INTRODUCTION.

ALGEBRA is a branch of the mathematics, which has for its object whatever can be expressed by number, either exactly or by approximation.

In this respect, and also in its employing arbitrary signs to denote the things of which it treats, it agrees with arithmetic. The analogy between the two sciences induced Sir Isaac Newton to denominate it Universal Arithmetic; but by the application of algebra to geometry, the science has acquired a new character and new powers, which render this appellation too limited, and not sufficiently descriptive of its nature. In its present state it is nearly alike related to arithmetic and geometry. In its application to both sciences, the reasoning is carried on by general symbols; its true character consists in this, that the results of its operations do not exhibit the individual values of the quantities which are the subject of investigation, such as we obtain in arithmetical calculations or geometrical constructions. They only indicate the operations, whether arithmetical or geometrical, which ought to be performed on the given quantities, to obtain the value of the quantities sought.

It has been a question much agitated, at what period and in what country was algebra invented? Who were the earliest writers on the subject? What was the progress of its improvement? And lastly, by what means, and at what period, was the science diffused over Europe?

It was a common opinion in the 17th century, that the ancient Greek mathematicians must have possessed an analysis of the nature of modern algebra, by which they discovered the theorems and solutions of the problems which we so much admire in their writings; but that they carefully concealed their instruments of investigation, and gave only the results, with synthetic demonstrations.

This opinion is, however, now exploded. A more intimate acquaintance with the writings of the ancient geometers has shown that they had an analysis, but that it was purely geometrical, and essentially different from our algebra.

Although there be no reason to suppose that the great geometers of antiquity derived any aid in their discoveries from the algebraic analysis, yet we find, that at a considerably later period it was known to a certain extent among the Greeks.

About the middle of the 4th century of the Christian era, a period when the mathematical sciences were on the decline, and their cultivators, instead of producing original works of genius, contented themselves with commentaries on the works of their more illustrious predecessors, there was a valuable addition made to the fabric of ancient learning. This was the treatise of Diophantus on arithmetic, which originally consisted of thirteen books, but of which only the first six, and an incomplete book on polygonal numbers, supposed to be the thirteenth, have descended to our times.

This precious fragment does not exhibit anything like a complete treatise on algebra. It rather is an application of its doctrines to a peculiar class of arithmetical questions, which belong to what is now called the indeterminate analysis.

Diophantus may have been the inventor of the Greek algebra, but it is more likely that its principles were not unknown before his time; and that, taking the science in the state he found it as the basis of his labours, he enriched it with new applications. The elegant solutions of Diophantus show that he possessed great address in the particular branch of which he treated, and that he was able to resolve determinate equations of the second degree. Probably this was the greatest extent to which the science had been carried among the Greeks. Indeed, in no country did it pass this limit, until it had been transplanted into Italy on the revival of learning.

The celebrated Hypatia, the daughter of Theon, composed a commentary on the work of Diophantus. This, however, is now lost, as well as a similar labour of this illustrious and ill-fated lady on the Conics of Apollonius. It is commonly known that she fell a sacrifice to the fury of a fanatical mob about the beginning of the 5th century.

About the middle of the 16th century, the work of Diophantus, written in the Greek language, was discovered at Rome in the Vatican library, where probably it had been carried from Greece when the Turks possessed themselves of Constantinople. A Latin translation, without the original text, was given to the world by Xylander in 1575; and a more complete translation, by Bachet de Mezeriac (one of the oldest members of the French academy), accompanied by a commentary, appeared in 1621. Bachet was eminently skilful in the indeterminate analysis, and therefore well qualified for the work he had undertaken; but the text of Diophantus was so much injured, that he was frequently obliged to divine the meaning of the author, or supply the deficiency. At a later period, the celebrated French mathematician Fermat, in addition to the commentary of Bachet, added notes of his own on the writings of the Greek algebraist. These are extremely valuable, on account of Fermat's profound knowledge of this particular branch of analysis. This edition, the best which exists, appeared in 1670.

Although the revival of the writings of Diophantus was an important event in the history of the mathematics, yet it was not from them that algebra became first known in Europe. This important invention, as well as the numeral characters and decimal arithmetic, was received from the Arabs. That ingenious people fully appreciated the value of the sciences; for at a period when all Europe was enveloped in the darkness of ignorance, they preserved from extinction the lamp of knowledge. They carefully collected the writings of the Greek mathematicians; they translated them into their language, and illustrated them with commentaries. It was through the medium of the Arabic tongue that the elements of Euclid were first introduced into Europe; and a part of the writings of Apollonius are only known at the present day by a translation from the Arabic, the Greek original being probably irrecoverably lost. The Arabs ascribe the invention of their algebra to one of their mathematicians, Mahomméd-Ben-Musa, or Moses, called also Mahomméd of Buziana, who flourished about the middle of the 9th century, in the reign of the Caliph Almamon.

It is certain that this person composed a treatise on Algebra, this subject, because an Italian translation was known at one time to have existed in Europe, although it be now lost. Fortunately, however, a copy of the Arabic original is preserved in the Bodleian Library at Oxford, bearing a date of transcription corresponding to the year 1342. The title-page identifies its author with the ancient Arabian. A marginal note concurs in this testimony, and farther declares the work to be the first treatise composed on algebra among the faithful; and the preface, besides indicating the author, intimates that he was encouraged by Almamon, commander of the faithful, to compile a compendious treatise of calculation by algebra.

The circumstance of this treatise professing to be only a compilation, and, moreover, the first Arabian work of the kind, has led to an opinion that it was collected from books in some other language. As the author was intimately acquainted with the astronomy and computations of the Hindoos, he may have derived his knowledge of algebra from the same quarter. Hence we may conclude, with some probability, that the Arabian algebra was originally derived from India.

The algebraic analysis having been once introduced among the Arabs, it was cultivated by their own writers. One of these, Mahomméd Abulwafa, who flourished in the last forty years of the 10th century, composed commentaries on the writers who had preceded him. He also translated the writings of Diophantus. Probably this was the first translation that was made of the Greek algebra into the Arabian tongue.

It is remarkable, that although the mathematical sciences were received with avidity, and sedulously cultivated during a long period, by the Arabians, yet in their hands they received hardly any improvement. It might have been expected that an acquaintance with the writings of Diophantus would have produced some change in their algebra. This, however, did not happen: their algebra continued nearly in the same state, from their earliest writer on the subject, to one of their latest, Behaudin, who lived between the years 953 and 1031.

Writers on the history of algebra were long under a mistake as to the time and manner of its introduction into Europe. It has now, however, been ascertained that the science was brought into Italy by Leonardo, a merchant of Pisa. This ingenious man resided in his youth in Barbary, and there learned the Indian method of accounting by the nine numeral characters. Commercial affairs led him to travel into Egypt, Syria, Greece, and Sicily, where we may suppose he made himself acquainted with everything known respecting numbers. The Indian mode of computation appeared to him to be by far the best. He accordingly studied it carefully; and, with this knowledge, and some additions of his own, and also taking some things from Euclid's Geometry, he composed a treatise on arithmetic. At that period algebra was regarded only as a part of arithmetic. It was indeed the sublime doctrine of that science; and under this view the two branches were handled in Leonardo's treatise, which was originally written in 1202, and again brought forward under a revised form in 1228. When it is considered that this work was composed two centuries before the invention of printing, and that the subject was not such as generally to interest mankind, we need not wonder that it was but little known; hence it has always remained in manuscript, as well as some others by the same author. Indeed it was not known to exist from an early period until the middle of the last century, when it was discovered in the Magliabecchian library at Florence.

The extent of Leonardo's knowledge was pretty much Algebra, the same as that of the preceding Arabian writers. He could resolve equations of the first and second degree, and he was particularly skilful in the Diophantine analysis. He was well acquainted with geometry, and he employed its doctrines in demonstrating his algebraic rules. Like the Arabian writers, his reasoning was expressed in words at length; a mode highly unfavourable to the progress of the art. The use of symbols, and the method of combining them so as to convey to the mind at a single glance a long process of reasoning, was an invention considerably later than Leonardo's time.

Considerable attention was given to the cultivation of algebra between the time of Leonardo and the invention of printing. It was publicly taught by professors. Treatises were composed on the subject; and two works of the oriental algebraists were translated from the Arabian language into Italian. One was entitled the Rule of Algebra, and the other was the oldest of all the Arabian treatises, that of Mahommed-Ben-Musa of Corsan.

The earliest printed book on algebra was composed by Lucas Paciolus, or Lucas de Burgo, a minorite friar. It was first printed in 1494, and again in 1523. The title is Summa de Arithmetica, Geometria, Proportioni, et Propor- tionalita.

This is a very complete treatise on arithmetic, algebra, and geometry, for the time in which it appeared. The author followed close on the steps of Leonardo; and, indeed, it is from this work that one of his lost treatises has been restored.

Lucas de Burgo's work is interesting, inasmuch as it shows the state of algebra in Europe about the year 1500; probably the state of the science was nearly the same in Arabia and Africa, from which it had been re- ceived.

The power of algebra as an instrument of research is in a very great degree derived from its notation, by which all the quantities under consideration are kept con- stantly in view; but in respect of convenience and brevity of expression, the algebraic analysis in the days of Lucas de Burgo was very imperfect: the only symbols employed were a few abbreviations of the words or names which occurred in the processes of calculation, a kind of short-hand, which formed a very imperfect substitute for that compactness of expression which has been attained by the modern notation.

The application of algebra was also at this period very limited; it was confined almost entirely to the resolution of certain questions of no great interest about numbers. No idea was then entertained of that extensive applica- tion which it has received in modern times.

The knowledge which the early algebraists had of their science was also circumscribed; it extended only to the resolution of equations of the first and second degree; and they divided the last into cases, each of which was resolved by its own particular rule. The important ana- lytical fact, that the resolution of all the cases of a prob- lem may be comprehended in a single formula, which may be obtained from the solution of one of its cases, merely by a change of the signs, was not then known: in- deed it was long before this principle was fully compre- hended. Dr Halley expresses surprise, that a formula in optics which he had found, should by a mere change of the signs give the focus of both converging and diverging rays, whether reflected or refracted by convex or concave spec- ula or lenses; and Molyneux speaks of the universality of Halley's formula as something that resembled magic.

The rules of algebra may be investigated by its own principles, without any aid from geometry; and although in some cases the two sciences may serve to illustrate the doctrines of each other, there is now not the least neces- sity in the more elementary parts to call in the aid of the latter to the former. It was otherwise in former times. Lucas de Burgo found it to be convenient, after the ex- ample of Leonardo, to employ geometrical constructions to prove the truth of his rules for resolving quadratic equations, the nature of which he did not completely com- prehend; and he was induced by the imperfect nature of his notation to express his rules in Latin verses, which will not now be read with the satisfaction we receive from the perusal of the well-known poem, "the Loves of the Triangles."

As it was in Italy that algebra became first known in Europe, so it was there that it received its earliest im- provements. The science had been nearly stationary from the days of Leonardo to the time of Paciolus, a period of three centuries; but the invention of printing soon excited a spirit of improvement in all the mathematical sciences. Hitherto an imperfect theory of quadratic equations was all the extent to which it had been carried. At last this boundary was passed, and about the year 1505 a particu- lar case of equations of the third degree was resolved by Scipio Ferreus, a professor of mathematics in Bononia. This was an important step, because it showed that the difficulty of resolving equations of the higher orders, at least in the case of the third degree, was not insurmount- able, and a new field was opened for discovery. It was then the practice among the cultivators of algebra, when they advanced a step, to conceal it carefully from their contemporaries, and to challenge them to resolve arithme- tical questions, so framed as to require for their solution a knowledge of their own new-found rules. In this spirit did Ferreus make a secret of his discovery: he com- municated it, however, to a favourite scholar, a Venetian named Florido. About the year 1535 this person, having taken up his residence at Venice, challenged Tartalea of Brescia, a man of great ingenuity, to a trial of skill in the resolution of problems by algebra. Florido framed his questions so as to require for their solution a knowledge of the rule which he had learned from his preceptor Fer- reus; but Tartalea had, five years before this time, ad- vanced farther than Ferreus, and was more than a match for Florido. He therefore accepted the challenge, and a day was appointed when each was to propose to the other thirty questions. Before the time came, Tartalea had re- sumed the study of cubic equations, and had discovered the solution of two cases in addition to two which he knew before. Florido's questions were such as could be resolved by the single rule of Ferreus; while, on the con- trary, those of Tartalea could only be resolved by one or other of three rules, which he himself had found, but which could not be resolved by the remaining rule, which was also that known to Florido. The issue of the contest is easily anticipated; Tartalea resolved all his adversary's questions in two hours, without receiving one answer from him in return.

The celebrated Cardan was a contemporary of Tartalea. This remarkable person was a professor of mathematics at Milan, and a physician. He had studied algebra with great assiduity, and had nearly finished the printing of a book on arithmetic, algebra, and geometry; but being desirous of enriching his work with the discoveries of Tartalea, which at that period must have been the object of considerable attention among literary men in Italy, he endeavoured to draw from him a disclosure of his rules. Tartalea resis- ted for a time Cardan's entreaties. At last, overcome by his importunity, and his offer to swear on the holy Evange- lists, and by the honour of a gentleman, never to publish them, and on his promising on the faith of a Christian to Algebra commit them to cypher, so that even after his death they would be unintelligible to any one, he ventured with much hesitation to reveal to him his practical rules, which were expressed by some very bad Italian verses, themselves in no small degree enigmatical. He reserved, however, the demonstrations. Cardan was not long in discovering the reason of the rules, and he even greatly improved them, so as to make them in a manner his own. From the imperfect essays of Tartalea, he deduced an ingenious and systematic method of resolving all cubic equations whatsoever; but with a remarkable disregard for the principles of honour, and the oath he had taken, he published, in 1545, Tartalea's discoveries, combined with his own, as a supplement to a treatise on arithmetic and algebra, which he had published six years before. This work is remarkable for being the second printed book on algebra known to have existed.

In the following year Tartalea also published a work on algebra, which he dedicated to Henry VIII. king of England.

It is to be regretted that in many instances the authors of important discoveries have been overlooked, while the honours due to them have been transferred to others having only secondary pretensions. The formulae for the resolution of cubic equations are now called Cardan's rules, notwithstanding the prior claim of Tartalea. It must be confessed, however, that he evinced considerable selfishness in concealing his discovery; and although Cardan cannot be absolved from the charge of bad faith, yet it must be recollected that by his improvements in what Tartalea communicated to him, he made the discovery in some measure his own; and he had moreover the high merit of being the first to publish this important improvement in algebra to the world.

The next step in the progress of algebra was the discovery of a method of resolving equations of the fourth order. An Italian algebraist had proposed a question which could not be resolved by the newly invented rules, because it produced a biquadratic equation. Some supposed that it could not be at all resolved; but Cardan was of a different opinion: he had a pupil named Lewis Ferrari, a young man of great genius, and an ardent student in the algebraic analysis: to him Cardan committed the solution of this difficult question, and he was not disappointed. Ferrari not only resolved the question, but he also found a general method of resolving equations of the fourth degree, by making them depend on the solution of equations of the third degree.

This was another great improvement; and although the precise nature of an equation was not then fully understood, nor was it indeed until half a century later, yet, in the general resolution of equations, a point of progress was then reached which the utmost efforts of modern analyses have never been able to pass.

There was another Italian mathematician of that period who contributed somewhat to the improvement of algebra. This was Bombelli. He published a valuable work on the subject in 1572, in which he brought into one view what had been done by his predecessors. He explained the nature of the irreducible case of cubic equations, which had greatly perplexed Cardan, who could not resolve it by his rule; he showed that the rule would apply sometimes to particular examples, and that all equations of this case admitted of a real solution; and he made the important remark, that the algebraic problem to be resolved in this case corresponds to the ancient problem of the trisection of an angle.

There were two German mathematicians contemporary with Cardan and Tartalea, viz. Stifelius and Scheubelius. Their writings appeared about the middle of the 16th century, before they knew what had been done by the Italians. Their improvements were chiefly in the notation. Stifelius, in particular, introduced for the first time the characters which indicate addition and subtraction, and the symbol for the square root.

The first treatise on algebra in the English language was written by Robert Recorde, teacher of mathematics and practitioner in physic at Cambridge. At this period it was common for physicians to unite with the healing art the studies of mathematics, astrology, alchemy, and chemistry. This custom was derived from the Moors, who were equally celebrated for their skill in medicine and calculation. In Spain, where algebra was early known, the title of physician and algebraist were nearly synonymous. Accordingly, in the romance of Don Quixotte, when the bachelor Sanson Carasco was grievously wounded in his encounter with the knight, an algebraist was called in to heal his bruises.

Recorde published a treatise on arithmetic, which was dedicated to Edward VI.; and another on algebra, with this title, "The Whetstone of Wit," &c. Here, for the first time, the modern sign for equality was introduced.

By such gradual steps did algebra advance in improvement from its first introduction by Leonardo, each succeeding writer making some change for the better; but with the exception of Tartalea, Cardan, and Ferrari, hardly any one rose to the rank of an inventor. At length came Vieta, to whom this branch of mathematical learning, as well as others, is highly indebted. His improvements in algebra were very considerable; and some of his inventions, although not then fully developed, have yet been the gems of later discoveries. He was the first that employed general characters to represent known as well as unknown quantities. Simple as this step may appear, it has yet led to important consequences. He must also be regarded as the first that applied algebra to the improvement of geometry. The older algebraists had indeed resolved geometrical problems, but each solution was particular; whereas Vieta, by introducing general symbols, produced general formulæ, which were applicable to all problems of the same kind, without the trouble of going over the same process of analysis for each.

This happy application of algebra to geometry has produced great improvements: it led Vieta to the doctrine of angular sections, one of the most important of his discoveries, which is now expanded into the arithmetic or calculus of sines. He also improved the theory of algebraic equations, and he was the first that gave a general method of resolving them by approximation. As he lived between the years 1540 and 1603, his writings belong to the latter period of the 16th century. He printed them at his own expense, and liberally bestowed them on men of science.

The Flemish mathematician Albert Girard was one of the improvers of algebra. He extended the theory of equations somewhat farther than Vieta, but he did not completely unfold their composition; he was the first that showed the use of the negative sign in the resolution of geometrical problems, and he also first spoke of imaginary quantities, a subject not yet completely cleared up; and be inferred by induction that every equation has precisely as many roots as there are units in the number that expresses its degree. His algebra appeared in 1629.

The next great improver of algebra was Thomas Harriot, an Englishman. As an inventor he has been the boast of this country. The French mathematicians have accused the British of giving discoveries to him which were really due to Vieta. It is probable that some of these may be justly claimed for both, because each may have made the discovery for himself, without knowing what had been done by the other. Harriot's principal discovery, and indeed the most important ever made in algebra, was, that every equation may be regarded as formed by the product of as many simple equations as there are units in the number expressing its order. This important doctrine, now familiar to every student of algebra, was yet slowly developed: it was quite within the reach of Vieta, who unfolded it in part, but left its complete discovery to Harriot.

We have seen the very unnatural form in which algebra first appeared in Europe. The improvements of almost 400 years had not given its notation that compactness and elegance of which it is susceptible. Harriot made several changes in the notation, and added some new signs; he thus gave to algebra greater symmetry of form. Indeed, as it came from his hands, it differed but little from its state at the present time.

Oughtred, another early English algebraist, was a contemporary with Harriot, but lived long after him. He wrote a treatise on the subject, which was long taught in the universities.

In tracing the history of algebra, we have seen, that in the form under which it was received from the Arabs, it was hardly distinguishable as a peculiar mode of reasoning, because of the want of a suitable notation; and that, poor in its resources, its applicability was limited to the resolution of a small number of uninteresting numeral questions. We have followed it through different stages of improvement, and we are now arrived at a period when it was to acquire additional power as an instrument of analysis, and to admit of new and more extended applications. Vieta saw the great advantage that might be derived from the application of algebra to geometry. The essay he made in his theory of angular sections, and the rich mine of discovery thus opened, proved the importance of his labours. He did not fully explore it, but it has seldom happened that one man began and completed a discovery. He had, however, an able and illustrious successor in Descartes, who, employing in the study of algebra that high power of intellect with which he was endowed, not only improved it as an abstract science, but, more especially by its application to geometry, he laid the foundation of the great discoveries which have since so much engaged mathematicians, and made the last two centuries ever memorable in the history of the progress of the human mind.

Descartes's grand improvement was the application of algebra to the doctrine of curve lines. As in geography we refer every place on the earth's surface to the equator, and to a determinate meridian, so he referred every point of a curve to some line given by position. For example, in a circle, every point in the circumference might be referred to the diameter. The perpendicular from any point in the curve, and the distance of that perpendicular from the centre or from the extremity of a diameter, were lines which, although varying with every change of position in the point from which the perpendicular was drawn, yet had a determinate relation to each other, which was the same for all points in the curve, which depended on its nature, and which, therefore, served as a characteristic to distinguish it from all other curves.

The relations of lines drawn in this way could be readily expressed in algebraic symbols; and the combination of these constituted what is called the equation of the curve. This might serve as its definition; and from the equation by the processes of algebra, all the properties of the curve could be investigated.

Descartes's geometry (or, as it might have been named, Algebra, the application of algebra to geometry) appeared first in 1637. This was six years after the publication of Harriot's discoveries, which was a posthumous work. Descartes availed himself of some of Harriot's views, particularly the manner of generating an equation without acknowledgement; and on this account Dr Wallis, in his algebra, has reflected with considerable severity on the French algebraist.

This spirit has engendered a corresponding eagerness in the French mathematicians to defend him. Montucla, in his history of the mathematics, has evinced a strong national prejudice in his favour; and, as usually happens, in order to exalt him, he hardly does justice to Harriot, the idol of his adversaries.

In treating of the claims of algebra and geometry to be considered as kindred sciences, a question arises, why was this relation not sooner perceived and appreciated? The sciences of geometry and algebra have each had a distinct origin. The former is the more ancient, and no doubt for this reason, that its principles are less removed from the ordinary affairs of men. The subjects of geometry, extension, and figure are continually presented to attention; and the elements of the science are to a certain extent employed in the most ordinary arts of life. We cannot sufficiently admire the ingenuity with which the natural geometry of the early times had been wrought up into a system more than two thousand years ago; but when we consider that its assistance was wanted in the partition of land, in the erection of houses and temples, and numberless other cases, we need not wonder at the early progress of the science among such an ingenious people as the ancient Greeks.

Algebra, however, is a more refined speculation. Its first object was number; but the properties of number are more recondite than those of extension and figure.

In geometry, the objects of our attention are the very figures themselves; but in algebra, the subjects of our reasonings are represented by symbols, which have no resemblance to the things they represent; hence it is not wonderful that algebra should have a later origin, and that it should have been slower in its progress towards perfection.

Notwithstanding the different origin of geometry and algebra, and their long-continued separate existence, like some chemical substances of different natures, they have a strong affinity; and, when united, their new properties are entirely different from those which belong to each apart. By their union, a new science was created, and new instruments of invention furnished, vastly more powerful than any possessed by the sciences apart.

The new views which the labours of Vieta, Harriot, and Descartes opened in geometry and algebra were seized with avidity by the powerful minds of men eager in the pursuit of real knowledge. Accordingly, we find in the seventeenth century a whole host of writers on algebra, or algebra combined with geometry.

Our limits will not allow us to enter minutely into the claims which each has on the gratitude of posterity. Indeed, in pure algebra the new inventions were not so conspicuous as the discoveries made by its applications to geometry, and the new theories which were suggested by their union. The refined speculations of Kepler concerning the solids formed by the revolutions of curvilinear figures, the Geometry of Indivisibles by Cavalieri, the Arithmetic of Infinites of Wallis, and, above all, the Method of Fluxions of Newton, and the Differential and Integral Calculus of Leibnitz, are fruits of the happy union. All these were agitated incessantly by their inventors and Algebra. contemporaries; such men as Barrow, James Gregory, Wren, Cotes, Taylor, Halley, De Moivre, MacLaurin, Stirling, and others, in this country; and abroad by Roberval, Fermat, Huygens, the two Bernoullis, Herman, Pascal, and many others.

It is at this period, then, that our sketch of the history of algebra, at least in Europe, must terminate, because of the great number of writers who have in one way or other elucidated or improved different parts of the subject, either directly, or when treating of collateral theories.

We have been as copious as our limits would permit on the early history, because it presents the interesting spectacle of the progress of a science from an almost imperceptible beginning, until it has attained a magnitude too great to be fully grasped by the human mind.

Of the Indian Algebra.

The attention of the learned has, within the last thirty years, been called to a branch of the history of algebra, in no small degree interesting; we mean the cultivation of the science to a considerable extent, and at a remote period, in India.

We are indebted, we believe, to Mr Reuben Burrow for some of the earliest notices which reached Europe on this very curious subject. His eagerness to illustrate the history of the mathematical sciences led him to collect oriental manuscripts, some of which, in the Persian language, with partial translations, were bequeathed to his friend Mr Dalby of the Royal Military College, who communicated them to such as took an interest in the subject, about the year 1800.

In the year 1813 Mr Edward Strachey published in this country a translation from the Persian of the Bija Ganita (or Vija Ganita), a Hindoo treatise on algebra; and in 1816 Dr John Taylor published at Bombay a translation of Lilavati (or Lilavati), from the Sanscrit original. This last is a treatise on arithmetic and geometry, and both are the production of an oriental algebraist, Bhascara Acharya. Lastly, in 1817 there came out a work entitled Algebra, Arithmetic, and Mensuration, from the Sanscrit of Brahmagupta and Bhascara, translated by Henry Thomas Colebrooke, Esq. This contains four different treatises, originally written in Sanscrit verse, viz., the Vija Ganita and Lilavati of Bhascara Acharya, and the Ganitadhyana and Cuttacodhyaya of Brahmagupta. The first two form the preliminary portion of Bhascara's Course of Astronomy, entitled Siddhanta Siromani, and the last two are the twelfth and eighteenth chapters of a similar course of astronomy, entitled Brahma-siddhanta.

The time when Bhascara wrote is fixed with great precision, by his own testimony and other circumstances, to a date that answers to about the year 1150 of the Christian era. The works of Brahmagupta are extremely rare, and the age in which he lived is less certain. Mr Davis, an oriental scholar, who first gave the public a correct view of the astronomical computations of the Hindoos, is of opinion that he lived in the 7th century; and Dr William Hunter, another diligent inquirer into Indian science, assigns the year 628 of the Christian era as about the time he flourished. From various arguments, Mr Colebrooke concludes that the age of Brahmagupta was antecedent to the earliest dawn of the culture of the sciences among the Arabians, so that the Hindoos must have possessed algebra before it was known to that nation.

Brahmagupta's treatise is not, however, the earliest work known to have been written on this subject. Ganessa, a distinguished astronomer and mathematician, and the most eminent scholiast of Bhascara, quotes a passage from a much older writer, Arya-Bhatta, specifying algebra under the designation of Vija, and making separate mention of Cuttaca, a problem subservient to the resolution of indeterminate problems of the first degree. He is understood by another of Bhascara's commentators to be at the head of the older writers. They appear to have been able to resolve quadratic equations, by the process of completing the square; and hence Mr Colebrooke presumes that the treatise of Arya-Bhatta then extant extended to quadratic equations in the determinate analysis, and to indeterminate equations of the first degree, if not to those of the second likewise, as most probably it did.

Considering the proficiency of Arya-Bhatta in astronomical science, and adverting to the fact of his having written on algebra, and being placed at the head of algebraists when the commentators of extant treatises have occasion to mention the early and original writers on this branch of science, he may be regarded as the great improver of the analytic art in India, and likely to have been the person by whom it was carried to the pitch it was found to have attained among the Hindoos, and at which it was observed to be nearly stationary through the long lapse of ages which have since passed; the later additions being few and unessential in the writings of Brahmagupta, of Bhascara, and of Jyamarija, though they lived at intervals of centuries from each other.

The exact period when Arya-Bhatta lived cannot be determined with certainty; but Mr Colebrooke thinks it probable that this earliest of known Hindoo algebraists wrote as far back as the fifth century of the Christian era, and perhaps earlier. He was therefore nearly as ancient as the Grecian algebraist Diophantus, who is reckoned to have flourished in the time of the emperor Julian, or about A.D. 360. Supposing then the Hindoo and Greek algebraists to be nearly of the same antiquity, it must be conceded in favour of the former, that he was farthest advanced in the science, since he knew how to resolve equations containing several unknown quantities; now it does not appear that Diophantus could do this. He also had a general method for indeterminate equations, of at least the first degree, to a knowledge of which the Grecian algebraist had certainly not attained.

It appears from the Hindoo treatises on algebra, that they understood well the arithmetic of surd roots; that they were aware of the infinite quotient resulting from the division of finite quantity by cypher; that they knew the general resolution of equations of the second degree, and had touched on those of higher denomination, resolving them in particular cases, and in those in which the solution may be effected in the manner of quadratics; that they had found a general solution of indeterminate equations of the first degree, and a method for deriving a multitude of answers to problems of the second degree, when one solution was obtained by trials; now this is as near an approach to a general solution of such problems as was made until the time of Lagrange. The Hindoos had also attempted to solve indeterminate equations of higher orders, but, as might be expected, with very little success.

We have seen how long it was before algebra was applied to geometry in Europe; but the Hindoos not only applied algebra both to astronomy and geometry, but conversely applied geometry to the demonstration of algebraic rules; and indeed they cultivated algebra much more and with greater success than geometry, as appears by the low state of their knowledge of the one and the high pitch of their attainments in the other.

Mr Colebrooke has instituted a comparison between the Indian algebraist and Diophantus, and found reason to conclude, that in the whole science the latter is very far behind the former. He says, the points in which the Hindoo algebra appears particularly distinguished from the Greek art, besides a better and more convenient algorithm; 1st, the management of equations of more than one unknown quantity; 2d, the resolution of equations of a higher order, in which, if they achieved little, they had at least the merit of the attempt, and anticipated a modern discovery in the resolution of biquadratics; 3d, general methods for the resolution of indeterminate problems of the first and second degrees, in which they went far indeed beyond Diophantus, and anticipated discoveries of modern algebraists; 4th, the application of algebra to astronomical investigations and geometrical demonstration, in which also they hit upon some matters which have been re-invented in modern times.

When we consider that algebra made little or no progress among the Arabians, a most ingenious people, and particularly devoted to the study of the sciences, and that centuries elapsed from its first introduction into Europe until it reached any considerable degree of perfection, we may reasonably conjecture, that it may have existed in one shape or other in India long before the time of Arya-Bhatta: indeed, from its close connection with their doctrines of astronomy, it may be supposed to have descended from a very remote period, along with that science. The late learned Professor Playfair took a great interest in this curious and interesting subject; and, adopting the opinion of Bailly, the eloquent author of the Astronomie Indienne, he with great ingenuity attempted to prove, in a Memoir on the Astronomy of the Brahmins, that the observations on which the Indian astronomy is founded were of great antiquity, indeed more than 3000 years before the Christian era. Again, in a later memoir, On the Trigonometry of the Brahmins, he endeavoured to establish, that the origin of the mathematical sciences in Hindostan must be referred to an equally remote period. The same judicious writer has further considered this most curious subject in a Review of Strachey's Translation of Bija Gamiita (Edinburgh Review, No. 42), and again, in a Review of Colebrooke's work on the Indian algebra, to which we have so frequently adverted (Edinburgh Review, No. 57). This last article, published in 1817, may be supposed to contain the matured opinions of one of the most ardent, able, and we must say most candid, inquirers into the history of Hindoo mathematical science. There is here certainly an abatement of his first confidence in the opinions of Bailly on the Indian astronomy, and a corresponding caution in his own opinion as to the antiquity of the mathematical sciences. The very remote origin of the Indian astronomy had been strongly questioned by many in this country, and also on the Continent; particularly by Laplace, also by Delambre in his Histoire de l'Astronomie Ancienne, tome i. p. 400, &c., and again in Histoire de l'Astronomie du Moyen Age, Discours Préliminaire, p. 18, &c. where he speaks slightingly of their algebra; and in this country, Professor Leslie, in his very learned work on The Philosophy of Arithmetic, p. 225 and 226, calls the Lilavati "a very poor performance, containing merely a few scanty precepts couched in obscure memorial verses." We shall conclude this slight sketch of the history of Indian algebra with the last recorded sentiments of Professor Playfair on the mathematical science of India. "Among many subjects of wonder which the study of these ancient fragments cannot fail to suggest, it is not one of the least that algebra has existed in India, and has been cultivated for more than 1200 years, without any signal improvement, or the addition of any material discovery. The works of the ancient teachers of science have been commented on, elucidated, and explained with skill and learning; but no new methods have been invented, nor any new principle introduced. The method of resolving indeterminate problems, that constitute the highest merit of their analytical science, were known to Brahmagupta hardly less accurately than to Bhaskara; and they appear to have been understood even by Arya-Bhatta, more ancient by several centuries than either. A long series of scholastics display in their annotations great acuteness, intelligence, and judgment; but they never pass far beyond the line drawn by their predecessors, which probably seemed even to those learned and intelligent men as the barrier within which it was to be confined. In India, indeed, everything seems equally insurmountable, and truth and error are equally assured of permanence in the stations they have once occupied. The politics, the laws, the religion, the science, and the manners, seem all nearly the same as at the remotest period to which history extends. Is it because the power which brought about a certain degree of civilisation, and advanced science to a certain height, has either ceased to act, or has met with such a resistance as it is barely able to overcome? or is it because the discoveries which the Hindoos are in possession of are an inheritance from some more inventive and more ancient people, of whom no memorial remains but some of their attainments in science?"

Writers on Algebra, with the years in which they wrote or flourished.

Diophantus, Arithmeticon Libri sex, flourished, A.C. 360

First edition of his writings, 1575; the best, 1670.

Leonardo Bonacci (his works described in Costali)...1202

Lucas Paciolus, or De Burgo, Summa de Arithmetica, &c. ..................................................1470

Rudolph, Algebra ..............................................1522

Stifelius, Arithmetica Integra, &c. ..................................................1544

Cardan, Ars Magna quam vulgo Cossam vocant ........1545

Ferreus ..........................................................1545

Ferrari, (first resolved biquadratic equations) ..........1545

Tartalea, Questi et Inventioni diversi .........................1546

Scheubelinus, Algebra Compendiosa .......................1551

Recorde, Whetstone of Wit ..................................1557

Peletarius, De Occulta parte Numerorum .................1558

Buteo, De Logistica ...........................................1559

Ramus, Arithmetice Libri duo et totidem Algebrae ....1560

Pedro Nunez or Nonius, Libro de Algebra, &c. ..........1567

Jossalin, De Occulta parte Mathematicorum ..........1576

Bombelli .....................................................1579

Clavius .......................................................1580

Bernard Solignac, Arith. Libri ii. et Algebrae totidem .1580

Stevinus, Arithmétique, &c. aussi l'Algèbre ..............1585

Vieta, Opera Mathematica ..................................1600

Folius, Algebra, sive Liber de Rebus Occultis ..........1619

Van Ceulen ..................................................1619

Bachet, Diophantus cum Commentariis ....................1621

Albert Girard, Invention Nouvelle en Algèbre ..........1629

Ghetaldus, De Resolutione et Compositione Mathematica ..................................................1630

Descartes .....................................................1630

Harriot, Artis Analyticae Praxis ............................1631

Oughtred, Clavis Mathematicae .............................1631

Herigonus, Cursus Mathematicus ..........................1634

Cavalieri, Geometria Indivisibilibus Continuorum ..1635

Descartes, Geometria .......................................1637

Commentators on Descartes.—Franciscus à Schooten,

Florimond de Beaune, Erasmus Bertholinus,

Joh. Hudde, F. Rabuel, James Bernoulli, John de Witt, &c.

Roberval, De Recognitione Æquationum, &c. ..........1640

De Billy, Nova Geometricæ Clavis Algebra ...............1643 Algebra. Renaldinus, Opus Algebraicum flourished a.c. 1644 In addition to the preceding list of writers, which contains Algebra almost all of an early date, we shall add the following.

Pascal, in his works..................................................1654 Wallis, Arithmeticae Infinitorum..................................1655 — Algebra.................................................................1685 Slusius, Mesolabum....................................................1659 Rhonius, Algebra (translated into English)......................1659 Kinckhausen, used as a text-book by Sir I. Newton...........1661 Sir Isaac Newton, The Binomial Theorem.......................1666 Frenicle, Various papers in Mem. of F. Academy..............1666 Pell, translated and improved Rhonius' Algebra.................1668 James Gregory, Exercitationes Geometricae.....................1668 Mercator, Logarithmotechnia......................................1668 Branner...............................................................1668 Barrow, in Lectiones Geometricae...............................1669 Kersey, Elements of Algebra......................................1673 Prescot, Nouveaux Éléments de Mathématiques...............1675 Leibnitz, in Leipsic Acts, &c.....................................1677 Fermat, in Varia Opera Mathematica............................1679 Bulliadd, Opus Novum ad Arithmeticam Infinitorum..........1682 Tscherinhausen, in the Leipsic Acts...........................1683 Baker, Geometrical Key, &c......................................1684 Dr Halley, in Phil. Trans........................................1687 and 1694 Rolle, Une Méthode pour la Résolution des Equations Indéterminées..................................................1690 Raphson, Analysis Æquationum Universalis......................1690 Deschales, Cursus seu Mundus Mathematicus....................1690 De Lagny, various pieces on Equations..........................1692 Alexander, Synopsis Algebraica.................................1693 Ward, Compendium of Algebra...................................1695 Young Mathematician's Guide....................................1706 De Moivre, various Memoirs in Phil. Trans......................1697-1730 Sault, New Treatise of Algebra..................................1698 Christopher, De Constructione Æquationum......................1702 Ozanam, Nouveaux Éléments d'Algèbre.........................1704 Harris, Lexicon Technicum......................................1704 Guissée, Application de l'Algèbre à la Géométrie.............1705 Jones, Synopsis Palmariorum Matheseos.........................1706 Newton, Arithmetica Universalis.................................1707 L'Hôpital, Traité Analytique de Sections Coniques............1707 Reyneau, Analyse Démontrée....................................1708 Brooke Taylor, Methodus Incrementorum.........................1715 Stirling, Linea Tertii Ordinis...................................1717 Methodus Differentialis...........................................1730 Nicole on Cubic Equations, in Mém. Acad. des Sciences.....1717 SGravesande, Algebra.............................................1727 Wolfius, Algebra : Cursus Mathematicus.........................1732 Kirby, Arithmetic and Algebra..................................1735 James Gregory.....................................................1736 Simpson, Algebra and various works............................1740, 1742 Saunders on Algebra, 2 vols. 4to................................1740 La Caille, Algèbre in Leçons de Mathématiques.................1741 De Guin on the Roots of Equations, in Mém. Acad. des Sciences..................................................1741 Clairaut, Éléments d'Algèbre.....................................1746 Maclaurin, Algebra................................................1747 Fontaine, L'Art de Résoudre les Equations.......................1747 Donna Maria Gaetana Agnesi, Instituzioni Analitiche........1748 Boscovich, in Elementa Universae Matheseos...................1754 Castillon, Arithmetica Universalis Newtoni cum Commentario...1761 Emerson, Algebra, &c.............................................1763 Landen, Residual Analysis, &c...................................1764 Lagrange, Traité de la Résolution des Equations Numériques..1767 Euler, Algèbre.....................................................1770 Waring, Meditationes Algebraicae, &c............................1770, 1776 Soladini, Compendio d'Analisi...................................1775 Paoli, Elementi d'Algèbre.......................................1794

Arbogast, Calcul des Dérivations. The Bernoullis, Begnalt, Bertrand, Bezout, Bossuet, Burja, Brunacci, Babbage, Bridges, Bland, Budan, Bonnycastle, Burdon, Barlow. Cousin, Cauchy, Coignet, Carnot. Degraave, Dodson, Ditton. Frisius, Francocur, Frend. Gauss, Disquisitiones Arithmeticae. Hemischius, Hales, Hirsch, Hutton, Holdred. Kuhnius, Krumpp, Kaestner. Laloubre, Lorgna, Le Blond, Lee, Lacroix, Ludlam, Legendre, L'Huillier, Leroy. Mescher, Malebranche, Manfredi, Mascrez. Nicholson, Nieuwentit Analysis Infinitorum. Polleti, Poignard (on Magic Squares), Playfair. Rowning, Reimer. Surenam-Missery (on Impossible Quantities), Schonerus, Salignat. Trail, Tedenat, Thacker. Vilent, Vandermonde. Wells, Wilson, Wood, Woodhouse, Warren.

Writers on the History of Algebra.

Wallis in his Algebra; Montucla in Histoire des Mathématiques; Bossuet, Histoire des Mathématiques; Cossali, Origine, Trasporto in Italia, Primi Progressi in Essa dell'Algebra, 2 vols. printed in 1797; Hutton in his Dictionary, and more diffusely in his Tracts, vol. ii.

For the titles of works on Algebra, consult Murhard, Bibliotheca Mathematica; and for memoirs on algebra, in Academical Collections, see Reuss, Repertorium Commemorationum, tom. vii.

NOTATION AND EXPLANATION OF THE SIGNS.

I. In arithmetic there are ten characters, which being variously combined, according to certain rules, serve to denote all magnitudes whatever. But this method of expressing quantities, although of the greatest utility in every branch of the mathematics (for we must always have recourse to it in the different applications of that science to practical purposes), is yet found to be inadequate, taken by itself, to the more difficult cases of mathematical investigation; and it is therefore necessary, in many inquiries concerning the relations of magnitude, to have recourse to that more general mode of notation, and more extensive system of operations, which constitute the science of algebra.

In algebra quantities of every kind may be denoted by any characters whatever, but those commonly used are the letters of the alphabet; and as in every mathematical problem there are certain magnitudes given, in order to determine other magnitudes which are unknown, the first letters of the alphabet \(a, b, c\), &c., are used to denote known quantities, while those to be found are represented by \(x, y, z\), &c., the last letters of the alphabet.

2. The sign \(+\) (plus) denotes that the quantity before which it is placed is to be added to some other quantity. Thus, \(a + b\) denotes the sum of \(a\) and \(b\); \(3 + 5\) denotes the sum of 3 and 5, or 8.

The sign \(-\) (minus) signifies that the quantity before which it is placed is to be subtracted. Thus, \(a - b\) denotes the excess of \(a\) above \(b\); 6 — 2 is the excess of 6 above 2 or 4.

3. Quantities which have the sign \(+\) prefixed to them are called positive or affirmative; and such as have the sign \(-\) are called negative. When quantities are considered abstractedly, the terms positive and negative can only mean that such quantities are to be added or subtracted; for as it is impossible to conceive a number less than 0, it follows, that a negative quantity by itself is unintelligible. But, in considering the affections of magnitude, it appears, that in many cases a certain opposition may exist in the nature of quantities. Thus, a person's property may be considered as a positive quantity, and his debts as a negative quantity. Again, any portion of a line drawn to the right hand may be considered as positive, while a portion of the same line, continued in the opposite direction, may be taken as negative.

When no sign is prefixed to a quantity, + is always understood, or the quantity is to be considered as positive.

Quantities which have the same sign, either + or —, are said to have like signs. Thus, +a and +b have like signs, but +a and —b have unlike signs.

4. A quantity which consists of one term, is said to be simple; but if it consist of several terms, connected by the signs + or —, it is then said to be compound. Thus, +a and —c are simple quantities; and b+c, also a+b—d, are compound quantities.

5. To denote the product arising from the multiplication of quantities. If they be simple, they are either joined together, as if intended to form a word, or else the quantities are connected together, with the sign × interposed between every two of them. Thus, ab, or a×b, denotes the product of a and b; also abc, or a×b×c, denotes the product of a, b, and c; the latter method is used when the quantities to be multiplied are numbers. If some of the quantities to be multiplied be compound, each of them has a line drawn over it called a vinculum, and the sign × is interposed, as before. Thus, a×c+d×e−f denotes that a is to be considered as one quantity, the sum of c and d as a second, and the difference between e and f as a third; and that these three quantities are to be multiplied into one another. Instead of placing a line over such compound quantities as enter a product, it is now common among mathematical writers to inclose each of them between two parentheses, so that the last product may be otherwise expressed thus, a(c+d)(e−f); or thus, a×(c+d)×(e−f).

6. A number prefixed to a letter is called a numerical coefficient, and denotes how often that quantity is to be taken. Thus, 3a signifies that a is to be taken three times. When no number is prefixed, the coefficient is understood to be unity.

7. The quotient arising from the division of one quantity by another is expressed by placing the dividend above a line, and the divisor below it. Thus, \( \frac{12}{3} \) denotes the quotient arising from the division of 12 by 3, or 4; \( \frac{b}{a} \) denotes the quotient arising from the division of b by a. This expression of a quotient is also called a fraction.

8. The equality of two quantities is expressed by putting the sign = between them. Thus, a + b = c − d denotes that the sum of a and b is equal to the excess of c above d.

9. Simple quantities, or the terms of compound quantities, are said to be like, which consist of the same letter or letters. Thus, +ab and −5ab are like quantities, but +ab and +abb are unlike.

There are some other characters, which will be explained when we have occasion to use them; and in what follows we shall suppose that the operations of common arithmetic are sufficiently understood; for algebra, being an expansion of that science, ought not to be embarrassed by Algebra, the demonstration of its elementary rules.

**Sect. I.—Fundamental Operations.**

The primary operations in algebra are the same as in common arithmetic; namely, addition, subtraction, multiplication, and division; and from the various combinations of these four, all the others are derived.

**Problem I.—To Add Quantities.**

10. In addition there may be three cases; the quantities to be added may be like, and have like signs; or they may be like, and have unlike signs; or, lastly, they may be unlike.

**Case 1.** To add quantities which are like, and have like signs.

**Rule.** Add together the co-efficients of the quantities, prefix the common sign to the sum, and annex the letter or letters common to each term.

**Examples.**

Add together \( \begin{array}{c} + 7a \\ + 3a \\ + a \\ + 2a \end{array} \)

Add together \( \begin{array}{c} - 2ax \\ - ax \\ - 5ax \\ - 12ax \end{array} \)

Sum, \( +13a \) Sum, \( -20ax \)

**Case 2.** To add quantities which are like, but have unlike signs.

**Rule.** Add the positive co-efficients into one sum, and the negative ones into another; then subtract the least of these sums from the greatest, prefix the sign of the greatest to the remainder, and annex the common letter or letters as before.

**Examples.**

Add together \( \begin{array}{c} + 2ax \\ - ax \\ - 3ax \\ + 9ax \end{array} \)

Add together \( \begin{array}{c} + 6ab + 7 \\ - 4ab + 9 \\ + ab - 5 \\ + 7ab - 13 \end{array} \)

Sum of the pos. \( +11ax \) Sum of the pos. \( -10ab + 16 \)

Sum of the neg. \( -4ax \) Sum of the neg. \( -4ab - 18 \)

Sum required, \( +7ax \) Sum required, \( +10ab - 2 \)

\( ax + 2ax - xx - 4ab \)

\( -2ax + 3ax - 4xx + ab \)

\( 6ax - 5ax + 11zx + 3ab \)

Sum, \( 5aa \) 0 \( + 6zx \) Sum, 0

**Case 3.** To add unlike quantities.

**Rule.** Put down the quantities, one after another, in any order, with their signs and co-efficients prefixed.

**Examples.**

\( \begin{array}{c} 2a \\ 3b \\ -4c \end{array} \)

\( \begin{array}{c} ax + 2ay \\ bb - 3bz \end{array} \)

Sum, \( 2a + 3b - 4c \) Sum, \( ax + 2ay + bb - 3bz \)

**Prob. II.—To Subtract Quantities.**

11. **General Rule.** Change the signs of the quantities to be subtracted, or suppose them changed, and then add them to the other quantities, agreeably to the rules of addition. Examples.

From $5a - 12b$ Subtract $2a - 5b$

Remainder $3a - 7b$ $5xy - 2 + 8x - y$ $3xy - 8 - 8x - 3y$

From $6x - 8y + 3$ Subtract $2x + 9y - 2$

Remainder $4x - 17y + 5$ $aa - ax - yy$ $bb - by + zz$

$2xy + 6 + 16x + 2y$ $aa - ax - yy - bb + by - zz$

The reason of the rule for subtraction may be explained thus. Let it be required to subtract $2p - 3q$ from $m + n$. If we subtract $2p$ from $m + n$, there will remain $m + n - 2p$; but if we are to subtract $2p - 3q$, which is less than $2p$, it is evident that the remainder will be greater by a quantity equal to $3q$; that is, the remainder will be $m + n - 2p + 3q$; hence the reason of the rule is evident.

Prob. III.—To Multiply Quantities.

12. General Rule for the Signs. If the quantities to be multiplied have like signs, the sign of the product is $+$; but if they have unlike signs, the sign of the product is $-$.

The examples of multiplication may be referred to two cases; the first is when both the quantities are simple, and the second when one or both of them are compound.

Case 1. To multiply simple quantities.

Rule. Find the sign of the product by the general rule, and annex to it the product of the numeral co-efficients; then set down all the letters, one after another, as in one word.

Examples.

| Multiply | $+a$ | |---------|-----| | By | $+c$ | | Product | $+ac$ |

| $+5b$ | |-------| | $-4a$ | | $-2ab$ |

| $-3ax$ | |--------| | $-20ab$ | | $-21aabx$ |

| $-2ab$ | |--------| | $-3cx$ | | $+6abcz$ |

Case 2. To multiply compound quantities.

Rule. Multiply every term of the multiplicand by all the terms of the multiplier, one after another, by the preceding rule, and collect their products into one sum, which will be the product required.

Examples.

Multiply $4a - 2b + c$ By $3a$

Product $12aa - 6ab + 3ac$

$2x + y$ $x - 2y$

$2xe + xy$ $-4xy - 2yy$

$aa - ab + bb$ $a + b$

$a - b + e$ $a + b - e$

$aa - ab + ae$ $aa - ab + ac$

$+ab - bb + bc$ $-ac + bc - cc$

$aaa + bbb$ $aa \cdot bb + 2bc - cc$

The reason of the rules for the multiplication of quantities may be explained in the following manner:—Let it be required to multiply $a - b$ by $c - d$; because multiplication is a repeated addition of the multiplicand as often as the multiplier contains unity; therefore, $a - b$ is to be taken as often as there are units in $c - d$, and the sum will be the product required. Now, if $a - b$ be taken as often as there are units in $c$, the result will evidently exceed the product required, and that by a quantity equal to $a - b$, taken as often as there are units in $d$. But, from the nature of addition, $a - b$ taken as often as there are units in $c$, is $ca - cb$, and for the same reason, $a - b$ taken as often as there are units in $d$, is $da - db$; therefore, to obtain the product required, we must subtract $da - db$ from $ca - cb$; but from what has been shown in subtraction, the remainder will be $ca - cb - da + db$; therefore the product arising from the multiplication of $a - b$ by $c - d$ is $ca - cb - da + db$; hence, the reason of the general rule for the signs, as well as the other rules, is manifest.

When several quantities are multiplied together so as to constitute a product, each of them is called a factor of that product: thus $a$, $b$, and $c$ are factors of the product $abc$; also, $a + x$ and $b - x$ are factors of the product $(a + x)(b - x)$.

The products arising from the continual multiplication of the same quantity are called powers of that quantity, which is called the root. Thus $aa$, $aaa$, $aaaa$, &c. are powers of the root $a$. These powers are commonly expressed by placing above the root, towards the right hand, a figure, denoting how often the root is repeated. This figure serves to denominate the power, and is called its index or exponent. Thus, the quantity $a$ being considered as the root, or as the first power of $a$, we have $aa$ or $a^2$ for its second power, $aaa$ or $a^3$ for its third power, $aaaa$ or $a^4$ for its fourth power, and so on.

The second and third powers of a quantity are generally called its square and cube; and the fourth, fifth, and sixth powers are sometimes respectively called its biquadratic, sursolid, and cubocube.

By considering the notation of powers, and the rules for multiplication, it appears that powers of the same root are multiplied by adding their exponents. Thus $a \times a^2 = a^3$, also $x^3 \times x^4 = x^7$; and in general $a^n \times a^m = a^{n+m}$.

Prob. IV.—To Divide Quantities.

13. General Rule for the Signs. If the signs of the divisor and dividend be like, the sign of the quotient is $+$; but if they be unlike, the sign of the quotient is $-$.

This rule is easily derived from the general rule for the signs in multiplication, by considering that the quotient must be such a quantity as, when multiplied by the divisor, shall produce the dividend, with its proper sign.

The quotient arising from the division of one quantity by another may be expressed by placing the dividend above a line and the divisor below it (sect. 25); but it may also be often expressed in a more simple manner by the following rules.

Case 1. When the divisor is simple, and a factor of every term of the dividend.

Rule. Divide the co-efficient of each term of the dividend by the co-efficient of the divisor, and expunge out of each term the letter or letters in the divisor: the result is the quotient.

Ex. 1. Divide $12abc$ by $3ac$.

From the method of notation, the quotient may be expressed thus, $\frac{12abc}{3ac}$; but the same quotient, by the rule just given, is more simply expressed thus, $4b$.

Ex. 2. Divide $16a^3xy - 28a^2xz^2 + 4a^2x^3$ by $4a^2x$.

The quotient is $4ay - 7z^2 + x^2$.

If the divisor and dividend be powers of the same quan- Algebra. tity, the division will evidently be performed by subtracting the exponent of the divisor from that of the dividend. Thus \(a^3\), divided by \(a^2\), has for a quotient \(a^{3-2} = a\).

Case 2. When the divisor is simple, but not a factor of the dividend.

Rule. The quotient is expressed by a fraction, of which the numerator is the dividend, and the denominator the divisor.

Thus the quotient of \(3ab^2\), divided by \(2abc\), is the fraction \(\frac{3ab^2}{2abc}\).

It will sometimes happen that the quotient found thus may be reduced to a more simple form, as shall be explained when we come to treat of fractions.

Case 3. When the divisor is compound.

Rule 1. The terms of this dividend are to be arranged according to the powers of some one of its letters, and those of the divisor according to the powers of the same letter.

2. The first term of the dividend is to be divided by the first term of the divisor, observing the general rule for the signs; and this quotient, being set down for a part of the quotient wanted, is to be multiplied by the whole divisor, and the product subtracted from the dividend. If nothing remain, the division is finished; but if there be a remainder, it is to be taken for a new dividend.

3. The first term of the new dividend is next to be divided by the first term of the dividend, as before, and the quotient joined to the part already found, with its proper sign. The whole divisor is also to be multiplied by this part of the quotient, and the product subtracted from the new dividend; and thus the operation is to be carried on till there be no remainder, or till it appear that there will always be a remainder.

To illustrate this rule, let it be required to divide \(8a^2 + 2ab - 15b^2\) by \(2a + 3b\), the operation will stand thus:

\[ \begin{align*} (2a + 3b) & (8a^2 + 2ab - 15b^2) \\ & \quad (4a - 5b) \end{align*} \]

Here the terms of the divisor and dividend are arranged according to the powers of the quantity \(a\). We now divide \(8a^2\), the first term of the dividend, by \(2a\) the first term of the divisor; and thus get \(4a\) for the first term of the quotient. We next multiply the divisor by \(4a\), and subtract the product \(8a^2 + 12ab\) from the dividend; we get \(-10ab - 15b^2\) for a new dividend.

By proceeding in all respects as before, we find \(-5b\) for the second term of the quotient, and no remainder: the operation is therefore finished, and the whole quotient is \(4a - 5b\).

The following examples will also serve to illustrate the manner of applying the rule.

Ex. 1.

\[ \begin{align*} (3a - b)(3a^2 - 12a^2 - ab + 10ab - 2b^2)(a^2 - 4a + 2b) \\ & \quad (6ab - 2b^2) \end{align*} \]

Ex. 2.

\[ \begin{align*} (a + b)(a^2 + b^2)(a^2 - ab + b^2) \\ & \quad (a^3 + a^2b) \\ & \quad (-a^2b + b^3) \\ & \quad (-a^2b - ab^2) \\ & \quad (+ab^2 + b^3) \\ & \quad (+ab^2 + b^3) \end{align*} \]

Ex. 3.

\[ \begin{align*} (a^3 - b^3)(a^2 - b^2)(a^3 + b^3) \\ & \quad (a^6 - a^3b^3) \\ & \quad (+a^3b^3 - b^6) \\ & \quad (+a^3b^3 - b^6) \end{align*} \]

Ex. 4.

\[ \begin{align*} (1-x)^1 & (1+x+x^2+\ldots) \\ & \quad (1-x) \\ & \quad (+x) \\ & \quad (+x-x^2) \\ & \quad (+x^2) \\ & \quad (+x^2-x^3) \\ & \quad (+x^3) \end{align*} \]

14. Sometimes, as in this last example, the quotient will never terminate: in such a case it may either be considered as an infinite series, the law according to which the terms are formed being in general sufficiently obvious; or the quotient may be completed as in arithmetical division, by annexing to it a fraction, the numerator of which is the remainder, and denominator the divisor. Thus the quotient in last example may stand thus:

\[ 1 + x + x^2 + \frac{x^3}{1-x} \]

The reason of the rule for division is sufficiently manifest. For, in the course of the operation, all the terms of the quotient obtained by it are multiplied by all the terms of the divisor, and the products successively subtracted from the dividend, till nothing remain; that therefore must evidently be the true quotient.

Sect. II.—Of Fractions.

15. In the operation of division, the divisor may be sometimes less than the dividend, or may not be contained in it an exact number of times: in either case the quotient is expressed by means of a fraction. There can be no difficulty, however, in estimating the magnitude of such a quotient; if, for example, it were the fraction \(\frac{4}{7}\), we may consider it as denoting either that some unit is divided into 7 equal parts, and that 5 of these are taken, or that 5 times the same unit is divided into 7 equal parts, and one of them taken.

16. In any fraction the upper number, or the dividend, is called the numerator, and the lower number or divisor is called the denominator. Thus, in the fraction \(\frac{a}{b}\), \(a\) is the numerator, and \(b\) the denominator.

17. If the numerator be less than the denominator, such a fraction is called a proper fraction; but if the numerator be either equal to, or greater than the denominator, it is called an improper fraction; and if a quantity be Algebra made up of an integer and a fraction, it is called a mixed quantity. Thus, \( \frac{a}{a+x} \) is a proper fraction; \( \frac{a+x}{a} \) also \( \frac{a+x}{a} \), are both improper fractions; and \( b + \frac{x}{a} \) is a mixed quantity.

18. The reciprocal of a fraction is another fraction, having its numerator and denominator respectively equal to the denominator and numerator of the former.

Thus, \( \frac{b}{a} \) is the reciprocal of the fraction \( \frac{a}{b} \).

19. The following proposition is the foundation of the operations relating to fractions.

If the numerator and denominator of a fraction be either both multiplied or both divided by the same quantity, the value of that fraction is the same as before.

For, let any fraction \( \frac{b}{a} = c \); then, because \( c \) is the quotient arising from the division of \( b \) by \( a \), it follows that \( b = ac \); and multiplying both by any quantity \( n \), we have \( nb = nc \); let these equals be both divided by the same quantity \( na \), and the quotients will be equal, that is, \( \frac{nb}{na} = \frac{nc}{na} \); hence the truth of the proposition is manifest.

From this proposition, it is obvious that a fraction may be very differently expressed, without changing its value, and that any integer may be reduced to the form of a fraction, by placing the product arising from its multiplication by any assumed quantity as the numerator, and the assumed quantity as the denominator of the fraction. It also appears that a fraction very complex in its form may often be reduced to another of the same value, but more simple, by finding a quantity which will divide both the numerator and denominator, without leaving a remainder. Such a common measure, or common divisor, may be either simple or compound; if it be simple, it is readily found by inspection, but if it be compound, it may be found as in the following problem.

20. Prob. I.—To find the greatest common Measure of two Quantities.

Rule 1. Range the quantities according to the power of some one of the letters, as taught in division, leaving out the simple divisors of each quantity.

2. Divide that quantity which is of most dimensions by the other one, and if there be a remainder, divide it by its greatest simple divisor; and then divide the last compound divisor by the resulting quantity, and if anything yet remain, divide it also by its greatest simple divisor, and the last compound divisor by the resulting quantity. Proceed in this way till nothing remains, and the last divisor shall be the common measure required.

Note. It will sometimes be necessary to multiply the dividends by simple quantities in order to make the divisions succeed.

Ex. 1. Required the greatest common measure of the quantities \( ax^2 - x^3 \) and \( ax^2 - 2ax + ax^2 \). The simple divisor \( x \) being taken out of the former of these quantities, and \( a \) out of the latter, they are reduced to \( ax^2 - x^2 \), and \( ax^2 - 2ax + x^2 \); and as the quantity \( a \) rises to the same dimensions in both, we may take either of them as the first divisor; let us take that which consists of several terms, and the operation will stand thus:

\[ \begin{align*} (ax^2 - x^2)(ax^2 - 2ax + x^2) & \\ (ax^2 - x^2) & \\ \end{align*} \]

\[ \begin{align*} -2ax + 2x^2 \text{ remainder,} \end{align*} \]

which divided by \( -2x \) is \( a - x \) \( a^2 - x^2(a + x) \)

\[ \begin{align*} a^2 - ax & \\ + ax - x^2 & \\ + ax - x^2 & \\ \end{align*} \]

Hence it appears that \( a - x \) is the greatest common measure required.

Ex. 2. Required the greatest common measure of \( 8a^2b - 10ab^2 + 2b^3 \) and \( 9a^2b - 9ab^2 + 3ab^2 - 3ab^3 \).

