The Project Gutenberg eBook ofEncyclopaedia Britannica, 11th Edition, "Equation" to "Ethics"This ebook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this ebook or online atwww.gutenberg.org. If you are not located in the United States, you will have to check the laws of the country where you are located before using this eBook.Title: Encyclopaedia Britannica, 11th Edition, "Equation" to "Ethics"Author: VariousRelease date: February 25, 2011 [eBook #35398]Most recently updated: January 7, 2021Language: EnglishCredits: Produced by Marius Masi, Don Kretz and the OnlineDistributed Proofreading Team at https://www.pgdp.net*** START OF THE PROJECT GUTENBERG EBOOK ENCYCLOPAEDIA BRITANNICA, 11TH EDITION, "EQUATION" TO "ETHICS" ***
This ebook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this ebook or online atwww.gutenberg.org. If you are not located in the United States, you will have to check the laws of the country where you are located before using this eBook.
Title: Encyclopaedia Britannica, 11th Edition, "Equation" to "Ethics"Author: VariousRelease date: February 25, 2011 [eBook #35398]Most recently updated: January 7, 2021Language: EnglishCredits: Produced by Marius Masi, Don Kretz and the OnlineDistributed Proofreading Team at https://www.pgdp.net
Title: Encyclopaedia Britannica, 11th Edition, "Equation" to "Ethics"
Author: Various
Author: Various
Release date: February 25, 2011 [eBook #35398]Most recently updated: January 7, 2021
Language: English
Credits: Produced by Marius Masi, Don Kretz and the OnlineDistributed Proofreading Team at https://www.pgdp.net
*** START OF THE PROJECT GUTENBERG EBOOK ENCYCLOPAEDIA BRITANNICA, 11TH EDITION, "EQUATION" TO "ETHICS" ***
Articles in This Slice
EQUATION(from Lat.aequatio,aequare, to equalize), an expression or statement of the equality of two quantities. Mathematical equivalence is denoted by the sign =, a symbol invented by Robert Recorde (1510-1558), who considered that nothing could be more equal than two equal and parallel straight lines. An equation states an equality existing between two classes of quantities, distinguished as known and unknown; these correspond to the data of a problem and the thing sought. It is the purpose of the mathematician to state the unknowns separately in terms of the knowns; this is called solving the equation, and the values of the unknowns so obtained are called the roots or solutions. The unknowns are usually denoted by the terminal letters, ... x, y, z, of the alphabet, and the knowns are either actual numbers or are represented by the literals a, b, c, &c...,i.e.the introductory letters of the alphabet. Any number or literal which expresses what multiple of term occurs in an equation is called the coefficient of that term; and the term which does not contain an unknown is called the absolute term. The degree of an equation is equal to the greatest index of an unknown in the equation, or to the greatest sum of the indices of products of unknowns. If each term has the sum of its indices the same, the equation is said to be homogeneous. These definitions are exemplified in the equations:—
(1) ax² + 2bx + c = 0,(2) xy² + 4a²x = 8a³,(3) ax² + 2hxy + by² = 0.
(1) ax² + 2bx + c = 0,
(2) xy² + 4a²x = 8a³,
(3) ax² + 2hxy + by² = 0.
In (1) the unknown is x, and the knowns a, b, c; the coefficients of x² and x are a and 2b; the absolute term is c, and the degree is 2. In (2) the unknowns are x and y, and the known a; the degree is 3,i.e.the sum of the indices in the term xy². (3) is a homogeneous equation of the second degree in x and y. Equations of the first degree are calledsimpleorlinear; of the second,quadratic; of the third,cubic; of the fourth,biquadratic; of the fifth,quintic, and so on. Of equations containing only one unknown the number of roots equals the degree of the equation; thus a simple equation has one root, a quadratic two, a cubic three, and so on. If one equation be given containing two unknowns, as for example ax + by = c or ax² + by² = c, it is seen that there are an infinite number of roots, for we can give x, say, any value and then determine the corresponding value of y; such an equation is calledindeterminate; of the examples chosen the first is a linear and the second a quadratic indeterminate equation. In general, an indeterminate equation results when the number of unknowns exceeds by unity the number of equations. If, on the other hand, we have two equations connecting two unknowns, it is possible to solve the equations separately for one unknown, and then if we equate these values we obtain an equation in one unknown, which is soluble if its degree does not exceed the fourth. By substituting these values the corresponding values of the other unknown are determined. Such equations are calledsimultaneous; and a simultaneous system is a series of equations equal in number to the number of unknowns. Such a system is not always soluble, for it may happen that one equation is implied by the others; when this occurs the system is calledporismaticorporistic. Anidentitydiffers from an equation inasmuch as it cannot be solved, the terms mutually cancelling; for example, the expression x² − a² = (x − a)(x + a) is an identity, for on reduction it gives 0 = 0. It is usual to employ the sign ≡ to express this relation.
An equation admits of description in two ways:—(1) It may be regarded purely as an algebraic expression, or (2) as a geometrical locus. In the first case there is obviously no limit to the number of unknowns and to the degree of the equation; and, consequently, this aspect is the most general. In the second case the number of unknowns is limited to three, corresponding to the three dimensions of space; the degree is unlimited as before. It must be noticed, however, that by the introduction of appropriate hyperspaces,i.e.of degree equal to the number of unknowns, any equation theoretically admits of geometrical visualization, in other words, every equation may be represented by a geometrical figure and every geometrical figure by an equation. Corresponding to these two aspects, there are two typical methods by which equations can be solved, viz. the algebraic and geometric. The former leads to exact results, or, by methods of approximation, to results correct to any required degree of accuracy. The latter can only yield approximate values: when theoretically exact constructions are available there is a source of error in the draughtsmanship, and when the constructions are only approximate, the accuracy of the results is more problematical. The geometric aspect, however, is of considerable value in discussing the theory of equations.
