit to the positive roots of the transformed equation. Limits to the negative roots of the equation f(x) = are found by finding limits to the positive roots of the equa- tion /(- a:) = (§ 466). Exercise 79. Find superior limits to the positive roots of the following equations : 1. a;'-2a;'-f4a: + 3 = 0. 2. 2x'-x'' — x + l = 0. 3. 3a:* + 5a;^-12a;^ + 10a:-18--=0. 4. 4:X* — Sx'-x''+7x+5 = 0. 5. x'-x' — 2x-'~4:X-24:=0. 6. 4:r'-Sx* + 223r' + 90x'--60x+l = 0. CHAPTER XXX. NUMERICAL EQUATIONS. 483. A real root of a numerical equation is either com- mensurable or incommensurable. Commensurable roots are either integers or fractions. Repeating decimals can be expressed as fractions (§ 231), and roots in that form are consequently commensurable. Incommensurable roots cannot be found exactly, but may be calculated to any desired degree of accuracy by the method of approximation explained in this chapter. COMMENSURABLE ROOTS. 484. Integral Eoots. The process of finding integral roots given in § 441 is long and tedious when there are many numbers to be tried. The number of divisors to be tried is diminished by the following theorem : Eve)y integral root of an equation with integral coefficients is a divisor of the last term. We shall prove this for an equation of the fourth degree, but the proof is perfectly general. Let h be an integral root of the equation ax^ + hx^ -\- ex'' -}- dx -{- e = 0, where the coefficients a, h, c, d, e are all integers. Since A is a root, ah' + bh' + ch' + dh + e = 0, (§ 432) or, e = — dh — cl^ — bh^ — ah'. NUMERICAL EQUATIONS. 437 Dividing by A, ^- = -d~ch-hh''-ah\ h Since tlie rigTit-member is an integer, the left-member must be an integer. That is, e is divisible by A. Similarly, for any equation with integral coefficients. Hence, in applying the method of § 441, we need try only divisors of the last term. The necessary labor may be still further reduced by the method of the following section. 485. Newton's Method of Divisors. In the equation above - is an integer. Put - = D, transpose d, and divide by h. h h Then, h Since the right-member is an integer, D -\- d must be divisible by h. Put — '^— ■= C, transpose — c, and divide by A. Then, h As before, C-\- c must be divisible by Ti. B, transp B±h_ Put 7"^ = B, transpose — h, and divide by h. Then, h a. h As before, B-\-h must be divisible by h. Transposing ■— a, we have B-\-b h provided A is a root. a = 0, 438 ALGEBRA. The preceding gives the following rule : Divide the last term hy]i\ if the quotient is an integer^ to it add the preceding coefficient, and again divide by h; if this quotient is an integer, to it add the preceding coefficient; and so on. If A is a root, the quotients will all be integral, and the last sum will be zero. A failure in either respect implies that h is not a root. From the above we also obtain Z>= _ (ah' + bh' + cA + d), C = - (ah' + bh + c), B = ~(ah-\- b), so that the successive quotients, with their signs changed, are (§ 436), in reversed order, the coefficients of the quo- tient obtained by dividing the left-member hy x — h. The above evidently applies to an equation of any de- gree. Ex. Find the integral roots of 3a;* - 23a;3 + ^2x' + 32a; - 96 = 0. By substitution neither + 1 nor — 1 is a root. The other divisors of — 96 are ± 2, ± 3, ± 4, ± 6, etc. Try +2: -96 +32 +42 -23 + 3 |_2 -48 - 8 +17 -3 16+34-6 Hence + 2 is a root. The coefficients of the depressed equation in reversed order are _.g _g ^,_ _o Try + 2 again : _48-8+17-3(_2 -24 -16 -32+1 Since 2 is not a divisor of + 1, + 2 is not again a root. NUMERICAL EQUATIONS. 439 Try - 2 : -48-8+17-3 | -2 + 24-8 + 16+9 and — 2 is not a root. Try + 3 : - 48 - 8 + 17 -^ 3 [_3 -16 - 8 +3 -24+9 Hence + 3 is a root. The depressed equation is 3a;2-8x-16 = 0, of which the roots are 4 and — f . Therefore the roots of the given equation are 2, 3, 4, — |. The advantage of this method over that of § 441 is that if the number tried is not a root, this fact is detected as soon as we come to a fractional quotient ; whereas, in § 441, we have to complete the division before we decide whether the number tried is a root or not. 486. Fractional Boots. A rational fraction cannot be a root of an equation with integral coefficients in the p form. If possible let -, where h and k are integers, and - is in rC rC its lowest terms, be a root. Then, Multiplying by ^"~^ and transposing, ^=- p.h'^-' -p^h^'-'k - -pjr^\ Now the right-member is an integer ; the left-member is a fraction in its lowest terms, since A" and k have no common divisor as h and k have no common divisor (§ 350, V.). But a fraction in its lowest terms cannot be equal to an integer. Hence -, or any other rational fraction, cannot be a root. k 440 ALGEBRA. The real roots of an equation with integral coefficients in the 'p form are, therefore, integral or incommensurable. In case an equation has fractional roots, we can find them as follows : Transform the equation into an equation with integral coefficients by multiplying the roots by some number m (§ 468). Find the integral roots of the transformed equa- tion, and divide each by m. Ex. Solve the equation 36a;''-55a:2-35a;-6 = 0. Write this Multiplying the roots by 6, we obtain a;*_55a;2-210a;-216 = 0, of which the roots are found to be — 2, — 3, — 4, 9. Hence, the roots of the given equation are ~~ f » ~~\i ~ f > f ; *^^' ~ ¥» ~ Y» ~" f » f • Exercise 80. Find the commensurable roots, and if possible all the roots, of each of the following equations : 1. a:* — 4.r' — 8^ + 32 = 0. 2. a;^-6a;'' + 10a;-8 = 0. 3. x'^2x'-nx^-^x-\-V2. = ^. 4. a;^ 3a;' -30a: + 36 = 0. . 5. a;^ - 12a;^ + 32a:' + 27a; -18 = 0. 6. a;^ - 9a:^ + 17a:' + 27a; -60 = 0. 7. a;^-5a:* + 3a;^ + 17a:'-28a:+12 = 0. 8. a;*-10a;H35a:'-50a; + 24--^0. NUMERICAL EQUATIONS. 441 9. o:^ - 8a:* + liar" + 29a;^- 36a; -45 = 0. 10. a^-x'--6x'-{-9x' + x-4: = 0. 11. 2a;* — Sar" — 20a7'' + 27a;+18 = 0. 12. 2a;* -9a:'-27a;'+ 134a; -120 = 0. 13. a;« + 3a;^ — 2a;*-15a:' — 15a;^ + 8a; + 20 = 0. 14. 18a;^ + 3a;^-7a;-2 = 0. 15. 24ar'-34a;'-5a;+3 = 0. 16. 273;^— 18a;'-3a; + 2 = 0. 17. 18a;* + 9ar'+10a;^-8a;+l = 0. 18. 36a;*H-48ar'-23a;'-17a; + 6 = 0. INCOMMENSURABLE ROOTS. 487. Location of the Boots. In order to calculate the value of an incommensurable root we must first find a rough approximation to the value of the root ; for example, two integers between which it lies. This can generally be accomplished by successive applications of the principle of § 477. In some equations the methods of § 479-482 may be useful. (1) Consider the equation We find {i 436), /(0)=+ 5 /(!) = + 3 /(2) = - 5 /(3) = -13 /(4) = -15 /(5) = - 5 /(6) = + 23 /(-I) = -6. All numbers above 6 give + ; all below — 1, give — . From the above (^ 477) the three roots are all real ; one between 1 and 2 ; one between 5 and 6 ; one between and — 1. 442 ALGEBRA. (2) The equation has, by Descartes's rule (§ 479), not more than two positive roots and not more than two negative roots. We find (^ 436), /(O) = + 2 ; /(5) = + 132 ; /(1) = - 4; /(-1) = - 12: /(2) = -30; /(-2) = - 22; /(3) = -52; /(-3) = + 20; /(4) = -22; /(-4) = + 186. Hence there are two positive roots, one between and 1, and one between 4 and 5 ; and two negative roots, one between and — 1, and one between — 2 and — 3. Let us find more closely a value for the root between and 1. We find /(0.5) = + 2.06+. Since /(I) = - 4, the root lies between 0.5 and 1. Try 0.8 : we find /(0.8) = - 0.9+. Hence the root lies between 0.5 and 0.8. We find /(0.7) = + 0.4+. Hence the root lies between 0.7 and 0.8. In' a similar manner we find the root between and — 1 to lie between - 0.2 and - 0.3. The first significant figures of the roots are accordingly 0.7, 4, -0.2, -2. Exercise 81. Determine the first significant figure of each real root of the following equations : 1. x^-x''-2x-}-l = 0. 5. x'-Qx'-Sx-i-5=^0. 2. x'-bx-S = 0. 6. :r^ + 9:r^ + 24a: +17 = 0. 3. x'-6x' + 7 = 0. 7. ^^-15^^ + 63:^-50 = 0. 4. x' + 2x''-S0x-JrS9 = 0. 8. x'-8x'-{-Ux''-\- 4:X-8 = 0. 488k Horner's Method, Positive Roots. Suppose the first figure of the root to have been found. Any number of remaining figures may be calculated by the method of approximation known as Horner's Method. NUMERICAL EQUATIONS. 