It is evident, from inspection, that \( b \) is a simple divisor of both quantities; it will therefore be a factor of the common measure required. Let the simple divisors be now left out of each quantity, and they are reduced to \( 4a^2 - 5ab + b^2 \) and \( 3a^2 - 3ab + ab^2 - b^3 \); but as the second of these is to be divided by the first, it must be multiplied by 4 to make the division succeed, and the operation will stand thus:

\[ \begin{align*} (4a^2 - 5ab + b^2)(12a^2 - 12ab + 4ab^2 - 4b^3(3a & \\ 12a^2 - 15ab + 3ab^2 & \\ \end{align*} \]

\[ \begin{align*} + 3ab^2 + ab^2 - 4b^3. & \end{align*} \]

This remainder is to be divided by \( b \), and the new dividend multiplied by 3, to make the division again succeed, and the work will stand thus:

\[ \begin{align*} 3a^2 + ab - 4b^3(12a^2 - 15ab + 3ab^2(4 & \\ 12a^2 + 4ab - 16b^3 & \\ \end{align*} \]

\[ \begin{align*} -19ab + 19b^2. & \end{align*} \]

This remainder is to be divided by \( -19b \), which being done, and the last divisor taken as a dividend as before, the rest of the operation will be as follows:

\[ \begin{align*} (a-b)(3a^2 + ab - 4b^3(3a + 4b & \\ 3a^2 - 3ab & \\ \end{align*} \]

\[ \begin{align*} + 4ab - 4b^2 & \\ + 4ab - 4b^2 & \\ \end{align*} \]

from which it appears that the compound divisor sought is \( a - b \), and remarking that the quantities proposed have also a simple divisor \( b \), the greatest common measure which is required will be \( b(a-b) \).

21. The reason of the rule given in this problem may be deduced from the following considerations.

1. If two quantities have a compound divisor common to both, and they be either multiplied or divided by any simple quantities, the results will each have the same compound divisor. Thus the quantities \( p(a-x) \) and \( q(a-x) \) have the common divisor \( a-x \), and the quantities \( n p(a-x), r q(a-x) \) have each the very same divisor.

2. In the operation of division, whatever quantity measures both the divisor and dividend, the same will also measure the remainder. For let \( x \) be such a quantity, then the divisor and dividend may be represented by \( ax \) and \( bx \); let \( q \) be the quotient, and the remainder will evidently be \( bx - qx \), which is evidently divisible by \( x \).

3. Whatever quantity measures both the divisor and remainder, the same will also measure the dividend; for, let the divisor be \( ax \), and the remainder \( rx \), then \( q \), denoting the quotient, the dividend will be \( aqx + rx \), which, as well as the divisor and dividend, is divisible by \( x \).

Let us apply these observations to the last example. From the first observation, the reason for leaving out the simple quantities in the course of the operation, as well as for multiplying by certain other quantities, to make the divisions succeed, is obvious; and from the second observation it appears, that whatever quantity measures $4a^2 - 5ab + b^2$, and $12a^2 - 12ab + 4b^2 - 4b^3$, the same must measure $3a^2 + ab - 4b^2$, the first remainder, as also $-19ab + 19ab^2$, the second remainder; but the only compound divisor which this last quantity can have is $a - b$, which is also found to be a divisor of $3a^2 + ab - 4b^2$, or of $3a^2 + ab - 4b^2$. The first remainder, therefore, by the third observation, $a - b$, must also be a divisor of $12a^2 - 15ab + 3b^2$, or of $4a^2 - 5ab + b^2$, the first divisor, and therefore also it must be a divisor of $12a^2 - 12ab + 4b^2 - 4b^3$, the first dividend; so that $a - b$ is the greatest common measure, as was required.

22. Prob. II.—To Reduce a Fraction to its lowest Terms.

Rule. Divide both numerator and denominator by their greatest common measure, which may be found by Prob. I.

Ex. 1. Reduce $\frac{56a^2bc}{24abc^2}$ to its lowest terms.

It appears from inspection, that the greatest common measure is $8ac$; and dividing both numerator and denominator by this quantity, we have $\frac{56a^2bc}{24abc^2} = \frac{7ab}{3c}$.

Ex. 2. Reduce $\frac{a^2x - x^2}{a^2 - 2ax + ax^2}$ to its lowest terms.

We have already found in the first example of Prob. I. that the greatest common measure of the numerator and denominator is $a - x$; and dividing both by this quantity, we have $\frac{a^2x - x^2}{a^2 - 2ax + ax^2} = \frac{ax + x^2}{a^2 - ax}$.

In like manner we find

$\frac{9a^2b - 9ab^2 + 3a^2b^2 - 3ab^3}{8a^2b^2 - 10ab^3 + 2b^4} = \frac{9a^2 + 3ab^2}{8ab^2 - 2b^3}$;

the common measure being $b(a - b)$, as was shown in Example 2, Problem I.

23. Prob. III.—To Reduce a mixed Quantity to an improper Fraction.

Rule. Multiply the integer by the denominator of the fraction, and to the product add the numerator; and the denominator being placed under this sum, the result will be the improper fraction required.

Ex. 1. Let $x + \frac{x^2}{a}$, and $x - \frac{a^2 - x^2}{x}$ be reduced to improper fractions.

First, $x + \frac{x^2}{a} = \frac{ax + x^2}{a}$, the answer.

And $x - \frac{a^2 - x^2}{x} = \frac{x^2 - a^2 + x^2}{x} = \frac{2x^2 - a^2}{x}$, Ans.

Ex. 2. Reduce $a - x + \frac{x^2}{a + x}$ to an improper fraction.

$a - x + \frac{x^2}{a + x} = \frac{(a + x)(a - x) + x^2}{a + x} = \frac{a^2}{a + x}$, Ans.

24. Prob. IV.—To Reduce an improper Fraction to a whole or mixed Number.

Rule. Divide the numerator by the denominator for the integral part, and place the remainder, if any, over the denominator, and it will be the mixed quantity required.

Ex. 1. Reduce $\frac{ax + a^2}{x}$ to a whole or mixed quantity.

$\frac{ax + a^2}{x} = a + \frac{a^2}{x}$, the answer required.

Ex. 2. Reduce $\frac{ax + 2x^2}{a + x}$, also $\frac{x^2 - y^2}{x - y}$, to whole or mixed quantities.

First $\frac{ax + 2x^2}{a + x} = x + \frac{x^2}{a + x}$, the answer.

And $\frac{x^2 - y^2}{x - y} = x + y$ a whole quantity, which is the answer.

25. Prob. V.—To Reduce Fractions of different Denominators to others of the same value which shall have a common Denominator.

Rule. Multiply each numerator separately into all the denominators except its own for the new numerators, and all the denominators together for the common denominator.

Ex. 1. Reduce $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$, to fractions of equal value which have a common denominator.

$\begin{align*} \frac{a \times d \times f}{b \times d \times f} &= \text{New numerators.} \\ \frac{c \times b \times f}{b \times d \times f} &= \text{Common denominator.} \end{align*}$

Hence we find $\frac{a}{b} = \frac{adf}{bd}$, $\frac{c}{d} = \frac{cbf}{bd}$, and $\frac{e}{f} = \frac{ebd}{bd}$, where the new fractions have a common denominator, as was required.

Ex. 2. Reduce $\frac{ax}{a - x}$ and $\frac{a^2 - x^2}{a + x}$ to fractions of equal value, and having a common denominator.

$\begin{align*} (ax + x)(a - x) &= a^2x + ax^2 \\ (a^2 - x^2)(a + x) &= a^2 - ax^2 - ax^2 + x^2 \end{align*}$

New numerators.

$\frac{ax}{a - x} = \frac{a^2x + ax^2}{a^2 - x^2}$ and $\frac{a^2 - x^2}{a + x} = \frac{a^2 - ax^2 - ax^2 + x^2}{a^2 - x^2}$.

26. Prob. VI.—To Add or Subtract Fractions.

Rule. Reduce the fractions to a common denominator, and add or subtract their numerators, and the sum or difference placed over the common denominator, is the sum or remainder required.

Ex. 1. Add together $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$.

$\begin{align*} \frac{a}{b} &= \frac{adf}{bd} \\ \frac{c}{d} &= \frac{bef}{bd} \\ \frac{e}{f} &= \frac{bd}{bd} \end{align*}$

Hence $\frac{a}{b} + \frac{c}{d} + \frac{e}{f} = \frac{adf + bef + bde}{bd}$, the sum required.

Ex. 2. From $\frac{a + x}{a}$ subtract $\frac{a}{a + x}$.

$\begin{align*} \frac{a + x}{a} &= \frac{a^2 + 2ax + x^2}{a^2 + ax} \\ \frac{a}{a + x} &= \frac{a^2}{a^2 + ax} \end{align*}$

Hence $\frac{a + x}{a} - \frac{a}{a + x} = \frac{2ax + x^2}{a^2 + ax}$.

Ex. 3. Add together $\frac{x + 2}{3}$, $\frac{x}{4}$, and $\frac{x - 5}{2}$.

$\frac{x + 2}{3} + \frac{x}{4} + \frac{x - 5}{2} = \frac{8x + 16 + 6x + 12x - 60}{24} =$ If it be required to add or subtract mixed quantities, they may either be reduced to the form of fractions by prob. 3, and then added or subtracted, or else these operations may be performed first on the integer quantities, and afterwards on the fractions.

27. PROB. VII.—To Multiply Fractions.

Rule. Multiply the numerators of the fractions for the numerator of the product, and the denominators for the denominator of the product.

Ex. 1. Multiply \( \frac{b}{a} \times \frac{d}{c} = \frac{bd}{ac} \), the product required.

Ex. 2. Multiply \( \frac{a+b}{c} \times \frac{a-b}{d} = \frac{(a^2-b^2)}{cd} \), the product.

If it be required to multiply an integer by a fraction, the integer may be considered as having unity for a denominator. Thus, \( (a+x) \times \frac{3d}{c} = \frac{a+3x}{c} \times \frac{3d}{c} = \frac{3ad+3adx}{c} \).

Mixed quantities may be multiplied after being reduced to the form of fractions by prob. 3. Thus, \( \left( b + \frac{bx}{a} \right) \times \frac{a}{x} = \frac{ab+bx}{a} \times \frac{a}{x} = \frac{ab+bx}{ax} = \frac{ab+bx}{x} \).

The reason of the rule for multiplication may be explained thus. If \( \frac{a}{b} \) is to be multiplied by \( c \), the product will evidently be \( \frac{ac}{b} \); but if it is only to be multiplied by \( \frac{c}{d} \), the former product must be divided by \( d \), and it becomes \( \frac{ac}{bd} \), which is the product required. Or let \( \frac{a}{b} = m \), and \( \frac{c}{d} = n \), then \( a = bm \) and \( c = dn \) and \( ac = bdmn \); hence, \( mn \) or \( \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \).

28. PROB. VIII.—To Divide Fractions.

Rule. Multiply the denominator of the divisor by the numerator of the dividend for the numerator of the quotient. Then multiply the numerator of the divisor by the denominator of the dividend for the denominator of the quotient.

Or, multiply the dividend by the reciprocal of the divisor, the product will be the quotient required.

Ex. 1. Divide \( \frac{a}{b} \) by \( \frac{c}{d} \)

\( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} \) as before,

\( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} \), the quotient.

Ex. 2. Divide \( \frac{a^2+ab}{2x} \) by \( \frac{a-b}{2x} \),

\( \frac{3a^2}{2x} \div \frac{a^2-ab}{6ax^2} = \frac{a^2-b^2}{6ax^2} \), the quotient.

If either the divisor or dividend be an integer quantity, it may be represented as a fraction, by placing unity for a denominator; or if it be a mixed quantity, it may be reduced to a fraction by prob. 3, and the operation of division performed agreeably to the rule.

The reason of the rule for division may be explained thus: Let it be required to divide \( \frac{c}{d} \) by \( \frac{a}{b} \). If \( \frac{c}{d} \) is to be divided by \( a \), the quotient is \( \frac{c}{ad} \); but if it is to be divided by \( \frac{a}{b} \), then the last quotient must be multiplied by \( b \); thus we have \( \frac{cb}{ad} \) for the quotient required. Or let \( \frac{a}{b} = m \), and \( \frac{c}{d} = n \), then \( a = bm \) and \( c = dn \); also \( ad = bdm \) and \( bc = bdn \); therefore \( \frac{bdn}{bdm} = \frac{n}{m} = \frac{bc}{ad} \).

Sect. III.—Involution and Evolution.

29. In treating of multiplication, we have observed, that when a quantity is multiplied by itself any number of times, the product is called a power of that quantity, while the quantity itself, from which the powers are formed, is called the root (sect. 12). Thus, \( a, a^2, \) and \( a^3 \) are the first, second, and third powers of the root \( a \); and in like manner \( \frac{1}{a}, \frac{1}{a^2}, \) and \( \frac{1}{a^3} \) denote the same powers of the root \( \frac{1}{a} \).

But before considering more particularly what relates to powers and roots, it will be proper to observe, that the quantities \( \frac{1}{a}, \frac{1}{a^2}, \frac{1}{a^3}, \) &c. admit of being expressed under a different form; for, like as the quantities \( a, a^2, a^3, \) &c. are expressed as positive powers of the root \( a \), so the quantities \( \frac{1}{a}, \frac{1}{a^2}, \frac{1}{a^3}, \) &c. may be respectively expressed thus, \( a^{-1}, a^{-2}, a^{-3}, \) &c. and considered as negative powers of the root \( a \).

30. This method of expressing the fractions \( \frac{1}{a}, \frac{1}{a^2}, \frac{1}{a^3}, \) as powers of the root \( a \), but with negative indices, is a consequence of the rule which has been given for the division of powers; for we may consider \( \frac{1}{a} \) as the quotient arising from the division of any power of \( a \) by the next higher power; for example, from the division of the 2d by the 3d, and so we have \( \frac{1}{a} = \frac{a^2}{a^3} \); but since powers of the same quantity are divided by subtracting the exponent of the divisor from that of the dividend (sect. 30), it follows, that \( \frac{a^2}{a^3} = a^{2-3} = a^{-1} \); therefore the fraction \( \frac{1}{a} \) may also be expressed thus, \( a^{-1} \). By considering \( \frac{1}{a} \) as equal to \( \frac{a^2}{a^3} \), it will appear in the same manner that \( \frac{1}{a^2} = \frac{a^3}{a^4} = a^{-2} \); and proceeding in this way, we get \( \frac{1}{a^3} = \frac{a^4}{a^5} = a^{-3}, \frac{1}{a^4} = \frac{a^5}{a^6} = a^{-4}, \) &c. and so on, as far as we please. It also appears that unity or 1 may be represented by \( a^0 \), where the exponent is a cypher, for \( 1 = \frac{a^2}{a^2} = a^{2-2} = a^0 \).

31. The rules which have been given for the multiplication and division of powers with positive exponents will apply in every case, whether the exponents be positive or negative; and this must evidently take place; for the mode of notation, by which we represent fractional quantities as the powers of integers, but with negative exponents, has been derived from those rules. Thus, \( \frac{1}{a^2} \times a^3 \) or \( a^{-2} \times a^3 \) Hence it appears that the powers of \(a + x\) differ from the powers of \(a - x\) only in this respect, that in the former the signs of the terms are all positive, but in the latter they are positive and negative alternately.

33. Besides the method of finding the powers of a compound quantity by multiplication, which we have just now explained, there is another more general, as well as more expeditious, by which a quantity may be raised to any power whatever without the trouble of finding any of the inferior powers, namely, by means of what is commonly called the binomial theorem. This theorem may be expressed as follows: Let \(a - x\) be a binomial quantity, which is to be raised to any power denoted by the number \(n\), then \((a + x)^n =\)

\[ \begin{align*} &= a^n + \frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 \\ &+ \frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} a^{n-4}x^4 + \cdots + \frac{n(n-1)(n-2)(n-3)(n-4)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^{n-5}x^5 + \cdots \end{align*} \]

This series will always terminate when \(n\) is any whole positive number, by reason of some one of the factors \(n-1, n-2, \ldots\) becoming \(0\); but if \(n\) be either a negative or fractional number, the series will consist of an infinite number of terms. As, however, we mean to treat in this section only of the powers of quantities when their exponents are whole positive numbers, we shall make no further remarks upon any other: we shall afterwards give a demonstration of the theorem, and show its application to fractional and negative powers, in treating of infinite series. The \(n\)th power of \(a - x\) will not differ from the same power of \(a + x\), but in the signs of the terms which compose it; for it will stand thus: \((a - x)^n = a^n - \frac{n}{1} a^{n-1}x + \cdots\)

\[ \begin{align*} &= a^n - \frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 - \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 \\ &- \frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} a^{n-4}x^4 - \cdots - \frac{n(n-1)(n-2)(n-3)(n-4)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^{n-5}x^5 - \cdots \end{align*} \]

and so on—alternately.

Ex. 1. Let it be required to raise \(a + x\) to the fifth power.

Here \(n\), the exponent of the power, being 5, the first term \(a^5\) of the general theorem will be equal to \(a^5\), the second \(na^{n-1}x = 5a^4x\), the third \(\frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 = \frac{5 \times 4}{1 \cdot 2} a^3x^2 = 10a^3x^2\), the fourth \(\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 = \frac{5 \times 4 \times 3}{1 \cdot 2 \cdot 3} a^2x^3 = 10a^2x^3\), the fifth \(\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} a^{n-4}x^4 = \frac{5 \times 4 \times 3 \times 2}{1 \cdot 2 \cdot 3 \cdot 4} ax^4 = 5ax^4\), and the sixth and last \(\frac{n(n-1)(n-2)(n-3)(n-4)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^{n-5}x^5 = \frac{5 \times 4 \times 3 \times 2 \times 1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} x^5 = x^5\); the remaining terms of the general theorem all vanish, by reason of the factor \(n - 5 = 0\) by which each of them is multiplied, so that we get \((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5\).

Ex. 2. It is required to raise \(2d - \frac{x}{3}\) to the third power.

In this case \(n = 3\), so that if we put \(a = 2d\) and \(x = \frac{x}{3}\), we have the first term of the general theorem, or \(a^3 = 8d^3\), the second \(\frac{n}{1} a^{n-1}x = 3 \times 4d^2 \times \frac{x}{3} = 4d^2x\), the third

\[ n(n-1) \cdot \frac{a^{n-2} x^2}{1 \cdot 2} = 3 \times 2d \times \frac{x^2}{y} = \frac{2d x^2}{3}, \text{ and the fourth} \]

and last term \( \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3} x^3 = \frac{x^3}{27}; \text{ and since the} \)

signs of the terms of any power of \( a - x \) are + and - alternately, we have \( (2d - \frac{5}{3})^5 = 8d^5 - 4d^2 x + \frac{2d x^2}{3} - \frac{x^3}{27}. \)

34. If the quantity to be involved consists of more than two terms, as if \( p + q - r \) were to be raised to the second power, put \( p = a \) and \( q - r = b \), then \( (p + q - r)^2 = (a + b)^2 = a^2 + 2ab + b^2 = p^2 + 2pq - 2pr + (q - r)^2, \)

but \( 2p(q - r) = 2pq - 2pr; \text{ by the general theorem} \)

\((q - r)^2 = q^2 - 2qr + r^2; \text{ therefore we get } (p + q - r)^2 = p^2 + 2pq - 2pr + q^2 - 2qr + r^2; \text{ and by a similar method of proceeding a quantity consisting of four or more terms may be raised to any power.} \)

Of Evolution.

35. Evolution is the reverse of involution, or it is the method of finding the root of any quantity, whether simple or compound, which is considered as a power of that root: hence it follows that its operations, generally speaking, must be the reverse of those of involution.

To denote that the root of any quantity is to be taken, the sign \( \sqrt{} \) (called the radical sign) is placed before it, and a small number placed over the sign to express the denomination of the root. Thus \( \sqrt{a} \) denotes the square root of \( a \), \( \sqrt[3]{a} \) its cube root, \( \sqrt[4]{a} \) its fourth root, and in general, \( \sqrt[n]{a} \) its \( n \)-th root. The number placed over the radical sign is called the index or exponent of the root, and is usually omitted in expressing the square root: thus, either \( \sqrt{a} \) or \( \sqrt[2]{a} \) denotes the square root of \( a \).

Case 1. When roots of simple quantities are to be found.

Rule. Divide the exponents of the letters by the index of the root required, and prefix the root of the numeral co-efficient; the result will be the root required.

Note 1. The root of any positive quantity may be either positive or negative, if the index of the root be an even number; but if it be an odd number, the root can be positive only.

2. The root of a negative quantity is also negative when the index of the root is an odd number.

3. But if the quantity be negative, and the index of the root even, then no root can be assigned.

Ex. 1. Required the square root of \( 36a^2x^4 \).

Here the index of the root is 2, and the root of the co-efficient 6, therefore \( \sqrt{36a^2x^4} = \pm 6ax^2 \) or \( \sqrt{36a^2x^4} = \pm 6ax^2 \); for either of these quantities, when multiplied by itself, produces \( 36a^2x^4 \); so that the root required is \( \pm 6ax^2 \), where the sign \( \pm \) denotes that the quantity to which it is prefixed may be considered either as positive or negative.

Ex. 2. Required the cube root of \( 125a^3x^9 \).

Here the index of the root is 3, and the root of the co-efficient 5, therefore \( \sqrt[3]{125a^3x^9} = 5a^x \), the root required; and in like manner the cube root of \( -125a^3x^9 \) is found to be \( -5a^x \).

If it be required to extract the square root of \( -a^2 \), it will immediately appear that no such root can be assigned; for it can neither be \( +a \) nor \( -a \), seeing that each of these quantities, when squared, produces \( +a^2 \): the root required is therefore said to be impossible, and may be expressed thus: \( \sqrt{-a^2} \).

The root of a fraction is found by extracting that root out of both numerator and denominator. Thus the square root of \( \frac{4a^2x^4}{9b^2y^2} \) is \( \frac{2ax^2}{3by^2} \).

Case 2. When the quantity of which the root is to be extracted is compound.

36. I. To extract the square root.

Range the terms of the quantity according to the powers of the letters, as in division.

Find the square root of the first term for the first part of the root sought, subtract its square from the given quantity, and divide the remainder by double the part already found, and the quotient is the second term of the root.

Add the second part to double the first, and multiply their sum by the second part; subtract the product from the remainder, and if nothing remain, the square root is obtained. But if there is a remainder, it must be divided by the double of the parts already found, and the quotient will give the third term of the root, and so on.

Ex. 1. Required the square root of \( a^2 + 2ax + x^2 \).

\[ \begin{array}{c} a^2 + 2ax + x^2 \\ \hline \end{array} \]

Ex. 2. Required the square root of \( x^4 - 2x^3 + \frac{3}{2}x^2 \).

\[ \begin{array}{c} x^4 - 2x^3 + \frac{3}{2}x^2 \\ \hline \end{array} \]

To understand the reason of the rule for finding the square root of a compound quantity, it is only necessary to involve any quantity, as \( a + b + c \), to the second power, and observe the composition of its square; for we have \( (a + b + c)^2 = a^2 + 2ab + b^2 + 2ac + 2bc + c^2; \text{ but } 2ab + b^2 = (2a + b)b \text{ and } 2ac + 2bc + c^2 = (2a + 2b + c)c; \text{ therefore,} \)

\((a + b + c)^2 = a^2 + (2a + b)b + (2a + 2b + c)c; \text{ and from this expression the manner of deriving the rule is obvious.} \)

As an illustration of the common rule for extracting the square root of any proposed number, we shall suppose that the root of 59049 is required.

Accordingly we have \( (a + b + c)^2 = 59049 \), and from hence we are to find the values of \( a, b, \) and \( c \). Hence 243 is the root required.

Add together three times the square of the part of the root already found, three times the product of that part and the second part of the root, and the square of the second part; multiply the sum by the second part, and subtract the product from the first remainder, and if nothing remain, the root is obtained; but if there is a remainder, it must be divided by three times the square of the sum of the parts already found, and the quotient is a third term of the root, and so on, till the whole root is obtained.

Ex. Required the cube root of \(a^3 + 3a^2x + 3ax^2 + x^3\).

The reason of the preceding rule is evident from the composition of a cube; for if any quantity, as \(a + b + c\) be raised to the third power, we have \((a + b + c)^3 = a^3 + (3a^2 + 3ab + b^2) + 3(a + b)c + c^3\), and by considering in what manner the terms \(a, b,\) and \(c\) are deduced from this expression for the cube of their sum, we also see the reason for the common rule for extracting the cube root in numbers. Let it be required to find the cube root of 13312053, where the root will evidently consist of three figures; let us suppose it to be represented by \(a + b + c\), and the operation for finding the numerical values of these quantities may stand as follows.

The operation as performed by the common rule (see Arithmetic) will stand thus:

38. III. To extract any other root.

Rule. Range the quantity of which the root is to be found, according to the powers of its letters, and extract the root of the first term, and that shall be the first member of the root required. In this example, the cube root of \( x^3 \) or \( x^2 \), is the first member of the root; and to find a second member, the first is raised to the power next lower, or to the second power, and also multiplied by 3, the index of the root required. Thus we get \( 3x^4 \) for a divisor, by which the second term \( 6x^3 \) being divided, we find \( 2x \) for the second member of the root. We must now consider \( x^2 + 2x \) as forming one term: accordingly, having subtracted its cube from the quantity of which the root is sought, we have \( -12x^4 \), &c., for a new dividend; and having also raised \( x^2 + 2x \) to the second power, and multiplied the result by 3, we find \( 3x^4 + \ldots \), &c. for a divisor. (As it is only the terms which contain the highest powers of the dividend and divisor that we have occasion for, the remaining terms are expressed by &c.) Having divided \( -12x^4 \) by \( 3x^4 \), we find \( -4 \) for the third term of the root; and because it appears that \( x^2 + 2x - 4 \), when raised to the third power, gives a result the very same with the proposed power, we conclude \( x^2 + 2x - 4 \) to be the root sought.

39. In the preceding examples, the quantities whose roots were to be found have been all such as could have their roots expressed by a finite number of terms; but it will frequently happen that the root cannot be otherwise assigned than by a series consisting of an infinite number of terms. The preceding rules, however, will serve to determine any number of terms of the series. Thus, the square root of \( a^2 + x^2 \) will be found to be \( a + \frac{x^2}{2a} - \frac{x^4}{8a^3} + \ldots \), and the cube root of \( a^3 + x^3 \) will stand thus, \( a + \frac{x^3}{3a^2} - \frac{x^6}{9a^3} + \frac{5x^9}{81a^4} - \frac{10x^{12}}{243a^5} + \ldots \). But as the extraction of roots in the form of series can be more easily performed by other methods, we shall refer the reader to sect. 19, which treats of series, where this subject is resumed.

Sect. IV.—Of Surds.

40. It has been already observed (35), that the root of any proposed quantity is found by dividing the exponent of the quantity by the index of the root; and the rule has been illustrated by suitable examples, in all which, however, the quotient expressing the exponent of the result is a whole number; but there may be cases in which the quotient is a fraction. Thus, if the cube root of \( a^2 \) were required, it might be expressed, agreeably to the method of notation already explained, either thus, \( \sqrt[3]{a^2} \), or thus, \( a^{\frac{2}{3}} \).

Quantities which have fractional exponents are called surds, or imperfect powers, and are said to be irrational, in opposition to others with integral exponents, which are called rational.

Surds may be denoted by means of the radical sign, but it will be often more convenient to use the notation of fractional exponents. The following examples will show how they may be expressed either way.

\[ \begin{align*} \sqrt[3]{a} &= a^{\frac{1}{3}}, \\ \sqrt[4]{4a^2} &= 2a^{\frac{1}{2}}, \\ \sqrt[5]{a^3b^2} &= a^{\frac{3}{5}}b^{\frac{2}{5}}, \\ \sqrt{a^2 + b^2} &= (a^2 + b^2)^{\frac{1}{2}}, \\ \sqrt{(a-b)^2} &= (a-b)^{\frac{1}{2}}, \\ \sqrt{ab} &= (a+b)^{\frac{1}{2}}a^{-\frac{1}{2}}b^{-\frac{1}{2}}. \end{align*} \]

The operations concerning surds depend on the following principle: If the numerator and denominator of a fractional exponent be either both multiplied or both divided by the same quantity, the value of the power is the same. Thus, \( a^n = a^n \). For let \( a^n = b^n \), then raising both to the power \( n \), \( a^{n^2} = b^{n^2} \); and further, raising both to the power \( e \), we get \( a^{ne} = b^{ne} \); let the root \( cn \) be now taken, and we find \( a^n = b^n \).

41. Prob. I.—To Reduce a Rational Quantity to the form of a Surd of any given denomination.

Rule. Reduce the exponent of the quantity to the form of a fraction of the same denomination as the given surd.

Ex. 1. Reduce \( a^2 \) to the form of the cube root.

Here the exponent 2 must be reduced to the form of a fraction having 3 for a denominator, which will be the fraction \( \frac{2}{3} \); therefore \( a^2 = a^{\frac{2}{3}} = \sqrt[3]{a^2} \).

Ex. 2. Reduce 5 to the form of the cube root, and \( 3a^2 \) to the form of the square root.

First, \( 5 = \frac{5}{1} = \sqrt[3]{5 \times 5 \times 5} = \sqrt[3]{125} \),

And \( 3a^2 = \frac{3}{1}a^2 = (3a^2)^{\frac{1}{3}} = \sqrt[3]{9a^2} \).

42. Prob. II.—To Reduce Surds of different denominations to others of the same value, and of the same denomination.

Rule. Reduce the fractional exponents to others of the same value, and having the same common denominator.

Ex. 1. Reduce \( \sqrt{a} \) and \( \sqrt{b} \), or \( a^{\frac{1}{2}} \) and \( b^{\frac{1}{2}} \) to other equivalent surds of the same denomination.

The exponents \( \frac{1}{2} \), \( \frac{1}{3} \), when reduced to a common denominator, are \( \frac{3}{6} \) and \( \frac{2}{6} \); therefore the surds required are \( a^{\frac{3}{6}} \) and \( b^{\frac{2}{6}} \), or \( \sqrt{a^3} \) and \( \sqrt{b^2} \).

Ex. 2. Reduce \( 3^{\frac{1}{2}} \) and \( 2^{\frac{1}{2}} \) to surds of the same denomination.

The new exponents are \( \frac{5}{6} \) and \( \frac{3}{6} \); therefore we have \( 3^{\frac{5}{6}} = \frac{6}{\sqrt{3^5}} = \sqrt[6]{27} \), and \( 2^{\frac{3}{6}} = \frac{6}{\sqrt{2^3}} = \sqrt[6]{8} \).

And in the same way the surds \( A^{\frac{1}{6}} \), \( B^{\frac{1}{6}} \), are reduced to these two, \( \sqrt{A^6} \) and \( \sqrt{B^6} \).

43. Prob. III.—To Reduce Surds to their most simple terms.

Rule. Reduce the surd into two factors, so that one of them may be a complete power, having its exponent divisible by the index of the surd. Extract the root of that power, and place it before the remaining quantities, with the proper radical sign between them.

Ex. 1. Reduce \( \sqrt{48} \) to its most simple terms.

The number 48 may be resolved into the two factors 16 and 3, of which the first is a complete square; therefore \( \sqrt{48} = (4^2 \times 3)^{\frac{1}{2}} = 4 \times 3^{\frac{1}{2}} = 4\sqrt{3} \).

Ex. 2. Reduce \( \sqrt{98a^2} \), and \( \sqrt{24a^2x + 40a^2x^2} \), each to its most simple terms.

First, \( \sqrt{98a^2} = (7a^2 \times 2)^{\frac{1}{2}} = 7a^2 \times (2x)^{\frac{1}{2}} = 7a^2\sqrt{2x} \).

Also \( \sqrt{24a^2x + 40a^2x^2} = (2a(3x + 5x^2))^{\frac{1}{2}} = 2a\sqrt{3x + 5x^2} \).

44. Prob. IV.—To Add and Subtract Surds.

Rule. If the surds are of different denominations, reduce them to others of the same denomination, by prob. 2. and then reduce them to their simplest terms by last problem. Then, if the surd part be the same in them all, annex it to the sum or difference of the rational parts, with the sign of multiplication, and it will give the sum or difference required. But if the surd part be not the same in all the quantities, they can only be added or subtracted by placing the signs + or − between them.

Ex. 1. Required the sum of \( \sqrt{27} \) and \( \sqrt{48} \).

By prob. 3 we find \( \sqrt{27} = 3\sqrt{3} \) and \( \sqrt{48} = 4\sqrt{3} \), therefore \( \sqrt{27} + \sqrt{48} = 3\sqrt{3} + 4\sqrt{3} = 7\sqrt{3} \).

Ex. 2. Required the sum of \( 3\sqrt{\frac{1}{4}} \) and \( 5\sqrt{\frac{1}{2}} \).

\( 3\sqrt{\frac{1}{4}} = 3\sqrt{\frac{1}{2}} = \frac{3}{2}\sqrt{2} \) and \( 5\sqrt{\frac{1}{2}} = 5\sqrt{\frac{1}{2}} = \frac{5}{2}\sqrt{2} \), therefore \( 3\sqrt{\frac{1}{4}} + 5\sqrt{\frac{1}{2}} = \frac{3}{2}\sqrt{2} + \frac{5}{2}\sqrt{2} = 4\sqrt{2} \).

Ex. 3. Required the difference between \( \sqrt{80a^2x} \) and \( \sqrt{20a^2x^3} \).

\( \sqrt{80a^2x} = (4a^2 \times 5x)^{\frac{1}{2}} = 4a^2\sqrt{5x} \), and \( \sqrt{20a^2x^3} = (2a^2x^2 \times 5x)^{\frac{1}{2}} = 2ax\sqrt{5x} \); therefore \( \sqrt{80a^2x} - \sqrt{20a^2x^3} = (4a^2 - 2ax)\sqrt{5x} \).

45. Prob. V.—To Multiply and Divide Surds.

Rule. If they are surds of the same rational quantity, add or subtract their exponents.

But if they are surds of different rational quantities, let them be brought to others of the same denomination, by prob. 2. Then, by multiplying or dividing these rational quantities, their product or quotient may be set under the common radical sign.

Note. If the surds have any rational co-efficients, their product or quotient must be prefixed.

Ex. 1. Required the product of \( \sqrt{\frac{5}{a^2}} \) and \( \sqrt{\frac{5}{a^2}} \).

\( \sqrt{\frac{5}{a^2}} \times \sqrt{\frac{5}{a^2}} = \frac{5}{a^2} \times \frac{5}{a^2} = \frac{25}{a^4} = \frac{15}{a^4} = \frac{15}{a^4} \), Ans.

Ex. 2. Divide \( \sqrt{a^2 - b^2} \) by \( \sqrt{a + b} \).

These surds, when reduced to the same denomination, are \( (a^2 - b^2)^{\frac{1}{2}} \) and \( (a + b)^{\frac{1}{2}} \). Hence \( \frac{\sqrt{a^2 - b^2}}{\sqrt{a + b}} = \frac{(a + b)(a - b)}{(a + b)^2} = \frac{\sqrt{(a + b)(a - b)}}{\sqrt{a + b}} \).

Ex. 3. Required the product of \( 5\sqrt{8} \) and \( 3\sqrt{5} \).

\( 5\sqrt{8} \times 3\sqrt{5} = 5 \times 3 \times \sqrt{8} \times \sqrt{5} = 15 \times \sqrt{40} = 15 \times \sqrt{10} = 30\sqrt{10} \).

Ex. 4. Divide \( 8\sqrt{56} \) by \( 4\sqrt{2} \).

\( \frac{8\sqrt{56}}{4\sqrt{2}} = 2\sqrt{56} = 2\sqrt{28} \).

Ex. 5. Required the product of \( \frac{1}{a^2} \) and \( \frac{1}{b^2} \); also the quotient arising from the division of \( \frac{1}{a^2} \) by \( \frac{1}{b^2} \).

First, \( \frac{1}{a^2} \times \frac{1}{b^2} = \frac{1}{a^2 + b^2} = \frac{1}{a^2 + b^2} = \frac{1}{a^2 + b^2} \),

And \( \frac{1}{a^2} \div \frac{1}{b^2} = \frac{a^2}{b^2} = \frac{a^2}{b^2} \).

46. Prob. VI.—To Involve and Evolve Surds.

Surds are involved or evolved in the same manner as any other quantities, namely, by multiplying or dividing their exponents by the index of the power or root required. Thus, the square of \( 3\sqrt{3} \) is \( 3 \times 3 \times (3)^{\frac{3}{2}} = 9\sqrt{9} \). The nth power of \( x^{\frac{1}{n}} \) is \( x^{\frac{n}{n}} \). The cube root of \( \frac{1}{8} \sqrt{2} \) is \( \frac{1}{2}(2)^{\frac{1}{3}} = \frac{\sqrt{2}}{2} \), and the nth root of \( x^{\frac{m}{n}} \) is \( x^{\frac{m}{n}} \).

47. If a compound quantity involve one or more surds, its powers may be found by multiplication. Thus, the square of \( 3 + \sqrt{5} \) is found as follows:

\[ \begin{array}{c} 3 + \sqrt{5} \\ 3 + \sqrt{5} \\ \hline 9 + 3\sqrt{5} + 3\sqrt{5} + 5 \\ \end{array} \]

\( 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5} \), the square required.

48. The square root of a binomial or residual surd, \( A + B \) or \( A - B \), may be found thus. Find \( D = \sqrt{A^2 - B^2} \), then \( \sqrt{A + B} = \sqrt{\frac{A + D}{2}} + \sqrt{\frac{A - D}{2}} \), and \( \sqrt{A - B} = \sqrt{\frac{A + D}{2}} - \sqrt{\frac{A - D}{2}} \).

Thus the square root of \( 8 + 2\sqrt{7} \) is \( 1 + \sqrt{7} \); and the square root of \( 3 - \sqrt{8} \) is \( \sqrt{2} - 1 \). With respect to the extraction of the cube or any higher root, no general rule can be given.

Sect. V.—Of Proportion.

49. In comparing together any two quantities of the same kind in respect of magnitude, we may consider how much the one is greater than the other, or else how many times the one contains either the whole or some part of the other; or, which is the same thing, we may consider either what is the difference between the quantities, or what is the quotient arising from the division of the one quantity by the other: the former of these is called their arithmetical ratio, and the latter their geometrical ratio. These denominations, however, have been assumed arbitrarily, and have little or no connection with the relations they are intended to express.

I. Of Arithmetical Proportion.

50. When of four quantities the difference between the first and second is equal to the difference between the third and fourth, the quantities are called arithmetical proportions. Such, for example, are the numbers 2, 5, 9, 12; and, in general, the quantities \( a, a+d, b+d \). If the two middle terms are equal, the quantities constitute an arithmetical progression.

51. The principal property of four arithmetical proportions is this:—If four quantities be arithmetically proportional, the sum of the extreme terms is equal to the sum of the means. Let the quantities be \( a, a+d, b, b+d \); where \( d \) is the difference between the first and second, and also between the third and fourth, the sum of the extremes is \( a+d+d \), and that of the means \( a+d+b \); so that the truth of the proposition is evident. Hence it follows, that if any three quantities be arithmetically proportional, the sum of the two extremes is double the mean.

52. If any three terms of four arithmetical proportions be given, the fourth may be found from the preceding Let \(a, b, c\) be the first, second, and fourth terms, and let \(x\) be the third term required; because \(a + c = b + x\), therefore \(x = a + c - b\). In like manner any two of three arithmetical proportions being supposed given, the remaining term may be readily found.

53. If a series of quantities be such, that the difference between any two adjacent terms is always the same, these terms form a continued arithmetical progression. Thus, the numbers 2, 4, 6, 8, 10, &c., form a series in continued arithmetical proportion, and, in general, such a series may be represented thus:

\[a, a+d, a+2d, a+3d, a+4d, a+5d, \ldots\]

where \(a\) denotes the first term, and \(d\) the common difference.

By a little attention to this series, we readily discover that it has the following properties:

1. The last term of the series is equal to the first term, together with the common difference taken as often as there are terms after the first. Thus, when the number of terms is 7, the last term is \(a + 6d\); and so on. Hence if \(z\) denote the last term, \(n\) the number of terms, and \(a\) and \(d\) express the first term and common difference, we have \(z = a + (n-1)d\).

2. The sum of the first and last term is equal to the sum of any two terms at the same distance from them. Thus, suppose the number of terms to be 7, then the last term is \(a + 6d\), and the sum of the first and last \(2a + 6d\); but the same is also the sum of the second and last but one, of the third and last but two, and so on till we come to the middle term, which, because it is equally distant from the extremes, must be added to itself.

From the last-mentioned property we derive a rule for finding the sum of all the terms of the series. For if the sum of the first and last be taken, as also the sum of the second and last but one, of the third and last but two, and so on along the series till we come to the sum of the last and first terms, it is evident that we shall have as many sums as there are terms, and each equal to the sum of the first and last terms; but the aggregate of those sums is equal to all the terms of the series taken twice, therefore the sum of the first and last term, taken as often as there are terms, is equal to twice the sum of all the terms; so that if \(s\) denote that sum, we have \(2s = n(a+z)\), and

\[s = \frac{n}{2}(a+z).\]

Hence the sum of the odd numbers 1, 3, 5, 7, 9, &c., continued to \(n\) terms, is equal to the square of the number of terms. For in this case \(a = 1, d = 2, z = 1 + (n-1)d = 2n-1\), therefore \(s = \frac{n}{2} \times 2n = n^2\).

II. Of Geometrical Proportion.

54. When, of four quantities, the quotient arising from the division of the first by the second is equal to that arising from the division of the third by the fourth, these quantities are said to be in geometrical proportion, or are called simply proportional. Thus, 12, 4, 15, 5, are four numbers in geometrical proportion; and, in general, \(na, nb, nc, nd\) may express any four proportional, for \(\frac{na}{nb} = n\), and also \(\frac{nc}{nd} = n\).

To denote that any four quantities \(a, b, c, d\) are proportional, it is common to place them thus, \(a : b :: c : d\); or thus, \(a : b = c : d\); which notation, when expressed in words, is read thus, \(a\) is to \(b\) as \(c\) to \(d\), or the ratio of \(a\) to \(b\) is equal to the ratio of \(c\) to \(d\).

The first and third terms of a proportion are called the antecedents, and the second and fourth the consequents.

When the two middle terms of a proportion are the same, the remaining terms, and that quantity, constitute three geometrical proportions; such as 4, 6, 9, and in general \(na, nc, nd\). In this case the middle quantity is called a mean proportional between the other two.

55. The principal properties of four proportional are the following:

1. If four quantities be proportional, the product of the extremes is equal to the product of the means. Let \(a, b, c, d\) be four quantities, such that \(a : b :: c : d\); then, from the nature of proportions, \(\frac{a}{b} = \frac{c}{d}\); let these equal quotients be multiplied by \(bd\), and we have \(\frac{abd}{b} = \frac{cdb}{d}\), or \(ad = bc\). Hence it follows, that when three quantities are proportional, the product of the extremes is equal to the square of the middle term. It also appears, that if any three of four proportional be given, the remaining one may be found. Thus, let \(a, b, c\), the first three, be given, and let it be required to find \(x\), the fourth term; because \(a : b :: c : x\), \(ax = bc\), and dividing by \(a\), \(x = \frac{bc}{a}\).

This conclusion may be considered as a demonstration of what is called the rule of three in arithmetic.

2. If four quantities be such, that the product of two of them is equal to the product of the other two, these quantities are proportional.

Let \(a, b, c, d\) be the quantities, which are such that \(ad = bc\); if these equals be divided by \(bd\), then \(\frac{ad}{bd} = \frac{bc}{bd}\) or \(\frac{a}{b} = \frac{c}{d}\); hence, from the definition given of proportional (sect. 54), \(a : b :: c : d\). From this property of proportional it appears, that if three quantities be such that the square of one of them is equal to the product of the other two, these quantities are three proportional.

If four quantities are proportional, that is, if \(a : b :: c : d\), then will each of the following combinations or arrangements of the quantities be also four proportional.

1st. By inversion, \(b : a :: d : c\). 2nd. By alternation, \(a : c :: b : d\).

Note.—The quantities in the second case must be all of the same kind.

3rd. By composition, \(a + b : a :: c + d : c\), or \(a + b : b :: c + d : d\).

4th. By division, \(a - b : a :: c - d : c\), or \(a - b : b :: c - d : d\).

5th. By mixing, \(a + b : a :: c + d : c - d\). 6th. By taking any equimultiples of the antecedents, and also any equimultiples of the consequents,

\[na : pb :: nc : pd.\]

7th. Or, by taking any parts of the antecedents and consequents,

\[\frac{a}{p} : \frac{b}{q} :: \frac{c}{r} : \frac{d}{s}.\]

That the preceding combinations of the quantities \(a, b, c, d\) are proportional, may be readily proved, by taking the products of the extremes and means; for from each of them we derive this conclusion, that \(ad = bc\), which is known to be true, from the original assumption of the quantities.

If four quantities be proportional, and also other four, the product of the corresponding terms will be proportional.

Let \(a : b :: c : d\), And \(e : f :: g : h\); Then \(ae : bf :: cg : dh\). Algebra. For \( ad = bc \), and \( ch = fg \) (sect. 55), therefore, multiplying together these equal quantities, \( adeh = bcfg \), or \( ae \times dh = bf \times cg \); therefore, by the second property (sect. 55), \( ae : bf :: cg : dh \).

Hence it follows, that if there be any number of proportions whatever, the products of the corresponding terms will still be proportional.

56. If a series of quantities be so related to each other, that the quotient arising from the division of any term by that which follows it is always the same quantity, these are said to be in continued geometrical proportion; such are the numbers 2, 4, 8, 16, 32, &c. also \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots \), &c. and in general, a series of such quantities may be represented thus, \( a, ar, ar^2, ar^3, ar^4, ar^5, \ldots \). Here \( a \) is the first term, and \( r \) the quotient of any two adjoining terms, which is also called the common ratio.

By inspecting this series, we find that it has the following properties:

1. The last term is equal to the first, multiplied by the common ratio raised to a power, the index of which is one less than the number of terms. Therefore, if \( z \) denote the last term, and \( n \) the number of terms, \( z = ar^{n-1} \).

2. The product of the first and last term is equal to the product of any two terms equally distant from them: thus, supposing \( ar^5 \) the last term, it is evident that \( a \times ar^5 = ar \times ar^4 = ar^2 \times ar^3, \ldots \).

The sum of \( n \) terms of a geometrical series may be found thus:

Let \( s = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1} \).

Then \( ss = ar + ar^2 + ar^3 + \ldots + ar^{n-1} + ar^n \).

Subtract, \( ss - s = ar^n - a \);

That is, \((r - 1)s = a(r^n - 1)\).

Hence \( s = \frac{r^n - 1}{r - 1}a \).

Sect. VI.—Of the Resolution of Equations involving one Unknown Quantity.

57. The general object of algebraic investigation is to discover certain unknown quantities, by comparing them with other quantities which are given, or supposed to be known. The relation between the known and unknown quantities is either that of equality, or else such as may be reduced to equality; and a proposition which affirms that certain combinations of quantities are equal to one another is called an equation. Such are the following:

\[ \frac{x}{2} + \frac{x}{3} = \frac{2x}{3}, \]

\(2x + 3y = xy\).

The first of these equations expresses the relation between an unknown quantity \( x \) and certain known numbers; and the second expresses the relation which the two indefinite quantities \( x \) and \( y \) have to each other.

58. When a quantity stands alone on one side of an equation, the terms on the other side are said to be a value of that quantity. Thus, in the equation \( x = ay + b - c \), the quantity \( x \) stands alone on one side, and \( ay + b - c \) is its value.

The conditions of a problem may be such as to require several equations and symbols of unknown quantities for their complete expression. These, however, by rules hereafter to be explained, may be reduced to one equation, involving only one unknown quantity and its powers, besides the known quantities; and the method of expressing that quantity by means of the known quantities constitutes the theory of equations, one of the most important as well as most intricate branches of algebraic analysis.

59. An equation is said to be resolved when the unknown quantity is made to stand alone on one side, and only known quantities on the other side; and the value of the unknown quantity is called a root of the equation.

60. Equations containing only one unknown quantity and its powers, are divided into different orders, according to the highest power of that quantity contained in any one of its terms. The equation, however, is supposed to be reduced to such a form that the unknown quantity is found only in the numerators of the terms, and that the exponents of its powers are expressed by positive integers.

61. If an equation contains only the first power of the unknown quantity, it is called a simple equation, or an equation of the first order. Such is \( ax + b = c \), where \( x \) denotes an unknown, and \( a, b, c \), known quantities.

If the equation contains the second power of the unknown quantity, it is said to be of the second degree, or is called a quadratic equation; such is \( 4x^2 + 3x = 12 \), and in general \( ax^2 + bx = c \). If it contains the third power of the unknown quantity, it is of the third degree, or is a cubic equation; such are \( x^3 - 2x^2 + 4x = 10 \), and \( ax^3 + bx^2 + cx = d \); and so on with respect to equations of the higher orders. A simple equation is sometimes said to be linear, or of one dimension. In like manner, quadratic equations are said to be of two dimensions, and cubic equations of three dimensions.

62. When in the course of an algebraic investigation we arrive at an equation involving only one unknown quantity, that quantity will often be so entangled in the different terms as to render several previous reductions necessary before the equation can be expressed under its characteristic form, so as to be resolved by the rules which belong to that form.

These reductions depend upon the operations which have been explained in the former part of this treatise, and the application of a few self-evident principles, namely, that if equal quantities be added to or subtracted from equal quantities, the sums or remainders will be equal; if equal quantities be multiplied or divided by the same quantity, the products or quotients will be equal; and, lastly, if equal quantities be raised to the same power, or have the same root extracted out of each, the results will still be equal.

From these considerations are derived the following rules, which apply alike to equations of all orders, and are alone sufficient for the resolution of simple equations.

63. Rule 1. Any quantity may be transposed from one side of an equation to the other, by changing its signs.

Thus, if \( x - 3 = 5 \),

Then \( x = 5 + 3 \),

Or \( x = 8 \).

And if \( 3x - 10 = 2x + 5 \),

Then \( 3x - 2x = 5 + 10 \),

Or \( x = 15 \).

Again, if \( ax + b = cx - dx + e \),

Then \( ax - cx + dx = e - b \),

Or \( (a - c + d)x = e - b \).

The reason of this rule is evident, for the transposing of a quantity from one side of an equation to the other is nothing more than adding the same quantity to each side of the equation, if the sign of the quantity transposed was — ; or subtracting it, if the sign was +.

From this rule we may infer, that if any quantity be found on each side of the equation with the same sign, it may be left out of both. Also, that the signs of all the terms of an equation may be changed into the contrary, without affecting the truth of the equation. Thus, if \(a + x = b + a - c\), Then \(x = b + c\); And if \(a - x = b - d\), Then \(x = a - d - b\).

64. Rule 2. If the unknown quantity in an equation be multiplied by any quantity, that quantity may be taken away, by dividing all the other terms of the equation by it.

If \(3x = 24\), Then \(x = \frac{24}{3} = 8\).

If \(ax = b - c\), Then \(x = \frac{b - c}{a} = \frac{b}{a} - \frac{c}{a}\).

Here equal quantities are divided by the same quantity, and therefore the quotients are equal.

65. Rule 3. If any term of an equation be a fraction, its denominator may be taken away, by multiplying all the other terms of the equation by that denominator.

If \(\frac{x}{5} = 7\), Then \(x = 35\).

If \(\frac{x}{a} = b - c + d\), Then \(x = ab - ac + ad\).

If \(a - \frac{b}{x} = c\), We have \(ax - b = cx\).

In these examples, equal quantities are multiplied by the same quantity, and therefore the products are equal.

66. The denominators may be taken away from several terms of an equation by one operation, if we multiply all the terms by any number which is a multiple of each of these denominators.

Thus, if \(\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 26\);

let all the terms be multiplied by 12, which is a multiple of 2, 3, and 4, and we have

\[ \frac{12x}{2} + \frac{12x}{3} + \frac{12x}{4} = 312; \]

Or \(6x + 4x + 3x = 312\); Hence \(13x = 312\).

Universally, if \(\frac{x}{a} - \frac{x}{b} + \frac{x}{c} = d - e\);

to take away the denominators \(a, b, c\), let the whole equation be multiplied by \(abc\), their product, and we have

\(bcx - acx + abx = abc(d - e)\), Or \((bc - ac + ab)x = abc(d - e)\).

From the last two rules it appears, that if all the terms of an equation be either multiplied or divided by the same quantity, that quantity may be left out of all the terms.

If \(ax = ab - ac\), Then \(x = b - c\); And if \(\frac{x}{a} = \frac{b}{a} + \frac{c}{a}\), Then \(x = b + c\).

67. Rule 4. If the unknown quantity is found in any term which is a surd, let that surd be made to stand alone on one side of the equation, and the remaining terms on the opposite side; then involve each side to a power denoted by the index of the surd, and thus the unknown quantity shall be freed from the surd expression.

If \(\sqrt{x + 6} = 10\), Then, by transposition, \(\sqrt{x} = 10 - 6 = 4\); And, squaring both sides, \(\sqrt{x} \times \sqrt{x} = 4 \times 4\), Or \(x = 16\).

Also, if \(\sqrt{a^2 + x^2} = b + x\), By trans. \(\sqrt{a^2 + x^2} = b + x\), And, squaring, \(a^2 + x^2 = (b + x)^2 = b^2 + 2bx + x^2\), Hence \(a^2 = b^2 + 2bx\).

And if \(\sqrt{a^2 - bx} = x\), Then \(a^2 - bx = x^2\).

68. Rule 5. If the side of the equation which contains the unknown quantity be a perfect power, the equation may be reduced to another of a lower order, by extracting the root of that power out of each side of the equation.

Thus, if \(x^3 = 64a^3\), Then, by extracting the cube root, \(x = 8a\); And if \((a + x)^2 = b^2 - a^2\), Then \(a + x = \sqrt{b^2 - a^2}\).

69. The use of the preceding rules will be further illustrated by the following examples:

Ex. 1. Let \(20 - 3x - 8 = 60 - 7x\), By rule 1, \(7x - 3x = 60 + 8 - 20\), Or \(4x = 48\), Therefore, by rule 2, \(x = 12\).

Ex. 2. Let \(ax - b = cx + d\), By rule 1, \(ax - cx = b + d\), Or \((a - c)x = b + d\), And by rule 2, \(x = \frac{b + d}{a - c}\).

Ex. 3. Let \(\frac{x + 1}{2} + \frac{x + 2}{3} = 16 - \frac{x + 3}{4}\), By rule 3, \[ \begin{align*} x + 1 + \frac{2x + 4}{3} &= 32 - \frac{2x + 6}{4}, \\ 3x + 3 + 2x + 4 &= 96 - 6x + 18, \\ 12x + 12 + 8x + 16 &= 384 - 6x - 18, \\ 20x + 28 &= 366 - 6x, \\ \end{align*} \] Or \(26x = 338\), Hence, by rule 1, \(x = 13\), And by rule 2, \(x = 13\).

In this example, instead of taking away the denominators one after another, they might have been all taken away at once, by multiplying the given equation by 12, which is divisible by the numbers 2, 3, and 4; thus we should have got \(6x + 6 + 4x + 8 = 192 - 3x - 9\), and hence, as before, \(x = 13\).

Ex. 4. Let \(6x^3 - 20x^2 = 16x^2 + 2x^3\); Then, dividing by \(2x^2\), \(3x - 10 = 8 + x\), And transposing, \(3x - x = 8 + 10\), Or \(2x = 18\), And therefore \(x = 9\).

Ex. 5. Let \(a - \frac{b^2}{x} = c\), Then \(ax - b^2 = cx\), And \(ax - cx = b^2\), Whence \(x = \frac{b^2}{a - c}\).

Ex. 6. Let \(x - 6 = \frac{x^2}{x + 24}\), Then \((x - 6)(x + 24) = x^2\); That is, \( x^2 + 18x - 144 = x^2 \), Therefore \( 18x = 144 \), And \( x = 8 \).

Ex. 7. Let \( ax + b^2 = \frac{ax^2 + ac^2}{a + x} \); Then \( (a + x)(ax + b^2) = ax^2 + ac^2 \), Or \( ax^2 + ab^2 + ax^2 + bx = ax^2 + ac^2 \), Hence \( ax^2 + bx = ac^2 - ab^2 \), And \( x = \frac{ac^2 - ab^2}{a^2 + b^2} \).

Ex. 8. Let \( \frac{1-x}{1+a} = a \); Then \( 1-x = a + ax \), And \( -x = a + ax - 1 \), Or, changing the signs, \( x + ax = 1 - a \); Hence \( x = \frac{1-a}{1+a} \).

Ex. 9. Let \( \sqrt{12 + x} = 2 + \sqrt{x} \); Then, by rule 4, \( 12 + x = 4 + 4\sqrt{x} + x \), And by transposition, \( 8 = 4\sqrt{x} \), And by division, \( 2 = \sqrt{x} \), And again, by rule 4, \( 4 = x \).

Ex. 10. Let \( x + \sqrt{a^2 + x^2} = \frac{2a^2}{\sqrt{a^2 + x^2}} \); Then, by rule 3, \( x \sqrt{a^2 + x^2} + a^2 + x^2 = 2a^2 \), And by transposition, \( x \sqrt{a^2 + x^2} = a^2 - x^2 \), Therefore, by rule 4, \( a^2 + x^2 = a^2 - 2a^2x^2 + x^4 \), Whence \( 3a^2x^2 = a^4 \), And \( x^2 = \frac{a^2}{3} \), therefore, by rule 5, \( x = \frac{a}{\sqrt{3}} \).

Ex. 11. Let \( \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = a \); Then \( 1-\sqrt{1-x^2} = a + a\sqrt{1-x^2} \), And \( 1-a = a\sqrt{1-x^2} + \sqrt{1-x^2} = (1+a)\sqrt{1-x^2} \), Whence \( \frac{1-a}{1+a} = \sqrt{1-x^2} \); And, taking the square of both sides, \( \frac{(1-a)^2}{(1+a)^2} = 1-x^2 \), Therefore, by transposition, \( x^2 = 1 - \frac{(1-a)^2}{(1+a)^2} \), That is, \( x^2 = \frac{(1+a)^2 - (1-a)^2}{(1+a)^2} = \frac{4a}{(1+a)^2} \), Therefore \( x = \frac{\sqrt{a}}{1+a} \).

Ex. 12. Let \( a + x = \sqrt{a^2 + x^2 + b^2 + x^2} \), Then \( (a + x)^2 = a^2 + x^2 + b^2 + x^2 \), That is, \( a^2 + 2ax + x^2 = a^2 + x^2 + b^2 + x^2 \), Therefore \( 2ax + x^2 = x^2 + b^2 + x^2 \), And dividing by \( x \), \( 2a + x = \sqrt{b^2 + x^2} \), Again, taking the squares of both sides, \( 4a^2 + 4ax + x^2 = b^2 + x^2 \), Whence \( 4a^2 + 4ax = b^2 \), And \( 4ax = b^2 - 4a^2 \); so that \( x = \frac{b^2 - 4a^2}{4a} \).

In all these examples we have been able to determine the value of the unknown quantity by the rules already delivered, because in every case the first, or at most the second power of that quantity, has been made to stand algebraically on one side of the equation, while the other consisted only of known quantities; but the same methods of reduction serve to bring equations of all degrees to a proper form for solution. Thus, if \( \frac{1-p+q+r}{x+1} = 1-p-x + \frac{r}{x} \); by proper reduction, we have \( x^3 + px^2 + qx = r \), a cubic equation, which may be resolved by rules to be afterwards explained.

Sect. VII.—Of the Reduction of Equations involving more than one Unknown Quantity.

70. Having shown in the last section in what manner an equation involving one unknown quantity may be resolved, or at least fitted for a final solution, we are next to explain the methods by which two or more equations, involving as many unknown quantities, may at last be reduced to one equation and one unknown quantity.

As the unknown quantities may be combined together in very different ways, so as to constitute an equation, the methods most proper for their elimination must therefore be various. The three following, however, are of general application, and the last of them may be used with advantage, not only when the unknown quantity to be eliminated rises to the same power in all the equations, but also when the equations contain different powers of that quantity.

71. Method 1. Observe which of the unknown quantities is the least involved, and let its value be found from each equation, by the rules of last section.

Let the values thus found be put equal to each other, and hence new equations will arise, from which that quantity is wholly excluded. Let this operation be now repeated with these equations, thus eliminating the unknown quantities one by one, till at last an equation be found which contains only one unknown quantity.

Ex. Let it be required to determine \( x \) and \( y \) from these two equations.

\[ \begin{align*} 2x + 3y &= 23, \\ 5x - 2y &= 10. \end{align*} \]

From the first equation, \( 2x = 23 - 3y \), And \( x = \frac{23 - 3y}{2} \).

From the second equation, \( 5x = 10 + 2y \), And \( x = \frac{10 + 2y}{5} \).

Let these values of \( x \) be now put equal to each other, And we have \( \frac{10 + 2y}{5} = \frac{23 - 3y}{2} \), Or \( 20 + 4y = 125 - 15y \), Therefore \( 19y = 95 \), And \( y = 5 \).

And since \( x = \frac{23 - 3y}{2} \), or \( x = \frac{10 + 2y}{5} \), from either of these values we find \( x = 4 \).