An equation admits of description in two ways:—(1) It may be regarded purely as an algebraic expression, or (2) as a geometrical locus. In the first case there is obviously no limit to the number of unknowns and to the degree of the equation; and, consequently, this aspect is the most general. In the second case the number of unknowns is limited to three, corresponding to the three dimensions of space; the degree is unlimited as before. It must be noticed, however, that by the introduction of appropriate hyperspaces,i.e.of degree equal to the number of unknowns, any equation theoretically admits of geometrical visualization, in other words, every equation may be represented by a geometrical figure and every geometrical figure by an equation. Corresponding to these two aspects, there are two typical methods by which equations can be solved, viz. the algebraic and geometric. The former leads to exact results, or, by methods of approximation, to results correct to any required degree of accuracy. The latter can only yield approximate values: when theoretically exact constructions are available there is a source of error in the draughtsmanship, and when the constructions are only approximate, the accuracy of the results is more problematical. The geometric aspect, however, is of considerable value in discussing the theory of equations.
History.—There is little doubt that the earliest solutions of equations are given, in the Rhind papyrus, a hieratic document written some 2000 years before our era. The problems solved were of an arithmetical nature, assuming such forms as “a mass and its1⁄7th makes 19.” Calling the unknown mass x, we have given x +1⁄7x = 19, which is a simple equation. Arithmetical problems also gave origin to equations involving two unknowns; the early Greeks were familiar with and solved simultaneous linear equations, but indeterminate equations, such, for instance, as the system given in the “cattle problem” of Archimedes, were not seriously studied until Diophantus solved many particular problems. Quadratic equations arose in the Greek investigations in the doctrine of proportion, andalthough they were presented and solved in a geometrical form, the methods employed have no relation to the generalized conception of algebraic geometry which represents a curve by an equation and vice versa. The simplest quadratic arose in the construction of a mean proportional (x) between two lines (a, b), or in the construction of a square equal to a given rectangle; for we have the proportion a:x = x:b;i.e.x² = ab. A more general equation, viz. x² − ax + a² = 0, is the algebraic equivalent of the problem to divide a line in medial section; this is solved inEuclid, ii. 11. It is possible that Diophantus was in possession of an algebraic solution of quadratics; he recognized, however, only one root, the interpretation of both being first effected by the Hindu Bhaskara. A simple cubic equation was presented in the problem of finding two mean proportionals, x, y, between two lines, one double the other. We have a:x = x:y = y:2a, which gives x² = ay and xy = 2a²; eliminating y we obtain x³ = 2a³, a simple cubic. The Greeks could not solve this equation, which also arose in the problems of duplicating a cube and trisecting an angle, by the ruler and compasses, but only by mechanical curves such as the cissoid, conchoid and quadratrix. Such solutions were much improved by the Arabs, who also solved both cubics and biquadratics by means of intersecting conics; at the same time, they developed methods, originated by Diophantus and improved by the Hindus, for finding approximate roots of numerical equations by algebraic processes. The algebraic solution of the general cubic and biquadratic was effected in the 16th century by S. Ferro, N. Tartaglia, H. Cardan and L. Ferrari (seeAlgebra:History). Many fruitless attempts were made to solve algebraically the quintic equation until P. Ruffini and N.H. Abel proved the problem to be impossible; a solution involving elliptic functions has been given by C. Hermite and L. Kronecker, while F. Klein has given another solution.
In the geometric treatment of equations the Greeks and Arabs based their constructions upon certain empirically deduced properties of the curves and figures employed. Knowing various metrical relations, generally expressed as proportions, it was found possible to solve particular equations, but a general method was wanting. This lacuna was not filled until the 17th century, when Descartes discovered the general theory which explained the nature of such solutions, in particular those wherein conics were employed, and, in addition, established the most important facts that every equation represents a geometrical locus, and conversely. To represent equations containing two unknowns, x, y, he chose two axes of reference mutually perpendicular, and measured x along the horizontal axis and y along the vertical. Then by the methods described in the articleGeometry:Analytical, he showed that—(1) a linear equation represents a straight line, and (2) a quadratic represents a conic. If the equation be homogeneous or break up into factors, it represents a number of straight lines in the first case, and the loci corresponding to the factors in the second. The solution of simultaneous equations is easily seen to be the values of x, y corresponding to the intersections of the loci. It follows that there is only one value of x, y which satisfies two linear equations, since two lines intersect in one point only; two values which satisfy a linear and quadratic, since a line intersects a conic in two points; and four values which satisfy two quadratics, since two conics intersect in four points. It may happen that the curves do not actually intersect in the theoretical maximum number of points; the principle of continuity (seeGeometrical Continuity) shows us that in such cases some of the roots are imaginary. To represent equations involving three unknowns x, y, z, a third axis is introduced, the z-axis, perpendicular to the plane xy and passing through the intersection of the lines x, y. In this notation a linear equation represents a plane, and two linear simultaneous equations represent a line,i.e.the intersection of two planes; a quadratic equation represents a surface of the second degree. In order to graphically consider equations containing only one unknown, it is convenient to equate the terms to y;i.e.if the equation be ƒ(x) = 0, we take y = ƒ(x) and construct this curve on rectangular Cartesian co-ordinates by determining the values of y which correspond to chosen values of x, and describing a curve through the points so obtained. The intersections of the curve with the axis of x gives the real roots of the equation; imaginary roots are obviously not represented.
In this article we shall treat of: (1) Simultaneous equations, (2) indeterminate equations, (3) cubic equations, (4) biquadratic equations, (5) theory of equations. Simple, linear simultaneous and quadratic equations are treated in the articleAlgebra; for differential equations seeDifferential Equations.