443 We proceed to illustrate the process by an example. Take the equation x'-6x'-{-Sx + b = 0. (1) By § 487, Ex. 1, one root of this equation lies between 1 and 2. We proceed to calculate that root. Diminish the roots by 1 (§ 471) : 1 -6 +3 +5 [1 •I- 1 - 5 - 2 - 5 - 2 +3 ±i zA -4 -6 + 1 -3 The transformed equation is, therefore, y3_33^2_6y + 3-0. (2) The roots of equation (2) are each less by 1 than the roots of equation (1). Equation (1) has a root between 1 and 2 ; equation (2) has, therefore, a root between and 1. Since this root is less than 1, y' and y^ are both less than y. Neglecting these terms, we have -6y + 3==0, or ?/ -= 0.5. At this stage of the process the figure thus obtained will not in general be the correct one. If, however, we neglect only the 3/^ term, we obtain -3y-^-6y + 3=:0, ^^ + 2^-1 = 0, of which one root is V2 — 1 = 0.4 +. We can also find the second figure of the root as follows : Take the first value 0.5. With this assumed value of y, computing the value of y' — 3y', and substituting, we obtain 63/ = 2.375; whence y = 0.4, approxi- mately. 444 ALGEBRA. We now diminish the roots of (2) by 0.4 : 1 - 3 - 6 +3 I 0.4 + 0.4 -1.04 - 2.816 — 2.6 -7.04 +0.184 + 0.4 -0.88 — 2.2 -7.92 + 0.4 -1.8 The second transformed equation is 2^- 1.82^ - 7.922 + 0.184 = 0. (^3) The roots of (3) are less by 0.4 than those of (2), and less by 1.4 than those of (1). Equation (2) has a root between 0.4 and 0.5 ; equation (3) has, therefore, a root between and 0.1. Since this root is much less than 1, we shall probably obtain a correct value for the next figure of the root by neglecting the z^ and z^ terms in equation (3). This gives ~ 7.92 z + 0.184 =- ; whence 2 = 0.02+. Diminish the roots of (3) by 0.02 : -1.8 -7.92 + 0.184 i 0.02 + 0.02 - 0.0356 -0.159112 -1.78 - 7.9556 + 0.024888 + 0.02 - 0.0352 -1.76 — 7.9908 + 0.02 -1.74 The third transformed equation is u^ ~ 1.74: u" - 7.99082^ + 0.024888 = 0. (4) The roots of (4) are less by 0.02 than those of (3), and less by 1.42 than those of (1). Neglecting the u^ and u^ terms, we obtain w = 0.0031 +, NUMERICAL EQUATIONS. 445 SO that to four places of decimals tlie root of (1) is 1.4231. The process may evidently be continued until the root is calculated to any desired degree of accuracy. 489. We shall now make some observations on the pre- ceding work. First : If we diminish the roots by a number less than the required root, as we do not pass through the root, the sign of the last term remains unchanged throughout the work. The last coefficient but one will always have a sign opposite to that of the last term. If, in (3), the signs of the last two terms were alike, the value of z would be — 0.02+. This would show that the value assumed for z was too great, and we should diminish the value of z and make the last transformation again. In beginning an example, one is very likely to assume too large a value for the next figure of the root ; in solving (2), for instance, the first solution gave y = 0,5, and had that value been tried, it would have proved to be too great. Remaek. The first transformation may, however, change the sign of the last term. Thus, if there had been a root between and 1 in equation (1), diminishing the roots by 1 would have changed the sign of the last term. Second : In finding the second figure of the root we make use of the last three terms of the first transformed equation instead of the last two terms. Or, we may use the alternative method. One of these methods will gener- ally give the correct figure. In any case we can find the correct figure by another trial. Any figure after the second is generally found correctly from the last two terms ; for, in this case, the root is small and its square and cube so much smaller than the root itself that the terms in which they appear have but slight influence upon the result. 490. It is not necessary to write out the successive trans- formed equations. When the coefficients of any transformed 446 ALGEBRA. equation have been computed, the next figure of the root may be found by dividing the last coefficient by the pre- ceding coefficient, and changing the sign of the quotient. Thus, in equation (4), the next figure of the root is obtained by dividing 0.024888 by 7.9908. On this account the last coefficient but one of each trans- formed equation is called a trial divisor. Sometimes the last coefiicient but one in one of the transformed equations is zero. To find the next figure of the root in this case follow the method given for finding the second figure of the root. Th«i work may now be collected and arranged as follows : 1 -6 +3 + 1 -5 + 5 1 1.423+ -2 -5 + 1 -2 -4 + 3 -2.816 -4 + 1 -6 -1.04 + 0.184 - 0.159112 -3 + 0.4 -7.04 -0.88 + 0.024888 -2.6 + 0.4 -2.2 + 0.4 -7.92 -0.0356 - 7.9556 -0.0352 -1.8 + 0.02 -1.78 + 0.02 -1.76 + 0.02 — 7.990 8 -1.74 NUMERICAL EQUATIONS. 447 The broken lines mark the cpnclusion of each transformation. The numbers in heavy type are the coefficients of the successive transformed equations, the first coefficient of each equation being the same as the first coefficient of the given equation. In this example the first coefficient is 1. When we have obtained the root to three places of decimals we fian generally obtain two or three more figures of the root by simple division. 491. In practice it is convenient to avoid the use of the decimal points. We can do this as follows : multiply the roots of the first transformed equation by 10, the roots of the second transformed equation by 100, and so on. In the last example the first transformed equation will now be f - mf -~ 600y + 3000 = 0, and this equation will have a root between 4 and 5. The second transformed equation will now be z' - 1802^ - 79,2002 + 184,000 = 0, and this equation will have a root between 2 and 3. And so on. Comparing these equations with the equations in § 488, we see that we can avoid the use of the decimal point by adopting the following rule : When the coefficients of a transformed equation have been obtained, add one cipher to the second coefficient, two ciphers to the third coefficient, and so on. The coefficients and the next figure of the root are now integers. The work proceeds as in § 490. If the root of the given equation lay between and 1, we should begin by multiplying the roots of the given equation by 10. 448 ALGEBRA. The complete work of the last example, for six figures of the root, will now be as follows : -6 + 1 -5 + 1 -4 + 1 -30 + 4 -26 + 4 -22 + 4 -180 + 2 -178 + 2 -176 + 2 -1740 + 3 - 1737 + 3 - 1734 + 3 - 17310 + 1 - 17309 + 1 - 17308 + 1_ - 17307 + 3 -5 -2 -4 -600 -104 -704 - 79200 - 356 - 79556 - 352 - 7990800 5211 -7996011 5202 - 800121300 17309 -800138609 17308 - 800155917 + 5 1 1.42311+ + 3000 -2816 184000 159112 + 24888000 - 23988033 + 899967000 - 800138609 + 99828391 NUMERICAL EQUATIONS. 449 We can find five more figures of the root by simple division. If we divide 99,828,391 by 800,155.917, we obtain 0.124761, so that the required root to ten places of decimals is 1.4231124761. The reason is seen by examining the last transformed equation. Write this 8.00155917^ = 0.000099828391 - 1.7307 w^ + w\ As w is about 0.00001, w"^ is about 0.0000000001, and w^ is still smaller. Hence the error in neglecting the w^ and w^ terms is in 8 w about 0.00000000017, and in w about 0.00000000002. The result obtained by division will therefore be true to ten places of decimals. 492. Negative Eoots. To avoid the- inconvenience of working with negative numbers, when we wish to calculate a negative root, we change the signs of the roots (§ 466), and calculate the corresponding positive roots of the trans- formed equation. Thus one root of the equation a^-6a;2 + 3a; + 5=.0 lies between and — 1 (§ 487). By Horner's Method we find the corresponding root of ic3 + 6a;2 + 3a;-5 = to be 0.6696+. Hence, the required root of the given equation is - 0.6696+. Exercise 82. Compute for each of the following equations the root of which the first figure is the number in parenthesis opposite the equation. Carry out the work to three places of deci- mals : 1. x^ + 2>x- 5 = (1). 2. a:'-6a:-12-:0 (3). 3. a;» + a;' + a7-100 = (4). 4. ar» + 10a;^ + 62r- 120 = (2). 5. a;'+92;2 + 24a: +17 = (-4). 6. a;*-12r'+12a:-3 = (-1). 7. a;*-8a;^ + 14^ + 4a;-8 = (-0.). 450 ALGEBRA. 