72. Method 2. Let the value of the unknown quantity which is to be eliminated be found from that equation wherein it is least involved. Let this value and its powers be substituted for that quantity, and its respective powers in the other equations; and with the new equations thus arising, let the operation be repeated till there remain only one equation and one unknown quantity. Ex. Let the given equations, as in last method, be

\[2x + 3y = 23,\] \[5x - 2y = 10.\]

From the first equation, \(x = \frac{23 - 3y}{2};\)

And this value of \(x\) being substituted in the second equation, we have \(5 \times \frac{23 - 3y}{2} - 2y = 10,\)

Or \(115 - 15y - 4y = 20;\) Therefore \(95 = 19y,\) And \(5 = y,\) And hence \(x = \frac{23 - 3y}{2} = 4,\) as before.

73. Method 3. Let the given equations be multiplied or divided by such numbers or quantities, whether known or unknown, that the term which involved the highest power of the unknown quantity may be the same in each equation.

Then, by adding or subtracting the equations, as occasion may require, that term will vanish, and a new equation emerge, wherein the number of dimensions of the unknown quantity in some cases, and in others the number of unknown quantities, will be diminished; and by a repetition of the same or similar operations, a final equation may be at last obtained, involving only one unknown quantity.

Ex. Let the same example be taken, as in the illustration of the former methods, namely,

\[2x + 3y = 23,\] \[5x - 2y = 10;\]

and from these equations we are to determine \(x\) and \(y.\)

To eliminate \(x,\) let the first equation be multiplied by 5, and the second by 2; thus we have

\[10x + 15y = 115,\] \[10x - 4y = 20.\]

Here the term involving \(x\) is the same in both equations; and it is obvious, that by subtracting the one from the other, the resulting equation will contain only \(y,\) and known numbers; for by such subtraction we find \(19y = 95,\) and therefore \(y = 5.\)

Having got the value of \(y,\) it is easy to see how \(x\) may be found, from either of the given equations; but it may also be found in the same manner as we found \(y.\) For let the first of the given equations be multiplied by 2, and the second by 3, and we have

\[4x + 6y = 46,\] \[15x - 6y = 30.\]

By adding these equations, we find

\[19x = 76,\] and \(y = 4.\)

74. The following examples will serve further to illustrate these different methods of eliminating the unknown quantities from equations.

Ex. 1. Given

\[\begin{align*} \frac{x}{2} + \frac{y}{3} &= 16, \\ \frac{x}{5} - \frac{y}{9} &= 2; \end{align*}\]

required \(x\) and \(y.\)

By Method 1.

From the first equation, \(x = 32 - \frac{2y}{3},\)

And from the second, \(x = 10 + \frac{5y}{9};\)

Therefore \(10 + \frac{5y}{9} = 32 - \frac{2y}{3},\)

Or \(90 + 5y = 288 - 6y;\) Hence \(11y = 198\) and \(y = 18.\)

The value of \(y\) being substituted in either of the values of \(x,\) namely, \(32 - \frac{2y}{3},\) or \(10 + \frac{5y}{9},\) we find \(x = 20.\)

By Method 2.

Having found from the first equation

\[x = 32 - \frac{2y}{3},\]

let this value of \(x\) be substituted in the second, and

\[\frac{1}{5}(32 - \frac{2y}{3}) - \frac{y}{9} = 2,\]

Or \(\frac{32}{5} - \frac{2y}{15} - \frac{y}{9} = 2;\)

Hence \(198 = 11y\) and \(y = 18.\)

The value of \(y\) being now substituted in either of the given equations, we thence find \(x = 20,\) as before.

By Method 3.

The denominators of the two given equations being taken away by rule 3 of last section, we have

\[3x + 2y = 96,\] \[9x - 5y = 90.\]

From three times the first of these, or \(9x + 6y = 288,\) let the second be subtracted, there remains

\[11y = 198\) and \(y = 18.\)

The value of \(y\) being now substituted in either of the equations \(3x + 2y = 96,\) \(9x - 5y = 90,\) we readily find \(x = 20.\)

75. Having shown in what way the different methods of eliminating the unknown quantities may be applied, we shall, in the remaining examples of this section, chiefly make use of the last method, because it is the most easy and expeditious in practice.

Ex. 2. Given

\[\begin{align*} \frac{x}{2} - 12 &= \frac{y}{4} + 8, \\ \frac{x}{5} + \frac{x}{3} - 8 &= \frac{2y - x}{4} + 27. \end{align*}\]

It is required to determine \(x\) and \(y.\)

From the 1st equation, \(4x - 96 = 2y + 64;\) From the 2d, \(12x + 12y + 20x = 480 = 30y - 15x + 1620.\)

These equations, when abridged, become

\[4x - 2y = 160,\] \[47x - 18y = 2100.\]

To eliminate \(y,\) from this last equation let 9 times the one preceding it be subtracted.

Thus we find \(11x = 660,\) and \(x = 60;\) And because \(2y = 4x - 160 = 80,\) Therefore \(y = 40.\)

Ex. 3. Given

\[\begin{align*} ax + by &= c, \\ dx + fy &= g; \end{align*}\]

to determine \(x\) and \(y.\)

To eliminate \(y,\) let the first equation be multiplied by \(f,\) and the second by \(b,\) and we have

\[afx + bfy = cf,\] \[bdx + bfy = bg.\]

Taking now the difference between these equations,

\[afx - bdx = cf - bg,\] Or \((af - bd)x = cf - bg,\) And therefore \(x = \frac{cf - bg}{af - bd}.\)

In the same manner may \(y\) be determined, by multiply- Next, to eliminate \( y \), let the first of these equations be multiplied by 3, and the second by 5; hence,

\[ 60x + 15y = 2340, \] \[ 50x + 15y = 2100. \]

Subtracting now the latter equation from the former,

\[ 10x = 240 \quad \text{and} \quad x = 24, \]

Therefore

\[ y = \frac{420 - 10x}{3} = 60, \]

And

\[ z = \frac{1448 - 12x - 8y}{6} = 120. \]

77. From the preceding examples, it is manifest in what manner any number of unknown quantities may be determined by an equal number of equations, which contain only the first power of those quantities, in the numerators of the terms. Such are the following:

\[ ax + bx + cx = n, \] \[ dx + ey + fz = p, \] \[ gx + hy + kz = q; \]

where \( a, b, c, \) &c. represent known, and \( x, y, z \), unknown quantities; and in every case the unknown quantities may be directly found, for they will be always expressed by whole numbers or rational fractions, provided that the known quantities, \( a, b, c, \) &c. are also rational.

78. We shall now add a few examples, in which the equations that result from the elimination of an unknown quantity rise to some of the higher degrees; and therefore their final solution must be referred to the sections which treat of those degrees.

Ex. 6. Let \( x - y = 2 \), and \( xy + 5x - 6y = 120 \); it is required to eliminate \( x \).

From the first equation \( x = y + 2 \); which value being substituted in the other equation according to the second general method (sect. 72) it becomes

\[ (y + 2)y + 5(y + 2) - 6y = 120, \]

that is, \( y^2 + 2y + 5y + 10 - 6y = 120 \);

therefore the equation required is \( y^2 + y = 110 \).

Ex. 7. There is given \( x + y = a \), and \( x^2 y^2 = b \), to eliminate \( x \).

From the first equation \( x = a - y \), and \( x^2 = (a - y)^2 \),

And from the second \( x^2 = b - y^2 \);

Therefore \( (a - y)^2 = b - y^2 \);

That is, \( a^2 - 2ay + y^2 = b - y^2 \).

Hence \( 2y^2 - 2ay = b - a^2 \); an equation involving only \( y \).

Ex. 8. Given \( \begin{cases} axy + bx + cy = d \\ fzy + gx + hy = k \end{cases} \) to eliminate \( y \).

From the first equation \( y = \frac{d - bx}{ax + c} \),

And from the second \( y = \frac{k - gx}{fx + h} \);

Therefore \( \frac{d - bx}{ax + c} = \frac{k - gx}{fx + h} \),

an equation in which the unknown quantity \( y \) is not found.

Ex. 9. Given \( \begin{cases} y^2 - 3xy + ay = x^2 \\ y^2 + 2ax - by = 4x^2 - b^2 \end{cases} \) to eliminate \( y \).

As the co-efficient of \( y^2 \) is unity in both equations, if their difference be taken, the highest power of \( y \) will vanish; but to give a general solution, let the terms of the equations be all brought to one side and made equal to 0, thus,

\[ y^2 - (3x - a)y - x^2 = 0, \] \[ y^2 - by + 2ax - 4x^2 + b^2 = 0. \] In the first equation put $1 = A$, $-(3x - a) = B$, $-x^2 = C$; and in the second, $1 = D$, $-b = E$, $2ax - 4x^2 + bx = F$; and the two become

$$Ay^2 + By + C = 0,$$ $$Dy^2 + Ey + F = 0.$$

To eliminate $y^2$, let the first equation be multiplied by $D$, and the second by $A$; then

$$ADy^2 + BDy + CD = 0,$$ $$ADy^2 + AEy + AF = 0.$$

Therefore, taking the difference of these equations,

$$(BD - AE)y + CD - AF = 0,$$ $$y = \frac{AF - CD}{BD - AE}.$$

Again, to find another value of $y$, multiply the first equation by $F$, and the second by $C$; then

$$AFy^2 + BFy + CF = 0,$$ $$CDy^2 + CEy + CF = 0.$$

Therefore, subtracting as before,

$$(AF - CD)y^2 + (BF - CE)y = 0,$$ $$y = \frac{CE - BF}{AF - CD}.$$

Let this value of $y$ be put equal to the former, then

$$\frac{AF - CD}{BD - AE} = \frac{CE - BF}{AF - CD};$$

And therefore $(AF - CD)^2 = (BD - AE)(CE - BF)$.

Now, as $y$ does not enter this equation, if we restore the values of $A$, $B$, $C$, &c., we have the following equation, which involves only $x$ and known quantities,

$$(b^2 + 2ax - 3x^2)^2 = (a + b - 3x)^2 [bx^2 - (a - 3x)(2ax - 4x^2 + b^2)].$$

This equation, when properly reduced, will be of the fourth order, and therefore its final resolution belongs not to this place.

**Sect. VIII.—Questions producing Simple Equations.**

79. When a problem is proposed to be resolved by the algebraic method of analysis, its true meaning ought in the first place to be perfectly understood, so that, if necessary, it may be freed from all superfluous and ambiguous expressions, and its conditions exhibited in the clearest point of view possible. The several quantities concerned in the problem are next to be denoted by proper symbols, and their relation to one another expressed agreeably to the algebraic notation. Thus we shall obtain a series of equations, which, if the question be properly limited, will enable us to determine all the unknown quantities required by the rules already delivered in the two preceding sections.

80. In reducing the conditions of a problem to equations, the following rule will be of service. Suppose that the quantities to be determined are actually found, and then consider by what operations the truth of the solution may be verified; then let the same operations be performed upon the quantities, whether known or unknown, and thus all the conditions of the problem will be reduced to a series of equations, such as is required. For example, suppose that it is required to find two numbers, such that their sum is 20, and the quotient arising from the division of their difference by the lesser 3; then if we denote the greater of the two numbers by $x$, and the lesser by $y$, and proceed as if to prove the truth of the solution, we shall have $x + y$ for the sum of the numbers, and $x - y$ for their difference. Now, as the former must be equal to 20, and the latter divided by $y$ equal to 3, the first condition of the problem will be expressed by this equation,

$$x + y = 20,$$ $$\frac{x - y}{y} = 3;$$

and from these the values of $x$ and $y$ may easily be found.

81. When the conditions of a problem have been expressed by equations, or translated from the common language into that of algebra, we must consider whether the problem be properly limited; for in some cases the conditions may be such as to admit of innumerable solutions, and in others they may involve an absurdity, and thus render the problem altogether impossible.

Now, by considering the examples of last section, it will appear, that to determine any number of unknown quantities, there must be given as many equations as there are unknown quantities. These, however, must be such as cannot be derived from each other, and they must not involve any contradiction; for in the one case the problem would admit of an unlimited number of answers, and in the other case it would be impossible. For example, if it were required to determine $x$ and $y$ from these two equations, $2x - 3y = 13$, $4x - 6y = 26$; as the latter equation is a consequence of the former (for each term of the one is the half of the corresponding term of the other), it is evident, that innumerable values of $x$ and $y$ might be found to satisfy both equations. Again, if $x$ and $y$ were to be determined from these equations, $x + 2y = 8$, $3x + 6y = 26$, it is easy to see that it is impossible to find such values of $x$ and $y$ as will satisfy both; for, from the first, we find $3x = 24 - 6y$; and from the second, $3x = 26 - 6y$; and therefore $24 - 6y = 26 - 6y$, or $24 = 26$, which is absurd; and so also must have been the conditions from which this conclusion is drawn.

82. But there is yet another case in which a problem may be impossible; and that is, when there are more equations than unknown quantities; for it appears, that in this case, by the rules of last section, we would at last find two equations, each involving the same unknown quantity. Now, unless these happened to agree, the problem would admit of no solution. On the whole therefore it appears, that a problem is limited when the conditions furnish just as many independent equations as there are known quantities to be determined; if there be fewer, the problem is indeterminate; but if there be more, the problem in general admits of no solution whatever.

83. In expressing the conditions of a problem by equations, it will sometimes be convenient to introduce as few symbols of unknown quantities as possible. Therefore, if two quantities be sought and their sum be given, suppose it $= s$; then if the one quantity be represented by $x$, the other may be denoted by $s - x$. If, again, their difference be given $= d$, the quantities may be denoted by $x$, and $d + x$, or by $x$, and $x - d$. If their product be given $= p$, the quantities are $x$, and $\frac{p}{x}$; and so on.

84. We shall now apply the preceding observations to some examples, which are so chosen as to admit of being resolved by simple equations.

**Ex. I.** What is that number, to which if there be added its half, its third, and its fourth parts, the sum will be 50?

Let $x$ denote the number sought; then its half will be $\frac{x}{2}$, its third $\frac{x}{3}$, and its fourth $\frac{x}{4}$;

Therefore $$x + \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 50.$$

Hence $$24x + 12x + 8x + 6x = 1200,$$ Or $$50x = 1200;$$ Therefore $$x = 24.$$ Thus it appears, that the number sought is 24, which upon trial will be found to answer the conditions of the question.

Ex. 2. A post is \( \frac{1}{4} \) of its length in the mud, \( \frac{1}{3} \) in the water, and 10 feet above the water; what is its whole length?

Let its length be \( x \) feet, then the part in the mud is \( \frac{x}{4} \), and that in the water \( \frac{x}{3} \); therefore, from the nature of the question,

\[ \frac{x}{4} + \frac{x}{3} + 10 = x. \]

From this equation we find \( 7x + 120 = 12x \), and \( x = 24 \).

Ex. 3. Two travellers set out at the same time from London and York, the distance between which places is 150 miles; the one goes 8 miles a day, and the other 7; in what time will they meet?

Suppose that they meet after \( x \) days.

Then the one traveller has gone 8\( x \) miles, and the other 7\( x \) miles; now the sum of the distances they travel is, by the question, equal to the distance from London to York.

Therefore \( 8x + 7x = 150 \),

That is, \( 15x = 150 \), and \( x = 10 \) days.

Ex. 4. A labourer engaged to serve for 40 days, upon these conditions: that for every day he worked he was to receive 20d., but for every day he played or was absent, he was to forfeit 8d.; now at the end of the time he had to receive Ll. 11s. 8d. It is required to find how many days he worked, and how many days he was idle.

Let \( x \) be the number of days he worked;

Then will \( 40 - x \) be the number of days he was idle,

Also \( 20 \times x = 20x \) is the sum he earned in pence,

And \( 8 \times (40 - x) = 320 - 8x \) is the sum he forfeited.

Now the difference of these two was Ll. 11s. 8d. or 380d.;

Therefore \( 20x - (320 - 8x) = 380 \);

That is, \( 28x = 700 \).

Hence \( x = 25 \) is the number of days he worked,

And \( 40 - x = 15 \) is the number of days he was idle.

Ex. 5. A market-woman bought a certain number of eggs at 2 a penny, and as many at 3 a penny, and sold them all out again at 5 for 2d.; but, instead of getting her own money for them, as she expected, she lost 4d.: what number of eggs did she buy?

Let \( x \) be the number of eggs of each sort;

Then will \( \frac{x}{2} \) be the price of the first sort,

And \( \frac{x}{3} \) is the price of the second sort.

Now, the whole number being \( 2x \), we have

\[ 5 : 2x :: 2 : \frac{4x}{5} = \text{price of both sorts at } 5 \text{ for } 2d.; \]

Therefore \( \frac{x}{2} + \frac{x}{3} - \frac{4x}{5} = 4 \), by the question.

Hence \( 15x + 10x - 24x = 120 \),

And \( x = 120 \), the number of each sort.

Ex. 6. A bill of L.120 was paid in guineas and moidores: the number of pieces of both sorts used was 100; how many were there of each?

Let the number of guineas be \( x \);

Then the number of moidores will be \( 100 - x \);

Also the value of the guineas, reckoned in shillings, will be 21s, and that of the moidores \( 27(100 - x) = 2700 - 27x \);

Therefore, by the question, \( 21x + 2700 - 27x = 2400 \).

Hence \( 6x = 300 \), and \( x = 50 \);

So that the number of pieces of each sort was 50.

Ex. 7. A footman agreed to serve his master for L.8 Algebra, a year and livery, but was turned away at the end of 7 months, and received only L.2. 13s. 4d. and his livery; what was its value?

Suppose \( x \) the value of the livery, in pence.

Then his wages for a year were to be \( x + 1920 \) pence.

But for 7 months he received \( x + 640 \) pence.

Now he was paid in proportion to the time he served;

Therefore \( 12 : 7 :: x + 1920 : x + 640 \).

And, taking the product of the extremes and means,

\( 12x + 7680 = 7x + 13440 \).

Hence \( 5x = 5760 \), and \( x = 1152 \) d. = L.4. 16s.

Ex. 8. A person at play lost \( \frac{1}{4} \) of his money, and then won 3s.; after which he lost \( \frac{1}{4} \) of what he then had, and then won 2s.; lastly, he lost \( \frac{1}{4} \) of what he then had; and, this done, found he had only 12s. left: what had he at first?

Suppose he began to play with \( x \) shillings.

He lost \( \frac{1}{4} \) of his money, or \( \frac{x}{4} \), and had left \( x - \frac{x}{4} = \frac{3x}{4} \).

He won 3s. and had then \( \frac{3x}{4} + 3 = \frac{3x + 12}{4} \).

He lost \( \frac{1}{4} \) of \( \frac{3x + 12}{4} \), or \( \frac{x + 4}{4} \), and had left

\( \frac{3x + 12}{4} - \frac{x + 4}{4} = \frac{2x + 8}{4} \).

He won 2s. and had then \( \frac{2x + 8}{4} + 2 = \frac{2x + 16}{4} \).

He lost \( \frac{1}{4} \) of \( \frac{2x + 16}{4} \) or \( \frac{2x + 16}{28} \), and had left

\( \frac{2x + 16}{4} - \frac{2x + 16}{28} = \frac{12x + 96}{28} \).

And because he had now 12s. left, we have this equation,

\( \frac{12x + 96}{28} = 12 \).

Hence \( 12x = 240 \), and \( x = 20 \).

Ex. 9. Two tradesmen, A and B, are employed upon a piece of work; A can perform it alone in 15 hours, and B in 10 hours: in what time will they do it when working together?

Suppose that they can do it in \( x \) hours, and let the whole work be denoted by 1.

Then \( 15 : x :: 1 : \frac{x}{15} \) is the part of the work done by A.

And \( 10 : x :: 1 : \frac{x}{10} \) is the part done by B.

Now, by the question, they are to perform the whole work between them;

Therefore \( \frac{x}{15} + \frac{x}{10} = 1 \).

Hence \( 25x = 150 \), and \( x = 6 \) hours.

Ex. 10. The sum of any two quantities being given \( = s \), and their difference \( = d \), it is required to find each of the quantities.

Let \( x \) denote the greater of the two quantities, and \( y \) the lesser.

Then \( x + y = s \), and \( x - y = d \).

Taking the sum of the equations, we get \( 2x = s + d \);

And, subtracting the second from the first, \( 2y = s - d \);

Therefore \( x = \frac{s + d}{2} \), and \( y = \frac{s - d}{2} \). Ex. 11. A gentleman distributing money among some poor people, found he wanted 10s. to be able to give each 5s.; therefore he gave only 4s. to each, and had 5s. left. Required the number of shillings and poor people.

Let the number of shillings be \( x \), and that of the poor people \( y \); then, from the nature of the question, we have these two equations,

\[ 5y = x + 10 \\ 4y = x - 5. \]

From the first equation, \( x = 5y - 10 \), And from the second, \( x = 4y + 5 \); Therefore \( 5y - 10 = 4y + 5 \); Hence \( y = 15 \), and \( x = 4y + 5 = 65 \).

Ex. 12. A farmer kept a servant for every 40 acres of ground he rented, and on taking a lease of 104 more acres, he engaged 5 additional servants, after which he had a servant for every 36 acres. Required the number of servants and acres.

Suppose that he had at first \( x \) servants, and \( y \) acres.

From the first condition of the question, \( x = \frac{y}{40} \);

And from the second, \( x + 5 = \frac{y + 104}{36} \).

By comparing the values of \( x \), as found from these equations, we have

\[ \frac{y + 104}{36} - 5 = \frac{y}{40}. \]

Hence \( 40y + 4160 - 7200 = 36y \), so that \( 4y = 3040 \); Therefore \( y = 760 \), and \( x = \frac{y}{40} = 19 \).

Ex. 13. Two persons, A and B, were talking of their ages; says A to B, seven years ago I was just three times as old as you were then, and seven years hence I shall be just twice as old as you will be. What is their present ages?

Let the ages of A and B be \( x \) and \( y \) respectively. Their ages seven years ago were \( x - 7 \) and \( y - 7 \), and seven years hence they will be \( x + 7 \) and \( y + 7 \). Therefore by the question

\[ x - 7 = 3(y - 7) \quad \text{and} \quad x + 7 = 2(y + 7). \]

From the first equation, \( x = 3y - 14 \), And from the second, \( x = 2y + 7 \); Therefore \( 3y - 14 = 2y + 7 \); hence \( y = 21 \). And because \( x = 2y + 7 \), therefore \( x = 49 \).

Ex. 14. A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound's 3, but 2 of the greyhound's leaps are as much as 3 of the hare's. How many leaps must the greyhound take to catch the hare?

In this example there is only one quantity required, it will, however, be convenient to make use of two letters; therefore let \( x \) denote the number of leaps of the greyhound, and \( y \) those of the hare; then, by considering the proportion between the number of leaps each takes in the same time, we have

\[ 3 : 4 :: x : y, \quad \text{hence} \quad 3y = 4x. \]

Again, by considering the proportion between the number of leaps each must take to run the same distance, we find

\[ 50 : y :: 2 : 3, \quad \text{hence} \quad 100 + 2y = 3x. \]

From the first equation we find \( 6y = 8x \), And from the second \( 6y = 9x - 300 \). Hence \( 9x - 300 = 8x \), and \( x = 300 \).

Ex. 15. To divide the number 90 into 4 such parts, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sum, difference, product, and quotient, shall be all equal to each other.

In this question there are four quantities to be determined; but instead of introducing several letters, having put \( x \) to denote the first of them, we may find an expression for each of the remaining ones, as follows:

Because \( x + 2 \) is the second quantity — 2, Therefore \( x + 4 \) is the second quantity; And because \( x + 2 \) is third \( \times 2 \), Therefore \( \frac{x + 2}{2} \) is the third quantity.

And in like manner \( 2(x + 2) \) is the fourth quantity. Now, by the question, the sum of all the four = 90; Therefore \( x + x + 4 + \frac{x + 2}{2} + 2(x + 2) = 90 \). Hence \( 9x = 162 \), and \( x = 18 \); Therefore the numbers required are 18, 22, 10, and 40.

Ex. 16. A and B together can perform a piece of work in 12 hours, A and C in 20, and B and C in 15 hours; in what time will each be able to perform it when working separately?

That we may have a general solution, let us suppose A and B can perform the work in \( a \) hours, A and C in \( b \) hours, and B and C in \( c \) hours. Let \( x, y, \) and \( z \), denote the times in which A, B, and C, could perform it respectively, if each wrought alone; and let the whole work be represented by 1.

Then \( x : a :: 1 : \frac{a}{x} \) is the part done by A in \( a \) hours. \( y : a :: 1 : \frac{a}{y} \) is the part done by B in \( a \) hours. Also \( x : b :: 1 : \frac{b}{x} \) is the part done by A in \( b \) hours. \( z : b :: 1 : \frac{b}{z} \) is the part done by C in \( b \) hours. And \( y : c :: 1 : \frac{c}{y} \) is the part done by B in \( c \) hours. \( z : c :: 1 : \frac{c}{z} \) is the part done by C in \( c \) hours.

The question gives the three following equations:

\[ \frac{a}{x} + \frac{a}{y} = 1, \quad \frac{b}{x} + \frac{b}{z} = 1, \quad \frac{c}{y} + \frac{c}{z} = 1. \]

Let the first equation be divided by \( a \), the second by \( b \), and the third by \( c \); thus we have

\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{a}, \quad \frac{1}{x} + \frac{1}{z} = \frac{1}{b}, \quad \frac{1}{y} + \frac{1}{z} = \frac{1}{c}. \]

If these be added, and their sum divided by 2, we find

\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}. \]

From this equation let each of the three preceding be subtracted in its turn; thus we get

\[ \frac{1}{z} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{-ab + ac + bc}{2abc}, \] \[ \frac{1}{y} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{ab - ac + bc}{2abc}, \] \[ \frac{1}{x} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{-ab + ac + bc}{2abc}. \] Hence \( z = \frac{2abc}{ab + ac - bc} = \frac{7200}{120} = 60 \),

\[ y = \frac{2abc}{ab - ac + bc} = \frac{7200}{360} = 20 \]

\[ x = \frac{2abc}{ab + ac + bc} = \frac{7200}{240} = 30 \]

**Sect. IX.—Of Quadratic Equations.**

85. We are next to explain the resolution of equations of the second degree, or quadratic equations. These involve the second power of the unknown quantity, and may be divided into two kinds, pure and affected.

I. Pure quadratic equations are such as after proper reduction have the square of the unknown quantity in one term, while the remaining terms contain only known quantities. Thus, \( x^2 = 64 \), and \( ax^2 + bx + c = 0 \) are examples of pure quadratics.

II. Affected quadratic equations contain the square of the unknown quantity in one term, and its first or simple power in another; the remaining terms consisting entirely of known quantities. Such are the following, \( x^2 + 3x = 28 \), \( 2x^2 - 33 - 5x \), \( ax^2 + bx + c = 0 \).

The manner of resolving a pure quadratic equation is sufficiently evident. If the unknown quantity be made to stand alone on one side, with unity as a co-efficient, while the other side consists entirely of known quantities, and the square root of each side be taken, we immediately obtain the value of the simple power of the unknown quantity as directed by rule 5th of sect. VI.

86. In extracting the square root of any quantity, it is necessary to observe, that the sign of the root may be either \( + \) or \( - \). This is an evident consequence of the rule for the signs in multiplication; for since by that rule any quantity, whether positive or negative, if multiplied by itself, will produce a positive quantity, and therefore the square of \( +a \), as well as that of \( -a \), is \( +a^2 \); so, on the contrary, the square root of \( +a^2 \) is to be considered either as \( +a \) or as \( -a \), and may accordingly be expressed thus, \( \pm a \).

87. Having remarked that the square of any quantity, whatever be its sign, is always positive, it evidently follows that no real quantity whatever, when multiplied by itself, can produce a negative quantity; therefore, if the square root of a negative quantity be required, no such root can be assigned. Hence it also follows, that if a problem requires for its solution the extraction of the square root of a negative quantity, some contradiction must necessarily be involved, either in the condition of the problem, or in the process of reasoning by which that solution has been obtained.

88. When an affected quadratic equation is to be resolved, it may always, by proper reduction, be brought to one or other of the three following forms:

1. \( x^2 + px = q \). 2. \( x^2 - px = q \). 3. \( x^2 - px = -q \).

But as the manner of resolving each of the three forms is the very same, it will be sufficient if we consider any one of them.

Resuming therefore the first equation, or \( x^2 + px = q \), let us compare the side of it which involves the unknown quantity \( x \) with the square of a binomial \( x + a \); that is, let us compare \( x^2 + px \) with \( x^2 + 2ax + a^2 = (x + a)^2 \), and it will presently appear, that if we suppose \( p = 2a \), or \( \frac{p}{2} = a \), the quantities \( x^2 + px \) and \( x^2 + 2ax \) will be equal; and as \( x^2 + 2ax \) is rendered a complete square, by adding to it \( a^2 \), so also may \( x^2 + px \) be completed into a square by adding to it \( \frac{p^2}{4} \), which is equal to \( a^2 \); therefore, let \( \frac{p^2}{4} \) be added to both sides of the equation \( x^2 + px = q \), and we have

\[ x^2 + px + \frac{p^2}{4} = \frac{p^2}{4} + q, \text{ or } \left( x + \frac{p}{2} \right)^2 = \frac{p^2}{4} + q; \]

and, extracting the square root of each side, \( x + \frac{p}{2} = \pm \sqrt{\frac{p^2}{4} + q}; \) hence \( x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} + q}. \)

89. From these observations we derive the following general rules for resolving affective quadratic equations.

1. Bring all the terms involving the unknown quantity to one side, and the known quantities to the other side, and so that the term involving the square of the unknown quantity may be positive.

2. If the square of the unknown quantity be multiplied by a co-efficient, let the other terms be divided by it, so that the co-efficient of the square of the unknown quantity may be 1.

3. Add to both sides the square of half the co-efficient of the unknown quantity itself, and the side of the equation involving the unknown quantity will now be a complete square.

4. Extract the square root of both sides of the equation, by which it becomes simple with respect to the unknown quantity; and by transposition, that quantity may be made to stand alone on one side of the equation, while the other side consists of known quantities; and therefore the equation is resolved.

Note. The square root of the first side of the equation is always equal to the sum or difference of the unknown quantity, and half the co-efficient of the second term. If the sign of that term be \( + \), it is equal to the sum, but if it be \( - \), then it is equal to the difference.

Ex. 1. Given \( x^2 + 2x = 35 \), to determine \( x \).

Here the co-efficient of the second term is 2; therefore, adding the square of its half to each side, we have

\[ x^2 + 2x + 1 = 35 + 1 = 36, \]

And, extracting the square root, \( x + 1 = \sqrt{36} = \pm 6. \)

Hence \( x = \pm 6 - 1 \), that is \( x = +5 \), or \( x = -7 \), and either of these numbers will be found to satisfy the equation, for \( 5 \times 5 + 2 \times 5 = 35 \), also \( -7 \times -7 + 2 \times -7 = 35. \)

Ex. 2. Given \( \frac{x^2}{6} - 12 = x \), to find \( x \).

This equation, when reduced, becomes \( x^2 - 6x = 72 \).

And, by completing the square, \( x^2 - 6x + 9 = 72 + 9 = 81. \)

Hence, by extracting the square root, \( x - 3 = \pm 9 \), and \( x = \pm 9 + 3; \)

Therefore \( x = +12 \), or \( x = -6 \); and upon trial we find that each of these values satisfies the original equation, for \( \frac{12 \times 12}{6} - 12 = 12 \), also \( \frac{-6 \times -6}{6} - 12 = -6. \)

Ex. 3. Given \( x^2 + 28 = 11x \), to find \( x \).

Then \( x^2 - 11x = -28. \)

And, by completing the square, \( x^2 - 11x + \frac{121}{4} = \frac{121}{4} - 28 = \frac{9}{4}. \) Therefore, by extracting the root, \( x = \frac{11}{2} = \frac{3}{2} \).

Hence \( x = \frac{11}{2} = \frac{3}{2} \); that is, \( x = +7 \), or \( x = +4 \).

In the first two examples, we found one positive value for \( x \) in each, and also one negative value; but in this example both the values of \( x \) are positive, and, upon trial, each of them is found to satisfy the equation; for \( 7 \times 7 + 28 = 11 \times 7 \), also \( 4 \times 4 + 28 = 11 \times 4 \).

90. As at first sight it appears remarkable, that in every quadratic equation the unknown quantity admits always of two distinct values or roots, it will be proper to consider a little further the circumstances upon which this peculiarity depends. This is the more necessary, as the property of the unknown quantity admitting of several values is not peculiar to quadratics, but takes place also in equations of the higher degrees, where the cause of the ambiguity requires an explanation somewhat different from that which we have already given in the present case.

91. Let us again consider the equation \( x^2 + 2x - 35 \), which forms the first of the three preceding examples. By bringing all the terms to one side, the same equation may be also expressed thus, \( x^2 + 2x - 35 = 0 \); so that we shall have determined \( x \), when we have found such a number as, when substituted for it in the quantity \( x^2 + 2x - 35 \), will render the result equal to 0. But \( x^2 + 2x - 35 \) is the product of these two factors \( x - 5 \) and \( x + 7 \), as may be proved by actual multiplication; therefore, to find \( x \), we have \( (x - 5)(x + 7) = 0 \); and as a product can only become 0 when one of its factors is reduced to 0, it follows that either of the two factors \( x - 5 \) and \( x + 7 \) may be assumed = 0. If \( x - 5 = 0 \), then \( x = 5 \); but if \( x + 7 = 0 \), then \( x = -7 \); so that the two values of \( x \), or two roots of the equation \( x^2 + 2x - 35 \), are \( +5 \) and \( -7 \), as we have already found in a different manner.

92. What has been shown in a particular case is true of any quadratic equation whatever; that is, if \( x^2 + px + q = 0 \), or, by bringing all the terms to one side, \( x^2 + px - q = 0 \), it is always possible to find two factors \( x - a \) and \( x + b \), such that

\[ x^2 + px + q = (x - a)(x + b), \]

where \( a \) and \( b \) are known quantities, which depend only upon \( p \) and \( q \), the given numbers in the equation; and since that to have \( (x - a)(x + b) = 0 \), we may either assume \( x - a = 0 \) or \( x + b = 0 \), it evidently follows that the conditions of the equation \( x^2 + px + q = 0 \), or \( x^2 + px - q = 0 \), are alike satisfied by taking \( x = a \) or \( x = -b \).

From these considerations it follows, that \( x \) can have only two values in a quadratic equation; for if it could be supposed to have three or more values, then it would be possible to resolve \( x^2 + px - q \) into as many factors, \( x - c, x - d, \ldots \); but the product of more than two factors must necessarily contain the third or higher powers of \( x \), and as \( x^2 + px - q \) contains no higher power than the second, therefore no such resolution can take place.

93. Since it appears that \( x^2 + px - q \) may be considered as the product of the two factors \( x - a \) and \( x + b \), let us examine the nature of these factors. Accordingly, taking their product, we find it \( x^2 + (b - a)x - ab \); and since this quantity must be equal to \( x^2 + px - q \), it follows that \( b - a = p \) and \( ab = q \), or, changing the signs of the terms of both equations, \( a - b = -p \), \( ab = -q \). Now, if we consider that \( +a \) and \( -b \) are the roots of the equation \( x^2 + px - q \), it is evident that \( a - b \) is the sum of the roots, and \( -ab \) their product. So that from the equations \( a - b = -p \), and \( -ab = q \), we derive the following proposition relating to the roots of any quadratic equation. The sum of the roots of any quadratic equation \( x^2 + px + q = 0 \) is equal to \( -p \), that is, to the coefficient of the second term, having its sign changed; and their product is equal to \( -q \), or to the latter side of the equation, having its sign also changed.

This proposition enables us to resolve several important questions concerning the roots of a quadratic equation, without actually resolving that equation. Thus we learn from it, that if \( q \), the term which does not involve the unknown quantity (called sometimes the absolute number), be positive, the equation has one of its roots positive and the other negative; but if that term be negative, the roots are either both positive or both negative. It also follows, that in the former case the root which is denoted by the least number will have the same sign with the second term; and in the latter case, the common sign of the roots will be the contrary to that of the second term.

94. From this property of the roots we may also derive a general solution to any quadratic equation \( x^2 + px + q = 0 \); for we have only to determine two quantities whose sum is \( -p \) and product \( -q \), and these shall be the two values of \( x \), or the two roots of the equation.

Without considering the signs of the roots, let us call them \( v \) and \( z \); then

\[ v + z = -p, \quad vz = -q. \]

From the square of each side of the first equation let four times the second be subtracted, and we have

\[ v^2 - 2vz + z^2 = p^2 + 4q, \quad \text{or} \quad (v - z)^2 = p^2 + 4q; \]

therefore, \( v - z = \pm \sqrt{p^2 + 4q} \).

From this equation, and from the equation

\[ v + z = -p, \]

we readily obtain

\[ v = \frac{-p + \sqrt{p^2 + 4q}}{2}, \quad z = \frac{-p - \sqrt{p^2 + 4q}}{2}; \]

that is, if \( v = \frac{-p + \sqrt{p^2 + 4q}}{2} \),

then \( z = \frac{-p - \sqrt{p^2 + 4q}}{2} \); and if \( v = \frac{-p - \sqrt{p^2 + 4q}}{2} \),

then \( z = \frac{-p + \sqrt{p^2 + 4q}}{2} \).

But the value of \( v \) upon the one supposition is the same as the value of \( z \) upon the other supposition, and vice versa; therefore, in reality, the only two distinct values of the roots \( v \) and \( z \) are \( \frac{-p + \sqrt{p^2 + 4q}}{2} \) and \( \frac{-p - \sqrt{p^2 + 4q}}{2} \), which agrees with the conclusion we have already found (sect. 89).

95. It appears, from what has been already shown, that the roots of a quadratic equation \( x^2 + px + q = 0 \) always involve the quantity \( \sqrt{p^2 + 4q} \); hence it follows, that \( p^2 + 4q \) must be a positive quantity; for if it were negative, as the square root of such a quantity could not be found, the value of \( x \) could not possibly be obtained. If, for example, the value of \( x \) were required from this equation, \( x^2 + 13 = 4x \), or \( x^2 - 4x = -13 \), we should find \( x = 2 \pm \sqrt{-9} \); and as this expression for the roots requires us to extract the square root of \( -9 \), the equation from which it is derived must necessarily have involved some contradiction. It is not difficult to see wherein the absurdity consists; for since in this case \( p = -4 \), and \( q = -13 \), the roots of the equation ought to be both positive (sect. 93), and such that their sum = 4, while their product = 13 (sect. 93), which is impossible.

96. Although imaginary quantities serve no other purpose in the resolution of quadratic equations than to show that a particular problem cannot be resolved, by reason of some want of consistency in its data, yet they are not upon that account to be rejected. By introducing them into mathematical investigations, many curious theories may be explained, and problems resolved, in a more concise way than can be done without the use of such quantities. This is particularly the case in the higher parts of the mathematics.

The method which has been applied to the resolution of quadratic equations properly so called, namely, such as are of this form, \(x^2 + px = q\), will also apply to all equations of this form,

\[x^n + px^n = q,\]

where the unknown quantity \(x\) is found only in two terms, and such that its exponent in the one term is double that in the other; for let us assume \(x^n = y\), then \(x^n = y^2\), and therefore the equation

\[y^2 + py = q;\]

a quadratic equation, from which \(x\) may be found, and thence \(x\), by considering that \(x = \sqrt{y}\).

97. Although every quadratic equation admits of two roots, yet it will frequently happen that only one of them can be of use, the other being excluded by the conditions of the question. This will often be the case with respect to the negative root; as, for example, when the unknown quantity denotes a number of men, a number of days, &c. And hence, in reckoning the cases of quadratic equations, it is common to neglect this one, \(x^2 + px = -q\), where the roots are both negative; for an equation of this form can only be derived from a question which has some fault in its enunciation, and which, by a proper change in its form, will produce another equation having both its roots positive.

98. The remainder of this section shall be employed in solving some questions which produce quadratic equations.

Ex. 1. It is required to divide the number 10 into two such parts that the sum of their squares may be 58.

Let \(x\) be the one number;

Then, since their sum is 10, we have \(10 - x\) for the other;

And by the question \(x^2 + (10 - x)^2 = 58;\)

That is, \(x^2 + 100 - 20x + x^2 = 58;\)

Or \(2x^2 - 20x = 58 - 100 = -42;\)

Hence \(x^2 - 10x = -21.\)

And completing the square, \(x^2 - 10x + 25 = 25 - 21 = 4;\)

Hence, by extracting the root, \(x - 5 = \pm \sqrt{4} = \pm 2\)

And \(x = 5 \pm 2,\)

That is, \(x = 7\) or \(x = 3.\)

If we take the greatest value of \(x\), viz. 7, the other number \(10 - x\) will be 3; and if we take the least value of \(x\), viz. 3, then the other number is 7. Thus it appears, that the greatest value of the one number corresponds to the least of the other; and indeed this must necessarily be the case, seeing that both are alike concerned in the question. Hence, upon the whole, the only numbers that will answer the conditions of the question are 7 and 3.

Ex. 2. What two numbers are those whose product is 28; and such, that twice the greater, together with thrice the lesser, is equal to 26?

Let \(x\) be the greatest and \(y\) the least number; then, from the nature of the question, we have these two equations,

\[xy = 28, \quad 2x + 3y = 26.\]

From the first equation we have \(y = \frac{28}{x},\)

And from the second \(y = \frac{26 - 2x}{3}.\)

Hence \(\frac{26 - 2x}{3} = \frac{28}{x};\)

And, reducing, \(26x - 2x^2 = 84;\)

Or \(2x^2 - 26x = -84;\)

Hence \(x^2 - 13x = -42;\)

And comp. the sq. \(x^2 - 13x + \frac{169}{4} = \frac{169}{4} - 42 = \frac{1}{4};\)

Hence, by extracting the root, \(x - \frac{13}{2} = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2};\)

Therefore \(x = \frac{15}{2} \pm \frac{1}{2};\)

That is, \(x = 7\) or \(x = 6.\)

And since \(y = \frac{28}{x},\) we have \(y = 4,\) or \(y = \frac{14}{3}.\)

Thus we have obtained two sets of numbers, which fulfil the conditions required, viz.

\[x = 7, \quad y = 4;\] Or \(x = 6, \quad y = \frac{14}{3}.\)

And besides these, there can be no other numbers.

Ex. 3. A company dining together at an inn, find their bill amount to 175 shillings; two of them were not allowed to pay, and the rest found that their shares amounted to 10 shillings a man more than if they had all paid. How many were in company?

Suppose their number to be \(x.\)

Then, if all had paid, the share of each would have been \(\frac{175}{x}.\)

But because only \(x - 2\) paid, the share of each was \(\frac{175}{x - 2}.\)

Therefore, by the question, \(\frac{175}{x - 2} - \frac{175}{x} = 10.\)

And, by reduction, \(175x - 175(x - 2) = 10x^2 - 20x;\)

That is, \(10x^2 - 20x = 350;\)

Or \(x^2 - 2x = 35;\)

And comp. the sq. \(x^2 - 2x + 1 = 35 + 1 = 36;\)

Hence, by extracting the root, \(x^2 + 1 = \pm 6;\)

Therefore \(x = 5\) or \(x = -7.\) But from the nature of the question, the negative root can be of no use; therefore \(x = 6.\)

Ex. 4. A mercer sold a piece of cloth for L.24, and gained as much per cent. as the cloth cost him. What was the price of the cloth?

Suppose that it cost \(x\) pounds;

Then the gain was \(24 - x;\)

And, by the question, \(100 : x = x : 24 - x.\)

Therefore, taking the product of the extremes and means,

\[2400 - 100x = x^2;\]

Or \(x^2 + 100x = 2400;\)

And comp. the sq. \(x^2 + 100x + 2500 = 4900;\)

Hence, taking the root, \(x + 50 = \pm 70;\)

And \(x = 20\) or \(-120.\)

Here, as in the last question, the negative root cannot apply; therefore \(x = 20\) pounds, the price required.

Ex. 5. A grazier bought as many sheep as cost him L.60, out of which he reserved 15, and sold the remainder for L.54, and gained 2s. each upon them. How many sheep did he buy, and what did each cost him? Suppose that he bought $x$ sheep; then each would cost him $\frac{1200}{x}$ shillings.

Therefore, after reserving 15, he sold each of the remaining $x-15$ for $\frac{1200}{x} + 2$ shillings.

Hence, he would receive for them $(x-15)\left(\frac{1200}{x} + 2\right)$ shillings. And, because L54 = 1080 shillings, we have by the question

$$(x-15)\left(\frac{1200}{x} + 2\right) = 1080;$$

Which, by proper reduction, becomes $x^2 + 45x = 9000$;

And, completing the square, $x^2 + 45x + \frac{2025}{4} = \frac{38025}{4}$.

Therefore, extracting the root, &c. $x = \pm \frac{195}{2} - \frac{45}{2}$.

And, taking the positive root, $x = 75$, the number of sheep; and consequently $\frac{1200}{75} = 16$ shillings, the price of each.

Ex. 6. What number is that which, when divided by the product of its two digits, the quotient is 3; and if 18 be added to it, the digits are inverted?

Let $x$ and $y$ denote the digits; then the number itself will be expressed by $10x+y$, and that number in which the digits are inverted, by $10y+x$. Thus the conditions of the problem will be expressed by these two equations,

$$\frac{10x+y}{xy} = 3, \quad 10x+y+18 = 10y+x.$$

From the first equation we have $y = \frac{10x}{3x-1}$,

And from the second, $y = x + 2$;

Therefore $x + 2 = \frac{10x}{3x-1}$;

And $3x^2 + 5x - 2 = 10x$.

Hence, $x^2 + \frac{5}{3}x - \frac{2}{3} = 0$;

And comp. sq. $x^2 + \frac{5}{3}x + \frac{25}{36} = \frac{39}{36}$;

Therefore, taking the root, $x = \frac{-5}{6} \pm \frac{\sqrt{39}}{6}$;

So that $x = 2$, or $x = -\frac{1}{2}$.

Here it is evident that the negative root is useless; hence, $y = x + 2 = 4$, and 24 is the number required.

Ex. 7. Find two numbers whose product is 100, and the difference of their square roots 3.

Let $x$ be the one number; then $\frac{100}{x}$ is the other.

Now by the question, $\frac{10}{\sqrt{x}} = 3$;

Hence we have $10 - 3\sqrt{x} = 3x^{\frac{1}{2}}$,

Or $x + 3x^{\frac{1}{2}} = 10$;

And comp. the sq. $x + 3x^{\frac{1}{2}} + \frac{9}{4} = 10 + \frac{25}{4}$,

and taking the root, $x^{\frac{1}{2}} + \frac{3}{2} = \frac{7}{2}$;

So that $x^{\frac{1}{2}} = +5$, or $x^{\frac{1}{2}} = -2$,

and therefore $x = 25$, or $x = 4$.

If $x = 4$, the other number is $\frac{100}{4} = 25$, and if $x = 25$, then the other number is 4; so that, in either case, the two numbers which answer the conditions of the question are 4 and 25.

Ex. 8. It is required to find two numbers, of which the product shall be 6, and the sum of their cubes 35.

Let $x$ be the one number; then $\frac{6}{x}$ will be the other.

Therefore, by the question, $x^3 + \frac{216}{x^3} = 35$;

Hence $x^3 + 216 = 35x^3$,

Or $x^3 - 35x^3 = -216$.

This equation, by putting $x^3 = y$, becomes

$y^2 - 35y = -216$;

Hence we find $y = 27$, or $y = 8$.

And since $x^3 = y$, therefore $x = 3$, or $x = 2$.

If $x = 3$, then the other number is 2, and if $x = 2$, the other number is 3; so that 2 and 3 are the numbers required.

In general, if it be required to find two numbers which are exactly alike concerned in a question that produces a quadratic equation, they will be the roots of that equation. A similar observation applies to any number of quantities which require for their determination the resolution of an equation of any degree whatever.

Sect. X.—Of Equations in General

99. Before we proceed to the resolution of cubic and the higher orders of equations, it will be proper to explain some general properties which belong to equations of every degree, and also certain operations which must frequently be performed upon equations before they be fitted for a final solution.

In treating of equations in general, we shall suppose all the terms brought to one side, and put equal to 0; so that an equation of the fourth degree will stand thus:

$$x^4 + px^3 + qx^2 + rx + s = 0,$$

where $x$ denotes an unknown quantity, and $p$, $q$, $r$, $s$, known quantities, either positive or negative. Here the co-efficient of the highest power of $x$ is unity, but had it been any other quantity, that quantity might have been taken away, and the equation reduced to the above form, by rules already explained (Sect. VI).

The terms being thus arranged, if such a quantity be found as, when substituted for $x$, will render both sides = 0, and therefore satisfy the equation, that quantity, whether it be positive or negative, or even imaginary, is to be considered as a root of the equation. But we have seen that every quadratic equation has always two roots, real or imaginary; we may therefore suppose that a similar diversity will take place in all equations of a higher degree; and this supposition appears to be well founded, by the following proposition, which is of great importance in the theory of equations.

If a root of any equation, as $x^4 + px^3 + qx^2 + rx + s = 0$, be represented by $a$, the first side of that equation is divisible by $x-a$;

For since $x^4 + px^3 + qx^2 + rx + s = 0$,

And also $a^4 + pa^3 + qa^2 + ra + s = 0$;

Therefore, by subtraction,

$$x^4 - a^4 + p(x^3 - a^3) + q(x^2 - a^2) + r(x - a) = 0.$$

But any quantity of this form $x^n - a^n$, where $n$ denotes a whole positive number, is equal to

$$(x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-2}x + a^{n-1}),$$

as may be proved by multiplication; therefore, putting $x = 4$, 3, and 2 successively, we have

$$x^4 - a^4 = (x-a)(x^3 + ax^2 + a^2x + a^3),$$

$$x^3 - a^3 = (x-a)(x^2 + ax + a^2),$$

$$x^2 - a^2 = (x-a)(x+a),$$

$$x - a = (x-a);$$

and by substitution, and collecting into one term the coefficients of the like powers of $x$, the equation becomes Algebra.

\[(x-a)(x^3 + (a+p)x^2 + (a^2+pa+q)x + a^3+pa^2+qa+r) = 0;\] so that, putting \(p' = a + p, q' = a^2 + pa + q, r' = a^3 + pa^2 + qa + r\), we have

\[x^4 + px^3 + qx^2 + rx + s = (x-a)(x^3 + p'x^2 + q'x + r').\]

Hence, if the proposed equation \(x^4 + px^3 + qx^2 + rx + s\) be divided by \(x-a\), the quotient will be \(x^3 + p'x^2 + q'x + r'\), an integer quantity; and since the same mode of reasoning will apply to any equation whatever, the truth of the proposition is evident.

We have found that \((x-a)(x^3 + p'x^2 + q'x + r') = 0\); and as a product becomes \(=0\), when any one of its factors \(=0\), therefore the equation will have its conditions fulfilled, not only when \(x-a = 0\), but also when \(x^3 + p'x^2 + q'x + r' = 0\).

Let us now suppose that \(b\) is a root of this equation; then, by reasoning exactly as in last article, and putting \(p'' = b + p', q'' = b^2 + pb + q'\), we shall have

\[x^3 + p''x^2 + q''x + r'' = (x-b)(x^2 + p''x + q'') = 0.\]

By proceeding in the same manner with the quadratic equation \(x^2 + p''x + q'' = 0\), we shall find that if \(c\) denote one of its roots, then

\[x^2 + p''x + q'' = (x-c)(x+c+p').\]

So that if we put \(d = -(c+p')\), we at last find

\[x^4 + px^3 + qx^2 + rx + s = (x-a)(x-b)(x-c)(x-d) = 0;\]

and since each of the factors \(x-a, x-b, x-c, x-d\) may be assumed \(=0\), it follows that there are four different values of \(x\), which will render the equation \(x^4 + px^3 + qx^2 + rx + s = 0\), namely, \(x = a, x = b, x = c, x = d\).

The mode of reasoning which has been just now employed in a particular case, may be applied to an equation of any order whatever; we may therefore conclude, that every equation may be considered as the product of as many simple factors as the number denoting its order contains unity, and therefore, that the number of roots in any equation is precisely equal to the exponent of the highest power of the unknown quantity contained in that equation.

100. By considering equations of all degrees as formed from the products of factors \(x-a, x-b, x-c, \ldots\), we discover curious relations, which subsist between the roots of any equation and its co-efficients. Thus, if we limit the number of factors to four, and suppose that \(a, b, c, d\) are the roots of this equation of the fourth degree,

\[x^4 + px^3 + qx^2 + rx + s = 0,\]

we shall also have \((x-a)(x-b)(x-c)(x-d) = 0\); and therefore, by actual multiplication,

\[\begin{align*} x^4 &- ax^3 - bx^3 - cx^3 - dx^3 \\ &+ ax^2 + bx^2 + cx^2 + dx^2 \\ &- ax - bx - cx - dx \\ &+ ab + ac + ad + bc + cd \\ &- abc - abd - acd - bcd \\ &+ abcd = 0. \end{align*}\]

If we compare together the co-efficients of the same powers of \(x\), we find the following series of equations:

\[\begin{align*} a + b + c + d &= -p, \\ ab + ac + ad + bc + cd &= +q, \\ abc + abd + acd + bcd &= -r, \\ abcd &= +s; \end{align*}\]

and as similar results will be obtained for equations of all degrees, we hence derive the following propositions, which are of great importance in the theory of equations.

1. The co-efficient of the second term of any equation, taken with a contrary sign, is equal to the sum of all the roots.

2. The co-efficient of the third term is equal to the sum of the products of the roots multiplied together two and two.

3. The co-efficient of the fourth term, taken with a contrary sign, is equal to the sum of the roots multiplied together three and three; and so on in the remaining coefficients, till we come to the last term of the equation, which is equal to the product of all the roots having their signs changed.

Instead of supposing an equation to be produced by multiplying together simple equations, we may consider it as formed by the product of equations of any degree, provided that the sum of their dimensions be equal to that of the proposed equation. Thus, an equation of the fourth degree may be formed either from a simple and cubic equation, or from two quadratic equations.

101. If \(n\) denote the degree of an equation, we have shown, that by considering it as the product of simple factors, that equation will have \(n\) divisors of the first degree; but if we suppose the simple factors to be combined two and two, they will form quantities of the second degree, which are also factors of the equation; and since there may be formed \(\frac{n(n-1)}{1 \cdot 2}\) such combinations, any equation will admit of \(\frac{n(n-1)}{1 \cdot 2}\) divisors of the second degree.

For example, the equation \(x^4 + px^3 + qx^2 + rx + s = 0\), which we have considered as equal to

\[(x-a)(x-b)(x-c)(x-d) = 0,\]

may be formed of the product of two factors of the second degree, in these six different ways.

By the product of \((x-a)(x-b)\) and \((x-c)(x-d)\),

\[\begin{align*} (x-a)(x-c) &- (x-a)(x-d) \\ (x-a)(x-d) &- (x-b)(x-c) \\ (x-b)(x-c) &- (x-a)(x-d) \\ (x-b)(x-d) &- (x-a)(x-c) \\ (x-c)(x-d) &- (x-a)(x-b). \end{align*}\]

Thus an equation of the fourth degree may have \(\frac{4 \cdot 3}{1 \cdot 2} = 6\) quadratic divisors.

By combining the simple factors three and three, we shall have divisors of the third degree, of which the number for an equation of the \(n\)th order will be \(\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}\); and so on.

102. When the roots of an equation are all positive, its simple factors will have this form, \(x-a, x-b, x-c, \ldots\); and if, for the sake of brevity, we take only these three, the cubic equation which results from their product will have this form,

\[x^3 - px^2 + qx - r = 0,\]

where \(p = a + b + c, q = ab + ac + bc, r = abc;\)

and here it appears that the signs of the terms are \(+\) and \(-\) alternately.

Hence we infer, that when the roots of an equation are all positive, the signs of its terms are positive and negative alternately.

If again the roots of the equation be all negative, and therefore its factors \(x+a, x+b, x+c\), then \(p, q, r\) being as before, the resulting equation will stand thus:

\[x^3 + px^2 + qx + r = 0.\]

And hence we conclude, that when the roots are all negative, there is no change whatever in the signs.

103. In general, if the roots of an equation be all real, that equation will have as many positive roots as there are changes of the signs from \(+\) to \(-\), or from \(-\) to \(+\); and the remaining roots are negative. This rule, however, does not apply when the equation has imaginary Algebra roots, unless such roots be considered as either positive or negative.

That the rule is true when applied to quadratic equations will be evident from Sect. IX. With respect to cubic equations, the rule also applies when the roots are either all positive, or all negative, as we have just now shown.

When a cubic equation has one positive root, and the other two negative, its factors will be \(x-a\), \(x+b\), \(x+c\), and the equation itself

\[ \begin{align*} (x-a) & \quad (x-b) \\ +ab & \quad +ac \\ +bc & \quad +be \\ \end{align*} \]

Here there must always be one change of the signs, since the first term is positive and the last negative; and there can be no more than one; so the second term is negative, or \(b+c\) less than \(a\); then \((b+c)^2\) will be less than \((b+c)a\); but \((b+c)^2\) is always greater than \(bc\), therefore \(bc\) will be much less than \((b+c)a\), or \(ab+ac\), so that the third term must also be negative, and therefore, in this case there can be only one change of the signs. If, again, the second term be positive, then because the sign of the last term is negative, whatever be the sign of the third term, there can still be no more than one change of the signs.

When the equation has two positive roots, and one negative, its factors are \(x-a\), \(x-b\), \(x+c\), and the equation

\[ \begin{align*} (x-a) & \quad (x-b) \\ +ab & \quad +ac \\ +bc & \quad +be \\ \end{align*} \]

Here there must always be two changes of the signs; for if \(a+b\) be greater than \(c\), the second term is negative, and the last term being always positive, there must be two changes, whether the sign of the third term be positive or negative. If, again, \(a+b\) be less than \(c\), and therefore the second term positive, it may be shown as before, that \(ab\) is much less than \(ac+bc\); and hence the third term will be negative; so that in either case there must be two changes of the signs. We may conclude, therefore, upon the whole, that in cubic equations there are always as many positive roots as changes of the signs from \(+\) to \(-\), or from \(-\) to \(+\); and, by the same method of reasoning, the rule will be found to extend to all equations whatever.

104. It appears, from the manner in which the co-efficients of an equation are formed from its roots, that when the roots are all real, the co-efficients must consist entirely of real quantities. But it does not follow, on the contrary, that when the co-efficients are real, the roots are also real; for we have already found, that in a quadratic equation, \(x^2+px+q=0\), where \(p\) and \(q\) denote real quantities, the roots are sometimes both imaginary.

When the roots of a quadratic equation are imaginary, they have always this form, \(a+\sqrt{-b}\), \(a-\sqrt{-b}\), which quantities may also be expressed thus, \(a+b\sqrt{-1}\), \(a-b\sqrt{-1}\); so that we have these two factors \(x-a-b\sqrt{-1}\), \(x-a+b\sqrt{-1}\), and taking their product,

\[ x^2-2ax+a+b^2=0. \]

Thus we see that two imaginary factors may be of such a form as to admit of their product being expressed by a real quantity; and hence the origin of imaginary roots in quadratic equations.

105. It appears by induction, that no real equation can be formed from imaginary factors, unless those factors be taken in pairs, and each pair have the form \(x-a-b\sqrt{-1}\), \(x-a+b\sqrt{-1}\); for the product of three, or any odd number of imaginary factors, whatever be their form, is still an imaginary quantity. Thus, if we take the product of any three of these four imaginary expressions, \(x+a-b\sqrt{-1}\), \(x+a+b\sqrt{-1}\), \(x+c-d\sqrt{-1}\), \(x+c+d\sqrt{-1}\), we may form four different equations, each of which will involve imaginary quantities. If, however, each equation be multiplied by the remaining factor, which had not previously entered into its composition, the product will be found to be rational, and the same for all the four.

Hence we may deduce the three following inferences respecting the roots of equations:

1. If an equation have imaginary roots, it must have two, or four, or some even number of such roots. 2. If the degree of an equation be denoted by an odd number, that equation must have at least one real root. 3. If the degree of an equation be denoted by an even number, and that equation have one real root, it will also have another real root.

106. We shall now explain some transformations which are frequently necessary to prepare the higher orders of equations for a solution.

Any equation may have its positive roots changed into negative roots of the same value, and its negative roots into such as are positive, by changing the signs of the terms alternately, beginning with the second. The truth of this remark will be evident, if we take two equations,

\[ \begin{align*} (x-a)(x-b)(x+c)=0, \\ (x+a)(x+b)(x-c)=0, \end{align*} \]

(which are such, that the positive roots of the one have the same values as the negative roots of the other), and multiply together their respective factors; for these equations will stand thus:

\[ \begin{align*} (x-a) & \quad (x-b) \\ +ab & \quad +ac \\ +bc & \quad +be \\ \end{align*} \]

where it appears that the signs of the first and third terms are the same in each, but the signs of the second and fourth are just the opposite of each other. And this will be found to hold true, not only of cubic equations, but of all equations, to whatever order they belong.

107. It will sometimes be useful to transform an equation into another that shall have each of its roots greater or less than the corresponding roots of the other equation, by some given quantity.

Let \((x-a)(x-b)(x+c)=0\) be any proposed equation which is to be transformed into another, having its roots greater or less than those of the proposed equation by the given quantity \(n\); then, because the roots of the transformed equation are to be \(+a+n\), \(+b+n\), and \(-c-n\), the equation itself will be

\[ (y-n-a)(y-n-b)(y-n+c)=0. \]

Hence the reason of the following rule is evident.

If the new equation is to have its roots greater than those of the proposed equation; instead of \(x\) and its powers, substitute \(y-n\) and its powers; but if the roots are to be less; then, instead of \(x\) substitute \(y+n\); and in either case, a new equation will be produced, the roots of which shall have the property required.