I.Simultaneous Equations.Simultaneous equations which involve the second and higher powers of the unknown may be impossible of solution. No general rules can be given, and the solution of any particular problem will largely depend upon the student’s ingenuity. Here we shall only give a few typical examples.1.Equations which may be reduced to linear equations.—Ex.To solve x(x − a) = yz, y (y − b) = zx, z (z − c) = xy. Multiply the equations by y, z and x respectively, and divide the sum by xyz; thena+b+c= 0zxy(1).Multiply by z, x and y, and divide the sum by xyz; thena+b+c= 0yzx(2).From (1) and (2) by cross multiplication we obtain1=1=1=1(suppose)y (b² − ac)z (c² − ab)x (a² − bc)λ(3).Substituting for x, y and z in x (x − a) = yz we obtain1=3abc − (a³ + b³ + c³);λ(a² − bc) (b² − ac) (c² − ab)and therefore x, y and z are known from (3). The same artifice solves the equations x² − yz = a, y² − xz = b, z² − xy = c.2.Equations which are homogeneous and of the same degree.—These equations can be solved by substituting y = mx. We proceed to explain the method by an example.Ex.To solve 3x² + xy + y² = 15, 31xy − 3x² − 5y² = 45. Substituting y = mx in both these equations, and then dividing, we obtain 31m − 3 − 5m² = 3 (3 + m + m²) or 8m² − 28m + 12 = 0. The roots of this quadratic are m = ½ or 3, and therefore 2y = x, or y = 3x.Taking 2y = x and substituting in 3x² + xy + y² = 0, we obtain y² (12 + 2 + 1) = 15; ∴ y² = 1, which gives y = ±1, x = ±2. Taking the second value, y = 3x, and substituting for y, we obtain x² (3 + 3 + 9) = 15; ∴ x² = 1, which gives x = ±1, y = ±3. Therefore the solutions are x = ±2, y = ±1 and x = ±1, y = ±3. Other artifices have to be adopted to solve other forms of simultaneous equations, for which the reader is referred to J.J. Milne,Companion to Weekly Problem Papers.II.Indeterminate Equations.1. When the number of unknown quantities exceeds the number of equations, the equations will admit of innumerable solutions, and are therefore said to beindeterminate. Thus if it be required to find two numbers such that their sum be 10, we have two unknown quantities x and y, and only one equation, viz. x + y = 10, which may evidently be satisfied by innumerable different values of x and y, if fractional solutions be admitted. It is, however, usual, in such questions as this, to restrict values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions,x = 1, 2, 3, 4, 5, 6, 7, 8, 9;y = 9, 8, 7, 6, 5, 4, 3, 2, 1;which indeed may be reduced to five; for the first four become the same as the last four, by simply changing x into y, and the contrary. This branch of analysis was extensively studied by Diophantus, and is sometimes termed the Diophantine Analysis.2. Indeterminate problems are of different orders, according to the dimensions of the equation which is obtained after all the unknown quantities but two have been eliminated by means of the given equations. Those of the first order lead always to equations of the formax ± by = ±c,where a, b, c denote given whole numbers, and x, y two numbers to be found, so that both may be integers. That this condition may be fulfilled, it is necessary that the coefficients a, b have no common divisor which is not also a divisor of c; for if a = md and b = me, then ax + by = mdx + mey = c, and dx + ey = c/m; but d, e, x, y are supposed to be whole numbers, therefore c/m is a whole number; hence m must be a divisor of c.Of the four forms expressed by the equation ax ± by = ±c, it is obvious that ax + by = −c can have no positive integral solutions. Also ax − by = −c is equivalent to by − ax = c, and so we have only to consider the forms ax ± by = c. Before proceeding to the general solution of these equations we will give a numerical example.To solve 2x + 3y = 25 in positive integers. From the given equationwe have x = (25 − 3y) / 2 = 12 − y − (y − 1) / 2. Now, since x must be a whole number, it follows that (y − 1)/2 must be a whole number. Let us assume (y − 1) / 2 = z, then y = 1 + 2z; and x = 11 − 3z, where z might be any whole number whatever, if there were no limitation as to the signs of x and y. But since these quantities are required to be positive, it is evident, from the value of y, that z must be either 0 or positive, and from the value of x, that it must be less than 4; hence z may have these four values, 0, 1, 2, 3.Ifz = 0,z = 1,z = 2,z = 3;Thenx = 11,x = 8,x = 5,x = 2,y = 1,y = 3,y = 5,y = 7.3. We shall now give the solution of the equation ax − by = c in positive integers.Convert a/b into a continued fraction, and let p/q be the convergent immediately preceding a/b, then aq − bp = ±1 (seeContinued Fraction).(α) If aq − bp = 1, the given equation may be writtenax − by = c (aq − bp);∴ a (x − cq) = b (y − cp).Since a and b are prime to one another, then x − cq must be divisible by b and y − cp by a; hence(x − cq) / b = (y − cq) / a = t.That is, x = bt + cq and y = at + cp.Positive integral solutions, unlimited in number, are obtained by giving t any positive integral value, and any negative integral value, so long as it is numerically less than the smaller of the quantities cq/b, cp/a; t may also be zero.(β) If aq − bp = −1, we obtain x = bt − cq, y = at − cp, from which positive integral solutions, again unlimited in number, are obtained by giving t any positive integral value which exceeds the greater of the two quantities cq/b, cp/a.If a or b is unity, a/b cannot be converted into a continued fraction with unit numerators, and the above method fails. In this case the solutions can be derived directly, for if b is unity, the equation may be written y = ax − c, and solutions are obtained by giving x positive integral values greater than c/a.4. To solve ax + by = c in positive integers. Converting a b into a continued fraction and proceeding as before, we obtain, in the case of aq − bp = 1,x = cq − bt, y = at − cp.Positive integral solutions are obtained by giving t positive integral values not less than cp/a and not greater than cq/b.In this case the number of solutions is limited. If aq − bp = −1 we obtain the general solution x = bt − cq, y = cp − at, which is of the same form as in the preceding case. For the determination of the number of solutions the reader is referred to H.S. Hall and S.R. Knight’sHigher Algebra, G. Chrystal’sAlgebra, and other text-books.5. If an equation