493. Oontraction of Horner's Method. In § 491 the student will see that if we seek only the first six figures of the root, the last six figures of the fourth coefficient of the last trans- formed equation may be rejected without affecting the result. Those figures of the second and third coefficients which enter into the fourth coefficient only in the rejected figures may also be rejected. Moreover, we may reject all the figures which stand in vertical lines over the figures already rejected. The work may now be conducted as follows : 1-6 - +3 +5 I 1.42311+ -5 -2 + 1 -5 + 1 174 2 800 80 3000 2816 -4 + 1 -600 -104 + 184000 - 159112 -30 + 4 -704 - 88 -f- 24888 - 23991 -26 + 4 - 79200 - 356 -h897 -800 -22 + 4 - 79556 - 352 + 97 - 80 -180 + 2 -178 - 79908 -7991 - 6 + 2 -176 - 7997 - 6 + 2 -8003 NUMERICAL EQUATIONS. 451 The double lines in the first column indicate that beyond this stage of the work the first column disappears alto- gether. In the present example we find three figures of the root before we begin to contract. We then contract the work as follows : Instead of adding ciphers to the coefficients of the trans- formed equation, we leave the last term as it is ; from the last coefficient but one we strike off the last figure ; from the last coefficient but two we strike off the last two figures ; and so on. In each case we take for the remainder the nearest integer. Thus, in the first column of the preceding example we strike off from 174 the last two figures, and take for the remainder 2 instead of 1. The contracted process soon reduces to simple division. Thus, in the last example, the last two figures of the root were found by simply dividing 897 by 800. To insure accuracy in the last figure, the last divisor must consist of at least two figures. Consider the trial divisor at any stage of the work. If we begin to contract, we strike off one figure from the trial divisor before finding the next figure of the root. Since the last divisor is to consist of two figures, the contracted process will give us two less figures than there are figures in the trial divisor. Thus, in § 491, if we begin to contract at the third trial divisor, — 79,908, we can obtain three more figures of the root ; if we begin to contract at the fourth trial divisor, — 8,001,213, we can obtain five more figures of the root ; and so on. The student should carefully compare the contracted process on page 450 with the un contracted on page 448. 494. When the root sought is a large number, we cannot find the successive figures of its integral portion by dividing the absolute term by the preceding coefficient, because 452 ALGEBRA. the neglect of the higher powers, which are in this case large numbers, leads to serious error. Let it be required to find one root of a;4 _ 3 a;2_^ 11 a, _ 4 842,624,131=0. (1) By trial, we find that a root lies between 200 and 300. Dimin- ishing the roots of (1) by 200, we have 2/* + 8002/3 + 239,9973/2 + 31,998,8112/ - 3,242,741,931 = 0. (2) If y = 60, f{y) = - 273,064,071. If 2/ = 70, f{y) = + 471,570,139. The signs oif{y) show that a root lies between 60 and 70. Dimin- ishing the roots of (2) by 60, we obtain 2* + 1040 z^ + 405,597 z" + 70,302,451 z - 273,064,071 = 0. (3) The root of this equation is found by trial to lie between 3 and 4. Diminishing the roots by 3, we may find the remaining figures of the root by the usual process. 495. Any root of a number can be extracted by Horner's Method Ex. Find the fourth root of 473. Here x^ = 473, or a;* + 0a;3 + 0a;2^0a;-473 = 0. Calculating the root, x = 4.66353+. If the number be a perfect power, the root will be obtained ex- actly. 496.* Boots nearly Equal, In the preceding examples the changes of sign in the value of f{x) enable us to deter- mine the situation of the roots. In rare cases two roots may be so nearly equal that they both lie between consecu- tive integers. In this case the existence of the roots will not be indicated by a change of sign in f(x), and we must resort to other means to detect their presence. NUMERICAL EQUATIONS. 463 Ex. Consider the equation 3^-515x2 + 1155 a; -649 = 0. (1) By Descartes' rule this equation has no negative root. It has therefore certainly one, and perhaps three, positive roots. We find /(~l) = -2320; /(O) =- 649; /(I) =- 8; /(2) =- 391; /(3) =-1792. The approach of f{x) towards indicates either that there are two roots near 1, or that the function approaches without reaching it ; the graph in the latter case being as here given. Let us proceed on the supposition that two roots near 1 do exist. Diminish the roots by 1. The transformed equation 3/3„512.y2 + 1282/ -8 = 0, (2) by Descartes' rule, still has either one or three positive roots, so that we have not passed the roots. If we had diminished the roots by 2, we should have obtained y3_509y2_g93y_ 391=0, which has but one positive root ; so that we have passed both roots. To find the second figure of the root, neglect the first term of equation (2). Since the roots are nearly equal, the expression 5122/2 -l28y + 8 must be nearly a perfect square. Comparing this with a{y — of, or 128 J 2x8 . , and ■ are approximate 2 X 512 128 ^^ ay"^ — 2 aay + aa?, we see that values for the roots ; these both give |, or 0.12. Diminish the roots by 0.1 ; the work is as before. Continue until the two quotients obtained as above give difi'erent numbers for the next figure of the root. In the present example this occurs when we come to the third decimal figure ; the transformed equation is V? - 51,164 u2 + 51,632w - 11,072 - 0. 454 ALGEBRA. and the two quotients are 0.5 + and 0.3 +. To separate the roots, try 0.4 ; the left-member of the last equation is found to be +. Since gives — and 1 gives — , there is one root between and 0.4, and one between 0.4 and 1. To calculate the first root, we try 0.3 ; as this gives a — sign we diminish the roots by 0.3, and proceed as in ^ 493. 515 1 514 1 513 1 5120 1 611631 5116 51 + 1155 - 514 + 641 - 513 + 12800 - 5119 + 7681 -5118 -649 1 1.1230907 + 641 -8000 + 7681 - 319000 + 307928 - 11072000 + 10885167 -5119 1 + 256300 - 102336 - 186833 + 184284 - 5118 1 + 153964 -102332 -1549 + 1400 - 51170 2 + 5163200 - 1534911 - 149 - 51168 2 + 3628389 - 1534902 - 51166 2 + 2093487 + 209349 - 511640 3 + 20935 - 459 + 20476 - 459 - 511637 3 - 511634 3 + 20017 2002 200 NUMERICAL EQUATIONS. 455 To calculate the second root, we return to the equation V? - 51,164^2 + 51,632 w - 11,072 = 0. We have /(0.4) = +, /(I) = - ; we find /(0.6) -■= +,/(0.7) = +0.383. Since /(0.7) is so small, /(0.8) is undoubtedly negative. Diminish the roots by 7 and proceed as follows : - 511640 7 + 5163200 -3581431 -11072000 11.1270002 + 11072383 -511633 7 + 1581769 -3581382 + 383 -511626 7 - 1999613 -200 -511619 Since the sum of the roots (| 442) is 515, we can find the third root by subtracting from 515 the sum of the two roots already found. 1st root, 1.1230907 2d root, 1.1270002 515 - 2.2500909 = 512.7499091, 3d root. 497. From the preceding sections we obtain the following general directions for solving a numerical equation : I. Find and remove commensurable roots by §§ 484-486, if there are any such roots in the equation. II. Determine the situation and thence the first figure of each of the incommensurable roots as in § 487. III. Calculate the incommensurable roots by Horner's Method. Exercise 83. Calculate to six places of decimals the positive roots of the following equations : 1. x^-^x-\=^0. 2. a;'-f-2a7'-4a;-43 = 0. 456 ALGEBRA. 3. ^x'-{-Sx' + 8x-S2=0. 4. 2:^^-2607^+131^-202 = 0. V 5. .x'-12x-\-1 = 0. 6. x'-5x' + 2x'~l3x + 5b = 0. Calculate, to six places of decimals where incommensura- ble, the real roots of the following equations : 7. x' = S5,^99. ■ 10. :^^ = 147,008,443. 8. a;' -- 242,970,624. 11. a;' + 2:r + 20 = 0. 9. :r* = 707,281. 12. a;^-10a;2 + 807+ 120 = 0. Each of the following equations has two roots nearly- equal. Calculate them to six places of decimals: 13.* x'-Sx^-4:X + lS = 0. 14.* 2o;* + 807^-35 07^-407+117 = 0. 15.* 07^+ llo7^- 10207+181 = 0. STURM'S THEOREM. 498. The problem of determining the number and situa- tion of the real roots of an equation is completely solved by Sturm's Theorem. In theory Sturm's method is per- fect ; in practice its application is long and tedious. For this reason, the situation of the roots is in general more easily determined by the methods already given. Before passing on to Sturm's Theorem itself we shall prove two preliminary theorems. 