108. By the preceding rule, an equation may be changed into another, which has its roots either all positive, or all negative; but it is chiefly used in preparing cubic and biquadratic equations for a solution, by transforming them Let \( x^3 + px^2 + qx + r = 0 \) be any cubic equation; if we substitute \( y + n \) for \( x \), the equation is changed into the following:

\[ \begin{align*} y^3 + 3ny^2 + 3n^2y + n^3 + p(y^2 + 2ny + n^2) + q(y + n) + r &= 0. \end{align*} \]

Now, that this equation may want its second term, it is evident that we have only to suppose \( 3n + p = 0 \), or \( n = -\frac{p}{3} \); for this assumption being made, and the value of \( n \) substituted in the remaining terms, the equation becomes

\[ y^3 + (q - \frac{p^2}{3})y + \frac{2p^3}{27} - \frac{pq}{3} + r = 0; \]

or, putting \( -\frac{p^2}{3} + q = q' \), and \( \frac{2p^3}{27} - \frac{pq}{3} + r = r' \), the same equation may also stand thus,

\[ y^3 + q'y + r' = 0. \]

109. In general, any equation whatever may be transformed into another, which shall want its second term, by the following rule.

Divide the co-efficient of the second term of the proposed equation by the exponent of the first term, and add the quotient, with its sign changed, to a new unknown quantity; the sum being substituted for the unknown quantity in the proposed equation, a new equation will be produced, which will want the second term, as required.

By this rule any affected quadratic equation may be readily resolved; for by transforming it into another equation which wants the second term, we thus reduce its solution to that of a pure quadratic. Thus, if the quadratic equation \( x^2 - 5x + 6 = 0 \) be proposed; by substituting \( y + \frac{5}{2} \) for \( x \), we find

\[ \begin{align*} y^2 + 5y + \frac{25}{4} - 3y - \frac{25}{4} &= 0, \\ \text{or } y^2 + 2y + \frac{9}{4} &= 0. \end{align*} \]

Hence \( y = \pm \frac{3}{2} \), and since \( x = y + \frac{5}{2} \), therefore \( x = \pm \frac{1}{2} + \frac{5}{2} = \pm 3, \text{ or } \pm 2 \).

110. It has been shown (sect. 109) that in any equation, the co-efficient of the second term, having its sign changed, is equal to the sum of all the roots; or, abstracting from their signs, it is equal to the difference between the sum of the positive and the sum of the negative roots; Therefore, if the second term be wanting, the sum of the positive roots in the equation must necessarily be equal to that of the negative roots.

111. Instead of taking away the second term from an equation, any other term may be made to vanish, by an assumption similar to that which has been employed to take away the second term. Thus, if in sect. 108 we assume \( 3n^2 + 2pn + q = 0 \), by resolving this quadratic equation, a value of \( n \) will be found which, when substituted in the equation, will cause the third term to vanish; and, by the resolution of a cubic equation, the third term may be taken away; and so on.

112. Another species of transformation, of use in the resolution of equations, is that by which an equation, having the co-efficients of some of its terms expressed by fractional quantities, is changed into another, the co-efficients of which are all integers.

Let \( x^3 + \frac{P}{a}x^2 + \frac{Q}{b}x + \frac{R}{c} = 0 \) denote an equation to be so transformed, and let us assume \( y = abcx \), and therefore \( x = \frac{y}{abc} \); then, by substitution, our equation becomes

\[ \frac{y^3}{a^3b^3c^3} + \frac{P}{a^2b^3c^3}y^2 + \frac{Q}{ab^2c^3}y + \frac{R}{abc^3} = 0; \]

and multiplying the whole equation by \( a^3b^3c^3 \), we have

\[ y^3 + bcpy^2 + a^2bc^2gy + a^3b^2c^3r = 0. \]

Thus we have an equation free from fractions, while at the same time the co-efficient of the highest power of the unknown quantity is unity, as before.

This transformation may always be performed by the following rule: Instead of the unknown quantity, substitute a new unknown quantity divided by the product of all the denominators; then, by proper reduction, the equation will be found to have the form required.

If, however, the equation have this form,

\[ x^3 + \frac{P}{a}x^2 + \frac{Q}{b}x + \frac{R}{c} = 0, \]

it will be sufficient to assume \( y = ax \), and therefore \( x = \frac{y}{a} \); for then we have

\[ \frac{y^3}{a^3} + \frac{P}{a^2}y^2 + \frac{Q}{ab}y + \frac{R}{ac} = 0, \]

and \( y^3 + py^2 + qy + r = 0 \);

which last equation has the form required.

Sect. XI.—Of Cubic Equations.

113. Cubic equations, as well as equations of every higher degree, are, like quadratics, divided into two classes: they are said to be pure when they contain only one power of the unknown quantity; and affected when they contain two or more powers of that quantity.

Pure cubic equations are therefore of this form, \( x^3 = 125 \), or \( x^3 = -27 \), or, in general, \( x^3 = r \); and hence it appears, that the 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; thus, from the first of the three preceding examples we find \( x = \sqrt[3]{125} \), from the second, \( x = \sqrt[3]{-27} \), and from the third, \( x = \sqrt[3]{r} \).

It would seem at first sight that the only value which \( x \) can have in the cubic equation \( x^3 = r \), or putting \( r = c^3 \), \( x^3 - c^3 = 0 \), is this one, \( x = c \); but since \( x^3 - c^3 \) may be resolved into these two factors, \( x - c \) and \( x^2 + cx + c^2 \), it follows, that besides the value of \( x \) already found, which results from making the factor \( x - c = 0 \), it has yet other two values, which may be found by making the other factor \( x^2 + cx + c^2 = 0 \); and accordingly, by resolving the quadratic equation \( x^2 + cx + c^2 = 0 \), we find these values to be

\[ \frac{-c + \sqrt{-3c^2}}{2} \quad \text{and} \quad \frac{-c - \sqrt{-3c^2}}{2}, \]

or \( \frac{-1 + \sqrt{-3}}{2}c \) and \( \frac{-1 - \sqrt{-3}}{2}c \).

Thus it appears, that any cubic equation of this form, \( x^3 = c^3 \), or \( x^3 - c^3 = 0 \), has these three roots,

\[ x = c, \quad x = \frac{-1 + \sqrt{-3}}{2}c, \quad x = \frac{-1 - \sqrt{-3}}{2}c; \]

the first of which is real, but the last two are imaginary.

If, however, each of the imaginary values of \( x \) be raised to the third power, the same results will be obtained as from the real value of \( x \): the original equation \( x^3 - c^3 = 0 \) may also be reproduced, by multiplying together the three factors \( x - c, x - \frac{-1 + \sqrt{-3}}{2}c, \text{ and } x - \frac{-1 - \sqrt{-3}}{2}c \). Let us now consider such cubic equations as have all their terms, and which are therefore of this form,

\[ x^3 + Ax^2 + Bx + C = 0, \]

where \( A, B, \) and \( C \) denote known quantities, either positive or negative.

It has been shown (sect. 108) how an equation having all its terms may be transformed into another which wants the second term; therefore, assume \( x = y - \frac{A}{3} \), as directed in that article; then, by proper substitution, the above equation will be changed into another of this form,

\[ y^3 + qy + r = 0, \]

where \( q \) and \( r \) denote known quantities, whether positive or negative; now the roots of this equation being found, it is evident that those of the former may be readily obtained by means of the assumed equation \( x = y - \frac{A}{3} \).

Resuming, therefore, the equation \( y^3 + qy + r = 0 \), let us suppose \( y = v + z \), and it becomes

\[ v^3 + 3v^2z + 3vz^2 + z^3 + qv + rz = 0. \]

Thus we have a new equation, which, as it involves two unknown quantities, \( v \) and \( z \), may be resolved into any two others, which will simplify the determination of those quantities.

Now it appears, that the only way in which we can divide that equation into two others, so as to simplify the question, is the following:

\[ 3v^2z + 3vz^2 + qv + rz = 0, \] \[ v^3 + z^3 + r = 0. \]

The first of these may also be expressed thus,

\[ (3v + q)(v + z) = 0. \]

Hence, we must either suppose that \( v + z = 0 \), or that \( 3v + q = 0 \); but the former supposition cannot be admitted without supposing also that \( y = 0 \), which does not agree with the hypothesis of the equation \( y^3 + qy + r = 0 \); therefore we must adopt the latter. So that to determine \( v \) and \( z \) we have these two equations,

\[ 3v + q = 0, \] \[ v^3 + z^3 + r = 0. \]

From the first, we find \( v = -\frac{q}{3} \), and \( v^3 = -\frac{r}{27} \); and from the second \( v^3 + z^3 = -r \); so that to determine the quantities \( v^3 \) and \( z^3 \), we have given their sum and product: now this is a problem which we have already resolved when treating of quadratic equations; and by proceeding in the same manner in the present case, we shall find

\[ v^3 = -\frac{1}{2}r + \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ z^3 = -\frac{1}{2}r - \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ v = \sqrt{-\frac{1}{2}r + \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}}, \] \[ z = \sqrt{-\frac{1}{2}r - \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}}, \] \[ y = v + z = \sqrt{-\frac{1}{2}r + \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}} + \sqrt{-\frac{1}{2}r - \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}}. \]

Thus we have at last obtained a value of the unknown quantity \( y \), in terms of the known quantities \( q \) and \( r \); therefore the equation is resolved.

But this is only one of three values which \( y \) may have. Let us, for the sake of brevity, put

\[ A = -\frac{1}{2}r + \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ B = -\frac{1}{2}r - \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ \alpha = \frac{-1 + \sqrt{-3}}{2}, \] \[ \beta = \frac{-1 - \sqrt{-3}}{2}. \]

Then, from what has been shown (sect. 113), it is evident that \( v \) and \( z \) have each these three values,

\[ v = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ v = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ z = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ z = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}. \]

To determine the corresponding values of \( v \) and \( z \), we must consider that \( vz = -\frac{q}{3} = \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2} \): Now if we observe that \( \alpha \beta = 1 \), it will immediately appear that \( v + z \) has these three values,

\[ v + z = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ v + z = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ v + z = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}. \]

Hence the three values of \( y \) are also these,

\[ y = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ y = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ y = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}. \]

The first of these formulae is commonly known by the name of Cardan's rule; but it is well known that Cardan was not the inventor, and that it ought to be attributed to Nicholas Tartalea and Scipio Ferreus, who discovered it much about the same time, and independently of each other. (See the Introduction.)

The formulae given above for the roots of a cubic equation may be put under a different form, and better adapted to the purposes of arithmetical calculation as follows.

Because \( vz = -\frac{q}{3} \), therefore \( z = -\frac{q}{3} \times \frac{1}{v} = -\frac{q}{3v} \);

hence \( v + z = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2} \): thus it appears that the three values of \( y \) may also be expressed thus:

\[ y = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ y = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}, \] \[ y = \frac{5}{3} \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}. \]

116. To show the manner of applying these formulae, let it be required to determine \( x \) from the cubic equation

\[ x^3 + 3x^2 + 9x - 13 = 0; \]

And as this equation has all its terms, the first step towards its resolution is to transform it into another which shall want the second term, by substituting \( y - 1 \) for \( x \) as directed (sect. 109). The operation will stand thus:

\[ x^3 = y^3 - 3y^2 + 3y - 1, \] \[ + 3x^2 = + 3y^2 - 6y + 3, \] \[ + 9x = + 9y - 9, \] \[ - 13 = - 13. \]

The transformed equation is \( y^3 + 6y - 20 = 0 \), which being compared with the general equation,

\[ y^3 + qy + r = 0, \]

gives \( q = 6, r = -20 \); hence

\[ A = \sqrt{-\frac{1}{2}r + \sqrt{\frac{1}{4}r^2 + \frac{1}{4}q^2}} = \sqrt{\frac{10}{3} + \sqrt{108}}, \]

therefore the second formula of last article gives \( y = \frac{5}{3} \sqrt{\frac{10}{3} + \sqrt{108}} - \frac{2}{3} \sqrt{\frac{10}{3} + \sqrt{108}} \); but as this expression involves a radical quantity, let the square root of 108 be taken and added to 10, and the cube root of the sum found; thus we have \( \sqrt[3]{10 + \sqrt{108}} = 2.732 \) nearly, and therefore \( \frac{2}{\sqrt[3]{10 + \sqrt{108}}} = 2.732 \); hence we at last find one of the values of \( y \) to be \( 2.732 - 0.732 = 2 \).

In finding the cube root of the radical quantity \( \sqrt[3]{10 + \sqrt{108}} \), we have taken only its approximate value, so as to have the expression for the root under a rational form, and in this way we can always find, as near as we please, the cube root of any surd of the form \( a + b\sqrt{b} \), where \( b \) is a positive number. But it will sometimes happen that the cube root of such a surd can be expressed exactly by another surd of the same form; and accordingly, in the present case, it appears that the cube root of \( 10 + \sqrt{108} \) is \( 1 + \sqrt{3} \), as may be proved by actually raising \( 1 + \sqrt{3} \) to the third power. Hence we find

\[ \frac{2}{\sqrt[3]{10 + \sqrt{108}}} = \frac{2}{1 + \sqrt{3}} = \frac{2(1 - \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = (1 - \sqrt{3}) \]

so that \( y = 1 + \sqrt{3} + 1 - \sqrt{3} = 2 \), as before.

The other two values of \( y \) will be had by substituting \( 1 + \sqrt{3} \) and \( 1 - \sqrt{3} \) for \( \sqrt[3]{A} \) and \( \frac{3}{\sqrt[3]{A}} \) in the second and third formulae of last article, and restoring the values of \( a \) and \( b \). We thus have

\[ y = \frac{-1 + \sqrt{-3}}{2} \times (1 + \sqrt{3}) + \frac{-1 - \sqrt{-3}}{2} \times (1 - \sqrt{3}) = -1 + \sqrt{-9} \]

So that the three values of \( y \) are

\[ +2, \quad -1 + \sqrt{-9}, \quad -1 - \sqrt{-9}; \]

and since \( x = y - 1 \), the corresponding values of \( x \) are

\[ +1, \quad -2 + \sqrt{-9}, \quad -2 - \sqrt{-9}. \]

Thus it appears that one of the roots of the proposed equation is real, and the other two imaginary.

The two imaginary roots might have been found otherwise, by considering that since one root of the equation is 1, the equation must be divisible by \( x - 1 \) (sect. 99). Accordingly, the division being actually performed, and the quotient put \( = 0 \), we have this quadratic equation,

\[ x^2 + 4x + 13 = 0; \]

which, when resolved by the rule for quadratics, gives \( x = -2 \pm \sqrt{-9} \), the same imaginary value as before.

117. In the application of the preceding formulae (sect. 114 and 115) to the resolution of the equation \( y^3 + qy + r = 0 \), it is necessary to find the square root of \( \frac{q^2}{4} + \frac{r^2}{4} \); now, when that quantity is positive, as in the equation \( y^3 + 6y - 20 = 0 \), which was resolved in last article, no difficulty occurs, for its root may be found either exactly or to as great a degree of accuracy as we please.

As, however, the co-efficients \( q \) and \( r \) are independent of each other, it is evident that \( q \) may be negative, and such, that \( \frac{q^2}{4} \) is greater than \( \frac{r^2}{4} \). In this case, the expression \( \frac{q^2}{4} + \frac{r^2}{4} \) will be negative, and therefore its square root an imaginary quantity. Let us take as an example, this equation, \( y^3 - 6y + 4 = 0 \); here \( q = -6 \), \( r = +4 \), \( \frac{q^2}{4} = 9 \), \( \frac{r^2}{4} = 4 \); \( \frac{q^2}{4} + \frac{r^2}{4} = \sqrt{-4} = 2\sqrt{-1} \); hence, by recurring to the formulae

\[ \text{sect. 115}, \text{we have } A = 2 + 2\sqrt{-1}, B = 2 - \sqrt{-1}, \text{ and therefore the three roots of the equation are} \]

\[ y = \sqrt[3]{2 + 2\sqrt{-1}} + \sqrt[3]{2 - 2\sqrt{-1}}, \]

\[ y = \sqrt[3]{2 + 2\sqrt{-1}} + \beta \sqrt[3]{2 - 2\sqrt{-1}}, \]

\[ y = \beta \sqrt[3]{2 + 2\sqrt{-1}} + \alpha \sqrt[3]{2 - 2\sqrt{-1}}. \]

Here all the roots appear under an imaginary form; but we are certain, from the theory of equations, as explained in Section X., that every cubic equation must have at least one real root. The truth is, as we shall show immediately, that in this case, so far from any of the roots being imaginary, they are all real; for it appears by actual involution, that the imaginary expression \( 2 + 2\sqrt{-1} \) is the cube of this other imaginary expression, \( -1 + \sqrt{-1} \); and in like manner, that \( 2 - 2\sqrt{-1} \) is the cube of \( -1 - \sqrt{-1} \), so that we have

\[ y = \sqrt[3]{2 + 2\sqrt{-1}} + \sqrt[3]{2 - 2\sqrt{-1}} = -1 + \sqrt{-1}, \]

\[ y = \frac{-1 + \sqrt{-3}}{2} \times (-1 + \sqrt{-1}) + \frac{-1 - \sqrt{-3}}{2} \times (-1 - \sqrt{-1}) = 1 + \sqrt{-3}, \]

\[ y = \frac{-1 - \sqrt{-3}}{2} \times (-1 + \sqrt{-1}) + \frac{-1 + \sqrt{-3}}{2} \times (-1 - \sqrt{-1}) = 1 - \sqrt{-3}. \]

118. We shall now prove, that as often as the roots of the equation \( x^3 + qx + r = 0 \) are real, \( q \) is negative, and \( \frac{q^2}{4} \) greater than \( \frac{r^2}{4} \); and, on the contrary, that if \( \frac{q^2}{4} \) be greater than \( \frac{r^2}{4} \), the roots are all real.

Let us suppose \( a \) to be a real root of the proposed equation;

Then \( x^3 + qx + r = 0 \),

And \( a^3 + qa + r = 0 \).

And therefore, by subtraction, \( x^3 - a^3 + q(x - a) = 0 \);

hence, dividing by \( x - a \), we have

\[ x^2 + ax + a^2 + q = 0. \]

This quadratic equation is formed from the two remaining roots of the proposed equation, and by resolving it we find

\[ x = -\frac{a}{2} \pm \sqrt{-\frac{3}{4}a^2 - q}. \]

And as, by hypothesis, all the roots are real, it is evident that \( q \) must necessarily be negative, and greater than \( \frac{3}{4}a^2 \); for otherwise the expression \( \sqrt{-\frac{3}{4}a^2 - q} \) would be imaginary. Let us change the sign of \( q \), and put \( q = \frac{3}{4}a^2 + d \); thus the roots of the equation \( x^3 + qx + r = 0 \) will be

\[ a, \quad -\frac{a}{2} + \sqrt{\frac{3}{4}a^2 + d}, \quad -\frac{a}{2} - \sqrt{\frac{3}{4}a^2 + d}, \]

and here \( d \) is a positive quantity.

To find an expression for \( r \) in terms of \( a \) and \( d \), let \( \frac{3}{4}a^2 + d \) be substituted for \( q \) in the equation \( a^3 + qa + r = 0 \); we thence find \( r = -\frac{3}{4}a^2 + ad \); so that to compare together the quantities \( q \) and \( r \), we have these equations,

\[ q = \frac{3}{4}a^2 + d, \quad r = -\frac{3}{4}a^2 + ad. \]

In order to make this comparison, let the cube of \( \frac{3}{4}q \) be taken, also the square of \( \frac{3}{4}r \), the results are

\[ \frac{3}{4}q^2 = \frac{3}{4}a^2 + \frac{3}{4}ad + \frac{3}{4}d^2, \]

\[ \frac{3}{4}r^2 = \frac{3}{4}a^2 - \frac{3}{4}ad + \frac{3}{4}d^2, \]

and therefore, by subtraction,

\[ \frac{3}{4}q^2 - \frac{3}{4}r^2 = \frac{3}{4}ad + \frac{3}{4}d^2, \]

\[ = 3d(\frac{1}{4}a^2 - \frac{1}{4}d^2). \] Algebra. Now the square of any real quantity being always positive, it follows that \(3d(4a^2 - b^2)^2\) will be positive when \(d\) is positive; hence it is evident that in this case \(\frac{1}{2}q^2\) must be greater than \(\frac{1}{2}r^2\), and that the contrary cannot be true, unless \(d\) be negative, that is, unless that \(-\frac{1}{2}a + \sqrt{d}, -\frac{1}{2}a - \sqrt{d}\), the two other roots of the equation, are imaginary. If we suppose \(d = 0\), then \(\frac{1}{2}q^2 = \frac{1}{2}r^2\); and the roots of the equations, which in this case are also real, are \(a, -\frac{1}{2}a, -\frac{1}{2}a\).

Upon the whole, therefore, we infer, that since a cubic equation has always one real root, its roots will be all real as often as \(q\) is negative, and \(\frac{1}{2}q^2\) greater than \(\frac{1}{2}r^2\); and consequently, that in this case the formulae for the roots must express real quantities, notwithstanding their imaginary form.

119. Let \(y^3 - qy + r = 0\) denote any equation of the form which has been considered in last article, namely, that which has its roots all real; then, if we put \(a = -\frac{1}{2}r\), \(b^2 = \frac{1}{4}q^2 - \frac{1}{4}r^2\), one of the roots, as expressed by the first formula, sect. 115, will be,

\[ y = \sqrt[3]{a + b\sqrt{-1}} + \sqrt[3]{a - b\sqrt{-1}}. \]

This expression, although under an imaginary form, must (as we have shown in last article) represent a real quantity. It may happen, as in last example, sect. 117, that the two surds which compose the root are perfect cubes of the form \((A + B\sqrt{-1})^3\), and \((A - B\sqrt{-1})^3\), and then the value of \(y\) becomes

\[ A + B\sqrt{-1} + A - B\sqrt{-1} = 2A. \]

But the rules for determining when this is the case depend upon trials, and are, besides, troublesome in the application; and if we attempt by a direct process to investigate the numerical values of \(A\) and \(B\), we are brought to a cubic equation of the very same form as that whose root is required.

This imaginary expression for a real quantity has greatly perplexed mathematicians; and much pains has been taken to obtain the root under another form, but without success. Accordingly, the case of cubic equations, in which the roots are all real, is now called the irreducible case.

120. It is remarkable that the expression

\[ \sqrt{a + b\sqrt{-1}} + \sqrt{a - b\sqrt{-1}}, \]

and in general,

\[ \sqrt{a + b\sqrt{-1}} + \sqrt{a - b\sqrt{-1}}, \]

where \(n\) is any power of 2, admits of being reduced to another form, in which no impossible quantity is found. Thus,

\[ \sqrt{a + b\sqrt{-1}} + \sqrt{a - b\sqrt{-1}} = \sqrt{2a + 2\sqrt{a^2 + b^2}}, \]

and

\[ \sqrt{a + b\sqrt{-1}} + \sqrt{a - b\sqrt{-1}} = \sqrt{\sqrt{2a + 2\sqrt{a^2 + b^2}}}. \]

as is easily proved by first raising the imaginary formulae to the second and fourth powers, and then taking the square and fourth root of each. But when \(n\) is 3, it does not seem that such reduction can possibly take place.

If each of the surds be expanded into an infinite series, and their sum be taken, the imaginary quantity \(\sqrt{-1}\) will vanish, and thus the root may be found by a direct process. There are, however, other methods which seem preferable. The following, which is derived from the calculus of sines, seems the best.

121. It will be demonstrated in Sect. XXV., that if \(a\) de-

note an arch of a circle, the relation between the cosine of the arch and the cosine of \(\frac{a}{3}\), one-third of that arch, is expressed by the following cubic equation:

\[ \cos^3 \frac{a}{3} - \frac{3}{4} \cos \frac{a}{3} = \frac{1}{4} \cos a. \]

Let us assume \(\cos \frac{a}{3} = \frac{y}{n}\); then, by substitution, the equation is transformed into the following:

\[ \frac{y^3}{n^3} - \frac{3y}{4n} = \frac{1}{4} \cos a, \]

Or

\[ y^3 - \frac{3n^2}{4} y = n^2 \times \frac{1}{4} \cos a; \]

and in this cubic equation, one of the roots is evidently \(y = n \times \cos \frac{a}{3}\). Now from the calculus of sines it appears that \(\cos a, \cos (360^\circ - a)\), and \(\cos (360^\circ + a)\), are all expressed by the same quantity; therefore the equation must have for a root not only \(n \times \cos \frac{a}{3}\) but also \(n \times \cos \frac{360^\circ - a}{3}\), and \(n \times \cos \frac{360^\circ + a}{3}\). But from the calculus of sines, \(\cos \frac{360^\circ - a}{3} = -\sin \frac{90^\circ - a}{3}\), and \(\cos \frac{360^\circ + a}{3} = -\sin \frac{90^\circ + a}{3}\); therefore the roots of the equation are

\[ n \times \cos \frac{a}{3}, \quad -n \times \sin \frac{90^\circ - a}{3}, \quad -n \times \sin \frac{90^\circ + a}{3}. \]

Let us next suppose that \(y^3 - qy + r = 0\) is a cubic equation whose roots are required, and let us compare it with the former equation \(y^3 - \frac{3n^2}{4} y = n^2 \times \frac{1}{4} \cos a\); then it is evident, that if we assume the quantities \(n\) and \(\cos a\), such that

\[ \frac{3n^2}{4} = q, \quad n^2 \times \frac{1}{4} \cos a = r, \]

the two equations will become identical, and thus their roots will be expressed by the very same quantities. But from these two assumed equations we find

\[ n = \sqrt{\frac{4q}{3}}, \quad \cos a = \frac{4r}{n^2} = \sqrt{\frac{27r^2}{4q^3}} = \frac{3r\sqrt{3}}{2q\sqrt{q}}; \]

and since the cosine of an arch cannot exceed unity, therefore \(\frac{27r^2}{4q^3}\) must be a proper fraction, that is, \(4q^3\) must exceed \(27r^2\), or \(\frac{1}{3}q^3\) must exceed \(\frac{1}{3}r^2\). If we now recollect that \(q\) is a negative quantity, it will immediately appear that the proposed equation must necessarily belong to the irreducible case.

The rule, therefore, which we derive from the preceding analysis for resolving that case is as follows:

Let \(y^3 - qy + r = 0\) be the proposed equation.

Find in the trigonometrical tables an arch \(a\), whose natural cosine \(= \frac{3r\sqrt{3}}{2q\sqrt{q}}\);

the roots of the equation are

\[ y = 2\sqrt{\frac{q}{3}} \times \cos \frac{a}{3}. \] \[ y = -2 \sqrt{\frac{q}{3}} \times \sin \frac{90^\circ - a}{3}, \] \[ y = -2 \sqrt{\frac{q}{3}} \times \sin \frac{90^\circ + a}{3}. \]

These formulae will apply, whether \( r \) be positive or negative, by proper attention to the signs. If, however, \( r \) be negative, or the equation have this form, \( y^2 - qy = -r \), the following will be more convenient:

Find in the tables an arch \( a \), whose sine \( = \frac{3r\sqrt{3}}{2q\sqrt{q}} \).

Then the roots of the equation are

\[ y = 2 \sqrt{\frac{q}{3}} \times \sin \frac{a}{3}, \] \[ y = 2 \sqrt{\frac{q}{3}} \times \cos \frac{90^\circ - a}{3}, \] \[ y = -2 \sqrt{\frac{q}{3}} \times \cos \frac{90^\circ + a}{3}. \]

The last formulae are derived from the equation

\[ \sin^2 \frac{a}{3} - \frac{3}{4} \sin \frac{a}{3} = -\sin a, \]

in the same manner as the former were found from the first equation of last article.

Ex. 1. It is required to find the roots of the equation \( x^3 - 3x = 1 \).

Here \( \frac{3r\sqrt{3}}{2q\sqrt{q}} = \frac{3 \times \sqrt{3}}{6 \times \sqrt{3}} = \frac{1}{2} = \cos 60^\circ = \cos a \), and

\[ x = 2 \cos \frac{60^\circ}{3} = 2 \cos 20^\circ = 1.8793852, \] \[ x = -2 \sin \frac{150^\circ}{3} = -2 \sin 50^\circ = -1.5320888, \] \[ x = -2 \sin \frac{30^\circ}{3} = -2 \sin 10^\circ = -3472964. \]

Ex. 2. It is required to find the roots of the equation \( x^3 - 3x = -1 \).

Here \( \frac{3r\sqrt{3}}{2q\sqrt{q}} = \frac{3 \times \sqrt{3}}{6 \times \sqrt{3}} = \frac{1}{2} = \sin 30^\circ = \sin a \), and

\[ x = 2 \sin \frac{30^\circ}{3} = 2 \sin 10^\circ = 3472964, \] \[ x = 2 \cos \frac{120^\circ}{3} = 2 \cos 40^\circ = 1.5320888, \] \[ x = -2 \cos \frac{60^\circ}{3} = -2 \cos 20^\circ = -1.8793852. \]

Sect. XII.—Of Biquadratic Equations.

122. When a biquadratic equation contains all its terms, it has this form,

\[ x^4 + Ax^3 + Bx^2 + Cx + D = 0, \]

where \( A, B, C, D \) denote any known quantities whatever.

We shall first consider pure biquadratics, or such as contain only the first and last terms, and therefore are of this form, \( x^4 = b^4 \). In this case it is evident that \( x \) may be readily had by two extractions of the square root; by the first we find \( x^2 = b^2 \), and by the second \( x = b \). This, however, is only one of the values which \( x \) may have; for since \( x^4 = b^4 \), therefore \( x^2 = b^2 = 0 \); but \( x^2 = b^2 \) may be resolved into two factors \( x^2 - b^2 \) and \( x^2 + b^2 \), each of which admits of a similar resolution; for \( x^2 - b^2 = (x - b)(x + b) \) and \( x^2 + b^2 = (x - b \sqrt{-1})(x + b \sqrt{-1}) \).

Hence it appears that the equation \( x^4 - b^4 = 0 \) may also be expressed thus:

\[ (x - b)(x + b)(x - b \sqrt{-1})(x + b \sqrt{-1}) = 0; \]

so that \( x \) may have these four values,

\[ + b, \quad - b, \quad + b \sqrt{-1}, \quad - b \sqrt{-1}, \]

two of which are real, and the others imaginary.

123. 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,

\[ x^4 + qx^2 + s = 0. \]

These may be resolved in the manner of quadratic equations; for if we put \( y = x^2 \), we have

\[ y^2 + qy + s = 0, \]

from which we find \( y = \frac{-q \pm \sqrt{q^2 - 4s}}{2} \), and therefore

\[ x = \pm \sqrt{\frac{-q \pm \sqrt{q^2 - 4s}}{2}}. \]

124. When a biquadratic equation has all its terms, the manner of resolving it is not so obvious as in the two former cases, but 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, which we select as one of the most ingenious, was first given by Euler in the Petersburg Commentaries, and afterwards explained more fully in his Elements of Algebra.

We have already explained, sect. 109, 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,

\[ y^4 + py^2 + qy + r = 0, \]

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 \( y = \sqrt{a} + \sqrt{b} + \sqrt{c} \), and also suppose that the letters \( a, b, c \) denote the roots of the cubic equation

\[ z^3 + Pz^2 + Qz - R = 0; \]

then, from the theory of equations we have

\[ a + b + c = -P, \quad ab + ac + bc = Q, \quad abc = R. \]

We square the assumed formula

\[ y = \sqrt{a} + \sqrt{b} + \sqrt{c}, \]

and obtain

\[ y^2 = a + b + c + 2(\sqrt{ab} + \sqrt{ac} + \sqrt{bc}), \]

or, substituting \( -P \) for \( a + b + c \), and transposing;

\[ y^2 + P = 2(\sqrt{ab} + \sqrt{ac} + \sqrt{bc}). \]

Let this equation be also squared, and we have

\[ y^4 + 2Py^2 + P^2 = 4(ab + ac + bc) + 8(\sqrt{abc} + \sqrt{ab} + \sqrt{ac} + \sqrt{bc}) + \sqrt{abc}; \]

and since \( ab + ac + bc = Q \), and

\[ \sqrt{abc} + \sqrt{ab} + \sqrt{ac} + \sqrt{bc} = \sqrt{abc}(\sqrt{a} + \sqrt{b} + \sqrt{c}) = \sqrt{Ry}, \]

the same equation may be expressed thus:

\[ y^4 + 2Py^2 + P^2 = 4Q + 8\sqrt{Ry}. \]

Thus we have the biquadratic equation

\[ y^4 + 2Py^2 - 8\sqrt{Ry} + P^2 - 4Q = 0, \]

one of the roots of which is \( y = \sqrt{a} + \sqrt{b} + \sqrt{c} \), and in which \( a, b, c \) are the roots of the cubic equation \( z^3 + Pz^2 + Qz - R = 0 \).

125. In order to apply this resolution to the proposed equation \( y^4 + py^2 + qy + r = 0 \), we must express the assumed co-efficients \( P, Q, R \), by means of \( p, q, r \), the co- Algebraic coefficients of that equation. For this purpose, let us compare the equations

\[ y^4 + py^2 + qy + r = 0, \] \[ y^4 + 2py^2 - 8\sqrt{R} y + P^2 - 4Q = 0, \]

and it immediately appears that \( 2P = p, -8\sqrt{R} = q, \) \[ P^2 - 4Q = r; \] and from these three equations we find

\[ P = \frac{p}{2}, Q = \frac{p^2 - 4r}{16}, R = \frac{q^2}{64}. \]

Hence it follows that the roots of the proposed equation are generally expressed by the formula \( y = \sqrt{a} + \sqrt{b} + \sqrt{c}; \) where \( a, b, c \) denote the roots of this cubic equation,

\[ z^3 + \frac{p}{2}z^2 + \frac{p^2 - 4r}{16}z - \frac{q^2}{64} = 0. \]

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 \( \sqrt{a}, \sqrt{b}, \sqrt{c}, \) 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 \( \sqrt{a}, \sqrt{b}, \sqrt{c}, \) must be equal to \( \sqrt{R} \) or to \( -\frac{q}{8} \) 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.

126. We shall now give the result of the preceding investigation in the form of a practical rule; and as the co-efficients of the cubic equation which has been found involve fractions, we shall transform it into another, in which the co-efficients are integers, by supposing \( z = \frac{v}{4}. \) Thus the equation

\[ z^3 + \frac{p}{2}z^2 + \frac{p^2 - 4r}{16}z - \frac{q^2}{64} = 0 \]

becomes, after reduction,

\[ v^3 + 2pv^2 + (p^2 - 4r)v - q^2 = 0; \] it also follows, that since the roots of the former equation are \( a, b, c, \)

the roots of the latter are \( \frac{a}{4}, \frac{b}{4}, \frac{c}{4} \) so that our rule may now be expressed thus:

Let \( y^4 + py^2 + qy + r = 0 \) be any biquadratic equation wanting its second term. Form this cubic equation,

\[ v^3 + 2pv^2 + (p^2 - 4r)v - q^2 = 0, \]

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,

\[ y = \frac{1}{2}(\sqrt{a} + \sqrt{b} + \sqrt{c}), \] \[ y = \frac{1}{2}(\sqrt{a} - \sqrt{b} - \sqrt{c}), \] \[ y = \frac{1}{2}(-\sqrt{a} + \sqrt{b} + \sqrt{c}), \] \[ y = \frac{1}{2}(-\sqrt{a} - \sqrt{b} - \sqrt{c}). \]

when \( q \) is positive,

\[ y = \frac{1}{2}(-\sqrt{a} - \sqrt{b} - \sqrt{c}), \] \[ y = \frac{1}{2}(-\sqrt{a} + \sqrt{b} + \sqrt{c}), \] \[ y = \frac{1}{2}(\sqrt{a} - \sqrt{b} - \sqrt{c}), \] \[ y = \frac{1}{2}(\sqrt{a} + \sqrt{b} + \sqrt{c}). \]

127. This resolution of biquadratic equations suggests the following general remarks upon the nature of their roots.

1. It is evident from the form of the roots, that if the cubic equation

\[ v^3 + 2pv^2 + (p^2 - 4r)v - q^2 = 0, \]

have all its roots real and positive, those of the biquadratic equation will be all real.

2. Since the last term of the cubic equation is negative; when its three roots are real, they must either be all positive, or two of them must be negative and one positive; for the last term is equal to the product of all the roots taken with contrary signs, sect. 100; so that in this last case, two of the three quantities \( a, b, c, \) must be negative, and therefore all the four roots of the biquadratic equation imaginary. If, however, the two negative roots be equal, they will destroy each other in two of the roots of the biquadratic equation, which will then become real and equal. Let us suppose, for example, that \( b \) and \( c \) are negative and equal; the first two values of \( y \) in each column become then imaginary, and the remaining values of \( y \) are in the first set of roots, \( y = -\frac{1}{2}\sqrt{a}, y = -\frac{1}{2}\sqrt{a}, \) and in the second, \( y = \frac{1}{2}\sqrt{a}, y = \frac{1}{2}\sqrt{a}. \)

3. When the cubic equation has only one real and two imaginary roots, its real roots must necessarily be positive; for the imaginary roots can only come from a quadratic equation having its last term positive, Sect. IX, and therefore are of this form, \( v^2 + Av + B = 0; \) hence the simple factor which contains the remaining root must have this form, \( v - \gamma, \) otherwise the last term of the cubic equation could not be negative.

By resolving the equation \( v^3 + Av + B = 0, \) we find

\[ v = -\frac{A}{2} \pm \sqrt{\frac{A^2}{4} - B}. \]

Here, the roots being supposed imaginary, \( \frac{A^2}{4} - B \) must be a negative quantity. That we may simplify the form of the roots, let us put \( -\frac{A}{2} = a, \) and \( \frac{A^2}{4} - B = -\beta^2, \) then

\[ v = -a + \beta\sqrt{-1}, \quad v = -a - \beta\sqrt{-1}, \] \[ v = -a + \beta\sqrt{-1}, \quad v = -a - \beta\sqrt{-1}. \]

Hence we have

\[ a = a + \beta\sqrt{-1}, \quad b = a - \beta\sqrt{-1}, \quad c = \gamma; \]

so that in two of the four values of \( y, \) we have a quantity of this form,

\[ \sqrt{a + \beta\sqrt{-1}} + \sqrt{a - \beta\sqrt{-1}}; \]

but this quantity, although it appears to be imaginary, is indeed real; for if we first square it, and then take its square root, it becomes

\[ \sqrt{2a + 2\beta\sqrt{-1}}, \]

which is a real quantity. The other two roots involve this other expression,

\[ \sqrt{a + \beta\sqrt{-1}} - \sqrt{a - \beta\sqrt{-1}}; \]

which being treated in the same manner as the former, becomes

\[ \sqrt{2a - 2\beta\sqrt{-1}}, \]

an imaginary quantity, and therefore the roots into which it enters are imaginary.

4. We may discover from the co-efficients of the proposed biquadratic equation in what case the roots of the cubic equation are all real. For this purpose, the latter is to be transformed into another which shall want the second term, by assuming \( v = u - \frac{2p}{3}; \) thus it becomes

\[ u^3 - \left(\frac{p^2}{3} + 4r\right)u - \frac{2p^3}{27} + \frac{8rp}{3} - q^2 = 0; \]

and in this equation the three roots will be real when

\[ \left(\frac{p^2}{3} + 4r\right)^3 \text{ is greater than } \frac{1}{4}\left(\frac{2p^3}{27} - \frac{8rp}{3} + q^2\right)^2. \]

128. As an example of the method of resolving a biquadratic equation, let it be required to determine the roots of the following,

\[ x^4 - 25x^2 + 60x - 36 = 0. \] Algebra. By comparing this equation with the general formula, we have \( p = -25, q = +60, r = -36 \); hence

\[ 2p = -50, \quad p^2 - 4r = 769, \quad q^2 = 3600, \]

and the cubic equation to be resolved is

\[ x^3 - 50x^2 + 769x - 3600 = 0; \]

the roots of which are found, by the rules for cubics, to be 9, 16, and 25, so that \( \sqrt{a} = 3, \sqrt{b} = 4, \sqrt{c} = 5 \). Now in this case \( q \) is positive, therefore

\[ x = \frac{(-3 - 4 - 5)}{-6}, \quad x = \frac{(-3 + 4 + 5)}{+3}, \quad x = \frac{(3 + 4 + 5)}{+2}, \quad x = \frac{(-3 + 4 - 5)}{-1}. \]

129. We have now explained the particular rules by which the roots of equations belonging to each of the first four orders may be determined; and this is the greatest length mathematicians have been able to go in the direct resolution of equations; for as to those of the fifth, and all higher degrees, no general method has hitherto been found, either for resolving them directly, or reducing them to others of an inferior degree.

It even appears that the formulæ which express the roots of cubic equations are not of universal application; for in one case, that is, when the roots are all real, they become illusory, so that no conclusion can be drawn from them. The same observation will also apply to the formulæ for the roots of biquadratic equations, because, before they can be applied, it is always necessary to find the roots of a cubic equation. But in either cubics or biquadratic equations, even when the formulæ involve no imaginary quantities, and therefore can be always applied, it is more convenient in practice to employ some other methods, which we are hereafter to explain.

Sect. XIII.—Of Reciprocal Equations.

130. Although no general resolution has hitherto been found of equations belonging to the fifth or any higher degree, yet there are particular equations of all orders, which, by reason of certain peculiarities in the nature of their roots, admit of being reduced to others of a lower degree; and thus, in some cases, equations of the higher orders may be resolved by the rules which have been already explained for the resolution of equations belonging to the first four orders.

When the co-efficients of the terms of an equation form the same numerical series, whether taken in a direct or an inverted order, as in this example,

\[ x^4 + px^3 + qx^2 + rx + 1 = 0, \]

it may always be transformed into another of a degree denoted by half the exponent of the highest power of the unknown quantity, if that exponent be an even number; or half the exponent diminished by unity, if it be an odd number.

The same observation will also apply to any equation of this form,

\[ x^4 + px^3 + qx^2 + rx + 1 = 0, \]

where the given quantity \( a \) and the unknown quantity \( x \) are precisely alike concerned; for by substituting \( ay \) for \( x \), it becomes

\[ ay^4 + py^3 + qy^2 + ry + 1 = 0; \]

and dividing by \( a \),

\[ y^4 + py^3 + qy^2 + ry + 1 = 0, \]

an equation of the same kind as the former.

131. That we may effect the proposed transformation upon the equation

\[ x^4 + px^3 + qx^2 + rx + 1 = 0, \]

let every two terms which are equally distant from the extremes be collected into one, and the whole be divided by \( x^2 \), then

\[ x^2 + \frac{1}{x^2} + p(x + \frac{1}{x}) + q = 0. \]

Let us assume \( x + \frac{1}{x} = z \);

then \( x^2 + \frac{1}{x^2} = z^2 \), and \( x^2 + \frac{1}{x^2} = z^2 - 2 \).

Thus the equation \( x^2 + \frac{1}{x^2} + p(x + \frac{1}{x}) + q = 0 \),

becomes \( z^2 + pz + q - 2 = 0 \);

and since \( x + \frac{1}{x} = z \), therefore \( x^2 - zx + 1 = 0 \).

Hence, to determine the roots of the biquadratic equation

\[ x^4 + px^3 + qx^2 + rx + 1 = 0, \]

we have the following rule:

Form this quadratic equation,

\[ z^2 + pz + q - 2 = 0, \]

and find its roots, which, let us suppose denoted by \( z' \) and \( z'' \); then the four roots of the proposed equation will be found by resolving two quadratic equations, viz.

\[ x^2 - zx + 1 = 0, \quad x^2 - zx + 1 = 0. \]

132. It may be observed, respecting these two quadratic equations, that since the last term of each is unity, if we put \( a, a' \) to denote the roots of the one, and \( b, b' \) those of the other, we have, from the theory of equations, \( a a' = 1 \), and therefore \( a' = \frac{1}{a} \); also \( b b' = 1 \), and \( b' = \frac{1}{b} \); now \( a, a', b, b' \) are also the roots of the equation

\[ x^4 + px^3 + qx^2 + rx + 1 = 0. \]

Hence it appears that the proposed equation has this peculiar property, that one half of its roots are the reciprocals of the other half; and to that circumstance we are indebted for the simplicity of its resolution.

133. The following equation,

\[ x^6 + px^5 + qx^4 + rx^3 + sx^2 + tx + 1 = 0, \]

which is of the sixth order, admits of a resolution in all respects similar to the former; for, by putting it under this form,

\[ x^3 + \frac{1}{x^3} + p(x^2 + \frac{1}{x^2}) + q(x + \frac{1}{x}) + r = 0, \]

and putting also \( x + \frac{1}{x} = z \), so that \( x^2 - zx + 1 = 0 \), we have \( x^2 + \frac{1}{x^2} = z^2 - 2 \),

\[ x^3 + \frac{1}{x^3} = z^3 - 3(z + \frac{1}{z}) = z^3 - 3z. \]

Hence, by substitution, the proposed equation is transformed into the following cubic equation,

\[ z^3 + pz^2 + (q - 3)z + r - 2p = 0; \]

therefore, putting \( z, z', z'' \), to denote its roots, the six roots of the proposed equation will be had by resolving these three quadratics,

\[ x^2 - zx + 1 = 0, \quad x^2 - zx + 1 = 0, \quad x^2 - zx + 1 = 0; \]

and here it is evident, as in the former case, that the roots of each quadratic equation are the reciprocals of each other, so that the one half of the roots of the proposed equation are the reciprocals of the other half.

The method of resolution we have employed in the two preceding examples is general for all equations whatever, in which the terms placed at equal distances from the first Algebra and last have the same co-efficients, and which are called reciprocal equations, because any such equation has the same form when you substitute for \( x \) its reciprocal, \( \frac{1}{x} \).

134. If the greatest exponent of the unknown quantity in a reciprocal equation is an odd number, as in this example,

\[ x^5 + px^4 + qx^3 + rx^2 + sx + t = 0, \]

the equation will always be satisfied by substituting \(-1\) for \( x \); hence, \(-1\) must be a root of the equation, and therefore the equation must be divisible by \( x+1 \). Accordingly, if the division be actually performed, we shall have in the present case

\[ x^5 + (p-1)x^4 - (p-q-1)x^3 + (p-1)x + 1 = 0, \]

another reciprocal equation, in which the greatest exponent of \( x \) is an even number, and therefore resolvable in the manner we have already explained.

135. As an application of the theory of reciprocal equation, let it be proposed to find \( x \) from this equation,

\[ \frac{x^5 + 1}{(x+1)^5} = a, \]

where \( a \) denotes a given number.

Every expression of the form \( x^n + 1 \) is divisible by \( x+1 \) when \( n \) is an odd number. In the present case, the numerator and denominator being divided by \( x+1 \), the equation becomes

\[ \frac{x^5 - x^4 + x^3 - x + 1}{x^5 + 4x^4 + 6x^3 + 4x^2 + 1} = a; \]

and this again, by proper reduction,

\[ (a-1)x^5 + (4a+1)x^4 + (6a-1)x^3 + (4a+1)x + a-1 = 0; \]

and, putting \( p = \frac{4a+1}{a-1} \), \( q = \frac{6a-1}{a-1} \),

\[ x^5 + px^4 + qx^3 + rx + s = 0; \]

a reciprocal equation, resolvable into two quadratics.

Sect.XIV.—Of Equations which have Equal Roots.

136. When an equation has two or more equal roots, these may always be discovered, and the equation reduced to another of an inferior degree, by a method of resolution which is peculiar to this class of equations.

Although the method of resolution we are to employ will apply alike to equations of every degree, having equal roots, yet, for the sake of brevity, we shall take a biquadratic equation,

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

the roots of which may be generally denoted by \( a, b, c \), and \( d \). Thus we have, from the theory of equations,

\[ (x-a)(x-b)(x-c)(x-d) = x^4 + px^3 + qx^2 + rx + s = 0. \]

Let us put

\[ A = (x-a)(x-b)(x-c)(x-d), A' = (x-a)(x-c)(x-d), \]

then, by actual multiplication, we have

\[ A = x^4 - ax^3 + bx^2 - acx + ab, \]

\[ A' = x^3 - ax^2 + bx - ac, \]

\[ A'' = x^2 - ax + b, \]

\[ A''' = x - a. \]

But since \( a, b, c, d \) are the roots of the equation

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

we have

\[ -(a+b+c+d) = 3p, \]

\[ 2(ab + ac + ad + bc + bd + cd) = 2q, \]

\[ -(abc + abd + acd + bcd) = r; \]

therefore, by substitution,

\[ A + A' + A'' + A''' = 4x^3 + 3px^2 + 2qx + r. \]

137. Let us now suppose that the proposed biquadratic equation has two equal roots, or \( a = b \); then \( x-a=x-b \), and since one or other of these equal factors enters each of the four products \( A, A', A'', A''' \), it is evident that \( A + A' + A'' + A''' \), or \( 4x^3 + 3px^2 + 2qx + r \) must be divisible by \( x-a \), or \( x-b \). Thus it appears that if the proposed equation

\[ x^4 + px^3 + qx^2 + rx + s = 0 \]

have two equal roots, each of them must also be a root of this equation,

\[ 4x^3 + 3px^2 + 2qx + r = 0; \]

for when the first of these equations is divisible by \((x-a)^2\), the latter is necessarily divisible by \(x-a\).

138. Let us next suppose that the proposed equation has three equal roots, or \( a = b = c \); then, two at least of the three equal factors \( x-a, x-b, x-c \), must enter each of the four products \( A, A', A'', A''' \); so that in this case, \( A + A' + A'' + A''' \), or \( 4x^3 + 3px^2 + 2qx + r \), must be twice divisible by \( x-a \). Hence it follows, that as often as the proposed equation has three equal roots, two of them must also be equal roots of the equation

\[ 4x^3 + 3px^2 + 2qx + r = 0. \]

139. Proceeding in the same manner, it may be shown, that whatever number of equal roots are in the proposed equation

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

they will remain, except one, in this equation,

\[ 4x^3 + 3px^2 + 2qx + r = 0, \]

which may be derived from the former, by multiplying each of its terms by the exponent of \( x \) in that term, and then diminishing the exponent by unity.

140. If we suppose that the proposed equation has two equal roots, or \( a = b \), and also two other equal roots, or \( c = d \), then, by reasoning as before, it will appear that the equation derived from it must have one root equal to \( a \) or \( b \), and another equal to \( c \) or \( d \); so that when the former is divisible both by \((x-a)^2\) and \((x-c)^2\), the latter will be divisible by \((x-a)(x-c)\).

141. The same mode of reasoning may be extended to all equations whatever; so that if we suppose

\[ x^n + Px^{n-1} + Qx^{n-2} + \ldots + Sx^2 + Tx + U = 0, \]

an equation of the \( n \)th degree, to have a divisor of this form,

\[ (x-a)^m (x-d)^p (x-f)^q \ldots \text{&c.} \]

the equation

\[ mx^{n-1} + (m-1) Px^{n-2} + (m-2) Qx^{n-3} + \ldots + 2Sx + T = 0, \]

which is of the next lower degree, will have for a divisor

\[ (x-a)^{n-1} (x-d)^{p-1} (x-f)^{q-1} \ldots \text{&c.} \]

and as this last product must be a divisor of both equa- tions, it may always be discovered by the rule which has been given (sect. 20) for finding the greatest common divisor of two algebraic quantities.

142. Again, as this last equation must, in the case of equal roots, have the same properties as the original equation; therefore, if we multiply each of its terms by the exponent of \( x \), and diminish that exponent by unity, as before, we have

\[ m(m-1)x^{m-2} + (m-1)(m-2)P_{x^{m-3}} + (m-2)(m-3)Q_{x^{m-4}} \ldots + 2S = 0, \]

a new equation, which has for a divisor

\[ (x-a)^{m-2}(x-b)^{m-2}(x-c)^{m-2}, \]

where the exponents of the factors are one less than those of the equation from which it was derived; and as this last divisor is also a divisor of the original equation, it may be discovered in the same manner as the former, namely, by finding the greatest common measure of both equations; and so on we may proceed, as far as we please.

143. As a particular example, let us take this equation,

\[ x^5 - 18x^4 + 67x^3 - 171x^2 + 216x - 108 = 0, \]

and apply to it the method we have explained, in order to discover whether it has equal roots, and if so, what they are. We must therefore seek the greatest common measure of the proposed equation and this other equation, which is formed agreeably to what has been shown, sect. 139,

\[ 5x^5 - 52x^4 + 201x^3 - 342x + 216 = 0; \]

and the operation being performed, we find that they have a common divisor, \( x^2 - 8x + 21x - 18 \), which is of the third degree, and consequently may have several factors. Let us therefore seek whether the last equation, and the following,

\[ 20x^3 - 156x^2 + 402x - 342 = 0, \]

which is derived from it, as directed in sect. 139, have any common divisor; and, by proceeding as before, we find that they admit of this divisor, \( x - 3 \), which is also a factor of the last divisor, \( x^2 - 8x + 21x - 18 \); and therefore the product of remaining factors is immediately found by division to be \( x^2 - 5x + 6 \), which is evidently resolvable into \( x - 2 \) and \( x - 3 \).

Thus it appears, upon the whole, that the common divisor of the original equation, and that which is immediately derived from it, is \( (x-2)(x-3)^2 \); and that the common divisor of the second and third equations is \( x - 3 \). Hence it follows that the proposed equation has \( (x-2)^2 \) for one factor, and \( (x-3)^2 \) for another factor; so that the equation itself may be expressed thus, \( (x-2)^2(x-3)^2 = 0 \); and the truth of this conclusion may be easily verified by multiplication.

**Sect. XV.—Resolution of Equations whose Roots are Rational.**

144. It has been shown in sect. 100, that the last term of any equation is always the product of its roots taken with contrary signs. Hence, when the roots are rational, they may be discovered by the following rule:

Bring all the terms of the equation to one side; find all the divisors of the last term, and substitute them successively for the unknown quantity. Then each divisor, which produces a result equal to 0, is a root of the equation.

**Ex. 1.** Let \( x^3 - 4x^2 - 7x + 10 = 0 \).

The divisors of 10, the last term, are 1, 2, 5, 10, each of which may be taken either positively or negatively; and these being substituted successively for \( x \), we obtain the following results:

By putting \( x + 1 \) for \( x \),

\[ \begin{align*} 1 & - 4 + 7 + 10 = 0, \\ -1 & -4 + 7 + 10 = 12, \\ +2 & -8 + 16 - 14 + 10 = 12, \\ -2 & -8 + 16 - 14 + 10 = 0, \\ +5 & 125 - 100 - 35 + 10 = 0. \end{align*} \]

Here the divisors which produce results equal to 0 are \( +1, -2, \) and \( +5 \); therefore these are the three roots of the proposed equation.

145. When the number of divisors to be tried is considerable, it will be convenient to transform the equation into another, in which the last term has fewer divisors. This may in general be done by forming an equation, the roots of which are greater or less than those of the proposed equation by some determinate quantity; as in the following example:

**Ex. 2.** Let \( y^4 - 4y^3 - 8y + 32 = 0 \) be proposed.

Here the divisors to be tried are 1, 2, 4, 8, 16, 32, each taken either positively or negatively; but to prevent the trouble of so many substitutions, transform the equation, by putting \( x + 1 \) for \( y \).

Then

\[ \begin{align*} y^4 &= x^4 + 4x^3 + 6x^2 + 4x + 1, \\ -4y^3 &= -4x^3 - 12x^2 - 12x - 4, \\ -8y &= -8x - 8, \\ +32 &= +32. \end{align*} \]

Therefore \( x^4 - 6x^2 - 16x + 21 = 0 \)

is the transformed equation, and the divisors of the last term are \( +1, -1, +3, -3, +7, -7 \). These being put successively for \( x \), we get \( +1 \) and \( +3 \) for two roots of the equation; as to the two remaining roots, it is easy to see that they must be imaginary. They may, however, be readily exhibited by considering that the equation \( x^4 - 6x^2 - 16x + 21 = 0 \) is divisible by the product of the two factors \( x - 1 \) and \( x - 3 \), and therefore may be reduced to a quadratic. Accordingly, by performing the division, and putting the quotient equal 0, we have this equation,

\[ x^2 + 4x + 7 = 0, \]

the roots of which are the imaginary quantities \( -2 + \sqrt{-3} \) and \( -2 - \sqrt{-3} \); so that, since \( y = x + 1 \), the roots of the equation \( y^4 - 4y^3 - 8y + 32 = 0 \) are these, \( y = +2, y = +4, y = -1 + \sqrt{-3}, y = -1 - \sqrt{-3} \).

If this literal equation were proposed,

\[ x^3 - (3a + b)x^2 + (2a^2 + 3ab)x - 2a^3b = 0, \]

by proceeding as before, we should find \( x = a, x = 2a, x = b \) for the roots.

146. To avoid the trouble of trying all the divisors of the last term, a rule may be investigated for restricting the number to very narrow limits as follows:

Suppose that the cubic equation \( x^3 + px^2 + qx + r = 0 \) is to be resolved. Let it be transformed into another, the roots of which are less than those of the proposed equation by unity. This may be done by assuming \( y = x - 1 \), and the last term of the transformed equation will be \( 1 + p + q + r \). Again, by assuming \( y = x + 1 \), another equation will be formed whose roots exceed those of the proposed equation by unity, and the last term of this other transformed equation will be \( -1 + p + q + r \). And it is to be observed, that these two quantities \( 1 + p + q + r \) and \( -1 + p + q + r \) are formed from the proposed equation \( x^3 + px^2 + qx + r \) by substituting successively \( +1 \) and \( -1 \) for \( x \). Now the values of \( x \) are some of the divisors of \( r \), which is the term left in the proposed equation, when \( x \) is supposed \( = 0 \); and the values of \( y \) are some of the divisors of \( 1 + p + q + r \) and \( -1 + p - q + r \) respectively; and these are in arithmetical progression, increasing by the common difference, unity; because \( x = 1, x = 2, x = 3 \), are in that progression; and it is obvious that the same reasoning will apply to an equation of any degree whatever.

Hence the following rule:

Substitute in place of the unknown quantity, successively, three or more terms of the progression \( 1, 0, -1, \ldots \), and find all the divisors of the sums that result; then take out all the arithmetical progressions that can be found among these divisors, whose common difference is 1, and the values of \( x \) will be among these terms of the progressions, which are the divisors of the result arising from the substitution of \( x = 0 \). When the series increases, the roots will be positive; and when it decreases, they will be negative.

**Ex. 1.** Let it be required to find a root of the equation

\[ x^3 - x^2 - 10x + 6 = 0. \]

| Substit. | Result | Divisors | Prog. | |---------|--------|----------|------| | \( x = +1 \) | \( x^3 - x^2 \) | \( -4, 1, 2, 4 \) | 4 | | \( x = 0 \) | \( -10x + 6 \) | \( +6, 1, 2, 3, 6 \) | 3 | | \( x = -1 \) | \( +14 \) | \( 1, 2, 7, 14 \) | 2 |

In this example there is only one progression, 4, 3, 2, the term of which opposite to the supposition of \( x = 0 \) being 3, and the series decreasing, we try it \( -3 \) substituted for \( x \) makes the equation vanish; and as it succeeds, it follows that \( -3 \) is one of its roots. To find the remaining roots, if \( x^3 - x^2 - 10x + 6 \) be divided by \( x + 3 \), and the quotient \( x^2 - 4x + 2 \) put \( = 0 \), they will appear to be \( 2 + \sqrt{2} \), and \( 2 - \sqrt{2} \).

**Ex. 2.** Let the proposed equation be

\[ x^4 + x^3 - 29x^2 - 9x + 180 = 0. \]

To find its roots.

Here there are four progressions, two increasing and two decreasing; hence, by taking their terms, which are opposite to the supposition of \( x = 0 \), we have these four numbers, \( +3, +4, -3, -5 \), to be tried as roots of the equation, all of which are found to succeed.

147. If any of the co-efficients of the proposed equation be a fraction, the equation may be transformed into another having the co-efficient of the highest power unity, and those of the remaining terms integers by sect. 112, and the roots of the transformed equation being found, those of the proposed equation may be easily derived from them.

For example, if the proposed equation be \( x^3 - \frac{3}{4}x^2 + \frac{3}{4}x - 6 = 0 \); let us assume \( x = \frac{y}{4} \). Thus the equation is transformed to

\[ \frac{y^3}{64} - \frac{7y^2}{64} + \frac{35y}{16} - 6 = 0; \]

or \( y^3 - 7y^2 + 140y - 384 = 0 \),

one root of which is \( y = 3 \); hence \( x = \frac{y}{4} = \frac{3}{4} \).

The proposed equation being now divided by \( x - \frac{3}{4} \), is reduced to this quadratic, \( x^2 - x + 8 = 0 \), the roots of which are both impossible.

148. When the co-efficients of an equation are integers, and unity that of the highest power of the unknown quantity, if its roots are not found among the divisors of the last term, we may be certain that, whether the equation be pure or affected, its roots cannot be exactly expressed either by whole numbers or rational fractions. This may be demonstrated by means of the following proposition.

If a prime number \( P \) be a divisor of the product of two numbers \( A \) and \( B \), it will also be a divisor of at least one of the numbers.