were proposed involving three unknown quantities, as ax + by + cz = d, by transposition we have ax + by = d − cz, and, putting d − cz = c′, ax + by = c′. From this last equation we may find values of x and y of this form,x = mr + nc′, y = mr + n′c′,or x = mr + n (d − cz), y = m′r + n′ (d − cz);where z and r may be taken at pleasure, except in so far as the values of x, y, z may be required to be all positive; for from such restriction the values of z and r may be confined within certain limits to be determined from the given equation. For more advanced treatment of linear indeterminate equations seeCombinatorial Analysis.6. We proceed to indeterminate problems of the second degree: limiting ourselves to the consideration of the formula y² = a + bx + cx², where x is to be found, so that y may be a rational quantity. The possibility of rendering the proposed formula a square depends altogether upon the coefficients a, b, c; and there are four cases of the problem, the solution of each of which is connected with some peculiarity in its nature.Case1. Let a be a square number; then, putting g² for a, we have y² = g² + bx + cx². Suppose √(g² + bx + cx²) = g + mx; then g² + bx + cx² = g² + 2gmx + m²x², or bx + cx² = 2gmx + m²x², that is, b + cx = 2gm + m²x; hencex =2gm − b, y = √(g² + bx + cx²)=cg − bm + gm².c − m²c − m²Case 2. Let c be a square number = g²; then, putting √(a + bx + g²x²) = m + gx, we find a + bx + g²x² = m² + 2mgx + g²x², or a + bx = m² + 2mgx; hence we findx =m² − a, y = √(a + bx + g²x²) =bm − gm² − ag.b − 2mgb − 2mgCase 3. When neither a nor c is a square number, yet if the expression a + bx + cx² can be resolved into two simple factors, as f + gx and h + kx, the irrationality may be taken away as follows:—Assume √(a + bx + cx²) = √{ (f + gx) (h + kx) } = m (f + gx), then (f + gx) (h + kx) = m² (f + gx)², or h + kx = m² (f + gx); hence we findx =fm² − h, y = √{ (f + gx) (h + kx) } =(fk − gh) m;k − gm²k − gm²and in all these formulae m may be taken at pleasure.Case 4. The expression a + bx + cx² may be transformed into a square as often as it can be resolved into two parts, one of which is a complete square, and the other a product of two simple factors; for then it has this form, p² + qr, where p, q and r are quantities which contain no power of x higher than the first. Let us assume √(p² + qr) = p + mq; thus we have p² + qr = p² + 2mpq + m²q² and r = 2mp + m²q, and as this equation involves only the first power of x, we may by proper reduction obtain from it rational values of x and y, as in the three foregoing cases.The application of the preceding general methods of resolution to any particular case is very easy; we shall therefore conclude with a single example.Ex.It is required to find two square numbers whose sum is a given square number.Let a² be the given square number, and x², y² the numbers required; then, by the question, x² + y² = a², and y = √(a² − x²). This equation is evidently of such a form as to be resolvable by the method employed in case 1. Accordingly, by comparing √(a² − x²) with the general expression √(g² + bx + cx²), we have g = a, b = 0, c = −1, and substituting these values in the formulae, and also −n for +m, we findx =2an, y =a (n² − 1).n² + 1n² + 1If a = n² + 1, there results x = 2n, y = n² − 1, a = n² + 1. Hence if r be an even number, the three sides of a rational right-angled triangle are r, (½ r)² − 1, (½ r)² + 1. If r be an odd number, they become (dividing by 2) r, ½ (r² − 1), ½ (r² + 1).For example, if r = 4, 4, 4 − 1, 4 + 1, or 4, 3, 5, are the sides of a right-angled triangle; if r = 7, 7, 24, 25 are the sides of a right-angled triangle.III.Cubic Equations.1. Cubic equations, like all equations above the first degree, are divided into two classes: they are said to bepurewhen they contain only one power of the unknown quantity; andadfectedwhen they contain two or more powers of that quantity.Pure cubic equations are therefore of the form x³ = r; and hence it appears that a value of the simple power of the unknown quantity may always be found without difficulty, by extracting the cube root of each side of the equation. Let us consider the equation x³ − c³ = 0 more fully. This is decomposable into the factors x − c = 0 and x² + cx + c² = 0. The roots of this quadratic equation are ½ (−1 ± √−3) c, and we see that the equation x³ = c³ has three roots, namely, one real root c, and two imaginary roots ½ (−1 ± √−3) c. By making c equal to unity, we observe that ½ (−1 ± √−3) are the imaginary cube roots of unity, which are generally denoted by ω and ω², for it is easy to show that (½ (−1 − √−3))² = ½ (−1 + √−3).2. Let us now consider such cubic equations as have all their terms, and which are therefore of this form,x³ + Ax² + Bx + C = 0,where A, B and C denote known quantities, either positive or negative.This equation may be transformed into another in which the second term is wanting by the substitution x = y − A/3. This transformation is a particular case of a general theorem. Let xn+ Axn−1+ Bxn−2... = 0. Substitute x = y + h; then (y + h)n+ A (y + h)n−1... = 0. Expand each term by the binomial theorem, and let us fix our attention on the coefficient of yn−1. By this process we obtain 0 = yn+ yn−1(A + nh) + terms involving lower powers of y.Now h can have any value, and if we choose it so that A + nh = 0, then the second term of our derived equation vanishes.Resuming, therefore, the equation y³ + qy + r = 0, let us suppose y = v + z; we then have y³ = v³ + z³ + 3vz (v + z) = v³ + z³ + 3vzy, and the original equation becomes v³ + z³ + (3vz + q) y + r = 0. Now v and z are any two quantities subject to the relation y = v + z, and if we suppose 3vz + q = 0, they are completely determined. This leads to v³ + z³ + r = 0 and 3vz + q = 0. Therefore v³ and z³ are the roots of the quadratic t² + rt − q²/27 = 0. Thereforev³ =−½ r + √(1⁄27q³ + ¼ r²); z³ = −½ r − √(1⁄27q³ + ¼r²);v =3√{−½ r + √(1⁄27q³ + ¼ r²) }; z =3√{ (−½ r − √(1⁄27q³ + ¼ r²) };and y =v + z =3√{−½ r + √(1⁄27q³ + ¼ r²) } +3√{−½ r − √(1⁄27q³ + ¼ r²) }.Thus we have obtained a value of the unknown quantity y, in terms of the known quantities q and r; therefore the equation is resolved.3. But this is only one of three values which y may have. Let us, for the sake of brevity, putA = −½ r + √(1⁄27q³ + ¼ r²), B = −½ r − √(1⁄27q³ + ¼ r²),and putα = ½ (−1 + √−3),β = ½ (−1 − √−3).Then, from what has been shown (§ 1), it is evident that v and z have each these three values,v =3√A, v = α3√A, v = β3√A;z =3√B, z = α3√B, z = β3√B.To determine the corresponding values of v and z, we must consider that vz = −1⁄3q =3√(AB). Now if we observe that αβ = 1, it will immediately appear that v + z has these three values,v + z =3√A +3√B,v + z = α3√A + β3√B,v + z = β3√A + α3√B,which are therefore the three values of y.The first of these formulae is commonly known by the name of Cardan’s rule (seeAlgebra:History).The formulae given above for the roots of a cubic equation may be put under a different form, better adapted to the purposes of arithmetical calculation, as follows:—Because vz = −1⁄3q, therefore z = −1⁄3q × 1/v = −1⁄3q /3√A; hence v + z =3√A −1⁄3q /3√A: thus it appears that the three values of y may also be expressed thus:y =3√A −1⁄3q /3√Ay = α3√A −1⁄3qβ /3√Ay = β3√A −1⁄3qα /3√A.See below,Theory of Equations, §§ 16 et seq.IV.Biquadratic Equations.1. When a biquadratic equation contains all its terms, it has this form,x4+ Ax³ + Bx² + Cx + D = 0,where A, B, C, D denote known quantities.We shall first consider pure biquadratics, or such as contain only the first and last terms, and therefore are of this form, x4= b4. In this case it is evident that x may be readily had by two extractions of the square root; by the first we find x² = b², and by the second x = b. This, however, is only one of the values which x may have; for since x4= b4, therefore x4− b4= 0; but x4− b4may be resolved into two factors x² − b² and x² + b², each of which admits of a similar resolution; for x² − b² = (x − b)(x + b) and x² + b² = (x − b√−1)(x + b√−1). Hence it appears that the equation x4− b4= 0 may also be expressed thus,(x − b) (x + b) (x − b√−1) (x + b√−1) = 0;so that x may have these four values,+b, −b, +b√−1, −b√−1,two of which are real, and the others imaginary.2. Next to pure biquadratic equations, in respect of easiness of resolution, are such as want the second and fourth terms, and therefore have this form,x4+ qx² + s = 0.These may be resolved in the manner of quadratic equations; for if we put y = x², we havey² + qy + s = 0,from which we find y = ½ {−q ± √(q² − 4s) }, and thereforex = ±√½ {−q ± √(q² − 4s) }.3. When a biquadratic equation has all its terms, its resolution may be always reduced to that of a cubic equation. There are various methods by which such a reduction may be effected. The following was first given by Leonhard Euler in thePetersburg Commentaries, and afterwards explained more fully in hisElements of Algebra.We have already explained how an equation which is complete in its terms may be transformed into another of the same degree, but which wants the second term; therefore any biquadratic equation may be reduced to this form,y4+ py² + qy + r = 0,where the second term is wanting, and where p, q, r denote any known quantities whatever.That we may form an equation similar to the above, let us assume y = √a + √b + √c, and also suppose that the letters a, b, c denote the roots of the cubic equationz³ + Pz² + Qz − R = 0;then, from the theory of equations we havea + b + c = −P, ab + ac + bc = Q, abc = R.We square the assumed formulay = √a + √b + √c,and obtain y² = a + b + c + 2(√ab + √ac + √bc);or, substituting −P for a + b + c, and transposing,y² + P = 2(√ab + √ac + √bc).Let this equation be also squared, and we havey4+ 2Py² + P² = 4 (ab + ac + bc) + 8 (√a²bc + √ab²c + √abc²);and since ab + ac + bc = Q,and √a²bc + √ab²c + √abc² = √abc (√a + √b + √c) = √R·y,the same equation may be expressed thus:y4+ 2Py² + P² = 4Q + 8√R·y.Thus we have the biquadratic equationy4+ 2Py² − 8√R·y + P² − 4Q = 0,one of the roots of which is y = √a + √b + √c, while a, b, c are the roots of the cubic equation z³ + Pz² + Qz − R = 0.4. In order to apply this resolution to the proposed equation y4+ py² + qy + r = 0, we must express the assumed coefficients P, Q, R by means of p, q, r, the coefficients of that equation. For this purpose let us compare the equationsy4+ py² + qy + r = 0,y4+ 2Py² − 8√Ry + P² − 4Q = 0,and it immediately appears that2P = p, −8√R = q, P² − 4Q = r;and from these equations we findP = ½ p, Q =1⁄16(p² − 4r), R =1⁄64q².Hence it follows that the roots of the proposed equation are generally expressed by the formulay = √a + √b + √c;where a, b, c denote the roots of this cubic equation,z³ +pz² +p² − 4rz −q²= 0.21664But to find each particular root, we must consider, that as the square root of a number may be either positive or negative, so each of the quantities √a, √b, √c may have either the sign + or − prefixed to it; and hence our formula will give eight different expressions for the root. It is, however, to be observed, that as the product of the three quantities √a, √b, √c must be equal to √R or to −1⁄8q; when q is positive, their product must be a negative quantity, and this can only be effected by making either one or three of them negative; again, when q is negative, their product must be a positive quantity; so that in this case they must either be all positive, or two of them must be negative. These considerations enable us to determine that four of the eight expressions for the root belong to the case in which q is positive, and the other four to that in which it is negative.5. We shall now give the result of the preceding investigation in the form of a practical rule; and as the coefficients of the cubic equation which has been found involve fractions, we shall transform it into another, in which the coefficients are integers, by supposing z = ¼ v. Thus the equationz³ +pz² +p² − 4rz −q²= 021664becomes, after reduction,v³ + 2pv² + (p² − 4r) v − q² = 0;it also follows, that if the roots of the latter equation are a, b, c, the roots of the former are ¼ a, ¼ b, ¼ c, so that our rule may now be expressed thus:Let y4+ py² + qy + r = 0 be any biquadratic equation wanting its second term. Form this cubic equationv³ + 2pv² + (p² − 4r) v − q² = 0,and find its roots, which let us denote by a, b, c.Then the roots of the proposed biquadratic equation are,when q is negative,when q is positive,y = ½ (√a + √b + √c),y = ½ (−√a − √b − √c),y = ½ (√a − √b − √c),y = ½ (−√a + √b + √c),y = ½ (−√a + √b − √c),y = ½ (√a − √b + √c),y = ½ (−√a − √b + √c),y = ½ (√a + √b − √c).See also below,Theory of Equations, § 17 et seq.