499. Situation of the Eoots of f (x) = 0. Between any two distinct real roots of the equation f (x) = there lies at least one real root of the equation f (x) = 0. Let a and /3 be two real roots of /(07) = 0, ^ being greater than a. Then /(a) = and /(^) = 0. As 07 in- Sturm's theorem. 457 creases continuously from a to ^, f{x) changes from to again ; and must first increase and then decrease, or first decrease and then increase. Hence, there must be some point at which f\x) changes from + to — , or vice versa. Therefore, for some value of x between a and )8, f(x) must be zero. Hence, at least one root of f\x) = must lie between a and y8. In the graph the curve will be horizontal where f\x) = 0. In the figure here given, A, B, Q D correspond to roots of f{x) = 0. Between A and B there is one root of f\x) = ; between B and C, three roots ; and between Cand D, one root. It is evident that if more than one root of f'(x) lies between a and (3, the number of roots must be an odd number. 500. Signs of f (x) and f '(x). Let a be any real root of an equation, f (x) = 0, which has no equal roots. Let X change continuously from a — h, a value a little less than a, /o a + h, a value a little greater than a. Then f (x) and f (x) will have unlike signs immediately before x passes through the root, and like signs immediately after x passes through the root. For /(a -h) = - hf\a) + |/"(a) - , and fXoi-h)= f\a)- hf"(a) + ; (§463) since /(a) = 0, as a is a root oif{x) = 0. When h is very small, the sign of each series on the right will be the sign of its first term (§ 475) ; and /(a — h) and /'(a — A) will evidently have opposite signs. Similarly, /(a -f h) and /'(a + h) will have like signs. The above is also evident from the graph off{x). 458 ALGEBRA. 501. Sturm's Punctions. The process of finding the H. C. F. of f{x) and f\x) has been employed (§ 462) in obtaining the multiple roots of the equation f(x) = 0. We use the same process in Sturm's Method. Let /(^) = be an equation which has no multiple roots ; let the operation of finding the H. 0. F. of f(x) and /'(x) be carried on until the remainder does not involve x, the sigii of each re^nainder obtained being changed before it is used as a divisor. If there is a H. C. F,, the equation has multiple roots. Remove them and proceed with the reduced equation. Represent by fix), f(x), /n(^) the several remain- ders with their signs changed. These expressions with /'(a;) are called Sturm's Punctions. Now, if D represent the dividend, d the divisor, q the quotient, and R the remainder, D=qd-^R. Consequently, / {x) = qj\x) —fix), fKx) = q,f(x)~f(x), f{x) = qj^{x) ~f(x), fn-2(^) = qn-Jn-l{x) —fn(x) ', where qi, q^, q^-i represent the several quotients, or the quotients multiplied by positive integers. From the above identities we have the following : I. Two consecutive functions cannot vanish for the same value of X. For example, suppose /j (;r) and/3(a;) to vanish for a par- ticular value of X. Give to x this value in all the identi- ties. By the third identity, f(x) will vanish; by the Sturm's theorem. 459 fourth, /s (a;) will vanish; finally, /„(-^) will vanish, which is contrary to the hypothesis that f(x) = has no multiple roots. II. When we give to rr a value which causes any one function to vanish, the adjacent functions have opposite signs. Thus, if/aC^) = 0, from the third identity /2(a;) = —/4(a:). 502. Sturm's Theorem. We are now in a position to enun- ciate Sturm's Theorem : If in the series of functions f(x), f'(x), f,(x) f„(x) we give to x any particular value a, and determine the num- ber of variations of sign; then give to x any greater value b, and determine the number of variations of sign ; the number of variations lost is the number of real roots of the equation f (x) = between a and h. For, let X increase continuously from a to b. First : Take the case in which x passes through a root of any of the functions /'(a;), /a (a;) f„-i(x), for example /^(a;). The adjacent functions in this case have opposite signs. fi(x) itself changes sign, but this has no effect on the num- ber of variations ; for if just before x passes through the root the signs are + H , just after x passes through the root they will be -| , and the number of variations is in each case one. Hence, there is no change in the number of variations of sign when x passes through a root of any of the functions Second : Take the case in which x passes through a root oif(x) = 0. Since /(:r) and /'(a:) have unlike signs just before x passes through the root, and like signs just after (§ 500), there is one variation lost for each root off(x) = 0. 460 ALGEBRA. Hence, the number of real roots between a and h is the number of variations of sign lost as x passes from a to h. To determine the total number of real roots, we take x first very large and negative, and then very large and pos- itive. The sign of each function is then the sign of its first term (§ 474). The student may not understand how it is that/(a;) and/^(a') always have unlike signs just before x passes through a root. Let o and j8 be two consecutive roots of f{x) = ; let A be very small. Suppose that at a f{x) changes from + to — ; then /'(a) is — (§ 460). When x = a — h, f{x) = +, f{x) is — ; a; = a, f{x) = 0, f\x) is — . As X changes from a to )8, f\x) passes through an odd number of roots (§ 499), and consequently changes sign. Hence, when x = — h, f{x) is — , f{x) is + ; and/^(a;) and /(a;) again have unlike signs. 503. Examples. (1) Determine the number and signs of the real roots of the equation x' — 4:x^-\-^x'' — l2x-{-l=0. Here f{x)^^x^ -I2x'' + I2x-12. Let us take ior f{x), however, the simpler expression a;3-3a;2 + 3a;-3. We proceed as if to find the H. C. F., changing the sign of each remainder before using it as a divisor. 1- 3+ 3- 3- 9+ 9- 3+ 1 -10+9 -30 + 27 -30-10 37- 9 111- 27 111 + 37 -64 + 64 l_4+6-12 + l l_3+3_ 3 1 + 3 1 + 3 9 + 1 3 + 3 - 6-2 3 + 1 1-1 10 + 37 Sturm's theorem. 461 The coefficients of the several functions are in heavy type. In the ordinary process of finding the highest common factor we can change signs at pleasure. In finding Sturm's functions we cannot do this as the sign is all important. We can, however, take out any positive factor. We now have /(x) = a;* — 4a::3 + Gx^ _ 12a; + 1, /^(x) = ar^-3a;2 + 3a; - 3, /2(a;) = 3a; + 1, f,{x) = + 64. When fix) fix) f,{x) Mx) a; == — 1000 + — — .+ 2 variations. a;=0 + — + + 2 variations. x= + 1000 + + + + variations. Hence, the equation has two real positive roots ; it must therefore have two imaginary roots. The real roots will be found by g 487 to lie one between and 1, and one between 3 and 4. (2) Investigate the reality of the roots of the equation We find fix) =oi^ + SHx+G, f\x) = 3ix' + H), fj,x) = -2Hx-G, f,ix) = -iG^ + AH^). If G^'^ + 4 H^ is positive, we have A^) /i(^) /2(^) Ai^) a; = — oo — + ± — 2 variations. a;== + oo + + T — 1 variation. Since H may be either + or — , the sign of/j (a;) is ambiguous. Hence, when O^ + 4: H'^ is positive, tliere is but one real root If G^ + 4 H^ is negative, H must be negative, and we have x= — Qo — + _ +3 variations. a; = + oo + + + +0 variation. Hence, when O"^ + 4 H^ is negative, there are three real roots. 462 ALGEBRA. Exercise 84. Determine by Sturm's Theorem the number and situation of the real roots of the following equations : 1. a;' — 4:^2 -11a; + 43 = 0. 2. x'-6x''-{-1x-S = 0. 3. x' — Ax'-\-x' + 6x + 2 = 0. 4. x*-^x'-\-10x'-6x-21 = 0. 5. x* — x'-x'' + 6 = 0. 6. x' — 2x^-Sx'' + 10x-4: = 0. 7. x' + 2x' + Sa^-\-Sx^—l = 0. 8. a^ + x^-2x'' + Sx-2 = 0, CHAPTEK XXXI. GENERAL SOLUTION OF EQUATIONS. 504. Numerical and Algebraic Solutions. By the methods of the preceding chapter we can find to any desired degree of accuracy the real roots of a numerical equation of any degree. The methods are theoretically complete, and the solution of a numerical equation becomes simply a question of the labor required for the necessary computations. In the case of a literal equation we have an entirely dif- ferent problem to solve. To solve a literal equation, we have to find in terms of the coefiicients expressions which will, when substituted for the unknown in the given equa- tion, reduce that equation to an identity. Thus, the roots of the general quadratic have been found ; they are given by ^ — Jrfc V^'"* — 4 ac 2a In the case of a particular quadratic with numerical co- efficients the roots can be found by putting for a, h, c in the above expression their particular values, and performing the indicated operations. Similar solutions have been obtained for the general equa- tions of the third and fourth degrees, and for certain special forms of equations of higher degrees. The solution of the general equation of the fifth degree involves expressions called elliptic functions, and is conse- quently beyond the scope of the present treatise. 