Let us suppose that it does not divide \( B \), and that \( B \) is greater than \( P \); then, putting \( q \) for the greatest number of times that \( P \) can be had in \( B \), and \( B' \) for the remainder, we have

\[ \frac{B}{P} = q + \frac{B'}{P}, \]

and therefore

\[ \frac{AB}{P} = qA + \frac{AB'}{P}. \]

Hence it appears, that if \( P \) be a divisor of \( AB \), it is also a divisor of \( AB' \). Now \( B' \) is less than \( P \), for it is the remainder which is found in dividing \( B \) by \( P \); therefore, seeing we cannot divide \( B' \) by \( P \), let \( P \) be divided by \( B' \), and \( q' \) put for the quotient, also \( B'' \) for the remainder. Again, let \( P \) be divided by \( B'' \), and \( q'' \) put for the quotient, and \( B''' \) for the remainder, and so on; and as \( P \) is supposed to be a prime number, it is evident that this series of operations may be continued till a remainder be found equal to unity, which will at last be the case; for the divisors are the successive remainders of the divisions, and therefore each is less than the divisor which preceded it. By performing these operations, we obtain the following series of equations;

\[ P = q'B' + B'', \quad P = q'B'' + B''', \]

and therefore

\[ B' = \frac{P - B'}{q'}, \quad B'' = \frac{P - B''}{q''}, \quad \text{etc.} \]

Hence we have

\[ AB' = \frac{AP - AB'}{q'}, \quad \text{and} \]

\[ \frac{q'AB'}{P} = \frac{AP - AB'}{P} = A - \frac{AB'}{P}. \]

Now, if \( AB \) be divisible by \( P \), we have shown that \( AB' \), and consequently \( q'AB' \), is divisible by \( P \); therefore, from the last equation, it appears that \( AB' \) must also be divisible by \( P \).

Again, from the preceding series of equations, we have

\[ AB'' = \frac{AP - AB''}{q''}, \quad \text{and therefore} \]

\[ \frac{q''AB''}{P} = \frac{AP - AB''}{P} = A - \frac{AB''}{P}; \]

hence we conclude that \( AB'' \) is also divisible by \( P \).

Proceeding in this manner, and observing that the series of quantities \( B', B'', B''', \ldots \) continually decrease till one of them \( = 1 \), it is evident that we shall at last come to a product of this form, \( A \times 1 \), which must be divisible by \( P \), and hence the truth of the proposition is manifest.

149. It follows from this proposition, that if the prime number \( P \), which we have supposed not to be a divisor of Let \( \frac{b}{a} \) be a fraction in its lowest terms, then the numbers \( a \) and \( b \) have no common divisor; but from what has been just now shown, it appears, that if a prime number be not a divisor of \( a \), it cannot be a divisor of \( a \times a \) or \( a^2 \); and in like manner, that if a prime number is not a divisor of \( b \), it cannot be a divisor of \( b \times b \), or \( b^2 \); therefore it is evident that \( a^2 \) and \( b^2 \) have no common divisor, and thus the fraction \( \frac{b^2}{a^2} \) is also in its lowest terms.

Hence it follows that the square of any fractional quantity is still a fraction, and cannot possibly be a whole number; and, on the contrary, that the square root of a whole number cannot possibly be a fraction; so that all such whole numbers as are not perfect squares can neither have their roots expressed by integers nor by fractions.

Seeing that if a prime number is not a divisor of \( a \), it is also not a divisor of \( a^2 \); therefore, if it is not a divisor of \( a \), it cannot be a divisor of \( a \times a^2 \) or \( a^3 \); and by reasoning in this way, it is obvious that if a prime number is not a divisor of \( a \), it cannot be a divisor of \( a^3 \); also, that if it is not a divisor of \( b \), it cannot be a divisor of \( b^3 \); therefore if \( \frac{b}{a} \) is a fraction in its lowest terms, \( \frac{b^3}{a^3} \) is also a fraction in its lowest terms; so that any power whatever of a fraction is also a fraction; and on the contrary, any root of a whole number is also a whole number. Hence it follows, that if the root of a whole number is not expressible by an integer, such root cannot be expressed by a fraction, but is therefore irrational or incommensurable.

150. Let us next suppose that

\[ x^n + Px^{n-1} + Qx^{n-2} + \ldots + Tx + U = 0 \]

is any equation whatever, in which \( P, Q, \ldots \) denote integer numbers; then, if its roots are not integers, they cannot possibly be rational fractions. For, if possible, let us suppose \( x = \frac{b}{a} \) a fraction reduced to its lowest terms; then, by substitution,

\[ \frac{a^n}{b^n} + P \frac{a^{n-1}}{b^{n-1}} + Q \frac{a^{n-2}}{b^{n-2}} + \ldots + T \frac{a}{b} + U = 0; \]

and, reducing all the terms to a common denominator,

\[ a^n + Pa^{n-1}b + Qa^{n-2}b^2 + \ldots + Tab + Ub^n = 0; \]

which equation may also be expressed thus,

\[ a^n + b(Pa^{n-1} + Qa^{n-2}b + \ldots + Tab + Ub^n) = 0, \]

where the equation consists of two parts, one of which is divisible by \( b \). But by hypothesis \( a \) and \( b \) have no common measure, therefore \( a^n \) is not divisible by \( b \), sect. 149; hence it is evident that the two parts of the equation cannot destroy each other as they ought to do; therefore \( x \) cannot possibly be a fraction.

Sect. XVI.—Resolution of Equations by Approximation.

151. When the roots of an equation cannot be accurately expressed by rational numbers, it is necessary to have recourse to methods of approximation; and by these we can always determine the numerical values of the roots to as great a degree of accuracy as we please.

The application of methods of approximation is rendered easy by means of the following propositions:

I. If two numbers, either whole or fractional, be found, which, when substituted for the unknown quantity in any equation, produce results with contrary signs, we may conclude that at least one root of the proposed equation is between those numbers, and is consequently real.

Let the proposed equation be

\[ x^3 - 5x^2 + 10x - 15 = 0, \]

which, by collecting the positive terms into one sum, and the negative into another, may also be expressed thus,

\[ x^3 + 10x - (5x^2 + 15) = 0; \]

then, to determine a root of the equation, we must find such a number as, when substituted for \( x \), will render

\[ x^3 + 10x = 5x^2 + 15. \]

Let us suppose \( x \) to increase and to have every degree of magnitude, from 0 upwards in the scale of number; then \( x^3 + 10x \) and \( 5x^2 + 15 \) will both continually increase, but with different degrees of quickness, as appears from the following table.

Successive values of \( x \): 0, 1, 2, 3, 4, 5, 6, &c.

of \( x^3 + 10x \): 0, 11, 28, 57, 104, 175, 276, &c.

of \( 5x^2 + 15 \): 15, 20, 35, 60, 95, 140, 195, &c.

By inspecting this table, it appears that while \( x \) increases from 0 to a certain numerical value, which exceeds 3, the positive part of the equation, or \( x^3 + 10x \), is always less than the negative part, or \( 5x^2 + 15 \); so that the expression

\[ x^3 + 10x - (5x^2 + 15) \text{ or } x^3 - 5x^2 + 10x - 15 \]

must necessarily be negative.

It also appears, that when \( x \) has increased beyond that numerical value, and which is evidently less than 4, the positive part of the equation, instead of being less than the negative part, is now greater, and therefore the expression

\[ x^3 - 5x^2 + 10x - 15 \]

is changed from a negative to a positive quantity.

Hence we may conclude that there is some real and determinate value of \( x \), which is greater than 3, but less than 4, and which will render the positive and negative parts of the equation equal to one another; therefore that value of \( x \) must be a root of the proposed equation; and as what has been just now shown in a particular case will readily apply to any equation whatever, the truth of the proposition is obvious.

152. Two limits, between which all the roots of any equation are contained, may be determined by this other proposition:

II. Let \( N \) be the greatest negative co-efficient in any equation. Change the signs of the terms taken alternately, beginning with the second, and let \( N' \) be the greatest negative co-efficient after the signs are so changed. The positive roots of the equation are contained between 0 and \( N + 1 \), and the negative roots between 0 and \(-N' - 1\).

Suppose the equation to be

\[ x^4 - px^3 + qx^2 - rx - s = 0, \]

which may be also expressed thus:

\[ x^4 \left(1 - \frac{p}{x} + \frac{q}{x^2} - \frac{r}{x^3} - \frac{s}{x^4}\right) = 0. \]

Then, whatever be the values of the co-efficients \( p, q, r, \ldots \), it is evident that \( x \) may be taken so great as to render each of the quantities \( \frac{p}{x}, \frac{q}{x^2}, \frac{r}{x^3}, \frac{s}{x^4} \) as small as we please, and therefore their sum, or \( -\frac{p}{x} + \frac{q}{x^2} - \frac{r}{x^3} - \frac{s}{x^4} \), less than 1; but in that case the quantity

\[ x^4 \left(1 - \frac{p}{x} + \frac{q}{x^2} - \frac{r}{x^3} - \frac{s}{x^4}\right), \]

or \( x^4 - px^3 + qx^2 - rx + s \),

will be positive, and such, that the first term \( x^4 \) is greater than the sum of all the remaining terms; therefore also \( x^4 + qx^2 \), the sum of the positive terms, will be much greater than \( px^3 + rx + s \), the sum of the negative terms alone.

Hence it follows, that if a number be found, which when substituted for \( x \), renders the expression \( x^4 - px^3 + qx^2 - rx - s \) positive, and which is also such, that every greater number has the same property, that number will exceed the greatest positive root of the equation.

Now, if we suppose $N$ to be the greatest negative coefficient, it is evident that the positive part of the equation, or $x^4 + qx^2 - rx + s$, is greater than $px^3 + qx^2 + rx + s$, provided that $x^4$ is greater than $N x^3 + Nx^2 + Nx + N$, or $N (x^3 + x^2 + x + 1)$; but $x^3 + x^2 + x + 1 = \frac{x^4 - 1}{x - 1}$, therefore a positive result will be obtained, if for $x$ there be substituted a number such that $\frac{N(x^3 - 1)}{x - 1} > N x^3 - N$. Now this last condition will evidently be fulfilled if we take $x^5 - N x^4$, and from this equation we find $x = N + 1$; but it further appears that the same condition will also be fulfilled as often as $x^5 - N x^4$, or $x - 1 > N$, that is, $x > N + 1$, therefore $N + 1$ must be a limit to the greatest positive root of the proposed equation, as was to be shown.

If $y$ be substituted for $x$, the equation $x^4 - px^3 + qx^2 - rx + s = 0$ will be transformed into $y^4 + py^3 + qy^2 + ry + s = 0$; which equation differs from the former only in the signs of the second, fourth, &c. terms; and as the positive roots of this last equation are the same as the negative roots of the proposed equation, it is evident that their limit must be such as has been assigned.

153. From the two preceding propositions it will not be difficult to discover, by means of a few trials, the nearest integers to the roots of any proposed numeral equation; and those being found, we may approximate to the roots continually, as in the following example:

\[ x^4 - 4x^3 - 3x + 27 = 0. \]

Here the greatest negative coefficient being 4, it follows, sect. 152, that the greatest positive root is less than 5. If $-y$ be substituted for $x$, the equation is transformed to

\[ y^4 + 4y^3 + 3y + 27 = 0, \]

an equation having all its terms positive; therefore it can have no positive roots, and consequently the proposed equation can have no negative roots; its real roots must therefore be contained between 0 and +5.

To determine the limits of each root in particular, let 0, 1, 2, 3, 4, be substituted successively for $x$; thus we obtain the following corresponding results:

| Substitutions for $x$ | Results | |----------------------|---------| | 0 | +27 | | 1 | +21 | | 2 | +9 | | 3 | +15 |

Hence it appears that the equation has two real roots, one between 2 and 3, and another between 3 and 4.

That we may approximate to the first root, let us suppose $x = 2 + y$, where $y$ is a fraction less than unity, and therefore its second and higher powers but small in comparison to its first power; hence, in finding an approximate value of $y$, they may be rejected. Thus we have

\[ x^4 = +16 + 32y, \text{ &c.} \] \[ -4x^3 = -32 - 48y, \text{ &c.} \] \[ -3x = -6 - 3y \] \[ +27 = +27 \]

Hence $0 = 5 - 19y$ nearly,

and $y = \frac{5}{19} = 0.26$; therefore, for a first approximation we have $x = 2.26$.

Let us next suppose $x = 2.26 + y$; then, rejecting as before the second and higher powers of $y$ on account of their smallness, and retaining three decimal places, we have

\[ x^4 = +26.087 + 46.172y, \text{ &c.} \] \[ -4x^3 = -46.172 - 61.291y, \text{ &c.} \] \[ -3x = -6.780 - 3y \] \[ +27 = +27 \]

\[ 0 = 1.35 - 18.119y \text{ nearly.} \]

Hence $y = \frac{1.35}{18.119} = 0.075$, and $x = 2.26 + y = 2.2675$. This value of $x$ is true to the last figure, but a more accurate value may be obtained by supposing $x = 2.2675 + y'$, and finding the value of $y'$ in the same manner as we have already found those of $y$ and $y'$; and thus the approximation may be continued till any required degree of accuracy be obtained.

The second root of the equation, which we have already found to be between 3 and 4, may be investigated in the same manner as the first, and will appear to be 3.6797, the approximation being carried on to the fourth figure of the decimal, in determining each root.

154. In the preceding example we have shown how to approximate to the roots of an affected equation; but the same method will also apply to pure equations.

For example, let it be required to determine $x$ from this equation, $x^2 = 2$.

Because $x$ is greater than 1 and less than 2, but nearer to the former number than to the latter, let us assume $x = 1 + y$; then, rejecting the powers of $y$ which exceed the first, we have $x^2 = 1 + 3y$, and therefore $2 = 1 + 3y$, and $y = \frac{1}{3} = 0.3$ nearly; hence $x = 1.3$ nearly.

Let us next assume $x = 1.3 + y$, then proceeding as before, we find $2 = 2.197 + 5.07y$, hence $y = \frac{-1.97}{5.07} = -0.39$, and $x = 1.3 - 0.39 = 1.26$ nearly.

To find a still nearer approximation, let us suppose $x = 1.26 + y$, then, from this assumption we find $y = -0.000079$, and therefore $x = 1.259921$, which value is true to the last figure.

155. By assuming an equation of any order with literal coefficients, a general formula may be investigated for approximating to the roots of equations belonging to that particular order.

Let us take for an example the cubic equation

\[ x^3 + px^2 + qx + r = 0, \]

and suppose that $x = a + y$, where $a$ is nearly equal to $x$, and $y$ is a small fraction. Then, by substituting $a + y$ for $x$ in the proposed equation, and rejecting the powers of $y$ which exceed the first, on account of their smallness, we have

\[ a^3 + pa^2 + qa + r + (3a^2 + 2pa + q)y = 0. \]

Hence $y = \frac{-a^3 - pa^2 - qa - r}{3a^2 + 2pa + q},$

and $x = a - \frac{a^3 + pa^2 + qa + r}{3a^2 + 2pa + q} = \frac{2a^3 + pa^2 - r}{3a^2 + 2pa + q}. $

Let it be required to approximate to a root of the cubic equation $x^3 + 2x^2 + 3x - 50 = 0$. Here $p = 2$, $q = 3$, and $r = -50$; and by trials it appears that $x$ is between 2 and 3, but nearest the latter number; therefore, for the first approximation, $a$ may be supposed = 3, hence we find

\[ x = \frac{2a^3 + pa^2 - r}{3a^2 + 2pa + q} = \frac{129}{42} = 3.1. \]

By substituting $\frac{1}{4}$ for $a$ in the formula, and proceeding as 156. The method we have hitherto employed for approximating to the roots of equations is known by the name of the method of successive substitutions, and was first proposed by Newton. It has been since improved by Lagrange, who has given it a form which has the advantage of showing the progress made in the approximation by each operation. This improved form we now proceed to explain.

Let \(a\) denote the whole number next less to the root sought, and \(\frac{1}{y}\) a fraction, which, when added to \(a\), completes the root; then \(x = a + \frac{1}{y}\). If this value of \(x\) be substituted in the proposed equation, a new equation involving \(y\) will be had, which, when cleared of fractions, will necessarily have a root greater than unity.

Let \(b\) be the whole number which is next less than that root; then, for a first approximation, we have \(x = a + \frac{1}{b}\). But \(b\) being only an approximate value of \(y\), in the same manner as \(a\) is an approximate value of \(x\), we may suppose \(y = b + \frac{1}{y}\); then, by substituting \(b + \frac{1}{y}\) for \(y\), we shall have a new equation, involving only \(y\), which must be greater than unity. Putting therefore \(b'\) to denote the next whole number less than the root of the equation involving \(y\), we have \(y = b + \frac{1}{b'} = \frac{bb' + 1}{b'}\); and substituting this value in that of \(x\), the result is

\[x = a + \frac{b}{bb' + 1}\]

for a second approximate value of \(x\).

To find a third value, we may take \(y = b' + \frac{1}{y'}\); then if \(b''\) denote the next whole number less than \(y'\), we have

\[y = b' + \frac{1}{b''} = \frac{bb' + b'' + 1}{bb' + b''},\]

and so on, to obtain more accurate approximations.

We shall apply this method to the following example:

\[x^3 - 7x + 7 = 0.\]

Here the positive roots must be between 0 and 8; let us therefore substitute successively, 0, 1, 2, ... to 8, and we obtain the following results:

| Substitutions | Results | |---------------|---------| | 0 | | | 1 | | | 2 | | | 3 | | | 4 | | | 5 | | | 6 | | | 7 | |

But as these results have all the same sign, nothing can be concluded respecting the magnitude of the roots from that circumstance alone. It is however observable, that while \(x\) increases from 0 to 1, the results decrease; but that whatever successive magnitudes \(x\) has greater than 2, the results increase. We may therefore reasonably conclude, that if the equation have any positive roots, they must be between 1 and 2. Accordingly, by substituting 1-2, 1-4, 1-6, and 1-8, successively for \(x\), we find these results, +328, -056, -104, +232; and as there are here two changes of the signs, it follows that the equation has two positive roots, one between 1-2 and 1-4, and another between 1-6 and 1-8.

Hence it appears, that to find either value of \(x\), we may assume \(x = 1 + \frac{1}{y}\); then, by substitution, we have

\[y^3 - 4y^2 + 3y + 1 = 0.\]

The limit of the positive root of this last equation is 5, and by substituting 0, 1, 2, 3, 4, successively for \(y\), it will be found to have two, one between 1 and 2, and the other between 2 and 3. Therefore, for a first approximation, we have

\[x = 1 + \frac{1}{y}, x = 1 + \frac{1}{2}, \text{ that is, } x = 2, x = 3.\]

To approach nearer to the first value of \(y\), let us take \(y = 1 + \frac{1}{y'}\), and therefore

\[y^3 - 2y^2 - y + 1 = 0.\]

This last equation will be found to have only one real root between 2 and 3; from which it appears that \(y = 1 + \frac{1}{2} = \frac{3}{2}\), and \(x = 1 + \frac{3}{2} = \frac{5}{2}\).

Let us next suppose \(y = 2 + \frac{1}{y''}\); hence we find

\[y^3 - 3y^2 - 4y' - 1 = 0,\]

and from this equation, \(y'\) is found to be between 4 and 5. Taking the least limit we have

\[y = 2 + \frac{1}{4}, y = 1 + \frac{1}{4}, x = 1 + \frac{1}{4} = \frac{5}{4}.\]

It is easy to continue this process, by assuming \(y = 4 + \frac{1}{y''}\), and so on, as far as may be judged necessary.

We return to the second value of \(x\), which was found = \(\frac{5}{2}\) by the first approximation, and which corresponds to \(y = 2\). Putting \(y = 2 + \frac{1}{y'}\), and substituting this value in the equation \(y^3 - 4y^2 + 3y + 1 = 0\), which was formerly found, we get

\[y^3 + y^2 - 2y - 1 = 0.\]

This, as well as the corresponding equation employed in determining the other value of \(x\), has only one root greater than unity, which root being between 1 and 2, let us take \(y = 1\), we thence find

\[y = 3, \text{ and } x = 1 + \frac{1}{3} = \frac{4}{3}.\]

Put \(y = 1 + \frac{1}{y''}\), and we thence find by substitution

\[y^3 - 3y^2 - 4y' - 1 = 0,\]

an equation which gives \(y'\) between 4 and 5; hence, as before,

\[y = \frac{5}{4}, y = \frac{1}{4}, x = \frac{5}{4}.\]

That we may proceed in the approximation, we have only to suppose \(y = 4 + \frac{1}{y''}\), and so on.

The equation \(x^3 - 7x + 7\) also has a negative root between -3 and -4; and to find a nearer value, put \(x = -3 - \frac{1}{y}\); hence, \(y^3 - 20y^2 - 9y - 1 = 0\), and \(y = 20\), \(y = 21\); and therefore, for the first approximation, \(x = -3 - \frac{1}{20} = -\frac{61}{20}\). By putting \(y = 20 + \frac{1}{y''}\), &c., we may obtain successive values of \(x\), each of which will be more exact than that which preceded it.

157. The successive equations which involve \(y, y', y''\), &c., have never more than one root greater than unity. Alg. unless two or more roots of the proposed equation are contained between the limits \(a\) and \(a+1\); but then, as in the preceding example, some one of the equations involving \(y\), \(y'\), &c. will have more than one root greater than unity, and from each root a series of equations may be derived, by which we may approximate to the particular roots of the proposed equation contained between the limits \(a\) and \(a+1\).

**Sect. XVII.—Of Infinite Series.**

158. The resolving of any proposed quantity into a series, is a problem of considerable importance in the application of algebra to the higher branches of the mathematics; and there are various methods by which it may be performed, suited to the particular forms of the quantities.

Any rational fraction may be resolved into a series, by the common operation of algebraic division, as in the following examples:

**Ex. 1.** To change \(\frac{ax}{a-x}\) into an infinite series.

\[ \begin{align*} \text{Operation.} \\ (ax - x^2) \div (x + \frac{x^2}{a} + \frac{x^3}{a^2} + \frac{x^4}{a^3}, & \text{&c.}) \\ \frac{ax}{a} - \frac{x^2}{a} + \frac{x^3}{a^2} - \frac{x^4}{a^3} + \frac{x^5}{a^4} - \frac{x^6}{a^5} + \cdots \end{align*} \]

Thus it appears that

\[ \frac{ax}{a-x} = x + \frac{x^2}{a} + \frac{x^3}{a^2} + \frac{x^4}{a^3} + \cdots \]

Here the law of the series being evident, the terms may be continued at pleasure.

**Ex. 2.** By a like process we find

\[ \frac{a^2}{(a+x)^2} = 1 - \frac{2x}{a} + \frac{3x^2}{a^2} - \frac{4x^3}{a^3} + \frac{5x^4}{a^4} - \cdots \]

the law of continuation being evident.

159. A second method by which algebraic quantities, whether rational or irrational, may be converted into series, and which is also of very extensive use in the higher parts of the mathematics, consists in assuming a series with indeterminate co-efficients, and having its terms arranged according to the powers of some quantity contained in the proposed expression.

That we may explain this method, let us suppose that the fraction \(\frac{a^2}{a^2 + ax + x^2}\) is to be converted into a series proceeding by the powers of \(x\). We are therefore to assume

\[ \frac{a^2}{a^2 + ax + x^2} = A + Bx + Cx^2 + Dx^3 + Ex^4 + \cdots \]

where \(A\) denotes those terms of the series into which \(x\) does not at all enter, \(Bx\) the terms which contain only the first power of \(x\), \(Cx^2\) the terms which contain only the second power, and so on. Let both sides of the equation be multiplied by \(a^2 + ax + x^2\), so as to take away the denominator of the fraction, and let the numerator \(a^2\) be transposed to the other side, so that the whole expression may be \(= 0\); then

\[ \begin{align*} a^2A + a^2B(x + aC) + a^2D(x^2 + aC + aD) + a^2E(x^3 + aC + aD + aE) + \cdots &= 0 \\ -a^2 + aA(x + aB) + a^2C(x^2 + aC + aD) + a^2D(x^3 + aC + aD + aE) + \cdots &= 0 \\ \end{align*} \]

Now the quantities \(A, B, C, D, &c.\) being supposed to be entirely independent of any particular value of \(x\), it follows that the whole expression can only be \(= 0\), upon the supposition that the terms which multiply the same powers of \(x\) are separately \(= 0\); for if that were not the case, it would follow that \(x\) had a determinate relation to the quantities \(A, B, C, &c.\) which is contrary to what we have supposed. To determine the quantities \(A, B, C, &c.\) therefore, we have this series of equations,

\[ \begin{align*} a^2A - a^2 &= 0, \quad \text{hence } A = 1; \\ a^2B + aA &= 0, \quad B = -\frac{A}{a} = -\frac{1}{a}; \\ a^2C + aB + A &= 0, \quad C = -\frac{B}{a} = -\frac{1}{a^2}; \\ a^2D + aC + B &= 0, \quad D = -\frac{C}{a} = -\frac{1}{a^3}; \\ a^2E + aD + C &= 0, \quad E = -\frac{D}{a} = -\frac{1}{a^4}; \\ \end{align*} \]

&c.

Here the law of relation which takes place among the quantities \(A, B, C, D, &c.\) is evident, viz. that if \(P, Q, R\) denote any three co-efficients which immediately follow each other,

\[ a^2R + aQ + P = 0; \]

and from this equation, by means of the co-efficients already determined, we find \(F = 0, G = \frac{1}{a^6}, H = -\frac{1}{a^7}, K = 0, &c.\)

Therefore, resuming the assumed equation, and substituting for \(A, B, C, &c.\) their respective values, we have

\[ \frac{a^2}{a^2 + ax + x^2} = 1 - \frac{2x}{a} + \frac{3x^2}{a^2} - \frac{4x^3}{a^3} + \frac{5x^4}{a^4} - \cdots \]

As a second example of the method of indeterminate co-efficients, let it be required to express the square root \(a^2 - x^2\) by means of a series. For this purpose we might assume

\[ \sqrt{a^2 - x^2} = A + Bx + Cx^2 + Dx^3 + Ex^4 + \cdots \]

but as we would then find the co-efficients of the odd powers of \(x\) to be each \(= 0\), let us rather assume

\[ \sqrt{a^2 - x^2} = A + Bx^2 + Cx^4 + Dx^6 + \cdots \]

then, squaring both sides, and transposing, we have

\[ 0 = \left\{ \begin{array}{c} A^2 + 2AB \\ - a^2 + 1 \\ \end{array} \right\} x^2 + \left\{ \begin{array}{c} 2AC + 2AD \\ + B^2 \\ \end{array} \right\} x^4 + \cdots \]

Hence \(A^2 - a^2 = 0\), and \(A = a\);

\[ \begin{align*} 2AB + 1 &= 0, \quad B = -\frac{1}{2a} = -\frac{1}{2a}; \\ 2AC + B^2 &= 0, \quad C = -\frac{B^2}{2A} = -\frac{1}{8a^2}; \\ AD + BC &= 0, \quad D = -\frac{BC}{A} = -\frac{1}{16a^3}; \\ \end{align*} \]

&c. Algebra, and substituting for A, B, C, &c. their values;

\[ \sqrt{a^2 - x^2} = \frac{a^2}{2x} - \frac{x^4}{8a^3} + \frac{x^6}{16a^5}, \ldots \]

This method of resolving a quantity into any infinite series will be found more expeditions than any other, as often as the operations of division and evolution are to be performed at the same time, as in these expressions,

\[ \frac{1}{\sqrt{a^2 + x^2}} \text{ or } \frac{\sqrt{a^2 - x^2}}{\sqrt{a^2 + x^2}} \]

160. The binomial theorem affords a third method of resolving quantities into series; but we must first show how the theorem itself may be investigated.

Let \(a + x\) be any binomial quantity which is to be raised to a power denoted by \(\frac{m}{n}\), where \(m\) and \(n\) are any numbers either positive or negative.

Because \(a + x = a(1 + \frac{x}{a})\), if we put \(\frac{x}{a} = y\), then \((a + x)^n = a^n \times (1 + y)^n\); therefore, instead of \(a + x\), we may consider \(1 + y\), which is somewhat more simple in its form.

By considering some integer powers of \(1 + x\), as

\[ (1 + x) = 1 + x, \] \[ (1 + x)^2 = 1 + 2x + x^2, \] \[ (1 + x)^3 = 1 + 3x + 3x^2 + x^3, \] \[ (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4, \] \[\ldots\]

it may be inferred that all powers of \(1 + x\) have this form,

\[ 1 + Ax + Bx^2 + Cx^3 + Dx^4 + Ex^5 + \ldots \]

where the co-efficients \(A, B, C, D, E, \ldots\) are numbers which are altogether independent of any particular value of \(x\). It also appears that the series cannot contain any negative power of \(x\); for if any of its terms had this form, \(\frac{Q}{x^n}\), then the supposition of \(x = 0\) would render that term indefinitely great; whereas the whole series ought in that case to be reduced to unity.

Let us therefore assume

\[ (1 + y)^n = 1 + Ay + By^2 + Cy^3 + Dy^4 + \ldots \]

Then we have also

\[ (1 + z)^n = 1 + Az + Bz^2 + Cz^3 + Dz^4 + \ldots \]

Let us put \((1 + y)^n = u\), \((1 + z)^n = v\), and therefore

\[ u^n - v^n = A(y - z) + B(y^2 - z^2) + C(y^3 - z^3) + D(y^4 - z^4) + \ldots \]

Because \(u^n = 1 + y\) and \(v^n = 1 + z\), by subtracting the latter equation from the former, we have \(u^n - v^n = y - z\); hence, and from the last series, it follows that

\[ \frac{u^n - v^n}{y - z} = \frac{A(y - z)}{y - z} + \frac{B(y^2 - z^2)}{y - z} + \frac{C(y^3 - z^3)}{y - z} + \ldots \]

But every expression of the form \(u^n - v^n\) is divisible by \(u - v\), when \(m\) is a whole number. Thus we have

\[ u^n - v^n = (u - v)(u^{n-1} + u^{n-2}v + \ldots + uv^{n-2} + v^{n-1}) \]

so that if we substitute for \(u^n - v^n\) its value as found from these equations, and divide each term of the series by the denominator \(y - z\), we have

\[ \frac{u^{n-1} + u^{n-2}v + \ldots + uv^{n-2} + v^{n-1}}{y - z} = \]

\[ A + B(y + z) + C(y^2 + yz + z^2) + D(y^3 + y^2z + yz^2 + z^3) + \ldots \]

Now, as this last equation must be true, whatever be the values of \(y\) and \(z\), we may suppose \(y = z\), but in that case \(1 + y = 1 + z\), or \(u^n = v^n\), and therefore \(u = v\). Thus the equation is reduced to

\[ \frac{u^{n-1} + u^{n-2}v + \ldots + uv^{n-2} + v^{n-1}}{y - z} = \]

\[ A + 2By + 3Cy^2 + 4Dy^3 + 5Ey^4 + \ldots \]

or to the following:

\[ \frac{m}{n} u^n = u^n(A + 2By + 3Cy^2 + 4Dy^3 + 5Ey^4 + \ldots) \]

so that, putting for \(u^n\) and \(u^n\) their values \((1 + y)^n\) and \(1 + y\), we have

\[ \frac{m}{n}(1 + y)^n = (1 + y)(A + 2By + 3Cy^2 + 4Dy^3 + 5Ey^4 + \ldots) \]

\[ = \left\{ \begin{array}{l} A + 2By + 3Cy^2 + 4Dy^3 + 5Ey^4 + \ldots \\ + Ay + 2By^2 + 3Cy^3 + 4Dy^4 + \ldots \end{array} \right. \]

But from the equation originally assumed we have

\[ \frac{m}{n}(1 + y)^n = \]

\[ \frac{m}{n} + \frac{m}{n}Ay + \frac{m}{n}By^2 + \frac{m}{n}Cy^3 + \frac{m}{n}Dy^4 + \ldots \]

Therefore

\[ \frac{m}{n} + \frac{m}{n}Ay + \frac{m}{n}By^2 + \frac{m}{n}Cy^3 + \frac{m}{n}Dy^4 + \ldots \]

\[ = \left\{ \begin{array}{l} A + 2By + 3Cy^2 + 4Dy^3 + 5Ey^4 + \ldots \\ + Ay + 2By^2 + 3Cy^3 + 4Dy^4 + \ldots \end{array} \right. \]

And as the co-efficients of the terms have no connection with any particular value of \(y\), it follows that the co-efficient of any power of \(y\) on the one side of the equation must be equal to the co-efficient of the same power of \(y\) on the other side. Therefore, to determine \(A, B, C, \ldots\) we have the following series of equations:

\[ A = \frac{m}{n}, \quad \text{hence } A = \frac{m}{n}; \]

\[ 2B + \frac{m}{n}A = \frac{A(m - n)}{2n}; \]

\[ 3C + 2B = \frac{B(m - 2n)}{3n}; \]

\[ 4D + 3C = \frac{C(m - 3n)}{4n}; \]

\[ 5E + 4D = \frac{D(m - 4n)}{5n}; \]

\[ \ldots \]

Or, substituting for \(A, B, C, \ldots\) their values as determined from the preceding equations:

\[ A = \frac{m}{n}, \]

\[ B = \frac{m(m - n)}{2n}; \]

\[ C = \frac{m(m - n)(m - 2n)}{3n}; \] Resuming now the assumed equation

\[ (1+y)^n = 1 + Ay + By^2 + Cy^3 + \ldots \]

and observing that \( \frac{x}{a} = y \), and \( (a+x)^n = a^n (1+y)^n \), we have \( (a+x)^n \) expressed by the series

\[ \begin{align*} & a^n \left( 1 + \frac{m}{n} x + \frac{A(m-n)}{2n} x^2 + \frac{B(m-2n)}{3n} x^3 + \ldots \right) \\ & + \frac{C(m-3n)}{4n} x^4 + \frac{D(m-4n)}{5n} x^5 + \ldots , \end{align*} \]

where A, B, C, &c. denote the co-efficients of the preceding terms, or

\[ \begin{align*} & a^n + \frac{m}{n} a^{n-1} x + \frac{m(m-n)}{1 \cdot 2 \cdot n^2} x^2 \\ & + \frac{m(m-n)(m-2n)}{1 \cdot 2 \cdot 3 \cdot n^3} x^3 \\ & + \frac{m(m-n)(m-2n)(m-3n)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot n^4} x^4 + \ldots , \end{align*} \]

and either of these formulae may be considered as a general theorem for raising a binomial quantity \( a+x \) to any power whatever.

161. In determining the value of the expression \( \frac{m-mn}{m-mn} \)

when \( u=v \), it has been assumed that \( \frac{m}{n} \) is positive; but the same conclusion will be obtained when \( \frac{m}{n} \) is negative.

For, changing \( +m \) into \( -m \), and observing that

\[ \begin{align*} & \frac{m-mn}{m-mn} = \frac{1}{m} - \frac{1}{v_m} = \frac{v_m-mn}{v_m}, \end{align*} \]

we have

\[ \begin{align*} & \frac{v_m-mn}{v_m} = \frac{1}{v_m} - \frac{1}{v_m} = \frac{v_m-mn}{v_m}. \end{align*} \]

Now we have already found, that when \( u=v \), the fraction \( \frac{m-mn}{m-mn} \) becomes \( \frac{mn-1}{mn-1} \); therefore, in the same case,

\[ \begin{align*} & \frac{u-mn}{u-mn} = \frac{1}{u} \times \frac{mn-1}{mn-1} = \frac{mn-1}{mn-1}; \end{align*} \]

and from this last expression we derive the same value for \( \frac{u-mn}{u-mn} \) or \( (1+y)^n \) as before, regard being had to the change of the sign of the exponent.

162. If we suppose \( m \) to be a positive integer, and \( n=1 \), the series given in last article for the powers of \( a+x \) will always terminate, as appears also from the operation of involution; but if \( m \) be negative, or \( \frac{m}{n} \) a fraction, the series will consist of an indefinite number of terms. Examples of the application of the theorem have been already given upon the first supposition, when treating of involution; we now proceed to show how it is to be applied to the expansion of algebraic quantities into series upon either of the last two hypotheses.

Ex. 1. It is required to express \( \frac{r^3}{(r+z)^3} \) by means of a series.

Because \( \frac{r}{r+z} = \frac{1}{1+\frac{z}{r}} \),

therefore \( \frac{r^3}{(r+z)^3} = \frac{1}{\left(1+\frac{z}{r}\right)^3} = \left(1+\frac{z}{r}\right)^{-3} \).

Let \( \left(1+\frac{z}{r}\right)^{-3} \) be compared with \( (a+x)^n \), and we have

\( a=1, x=\frac{z}{r}, m=-3, n=1 \).

Hence, by substituting these values of \( a, x, m, n \), in the first general formula of sect. 160, we have

\[ \begin{align*} & r^3 = 1 - \frac{3z}{r} + \frac{3 \cdot 3z^2}{1 \cdot 2 \cdot r^2} - \frac{3 \cdot 3 \cdot 3z^3}{1 \cdot 2 \cdot 3 \cdot r^3} + \ldots , \\ & (r+z)^3 = 1 - \frac{3z}{r} + \frac{6z^2}{r^2} - \frac{10z^3}{r^3} + \frac{15z^4}{r^4} - \ldots . \end{align*} \]

Ex. 2. It is required to express \( \sqrt{a+b} \) in the form of a series.

Because \( a+b=a\left(1+\frac{b}{a}\right) \),

therefore \( \sqrt{a+b} = \sqrt{a} \times \sqrt{1+\frac{b}{a}} = a^{\frac{1}{2}}\left(1+\frac{b}{a}\right)^{\frac{1}{2}} \).

By comparing \( \left(1+\frac{b}{a}\right)^{\frac{1}{2}} \) with \( (a+x)^n \), we have \( a=1 \),

\( x=\frac{b}{a}, m=1, n=3 \), and \( \sqrt{a+b} \) is

\[ \begin{align*} & a^{\frac{1}{2}} \left(1 + \frac{b}{3a} - \frac{1 \cdot 2 \cdot b^2}{3 \cdot 6 \cdot a^2} + \frac{1 \cdot 2 \cdot 3 \cdot b^3}{3 \cdot 6 \cdot 9 \cdot a^3} - \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot b^4}{3 \cdot 6 \cdot 9 \cdot 12 \cdot a^4} + \ldots \right) \\ & = a^{\frac{1}{2}} \left(1 + \frac{b}{3a} - \frac{b^2}{9a^2} + \frac{5b^3}{81a^3} - \frac{10b^4}{243a^4} + \ldots \right). \end{align*} \]

Ex. 3. It is required to resolve \( \frac{r^2}{(r^2+z^2)^{\frac{3}{2}}} \) into a series.

Because \( \frac{r^2}{(r^2+z^2)^{\frac{3}{2}}} = r^2(-z)^{-\frac{3}{2}} \) if we raise \( r^2+z^2 \) to the \( -\frac{3}{2} \) power, and multiply the resulting series by \( r^2 \), we shall have the series required. Or the given quantity may be reduced to a more simple form thus: because

\[ \begin{align*} & r^2 + z^2 = r^2 \left(1 + \frac{z^2}{r^2}\right), \end{align*} \]

therefore \( \left(r^2 + z^2\right)^{\frac{3}{2}} = r^2 \left(1 + \frac{z^2}{r^2}\right)^{\frac{3}{2}} \),

and \( \frac{r^2}{\left(r^2 + z^2\right)^{\frac{3}{2}}} = \frac{1}{\left(1 + \frac{z^2}{r^2}\right)^{\frac{3}{2}}} = \left(1 + \frac{z^2}{r^2}\right)^{-\frac{3}{2}} \).

Hence \( \frac{r^2}{\left(r^2 + z^2\right)^{\frac{3}{2}}} = \left(1 + \frac{z^2}{r^2}\right)^{-\frac{3}{2}} \),

\[ \begin{align*} & = 1 - \frac{2z^2}{3r^2} + \frac{2 \cdot 5z^4}{3 \cdot 6 \cdot r^4} - \frac{2 \cdot 5 \cdot 8 \cdot 11z^6}{3 \cdot 6 \cdot 9 \cdot 12 \cdot r^6} + \ldots , \\ & = 1 - \frac{2z^2}{3r^2} + \frac{5z^4}{9r^4} - \frac{40z^6}{81r^6} + \frac{110z^8}{243r^8} - \ldots . \end{align*} \]

Ex. 4. It is required to find a series equal to \( \frac{\sqrt{a^2+x^2}}{\sqrt{a^2-x^2}} \)

By the binomial theorem, we have

\[ \begin{align*} & \sqrt{a^2+x^2} = \left(a^2+x^2\right)^{\frac{1}{2}} = a + \frac{x^2}{2a} - \frac{x^4}{8a^3} + \frac{x^6}{16a^5} - \ldots , \\ & \sqrt{a^2-x^2} = \left(a^2-x^2\right)^{\frac{1}{2}} = a - \frac{x^2}{2a} + \frac{x^4}{8a^3} - \frac{x^6}{16a^5} + \ldots . \end{align*} \] Therefore, by taking the product of the two series, and proceeding in the operation only to such terms as involve the 6th power of \( x \), we find

\[ \frac{\sqrt{a^2 + x^2}}{\sqrt{a^2 - x^2}} = 1 + \frac{x^2}{a^2} - \frac{x^4}{2a^4} + \frac{x^6}{2a^6}, \text{ &c.} \]

**Sect. XVIII.—Of the Reversion of Series.**

163. The method of indeterminate co-efficients, which we have already employed when treating of infinite series, may also be applied to what is called the reverting of series; that is, having any quantity expressed by an infinite series composed of the powers of another quantity, to express, on the contrary, the latter quantity by means of an infinite series composed of the powers of the former.

Let \( y = n + ax + bx^2 + cx^3 + dx^4 + \text{&c.} \)

Then, to revert the series, we must find the value of \( x \) in terms of \( y \). For this purpose, we transpose \( n \) and put \( z = y - n \); then

\[ z = ax + bx^2 + cx^3 + dx^4 + \text{&c.} \]

Now, when \( z = 0 \), it is evident that \( z = 0 \); therefore we may assume for \( x \) a series of this form,

\[ x = Az + Bz^2 + Cz^3 + Dz^4 + \text{&c.} \]

where the co-efficients \( A, B, C, D, \text{&c.} \) denote quantities as yet unknown, but which are entirely independent of the quantity \( x \). To determine these, let the first, second, third, &c. powers of the series

\[ Az + Bz^2 + Cz^3 + Dz^4 + \text{&c.} \]

be found by multiplication, and substituted for \( x, x^2, x^3, \text{&c.} \) respectively, in the equation

\[ 0 = -z + ax + bx^2 + cx^3 + \text{&c.} \]

thus we have

\[ \begin{align*} -1 &= -1 \\ +ax &= aAz + aBz^2 + aCz^3 + aDz^4 + \text{&c.} \\ +bx^2 &= bAz^2 + 2bABz^3 + 2bACz^4 + \text{&c.} \\ +cx^3 &= cAz^3 + 3cABz^4 + \text{&c.} \\ +dx^4 &= dAz^4 + \text{&c.} \end{align*} \]

Hence, putting the co-efficients of \( x, x^2, x^3, \text{&c.} \) each \( = 0 \),

\[ \begin{align*} aA - 1 &= 0 \\ aB + bA^2 &= 0 \\ aC + 2bAC + cA^3 &= 0 \\ aD + 2bAC + bB^2 + 3cA^2B + dA^4 &= 0, \text{&c.} \end{align*} \]

these equations give

\[ \begin{align*} A &= \frac{1}{a} \\ B &= \frac{b}{a^2} \\ C &= \frac{2b^2 - ac}{a^3} \\ D &= \frac{-5b^3 - 5abc + a^2d}{a^7} \\ &\text{&c.} \end{align*} \]

Therefore \( x = \frac{1}{a} - \frac{b}{a^2} z^2 + \frac{2b^2 - ac}{a^3} z^3 - \frac{5b^3 - 5abc + a^2d}{a^7} z^4 + \text{&c.} \)

As an example of the application of this formula, let it be required to determine \( x \) from the equation

\[ y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \text{&c.} \]

In this case we have

\[ z = y, \quad a = 1, \quad b = -\frac{1}{2}, \quad c = \frac{1}{3}, \quad d = -\frac{1}{4}, \text{&c.} \]

Therefore, substituting these values, we have

\[ x = y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24} + \text{&c.} \]

In the equation

\[ ay + by^2 + cy^3 + \text{&c.} = ax + bx^2 + cx^3 + \text{&c.} \]

in which both sides are expressed by series, and it is required to find \( y \) in terms of \( x \), we must assume, as before,

\[ y = Ax + Bx^2 + Cx^3 + Dx^4 + \text{&c.} \]

and substitute this series and its powers for \( y \) and its powers in the proposed equation; afterwards, by bringing all the terms to one side, and making the co-efficients of each power of \( y = 0 \), a series of equations will be had by which the quantities \( A, B, C, D, \text{&c.} \) may be determined.

**Sect. XIX.—Of Logarithms and Exponential Quantities.**

164. All positive numbers may be considered as powers of any one given affirmative number. The powers of 2, for instance, may become equal, either exactly, or nearer than by any assignable difference, to all numbers whatever, from 1 upwards. If the exponents be integers, we shall have only the numbers which form the geometrical progression, 1, 2, 4, 8, 16, &c.; but the intermediate numbers may be expressed, at least nearly, by means of fractional exponents. Thus, the numbers from 1 to 10 may be expressed by the powers of 2 as follows:

\[ \begin{align*} 2^0 &= 1 \\ 2^1 &= 2 \\ 2^2 &= 4 \\ 2^3 &= 8 \\ 2^4 &= 16 \\ 2^5 &= 32 \\ 2^6 &= 64 \\ 2^7 &= 128 \\ 2^8 &= 256 \\ 2^9 &= 512 \\ 2^{10} &= 1024 \end{align*} \]

In like manner may fractions be expressed by the powers of 2. Thus,

\[ \begin{align*} \frac{1}{2} &= 2^{-1} \\ \frac{1}{4} &= 2^{-2} \\ \frac{1}{8} &= 2^{-3} \\ \frac{1}{16} &= 2^{-4} \\ \frac{1}{32} &= 2^{-5} \\ \frac{1}{64} &= 2^{-6} \\ \frac{1}{128} &= 2^{-7} \\ \frac{1}{256} &= 2^{-8} \\ \frac{1}{512} &= 2^{-9} \\ \frac{1}{1024} &= 2^{-10} \end{align*} \]

where it is observable that the exponents are now negative.

In the same manner may all numbers be expressed by the powers of 10.

\[ \begin{align*} 10^0 &= 1 \\ 10^1 &= 10 \\ 10^2 &= 100 \\ 10^3 &= 1000 \\ 10^4 &= 10000 \\ 10^5 &= 100000 \\ 10^6 &= 1000000 \\ 10^7 &= 10000000 \\ 10^8 &= 100000000 \\ 10^9 &= 1000000000 \\ 10^{10} &= 10000000000 \end{align*} \]

Even a fraction might be taken in place of 2, or 10, in the preceding examples; and such exponents might be found as would give its powers equal to all numbers, from 0 upwards. There are therefore no limitations with respect to the magnitude of the number, by the powers of which all other numbers may be expressed, except that it must neither be unity nor a negative quantity. If it were \( = 1 \), then all its powers would also be \( = 1 \); and if it were negative, there are numbers to which none of its powers could possibly be equal.

165. If therefore \( y \) denote any number whatever, and \( r \) a given number, a number \( x \) may be found, such, that \( r^x = y \), and \( x \), that is, the exponent of \( r \) which gives a number equal to \( y \), is called the logarithm of \( y \).

The given number \( r \), by the powers of which all other Algebra. numbers are expressed, is called the radical number of the logarithms, which are the indices of those powers.

166. From this definition of logarithms their properties are easily deduced as follows:

1. The sum of two logarithms is equal to the logarithm of their product. Let \( y \) and \( y' \) be two numbers, and \( x \) and \( x' \) their logarithms, so that \( r = \log_y \) and \( r' = \log_{y'} \); then \( r + r' = \log_{yy'} \), or \( r + r' = \log_{yy'} \); hence, from the definition, \( x + x' \) is the logarithm of \( yy' \), that is, the sum of the logarithms of \( y \) and \( y' \) is the logarithm of \( yy' \).

2. The difference of the logarithms of two numbers is equal to the logarithm of their quotient; for if \( r = \log_y \) and \( r' = \log_{y'} \), then \( r - r' = \log_{\frac{y}{y'}} \) or \( r - r' = \log_{\frac{y}{y'}} \), therefore, by the definition, \( x - x' \) is the logarithm of \( \frac{y}{y'} \); that is, the difference of the logarithms of \( y \) and \( y' \) is the logarithm of \( \frac{y}{y'} \).

3. Let \( n \) be any number whatever, then \( \log_y^n = n \times \log_y \). For \( y^n \) is \( y \) multiplied into itself \( n \) times, therefore the logarithm of \( y^n \) is equal to the logarithm of \( y \) added to itself \( n \) times, or to \( n \times \log_y \).

167. From these properties of logarithms it follows, that if we possess tables by which we can assign the logarithm corresponding to any given number, and also the number corresponding to any given logarithm, the operations of multiplication and division of numbers may be reduced to the addition and subtraction of their logarithms, and the operations of involution and evolution, to the more simple operations of multiplication and division. Thus, if two numbers \( x \) and \( y \) are to be multiplied together, by taking the sum of their logarithms, we obtain the logarithm of their product, and, by inspecting the table, the product itself. A similar observation applies to the quotient of two numbers, and also to any power or to any root of a number.

168. The general properties of logarithms are independent of any particular value of the radical number, and hence there may be various systems of logarithms, according to the radical number employed in their construction. Thus, if the radical number be 10, we shall have the common or Briggs's system of logarithms; but if it were 2^7182818, we should have the logarithms first constructed by Lord Napier, which are sometimes called hyperbolic logarithms.

We have already observed (sect. 165), that the relation between any number and its logarithm is expressed by the equation \( r = \log_y \), where \( y \) denotes a number, \( x \) its logarithm, and \( r \) the radical number of the system; and any two of these three quantities being given, the remaining one may be found. If either \( y \) or \( r \) were the quantity required, the problem would involve no difficulty; if, however, the exponent \( x \) were considered as the unknown quantity, while \( r \) and \( y \) were supposed given, the equation to be resolved would be of a different kind from any we have hitherto considered. Equations of this form are called exponential equations. To resolve such an equation is evidently the same thing as to determine the logarithm of a given number.

169. We therefore resume the equation \( r = \log_y \), where \( r, x, \) and \( y \) denote as before. We are now to find a value of \( x \) in terms of \( r \) and \( y \). Let us suppose \( r = 1 + a \) and \( y = 1 + v \), then our equation will stand thus:

\[ (1 + a)^r = 1 + v. \]

So that, by raising both sides to a power \( n \), where \( n \) denotes an indeterminate number, which is to disappear in the course of the investigation, we have \( (1 + a)^n = (1 + v)^n \); let us now put \( A \) to denote the constant multiplier and resolving both sides of the equation into series by means of the binomial theorem,

\[ 1 + nx + \frac{n(n-1)}{1 \cdot 2}x^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}x^3 + \cdots \]

\[ + \frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}x^4 + \cdots \]

\[ = 1 + nv + \frac{n(n-1)}{1 \cdot 2}v^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}v^3 + \cdots \]

\[ + \frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}v^4 + \cdots \]

Therefore, subtracting unity from both sides, and dividing by \( n \), we have

\[ xa + \frac{x(nx-1)}{1 \cdot 2}a^2 + \frac{x(nx-1)(nx-2)}{1 \cdot 2 \cdot 3}a^3 + \cdots \]

\[ + \frac{x(nx-1)(nx-2)(nx-3)}{1 \cdot 2 \cdot 3 \cdot 4}a^4 + \cdots \]

\[ = v + \frac{n-1}{2}v^2 + \frac{(n-1)(n-2)}{2 \cdot 3}v^3 + \cdots \]

\[ + \frac{(n-1)(n-2)(n-3)}{2 \cdot 3 \cdot 4}v^4 + \cdots \]

and by supposing the factors which constitute the terms of each series to be actually multiplied, and the products arranged according to the power of \( n \), the last equation will have this form,

\[ xa + \left( Pn - \frac{x}{2} \right)a^2 + \left( Pn + Qn^2 + \frac{x}{3} \right)a^3 + \cdots \]

\[ + \left( Pn + Qn^2 + Rn^3 - \frac{x}{4} \right)a^4 + \cdots \]

\[ = v + \left( pn - \frac{1}{2} \right)v^2 + \left( pn + qv^2 + \frac{1}{3} \right)v^3 + \cdots \]

\[ + \left( pn + qv^2 + rv^3 - \frac{1}{4} \right)v^4 + \cdots \]

Here the co-efficients of the power \( n \), viz. \( P, P', P'', \ldots Q, Q', \ldots R, \ldots \) etc., also \( p, p', p'', \ldots q, q', \ldots r, \ldots \) etc., are expressions which denote certain combinations of the powers of \( x \) in the first series, and certain numbers in the second; but as they are all to vanish in the course of the investigation, it is not necessary that they should be expressed in any other way than by a single letter.

Now each side of this last equation may evidently be resolved into two parts, one of which is entirely free from the quantity \( n \), and the other involves that quantity; hence the same equation may also stand thus:

\[ xa - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]

\[ + Pna^2 + \left( Pn + Qn^2 \right)a^3 + \left( Pn + Qn^2 + Rn^3 \right)a^4 + \cdots \]

\[ = v - \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \cdots \]

This equation must hold true, whatever be the value of \( n \), which is a quantity entirely arbitrary, and therefore ought to vanish from the equation expressing the relation between \( x \) and \( e \); hence it follows that the terms on each side of the equation, which involve \( n \), ought to destroy each other, and thus there will remain only the part of each side which does not involve \( n \); that is,

\[ xa - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]

\[ = v - \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \cdots \]

or \((a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \cdots)x\)

\[ = v - \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \cdots \] \[ a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \ldots \]

\[ = (r-1) - \frac{(r-1)^2}{2} + \frac{(r-1)^3}{3} - \frac{(r-1)^4}{4} + \ldots \]

and substitute for \( r \) its value, \( y = 1 \); thus we find

\[ \log y = \frac{1}{A} \left( (y-1) - \frac{(y-1)^2}{2} + \frac{(y-1)^3}{3} - \frac{(y-1)^4}{4} + \ldots \right) \]

and by this formula, the logarithm of any number a little greater than unity may be readily found.

170. If \( y \) be nearly \( = 2 \), the series will converge too slowly to be of use, and if it exceed 2, the series will diverge, and therefore cannot be directly applied to the finding of its logarithm. But a series which converges faster, and is applicable to every case, may be investigated as follows:

Because

\[ \log (1+v) = \frac{1}{A} \left( v - \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \ldots \right) \]

by substituting \(-v\) for \(+v\), we have

\[ \log (1-v) = \frac{1}{A} \left( -v - \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \ldots \right) \]

Now,

\[ \log (1+v) - \log (1-v) = \log \frac{1+v}{1-v} \]

therefore, subtracting the latter series from the former, we have

\[ \log \frac{1+v}{1-v} = \frac{2}{A} \left( v + \frac{v^3}{3} + \frac{v^5}{5} + \ldots \right) \]

Put \( \frac{1+v}{1-v} = y \), then \( v = \frac{y-1}{y+1} \), and the last series becomes

\[ \log y = \frac{2}{A} \left( \frac{y-1}{y+1} + \frac{1}{3} \left( \frac{y-1}{y+1} \right)^3 + \frac{1}{5} \left( \frac{y-1}{y+1} \right)^5 + \ldots \right) \]

This series will always converge, whatever be the value of \( y \); and by means of it the logarithms of small numbers may be found with great facility.

171. When a number is composite, its logarithm will most easily be found, by adding together the logarithms of its factors; but if it be a prime number, its logarithm may be derived from that of some convenient composite number, either greater or less, and an infinite series. Let \( n \) be a number of which the logarithm is already found; then, substituting \( \frac{n+z}{n} \) for \( y \) in the last formula, we have

\[ \log \frac{n+z}{n} = \frac{1}{A} \left( \frac{2z}{2n+z} + \frac{1}{3} \left( \frac{2z}{2n+z} \right)^3 + \frac{1}{5} \left( \frac{2z}{2n+z} \right)^5 + \ldots \right) \]

But \( \log \frac{n+z}{n} = \log (n+z) - \log n \), therefore \( \log (n+z) = \log n + \frac{1}{A} \left( \frac{2z}{2n+z} + \frac{1}{3} \left( \frac{2z}{2n+z} \right)^3 + \frac{1}{5} \left( \frac{2z}{2n+z} \right)^5 + \ldots \right) \)

This series gives the logarithm of \( n+z \) by means of the logarithm of \( n \), and converges very fast when \( n \) is considerable.

172. It appears, from the series which have been found for \( \log y \), that the logarithm of a number is always the product of two quantities: one of these is variable, and depends upon the number itself; but the other, viz. \( \frac{1}{A} \), is constant, and depends entirely on the radical number of the system. This quantity has been called by writers on logarithms the modulus of the system.

173. The most simple system, in respect to facility of computation, is that in which \( \frac{1}{A} = 1 \) or \( A = 1 \). The logarithms of this system are the same as those first invented by Napier. They have been called Hyperbolic Logarithms, because they serve to express the area of an hyperbola; but this may be done by logarithms of any system. We shall therefore distinguish them by calling them Napierian Logarithms. The Napierian logarithm of any number \( y \) is therefore

\[ (y-1) - \frac{1}{2} (y-1)^2 + \frac{1}{3} (y-1)^3 - \frac{1}{4} (y-1)^4 + \ldots \]

and that of \( r \), the radical number of any system, is

\[ (r-1) - \frac{1}{2} (r-1)^2 + \frac{1}{3} (r-1)^3 - \frac{1}{4} (r-1)^4 + \ldots \]

But this last series is the same as we have denoted by \( A \); hence it follows, that the modulus of any system is the reciprocal of the Napierian logarithm of the radical number of that system. Thus it appears that the logarithms of numbers, according to any proposed system, may be readily found from the Napierian logarithm of the same numbers, and the Napierian logarithm of the radical number of that system.

174. Let \( L \) denote the Nap. log. of any number, and \( l, l' \) the logarithms of the same number according to two other systems whose moduli are \( m \) and \( m' \); then

\[ l = mL, l' = m'L; \]

therefore, \( \frac{l}{m} = \frac{l'}{m'} \) and \( m : m' :: l : l' \).

That is, the logarithms of the same number, according to different systems, are directly proportional to the moduli of these systems, and therefore have a given ratio to one another.

175. We shall now apply the series here investigated to the calculation of Napier's logarithm of 10, the reciprocal of which is the modulus of the common system of logarithms; and also to the calculation of the common logarithm of 2. The Nap. log. of 10 may be obtained by substituting 10 for \( y \) in the formula

\[ \text{Nap. log. } y = \frac{2(y-1)}{y+1} + \frac{2(y-1)}{3(y+1)} + \frac{2(y-1)}{5(y+1)} + \ldots \]

but the resulting series

\[ \frac{2}{11} + \frac{2}{3 \cdot 11} + \frac{2}{5 \cdot 11} + \ldots \]

converges too slowly to be of any practical utility; it will therefore be better to derive the logarithm of 10 from those of 2 and 5. By substituting 2 in the formula, we have

\[ \text{Nap. log. } 2 = \frac{2}{3} + \frac{1}{3 \cdot 3} + \frac{1}{5 \cdot 3} + \frac{1}{7 \cdot 3} + \ldots \]

This series converges very fast, so that by reducing its terms to decimal fractions, and taking the sum of the first seven terms, we find the Nap. log. of 2 to be .9031472.

The Nap. log. of 5 may be found in the same manner, but more easily from the formula given in sect. 171. For the log. of 2 being given, that of \( \frac{4}{2} = 2 \) is also given (sect. 166); therefore, substituting \( \log_4 = 2 \log_2 \) for \( \log_2 \), and 1 for \( z \), in the series

\[ \text{Nap. log. } (n+z) = \text{Nap. log. } n + \frac{2}{3} \left( \frac{z}{2n+z} + \frac{1}{3} \left( \frac{z}{2n+z} \right)^3 + \frac{1}{5} \left( \frac{z}{2n+z} \right)^5 + \ldots \right) \]

we have

\[ \text{Nap. log. } 5 = \text{Nap. log. } 2 + \frac{2}{3} \left( \frac{1}{3} + \frac{1}{5 \cdot 3} + \frac{1}{7 \cdot 3} + \ldots \right) \]

The first three terms of this series are sufficient to give the result true to the seventh decimal, so that we have

\[ \text{Nap. log. } 5 = 1.6094379, \]

and

\[ \text{Nap. log. } 10 = \text{Nap. log. } 2 + \text{Nap. log. } 5 = 2.3025851. \]

Hence the modulus of the common system of logarithms, or \( \frac{1}{\text{Nap. log. } 10} \), is found \( = 4342945 \). The same number, because of its great utility in the construction of tables of Algebra. logarithms, has been calculated to a much greater number of decimals. A celebrated calculator of the last century, Mr. A. Sharp, found it to be

\[0.43429448190325182765112891891660508229439700\]

Having found the Nap. log. of 2 to be -6931472, the common logarithm of 2 is got immediately by multiplying the Nap. log. of 2 by the modulus of the system; thus, we find

\[ \text{com. log. } 2 = 4.342945 \times 6931472 = 3010300. \]

We have seen, sect. 169, that to determine the logarithm of a given number, is the same problem as to determine the value of \( x \) in an equation of this form, \( a^x = b \), where the unknown quantity is an exponent. But in order to resolve such an equation, it is not necessary to have recourse to series; for a table of logarithms being once supposed constructed, the value of \( x \) may be determined thus: It appears, from sect. 166, that \( x \times \log_a = \log_b \); hence it follows, that \( x = \frac{\log_b}{\log_a} \). The use of this formula will appear in next Section, which treats of computations relative to interest and annuities.