I.Simultaneous Equations.
Simultaneous equations which involve the second and higher powers of the unknown may be impossible of solution. No general rules can be given, and the solution of any particular problem will largely depend upon the student’s ingenuity. Here we shall only give a few typical examples.
1.Equations which may be reduced to linear equations.—Ex.To solve x(x − a) = yz, y (y − b) = zx, z (z − c) = xy. Multiply the equations by y, z and x respectively, and divide the sum by xyz; then
(1).
Multiply by z, x and y, and divide the sum by xyz; then
(2).
From (1) and (2) by cross multiplication we obtain
(3).
Substituting for x, y and z in x (x − a) = yz we obtain
and therefore x, y and z are known from (3). The same artifice solves the equations x² − yz = a, y² − xz = b, z² − xy = c.
2.Equations which are homogeneous and of the same degree.—These equations can be solved by substituting y = mx. We proceed to explain the method by an example.
Ex.To solve 3x² + xy + y² = 15, 31xy − 3x² − 5y² = 45. Substituting y = mx in both these equations, and then dividing, we obtain 31m − 3 − 5m² = 3 (3 + m + m²) or 8m² − 28m + 12 = 0. The roots of this quadratic are m = ½ or 3, and therefore 2y = x, or y = 3x.
Taking 2y = x and substituting in 3x² + xy + y² = 0, we obtain y² (12 + 2 + 1) = 15; ∴ y² = 1, which gives y = ±1, x = ±2. Taking the second value, y = 3x, and substituting for y, we obtain x² (3 + 3 + 9) = 15; ∴ x² = 1, which gives x = ±1, y = ±3. Therefore the solutions are x = ±2, y = ±1 and x = ±1, y = ±3. Other artifices have to be adopted to solve other forms of simultaneous equations, for which the reader is referred to J.J. Milne,Companion to Weekly Problem Papers.
II.Indeterminate Equations.
1. When the number of unknown quantities exceeds the number of equations, the equations will admit of innumerable solutions, and are therefore said to beindeterminate. Thus if it be required to find two numbers such that their sum be 10, we have two unknown quantities x and y, and only one equation, viz. x + y = 10, which may evidently be satisfied by innumerable different values of x and y, if fractional solutions be admitted. It is, however, usual, in such questions as this, to restrict values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions,
x = 1, 2, 3, 4, 5, 6, 7, 8, 9;y = 9, 8, 7, 6, 5, 4, 3, 2, 1;
x = 1, 2, 3, 4, 5, 6, 7, 8, 9;
y = 9, 8, 7, 6, 5, 4, 3, 2, 1;
which indeed may be reduced to five; for the first four become the same as the last four, by simply changing x into y, and the contrary. This branch of analysis was extensively studied by Diophantus, and is sometimes termed the Diophantine Analysis.
2. Indeterminate problems are of different orders, according to the dimensions of the equation which is obtained after all the unknown quantities but two have been eliminated by means of the given equations. Those of the first order lead always to equations of the form
ax ± by = ±c,
where a, b, c denote given whole numbers, and x, y two numbers to be found, so that both may be integers. That this condition may be fulfilled, it is necessary that the coefficients a, b have no common divisor which is not also a divisor of c; for if a = md and b = me, then ax + by = mdx + mey = c, and dx + ey = c/m; but d, e, x, y are supposed to be whole numbers, therefore c/m is a whole number; hence m must be a divisor of c.
Of the four forms expressed by the equation ax ± by = ±c, it is obvious that ax + by = −c can have no positive integral solutions. Also ax − by = −c is equivalent to by − ax = c, and so we have only to consider the forms ax ± by = c. Before proceeding to the general solution of these equations we will give a numerical example.
To solve 2x + 3y = 25 in positive integers. From the given equationwe have x = (25 − 3y) / 2 = 12 − y − (y − 1) / 2. Now, since x must be a whole number, it follows that (y − 1)/2 must be a whole number. Let us assume (y − 1) / 2 = z, then y = 1 + 2z; and x = 11 − 3z, where z might be any whole number whatever, if there were no limitation as to the signs of x and y. But since these quantities are required to be positive, it is evident, from the value of y, that z must be either 0 or positive, and from the value of x, that it must be less than 4; hence z may have these four values, 0, 1, 2, 3.
3. We shall now give the solution of the equation ax − by = c in positive integers.
Convert a/b into a continued fraction, and let p/q be the convergent immediately preceding a/b, then aq − bp = ±1 (seeContinued Fraction).
(α) If aq − bp = 1, the given equation may be written
ax − by = c (aq − bp);∴ a (x − cq) = b (y − cp).
Since a and b are prime to one another, then x − cq must be divisible by b and y − cp by a; hence
(x − cq) / b = (y − cq) / a = t.
That is, x = bt + cq and y = at + cp.
Positive integral solutions, unlimited in number, are obtained by giving t any positive integral value, and any negative integral value, so long as it is numerically less than the smaller of the quantities cq/b, cp/a; t may also be zero.
(β) If aq − bp = −1, we obtain x = bt − cq, y = at − cp, from which positive integral solutions, again unlimited in number, are obtained by giving t any positive integral value which exceeds the greater of the two quantities cq/b, cp/a.
If a or b is unity, a/b cannot be converted into a continued fraction with unit numerators, and the above method fails. In this case the solutions can be derived directly, for if b is unity, the equation may be written y = ax − c, and solutions are obtained by giving x positive integral values greater than c/a.
4. To solve ax + by = c in positive integers. Converting a b into a continued fraction and proceeding as before, we obtain, in the case of aq − bp = 1,
x = cq − bt, y = at − cp.
Positive integral solutions are obtained by giving t positive integral values not less than cp/a and not greater than cq/b.