464 ALGEBRA. In many cases, however, the numerical values of the roots of a particular equation are not easily obtained from the general solution, and for numerical equations the gen- eral solutions are in such cases of little value. A general solution differs from the solutions obtained in the last chapter in that a general solution represents not one particular root but all the roots indiscriminately. We shall first consider equations of two special forms, reciprocal and binomial equations. 505. Eeciprocal Equations. Keciprocal equations (§ 470), called also recurring equations, are of four forms : (1) Degree even ; corresponding coefficients equal with like signs. (2) Degree even ; corresponding coefficients numerically equal but with unlike signs. (3) Degree odd ; corresponding coefficients equal with like signs. (4) Degree odd ; corresponding coefficients numerically equal but with unlike signs. The following are examples of the four forms : (1) 2a;*-3a:3+4tc2-3a; + 2 = 0; (2) 3a^-a^ + 2a;*-2a;2 + a;-3 = 0; (3) a^ + 3a;*-2a^-2a;2 + 3a; + l=0; (4) 2:f^ + bx^-\-s?-x''-bx-2 = 0. Every equation of the second form will evidently want the mid- dle term. Every reciprocal equation of the second, third, or fourth form can be depressed to an equation of the first form. GENERAL SOLUTION OP EQUATIONS. 465 Second Form : Consider the equation ax^ -j- ^^^ + ^^* — cx^ — bx — a — O. Writing this a{x^-l) + hx{x'~-l) + cx''{x^ - 1) = 0, we see that the equation is divisible by x^ — \; conse- quently 1 and — 1 are both roots. The depressed equation is evidently of the first form. Similarly for any equation of the second form. Third Form : Consider the equation ax^ + bx^ -\- co(^-\-cx'^-\-hx-\-a=^0, "Writing this aix' + 1) + bx{x^ + 1) + cx\x + 1) = 0, we see that the equation is divisible by :r+l ; consequently — 1 is a root. The depressed equation is evidently of the first form. Similarly for any equation of the third form. Fourth Form : Consider the equation ao(^ + bx*' -\- C3^ — cx^ — bx — a = 0. Writing this a{a^ - 1) + bx{x' - 1) + cx\x - 1) = 0, we see that the equation is divisible by a;— 1; consequently + 1 is a root. The depressed equation is evidently of the first lorm. Similarly for any equation of the fourth form. By the preceding, to solve any reciprocal equation, it is only necessary to solve one of the first form. 466 ALGEBRA. 506. Any reciprocal equation of the first form can be depressed to an equation of half the degree. We proceed to illustrate by examples : (1) Solve the equation yA _ Y^x" + 29ri;« - 12:i; + 1 = 0. Divide by x^ a;^ + 1 - 12 /'a? + - V 29 = 0. x^ \ xj Put 1. X ' then 22 _2_ 120 + 29 = 0, or 22 _ 122 + 36 = 9, whence z = 9 or 3. * Solving the equations a^ + i = 9, «' + ^ = 3. we find X = , and x == 2 2 The first two roots will be found to be reciprocals each of the other ; also the second two roots. (2) Solve the equation This is of the fourth form ; dividing by re — 1 we find the depressed equation to be a;*-2a;3 + 3a;2-2a; + l = 0. This may be written x^ \ X j a;2 + 2 + -:!^-2fa; + -l + l-0, or 22-22 + 1 = 0, whence 2=1. Solving the equation a; + - = 1, we find X = t 2 these expressions being double roots. GENERAL SOLUTION OF EQUATIONS. 467 Exercise 85. Solve the equations : 1. x*-\-7x'-1x-l = 0. 2. x'-i-2x'-^x' + 2x+1^0. 3. x'-Sx^ + 5x' — bx^ + Sx—l = 0. 4. x' — 5x' + 6x' — 5x+l = 0. 5. 2x* — 5x'-i-6x''-bx + 2 = 0. 6. x^~4:x' + x^-i-x^ — 4.x+l = 0. 7. a:*— 10^-' + 26:r^ — 1007+1 = 0. 8. x^ + mx^ + mx + 1 = 0. 9. x^ + x' — x'-x' + x-\-l = 0. 10. Sx'-2x' + 6x'-5x' + 2x-S = 0. 507. Binomial Equations. An equation of the form a;" rb a = is called a binomial equation. We shall first consider the two equations .^r**— 1 = 0, a;'»+l = 0. If n is even, the equation a;" + 1 = 0, by Descartes' Rule (§ 479), has no real roots ; the equation ^i;" — 1 = has two real roots, +1 and —1, the remaining n — 2 roots being imaginary. If n is odd, the equation a;" + 1 = has one real root, — 1 ; the equation a;" — 1 = has one real root, + 1 ; the remaining n—1 roots being in each case imaginary. 468 ALGEBRA. 608. Now consider the equation ^" =fc a = 0, where a is positive. Represent by Va the real positive nth root of a. Then, if a is any root of a;" ± 1 = 0, aVa will be a root of ^r"* ± a = 0. For (a Vay = a"a = =F 1 X a = ^ <2. Since a is any root of a;" it 1 = 0, the n roots of ^'^i a = are found by multiplying each of the n roots of :r" ± 1 = by Va. The roots of a binomial equation are all different. For x"" ±a and its derivative nx^~'^, can have no common factor involving x (§ 462). 509. If a is a root of the equation .t"— 1 = 0, a*, where h is any integer, is also a root. For, if a is a root, a**=: 1. But(a^)'*=(a»7-=(1)''-=1. Therefore a* is a root of :r" = 1, or of re" — 1 = 0. Similarly for a root of x^-{- 1 = 0, provided h is an odd integer. 510. The Cube Eoots of Unity. The equation x^ = \, or a;' — 1 = 0, may be written (.'.-l)(a;^ + ^+l) = 0, of which we find the three roots to be : 1. -i+iV^, -i-i-V^- If either of the imaginary roots be represented by w, the other is found by actual multiplication to be cx-\-d = Q. (1 ) Before attempting to solve this equation we shall trans- form it into an equation in which the second term is wanting. Put z=^ax-\-h\ :. x =^ . Substituting this ex- a pression for x, and reducing, we obtain z^ + 3(«c - h") z + {a^d- Zabc + 2^>^) = 0, ' or, putting ^= ac — l>\ G = a^d — 3 ahc + 2 W, 2^ + 3^2+6^ = 0. (2) In the transformed equation put 2; = w* + v^. We obtain (t^^ + v*)^ + 3 ^(2^* + t;^) + G^ = 0, which reduces to % + v + 3(wM + ^)(2^^ + v*)+G^ = 0. (3) Since we have assumed but one relation between u and V, we are at liberty to assume one more relation. Let us assume u^v^ = -H. (4) Equation (3) now reduces to u^v^-0. (5) And (4) may be written uv = -B:\ (6) Eliminating v, we obtain the quadratic u'^-Gu^IP, (7) called the reducing quadratic of the cubic. GENERAL SOLUTION OF EQUATIONS. 471 Solving this quadratic, we find P (8) "" 2 ^ u 2 Since ax-\-h=z = u^ -\-v^, the three values of z are where u^ is any one of the three cube roots of u. Since there is the sign db before the radical, we have apparently six values of z. From (4) it is seen, however, that there are really but three different values of z. The above solution is known as Cardans. Ex. Solve, by Cardan's method, 2r»- 6^2 + 12a;- 11=0. Here a = 2, b = -2. Putting 2 = 2a; — 2, we obtain 23 + 122-12 = 0. .*. 5"= 4, = — 12, and the reducing quadratic is ^2- 12m = 64. Solving, w = 6±10= 16 or - 4; .; v = = - 4 or + 16. u Henoe the values of 2 are 2v^-v^; .2«v^2-«2v/i; 2 a^) = 0. 11. Prove that J" vanishes for the biquadratic ^a{x — 2ay = 2a{x-^ay. 12. Let a, p, y, 8 be the distances of four points A, B, C, D, on a straight line from a fixed point on that line. Prove that when the line is harmonically divided at ^, B, C, D the roots of Euler's cubic are in arithmetical progres- sion. The student who wishes to pursue the subject of this chapter farther is referred to Burnside and Panton's Theory of Equations, published by Longmans, London. CHAPTER XXXII. COMPLEX NUMBERS. 518. Eepresentation of Eeal Numbers. Let XX^ be a straight line of unlimited length. Let be a fixed point on that line. With any convenient unit of length measure off along the line from to the right and left a series of equal distances. X' — J I ■ I ■ rf ■ r? ■ ? f^f ■ i^-? I I ■ I ■ X Each of the points of division thus obtained will repre- sent an integer (§ 8). If the points to the right represent positive integers, those to the left will represent negative integers. The point will represent 0. To represent a rational fraction -, where a and h are integers, h being positive and a either positive or negative, we divide the unit into h equal parts, and then measure off a of these parts. The point obtained will lie between two of the points which represent integers. We cannot find exactly the point which represents a given incommensurable number. We can, however, always find two fractions between which the given incommensurable number lies ; and the point which represents the incommen- surable number will lie between the points which represent the two fractions. Since the difference between the fractions can be made COMPLEX NUMBERS. 