176. The theory of logarithms requires the solution of this other problem. Having given the radical number of a system, and a logarithm, to determine the corresponding number. Or, having given the equation \( r = y \) (where \( r, x, \) and \( y \) denote, as in sect. 165), to find a series which shall express \( y \) in terms of \( r \) and \( x \).

For this purpose, let us suppose \( r = 1 + a \), then our equation becomes \( y = (1 + a)^n \), which may also be expressed thus:

\[ y = [(1 + a)^n]^{\frac{n}{n}}, \]

where \( n \) is an arbitrary quantity, which is to disappear in the course of the investigation.

By the binomial theorem we have

\[ (1 + a)^n = 1 + na + \frac{n(n-1)}{1 \cdot 2}a^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}a^3 + \ldots \]

This equation, by multiplying together the factors which compose the terms of the series, and arranging the results according to the powers of \( n \), may also be expressed thus:

\[ (1 + a)^n = 1 + An + Bn^2 + Cn^3 + \ldots \]

where it will readily appear that

\[ A = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \ldots \]

As to the values of \( B, C, \ldots \), it is of no importance to know them, for they will all disappear in the course of the investigation. Hence, by substituting for \( (1 + a)^n \) its value, as expressed by this last series, we have

\[ y = (1 + An + Bn^2 + Cn^3 + \ldots)^{\frac{n}{n}}; \]

and expanding the latter part of this equation by means of the binomial theorem, it becomes

\[ y = 1 + \frac{x}{n}(An + Bn^2 + \ldots) + \frac{x(x-n)}{1 \cdot 2 \cdot n^2}(An + Bn^2 + \ldots)^2 + \ldots \]

But \( An + Bn^2 + \ldots = n(A + Bn + \ldots) \),

\[ (An + Bn^2 + \ldots)^2 = n^2(A + Bn + \ldots)^2, \]

\[ (An + Bn^2 + \ldots)^3 = n^3(A + Bn + \ldots)^3, \ldots \]

therefore, by leaving out of each term of the series the powers of \( n \), which are common to the numerator and denominator, the equation will stand thus:

\[ y = 1 + x(A + Bn + \ldots) + \frac{x(x-n)}{1 \cdot 2}(A + Bn + \ldots)^2 + \ldots \]

Now \( n \) is here an arbitrary quantity, and ought, from the nature of the original equation, to disappear from the value of \( y \); the terms of the equation which are multiplied by \( n \) ought therefore to destroy each other, and then the equation is reduced to

\[ r = y = 1 + \frac{x}{1} + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \ldots \]

and since we have found

\[ A = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \ldots \]

it is evident, from sect. 173, that \( A \) is Napier's logarithm of the radical number of the system.

177. If, in the equation \( r = y \), we suppose \( x = 1 \), the value of \( y \) becomes

\[ r = 1 + \frac{A}{1} + \frac{A^2}{1 \cdot 2} + \frac{A^3}{1 \cdot 2 \cdot 3} + \ldots \]

Here the radical number is expressed by means of its Napierian logarithm. Again, if we suppose \( x = \frac{1}{A} \), then

\[ r = 1 + \frac{1}{1} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2 \cdot 3} + \ldots \]

Thus it appears that the quantity \( r \) is equal to a constant number, which, by taking the sum of a sufficient number of terms of the series, will be found \( = 2.718281828459045 \ldots \). Let us denote this number by \( e \), then \( r = e \), and hence \( r = e^A \). Now, if we remark that \( A \) is the Nap. log. of \( r \), it must be evident (sect. 165 and 173), that \( e \) is the radical number of Napier's system of logarithms.

Again, since \( r = e \), therefore \( \frac{1}{A} \times \log_r = \log_e \), and

\[ A = \frac{\log_r}{\log_e}. \]

Here \( \log_r \) and \( \log_e \) denote logarithms taken according to any system whatever.

178. If we now resume the equation

\[ r = y = 1 + \frac{x}{1} + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \ldots \]

and substitute for \( A \) its value \( \frac{\log_r}{\log_e} \), we shall have the following general expression for any exponential quantity whatever:

\[ r = 1 + \frac{x(\log_r)}{1(\log_e)} + \frac{x^2(\log_r)^2}{1 \cdot 2(\log_e)^2} + \frac{x^3(\log_r)^3}{1 \cdot 2 \cdot 3(\log_e)^3} + \ldots \]

which, by supposing \( r = e \), becomes

\[ e = 1 + \frac{x}{1} + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \ldots \]

Sect. XX.—Of Interest and Annuities.

179. The theory of logarithms admits of extensive application to calculations relating to interest and annuities; these we now proceed to explain. There are two hypotheses, according to either of which money put out at interest may be supposed to be improved. We Algebra may suppose that the interest, which is always proportional to the sum lent, or principal, is also proportional to the time during which the principal is employed; and on this hypothesis, the money is said to be improved at simple interest. Or we may suppose that the interest which ought to be paid to the lender at successive stated periods is added to the principal, instead of being actually paid, and thus their amount converted into a new principal. When money is lent according to this second hypothesis, it is said to be improved at compound interest.

180. In calculations relating to interest, the things to be considered are the principal, or sum lent; the rate of interest, or sum paid for the use of L.100 for one year; the time during which the principal is lent; and the amount, or sum of the principal and interest, at the end of that time.

Let \( p \) denote the principal, L.1 being the unit; \( r \) the interest of L.1 for one year, at the given rate; \( t \) the time, one year being the unit; \( a \) the amount.

We shall now examine the relations which subsist between these quantities, according to each of the two hypotheses of simple and compound interest.

**Simple Interest.**

181. Because the interest of L.1 for one year is \( r \), the interest of L.1 for \( t \) years must be \( rt \); and the interest of \( p \) pounds for the same time \( prt \); hence we have this formula,

\[ p + prt = a, \]

from which we find

\[ P = \frac{a}{1+rt}, \quad r = \frac{a-p}{pt}, \quad t = \frac{a-p}{pr}. \]

As the manner of applying these formulas to questions relating to simple interest is sufficiently obvious, we proceed to consider compound interest.

**Compound Interest.**

182. In addition to the symbols already assumed, let \( R = 1+r \) = amount of L.1 in one year; then, from the nature of compound interest, \( R \) is also the principal at the beginning of the second year. Now, interest being always proportional to the principal, we have

\[ 1 : r :: R : rR = \text{the interest of } R \text{ for a year}, \]

and \( R+rR = (1+r)R = R^2 = \text{amount of } R \text{ in a year}; \)

therefore \( R^2 \) is the amount of L.1 in two years, which sum being assumed as a new principal, we find, as before, its interest for a year to be \( rR^2 \), and its amount \( R^2+rR^2 = (1+r)R^2 = R^3 \); so that \( R^3 \) is the amount of L.1 in three years. Proceeding in this manner, we find, in general, that the amount of L.1 in \( t \) years is \( R^t \), and of \( p \) pounds, \( pR^t \); hence we have this formula,

\[ pR^t = a; \]

which, from the nature of logarithms, may also be expressed thus:

\[ \log_p + t \times \log_R = \log_a. \]

We have also

\[ n = \frac{a}{R} \quad R = \sqrt[n]{a}; \]

or, by logarithms,

\[ \log_p = \log_a - t \times \log_R. \quad \log_R = \log_a - \log_p \]

\[ t = \frac{\log_a - \log_p}{\log_R}. \]

**Ex. 1.** As an example of the use of these formulas, let it be required to determine what sum improved at 5 per cent. compound interest will amount to L.500 in 42 years. In this case we have given \( a = 500, r = 0.05, R = 1.05, t = 42 \), to find \( p \).

From

\[ \log_a = \log_500 = 2.6989700 \]

subtract

\[ t \times \log_R = 42 \times \log_1.05 = 0.8899506 \]

remains \( \log_p \).

\[ 1.8090194 \]

therefore \( p = L.6442 = L.64.8s. 5d. \) the sum required.

**Ex. 2.** In what time will a sum laid out at 4 per cent. compound interest be doubled?

Let any sum be expressed by unity; then we have given \( p = 1, r = 0.04, R = 1.04, a = 2 \), to find \( t \).

From the formula,

\[ t = \frac{\log_a - \log_p}{\log_R} = \frac{\log_2}{\log_1.04} = \frac{0.3010300}{0.0170333} = 17.7 \text{ years nearly}. \]

In treating of compound interest, we have supposed the interest to be joined to the principal at the end of the year. But we might have supposed it to be added at the end of every half-year, or every quarter, or even every instant; and suitable rules might have been found for performing calculations, according to each hypothesis. As such suppositions are, however, never made in actual business, we shall not at present say anything more of them.

**Annuities.**

183. An annuity is a payment made annually for a term of years; and the chief problem relating to it is to determine its present worth, that is, the sum a person ought to pay immediately to another, upon condition of receiving from the latter a certain sum annually for a given time. In resolving this problem, it is supposed that the buyer improves his annuity from the time he receives it, and the seller the purchase-money, in a certain manner, during the continuance of the annuity, so that at the end of the time the amount of each may be the same. There may be various suppositions as to the way in which the annuity and its purchase-money may be improved; but the only one commonly applied to practice is the highest improvement possible of both, viz. by compound interest. As the taking of compound interest is, however, prohibited by law, the realizing of this supposed improvement requires punctual payment of interest, and therefore the interest in such calculations is usually made low.

184. Let \( A \) denote the annuity, \( P \) the present worth, or purchase-money, \( t \) the time of its continuance; let \( r \) and \( R \) denote as before.

The seller, by improving the price \( P \) at compound interest during the time \( t \), has \( PR \).

The purchaser is supposed to receive the first annuity \( A \) at the end of one year, which, being improved for \( t-1 \) years, amounts to \( AR^{t-1} \). He receives the second year's annuity at the end of the second year, which, being improved for \( t-2 \) years, amounts to \( AR^{t-2} \). In like manner, the third year's annuity becomes \( AR^{t-3} \), and so on, to the last year's annuity, which is simply \( A \). Therefore, the whole amount of the improved annuities is the geometrical series A + AR + AR^2 + AR^3 ... + AR^{n-1},

the sum of which, by sect. 56, is \( A \frac{R^n - 1}{R - 1} = A \frac{R^n - 1}{r} \);

and since this sum must be equal to the amount of the purchase-money, or PR', we have

\[ PR' = A \frac{R^n - 1}{r} \]

and from this equation we find

\[ P = \frac{A}{r} \left( 1 - \frac{1}{R^n} \right), \quad A = rPR' \frac{R^n - 1}{R - 1}, \quad t = \log A - \log (A - rP) \]

As to \( r \), it can only be found by the resolution of an equation of the \( t \) order.

185. To find the present value of an annuity in reversion, that is, an annuity which is to commence at the end of \( n \) years, and continue during \( t \) years; first find its value for \( n + t \) years, and then for \( n \) years, and subtract the latter from the former; we thus obtain the following formula:

\[ P = \frac{A}{rR^n} \left( 1 - \frac{1}{R} \right). \]

186. If the annuity is to commence immediately, and to continue for ever, then, because in this case \( R^n \) is infinitely great, and therefore \( \frac{1}{R^n} = 0 \), the formula \( P = \frac{A}{r} \left( 1 + \frac{1}{R} \right) \) becomes simply \( P = \frac{A}{r} \).

And if the annuity is to commence after \( n \) years, and continue for ever, the formula \( P = \frac{A}{rR^n} \left( 1 - \frac{1}{R} \right) \) becomes

\[ P = \frac{A}{rR^n}. \]

Note. The subject of life annuities forms by itself a distinct article.

Sect. XXI.—Of Continued Fractions.

187. Every quantity which admits of being expressed by a common fraction may also be expressed in the form of what is called a continued fraction. The nature of such fractions will be easily understood by the following example.

Let the common fraction be \( \frac{314159}{100000} \) or, which is the same, \( 3 + \frac{14159}{100000} \). Since \( 100000 = 7 \times 14159 + 887 \), therefore

\[ \frac{14159}{100000} = \frac{14159}{7 \times 14159 + 887} = \frac{1}{7 + \frac{887}{14159}}. \]

Now \( \frac{887}{14159} = \frac{887}{15 \times 887 + 854} = \frac{1}{15 + \frac{854}{887}} \), and substituting this for \( \frac{887}{14159} \) in the value of \( \frac{314159}{100000} \) already found, we have

\[ \frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{854}}}. \]

Again, \( \frac{854}{887} = \frac{854}{854 + 33} = \frac{1}{1 + \frac{33}{854}} \), which being substituted as before, gives \( \frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{33}{854}}}} \).

By operations similar to the preceding, we find

\[ \frac{1}{25 + \frac{29}{33}} = \frac{1}{25 + \frac{29}{33}} = \frac{1}{4 + \frac{29}{7 + \frac{1}{4}}} \]

therefore, by substitution,

\[ \frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{25 + \frac{1}{1 + \frac{1}{7 + \frac{1}{4}}}}}}. \]

By an operation in all respects the same as has been just now performed, may any fraction whatever be reduced to the form

\[ a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \ldots}}} \]

and it is then called a continued fraction.

188. It is easy to see in what manner the inverse of the preceding operation is to be performed, or a continued fraction reduced to a common fraction.

Thus, if the continued fraction be

\[ a + \frac{1}{b + \frac{1}{c + \frac{1}{d}}} \]

it will evidently be reduced to a common fraction by adding the reciprocal of \( d \) to \( c \), and the reciprocal of that sum to \( b \), and again the reciprocal of this last sum to \( a \); now the reciprocal of \( d \), or \( \frac{1}{d} \), added to \( c \), is \( c + \frac{1}{d} = \frac{cd + 1}{d} \); again, the reciprocal of this sum, or \( \frac{d}{cd + 1} \), added to \( b \), is \( b + \frac{d}{cd + 1} = \frac{bcd + b + d}{cd + 1} \), and the reciprocal of this last quantity, viz., \( \frac{cd + 1}{bcd + b + d} \), when added to \( a \), gives

\[ abcd + ab + ad + cd + 1 = a + \frac{1}{b + \frac{1}{c + \frac{1}{d}}}. \]

189. This manner of expressing a fraction enables us to find a series of other fractions, that approach in value to any given one, and each of them expressed in the smallest numbers possible. Thus, in the example \( \frac{314159}{100000} \) which has been resolved into a continued fraction, sect. 187, and which is known to express nearly the proportion of the diameter of a circle to its circumference, if we take only the first two terms of the continued fraction, and put $\pi$ for $\frac{314159}{100000}$, we shall have $\pi = 3 + \frac{1}{7} = 3.142857$ nearly; and this is the proportion which was found by Archimedes.

Again, by taking the first three terms, we have

$$\pi = 3 + \frac{1}{7} + \frac{1}{15} = 3 + \frac{15}{106} = \frac{333}{106},$$

which is nearer the truth than the former.

And by taking the first four terms, we have

$$\pi = 3 + \frac{1}{7} + \frac{1}{15} + \frac{1}{113} = \frac{355}{113},$$

which is the proportion assigned by Metius, and is more exact than either of the preceding. The results are alternately greater and less than the truth.

190. Among continued fractions, those have been particularly distinguished in which the denominators, after a certain number of changes, are continually repeated in the same order. Such, for example, is the fraction

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3} + \ldots,$$

The value of this fraction, though continued ad infinitum, may be easily found; for leaving out the first term, which is an integer, let us suppose

$$x = \frac{1}{2} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3} + \ldots,$$

Then, since after the second, all the terms return in the same order, it follows that their amount is also $= x$. Thus we have

$$x = \frac{1}{2} + \frac{1}{3 + x};$$

hence $x = \frac{3 + x}{6 + 2x + 1}$, and $x^2 + 3x = \frac{3}{2}$ and $x = \frac{-3 + \sqrt{15}}{2}$;

therefore $x + 1$, or the sum of the series, $= \frac{-1 + \sqrt{15}}{2}$.

In general, if $x = \frac{1}{a + b + \frac{1}{a + b + \ldots}}$, &c.

we find $x = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} + \frac{b}{a}}$. Though the denominators did not return in the same order till after a greater interval, the value of the fraction would still be expressed by the root of a quadratic equation. And conversely, the roots of all quadratic equations may be expressed by periodical continued fractions, and may often by that means be very readily approximated in numbers, without the trouble of extracting the square root.

191. The reduction of a decimal into the form of a continued fraction sometimes renders the law of its continuation evident. Thus we know that $\sqrt{2} = 1.41421356\ldots$; but from the bare inspection of this decimal we discover no rule for its further continuation. If, however, it be reduced into a continued fraction, it becomes

$$1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \ldots}}},$$

and hence we see in what way it may be continued to any degree of accuracy.

When the root of any equation is found by the method explained in sect. 156, the value of the unknown quantity is evidently expressed by a continued fraction

For if $x$ be the root sought, we have $x = a + \frac{1}{y}$, $y = b + \frac{1}{y'}$, $y' = b' + \frac{1}{y''}$, &c. where $a, b, b', b''$, &c. denote the whole numbers, which are next less than the true values of $x, y, y', y''$, &c. If, therefore, in the value of $x$ we substitute $b + \frac{1}{y}$ for $y$, it becomes

$$x = a + \frac{1}{b + \frac{1}{y'}}.$$

Again, if in this second value of $x$ we substitute $b' + \frac{1}{y''}$ for $y'$, it becomes

$$x = a + \frac{1}{b + \frac{1}{b' + \frac{1}{y''}}}.$$

The next value of $x$ is in like manner found to be

$$x = a + \frac{1}{b + \frac{1}{b' + \frac{1}{b'' + \frac{1}{y''}}}},$$

and so on continually.

Sect. XXII.—Of Indeterminate Problems.

192. When the conditions of a question are such that the number of equations exceeds the number of unknown quantities, that question will admit of innumerable solutions, and is therefore said to be indeterminate. Thus, if it be required to find two numbers subject to no other limitation than 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.

193. 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 exterminated by means of the given equations. Those of the first order lead always to equations of this 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 $b = mc$, then $ax + by = m(ax + by)$. to be whole numbers, therefore \( \frac{c}{m} \) is a whole number; hence \( m \) must be a divisor of \( c \).

We proceed to illustrate the manner of resolving indeterminate equations of the first order, by some numerical examples.

**Ex. 1.** Given \( 2x + 3y = 25 \), to determine \( x \) and \( y \) in whole positive numbers.

From the given equation we have \( x = \frac{25 - 3y}{2} = 12 - y + \frac{1-y}{2} \). Now, since \( x \) must be a whole number, it follows that \( \frac{1-y}{2} \) must be a whole number. Let us assume \( \frac{1-y}{2} = z \), then \( 1 - y = 2z \) and \( y = 1 - 2z \); and since \( x = 12 - y + \frac{1-y}{2} = 12 - y + z \), therefore \( x = 12 - 1 + 2z + z \); hence we have

\[ x = 11 + 3z, \quad y = 1 - 2z, \]

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 negative, and from the value of \( x \), that, abstracting from the sign, it must be less than 4; hence \( z \) may have these three values, 0, -1, -2, -3.

If \( z = 0 \), \( z = -1 \), \( z = -2 \), \( z = -3 \);

Then \( x = 11 \), \( x = 8 \), \( x = 5 \), \( x = 2 \),

\( y = 1 \), \( y = 3 \), \( y = 5 \), \( y = 7 \).

**Ex. 2.** It is required to divide 100 into such parts that the one may be divisible by 7 and the other by 11.

Let \( 7x \) be the first part, and \( 11y \) the second; then, by the question, \( 7x + 11y = 100 \), and

\[ x = \frac{100 - 11y}{7} = 14 - y + \frac{2 - 4y}{7}. \]

Hence it appears, that \( \frac{2 - 4y}{7} \) must be a whole number.

Let us assume \( \frac{2 - 4y}{7} = z \), then \( x = 14 - y + z \), and

\[ 4y = 2 - 7z, \text{ or } y = \frac{2 - 7z}{4} = \frac{2 - 3z}{4} - z; \text{ therefore } \frac{2 - 3z}{4} \]

must be a whole number. Assume \( \frac{2 - 3z}{4} = t \), then \( y = t - z \), and \( 3z = 2 - 4t \), or \( z = \frac{2 - 4t}{3} = \frac{2 - t}{3} - t \); therefore \( \frac{2 - t}{3} \) must be a whole number.

Assume now \( \frac{2 - t}{3} = v \), then \( z = v - t \), and \( t = 2 - 3v \).

Here it is evident that \( v \) may be any whole number taken at pleasure; so that to determine \( x \) and \( y \) we have the following series of equations:

\[ t = 2 - 3v, \] \[ z = v - t = 4v - 2, \] \[ y = t - z = 4 - 7v, \] \[ x = 14 - y + z = 11v + 8. \]

From the value of \( y \), it appears that \( v \) must either be 0 or negative; but from the value of \( x \), \( v \) cannot be a negative whole number; therefore \( v \) can only be 0;

hence the only values which \( a \) and \( y \) can have in whole Algebraic numbers, are \( x = 8, y = 4 \).

**Ex. 3.** It is required to find all the possible ways in which L60 can be paid in guineas and moidores only.

Let \( x \) be the number of guineas, and \( y \) the number of moidores. Then the value of the guineas, expressed in shillings, is \( 21x \), and that of the moidores \( 27y \); therefore, from the nature of the question, \( 21x + 27y = 1200 \), or, dividing the equation by 3, \( 7x + 9y = 400 \); hence

\[ x = \frac{400 - 9y}{7} = 57 - y + \frac{1 - 2y}{7}, \text{ so that } \frac{1 - 2y}{7} \text{ must be a whole number.} \]

Assume \( \frac{1 - 2y}{7} = z \), then \( x = 57 - y + z \), and \( 2y = 1 - 7z \) or \( y = \frac{1 - 7z}{2} = \frac{1 - z}{2} - 3z; \text{ therefore } \frac{1 - z}{2} \text{ must be a whole number.} \)

Assume \( \frac{1 - z}{2} = v \), then \( y = v - 3z \), and \( z = 1 - 2v; \)

therefore \( v \) may be taken any whole number at pleasure, and \( x \) and \( y \) may be determined by the following equations:

\[ x = 1 - 2v, \] \[ y = v - 3z = 7v - 3, \] \[ x = 57 - y + z = 61 - 9v. \]

From the value of \( x \), it appears that \( v \) cannot exceed 6, and from the value of \( y \), that it cannot be less than 1.

Hence if \( v = 1, 2, 3, 4, 5, 6, \)

we have \( x = 52, 43, 34, 25, 16, 7, \) \[ y = 4, 11, 18, 25, 32, 39. \]

**194.** In the foregoing examples the unknown quantities \( x \) and \( y \) have each a determinate number of positive values; and this will evidently be the case as often as the proposed equation is of this form, \( ax + by = c \). If, however, \( b \) be negative, that is, if the equation be of this form, \( ax - by = c \), or \( ax = by + c \), we shall have questions of a different kind, admitting each of an infinite number of solutions; these, however, may be resolved in the same manner as the preceding, as will appear from the following example.

**Ex. 4.** A person buys some horses and oxen: he pays 31 crowns for each horse, and 20 crowns for each ox, and he finds that the oxen cost him seven crowns more than the horses. How many did he buy of each?

Let \( x \) be the number of horses, and \( y \) that of the oxen; then, by the question,

\[ 20x = 31y + 7, \text{ and } x = \frac{31y + 7}{20} = y + \frac{11y + 7}{20}. \]

Therefore \( \frac{11y + 7}{20} \) must be a whole number.

Let \( \frac{11y + 7}{20} = v \), then \( x = y + v \), and \( y = \frac{20v - 7}{11} = v + \frac{9v - 7}{11}; \text{ hence } \frac{9v - 7}{11} \text{ must be a whole number.} \)

Let \( \frac{9v - 7}{11} = t \), then \( y = v + t \), and \( v = \frac{11t + 7}{9} = t + \frac{2t + 7}{9}; \text{ therefore } \frac{2t + 7}{9} \text{ is a whole number.} \)

Let \( \frac{2t + 7}{9} = s \), then \( v = t + s \), and \( t = \frac{9s - 7}{2} = 4s + \frac{s - 7}{2}; \text{ therefore } \frac{s - 7}{2} \text{ is a whole number.} \) Put \( \frac{s-7}{2} = r \), then \( t = 4s + r \) and \( s = 2r - 7 \).

Having now no longer any fractions, we return to the values of \( x \) and \( y \) by the following series of equations:

\[ \begin{align*} s &= 2r + 7, \\ t &= 4s + r = 9r + 28, \\ v &= t + s = 11r + 35, \\ y &= v + t = 20r + 63 = \text{number of oxen}, \\ z &= y + t = 31r + 98 = \text{number of horses}. \end{align*} \]

The least positive values of \( x \) and \( y \) will evidently be obtained by making \( r = -3 \), and innumerable other values may be had by putting \( r = -2, r = -1, r = 0, r = +1, \ldots \). Thus we have

\[ \begin{align*} x &= 5, 36, 67, 98, 129, 160, 191, 222, \ldots \\ y &= 3, 23, 43, 63, 83, 103, 123, 143, \ldots \end{align*} \]

each series forming an arithmetical progression, the common difference in the first being 31, and in the second 20.

195. If we consider the manner in which the numbers \( x, y \), in this example, are determined from the succeeding quantities \( r, t, \ldots \), we shall immediately perceive that the co-efficients of those quantities are the same as the successive quotients which arise in the arithmetical operation for finding the greatest common measure of 20 and 31, the co-efficients of the given equation \( 20x = 31y + 7 \). The operation performed at length will stand thus:

\[ \begin{array}{c|c} 20 & 31 \\ \hline 11 & 20 \\ \hline 9 & 11 \\ \hline 2 & 9 \\ \hline 1 & 2 \\ \hline 0 & 0 \end{array} \]

Hence we may form a series of numeral equations which, when compared with the series of literal equations expressing the relations between \( x, y, r, \ldots \), as put down in the following table, will render the method of determining the latter from the former sufficiently obvious.

\[ \begin{align*} 31 &= 1 \times 20 + 11 \\ 20 &= 1 \times 11 + 9 \\ 11 &= 1 \times 9 + 2 \\ 9 &= 4 \times 2 + 1 \\ 2 &= 2 \times 1 + 0 \\ &= 2r + 2 \\ &= s + r = 3r + 2 \\ &= 2t + s = 8r + 6 \\ &= v + t = 11r + 8 \\ &= y + v = 19r + 14; \end{align*} \]

and as every question of this kind may be analyzed in the same manner, we may hence form the following general rule for resolving indeterminate problems of the first order.

196. Let \( bx = ay + n \) be the proposed equation, in which \( a, b, n \) are given integers, and \( x, y \), numbers to be found. Let \( a \) be the greatest of the two numbers \( a, b \), and let \( A \) denote the greatest multiple of \( b \) which is contained in \( a \), and \( c \) the remainder; also let \( B \) denote the greatest multiple of \( c \) contained in \( b \), and \( d \) the remainder; and \( C \) the greatest multiple of \( d \) contained in \( c \), and \( e \) the remainder; and so on, till one of the remainders be found equal to 0. The numbers \( A, B, C, \ldots \) afford a series of equations from which another series may be derived, as in the following table:

\[ \begin{align*} a &= Ab + c, \\ b &= Be + d, \\ c &= Cd + e, \\ d &= De + f, \\ e &= Ef + g, \\ f &= Fg + 0 \end{align*} \]

And in the last equation of the second series any number whatever may be put for \( q \). It is also to be observed, that the given number \( n \) is to have the sign \( + \) prefixed to it, if the number of equations be odd, but — if that number be even. Having formed the second series of equations, the values of \( x \) and \( y \) may be thence found as in the foregoing examples. We proceed to show the application of the rule.

Ex. 5. Required a number which, being divided by 11, leaves the remainder 3, but being divided by 19, leaves the remainder 5.

Let \( N \) be the number, and \( x, y \), the quotients which arise from the respective divisions; then we have \( N = 11x + 3 \), also \( N = 19y + 5 \); hence \( 11x + 3 = 19y + 5 \), and \( 11x = 19y + 2 \), an equation which furnishes the following table:

\[ \begin{align*} 19 &= 1 \times 11 + 8 \\ 11 &= 1 \times 8 + 3 \\ 8 &= 2 \times 3 + 2 \\ 3 &= 1 \times 2 + 1 \\ 2 &= 2 \times 1 + 0 \\ &= 2r + 2 \\ &= s + r = 3r + 2 \\ &= 2t + s = 8r + 6 \\ &= v + t = 11r + 8 \\ &= y + v = 19r + 14; \end{align*} \]

Here \( r \) may be assumed of any value whatever. Hence we have

\[ \begin{align*} x &= 1500 - 5e, \\ y &= 6e - 500. \end{align*} \]

Let these values of \( x \) and \( y \) be substituted in either of the original equations; in the first, for example, as being the most simple, and we find \( 7z + 15e = 1560 \). This last equation being resolved in the same manner, we find

\[ \begin{align*} z &= 1560 - 7t, \\ t &= 15t - 3120, \\ y &= 8860 - 42t, \\ x &= 35t - 7300; \end{align*} \]

and hence it appears that the only values which \( t \) can have, so as to give whole positive numbers for \( x, y, z \), are 209 and 210; thus we have

\[ \begin{align*} x &= 15 \\ y &= 82 \\ z &= 15, \\ \text{or } x &= 50 \\ y &= 40 \\ z &= 30. \end{align*} \]

197. If an equation were proposed involving three unknown quantities, as \( ax + by + cz = d \), by transposition we have \( ax + by = d - cz \), and putting \( d - cz = e' \), \( ax + by = e' \). From this last equation we may find values of \( x \) and \( y \) of this form,

\[ \begin{align*} x &= mr + nc', \\ y &= mr + nc', \end{align*} \]

or \( x = mr + n(d - cz) \), \( y = mr + n(d - cz) \); 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.

198. We proceed to indeterminate problems of the second degree. These produce equations of the three following forms:

I. \( y = \frac{a}{b + cx} \) II. \( y = \frac{a + bx}{c + dx} \) III. \( y = \sqrt{a + bx + cx^2} \)

In all these equations \( a, b, c \) denote given numbers. In the first two, \( x \) is to be determined so that \( y \) may be an integer; and in the third, \( x \) is to be determined so that \( y \) may be a rational quantity.

In the equation \( y = \frac{a}{b + cx} \), it is evident, \( b + cx \) must be a divisor of \( a \); let \( d \) be one of its divisors, then \( b + cx = d \), and \( x = \frac{d - b}{c} \); hence, to find \( x \) we must search among the divisors of \( a \) for one such, that if \( b \) be subtracted from it, the remainder may be divisible by \( c \), and the quotient will be such a value of \( x \) as is required.

When \( y = \frac{a + bx}{c + dx} \), if \( d \) be a divisor of \( b \), \( x \) will be taken out of the numerator if we divide it by \( c + dx \), and this form is then reduced to the preceding. But if \( d \) is not a divisor of \( b \), multiply both sides by \( d \), then \( dy = \frac{da + dbx}{c + dx} \) or \( dy = b + \frac{ad - bc}{c + dx} \), and so \( x \) is found by making \( c + dx \) equal to a divisor of \( ad - bc \).

Example. Given \( x + y + 2xy = 195 \), to determine \( x \) and \( y \) in whole numbers.

From the given equation, \( y = \frac{195 - x}{1 + 2x} \), therefore

\[ y = \frac{390 - 2x}{1 + 2x} = -1 + \frac{391}{1 + 2x} \]

Now \( 391 = 17 \times 23 \), hence we must assume \( 1 + 2x = 17 \), or \( 1 + 2x = 23 \); the first supposition gives us \( x = 8 \), \( y = 11 \); and the second \( x = 11 \), \( y = 8 \), the same result in effect as the former.

199. We are next to consider the formula \( y = \sqrt{a + bx + cx^2} \), where \( x \) is to be found, so that \( y \) may be a rational quantity; but as the condition of having \( x \) and \( y \) also integers would add greatly to the difficulty of the problem, and produce researches of a very intricate nature, we must be satisfied for the most part with fractional values. The possibility of rendering the proposed formula a square depends altogether upon the co-efficients \( a, b, c \); and there are four cases of the problem, the solution of each of which is connected with some peculiarity in its nature.

Case 1. Let \( a \) be a square number; then, putting \( g^2 \) for \( a \), we have \( y = \sqrt{g^2 + bx + cx^2} \). Suppose \( \sqrt{g^2 + bx + cx^2} = g + mx \); then \( g^2 + bx + cx^2 = g^2 + 2gmx + m^2x^2 \), or \( bx + cx^2 = 2gmx + m^2x^2 \); hence \( x = \frac{2gm - b}{c - m^2} \), \( y = \sqrt{g^2 + bx + cx^2} = \frac{cg - bm + gm^2}{c - m^2} \).

Here \( m \) may be any rational quantity, either whole or fractional.

Case 2. Let \( c \) be a square number \( = g^2 \); then, putting \( \sqrt{a + bx + g^2x^2} = m + gx \), we find \( a + bx + g^2x^2 = m^2 + 2mgx + g^2x^2 \), or \( a + bx = m^2 + 2mgx \); hence we find \( x = \frac{m^2 - a}{b - 2mg} \), \( y = \sqrt{a + bx + g^2x^2} = \frac{bm - gm^2 - ag}{b - 2mg} \).

Here \( m \), as before, may be taken at pleasure.

Case 3. When neither \( a \) nor \( c \) is a square number, yet if the expression \( a + bx + cx^2 \) can be resolved into two simple factors, as \( f + gx \) and \( h + kx \), the irrationality may be taken away as follows.

Assume \( \sqrt{a + bx + cx^2} = \sqrt{(f + gx)(h + kx)} = m(f + gx) \), then \( (f + gx)(h + kx) = m^2(f + gx)^2 \), or \( h + kx = m^2(f + gx) \); hence we find

\[ x = \frac{fm^2 - h}{k - gm^2}, \quad y = \sqrt{(f + gx)(h + kx)} = \frac{(fk - gh)m}{k - gm^2} \]

and in these formulæ \( m \) may be taken at pleasure.

Case 4. The expression \( a + bx + cx^2 \) 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, \( p^2 + qr \), where \( p, q, r \) are quantities which contain no power of \( x \) higher than the first. Let us assume \( \sqrt{p^2 + qr} = p + mq \); thus we have \( p^2 + qr = p^2 + 2mpq + m^2q^2 \) and \( r = 2mp + m^2q \), 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.

200. If we can by trials discover any one value of \( x \) which renders the expression \( \sqrt{a + bx + cx^2} \) rational, we may immediately reduce the quantity under the radical sign to the above-mentioned form, and thence find a general expression from which as many more values of \( x \) may be determined as we please. Thus, let us suppose that \( p \) is a value of \( x \) which satisfies the condition required, and that \( q \) is the corresponding value of \( y \); then

\[ y^2 = a + bx + cx^2 \]

Therefore, by subtraction,

\[ y^2 - q^2 = b(x - p) + c(x^2 - p^2) = (b + cp + cx)(x - p) \]

and \( y = \sqrt{q^2 + (b + cp + cx)(x - p)} \). The quantity under the radical being now reduced to the prescribed form, it may be rendered rational by the substitution pointed out in the last article.

The application of the preceding general methods of resolution to any particular case is very easy; we shall therefore conclude with a very few examples.

Ex. 1. It is required to find two square numbers whose sum is a given square number.

Let \( a^2 \) be the given square number, and \( x^2, y^2 \), the numbers required; then, by the question, \( x^2 + y^2 = a^2 \), and \( y = \sqrt{a^2 - x^2} \). This equation is evidently of such a form as to be resolvable by the method employed in case 1. Accordingly, by comparing \( \sqrt{a^2 - x^2} \) with the general expression \( \sqrt{g^2 + bx + cx^2} \), we have \( g = a, b = 0, c = -1 \), and substituting these values in the formulæ of sect. 199, also \( -n \) for \( +m \), we find

\[ x = \frac{2an}{n^2 + 1}, \quad y = \frac{a(n^2 - 1)}{n^2 + 1} \]

hence the numbers required are

\[ x^2 = \frac{4a^2n^2}{(n^2 + 1)^2}, \quad y^2 = \frac{a^2(n^2 - 1)^2}{(n^2 + 1)^2} \]

If \( a = n^2 + 1 \), where \( n \) is any number whatever, the square numbers \( x^2 \) and \( y^2 \) will both be integers, viz. \( x^2 = 4n^2 \) and \( y^2 = (n^2 - 1)^2 \). Let us suppose \( n = 2 \), then \( a = n^2 + 1 = 5 \), and \( a^2 = 25 \), hence \( x^2 = 4n^2 = 16 \), \( y^2 = (n^2 - 1)^2 = 9 \). Thus it appears that the square number 25 may be resolved into two other square numbers, 9 and 16. Ex. 2. It is required to find two square numbers whose difference shall be equal to a given square number $b^2$.

This question may be resolved in the same manner as the last. Or, without referring to any former investigation, let $(x+n)^2$ and $x^2$ be the numbers sought, then $(x+n)^2 - x^2 = b^2$; that is, $2nx + n^2 = b^2$, hence

$$x = \frac{b^2 - n^2}{2n} \quad \text{and} \quad x + n = \frac{b^2 + n^2}{2n}.$$

So that the numbers sought are

$$\left(\frac{(b^2 + n^2)^2}{4n^2}\right), \quad \left(\frac{(b^2 - n^2)^2}{4n^2}\right),$$

where $n$ may be any number whatever. If, for example, $b^2 = 25$ and $n = 1$, then $x = 12$ and $x + n = 13$; so that the numbers required are 144 and 169.

Ex. 3. It is required to determine $x$, so that $\frac{x^2 + x}{2}$ may be a rational square.

Let $y$ be the side of the square required; then $\frac{x^2 + x}{2} = y^2$ and $4x^2 + 4x = 8y^2$. Let the first part of this equation be completed into a square by adding 1 to each side, then $4x^2 + 4x + 1 = 1 + 8y^2$; and taking the root,

$$2x + 1 = \sqrt{1 + 8y^2},$$

so that we have to make $1 + 8y^2$ a square. Assume

$$1 + 8y^2 = \left(1 + \frac{p}{q}\right)^2 = 1 + \frac{2p}{q}y + \frac{p^2}{q^2}y^2,$$

then $8y = \frac{2p}{q} + \frac{p^2}{q^2}y$. Hence, by proper reduction, $y = \frac{2pq}{8q^2 - p^2}$, and since $2x + 1 = \sqrt{1 + 8y^2} = \frac{8q^2 + p^2}{8q^2 - p^2}$, therefore $x = \frac{p^2}{8q^2 - p^2}$, and $\frac{x^2 + x}{2} = \frac{4p^2 q^2}{(8q^2 - p^2)^2}$, a rational square, as was required.

Sect. XXIII.—Of the Resolution of Geometrical Problems.

201. When a geometrical problem is to be resolved by algebra, the figure which is to be the subject of investigation must be drawn, so as to exhibit as well the known quantities connected with the problem, as the unknown quantities which are to be found. The conditions of the problem are next to be attentively considered, and such lines drawn, or produced, as may be judged necessary to its resolution. This done, the known quantities are to be denoted by symbols in the usual manner, and also such unknown quantities as can most easily be determined; which may be either those directly required, or others from which they can be readily found. We must next proceed to deduce from the known geometrical properties of the figure a series of equations, expressing the relations between the known and unknown quantities; these equations must be independent of each other, and as many in number as there are unknown quantities. Having obtained a suitable number of equations, the unknown quantities are to be determined in the same manner as in the resolution of numerical problems.

No general rule can be given for drawing the lines, and selecting the quantities most proper to be represented by symbols, so as to bring out the simplest conclusion; because different problems require different methods of solution. The best way to gain experience in this matter, is to try the solution of the same problem in different ways, and then apply that which succeeds best to other cases of the same kind, when they afterwards occur. The following particular directions, however, may be of some use:

1. In preparing the figure by drawing lines, let them be either parallel or perpendicular to other lines in the figure, so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.

2. In selecting the quantities for which symbols are to be substituted, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adjacent parts may be expressed by addition and subtraction only, without the intervention of surds.

3. When two lines, or quantities, are alike related to other parts of the figure, or problem, the best way is to substitute for neither of them separately, but to substitute for their sum, or difference, or rectangle, or the sum of their alternate quotients, or some line or lines in the figure, to which they have both the same relation.

4. When the area or the perimeter of a figure is given, or such like parts of it as have only a remote relation to the parts required, it is sometimes of use to assume another figure similar to the proposed one, having one side equal to unity, or some other known quantity. For thence the other parts of the figure may be found by the known proportions of like sides or parts, and so an equation will be obtained.

202. We shall now give the algebraical solutions of some geometrical problems.

Prob. 1. In a right-angled triangle, having given the base, and the sum of the hypothenuse and perpendicular, to find both these sides.

Let ABC (Plate XVIII, fig. 1) represent the proposed triangle, right-angled at B. Let AB, the given base, be denoted by $b$, and AC + BC, the sum of the hypothenuse and perpendicular, by $s$; then if $x$ be put for BC the perpendicular, the hypothenuse AC will be $s - x$. But from the nature of a right-angled triangle, AC$^2$ = AB$^2$ + BC$^2$, that is,

$$b^2 + x^2 = (s - x)^2 = s^2 - 2sx + x^2.$$

Hence $b^2 = s^2 - 2sx$, and $x = \frac{s^2 - b^2}{2s} = BC$.

Also $s - x = s - \frac{s^2 - b^2}{2s} = \frac{s^2 + b^2}{2s} = AC$. Thus the perpendicular and hypothenuse are expressed by means of the known quantities $b$ and $s$, as required.

If a solution in numbers be required, we may suppose $AB = b = 3$, and $AC + CB = s = 9$; then

$$BC = \frac{s^2 - b^2}{2s} = 4,$$ and $$AC = \frac{s^2 + b^2}{2s} = 5.$$ Algebra.

\[ x^2 - sx = \frac{a^2}{2} - s^2, \]

which being resolved, by completing the square, we find

\[ x = \frac{s + \sqrt{2a^2 - s^2}}{2} = AB, \]

and

\[ s - x = \frac{s - \sqrt{2a^2 - s^2}}{2} = BC. \]

Thus it appears, that either of the two quantities

\[ \frac{s + \sqrt{2a^2 - s^2}}{2}, \quad \frac{s - \sqrt{2a^2 - s^2}}{2}, \]

may be taken for \(AB\); but whichever of the two be taken, the remaining one is necessarily equal to \(BC\).

**Prob. 3.** It is required to inscribe a square in a given triangle.

Let \(ABC\) (fig. 2) be the given triangle, and \(EFHG\) the inscribed square. Draw the perpendicular \(AD\), cutting \(EF\), the side of the square, in \(K\); then, because the triangle is given, the perpendicular \(AD\) may be considered as given. Let \(BC = b\), \(AD = p\), and considering \(AK\) as the unknown quantity (because from it the square may be readily determined), let \(AK = x\); then \(KD = EF = p - x\).

The triangles \(ABC\), \(AEF\), are similar; therefore

\[ AD : BC :: AK : EF, \]

that is, \(p : b :: x : p - x\). Hence, by taking the product of the extremes and means,

\[ \frac{p^2}{p - x} = bx, \]

and

\[ x = \frac{p^2}{p + b} = AK. \]

If the side of the square be required, it may be immediately found by subtracting \(AK\) from \(AD\) the perpendicular. Thus we have

\[ \frac{p^2}{p + b} = KD = EF. \]

Hence it appears that we may either take \(AK\), a third proportional to \(AD + BC\) and \(AD\), or take \(DK\), a fourth proportional to \(AD + BC\), \(AD\) and \(BC\); and the point \(K\) being found, the manner of describing the square is sufficiently obvious.

**Prob. 4.** Having given the area of a rectangle inscribed in a given triangle; it is required to determine the sides of the rectangle.

Let \(ABC\) (fig. 3) be the given triangle, and \(EDGF\) the rectangle whose sides are required. Draw the perpendicular \(CI\), cutting \(DG\) in \(H\). Put \(AB = b\), \(CI = p\), \(DG = EF = x\), \(DE = HI = y\); then \(CH = p - y\). Let \(a^2\) denote the given area.

The triangles \(CDG\), \(CAB\), are similar; hence

\[ CH : DG :: CI : AB, \text{ or } p - y : x :: p : b. \]

So that to determine \(x\) and \(y\), we have these two equations,

\[ xy = a^2, \quad bp - by = px. \]

From the first equation we find \(y = \frac{a^2}{x}\), and from the second, \(y = \frac{bp - px}{b}\), therefore

\[ \frac{bp - px}{b} = \frac{a^2}{x}; \]

hence \(x^2 - bx = \frac{a^2b}{p}\), and from this quadratic equation, by completing the square, &c., we find

\[ x = \frac{b}{2} + \frac{\sqrt{b^2 - a^2b}}{2p}, \quad \text{and} \quad y = \frac{a^2}{x} = \frac{p}{2} - \frac{\sqrt{p^2 - pa^2}}{b}. \]

Hence it appears, that if \(\frac{a^2b}{p}\) be less than \(\frac{b^2}{4}\), that is, if \(a^2\) be less than \(\frac{pb}{4}\), there are two different rectangles, having the same area, which may be inscribed in the given triangle. It also appears that, to render the problem possible, the given space \(a^2\) must not be greater than \(\frac{pb}{4}\), that is, than half the area of the given triangle.

**Prob. 5.** In a triangle, there are given the base, the vertical angle, and the sum of the sides about that angle; to determine each of these sides.

Let us suppose that \(ABC\) (fig. 4) is the triangle, of which there is given the base \(AC\), the vertical angle \(ABC\), and the sum of the sides \(AB, BC\). Put \(AC = a\), \(AB + BC = b\), cosine of \(\angle ABC = c\); and let \(AB, BC\), the sides required, be denoted by \(x\) and \(y\).

Let \(CD\) be drawn from either of the angles at the base perpendicular to the opposite side \(AB\); then, rad. : cos. \(B :: CB : BD\); therefore \(BD = \cos B \times CB = cy\).

Now, from the principles of geometry, \(AC^2 = AB^2 + BC^2 - 2AB \times BD\). Hence, and from the question, we have these two equations,

\[ x + y = b, \quad x^2 - 2cxy + y^2 = a^2. \]

From the square of the first of these equations, viz.

\[ x^2 + 2xy + y^2 = b^2, \]

let the second be subtracted; thus we have

\[ 2(1 + c)xy = b^2 - a^2, \]

and

\[ 2xy = \frac{b^2 - a^2}{1 + c}. \]

Again, from the square of the first equation let the double of this last equation, viz.

\[ 4xy = \frac{2(b^2 - a^2)}{1 + c}, \]

be subtracted, and the result is

\[ 2a^2 - 2xy + y^2 = \frac{2a^2 - (1 - c)b^2}{1 + c}; \]

so that by taking the square root of this last equation, we obtain

\[ x - y = \sqrt{\frac{2a^2 - (1 - c)b^2}{1 + c}}. \]

Thus we have found the difference between the sides, now their sum is given \(= b\); hence, by adding \(\frac{1}{2}\) the difference to \(\frac{1}{2}\) the sum, we find

\[ x = \frac{b}{2} + \frac{1}{2} \sqrt{\frac{2a^2 - (1 - c)b^2}{1 + c}}; \]

and subtracting \(\frac{1}{2}\) the difference from \(\frac{1}{2}\) the sum,

\[ y = \frac{b}{2} - \frac{1}{2} \sqrt{\frac{2a^2 - (1 - c)b^2}{1 + c}}. \]

If the angle at \(B\) be a right angle, this problem becomes the same as prob. 2.

**Prob. 6.** To draw a straight line through a given point \(P\) (fig. 5), so as to form with \(AB, AC\), two straight lines given in position, a triangle \(DAE\) equal to a given space.

Draw \(PF\) parallel to \(AC\), one of the lines; and \(DH, PG\), perpendicular to \(AB\), the other line; then \(PF\) will be given in position; and \(AF, PG\), will be given in magnitude.

Put the known magnitudes \(AF = a\), \(PG = b\), and \(AE\), a side of the triangle which is to be determined, \(= x\). Also, let the given space to which the triangle \(ADE\) is to be equal, be \(c^2\).

By similar triangles, \(FE : PG :: AE : DH\);

\[ \text{that is, } x - a : b :: x : DH; \]

hence \(DH = \frac{bx}{x - a}\); and triangle \(ADE = \frac{bx^2}{2x - 2a}\).

Therefore \(\frac{bx^2}{2x - 2a} = c^2\), and \(bx^2 = 2cx - 2ac^2\),

and \(x^2 - \frac{2a}{b}x = -\frac{2ac^2}{b}\). This is a quadratic equation, which, by completing the square, becomes

\[ x^2 - \frac{2c}{b}x + \frac{c^2}{b^2} = \frac{c^2 - 2ab}{b^2}. \]

Hence, by taking the square root and transposing,

\[ x = \frac{c \pm c\sqrt{c^2 - 2ab}}{b}. \]

It appears, from this expression, that \( x \) will have two values, viz.

\[ x = \frac{c + c\sqrt{c^2 - 2ab}}{b}, \quad x = \frac{c - c\sqrt{c^2 - 2ab}}{b}; \]

therefore there are two positions of the line DE, which will satisfy the problem. It further appears, that the problem will only be possible when \( c^2 > 2ab \); for if \( c^2 < 2ab \), then \( c^2 - 2ab \) will be a negative quantity, and can have no real square root. Since, then, \( c^2 \) cannot be less than \( 2ab \), it follows that the least triangle which can be formed by drawing a line throughout P, a point in the angle CAB, is equal to \( 2ab \), that is, to \( 2AF \times PG \).

If the point P were in the angle DAB, adjacent to DAE (fig. 6), the solution would be the same; but in this case, the same notation being employed,

\[ x = \frac{c \pm c\sqrt{c^2 + 2ab}}{b}. \]

The quantity under the radical sign is always positive, therefore there is no limitation to the problem.

**Prob. 7.** To find the area of a triangle whose sides are given.

In the triangle ABC (fig. 7) draw the perpendicular CD; put BC = \( a \), AC = \( b \), AB = \( c \), CD = \( p \).

By the elements of Geometry (see Geometry),

\[ CB^2 = BA^2 + CA^2 - 2AB \cdot AD; \]

that is,

\[ a^2 = c^2 + b^2 - 2c \cdot AD; \]

hence

\[ AD = \frac{c^2 + b^2 - a^2}{2c}; \]

and since

\[ p^2 = AC^2 - AD^2 = b^2 - \left( \frac{c^2 + b^2 - a^2}{2c} \right)^2, \]

therefore \( 4c^2p^2 = 4c^2b^2 - (c^2 + b^2 - a^2)^2 \).

The second side of this equation is the difference of two squares; therefore it admits of being resolved into the product of the sum and difference of the roots, thus,

\[ [c^2 + 2cb + b^2 - a^2] \left[ a^2 - (c^2 - 2cb + b^2) \right]; \]

or,

\[ [(c + b)^2 - a^2] \left[ a^2 - (c - b)^2 \right]. \]

Again, \( (c + b)^2 - a^2 = (a + b + c)(a + b - c) \);

and \( a^2 - (c - b)^2 = (a + c - b)(a - c + b) \);

therefore,

\[ 4c^2p^2 = (a + b + c)(a + b - c)(a - b + c)(b + c - a). \]

If we put \( s = \frac{1}{2}(a + b + c) \), then

\[ a + b + c = 2s, \quad a + b - c = 2(s - e), \]

\[ a - b + c = 2(s - b), \quad b + c - a = 2(s - a). \]

We have now

\[ \frac{c^2p^2}{4} = s(s - a)(s - b)(s - c); \]

and hence, observing that \( \frac{cp}{2} = \text{area of triangle} \),

\[ \text{area} = \sqrt{s(s - a)(s - b)(s - c)}. \]

203. By a method of investigation in all respects similar to that which has been employed in these examples, any proposed geometrical problem may be reduced to an algebraic equation, the roots of which will exhibit arithmetical values of that geometrical magnitude which constitutes the unknown quantity in the equation. But the roots of algebraic equations may also be expressed by geometrical magnitudes, and hence a geometrical construction of a problem may be derived from its algebraic solution. For example, quadratic equations, which all belong to one or other of these three forms,

\[ x^2 + ax = bc, \quad x^2 - ax = bc, \quad x^2 - ax = -bc, \]

or, \( x(x + a) = bc, \quad x(x - a) = bc, \quad x(a - x) = bc; \)

may be constructed as follows.

**Construction of the first and second forms.**—Let a circle EABD (fig. 8) be described with a radius \( \frac{1}{2}a \), in which, from any point A in the circumference, apply a chord AB \( = b - c \) (b being supposed greater than c) and produce AB so that BC \( = c \); then AC \( = b \).

Let H be the centre of the circle; join CH cutting the circumference in D and E, then, in the first case, the positive value of \( x \) will be represented by CD, and in the second by CE; for by construction DE \( = a \); therefore, if CD be called \( x \), then CE \( = x + a \); but if CE \( = x \), then CD \( = x - a \).

Now, by the elements of geometry, \( EC \times CD = AC \times CB \); that is, \( x(x + a) = bc, \) or \( x(x - a) = bc, \) which equation comprehends the first and second cases.

If the negative roots be required, that of the first case will be CE and that of the second CD.

When \( b \) and \( c \) are equal, the construction will be rather more simple; for then AB vanishing, AC will coincide with the tangent CF. Therefore, if a right-angled triangle HFC be constructed, whose legs HF and FC are equal respectively to \( \frac{1}{2}a \) and \( b \), then will CD, the value of \( x \) in the first case, be equal to CH — HF, and CE, the value of \( x \) in the latter, \( = CH + HF \).

**Construction of the third form.**—Let a circle EADB (fig. 9) be described with a radius \( \frac{1}{2}a \) as before, in which apply a chord AB \( = b + c \), and take AC \( = b \). Through C draw the diameter DCE, then either DC or EC will be positive roots of the equation. For since ED \( = a \), if either EC or CD \( = x \), the remaining part of the diameter will be \( a - x \); now by the nature of the circle \( EC \times CD = AC \times CB \), that is, \( x(a - x) = bc, \) or \( x^2 - ax = -bc \); hence it is evident that the roots are rightly determined.

If \( b \) and \( c \) are equal, the construction will be the same, only it will then not be necessary to describe the whole circle; for since AC will be perpendicular to the diameter, if a right-angled triangle HCA be constructed, having its hypothenuse HA \( = \frac{1}{2}a \) and base AC \( = b \), the roots of the equation will be expressed by AH \( + HC \) and AH \( - HC \).

If \( b \) and \( c \) be so unequal that \( b - c \) in the first two cases, or \( b + c \) in the third, is greater than \( a \), then, instead of these quantities, \( \frac{b}{2} \) and \( 2c \), or in general \( \frac{b}{n} \) and \( nc \) (where \( n \) is any number whatever) may be used. Or a mean proportional may be found between \( b \) and \( c \), and the construction performed as directed in each case when \( b \) and \( c \) are equal.

It appears from this section, that every geometrical problem which produces a quadratic equation may be constructed by means of a straight line and a circle, or is a plane problem; hence, on the contrary, if a problem can be constructed by straight lines and circles, its algebraic resolution will not produce an equation higher than a quadratic. Cubic and biquadratic equations may be constructed geometrically by means of any two conic sections; hence it follows that every geometrical problem which requires for its construction two conic sections, will, when resolved by algebra, produce a cubic or biquadratic equation.

**Sect. XXIV. Of the Loci of Equations.**

204. When an equation contains two indeterminate quantities \( x \) and \( y \), then for each particular value of \( x \) Algebra. there may be as many values of \( y \) as it has dimensions in that equation. So that if in an indefinite line \( AE \) (fig. 10) there be taken a part \( AP \) to represent \( x \), and a perpendicular \( PM \) be drawn to represent \( y \), there will be as many points \( M, M', \ldots \) the extremities of these perpendiculars, as there are dimensions of \( y \) in the proposed equation; and the values of \( PM, PM', \ldots \) will be the roots of the equation, which are found by substituting for \( x \) its value in any particular case. Hence it appears that in any particular equation we may determine as many points \( M \) as we please; and a line which passes through all these points is called the locus of the equation. The line \( AP \), which expresses any value of \( x \), is called an abscissa; and \( PM \), which expresses the corresponding value of \( y \), is called an ordinate. Any two corresponding values of \( x \) and \( y \) are also called co-ordinates.

205. When the equation that arises by substituting for \( x \) any particular value \( AP \) has all its roots positive, the points \( M, M', \ldots \) will lie all on one side of \( AE \); but if any of them be negative, these must be set off on the other side of \( AE \) towards \( m \).

If \( x \) be supposed to become negative, then the line \( Ap \), which represents it, is to be taken in a direction opposite to that which represents the positive values of \( x \); the points \( M, m \), are to be taken as before, and the locus is only complete when it passes through all the points \( M, m \), so as to exhibit a value of \( y \) corresponding to every possible value of \( x \).

If in any case one of the values of \( y \) vanish, then the point \( M \) coincides with \( P \), and the locus meets \( AE \) in that point. If one of the values of \( y \) becomes infinite, then it shows that the curve has an infinite arc, and in that case the line \( PM \) becomes an asymptote to the curve, or touches it at an infinite distance, if \( AP \) itself is finite.

If, when \( x \) is supposed infinitely great, a value of \( y \) vanish, then the curve approaches to \( AE \) as an asymptote.

If any values of \( y \) become impossible, then so many points \( M \) vanish.

206. From these observations, and the theory of equations, it appears that when an equation is proposed involving two indeterminate quantities \( x \) and \( y \), there may be as many intersections of the curve which is the locus of the equation, and of the line \( PM \), as there are dimensions of \( y \) in the equation; and as many intersections of the curve and the line \( AE \) as there are dimensions of \( x \) in the equation.

207. A curve line is called geometrical, or algebraic, when the equation which expresses the relation between \( x \) and \( y \) (any abscissa and its corresponding ordinate), consists of a finite number of terms, and contains, besides these quantities, only known quantities. Algebraic curves are divided into orders, according to the dimensions of the equations which express the relations between their abscissas and ordinates, or according to the number of points in which they can intersect a straight line.

Straight lines themselves constitute the first order of lines; and when the equation expressing the relation between \( x \) and \( y \) is only of one dimension, the points \( M, M', \ldots \) must be all found in a straight line which contains with \( AE \) a given angle. Suppose, for example, that the given equation is \( ay - bx - cd = 0 \), and that its locus is required.

Since \( y = \frac{bx + cd}{a} \), it follows that \( APM \) (fig. 11) being a right angle, if \( AN \) be drawn making the angle \( NAP \) such that its cosine is to its sine as \( a \) to \( b \), and drawing \( AD \) parallel to the ordinates \( PM \), and equal to \( \frac{cd}{a} \), if \( DF \) be drawn parallel to \( AN \), then will \( DF \) be the locus required; where it is to be observed that \( AD \) and \( PN \) are to be taken on the same side of \( AE \) if \( bx \) and \( cd \) have the same sign, but on opposite sides of \( AE \) if they have contrary signs.

Curves whose equations are of two dimensions constitute the second order of lines, and the first kind of curves. Their intersections with a straight line can never exceed two (sect. 204).

Curves whose equations are of three dimensions form the third order of lines and the second kind of curves, and their intersections with a straight line can never exceed three; and after the same manner curves of the higher orders are denominated.

Some curves, if they were completely described, would cut a straight line in an infinite number of points; but these belong to none of the orders we have mentioned, for the relation between their ordinates and abscissas cannot be expressed by a finite equation, involving only ordinates and abscissas with determinate quantities. Curves of this kind are called mechanical or transcendental.

208. As the roots of an equation become impossible always in pairs, so the intersections of a curve and its ordinate \( PM \) must vanish in pairs, if any of them vanish. Let \( PM \) (fig. 12) cut the curve in the points \( M \) and \( m \), and, by moving parallel to itself, come to touch it in the point \( N \), then the two points of intersection \( M \) and \( m \) go to form one point of contact \( N \). If \( PM \) still move on, parallel to itself, the points of intersection will, beyond \( N \), become imaginary; as the two roots of an equation first become equal, and then imaginary.

The curves of the 3d, 5th, 7th orders, and all whose dimensions are odd numbers, have always one real root at least; and consequently for every value of \( x \) the equation by which \( y \) is determined must have at least one real root; so that as \( x \) or \( AP \) may be increased in infinitum on both sides, it follows that \( M \) must go off on both sides without limit.

In curves whose dimensions are even numbers, as the roots of their equations may become all impossible, it follows that the figure of the curve may be like a circle, or oval, that is, limited within certain bounds, beyond which it cannot extend.

209. When two roots of the equation by which \( y \) is determined become equal, either the ordinate \( PM \) touches the curve, two points of intersection in that case going into a point of contact; or the point \( M \) is a punctum duplex in the curve, two of its arcs intersecting each other there; or some oval that belongs to that kind of curve becoming infinitely little in \( M \), it vanishes into what is called a punctum conjugatum.

If, in the equation, \( y \) be supposed \( = 0 \), then the roots of the equation by which \( x \) is determined will give the distances of the points where the curve meets \( AE \) from \( A \); and if two of those roots be found equal, then either the curve touches the line \( AE \), or \( AE \) passes through a punctum duplex in the curve. When \( y \) is supposed \( = 0 \), if one of the values of \( x \) vanish, the curve in that case passes through \( A \); if two vanish, then either \( AE \) touches the curve in \( A \), or \( A \) is a punctum duplex.

As a punctum duplex is determined from the equality of two roots, so is a punctum tripulum from the equality of three roots.