In this case the number of solutions is limited. If aq − bp = −1 we obtain the general solution x = bt − cq, y = cp − at, which is of the same form as in the preceding case. For the determination of the number of solutions the reader is referred to H.S. Hall and S.R. Knight’sHigher Algebra, G. Chrystal’sAlgebra, and other text-books.
5. If an equation were proposed involving three unknown quantities, as ax + by + cz = d, by transposition we have ax + by = d − cz, and, putting d − cz = c′, ax + by = c′. From this last equation we may find values of x and y of this form,
x = mr + nc′, y = mr + n′c′,or x = mr + n (d − cz), y = m′r + n′ (d − cz);
where z and r may be taken at pleasure, except in so far as the values of x, y, z may be required to be all positive; for from such restriction the values of z and r may be confined within certain limits to be determined from the given equation. For more advanced treatment of linear indeterminate equations seeCombinatorial Analysis.
6. We proceed to indeterminate problems of the second degree: limiting ourselves to the consideration of the formula y² = a + bx + cx², where x is to be found, so that y may be a rational quantity. The possibility of rendering the proposed formula a square depends altogether upon the coefficients a, b, c; and there are four cases of the problem, the solution of each of which is connected with some peculiarity in its nature.
Case1. Let a be a square number; then, putting g² for a, we have y² = g² + bx + cx². Suppose √(g² + bx + cx²) = g + mx; then g² + bx + cx² = g² + 2gmx + m²x², or bx + cx² = 2gmx + m²x², that is, b + cx = 2gm + m²x; hence
Case 2. Let c be a square number = g²; then, putting √(a + bx + g²x²) = m + gx, we find a + bx + g²x² = m² + 2mgx + g²x², or a + bx = m² + 2mgx; hence we find
Case 3. When neither a nor c is a square number, yet if the expression a + bx + cx² can be resolved into two simple factors, as f + gx and h + kx, the irrationality may be taken away as follows:—
Assume √(a + bx + cx²) = √{ (f + gx) (h + kx) } = m (f + gx), then (f + gx) (h + kx) = m² (f + gx)², or h + kx = m² (f + gx); hence we find
and in all these formulae m may be taken at pleasure.
Case 4. The expression a + bx + cx² may be transformed into a square as often as it can be resolved into two parts, one of which is a complete square, and the other a product of two simple factors; for then it has this form, p² + qr, where p, q and r are quantities which contain no power of x higher than the first. Let us assume √(p² + qr) = p + mq; thus we have p² + qr = p² + 2mpq + m²q² and r = 2mp + m²q, and as this equation involves only the first power of x, we may by proper reduction obtain from it rational values of x and y, as in the three foregoing cases.
The application of the preceding general methods of resolution to any particular case is very easy; we shall therefore conclude with a single example.
Ex.It is required to find two square numbers whose sum is a given square number.
Let a² be the given square number, and x², y² the numbers required; then, by the question, x² + y² = a², and y = √(a² − x²). This equation is evidently of such a form as to be resolvable by the method employed in case 1. Accordingly, by comparing √(a² − x²) with the general expression √(g² + bx + cx²), we have g = a, b = 0, c = −1, and substituting these values in the formulae, and also −n for +m, we find
If a = n² + 1, there results x = 2n, y = n² − 1, a = n² + 1. Hence if r be an even number, the three sides of a rational right-angled triangle are r, (½ r)² − 1, (½ r)² + 1. If r be an odd number, they become (dividing by 2) r, ½ (r² − 1), ½ (r² + 1).
For example, if r = 4, 4, 4 − 1, 4 + 1, or 4, 3, 5, are the sides of a right-angled triangle; if r = 7, 7, 24, 25 are the sides of a right-angled triangle.
III.Cubic Equations.
1. Cubic equations, like all equations above the first degree, are divided into two classes: they are said to bepurewhen they contain only one power of the unknown quantity; andadfectedwhen they contain two or more powers of that quantity.
Pure cubic equations are therefore of the form x³ = r; and hence it appears that a value of the simple power of the unknown quantity may always be found without difficulty, by extracting the cube root of each side of the equation. Let us consider the equation x³ − c³ = 0 more fully. This is decomposable into the factors x − c = 0 and x² + cx + c² = 0. The roots of this quadratic equation are ½ (−1 ± √−3) c, and we see that the equation x³ = c³ has three roots, namely, one real root c, and two imaginary roots ½ (−1 ± √−3) c. By making c equal to unity, we observe that ½ (−1 ± √−3) are the imaginary cube roots of unity, which are generally denoted by ω and ω², for it is easy to show that (½ (−1 − √−3))² = ½ (−1 + √−3).
2. Let us now consider such cubic equations as have all their terms, and which are therefore of this form,
x³ + Ax² + Bx + C = 0,
where A, B and C denote known quantities, either positive or negative.
This equation may be transformed into another in which the second term is wanting by the substitution x = y − A/3. This transformation is a particular case of a general theorem. Let xn+ Axn−1+ Bxn−2... = 0. Substitute x = y + h; then (y + h)n+ A (y + h)n−1... = 0. Expand each term by the binomial theorem, and let us fix our attention on the coefficient of yn−1. By this process we obtain 0 = yn+ yn−1(A + nh) + terms involving lower powers of y.
Now h can have any value, and if we choose it so that A + nh = 0, then the second term of our derived equation vanishes.
Resuming, therefore, the equation y³ + qy + r = 0, let us suppose y = v + z; we then have y³ = v³ + z³ + 3vz (v + z) = v³ + z³ + 3vzy, and the original equation becomes v³ + z³ + (3vz + q) y + r = 0. Now v and z are any two quantities subject to the relation y = v + z, and if we suppose 3vz + q = 0, they are completely determined. This leads to v³ + z³ + r = 0 and 3vz + q = 0. Therefore v³ and z³ are the roots of the quadratic t² + rt − q²/27 = 0. Therefore
Thus we have obtained a value of the unknown quantity y, in terms of the known quantities q and r; therefore the equation is resolved.
3. But this is only one of three values which y may have. Let us, for the sake of brevity, put
A = −½ r + √(1⁄27q³ + ¼ r²), B = −½ r − √(1⁄27q³ + ¼ r²),
Then, from what has been shown (§ 1), it is evident that v and z have each these three values,
v =3√A, v = α3√A, v = β3√A;z =3√B, z = α3√B, z = β3√B.