481 as small as we please, the distance between the two points representing the fractions can be made as small as we please, and the position of the point which represents the given incommensurable number can therefore be deter- mined to any desired degree of accuracy. It appears, then, that all real numbers may be represented by points in the line XX\ Conversely, every point in the line XX^ will represent some real number which may be integral, fractional, or incommensurable, and either positive or negative. Instead of the representative points A, B, etc., we shall generally use the representative lines OA and OB. 519. Eemarks on Imaginaries. Imaginary expressions are not numbers in the ordinary arithmetical sense. We per- form upon them, however, the operations which we perform upon numbers, subject to the four fundamental laws which govern all algebraical operations (§ 47), viz. : the commu- tative, associative, distributive, and index laws. In finding the product of two imaginaries, however, the operation must be performed in a particular way (§ 168). We shall in this chapter often extend the term number to include imaginary expressions. When we are considering imaginary expressions without attempting to give them any arithmetical interpretation, there is nothing " imaginary " about the so-ca lled imagina- ries. The collection of symbols 3 + 4 V^ is, as far as symbols go, as " real " as the collection of symbols 34-4V2. It is only when we wish to obtain a result arithmetically interpretable, and arrive at an imaginary expression, that the latter can be called in a strict sense " imaginary." On this account the term complex number is preferable to im^aginary number. 482 ALGEBRA. 520. Pure Imaginaries. A pure imaginary cannot be rep- resented by a point on the line JTX' (§ 518), since all points on that line represent real numbers. We must there- fore seek elsewhere for its representative point. Represent V~ 1 by i. Assuming the commutative and associative laws, we have (§ 169) : ix a = ai; i X i X a = i"^ X a = (— 1) a = — a; ixixiXa — i^Xa = (—i)a = — ai] ixixixiXa = i^Xa = (+l)a = -i-a; i X i X i X i X i X a == i^ X a = i X a = ai ) and so on. From the above we see that the effect of multiplying by i twice is to change a to — a; twice more is to change — a back to + a. That is, two multiplications hy i reverse the sign of the multiplicand. Hence, two multiplications by i turn the representative line through 180° ; four multiplications by i through 360° ; and so on. We may, therefore, consistently assume that one multipli- cation by i turns the representative line through 90° ; three multiplications by i through 270° ; and so on. If, then, we draw through a line YY^ perpendicular to XX^, all pure imaginaries will be represented by points on this line, just as all real numbers are represented by points on XX'. 521. The lines XX* and FF' are called axes, XX' the axis of reals, and YY' the axis of pure imaginaries. is called the origin. It is customary to regard rotation opposite to that of the hands of a clock as positive. With this convention the COMPLEX NUMBERS. 483 point M, or the line OM, in the figure will represent + 5i or H-5V— 1; the point iV, or the line ON, — 6z or - 6V=n[. The only point which is on both axes is 0. This agrees X^ M I I I I I I N F' with the fact that is the only number which may be con- sidered either real or imaginary. Again, a and ai are measured on different lines. This agrees with the fact that a and ai are different in hind. 522. Vectors. A directed straight line of definite length is called a vector. Thus, the lines used to represent real numbers, and those used to represent pure imaginaries, are all vectors. Vectors need not, however, be parallel to either of the axes ; they may have any direction. The line AB, considered as a vector beginning at A and ending at B, is generally written AB. Tv/o parallel vectors which have the same length, and extend in the same direction, are said to be equal vectors. 484 ALGEBRA. 523. Vector Addition. D To add a vector CD to a vector AB we place on B, keeping CD parallel to its original position, and draw AD. Then, AD = AB + BD = AB+GD. The addition here meant by the sign + is not addition of numbers, but addition of vectors, generally called geometric addition. It is evidently identical with the composition of forces. From the dotted lines in the figure, and the known prop- erties of a parallelogram, it is easily seen that AD^GD-\-AB. :. AB+GD=GDi- AB. Consequently, vector addition is commutative (§ 21). It is easily seen that it is also associative (§ 27). 524. Complex Numbers. A complex number in general consists of a real part and an imaginary part, and may be written (§ 172) in the typical form x + yi, where :^ and y are both real. If we understand the sign -f to indicate geometric addi- tion, we shall obtain the vector which represents x -{- yi as follows : Lay off X on the axis of reals from to M. From M draw the vector MP to represent yi. Then the vector OP is the geometric sum of the vectors OM and MP, and represents the complex number x + yi. COMPLEX NUMBERS. 485 Instead of the vector OF we sometimes use the point P to represent the complex number. Thus, in the figure, the vectors OF, OQ, OF, OS, or the points F, Q, F, S, respectively, represent the^ complex numbers 6 + 4 ^, -6 + 5«, -5-3z, 3-5^■. In the complex number x-j-yi, a: and yi are repre- sented by vectors. Now vector addition is commutative. Consequently, x-{-yi = yi -f x. This is also evident from the preceding figure. The expression x-\-yi is the general expression for all numbers. This expression includes zero when x=0 and y = ; includes all real numbers when y = ; all pure imaginaries when a; = ; all complex numbers when x and y both differ from 0. 525. Addition of Complex Numbers. Let x + yi and x'-{-y'i be two complex numbers. Their sum, x -\- yi -\- x^ -\- yH, may by the commutative law be written a: + a;' + (3/ + 2/')^- LetOZ and OS be the representative vectors of X 4" yi and x^ + yH. Take 2^= UB; then, OC=OA + OB. Draw the other lines in the figure. Then, 0H= OF-h FH = OF^OE=^x^x\ 486 ALGEBRA. and HQ= FA + KC= FA^EB = y-\- y\ ,'. OC=x-\-x^ + {y-\- y^ = {x -\- yi) + (a;' + yH). But OC=OA + aB. Consequently, the geometric sum of the vectors of two com- plex numbers is the vector of their sum. Since vector addition is commutative, it follows that the addition of complex numbers is commutative. The sum of two complex numbers is the geometric sum of the sum of the real and the sum of the imaginary parts of the two numbers. The preceding may be made clearer by a particular example. Find the sum of 2 + 3 1 and — 4 + i 2 + 3i - OM and - 4 + i = OM^. Y M If now we proceed from M, the extremity of OM, in the direction of OM^ as far as the absolute value of 0M\ we reach the point M^^. Hence, OM^^ = - 2 + 4i, the sum of the two given complex numbers. The same result is reached if we first find the value of 2 + (- 4) = - 2. That is, if we count from two real units to A^^, and add to this sum 3 i + i = 4 i ; that is, count four imaginary units from A^^ on the perpendicular A'^M^^. A X 526. Modulus and Amplitude. Any complex number, x-\-yi, can be written in the form Wa;2 + 2/' s/x' + y'' J The expressions and ^ may be taken -y/^2 _j_ yi -yjx^ -f y COMPLEX NUMBERS. 487 as the sine and cosine of some angle <^, since they satisfy the equation cos'^i^ + sin^<^ = l. If we put r = V:r^ + 2/', the complex numj|er may be written r (cos -\-i sin ). Since r = V^M-y^, the sign of r is indeterminate. We shall, however, take r sAwSijs positive. The positive number r is called the modulus, the angle the amplitude, of the complex number x + yi. Let OF be the representative vector of a; + yi Since r is the positive value of Vx"^ + 2/^ it is evident that r is the length of OP. Since K X X OM COB V^M^' r OP and . , V y MP V^TifP r OP it follows that is the angle MOP. The above is easily seen to hold true when x and y are one or both negative. The modulus of a real number is its absolute value ; the amplitude is if the number is positive, 180* if the num- ber is negative. The modulus of a pure imaginary ai is a ; the amplitude is 90° if a is positive, 270° if a is negative. 