210. To illustrate these observations, we shall take a few examples.

Ex. 1. It is required to describe the line that is the locus of this equation, \( y^2 = ax + ab \), or \( y^2 = ax - ab = 0 \), where \( a \) and \( b \) denote given quantities. Since \( y^2 \) Algebra.

\[ \sqrt{ax + ab} \] if \( AP = x \) (fig. 13) be assumed of a known value, and PM, Pm, set off on each side equal to \( \sqrt{ax + ab} \), the points M, m will belong to the locus required; and for every positive value of AP there may thus be found a point of the locus on each side. The greater that AP or x is taken, the greater does \( \sqrt{ax + ab} \) become, and consequently PM and Pm the greater; and if AP be supposed infinitely great, PM and Pm will also become infinitely great; therefore the locus has two infinite arcs, that go off to an infinite distance from AE and from AD. If x be supposed to vanish, then \( y = \pm \sqrt{ab} \), so that y does not vanish in that case, but passes through D and d, taking AD and Ad each \( = \sqrt{ab} \).

If P be supposed to move on the other side of A, then x becomes negative, and \( y = \pm \sqrt{ab - ax} \), so that y will have two values as before, while x is less than b; but if \( AB = b \), and the point P be supposed to come to B, then \( ab = ax \), and \( y = \pm \sqrt{ab - ax} = 0 \); that is, PM and Pm vanish, and the curve there meets the line AE. If P be supposed to move from A beyond B, then x becomes greater than b, and ax greater than ab, so that \( ab - ax \) being negative, \( \sqrt{ab - ax} \) becomes imaginary; that is, beyond B there are no ordinates which meet the curve, and consequently, on that side the curve is limited at B.

All this agrees very well with what is known by other methods, that the curve whose equation is \( y^2 = ax + ab \) is a parabola whose vertex is B, axis BE, and parameter equal to a. For since \( b = BP \), and \( y = PM \), from the equation \( ab = ax = y^2 \), or \( a(b - x) = y^2 \), we have \( a \times BP = PM^2 \), which is the well-known property of the parabola.

Ex. 2. It is required to describe the line that is the locus of the equation \( xy + ay + cy = bc + bx \), or

\[ y = \frac{bc + bx}{a + c + x} \]

Here it is evident (fig. 14) that the ordinate PM can meet the curve in one point only, there being but one value of y corresponding to each value of x. When \( x = 0 \), then \( y = \frac{bc}{a + c} \); so that the curve does not pass through A. If x be supposed to increase, then y will increase, but will never become equal to b, since \( y = \frac{c + x}{a + c + x} \) and \( a + c + x \) is always greater than \( c + x \). If x be supposed infinite, then the terms a and c vanish compared with x, and consequently \( y = \frac{x}{x} = b \); from which it appears, that taking AD = b, and drawing GD parallel to AE, it will be an asymptote, and touch the curve at an infinite distance. If x be now supposed negative, and AP be taken on the other side of A, then \( y = \frac{c - x}{a + c - x} \); and if x be taken on that side = c, then \( y = \frac{c - c}{a} = 0 \); so that the curve must pass through B, if \( AB = c \). If x be supposed greater than c, then will \( c - x \) become negative, and the ordinate will become negative, and lie on the other side of AE, till x become equal to \( a + c \), and then \( y = \frac{b - a}{0} \), that is, because the denominator is 0, x becomes infinite; so that if AK be taken \( = a + c \), the ordinate KF Algebra. will be an asymptote to the curve.

If x be taken greater than \( a + c \), or AP greater than AK, then both \( c - x \) and \( a + c - x \) become negative, and consequently \( y = \frac{b \times \frac{x - c}{x - a - c}}{x - a - c} \) becomes a positive quantity; and since \( x - c \) is always greater than \( x - a - c \), it follows that y will be always greater than \( b \) or KG, and consequently the rest of the curve lies in the angle FGH. And as x increases, since the ratio of \( x - c \) to \( x - a - c \) approaches still nearer to a ratio of equality, it follows that PM approaches to an equality with PN, therefore the curve approaches to its asymptote GH on that side also.

The curve is the common hyperbola, for since \( b(c + x) = y(a + c + x) \), by adding \( ab \) to both sides, \( b(c + x) + ab \) and \( (b - y)(a + c + x) = ab \), that is, \( NM \times GN = GC \times BC \), which is the property of the common hyperbola.

Ex. 3. It is required to describe the locus of the equation \( ay^2 - xy^2 = x^3 + bx^2 \).

Here \( y^2 = \frac{x^3 + bx^2}{a - x} \), and therefore \( y = \pm \sqrt{\frac{x^3 + bx^2}{a - x}} \), hence PM and PM (fig. 15) are to be taken on each side, and equal to \( \sqrt{\frac{x^3 + bx^2}{a - x}} \). This expression, by supposing \( x = a \), becomes infinite, because its denominator is then = 0; therefore if AB be taken \( = a \), and BK be drawn perpendicular to AB, the line BK will be an asymptote to the curve. If x be supposed greater than a, or AP greater than AB, then \( a - x \) being negative, the fraction \( \frac{x^3 + bx^2}{a - x} \) will become negative, and its square root impossible; so that no part of the locus can lie beyond B. If x be supposed negative, or P taken on the other side of A, then \( y = \pm \sqrt{\frac{-x^3 + bx^2}{a + x}} \); hence the values of y will be real and equal as long as x is less than b; but if \( x = b \), then \( y = \sqrt{\frac{-x^3 + bx^2}{a + x}} = \sqrt{\frac{-b^3 + b^2}{a + b}} = 0 \), and consequently if AD be taken \( = b \), the curve will pass through D, and there touch the ordinate. If x be taken greater than b, then \( \pm \sqrt{\frac{-x^3 + bx^2}{a + x}} \) becomes imaginary, so that no part of the curve is found beyond D. The portion between A and D is called a nodus. If y be supposed = 0, then will \( x^3 + bx^2 = 0 \) be an equation whose roots are \( -b, 0, 0 \); from which it appears that the curve passes twice through A, and has in A a punctum duplex. This locus is a line of the third order.

If b is supposed to vanish in the supposed equation, so that \( ay^2 - xy^2 = x^3 \), then will A and D coincide (fig. 16) and the nodus vanish, and the curve will have in the point A a cuspis, the two arcs AM and Am, in this case, touching one another in that point. This is the same curve which the ancients called the Cissoid of Diocles.

If, instead of supposing b positive, or equal to 0, we suppose it negative, the equation will be \( ay^2 - xy^2 = x^3 - bx^2 \); the curve will in this case pass through D as before (fig. 17), and taking \( AB = a \), BK will be its asymptote. It will have a punctum conjugatum in A, because when y vanishes, two values of x vanish, and the third becomes \( = b \). The whole curve, except this point, lies between DQ and BK. These remarks are demonstrated after the same manner as in the first case.

211. If an equation have this form,

\[ y = ax^n + bx^{n-1} + cx^{n-2} + \ldots \]

and \( n \) is an even number, then will the locus of the equation have two infinite arcs lying on the same side of AE (fig. 18); for if \( x \) become infinite, whether positive or negative, \( x^n \) will be positive and \( ax^n \) have the same sign in either case; and as \( ax^n \) becomes infinitely greater than the other terms \( bx^{n-1}, \ldots \), it follows that the infinite values of \( y \) will have the same sign in these cases, and consequently the two infinite arcs of the curve will lie on the same side of AE.

But if \( n \) be an odd number, then when \( x \) is negative, \( x^n \) will be negative, and \( ax^n \) will have the contrary sign to what it has when \( x \) is positive; and therefore the two infinite arcs will in this case lie on different sides of AE, as in fig. 19, and tend towards parts directly opposite.

212. If an equation have this form,

\[ y^2 = ax^{n+1} \]

and \( n \) be an odd number, then when \( x \) is positive, \( y = \frac{ax^{n+1}}{x^n} \); but when \( x \) is negative, \( y = -\frac{ax^{n+1}}{x^n} \); so that this curve must all lie in the vertically opposite angles KAE, FAE (fig. 20), as the common hyperbola, FK, EC being asymptotes.

But if \( n \) be an even number, then \( y \) is always positive, whether \( x \) be positive or negative, because \( x^n \) in this case is always positive; and therefore the curve must all lie in the two adjacent angles KAE and KAE (fig. 21), and have AK and AE for its asymptotes.

213. If an equation be such as can be reduced into two other equations of lower dimensions, without affecting \( y \) or \( x \) with any radical sign, then the locus will consist of the two loci of those inferior equations. Thus, the locus of the equation

\[ y^2 - 2xy + bx^2 - bx = 0 \]

which may be resolved into these two, \( x - y = 0, y - x + b = 0 \), is found to be two straight lines cutting the absciss AE (fig. 22) in angles of 45° in the points A, B, whose distance AB = \( b \). In like manner some cubic equations can be resolved into three simple equations, and then the locus is three straight lines; or may be resolved into a quadratic and simple equation, and then the locus is a straight line and a conic section. In general, curves of the superior orders include all the curves of the inferior orders, and what is demonstrated generally of any one order is also true of the inferior orders. Thus, for example, any general property of the conic sections holds true of two straight lines as well as a conic section, particularly, that the rectangles of the segments of parallels bounded by them will always be to one another in a given ratio.

214. From the analogy which subsists between algebraic equations and geometrical curves, it is easy to see that the properties of the former must suggest corresponding properties of the latter. Hence the principles of algebra admit of the most extensive application to the theory of curve lines. It may be demonstrated, for example, that the locus of every equation of the second order is a conic section; and, on the contrary, the various properties of the diameters, ordinates, tangents, &c. of the conic sections may be readily deduced from the theory of equations.

Sect. XXV. Arithmetic of Sines.

215. The calculus of sines is one of the mathematical theories which have been produced by the application of algebra to geometry. It treats of the relations which sines, cosines, tangents, &c. of angles have one to another.

216. The geometrical principles of this theory were known to the ancients. They may be deduced from the beautiful property of a quadrilateral inscribed in a circle, namely, that the rectangle contained by its diagonals is equal to the sum of the rectangles contained by its opposite sides; which we find in the writings of Ptolemy, and which was employed in the trigonometry of the Greeks. In comparatively modern times we find the same propositions distinctly recognised in the Opus Palatinum, the great work on triangles begun by Rhetius, and finished by Otho, who published it in 1596; also in the trigonometry of Pitiscus, first printed in 1599; and probably they may be found in still earlier works. Montucla, in his Histoire des Mathématiques, says, "I do not see that any one before the beginning of this century (the 18th) had thought of seeking formulæ proper to express the sines or cosines, tangents or co-tangents, of the sum or difference of arcs of a circle, or their powers, &c. It seems to me to have been very natural; and that there must have been frequently occasion to know what was the sine or cosine, the tangent or co-tangent, of an arc, that was the sum or difference of two arcs of which the sines or cosines, or tangents or co-tangents, were known. The first theorems on this subject appear to be the work of Frederic-Christian Mayer, one of the first members of the Petersburg Academy." Montucla, however, seems to have overlooked the trigonometry of Pitiscus, who, in problems 8 and 9 of his second book, gives rules for finding the sines and cosines of the sum and difference of two arcs, when the sines and cosines of the arcs are given. These are in the edition of 1612. The formula which expresses the tangent of the sum of several arcs by the tangent of the arcs themselves, was given, for the first time, we believe, by John Bernoulli, in the Leipsic Acts for July 1722. Both these are prior to Mayer's Memoir, which is in the Petersburg Commentaries with the date 1727. It was probably here, however, that the first essay of analytical trigonometry was given. Euler, who stands pre-eminent in every branch of the mathematics, has contributed more especially to this doctrine, as in his Subsidium Calculi Sinuum, in the New Petersburg Commentaries, vol. v. (for 1754 and 1755), and his Introductio in Analysis Infinitorum. The doctrine of spherical trigonometry was given in an analytic form by the same writer, in a memoir entitled Trigonometria Sphaerica universa ex primis principiis derivata, in the Petersburg Acts for 1779; and also by De Guan in the volume of the Academy of Sciences for 1783; and, lastly, by Lagrange in a memoir of the Journal Polytechnique, vol. ii.

217. The calculus of sines is of great importance as a branch of analysis. It enables us in a great measure to dispense with the complicated diagrams which the earlier writers on geometry, and on the physico-mathematical theories, employed in their investigations and demonstrations. In geometry, when combined with the modern analysis, it gives the power of expressing the most complicated relations of figure, and magnitude, and position, almost without the help of graphic representations. As an example, we may give the fine discovery by Gauss, of formulæ by which a regular polygon of seventeen sides may be inscribed in a circle by a geometrical construction. In algebra it has served to extend greatly the theory of equations; in astronomy, when applied to the theories of the planets and comets, it gives elegant and convenient expressions for their angular motions, and compact formulæ for computing the elements of their orbits, and their true and apparent positions; and in statics and dynamics, it gives, without diagrams, the relations of forces which produce rest or motion, and the laws of the motions. The precious invention of logarithms abridged the irksome Algebra. labour of calculation, even in the simplest applications of geometry; but this, combined with the trigonometrical tables, and the theory of the calculus of sines, affords a still more powerful instrument for abridging labour; just as two mechanical inventions, which apart can overcome considerable resistance, yet are vastly more potent when united.

218. In the calculus of sines, as well as in other applications of algebra to geometry, all quantities, whether lines or angles, are considered as expressed by numbers; some line or angle is assumed as a unit, and the number of times that unit is contained in the line or angle is its numerical value.

The magnitude of the unit is altogether arbitrary; and since, in general, the magnitude to be expressed in numbers may not contain the unit an exact number of times, it ought to be so small, that the remainders may, in respect of the wholes, be rejected.

219. The primary unit, by which angles are expressed in numbers, is one ninetieth part of a right angle, that is, one 360th part of four right angles. Each of these is called a degree, and is conceived to contain 60 units of a lower order, called minutes; and each minute, again, 60 units of the next lower order, called seconds; and so on to thirds, &c.

The degrees, minutes, &c. in an angle, are usually written thus: $12^\circ 15' 10'' 25''$. This means an angle of 12 degrees, 15 minutes, 10 seconds, and 25 thirds. It is not common to estimate angles to a greater degree of accuracy than seconds and tenths of a second; thus $3^\circ 4'$ means three seconds and four tenths of a second.

The French mathematicians attempted, in the time of the Revolution, to introduce a new unit for angles. They supposed a right angle to be divided into 100 equal parts, called degrees; and each degree to be subdivided into 100 minutes; and each minute into 100 seconds; and so on. This division, however, was not adopted out of France. Unfortunately, Laplace had employed it in his Mécanique Céleste, to the inconvenience of its readers; but now the French mathematicians themselves have almost all returned to the old sexagenary division.

220. Although an angle and a circle are not necessarily related, because we can form a distinct notion of each independently of the other; yet there is such an analogy between them, as to make it convenient to consider them together. Therefore (Plate XVIII. fig. 23), let PO, QO, be straight lines which contain an angle whose vertex is O. Let QO, one of the lines, be given in position, and let OE, a segment of the other line adjacent to O, be an invariable magnitude. Let us now suppose the line PO to coincide at first with QO, and, departing from that position, to turn about O as a centre; the line PO will now generate an angle POQ, and E, the extremity of the constant radius, will describe an arc AE. When the line has made a complete revolution about the centre O, it will have generated four right angles, and the point E will have described a complete circle; and it is manifest that any angle POQ will have the same ratio to four right angles that the arc AE has to the whole circumference. Hence it follows that the arc may conveniently serve as a geometrical measure of the angle; and if we suppose the circumference to be divided into 360 equal parts, called degrees, and each of these to be subdivided into 60 minutes, and so on, then the angle and arc will be expressed by the same number of degrees, minutes, &c.

221. Let the diameters AC, BD, be perpendicular to each other. These divide the circumference into four quadrants, and the circle into four regions AOB, BOC, COD, DOA. The difference between an angle and a right angle, or between an arc and a quadrant, is called the complement of the angle or arc; and the difference between an angle and two right angles, or between an arc and a semicircle, is called the supplement of the angle or arc. Thus, the angle POB is the complement of POQ, and POC is its supplement; and the arc BE is the complement of AE, and CE its supplement.

222. Assuming A as the origin of the variable arc AE, a straight line EF, drawn from its extremity E, perpendicular to the diameter AC, is called the sine of the arc AE; and a straight line EG, drawn perpendicular to BD (which is the sine of the arc EB, the complement of AE), is called the cosine of the arc AE. Hence it appears that the cosine of an arc is equal to the distance of the sine from the centre of the circle.

223. Supposing now the arc AE to increase continually by the revolving of the radius OE, departing from the position OA, and moving in the direction ABCD, &c. the sine EF, which at first is =0, will increase until the arc become a quadrant; afterwards, while the arc increases from one to two quadrants, the sine EF will decrease, and at last again become =0. The arc continuing to increase, the sine will be reproduced; but its direction FE will now be the opposite to its first direction FE. It will increase in magnitude, until the arc become three quadrants ABCD; and it will again decrease, while the arc is increasing from three to four quadrants: and when the radius has made a complete revolution, the sine will have decreased to zero. We may, however, suppose the revolving radius OE still to proceed; there will thus be generated an arc, ABCDAE, greater than a complete circumference, of which EF will be the sine; and, in the second and every succeeding revolution, the sine will pass through the same changes as in the first.

224. We may trace the changes in the magnitude and direction of the cosine just in the same way. While the arc AE increases from zero to a quadrant, the cosine OF, at first equal to the radius OA, decreases, and at last vanishes; and while the arc continues to increase, and has any value ABE between one and two quadrants, the cosine OP is reproduced, with a direction the opposite to the first. When the arc is exactly two quadrants, the cosine becomes in magnitude OC, equal to OA, but opposite in direction. The arc still increasing to ACE', the cosine decreases to OF'; and when the arc becomes ABCD, three quadrants, the cosine again becomes =0; between three and four quadrants the cosine OE" reappears in its original direction; and when the revolution is completed, the cosine acquires its first value OA. In a second revolution the same changes are repeated, and so on continually.

We may suppose the radius to turn round O in the contrary direction ADC, and the very same changes will be produced on the direction and magnitude of the sine and cosine of the generated arc AE'.

225. In conformity to the principles of analytic geometry, the contrary position of the arcs AE, AE', and the sines EF, EF', in respect of the position of the arcs AE", AE', and the sines E"F', E"F", may be indicated by the signs + and —; and the same is true of the cosines OF, OF', which are opposed in direction to the cosines OF, OF', the arcs and sines on opposite sides of the diameter AC being regarded as having opposite signs, and a like contrariety being supposed in the cosines on opposite sides of the diameter BD, which is perpendicular to AC. The arc AE being regarded as positive, it does not signify whether we reckon the sines and cosines positive or negative in the beginning, provided we change the signs, Let us suppose the sine and cosine of any positive arc AE in the first quadrant to be positive; then, from what has been explained, the sine will continue to be positive in the second quadrant, but the cosine will have become negative; in the third the sine and cosine will both be negative; and in the fourth the sine will be negative, but the cosine positive.

Again, if the radius OE turn about O in the contrary direction, ADCB, &c., the arc AE', &c., thus generated will be negative; the sines will also be negative in the first semicircle, but the cosines positive in the quadrant AD, and negative in DC, and so on; the changes from + to —, and from — to +, following each other exactly as in the generation of a positive arc.

226. Let a straight line AH, perpendicular to the diameter AC, meet the radius OE produced in H (fig. 24): the line AH is called the tangent of the arc AE, and the line OH the secant of that arc.

227. A line BK, perpendicular to OB (which is at right angles to OA), and which meets OH in K, will be the tangent of the complement of the arc AE; and OK will be its secant. The former is therefore called the cotangent of the arc AE, and the latter its co-secant.

228. The segment AF of the diameter, between the beginning of the arc and its sine, is called the versed sine of the arc; and, similarly, the segment BG, the versed sine of the complement, is its co-versed sine.

229. Since the arc AE is assumed as a geometrical measure of the angle POQ, we may regard the lines which are the sine, tangent, secant, &c. of the arc, as also the sine, tangent, secant, &c. of the angle; or, since all quantities in this calculus are to be expressed in numbers, we may take any numbers proportional to these lines. Hence we may assume the fraction \(\frac{\text{arc AE}}{\text{rad. OA}}\) as the measure of the angle POQ, and, similarly, \(\frac{\text{OE}}{\text{OE}}\) as the sine of the angle POQ, \(\frac{\text{OF}}{\text{OB}}\) its cosine, \(\frac{\text{HA}}{\text{OA}}\) its tangent, and \(\frac{\text{OH}}{\text{OA}}\) its secant; then \(\frac{\text{BK}}{\text{OB}} = \frac{\text{AO}}{\text{AH}}\) will be the cotangent, and \(\frac{\text{OB}}{\text{AH}}\) will be the co-secant of the angle. These quantities will have among themselves the same ratios as the lines which bear the same designations; and it is evident, from the nature of similar figures, that supposing the angle POQ to remain of the same magnitude, its sine, tangent, secant, &c. thus defined, are the same, whatever be the magnitude of the circle, which now serves merely to connect them as related quantities.

We may consider OA, the radius of the circle, as a unit, and then the sines, tangents, &c. will be expressed in parts of that unit, according to the decimal notation. This is the common form in which they are given in the trigonometrical tables.

230. The power of algebra, as an instrument of investigation, depends greatly on the proper employment of conventional signs: it is common to express the sine of an arc \(a\) by the abbreviation \(\sin a\), its cosine by \(\cos a\), its tangent by \(\tan a\), its secant by \(\sec a\), its co-tangent by \(\cot a\), its co-secant by \(\csc a\), and its versed sine by \(\ver \sin a\).

231. From the definition given of an angle by the ancient geometers, its measure cannot exceed half the circumference of a circle. When, however, we regard an angle as generated by the rotation of a line about a fixed point, the arc, which is its measure, may be regarded as capable of continual increase. In astronomy, the angular motion of the sun and planets is reckoned from an angle or arc of 0° forward to 360°. We are thus led to consider the sines of arcs which exceed 180°, and even which exceed a whole circumference, or any number of circumferences.

232. Let \(x\) denote half the circumference of the circle which is assumed as a scale for measuring angles. It is easy, by induction, to infer from what has been explained, that \(a\) being any arc,

\[ \begin{align*} \cos (\pi + a) &= -\cos a, \\ \sin (\pi + a) &= -\sin a, \\ \cos (2\pi + a) &= +\cos a, \\ \sin (2\pi + a) &= +\sin a, \\ \cos (3\pi + a) &= -\cos a, \\ \sin (3\pi + a) &= -\sin a, \\ \cos (4\pi + a) &= +\cos a, \\ \sin (4\pi + a) &= +\sin a, \\ &\vdots \end{align*} \]

And in general, \(n\) being 0, or any whole number,

\[ \begin{align*} \cos [(2n+1)\pi + a] &= -\cos a, \\ \cos [2n\pi + a] &= +\cos a, \\ \sin [(2n+1)\pi + a] &= -\sin a, \\ \sin [2n\pi + a] &= +\sin a. \end{align*} \]

It is also evident that

\[ \begin{align*} \cos (\pi - a) &= -\cos a, \\ \sin (\pi - a) &= +\sin a, \\ \cos (2\pi - a) &= +\cos a, \\ \sin (2\pi - a) &= -\sin a, \\ \cos (3\pi - a) &= -\cos a, \\ \sin (3\pi - a) &= +\sin a, \\ \cos (4\pi - a) &= +\cos a, \\ \sin (4\pi - a) &= -\sin a, \\ &\vdots \end{align*} \]

And in general that

\[ \begin{align*} \cos [(2n+1)\pi - a] &= -\cos a, \\ \cos [2n\pi - a] &= +\cos a, \\ \sin [(2n+1)\pi - a] &= +\sin a, \\ \sin [2n\pi - a] &= -\sin a. \end{align*} \]

When the arc \(a = 0\), we have

\[ \begin{align*} \cos (2n+1)\pi &= -1, \\ \sin (2n+1)\pi &= 0, \\ \cos 2n\pi &= +1, \\ \sin 2n\pi &= 0. \end{align*} \]

233. If \(a\) and \(a'\) be arcs which differ by a quadrant, that is, if \(a' = \frac{\pi}{2} + a\); then it will appear by inspection of figure 23, and a little consideration, that

\[ \begin{align*} \cos a &= -\sin a, \\ \sin a &= +\cos a. \end{align*} \]

Now, by changing \(a\) into \(a'\), formulae (1) of last section become

\[ \begin{align*} \cos [(4n+3)\frac{\pi}{2} + a] &= +\sin a, \\ \cos [(4n+1)\frac{\pi}{2} + a] &= -\sin a, \\ \sin [(4n+3)\frac{\pi}{2} + a] &= -\cos a, \\ \sin [(4n+1)\frac{\pi}{2} + a] &= +\cos a. \end{align*} \]

If, again, there be two arcs \(a\) and \(a'\), whose sum is a quadrant, that is, \(a' = \frac{\pi}{2} - a\), then

\[ \begin{align*} \cos a &= \sin a, \\ \sin a &= \cos a. \end{align*} \]

If, in the above formulae, we now put \(\frac{\pi}{2} - a\) for \(a\), and \(\sin a\) for \(\cos a\), and \(\cos a\) for \(\sin a\), we shall have

\[ \begin{align*} \cos [(4n+3)\frac{\pi}{2} - a] &= -\sin a, \\ \cos [(4n+1)\frac{\pi}{2} - a] &= +\sin a, \\ \sin [(4n+3)\frac{\pi}{2} - a] &= -\cos a, \\ \sin [(4n+1)\frac{\pi}{2} - a] &= +\cos a. \end{align*} \] 234. The formulæ investigated in the last two sections may be brought together in an abridged form, as in the following table:

\[ \begin{align*} \text{Cos. } [(2n+1)\pi \pm a] &= -\cos a, \\ \text{Cos. } [2n\pi \pm a] &= +\cos a, \\ \text{Sin. } [(2n+1)\pi \pm a] &= -\sin a, \\ \text{Sin. } [2n\pi \pm a] &= +\sin a; \\ \text{Cos. } [(4n+1)\frac{\pi}{2} \pm a] &= -\sin a, \\ \text{Cos. } [(4n+3)\frac{\pi}{2} \pm a] &= +\sin a, \\ \text{Sin. } [(4n+1)\frac{\pi}{2} \pm a] &= +\cos a, \\ \text{Sin. } [(4n+3)\frac{\pi}{2} \pm a] &= -\cos a. \end{align*} \]

235. The perfection of the modern analysis consists in the employment of the fewest possible principles, and in deducing from these all the truths which can be extracted by the power of the analysis itself. We have therefore occasion for very few of the simplest propositions in geometry, and these only at the outset.

By the nature of a right-angled triangle (fig. 24), \(OF^2 + FE^2 = OE^2; OA^2 + AH^2 = OH^2; OB^2 + BK^2 = OK^2.\)

The triangles OFE, OAH, OBK, being similar,

\[ \begin{align*} OF : FE &:: OA : AH : BK : BO; \\ FE : EO &:: AH : HO : BO : OK; \\ EO : OF &:: HO : OA :: OK : KB. \end{align*} \]

Hence, observing that \(OA = OB = OE = 1,\) and employing our premised notation, we have

\[ \begin{align*} \text{Cos. } a &:: \sin a :: 1 :: \tan a :: \cot a :: 1; \\ \text{Sin. } a &:: 1 :: \tan a :: \cot a :: 1 :: \csc a; \\ 1 &:: \cos a :: \sec a :: 1 :: \csc a :: \cot a. \end{align*} \]

From these we obtain the following table of formulæ, exhibiting the relations among trigonometrical quantities, where \(\cos^2 a\) and \(\sin^2 a\) denote the squares of the cosine and sine of the arc or angle \(a.\)

(A)

\[ \begin{align*} \text{Cos. } a + \sin^2 a &= 1, \\ \text{Sec. } a - \tan^2 a &= 1, \\ \text{Cosec. } a - \cot^2 a &= 1, \\ \text{Cos. } a \sec a &= 1, \\ \text{Sin. } a \csc a &= 1, \\ \text{Tan. } a \cot a &= 1, \\ \text{Tan. } a \cos a &= \sin a, \\ \text{Cot. } a \sin a &= \cos a, \\ \text{Sin. } a \sec a &= \tan a, \\ \text{Cos. } a \csc a &= \cot a, \\ \text{Tan. } a \csc a &= \sec a, \\ \text{Cotan. } a \sec a &= \csc a. \end{align*} \]

It appears from formulæ 7, 8, 4, 5, that

\[ \begin{align*} \tan a &= \frac{\sin a}{\cos a}, \\ \cot a &= \frac{\cos a}{\sin a}, \\ \sec a &= \frac{1}{\cos a}, \\ \csc a &= \frac{1}{\sin a}. \end{align*} \]

Hence it follows that the tangent and co-tangent must be regarded as positive quantities when the sine and cosine are both positive, or both negative quantities; but as negative when, of the sine and cosine, one is positive and the other negative; and that the secant and cosine, also the co-secant and sine, must always have the same sign. Therefore the tangent will be positive in the first quadrant, negative in the second, positive in the third, negative in the fourth, and so on; and the secant will be positive in the first, negative in the second and third, positive in the fourth and fifth, and so on to any number of quadrants.

236. We have seen that the sines and cosines, when changing their sign from \(+\) to \(-,\) or from \(-\) to \(+,\) in the transition become \(=0.\) Now, since \(\tan a = \frac{\sin a}{\cos a},\) and \(\sec a = \frac{1}{\cos a};\) when \(a\) is a quadrant, and therefore \(\cos a = 0,\) and \(\sin a = 1,\) then both \(\tan a\) and \(\sec a\) become infinite; and when \(a\) is between a quadrant and a semicircumference, \(\cos a\) being then negative, both \(\tan a\) and \(\sec a\) are negative. Hence it appears that the tangent and secant become both infinite in changing their sign from \(+\) to \(-.\) Is passing from the second to the third quadrant, the tangent, from being negative, becomes positive, because the numerator of the fraction \(\frac{\sin a}{\cos a}\) changes from \(+\) to \(-.\) In this transition the tangent becomes \(=0.\) The secant, however, continues negative; and when the arc has attained a magnitude between three and four quadrants, both tangent and secant change their signs, after having become infinite, the tangent being now negative, and the secant positive. When the circumference is completed, the tangent is reduced to 0, and the secant to the radius; and the arc being supposed to increase, the same changes will be repeated.

237. Let the sides of any triangle \(A B C\) (Fig. 7. Plate XVIII.) be \(a, b, c,\) and the opposite angles \(A, B, C.\)

From \(C,\) any one of the angles, draw \(C D\) perpendicular to the opposite side \(c,\) dividing it into the two segments \(AD, BD;\) then, by the definitions (sect. 229),

\[ \cos B = \frac{BD}{BC} = \frac{BD}{a}, \]

and hence

\[ BD = a \cos B,\] and, similarly, \(AD = b \cos A;\) and \(c = AB = BD + AD = a \cos B + b \cos A.\)

Hence we have this property of a triangle:

If each of any two of its sides be multiplied by the cosine of the angle it makes with the third side, the sum of the products is equal to that third side. This theorem, combined with these formulæ,

\[ \begin{align*} \cos^2 A + \sin^2 A &= 1, \\ \cos (\pi - A) &= -\cos A, \\ \sin (\pi - A) &= \sin A, \end{align*} \]

is to serve as the basis of the whole calculus of sines.

From the preceding theorem, we obtain the following system of formulæ:

I.

\[ \begin{align*} a &= b \cos C + c \cos B, \\ b &= a \cos C + c \cos A, \\ c &= a \cos B + b \cos A. \end{align*} \]

From the first of these we get

\[ \begin{align*} a \cos C &= b \cos^2 C + c \cos C \cos B, \\ a \cos B &= b \cos B \cos C + c \cos^2 B; \end{align*} \]

and from these again, by transposing,

\[ \begin{align*} b \cos^2 C &= a \cos C - c \cos B \cos C, \\ c \cos^2 B &= a \cos B - b \cos B \cos C. \end{align*} \]

Subtract each side of the first of these two from the corresponding side of 2 of system I, and each side of the second from the corresponding side of 3, and write in the Algebra results $\sin^2 C$ for $1 - \cos^2 C$, and $\sin^2 B$ for $1 - \cos^2 B$; we then get

$$b \sin^2 C = c (\cos A + \cos B \cos C),$$ $$c \sin^2 B = b (\cos A + \cos B \cos C).$$

Take now the product of the sides of these equations, and leave out the common factors $b, c$ of the result; we thus obtain

$$\sin^2 B \sin^2 C = (\cos A + \cos B \cos C)^2.$$

Hence, taking the square roots,

$$\sin B \sin C = \cos A + \cos B \cos C.$$

Here we have a property of the angles of any triangle, which is independent of its sides; this, by interchanging the angles, gives the following system of formulae:

II.

$$\begin{align*} \cos A &= \cos B \cos C + \sin B \sin C, \\ \cos B &= \cos A \cos C + \sin A \sin C, \\ \cos C &= \cos A \cos B + \sin A \sin B. \end{align*}$$

Again, from the first of these we have

$$\cos A \cos C = \cos B \cos^2 C + \sin B \sin C \cos C,$$ and hence, by transposition, $$\cos B \cos^2 C = \cos A \cos C + \sin B \sin C \cos C.$$

Subtract now the sides of this equation from the corresponding sides of 2; the result is

$$\cos B (1 - \cos^2 C) = \sin C (\sin A - \sin B \cos C).$$

Substitute $\sin^2 C$ for $1 - \cos^2 C$, and divide by $\sin C$; then,

$$\cos B \sin C = \sin A - \sin B \cos C.$$

This gives us a third system of formulae; viz.

III.

$$\begin{align*} \sin A &= \sin B \cos C + \cos B \sin C, \\ \sin B &= \sin A \cos C + \cos A \sin C, \\ \sin C &= \sin A \cos B + \cos A \sin B. \end{align*}$$

If it be now considered that $A + B + C$ are two right angles $= \pi$, so that $A + B = \pi - C$, and therefore

$$\sin(A + B) = \sin C, \quad \cos(A + B) = -\cos C,$$

we obtain, from the third formulae of systems II and III,

$$\begin{align*} \cos(A + B) &= \cos A \cos B - \sin A \sin B, \\ \sin(A + B) &= \sin A \cos B + \cos A \sin B. \end{align*}$$

238. In this investigation, we have supposed that $A$ and $B$ are two angles of a triangle,—an hypothesis which requires that their sum shall not exceed two right angles, and that the sum of the arcs which measure them do not exceed a semicircumference. The formulae, however, hold true, whatever be the magnitude of the arcs: for, let $A, a$, and $B, b$, denote arcs, such that $a + A = \pi$, $b + B = \pi$; and therefore $(a + b) + (A + B) = 2\pi$; then,

$$A = \pi - a, \quad B = \pi - b,$$

and (sect. 232)

$$\cos A = -\cos a, \quad \sin A = \sin a,$$ $$\cos B = -\cos b, \quad \sin B = \sin b.$$

Hence it follows, that

$$\cos A \cos B = \sin A \sin B,$$ $$\cos(a + b) = \cos(A + B) = \cos A \cos B - \sin A \sin B,$$ $$\sin(a + b) = -\sin(A + B) = -\sin A \cos B + \cos A \sin B,$$ $$\cos(a + b) = \cos a \cos b - \sin a \sin b,$$ $$\sin(a + b) = \sin a \cos b + \cos a \sin b.$$

And since (sect. 232)

$$\cos(a + b) = \cos(A + B) = \cos A \cos B - \sin A \sin B,$$ $$\sin(a + b) = -\sin(A + B) = -\sin A \cos B + \cos A \sin B,$$ $$\cos(a + b) = \cos a \cos b - \sin a \sin b,$$ $$\sin(a + b) = \sin a \cos b + \cos a \sin b.$$

Now, on the hypothesis that $A + B < \pi$; then, because $A + B + a + b = 2\pi$, it follows that $a + b > \pi$. Hence it appears that the formulae for the cosine and sine of the sum of $A, B$, arcs less than a semicircle, apply equally to arcs $a, b$, whose sum is greater than the half, but less than the whole circumference.

Again, if we suppose that $a, a'$, and $b, b'$, are such arcs, that

$$a + a' = 2\pi, \quad b + b' = 2\pi,$$ then we shall have $a = 2\pi - a', \quad b = 2\pi - b'$, $$\cos a = \cos a', \quad \sin a = -\sin a',$$ $$\cos b = \cos b', \quad \sin b = -\sin b'.$$ $$\cos(a + b') = \cos(a + b'), \quad \sin(a + b') = -\sin(a + b').$$

Here, again, we shall have

$$\cos(a + b') = \cos a \cos b' - \sin a \sin b',$$ $$\sin(a + b') = \sin a \cos b' + \cos a \sin b';$$ and because $a + b'$ is greater than $\pi$, but less than $2\pi$, therefore $a + b'$ is an arc between two and three semicircumferences.

It is evident we may proceed in this way continually, and thus be assured that the formulae are true, when $A$ and $B$ are any arcs of a circle, even although they should exceed a circumference, or any number of circumferences.

239. If, in these formulae, we substitute $a - b$ for $A$, and $b$ for $B$, they become

$$\cos a = \cos b \cos(a - b) - \sin b \sin(a - b),$$ $$\sin a = \sin b \cos(a - b) + \cos b \sin(a - b).$$

To abridge, let us put

$$P = \cos(a - b), \quad Q = \sin(a - b);$$ we have now, by transposition,

$$P \cos b - Q \sin b = \cos a,$$ $$P \sin b + Q \cos b = \sin a.$$

From these equations, by the ordinary process of elimination, we can determine $P$ and $Q$; thus, proceeding by sect. 73, we get

$$\begin{align*} P \cos^2 b - Q \sin b \cos b &= \cos a \cos b, \\ P \sin^2 b + Q \sin b \cos b &= \sin a \sin b, \\ P \cos b \sin b + Q \cos^2 b &= \cos a \sin b, \\ P \cos b \sin b + Q \sin^2 b &= \sin a \cos b. \end{align*}$$

By adding equations (1) and (2), and substituting 1 for $\cos^2 b + \sin^2 b$, we obtain

$$P = \cos(a - b) = \cos a \cos b + \sin a \sin b.$$

And again, by subtracting (3) from (4), we get

$$Q = \sin(a - b) = \sin a \cos b - \cos a \sin b.$$

For the sake of reference, and a uniform notation, we shall now bring together the formulae of this and sect. 237 into one group.

(B)

$$\begin{align*} \cos(a + b) &= \cos a \cos b - \sin a \sin b, \\ \cos(a - b) &= \cos a \cos b + \sin a \sin b, \\ \sin(a + b) &= \sin a \cos b + \cos a \sin b, \\ \sin(a - b) &= \sin a \cos b - \cos a \sin b. \end{align*}$$

These constitute what we have termed the fundamental theorems of this calculus.

240. By adding and subtracting these formulae, we obtain others equivalent to them, but under a different form, viz. these:

(C)

$$\begin{align*} \cos a \cos b &= \frac{1}{2} \cos(a - b) + \frac{1}{2} \cos(a + b), \\ \sin a \sin b &= \frac{1}{2} \cos(a - b) - \frac{1}{2} \cos(a + b), \\ \sin a \cos b &= \frac{1}{2} \sin(a + b) + \frac{1}{2} \sin(a - b), \\ \cos a \sin b &= \frac{1}{2} \sin(a + b) - \frac{1}{2} \sin(a - b). \end{align*}$$

If we now put $a + b = A$, and $a - b = B$, so that $a = \frac{1}{2}(A + B)$, $b = \frac{1}{2}(A - B)$, and substitute these values of $a, b, a + b, a - b$, in the above formulae, and, for the sake of uniformity, write the letters $a$ and $b$ instead of $A$ and $B$, we obtain Cos. \( b + \cos. a = 2 \cos. \frac{1}{2} (a + b) \cos. \frac{1}{2} (a - b), \quad (1) \) Cos. \( b - \cos. a = 2 \sin. \frac{1}{2} (a + b) \sin. \frac{1}{2} (a - b), \quad (2) \) Sin. \( a + \sin. b = 2 \sin. \frac{1}{2} (a + b) \cos. \frac{1}{2} (a - b), \quad (3) \) Sin. \( a - \sin. b = 2 \cos. \frac{1}{2} (a + b) \sin. \frac{1}{2} (a - b). \quad (4) \)

In formulae (C), let \( nA \) be written instead of \( a \), and \( A \) instead of \( b \); and consequently \((n-1)A\) instead of \( a-b \), and \((n+1)A\) instead of \( a+b \); the result, after putting \( a \) and \( b \) for \( A \) and \( B \), will be

\[ \begin{align*} 2 \cos. a \cos. na &= \cos. (n-1)a + \cos. (n+1)a, \quad (1) \\ 2 \sin. a \sin. na &= \cos. (n-1)a - \cos. (n+1)a, \quad (2) \\ 2 \cos. a \sin. na &= \sin. (n+1)a + \sin. (n-1)a, \quad (3) \\ 2 \sin. a \cos. na &= \sin. (n+1)a - \sin. (n-1)a. \quad (4) \end{align*} \]

241. Resuming the first system of formulae, viz.

\[ \begin{align*} \sin. (a+b) &= \sin. a \cos. b + \cos. a \sin. b, \\ \cos. (a+b) &= \cos. a \cos. b - \sin. a \sin. b \end{align*} \]

let each side of the first be divided by the corresponding side of the second; then we have

\[ \begin{align*} \frac{\sin. (a+b)}{\cos. (a+b)} &= \frac{\sin. a \cos. b + \cos. a \sin. b}{\cos. a \cos. b - \sin. a \sin. b} \\ &= \frac{\sin. a + \sin. b}{\cos. a \cos. b} \\ &= \frac{1 - \sin. a \sin. b}{\cos. a \cos. b}. \end{align*} \]

Hence, observing that tan. \( A = \frac{\sin. A}{\cos. A} \), we obtain

\[ \tan. (a+b) = \tan. a + \tan. b - \tan. a \tan. b. \]

By a like process we obtain from the formulae for sin. \((a-b)\), and cos. \((a-b)\), a formula for tan. \((a-b)\).

The two, for the sake of reference, may be put together.

\[ \begin{align*} \tan. (a+b) &= \frac{\tan. a + \tan. b}{1 - \tan. a \tan. b}, \quad (1) \\ \tan. (a-b) &= \frac{\tan. a - \tan. b}{1 + \tan. a \tan. b}. \quad (2) \end{align*} \]

242. The expressions which have been found for tan. \((a+b)\), tan. \((a-b)\), give like formulae for the co-tangent of the sum and difference of two arcs; we have only to substitute in them

\[ \frac{1}{\cot. (a+b)}, \quad \frac{1}{\cot. a}, \quad \text{and} \quad \frac{1}{\cot. b} \]

for tan. \( b \). The results are,

\[ \begin{align*} \cot. (a+b) &= \frac{1 - \cot. a \cot. b}{\cot. a + \cot. b}, \quad (1) \\ \cot. (a-b) &= \frac{1 + \cot. a \cot. b}{\cot. a - \cot. b}. \quad (2) \end{align*} \]

243. We have found, formulae (C), that

\[ \begin{align*} 2 \sin. a \sin. b &= \cos. (a-b) - \cos. (a+b), \\ 2 \cos. a \cos. b &= \cos. (a-b) + \cos. (a+b); \end{align*} \]

therefore, \(\cos. a \cos. b = \cos. (a-b) + \cos. (a+b)\).

From these formulae we also obtain like expressions for

\[ \begin{align*} \frac{\sin. a \cos. b}{\cos. a \sin. b} &= \frac{\cos. a \sin. b}{\sin. a \cos. b} = \frac{\cos. a \cos. b}{\sin. a \sin. b}; \end{align*} \]

and from these, putting tan. \( a \) for \(\frac{\sin. a}{\cos. a}\) and cot. \( a \) for \(\frac{\cos. a}{\sin. a}\), &c., we deduce this new set of formulae:

\[ \begin{align*} \tan. a \tan. b &= \frac{\cos. (a-b) - \cos. (a+b)}{\cos. (a-b) + \cos. (a+b)}, \quad (1) \\ \tan. a \cot. b &= \frac{\sin. (a+b) + \sin. (a-b)}{\sin. (a+b) - \sin. (a-b)}, \quad (2) \\ \cot. a \tan. b &= \frac{\sin. (a+b) + \sin. (a-b)}{\sin. (a+b) - \sin. (a-b)}, \quad (3) \\ \cot. a \cot. b &= \frac{\cos. (a-b) + \cos. (a+b)}{\cos. (a-b) - \cos. (a+b)}. \quad (4) \end{align*} \]

244. Because, by formulae (B),

\[ \sin. a \cos. b + \cos. a \sin. b = \sin. (a+b); \]

therefore, \(\frac{\sin. a}{\cos. a} + \frac{\sin. b}{\cos. b} = \frac{\sin. (a+b)}{\cos. a \cos. b}\).

By treating the remaining formulae in the same way, and substituting the tangent and co-tangent for their values, we obtain

\[ \begin{align*} \tan. a + \tan. b &= \frac{\sin. (a+b)}{\cos. a \cos. b}, \quad (1) \\ \tan. a - \tan. b &= \frac{\sin. (a-b)}{\cos. a \cos. b}, \quad (2) \\ \cot. a + \cot. b &= \frac{\sin. (a+b)}{\sin. a \sin. b}, \quad (3) \\ \cot. a - \cot. b &= \frac{\sin. (a-b)}{\sin. a \sin. b}, \quad (4) \\ \tan. a + \cot. b &= \frac{\cos. (a-b)}{\cos. a \sin. b}, \quad (5) \\ \cot. a - \tan. b &= \frac{\cos. (a+b)}{\sin. a \cos. b}. \quad (6) \end{align*} \]

245. Because, by formulae (D),

\[ \begin{align*} \sin. a + \sin. b &= 2 \sin. \frac{1}{2} (a+b) \cos. \frac{1}{2} (a-b), \\ \sin. a - \sin. b &= 2 \cos. \frac{1}{2} (a+b) \sin. \frac{1}{2} (a-b). \end{align*} \]

Hence, multiplying equals by equals, and considering that by formula (3) of (B)

\[ \begin{align*} 2 \sin. \frac{1}{2} (a+b) \cos. \frac{1}{2} (a+b) &= \sin. [\frac{1}{2} (a+b) + \frac{1}{2} (a+b)], \\ 2 \sin. \frac{1}{2} (a-b) \cos. \frac{1}{2} (a-b) &= \sin. [\frac{1}{2} (a-b) + \frac{1}{2} (a-b)], \end{align*} \]

we find

\[ (\sin. a + \sin. b)(\sin. a - \sin. b) = \sin. (a+b) \sin. (a-b). \]

If we write \(90-a\) for \(a\), the formula becomes

\[ (\cos. a + \sin. b)(\cos. a - \sin. b) = \cos. (a+b) \cos. (a-b). \]

The two may be written thus:

\[ \begin{align*} \sin^2 a - \sin^2 b &= \sin. (a+b) \sin. (a-b), \\ \cos^2 a - \cos^2 b &= \cos. (a+b) \cos. (a-b). \end{align*} \]

246. We have found, formulae (D),

\[ \begin{align*} \sin. a + \sin. b &= 2 \sin. \frac{1}{2} (a+b) \cos. \frac{1}{2} (a-b), \\ \sin. a - \sin. b &= 2 \cos. \frac{1}{2} (a+b) \sin. \frac{1}{2} (a-b). \end{align*} \]

Therefore, dividing equals by equals, and putting

\[ \tan. \frac{1}{2} (a+b) \text{ for } \sin. \frac{1}{2} (a+b), \text{ we obtain} \]

\[ \frac{\sin. a + \sin. b}{\sin. a - \sin. b} = \tan. \frac{1}{2} (a+b). \]

By a similar process we may deduce other five formulæ Algebra of the same nature from formulae (D). The whole are contained in this table.

(M)

\[ \begin{align*} \sin a + \sin b &= \tan \frac{1}{2} (a + b), \\ \sin a - \sin b &= \tan \frac{1}{2} (a - b), \\ \cos a + \cos b &= \cot \frac{1}{2} (a + b), \\ \cos a - \cos b &= \cot \frac{1}{2} (a - b), \\ \sin a + \sin b &= \tan \frac{1}{2} (a + b), \\ \sin a - \sin b &= \cot \frac{1}{2} (a + b), \\ \cos a + \cos b &= \cot \frac{1}{2} (a - b), \\ \cos a - \cos b &= \cot \frac{1}{2} (a - b). \end{align*} \]

247. We have found (formula K, No. 4) that

\[ \begin{align*} \sin (a-b) &= \cot a - \cot b, \\ \sin (b-c) &= \cot b - \cot c, \\ \sin (c-a) &= \cot c - \cot a. \end{align*} \]

Hence, by adding, we obtain

\[ \begin{align*} \sin (a-b) + \sin (b-c) + \sin (c-a) &= 0. \end{align*} \]

In like manner, it appears, from No. 2, that

\[ \begin{align*} \cos (a-b) + \cos (b-c) + \cos (c-a) &= 0. \end{align*} \]

Therefore, taking away the denominators, and observing that \(\sin (c-a) = -\sin (a-c)\), we have

(N)

\[ \begin{align*} \sin b \sin (a-c) &= \sin a \sin (a-b) + \sin a \sin (b-c), \\ \cos b \sin (a-c) &= \cos a \sin (a-b) + \cos a \sin (b-c). \end{align*} \]

These formulae are true, \(a, b, c\) being any arcs whatever; and they are sufficient to indicate how innumerable others of the same kind may be obtained.

248. If in the formula for \(\cos (a+b)\) we suppose \(a = b\), it becomes \(\cos 2a = \cos^2 a - \sin^2 a\); and, since \(\cos^2 a = 1 - \sin^2 a\), and \(\sin^2 a = 1 - \cos^2 a\), we have also

\[ \cos 2a = 1 - 2 \sin^2 a = 2 \cos^2 a - 1. \]

By assuming \(b = a\), we get from the formulae for \(\sin (a+b)\), \(\tan (a+b)\), \(\cot (a+b)\), other formulae for the sine, tangent, and co-tangent, of the double of an arc. These, being frequently wanted, may be put together in a table.

(O)

\[ \begin{align*} \cos 2a &= \cos^2 a - \sin^2 a = 2 \cos^2 a - 1 = 1 - 2 \sin^2 a, \\ \sin 2a &= 2 \sin a \cos a, \\ \tan 2a &= \frac{2 \tan a}{1 - \tan^2 a}, \\ \cot 2a &= \frac{1}{2} (\cot a - \tan a). \end{align*} \]

249. From the first formula we may get formulae for the sine and cosine of the half of an arc in terms of the cosine of the whole arc. But we may also express the sine and cosine of the half arc by the sine of the whole arc; thus,

\[ \cos \frac{1}{2} a + \sin \frac{1}{2} a = 1, \]

and \(2 \cos \frac{1}{2} a \sin \frac{1}{2} a = \sin a\),

we have \(\cos^2 \frac{1}{2} a + 2 \cos \frac{1}{2} a \sin \frac{1}{2} a + \sin^2 \frac{1}{2} a = 1 + \sin a\),

and \(\cos^2 \frac{1}{2} a - 2 \cos \frac{1}{2} a \sin \frac{1}{2} a + \sin^2 \frac{1}{2} a = 1 - \sin a\);

whence, taking the square roots,

\[ \begin{align*} \cos \frac{1}{2} a + \sin \frac{1}{2} a &= \sqrt{1 + \sin a}, \\ \cos \frac{1}{2} a - \sin \frac{1}{2} a &= \sqrt{1 - \sin a}; \end{align*} \]

which, by addition and subtraction, give

\[ \begin{align*} \cos \frac{1}{2} a &= \frac{1}{2} [\sqrt{1 + \sin a} + \sqrt{1 - \sin a}], \\ \sin \frac{1}{2} a &= \frac{1}{2} [\sqrt{1 + \sin a} - \sqrt{1 - \sin a}]. \end{align*} \]

Again, because \(\sin a = 2 \sin \frac{1}{2} a \cos \frac{1}{2} a\), therefore

\[ \sin a = \frac{\sin \frac{1}{2} a}{\cos \frac{1}{2} a} \times 2 \cos^2 \frac{1}{2} a; \]

but \(\frac{\sin \frac{1}{2} a}{\cos \frac{1}{2} a} = \tan \frac{1}{2} a\), and \(2 \cos^2 \frac{1}{2} a = 1 + \cos a\) (by No. 1 of O),

therefore \(\sin a = \tan \frac{1}{2} a (1 + \cos a)\),

\[ \begin{align*} \tan \frac{1}{2} a &= \frac{\sin a}{1 + \cos a} = \frac{1 - \cos a}{\sin a}, \\ \cot \frac{1}{2} a &= \frac{1 + \cos a}{\sin a} = \frac{1 - \cos a}{\sin a}; \end{align*} \]

Again, since \(\frac{1 - \cos a}{\sin a} = \frac{1}{\sin a} - \frac{\cos a}{\sin a} = \csc a - \cot a\),

and \(\frac{1 + \cos a}{\sin a} = \frac{1}{\sin a} + \frac{\cos a}{\sin a} = \csc a + \cot a\),

therefore \(\tan \frac{1}{2} a = \csc a - \cot a\),

and \(\cot \frac{1}{2} a = \csc a + \cot a\).

The formulæ investigated in this article may now be brought into one table, viz.

(P)

\[ \begin{align*} \cos \frac{1}{2} a &= \frac{\sqrt{1 + \cos a}}{2} = \frac{1}{2} [\sqrt{1 + \sin a} + \sqrt{1 - \sin a}], \\ \sin \frac{1}{2} a &= \frac{\sqrt{1 - \cos a}}{2} = \frac{1}{2} [\sqrt{1 + \sin a} - \sqrt{1 - \sin a}], \\ \tan \frac{1}{2} a &= \frac{\sin a}{1 + \cos a} = \csc a - \cot a, \\ \cot \frac{1}{2} a &= \frac{\sin a}{1 - \cos a} = \csc a + \cot a. \end{align*} \]

250. We shall now investigate formulæ for the cosines and sines of the multiples of any arc.

By the first of formulæ (E) we find that

\[ \cos (n+1)a = 2 \cos a \cos na - \cos (n-1)a. \]

Let \(x\) denote \(a\); then,

\[ \begin{align*} \cos x &= x, \\ \cos 2x &= 2x \cos x - \cos 0a, \\ \cos 3x &= 2x \cos 2x - \cos x, \\ \cos 4x &= 2x \cos 3x - \cos 2x, \\ &\text{etc.} \end{align*} \]

Hence, by substituting successively for \(\cos a, \cos 2a, \ldots\), their values in the second members of these formulæ,

(Q)

\[ \begin{align*} \cos x &= x, \\ \cos 2x &= 2x^2 - 1, \\ \cos 3x &= 4x^3 - 3x, \\ \cos 4x &= 8x^4 - 8x^2 + 1, \\ \cos 5x &= 16x^5 - 20x^3 + 5x, \\ \cos 6x &= 32x^6 - 48x^4 + 18x^2 - 1, \\ \cos 7x &= 64x^7 - 112x^5 + 56x^3 - 7x, \\ &\text{etc.} \end{align*} \] Algebra. The law of the series is \(2 \cos na =\)

\[ (2x)^n - n(2x)^{n-2} + \frac{n(n-3)}{1 \cdot 2} (2x)^{n-4} \\ - \frac{n(n-4)(n-5)}{1 \cdot 2 \cdot 3} (2x)^{n-6} + \ldots \]

251. The third formula of (E) gives

\[ \sin (n+1)a = 2 \cos a \sin na - \sin (n-1)a. \]

Let \(x\) denote the cosine, and \(y\) the sine of \(a\); then

\[ \begin{align*} \sin a &= y, \\ \sin 2a &= 2x \sin a - \sin 0a, \\ \sin 3a &= 2x \sin 2a - \sin a, \\ \sin 4a &= 2x \sin 3a - \sin 2a, \\ &\vdots \end{align*} \]

Therefore, proceeding as in last article,

\[ (R) \]

\[ \begin{align*} \sin a &= y, \\ \sin 2a &= 2xy, \\ \sin 3a &= y(4x^2 - 1), \\ \sin 4a &= y(8x^2 - 4x), \\ \sin 5a &= y(16x^2 - 12x^2 + 1), \\ \sin 6a &= y(32x^2 - 32x^2 + 6x), \\ \sin 7a &= y(64x^2 - 80x^2 + 24x^2 - 1), \\ &\vdots \end{align*} \]

Here the law of the series is \(\sin na =\)

\[ y[(2x)^n - (n-2)(2x)^{n-2} + \frac{n(n-3)(n-4)}{1 \cdot 2} (2x)^{n-4} \\ - \frac{n(n-4)(n-5)(n-6)}{1 \cdot 2 \cdot 3} (2x)^{n-6} + \ldots]. \]

252. The series for the cosines and sines of the multiples of an arc are arranged according to the descending powers of the cosine of the arc; but we may find others which proceed according to the ascending powers. It is, however, then necessary to distinguish between the even and odd multiples of the arc.

Let \(n\) be odd, and we have, from table (Q),

\[ (S) \]

\[ \begin{align*} \cos a &= x, \\ \cos 3a &= -(3x - 4x^3), \\ \cos 5a &= 5x - 20x^3 + 16x^5, \\ &\vdots \end{align*} \]

and in general, \(\cos na =\)

\[ \pm \left[ nx - \frac{n(n^2 - 1)}{2 \cdot 3} x^3 + \frac{n(n^2 - 1)(n^2 - 9)}{2 \cdot 3 \cdot 4 \cdot 5} x^5 - \ldots \right]. \]

The upper sign is to be taken when \(n\) is any number in the series 1, 5, 9, &c.; that is, when \(n\) has the form \(4m + 1\); and the lower, when \(n\) is one of these, 3, 7, 11, &c., or when it is of the form \(4m + 3\).

Next, for the even multiples of the arc we find

\[ (2S) \]

\[ \begin{align*} \cos 2a &= -(1 - 2x^2), \\ \cos 4a &= + (1 - 8x^2 + 8x^4), \\ \cos 6a &= -(1 - 18x^2 + 48x^4 - 32x^6), \\ &\vdots \end{align*} \]

and in general, \(\cos na =\)

\[ \pm \left[ 1 - \frac{n^2}{2} x^2 + \frac{n(n^2 - 4)}{2 \cdot 3 \cdot 4} x^4 \\ - \frac{n(n^2 - 4)(n^2 - 16)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} x^6 + \ldots \right]. \]

Here the upper sign is to be taken when \(n\) has the form \(4m + 2\), and the lower when it has the form \(4m\).

253. The corresponding expressions for the sines are obtained from table (R), viz.

\[ \text{(T)} \]

\[ \begin{align*} \sin a &= y, \\ \sin 3a &= -(y(1 - 4x^2)), \\ \sin 5a &= y(1 - 12x^2 + 16x^4), \\ &\vdots \end{align*} \]

and in general, \(\sin na =\)

\[ \pm y \left\{ 1 - \frac{n^2 - 1}{2} x^2 + \frac{n(n^2 - 1)(n^2 - 9)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} x^6 + \ldots \right\}; \]

the upper sign to be taken when \(n\) has the form \(4m + 1\), and the lower when it has the sign \(4m + 3\).

Also we have

\[ (2T) \]

\[ \begin{align*} \sin 2a &= 2xy, \\ \sin 4a &= -(y(4x - 8x^3)), \\ \sin 6a &= y(6x - 32x^3 + 32x^5), \\ &\vdots \end{align*} \]

and in general, \(\sin na =\)

\[ \pm \left[ nx - \frac{n(n^2 - 4)}{2 \cdot 3} x^3 + \frac{n(n^2 - 4)(n^2 - 16)}{2 \cdot 3 \cdot 4} x^5 - \ldots \right]; \]

the upper sign to be taken when \(n\) has the form \(4m + 2\), and the lower when it has the form \(4m\).

254. Because \(x^2 = 1 - y^2\), we have, from table (Q),

\[ (V) \]

\[ \begin{align*} \cos a &= x, \\ \cos 3a &= x(1 - 4y^2), \\ \cos 5a &= x(1 - 12y^2 + 16y^4), \\ &\vdots \end{align*} \]

and in general, when \(n\) is an odd number, \(\cos na =\)

\[ x \left\{ 1 - \frac{n^2 - 1}{2} y^2 + \frac{n(n^2 - 1)(n^2 - 9)}{2 \cdot 3 \cdot 4} y^4 \\ - \frac{n(n^2 - 1)(n^2 - 9)(n^2 - 25)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} y^6 + \ldots \right\}; \]

and when \(n\) is even,

\[ (2V) \]

\[ \begin{align*} \cos 2a &= 1 - 2y^2, \\ \cos 4a &= 1 - 8y^2 + 8y^4, \\ \cos 6a &= 1 - 18y^2 + 48y^4 - 32y^6, \\ &\vdots \end{align*} \]

and in general, \(\cos na =\)

\[ 1 - \frac{n^2}{2} y^2 + \frac{n(n^2 - 4)}{2 \cdot 3 \cdot 4} y^4 - \frac{n(n^2 - 4)(n^2 - 16)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} y^6 + \ldots; \]

255. In like manner, for the sines, when \(n\) is odd, we find

\[ (X) \]

\[ \begin{align*} \sin a &= y, \\ \sin 3a &= 3y - 4y^3, \\ \sin 5a &= 5y - 20y^3 + 16y^5, \\ &\vdots \end{align*} \]

and in general, \(\sin na =\)

\[ ny \left\{ 1 - \frac{n^2 - 1}{2 \cdot 3} y^2 + \frac{n(n^2 - 1)(n^2 - 9)}{2 \cdot 3 \cdot 4 \cdot 5} y^4 - \ldots \right\}; \]

Again, when \(n\) is even,

\[ (2X) \]

\[ \begin{align*} \sin 2a &= 2xy, \\ \sin 4a &= x(4y - 8y^3), \\ \sin 6a &= x(6y - 32y^3 + 32y^5), \\ &\vdots \end{align*} \] And in general, \(\sin na = \frac{n(n^2 - 4)}{2 \cdot 3} y^3 + \frac{n(n^2 - 4)(n^2 - 16)}{2 \cdot 3 \cdot 4 \cdot 5} y^5 + \ldots\).