To determine the corresponding values of v and z, we must consider that vz = −1⁄3q =3√(AB). Now if we observe that αβ = 1, it will immediately appear that v + z has these three values,
v + z =3√A +3√B,v + z = α3√A + β3√B,v + z = β3√A + α3√B,
which are therefore the three values of y.
The first of these formulae is commonly known by the name of Cardan’s rule (seeAlgebra:History).
The formulae given above for the roots of a cubic equation may be put under a different form, better adapted to the purposes of arithmetical calculation, as follows:—Because vz = −1⁄3q, therefore z = −1⁄3q × 1/v = −1⁄3q /3√A; hence v + z =3√A −1⁄3q /3√A: thus it appears that the three values of y may also be expressed thus:
y =3√A −1⁄3q /3√Ay = α3√A −1⁄3qβ /3√Ay = β3√A −1⁄3qα /3√A.
See below,Theory of Equations, §§ 16 et seq.
IV.Biquadratic Equations.
1. When a biquadratic equation contains all its terms, it has this form,
x4+ Ax³ + Bx² + Cx + D = 0,
where A, B, C, D denote known quantities.
We shall first consider pure biquadratics, or such as contain only the first and last terms, and therefore are of this form, x4= b4. In this case it is evident that x may be readily had by two extractions of the square root; by the first we find x² = b², and by the second x = b. This, however, is only one of the values which x may have; for since x4= b4, therefore x4− b4= 0; but x4− b4may be resolved into two factors x² − b² and x² + b², each of which admits of a similar resolution; for x² − b² = (x − b)(x + b) and x² + b² = (x − b√−1)(x + b√−1). Hence it appears that the equation x4− b4= 0 may also be expressed thus,
(x − b) (x + b) (x − b√−1) (x + b√−1) = 0;
so that x may have these four values,
+b, −b, +b√−1, −b√−1,
two of which are real, and the others imaginary.
2. Next to pure biquadratic equations, in respect of easiness of resolution, are such as want the second and fourth terms, and therefore have this form,
x4+ qx² + s = 0.
These may be resolved in the manner of quadratic equations; for if we put y = x², we have
y² + qy + s = 0,
from which we find y = ½ {−q ± √(q² − 4s) }, and therefore
x = ±√½ {−q ± √(q² − 4s) }.
3. When a biquadratic equation has all its terms, its resolution may be always reduced to that of a cubic equation. There are various methods by which such a reduction may be effected. The following was first given by Leonhard Euler in thePetersburg Commentaries, and afterwards explained more fully in hisElements of Algebra.
We have already explained how an equation which is complete in its terms may be transformed into another of the same degree, but which wants the second term; therefore any biquadratic equation may be reduced to this form,
y4+ py² + qy + r = 0,
where the second term is wanting, and where p, q, r denote any known quantities whatever.
That we may form an equation similar to the above, let us assume y = √a + √b + √c, and also suppose that the letters a, b, c denote the roots of the cubic equation
z³ + Pz² + Qz − R = 0;
then, from the theory of equations we have
a + b + c = −P, ab + ac + bc = Q, abc = R.
We square the assumed formula
y = √a + √b + √c,
and obtain y² = a + b + c + 2(√ab + √ac + √bc);
or, substituting −P for a + b + c, and transposing,
y² + P = 2(√ab + √ac + √bc).
Let this equation be also squared, and we have
y4+ 2Py² + P² = 4 (ab + ac + bc) + 8 (√a²bc + √ab²c + √abc²);
and since ab + ac + bc = Q,
and √a²bc + √ab²c + √abc² = √abc (√a + √b + √c) = √R·y,
the same equation may be expressed thus:
y4+ 2Py² + P² = 4Q + 8√R·y.
Thus we have the biquadratic equation
y4+ 2Py² − 8√R·y + P² − 4Q = 0,
one of the roots of which is y = √a + √b + √c, while a, b, c are the roots of the cubic equation z³ + Pz² + Qz − R = 0.
4. In order to apply this resolution to the proposed equation y4+ py² + qy + r = 0, we must express the assumed coefficients P, Q, R by means of p, q, r, the coefficients of that equation. For this purpose let us compare the equations
y4+ py² + qy + r = 0,y4+ 2Py² − 8√Ry + P² − 4Q = 0,
and it immediately appears that
2P = p, −8√R = q, P² − 4Q = r;
and from these equations we find
P = ½ p, Q =1⁄16(p² − 4r), R =1⁄64q².
Hence it follows that the roots of the proposed equation are generally expressed by the formula
y = √a + √b + √c;
where a, b, c denote the roots of this cubic equation,
But to find each particular root, we must consider, that as the square root of a number may be either positive or negative, so each of the quantities √a, √b, √c may have either the sign + or − prefixed to it; and hence our formula will give eight different expressions for the root. It is, however, to be observed, that as the product of the three quantities √a, √b, √c must be equal to √R or to −1⁄8q; when q is positive, their product must be a negative quantity, and this can only be effected by making either one or three of them negative; again, when q is negative, their product must be a positive quantity; so that in this case they must either be all positive, or two of them must be negative. These considerations enable us to determine that four of the eight expressions for the root belong to the case in which q is positive, and the other four to that in which it is negative.
5. We shall now give the result of the preceding investigation in the form of a practical rule; and as the coefficients of the cubic equation which has been found involve fractions, we shall transform it into another, in which the coefficients are integers, by supposing z = ¼ v. Thus the equation
becomes, after reduction,
v³ + 2pv² + (p² − 4r) v − q² = 0;
it also follows, that if the roots of the latter equation are a, b, c, the roots of the former are ¼ a, ¼ b, ¼ c, so that our rule may now be expressed thus:
Let y4+ py² + qy + r = 0 be any biquadratic equation wanting its second term. Form this cubic equation
v³ + 2pv² + (p² − 4r) v − q² = 0,
and find its roots, which let us denote by a, b, c.
Then the roots of the proposed biquadratic equation are,
See also below,Theory of Equations, § 17 et seq.