527. Since the sum of the lengths of two sides of a triangle is greater than the length of the third side, it follows from §§ 523, 525 that the modulus of -the sum of two cowiplex nuTnbers is less than the sum, of the moduli. 488 ALGEBRA. In one case, however, that in which the representative vectors are collinear, the modulus of the sum is equal to the sum of the moduli. 528. Multiplication by Eeal Numbers. Let x -\- yi be any- complex number. If the representative vector be multi- plied by any real number a, it is easily seen from a figure that the product is ax -f- ayi. Therefore, a(x -f yi) — ao; + ayi. It follows that the multiplication of a complex number by a real number is distributive. Y Ex. To multiply - 2 +i by 3 : Take OA = — 2 on OX^, and erect at A the per- pendicular AM= 1. Then OM^ - 2 + i ; and, by taking OM three times, the result is OM^ = — 6 + 3 i, the product of (-2 + t)by3. X' A' ' ' A O 629. Multiplication by Pure Imaginaries. We have seen (§ 520) that multiplying a real number, or a pure imaginary, by i turns that number through 90°. Let us consider the effect of multiplying a complex number by i. By the commutative, associative, and distributive laws, i X r (cos <^ + «sin <^) ■= r(i cos <^ — sin <^) = r (— sin <;^ + i cos <^) = r [cos (90° + <^) -I- ^■ sin(90° + <^)]. Here, also, the effect of multiplying by i is to increase <^ to <^ + 90° ; that is, to turn the representative vector in the positive direction through an angle of 90°. The effect of multiplying by a pure imaginary ai will be to turn the complex number through a positive angle of 90°, and also to multiply the modulus by a. COMPLEX NUMBERS. ' 489 530. Multiplication by a Complex Number. We come now to the general problem of the multiplication of one complex number by another. This includes all other cases as par- ticular cases. Let r (cos <^ + z sin <^) and r'(cos<^' + *sin<^') be two complex numbers. By actual multiplication their product is rr'[cos cos ' — sin <^ sin <^' + *(sin <^ cos <^' + cos <^ sin (/>')]. By Trigonometry, this may be written rr'[cos(<^ + ') + i sin (<^ -f- <^')]. Therefore, the modulus of the product of two complex numbers is the product of their moduli ; and the amplitude of the product is the sum. of the amplitudes. Consequently, the effect of multiplying one complex number by another is to multiply the modulus of the first hy the modulus of the second ; and to turn the representative vector of the first through the amplitude of the second. 631. Division by a Complex Number. The quotient r(cos<^4-^sin ^) r'(cos<^'4-^sin <^') becomes, multiplying numerator and denominator by cos ^' — i sin ', ^,[cos (<^ - <^') + * sin (<^ -). 533. Boots of a Complex Number. From § 632, putting <^ for n(f>, and r for r"', we obtain + ^ sm — n r (cos <^ + ?i sin <^) ; or [Vr ( cos - \ n [r (cos <^ + ^ sin <^)]» = Vr ( cos ^ + i sin ^ )> \ n nj where by Vr is meant the real positive value of the root. The last expression gives apparently but one value for the 92th root of a complex number. But we must remem- ber that there are an unlimited number of angles which have a given sine and cosine. Thus the angles <^, <^ + 360°, (^ + 720°, (^4-^(360°), all have the same sine and cosine. We have, therefore, the following n\h roots of r (cos <;^ -j- i sin ^ V^^^cos + zsm j] {n) ( (^ + ^(360°) , . . <^ + 7z360° >^ Vrfcos ^ f-^sin 1; (^ + 1); COMPLEX NUMBERS. 491 In this series the (n -f l)th expression is the same as the ^rs^; the (n-\-2)th the same as the second; and so on. Consequently, there are but n different nth roots, those numbered (1) to (n). From this and the preceding section we can obtain an expression for [r(cos<^ + isin<^)]'*, where — is a rational fraction. n Ex. The 12 twelfth roots of 1 are: cos0° + i8in0° = l; (1) cos 30° + i sin 30° - ^^^^-^ ; (2) A COS 60° + i sin 60° = ^+^^ ; (3) cos90° + isin90° = i; (4) cos 330° + % sin 330° = ^"^ (12) 634. Complex Exponents. The meaning of a complex exponent is determined by subjecting it to the same opera- tions as a real exponent. It follows that such an expression as a*+^*, where a is a real number and x-^yi a complex exponent, may be sim- plified by resolving it into two factors, one of which is a real number, and the other an imaginary power of e (§ 392). From the ordinary rules for exponents, Put a» = e" ; then, u = log, a" = y log, a. 492 ALGEBRA. Since ^^l + ^+|| + | + g (§392) therefore, e''' = 1 + ui + '^ + ^ + -—■+—- -{- [2 |3 [4 |5_ By the Differential Calculus it is proved that when u is the circular measure of an angle, co.«=l-- + - + -+ sm« = «-- + j^- each series being an infinite series. Consequently, e"** = cos u + i sin u, and e^+"' = ^ (cos u-\-i sin u). Also, a^^^^ = a" (cos u-\-i sin w) = a*[cos(2/log,a) + ^sin(3/logea)]. 535. Trigonometric Solution of Cubic Equations. In the irreducible case (§ 513, 1.) the numerical values of the roots of a cubic equation may be found by the Trigonometric tables. We have (§ 513, III.), a. + 5^-(^ 2 ) +( 2 ) ' In the case to be considered G"^ -}- 4: H^ is negative (§ 513, L). Put _^=iJcos^, y^^^^^^iRAni,. A A Then, i?^ = (-//)^ R = (,-Hf, COMPLEX NUMBERS. 493 And, by § 533, ax -{-b = (— Hy [cos <^ + *' sin <^)* + (cos <^ — t sin <^)^]. The cube roots in the right member must be so taken that their product is 1, since in § 512 u^v^ = — IT. The three values of ax-{-h are : 2(-^/cos|; 2(-^)*cosC| + 120°') 2(-^f cos/'|4-240°\ IS given by tan cf> = ^> — — ^ ^• Ex. Take the equation s:' — 63 + 2 = 0. Here 0=^2, H^-2, G^ + 4H^ = -~28. tan0 = :^=V7. 1= 23°5/54/^ \/28 2 log 7 = 0.84510. I + 120° = 143° 5^ 54^^ log tan (() = 0.42255. ? + 240° = 263° 5^ 64^^. <^ = 69° 17M2^^. 3 + ^^'' And the three values of z are found by logarithms to be 2.6016, -2.2618, -0.3399. Check: 2.6016 - 2.2618 -0.3399 • 0. Horner's method is, however, to be preferred to the method of the present section. 494 ALGEBRA. Exercise 89. Express in the typical form : 1. (a + biy-\-{a-bi)\ 2. M:i,+ 1-^ 1 + 2^ l-2^ 2 + 36z ■ 7-26z * 6 + 8z 3-4^' 4. Show that [(V3 + 1) + (V3 -!)(]» = 16 + 16 ^. 5. If -wx -\-yi — a-\- hi, show that 6. F.nd the modulus of |f^|±M. 7. Find the three cube roots of \-\-i, 8. Find the five fifth roots of 1. 9. Find the four fourth roots of 3 -|- 4z. 10. Solve the equation z' — 122 + 3 = 0. 11. 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The first two chapters which give the theory and fundamental ideas and processes of his method, will enable students to master the remaining chapters, containing applications to Plane and Solid Geometry and Mechanics ; or to read Grassmann's original works. A very elementary knowledge of Trigonometry, the Differ- ential Calculus and Determinants, will be sufficient as a preparation for reading this book. Daniel Carhart, Prof, of Mathe- matics, Western University of Penn- sylvania: I am pleased to note the success which has attended Professor Hyde's efforts to bring into more popular form a branch of mathemat- ics which is at once so abbreviated in form and so comprehensive in results. Elements of the Differential and Integral Calculus. With Examples and Applications. By J. M. Taylor, Professor of Mathematics in Madison University. 8vo. Cloth. 249 pages. Mailing price, $1.95; for introduction, $1.80; allowance for old book, 40 cents. rpHE aim of this treatise is to present simply and concisely the fundamental problems of the Calculus, their solution, and more common applications. Many theorems are proved both by the method of rates and that of limits, and thus each is made to throw light upon the other. The chapter on differentiation is followed by one on direct integra- tion and its more important applications. Throughout the work there are numerous practical problems in Geometry and Mechanics, which serve to exhibit the power and use of the science, and to excite and keep alive the interest of the student. In February, 1891, Taylor's Calculus was found to be in use in about sixty colleges. The Nation, New York : In the first place, it is evidently a most carefully written book We are acquainted with no text-book of the Calculus which compresses so much matter into so few pages, and at the same time leaves the impression that all that is necessary has been said. In the second place, the number of carefully selected examples, both of those worked out in full in illustra- tion of the text, and of those left for the student to work out for himself, is extraordinary. MATHEMATICS. 