256. We owe to Vieta the formulae in tables (Q), (R), (2 V), and (X). He, however, enunciated them as properties of the chords of arcs, to which they may be transformed, by considering that chord \(a = 2 \sin \frac{1}{2} a\), and chord \((\pi - a) = \sin\) chord \(a = 2 \cos \frac{1}{2} a\); he did not indicate the general law of the series, but merely showed how the cosines and sines of the multiple arcs might be formed one from another. John Bernoulli gave, in the Leipsic Acts for 1701, a general formula for the chords of multiple arcs, but without demonstration. This is equivalent to table (R). Afterwards, James Bernoulli gave, in the Memoirs of the Academy of Sciences for 1702, two formulae for the chords of multiple arcs; these answer to the general formulae of tables (2 V) and (X); but the latter had previously been given by Newton in his first letter to Oldenburg, the secretary of the Royal Society.

The general formulae of tables (Q), (R), (X), and (2 X), and these only, are given in the Introductio in Analysis Infinitorum of Euler; and the whole are given by Lagrange, in his Lecons sur le Calcul des Fonctions, where they are also strictly demonstrated by his calculus, which is equivalent to the differential calculus.

257. If in the formula

\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]

we make \(b = na\), it becomes

\[ \tan(n + 1)a = \frac{\tan na + \tan a}{1 - \tan na \tan a} \]

Let us make \(\tan a = t\), then, making \(a\) equal to the numbers 2, 3, 4, &c., successively, and substituting for \(\tan na\) its value in \(\tan(n + 1)a\), we get

\[ \tan a = t, \] \[ \tan 2a = \frac{t^2}{1 - t^2}, \] \[ \tan 3a = \frac{3t - t^3}{1 - 3t^2}, \] \[ \tan 4a = \frac{4t - 4t^3}{1 - 6t^2 + t^4}, \] \[ \tan 5a = \frac{5t - 10t^3 + 5t^5}{1 - 10t^2 + 5t^4}, \] \[ \text{&c.} \]

In general, if we put \(a = \frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}\), &c.; that is, if \(\alpha, \beta, \gamma, \delta, \epsilon, \&c.\) denote the co-efficients of the second, third, fourth, and following terms of a binomial raised to the \(n\)th power, we have

\[ \tan na = \frac{at - \beta t^3 + \gamma t^5 - \delta t^7 + \&c.}{1 - \beta t^2 + \gamma t^4 - \delta t^6 + \&c.} \]

This formula was first given by John Bernoulli, in the Leipsic Acts for 1722.

258. We shall next investigate formulæ for the successive integer powers of the cosine and sine of an arc.

By formula (1) of (E) we have, making \(n = 1\),

\[ 2 \cos^2 a = \cos 2a + 1. \]

Let both sides be multiplied by \(2 \cos a\), the result is

\[ 4 \cos^3 a = 2 \cos a \cos 2a + 2 \cos a; \]

but, by the same formulæ (E),

\[ 2 \cos a \cos 2a = \cos 3a + \cos a; \]

therefore, \(4 \cos^3 a = \cos 3a + 3 \cos a\).

Again, multiplying both sides by \(2 \cos a\), we have

\[ 8 \cos^4 a = 2 \cos a \cos 3a + 6 \cos a \cos a. \]

Now, \(2 \cos a \cos 3a = \cos 4a + \cos 2a\),

and \(6 \cos a \cos a = 3 \cos 2a + 3\);

therefore, \(8 \cos^4 a = \cos 4a + 4 \cos 2a + 3\).

By proceeding in this way, multiplying both sides of each new formula by \(2 \cos a\), and substituting \(\cos(n + 1)a\),

\[ \cos(n - 1)a\) for \(2 \cos a \cos na\), there is obtained a series of formulæ for the successive powers of the cosine of an arc, in terms of the cosines of the multiples of the arc.

These are given in the following table:

\[ \begin{align*} \cos a &= \cos a, \\ 2 \cos^2 a &= \cos 2a + 1, \\ 4 \cos^3 a &= \cos 3a + 3 \cos a, \\ 8 \cos^4 a &= \cos 4a + 4 \cos 2a + 3, \\ 16 \cos^5 a &= \cos 5a + 5 \cos 3a + 10 \cos a, \\ 32 \cos^6 a &= \cos 6a + 6 \cos 4a + 15 \cos 2a + 10, \\ 64 \cos^7 a &= \cos 7a + 7 \cos 5a + 21 \cos 3a + 35 \cos a, \\ &\text{&c.} \end{align*} \]

The general law of the series is \(2^{n-1} \cos^n a = \cos na + \frac{n(n-1)}{1 \cdot 2} \cos(n-2)a + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} \cos(n-4)a + \&c.\)

The series is to be continued until we come to a negative arc, and if \(n\) be an even number, the half of the coefficient of the last term (viz. that in which the arc \(= 0\)) is to be taken.

259. Next, for the powers of the sine; from (2) of (E), we have \(2 \sin^2 a = \cos 2a + 1\);

multiply both sides by \(2 \sin a\); then,

\[ 4 \sin^3 a = 2 \sin a \cos 2a + 2 \sin a. \]

Now, by formula (4) of (E),

\[ 2 \sin a \cos 2a = \sin 3a - \sin a; \]

therefore \(4 \sin^3 a = \sin 3a + 3 \sin a\).

Again, multiplying both sides by \(2 \sin a\),

\[ 8 \sin^4 a = 2 \sin a \sin 3a + 6 \sin a \sin a. \]

But by formula (2) of (E),

\[ 2 \sin a \sin 3a = \cos 4a + \cos 2a, \]

and \(6 \sin a \sin a = 3 \cos 2a + 3\);

therefore \(8 \sin^4 a = \cos 4a + 4 \cos 2a + 3\).

In this way we may form the following table:

\[ \begin{align*} \sin a &= \sin a, \\ 2 \sin^2 a &= \cos 2a + 1, \\ 4 \sin^3 a &= \sin 3a + 3 \sin a, \\ 8 \sin^4 a &= \cos 4a + 4 \cos 2a + 3, \\ 16 \sin^5 a &= \sin 5a + 5 \sin 3a + 10 \sin a, \\ 32 \sin^6 a &= \cos 6a + 6 \cos 4a + 15 \cos 2a + 10, \\ 64 \sin^7 a &= \sin 7a + 7 \sin 5a + 21 \sin 3a + 35 \sin a. \end{align*} \]

In general, when \(n\) is an odd number, \(2^{n-1} \sin^n a = \sin na + \frac{n(n-1)}{1 \cdot 2} \sin(n-2)a + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} \sin(n-4)a + \&c.\)

The upper sign is to be taken when \(n\) is a number in the series 1, 5, 9, &c.; that is, when \(n\) has the form \(4m + 1\); and the lower when \(n\) is one of the intermediate odd numbers, 3, 7, 11, &c. In either case the series terminates with a multiple of sin. \(a\).

When \(n\) is an even number, then \(2^n - 1 \sin^a = \cos. na - n \cos. (n - 2) a + \frac{n(n-1)}{1 \cdot 2} \cos. (n - 4) a - \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} \cos. (n - 6) a + \ldots\)

Here the upper sign applies when \(n\) is an even number in the series 2, 6, 10, &c.; that is, when \(\frac{n}{2}\) is an odd number; and the lower when \(n\) has the form \(4m\). In either case the series is to be continued until a term contain cos. \((0a)\) (which is = 1), and then the numeral co-efficient must be divided by 2.

260. The same formula (E) serve to resolve any expression of the form \(\cos^a \sin^a\) into sines and cosines of multiples of \(a\); thus,

\[2 \cos. a \sin. a = \sin. 2a;\] \[4 \cos^2 a \sin. a = 2 \cos. a \sin. 2a,\] \[= \sin. 3a + \sin. a;\] \[4 \cos. a \sin^2 a = 2 \sin. a \sin. 2a,\] \[= - \cos. 3a + \cos. a.\]

In like manner we find

\[8 \cos. a \sin^3 a = - \sin. 4a + 2 \sin. 2a,\] \[8 \cos^2 a \sin^2 a = - \cos. 4a + 1,\] \[8 \cos^3 a \sin. a = + \sin. 4a + 2 \sin. 2a.\]

\[16 \cos. a \sin^4 a = \cos. 5a - 3 \cos. 3a + 2 \cos. a,\] \[16 \cos^2 a \sin^3 a = - \sin. 5a + \sin. 3a + 2 \sin. a,\] \[16 \cos^3 a \sin^2 a = - \cos. 5a - \cos. 3a + 2 \cos. a,\] \[16 \cos^4 a \sin. a = \sin. 5a + 3 \sin. 3a + 2 \sin. a.\]

261. The introduction of the imaginary symbol \(\sqrt{-1}\) into analysis, has given great assistance in all investigations connected with the calculus of sines. Let \(x\) denote the cosine of any arc \(a\), and \(y\) its sine; then, by formulæ (B),

\[\cos. (n+1)a = x \cos. na - y \sin. na,\] \[\sin. (n+1)a = y \cos. na + x \sin. na.\]

Let \(v\) denote a quantity at present indefinite, but to be determined in the course of the investigation; then, from formula (2),

\[v \sin. (n+1)a = vy \cos. na + vx \sin. na.\]

The sum and difference of formulæ (1) and (3) give us

\[\cos. (n+1)a + v \sin. (n+1)a = (x + vy) \cos. na + \left(\frac{x}{v} - y\right) v \sin. na,\] \[\cos. (n+1)a - v \sin. (n+1)a = (x - vy) \cos. na - \left(\frac{x}{v} + y\right) v \sin. na.\]

Let us now assume that \(v^2 = -1\), so that \(v = \sqrt{-1}\), and \(v = -\frac{1}{v}\); the two last formulæ will become

\[\cos. (n+1)a + v \sin. (n+1)a = (x + vy) \cos. na + \left(\frac{x}{v} - y\right) v \sin. na,\] \[\cos. (n+1)a - v \sin. (n+1)a = (x - vy) \cos. na - \left(\frac{x}{v} + y\right) v \sin. na.\]

By giving to \(n\) the values 0, 1, 2, 3, &c. in succession, we obtain from the first of these,

\[\cos. a + v \sin. a = x + vy,\] \[\cos. 2a + v \sin. 2a = (x + vy)^2,\] \[\cos. 3a + v \sin. 3a = (x + vy)^3,\]

and, in general, \(m\) being any whole number,

\[\cos. ma + v \sin. ma = (x + vy)^m.\]

The other equation gives us

\[\cos. a - v \sin. a = x - vy,\] \[\cos. 2a - v \sin. 2a = (x - vy)^2,\] \[\cos. 3a - v \sin. 3a = (x - vy)^3,\]

and in general,

\[\cos. ma - v \sin. ma = (x - vy)^m.\]

Formulæ (4) and (5), when the quantities denoted by \(x, y,\) and \(v\), have been restored, may stand thus:

(B 2)

\[\cos. na + \sqrt{-1} \sin. na = (\cos. a + \sqrt{-1} \sin. a)^n,\] \[\cos. na - \sqrt{-1} \sin. na = (\cos. a - \sqrt{-1} \sin. a)^n.\]

These have been investigated on the hypothesis that \(n\) is a whole positive number, but they are also true when \(n\) is a fraction, or negative; for we have manifestly

\[\cos. a + \sqrt{-1} \sin. a = (\cos. na + \sqrt{-1} \sin. na)^{\frac{1}{n}},\] \[\cos. a - \sqrt{-1} \sin. a = (\cos. na - \sqrt{-1} \sin. na)^{\frac{1}{n}};\]

therefore, raising these equal expressions to the power \(m\), we have

\[\cos. ma + \sqrt{-1} \sin. ma = (\cos. na + \sqrt{-1} \sin. na)^{\frac{m}{n}},\] \[\cos. ma - \sqrt{-1} \sin. ma = (\cos. na - \sqrt{-1} \sin. na)^{\frac{m}{n}};\]

and, putting \(na = a'\), and therefore \(ma = \frac{m}{n} a'\),

\[\cos. \frac{m}{n} a' + \sqrt{-1} \sin. \frac{m}{n} a' = (\cos. a' + \sqrt{-1} \sin. a')^{\frac{m}{n}},\] \[\cos. \frac{m}{n} a' - \sqrt{-1} \sin. \frac{m}{n} a' = (\cos. a' - \sqrt{-1} \sin. a')^{\frac{m}{n}}.\]

Exactly in the same way it may be proved that

\[\cos. \frac{m}{n} a' - \sqrt{-1} \sin. \frac{m}{n} a' = (\cos. a' - \sqrt{-1} \sin. a')^{\frac{m}{n}}.\]

Thus it appears that formulæ (B 2) are true, whether \(n\) be whole or fractional.

Again, we have manifestly

\[\cos. na + \sqrt{-1} \sin. na = (\cos. a + \sqrt{-1} \sin. a)^n.\]

Let the numerator and denominator of the first member of this equation be multiplied by \(\cos. na - \sqrt{-1} \sin. na\); then, observing that

\[(\cos. na + \sqrt{-1} \sin. na) (\cos. na - \sqrt{-1} \sin. a) = \cos^2 na + \sin^2 na = 1,\]

also, that \(\cos. na = \cos. (-na)\), and \(-\sin. na = \sin. (-na)\), we have

\[\cos. (-na) - \sqrt{-1} \sin. (-na) = (\cos. a + \sqrt{-1} \sin. a)^n.\]

Exactly in the same way it may be shown, that

\[\cos. (-na) - \sqrt{-1} \sin. (-na) = (\cos. a - \sqrt{-1} \sin. a)^n;\]

so that the formulæ are universally true, \(n\) being either a positive or negative whole number or fraction. We may even extend the proof to the case of \(n\), an irrational quantity; for every such quantity may be expressed to any degree of nearness by numbers.

262. These very remarkable and important expressions (B 2) were first found by Abraham de Moivre, and appeared in his Miscellanea Analytica, which was published in 1730. He was led to them by considering the analogy between the circle and hyperbola; but we owe to Euler their introduction into analysis as elementary formulæ. By taking their sum and difference, we obtain them in a different shape, as follows:

(C 2)

\[\cos. na = (\cos. a + \sqrt{-1} \sin. a)^n + (\cos. a - \sqrt{-1} \sin. a)^n,\] \[\sin. na = \frac{(\cos. a + \sqrt{-1} \sin. a)^n - (\cos. a - \sqrt{-1} \sin. a)^n}{2\sqrt{-1}}.\] Algebra. In either case, they are purely symbolical expressions of the kind found by Cardan's rule for the roots of a cubic equation belonging to the irreducible case. (Sect. XL.) Like them, they cannot be immediately applied to calculation; but when treated by the rules of analysis, they reveal some of the most elegant and recondite theorems in geometry.

263. The formulae (C 2) admit of an immediate application to the determination of the cosine and sine of any multiple of an arc. Let \( x = \cos a, y = \sin a \), so that

\[ \begin{align*} \cos na &= \frac{(x + \sqrt{-1}y)^n + (x - \sqrt{-1}y)^n}{2}, \\ \sin na &= \frac{(x + \sqrt{-1}y)^n - (x - \sqrt{-1}y)^n}{2\sqrt{-1}}. \end{align*} \]

Now, by the binomial theorem,

\[ (x + \sqrt{-1}y)^n = x^n + nx^{n-1}y\sqrt{-1} - \frac{n(n-1)}{1 \cdot 2}x^{n-2}y^2 + \cdots, \]

\[ (x - \sqrt{-1}y)^n = x^n - nx^{n-1}y\sqrt{-1} - \frac{n(n-1)}{1 \cdot 2}x^{n-2}y^2 + \cdots, \]

Hence, taking half the sum and difference of these series, and in the latter case dividing by \(\sqrt{-1}\), we obtain

\[ \begin{align*} \cos na &= \frac{n(n-1)}{1 \cdot 2}x^{n-2}y^2 + \cdots, \\ \sin na &= \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}x^{n-3}y^3 + \cdots. \end{align*} \]

These series terminate as often as \( n \) is a whole positive number. They were first given in 1701, by John Bernoulli, in the Leipsic Acts, but without demonstration. They also appear in letter 129 of the Commercium Epistolicum of Leibnitz and Bernoulli, and in the Sections Coniques of M. de l'Hôpital. It is remarkable that John Bernoulli should have fallen on these beautiful theorems, and yet missed the formulae of De Moivre, which are easily deducible from them, but which were not discovered until twenty years later.

If it be observed that \(\sin a = \tan a \cos a\), and \(\sin na = \tan na \cos na\); and if, after substitution, the second formula be divided by the first, the result, after due reduction, will be the expression for the tangent which was given in sect. 257.

264. The same powerful instrument of analysis might enable us to investigate all the general expressions for the cosines and sines of multiple arcs from (R) to (2 X); but we shall rather reserve them to show the application of the differential calculus, or method of fluxions. We shall, however, now exemplify their use in the investigation of general series for the powers of the cosine and sine of an arc, in terms of the cosines and sines of its multiples.

Let us put \( p = \cos a + \sqrt{-1} \sin a, q = \cos a - \sqrt{-1} \sin a \);

then \( pq = \cos^2 a + \sin^2 a = 1 \), and \( q = \frac{1}{p} \).

De Moivre's formulae may now be expressed thus:

\[ \begin{align*} 2 \cos na &= p^n + q^n = p^n + \frac{1}{p^n}, \\ 2 \sqrt{-1} \sin na &= p^n - q^n = p^n - \frac{1}{p^n}; \end{align*} \]

and since \( 2 \cos a = p + q \), therefore, putting \( c, c', c'' \), &c., for the numeral co-efficients of the second, third, fourth, &c., terms of a binomial, we have

\[ (2 \cos a)^n = p^n + c' p^{n-1} q + c'' p^{n-2} q^2 + c''' p^{n-3} q^3 + \cdots. \]

Because \( 2 \cos a = q + p \), we have also

\[ (2 \cos a)^n = q^n + c' q^{n-1} p + c'' q^{n-2} p^2 + c''' q^{n-3} p^3 + \cdots. \]

Taking now the sum of both series, and observing that \( pq = 1 \), we obtain

\[ 2(2 \cos a)^n = p^n + q^n + c'(p^{n-2} + q^{n-2}) + c''(p^{n-4} + q^{n-4}) + \cdots. \]

Now,

\[ \begin{align*} p^n + q^n &= 2 \cos na, \\ p^{n-2} + q^{n-2} &= 2 \cos (n-2)a, \\ p^{n-4} + q^{n-4} &= 2 \cos (n-4)a, \\ p^{n-6} + q^{n-6} &= 2 \cos (n-6)a, \\ &\vdots \end{align*} \]

Therefore, substituting and dividing by 2,

\[ (2 \cos a)^n = \cos na + c' \cos (n-2)a + c'' \cos (n-4)a + \cdots. \]

By continuing the series, we come to the cosines of negative arcs, which are exactly the same as the cosines of positive arcs having the same co-efficient: Thus, \(\cos (m-n)a = \cos (n-m)a\), for each is equal to \(\cos ma \cos na + \sin ma \sin na\). Now, in the series which is the expansion of \((p+q)^n\), the terms at equal distances from the first and last have the same co-efficient; therefore the series for \((2 \cos a)^n\) must have the same property; and further, the cosines of the arcs at equal distances from the extremes are, as has been just proved, equal; we may therefore omit the terms which are the cosines of negative arcs, and in their stead take the doubles of these, which are the cosines of positive arcs. The above formula will then be abbreviated to

\[ (2 \cos a)^n = 2[\cos na + c' \cos (n-2)a + c'' \cos (n-4)a + \cdots]. \]

When \( n \) is an even number, the series has a term at equal distances from both extremes, which, being single in the expansion, must not be doubled; but when \( n \) is an odd number, the cosines of the negative arcs are equal in number to those of the positive arcs. Lastly, by considering that the cosine of \((n-n)a = 1\), the truth of the general formulae of table (Z) will be sufficiently obvious.

265. The general formulae of table (A 2) may be established by a process quite similar to the above, by the expression \( 2\sqrt{-1} \sin a = p - q \); but more easily by substituting \( \pi - a' \) instead of \( a \), and remarking that \(\cos (\frac{1}{2}\pi - a') = \sin a'\); we have thus

\[ (2 \sin a')^n = 2 \cos m(\frac{1}{2}\pi - a') + 2c' \cos (n-2)(\frac{1}{2}\pi - a') + 2c'' \cos (n-4)(\frac{1}{2}\pi - a') + \cdots. \]

The terms of this series being all of the form

\[ \cos m(\frac{1}{2}\pi - a') = \cos m\frac{\pi}{2} \cos ma' + \sin m\frac{\pi}{2} \sin ma' \]

will have the form \( \pm \sin ma' \), when \( m \) is an odd number; and the form \( \pm \cos ma' \), when \( m \) is an even number.

266. Before we proceed to other applications of De Moivre's formula, it will be convenient to establish some analytic principles of frequent application, regarding the limits between which certain expressions involving arcs and their sines, tangents, &c. are always contained.

It is an axiom in geometry, that any arc is less than its tangent (as defined sect. 223), but greater than its sine, therefore, \( \frac{a}{n} \) being any arc,

\[ \tan \frac{a}{n} > \frac{a}{n}, \quad \sin \frac{a}{n} < \frac{a}{n}. \]

Multiplying the first of these by \( n \), and the second by sec. \( \frac{a}{n} \), and observing that sec. \( \frac{a}{n} \cdot \sin \frac{a}{n} = \tan \frac{a}{n} \), we have

\[ n \tan \frac{a}{n} > a, \quad n \tan \frac{a}{n} < a \sec \frac{a}{n}. \]

Suppose now \( n \) to increase continually, while the arc \( a \) is invariable, it is manifest, that since \( \frac{a}{n} \) will decrease, and may become less than any assignable arc, sec. \( \frac{a}{n} \) will continually approach to the radius \( = 1 \), and may differ from it by less than any assignable quantity. Hence it follows that \( n \) being supposed to increase continually, the expression \( n \tan \frac{a}{n} \) approaches continually to \( a \), which is its limit.

In like manner, because

\[ n \sin \frac{a}{n} < a, \quad n \sin \frac{a}{n} > a \cos \frac{a}{n}; \]

and because, while \( n \) increases, \( \cos \frac{a}{n} \) approaches to radius \( = 1 \), which is its limit; therefore the limit of \( n \sin \frac{a}{n} \) is the arc \( a \).

Hence, it appears that the arc \( a \) being supposed to decrease continually,

\[ \text{limit of } \frac{a}{\sin a} = 1, \quad \text{and limit of } \frac{a}{\tan a} = 1. \]

267. It is sufficiently evident that the limit of \( \cos \frac{a}{n} \) is \( 1 \); but, in what is to follow, it will be necessary to find the limit of \( \cos^n \frac{a}{n} \) when \( n \) increases indefinitely.

By a known formula, \( \cos \frac{a}{n} = \sqrt{1 - \sin^2 \frac{a}{n}} \). Let us put \( w = n \sin \frac{a}{n} \); then \( w \) will be less than \( a \), and

\[ \cos^n \frac{a}{n} = \left(1 - \frac{w^2}{n^2}\right)^{\frac{n}{2}}. \]

By the binomial theorem, the second member of this equation is

\[ 1 - \frac{w^4}{2n^2} + \left(1 - \frac{2}{n}\right) \frac{w^4}{2^2 \cdot 4 \cdot n^2} - \left(1 - \frac{2}{n}\right) \left(1 - \frac{4}{n}\right) \frac{w^6}{2^3 \cdot 4 \cdot 6 \cdot n^3} + \ldots \]

But since, by hypothesis, \( n \) may be greater than any assignable number, and \( w \) is not greater than \( a \), every term of this series, except the first, will evidently be \( = 0 \), when \( n \) is indefinitely great; therefore, in that case, \( \cos^n \frac{a}{n} \) accurately \( = 1 \).

268. Resuming now De Moivre's formulæ (261), we have

\[ \cos a + \sqrt{-1} \sin a = \left(\cos \frac{a}{n} + \sqrt{-1} \sin \frac{a}{n}\right)^n, \]

\[ \cos a - \sqrt{-1} \sin a = \left(\cos \frac{a}{n} - \sqrt{-1} \sin \frac{a}{n}\right)^n. \]

Let \( t \) denote the tangent of the arc \( \frac{a}{n} \); then observing that \( \sin \frac{a}{n} = \cos \frac{a}{n} \tan \frac{a}{n} \), we have

\[ \cos a + \sqrt{-1} \sin a = \cos^n \frac{a}{n} \left(1 + \sqrt{-1} t\right)^n, \]

\[ \cos a - \sqrt{-1} \sin a = \cos^n \frac{a}{n} \left(1 - \sqrt{-1} t\right)^n. \]

By the binomial theorem \( \left(1 + \sqrt{-1} t\right)^n = 1 - \left(1 - \frac{1}{n}\right) \frac{n^2}{1^2} + \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \frac{n^4}{1^2 \cdot 2^2} - \ldots \)

\[ + \sqrt{-1} \left\{nt - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \frac{n^3}{1^2 \cdot 2^2} + \ldots \right\}. \]

(Here we have resolved the whole expansion into two parts; one entirely free from the imaginary sign, and the other having that sign as a common factor of all its terms.)

Let us denote the first of these series by \( P \), and the second, setting aside its imaginary factor \( \sqrt{-1} \), by \( Q \); then

\[ \left(1 + \sqrt{-1} t\right)^n = P + \sqrt{-1} Q. \]

By a like process we find

\[ \left(1 - \sqrt{-1} t\right)^n = P - \sqrt{-1} Q; \]

and hence again

\[ \cos a + \sqrt{-1} \sin a = \cos^n \frac{a}{n} \left(P + \sqrt{-1} Q\right); \]

\[ \cos a - \sqrt{-1} \sin a = \cos^n \frac{a}{n} \left(P - \sqrt{-1} Q\right). \]

and from these, by adding and subtracting,

\[ \cos a = \cos^n \frac{a}{n} P, \quad \sin a = \cos^n \frac{a}{n} Q; \]

or, substituting for \( P, Q, \) and \( t \), their values,

\[ \cos a = \cos^n \frac{a}{n} \left\{1 - \left(1 - \frac{1}{n}\right) \frac{n^2}{1^2} + \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \frac{n^4}{1^2 \cdot 2^2} - \ldots \right\} \]

\[ + \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \frac{n^3}{1^2 \cdot 2^2} + \ldots \]

\[ \sin a = \cos^n \frac{a}{n} \left\{nt - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \frac{n^3}{1^2 \cdot 2^2} + \ldots \right\} \]

These formulæ must hold true, whatever value we give to \( n \). But let us now suppose \( n \) indefinitely great; then, from what was proved in last section,

\[ \cos^n \frac{a}{n} = 1, \quad nt = n \tan \frac{a}{n} = a, \quad \frac{1}{n} = 0, \quad \frac{2}{n} = 0, \quad \ldots \]

We have therefore, by substitution,

\[ \cos a = 1 - \frac{a^2}{1^2} + \frac{a^4}{1^2 \cdot 2^2} - \ldots \]

\[ \sin a = a - \frac{a^3}{1^2 \cdot 2^2} + \frac{a^5}{1^2 \cdot 2^2 \cdot 3^2} - \ldots \]

These elegant theorems, discovered by Newton, were among the first fruits of his analytical methods.

269. It has been shown (sect. 177) that if \( e \) denote the number \( 2.7182818 \), viz. that whose Napierian logarithm is \( 1 \), then

\[ e^{x} = 1 + x + \frac{x^2}{1^2} + \frac{x^3}{1^2 \cdot 2^2} + \frac{x^4}{1^2 \cdot 2^2 \cdot 3^2} + \ldots \]

Now, \( a \) being any arc, let \( x = a \sqrt{-1} \); then, by substituting and separating the real and imaginary terms, so as to form two series, we have

\[ e^{a \sqrt{-1}} = \left\{1 - \frac{a^2}{2} + \frac{a^4}{1^2 \cdot 2^2 \cdot 3^2} - \ldots \right\} \]

\[ + \sqrt{-1} \left\{a - \frac{a^3}{1^2 \cdot 2^2} + \frac{a^5}{1^2 \cdot 2^2 \cdot 3^2} - \ldots \right\} \]

Again, putting \( x = -a \sqrt{-1} \), we obtain, in like manner, But it was found that

\[ \cos a = 1 - \frac{a^2}{1 \cdot 2} + \frac{a^4}{1 \cdot 2 \cdot 3 \cdot 4}, \ldots \]

\[ \sin a = a - \frac{a^3}{1 \cdot 2 \cdot 3} + \frac{a^5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}, \ldots \]

Hence it follows that

\[ e^{i\sqrt{-1}} = \cos a + i \sin a, \]

\[ e^{-i\sqrt{-1}} = \cos a - i \sin a; \]

and consequently that

\[ \cos a = \frac{e^{i\sqrt{-1}} + e^{-i\sqrt{-1}}}{2}, \quad \sin a = \frac{e^{i\sqrt{-1}} - e^{-i\sqrt{-1}}}{2i}. \]

Here the cosine and sine are expressed by imaginary exponentials. Lagrange considered these formulae to be the happiest analytical inventions of the age. The series for the cosine and sine from which they have been here deduced had been given by Newton before the end of the 17th century. They might therefore then have been found, and a perfection given to this subject which it did not attain until fifty years later, by the labours of Euler.

270. From formulae (G 2) we find

\[ \cos a + i \sin a = e^{i\sqrt{-1}}, \]

\[ \cos a - i \sin a = e^{-i\sqrt{-1}} = e^{i\sqrt{-1}}; \]

or, putting, instead of sin \( a \), its equivalent tan \( a \) cos \( a \), and leaving out the common factor cos \( a \),

\[ \frac{1 + i \tan a}{1 - i \tan a} = e^{i\sqrt{-1}}. \]

Now, observing that \( e \) is the base or radical number of Napier's logarithms, it follows, from the theory of logarithms (Sect. XIX.), that

\[ 2a\sqrt{-1} = \text{Nap. log. } \frac{1 + i \tan a}{1 - i \tan a}. \]

But it was shown, sect. 167, that \( e \) being any number,

\[ \log \frac{1 + r}{1 - r} = 2 \left( \frac{r^3}{3} + \frac{r^5}{5} + \frac{r^7}{7} + \ldots \right). \]

Put now \( i \tan a \) instead of \( r \), and the result equal to \( 2a\sqrt{-1} \), and divide both sides by \( 2\sqrt{-1} \), and we have

\[ a = \tan a - \frac{1}{2} \tan^3 a + \frac{1}{3} \tan^5 a - \frac{1}{4} \tan^7 a + \ldots \]

This elegant expression for an arc of a circle was first found by James Gregory, from whom it was received by Collins, an eminent mathematician of that period, in the beginning of 1671. It was sent to Leibnitz in 1675 (see Commercium Epistolicum, p. 98 and 120), and it appears that this celebrated person had communicated the same series to his friends on the Continent as his own discovery, and even sent it to England in 1676. (Com. Ep. 133.) He was accused by the English mathematicians of appropriating to himself the discovery of Gregory; but this, of course, he denied. (Com. Ep. Leibnitz et Bernoulli, tom. ii. p. 313.) We have been thus particular, because Lagrange has given it to Leibnitz. (Calcul des Fonctions, p. 68.) John Bernoulli, however, in expressing his belief that Leibnitz had found the series himself, admits that it was first discovered by Gregory.

271. If we suppose \( a \) to be an arc of 45°, the Gregorian Algebra series gives

\[ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots \]

for one eighth of the circumference. This, however, converges too slowly to be of any practical use. Newton found a different series, viz.

\[ 1 + \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \frac{1}{6} - \frac{1}{7} - \frac{1}{8} + \ldots \]

for the length of the quadrantial arc of a circle of which the chord \( = 1 \). This converges somewhat faster than Gregory's series; but Newton says that the addition of no fewer than 5000,000,000 of the terms would be required to give the length of the quadrant true to twenty decimal places of figures; a labour which would require about one thousand years.

272. The simplicity of Gregory's series is a great recommendation; and as the determination of the ratio of the diameter of a circle to the circumference is a problem of primary importance, we shall investigate an auxiliary formula, which will make it the fittest of any for the solution.

Let \( n, x, y \) be three whole numbers, such, that the arc whose tangent is \( \frac{1}{n} \) is equal to the sum of the arcs whose tangents are \( \frac{1}{x} \) and \( \frac{1}{y} \); then, by formulae (F),

\[ \frac{1}{n} = \frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy-1}; \]

hence we find \( y = \frac{nx+1}{x-n} = n + \frac{n^2+1}{x-n}. \)

Now as \( y \) is a whole number, \( n^2+1 \) must be divisible by \( x-n \). Let \( p \) be any divisor of \( n^2+1 \), and \( q \) the quotient, so that \( pq = n^2+1 \); then \( x-n \) may be assumed \( = p \), and \( x = n+p \), and therefore

\[ y = n + \frac{n^2+1}{x-n} = n + p. \]

Hence we have the following theorem:

Let \( n \) be any whole number, and let \( n^2+1 \) be resolved into any two factors \( p \) and \( q \), one of which may be unity.

The arc whose tangent is \( \frac{1}{n} \) is equal to the sum of the arcs whose tangents are \( \frac{1}{n+p} \) and \( \frac{1}{n+q} \).

As a convenient notation, let \( A_n \) denote the arc whose tangent is \( \frac{1}{n} \), and similarly, \( A_{n+p} \) and \( A_{n+q} \), the arcs whose tangents are \( \frac{1}{n+p} \) and \( \frac{1}{n+q} \); our theorem will then stand thus:

\[ A_n = A_{n+p} + A_{n+q}. \]

By giving to \( n \) different values, we form the following table:

| \( n \) | \( n^2+1 \) | \( A_n \) | \( A_{n+p} \) | \( A_{n+q} \) | |-------|-------------|----------|--------------|--------------| | 1 | 2 | \( A_1 \) | \( A_2 \) | \( A_1 \) | | 2 | 5 | \( A_2 \) | \( A_5 \) | \( A_3 \) | | 3 | 10 | \( A_3 \) | \( A_{10} \) | \( A_7 \) | | 4 | 17 | \( A_4 \) | \( A_{17} \) | \( A_{13} \) | | 5 | 26 | \( A_5 \) | \( A_{26} \) | \( A_{21} \) | | 6 | 37 | \( A_6 \) | \( A_{37} \) | \( A_{33} \) | | 7 | 50 | \( A_7 \) | \( A_{50} \) | \( A_{43} \) |

We may proceed with this series to any extent; and from these formulae, by elimination, we obtain the following: \[ \frac{1}{n \tan \frac{a}{n}} = \frac{1}{\tan a} + \frac{1}{2} \tan \frac{a}{2} + \frac{1}{3} \tan \frac{a}{3} + \cdots + \frac{1}{n} \tan \frac{a}{n} \]

Suppose now \( n \) to be increased indefinitely, the series will then consist of an infinite number of terms; and since it was proved in sect. 266 that \( n \tan \frac{a}{n} = a \), we have

\[ \frac{1}{a} = \frac{1}{\tan a} + \frac{1}{2} \tan \frac{a}{2} + \frac{1}{3} \tan \frac{a}{3} + \cdots \]

This very simple and neat expression for the reciprocal of an arc was found by Professor Wallace, and given along with various others in the sixth volume of the *Transactions of the Royal Society of Edinburgh*, about the year 1812. He believed it to be new, and indeed it was not then known in this country; but it had been given before by Euler, in his *Opuscula Analytica*, tom. i. p. 350.

This series converges very fast; for the tangent of the half of an arc being less than half the tangent of the whole arc, as is easily proved, each term is less than one fourth of the term before it. As, however, the series proceeds, the ratio of any two consecutive terms approaches continually to that of 4 to 1; and hence any term somewhat advanced in the series will be nearly three times the sum of all that follow it; and, by this property, as soon as a term is found to be nearly one fourth of that before it, one third of that term will give the sum of all that follow it, very near.

From the formula \( \tan \frac{a}{2} = \cot a - \cot \frac{a}{2} \) (sect. 249), we get

\[ \tan \frac{a}{2} = \sqrt{1 + \cot^2 a} - \cot a = \cot a (\sqrt{1 + \tan^2 a} - 1) \]

But by the binomial theorem

\[ \sqrt{1 + \tan^2 a} = 1 + \frac{1}{2} \tan^2 a - \frac{1}{8} \tan^4 a + \cdots \]

therefore, \( \tan \frac{a}{2} = \frac{1}{2} \tan a - \frac{1}{8} \tan^3 a + \cdots \)

By the first of these expressions we may compute the terms of the series until the seventh or fifth power may be neglected, and then the remaining terms may be more readily computed by the second.

Let \( a \) be a quadrant \( = \frac{1}{2} \pi \). In this case, \( \frac{1}{\tan a} = 0 \).

The calculation of the length of the arc will stand thus:

\[ \begin{align*} \tan \frac{a}{2} &= 5000000000 \\ \tan \frac{a}{4} &= 41421356237 \\ \tan \frac{a}{8} &= 19891236738 \\ \tan \frac{a}{16} &= 0894914036 \\ \tan \frac{a}{32} &= 04912684977 \\ \tan \frac{a}{64} &= 02454862211 \\ \tan \frac{a}{128} &= 01227246238 \\ \tan \frac{a}{256} &= 00613600016 \\ \tan \frac{a}{512} &= 00306797120 \\ \end{align*} \]

\[ \begin{align*} \frac{1}{2} \tan \frac{a}{2} &= 50000000000 \\ \frac{1}{8} \tan \frac{a}{4} &= 10355339059 \\ \frac{1}{32} \tan \frac{a}{8} &= 02486404592 \\ \frac{1}{128} \tan \frac{a}{16} &= 006153571271 \\ \frac{1}{512} \tan \frac{a}{32} &= 00153521406 \\ \frac{1}{2048} \tan \frac{a}{64} &= 00038357222 \\ \frac{1}{8192} \tan \frac{a}{128} &= 00009587861 \\ \frac{1}{32768} \tan \frac{a}{256} &= 00000396875 \\ \frac{1}{131072} \tan \frac{a}{512} &= 000000599213 \\ \frac{1}{524288} \tan \frac{a}{1024} &= 000000199737 \\ \end{align*} \]

\[ \begin{align*} \frac{1}{2} \tan \frac{a}{2} &= 50000000000 \\ \frac{1}{8} \tan \frac{a}{4} &= 10355339059 \\ \frac{1}{32} \tan \frac{a}{8} &= 02486404592 \\ \frac{1}{128} \tan \frac{a}{16} &= 006153571271 \\ \frac{1}{512} \tan \frac{a}{32} &= 00153521406 \\ \frac{1}{2048} \tan \frac{a}{64} &= 00038357222 \\ \frac{1}{8192} \tan \frac{a}{128} &= 00009587861 \\ \frac{1}{32768} \tan \frac{a}{256} &= 00000396875 \\ \frac{1}{131072} \tan \frac{a}{512} &= 000000599213 \\ \frac{1}{524288} \tan \frac{a}{1024} &= 000000199737 \\ \end{align*} \]

\[ \begin{align*} \frac{1}{2} \tan \frac{a}{2} &= 50000000000 \\ \frac{1}{8} \tan \frac{a}{4} &= 10355339059 \\ \frac{1}{32} \tan \frac{a}{8} &= 02486404592 \\ \frac{1}{128} \tan \frac{a}{16} &= 006153571271 \\ \frac{1}{512} \tan \frac{a}{32} &= 00153521406 \\ \frac{1}{2048} \tan \frac{a}{64} &= 00038357222 \\ \frac{1}{8192} \tan \frac{a}{128} &= 00009587861 \\ \frac{1}{32768} \tan \frac{a}{256} &= 00000396875 \\ \frac{1}{131072} \tan \frac{a}{512} &= 000000599213 \\ \frac{1}{524288} \tan \frac{a}{1024} &= 000000199737 \\ \end{align*} \] It appears from Sect. X. that the resolution of any equation depends on our being able to resolve it into its factors, whether of the first or second degree. In one class of equations this can be effected by the calculus of sines, as will appear from the following analysis:

Since $\cos(m+1)x = 2\cos x \cos mx - \cos(m-1)x$,

let $2\cos x = v + \frac{1}{v}$;

then $2\cos 2x = 4\cos x \cos x - 2\cos 0x = v^2 + \frac{1}{v^2}$,

$2\cos 3x = 4\cos x \cos 2x - 2\cos x = v^3 + \frac{1}{v^3}$,

$2\cos 4x = 4\cos x \cos 3x - 2\cos 2x = v^4 + \frac{1}{v^4}$,

&c.

and in general, $2\cos mx = v^m + \frac{1}{v^m}$;

hence we get

$v^{2m} - 2v^m \cos mx + 1 = 0$.

The equation $v + \frac{1}{v} = 2\cos x$ gives

$v^2 - 2v \cos x + 1 = 0$.

Now, since the two equations

$v^{2m} - 2v^m \cos mx + 1 = 0$,

$v^2 - 2v \cos x + 1 = 0$,

require to be both satisfied at the same time, they must have at least one common root. Let $a$ be that root, then $\frac{1}{a}$ will also be a root of both; for putting them under the form

$v^m + \frac{1}{v^m} - 2\cos mx = 0$,

$v + \frac{1}{v} - 2\cos x = 0$,

the results will be the very same whether we substitute $a$ or $\frac{1}{a}$ for $v$.

Now the quadratic equation

$v^2 - 2v \cos x + 1 = 0$

can only have two roots, therefore these must also be roots of the equation

$v^{2m} - 2v^m \cos mx + 1 = 0$;

hence it follows that the trinomial

$v^{2m} - 2v^m \cos mx + 1$

must be divisible by this other expression,

$v^2 - 2v \cos x + 1$;

or, putting $mx = \phi$, and therefore $x = \frac{\phi}{m}$, that

$v^{2m} - 2v^m \cos \phi + 1$

must have for a divisor

$v^2 - 2v \cos \frac{\phi}{m} + 1$.

Now we found, sect. 232, that $\sigma$ being put for half the circumference,

$\cos \phi = \cos(zm\pi + \phi)$;

hence it follows, that instead of $\frac{\phi}{m}$, we may take any one of this series of arcs,

$\frac{2\pi}{m} + \frac{\phi}{m}, \frac{4\pi}{m} + \frac{\phi}{m}, \frac{6\pi}{m} + \frac{\phi}{m}, \ldots \text{to } \left\{ \frac{2(m-1)\pi}{m} + \frac{\phi}{m} \right\}$,

for the cosine of $m$ times any one of these is the very same quantity; we may therefore infer that the trinomial

$v^{2m} - 2v^m \cos \phi + 1$

is divisible by each of these trinomial expressions of the second degree, viz.

$v^2 - 2v \cos \frac{\phi}{m} + 1$,

$v^2 - 2v \cos \left( \frac{2\pi}{m} + \frac{\phi}{m} \right) + 1$,

$v^2 - 2v \cos \left( \frac{4\pi}{m} + \frac{\phi}{m} \right) + 1$,

&c.

It is needless to carry the series farther, because the next divisor would involve $\cos \left( \frac{2m}{m} \pi + \frac{\phi}{m} \right)$, which is the same as $\cos \left( 2\pi + \frac{\phi}{m} \right) = \cos \frac{\phi}{m}$, a repetition of the first, and so on. As the number of divisors which differ from one another is $m$, and the trinomial is of the $2m$th degree, it follows from the theory of equations, that the trinomial is the continual product of all the divisors.

This elegant theorem is due to De Moivre, as appears from his Miscellanea Analytica.

If we suppose $\phi = 0$, then $\cos \phi = 1$, and the trinomial $v^{2m} - 2v^m \cos \phi + 1$ becomes in this case

$v^{2m} - 2v^m + 1 = (v^m - 1)^2$, and the factors are

$v^2 - 2v + 1 = (v - 1)^2$,

$v^2 - 2v \cos \frac{2\pi}{m} + 1$,

$v^2 - 2v \cos \frac{4\pi}{m} + 1$,

&c.

If, again, $\phi = \pi$, then $\cos \phi = -1$, and the trinomial

$v^{2m} - 2v^m \cos \phi + 1 = v^{2m} + 2v^m + 1 = (v^m + 1)^2$;

therefore the factors of this quantity are

$v^2 - 2v \cos \frac{\pi}{m} + 1$,

$v^2 - 2v \cos \frac{3\pi}{m} + 1$,

$v^2 - 2v \cos \frac{5\pi}{m} + 1$,

&c.

The last two formulae are the analytic expressions of a very remarkable property of the circle discovered by Cotes, the friend and contemporary of Newton, and called the Cotesian Theorem. It consists of two parts, and may be thus enunciated (see Plate XVIII. fig. 25 and 26):

Let the circumference of any circle be divided into $2m$ equal parts (for example ten), at the points $A_0, A_1, A_2, A_3, A_4, \ldots A_{9}$; and let $A_0C$ be a diameter drawn through $A_0$, one of the points of division; and from any point $P$ in the diameter (fig. 25), or the diameter produced (fig. 26), let straight lines $PA_1, PA_2, PA_3, PA_4, PA_5, PA_6$, be drawn to the points in the circumference which are the first, third, fifth, &c. from $A_0$; and also straight lines $PA_7, PA_8, PA_9$, to the points which are the second, fourth, The continual product of the lines drawn from P to \( A_1, A_3, A_5, \ldots \) (the points marked with odd numbers), is equal to \( r^n + r^m \).

II. The continual product of the lines drawn from P to \( A_0, A_2, A_4, A_6, \ldots \) (the points marked with even numbers), is equal to \( r^n - r^m \), when the point is within the circle (fig. 25), or to \( r^n - r^m \), when it is without (fig. 26).

Join CA_1, and, to simplify, let \( r = 1 \); because the whole circumference is divided into \( 2m \) parts, the arcs \( A_0 A_1 = \frac{\pi}{m}, A_0 A_2 = \frac{2\pi}{m}, A_0 A_3 = \frac{3\pi}{m} \), and so on. Now by the elements of Geometry,

\[ PA_1^2 = PC^2 - 2PC \times CA_1 \cos \frac{\pi}{m} + CA_1^2 = v^2 - 2v \cos \frac{\pi}{m} + 1. \]

In like manner, it will appear that

\[ PA_2^2 = v^2 - 2v \cos \frac{2\pi}{m} + 1, \] \[ PA_3^2 = v^2 - 2v \cos \frac{3\pi}{m} + 1, \] and so on, with respect to the lines drawn to the remaining points \( PA_4, PA_5 \).

Now, remarking that \( PA_0 = v - 1 \), from what has been proved, it appears that the product

\[ PA_1 \times PA_2 \times PA_3 \ldots \text{to } PA_{2n-1} \]

is equal to the product of the trinomials

\[ v^2 - 2v \cos \frac{\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{2\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{3\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{(2m-1)\pi}{m} + 1; \]

and that the product

\[ PA_1^2 \times PA_2^2 \times PA_3^2 \ldots \text{to } PA_{2n-1}^2 \]

is equal to the product of the trinomials,

\[ (v - 1)^2 \text{ or } (1 - v)^2, \] \[ v^2 - 2v \cos \frac{\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{2\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{(2m-1)\pi}{m} + 1. \]

But we have proved that the first of these products is equal to \( (v^m + 1)^2 \), and the second to \( (1 - v^m)^2 = (v^m - 1)^2 \); therefore, taking the square roots, we have

\[ PA_1 \times PA_2 \times PA_3 \times PA_4, \ldots \text{etc.} = v^m + 1, \] \[ PA_0 \times PA_2 \times PA_4 \times PA_6, \ldots \text{etc.} = 1 - v^m \text{ or } -v^m - 1. \]

Thus both parts of the theorem are demonstrated.

This elegant geometrical theorem was found among the papers of Mr Cotes after his death. It had no demonstration, but that was soon supplied by his contemporaries, particularly De Moivre, who gave it the more general form which is expressed analytically in last article. At the time of its discovery, it was greatly esteemed; but now, analytical formulae are found by experience more convenient than geometrical diagrams, and therefore are employed instead of them.

277. If we consider that

\[ \cos \frac{2(m-n)\pi}{m} = \cos(2\pi - \frac{n\pi}{m}) = \cos \frac{2n\pi}{m}, \]

it will readily appear that the factors of the quantity \( (v^m - 1)^2 \), which have been given in sect. 275, may be written thus:

\[ v^2 - 2v + 1 = (v - 1)^2, \] \[ v^2 - 2v \cos \frac{2\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{4\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{2(m-1)\pi}{m} + 1; \]

where the second and last are identical; and the same is true of the third and last but two, the fourth and last but three, and so on, until, in the case of \( m \) an even number, we come to two equal factors, one on each side of the \( (\frac{1}{2}m+1) \)th factor, which being \( v^2 - 2v \cos \frac{m\pi}{m} + 1 \), will be \( v^2 + 2v + 1 = (v + 1)^2 \); but in the case of \( m \) an odd number, we come to two adjoining equal factors, the first of which is \( v^2 - 2v \cos \frac{m-1}{m} \pi + 1 \); therefore, taking the square root of \( (v^m - 1)^2 \), and at the same time rejecting one of each pair of equal factors, we find that when \( m \) is an even number, \( v^m - 1 \) is equal to the product of these quantities,

\[ v - 1, \] \[ v + 1, \] \[ v^2 - 2v \cos \frac{2\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{4\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{m-2}{m} \pi + 1. \]

But when \( m \) is an odd number, the factors are

\[ v - 1, \] \[ v^2 - 2v \cos \frac{2\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{4\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{m-1}{m} \pi + 1. \]

278. Next let us consider the expression \( (v^m + 1)^2 \) and its factors, as given sect. 275. We have for this case

\[ \cos \frac{2m-n\pi}{m} = \cos(2\pi - \frac{n\pi}{m}) = \cos \frac{n\pi}{m}. \]

The factors are therefore

\[ v^2 - 2v \cos \frac{\pi}{m} + 1, \] Here it appears that the first and last factors are identical, also the second and last but one, and so on, until in the case of \( m \) an even number, we come to two adjoining equal factors, the first of which is

\[ v^2 - 2v \cos \frac{m-1}{m} \sigma + 1. \]

But if \( m \) be an odd number, there are two equal factors, and a single one between them, viz.

\[ v^2 - 2v \cos \frac{m}{m} \sigma + 1 = v^2 + 2v + 1 = (v + 1)^2; \]

therefore, taking the square roots of \((v + 1)^2\) and \((v + 1)^2\), and rejecting one of each pair of equal factors, we obtain, in the case of \( m \) an even number, \( v^2 + 1 \), equal to the product of these factors:

\[ v^2 - 2v \cos \frac{\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{3\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{5\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{m-1}{m} \sigma + 1. \]

But when \( m \) is an odd number, the factors are,

\[ v + 1, \] \[ v^2 - 2v \cos \frac{\pi}{m} + 1, \] \[ v^2 - 2v \cos \frac{3\pi}{m} + 1, \] \[ \ldots \] \[ v^2 - 2v \cos \frac{m-2}{m} \sigma + 1. \]

These formulae, exhibiting the decomposition of the expressions \( e^{2m} = 2e^m \cos \phi + 1 \) and \( e^{2m} = 1 \), are of great importance in the Differential Calculus or doctrine of Fluxions, and in the more recondite theories of Analysis.

279. In concluding the calculus of sines, we shall yet give three other examples of its application: the first shall be to the resolution of quadratic equations.

(1.) Let the equation be

\[ x^2 + px = q, \]

where \( p \) and \( q \) are both positive numbers. This equation has two roots, a positive and a negative, of which \( -p \) is the sum, and \( -q \) the product. (Sect. 93.) Let \( \tan \frac{1}{2} v \sqrt{q} \) be the positive root, where \( v \) is an angle to be determined; then \( -\cotan \frac{1}{2} v \sqrt{q} \times \cotan \frac{1}{2} v \sqrt{q} = -q \). To determine \( v \), we have this equation,

\[ \sqrt{q} (\cotan \frac{1}{2} v - \cotan \frac{1}{2} v) = p; \]

but by (3) and (4) of (Q),

\[ \cotan \frac{1}{2} v - \cotan \frac{1}{2} v = 2 \cotan \frac{1}{2} v = \frac{2}{\tan v}; \]

therefore, \( \tan v = \frac{\sqrt{q}}{\frac{1}{2} p} \). Hence we have this rule:

To resolve the equation \( x^2 + px = q \),

find an angle \( v \), such that \( \tan v = \frac{\sqrt{q}}{\frac{1}{2} p} \times \text{radius}; \)

then the roots of the equation are,

\[ x = + \tan \frac{1}{2} v \sqrt{q}, \quad \text{and} \quad x = - \tan \frac{1}{2} v \sqrt{q}. \]

(2.) In like manner we find that in the equation

\[ x^2 - px = q, \]

the angle \( v \) being found as before, the roots of the equation are,

\[ x = + \tan \frac{1}{2} v \sqrt{q}; \quad \text{and} \quad x = - \tan \frac{1}{2} v \sqrt{q}. \]

(3.) When the equation is

\[ x^2 - px = -q, \]

which has two positive roots whose sum is \( p \), and product \( q \) (sect. 93), these may be expressed by \( \tan \frac{1}{2} v \sqrt{q} \), and \( \cotan \frac{1}{2} v \sqrt{q} \). In this case,

\[ \sqrt{q} (\cotan \frac{1}{2} v + \cotan \frac{1}{2} v) = p; \]

but by (3) and (4) of formulae (Q),

\[ \cotan \frac{1}{2} v + \cotan \frac{1}{2} v = 2 \cosec v = \frac{2}{\sin v}; \]

therefore, \( \sin v = \frac{\sqrt{q}}{\frac{1}{2} p} \). Hence, to resolve this case, find an angle \( v \), such that

\[ \sin v = \frac{\sqrt{q}}{\frac{1}{2} p} \times \text{radius}; \]

then \( x = + \sqrt{q} \tan \frac{1}{2} v, \quad x = + \tan \frac{1}{2} v \sqrt{q}. \)

280. The second example is the resolution of a problem in pure algebra, said to have been proposed by the late Professor Porson.

Find \( u, x, y, z \), from these equations,

\[ ux + yz = a, \quad \ldots \quad (A) \] \[ uz + xy = b, \quad \ldots \quad (B) \] \[ uy + zx = c, \quad \ldots \quad (C) \] \[ uz + yz = d, \quad \ldots \quad (D) \]

Solution.

Let \( p, q, r \) be such angles that

\[ uz = \sqrt{a} \tan p, \quad \ldots \quad 1 \] \[ uz = \sqrt{a} \tan q, \quad \ldots \quad 2 \] \[ uz = \sqrt{a} \tan r, \quad \ldots \quad 3 \]

then from the equation (A),

\[ xy = \sqrt{a} \cotan p, \quad \ldots \quad 4 \] \[ xz = \sqrt{a} \cotan q, \quad \ldots \quad 5 \] \[ yz = \sqrt{a} \cotan r, \quad \ldots \quad 6 \]

and hence by substitution in the other equations

\[ \sqrt{a} (\cotan p + \cotan q) = b, \] \[ \sqrt{a} (\cotan q + \cotan r) = c, \] \[ \sqrt{a} (\cotan r + \cotan p) = d; \]

and from these again (because \( A \) being any arc,

\[ \tan A + \cotan A = 2 \cosec A = \frac{2}{\sin 2A}, \text{sect. 250); } \]

\[ \frac{2 \sqrt{a}}{\sin 2p} = b, \quad \frac{2 \sqrt{a}}{\sin 2q} = c, \quad \frac{2 \sqrt{a}}{\sin 2r} = d, \]

and \( \sin 2p = \frac{2 \sqrt{a}}{b}, \quad \ldots \quad 7 \) \[ \sin 2q = \frac{2 \sqrt{a}}{c}, \quad \ldots \quad 8 \] \[ \sin 2r = \frac{2 \sqrt{a}}{d}. \quad \ldots \quad 9 \] By these equations the angles \( p, q, r \) are determined.

By multiplying the corresponding sides of (1) (2) (3) we get

\[ \begin{align*} wxyz &= (\sqrt{a})^3 \tan p \tan q \tan r : \\ u^2 &= \sqrt{a} \tan p \tan q \tan r . \end{align*} \]

From this and equations 3, 2, and 1,

\[ \begin{align*} x^2 &= \frac{\sqrt{a}}{\tan p \tan q} = \sqrt{a} \cot p \cot q \tan r , \\ y^2 &= \frac{\sqrt{a}}{\tan q \tan r} = \sqrt{a} \cot p \tan q \cot r , \\ z^2 &= \frac{\sqrt{a}}{\tan r \tan p} = \sqrt{a} \tan p \cot q \cot r ; \end{align*} \]

so that, on the whole, to determine \( u, x, y, z \), we first find the angles \( p, q, r \), from these formulae,

\[ \begin{align*} \sin 2p &= \frac{2\sqrt{a}}{b}, \\ \sin 2q &= \frac{2\sqrt{a}}{c}, \\ \sin 2r &= \frac{2\sqrt{a}}{d}; \end{align*} \]

and then

\[ \begin{align*} u &= \sqrt{a} \tan p \tan q \tan r , \\ x &= \sqrt{a} \cot p \cot q \tan r , \\ y &= \sqrt{a} \cot p \tan q \cot r , \\ z &= \sqrt{a} \tan p \cot q \cot r . \end{align*} \]

281. The last example shall be the manner of inscribing a regular polygon of 17 sides in a circle,—a discovery due to Mr Gauss of Brunswick, and one of the most remarkable that has been made in geometry. We take it as given in the excellent Treatise on Geometry by Legendre.

Let the arc \( \frac{\pi}{17} = \varphi \); in the first place we have this equation,

\[ \cos \varphi + \cos 3\varphi + \cos 5\varphi + \cos 7\varphi + \cos 9\varphi + \cos 11\varphi + \cos 13\varphi + \cos 15\varphi = \frac{1}{2}. \]

For, putting \( P \) for the first member, and multiplying all the terms by \( 2 \cos \varphi \), and transforming the results by the formula \( 2 \cos a \cos b = \cos (a-b) + \cos (a+b) \)

(sect. 238), we shall have \( 2P \cos \varphi \)

\[ = \left\{ \begin{array}{l} 1 + 2\cos 2\varphi + 2\cos 4\varphi + 2\cos 6\varphi + 2\cos 8\varphi \\ + 2\cos 10\varphi + 2\cos 12\varphi + 2\cos 14\varphi + 2\cos 16\varphi. \end{array} \right. \]

But since \( 17\varphi = \pi \), therefore \( 2\varphi = \cos (\pi - 15\varphi) \)

\( = -\cos 15\varphi; \quad 4\varphi = \cos (\pi - 13\varphi) = -\cos 13\varphi \),

and so on, to \( 16\varphi = -\cos \varphi \); therefore

\[ 2P \cos \varphi = 1 - 2\cos 15\varphi - 2\cos 13\varphi - 2\cos 3\varphi - \cos \varphi; \]

or \( 2P \cos \varphi = 1 + \cos \varphi - 2P \),

or \( 2P (1 + \cos \varphi) = 1 + \cos \varphi \), hence \( P = \frac{1}{2} \).

This being proved, we now divide the terms which compose \( P \) into two parts,

\[ \begin{align*} x &= \cos 3\varphi + \cos 5\varphi + \cos 7\varphi + \cos 11\varphi, \\ y &= \cos \varphi + \cos 9\varphi + \cos 13\varphi + \cos 15\varphi. \end{align*} \]

We have therefore \( x + y = \frac{1}{2} \). We next multiply the four terms of \( x \) by the four terms of \( y \), and changing the products of cosines into the cosines of simple arcs, we obtain, all reductions being made,

\[ xy = 2(\cos 2\varphi + \cos 4\varphi + \cos 6\varphi + \cdots + \cos 16\varphi), \]

or \( xy = -2(\cos 15\varphi + \cos 13\varphi + \cos 11\varphi + \cdots + \cos \varphi) \),

or at last \( xy = -1 \).

By these equations we obtain

\[ x = \frac{1}{2} + \frac{1}{2}\sqrt{17}, \quad y = \frac{1}{2} - \frac{1}{2}\sqrt{17}. \]

Now, if we divide anew the sums \( x \) and \( y \) into two parts, viz.

\[ \begin{align*} x &= s + t, \\ y &= u + z, \end{align*} \]

\[ \begin{align*} s &= \cos 3\varphi + \cos 5\varphi, \\ t &= \cos 7\varphi + \cos 11\varphi, \\ u &= \cos 9\varphi + \cos 13\varphi, \\ z &= \cos 15\varphi. \end{align*} \]

We obtain in like manner,

\[ st = -\frac{1}{2}, \quad uz = -\frac{1}{2}; \]

so that we may now determine the four numbers \( s, t, u, z \), by means of two new equations of the second degree.

Lastly, knowing \( \cos \varphi + \cos 13\varphi = u \), and \( \cos \varphi \cos 13\varphi = \frac{1}{2}(\cos 12\varphi + \cos 14\varphi) = -\frac{1}{2}(\cos 3\varphi + \cos 5\varphi) = -\frac{1}{2}s \),

we may obtain by a fourth equation of the second degree the value of \( \cos \varphi \), and thence the side of the polygon, which is \( 2 \sin \varphi = 2\sqrt{1 - \cos^2 \varphi} \).

As to the method which has guided us in forming these equations, it depends on a very delicate theory, founded on the Indeterminate Analysis. For this we must refer to the work of Gauss, Disquisitiones Arithmeticae; or the French translation, Recherches Arithmétiques. (w. w.)