93 Elementary Co-ordinate Geometry. By W. B. Smith, Professor of Math., Missouri State University. 8vo. Cloth. 312 pages. Mailing Price, $2.15; for introduction, $2.00. VITHILE in the study of Analytic Geometry either gain ol '* knowledge or culture of mind may be sought, the latter object alone can justify placing it in a college curriculum. Yet the subject may be so pursued as to be of no great educational value. Mere calculation, or the solution of problems by algebraic processes, is a very inferior discipline of reason. Even geometry is not the best discipline. In all thinking, the real difficulty lies in forming clear notions of things. In doing this all the higher faculties are brought into play. It is this formation of concepts, therefore, that is the essential part of mental training. And it is in line with this idea that the present treatise has been composed. Professors of mathematics speak of it as the most exhaustive work on the sub- ject yet issued in America ; and in colleges where an easier text- book is required for the regular course, this will be found of great value for post-graduate study. Wm. G. Peck, Prof, of Mathe- matics and Astronomy, Columbia College : I have read Dr. Smith's Co- ordinate Geometry from begintiing to end with unflagging interest. Its well compacted pages contain an im- mirably arranged. It is an excellent book, and the author is entitled to the thanks of every lover of mathe- matical science for this valuable con- tribution to its literature. I shall recommend its adoption as a text- mense amount of matter, most ad- i book in our graduate course. Elements of the Theory of the Newtonian Poten- tial Function. By B. O. Peirce, Professor of Mathematics and Physics, in Harvard University. 8vo. Cloth. 154 pages. Mailing price, $1.60; for intro- duction, $1.50. npmS book was written for the use of Electrical Engineers and students of Mathematical Physics because there was in English no mathematical treatment of the Theory of the Newtonian Poten- tial Function in sufficiently simple form. It gives as briefly as is consistent with clearness so much of that theory as is needed be- fore the study of standard works on Physics can be taken up with advantage. In the second edition a brief treatment of Electro- kinematics and a large number of problems have been added. MATHEMATICS. 95 Analytic Geometry. By A. S. Hardy, Ph.D., Professor of Mathematics in Dartmouth College, and author of Elements of Quaternions. 8vo. Cloth, xiv + 239 pages. Mailing Price, $1.60 j for "introduction, $1.50. rpmS work is designed for the student, not for the teacher. Particular attention has been given to those fundamental con- ceptions and processes which, in the author's experience, have been found to be sources of difficulty to the student in acquiring a grasp of the subject as a method of research. The limits of the work are fixed by the time usually devoted to Analytic Geometry in our college courses by those who are not to make a special study in mathematics. It is hoped that it will prove to be a text-booh which the teacher will wish to use in his class-room, rather than a hook oj reference to be placed on his study shelf. Oren Boot, Professor of Mathemat- ics, Hamilton College: It meets quite fully my notion of a text for our classes. I have hesitated somewhat about introducing a generalized dis- cussion of the conic in required work. I have, however, read Mr. Hardy's discussion carefully twice; and it seems to me that a student who can get the subject at all can get that. It is my present purpose to use the work next year. John E. Clark, Professor of Mathe- matics, Sheffield Scientific School of Yale College : I need not hesitate to say, after even a cursory examina- tion, that it seems to me a very at- tractive book, as I anticipated it would be. It has evidently been pre- pared with real insight alike into the nature of the subject and the difficul- ties of beginners, and a very thought- ful regard to both; and I think its aims and characteristic features will meet with high approval. While leading the student to the usual use- ful results, the author happily takes especial pains to acquaint him with the character and spirit of analytical methods, and, so far as practicable, to help him acquire skill in using them. John R. French, Dean of College of Liberal Arts, Syracuse Univer- sity : It is a very excellent work, and well adapted to use in the reci- tation room. Elements of Quaternions. By A. S. Hardy, Ph.D., Professor of Mathematics, Dartmouth College. Second edition, revised. Crown 8vo. Cloth, viii + 234 pages. Mailing Price, $2.15; Introduction, $2.00. nPHE chief aim has been to meet the wants of beginners in the class-room, and it is believed that this work will be found superior in fitness for beginners in practical compass, in explana- tions and applications, and in adaptation to the methods of instruc- tion common in this country. 96 MATHEMATICS. Elements of the Calculus. By A. S. Hardy, Professor of Mathematics in Dartmouth College. 8vo. Cloth, xi + 239 pages. By mail, f 1.60; for introduction, $1.50. rFHIS text-book is based upon the method of rates. The object of the Differential Calculus is the measurement and comparison of rates of change when the change is not uniform. Whether a quantity is or is not changing uniformly, however, its rate at any- instant is determined essentially in the same manner, viz. : by let- ting it change at the rate it had at the instant in question and observing what this change is. It is this change which the Cal- culus enables us to determine, however complicated the law of variation may be. From the author's experience in presenting the Calculus to beginners, the method of rates gives the student a mcr intelligent, that is, a less mechanical, grasp of the problems witl.i its scope than any other. No comparison has been made between this method and those of limits and of infinitesimals. This larger view of the Calculus is for special or advanced students, for which this work is not intended ; the space and time which would be required by such general comparison being devoted to the applica- tions of the method adopted. Part I., Differential Calculus, occupies 166 pages. Part II., Inte- gral Calculus, 73 pages. George B. Merriman, Prof, of Mathematics and Astronomy, Rut- gers College: I am glad to observe that Professor Hardy has adopted the method of rates in his new Calcu- lus, a logical and intelligent method, which avoids certain difficulties in- volved in the usual methods. J. B. Colt, Prof, of Mathematics, Boston University : It pleases me very much. The treatment of the first principles of Calculus by the method of rates is eminently clear. Its use next year is quite probable. Ellen Hayes, Prof, of Mathemat- ics, Wellesley College : I have found it a pleasure to examine the book. It must commend itself in many respects to teachers of Calculus. W. K. McDaniel, Prof, of Mathe- matics, Westei'n Maryland College: Hardy's Calculus and Analytic Ge- ometry are certainly far better books for the college class-room than any others I know of. The feature of both books is the directness with which the author gets right at the very fact that he intends to convey to the student, and the force of his presentation of the fact is greatly augmented by the excellent arrange- ment of type and other features of the mechanical make-up. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. MAR 6 194a ,iFeb'50R' REC'D LD MAY 27 1959 2SiuV b^ JUN 111959 ? REC'D LD JUN 9 ibb9 REC'D LDL_-^^^^^i5eoi^ o#^ REC'D LD AUG 2 7 m SEP 25 1951 LD 21-100m.9 •47(A5702sl6)47CIRCULATION DEPT, U.C.BERKELEY LIBRARIES comnDShT / 80056 >o (Dih ' ''- UNIVERSITY OF CALIFORNIA LIBRARY ^^.